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ENGINEERING – CHEMICAL Focusing on the application of mathematics to chemical engineering, Applied Mathematical Methods for Chemical Engineers addresses the setup and verification of mathematical models using experimental or other independently derived data. The book provides an introduction to differential equations common to chemical engineering, followed by examples of first-order and linear second-order ordinary differential equations. Later chapters examine Sturm–Liouville problems, Fourier series, integrals, linear partial differential equations, regular perturbation, combination of variables, and numerical methods emphasizing the method of lines with MATLAB® programming examples. Fully revised and updated, this Third Edition: • Includes additional examples related to process control, Bessel Functions, and contemporary areas such as drug delivery • Introduces examples of variable coefficient Sturm–Liouville problems both in the regular and singular types • Demonstrates the use of Euler and modified Euler methods alongside the Runge–Kutta order-four method • Inserts more depth on specific applications such as nonhomogeneous cases of separation of variables • Adds a section on special types of matrices such as upper- and lower-triangular matrices • Presents a justification for Fourier–Bessel series in preference to a complicated proof • Incorporates examples related to biomedical engineering applications • Illustrates the use of the predictor-corrector method • Expands the problem sets of numerous chapters APPLIED MATHEMATICAL METHODS for CHEMICAL ENGINEERS Third Edition Applied Mathematical Methods for Chemical Engineers, Third Edition uses worked examples to expose several mathematical methods that are essential to solving real-world process engineering problems. K15249 ISBN: 978-1-4665-5299-9 Norman W. Loney 90000 9 781466 552999 K15249_Cover_PubGr.indd All Pages 9/1/15 1:52 PM APPLIED MATHEMATICAL METHODS for CHEMICAL ENGINEERS Third Edition This page intentionally left blank APPLIED MATHEMATICAL METHODS for CHEMICAL ENGINEERS Third Edition Norman W. Loney MATLAB® is a trademark of The MathWorks, Inc. and is used with permission. The MathWorks does not warrant the accuracy of the text or exercises in this book. This book’s use or discussion of MATLAB® software or related products does not constitute endorsement or sponsorship by The MathWorks of a particular pedagogical approach or particular use of the MATLAB® software. 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Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com Dedication To my sons—Alexander David and Michael Oliver—and to their sons—Isaiah, Ishmael, and Michael Norman—as well as to my parents—Ella Esedora and David Alexander This page intentionally left blank Contents Preface to the First Edition���������������������������������������������������������������������������������������xi Preface to the Second Edition�������������������������������������������������������������������������������� xiii Preface to the Third Edition������������������������������������������������������������������������������������ xv Acknowledgments��������������������������������������������������������������������������������������������������xvii Author���������������������������������������������������������������������������������������������������������������������xix Chapter 1 Differential Equations........................................................................... 1 Introduction................................................................................1 Ordinary Differential Equations................................................2 Model Development....................................................................3 1.3.1 Turbulent Core Region (0 ≤ r ≤ Rb)................................4 1.3.2 Laminar Sublayer Region (0 < x < δ )...........................5 1.3.3 Outline of Model Development.....................................7 1.3.3.1 Tube Side (Equation of Continuity for Species A)......................................................8 1.3.3.2 Shell Side....................................................... 9 References........................................................................................... 10 1.1 1.2 1.3 Chapter 2 First-Order Ordinary Differential Equations...................................... 11 2.1 2.2 Linear Equations...................................................................... 11 Additional Information on Linear Equations........................... 18 2.2.1 Proof (non-rigorous).................................................... 19 2.3 Nonlinear Equations................................................................. 22 2.3.1 Separable Equations.................................................... 22 2.3.2 Exact Equations...........................................................24 2.3.3 Homogeneous Equations.............................................26 2.4 Problem Setup..........................................................................26 2.4.1 Problem Statement......................................................26 2.5 Problems................................................................................... 32 References........................................................................................... 35 Chapter 3 Linear Second-Order and Systems of First-Order Ordinary Differential Equations......................................................................... 37 3.1 3.2 Introduction.............................................................................. 37 Fundamental Solutions of Homogeneous Equations.................................................................................. 39 3.3Homogeneous Equations with Constant Coefficients............................................................................... 41 vii viii Contents 3.4 Nonhomogeneous Equations.................................................... 45 3.4.1 Method of Variation of Parameters............................. 52 3.5 Variable Coefficient Problems.................................................. 54 3.5.1 Series Solutions Near a Regular Singular Point.......... 56 3.6 Alternative Methods................................................................. 61 3.6.1 Summary..................................................................... 62 3.6.2 Initial Value Problems.................................................64 3.6.3 Some Useful Properties of Laplace Transforms......... 68 3.6.4 Inverting the Laplace Transform................................. 70 3.6.5 Taylor Series Solution of Initial Value Problems........ 75 3.7Applications of Second-Order Differential Equations............. 77 3.7.1 Problem Statement...................................................... 77 3.8Systems of First-Order Ordinary Differential Equations....... 103 3.8.1 Nonhomogeneous Linear Systems............................ 115 3.9 Problems................................................................................. 119 References......................................................................................... 126 Chapter 4 Sturm–Liouville Problems................................................................ 129 4.1 4.2 Introduction............................................................................ 129 Classification of Sturm–Liouville Problems.......................... 130 4.2.1Properties of the Eigenvalues and Eigenfunctions of a Sturm–Liouville Problem�������� 140 4.3 Eigenfunction Expansion........................................................ 143 4.4 Problems................................................................................. 145 References......................................................................................... 148 Chapter 5 Fourier Series and Integrals.............................................................. 149 5.1 Introduction............................................................................ 149 5.2 Fourier Coefficients................................................................ 151 5.3 Arbitrary Interval................................................................... 155 5.4 Cosine and Sine Series........................................................... 156 5.5 Convergence of Fourier Series............................................... 161 5.6 Fourier Integrals..................................................................... 167 5.7 Problems................................................................................. 173 References......................................................................................... 174 Chapter 6 Partial Differential Equations........................................................... 175 6.1 6.2 6.3 6.4 Introduction............................................................................ 175 Separation of Variables........................................................... 176 6.2.1 Boundary Conditions................................................ 181 Nonhomogeneous Problem and Eigenfunction Expansion............................................................................... 211 Laplace Transform Methods................................................... 219 ix Contents 6.5 Combination of Variables.......................................................240 6.6 Fourier Integral Methods........................................................ 247 6.7 Regular Perturbation Approaches.......................................... 250 6.8 Problems................................................................................. 263 References......................................................................................... 267 Chapter 7 Applications of Partial Differential Equations in Chemical Engineering....................................................................... 269 7.1 Introduction............................................................................ 269 7.2 Heat Transfer.......................................................................... 269 7.3 Mass Transfer......................................................................... 285 7.4 Comparison between Heat and Mass Transfer Results..........300 7.5 Simultaneous Diffusion and Convection................................302 7.6 Simultaneous Diffusion and Chemical Reaction................... 317 7.7Simultaneous Diffusion, Convection, and Chemical Reaction���������������������������������������������������������������� 330 7.8 Viscous Flow.......................................................................... 343 7.9 Problems................................................................................. 358 References......................................................................................... 365 Chapter 8 Dimensional Analysis and Scaling of Boundary Value Problems...... 369 8.1 Introduction............................................................................ 369 8.2 Classical Approach to Dimensional Analysis........................ 371 8.3 Finding the Πs........................................................................ 372 8.4 Scaling Boundary Value Problems......................................... 378 8.5 Problems................................................................................. 389 References......................................................................................... 390 Chapter 9 Selected Numerical Methods and Available Software Packages........ 393 9.1 9.2 Introduction and Philosophy.................................................. 393 Solution of Nonlinear Algebraic Equations........................... 393 9.2.1 Newton–Raphson Method......................................... 398 9.2.2 Modified Newton–Raphson Method......................... 399 9.3Solution of Simultaneous Linear Algebraic Equations..........400 9.3.1 Error Estimate........................................................... 411 9.3.2 Special Types of Matrices......................................... 416 9.4 Solution of Ordinary Differential Equations.......................... 423 9.4.1 Initial Value Problems............................................... 424 9.4.2 Boundary Value Problems......................................... 437 9.4.3 Systems of Ordinary Differential Equations............. 443 9.5Numerical Solution of Partial Differential Equations............446 9.5.1 Explicit and Implicit Finite Difference Methods......446 9.5.2 Method of Lines........................................................449 x Contents 9.5.3 Selected Applications Using the Method of Lines��������������������������������������������������������������������� 451 9.6 Summary................................................................................ 458 9.7 Problems................................................................................. 458 9.7.1 Summary...................................................................464 References.........................................................................................464 Appendix A: Elementary Properties of Determinants and Matrices....................... 467 Appendix B: Numerical Method of Lines Example Using MATLAB®.................481 Appendix C: Program for a Transport and Binding Kinetics Model of an Analyte...........................................................................489 Appendix D: Programmed Model of a Drug Delivery System............................... 511 Index���������������������������������������������������������������������������������������������������������������������� 539 Preface to the First Edition The purpose of this book is to introduce students of chemical engineering to several mathematical methods that are often essential to successfully solve real process engineering problems. The book emphasizes analytical methods, even though most realistic models will be solved using numerical methods. However, prior to an extensive and expensive numerical analysis of a model, it is very useful to develop some understanding of the gross tendencies of the model. This type of understanding usually comes from the derivation of analytical solutions of a modified version of the problem under consideration. Typical chemical engineering curriculums consist of the equivalent of three semesters of calculus capped off by a course in elementary ordinary differential equations. This usually occurs within the first 2 years of a 4-year program (5 years, if co-op is an option). The next 2 or 3 years are usually dedicated to solving unit operation problems using prederived formulae. The point being, with few exceptions, the use of the four semesters’ worth of mathematics is not applied until the first year of graduate school. Those graduates who do go on to industry and later encounter the need to understand and apply the results of computer algebraic systems—indeed, to choose a software package for their own applications—have to rely heavily on the salesperson’s judgment. This book provides worked-out examples using a number of solution techniques while exposing the use of mathematics in chemical engineering to the reader. The first chapter provides an introduction to the three classes of transport common to chemical engineering. Chapter 2 deals with select first-order ordinary differential equations and provides chemical engineering examples that demonstrate the use of solution techniques. A section addressing the formulation of some physically applicable first-order ordinary differential equations (problem setup) is included. The third chapter addresses linear second-order ordinary differential equations. A brief discourse, it reviews elementary differential equations, and the chapter serves as an important basis to the solution techniques of partial differential equations discussed in Chapter 6. An applications section is also included with 10 worked-out examples covering heat transfer, fluid flow, and simultaneous diffusion and chemical reaction. In addition, the residue theorem as an alternative method for Laplace transform inversion is introduced. Chapter 4 and Chapter 5 introduce Sturm–Liouville problems and Fourier series and integrals, respectively. These topics contain essential background material for use in solving linear partial differential equations. Applications of these are postponed until partial differential equations are discussed. The sixth chapter provides instruction in a number of solution techniques for linear partial differential equations. Also included is a section introducing regular perturbation, a common approach to solving some nonlinear differential equations. xi xii Preface to the First Edition As the material in this chapter will be a new experience for a large segment of the readership, a substantial number of drill-type examples are included. Chapter 7 is dedicated entirely to worked-out examples taken from the chemical engineering research literature. This chapter relies on the mathematics of the earlier six chapters to solve problems in heat transfer; mass transfer; simultaneous diffusion and convection; simultaneous diffusion and chemical reaction; simultaneous diffusion, convection, and chemical reaction; and viscous flow. The eighth chapter briefly discusses dimensional analysis and scaling of boundary value problems. This is an important topic in chemical engineering. The practicing engineer is continually faced with justifying the simplifying assumptions invoked in deriving a solution to some process model of concern to him or her. Chapter 9 introduces selected numerical methods and available software packages. Because methods that were too effort-consuming earlier are now commercially available in many software packages, it is more important to mention those packages and leave the algorithmic details to the numerical analysis literature. Here, it is hoped that enough of an introduction to numerical methods is made so that the interested reader can independently pursue the subject. As a goal of this text is to remove the mathematics phobia that usually exists among some of our bright young chemical engineers, rigor is sacrificed in favor of exposition. Therefore, the references at the end of each chapter have been carefully selected to aid the reader who wishes to pursue further study in the discussed subject matter. However, I do wish to point out that those references are not in any way exhaustive. Bold or italic type is used to draw attention to a term or statement that is significant to the concept under discussion. Others and I have successfully used this book as a text for both undergraduate and first-year graduate courses. Most of the students in the graduate course have been chemical engineers with varying backgrounds in elementary differential equations. For an undergraduate one-semester course, the applications in Chapter 2 and Chapter 3 are emphasized, and knowledge of those solution techniques is treated as a prerequisite. Chapters 4 through 6 are covered in their entirety with some applications taken from Chapter 7. Also, parts of Chapter 9 are covered based on the audience needs. The graduate class uses the entire book in a one-semester course. Preface to the Second Edition Because the purpose of this book has not changed, the users of the first edition will hopefully find the basic content and structure of this edition familiar. Two important aspects of the application of mathematics to chemical engineering are (1) setting up a mathematical model and (2) verification of the mathematical model with experimental or other independently derived data. In this edition of the text, I have attempted to address these two aspects. Chapter 1 has been expanded to include two popular approaches to model development. Each approach is discussed in the format of an example while using a real application from research. Both applications are solved as examples in Chapter 7. Further, a model of a one-dimensional rod is introduced in Chapter 6 and a planar model of heat conduction in one direction is introduced in Chapter 7. The solutions to the two examples of modeling approaches developed in Chapter 1 are verified with independently derived experimental data in Chapter 7. In addition, a figure comparing model results to independently derived experimental data is included for one example of mass transfer in a membrane separator discussed in Chapter 7. Additional changes have been made in Chapter 3 and Chapter 9. In particular, Chapter 3 has been expanded to include systems of first-order differential equations. Chapter 9 has been expanded to include the numerical method of lines. As usual, an example using the numerical method of lines is provided, which extends into Appendix B, where an actual MATLAB® program is given. xiii This page intentionally left blank Preface to the Third Edition Because the purpose of this book has not changed, the users of the first two editions will hopefully find the basic content and structure of this edition familiar. Feedback from students motivated the addition of more examples into this edition. In this edition of the text, changes have been made in Chapters 3, 4, 6, 7, and 9. More examples related to process control and an expanded problem set constitute the modification of Chapter 3. Chapter 4 has been modified to introduce examples of variable coefficient Sturm–Liouville problems both in the regular and singular types. Also, the examples are renumbered to emphasize the related procedures introduced earlier and the problem set is expanded. Examples in Chapter 6 are renumbered to insert more depth on specific applications such as nonhomogeneous cases of separation of variables. A justification for Fourier–Bessel series is now provided in preference to a complicated proof, along with five additional examples involving Bessel Functions. This chapter’s problem set is also expanded. Chapter 7 has been expanded to include additional examples on contemporary areas: absorption with chemical reaction, laminar flow in flat duct with permeable walls, drug delivery, and wall region velocity profile analysis. Chapter 9 has been expanded to include examples relating to biomedical engineering applications and an expanded problem set. These examples are solved using the numerical method of lines. As usual, examples using the numerical method of lines are provided, which extends into the appendices, where actual MATLAB® programs are provided. A new section is included in this chapter on special types of matrices, such as upper- and lower-triangular matrices, how to generate them, and factorization methods (Doolittle, Choleski, and Crout) that exploit their application to solve systems of equations. An example introduced under the explicit and implicit finite difference methods incorporates the notion of tridiagonal matrices, which was discussed earlier but not demonstrated. In addition, examples are introduced that demonstrate the use of Euler and modified Euler methods alongside the popular Runge–Kutta order-four method. Also included is an example demonstrating the use of the predictor-corrector procedure using the four-step Adams–Bashforth as the predictor and the three-step Adams– Moulton procedure as the corrector. xv xvi Preface to the Third Edition MATLAB ® are registered trademarks of The MathWorks, Inc. For product information, please contact: The MathWorks, Inc. 3 Apple Hill Drive Natick, MA 01760-2098 USA Tel: 508 647 7000 Fax: 508-647-7001 E-mail: info@mathworks.com Web: www.mathworks.com Acknowledgments I am indebted to many who have encouraged me on this project and to all my Mathematical Methods (ChE626) students for their contributions and patience and understanding since the printing of the first edition. I thank especially Professor Ali Elkamel of the University of Waterloo for his very timely suggestions and Professor William E. Schiesser of Lehigh University for his generosity in sharing his computer programs with me. xvii This page intentionally left blank Author Norman W. Loney is professor and department chair of the Otto H. York Department of Chemical, Biological and Pharmaceutical Engineering at New Jersey Institute of Technology (NJIT). He has authored or coauthored more than 70 publications and presentations related to the use of applied mathematics to solve transport phenomena-related problems in chemical engineering since joining the depart­ ment in 1991. Dr. Loney has been awarded several certificates of recognition from the National Aeronautics and Space Administration and the American Society for Engineering Education for research contributions. He has also been honored with the Newark College of Engineering Teaching Excellence award, the Saul K. Fenster Innovation in Engineering Education award, and the Excellence in Advising award. Dr. Loney is a fellow of the American Institute for Chemical Engineers. Prior to joining NJIT, Dr. Loney, a licensed professional engineer, practiced engineering at Foster Wheeler, M.W. Kellogg Company, Oxirane Chemical Company, and Exxon Chemical Company. xix This page intentionally left blank 1 1.1 Differential Equations INTRODUCTION There are several significant problems in chemical engineering that require a fundamental understanding of differential equations to fully appreciate the underlying transport phenomena. In this book, differential equation means an equation containing derivatives of an unknown function to be determined [1]. For example, Fourier’s law [2–4] for the molecular transport of heat in a fluid or a solid can be written as a first-order differential equation qz d(ρCP T ) = −α A dz (1.1) for constant density ρ and heat capacity CP. In this equation, qz/A represents the heat flux (J/s·m2), α the thermal diffusivity (m2/s), and ρ CPT the concentration of heat (J/m3), with the subscript z indicating that energy is transferred in the z direction. The unknown function is the temperature T(z). A second example that is familiar to chemical engineers is Fick’s law [2–4] for the molecular transport of mass in a fluid or a solid for constant total concentration in the fluid. This fundamental transport process can be written as J AZ = − DAB dCA dz (1.2) where JAZ is the flux of species A (kmol/s·m2), DAB is the molecular diffusivity (m2/s) of species A in B, and CA is the concentration of A (kmol/m3). In this case, the unknown function to be determined is CA(z). A third example is Newton’s law [2–4] of viscosity, written as follows for constant density ρ : τ zx = −γ d( ν x ρ) dz (1.3) where τ zx is the flux of x-directed momentum in the z direction [(kg·m/s)/s·m2] and γ is the kinematic viscosity (μ/ρ ) or momentum diffusivity. Transport or diffusion takes place in the z direction, and μ is the viscosity (kg/m·s). In this equation, the unknown function to be determined is the x component of velocity ν x(z). Differential equations are usually divided into two classes. If the unknown function depends on a single independent variable, then the differential equation is classified as an ordinary differential equation (ODE); if there are two or more independent variables, then the equation is called a partial differential equation (PDE). 1 2 Applied Mathematical Methods for Chemical Engineers Equations 1.1 through 1.3 are examples of ODEs. A general equation for the conservation of momentum, thermal energy, or mass can be written as ∂Γ ∂2 Γ −δ 2 = R ∂t ∂z (1.4) where Γ represents the concentration of the property (momentum, energy, or mass), δ is a proportionality constant (e.g., diffusivity), t is time, z indicates the distance in the direction of flow, and R is a source term (generation). In this differential equation, the unknown function to be determined is Γ (z, t), which depends on both distance and time. This equation is an example of a PDE. Several other examples of PDEs are given in Chapter 6. 1.2 ORDINARY DIFFERENTIAL EQUATIONS In discussions involving differential equations, the word order is very prevalent. The working definition of the order of an ODE is the order of the highest derivative that appears in that equation. Hence, the equation f [t , ρ(t ), ρ′(t ), L , ρ( n ) (t )] = 0 (1.5) is an ODE of the nth order. Equation 1.5 represents a relation between the variable t (independent) and the values of the dependent variable ρ and its first n derivatives, ρ ′ , ρ ″ , … , ρ (n). An explicit example of Equation 1.5 is ρ′′′ + 2et ρ′′ + ρρ′ = t 4 (1.6) which is a third-order differential equation for ρ = ρ (t). In this book, we will avoid the common assumption that it is always possible to solve a given ODE for the highest derivative. However, most of the discussion will be expedited when the form ρ( n ) = f (t , ρ, ρ′, ρ′′, L , ρ( n −1) ) (1.7) can be obtained. It should also be noted that even when the form given by Equation 1.7 is achievable, it does not generally mean that there is a function ρ = ɸ (t) that s­ atisfies it. Thus, a solution of the ODE 1.7 on α < t < β is a function ɸ such that ɸ ″ , ɸ ″ ′, … , ɸ (n − 1) exist and satisfy φ( n ) (t ) = f [t , φ(t ), φ ′(t ), L , φ( n −1) (t )] (1.8) for all t in α < t < β . In other words, a solution of a differential equation is a function that satisfies the differential equation and the domain of definition of the differential equation. By way of direct substitution into the first-order equation dQ = kQ dt (1.9) 3 Differential Equations it can be shown that Q(t ) = ce kt, −∞ < t < ∞ (1.10) is a solution, where c is an arbitrary constant and k is a given constant. There are three very important issues to be resolved for a given differential equation: 1.Existence of a solution 2.Uniqueness of the solution 3.How to determine a solution These issues are covered in detail by Ince 5 and some standard differential equation texts. However, if an engineering problem is to be formulated as a differential equation it is expected to have a solution. This solution provides one way of verifying the correctness of the mathematical formulation. The remaining chapters of this book are dedicated to the exposition of some of the common mathematical techniques that are useful in chemical engineering. However, before we address the mathematical techniques it may be helpful to review some modeling examples that lead to differential equations. In Section 1.3, examples are provided that use the conservation laws to set up differential equations. 1.3 MODEL DEVELOPMENT Regardless of the specific techniques used to derive the mathematical model of chemical processes, the conservation laws must be taken into consideration. This is the thesis of transport phenomena, and such conservation laws are well documented in standard textbooks. Examples 1.1 and 1.2 are provided as reminders of two popular approaches that lead to differential equations. The techniques used in these examples are the shell balance and the balance based on a previous mathematically formulated conservation law. Both approaches are acceptable in practice, but the second requires good judgment and experience to combine the conservation laws and appropriately eliminate the terms that may not be reflecting the current mechanisms under consideration. In Example 1.1, the shell balance [4, 6] is illustrated. Example 1.1 One approach to deriving correlations for mass transfer coefficients in process systems is to generate experimental data in momentum transport studies. In this approach, it is assumed that both molecular and eddy diffusions play a role in the intermediate region. Then, at any distance y from the wall the rate of mass transfer can be expressed as a function of both molecular and eddy diffusivities. However, applications of these models rely on knowledge of eddy diffusivity, ED, as a function of y, a relationship that is usually inferred from the experimental data [7–10]. There, the eddy diffusivity can be inferred from the eddy viscosity by similarity arguments. A substantial amount of published works is along this line [11–13]. 4 Applied Mathematical Methods for Chemical Engineers Consider an alternative approach that does not rely on the knowledge of the eddy diffusivity as a function of the distance y from the wall. Here, we examine the mass transfer for a turbulent flowing fluid in a smooth tube. In the tube, a turbulent core region and a laminar sublayer region are considered separately as contributing to the total mass transfer of the transferring species from the fluid toward the wall as well as away from the wall. 1.3.1 TURBULENt CORE REGION (0 ≤ r ≤ Rb) In this region, the velocity Vb is assumed to be independent of both r and z (Figure 1.1). The concentration of the transferring species Cb is independent of r but depends on z. For a core region of radius Rb in a smooth tube of radius R, a mass balance in a shell of volume πRb2 ∆z is N Az πRb2 z − N Az πRb2 z +∆z − N Ar Rb 2 πRb ∆z = 0 (1.11) where NAz and NAr are the fluxes of the diffusing species in the z and r directions, respectively [4, 14]. Dividing by πRb2 ∆z and letting Δ z → 0 in the limit, we get dN Az 2 N Ar =− dz Rb (1.12) Rb Further, assuming that diffusion in the z direction is negligible in comparison to bulk flow gives N Az = − D dCb + x A ( N Az + N Bz ) ≅ CbVb dz (1.13) By combining Equations 1.12 and 1.13, we get Vb dCb 2 N Ar =− dz Rb (1.14) Rb where Vb is the bulk velocity whose profile is sketched in Figure 1.2. Also shown in Figure 1.2 is an expanded view of the laminar sublayer in relation to the turbulent core. Rb x z R d Turbulent core region Z FIGURE 1.1 Turbulent core. (From Huang et al., Chem. Eng. Series, 59, 1191–1197, 2004. With permission.) 5 Differential Equations Vb Rb z vz R d x FIGURE 1.2 Laminar sublayer. (From Huang et al., Chem. Eng. Series, 59, 1191–1198, 2004. With permission.) 1.3.2 LAMINAR SUBLAYER REGION (0 < x < δ ) In the laminar sublayer region, the velocity of the fluid, vz, is assumed to be linear with respect to x. At the interface of the two regions, x = δ , the velocities are equal, that is, Vb = vz|x = δ . The concentration of the transferring species, CA, is expected to depend on both x and z. At the interface, Cb = CA|x = δ . Consider now a mass balance in a shell of volume Δ x Δ zw and length L over the laminar sublayer: w∆xN Az z − w∆x ⋅ N Az z + ∆z + w∆z ⋅ N Ax x − w∆z ⋅ N Ax x + ∆x =0 (1.15) Division by Δ x Δ zw and allowing Δ x and Δ z to go to zero in the limiting process results in ∂ N Az ∂ N Ax + =0 ∂x ∂z (1.16) Since there is no bulk flow in the x direction N Ax = − D ∂CA ∂C + x A ( N Ax + N Bx ) = − D A ∂x ∂x (1.17) Also, assuming that the mass flux of A in the z direction is controlled by bulk flow N Az = − D ∂CA + x A ( N Az + N Bz ) ≅ vz CA ∂z (1.18) Therefore, Equation 1.16 can be restated as vz ∂CA ∂2 CA =D ∂z ∂x 2 (1.19) where vz is the velocity profile in the laminar sublayer, as shown in Figure 1.2. Since we assumed that vz is proportional to the distance from the wall, vz = hx (1.20) 6 Applied Mathematical Methods for Chemical Engineers The proportionality constant h is defined as h= Vb δ (1.21) since at x = δ , vz = Vb. Then, Equation 1.20 becomes vz = Vb x δ (1.22) and Equation 1.19 becomes Vb ∂CA ∂2 CA =D x ∂z ∂x 2 δ (1.23) where CA is the concentration of the diffusing species in the laminar sublayer and D is the diffusion coefficient of that species. Equations 1.14 and 1.23 may be subjected to the conditions CA = 0 at x = 0 (1.24) CA = Cb ( z ) at x = δ (1.25) CA = C0 at z = 0 (1.26) when the mass transfer takes place from the fluid through the wall. The final problem to be solved (Example 7.12) consists of Vb dCb 2 N Ar =− dz Rb Rb Vb ∂CA ∂2 CA =D x ∂z ∂x 2 δ (1.27) (1.28) CA = 0 at x = 0 (1.29) CA = Cb ( z ) at x = δ (1.30) CA = C0 at z = 0 (1.31) The next example uses a previous mathematically formulated conservation law to develop a model for a mass exchanger. 7 Differential Equations CAS0 r CASL z CA0 CAB CASL CAS0 0 L FIGURE 1.3 Tubes-in-shell mass exchanger. Example 1.2 Large surface area membrane modules, such as hollow fiber units, are often used in the production of low-alcohol beer, hemodialysis, or desalination. In all these processes, the rate of transfer is believed to be governed by the concentration difference across the membrane, molecular size, and permeability characteristics of the membrane. A model that has shown some promise in correlating the amount removed as a function of flow rate is discussed here. The hollow fiber system consists of a shell, which houses a bundle of fibers. These fibers are grouped together in a parallel array. Both ends terminate in a tube sheet similar to a shell-and-tube heat exchanger. The length to diameter ratio of a typical channel in a well-packed hollow fiber system can be as large as 103 –104. The Reynolds number may be very low, and the system is expected to operate in a laminar flow regime. The entry region effect, which is important in traditional heat exchangers, is ­negligible for this type of unit that contains on the order of 102–104 fibers. Figure 1.3 shows the shell-in-tube arrangement of a typical unit. Notice that only one of the many tubes is shown to emphasize the orientation of solute concentrations in both shell and tube sides. The following assumptions are made in the development of a proposed mathematical model: • • • • Steady-state conditions prevail. Fully developed laminar flow on the tube side. Fick’s law can describe the diffusion process. Physical properties such as density, diffusivity, and overall mass transfer coefficient are the constants, independent of concentration. • Dialyzate-side mass transfer resistance is independent of position. • Plug flow occurs on the dialyzate side. 1.3.3 OUtLINE OF MODEL DEVELOPMENt Subdivide the mass exchanger into two subsystems: tube side and shell side. 8 Applied Mathematical Methods for Chemical Engineers 1.3.3.1 Tube Side (Equation of Continuity for Species A) Following Bird et al. [4], the equation of continuity for species A in terms of wA for constant ρ and DAB in cylindrical coordinates is ∂ wA ν θ ∂ wA ∂ wA ∂w ρ A + νr + + νz ∂t r ∂θ ∂r ∂z 1 ∂ ∂ wA 1 ∂ 2 wA ∂ 2 wA + rA = ρDAB + r + 2 2 ∂ z 2 r ∂r ∂r r ∂θ (1.32) where wA is the mass fraction of species A, and ρ and DAB represent the mass density and diffusion coefficient, respectively. Equation 1.32 can be reduced to 1 ∂ ∂ wA ∂ 2 wA ∂ wA ρ νz = ρDAB r + ∂z ∂ z 2 r ∂r ∂r (1.33) We make the following observations: ∂ wA =0 ∂t Because we assumed steady state, the radial and θ components of velocity, ν r and ν θ , are both zero in this system and there is no diffusion in the θ direction, that is, ρDAB 1 ∂ 2 wA ≡0 r 2 ∂θ2 Also, there is no chemical reaction indicated in this process; therefore, there is no generation or consumption of species A (rA ≡ 0). Based on the experience, we can anticipate that the convection contribution in the z direction will be much larger than the diffusion contribution, that is, νz ∂ wA ∂ 2 wA > DAB ∂z ∂z 2 Therefore, the equation of continuity for species A in this system is 1 ∂ ∂ wA ∂ wA ρ νz = ρDAB r ∂z r ∂ r ∂r Finally, if we move ρ inside the derivatives and divide each term by the molecular weight of species A and observe that ρwA ρ = A = CA MA MA 9 Differential Equations is the concentration of species A, we get the result ν z (r ) ∂CA 1 ∂ ∂CA = DA r ∂z r ∂r ∂r (1.34) subject to the boundary conditions − DA CA (0, r ) = CA0 (1.35) ∂CA ( z , 0) =0 ∂r (1.36) ∂CA ( z , R) = K (CA r = R − CAS ) ∂r (1.37) where CAS is the concentration of species A on the shell side and is a function of z only. CA(z, r) is the local solute concentration in the stream, CA0 is the inlet solute concentration for the tube side, DA is the diffusivity, and K is an overall mass transfer coefficient. Since the flow is fully developed, we can replace ν z(r) by the parabolic velocity profile r2 ν z (r ) = νmax 1 − 2 R (1.38) such that Equation 1.34 becomes r 2 ∂C 1 ∂ ∂CA νmax 1 − 2 A = DA r R ∂z r ∂r ∂r (1.39) 1.3.3.2 Shell Side The material balance on the shell side results in Q dCAS = −2 πRNK (CA dz r=R − CAS ) (1.40) subject to the condition that CAS ( L ) = CASL (1.41) where N is the number of fibers and Q is the volumetric flow rate of the shell side (sweep) stream. Also, a total mass balance of species A between the two streams results in Q(CAS0 − CASL ) = NπR 2 νmax 2 r=R 4 C − CA A0 R 2 r ∫= 0 z=L r2 1 − 2 r dr R (1.42) 10 Applied Mathematical Methods for Chemical Engineers where CAS0 is the outgoing sweep stream concentration of species A (Figure 1.3). Equations 1.39 through 1.42 along with the conditions given by Equations 1.35 through 1.37 represent the model. The solution to this problem is developed in Example 7.12. It is well known that mathematical models are useful tools that contribute to the understanding of underlying mechanisms occurring in a given process. It is also known that the most useful models are the ones benchmarked with experimental data. Although the benchmarking step can be challenging, it does guide the model development toward the actual physicochemical realities existing in a given system. Both Examples 1.1 and 1.2 are favorably compared with independently derived and published experimental data. REFERENCES 1.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley & Sons, New York, 2005, Chap. 1. 2.Geankopolis, C.J. Transport Processes and Unit Operations, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1978. 3.Bennett, C.O. and Myers, J.E. Momentum, Heat, and Mass Transfer, McGraw-Hill, New York, 1962. 4. Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena, John Wiley, New York, 2002. 5. Ince, E.L. Ordinary Differential Equations, Dover, New York, 1956. 6. Huang, C.R., Denny, A.F., and Loney, N.W. Molecular diffusion in the laminar ­sub-layer during turbulent flow in a smooth tube, Chem. Eng. Sci., 59, 1191 (2004). 7. Van Shaw, P. and Hanratty, T.J. Fluctuations in the local rate of turbulent mass transfer to a pipe wall, AIChE J., 10, 475 (1964). 8.Son, J.S. and Hanratty, T.J. Limiting relation for the eddy diffusivity close to a wall, AIChE J., 13, 689 (1967). 9. Hughmark, G.A. Wall region mass transfer for large Schmidt numbers in turbulent pipe flow, AIChE J., 23, 601 (1977). 10. Shaw, D.A. and Hanratty, T.J. Turbulent mass transfer rates to a wall for large Schmidt numbers, AIChE J., 23, 28 (1977). 11. Campbell, J.A. and Hanratty, T.J. Mass transfer between a turbulent fluid and a solid boundary: linear theory, AIChE J., 28, 988 (1982). 12.Campbell, J.A. and Hanratty, T.J. Turbulent velocity fluctuations that control mass transfer to a solid boundary, AIChE J., 29, 215 (1983). 13.Na, Y., Papavassiliou, D.V., and Hanratty, T.J. Use of direct numerical simulation to study the effect of Prandtl number on temperature fields, Int. J. Heat Fluid Flow, 20, 187 (1999). 14. Plawsky, J.L. Transport Phenomena Fundamentals, Marcel Dekker, New York, 2001. 2 First-Order Ordinary Differential Equations 2.1 LINEAR EQUATIONS The examples of linear first-order differential equations occur frequently in c­ hemical engineering practice through unsteady-state mass balances or first-order chemical reaction problems. Here, we review a few methods for solving first-order ordinary differential ­equations. Following each method are examples demonstrating the application of that method. Also, the notion of translating prose into mathematical symbolism is introduced in Section 2.4. A brief recap of the definition of linear equations in the context of differential equations is presented here. Following the recap are the examples of unsteady mass balances, which lead to linear first-order problems. Also presented are examples involving chemical reactions that can be treated as linear first-order problems. In this chapter, attention is focused on differential equations of the form ρ′ = f (t , ρ) (2.1) where f is a given function of t and ρ . By linear equations, we mean any equation that can be expressed in the polynomial form an (t )ρ( n ) + an −1 (t )ρ( n −1) + + a2 (t )ρ′ + a1 (t )ρ(0) + a0 (t ) = g(t ) (2.2) where ρ (.) symbolizes the derivative of ρ with respect to t. Consequently, the equation a2 (t )ρ(1) + a1 (t )ρ(0) + a0 = g(t ) (2.3) is a linear first-order differential equation and is more familiar in the following form: a2 (t )ρ′ + a1 (t )ρ + a0 (t ) = g(t ) (2.4) 11 12 Applied Mathematical Methods for Chemical Engineers The general solution of Equation 2.4 can be obtained by the following steps: 1. Rewrite Equation 2.4 as follows: ρ′ + a1 (t ) g(t ) − a0 (t ) , a2 (t ) ≠ 0 for all t ρ= a2 (t ) a2 (t ) (2.5) 2.Determine a (t ) µ(t ) = exp ∫ 1 dt a2 (t ) (2.6) where μ(t) is called an integrating factor. 3. Multiply both sides of Equation 2.5 by μ(t) a1 (t ) g(t ) − a0 (t ) µ(t ) ρ′ + a (t ) ρ µ(t ) = a2 (t ) 2 (2.7) and observe that the left-hand side of Equation 2.7 can be written as d [ρµ(t )] dt or a (t ) d a (t ) a (t ) ρ′ exp ∫ 1 dt + ρ 1 exp ∫ 1 dt = [ρµ(t )] a2 (t ) a2 (t ) dt a2 (t ) (2.8) Thus, Equation 2.7 can be recast as a1 (t ) g(t ) − a0 (t ) a (t ) d dt = exp ∫ 1 dt (2.9) ρ exp ∫ dt a2 (t ) a2 (t ) a2 (t ) 4.Integrate both sides of Equation 2.9 with respect to the independent variable to get a (t ) a (t ) g(t ) − a0 (t ) exp 1 dt dt + c ρ exp ∫ 1 dt = ∫ a2 (t ) a2 (t ) a2 (t ) or a (t ) a (t ) g(t ) − a0 (t ) exp 1 dt dt ρ(t ) = exp − ∫ 1 dt ∫ a2 (t ) a2 (t ) a2 (t ) a (t ) + c exp − ∫ 1 dt a2 (t ) where c is the constant of integration. (2.10) 13 First-Order Ordinary Differential Equations Example 2.1 Water containing 0.5 kg of salt per liter is poured into a tank at a rate of 2 L/min, and the well-stirred mixture leaves at the same rate [1]. After 10 minutes, the process is stopped and freshwater is poured into the tank at a rate of 2 L/min, with the new mixture leaving at 2 L/min. Determine the amount (in kilograms) of salt in the tank at the end of 20 minutes if there were 100 L of pure water initially in the tank. Solution Let CA (kg/L) be the concentration of salt in the tank at any time t. Then, from material balance [5] a salt balance gives 2 L/min, 1/2 kg salt/L CA 2 L/min, CA(kg/L) Rate of accumulation = rate of input - rate of output (2.11) or in symbols, 100 L dCA 1/2 kg 2 L 2L = − CA dt L min min (2.12) with the following initial condition: CA (0) = 0 (2.13) dCA 1 1 + CA = dt 50 100 (2.14) Equation 2.12 can be rewritten as as suggested in step 1. Then, following step 2, t 1 µ(t ) = exp ∫ dt = e 50 50 and Equation 2.14 is recast as (e t 50 1 50t e CA ′ = 100 ) which is in the form of Equation 2.9, as given in step 3. Following step 4, we get (2.15) 14 Applied Mathematical Methods for Chemical Engineers CA (t ) = 50 1 + ae − 50 t 100 (2.16) where a is an arbitrary constant. Using the given initial condition, Equation 2.13, we get 50 1 1 − e − 50 t 100 ( CA (t ) = ) (2.17) Equation 2.17 is the salt concentration profile for the first 10 minutes of the process. For the subsequent time during which pure water is added, Equation 2.11 reduces to Rate of accumulation = − rate of output (2.18) and the new initial condition is CA (10) = 50 10 1 − e − 50 100 ( ) Thus, dC A 2 =− CA dt 100 (2.19) and CA (10) = 50 10 1 − e − 50 100 ( ) (2.20) Equations 2.19 and 2.20 describe the process where no salt (pure water) is being poured in. The solution of Equations 2.19 and 2.20 gives CA (t ) = be − 50 t 1 (2.21) where b= 1 10 10 1 − e − 50 e 50 2 ( ) Then, at t = 20 minutes or after the second 10-minute period ( CA (20) = 1/2e − 50 1 − e − 50 10 10 ) and the amount of salt in the tank at the end of this period is 100 CA (20) = 50e −0.2 (1 − e −0.2 ) kg Example 2.2 Consider a tank with 500-L capacity that initially contains 200 L of water with 100 kg of salt in solution. Water containing 1 kg of salt per liter is entering at a rate of 3 L/min, and the mixture is allowed to flow out of the tank at a rate of 2 L/min. Determine the amount (in kilograms) of salt in the tank at any time prior to the instant when the solution begins to overflow. Determine the concentration (in kilograms per liter) of salt in the tank when 15 First-Order Ordinary Differential Equations it is at the point of overflowing. Compare this concentration with the theoretical limiting concentration if the tank had infinite capacity [1]. Solution Let CA (t) (kg/L) be the concentration in the tank at any time t, and let V(t) (L) be the volume of the tank contents, with V0 as the initial volume. Then, Equation 2.11 becomes d(VCA) 1 kg 3 L 2 L = − CA dt L min min (2.22) but V (t ) = V0 + (volumetric rate in − volumetric rate out) t (2.23) and dV = Volumetric rate in − volumetric rate out dt Therefore, Equation 2.22 becomes CA dV dC A +V = 3 − 2CA dt dt (2.24) or CA + (200 + t ) dCA = 3 − 2CA dt subject to CA (0) = 1/2 kg L Equation 2.25 can be solved using the previously given four steps as follows: Step 1: dC A 3 3 + CA = dt 200 + t 200 + t Step 2: 3 µ(t ) = exp ∫ dt = (200 + t )3 200 + t Step 3: [(200 + t )3 CA ]′ = 3(200 + t )2 (200 + t )3 CA = 3 ∫ (200 + t )2 dt + k (2.25) 16 Applied Mathematical Methods for Chemical Engineers or Step 4: CA (t ) = 1 + k (200 + t )3 at t = 0, CA = 1 k = 1+ 2 (200)3 thus 1 k = − (200)3 2 1 200 CA (t ) = 1 − 2 200 + t 3 Then, the amount of salt in the tank at any time t prior to the instant when the solution begins to overflow is V(t)CA. That is, V (t )CA (t ) = 200 + t − 100(200) (200 + t )2 for t < instant of overflow Noting that the capacity of the tank is 500 L, at the instant of overflow 500 = 200 + t, t = 300. At t = 300, CA (300) = 1 − 1/2(200)3 121 = (500)3 125 in comparison with the theoretical limiting concentration of lim CA (t ) = 1 t →∞ Example 2.3 Consider the consecutive second-order, irreversible reactions occurring in a batch reactor [6]: 1 A + S k →X 2 X + S k →Y If one mole of A and two moles of S are initially added, determine the mole fraction of X remaining after half of A is consumed. Assume that k2/k1 = 2. 17 First-Order Ordinary Differential Equations Solution dCX = k1CACS − k 2CXCS dt (2.26) is the net rate of formation of X in terms of the appropriate concentrations. dCA = − k1CACS dt (2.27) dCY = k 2CXCS dt (2.28) is the rate of disappearance of A. is the rate of formation of Y. Dividing Equation 2.26 by Equation 2.27 results in dC X k C = −1 + 2 X dC A k1 CA Equation 2.29 is a linear first-order differential equation. For an integrating factor µ(CA) = CA−2 the differential equation can be represented as follows: (CXCA−2 )′ = −CA−2 This integrates to CX = CA + m1CA2 subject to the following: CA = 1, CX = 0 at t = 0 Therefore, CX = CA − CA2 Similarly, dividing Equation 2.28 by 2.27 gives dC Y = −2 + 2CA dC A which integrates to CY = 1 − 2CA + CA2 (2.29) 18 Applied Mathematical Methods for Chemical Engineers based on the initial condition CA = 1, CY = 0 at t = 0 Finally, the mole fraction of X is 1 CA − CA2 = C A + CS + C X + C Y 9 when half of A is consumed. Examples 2.1 through 2.3 demonstrate a technique to solve linear first-order differential equations of the type given by Equation 2.5. Although this method is straightforward, there are three things to note. First, the form given by Equation 2.5 is required, that is, the coefficient of the derivative term, ρ ′, must be one. Second, the functions a1(t), b2(t), and [g(t) – a0(t)]/a2(t) must be continuous. Third, Equation 2.8 provides a check as to whether the derivative of the product of μ and ρ is in fact the appropriate left-hand side of Equation 2.7. It should also be noted that each example problem was stated in prose and required transformation to mathematical symbolism. This transformation or problem setup is an important step and is usually where most students are left behind. However, in this book, whenever the demonstration involves physical phenomena such as those encountered in chemical engineering, the formats of Examples 2.1 and 2.2 will be followed. As an aid to this step, it is suggested that the student invest some time in reviewing the laws of conservation of mass and energy, as well as the unit operations principles discussed in undergraduate chemical engineering courses. 2.2 ADDITIONAL INFORMATION ON LINEAR EQUATIONS In this section, a very important fundamental theorem is discussed. This theorem is important because it resolves two of the issues raised at the end of Section 1.1. Specifically, the theorem addresses the existence and uniqueness of a solution. An initial value problem for a first-order linear equation will always have a unique solution if the conditions of the following theorem are satisfied [1,2]. Theorem 2.1 If the functions p and g are continuous on an open interval α < x < β containing the point x = x0, then there exists a unique function y = φ (x) that satisfies the differential equation y′ + p( x ) y = g( x ) (2.30) for α < x < β and that also satisfies the initial condition y( x 0 ) = y0 where y0 is an arbitrary prescribed initial value. (2.31) 19 First-Order Ordinary Differential Equations 2.2.1 PROOF (non-rigorous) We seek a function μ such that if Equation 2.30 is multiplied by μ then the left-hand side of Equation 2.30 can be written as the derivative of the single function μ(x)y, that is, μ(x) [y ′ + p(x)y] = [μ(x)y] ′ = μ(x)y ′ + μ ′(x)y. Thus, μ(x) must satisfy µ( x ) p( x ) y = µ′( x ) y or µ ′( x ) = p( x ), µ( x ) > 0 µ( x ) Then ln µ( x ) = ∫ x p(t ) dt or x µ( x ) = exp ∫ p(t ) dt (2.32) [µ( x ) y]′ = µ( x ) g( x ) (2.33) Therefore, following the multiplication of Equation 2.30 by μ(x). Integrating both sides of Equation 2.33 with respect to x and solving for y give y= 1 x µ(s) g(s) d s + c µ( x ) ∫ (2.34) Further, since p is continuous for α < x < β , it follows that μ is defined in this i­ nterval and is a nonzero differentiable function. Thus, the conversion of Equation 2.30 into the form of Equation 2.33 is justified. Also, the function μg has an a­ ntiderivative because μ and g are continuous and Equation 2.34 follows from Equation 2.33. The assumption that there is at least one solution of Equation 2.30 is verifiable by ­substituting Equation 2.34 into Equation 2.30. The initial condition, Equation 2.31, determines the integration constant c uniquely. Sometimes, nonlinear equations can be reduced to linear ones by a substitution. One example where such a substitution is helpful is in solving the Bernoulli equations. The form of the Bernoulli equations is y′ + p( x ) y = q( x ) y n (2.35) v ( x ) = y1− n ( x ) (2.36) and if n ≠ 0, 1 then reduces Equation 2.35 to a linear equation. 20 Applied Mathematical Methods for Chemical Engineers Example 2.4 Solve: x2y ′ + 2xy − y 3 = 0 Solution By comparing with Equation 2.35, n = 3, that is, x 2 y′ + 2 xy = y3 or y′ + Let v = y1− 3 = y −2 = 2 1 y = 2 y3 x x (2.37) 1 . y2 dv −2 dy = dx y3 dx Solving for dy to get dx dy − y3 dv 1 dv = = − v −3/ 2 dx 2 dx 2 dx Substituting for dy and y in Equation 2.37 gives dx 1 dv 2 −1/ 2 1 −3/ 2 − v −3/ 2 + v = 2v 2 dx x x Following simplification, the differential equation becomes dv 4 2 − v=− 2 dx x x (2.38) Equation 2.38 is a linear first-order differential equation with a new dependent variable, v(x). Equation 2.38 can now be solved using the method of Section 2.1. A more engineering-type example is demonstrated in Example 2.5. Example 2.5 Suppose that in a certain autocatalytic chemical reaction a compound A reacts to form a compound B. Further, suppose that the initial concentration of A is CA0 and that CB(t) is the concentration of B at time t. Then, CA0 – CB(t) is the concentration of A at time t. Determine CB(t) if CB(0) = CB0. 21 First-Order Ordinary Differential Equations Solution Note that in an autocatalytic reaction the substance produced stimulates the reaction; dCB is proportional to both CB(t) and CA0 − CB(t), that is, thus, the reaction rate dt dCB (t ) = kCB (t )(CA0 − CB (t )) dt (2.39) subject to CB (0) = CB0 where k is the reaction rate coefficient. Equation 2.39 can be restated as dC B − kCBCA0 = − kCB2 dt By comparing with Equation 2.35, n = 2. 1. Let v (t ) = CB1− 2 = CB Then, dCB dv dv = −CB2 = − v −2 dt dt dt Substitute dCB and v(t) into Equation 2.40 to give dt − v −2 dv − kv −1CA0 = − kv −2 dt Then, multiplying both sides of the resulting equation by −v2 gives dv + kCA0 v = k dt µ(t ) = exp ( ∫ kC dt ) A0 Then, [ v exp ( kCA0t )]′ = k exp( kCA0t ) (2.40) 22 Applied Mathematical Methods for Chemical Engineers Integrating both sides with respect to t gives v exp ( kCA0 t ) = 1 exp ( kCA0 t ) + m1 CA0 where m1 is an arbitrary constant. Therefore, υ= 1 1 + m1 exp (− kCA0 t ) = CA0 CB such that CB (t ) = 1 1 / CA0 + m1 exp(− kCA0 t ) with m1 = 2.3 CA0 − CB0 CA0CB0 NONLINEAR EQUATIONS For those first-order equations that cannot be expressed in polynomial form, there is no single analytical method to produce a solution, as seen in Section 2.1. This difficulty increases the importance of the issues of existence and uniqueness of a solution. For a very lucid discussion on the existence and uniqueness theorem for nonlinear first-order differential equations, many excellent texts are available [1,2]. In this section, a few standard methods are presented for use on those first-order nonlinear differential equations that can be solved analytically. Even though the form dρ = f (t , ρ) dt (2.41) is common, it is sometimes more convenient to rewrite Equation 2.41 in an alternate form: M (t , ρ) + N (t , ρ) dρ =0 dt (2.42) 2.3.1 SEPARABLE EQUAtIONs Suppose M is a function of t only, and N is a function of ρ only; then Equation 2.42 becomes M (t ) + N (ρ) dρ =0 dt (2.43) First-Order Ordinary Differential Equations 23 which can be written as N (ρ)dρ = − M (t ) dt (2.44) Whenever a first-order differential equation can be written in either of the forms of Equation 2.43 or Equation 2.44, the equation is said to be separable. We reconsider Equation 2.39 dCB = kCB (CA0 − CB ) dt subject to CB (0) = CB0 Then, this differential equation is separable and results in dC B = k dt CB (CA0 − CB ) (2.45) To solve Equation 2.45, the left-hand side must first be simplified. Consider now the fraction 1 CB (CA0 − CB ) = γ α + CB CA0 − CB where α and γ are constants to be determined. Then, α (CA0 − CB ) + γCB = 1 If we put CB = 0: αCA0 = 1 then, α= 1 CA0 If we put CB = CA0: γCA0 = 1 then, γ= 1 CA0 Equation 2.46 can now be expressed as 1 CB (CA0 1 1 CA0 CA0 = + − CB ) CB CA0 − CB (2.46) 24 Applied Mathematical Methods for Chemical Engineers and Equation 2.45 becomes 1 CA0 1 1 C + C − C dCB = kdt A0 B B (2.47) which integrates to 1 CB CA 0 = m1 exp(kt ) C − C B A0 where m1 is an arbitrary constant to be determined with the given initial condition. At t = 0, CB = CB0; then, 1 CB0 CA 0 = m1 C − C B A0 CB (t ) = (CA0 CB0 CA0 − CB0 ) exp(− kCA0 t ) + CB0 following simplification. As evidenced in this later illustration, the potential difficulty in applying this separation of variable technique lies in one’s ability to carry out the resulting integration that may arise, such as in Equation 2.45. Another nonlinear problem that is not of the variable-separable type may be solvable if it is exact or can be made exact by use of an appropriate factor. 2.3.2 EXAct EQUAtIONs Suppose Equation 2.42 is given; then, if a function w(t, ρ ) exists such that ∂ w(t , ρ) = M (t , ρ), ∂t ∂ w(t , ρ) = N (t , ρ) ∂ρ (2.48) and such that w(t, ρ ) = constant defines ρ = φ (t) implicitly as a differentiable function of t [1,3], then M (t , ρ) + N (t , ρ) dρ ∂ w ∂ w dρ d = + = w(t , ρ(t )) dt ∂t ∂ρ dt dt (2.49) By comparing Equation 2.49 with Equation 2.42, we get d w(t , ρ(t )) = 0 dt (2.50) 25 First-Order Ordinary Differential Equations When Equations 2.48 through 2.50 hold, Equation 2.50 is called an exact differential equation. To determine whether an equation is exact in a given region R, the following criteria are essential: 1. The functions M, N, ∂ M , and ∂ N must be continuous in the given region. ∂t ∂ρ ∂M ∂N must hold at each point in the region. 2. = ∂ρ ∂t 3.The region R must be simply connected, that is, a single closed curve that does not cross itself or a region without holes. Sometimes, criterion 2 is not immediately satisfied, and an adjustment can be made that will remedy such occurrences. Whenever such an adjustment is possible, the differential equation, Equation 2.42, will become exact. To determine this adjustment, consider ∂ ∂ (µM ) = (µN ) ∂t ∂ρ (2.51) where μ is the adjustment to be determined and can be a function of both t and ρ . Then, ∂µ ∂ M ∂ N 1 ∂µ −M N = − µ ∂t ∂ρ ∂ρ ∂t (2.52) Equation 2.52 is not easy to solve in its present form; however, if either μ = μ(t) or μ = μ(ρ ) then Equation 2.52 simplifies to 1 dµ 1 ∂ M ∂ N = − µ dt N ∂ρ ∂t (2.53) 1 dµ 1 ∂ N ∂ M = − µ dρ M ∂t ∂ρ (2.54) or Either Equation 2.53 or Equation 2.54 gives a formula to determine ρ μ. If μ = μ(t, ρ ), Equation 2.52 must be solved directly. Equation 2.39 can be solved by first finding μ = μ(CB) and multiplying both sides with μ to get an exact differential equation: µ(CB ) dCB − kµ(CB )CB (CA0 − CB ) = 0 dt (2.55) 26 Applied Mathematical Methods for Chemical Engineers Homogeneous equations comprise a third group of non linear-type problems that usually do not yield to either the variable-separable or exact solution techniques. An equation of this type, however, may yield a solution if a new variable can be introduced. 2.3.3 HOMOGENEOUs EQUAtIONs Whenever Equation 2.41 can be rewritten in the form dρ = h(ρ / t ) dt (2.56) then Equation 2.56 is said to be homogeneous. The quantity ρ /t can now be treated as a new variable, and one of the solution techniques of the previous sections may now be applicable. So far, the examples have assumed that the differential equations are given. However, as chemical engineers, we know that more often than not the main problem is in the derivation of the differential equation and the associated conditions. To address that aspect of mathematical methods in this chapter, a problem setup section follows. 2.4 PROBLEM SETUP The traditional approach of outlining the theory and presenting some supporting examples has been followed up to now. However, a needed deviation from tradition is a “how to” or a problem setup section. This section is included to demonstrate one approach to formulating a physically applicable first-order ordinary differential equation. 2.4.1 PROBLEM StAtEMENt Consider the continuous extraction of benzoic acid from a mixture of benzoic acid and toluene, using water as the extracting solvent [4]. Both streams (acidic mixture and water) are fed into a tank where they are stirred efficiently, and the mixture is then pumped into a second tank where it is allowed to settle into two layers. The upper organic phase and the lower aqueous phase are removed separately, and the problem is to determine what proportion of the acid has passed into the solvent phase. A list of simplifications for the idealized problem (model) follows: 1.Combine the two tanks into a single stage (Figure 2.1). 2.Express stream-flow rates on a solute-free basis. 3.Assume a steady flow rate for each phase. 4.Assume that toluene and water are immiscible. 5.Assume that feed concentration is constant. 6.Assume that the mixing is efficient enough such that the two streams leaving the stage (Figure 2.1) are always in equilibrium with each other and can be expressed as y = mx (2.57) 27 First-Order Ordinary Differential Equations R L/min, toluene CA0 kg/L acid V1 S L/min, water x x kg/L, acid V2 , y y kg/L, benzoic acid FIGURE 2.1 R L/min, toluene S L/min, water Equilibrium stage. where m is the distribution coefficient, x is the concentration of benzoic acid leaving the stage in the organic phase, and y is the aqueous phase benzoic mass concentration. 7. Assume that the composition of a stream leaving the stage is the same composition as that phase in the stage. 8.Assume that the stage initially contains V1 L of toluene, V2 L of water, and no benzoic acid. Then, using Equation 2.11, that is, Rate of accumulation = rate of input − rate of output The quantities for any time t can be derived. A helpful procedure [4] is to ­tabulate the quantities for any time t and for a small change in time ∆t. A typical table, Table 2.1, is given below. Then, during a time interval ∆t, input of acid = RCA0∆t while output of acid = dx dy dx dy R x + ∆t ∆t + S y + ∆t ∆t and accumulation of acid = V1 ∆t + V2 ∆t. dt dt dt dt Therefore, dy dx dy dx + V2 ∆t = RCA0 ∆t − R x + ∆t + S y + ∆t ∆t V1 dt dt dt dt which simplifies to V1 dx dy dx dy + V2 = RCA0 − R x + ∆t − S y + ∆t dt dt dt dt Then, dx dy dx dy lim V1 + V2 = lim RCA0 − R x + ∆t − S y + ∆t ∆ t → 0 dt dt dt dt ∆t → 0 gives V1 dx dy + V2 = RCA0 − Rx − Sy dt dt (2.58) 28 Applied Mathematical Methods for Chemical Engineers TABLE 2.1 Quantities for t and Δt t System Property t + Δt Flow rate of organic phase Flow rate of aqueous phase Volume of organic phase in stage Volume of aqueous phase in stage Input acid concentration in organic phase R S V1 V2 CA0 R S V1 V2 CA0 Output acid concentration in organic phase x x+ dx ∆t dt Output acid concentration in aqueous phase y y+ dy ∆t dt Amount (mass) of acid in organic phase V1x V1 x + V1 dx ∆t dt Amount (mass) of acid in aqueous phase V2y V2 y + V2 dy ∆t dt Input acid concentration in aqueous phase 0 0 Source: Jenson, V.G. and Jeffreys, G.V., Mathematical Methods in Chemical Engineering, Academic Press, London, United Kingdom, 1963. Equation 2.58 reduces to V1 dx dx + V2 m = RCA0 − Rx − Smx dt dt or (V1 + mV2 ) dx = RCA0 − ( R + Sm) x dt (2.59) a linear first-order ordinary differential equation subject to the initial condition (state) of the system (assumption 8) x = 0 when t = 0 (2.60) The solution of Equations 2.59 and 2.60 will give the organic phase acid concentration profile as a function of time. Both the organic and the aqueous phase acid ­concentration profiles can be used to forecast the behavior of a single stage liquid– liquid e­ xtraction unit during start-up. In the derivation of Equation 2.59, eight assumptions were listed. No general rule governs the number of assumptions that will result in a perfect model. However, a balance between too many and too few assumptions must be found if a workable First-Order Ordinary Differential Equations 29 solution is to be expected. If too many assumptions are made, the result will be in gross error, while too few assumptions can result in a mathematical problem that is not tractable. The earlier format may be modified according to the situation under consideration. The reader should keep in mind that there is no single way to set up problems. However, there are some key items to pay close attention to: • Always check for the involvement of some physical law or principle (mass, momentum, or energy balances). • Include a consistent set of units (same units in each term). • Appropriately include given conditions (initial or boundary). • Check the mode of operation (steady or unsteady process). Below are some examples (Examples 2.6 through 2.9) for which the straightforward application of mass or energy balance is sufficient to set up the differential equation. Example 2.6 Transient Behavior of an Air-Cooling System Consider an engine that generates heat at a rate of 8530 Btu/min [5]. Suppose this engine is cooled with air, and the air in the engine housing is circulated rapidly enough so that the air temperature can be assumed uniform and is the same as that of the outlet air. The air is fed to the housing at 6.0 lb·mol/min and 65°F. Also, an average of 0.20 lb·mol of air is contained within the engine housing and its temperature variation can be neglected. If heat is lost from the housing to its surroundings at a rate of Ql (Btu/min) = 33.0 (T − 65ºF) and the engine is started with the inside air temperature equal to 65ºF, 1. Derive a differential equation for the variation of the outlet temperature with time. 2. Calculate the steady-state air temperature if the engine runs continuously for an indefinite period of time, using Cv = 5.00 Btu/lb·molºF. Solution The unsteady-state balance equation for this system (air within the engine housing) is the first law of thermodynamics for open systems with changes in kinetic and potential energies neglected. Also, the temperature and composition of the system contents are assumed independent of position and no phase changes occur. This gives the equation MCv dTsys = mCp (Tm − Tsys ) + Q + Ws dt where Ws is the rate of transfer of shaft work, Q is the rate of heat transfer, M is the mass (or number of moles) of the system contents, and m is the mass flow rate. Taking Cp = Cv + R = 6.99 Btu/lb·mol°F 30 Applied Mathematical Methods for Chemical Engineers then mCp = (6.0 lb·mol/ min) (6.99 Btu/lb·mol°F) = 41.9 Btu min°F For this problem, Ws = 0, as there are no moving parts and Q = Qgen – Ql. Therefore, the differential equation becomes (0.2) (5.0) dT = 41.9(65 − T ) + 8530 − 33(T − 65) dt which reduces to dT = −74.9T + 13398 dt subject to t = 0, T = 65°F The solution to the differential equation and initial condition is T (t ) = 179 − 112e −74.9t The steady-state temperature is lim T (t ) = 179°F t →∞ Example 2.7 Comminution Operation Suppose a copper ore is fed to a ball mill at a steady rate Q (kg/h), and crushed ore is withdrawn at the same rate. At an initial time (t = 0), the total mass of ore in the mill is M (kg). Further suppose that a series of 10 particle-size ranges is defined and xi represents the mass fraction of particles in the ith size range, where i = 1 is the largest size range and i = 10 is the smallest. The rate at which particles are broken out of the ith size range is ri = ki mi where mi is the mass of particles in this size range. Further suppose that of the particles broken out of size range i in a differential time interval, a fraction bij go into size range j. Assume that the size distribution of particles is uniform throughout the mill and equals that of the product, and let xif be the mass fraction of the feed that falls in the ith size range. Show that a mass balance on the jth size fraction in the tank yields dx j dt j −1 = Q / M x jf − x j − k j x j + ∑ ki xi bij i =1 First-Order Ordinary Differential Equations Solution Accumulation = input + generation − output − consumption dx j d ( Mx j ) = M Accumulation = dt dt Input = Qxif Generation: The rate at which particles enter the jth size fraction from the ith size fraction by breakage is kimibif. Output = Qxj Consumption = kjmj Therefore, the balanced equation becomes M 10 dx j = Qx jf − Qx j − k j m j + ∑ ki mi bij dt i =1 Note that bij = 0 for j ≤ i (particle size cannot increase, and breakage within a size range i = j does not count as an event); then, dividing by M gives j −1 dx j Q = ( x jf − x j ) + ∑ ki xi bij − k j x j dt M i =1 subject to t = 0, xj = xj0 (jth-size fraction of initial contents of the mill) [5]. Example 2.8 Semi-Batch Reacting System A liquid-phase chemical reaction with the stoichiometry A → B takes place in a semibatch reactor [5]. The rate of consumption of A per unit volume of the reactor is given by the first-order rate expression rA (mol/L ⋅ s) = kC A where CA (mol/L) is the reactant concentration. The tank is initially empty. At time t = 0, a solution containing A at a concentration CA0 (mol/L) is fed to the tank at a steady rate φ (L/s). Develop differential balances on the total volume of the tank contents, V, and on the moles of A in the tank, nA. Solution Total volume balance: accumulation = input dV =φ dt t = 0, V = 0 31 32 Applied Mathematical Methods for Chemical Engineers Moles of A balance: accumulation = input − consumption CA = nA V dnA = CA 0φ − knA dt t = 0, nA = 0 Example 2.9 Water containing 2 oz. of pollutant per gallon flows through a treatment tank at a rate of 500 gal/min [7]. In the tank, the treatment removes 2% of the pollutant per minute and the water is thoroughly stirred. The tank holds 10,000 gal of water. On the day the treatment plant opens, the tank is filled with pure water. Determine the concentration profile of the tank effluent. Solution Let P(t) be the amount of pollutant in the tank at any time t. Then, dP = input − output dt oz gal oz gal = 2 500 − P 500 gal min 10,000 gal min input effluent oz − 0.02 P min treatment subject to P = 0 at t = 0. Therefore, P(t ) = (100,000 /7)(1− e −0.07t ) 2.5 PROBLEMS 1.a.Derive the organic phase benzoic acid concentration profile as a function of time (Section 2.4). b. Determine the steady-state solution to (V1 + mV2 ) dx = RCA0 − ( R + Sm) x dt First-Order Ordinary Differential Equations 33 i.Directly. ii. By taking the limit as t → ∞. c.If E is the proportion of benzoic acid extracted and α = R/ms, what is the relationship between E and α for the steady-state process? 2.Consider the two-stage solvent extraction of benzoic acid with the previously made assumptions and yi = mxi , i = 1,2 where i denotes the stage [4]. a. Develop a table similar to Table 2.1. b. Use your table to write the time-dependent mass balance of acid for each stage. c. Find the steady-state organic (x2) and aqueous (y1) acid concentration profiles. d.If E is the proportion of acid extracted and α = R/ms, what is the relationship between E and α for the steady-state process? 3.Give suitable initial conditions for the two first-order differential equations and outline a solution. 4.Phosgene (COCl2) is formed by reacting CO and Cl2 in the presence of activated charcoal [5]: CO 2 + Cl 2 → COCl 2 At a temperature of 303.8 K in the presence of 1 g of charcoal, the rate of f­ ormation of phosgene is R f (mol/min) = 8.75[CO][Cl 2] (1 + 58.6[Cl 2 ] + 34.3[COCl 2 ])2 where “[ ]” denotes concentration (mol/L). a. Given that the input to a 3.00-L batch reactor is 1.00 g of charcoal and a gas containing 60 mol% CO and 40 mol% Cl2 and that the initial reactor conditions are 303.8 K and 1 atm, determine the initial concentrations (in moles per liter) of both reactants. The volume occupied by the charcoal may be neglected. b. Write a differential balance on phosgene and show that it simplifies to d[COCl 2 ] 2.92(0.02407 − [COCl 2 ])(0.01605 − [COCl 2 ]) = dt (1.941 − 24.3[COCl 2 ])2 5.A gas that contains CO2 is contacted with liquid water in an agitated batch absorber [5]. Henry’s law gives the equilibrium solubility of CO2 in water CA = pA /H A 34 Applied Mathematical Methods for Chemical Engineers where CA (mol/L) is the CO2 concentration in solution, pA (atm) is the partial pressure of CO2 in the gas phase and HA is Henry’s Law constant. The rate of transfer of CO2 from the gas to the liquid per unit area of gas–liquid interface is given by = k ( CA* − CA ) ra (mol/cm 2 s) where CA* is the concentration of CO2 that would be in equilibrium with the CO2 in the gas phase (CA* = PA/ H A). Suppose the gas-phase total pressure is P (atm) and contains yA mol fraction of CO2, and the liquid phase initially has V (cm3) of pure water with the agitation of the liquid phase sufficient to neglect spatial composition dependency, and if the amount of absorbed CO2 is low enough for P, V, and yA to be assumed constant, write a differential balance on CO2 in the liquid phase and solve the differential equation to show that CA (t ) = CA* [1 − exp(− kSt /V )] 6. An iron bar 2 cm × 3 cm × 10 cm at a temperature of 95 ºC is dropped into a barrel of water at 25 ºC. The barrel is large enough that the water temperature rises negligibly as the bar cools. The rate at which heat is transferred from the bar to the water is given by Q( J /min) = UA(Tb − TW ) where U (= 0.050 J/min·cm2·ºC) is a heat-transfer coefficient, A (cm2) is the exposed area of the bar, Tb is the surface temperature of the bar, and TW is the water temperature. Given that the heat capacity of the bar is 0.460 J/g ºC, and heat conduction in iron is fast enough to assume that the temperature T b(t) is uniform throughout the bar, write an energy balance on the bar and determine the steady-state temperature of the bar. Also, calculate the time required for the bar to cool to 30 ºC. 7.A steam coil is immersed in a stirred heating tank [5]. Saturated steam at 7.50 bar condenses within the coil, and the condensate emerges at its saturation temperature. A solvent with a heat capacity of 2.30 kJ/kg·ºC is fed to the tank at a steady rate of 12.0 kg/min and a temperature of 25 ºC, and the heated solvent is discharged at the same rate. The tank is initially filled with 760 kg of solvent at 25 ºC, at which point the flows of both steam and solvent commence. The rate at which heat is transferred from the coil to the solvent is given by Q(kJ/ min) = UA(Tsteam − T ) where UA = 11.5 kJ/min ºC. The tank is stirred well such that the temperature can be considered as spatially uniform and is the same as the outlet temperature. Derive a differential energy balance on the tank contents, and calculate the time required to heat the solvent to an arbitrary temperature Tf (ºC). First-Order Ordinary Differential Equations 35 REFERENCES 1.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 3rd ed., John Wiley & Sons, New York, 1977. 2. Giordano, F.R. and Weir, M.D. Differential Equations, A Modeling Approach, AddisonWesley, New York, 1991. 3.Thomas, G.B. and Finney, R.L. Calculus and Analytic Geometry, 6th ed., AddisonWesley, Reading, MA, 1984. 4.Jenson, V.G. and Jeffreys, G.V. Mathematical Methods in Chemical Engineering, Academic Press, London, UK, 1963. 5.Felder, R.M. and Rousseau, R.W. Elementary Principles of Chemical Processes, 2nd ed., John Wiley & Sons, New York, 1976. 6. Rice, R.G. and Do, D.D. Applied Mathematics and Modeling for Chemical Engineers, John Wiley & Sons, New York, 1995. 7.West, B.H. Setting up differential equations from word problems, In Modules in Applied Mathematics, Vol. 1, Differential Equations, Lucas, W.F., ed., Springer-Verlag, New York, 1983, Chap. 1. This page intentionally left blank 3 Linear Second-Order and Systems of FirstOrder Ordinary Differential Equations 3.1 INTRODUCTION Chemical engineers working in the area of transport phenomena must frequently solve problems that involve linear second-order differential equations. These may occur as boundary value problems in diffusional systems or initial value problems in process control or reacting systems, but most frequently, they are the result of reduction of partial differential equations. In this chapter, linear second-order ordinary differential equations will be reviewed. There will be examples on applications; however, Chapters 6 and 7 emphasize both mathematical and engineering applications. Herein, the meaning of linearity is the same as given in Chapter 2. Attention is now focused on equations that can be written as follows: ζ = f (t , ζ, ζ′) (3.1) Previously, it was observed that in the case of first-order equations, the integrated solutions contained one arbitrary constant. This constant could be determined by a given condition at an initial time. In the case of second-order equations, two constants of integration are expected to occur, and therefore two conditions at an initial time will be needed for the so-called initial value problems, or two conditions at separate locations for the so-called boundary value problems [1,4,6]. In the discussion of the theory of second-order linear ordinary differential equations, the standard mathematical symbols are employed. In chemical engineering applications, the usual chemical engineering nomenclature is used whenever there is no conflict. The general second-order linear differential equation is of the form P( x ) d2 y dy + Q ( x ) + R( x ) y = G ( x ) 2 dx dx (3.2) where P, Q, R, and G are the given functions. Three important examples of second-order linear differential equations that ­frequently occur in chemical engineering are Legendre’s equation [3,4] of order α: (1 − x 2 ) y ′′ − 2 xy ′ + α (α + 1) y = 0 (3.3) 37 38 Applied Mathematical Methods for Chemical Engineers Bessel’s equation [6,9,10,11] of order n: x 2 y′′ + xy′ + ( x 2 − n 2 ) y = 0 (3.4) and the confluent hypergeometric equation [5,7,8,12]: x d2 y dy + (c − x ) − ay = 0 2 dx dx (3.5) where the quantities a, n, and c are constants. In the discussion to follow, unless otherwise stated, the functions P, Q, R, and G in Equation 3.2 are taken to be continuous on some interval a < x < b (a may be −∞ and b may be +∞). If the function P(x) ≠ 0 everywhere on the interval, then Equation 3.2 can be rewritten as d2 y dy (3.6) + p( x ) + q( x ) y = g( x ) 2 dx dx following division by P(x). Similar to first-order equations, the issue of existence and uniqueness of a solution to second-order equations must be dealt with. Theorem 3.1 [1,4] addresses the existence and uniqueness of solutions of second-order differential equations. Theorem 3.1 If the functions p, q, and g are continuous on the open interval a < x < b, then there exists one and only one function y = φ(x) satisfying the differential Equation 3.6, y′′ + p( x ) y′ + q( x ) y = g( x ) on the entire interval a < x < b and the given initial conditions, y( x 0 ) = y0 , y′( x 0 ) = y′0 at a particular point x0 in the interval. It is important to note that this theorem does not address the issue of existence and uniqueness of solutions for boundary value problems. Boundary value problems are discussed in Chapter 4. The following three elementary examples serve to demonstrate how the earlier interval can be determined for a unique solution to exist. Example 3.1 Given: xy′′ + 3 y = x , y( x 0 ) = y0 ; y′( x 0 ) = y0′ rewrite as y′′ + 3 y =1 x then the interval consists of all points not including the origin. 39 Linear Second-Order and Systems of First-Order Example 3.2 Given: y′′ + by′ + 7 y = 2sin x; y( x 0 ) = y0 ; y′( x 0 ) = y0′ Since the differential equation is defined on −∞ < x < + ∞, the whole real line is the interval. Example 3.3 Given: x ( x − 1) y′′ + 3 ξx y′ + 4 y = 2, y( x 0 ) = y0 ; y′( x 0 )′ = y0′ rewritten as 3x 4 2 y′′ + y′ + y= x ( x − 1) x ( x − 1) x ( x − 1) This differential equation is defined everywhere except at 0 and 1. Thus, the interval is any interval excluding the points 0 and 1. Essentially, to determine the interval or domain of definition of the differential equation, one elementary procedure is to look for points where division by zero will occur or points where the equation becomes unbounded. These so-called singular points will be classified later, but for now, these are to be excluded from the interval over which Theorem 3.1 is applied. To discuss the methods used to solve linear second-order differential equations, it is necessary to reintroduce the term homogeneous, or complementary, but with a meaning unrelated to previous usage in this book. When the forcing function g(x) is set to zero in Equation 3.6, the reduced equation d2y dy + p( x ) + q( x ) y = 0 dx 2 dx (3.7) results. This is the homogeneous form of the second-order linear differential equation. Any equation of second-order or higher with right-hand side identically zero is termed homogeneous as opposed to nonhomogeneous. 3.2 FUNDAMENTAL SOLUTIONS OF HOMOGENEOUS EQUATIONS Suppose p and q in Equation 3.7 are continuous on a < x < b then for any twicedifferentiable function φ on a < x < b, the linear differential operator L is defined to mean L[φ] = φ′′ + pφ′ + qφ (3.8) L ≡ D 2 + pD + q (3.9) for where D is the derivative operator. Then Equation 3.7 becomes L( y) = y′′ + p( x ) y′ + q( x ) y = 0 (3.10) 40 Applied Mathematical Methods for Chemical Engineers The use of the operator L reduces the task of integrating the linear second-order differential equation as given below. However, before the employing the integration process, a few more definitions are needed. Theorem 3.2 If y = y1(x) and y = y2(x) are solutions of the differential Equation 3.10, L[ y] = y′′ + p( x ) y′ + q( x ) y = 0 then the linear combination of y = c1 y1(x) + c2 y2(x), with c1 and c2 being arbitrary constants is also a solution. Proof Since y = y1(x) is a solution set of Equation 3.10, then L[y1] = y1′′+ p( x ) y1′ + q(x)y1 = 0, and since y = y2(x) is also a solution set of Equation 3.10, then L[y2] = y′′2 + y′2 + q(x)y2 = 0. But L[c1 y1 + c2 y2 ] = L[c1 y1 ] + L[c2 y2 ] = c1 L[ y1 ] + c2 L[ y2 ] =0 Theorem 3.2 is a statement of the superposition principle [1,2,4], which is also applicable to higher-order linear differential equations. The two solutions y1 and y2 form what is called a fundamental set of solutions for Equation 3.10. In general, two solutions y1 and y2 of Equation 3.10 are said to form a fundamental set of solutions if every solution of Equation 3.10 can be expressed as a linear combination of y1 and y2. In particular; Theorem 3.3 If the functions p and q are continuous on the interval a < x < b and if y1 and y2 are solutions of the differential Equation 3.10, L [ y] = y ′′ + p( x ) y ′ + q( x ) y = 0 satisfying the condition y1 ( x ) y2′ ( x ) − y1′ ( x ) y2 ≠ 0 (3.11) at every point in a < x < b, then any solution of Equation 3.10 on the interval a < x < b can be expressed as a linear combination of y1 and y2. The condition described by Equation 3.11 is called the Wronskian and is commonly written in the determinant form as [1] W ( y1 , y2 ) = y1 y2 y1′ y2′ (3.12) Linear Second-Order and Systems of First-Order 41 The linear combination mentioned in Theorem 3.3 is usually called the general, complementary, or homogeneous solution of Equation 3.10. The following theorem should help to clarify when Equation 3.11 or 3.12 is expected to be different from zero. Theorem 3.4 If the function p and q are continuous on a < x < b and the function y1 and y2 are linearly independent solutions of the differential Equation 3.10, L[ y] = y ′′ + p( x ) y ′ + q( x ) y = 0 then the W(y1, y2) is nonzero on a < x < b, and hence any solution of Equation 3.10 can be expressed as a linear combination of y1 and y2. Linear independence is defined as two or more functions on the interval a < x < b that are not linearly dependent [1,2,4]. For example, if f and g are two functions on a < x < b, such that c1 f ( x ) + c2 g( x ) = 0 for all x in a < x < b is a true statement for c1 and c2 not both zero, then f and g are linearly dependent; otherwise they are linearly independent. In other words, if f and g are a constant multiple of each other on the interval, then they are linearly dependent; otherwise, they are linearly independent. Finally, we come to the question “how to” find solution for linear secondorder differential equations. Specifically, the cases where p and q are constants in Equation 3.10. 3.3 HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS The most general constant coefficient, linear, second-order, ordinary, homogeneous differential equation is ay′′ + by′ + cy = 0 (3.13) where a, b, and c are real constants and a ≠ 0. Equation 3.13 in operator notation is L[ y] = ay ′′ + by ′ + cy = (aD 2 + bD + c) y = 0 (3.14) Now, we are interested in integrating Equation 3.14 and expressing our findings as y = φ(x) and, in this case, include the two integration constants as implied by the ­presence of the second derivative term. To solve Equation 3.14, we seek a function φ(x) such that a times its second derivative added to b times its first derivative added to c times the function itself results in zero. Among all possible candidate functions, the function φ(x) = erx, where r is a constant, turns out to be the best candidate. Thus L[erx ] = a(erx )′′ + b(erx )′ + c(erx ) = erx (ar 2 + br + c) = 0 (3.15) 42 Applied Mathematical Methods for Chemical Engineers Since erx is never zero, then ar 2 + br + c = 0 (3.16) Equation 3.16 is called the characteristic or auxiliary equation and the resulting roots r1 and r2 are called the characteristic roots or the eigenvalues. The roots r1 and r2 are given by r1 = − b + b 2 − 4 ac − b − b 2 − 4 ac , r2 = 2a 2a (3.17) If the discriminant b2 − 4ac > 0, then r1 and r2 are real and unequal. Then y1 ( x ) = er1 x , y2 ( x ) = er2 x (3.18) and Theorem 3.2, the superposition principle, tells us that the linear combination y = c1er1 x + c2 er2 x (3.19) with c1 and c2 being arbitrary (integration) constants is also a solution. It is easy to check that Theorem 3.3 and Theorem 3.4 both hold; that is, er1x and er2 x are linearly independent functions and thus form a fundamental set. If the discriminant b2 − 4ac = 0, then r1 and r2 are identical and there only results one solution, namely e−(b/2a)x, that is, y1 = e − ( b / 2 a ) x (3.20) To find y2, consider the following procedure. Let y = υ ( x )e−( b/ 2 a ) x (3.21) then y′ = υ ′( x )e−( b/ 2 a ) x − b υ ( x )e−( b/ 2 a ) x 2a b b2 y′′ = υ ′′( x ) − υ′( x ) + 2 υ ( x ) e−( b / 2 a ) x a 4a (3.22) (3.23) Substitute Equations 3.21 and 3.23 into Equation 3.13 to get b b2 b a υ ′′ − υ′ + 2 υ + b υ′ − υ + c υ = 0 a 4a 2a (3.24) and simplify further to get υ ′′ = 0 (3.25) υ ( x ) = (c1 x + c2 ) (3.26) Integrating Equation 3.25 twice gives Linear Second-Order and Systems of First-Order 43 Then Equation 3.21 becomes y = (c1 x + c2 )e − ( b / 2 a ) x (3.27) y2 = xe − ( b / 2 a ) x (3.28) Therefore, and Equation 3.27 is the general solution when r1 = r2, that is, y = c1 xe(r1 x ) + c2 e(r1 x ) (3.29) If the discriminant b – 4ac < 0 then r1 and r2 are complex numbers. Furthermore, since a, b, and c are real, then r1 and r2 will be a conjugate pair. That is, 2 r1 = − b i b 2 − 4 ac + = λ + iµ 2a 2a (3.30) r2 = − b i b 2 − 4 ac − = λ − iµ 2a 2a (3.31) where the real numbers λ and μ are introduced for convenience. Similar to the results obtained in Equation 3.19, the general solution is y = k1e( λ+ iµ ) x + k2 e( λ− iµ ) x (3.32) where k1 and k2 are arbitrary constants. Equation 3.32 can be rewritten as y = [ k1e iµx + k 2 e − iµx ]e λx = [ k1 (cos µx + i sin µx ) + k2 (cos µx − i sin µx )]e λx = [( k1 + k2 ) cos µx + i( k1 − k2 ) sin µx ]e λx (3.33) = [c1 cos µx + c 2 sin µx ]e λx where c1 and c2 are arbitrary constants since the sum or difference of two arbitrary constants is still an arbitrary constant. In going from Equation 3.32 to Equation 3.33, the following concept was employed to define eiμx and e−iμx: The function eμx has a Taylor series expansion of eµx = 1 + µx + (µx )2 (µx )3 (µx )4 (µx )5 + + + + 2! 3! 4! 5! about the origin. Then (iµx )2 (iµx )3 (iµx )4 (iµx )5 + + + + 2! 3! 4! 5! (µx )3 (µx )5 (µx )2 (µx )4 = 1 − + − + i µx − + − 2! 4! 3! 5! e iµx = 1 + iµx + 44 Applied Mathematical Methods for Chemical Engineers It is also known that the functions cos μx and sin μx each have Taylor series expansion of cos µx = 1 − (µx )2 (µx )4 + − 2! 4! sin µx = µx − (µx )3 (µx )4 + − 3! 5! and about the origin. Therefore eiμx is defined as e iµx = cos µx + i sin µx (3.34) e − iµx = cos µx − i sin µx (3.35) and Equations 3.19, 3.29, and 3.33 give the general solutions of the second order, constant coefficient, homogeneous, and linear differential equation for the respective cases of real unequal, repeated, and complex characteristic roots. However, the actual steps that are used in deriving a solution to the homogeneous problem are as follows: Step 1: Put differential equation into operator form. Step 2: Identify the characteristic equation and roots. Step 3: Use Equations 3.19, 3.29, or 3.33 to express the general solution. The following examples should clarify the three steps given above. Example 3.4 Given: 2y″ − 3y′ + y = 0 Step 1: Operator form: (2D2 − 3D + 1)y = 0 Step 2: Characteristic equation: (2r2 − 3r + 1) = 0 Characteristic roots: r = 1, 1/2 Step 3: Since the roots are real and distinct, Equation 3.19 gives y = c1e x + c2 e1/2 x Example 3.5 Given: y″ + 2y′ + y = 0 Step 1: Operator form: (D2 + 2D + 1)y = 0 Linear Second-Order and Systems of First-Order 45 Step 2: Characteristic equation: r2 + 2r + 1 = 0 Characteristic roots: r = −1, −1 Step 3: Since roots are repeated, Equation 3.29 gives y = c1 xe − x + c2 e − x Example 3.6 Given: y″ + 6y′ + 13y = 0 Step 1: Operator form: (D2 + 6D + 13)y = 0 Step 2: Characteristic equation: r2 + 6r + 13 = 0 Characteristic roots: r = −3 + 2i, −3 − 2i Step 3: Since roots are complex, Equation 3.33 gives y = [c1 cos 2 x + c2 sin 2 x ]e −3 x where λ = −3 and μ = 2. 3.4 NONHOMOGENEOUS EQUATIONS Previously, some brief theory and a few examples were given that demonstrate the approach necessary to construct solutions to the homogeneous problem. The summarized procedure for the constant coefficient case is very simple, and one would like to maintain that simplicity for the case of at least the nonhomogeneous constant coefficient problem. The nonhomogeneous differential equation can be written as L[ y] = y ′′ + p( x ) y ′ + q( x ) y = g( x ) (3.36) where p, q, and g are continuous on the interval of interest. The following two theorems are needed to establish the procedure for solving Equation 3.36. Theorem 3.5 The difference of any two solutions of the differential Equation 3.36 L[ y] = y ′′ + p( x ) y ′ + q( x ) y = g( x ) 46 Applied Mathematical Methods for Chemical Engineers is a solution of the corresponding homogeneous differential equation L[ y] = y ′′ + p( x ) y ′ + q( x ) y = 0 (3.37) Proof Suppose the two functions u1 and u2 are solutions of Equation 3.36; then L[u1 ] = g and L[u2 ] = g Therefore, L[u1] − L[u2] = 0. But L is a linear operator; thus L[u1] − L[u2] = L[u1 − u2] = 0. Theorem 3.6 Given one solution yp of the nonhomogeneous linear differential Equation 3.36 L[ y] = y ′′ + p( x ) y ′ + q( x ) y = g( x ) then any solution y = f(x) of this equation can be expressed as f ( x ) = y p ( x ) + c1 y1 ( x ) + c2 y2 ( x ) (3.38) where y1 and y2 are linearly independent solutions of the corresponding homogeneous equation. The above equation is the general solution of the nonhomogeneous Equation 3.36. That is, the general solution yg of the nonhomogeneous equation is yg = yc + yp (3.39) where yc is the general solution of the associated homogeneous equation and is termed the complementary solution. The function yp is the particular solution satisfying the nonhomogeneous differential Equation 3.36. A relatively simple procedure for finding the most suitable candidate for yp is demonstrated below for the constant coefficient case. For example, given y ′′ − 3 y ′ − 4 y = 2sin x (3.40) then the associated homogeneous problem is y ′′ − 3 y ′ − 4 y = 0 which can be put in the operator form ( D 2 − 3 D − 4) y = 0 In fact, the nonhomogeneous problem can be put in the operator form as well: ( D 2 − 3 D − 4) y = 2sin x (3.41) Linear Second-Order and Systems of First-Order 47 Then one can see that if the RHS or nonhomogeneous part is differentiated two times, the result is −2 sin x. Further, if the nonhomogeneous part is added to the twice-differentiated result, one gets a final result of zero. That is, d2 (2 sin x ) + 2 sin x = 0 dx 2 or ( D 2 + 1)2 sin x = 0 Then, since the RHS of Equation 3.41 is operated on by D2 + 1, the same operation must be carried out on the LHS, giving ( D 2 + 1)( D 2 − 3 D − 4) y = 0 (3.42) a homogeneous equation of the fourth order. Following steps 2 and 3 of Section 3.3, the characteristic equation is (r 2 + 1) (r 2 − 3r − 4) = 0 with characteristic roots r1 = −1, r2 = 4, r3 = +i, r4 = −i. Then Equation 3.42 and Equation 3.33 give yg = c1e − x + c2 e 4 x + c3 cos x + c 4 sin x (3.43) where yc = c1e − x + c2 e 4 x which comes from the associated homogeneous equation, and therefore yp = c3 cos x + c 4 sin x is the best candidate for the particular solution. The constants c3 and c4 are not arbitrary and are determined by substituting yp into Equation 3.40 yp″ − 3 yp′ − 4 yp = 2 sin x that is, (c3 cos x + c4 sin x )′′ − 3(c3 cos x + c4 sin x )′ −4(c3 cos x + c4 sin x ) = 2sin xs which results in an identity (−c3 cos x − c4 sin x ) − 3(c4 cos x + c3 sin x )′ − 4(c3 cos x + c4 sin x ) = 2sin x Following simplifications and equating coefficients, one gets −(5c3 + 3c4 ) cos x = 0 (3c3 − 5c4 )sin x = 2sin x 48 Applied Mathematical Methods for Chemical Engineers or −5c3 − 3c4 = 0 and 3c3 − 5c4 = 2 giving c4 = −5/17, c3 = 3/17, such that yp = 3 / 17 cos x − 5 / 17sin x Therefore, the problem of finding the general solution to Equation 3.36 is reduced to find a particular solution to Equation 3.36. It should be noted that the method works when the nonhomogeneous part can be differentiated to zero or when an appropriate number of differentiations will bring back some constant multiple of the original function. For example, this method would fail for RHS of the types tan x or xm/n, where the number m/n is not an integer, and for combinations of these. In other words, the method works in theory for g(x) of the type g( x ) = Pn ( x ) = a0 x n + a2 x n−2 + a2 x n−2 + + an = e ax Pn ( x ) = e ax Pn ( x )sin βx = e ax Pn ( x ) cos βx In general, given a linear ordinary differential equation P ( D ) y = R( x ) (3.44) the general solution is y = yc + yp where P(D)yc = 0 and P(D)yp = R(x). Suppose there is a differential operator “Annihilator,” A(D), that is linear with constant coefficients such that A( D) R( x )0 If we operate on both sides of Equation 3.44 with A(D) we would get A( D) P( D) y = A( D) R( x ) = 0 Now, consider this new equation A( D) P( D) y = 0 (3.45) To find the general solution of Equation 3.44, we need the roots of the polynomial A(D)P(D) and they are r1, r2,…, rj; q1, q2,…, qk where the r1, r2,…, rj come from P(D) and q1, q2,…, qk come from A(D). Linear Second-Order and Systems of First-Order 49 The general solution of Equation 3.45 can be written as y = yc + yq But A( D) P( D)[ yc + yp ] = A( D) R( x ) = 0 thus yc + yp is also a solution of Equation 3.45. Since yc + yq is the general solution of Equation 3.45, it contains yc + yp. Therefore yc + yq ⊂ yc + yq or yp ⊂ yq The function yq can be thought of as the best candidate for the particular solution, and the method of undetermined coefficients can be used to find yp. Below are three more elementary examples that demonstrate successful application of the method, and a fourth example in which the method fails. Example 3.7 Find the general solution of y″ − y′ − 6y = 12xex. Solution Operator form: (D2 − D − 6)y = 12xex Determine the operator that will annihilate the nonhomogeneous part: 12xex d (12 xe x ) = 12e x + 12 xe x dx Annihiliate RHS: d2 d (12 xe x ) = (12e x + 12 xe x ) = 24e x + 12 xe x dx 2 dx then the ­combination of d2/dx (12xex), −2 d/dx (12xex), and the function itself adds to zero. That is, ( D 2 − 2 D + 1)12 xe x = 0 operates on the LHS of the given equation with the newly found operator to get ( D 2 − 2 D + 1)( D 2 − D − 6) y = 0 Solve using the procedure for the homogeneous problem. (r − 1)2 (r − 3) (r + 2) = 0 is the characteristic equation; r = 1, 1, −2, 3 are the characteristic roots. Therefore, yg = c1e x + c2 xe x + c4 e −2 x + c5e3 x yc = c4 e −2 x + c5e3 x where yc is the solution to the associated homogeneous problem, and the ­candidate for yp is yp = c1e x + c2 xe x 50 Applied Mathematical Methods for Chemical Engineers Determine c1 and c2 by substituting yp into the given differential equation (c1e x + c2 xe x )′′ − (c1e x + c2 xe x )′ − 6(c1e x + c2 xe x ) = 12 xe x which results in c1e x + 2c2 e x + c2 xe x − c1e x − c2 e x − c2 xe x − 6c1e x − 6c2 xe x = 12 xe x an identity, which simplifies to c2 − 6c1 = 0 −6c2 = 12 Therefore, 1 yp = − e x − 2 xe x 3 and 1 yg = c4 e −2 x + c5e3 x − e x − 2 xe x 3 where c4 and c5 are the arbitrary constants. Example 3.8 Find the general solution of y″− 6y′ + 8y = 3ex. ( D 2 − 6 D + 8) = 3e x becomes ( D − 1)( D 2 − 6 D + 8) y = 0 ( D − 1)( D − 4)( D − 2) y = 0 with characteristic roots: 1, 2, 4 yg = c1e 2 x + c2 e 4 x + c3e x = yc + yp where yp = c3e x yp′ = yp′′ = c3e x That is, c3e x − 6c3e x + 8c3e x = 3e x Therefore, 3c3 = 3 giving c3 = 1 51 Linear Second-Order and Systems of First-Order and yp = e x such that yg = c1e 2x + c2 e 4x + e x Example 3.9 Find the general solution of y″ − 2y′ + 10y = 20x2 + 2x − 8. Solution ( D 2 − 2 D + 10) y = 20 x 2 + 2 x − 8 D 3 ( D 2 − 2 D + 10) y = 0 Characteristic roots: 0,0,0,1 + 3i,1 − 3i yg = (c1 cos 3 x + c2 sin 3 x )e x + c4 + c5 x + c6 x 2 yc yp yp′ = 2c6 x + c5 yp′′ = 2c6 2c6 − 2(2c6 x + c5 ) + 10c4 + 10c5 x + 10c6 x 2 = 20 x 2 + 2 x − 8 Equating coefficients x 0 : 2c6 − 2c5 + 10c4 = −8 x : − 4c6 + 10c5 = 2 x 2 : 10c6 = 20 from which c6 = 2, c5 = 1, c4 = −1 and yp = −1 + x + 2 x 2 yg = [c1 cos (3 x ) + c2 sin (3 x )]e x + 2 x 2 + x − 1 Example 3.10 Find the general solution of y′′ + y = tan x , 0 < x < ( D 2 + 1) y = tan x = π 2 sin x cos x There is no combination of derivatives that will give the appropriate multiple of sin x and cos x to drive the RHS to zero. Therefore, a more general method is needed. 52 Applied Mathematical Methods for Chemical Engineers 3.4.1 MEthOD OF VARIAtION OF PARAMEtERs Given Equation 3.36, y ′′ + p( x ) y ′ + q( x ) y = g( x ) a general method known as the method of variation of parameters where p, q, and g are continuous on the interval of interest can be used to find particular solutions of the nonhomogeneous differential equation. To use this method, we must know a fundamental set of solutions (for the secondorder case y1 and y2) for the homogeneous equation. In the homogeneous case, we found that yH = c1 y1 + c2 y2 where y1 and y2 are the fundamental solutions. Then if we replace c1 and c2 with two functions, u1 and u2, such that a candidate for the particular solution yp = u1 ( x ) y1 + u2 ( x ) y2 (3.46) satisfies the nonhomogeneous Equation 3.36 (two conditions on u1 and u2 are required), then yp′ = (u1′ y1 + u2′ y2 ) + (u1 y1′ ) + u2 y2′ ) (3.47) To simplify the computation, let u1′ y1 + u2′ y2 = 0 (3.48) which is one condition on u1 and u2, thus yp′ = u1 y1′ + u2 y2′ and yp′′ = u1′ y1′′+ u1 y1′ + u2′ y2′ + u2 y2′′ Substituting into Equation 3.36 gives u1′y1′ + u1 y1′′+ u2′ y2′ + u2 y2′′+ p(u1 y1′ + u2 y2′ ) + q(u1 y1 + u2 y2 ) = g u1 ( y1″+ py1′ + qy1 ) + u2 ( y2″ + py2′ + qy2 ) + u1′y1′ + u2′ y2′ = g but y1′′+ py1′ + qy1 = 0 y2′′ + py2′ + qy2 = 0 both of which satisfy the homogeneous equation; therefore u1′ y1′ + u2′ y2′ = g is the second condition on u1 and u2. That is, for second order the two conditions that must be met are u1′ y1 + u2′ y2 = 0 u1′ y1′ + u2′ y2′ = g Linear Second-Order and Systems of First-Order For third order, the three conditions are u1′ y1 + u2′ y2 + u3′ y3 = 0 u1′ y1′ + u2′ y2′ + u3′ y3′ = 0 u1′ y1′′+ u2′ y2′′ + u3′ y3′′ = g As a demonstration of the procedure, consider the following examples. Example 3.11 y′′ − 5 y′ + 6 y = 2e x Consider y″ − 5y′ + 6y = 0 operator form: (D2 − 5D + 6)y = 0 then (r − 3) (r − 2) = 0. Thus yc = c1e3x + c2e2x Let yp = u1e3 x + u2 e 2 x u1′e3 x + u2′ e 2 x = 0 3u1′e3 x + 2u2′ e 2 x = 2e x 0 e2 x 2e x 2e 2 x u1′ = e3 x e 2 x 3e3 x 2e 2 x = −2e3 x 2e − 3e5 x 5x u1′ = 2e −2 x 1 u1 = 2 ∫ e −2 x d x = 2 − e −2 x = − e −2 x 2 and u2′ = − u1′e3 x = (−2e −2 x )e3 x e −2 x = −2e − x e2 x such that u2 = 2e−x. Therefore yp = − e −2 x e3 x + 2e − x e 2 x = − e x + 2e x = e x and yg = c1e3 x + c2 e 2 x + e x 53 54 Applied Mathematical Methods for Chemical Engineers Example 3.12 Find a particular solution of y″ + y = tan x. Solution ( D 2 + 1) y = tan x yc = c1 cos x + c2 sin x y1 = cos x , y2 = sin x , y1′ = − sin x , y2′ = cos x the candidate for yp is yp = u1 ( x ) y1 + u2 ( x ) y2 then u1′ cos x + u2′ sin x = 0 − u1′ sin x + u2′ cos x = u1′ = 0 sin x sin x cos x cos x cos x sin x − sin x cos x = sin x cos x − sin 2 x cos x and u1 ( x ) = − ∫ also u2′ = sin 2 x dx cos x − u1′ cos x = sin x sin x such that u2(x) = −cos x yp = cos x ∫ sin 2 x d x − sin x cos x cos x . As is evident in Example 3.12, challenging integration problems can arise when the variation of parameter method is used. 3.5 VARIABLE COEFFICIENT PROBLEMS Both of the previously discussed methods are applicable to constant coefficient as well as variable coefficient problems. However, for some variable coefficient problems, substitutions are possible that will reduce variable coefficient problems to constant coefficient ones. Linear Second-Order and Systems of First-Order 55 One such class of variable coefficient problems is the Euler or Equidimensional ­differential equation x 2 y′′ + bxy′ + cy = 0 (3.49) for the second-order case in which b and c are constants. If one lets x = ez (3.50) dy dy = e− z dx dz (3.51) d2 y d 2 y dy = e −2 z 2 − 2 dz dx dz (3.52) then and where Equation 3.51 and Equation 3.52 are derived by applying the chain rule. Substituting Equation 3.50 through 3.52 into Equation 3.49 gives d2 y dy + (b − 1) + cy = 0 2 dz dz (3.53) a constant coefficient equation. For the case y ′′ + p( x ) y ′ + q( x ) y = 0 (3.54) a change of independent variable can sometimes be made that will transform the above equation into a constant coefficient problem. That is, if x z = ∫ [q(t )]1/ 2 dt (3.55) then dy dz dy = dx dx dz and d 2 y dz 2 d 2 y d 2 z dy + = dx 2 dx dz 2 dx 2 dx transform Equation 3.54 into 2 dz dy dz d 2 y d 2 z + + p( x ) + q( x ) y = 0 dx dz 2 dx 2 dx dz (3.56) 56 Applied Mathematical Methods for Chemical Engineers provided that [q ′( x ) + 2 p( x )q( x )] = constant 2[q( x )]3/ 2 (3.57) If the procedure of changing the independent variable is too work intensive, then a simple substitution can be made for the Euler equation case. That is, if y = xr (3.58) where r is a constant to be determined, then y′ = rx r −1 and y′′ = r (r − 1) x r −2 can be substituted into Equation 3.49 to get [r(r − 1) + br + c]x r = 0 (3.59) Then for any interval not containing the origin, the following statements hold: y = c1 | x |r1 +c2 | x |r2 (3.60) y = (c1 + c2 ln( x )) | x |r1 (3.61) y =| x |λ [c1 cos(µ ln | x |) + c2 sin(µ ln | x |)] (3.62) for real and unequal roots for real and equal roots where r1, r2 = λ ± iμ are the complex roots. For the class of problems for which convenient simplifying substitutions are not available, infinite series methods may be successfully applied. The following section discusses such class of variable coefficient problems. 3.5.1 SERIEs SOLUtIONs NEAR A REGULAR SINGULAR POINt Consider the variable coefficient, linear second-order, and homogeneous differential equation P( x ) y ′′ + Q( x ) y ′ + R( x ) y = 0 (3.63) in the neighborhood of a regular singular point x = x0. By regular singular point (as opposed to irregular) we mean that both lim ( x − x 0 ) x → x0 Q( x ) P( x ) (3.64) 57 Linear Second-Order and Systems of First-Order and lim ( x − x 0 )2 x → x0 R( x ) P( x ) (3.65) are finite. Then one can assume a solution to Equation 3.63 of the form ∞ y = ∑ an x n + r (3.66) n=0 where the values of r are to be determined and a relation for the an is to be established. The singular points of Equation 3.63 are exactly those for which P(x) = 0, if P, Q, and R are polynomials without common factors. For a more in-depth discussion of this theory, standard differential equation text [1,4] should be consulted. Presented below is an example that illustrates the essential steps needed to successfully solve Equation 3.63 using Equation 3.66. Consider the differential equation 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 by comparison with Equation 3.63 P ( x ) = 2 x 2 , Q ( x ) = − x , R( x ) = 1 + x then −x 1 lim x 2 = − x→0 2 x 2 and 1+ x 1 lim x 2 2 = 2x 2 x→0 are both finite; therefore, x = 0 is a regular singular point. Let ∞ y = a0 x r + a1 x1+r + a2 x 2+r + + an x n+r + = ∑ an x n+r n=0 then ∞ y ′ = ∑ (n + r )an x n + r −1 n=0 and ∞ y ′′ = ∑ (n + r )(n + r − 1)an x n + r − 2 n=0 thus, the given differential equation 2 x 2 y ′′ − xy ′ + (1 + x ) y = 0 (3.67) 58 Applied Mathematical Methods for Chemical Engineers becomes the identity ∞ ∞ 2 x 2 ∑ (n + r )(n + r − 1)an x n+r − 2 − x ∑ (n + r )an x n+r −1 n= 0 n= 0 ∞ +(1 + x )∑ an x n+r = 0 n= 0 and can be recast as ∞ ∞ ∑ 2(n + r )(n + r − 1)an x n+r − ∑ (n + r )an x n+r n=0 n=0 ∞ ∞ n=0 n=0 (3.68) + ∑ an x n + r + ∑ an x n + r +1 = 0 but ∞ ∞ n=0 n =1 ∑ an x n+r +1 = ∑ an−1 x n+r by shifting the index of summation [4] or replacing n by n − 1 everywhere in the fourth term of Equation 3.68. Therefore, Equation 3.68 becomes ∞ ∞ n=0 n =1 ∑ [(n + r )(2n + 2r − 3) + 1]an x n+r + ∑ an−1 x n+r = 0 after factoring. By expanding the summation for n = 0, we get ∞ [(r )(2r − 3) + 1]a0 x r + ∑ {[(n + r )(2n + 2r − 3) + 1]an + an −1}x n + r = 0 n =1 for which a0 is arbitrary and x ≠ 0. Therefore, r (2r − 3) + 1 = 0 (3.69) [(n + r )(2n + 2r − 3) + 1]an + an −1 = 0 for n ≥ 1 (3.70) and are the so-called indicial equation and recurrence relationship, respectively. The roots of Equation 3.69 are r1 = 1 and r2 = 1/2. Then the recurrence relation for r1 is given by an = −1 an −1 , n ≥ 1 n(2n + 1) (3.71) bn = −1 bn −1 , n ≥ 1 n(2n − 1) (3.72) and for r2, by where the switch from an to bn is introduced to reduce confusion. 59 Linear Second-Order and Systems of First-Order A few constants for Equation 3.71 and Equation 3.72 are evaluated as follows: − a0 −b ; b1 = 0 1(3) 1(1) − a1 a0 −b b0 ; b2 = 1 = n = 2, a2 = = 2(5) 1 ⋅ 2 ⋅ 3 ⋅ 5 2(3) 3 ⋅ 2 ⋅ 1 −b − b0 − a2 − a0 ; b3 = 2 = n = 3, a3 = = 2 3(7) 2 ⋅ 3 ⋅ 5 ⋅ 7 3(5) 5 ⋅ 32 ⋅ 2 ⋅ 1 n = 1, a1 = Both sets of constants exhibit a pattern that can be generalized to an = (−1)n a0 , n ≥ 1 [(2n + 1)(2n − 1)5 ⋅ 3]n! (3.73) bn = (−1)n b0 , n ≥ 1 [(2n − 1)(2n − 3)3 ⋅ 1]n! (3.74) in the case of r1, and in the case of r2. Therefore, the two linearly independent solutions that are expected are ∞ y = a0 x + a1 x 2 + a2 x 3 + + an x n +1 + = ∑ an x n + r n=0 a x2 a x3 a0 x 4 = a0 x − 0 + 0 − + 3 5 ⋅ 3 ⋅ 2 7 ⋅ 5 ⋅ 32 ⋅ 2 ∞ (−1)n x n = x 1 + ∑ a0 n =1 [(2n + 1)(2n − 1)5 ⋅ 3]n! for the r1 case, that is, ∞ (−1)n x n y1 = x 1 + ∑ a0 + − ⋅ [(2 n 1)(2 n 1) 5 3] n ! n =1 (3.75) and the r2 case gives y = b0 x r − b0 x1+ r + b0 x 2+ r b0 x 3+ r − + 3 ⋅ 2 ⋅ 1 5 ⋅ 32 ⋅ 2 ∞ (−1)n x n = x1/ 2 1 + ∑ b0 n =1 [(2n − 1)(2n − 3)3 ⋅ 1]n! such that ∞ (−1)n x n y2 = x1/ 2 1 + ∑ [(2 n − 1)(2 n − 3) 3 ⋅ 1] n ! n =1 is the second linearly independent solution of the given differential equation. (3.76) 60 Applied Mathematical Methods for Chemical Engineers Before the general solution can be stated, it is important to determine if the series obtained in Equations 3.75 and 3.76 are convergent. One way is to apply a convergence test[14]. For example, the ratio test lim n→∞ an +1 x n +1 an x n gives lim n →∞ | x | [(2n + 1)(2n − 1)5 ⋅ 3]n! |x| = lim =0 [2(n + 1) + 1(2(n + 1) − 1)5 ⋅ 3](n + 1)! n→∞ (2n + 3)(n + 1) Therefore, the series Equation 3.75 converges for all x. It can also be shown that Equation 3.76 converges for all x. Then the general solution is given by y = a0 y1 + b0 y2 and the reader should verify that y1 and y2 are, in fact, linearly independent. There are exceptional cases [4,6] that occur, and the following theorem is useful in dealing with those cases. Theorem 3.7 Let r1 and r2 be roots of the indicial equation for x2 y″ + x[xq(x)]y′ + x2g(x)y = 0, which is assumed to have a regular singular point at the origin. Then 1. If r1 ≠ r 2 and r1 − r 2 is not an integer, then there are two linearly independent solutions: ∞ y1 = ∑ an x n + r1 and n=0 ∞ y2 = ∑ bn x n + r2 , x > 0 n=0 2. If r1 − r 2 is a positive integer, then there are linearly independent solutions of the form ∞ y1 = ∑ a n x n + r1 and n=0 ∞ y2 = Ay1 ln( x ) + ∑ bn x n + r2 , x > 0 n=0 where A is a constant that may turn out to be zero.** 3. If r1 = r 2 , then there are linearly independent solutions ∞ y1 = ∑ an x n + r1 n=0 and ∞ y2 = y1 ln( x ) + ∑ bn x n + r1 , x > 0 * Sometimes the smaller root will give the general solution or no solution. n=0 Linear Second-Order and Systems of First-Order 61 3.6 ALTERNATIVE METHODS So far, the methods discussed can be classified as standard. However, there are other techniques, some of which may even be applicable to a few nonlinear problems. For example, the problem y ′′ + x ( y ′)2 = 0 can be solved by making the change of variable y′ = v, y ′′ = v ′ such that the problem is reduced to v ′ + xv 2 = 0 a separable first-order differential equation. Here one takes advantage of the missing dependent variable, y, in y ′′ = f ( x , y ′) Another example of an otherwise difficult problem is yy ′′ + ( y ′)2 = 0 Here the independent variable, x, is missing from the equation. By making the substitutions v = y′ and using the chain rule dv dv dy dv = ⋅ =v dx dy dx dy a new independent variable, y, can be defined. The problem is now reduced to yv dv dv + v2 = y + v = 0 dy dy a linear first-order equation. Hence, taking advantage of the missing independent variable, x, in y ′′ = f ( y, y ′) can reduce the second-order problem to a first-order one that we know how to solve. In engineering, prototype differential equations often occur as a result of some peculiarity of the system under investigation. Then the approach is to compare one’s derived problem with a prototype, and extract and modify the pertinent result. For example, the problem 1 d dy x − (1 − x 2 )λy = 0 x d x dx (3.77) d2 f df 1 λ + (1 − y) − − f = 0 2 dy dy 2 4 (3.78) can be transformed into y 62 Applied Mathematical Methods for Chemical Engineers The above equation is the confluent hypergeometric equation (the prototype) with linearly independent solutions tabulated in the literature [12] whereas Equation 3.77 resulted from a fluid flow problem with a parabolic velocity profile. Another example of a prototype equation that should be familiar to most chemical engineers is Bessel’s differential equation of order n: x 2 y ′′ + xy ′ + ( x 2 − n 2 ) y = 0 (3.79) There are tabulated solutions of Bessel’s equation [6,9–11], and the standard approach is to solve by comparison. Equations 3.77 to 3.79 could be solved by using the Frobenius series (Equation 3.65), but that approach may be too work intensive. To solve Bessel’s differential equation by comparison one needs a standard form of the equation to compare to. For example, the form a 2 − γ 2c 2 2a − 1 y′′ − y′ + b 2c 2 x 2c− 2 + y = 0 x x2 (3.80) has linearly independent solutions y1 = x a J γ (bx c ) (3.81) y2 = x a J− γ (bx c ) (3.82) and based on the fact that J γ is a solution of Bessel’s equation of order γ. Thus, the problem y ′′ + 1 4 y′ + 4 x 2 − 2 y = 0 x 9x can be compared to Equation 3.80 to give 2a − 1 = −1 ⇒ a = 0 2c − 2 = 2 ⇒ c = 2 b 2c 2 = 4 ⇒ b = ±1 4 1 ⇒γ =± 9 3 y1 = J1/3 ( x 2 ), y2 = J−1/3 ( x 2 ) a 2 − γ 2c 2 = − yg = c1 J1/3 ( x 2 ) + c2 J−1/3 ( x 2 ) where J γ (•) and J−γ(•) are Bessel functions of the first kind of order γ. 3.6.1 SUMMARY In this chapter, a few methods were presented for obtaining a solution to the linear second-order (or higher) ordinary differential equations. To the inexperienced practitioners, these many options could present a dilemma; that is, given a problem, which method should one use? 63 Linear Second-Order and Systems of First-Order For example, consider the constant coefficient linear differential equation y ′′ + y ′ = xe − x sin 3 x with the intent to determine a general solution. One could attempt to annihilate the nonhomogeneous portion, but that would require some clever algebra to be successful without a lot of labor. Therefore, it is not the recommended procedure for this problem. A second alternative is to let v = y′; then v′ = y″ and we now have v ′ + v = xe − x sin 3 x a linear first-order differential equation, which can be solved by the method given in the previous chapter for such equation. That is, µ( x ) = e x ( v µ( x ))′ = ( ve x )′ x sin 3 x such that v= dy 1 x = − e − x cos3 x + e − x sin 3 x + c1e − x dx 3 9 Therefore y= 1 1 xe − x cos3 x d x + ∫ e − x sin 3 x d x + c1e − x + c2 3∫ 9 where the integration (arbitrary) constant c1 absorbs the change in sign resulting from an integration of e−x. With the aid of a good set of integral tables [13], the first term on the right-hand side can be quickly evaluated. After some simplification the result is yg = 1 1 (5 x − 9)e− x cos3 x − (45 x + 14)e− x sin 3 x + c1e− x + c2 150 450 y c yp As a third alternative, y ′′ + y ′ = xe − x sin 3 x can be solved using the method of variation of parameters. That is, yc = c1e − x + c2 where y1 = e − x y1′ = − e −x and y2 = 1 and y2′ = 0 Then a candidate for the particular solution is yp = u1 ( x ) y1 + u2 ( x ) y2 subject to 64 Applied Mathematical Methods for Chemical Engineers u1′ y1 + u2′ y2 = 0 u1′ y1′ + u2′ y2′ = xe − x sin 3 x or e − x u1′ + u2′ = 0 − e − x u1′ = xe − x sin 3 x such that u1 = − ∫ x sin 3 x d x = x sin 3 x cos3 x − 3 9 and u2 = ∫ xe − x sin 3 x d x = − e− x xe − x (6 cos3 x − 8sin 3 x ) + (− sin 3 x − 3cos3 x ) 100 10 where this integral was evaluated using a set of tables [13]. Therefore sin 3 x xe − x x (sin 3 x + 3cos3 x ) yp = e − x cos3 x − − 3 9 10 e− x + (8sin 3 x − 6 cos3 x ) 100 which simplifies to yp − 1 1 (5 x − 9)e − x cos3 x − (45 x + 14)e − x sin 3 x 150 450 From a cursory glance, it appears that both the second and third alternates are equivalent in expediency; however, the underlying algebra is more work intensive for the second alternate. Therefore, one should make a choice based on all the factors, including the amount of time that can be spent on the overall problem. Bear in mind that the more steps that have to be executed, the greater the chance to introduce errors in an otherwise complicated analysis. 3.6.2 INItIAL VALUE PROBLEMs Theorem 3.1 guarantees a unique solution to the initial value problem, and the subsequent methods that are discussed can be used to derive general solutions. Given a general solution, the integration constants can be evaluated with the use of the given initial conditions. Another procedure, which transforms a differential equation into an algebraic equation, is given below. This is the Laplace transform method. For example, the problem y ′′ − y ′ − 2 y = 0 y(0) = 1, y ′(0) = 0 Linear Second-Order and Systems of First-Order 65 can be solved using Laplace transform as follows: L{y ′′ − y ′ − 2 y} = L{y ′′} − L{y ′} − 2L{y} = 0 giving s 2Y (s) − sy(0) − y ′(0) − sY (s) + y(0) − 2Y (s) = 0 where L{y} = Y (s) Then, following simplification, we get using partial fractions. s −1 (s − 2)(s + 1) 1/ 3 2 / 3 = + s − 2 s +1 Y (s ) = To recover the result in terms of the variables with which we started, an inversion of Y(s) is needed. The inverse Laplace transform, L−1{y(s)}, is given as { } 1 −1 1 1 L = e 2t s−2 3 3 and { } 2 −1 1 2 L = e−t s +1 3 3 Therefore, the complete solution to the given initial value problem is 1 2 y(t ) = e 2t + e − t = L−1{Y (s)} 3 3 This method is very efficient when it is applicable, and is one of the many integral transformations that are useful for solving initial value problems. Integral transforms are certain functions that are defined by an integral, such as F (s ) = β ∫α K (s, t ) f (t ) dt (3.83) where the function f(t) is transformed into F(s), and K(s, t) is called the kernel of the transform. In this case, the Laplace transform of f(t) is defined to be ∞ L{ f (t )} = F (s) = ∫ e − st f (t ) dt 0 (3.84) where e−st is the kernel. Laplace transform is most efficient in solving problems with nonhomogeneous terms that are discontinuous or impulsive. These types of problems would at best be awkward if the previously discussed methods were attempted. When can this method be expected to work successfully? Below are some definitions that address this issue. 66 Applied Mathematical Methods for Chemical Engineers Theorem 3.8 Suppose that f is piecewise continuous on the interval 0 ≤ t ≤ A for any positive A and |f(t)| ≤ keat when t ≥ M, where k, a, and M are real constants. Then the Laplace transform L{f(t)} = F(s) exists for s > a. By piecewise continuous one means, given a function on an interval α ≤ t ≤ β, where the interval can be subdivided by a finite number of points in the following way, α = t0 < t1 < ⋯ < tm = β such that f is continuous on each open subinterval ti−1 < t < ti and f approaches a finite limit as the endpoints of each subinterval are approached from inside. Essentially, Theorem 3.8 says that if a function is piecewise continuous and is exponentially bounded, then one can expect it to have a Laplace transform. The theory of Laplace transform is well documented [1,4,6,15,16] and should be consulted for a deeper discussion. Also available are convenient tables of Laplace transforms [13]. Equation 3.84 can be used to determine Laplace transforms of derivatives provided that the derivatives satisfy the appropriate conditions of continuity and boundedness as required by Theorem 3.8. For example, ∞ L{y ′(t )} = ∫ e − st y ′(t ) dt 0 By definition, the improper integral is B lim ∫ e − st y ′(t ) dt B→∞ 0 then applying integration by partslet u = e − st then and dv = y ′(t ) dt du = − se − st dt and v = y(t ) therefore B B B ∫0 e− st y′(t ) dt = e− st y(t ) + s ∫ e− st y(t ) dt 0 = e− sB y( B) − y(0) + s ∫0 e− st y(t ) dt then B { B lim ∫ e − st y ′(t ) dt = lim e − sB y( B) − y(0) + s ∫ e − st y(t ) dt n→∞ 0 B→∞ 0 } ∞ = − y(0) + s ∫ e − st y(t ) dt (3.85) 0 = − y(0) + sY (s) where Equation 3.84 was used in Equation 3.85. This integration procedure can be extended to finding the Laplace transform of an nth-order derivative of a given function. That is, L{ f ( n ) (t )} = s n L { f (t ) − s n −1 f (0) − L − sf ( n − 2) (0) − f ( n −1) (0) (3.86) 67 Linear Second-Order and Systems of First-Order and for n = 2, d2 f L 2 = s 2 F (s) − sf (0) − f ′(0) dt (3.87) which is the Laplace transform of the second derivative that was used in the above example. As mentioned previously, Equation 3.86 can be found in tables of Laplace transform in most mathematical handbooks [13]. Equation 3.84 can also be used to derive the Laplace transforms of most elementary functions, such as L{1} = 1 s 1 1 = 1+1 2 s s 2 2 2 L{t } = 3 = 2 s s +1 n! L{t n } = n +1 ; n = 1,2, s L{t} = Now suppose n is not a positive integer; for example, what is L{t−1/2}? Since for any n, ∞ L{t n } = ∫ t n e − st dt , s > 0 0 letting u = st; du = s dt ∞ ∞ n gives n −u u e − u du = ∞ u e du = 1 s ∫0 s n + 1 s n +1 ∫0 t n e− st dt = ∫0 s ∞ ∫0 e − u u n d u ∞ If we denote ∫ e − u u n du by Γ (n + 1), that is, 0 ∞ ∫0 e− u un du = Γ(n + 1) (3.88) Γ (n + 1) then L{tn} becomes where Γ (n + 1) is the so-called gamma function or factorial s n +1 ∞ − u n function. Also, the quantity ∫ e u du can be evaluated using integration by parts to give 0 ∞ ∞ ∞ ∫0 e− uu n du = −u n e− u 0 + n ∫0 e− uu n−1 du = nΓ( n) Therefore Γ (n + 1) = nΓ (n); n ≠ 0 However, for n = 0, Equation 3.88 gives ∞ Γ (1) = ∫ e − u du = − e − u 0 ∞ 0 =1 (3.89) 68 Applied Mathematical Methods for Chemical Engineers Hence, 0! = 1. The gamma function, or factorial function, can now be evaluated for other values of n using Equation 3.89. For example, n = 1: Γ (2) = 1Γ (1) = 1 n = 2 : Γ (3) = 2Γ (2) = 2 ⋅ 1 n = 3 : Γ (4) = 3Γ (3) = 3 ⋅ 2Γ (2) = 3 ⋅ 2 ⋅ 1 and in general Γ (n + 1) = n! Now then, for n = −1/2, ∞ ∫0 e − uu −1/2 du = I Γ ( −1/2 + 1) = Γ (1/2) = Thus, by substituting u = x2, du = 2x dx, I becomes I = Γ (1/2) = ∞ ∞ 1 − x2 2 e 2x d x = 2∫ e− x d x 0 x ∞ ∫0 e − uu −1/2 du = ∫0 To evaluate I, the following device is employed: ( ∞ )( ∞ ) I2 = 2 ∫ e− x d x 2 ∫ e− y d y = 4 ∫ 0 2 2 0 ∞ 0 ∞ ∫0 e − ( x + y ) d x d y ∂x ∂( x , y ) ∂r dr dθ = But x = r cos θ, y = r sin θ dx dy = ∂x ∂(r , θ) ∂θ I2 = 4 ∫ π/2 0 (∫ ∞ 0 2 ∂y ∂r dr dθ = r dr dθ ∂y ∂θ ) e − r r dr dθ = π 2 such that I = Γ (1/2) = π and L{t −1/ 2} = 3.6.3 π s 1/ 2 = π s SOME UsEFUL PROPERtIEs OF LAPLAcE TRANsFORMs 1.If the Laplace transform of f(t) is F(s), then L{e − at f (t )} = F (s + a) where a is a constant. For example, L{e − at cos bt} = ? 2 69 Linear Second-Order and Systems of First-Order from a table of Laplace transforms L{cos bt} = therefore L(e − at cos bt ) = s (s + b 2 ) 2 s+a [(s + a)2 + b 2 ] 2.The transform of an integral is given by L {∫ f (t) dt} = 1s F (s) t 0 3.The derivative of a transform is given by d n F (s ) = L{(−t )n f (t )} ds n 4.Transform of a step function: if U(t) is a unit step function U (t ) = 0 =1 t<0 t>0 (3.90) and if U(t − τ) is a unit step function starting at t = τ U (t − τ) = 0 =1 t<τ t>τ (3.91) then L{U (t )} = 1 s and e sτ s f (t − τ) = 0 for L{U (t − τ)} = L{ f (t − τ)} = e −τs F (s); if 0<t <τ and L{ f (t − τ)U (t )} = e − sτ F (s) 5.Transform of an impulse function: the Dirac delta or unit impulse function is given as 0 if t < t0 − ε 1 δ ε (t − t0 ) = if t0 − ε ≤ t0 + ε 2ε 0 if t ≥ t0 + ε where ε > 0 is a small number and t0 > 0. 70 Applied Mathematical Methods for Chemical Engineers 1 2e de(t – t0) 0 t0 – e t0 + e Sketch of the Delta Function As can be seen from the sketch, as ε → 0, the rectangular pulse δ ε(t − t0) gets taller and thinner, but the limit does not exist. Even though the limit does not exist in the usual sense, one can still derive useful properties of the impulse function. For example, ∞ 1. ∫ δ ε (t − t0 ) dt = −∞ t0 +ε 1 ∫t −ε 2ε dt = 1 0 since δ ε (t − t0) = 0 for all t outside of the interval t0 − ε ≤ t ≤ t0 + ε. That is, if δ ε(t − t0) is considered as a force, then property (1) says the total impulse is unity. ∞ t0 +ε 1 1 t0 +ε 2. ∫−∞ δ ε (t − t0 ) f (t ) dt = ∫t0 −ε 2ε f (t ) dt = 2ε ∫t0 −ε f (t ) dt = f (τε ) for some τ ε in t0 − ε ≤ t ≤ t0 + ε and any continuous function f(t) defined on the interval −∞ < t < ∞. Further, if ε → 0, then ∞ ∫−∞ δ(t − t0 ) dt = 1 ∞ ∫−∞ δ(t − t0 ) f (t ) dt = f (t0 ) ∞ ∫−∞ δ n (t − t0 ) f (t ) dt = (−1)n f (n) (t0 ) 3.The Laplace transform of the impulse function is given by L{δ ε (t − t0 )} = e − st0 for t0 > 0 and L{δ(t )} = 1 if t0 → 0 3.6.4 INVERtING thE LAPLAcE TRANsFORM In order for this operational method to be useful to chemical engineers, there must be convenient ways to invert the transform. That is, it is necessary to be able to perform the operation Linear Second-Order and Systems of First-Order 71 L−1{F (s)} = f (t ) (3.92) to complete a problem. There are three standard ways to invert the Laplace transform and each is outlined below. The Laplace transform of a function can be arranged to be of the form F (s ) = P (s ) Q(s ) (3.93) where P and Q are polynomials in s and Q(s) is of higher degree than P(s). The function F(s) can then be expanded into its partial fractions [14] such that the form n Ai ( − s r )i i =1 F (s ) = ∑ (3.94) is obtained, where the quantity (s − r) is a linear factor of Q(s) and the Ai are determined constants. Once the form Equation 3.94 is obtained, then Equation 3.92 can be applied to each term with the aid of a table of transforms [13]. The second standard method of inverting the Laplace transform is by convolution. This method is most effective when F(s) is the product of two other transforms, H(s) and G(s), where transforms H(s) and G(s) are respectively transforms of the functions h(t) and g(t). The following theorem defines what is meant by convolution [1, 17]. Theorem 3.9 If H(s) = L{h(t)} and G(s) = L{g(t)} both exist for s > a ≥ 0, then F (s) = H (s)G (s) = L{ f (t )}s > a (3.95) ∫ 0 h(t − τ)g( τ) d τ = ∫ 0 h( τ)g(t − τ) d τ (3.96) where f (t ) = t t The function f is called the convolution of h and g and the integrals (Equation 3.96) are called convolution integrals. The convolution obeys the following rules: Commutative law: h * g = g * h Distributive law: h * ( g1 + g2 ) = h * g1 + h * g2 Associative law: h * ( g * h) = (h * g) * h h*0 = 0*h = 0 However, h *1 = h is not generally a true statement. That is, if h(t) = sin t, then t h * 1 = ∫ sin(t − τ)d τ = cos(t − τ) 0 = 1 − cos t ≠ sin t 0 t 72 Applied Mathematical Methods for Chemical Engineers Also h*h is not necessarily nonnegative, that is, t t h * h = ∫ sin(t − τ)sin τ d τ = ∫ (sin t cos τ − cos t sin τ)sin τ d τ 0 0 t t 0 0 = sin t ∫ sin τ cos τ d τ − cos t ∫ sin 2 τ d τ 1 = (sin t − t cos t ) 2 Essentially, the convolution integrals do not have all the properties of ordinary multiplication. This example illustrates inversion using the convolution integral. Consider F (s ) = s 1 s = 2 (s + 1)(s + 4) s + 1 s 2 + 4 From a set of tables of Laplace transforms [13] one can find L−1 = and L−1 = { } 1 = e−t s +1 { } s = cos 2t s +4 2 then t e − t * cos 2t = ∫ e − (t −τ ) cos 2τ d τ 0 t = e − t ∫ e τ cos 2τ d τ 0 t e τ (cos 2τ + 2sin 2τ) = e−t 5 0 1 = (cos 2t + 2sin 2t − e − t ) 5 Therefore s 1 L−1 = (cos 2t + 2sin 2t − e − t ) 2 (s + 1)(s + 4) 5 The third standard method of inverting a Laplace transform is by making use of the Residue theorem [6,15,16,18,22,30]. The transform function F(s) is analytic, except for singularities. In this discussion, when F(s) is analytic, the inverse transform of F(s) is given by ∞ f (t ) = L−1{F (s)} = ∑ ρn (t ) (3.97) 1 where ρ n(t) is termed the residue of F(s) at the singularities (poles) Sn. The residues of F(s) may be determined by employing the form given in Equation 3.93, that is, Linear Second-Order and Systems of First-Order F (s ) = 73 P (s ) Q(s ) If Sn is a simple pole of F(s), then ρ n(t) is given by ρn (t ) = P ( S n ) Sn t e Q ′ ( Sn ) (3.98) where Q′(Sn) is the value of dQ/ds evaluated at the singular point of interest. Recall that P (s) P ( sn ) P (s ) = lim = lim(s − sn ) s s s s → → n Q(s ) − Q(s ) n Q(s) Q ′ ( sn ) n − s s n The form lim(s − sn ) s → sn P (s) Q(s) is convenient for applications, especially when sn = 0. When Sn is a multiple pole of order m of F(s), then t2 t m −1 Am ρn (t ) = e Snt A1 + tA2 + A3 + + 2! (m − 1)! m t i −1 = e Snt ∑ Ai (i − 1)! i =1 (3.99) where 1 dm−i [(s − sn )m F (s)] s → sn (m − i )! ds m − i Ai = lim The following example illustrates the use of the residue theorem and should serve to clarify certain new terminology: Consider 1 P (s) F (s ) = = 2 s (s − a) Q(s) then the polynomial Q(s) has singularities at s = 0 and s = a. These singularities are termed poles [16,18] and in particular, s = 0 is termed a simple pole whereas s = a is a double pole (pole of order two). For the simple pole, Equation 3.98 gives ρ 0 (t ) = 1 1 1 e0t = = 2 Q ′(0) Q ′(0) a since Q′(s) = (s − a)2 + 2s(s − a). Alternatively, we could first determine the quantity P(0)/Q′(0) by taking the limit lim(s − 0) s→ 0 1 1 = s (s − a) 2 a 2 74 Applied Mathematical Methods for Chemical Engineers Then, multiply the result by e0t to get the residue at s = 0. The residue at the double pole, s = a, is given by ρa (t ) = e at ( A1 + tA2 ) since m = 2 where A1 = = 1 d 2 −1 1 (s − a) 2 (2 − 1)! ds 2−1 s(s − a)2 s = a d 1 1 =− 2 ds s s = a a and A2 = 1 d 2− 2 (2 − 2)! ds 2− 2 1 2 (s − a) s (s − a) 2 a 1 1 = = s s = a a then t 1 ρa (t ) = e at − 2 a a and f (t ) = L−1{F (s)} = 1 t 1 + e at − 2 a a a2 The previous example is very straightforward and could easily be inverted by use of partial fractions and a table of Laplace transforms. This next example may be tabulated. However, it is used here to demonstrate more clearly how the residue theorem may be useful for similar or more complicated inversions. Consider F (s ) = sinh x s sinh a s Then Q(s) = sinh a s If we let a s = iλ such that s = −λ 2/a2 then F(s) transforms to sin (x λ /a)/sin λ such that dQ dQ dλ cos λ = = − a 2i ds dλ ds 2λ where Q(λ) is given by isinλ. Then application of Equations 3.97 and 3.98 result in ∞ ∑− n= 0 2λ n sin( xλ n / a) −( λ 2n / a2 )t e a 2 cos λ n Linear Second-Order and Systems of First-Order 75 But the poles of sinh a s are the zeros of sin λ and these occur at λ = n π, n = 0, 1, 2, …. Therefore replacing λ n by the quantity n π and observing that cos(nπ) = (−1)n, we get the result 2π ∞ 2 2 2 (−1)n+1 n sin( xnπ/a)e − ( n π t / a ) 2 ∑ a n=1 More detailed usage of the residue theorem to invert Laplace transforms is discussed in Chapter 6 and in the literature [16, 30]. 3.6.5 TAYLOR SERIEs SOLUtION OF INItIAL VALUE PROBLEMs In Section 3.5.1, the Frobenius series method was discussed with regard to differential equations with regular singular points. In this section, a method is given that effectively deals with differential equations with ordinary points. To illustrate the importance of this method consider the relatively harmless-looking first-order initial value problem y′ + e x y = x 2 ; y(0) = 4 (3.100) This problem has a unique solution (Theorem 2.1) for all x and is given by y( x ) = −x 1 x 2 eξ ξ e d ξ + 4e x ∫ e e 0 (3.101) However, Equation 3.101 is not in closed form. By closed form we mean y ′ + 2 y = 1; y(0) = 3 has 1 y( x ) = (1 + 5e −2 x ); − ∞ < x < ∞ 2 as its solution in closed form. Equation 3.101 is difficult to graph or to use in forecasting results, and these are important aspects of engineering practice. Reconsider Equation 3.100, but this time, let y( x ) = y(0) + y′(0) x + 1 1 y′′(0) x 2 + + y( n ) (0) x n + 2! n! ∞ (3.102) 1 (n) y (0) x n ! n n=0 =∑ then each derivative can be evaluated at x = 0 from y ′(0) + e 0 y(0) = 0 or y′(0) = −4 76 Applied Mathematical Methods for Chemical Engineers to obtain the value of the second derivative at x = 0, one differentiates Equation 3.100 thus y ′′ + e x y ′ + e x y = 2 x (3.103) then at x = 0, Equation 3.103 gives y ′′(0) + y ′(0) + y(0) = 0 or y ′′(0) = 0 Differentiate Equation 3.103 to get y(3) + e x y ′′ + 2e x y ′ + e x y = 2 (3.104) then y(3) (0) + y ′′(0) + 2 y ′(0) + y(0) = 2 or y(3) (0) = 6 Therefore, the solution of Equation 3.100 using Equation 3.102 is y( x ) = 4 − 4 x + x 3 + (3.105) Notice that one can take this process up to any desired amount of terms. Equation 3.105 is more manageable than Equation 3.101. An appropriate question is when is this approach mathematically legal? The answer is whenever the initial value problem has analytic coefficients and forcing functions. The term analytic can be defined as any function having a Taylor series representation in some open interval about a given point. Theorem 3.10 If p, q, and g are analytic at x0, then the initial value problem y′′ + p( x ) y′ + q( x ) y = g( x ); y( x 0 ) = A, y′( x 0 ) = B (3.106) has a unique solution, which is also analytic at x0. If the initial value is not given at the origin, a shift to the origin can be made as given in the example that follows. Given y′′ + xy = 4; y(1) = 2, y′(1) = 0 let t = x − 1 and Y (t ) = y(1 + t ) then Y ′(t ) = dy dx dx dt but dx =1 dt dy Y ′(t ) = dx x = 1 + t, Linear Second-Order and Systems of First-Order 77 similarly Y ′′(t ) = y′′( x ) then Equation 3.106 becomes Y ′′(t ) + (1 + t )Y (t ) = 4; Y (0) = 2, Y ′(0) = 0 (3.107) Equation 3.107 can now be treated similar to the previous example to get 1 1 1 1 Y (t ) = 2 + t 2 − t 3 − t 4 − t 5 + t 6 + 3 12 30 72 such that 1 1 1 y( x ) = 2 + ( x − 1)2 − ( x − 1)3 − ( x − 1)4 − ( x − 1)5 3 12 30 1 + ( x − 1)6 + 72 Even though the above discussion is focused on linear problems, the Taylor series approach may be applied to some nonlinear problems. 3.7 APPLICATIONS OF SECOND-ORDER DIFFERENTIAL EQUATIONS In this section, a few applications of the theory and methods that were previously outlined will be illustrated. However, it should be noted that a substantial percentage of the application of second- (and higher) order ordinary differential equations is in association with solving partial differential equations, a topic discussed in Chapter 6. As in Section 2.4, some problem setup as well as solution techniques will be demonstrated. For example, consider the following problem in heat transfer through a cylindrical conductor [15]. 3.7.1 PROBLEM StAtEMENt Two concentric cylindrical metallic shells are separated by a solid material. If the two metal surfaces are maintained at different constant temperatures, what is the steady-state temperature distribution within the separating material? In this problem, temperature, T, and the heat flow per unit area, Q, depend on the radius, r. Then if we consider the condition of the system about an element of thickness, ∆r and with the aid of Figure 3.1, the important quantities can be organized according to the procedure of Section 2.4. Therefore, dQ 2πrQ − 2π(r + ∆r ) Q + ∆r = 0 dr or 78 Applied Mathematical Methods for Chemical Engineers r + Δr r a R FIGURE 3.1 Radial heat flow through cylindrical conductor. r System property T Temperature Heat transfer area per unit length 2πr Q Radial heat flux density 2πrQ Total radial heat flow Heat input to inner surface = 2πr.Q ( Heat output from outer surface = 2π(r + ∆r ) Q + Accumulation of heat = 0 −r dQ dr ∆r r + Δr T+ dT ∆r dr 2π (r + ∆r) dQ Q+ ∆r dr dQ 2 π(r + ∆r ) Q + ∆r dr ) dQ dQ −Q− ∆r = 0 dr dr Then, limit as ∆r → 0 r dQ +Q = 0 dr (3.108) But Q is related to T by Q = −k dT dr (3.109) where k is the thermal conductivity. Substituting Equation 3.109 into 3.108 gives r for a constant thermal conductivity. d 2 T dT + =0 dr 2 dr (3.110) Linear Second-Order and Systems of First-Order 79 Since the two metal surfaces are maintained at different constant temperatures, say T = T0 at r = a (3.111) T = T1 at r = R (3.112) and one can solve the second-order, linear, variable coefficient, and homogeneous differential equation subject to Equations 3.111 and 3.112. That is, d dT r =0 dr dr gives r dT = c1 dr and T (r ) = c1 ln r + c2 then T (a) = T0 = c1 ln a + c2 T ( R) = T1 = c1 ln R + c2 therefore R T1 − T0 = c1 ln a such that c1 = T1 − T0 ln( R/a) and c2 = T0 ln R − T1 ln a ln( R/a) Finally T (r ) − T0 ln(r /a) = T1 − T0 ln( R/a) As can be expected, there are other ways to set up physical problems that may be of interest. One approach that is very prominent in chemical engineering is to use the equations of change [19, 20]. That is, to set up constant density, 80 Applied Mathematical Methods for Chemical Engineers constant viscosity flow problems, one needs the equation of continuity and the equation of motion. Example 3.13 Consider the axial flow of an incompressible fluid in a circular tube of radius R. By considering a long tube and assuming that the θ-component and the r-component of velocities are negligible, one can reduce the z-component of motion for constant ρ and μ [19] to ρvz ∂ vz ∂P 1 ∂ ∂ vz ∂ 2 vz =− +µ r + 2 ∂z ∂z r ∂r ∂r ∂ z (3.113) Also, the equation of continuity [19] reduces to ∂ vz =0 ∂z (3.114) which further reduces Equation 3.113 to 0=− dP 1 1 d dvz + r dz µ r dr dr (3.115) subject to vz is finite at r = 0 (3.116) vz = 0 at r = R (3.117) and at the wall of the cylinder Equation 3.115 through 3.117 integrate to υz = ( P0 − PL ) R 2 r 2 1 − 4µL R where P0 and PL are the pressures at the entrance and exit of the cylinder, respectively. Example 3.14 Derive the temperature profile T(r) in a solid cylinder with heat generation if the g­ overning differential equation is 1 ∂ ∂ T 1 ∂ ∂ T ∂ ∂T ∂T + k k + q = ρCp kr + 2 r ∂r ∂z ∂t ∂r r ∂φ ∂φ ∂ z (3.118) 81 Linear Second-Order and Systems of First-Order where the coordinate system indicates the independent variables: ρ is mass density and Cp is specific heat. For a long solid cylinder as shown in Figure 3.2, with uniform heat generation q and at steady-state conditions, the rate at which heat is generated within the cylinder will be equal to the rate at which heat is convected from the surface of the cylinder to a moving fluid. This condition allows the surface temperature (Ts) to be maintained at a fixed value. Then, for constant thermal conductivity k, Equation 3.118 reduces to 1 d dT q r + =0 r dr dr k or q dT dr = − rdr dr k (3.119) such that T (r ) = − q 2 r + C1 ln r + C2 4k (3.120) Along the centerline of the solid cylinder (r = 0) the temperature distribution is symmetric and the temperature gradient (dT/dr) must be zero there. That is, at r = 0, dT =0 dr therefore, C1 = 0 at r = r0 , T = TS Z T (r, f, z) r x f y Cold fluid T∞ r qr h q L FIGURE 3.2 Solid cylinder with heat generation. 82 Applied Mathematical Methods for Chemical Engineers and C2 = TS + T (r ) − TS = q 2 r0 4k q 2 2 (r0 − r ) 4k (3.121) From an overall energy balance q ( πr02 L ) = h(2πr0 L )(TS − T∞ ) which simplifies to TS = T∞ + 0 qr 2h (3.122) Example 3.15 Consider a long solid tube, insulated at the outer radius r0 and cooled at the inner radius . ri with uniform heat generation q within the solid (as shown in Figure 3.3) [21]. 1. Determine the general solution for the temperature profile in the tube. 2. Suppose the maximum permissible temperature at the insulated surface r0 is T0. Identify appropriate boundary conditions that could be used to determine the arbitrary constants appearing in the general solution and find the temperature distribution. ri FIGURE 3.3 r0 Solid tube with heat generation. 83 Linear Second-Order and Systems of First-Order 3. What is the heat removal rate per unit length of tube? 4. If the coolant is available at a temperature T∞, obtain an expression for the convection coefficient that would have to be maintained at the inner surface . to allow for operation at the prescribed values of T0 and q. Solution Assumptions: 1. Steady-state conditions prevail. 2. One-dimensional radial conduction is reasonable. 3. Physical properties are constant. 4. Volumetric heat generation is uniform. 5. Outer surface is adiabatic. a. Equation 3.118 in Example 3.14 is the governing differential equation and it is reducible to Equation 3.119 by using the reasoning per Example 3.14. Therefore, the general solution is given in Equation 3.120: T (r ) = − q 2 r + C1 ln r + C2 4k b. Two boundary conditions are needed to determine C1 and C2. In this problem T (r0 ) = T0 (3.123) and Fourier’s law Qr = − kA dT dT = − k (2πrL ) dr dr expresses the rate at which heat is conducted across any cylindrical surface in the solid. In particular, at the adiabatic outer surface dT =0 dr Qr = 0 ⇒ at r = r0 , (3.124) Therefore, C1 = q 2 r0 2k and C2 = T0 + q 2 q 2 r0 − r0 ln r0 4k 2k Finally, the temperature distribution is T (r ) = T0 + q 2 2 q r (r0 − r ) − r02 ln 0 4k 2k r (3.125) 84 Applied Mathematical Methods for Chemical Engineers c. The heat removal rate may be determined by obtaining the conduction rate at r i or by evaluating the total generation rate for the tube. Since Qr = −2πkr dT dr represents the rate of conduction, then from Equation 3.125 dT q r02 = − r dr 2 k r and at r = r i dT q r02 = − ri dr 2 k ri Therefore, Qr (ri ) = −2πkri q r02 − ri 2 k ri = −πq ( r02 − ri2 ) (3.126) is the heat removal rate. d. Application of the energy conservation principle to the inner surface results in Qr,cond = Qr,conv or πq ( r02 − ri2 ) = h 2πri (Ti − T∞) (3.127) where −Qr(r i) accounts for heat flowing out of the wall, and Qr,conv is expressed as (Ti − T8) rather than (T8 − Ti). Therefore h= q ( r02 − ri2 ) , where Ti = T (ri ) 2ri (Ti − T∞) Example 3.16 This example covers simultaneous diffusion and chemical reaction in a tubular reactor [15]. A tubular reactor of length L and cross-sectional area 1.0 m2 is used to carry out a first-order chemical reaction of the type A→B The rate coefficient is k (sec−1). In a given feed rate of u m3/sec, the initial feed concentration of component A is C0 and the diffusivity of A is D m2/sec. What is the concentration of A as a function of the reactor length? It may be assumed that during the 85 Linear Second-Order and Systems of First-Order reaction the volume remains constant and that steady-state conditions are established. Also there is no concentration variation in the section following the reactor. Solution Let x represent the distance of any point from the beginning of the reaction section (0 < x < L), C represent the concentration profile of species A in the entry section (x < 0), and y be the concentration profile of species A in the reaction section as shown in Figure 3.4. Consider a material balance over the element of length ∆x at a distance x from the inlet. Then the bulk flow of A at x is (uy) and at x + ∆x is uy + u dy ∆x dx Also, the diffusion of A at x is − D dy dx and at x + ∆x is dy d dy − D + − D ∆x dx dx dx The accumulation rate is zero (assumed steady-state conditions given in problem statement). The rate of removal of component A by chemical reaction is given by ky∆x (cross-sectional area). Therefore, uy − D dy dy dy d dy − uy + u ∆x − − D + − D ∆x = ky∆x dx dx dx dx dx Entry section u, C0 L u, y x FIGURE 3.4 Tabular reactor. Δx (3.128) 86 Applied Mathematical Methods for Chemical Engineers Following simplification and division by ∆x we get D d2 y dy − u − ky = 0 dx 2 dx (3.129) Similarly, the material balance in the entry section gives D d 2C dC −u =0 2 dx dx (3.130) The characteristic equation for Equation 3.129 is Dr 2 − ur − k = 0 such that r1 = u(1 + m) / 2 D and 4 kD r2 = u(1 − m) / 2 D, where m = 1 + 2 u Therefore, ux ux y = h1 exp (1 + m) + h2 exp (1 − m) 2D 2D and the general solution for Equation 3.130 is ux C = h3 + h4 exp D The appropriate boundary conditions for Equation 3.129 are, at x = 0 C = y (contidC dy = also at x = L, dy/dx = 0 (no concentration variation dx dx following the reaction section). Appropriate boundary conditions for Equation 3.130 are, at x = −∞, C = C0 and the condition at x = 0 can be reused (i.e., C = y). Following application of these boundary conditions we get nuity of composition) or 2 y ux um um = exp (m + 1) exp ( L − x ) + (m − 1) exp − ( L − x ) D C0 K 2D 2D where K is given by uLm − uLm K = (m + 1)2 exp − (m − 1)2 exp 2D 2D Linear Second-Order and Systems of First-Order Example 3.17 This example covers the continuous hydrolysis of fat compound in a spray column [15]. A fat compound mixed with high-pressure hot water is fed to the bottom of a spray column. Water at the column operating conditions is sprayed into the top and descends in droplets through the rising fat phase. The hydrolysis reaction generates glycerine that is extracted by the descending water phase. Further, the hydrolysis reaction is first order with a specific rate constant of 0.17 sec−1. Estimate the concentration of glycerine in each phase as a function of column height. Also determine what fraction of the tower height H is required for the chemical reaction. Data: Rate of fat compound input = 8070 lb/h Rate of high-pressure hot water = 2270 lb/h Rate of water spray = 4120 lb/h Column operating conditions: 450ºF and 600 psia Height = 72 ft, Diameter = 2 ft 2 in Rate of final extract = 5560 lb/h containing 12.16% glycerine Rate of fatty acid raffinate = 8900 lb/h containing 0.24% raffinate Glycerine distribution ratio between the water and the fat phase is 10.32 at the given column temperature and pressure. Solution As shown in Figure 3.5, L represents the mass flow of raffinate and G the mass flow of extract. In addition, the following symbols are used to designate the other quantities: x = mass fraction of glycerine in raffinate y = mass fraction of glycerine in extract y* = equilibrium mass fraction of glycerine in extract z = mass fraction of hydrolyzable fat in raffinate S = cross-sectional area of tower a = interfacial area per unit volume of tower K = overall mass transfer coefficient expressed in terms of extract compositions m = distribution ratio k = reaction rate coefficient ρ = mass of fat per unit volume h = distance coordinate from base of column w = mass of fat per unit mass of glycerine H = effective height of column Then the changes occurring in the element of column of height ∆h are • Glycerine transferred from fat to water phase: KaS(y* − y)∆h. • Rate of destruction of fat by hydrolysis: kρSz∆h, thus giving the rate of ­production of glycerine as kρSz∆h/w. 87 88 Applied Mathematical Methods for Chemical Engineers L lb/h G lb/h xH yH zH x + Δx z + Δz y + Δy Δh x, z H y h x0, z0 FIGURE 3.5 y0 Continuously operating fat-hydrolyzing column under steady state. A glycerine balance over the element ∆h (Figure 3.5) is Lx + ( kρSz∆h) dx dy − L x + ∆h = Gy − G y + ∆h = KaS ( y* − y) ∆h w dh dh (3.131) y* = mx (3.132) where Also a glycerine balance between the element and the base of the tower is Lz 0 Lz + Gy = Lx + + Gy0 (3.133) w w Using Equation 3.132 and the last two parts of Equation 3.131 give KaSmx = KaSy − G dy dh (3.134) Substitution of Equation 3.133 into the first two parts of Equation 3.131 gives z G ( y − y0 ) dx dy kρS 0 + − x ∆h − L ∆h = − G ∆h L dh dh w (3.135) 89 Linear Second-Order and Systems of First-Order Multiplication of Equation 3.135 by (KaSm/LG) and substituting for x from Equation 3.134 gives kρS 2 Ka mz 0 mG dy kρS KaS + ( y − y0 ) − y − LG w L dh L G KaS dy d 2 y KaSm dy + =0 − − L dh G dh dh 2 and, using the following parameters r= mG ; L p= kρS KaS ; q= (r − 1) L G reduce the above equation to d2 y dy pq mz + ( p + q) + pqy = ry0 − 0 dh 2 dh r − 1 w (3.136) subject to the boundary conditionsat h = 0, x = 0 and at h = H, y = 0. Equation 3.136 is a linear constant coefficient equation and can be solved in the following way: pq mz Put in operator form: ( D 2 + [ p + q]D + pq) y = ry0 − 0 r − 1 w Annihilate in right-hand side: D(D2 + [p + q] D + pq)y = 0 Write down the characteristic equation: t(t2 + [p + q]t + pq) = 0. Such that t1 = 0, t2 = −p, and t3 = −q are the characteristic roots. Then the general solution is y = c1 + c2 e − ph + c3e − qh where c2 and c3 are arbitrary and c1 is a particular constant to be determined through substitution into Equation 3.136. That is, yp is given as yp = c1 , yp′ = yp′′ = 0 Then pqc1 = pq mz ry0 − 0 r − 1 w or c1 = 1 mz ry0 − 0 r − 1 w Finally, the general solution becomes yg = 1 mz ry0 − 0 + c2 e − ph + c3e − qh r − 1 w 90 Applied Mathematical Methods for Chemical Engineers and substitution of the boundary conditions into the general solution gives y= mz 0 − ph e − pH − v − qh ve − qH − re − pH e + e + r − e − qH w(r − v ) r − e qH (3.137) where v = 1+ y0 = kρG q + rp − p = KaL q mz 0 r − 1 − pH v − 1 − qH + 1− e e r − v w(r − e − qH ) r − v (3.138) The condition y = y0 at h = 0 was used to derive Equation 3.138. Equation 3.137 gives the weight fraction of glycerine in the extract phase as a function of column height h. Taking the solubility of water in tallow into consideration and using mean flow rates together with the given data result in L = 8540 G = 3760 y0 = 0.188 r = 4.544 p = 0.198 q = 0.00348Ka v = 1 + 201.6/(Ka) Solving Equation 3.137 with H = 72 ft, gives Ka = 14.2 lb/(h ft3). With Ka known, the values of y, y*, and z can be determined as functions of column height using Equations 3.137, 3.134, 3.132, and 3.133, and the fraction of tower height principally required for chemical reaction can be determined. Example 3.18 This example covers heat loss through pipe flanges [15]. Two thin-wall metal pipes of 1-in. outside diameter are connected by ½-in. thick and 4-in. diameter flanges that are carrying steam at 250ºF. Determine the rate of heat loss from the pipe and the proportion that leaves the rim of the flange. Thermal conductivity of the flange metal is k = 220 Btu/h ft2 ºF ft−1. The exposed surfaces of the flanges lose heat to the surroundings at T1 = 60ºF according to a heat transfer coefficient h = 2 Btu/h ft2 ºF. Solution Consider one flange with one exposed circular face and an exposed rim with radial coordinate r measured in inches from the axis of the pipe. Then the heat balance over an element of width ∆r, as shown in Figure 3.6, gives k dT Input = −2π (1/2)r 12 dr 91 Linear Second-Order and Systems of First-Order r + Δr r 2 in. FIGURE 3.6 1/2 in. Pipe flange. Output = − πkr dT d πkr dT 2πrh∆r + − ∆r + 144 (T − T1 ) dr r 12 dr 12 d flange surface heat lost to surroundings Accumulation = 0 Simplification of the heat balance leads to r d 2T dT h + − r (T − T1 ) = 0 dr 2 dr 6 k which can be reduced to x2 d2 y dy + x − x2y = 0 2 dx dx (3.139) for y = T − T1, and x = r h/6 k . Equation 3.139 is a modified Bessel equation of zero order. By comparison with Equation 3.80, we get y1 = J 0 (ix ) and y2 = Y0 (ix ) but J0(ix) is usually denoted as I0(x) and Y0(ix) as K0(x), where I0 (x) = 1 + x2 x4 x6 + 2 2 + 2 2 2 +L 2 2 24 246 and 1 2 x + L are real forms. The number γ is Euler’s 4 constant. Therefore, the general solution of Equation 3.139 is K 0 ( x ) = [ln 2 − γ ]I 0 ( x ) − I 0 ( x ) ln( x ) + yg = c1I 0 ( x ) + c2 K 0 ( x ) 92 Applied Mathematical Methods for Chemical Engineers Application of the boundary conditions at r = 1 / 2, T = 250 ⇒ x = 0.0195, y = 190 and at r = 2, − h k dT dy = (T − T1 ) ⇒ x = 0.078, = −0.0195 y 12 dr 144 dx gives T = 60 + 186.5I 0 (0.039r ) + 0.8636 K 0 (0.039r ) where the values of the pertinent Bessel function are taken from standard math tables [13]. The heat conducted from the pipe by the flange is given by Q1 = − πk dT = 47.81 Btu/h 24 dr r =1/2 and the heat lost through the rim is given by Q2 = − πk dT = 16.62 Btu/h 6 dr r = 2 Therefore, the pipe loses 47.8 Btu/h through each flange, of which 35% is lost through the rim. Example 3.19 This example covers a tubular gas preheater [15]. 1000 ft3/h of 70ºF air is drawn through a 4-ft long and 4-in. diameter heated cylindrical pipe. The wall temperature, Tw, of the pipe is maintained at 600ºF for the total length, and the overall heat transfer coefficient, h, as a function of x (distance measured from pipe inlet) is h = 5 x −1/2 Btu/h ft 2 °F The air properties are as follows: Specific heat (Cp) = 0.24 Btu/lbºF Thermal conductivity (k) = 0.020 Btu/h ft2 ºF ft−1 Density (ρ) = 0.050 lb/ft3 Assumptions: Heat transfer occurs by conduction within the gas in an axial direction. Mass flow of the gas is in an axial direction. Solution With the aid of Figure 3.7, the following heat balance can be established. Input dT dx Conduction − kA Mass flow uρCpT Wall heat transfer πDh(Tw – T)∆x Output − kA dT d dT + − kA ∆x dx dx dx dT ∆ uρC p T x dx 93 Linear Second-Order and Systems of First-Order 4 ft 70°F, u, Cp x FIGURE 3.7 Tw h Δx Gas preheater. where A is the pipe cross-sectional area and D is its diameter. Then dT + uρC pT + πDh(Tw − T ) ∆x dx dT d dT dT = − kA + − kA ∆x + uρC p T + ∆x dx dx dx dx − kA simplifies to πDh(Tw − T ) = − kA d 2T dT + uρC p dx dx 2 and rearranging gives d 2T uρC p dT πDh (Tw − T ) = 0 − + dx 2 kA dx kA (3.140) Inserting the numerical values and substituting x = z2, and t = Tw − T, results in z dt d 2t − (1 + 13760 z 2 ) − 12000 z 2t = 0 dz dz 2 Attempting to solve this differential equation by the method of Frobenius gives c = 0 and c = 2 as roots of the indicial equation. Then, for c = 2 we get a1 = 0 a3 = 800 a0 a2 = 3440 a0 a4 = 7.9 × 10 6 a0 The coefficients are increasing, even though a convergence test would suggest that this series is convergent. This is a case where a convergent series is not useful, since more than 100 terms would be required to determine the first solution. This dilemma occurs due to the much larger coefficients of dt/dz and t in comparison to that of d2t/dz2. By neglecting the gas conduction term, we get 94 Applied Mathematical Methods for Chemical Engineers dt + 0.436 x −1/2t = 0 dx which can be solved to give t = α exp(−0.872 x 1/2 ) Then at x = 0, t = 530, such that t = 530 exp(−0.872 x 1/2 ) (3.141) or T = 600 − 530 exp(−0.872 x 1/2 ) resulting in an exit gas temperature of 507ºF. A check on the accuracy of this approximation shows an error of 7.5% for an x = 10 −3. That is, Equation 3.141 can be differentiated twice to show that d 2t dx 2 1 1 688 dt = + dx 13760 x 15780 x 1/2 which is small, except when x is small. Example 3.20 A control valve of the type shown in Figure 3.8 is actuated by air pressure ranging from 3 to 15 psig operating on a 16-in. diameter diaphragm whose effective area is equivalent to 100 in.2. The effective dead weight of the moving parts of the valve, allowing for the friction in the gland, etc., is estimated to be 300 lbf, the stiffness of the spring is 600 lbf/in., and the damping constant is estimated to be 17 lbf-s/in. If the total lift Air Spring x Diaphragm Backing plate Valve stem FIGURE 3.8 Control valve. 95 Linear Second-Order and Systems of First-Order of the valve is 2.0 in., predict the response of the valve if the controlling air pressure suddenly changes from 6 to 12 psig. Solution A force balance on the valve is input force = ∆PAU(t), U(t) is the unit step function. Output force consists of • Force to overcome inertia of movable parts = m • Force to overcome resistance of the spring = kx dx • Force to overcome damping resistance = c dt d2 x dt 2 Therefore, at equilibrium m d2 x dx + c + kx = ∆PAU (t ) dt 2 dt By letting α = c/2mω where ω = k /m, simplify the differential equation to d2 x dx ∆P + 2αω + ω 2 x = U (t ) dt 2 dt m (3.142) Laplace transform of Equation 3.142 gives ( S 2 + 2αωS + ω 2 ) x = ∆PA + Sx (0) + 2αωx (0) + x ′(0) mS where x = L{x (t )} Then, assuming that the origin of the displacement of the moving parts of the valve corresponds to an air pressure of 6 psig gives x (0) = x ′(0) = 0 and x becomes x= ∆PA mS ( S 2 + 2αωS + ω 2 ) (3.143) By the method of partial fractions we get ∆PAS /mω 2 2α∆PA/mω ∆PA ∆PA = − 2 − 2 2 mS ( S + 2αωS + ω ) mω S S + 2αωS + ω 2 S 2 + 2αωS + ω 2 2 Then, inverting each term on the right-hand side and simplifying the algebra results in L−1{x (s)} = x (t ) = ∆PA α sin ω 1 − α 2 t e −αωt 1 − cos ω 1 − α 2 t + k 1 − α2 96 Applied Mathematical Methods for Chemical Engineers Alternatively, the inversion can be carried out in the following way: consider ∆PA P (s ) = mS ( S 2 + 2αωS + ω 2 ) Q(s) where the polynomial Q(s) has singularities at S0 = 0, S1 = −αω + iω 1 − α 2 , and S2 = −αω − iω 1 − α 2 , then ρ 0 (t ) = ∆PA 0 t ∆PA e = Q′(0) mω 2 ∆PA S1t ∆PAe −αωt e − iω 1−α t e = Q′( S1 ) 2mω 2 (α 2 − 1 − iα 1 − α 2 } 2 ρS1 (t ) = ∆PA S2t ∆PAe −αωt e − iω 1−α t e = e − iω Q ′( S2 ) 2mω α 2 − 1 + iα 1 − α 2 2 ρS2 (t ) = ( ) 1−α 2 t Then application of Equation 3.97 gives −αωt e ∆PA α sin ω 1 − α 2 t e −αωt = 1 − cos ω 1 − α 2 t + 2 k 1− α x (t ) = ∆PA ∆PA + mω 2 2mω 2 2 2 e − iω 1−α t e − iω 1−α t + 2 2 2 α − 1 − iα 1 − α 2 α − 1 − iα 1 − α where the Euler formulas {e iθ = cos θ + i sin θ} {e − iθ = cos θ − i sin θ} have been used. Therefore, x (t ) = 1 − (cos 25.5t + 0.427sin 25.5t )e −10.9t Example 3.21 This example covers an application to process control. Consider the differential equation ay′′ + by′ + cy = u(t ); y(0) = y0 and y′(0) = y0′ (3.144) where a, b, and c are constants and u(t) is a function that has a Laplace transform. Then in terms of process control nomenclature, u(t) is the input, while the solution y(t) is the output or response function. Further, y(t ) = yc (t ) + yp (t ) 97 Linear Second-Order and Systems of First-Order where yc(t) is the solution to the associated homogeneous equation ay′′ + by′ + cy = 0 and yp is chosen such that yp(0) = yp′(0) = 0, whereas yc(0) = y0 and yc′(0) = y0′. Then L{ay′′ + by′ + cy = 0} = 0 gives Yc (s) = 1 [a(sy0 + y0′ ) + by0 ] as 2 + bs + c Also L{ayp′ + byp′ + cyp = u(t )} with yp (0) = yp′ (0) = 0 gives Yp (s) = 1 U (s), U (s) = L{u(t )} as 2 + bs + c The factor 1/(as2 + bs + c) is denoted as H(s) and is called the transfer function of Equation 3.144. Notice here that H(s) and the initial conditions y(0) = y0 and y′(0) = y0′ together produce the complementary solution yc (t ) = L−1{Yc (s)} while the transfer function and the input function u(t) produce the particular solution yp (t ) = L−1{Yp (s)} Therefore, u(t) is said to control the response yp(t); that is, u(t) is the control function. From Figure 3.9, if H is considered fixed, then u(t) controls the output yp(t). Further, since H(s) = L{h(t)}, where h(t) is some function, then the transfer function is the Laplace transform of the impulse response. That is, ay′′ + by′ + cy = δ(t ), y(0) = y′(0) = 0 y0 , y09 u H yp + yc FIGURE 3.9 A control system with transfer function H(s). y 98 Applied Mathematical Methods for Chemical Engineers transforms to as 2Y (s) + bsY (s) + cY (s) = 1 or Y (s ) = 1 as 2 + bs + c If we recall the following three cases for positive coefficients of Equation 3.144, that is, a, b, and c are all-positive, then Case I: for b2 − 4ac > 0 m1 = − b + b 2 − 4 ac 2a and m2 = − b − b 2 − 4 ac 2a such that yc = c1e m1t + c2 e m2t Case II: for b2 − 4ac = 0 m1 = m2 = −b 2a and yc = c1e m1t + c2te m1t and Case III: for b2 − 4ac < 0 m1 = α + iβ and m2 = α − iβ and yc = eαt (c1 cos βt + c2 sin βt ) In each of the above three cases, the solution yc(t) → 0 as t → ∞. Therefore, yc(t) and the initial conditions y0 and y0′ become less important to the response function (y = yc + yp) as t gets larger and larger. That is, y → yp as t → ∞, thus yp is the steady-state solution. Also, yp = L−1{H (s)U (s)} = h(t ) * u(t ) That is, for positive coefficients, the response tends to the steady state given by the convolution of the impulse response with the control function. Example 3.22 Consider the steady flow between parallel planes [24]. Two planes are a distance 2b apart in the y direction and extend to infinity in the z direction. Assume that the fluid flow between them is steady. Derive the average velocity, ū, in terms of the impressed Linear Second-Order and Systems of First-Order pressure gradient, the distance between the plates, and the fluid viscosity. Also write down the ratio of the velocity to that of the average. Solution Starting with the general equation in Cartesian coordinates for incompressible fluid with viscosity independent of position [19], ∂u/∂z = 0, ∂2u/∂z2 = 0, and the pressure gradient can be taken as constant, that is, dp =A dx Then d2 y A = dy 2 µ which integrates to u= Ay 2 + c1 y + c2 2µ Applying the conditions u = 0 when y = ±b results in u= 1 dP 2 ( y − b2 ) 2µ dx (What is the sign of the pressure gradient when flow is to the right?) The average velocity is determined by integrating over the cross-section: u= 1 b 2 Ab 2 u dy = − ∫ 2b − b 3µ and u 3 y 2 = 1 − u 2 b Example 3.23 This example considers the dynamic behavior of the liquid level in a given cylindrical tank is described by A dh = Fi − k h 2/3 dt where A (m2) is the uniform cross-sectional area, Fi is the inlet flow rate (m3/min), h (m) is the liquid level, and k is a given constant. a.Linearize this process model around a steady-state level hSS and obtain an appropriate transfer function model relating the deviation variables h (s) and Fi (s). Explicitly specify what the steady-state gain and time constant of the approximate transfer function are. b.Suppose the given tank area is A = 0.25 m 2 and the height of the tank is 1m with a steady-state liquid level of 0.512 m while operating at an initial 99 100 Applied Mathematical Methods for Chemical Engineers steady-state flow rate of 0.32 m 3 min. If the flow rate were to be increased suddenly from 0.32 m 3 min to a new value of 0.52 m 3 min and remained there, the approximate linear model would (misleadingly) predict that the liquid level would arrive at a new steady-state level within the limits of the total tank height and would not result in an overflow; however, the nonlinear model would predict that this change in flow rate will in fact cause the tank to overflow. Confirm or refute this prediction. Solution 2/3 =0 SS-balance: Fi 0 − k hSS dh 2/3 US-balance: Fi − k h = A dt Purturbation: US-SS: 2/3 Fi − Fi ,SS − k h 2/3 − k hSS =A dh d ( h − hSS ) 2/3 ≡ F i − k h 2/3 − k hSS =A dt dt Let f (h) = h 2/3 Then f (h) = f (0) + df 2 ( h − hSS ) = hSS2/3 + hSS−1/3h dh SS 3 Therefore dh dh 2 −1/3 2 −1/3 Fi − k hSS h=A or Fi = k hSS h+A dt dt 3 3 Resulting in dh 1 A F =h+ −1/3 i −1/3 2 / 3k hSS 2 / 3k hSS dt τ K Taking Laplace transform results in K Fi ( s ) = h ( s ) + τ s h ( s ) = h ( s ) ( τ s + 1) such that h (s) = K Fi ( s ) ( τ s + 1) Using the given values result in Fi = 0.20 m 3 min; Fi ( s ) = τ= F ,SS 0.2 1 ; k = i2/3 = 0.5m 7/3 ; K = = 2.4 −1/3 s hSS 2 3 k hSS A A B 2.4 0.2 0.8 = 0.6; h ( s ) = = + ⇒ h ( s ) = −1/3 2 3 k hSS ( 0.6s + 1) s ( s + 5/3) s s s + 5/3 101 Linear Second-Order and Systems of First-Order Following inversion, we get h ( t ) = 0.48 − 0.48e −5t 3 Applying the final value theorem+ h ( ∞ ) = lim 0.48 (1 − e −5 t /3 ) = 0.48 ⇒ h ( ∞ ) + hSS = 0.992 < 1m t →∞ However, the given nonlinear system at a new steady state is given by F 2/3 Fi ,SS = k hSS ⇒ hSS = i ,SS k 3/2 0.52 = 0.5 3/2 = 1.06m > 1m which confirms that the system would overflow. + Terminal Value Theorem (AKA final value theorem) is one of two asymptotic values that are commonly employed in the solution of Process Control problems [30– 33]. The essential feature of the “final value theorem” rests on the following notion: Suppose that a function f (t ) is continuous on the open interval ( 0, ∞ ) and is of expodf nential order β and is piecewise continuous on ( 0, ∞ ); then dt L { } df = sL { f (t ) − f (0 + )} , Re s > β dt Furthermore if lim f (t ) exists, then lim f (t ) = lim s F (s) for real s. t →∞ s→ 0 t →∞ Example 3.24 This example covers an application of the Routh stability criterion [32]. Given the series reaction k3 k1 k2 A → B → C →D and the corresponding model equations: dC A = − k1C A dt (3.145) dC B = k1C A − k 2C B dt (3.146) dCc = k 2C B − k3Cc dt (3.147) 102 Applied Mathematical Methods for Chemical Engineers a. Show that the third order ordinary differential equation describing the concentration of component C is d 3Cc d 2C dC + k1 + k2 + k3 2 c + k1k2 + k1k3 + k2 k3 c + k1k2 k3Cc = 0 (3.148) 3 dt dt dt b. Assuming that all the kinetic parameters are positive, show that the s­ ystem is stable. Solution Solving for species A concentration in Equation 3.146 CA = 1 dC B k 2 + CB k1 dt k1 (3.149) Followed by differentiation with respect to time results in dC A 1 d 2C B k2 dC B = + dt k1 dt 2 k1 dt Substituting C A , (3.150) dC A into Equation 3.145 results in dt 1 dC B k 2 1 d 2C B k2 dC B = − k1 + + CB k1 dt 2 k1 dt k1 k1 dt (3.151) Similarly from Equation 3.147 we can solve for C B to get CB = 1 dCc k3 + C k2 dt k2 c (3.152) Then dC B 1 d 2Cc k3 dCc = + dt k2 dt 2 k2 dt (3.153) d 2C B 1 d 3Cc k3 d 2Cc = + dt 2 k 2 dt 3 k 2 dt 2 (3.154) and Upon substitution of Equations 3.152, 3.153, and 3.154 into Equation 3.151 result in 1 1 d 3Cc k3 d 2Cc k2 + k1 1 d 2Cc k3 dCc + + + k1 k2 dt 3 k2 dt 2 k1 k2 dt 2 k2 dt 1 dCc k3 + k2 + Cc = 0 k dt k 2 2 (3.155) 103 Linear Second-Order and Systems of First-Order Further simplification results in Equation 3.148. According to the Routh Stability Criterion, the characteristic polynomial is λ 3 + k1 + k2 + k3 λ 2 + k1k2 + k1k3 + k2 k3 λ + k1k2 k3 = 0 In this case a0 = 1, a1 = k1 + k2 + k3 , a2 = k1k2 + k1k3 + k2 k3 and a3 = k1k2 k3 a a −a a k1k2 k3 b1 = 1 2 0 3 = k1k2 + k1k3 + k2 k3 − > 0, b2 = 0 a1 k1 + k2 + k3 ba −a b c1 = 1 3 1 2 = k1k2 k3 , c2 = 0 b1 Such that Row 1 2 1 k1 + k2 + k3 3 k1k2 + k1k3 + k2 k3 − 4 k1k2 + k1k3 + k2 k3 k1k2 k3 k1k2 k3 k1 + k2 + k3 0 0 k1k2 k3 3.8 SYSTEMS OF FIRST-ORDER ORDINARY DIFFERENTIAL EQUATIONS There is an important connection between systems of first-order equations and single equations of higher orders. Indeed, an nth-order equation y( n ) = F (t , y, y′,, y( n−1) ) (3.156) can always be reduced to a system of n first-order equations. We can do this by introducing the variables x1, x2, x3, …, xn defined by x1 = y, x 2 = y′, x3 = y′′,, x n = y( n−1) Then (3.157) x1′ = x 2 x 2′ = x3 x3′ = x 4 x n′ −1 = x n (3.158) 104 Applied Mathematical Methods for Chemical Engineers For example, given the linear second-order equation x ′′ + 4 x = 0 if we let x ′ = y, then y ′ = −4 x such that our system of 2-first-order equations becomes x′ = y y ′ = −4 x Another example consists of the linear second-order nonhomogeneous equation y ′′ + 3 y ′ + 3 y = sin t The system becomes y′ = x x ′ = −3 x − 3 y + sin t upon substitution of x = y′ As a third illustration, consider y ′′′ + y = 0 Letting y′ = x; x′ = z results in the system y′ = x x′ = z z′ = − y As illustrated in these three examples, using the notion given by Equation 3.157, Equation 3.156 can be restated as x n′ = F (t , x1 , x 2 ,, x n ) (3.159) It is also important to note that Equation 3.158 and 3.159 represent a special case of the more general system: x1′ = F1 (t , x1 , x 2 ,…, x n ) x 2′ = F2 (t , x1 , x 2 ,…, x n ) x3′ = F3 (t , x1 , x 2 ,…, x n ) x n′ −1 = Fn (t , x1 , x 2 ,…, x n ) (3.160) Linear Second-Order and Systems of First-Order 105 The form given in Equation 3.160 is most suitable to begin our examples of how to solve systems of first-order differential equations. Example 3.25 Consider the given system of two equations. This is an example of a homogeneous system. x1′ = x1 + x 2 (3.161) x 2′ = 4 x1 − 2 x 2 (3.162) One approach to solving this system is to rearrange Equation 3.161 in terms of x2, that is, x 2 = x1′ − x1 and differentiate the rearranged equation to get x 2′ = x1′′− x1′ Then substitute both the differentiated and the rearranged equations into Equation 3.162 to get x1′′− x1′ = 4 x1 − 2( x1′ − x1 ) = 6 x1 − 2 x1′ or x1′′+ x1′ − 6 x1 = 0 This is now a linear, second-order, constant coefficient homogeneous equation (Section 3.3) whose general solution is x1 (t ) = c1e 2t + c2 e −3t Substitution into Equation 3.161 followed by simplification, we get x 2 (t ) = c1e 2t − 4c2 e −3t To define the arbitrary constants, c1 and c2, we would need conditions at some value of t. Example 3.26 In this next example, we will solve a nonhomogeneous system of two equations. Consider x1′ = 2 x1 − 5 x 2 − sin 2t; x1 (0) = 0 (3.163) x 2′ = x1 − 2 x 2 + t; x 2 (0) = 1 (3.164) 106 Applied Mathematical Methods for Chemical Engineers Using the same approach as in the previous example we start out by rearranging Equation 3.164: x 2′ + 2 x 2 − t = x1 Differentiate the rearranged equation: x 2′′ + 2 x 2′ − 1 = x1′ Then substitute both the rearranged and differentiated equations into Equation 3.163 and simplify to get x 2′′ + x 2 = 1 − 2t − sin 2t which is a linear, second-order, and nonhomogeneous differential equation. Applying the appropriate method from Section 3.4 leads to the general solution x 2 (t ) = c1 cos t + c2 sin t + 1 − 2t + 1 / 3sin 2t Substituting this result into Equation 3.164 and carrying out the necessary simplification results in 2 x1 (t ) = (2c1 + c2 ) cos t + (2c2 − c1 )sin t − 5t + (sin 2t + cos 2t ) 3 Following substitution of the initial conditions we arrive at the final result 2 4 2 x1 (t ) = − cos t − sin t − 5t + (sin 2t + cos 2t ) 3 3 3 2 2 x 2 (t ) = − sin t + 1 − 2t + sin 2t 3 3 The procedure that was used to arrive at the solutions in Examples 3.25 and 3.26 is called the Method of Elimination. Alternatively, one could have used Laplace transform methods to solve Example 3.26. To demonstrate the use of Laplace transform, we will examine a less cumbersome example; that is, consider x1′ = x1 − 5 x 2 , x1 (0) = 1 (3.165) x 2′ = 2 x1 − 5 x 2 , x 2 (0) = 0 (3.166) Then letting L{x1(t)} = X1(s) and L{x2 (t)} = X2(s) we transform Equation 3.165 into sX1 (s) − x1 (0) = X1 (s) − 5 X 2 (s) (3.167) 107 Linear Second-Order and Systems of First-Order and Equation 3.166 into sX 2 (s) − x 2 (0) = 2 X1 (s) − 5 X 2 (s) (3.168) We now have two simultaneous algebraic equations in the unknowns X1(s) and X2(s). Following insertion of the given initial values and simplifying, the transformed system becomes (s − 1) X1 + 5 X 2 = 1 −2 X1 + (s + 5) X 2 = 0 If we multiply the second equation by 12 (s − 1) and add the result to the first equation, we get an equivalent system [5 + 1 / 2(s − 1)(s + 5)]X 2 = 1 −2 X1 + (s + 5) X 2 = 0 Now the first equation when solved for X2(s) gives X 2 (s ) = 2 s 2 + 4s + 5 X1 ( s ) = s+5 s 2 + 4s + 5 whereas We now invert both X1(s) and X2(s) to recover the sought-after solution for the given system, that is, x1 (t ) = L−1{X1 (s)} = L−1 { s+5 s + 4s + 5 2 } (s + 2) + 3 = L−1 2 (s + 2) + 1 3 −2t (s + 2) −1 = L−1 + L = e cos t + 3e −2t sin t 2 2 (s + 2) + 1 (s + 2) + 1 Similarly 2 x 2 (t ) = L−1 = 2e −2t sin t 2 (s + 2) + 1 Formally, we say that the system of differential equations (Equation 3.160) is linear if all of the Fis are linear functions in the xis; otherwise, the system is nonlinear. A linear system is called homogeneous if all the Fis are independent of t and nonhomogeneous when at least one Fi depends explicitly on t. A compact way of writing the system given by Equation 3.160 is to introduce the vectors F = ⟨ F1, F2, …, Fn ⟩ and x = ⟨ x1, x2, …, xn⟩ such that 108 Applied Mathematical Methods for Chemical Engineers x1′ = F1 (t , x1 , x 2 ,…, x n ) x 2′ = F2 (t , x1 , x 2 ,…, x n ) x3′ = F3 (t , x1 , x 2 ,…, x n ) ⇔ x′ = F(t , x) (3.169) x n′ −1 = Fn (t , x1 , x 2 ,…, x n )} This compact notation reminds us of some of the notions about first-order linear differential equations that we have used. For example, the notion of existence and uniqueness still applies. Theorem 3.11 Existence and Uniqueness [25] Consider the initial value problem given by x′ = F(t , x), x(t0 ) = x 0 (3.170) Suppose there is a neighborhood of t0 and x0 in which F is continuous in t and continuously differentiable in x at x0. Then there is a δ > 0 for which there is a unique solution to x′ = F(t, x) , x(t0 ) = x 0 with t ε (t0 − δ, t0 + δ). The proof of this theorem is dealt with in the given reference. The form given by Equation 3.169 is often expressed more explicitly as x′ = Ax + f (t ) (3.171) where A is an n × n matrix (see Section A.2 in Appendix A) and f(t) is an n × 1 vector-valued function. For example, the initial value problem x1′ = 2 x1 − 3 x 2 , x 2′ = 3 x1 + 2 x 2 , x1 (0) = 1, x 2 (0) = −1 can be restated as x1′ 2 −3 x1 0 1 = x + 0 , x 0 = −1 3 2 x 2′ 2 Here 0 f (t ) = 0 which means that we are working with the homogeneous form x′ = Ax (3.172) Linear Second-Order and Systems of First-Order 109 In practice, this form is most conducive to the applications of matrix theory [28,29]. To construct a solution to the homogeneous case, we begin with an approach similar to that used in the discussion of linear second-order constant coefficient differential equations. We seek solutions to Equation 3.172 of the form x = pe λ t (3.173) where λ and the constant vector p are to be determined. Substituting Equation 3.173 into Equation 3.172 results in λpe λt = Ape λt which reduces to λp = Ap or [ A − λI]p = 0 (3.174) where I is the n × n identity matrix. Now, to solve the system Equation 3.172, we must solve the system of algebraic equations 3.174. It is important to note that solving the system given by Equations 3.174 is exactly solving for the eigenvalues and eigenvectors of the matrix A. Therefore the vector x is a solution of Equation 3.172 if λ is an eigenvalue and p is an associated eigenvector of the coefficient matrix A. For example, let us reconsider Example 3.25: x1′ = x1 + x 2 x 2′ = 4 x1 − 2 x 2 Step 1: Rewrite the given system as x1′ 1 1 x1 x′ = = x; x = x 2′ 4 −2 x 2 Step 2: Assume x = peλt and substitute for x in the given problem. This leads to the system of algebraic equations (Equation 3.174) 1− λ 1 4 −2 − λ p1 0 = p2 0 (3.175) which will have a nontrivial solution if and only if the determinant (Appendix A) of coefficients vanishes. That is, 1− λ 1 = (1 − λ )(−2 − λ ) − 4 = 0 4 −2 − λ 110 Applied Mathematical Methods for Chemical Engineers Step 3: Solve the characteristic equation for the eigenvalues and determine the associated eigenvectors, that is, λ = 2, −3. For λ = 2, we solve the system given in Equation 3.175 to get − p1 + p2 = 0 ⇒ p2 = p1 Therefore, the eigenvector corresponding to λ = 2 can be taken as 1 p(1) = 1 For λ = −3, substitution into Equation 3.175 produces 4 p1 + p2 = 0 ⇒ p2 = −4 p1 This produces the corresponding eigenvector 1 p(2) = −4 Finally, we can state the solution in the form 1 1 x (1) (t ) = e 2t ; x (2) (t ) = e−3t 1 −4 Step 4: Determine if the solutions form a fundamental set by examining the Wronskian (Section 3.2) of these solutions: W [ x (1) , x (2) ](t ) = e 2t e −3t = −5e − t ≠ 0 e 2t −4e −3t Because the Wronskian is different from zero for all finite values of t, the solutions x(1) and x(2) form a fundamental set and the general solution of the given system, Equation 3.175 is x = c1x (1) (t ) + c2 x (2) (t ) 1 1 = c1 e 2t + c2 1 −4 −3t e which is the identical result obtained previously. This later method, although straightforward, can become complicated for larger systems. There is a good number of computer algebraic software [29] available and Linear Second-Order and Systems of First-Order 111 it is recommended that the reader explore the use of those tools when faced with a system of three or more equations. Similar to linear second-order problems, there are cases of complex eigenvalues and repeated eigenvalues. First, we will explore an example having complex eigenvalues and illustrate how to obtain real (as opposed to imaginary) solutions. Then we will solve an example that has repeated eigenvalues and illustrate how to obtain linearly independent solutions. Example 3.27 This example demonstrates the complex eigenvalue case [1]. Consider the linear system given by 0 1 x′ = x −1 0 Determine the two real linearly independent solutions. Solution Step 1: Construct the characteristic polynomial of A by taking the determinant of −λ 1 , that is, −1 −λ −λ 1 = λ2 + 1 = 0 −1 −λ The roots of this polynomial are λ 1 = i and λ 2 = −i. Step 2: Determine the associated eigenvectors by solving the system of algebraic equations −λ 1 p1 0 = for each λ −1 −λ p2 0 This reduces to the equation p2 = ip1 for λ1 = i If p2 = 1, then p1 = 1/i = −i such that −i p(1) = 1 is the associated eigenvector for the eigenvalue λ 1 = i. Similarly, the eigenvalue λ 2 = −i when substituted into −λ 1 p1 0 = −1 −λ p2 0 112 Applied Mathematical Methods for Chemical Engineers produces i p(2) = 1 This would mean that the general solution for the given system is −i i x(t ) = c1 e it + c2 e − it 1 1 Step 3: Expand the general solution and rename the constants x1 (t ) = − ic1e it + ic2 e − it = − ic1 (cos t + i sin t ) + ic2 (cos t − i sin t ) = − i(c1 − c2 ) cos t + (c1 + c2 )sin t and x 2 (t ) = c1e it + c2 e − it = c1 (cos t + i sin t ) + c2 (cos t − i sin t ) = i(c1 − c2 )sin t + (c1 + c2 ) cos t If we make the substitutions k1 = (c1 + c2 ) and k 2 = − i(c1 − c2 ) we now have x1 (t ) = − i(c1 − c2 ) cos t + (c1 + c2 )sin t = k1 sin t + k 2 cos t and x 2 (t ) = i(c1 − c2 )sin t + (c1 + c2 ) cos t = k1 cos t − k 2 sin t Therefore, the general solution can be restated in the form sin t x(t ) = k1 cos t cos t + k 2 − sin t In general, if the n × n matrix A is real, then the coefficients in the polynomial det( A − λI) = | A − λI | = 0, for λ is real and any complex eigenvalues must occur in conjugate pairs. Further, the corresponding eigenvectors are also complex conjugates. The following example illustrates the key steps to developing a solution when the eigenvalues repeat and it is possible to find linearly independent eigenvectors. 113 Linear Second-Order and Systems of First-Order Example 3.28 This example demonstrates the repeated eigenvalues case [1]. Consider the system 11 x′ = x 01 (3.176) Then the characteristic equation is (1 − λ )2 = 0 which results in λ = 1,1. For λ 1 = 1, the associated eigenvector turns out 1 to be . 0 Therefore one solution is 1 x (1) (t ) = et 0 To determine a second linearly independent solution we will assume that u1 1 x (2) (t ) = tet + u2 0 et (3.177) u1 Notice that the 2 × 1 vector u2 is unknown, but we will determine the components of this vector knowing that x(2)(t) must satisfy Equation 3.176. Substitution of Equation 3.177 into 3.176 gives dx (2) 1 t 1 t u1 = e + te + dt u2 0 0 et which must be equated to 1 1 1 t 1 1 u1 1 1 (2) 0 1 x (t ) = 0 1 0 te + 0 1 u 2 et That is, 1 t 1 t u1 0 e + 0 te + u 2 1 1 1 t 1 1 u1 et = te + 0 1 u 0 1 0 2 Because 1 1 1 1 0 1 0 = 0 et 114 Applied Mathematical Methods for Chemical Engineers the terms containing tet cancel out and we have 1 t u1 0 e + u 2 1 1 u1 et = 0 1 u2 et which can be restated in the form u1 1 [ A − I] = u2 0 that is, 1 1 1 0 u1 0 1 − 0 1 u 2 01 0 0 1 = 0 0 1 u1 1 0 0 u = 0 ⇒ u2 = 1 2 and the unknown vector is 0 1 Therefore, the second solution is 1 0 x (2) (t ) = tet + et 0 1 We now need to check to see if the two solutions x(1)(t) and x(2)(t) are linearly independent by calculating the Wronskian. et tet 2t W [ x (1) , x (2) ](t ) = =e ≠0 t 0 e Therefore, the two linearly independent solutions can be combined as x = c1x (1) (t ) + c2 x (2) (t ) to give the general solution. Notice that x(1)(t) and x(2)(t) form a fundamental set or fundamental solutions. 115 Linear Second-Order and Systems of First-Order 3.8.1 NONhOMOGENEOUs LINEAR SYstEMs In Section 3.4, we employed the undetermined coefficients (equivalent—­annihilation) or variation of parameters methods to solve the nonhomogeneous problem. In this section we will suggest the use of the same two methods to determine particular solutions of the nonhomogeneous system of first-order differential equations. An example will be provided to illustrate the necessary steps to follow when constructing the general solution to Equation 3.171, x ′ = Ax + f (t ) using the method of undetermined coefficients. Although the necessary theory to generalize the method of variation of parameters is beyond the scope of this book, we will provide an example outlining the method. Example 3.29 This is an example using the undetermined coefficient method [1]. Consider the system −4e −3t 0 1 x′ = x + −2t −1 0 e Step 1: Determine the complementary solution. In solving Example 3.27, we, in fact, derived the solution to the homogeneous part of this new problem. This means that we have the complementary solution, that is, sin t cos t x c (t ) = k1 + k 2 − sin t cos t Step 2: Select the candidate particular solution. Recall that we selected candidate particular solution based on the complementary solutions being linearly independent of the nonhomogeneous part of the given differential equation. In this example, the complementary solution is a linear combination of sin t and cos t. Therefore, the candidate particular solution is a combination of the exponential functions appearing in the nonhomogeneous part of the given system x p = αe −3t + βe −2t (3.178) The unknown n × 1 vectors α and β are to be determined by substituting into the given system. Step 3: Determine the unknown vectors. Upon substitution into the given system we get 1 0 −3αe −3t − 2βe −2t = Aαe −3t + Aβe −2t − 4 e −3t + e −2t 0 1 116 Applied Mathematical Methods for Chemical Engineers or 1 0 − (3I + A)αe −3t − (2I + A)βe −2t = −4 e −3t + e −2t 0 1 Then, equating coefficients of the exponential functions results in α1 e −3t : (3I + A) α 2 β1 e −2t : (2I + A) β 2 4 = 0 0 = − 1 Solving these two systems results in 6/5 1/ 5 α= and β = −2 / 5 2 / 5 Therefore the particular solution is 6 / 5 −3t 1 / 5 −2t xp = e + −2 / 5 e 2/5 and the general solution is xg = xc + xp The method of undetermined coefficients is effective when the matrix, A in Equation 3.171 is constant and the vector f has a special form (exponentials, polynomials, sines, and cosines). However, as we know from our experience with second-order equations, the method of variation of parameters is more general. This method is expected to work even when A depends on t and f belongs to a much larger class of vector-valued functions. Recall that the method of variation of parameters requires that we already have a set of linearly independent complementary solutions. Following that, we choose the candidate for the particular solution by allowing the arbitrary constants in the complementary solution to be functions of t. For example, consider the linear system given by e −2t 13 x′ = x+ 31 0 The first step is to solve the associated homogeneous problem using the approach already illustrated in the previous examples. That is, to determine the characteristic equation 1− λ 3 = (1 − λ)2 − 9 = 0 3 1− λ 117 Linear Second-Order and Systems of First-Order This leads to the eigenvalues λ 1 = − 2 and λ 2 = 4. Using each of these eigenvalues in turn, find the associated eigenvectors, that is, 3 3 p1 0 3 3 p = 0 ⇒ p2 = − p1 2 or 1 p(1) = for λ1 = −2 −1 Similarly, −3 3 p1 0 3 −3 p = 0 ⇒ p2 = p1 2 or 1 p(2) = for λ 2 = 4 1 Therefore the complementary solution is 1 −2t 1 4t x c = c1 e + c2 1 e −1 From this complementary solution one can form the matrix e −2t e 4 t X(t ) = −2t 4 t −e e which is known as a fundamental matrix, consisting of the fundamental solutions. The candidate for the particular solution is now given as t x p (t ) = X(t ) ∫ X −1 (s) f (s) d s (3.179) where X−1 is the inverse of the fundamental matrix. To determine X−1, follow the procedure given in Section A.2 of Appendix A, using elementary operations: 1. Form the augmented matrix. e −2t e 4 t 1 0 −2t 4 t −e e 0 1 118 Applied Mathematical Methods for Chemical Engineers 2. Add row 1 to row 2 followed by adding − 12 row 2 to row 1 to get e −2t 0 1/2 −1/2 0 2e 4 t 1 1 1 −4t 3. Multiply row 1 by e2t and then multiply row 2 by 2 e to get 1 0 1/2e 2t −1/2e 2t 0 1 1/2e −4 t 1 /2e −4 t Therefore, 1/2e 2t −1/2e 2t X −1 (t ) = 1/2e −4 t 1/2e −4 t Now ∫ t 2s −1/2e 2 s t 1/2e X −1 (s) f (s) d s = ∫ −4 s 1/2e 1/2e −4 s 1/2 t =∫ 1/2e −6 s ds 1/2t = −1/12e −6t −2 s e 0 ds such that t x p (t ) = X(t ) ∫ X −1 (s) f (s) d s e −2t e 4 t 1/2t = −2t 4 t −6 t − e e −1/12e 1 /2t = −1/2t 1/12 e −2t e −2t − 1/12 Notice that the last column is a multiple of c1 in the complementary solution and can be dropped from the particular solution, that is, 1 −2t 1 4t x(t ) = c1 e + c2 1 e + x p − 1 1 / 2t −2t 1 −2t 1 = c1 e + c2 e 4 t + e −1 1 −1 / 2t Linear Second-Order and Systems of First-Order 119 As usual, to determine the arbitrary constants, initial conditions are needed. In summary, the following theorem provides us with the formula to solve the nonhomogeneous initial value problem. Theorem 3.12 The unique solution to the initial value problem x ′ = A(t ) x + f (t ), x(0) = x 0 (3.180) is given by t x(t ) = X(t ) X −1 (0) x 0 + X(t ) ∫ X −1 (s) f (s) d s 0 (3.181) where X(t) is a fundamental matrix solution of the homogeneous system. 3.9 PROBLEMS 1. Two thin-wall pipes of 1-in. outside diameter have flanges 1/2-in. thick and 4-in. diameter on the ends joining them together. If the conductivity of the flange metal is k Btu/h ft2 ºF ft−1, and the exposed surfaces of the flange lose heat to the surroundings, which are at T1ºF by means of a heat transfer coefficient, h Btu/h ft2 ºF, show that the equation giving the temperature distribution in the flange is d 2 T dT 6k r 2 + = hr (T − T1 ) dr dr where r is the radial distance coordinate in inches. If the circular faces of the flanges are thermally insulated and the flanges lose heat only through the rim where h = 20 Btu/h ft2 ºF, solve the differential equation and determine the temperature of the rim for a pipe temperature of 200ºF, T1 = 60ºF, and k = 200 Btu/h ft2 ºF ft−1. 2.Mass transfer: As liquid flows across any plate of a distillation column its composition, x, changes from the entry concentration, x0, to the exit concentration, x1[15]. The composition at any point on the plate is influenced by the passage of the stripping gas at a rate G, the bulk flow at a rate L, and the mixing on the plate by the eddy diffusivity, DE. If a constant Murphree point efficiency * = Emv y − y1 y* − y1 120 Applied Mathematical Methods for Chemical Engineers can be assumed and a straight line equilibrium curve given by y* = mx + b show that the liquid composition satisfies the equation DE d2 x dx mGC * −V − Emv ( x − x * ) = 0 dz 2 dz Lz1 where z is the distance measured along the plate from the inlet weir, z1 is the distance between the weirs, V is the linear velocity of the liquid, and x * is the liquid composition in equilibrium with the entering gas, which is constant across the plate. 3.Consider the problem of heat loss from the surface of an oven wall due to “through metal,” which conducts heat from the inside. The heat is dissipated to the air from the sheet-metal protective covering of the insulated housing. The metal covering consists of 0.005-ft-thick steel having thermal conductivity of 25 Btu/h ft2 ºFft−1. The surface coefficient of heat transfer is 2.5 Btu/h.ft2 ºF and the head of the bolt is 5/8-in. diameter. The temperature of the room is 70ºF and the bolt head temperature is constant at 150ºF. Neglecting heat loss, except by conduction along the bolt, determine temperatures of the outer metal wall at several points up to 1 ft from the bolt. (Hint: Since the temperature is symmetrical about the bolt, then T can be assumed to be a function of r only.) 4.A copper fin L-ft long is triangular in cross-section [6]. It is w-ft thick at the base and tapers off to a line (see Figure 3.10). The base of this wedgeshaped piece of metal is maintained at a constant temperature TA, and the fin loses heat by convection to the surrounding air, which is at a temperature TB. The surface coefficient of heat transfer is h Btu/h ft2 ºF. Derive the relationship between the temperature, T, of the metal fin and the distance from the base, L − x. (Hint: Assume T is a function of x only.) Answer : d w dT k x = h 2sec θ(T − TA ) dx L dx L Base at TB Air at TA x dx FIGURE 3.10 Wedge-shaped fin. 121 Linear Second-Order and Systems of First-Order y = +h, u = U, T = T1 Moving r = const. y y = –h, u = 0, T = T0 FIGURE 3.11 z Couette flow between parallel plates. 5. Steady flow between a fixed and a moving plate [26]: Assume that (1) two infinite plates are 2h apart, and the upper plate moves at speed U relative to the lower; (2) the pressure p is constant; and (3) the upper plate is held at temperature T1 and the lower plate at T0 (see Figure 3.11). Derive the velocity and temperature profiles assuming that u = u(y) and T = T(y). Also, find the shear stress. µU 2h ∂u =0 Continuity : ∂x d2u Momentum : µ 2 = 0 dy Answer : 2 Energy : k du d2T + µ = 0 dy dy 2 6.Diffusion with chemical reaction: A gas is absorbed by a solution with which it reacts chemically [6]. The rate of diffusion in the liquid can be assumed proportional to the concentration gradient, and the diffusing gas is eliminated as it diffuses by a first-order chemical reaction. The reaction rate is proportional to the concentration of the solute gas in the liquid. Obtain an expression for the concentration in the liquid as a function of the distance from the gas–liquid interface (see Figure 3.12). Answer : d dc D = kc dx dx Interface dx x FIGURE 3.12 Diffusion with chemical reaction. 122 Applied Mathematical Methods for Chemical Engineers 7.Consider the set of reversible reactions [6]. 1 A k →B 2 →A B k 3 →C B k 4 →B C k Suppose the initial amount of A is 1 mol and NA, NB, and NC denote the moles of A, B, and C, respectively, at any time t. Derive a second-order constant coefficient linear differential equation relating NA, t, and the rate constants. 8.Oxygen dissolves into and reacts irreversibly with aqueous sodium sul* fite solutions [23]. If the gas solubility is denoted as CA at the liquid–gas ­interface, develop a differential equation that describes the steady-state composition profiles of O2 in the liquid phase when kCAn is the O2 reaction rate, and J A = − DA dCA dz describes the local O2 diffusion flux. Answer : DA d 2C A = kCAn dz 2 9.If one of the planes of Example 3.22 is moving with speed V, what are the boundary conditions? Derive the velocity profile for this case. V y A Answer : u = + 1 + ( y 2 − b 2 ) 2 b 2µ 10.Consider the steady flow of a fluid through a circular pipe of radius R. Suppose the fluid flows with a velocity uz = U in the z direction, starting with the continuity equation in polar coordinates, and the Navier–Stokes equations [19] show that 0=− A µ 1 d duz + r ρ ρ r dr dr where A is the pressure gradient, which is taken as constant. Why is the quantity Linear Second-Order and Systems of First-Order 123 ∂ 2 uz = 0? ∂θ2 Also, show that dP = − 8µuz µu dz = −32 2z dz R2 D 11. Reduction of linear boundary value problems to initial value problems [27]: Consider the problem given by the differential equation d2 y dy + f1 ( x ) + f2 ( x ) y = r ( x ) dx 2 dx (3.182) and the boundary conditions y(a) = ya and y(b) = yb (3.183) This problem can be transformed into a set of two initial value problems. Transform Equation 3.182 into a set of two differential equations by ­assuming that y( x ) = y1 ( x ) + µy2 ( x ) (3.184) is a linear combination of solutions to the desired differential equations, where the quantity μ is a constant. Use the boundary conditions given by Equation 3.183 to derive the initial conditions y1 (a) = ya and y2 (a) = 0 (3.185) dy2 (a) =1 dx (3.186) and set dy1 (a) = 0 and dx to identify the missing slope. Transform the boundary condition at the ­second point to define μ. 124 Applied Mathematical Methods for Chemical Engineers Answer: Initial value problem-1: d 2 y1 dy + f1 ( x ) 1 + f2 ( x ) y1 = r ( x ) 2 dx dx dy1 (a) y1 (a) = ya , =0 dx Initial value problem-2: d 2 y2 dy + f1 ( x ) 2 + f2 ( x ) y2 = 0 dx 2 dx dy2 (a) y2 (a) = 0, =1 dx µ = [ yb − y1 (b)] / y2 (b) 12.Find a fundamental set of solutions of 1 −1 x ′ = Ax; A = 1 3 13. Use the method of undetermined coefficients to find a particular solution of 2e − t −2 1 x′ = x+ 1 −2 3t 14.Show that Equation 3.179 is a valid candidate particular solution when the method of variation of parameters is used. 15.Use the method of variation of parameters to find the general solution to 2e − t −2 1 x′ = x+ 1 −2 3t 16.Use the method of Laplace transforms to solve the system 2e − t −2 1 x′ = x+ 1 −2 3t 0 ; x0 = 0 17.Given the system of first-order ordinary differential equations below dx1 = −4 x1 + x 2 ; x1 ( 0 ) = 100 dt dx 2 = 4 x1 − 4 x 2 ; x 2 ( 0 ) = 0 dt Linear Second-Order and Systems of First-Order 125 a. Use the method of Laplace transform to solve the system. b. Solve the system by first identifying the eigenvalues followed by inverting the coefficient matrix. 18. Given a batch reactor with a series reaction in which species A reacts reversibly to form the desired specie B. In this reaction, it is observed that specie B react to form an undesired specie C kif k2 B A →C k (3.187) 1r where k1 f and k1rrepresent the rate constants for the forward and reverse reactions for the conversion of species A and B, while k2 is the rate constant for the conversion of species B to species C. If it is assumed that each of the reactions is of first-order then the modeling equations are given by dC A = − k1 f C A + k1r C B dt dC B = k1 f C A − k1r C B − k2C B dt dCC = k 2C B dt (3.188) (3.189) (3.190) where C A , C B and CC are the concentrations (mol/volume) of components A, B, and C. a. Given the following dimensionless quantities: The dimensionless time: τ = k1t Conversion of species A: x1 = (C A0 − C A ) C A0 Dimensionless concentration of species B: x2 = C B C A 0 Ratio of rate coefficients: α = k2 k1 f Ratio of forward and reverse rate coefficients: β = k1r k1 f Show that the equation for the dimensionless concentration of species B is d 2 x2 dx + (α + β + 1) 2 + αx 2 = 0 2 dτ dτ and that the roots of the characteristic equation can never be complex or unstable for positive rate coefficients. b.Determine x 2 ( α , β, τ ) . c.Given k1f = 2, k1r = 1, and k2 = 1.25 hr −1 use a computer algebraic system (e.g., Mathematica®) to determine the maximum conversion of species A to species B and the reaction time required for this conversion. d. Suppose the real value of k2 is 1.5hr −1 instead of the previous (1.25 hr –1) and if the reaction is run for the time found in part c), what will be the actual conversion of A to B? 126 Applied Mathematical Methods for Chemical Engineers REFERENCES 1.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley & Sons, New York, 2005. 2. Giordano, F.R. and Weir, M.D. Differential Equations, a Modeling Approach, AddisonWesley, Reading, MA, 1988. 3.Greenberg, M.D. Advanced Engineering Mathematics, Prentice Hall, Englewood Cliffs, NJ, 1988. 4. O’Neil, P.V. Advanced Engineering Mathematics, 4th ed., PWS-Kent, Boston, 1995. 5.Hildebrand, F.B. Advanced Calculus for Applications, 2nd ed., Prentice Hall, Englewood Cliffs, NJ, 1976. 6.Mickley, H.S., Sherwood, T.K., and Reid, C.E. Applied Mathematics in Chemical Engineering, McGraw-Hill, New York, 1957. 7. Loney, N.W. Analytical solution to mass transfer in laminar flow in hollow fiber with heterogeneous chemical reaction, Chem. Eng. Sci., 51, 3995, 1995. 8. Huang, C.R. Heat transfer to a laminar flow fluid in a circular tube, AIChE J., 30, 833, 1984. 9.Watson, G.N. A Treatise on the Theory of Bessel Functions, 2nd ed., Cambridge University Press, London, 1966. 10. Gray, A. and Matthews, G.B. A Treatise on Bessel Functions and Their Applications to Physics, 2nd ed., Dover, New York, 1966. 11.Tranter, C.J. Bessel Functions with Some Physical Applications, English University Press, London, 1968. 12.Slater, L.C.J. Confluent Hypergeometric Functions, Cambridge University Press, London, 1960. 13. Spiegel, M.R. Mathematical Handbook, McGraw-Hill, New York, 1968. 14.Thomas, G.B. and Finney, R.L. Calculus and Analytic Geometry, 6th ed., AddisonWesley, Reading, MA, 1984. 15.Jenson, V.G. and Jeffreys, G.V. Mathematical Methods in Chemical Engineering, Academic Press, London, 1963. 16. Churchill, R.V. and Brown, J.W. Complex Variables and Applications, 4th ed., McGrawHill, New York, 1984. 17.Kreyszig, E. Advanced Engineering Mathematics, 7th ed., John Wiley & Sons, New York, 1993. 18.Saff, E.B. and Snider, A.D. Fundamentals of Complex Analysis for Mathematics, Science and Engineering, Prentice Hall, Englewood Cliffs, NJ, 1976. 19.Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena, 2nd ed., John Wiley & Sons, New York, 2002. 20. Schlichting, H. Boundary-Layer Theory, 7th ed., McGraw-Hill, New York, 1979. 21.Incropera, F.P. and DeWitt, D.P. Fundamentals of Heat and Mass Transfer, 2nd ed., John Wiley & Sons, New York, 1985. 22. Dettman, J.W. Applied Complex Variables, Dover, New York, 1984. 23. Rice, R.G. and Do, D.D. Applied Mathematics and Modeling for Chemical Engineers, John Wiley & Sons, New York, 1995. 24.Walas, S.M. Modeling with Differential Equations in Chemical Engineering, Butterworth-Heinemann, Boston, 1991. 25. Hale, J.K. Ordinary Differential Equations, John Wiley & Sons, New York, 1974. 26. White, F.M. Viscous Fluid Flow, McGraw-Hill, New York, 1991. 27. Na, T.Y. Computational Methods in Engineering Boundary Value Problems, Academic Press, New York, 1979. 28. Strang, G. Linear Algebra and Its Applications, 2nd ed., Academic Press, New York, 1980. Linear Second-Order and Systems of First-Order 127 29.Sewell, G. Computational Methods of Linear Algebra, 2nd ed., John Wiley & Sons, New Jersey, 2005. 30. Schiff, J. L. The Laplace Transform: Theory and Applications, Springer Science, 1999. 31. Brosilow, C. and Joseph, B. Techniques of Model-Based Control, Prentice Hall, 2002. 32. Bequette, B.W. Process Dynamics: Modelling, Analysis and Simulation, Prentice Hall, 1998. 33.Stephanopoulos, G. Chemical Process Control: An Introduction to Theory and Practice, 1984. This page intentionally left blank 4 Sturm–Liouville Problems 4.1 INTRODUCTION In the previous chapters, the discussion was primarily about the initial value problems, for which one seeks a solution to a differential equation subject to conditions on the dependent variable and its derivatives specified at only one value of the independent variable. In this chapter, the discussion will be about the class of problems, for which one seeks a solution to a differential equation subject to conditions on the dependent variable specified at two or more values of the independent variable, namely boundary value problems. Among the many problems encountered in chemical engineering science, those that appear very frequently involve the Sturm–Liouville boundary value problems. Following is an illustration of what is meant by a Sturm–Liouville problem. Consider the problem y ′′ + 2 y ′ + ( x 2 − λx ) y = 0 x (4.1) then an integrating factor for the Equation 4.1 is µ( x ) = e ∫ x 2 dx = x2 If Equation 4.1 is multiplied by x2, we get x 2 y ′′ + 2 xy ′ + ( x 4 − λx 3 ) y = 0 or ( x 2 y ′) ′ + ( x 4 − λx 3 ) y = 0 (4.2) Equation 4.2 is considered to be in the Sturm–Liouville form. That is, ( p( x ) y ′) ′ + [q( x ) + λr ( x )] y = 0 (4.3) is the Sturm–Liouville differential equation. It is also helpful to distinguish between two types of linear boundary value ­problems, homogeneous and nonhomogeneous. The equation p( x ) y ′′ + Q( x ) y ′ + R( x ) y = G ( x ) (4.4) is a general, linear, second-order, and nonhomogeneous differential equation and was discussed in Chapter 3. The condition a1 y( x 0 ) + a2 y ′( x 0 ) = c (4.5) 129 130 Applied Mathematical Methods for Chemical Engineers is a general, linear, and nonhomogeneous boundary condition. Both Equations 4.4 and 4.5 are considered homogeneous if their right-hand sides are identically zero. Therefore, we are using the same definition of “homogeneous” as introduced in the discussion on second-order differential equations. However, it is very important to note what is meant by a homogeneous boundary value problem as opposed to a differential equation. A boundary value problem is homogeneous when both the differential equation and all boundary conditions are homogeneous [1–4] and nonhomogeneous otherwise. For example, the problem P ( x ) y ′′ + Q( x ) y ′ + R( x ) y = 0,0 < x < 1 a1 y(0) + a2 y ′(0) = 0 b1 y(1) + b2 y ′(1) = 0 is a second-order linear homogeneous boundary value problem. Sturm–Liouville problems are categorized according to the type of boundary conditions that the differential equation must satisfy. 4.2 CLASSIFICATION OF STURM–LIOUVILLE PROBLEMS A general homogeneous second-order differential equation of the form a1 ( x ) y ′′ + a2 ( x ) y ′ + [a3 ( x ) + λ] y = 0 (4.6) can be recast in the Sturm–Liouville form d dy p( x ) + [q( x ) + λr ( x )] y = 0 dx dx (4.7) with the aid of the following device x a (t ) p( x ) = exp ∫ 2 dt a1 (t ) q( x ) = a3 ( x ) p( x ) a1 ( x ) r (x) = p( x ) a1 ( x ) (4.8) (4.9) and (4.10) where p, q, and r are real-valued functions of x, and λ is a parameter. Further, to ensure the existence of solutions on a closed finite interval [a, b], q and r are to be continuous on [a, b], whereas p is to be continuously differentiable on [a, b] [2, 4]. 131 Sturm–Liouville Problems Generally, there are three classes of Sturm–Liouville problems: 1.Regular 2.Periodic 3.Singular Each class is discussed later with an illustrative problem. Again the boundary conditions are very influential as to what class of boundary value problem one has to solve. Class 1: The Regular Sturm–Liouville Problem on [a, b] If p(x) > 0 and r(x) > 0 on a ≤ x ≤ b, one seeks numbers λ and nontrivial solutions of ( p( x ) y ′) ′ + [q( x ) + λr ( x )] y = 0 (4.11) A1 y(a) + A2 y ′(a) = 0 (4.12) B1 y(b) + B2 y ′(b) = 0 (4.13) subject to in which A1 and A2 are given and are not both zero, and B1 and B2 are given and are not both zero. Equation 4.11 through Equation 4.13 constitute a regular Sturm– Liouville boundary value problem. Example 4.1 This example demonstrates the procedure to determine the eigenvalues and eigenfunctions of a Regular Sturm-Liouville problem. π y′′ + λy = 0; 0 ≤ x ≤ 2 y(0) = 0 y π = 0 2 where p(x) = 1, r(x) = 1, and q(x) = 0 is a regular Sturm–Liouville problem with A1 = B1 = 1 and A2 = B2 = 0. There are three cases to consider, depending on the parameter λ, that is, λ = 0, λ > 0, and λ < 0. Case I Suppose λ = 0, then y″ = 0 and yg ( x ) = k1 x + k 2 yg (0) = 0 = k 2 π π y = 0 = k1 , k1 = 0 2 2 giving the trivial solution. Therefore, λ ≠ 0. 132 Applied Mathematical Methods for Chemical Engineers Case II Suppose λ > 0, say λ = β2 (for convenience), then y′′ + β 2 y = 0 y(0) = 0, y(π /2) = 0 has general solution y( x ) = k1 cos βx + k 2 sin βx and y(0) = 0 = k1 π π y = 0 = k 2 sin β 2 2 therefore, either k2 = 0 or sin β π =0 2 If k2 = 0, then we get a trivial solution. Then, for nontrivial solution to exist sin β π π = 0 ⇒ β = nπ 2 2 or β = 2n, n = 1,2,3, Therefore, λn = 4n2 defines the so-called eigenvalues of this problem. Corresponding to each eigenvalue is an eigenfunction φn ( x ) = k 2 n sin(2nx ) with k2n an arbitrary constant. Case III Suppose λ < 0, say λ = −α2, α > 0, then y′′ − α 2 y = 0 π y(0) = y = 0 2 Therefore, yg(x) = c1eαx + c2e−αx and 133 Sturm–Liouville Problems y(0) = 0 = c1 + c2 ⇔ c2 = − c1 π y = 0 = c1eα ( π / 2) + c2 e −α ( π / 2) ⇒ c1 (eα ( π / 2) − e −α ( π / 2) ) = 0 2 from which one can conclude that c1 = 0 = c2, and there are no negative eigenvalues. In summary, the regular Sturm–Liouville problem y′′ − λy = 0, π y(0) = y = 0 2 has eigenvalues λ n = 4 n 2 , n = 1,2,3, and corresponding eigenfunctions φn ( x ) = k 2 n sin(2nx ) with k2n nonzero but otherwise arbitrary. It is important to note that although constant coefficient differential equations are used to demonstrate how to identify the eigenvalues and eigenfunctions of a regular Sturm–Liouville problem, there are also variable coefficient problems as well. Example 4.2 The example demonstrates two approaches to identify the eigenvalues and eigenfunctions of A Variable Coefficient Sturm–Liouville Problem. Given x 2 y′′ + xy′ + λy = 0 (4.14) Subject to: y′ ( e −1 ) = 0; y (1) = 0 Determine the eigenvalues and corresponding eigenfunctions. Method I – Convert to constant coefficient Let x = et ⇔ t = ln x Then dt 1 dy dy dt 1 dy = and = = dx x dx dt dx x dt (4.15 and 4.16) 134 Applied Mathematical Methods for Chemical Engineers Similarly d 2 y d dy d 1 dy = = dx 2 dx dx dx x dt −1 dy 1 d 2 y dt = 2 + x dt x dt 2 dx 1 d 2 y dy = 2 2 − x dt dt Substituting into Equation 4.14 earlier results in d2y + λy = 0 dt 2 now subject to y ( 0 ) = 0, y′ (1) = 0 One can now apply the procedure illustrated in Class 1 under “The Regular Sturm– Liouville Problem on [a, b]” to arrive at the case: λ > 0, say λ = β 2 Then y ( t ) = b1 sin αt + b2 cos αt and y′ ( t ) = b1 α cos αt − b2 α sin αt Applying the boundary conditions leads to the conclusions: b2 = 0 and b1 α cos α = 0 However, for nontrivial solution to exist b1 α cos α = 0 ⇒ b1 ≠ 0 and cos α = 0 such that π α = ( 2n − 1) , n = 1,2,3, 2 Therefore, the eigenvalues are given by λ n = ( 2n − 1)2 π2 , n = 1,2,3, 4 and the corresponding eigenfunctions are yn = sin ( 2n − 1) π 2 t ≡ sin ( 2n − 1) π 2 ln x 135 Sturm–Liouville Problems Method II Alternatively, one could have made the substitution y = xm into Equation 4.14, resulting in Linear Independent solutions y1 = x i λ and y2 = x − i λ See Problem 11. Class 2: The Periodic Sturm–Liouville Problem on [a, b] If p(x) > 0 and r(x) > 0 on a ≤ x ≤ b, one seeks numbers λ and nontrivial solution of ( p( x ) y′)′ + [q( x ) + λr ( x )] y = 0 (4.17) y( a ) = y( b ) (4.18) y′(a) = y′(b) (4.19) subject to Example 4.3 This example demonstrates the procedure to determine the eigenvalues and eigenfunctions of a Periodic Sturm-Liouville problem, y′′ + λy = 0 on [−π , π ] y(−π ) = y(π ) y′(−π ) = y′(π ) is a periodic Sturm–Liouville problem. Again, there are three cases to consider. Case I Suppose λ = 0, then yg ( x ) = k1 x + k 2 and y(−π) = − k1π + k 2 = y(π) = k1π + k 2 which results in y = k2 Thus, λ = 0 is an eigenvalue for this problem with the corresponding eigenfunction y = k2, which is an arbitrary constant. Case II Suppose λ > 0, say λ = β2, then yg = c1 cos βx + c2 sin βx 136 Applied Mathematical Methods for Chemical Engineers and y(−π ) = c1 cos βπ − c2 sin βπ = y(π) = c1 cos βπ + c2 sin βπ and y′(−π) = c1 β sin βπ + c2 β × = − c1 β × + c2 cos βπ = y′(π ) Therefore, 2c2 sin βπ = 0 and 2c1 sin βπ = 0 ⇒ sin βπ = 0 ⇒ βπ = nπ , n = 1,2,3, for nontrivial solution. Therefore, λ = n2 is an eigenvalue with corresponding eigenfunctions: φn ( x ) = c1n cos( nx ) + c2 n sin(nx ) in which c1n and c2n are arbitrary constants, which cannot both be zero but are ­otherwise arbitrary. For the case with λ < 0, it can be shown that only trivial solutions result. Therefore, the periodic Sturm–Liouville problem y′′ + λy = 0, y(−π) = y(π ) and y′(−π) = y′(π ) has eigenvalues λ = n 2 , n = 0,1,2, and eigenfunction φn ( x ) = c1n cos( nx ) + c2 n sin(nx ) with not both c1n and c2n zero. Class 3: The Singular Sturm–Liouville Problem on [a, b] If p(x) > 0 and r(x) > 0 on a ≤ x ≤ b, one seeks number λ and nontrivial solution of ( p( x ) y′)′ + [q( x ) + λr ( x )] y = 0 (4.20) satisfying one of the following three types of boundary conditions: 1.If p(a) = 0, then B1 y(b) + B2 y′(b) = 0 with B1 and B2 given and are nonzero, and also the solutions must be bounded at a. (4.21) 137 Sturm–Liouville Problems 2.If p(b) = 0, then A1 y(a) + A2 y′(a) = 0 (4.22) with A1 and A2 given and are nonzero, also solutions must be bounded at b. 3.If p(a) = p(b) = 0, we have no boundary conditions specified at either a or b but require that solutions be bounded on [a, b]. Example 4.4 This example demonstrates the procedure to determine the eigenvalues and eigenfunctions of a Singular Sturm–Liouville problem, such as Legendre’s equation [2] (1 − x 2 ) y′′ − 2 xy′ + ( + 1) y = 0 recasted as [(1 − x 2 ) y′]′ + λy = 0 − 1 ≤ x ≤ 1 with λ = ( + 1), p( x ) = 1 − x 2 such that p(−1) = p(1) = 0 This results in the class of singular Sturm–Liouville problems having type 3 boundary condition. Further, since (1 − x 2 ) y′′ − 2 xy′ + ( + 1) y = 0 can be rewritten as y′′ − 2x ( + 1) y′ + y=0 1− x2 1− x2 with singular points at x = ±1, but both −2x/(1 − x2) and ℓ(ℓ+1)/(1 − x2) have Taylor Series expansion about the origin in −1 < x < 1 (x = 0 is an ordinary point), then ∞ y = ∑ an x n n=0 can be assumed as a solution of the given differential equation. Proceeding formally, as in the Frobenius series method, results in 2a2 + ( + 1)a0 + [3 ⋅ 2a3 + ( + 1)a1 − 2a1 ]x ∞ + ∑ {(n + 2)(n + 1)an + 2 + [( + 1) − n(n − 1) − 2n]an }x n = 0 n=2 138 Applied Mathematical Methods for Chemical Engineers from which ( + 1) a0 2 ( + 2)( + 1) a3 = − a1 3⋅2 (n + + 1)( − n) =− an , n ≥ 2 (n + 2)(n + 1) a2 = − an + 2 where the latter equation is the recurrence relation. Following the determination of a few more constants, one can observe the patterns: a2 n = (−1)n ( + 2n − 1)( + 2n − 3)… ( + 1)( − 2)… ( − 2n + 4)( − 2n + 2) a0 (2n)! for even indexed coefficients and a2 n +1 = (−1)n ( + 2n)… ( + 4)( + 2)( − 1)( − 3)… ( − 2n + 1) a1 (2n + 1)! for odd indexed coefficients. Therefore, one can obtain two linearly independent solutions of Legendre’s equation. Say for a0 = 1 and a1 = 0 ∞ y1 = ∑ (−1)n n=0 ( + 2n − 1)( + 2n − 3)… ( + 1)( − 2)… ( − 2n + 4)( − 2n + 2) 2 n x (2n)! and for a 0 = 0 and a1 = 1 ∞ y2 = ∑ (−1)n n=0 ( + 2n)… ( + 4)( + 2)( − 1)( − 3)… ( − 2n + 1) 2 n +1 x (2n + 1)! By making appropriate choices of ℓ, the linearly independent solutions can be reduced to polynomials (Legendre’s polynomials). A frequently occurring example of a Variable Coefficient Sturm–Liouville Problem of the singular type is given below. Example 4.5 This example demonstrates the procedure to determine the eigenvalues and eigenfunctions of a Singular Sturm–Liouville problem Given the singular boundary value problem described by 1 R′′ + R′ − λR = 0 r 139 Sturm–Liouville Problems and subject to R(r ) being bounded at the origin (r = 0), and R(d) = 0; determine the eigenfunction and the equation that defines the eigenvalues. Solution By comparing the R-equation with the prototype Bessel Equation (Equation 3.80), y′′ − ( 2a − 1) x a 2 − γ 2c 2 y′ + b 2c 2 x 2 c − 2 + y = 0 x2 whose linearly independent solutions are y1 = x a J γ (bx c ), y2 = x a J−γ (bx c ) where the J γ and J−γ are Bessel functions of order γ as described in Section 3.6; we find that 2a − 1 = −1⇒ a = 0; 2c − 2 = 0 ⇒ c = 1; b 2c 2 = −λ ⇒ b = ± i λ ; and γ = 0 Therefore R(r ) = c1J0 (i λ r ) + c2 J−0 (i λ r ) = c1J0 (i λ r ) + c2Y0 (i λ r ) where J0 ( x ) = 1 − x2 x4 x6 + − + 2 2 2 2 2 2 i4 2 i 4 2 i 62 is the zero order Bessel function of the first kind and 2 x 2 x4 1 x6 2 1 + 1 + 1 − 1+ + Y0 ( x ) = ln ( x 2 ) + γ J0 ( x ) + − ; π π 22 22 i 4 2 2 22 i 4 2 i 62 2 3 { } γ = 0.5772156… is Euler ' s const . is the zero order Bessel function of the second kind. Applying the boundary condition; R(r ) is bounded at r = 0, then c2 must be chosen as zero because Y0 (βr ) → ∞ as r → 0 where β = i λ or λ = −β 2 r = d ⇒ R(d) = 0 = c1J0 (d β) 140 Applied Mathematical Methods for Chemical Engineers However, for a nontrivial solution to exist, c1 ≠ 0 such that 0 = J0 (d β) which defines the Eigenvalues. That is, J0 (d β n ) = 0, n = 1,2,3, and the eigenfunction is given by Rn (r ) = J0 (i λ n r ) ≡ J0 (β nr ) A Procedure to determine β n values: Given the eigenvalues, β n defined by the Bessel function: J0 (2β n ) = 0, n = 1,2,3, 1. Locate a table of values for the zeros of Bessel Functions. 2. Identify the case n = 1 (in this case) and locate the listed zero noting the order in which they are listed (most tables list smallest root first); in this specific case 2β1 = 2.4048, 2β 2 = 5.5201, 2.4048 = 1.2024. 3. Solve for β1 from step 2, that is, β1 = 2 4. Repeat steps 2 and 3 as needed. Remark: Knowledge of the properties of the eigenvalues and eigenfunctions can dramatically reduce the labor typically required to solve a Sturm–Liouville problem. These properties also provide a check on whether or not one’s solution is reasonable. 4.2.1 PROPERtIEs OF thE EIGENVALUEs AND EIGENFUNctIONs OF A StURM –LIOUVILLE PROBLEM The following theorem summarizes the important properties of the Sturm–Liouville problem and its eigenvalues [1, 2]. Theorem 4.1 1. For the regular and periodic Sturm–Liouville problems, there exist an ­infinite number of eigenvalues. Further, these eigenvalues can be labeled λ1, λ2, … so that λn < λm if n < m and nlim λ = ∞. →∞ n 2. If λn and λm are distinct eigenvalues of any of the three types of Sturm– Liouville problems, with corresponding eigenfunctions ɸ n and ɸ m, then ɸ n and ɸ m are orthogonal on [a, b] with weight function r(x). That is, b ∫ r ( x )φn ( x )φm ( x ) d x = 0 a if n ≠ m. 141 Sturm–Liouville Problems 3. For all three classes of Sturm–Liouville problems, all eigenvalues are real. 4. For a regular Sturm–Liouville problem, any two eigenfunctions corresponding to a given eigenvalue are linearly dependent. 5. The Laplace transform, F(s), of a solution to a Sturm–Liouville problem is analytic for all finite s except for poles, which correspond to the eigenvalues of the system. Sometimes it is helpful to approximate the value of the smallest eigenvalue when one is faced with a computationally difficult engineering problem. An iterative procedure known as the method of Stodola and Vianello [5] is outlined as follows. Consider the regular Sturm–Liouville problem d2 y + λy = 0 dx 2 subject to y(0) = 0, y( L ) = 0 (4.23) (4.24) This problem can be easily shown to have eigenvalues n2 π2 , n = 1,2, L2 λn = (4.25) with corresponding eigenfunctions φn ( x ) = sin nπx L (4.26) However, the method can be illustrated using this familiar problem: 1.Rewrite Equation 4.23 as d2 y = −λy dx 2 (4.27) 2.Replace the unknown function y on the right-hand side of the above equation by a conveniently chosen first approximation y1(x) giving d2 y = −λy1 ( x ) dx 2 (4.28) 3.Solve Equation 4.28 and Equation 4.24 formally to get y = λf1 ( x ) If y1(x) were actually an eigenfunction of the problem, then y = y1 and λ would be the eigenvalue given by the constant ratio (4.29) 142 Applied Mathematical Methods for Chemical Engineers y1 ( x ) f1 ( x ) λ= However, y1(x) is generally not an eigenfunction and therefore y1(x) and f1(x) generally will not be in a constant ratio. To illustrate further, suppose y1(x) is chosen to be y1 ( x ) = x ( L − x ) then d2 y = −λx ( L − x ) = λ( x 2 − Lx ) dx 2 whose solution satisfying Equation 4.24 is y= λ 4 ( x − 2 Lx 3 + L3 x ) ≡ λf1 ( x ) 12 Observe that y ≠ y1, but a first approximation of the smallest eigenvalue, λ 1, can be obtained from L L 0 0 ∫ y1 ( x ) d x = λ ∫ f1 ( x ) d x That is, L ∫ ( Lx − x 2 ) d x = 0 L λ ( x 4 − 2 Lx 3 + L3 x ) d x 12 ∫0 or L λ = 12 ∫ ( Lx − x 2 ) d x 0 L ∫ ( x 4 − 2 Lx 3 + L3 x ) d x 0 such that λ1(1) = 10 L2 is the first approximation to the smallest eigenvalue. By taking y2 ( x ) = x 4 − 2 Lx 3 + L3 x as the next approximation, and formally solving d2 y = −λy2 ( x ) dx 2 subject to Equation 4.24 gives (4.30) 143 Sturm–Liouville Problems y = λ f2 ( x ) = − λ 6 ( x − 3 Lx 5 + 5 L3 x 3 − 3 L5 x ) 30 then Equation 4.30 becomes L L 0 0 ∫ y2 ( x ) d x = λ ∫ f2 ( x ) d x such that the second approximation to λ 1 is λ1(2) = 168 9.882 = 2 17 L2 L In general, successive estimates of the eigenvalues λ 1 may be obtained after each iteration by requiring that the functions yn(x) and y = λ f n(x) agree as well as possible over the interval (0, L). In particular, the nth approximation to be the smallest value of λ is given by L λ (n) 1 = ∫ yn ( x ) d x 0 L (4.31) ∫ fn ( x ) d x 0 and the corresponding eigenfunctions is yn(x). An improved version of the above equation is given by b λ (n) 1 = ∫ r ( x ) fn ( x ) yn ( x ) d x a b ∫ r ( x )[ fn ( x )] 2 (4.32) dx a The above equation is a consequence of point 2 of Theorem 4.1. The next section briefly introduces an application of property 2 given in the theorem. This is a very important property that will be used repeatedly in solving application problems in both Chapter 6 and Chapter 7. 4.3 EIGENFUNCTION EXPANSION Consider a Sturm–Liouville problem on [a, b], that is, ( p( x ) y ′) ′ + [q( x ) + λr ( x )] y = 0 (4.33) A1 y(a) + A2 y ′(a) = 0 (4.34) B1 y(b) + B2 y ′(b) = 0 (4.35) subject to 144 Applied Mathematical Methods for Chemical Engineers Then, if the eigenvalues λ 1, λ 2, …, and corresponding eigenfunctions φ 1(x), φ 2(x), … have been found, one can sometimes expand a given function f(x) in a series of these eigenfunctions. That is, ∞ f ( x ) = ∑ cn φn ( x ), a ≤ x ≤ b (4.36) n =1 The series in the above equation is called an eigenfunction expansion of f(x) on [a, b]. To determine the cn′s, property 2 of Theorem 4.1 will be used in the ­following way: Multiply Equation 4.36 by r(x)φ k(x) and integrate the result from a to b. b ∞ b a n =1 a ∫ r ( x ) f ( x )φk ( x ) d x = ∑ cn ∫ r ( x )φn ( x )φk ( x ) d x (4.37) assuming that term-by-term integration of the series is permissible. Then property 2 of Theorem 4.1 gives b ∫ r ( x )φn ( x )φk ( x ) d x = 0 if n ≠ k a Therefore, Equation 4.37 reduces to b b a a ∫ r ( x ) f ( x )φk ( x ) d x = ck ∫ r ( x )[φk ( x )]2 d x for n = k Finally, b ck = ∫ r ( x ) f ( x )φk ( x ) d x a b (4.38) ∫ r ( x )[φk ( x )]2 d x a These ck are the so-called Fourier coefficients of f(x) with respect to the eigenfunctions of the given Sturm–Liouville problem. A more comprehensive discussion on Fourier coefficients will be given in the next chapter. Also, the conditions that a function must satisfy to have a series expansion, as given in Equation 4.36, will be discussed in Chapter 5. Following is an example demonstrating the computation of Fourier coefficients. Example 4.6 This example demonstrates the procedure to expand a given function into a series of eigenfunctions. Consider y ″ + λy = 0 subject to y(0) = y(π/2) = 0, a regular Sturm–Liouville problem on [0, π/2] with eigenvalues λ n = 4n2, n ≥ 1 and corresponding eigenfunctions φn(x) = sin 2nx. 145 Sturm–Liouville Problems Then, if f(x) = x2 on [0, π/2] satisfies all the required conditions to be given in Chapter 5, one can proceed to find the cn’s as follows: π /2 π /2 ∫ r ( x ) f ( x )φn( x ) dx = ∫ x 2 sin 2nx dx 0 0 −π2 1 = [cos nπ − 1] cos nπ + 8n 4n3 and π /2 π /2 0 0 ∫ r (x )[φn(x )]2 d x = ∫ sin 2 2nx d x = π/4 such that cn = = 4 −π2 1 cos nπ + (cos nπ − 1) π 8n 4n3 −π 1 cos nπ + (cos nπ − 1) 2n πn3 Finally, ∞ 1 π x2 = ∑ (cos nπ − 1) − cos nπ sin2nx 3 n 2 n =1 πn Again, eigenfunction expansion will be useful in solving certain types of boundary value problems to be discussed in Chapter 6. 4.4 PROBLEMS 1.Find all the values of λ, satisfying the boundary value problem X ′′ =λ X X (0) = 0, X (1) = 0 2.Find all the values of λ, satisfying the boundary value problem X ′′ =λ X X (0) = 0, X ′( L ) = 0 146 Applied Mathematical Methods for Chemical Engineers 3.Find all the values of λ, satisfying the boundary value problem X ′′ =λ X X ( − L ) = X ( L ), X ′( − L ) = X ′( L ) 4.Find all the values of λ, satisfying the boundary value problem X ′′ =λ X X ′(0) = 0, X ′( L ) = 0 5.Use property 2 of Theorem 4.1 to determine Bn in the series ∞ ∑ Bn sin n =1 nπx = f (x) L 1, 0 < x < 0.5 f (x) = 0, 0.5 < x < L 6.Use property 2 of Theorem 4.1 to determine An in the series ∞ ∑ An cos n= 0 nπx = f (x) L 1, 0 < x < 0.5 f (x) = 0, 0.5 < x < L 7.Use property 2 of Theorem 4.1 to determine An and Bn in the series ∞ ∑ An cos n= 0 nπx nπx + Bn sin = f (x) 2 2 − x , −2 < x < 0 f (x) = 2, 0 < x < 2 8.Derive the equation that is required to yield the values of λ in the singular Sturm–Liouville problem ρR ′′(ρ) + R ′(ρ) + λ 2 ρR(ρ) = 0 subject to R(c) = 0 Hint: Apply a boundednes condition after using Equation 3.80 to reduce labor. 147 Sturm–Liouville Problems 9.Find all the values of λ, satisfying the boundary value problem y ′′ + λy = 0; y(0) = 0, y( L ) = 0 10.Find all the values of λ, satisfying the boundary value problem y ′′ + λy = 0; y ′(0) = 0, y ′( L ) = 0 11.Complete the Variable Coefficient Sturm–Liouville Problem using the suggested Method II given in the solved Example 4.2 in Section 4.2. 12.Find all the values of λ, satisfying the boundary value problem ( xy′)′ + λx −1 y = 0 y(1) = 0, y(e) = 0 m a. Use the substitution y = x b. Use the substitution x = et 13.Find all the values of (or the equation defining) λ, satisfying the boundary value problem y′′ + λy = 0 y(0) − y′(0) = 0 y(0) + y′(1) = 0 14 a.Given the linear variable coefficient differential equation, determine its two linearly independent solutions x 2 y ′′ + xy ′ + ( x 2 − 1 4 ) y = 0 Hint: Consider comparing the given differential equation to a2 − γ 2c2 2a − 1 y ′′ − y ′ + b 2 c 2 x 2c − 2 + y = 0 x x2 (Recall Equation 3.80) y1 ( x ) = x J γ ( bx ) , y2 ( x ) = x J− γ ( bx a c a c ) b.Express the two linearly independent solutions in terms of sin x and cos x by employing the formulae J1/ 2 ( x ) = 2 sin x; J −1/ 2 ( x ) = πx 2 cos x πx 15.Given x 2 y ′′ + xy ′ + λy = 0; y (1) = 0, y ( e π ) = 0 148 Applied Mathematical Methods for Chemical Engineers Determine the eigenvalues and define the eigenfunctions. 16.Given the singular differential equation, along with appropriate boundary condition, derive the equation that defines the eigenvalues. 1 R ′′ + R ′ + λ 2 R = 0 r subject to R(0) < ∞ and R(r0 , t ) = 0 17.Given y ′′ + � y = 0; y(0) = 0, y(1) + y ′ (1) = 0 a. Determine the equation that defines the eigenvalues. b. Sketch the graph of the equation derived in part a identifying the first four eigenvalues. REFERENCES 1.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley & Sons, Inc., New York, 2005. 2. O’Neil, P.V. Advanced Engineering Mathematics, 4th ed., PWS, Boston, 1995. 3.Humi, M. and Miller, W.B. Boundary Value Problems and Partial Differential Equations, PWS, Boston, 1988. 4.Myint-U, T. and Debnath, L. Partial Differential Equations for Scientists and Engineers, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1987. 5.Hildebrand, F.B. Advanced Calculus for Applications, 2nd ed., Prentice Hall, Englewood Cliffs, NJ, 1976. 5 Fourier Series and Integrals 5.1 INTRODUCTION One of the most common solution techniques applicable to linear homogeneous partial differential equation problems involves the use of Fourier series. A discussion of the methods of solution of linear partial differential equations will be the topic of the next chapter. In this chapter, a brief outline of Fourier series is given. The primary concerns in this chapter are to determine when a function has a Fourier series expansion and then, does the series converge to the function for which the expansion was assumed? Also, the topic of Fourier transforms will be briefly introduced, as it can also provide an alternative approach to solve certain types of linear partial differential equations. To establish the conditions for a function to have a Fourier series expansion, the following definitions are necessary. A function is said to be piecewise continuous in an interval a ≤ x ≤ b if there exist finitely many points a = x1 < x2 < ⋯ < xn = b, such that f(x) is continuous in xj < x < xj+1 and the one-sided limits f ( x +j ) and f ( x −j +1 ) exist for all j = 1, 2, 3, …, n−1 [1–4]. Note that a function is piecewise continuous on the closed interval a ≤ x ≤ b; however, continuity on the open interval a < x < b does not imply piecewise continuity there. For example, f (x) = 1 x is continuous on 0 < x < 1, but it is not piecewise continuous as f(0+) fails to exist. When a function f(x) is piecewise continuous on an interval a < x < b, the integral of f(x) from x = a to x = b always exists. That integral is the sum of the integrals of f(x) over the open subintervals on which f is continuous, that is, b ∫a x1 x2 a x1 f (x) d x = ∫ f (x) d x + ∫ f (x) d x + + ∫ b x n −1 f (x) d x (5.1) For example, f (x) = x, 0 ≤ x < 1 = −1, 1 < x < 2 = 1, 3 1 2<x <3 2 3 1 ∫0 f ( x ) d x = ∫0 x d x + ∫1 − d x + ∫2 d x = 2 149 150 Applied Mathematical Methods for Chemical Engineers If two functions f1 and f 2 are piecewise continuous on an internal a < x < b, then there is a finite subdivision of the interval such that both functions are continuous on each closed subinterval whenever the functions are given their limiting values from the interior at the endpoints. This means that a linear combination c1f1 + c2f 2 or the product f1f 2 has the continuity on each subinterval and is, itself, piecewise continuous on a < x < b. As a consequence ∫ [c1 f1 + c2 f2 ] d x , ∫ f1 f2 d x , ∫ [ f1 ( x )]2 d x and all exist on that interval [1]. If f(x) is piecewise continuous on a ≤ x ≤ b and if the first derivative f′(x) is continuous on each of the subintervals xj < x < xj+1 and the limits f ( x +j ) and f ( x −j ) exist, then f is said to be piecewise smooth. The functions of a sequence {φn(x)} are said to be orthogonal with respect to the weight function r(x) on a ≤ x ≤ b if b ∫a r ( x )φn ( x )φm ( x ) d x = 0, m≠n (5.2) and if m = n, then b φ2 r ( x ) d x = φ n ∫a n (5.3) which is called the norm of the orthogonal system {φn}. For example, the sequence of function 1, cos x, sin x, …, cos nx, sin nx forms an orthogonal system on −π < x < π since π ∫− π sin mx sin nx d x = 0 π ∫− π sin mx sin nx d x = π π ∫− π sin mx cos nx d x = 0 π ∫− π cos mx cos nx d x = 0 π ∫− π cos mx cos nx d x = π m≠n m=n for all m, n (5.4) m≠n m=n for positive integers m and n. Note that an orthonormal system, one that is both orthogonal and normalized, for the given sequence is 1 cos x sin x cos nx sin nx , , , , , π π π π 2π where each element of the sequence is divided by its norm. It is also important to note that the elements of the sequence 1, cos x, sin x, …, cos nx, sin nx are periodic. In general, a piecewise continuous function f(x) in an interval a ≤ x ≤ b is said to be periodic if there exists a real positive number p such that 151 Fourier Series and Integrals f ( x + np) = f ( x ) (5.5) for any integer n. Further if f1, f 2, …, f k have the period p, then the linear combination f = c1 f1 + c2 f2 + + ck fk (5.6) has the period p. Also, a constant function is a periodic function with an arbitrary period p. 5.2 FOURIER COEFFICIENTS Since the linear independent functions 1, cos x, sin x, cos 2x, sin 2x, … are mutually orthogonal to each other −π < x < π, one can form the series f (x) ~ A0 ∞ + ∑ ( An cos nx + Bn sin nx ) 2 n =1 (5.7) where the symbol ~ indicates an association of A0, An, and Bn to f(x) in some unique manner. The series may or may not be convergent. Suppose f(x) is a Riemann integrable function that is defined on −π ≤ x ≤ π. Then one can define the kth partial sum Sk ( x ) = k A0 + ∑ ( An cos nx + Bn sin nx ) 2 n =1 (5.8) to represent f(x) on −π ≤ x ≤ π. We now seek the numbers A0, An, and Bn such that Sk(x) is the best approximation to f(x) in the sense of least squares; that is, we need to minimize the integral π I ( A0 , An , Bn ) = ∫ [ f ( x ) − Sk ( x )]2 d x −π (5.9) The necessary condition for I(A0, An, Bn) to be a minimum is ∂I ∂I ∂I =0= = ∂ A0 ∂ An ∂ Bn (5.10) k π ∂I A = − ∫ f ( x ) − 0 − ∑ ( A j cos jx + B j sin jx ) d x π − ∂ A0 2 j =1 (5.11) k π ∂I A = −2 ∫ f ( x ) − 0 − ∑ ( A j cos jx + B j sin jx ) cos nx d x −π ∂ An 2 j =1 (5.12) Thus 152 Applied Mathematical Methods for Chemical Engineers k π ∂I A = −2 ∫ f ( x ) − 0 − ∑ ( A j cos jx + B j sin jx ) sin nx d x −π ∂ Bn 2 j =1 (5.13) From the orthogonality of the trigonometric function (Equation 5.4) and noting that π π ∫− π cos mx d x = ∫− π sin mx d x = 0, m an integer Equations 5.11 through 5.13 become π ∂I = πA0 − ∫ f ( x ) d x − π ∂ A0 (5.14) π ∂I = 2πAn − 2 ∫ f ( x ) cos nx d x − π ∂ An (5.15) π ∂I = 2πBn − 2 ∫ f ( x )sin nx d x −π ∂ Bn (5.16) Based on Equation 5.10, there results 1 π f (x) d x π ∫− π (5.17) An = 1 π f ( x ) cos nx d x π ∫− π (5.18) Bn = 1 π f ( x )sin nx d x π ∫− π (5.19) A0 = Further, ∂2 I =π ∂ A02 ∂2 I ∂2 I = = 2π ∂ An2 ∂ Bn2 and ∂( m ) I ∂( m ) I ∂2 I ∂2 I = = = = 0, m ≥ 3, n = 1,2, ∂ A0 An ∂ A0 Bn ∂ An( m ) ∂ Bn( m ) By expanding I(A0, An, Bn) in a Taylor series about (A0, A1, …, An, B1, B2, …, Bn) one gets Fourier Series and Integrals I ( A0 + ∆A0 ,, Bn + ∆Bn ) = I ( A0 ,, Bn ) + ∆I 153 (5.20) where ∆I represents the remaining terms. But ∆I can be written as ∆I = k 1 ∂2 I ∂2 I ∂2 I 2 2 2 ∆ A + 2 0 ∑ 2 ∆An + 2 ∆Bn 2! ∂ A0 ∂ Bn n =1 ∂ An (5.21) since the first derivative, all mixed second derivatives, and all remaining higher order derivatives are zero. Then ∆I = k 1 2 2 2 π∆A0 + ∑ ( 2π∆An + 2π∆Bn ) > 0 2! n =1 Therefore, in order for I to have a minimum value, the coefficients A0, An, and Bn must be given by Equations 5.17 through 5.19, respectively. These coefficients are called the Fourier Coefficients of f(x), and the series given by Equation 5.7 represent the Fourier series. According to Equation 5.6, the Fourier series Equation 5.7, is periodic with period 2π, if it converges. The following are a few examples on calculating Fourier series of a given function. Example 5.1 Find the Fourier series expansion for the function f(x) = x + x2 on −π < x < π. Solution 1 π 1 π 2π 2 f (x) d x = ∫ (x + x 2 ) d x = ∫ π π − − π π 3 1 π 4 2 f ( x ) cos nx d x = ∫ ( x + x ) cos nx d x = 2 cos nπ n π −π 4 = 2 (−1)n , n ≥ 1 n A0 = An = 1 π π ∫− π Bn = 1 π 1 π 2 f ( x )sin nx d x = ∫ ( x + x 2 )sin nx d x = cos nπ ∫ − π − π n π π 2 n = (−1) , n ≥ 1 n Therefore, the Fourier series expansion for f is f (x) = π2 ∞ 4 2 +∑ (−1)n cos nx − (−1)n sin nx n 3 n =1 n 2 154 Applied Mathematical Methods for Chemical Engineers Example 5.2 Find the Fourier series expansion of the function f(x) = x in the interval −π < x < π. Solution A0 = 1 π 1 π f (x) d x = ∫ x d x = 0 ∫ π − π π −π Also, for n ≥ 1 An = Bn = 1 π 1 π f ( x ) cos nx d x = ∫ x cos nx d x = 0 π ∫− π π −π 1 π 1 π 2 f ( x )sin nx d x = ∫ x sin nx d x = (−1)n +1 ∫ − π n π π −π Therefore, ∞ (−1)n +1 sin nx n n =1 f ( x ) = 2∑ Example 5.3 Find the Fourier series of the periodic function shown in Figure 5.1. f ( x ) = − π, − π < x < 0 = x, 0 < x < π Solution A0 = π 1 π 1 0 −π f ( x ) d x = ∫ − π d x + ∫ x d x = 0 2 π ∫− π π −π π 1 π 1 0 f ( x ) cos nx d x = ∫ − π cos nx d x + ∫ x cos nx d x 0 π ∫− π π −π 1 = 2 [(−1)n − 1], n ≥ 1 πn An = f (x) –2π –π FIGURE 5.1 Periodic Function. 0 π 2π 155 Fourier Series and Integrals 1 π f ( x )sin nx d x π ∫− π π 1 0 1 = ∫ − π sin nx d x + ∫ x sin nx d x = [1 − 2(−1)n ] π 0 − n π Bn = Therefore, the Fourier series is f (x) = − 5.3 { π ∞ 1 1 + ∑ 2 [(−1)n − 1]cos nx + [1 − 2(−1)n ]sin nx n 4 n =1 n π } ARBITRARY INTERVAL In the previous two sections, the discussion was primarily based on the interval [−π, π]. However, in many applications, as will be observed in the next chapter, this interval is restrictive. Suppose the interval of interest is [a, b]. Then the interval a ≤ x ≤ b can be transformed to −π < x < π by using 1 b−a x = (b + a) + t 2π 2 (5.22) Therefore, the function (b + a) b − a f + t = F (t ) 2π 2 has period 2π. When F(t) is expanded in a Fourier series, one obtains F (t ) = A0 ∞ + ∑ ( An cos nt + Bn sin nt ) 2 n =1 (5.23) where for n ≥ 0 An = 1 π F (t ) cos nt dt π ∫− π Bn = 1 π F (t )sin nt dt π ∫− π and Upon resubstituting x, the expansion for f(x) in a ≤ x ≤ b is f (x) = A0 ∞ 2x − b − a 2x − b − a + Bn sin nπ + ∑ An cos nπ b − a b − a 2 n =1 (5.24) where b An = 2 2x − b − a f ( x ) cos nπ d x, n ≥ 0 ∫ b − a b−a a (5.25) 156 Applied Mathematical Methods for Chemical Engineers Bn b 2 2x − b − a f ( x )sin nπ d x, n ≥ 1 ∫ b − a b−a a (5.26) To illustrate the change of interval, consider the function f(x) = x, −2 < x < 2, then A0 = 1 2 x dx = 0 2 ∫−2 also for n ≥ 1, An = 0 and Bn = 1 2 nπx −4 x sin dx = ( −1)n , n ≥ 1 nπ 2 ∫−2 2 Therefore, the Fourier series for f(x) is ∞ nπx 4 ( −1)n +1 sin 2 n =1 nπ f (x) = ∑ a sine series. An alternate representation of Equations 5.25 and 5.26 can be achieved if the convenient interval [−L, L] is used [5, 6]. That is, let a = −L, and b = L, then f (x) = nπx A0 ∞ nπx + ∑ An cos + Bn sin L L 2 n =1 (5.27) where 5.4 An = 1 L nπx f ( x ) cos d x, n ≥ 0 ∫ − L L L (5.28) Bn = 1 L nπx f ( x )sin d x, n ≥ 1 ∫ − L L L (5.29) COSINE AND SINE SERIES Suppose f(x) is an even function defined on the interval [−π, π]; then since cos nx is an even function and sin nx is an odd function, the product f(x)cos nx and f(x) sin nx are even and odd functions, respectively. Then the Fourier coefficients of f(x) are An = 1 π 2 π f ( x ) cos nx d x = ∫ f ( x ) cos nx d x , n ≥ 0 ∫ − π π π 0 (5.30) 1 π f ( x )sin nx d x = 0, n ≥ 1 π ∫− π (5.31) and Bn = Therefore, the Fourier series representation of f(x) is 157 Fourier Series and Integrals f (x) ~ A0 ∞ + ∑ An cos nx 2 n =1 (5.32) On the other hand, if f(x) is an odd function defined on [−π, π], then f(x) cos nx and f(x) sin nx are odd and even functions, respectively. Then 1 π f ( x ) cos nx d x = 0, n ≥ 0 π ∫− π (5.33) 1 π 2 π f ( x )sin nx d x = ∫ f ( x )sin nx d x π ∫− π π 0 (5.34) An = and Bn = Therefore, the Fourier series representation of f(x) becomes ∞ f ( x ) ~ ∑ Bn sin nx (5.35) n =1 In practice, one frequently encounters problems in which a function is defined on the interval [−π, π]. In these cases, a periodic extension as shown in Figure 5.2 can be made, and those functions can be represented by the Fourier series expansion even though one is interested only in the expansion on [−π, π]. However, if a function f(x) is defined only on [0, π], then f(x) may be extended in either an even extension or an odd extension. By even extension of f(x), one means Fe ( x ) = f ( x ), 0 < x < π f (− x ), − π < x < 0 as shown in Figure 5.3. f (x) x –π 0 π FIGURE 5.2 Periodic extension. f (x) –π FIGURE 5.3 Even extension. 0 π x 158 Applied Mathematical Methods for Chemical Engineers f (x) –π 0 π x FIGURE 5.4 Odd extension. By odd extension of f(x), one means F0 ( x ) = f ( x ), 0 < x < π f (− x ), − π < x < 0 as shown in Figure 5.4. These functions Fe(x) and F0(x) with period 2π, both have Fourier series expansions given by ∞ A Fe ( x ) ~ 0 + ∑ An cos nx 2 n =1 (5.36) ∞ F0 ( x ) ~ ∑ Bn sin nx n =1 (5.37) with the Fourier coefficients as given by Equation 5.30 and 5.34. Following are the four examples of the expediency that is achievable when the concept of odd or even function is used in the context of Fourier series. Example 5.4 Show that an even function can have no sine terms in its Fourier expansion. Solution No sine terms if Bn = 0, n ≥ 1. Consider f(x) on [−L, L], then 1 L nπx f ( x )sin dx L ∫− L L 1 0 1 L nπx nπx d x + ∫ f ( x )sin dx = ∫ f ( x )sin L −L L L 0 L Bn = But if f(x) is even, then f(x) = f(−x). Specifically, let x = −y, dx = −dy then 1 0 nπx 1 0 nπy d x = ∫ f (− y)sin dy f ( x )sin L ∫− L L L L L 0 1 nπy = − ∫ f (− y)sin dy LL L 159 Fourier Series and Integrals Therefore, Bn = − 1 0 nπy 1 L nπx dx = 0 f (− y)sin d y + ∫ f ( x )sin ∫ L L L 0 L L in which case the knowledge that the function is even would eliminate the need to calculate Bn. Example 5.5 Expand f(x) = sin x, 0 < x < π in a Fourier cosine series. Solution Bn = 0, n ≥1 Since a Fourier series consisting of cosine terms alone is obtained only for an even function, then f(x) is redefined as an even extension. A sketch of the even extension of sin x is shown in Figure 5.5. An = 2 π sin x cos nx d x , n ≥ 0 π ∫0 for n = 0 A0 = 4 2 π sin x d x = π π ∫0 for n = 1, A1 = 0 (why?) for n > 1 An = − 2(1 + cos nπ) π(n 2 − 1) Therefore, sin x = 2 2 ∞ (1 + cos nπ) − ∑ cos nx π π n=2 n2 − 1 Example 5.6 Expand f(x) = x, 0 < x < 2 in a sine series. Solution We need the odd periodic extension of f(x) shown in Figure 5.6, which follows. f (x) 1– x FIGURE 5.5 Even extension of sin x. 160 Applied Mathematical Methods for Chemical Engineers f (x) FIGURE 5.6 Odd extension of f(x) = x. An = 0 nπx 2 2 x sin dx ∫ 0 2 2 −4 = (−1)n , n ≥ 1 nπ Bn = Therefore, ∞ f (x) = ∑ n =1 nπx 4 (−1)n +1 sin nπ 2 Example 5.7 Given 1, 0 < x < 1 / 2 f (x) = 0, 1 / 2 < x < 1 what is the cosine series expansion of f(x)? Solution An even extension of f(x) is needed for a period 2l = 2. For an even extension Bn = 0, n ≥ 1 1/2 2 L A0 = ∫ f ( x ) d x = 2 ∫ d x = 1 0 L 0 also An = 1/2 nπ 2 L nπx 2 f ( x ) cos d x = ∫ cos nπx dx = sin 0 L ∫0 nπ L 2 161 Fourier Series and Integrals Therefore, f (x) = 1 ∞ 2 +∑ (−1)n −1 cos(2n − 1)πx 2 n =1 (2n − 1)π 5.5 CONVERGENCE OF FOURIER SERIES If f(x) is piecewise continuous and periodic with period 2π, then π ∫− π [ f ( x ) − Sk ( x )]2 d x ≥ 0 (5.38) where k A0 + ∑ ( An cos nx + Bn sin nx ) 2 n =1 Sk ( x ) = Upon expansion of Equation 5.38, one obtains π π π π ∫− π [ f ( x ) − Sk ( x )2 ] d x = ∫− π [ f ( x )]2 d x − 2 ∫− π f ( x )Sk ( x ) d x + ∫− π [Sk ( x )]2 d x which reduces to π ∫− π f ( x ) Sk ( x ) d x = ∫ π −π k A f ( x ) 0 + ∑ ( An cos nx + Bn sin nx ) dx 2 n =1 k πA2 = 0 + π∑ ( An2 + Bn2 ) 2 n =1 (5.39) also 2 k π A 0 ( An cos nx + Bn sin nx ) d x ∫− π [Sk ( x )]2 d x = ∫− π 2 + ∑ n =1 k πA02 = + π∑ ( An2 + Bn2 ) 2 n =1 π (5.40) Therefore, k π πA02 2 2 [ f ( x ) − S ( x )] d x = [ f ( x )] d x − + π ( An2 + Bn2 ) ≥ 0 ∑ k ∫− π ∫− π n =1 2 π (5.41) from which it follows that k π A02 + ∑ ( An2 + Bn2 ) ≤ ∫ [ f ( x )]2 d x −π 2 n =1 (5.42) 162 Applied Mathematical Methods for Chemical Engineers for all k. Further, as 1 π [ f ( x )]2 d x π ∫− π is independent of k, then 1 π A02 ∞ + ∑ ( An2 + Bn2 ) ≤ ∫ [ f ( x )]2 d x 2 n =1 π −π (5.43) which is Bessel’s inequality. Therefore, the series A02 ∞ + ∑ ( An2 + Bn2 ) 2 n =1 (5.44) converges, since it is nondecreasing and is bounded from above. That is, lim An = 0 = lim Bn n →∞ (5.45) n →∞ is the necessary condition for convergence of the series (Equation 5.7). Finally, the Fourier series is said to converge in the mean to f(x) when 2 k π A lim ∫ f ( x ) − 0 + ∑ ( An cos nx + Bn sin nx ) dx = 0 k →∞ − π 2 n =1 (5.46) If the above equation holds, then one gets 1 π A02 ∞ + ∑ ( An2 + Bn2 ) = ∫ [ f ( x )2 ] d x 2 n =1 π −π (5.47) which is known as Parseval’s relation, and the set of functions 1, cos x, sin x, cos 2x, sin 2x, … is said to be complete. The following theorem can be used to establish point convergence of a given Fourier series [1, 4]. Theorem 5.1 If f(x) is piecewise smooth and periodic with period 2π in −π < x < π, then for any x 1 A0 ∞ + ∑ ( An cos nx + Bn sin nx ) = [ f ( x + ) + f ( x − )] 2 n =1 2 where An = 1 π f ( x ) cos nx d x , n ≥ 0 π ∫− π (5.48) 163 Fourier Series and Integrals and Bn = 1 π f ( x )sin nx d x , n ≥ 1 π ∫− π Proof of Theorem 5.1 can be found elsewhere [1], however, an example of its usage is given below. Previously (see Example 5.1), the Fourier series of x + x2 on (−π, π) was found to be π2 ∞ 4 2 +∑ (−1)n cos nx − (−1)n sin nx 3 n =1 n 2 n as f(x) is piecewise smooth, the series converges and x + x2 = π2 ∞ 4 2 + ∑ 2 (−1)n cos nx − (−1)n sin nx 3 n =1 n n at points of continuity. However, at points of discontinuity, such as x = π, Theorem 5.1 gives 1 π2 ∞ 4 [(π + π 2 ) + (− π + π 2 )] = + ∑ (−1)n cos nx 2 3 n =1 n 2 where f (π − ) = π + π 2 : approaching from the left and f (π + ) = f (− π + ) = − π + π 2 : approaching from the right Therefore, the series converges to ∞ π2 1 =∑ 2 6 n =1 n (5.49) at the points of discontinuity. Another useful result regarding convergence is given below. This result is known as the Riemann–Lebesgue lemma. Lemma 5.1 If g(x) is piecewise continuous on a ≤ x ≤ b, then b lim ∫ g( x )sin λx d x = 0 λ→∞ a (5.50) Proof Consider b I (λ) = ∫ g( x )sin λx d x a (5.51) 164 Applied Mathematical Methods for Chemical Engineers then, if π λ π sin λx = sin λ t + = − sin λt λ x=t+ and I(λ) becomes I (λ ) = − ∫ b − (π / λ ) a − (π / λ ) π g t + sin λt dt λ (5.52) But t is a dummy variable in the above equation; therefore, I(λ) can be rewritten as I (λ ) = − ∫ b − (π / λ ) a − (π / λ ) π g x + sin λx d x λ (5.53) Then adding Equation 5.51 and 5.53 gives b b − (π / λ ) a a − (π / λ ) 2 I (λ) = ∫ g( x )sin λx d x − ∫ =∫ π g x + sin λx d x λ b π g x + sin λx d x + ∫ g( x )sin λx d x a − (π / λ ) b − (π / λ ) λ a +∫ b − (π / λ ) a π g( x ) − g x + λ sin λx d x If g(x) is continuous in a ≤ x ≤ b, then g(x) is necessarily bounded; that is, there exists a number M such that |g(x)| ≤ M. Therefore a π a − (π / λ ) ∫a−(π / λ ) g x + λ sin λx d x = ∫a g( x )sin λx d x ≤ πM λ and b ∫b−(π / λ ) g( x )sin λx d x ≤ πM λ giving | I (λ ) | ≤ b − (π / λ ) Mπ π +∫ g( x ) − g x + d x a λ λ (5.54) as a consequence. Further, since g(x) is a continuous function on a closed interval a ≤ x ≤ b, it is uniformly continuous on a ≤ x ≤ b such that π ε g( x ) − g x + < λ b−a 165 Fourier Series and Integrals for all λ > Λ and all x in a ≤ x ≤ b. Choosing λ such that πM/λ < ε/2 whenever λ > Λ gives | I (λ ) | < ε ε + =ε 2 2 The second result makes use of the Dirichlet kernel, Dn(x), and is given here without proof. Lemma 5.2 If a given function g(x) is piecewise continuous on the interval 0 < x < π and the ­right-hand derivative g’R (0) exists, then π lim ∫ g( x ) Dn ( x ) d x = n→∞ 0 π g(0 + ) 2 where Dn(x) is given as Dn ( x ) + 1 N + ∑ cos nx 2 n =1 or Dn ( x ) = sin[2n + 1]x /2 x 2sin 2 Example 5.8 This example considers the least-square properties of Fourier series. Quite often one is interested in approximating a periodic function, f, over a period. Therefore, the trigonometric partial sum SK , which best approximates such a p­ eriodic function over one full period, p, is the one for which the total squared error p E = ∫ [ f ( x ) − SK ( x )]2 d x 0 (5.55) is a minimum. This example shows that among all trigonometric sums SK ( x ) = A0 K nπx nπx + ∑ An cos + Bn sin 2 n =1 p p SK ( x ) = A0 K nπx nπx + ∑ An cos + Bn sin 2 n =1 p p the Kth partial sum of the Fourier series of a periodic function is for every value K the best least-square approximation to f over one period (and can be extended to any number of periods). 166 Applied Mathematical Methods for Chemical Engineers Solution Let p E = ∫ [ f ( x ) − SK ( x )]2 d x 0 be the error _ associated with the particular partial sum Sk(x). Then it is important to compare E with E. Expansion of Equation 5.55 gives p E = ∫ [ f 2 x − 2 f ( x ) SK ( x ) + SK2 ( x )] d x 0 K p p A nπx nπx = ∫ f 2 ( x ) d x − 2 ∫ f ( x ) 0 + ∑ An cos + Bn sin dx 0 0 p p 2 n =1 2 K p A nπx nπx + ∫ 0 + ∑ An cos + Bn sin dx 0 2 p p n =1 The integral in the second term simplifies to K A A −2 p 0 0 + ∑ ( An An + Bn Bn ) 2 n =1 The third term reduces to A2 K p 0 + ∑ An2 + Bn2 2 n =1 ( ) such that K p A A E = ∫ f 2 ( x ) d x − 2 p 0 0 + ∑ ( An An + Bn Bn ) 0 n =1 2 2 A2 K + p 0 + ∑ An2 + Bn2 dx 2 n =1 ( ) By making the following substitutions A0 = A0 , An = An , and Bn = Bn the quantity E becomes p A2 K E = ∫ f 2 ( x ) d x − p 0 + ∑ ( An2 + Bn2 ) 0 2 n =1 _ Subtracting E from E results in A − A0 2 K E − E = p 0 + ∑ ( An − An )2 + ( Bn − Bn )2 2 n =1 _ From this result, one can conclude that for every possible set of coefficients ­(A n), ( Bn ), _ E ≥_ E. Further, _ equality holds only if each squared difference is zero, that is, if and only if A0 = A0, A n = An, and Bn = Bn, which means that SK ( x ) = Sk(x). 167 Fourier Series and Integrals 5.6 FOURIER INTEGRALS In the previous sections, some theory and a few applications involving the expansion of a function f(x) into a Fourier series were discussed. The function f(x) had period 2 L, with L assumed finite. An important question is what happens when L approaches infinity? Below, this question is explored. Suppose f(x) satisfies the following conditions: f(x) and f ′(x) are piecewise continuous in every finite interval (I). ∞ Then ∫−∞ | f ( x ) | d x converges, that is, is absolutely integrable in (−∞, ∞) (II). ∞ f ( x ) = ∫ [ A(ω ) cos ωx + B(ω )sin ωx ]dω (5.56) 1 ∞ f ( x ) cos ωx d x π ∫−∞ 1 ∞ B(ω ) = ∫ f ( x )sin ωx d x π −∞ (5.57) 0 where A(ω ) = The condition (I) is true if x is a point of continuity of f(x), and f(x) must be replaced by [f(x+) + f(x−)]/2 if x is a point of discontinuity. Equation 5.56 is referred to as a Fourier integral expansion of f(x). Example 5.9 Find the Fourier integral expansion of 0, x < 0 f ( x ) = − ax , a>0 e , x > 0 Solution Use Equation 5.52 with 1 ∞ f ( x ) cos ωx d x π ∫−∞ ∞ 1 = ∫ e − ax cos ωx d x π −∞ A(ω ) = = and 1 a π a 2 + ω 2 1 ∞ − ax e sin ωx d x π ∫−∞ 1 ω = 2 π a + ω 2 B(ω ) = 168 Applied Mathematical Methods for Chemical Engineers Therefore, the given function becomes ∞ f ( x ) = ∫ [ A(ω ) cos ωx + B(ω )sin ωx ] dω 0 ∞ =∫ 0 1 a ω cos ωx + 2 sin ωx dω π a 2 + ω 2 a + ω2 Some other useful forms of Fourier’s integral are f (x) = and 1 ∞ ∞ f (u) cos ω ( x − u) du dω π ∫0 ∫−∞ (5.58) 1 ∞ ∞ f (u)e iω ( x − u ) du dω π ∫0 ∫−∞ (5.59) f (x) = Example 5.10 Use Equation 5.59 to find the Fourier representation of 0, x < 0 f ( x ) = − ax e , x > 0, a>0 Solution 1 ∞ ∞ f (u)e iω ( x − u ) du dω 2π ∫−∞ ∫−∞ 1 ∞ ∞ − ax − iωx iωx e e e d x dω = 2π ∫−∞ ∫−∞ f (x) = where the substitution f (u)e − iωu du ≡ e − ax e − iωx dx has been made. However, ∞ ∞ 1 ∫−∞ e− (a+ iω ) x d x = ∫0 e− (a+ iω ) x d x = a + iω Therefore, f (x) = 1 ∞ 1 e iωt dω 2π ∫−∞ a + iω If f(x) is either an odd or an even function, the Fourier integral can be reduced to a sine or cosine representation. That is, for f(x) odd f (x) = ∞ 2 ∞ sin ωx dω ∫ f (u)sin ωu du ∫ 0 0 π (5.60) 169 Fourier Series and Integrals and for f(x) even f (x) = ∞ 2 ∞ cos ωx dω ∫ f (u) cos ωu du ∫ 0 0 π (5.61) Equations 5.60 and 5.61 are the Fourier sine and cosine integral representations of f(x), respectively. Example 5.11 Find the Fourier cosine and sine representation of f(t) = e−at, 0 < t < ∞, a > 0. Solution Equation 5.61 suggests that ∞ ∫0 ∞ f (u) cos ωu du ≡ ∫ e − at cos ωt dt 0 = a a2 + ω 2 Therefore, the cosine representation of the given function is f (t ) = 2 ∞ a cos ωt dω π ∫0 a 2 + ω 2 Similarly, the sine representation of e−at is f (t ) = ∞ 2 ω sin ω t dω π ∫0 a 2 + ω 2 Example 5.12 Find the Fourier cosine integral representation of the function 1, 0 < x < 1 f (x) = 0, x ≥ 1 Solution From Equation 5.61, f (x) = ∞ 2 ∞ cos ωx dω ∫ f (u) cos ωu du 0 π ∫0 Therefore, 1 2 ∞ cos ωx dω ∫ cos ωu du ∫ 0 0 π 2 ∞ sin ω cos ωx dω = ∫ π 0 ω 1 = f (x) = 170 Applied Mathematical Methods for Chemical Engineers The form given by Equation 5.59 is particularly useful, in that if F (ω ) = ∫ ∞ −∞ then f (x) = f (u)e − iωu du 1 ∞ F (ω )e iωx dω 2π ∫−∞ (5.62) (5.63) Equations 5.62 and 5.63 are the Fourier transform pair, where F(ω) is the Fourier transform of f(x) and Equation 5.63 is the inverse transform. The customary notation for the Fourier transform is F{f(x)} and its inverse is denoted by F−1{F(ω)}. These notations will be used in this book. Again, if f(x) is an odd function, then Equation 5.60 can be reinterpreted as ∞ Fs (ω ) = ∫ f (u)sin ωu du 0 (5.64) with inverse f (x) = 2 ∞ Fs (ω )sin ωx dω π ∫0 (5.65) Here, Fs(ω) is the Fourier sine transform of f(x). If f(x) is an even function, Equation 5.60 can be reinterpreted as ∞ Fc (ω ) = ∫ f (u) cos ωu du 0 (5.66) with inverse transform given by f (x) = 2 ∞ Fc (ω ) cos ωx dω π ∫0 (5.67) where Fc(ω) is the Fourier cosine transform of f(x). Similar to Laplace transforms, there is a convolution theorem for Fourier transforms [2],[4],[7–9], which states that the Fourier transform of the convolution of two functions f(x) and g(x) is equal to the product of their Fourier transforms. That is, F{ f * g} = F{ f ( x )}F{g( x )} (5.68) and the convolution obeys the commutative, associative, and distributive laws of algebra. The convolution of the functions f(x) and g(x) is defined to be f *g = ∫ ∞ −∞ f (u) g( x − u) du Example 5.13 Solve for f(x) in the integral equation given by ∞ ∫0 1 − α , 0 ≤ α ≤ 1 f ( x )sin αx d x = α >1 0, (5.69) 171 Fourier Series and Integrals Solution According to Equation 5.64, 1 − α , 0 ≤ α ≤ 1 ∞ Fs (α ) = ∫ f ( x )sin αx d x = 0 α >1 0, is the Fourier sine transform of f(x). Then, by Equation 5.65, f (x) = 2 ∞ 2 1 Fs (α )sin αx dα = ∫ (1 − α )sin αx dα π ∫0 π 0 2( x − sin x ) = πx 2 Example 5.14 If the Fourier transforms of the functions f1(x) and f 2(x) exist, what is the Fourier transform of a1f1 + a2f 2 for a1, a2 constants? Solution ∞ F{a1 f1 + a2 f2} = ∫ [a1 f1 (u) + a2 f2 (u)]e − iωu du −∞ = a1 ∫ ∞ f1 (u)e − iωu du + a2 ∫ −∞ ∞ −∞ f2 (u)e − iωu du (5.70) a1F{ f1} + a2 F{ f2} The linearity property displayed by Example 5.14 is generally true for Fourier transforms and their inverses. Example 5.15 If f(x) is continuous and f′(x) is at least piecewise continuous on (−∞, ∞), and if ∞ ∫−∞ | f ( x ) | d x and ∞ ∫−∞ | f ′( x ) | d x exist, then show that F{f′(x)} = iωF{f(x)}. Solution F{ f ′(u)} = ∫ ∞ −∞ f ′(u)e − iωu du = F (ω ) then using integration by parts with p = e − iωu and dq = f ′(u) du 172 Applied Mathematical Methods for Chemical Engineers results in F{ f ′(u)} = lim [e iωu f (u)]−HH + iω ∫ ∞ −∞ H →∞ f (u)e − iωu du However, f(x) is given as absolutely integrable; therefore, it must vanish at both −∞ and ∞, such that, the final result is F{ f ′(u)} = iω ∫ ∞ −∞ f (u)e − iωu du = iωF (ω ) (5.71) The result of this latter example can be extended to the general result F{ f ( n ) ( x )} = (iω )n F (ω ) (5.72) as long as successive derivatives of f(x) are continuous and absolutely integrable on (−∞, ∞). Additional properties of Fourier transforms can be found in most advanced engineering texts [2],[7],[8]. Example 5.16 Obtain the solution of the initial-value problem of heat conduction in an infinite rod as described by ∂T ∂2 T = k 2 , −∞ < x < ∞, t > 0 ∂t ∂x subject to T ( x ,0) = f ( x ) where T is the temperature distribution and is bounded, and k is a constant representing diffusivity. Solution ∞ Let U (ω , t ) = F{T ( x , t )} = ∫ T ( x , t )e − iωx d x −∞ then the inverse transform is T (x, t) = 1 ∞ U (ω , t )e − iωx dω 2π ∫−∞ Therefore, the differential equation becomes dU (ω , t ) + kω 2U = 0 dt and the given initial condition becomes U (ω , 0) = F (ω ) 173 Fourier Series and Integrals This is now a first-order linear ordinary differential equation with U (ω , t ) = Ce − kω t 2 as its general solution while treating ω as a parameter. The particular solution is U (ω , t ) = F (ω )e − kω t 2 Using Equation 5.63, the inverse transform is 1 ∞ U (ω , t )e iωx dω 2π ∫−∞ 1 ∞ 2 = F (ω )e − kω t e iωx dω 2π ∫−∞ T (x, t) = This is now in the appropriate form for which the convolution theorem for Fourier transform is useful to carry out the inversion. That is, f *g = ∫ ∞ −∞ f (u) g( x − u) du By comparison, g(x) is the inverse of G (ω ) = e − kω t, which has the form 2 1 ∞ G (ω )e iωx dω 2π ∫−∞ 1 ∞ − kω 2t iωx = e e dω 2π ∫−∞ g( x ) = = 1 π − x 2 /4 kt e 2π kt = 1 2 e − x /4 kt 4πkt Consequently, the final solution is T (x, t) = 1 ∞ 1 ( x − u) 2 f (u) exp − du ∫ −∞ 2π 4 πkt 4 kt 5.7 PROBLEMS 1.Assuming that f(t), f ′(t), f ″(t), and f ‴(t) are continuous and absolutely integrable on (−∞, ∞), show that a. F {f ′′ (t)} = iωF {f ′(t)} = (iω)2F(ω) b. F {f ‴(t)} = iωF {f ″(t)} = (iω)3F(ω) 174 Applied Mathematical Methods for Chemical Engineers 2.If F(ω) is the Fourier transform of f(t), show that F″(ω) = F{−f(t)}. Hint: Assume that f(t) is absolutely integrable on (−∞, ∞). 3. What error is made in approximating f(t) = e−t on 0 < t < 1 by the sum of the first three nonzero terms of its Fourier series? 4.Determine the values of a and b, which make the line y = a + bx the best least-square approximation to ex on 0 < x < 1, using the least-square crite1 rion given in Example 5.8 E (a, b) = ∫ [e x − (a + bx )]2 d x as the quantity to 0 be minimized. Recall that for E to be a minimum, the quantities ∂E/∂a and ∂E/∂b must both be zero. REFERENCES 1. Churchill, R.V. and Brown, J.W. Fourier Series and Boundary Value Problems, 4th ed., McGraw-Hill, New York, 1987. 2. O’Neil, P.V. Advanced Engineering Mathematics, 4th ed., PWS-Kent, Boston, 1995. 3.Spiegel, M.R. Fourier Analysis with Applications to Boundary Value Problems, McGraw-Hill, New York, 1974. 4. Myint-U.T. and Debnath, L. Partial Differential Equations for Scientists and Engineers, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1987. 5. Giordano, F.R. and Weir, M.D. Differential Equations, a Modeling Approach, AddisonWesley, Reading, MA, 1991. 6.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 3rd ed., John Wiley, New York, 1977. 7. Wylie, C.R. and Barrett, L.C. Advanced Engineering Mathematics, 6th ed., McGrawHill, New York, 1995. 8.Greenberg, M.D. Advanced Engineering Mathematics, Prentice Hall, Englewood Cliffs, NJ, 1988. 9. Zauderer, E. Partial Differential Equations of Applied Mathematics, John Wiley, New York, 1983. 6 Partial Differential Equations 6.1 INTRODUCTION In the previous five chapters, the essential tools necessary to tackle this chapter have been outlined. Herein, we will follow the “standard mathematician’s MO,” that is, I will attempt to reduce these new problems (solution of partial differential equations) to those we already know how to solve (ordinary differential equations). In the practice of chemical engineering science, there are many problems that must be described by two or more independent variables. For example, the equations of change for isothermal systems [1] are ∂ ∂ρ ∂ ∂ = − ρv x + ρv y + ρvz ∂x ∂t ∂y ∂z Dv ρ = −∇p − [∇ ⋅ τ] + ρg Dt Dv ρ = −∇p + µ∇ 2 v + g Dt (6.1) (6.2) (6.3) Equation 6.1 is the equation of continuity, and it describes the rate of change of density at a fixed point resulting from the changes in the mass velocity vector ρv . Equation 6.2 is the equation of motion, which states that a small volume element moving with the fluid is accelerated due to the forces acting on it. Equation 6.3 is the Navier–Stokes equation. These equations together with appropriate boundary and initial conditions make up a large portion of the research problems that are encountered in chemical engineering. Equation 6.1 through Equation 6.3 are partial differential equations (PDEs), as opposed to ordinary differential equations, but the definitions of linearity and homogeneity remain the same as those given for second-order ordinary differential equations. As a reminder, a linear operator L [2–5] must satisfy L (c1u1 + c2 u2 ) = c1 L (u1 ) + c2 L (u2 ) (6.4) for any two functions u1 and u2 where c1 and c2 are arbitrary constants. For example, ∂/∂t and ∂2/∂x2 are linear operators since they satisfy ∂ ∂u ∂u (c1u1 + c2 u2 ) = c1 1 + c2 2 ∂t ∂t ∂t 175 176 Applied Mathematical Methods for Chemical Engineers and ∂2 ∂2 u1 ∂2 u2 ( c u + c u ) = c + c 2 1 1 2 2 1 ∂x 2 ∂x 2 ∂x2 The general linear second-order PDE in one dependent variable u and independent variables x and y is Auxx + Buxy + Cu yy + Dux + Eu y + Fu = G (6.5) where the coefficients are functions of x and y and do not vanish simultaneously. Also, in Equation 6.5 it is assumed that u and the coefficients are twice continuously differentiable in some domain R. If G = 0 in Equation 6.5, then the equation is homogeneous. A fundamental property of linear operators (Equation 6.4) allows solutions of linear equations to be added together. This property is the principle of superposition and can be stated as follows: Given that u1, u2, …, uk satisfy a linear homogeneous equation, then an arbitrary linear combination of these, c1u1 + c2u2 + ⋯ + ckuk also satisfies the same linear homogeneous equation. The concepts of linearity and homogeneity also apply to boundary conditions (see definitions in Chapter 4). In the sections that follow, solution techniques for linear boundary value problems are developed. Specifically, the method of separation of variables in Section 6.2 is illustrated. In Section 6.3, the method of eigenfunction expansion is outlined. In Section 6.4, the method of Laplace transform is illustrated. The method of combination of variables is outlined in Section 6.5. In Section 6.6, Fourier integrals are introduced. Each method is accompanied by at least two worked-out examples demonstrating the important steps in the construction of a solution to a given problem. Also, the method of regular perturbation, a technique that can be very helpful in estimating a solution to some nonlinear problems, is briefly introduced in this chapter. 6.2 SEPARATION OF VARIABLES Separation of variables is one of the most widely used solution techniques for a system consisting of a second-order PDE with boundary and/or initial conditions. This solution technique requires that the PDE be reduced to a system of ordinary differential equations. The spatial part of the PDE forms the ordinary differential equation in the resulting boundary value problem. Typically, these boundary value problems are precisely those Sturm–Liouville problems discussed in Chapter 4. As a reminder, a boundary value problem consists of a differential equation together with appropriate boundary conditions. In separating the variables, a substitution is made that transforms the PDE to an ordinary differential equation. This same substitution is used to transform the boundary conditions accompanying the PDE into appropriate boundary conditions for the ordinary differential equation. The time or time-like part of the PDE usually results in an initial value type problem. The general solution of this initial value problem is then combined with the 177 Partial Differential Equations eigenfunctions resulting from solving the boundary value problem. Application of the initial condition usually results in a Fourier series. In order for this technique to be successfully applied, both the PDE and accompanying boundary conditions must be linear and homogeneous. To demonstrate the technique, we will consider the following model: 1.One-dimensional rod of length, L, which is laterally insulated. 2.Heat is not internally generated. 3.Rod is uniform with constant density throughout its length. 4.Material of construction has constant specific heat and constant thermal conductivity. 5.Ends of the rod are kept at a fixed temperature, 0ºK in this case. 6.Rod has an initial temperature distribution that is a function of its length. This model can be mathematically described for the temperature distribution, u, by conducting an energy balance over a finite segment of a one-dimensional rod: heat energy Let ϕ( x , t ) = heat flux surface area ⋅ time heat energy Q( x , t ) = heat source volume ⋅ time f (b, t) f (a, t) x=a L x=b x u(x, t) = temperature mass ρ( x ) = mass density volume heat energy c( x ) = specific heat mass ⋅ degree A = cross-sectional area b Then the total heat energy is ∫ c ρ u A d x. a 178 Applied Mathematical Methods for Chemical Engineers Since change in flow through side = total heat energy edges at x = a and x = b heat generated + inside the region (6.6) d dt (6.7) then we have ∫ a c ρ u A d x = φ(a, t ) A − φ(b, t ) A + ∫ a AQ d x b b b An ordinary derivative, d/dt is used in Equation 6.7 because ∫ c ρ u A d x depends on a t only. Also notice that A is constant and can be divided out. From calculus, the quantity b d b ∂u c ρ u A d x ≡ ∫ cρ A d x ∫ a a dt ∂t (6.8) as long as u is continuous and a and b are constants. Also Equation 6.8 holds because the derivative is taken with x held constant, which means the ordinary derivative is to be replaced by a partial derivative. Furthermore, we can observe that φ(a, t ) − φ(b, t ) = − ∫ b a ∂φ dx ∂x since φ (x, t) is being taken as continuously differentiable. Therefore, Equation 6.7 can be restated as b ∂u ∂φ ∫a cρ ∂t + ∂ x − Q d x = 0 (6.9) From which one can conclude that cρ ∂u ∂φ = − + Q( x , t ) ∂t ∂x (6.10) But ϕ( x , t ) = −κ ∂u (Fourier’s law) ∂x Therefore, upon substitution, we get ∂u κ ∂2 u ∂2 u Q x t k = + ( , ) = + Q( x , t ); ∂t cρ ∂ x 2 ∂x 2 κ =k cρ (6.11) In the present model, we are assuming that heat is not generated, therefore, the term Q(x, t) = 0, such that the model becomes l ∂u ∂2 u = , 0 < x < L, t > 0 k ∂t ∂ x 2 (6.12) 179 Partial Differential Equations u(0, t ) = 0 for t > 0 BC : (6.13) u( L , t ) = 0, IC : u( x , 0) = f ( x ), 0 < x < L (6.14) where k is thermal diffusivity (L2/time), and these boundary conditions are described as fixed homogeneous. Solution Assume that u( x , t ) = X ( x )T (t ) (6.15) where X(x) represents a function of x only and T(t) represents a function of t only. Then Equation 6.15 can be substituted into Equation 6.12 such that ∂u dT = X (x) = X ( x )T ′(t ) ∂t dt ∂u dX = T (t ) = X ′( x )T (t ) ∂ x dx ∂2 u d 2 X = T (t ) = X ′′( x )T (t ) ∂ x 2 dx 2 giving 1 X ( x )T ′(t ) = X ′′( x )T (t ), k which can be rewritten as 1 T ′ X ′′ = k T X (6.16) Now, observe that the left-hand side of Equation 6.16 is a function of t only, whereas the right-hand side is a function of x only. This can be a true statement only if Equation 6.16 is equal to a constant. That is, 1 T ′ X ′′ = =λ k T X (6.17) where the number λ is called the separation constant. Equation 6.17 can now be recasted as a system of two ordinary differential equations 1 T′ =λ k T (6.18) X ′′ =λ X (6.19) and 180 Applied Mathematical Methods for Chemical Engineers To further define λ, we sequentially substitute the BC, Equation 6.13, into 6.15, thus u(0, t ) = X (0)T (t ) ⇒ X (0) = 0 where the argument that T(t) is an arbitrary function independent of x has been applied to conclude that X(0) = 0. The result, u( L , t ) = X ( L )T (t ) ⇒ X (L) = 0 is arrived at using the same reasoning. Therefore, Equation 6.19 is to be subjected to the boundary conditions X(0) = 0 (6.20) X (L) = 0 (6.21) and Equation 6.19 together with Equations 6.20 and 6.21 constitute a homogeneous boundary value problem as defined in Chapter 4. Furthermore, Equation 6.19, together with Equations 6.20 and 6.21, is an example of a regular Sturm–Liouville problem, where p(x) = 1, r(x) = 1, and q(x) = 0. Also, A1 = B1 = 1 and A2 = B2 = 0 for the boundary conditions (Chapter 4). The three cases of λ are examined in the way of Chapter 4, that is, λ = 0 and λ > 0 both produce the trivial solution. The third case, λ < 0, say λ = −β 2 gives X ′′ + β 2 X = 0 X (0) = 0 X (L) = 0 which solves to λn = − n2π 2 , n = 1, 2, L2 and X n ( x ) = A1,n sin nπx L The numbers defined by λ n are the eigenvalues and the functions Xn(x) are the corresponding eigenfunctions as discussed in Chapter 4. Now, attention is given to Equation 6.18, which solves to Tn (t ) = ce λ n kt where c is an arbitrary constant. Therefore, Equations 6.12 through 6.15 result in nπx λ n kt un ( x , t ) = A1,n sin ce L 181 Partial Differential Equations for each n. However, by the principle of superposition the solution of Equations 6.12 and 6.13, using Equation 6.15, can be represented as ∞ u( x , t ) = ∑ Bn sin n =1 nπx n2π 2 exp − 2 kt L L (6.22) We now use the given initial condition, Equation 6.14, to define Bn in the following way: ∞ u( x ,0) = ∑ Bn sin n =1 nπx = f (x) L (6.23) Equation 6.23 can be recognized as the Fourier sine series representation of f(x). That is, Bn = nπx 2 L f ( x )sin d x for n ≥ 1 ∫ 0 L L (6.24) Therefore, the described model has Equations 6.22 and 6.24 as its solution. 6.2.1 BOUNDARY CONDItIONs In this section, we will use examples to demonstrate the four standard types (classes) of boundary conditions that usually accompany a PDE. Class 1: Bar with zero temperature at both ends ∂u ∂2 u = ,0 < x < 1, t > 0 ∂t ∂ x 2 (6.25) BC : u(0, t ) = u(1, t ) = 0 (6.26) 1, 0 < x < 0.5 IC : u( x , 0) = 0, 0.5 < x < 1 (6.27) Solution Step 1: Substitute u = XT into Equation 6.25 to obtain and X ′′ = constant X (6.28) T′ = same constant T (6.29) Interpret the boundary conditions, Equation 6.26, in terms of the substitution u = XT: u(0, t ) = X (0)T (t ) = 0 and u(1, t ) = X (1)T (t ) = 0 182 Applied Mathematical Methods for Chemical Engineers Then, for nontrivial solution to exist, we get X (0) = 0 and X (1) = 0 (6.30) Step 2: Usually there are three cases of the constant to consider in solving Equation 6.28. However, in the general model considered in Equation 6.12 through Equation 6.14, we found out that only the trivial solution resulted if the constant was zero or positive. Nontrivial solution resulted only for a negative constant, say −λ2, that is, designating the constant to be −λ 2, Equation 6.28 has general solution X ( x ) = c1 cos λx + c2 sin λx Then applying the conditions given by Equation 6.30, we get λ = nπ, n = 1, 2, 3, and X n ( x ) = c2 sin nπx Step 3: Solve Equation 6.29, T′ = −λ 2 = − n 2 π 2 T to get Tn (t ) = c3 exp(− n 2 π 2t ) Therefore, for each n we have the product solution un ( x , t ) = X n Tn = Bn sin nπx exp(−n 2 π 2t ) Step 4: By the superposition principle we get ∞ u( x , t ) = ∑ Bn e− n π t sin nπx 2 2 n=1 as the solution to Equation 6.25 and Equation 6.26, with Bn to be determined. Step 5: Here we determine the constants Bn with the aid of Equation 6.27. That is, ∞ 1, 0 < x < 0.5 ∑ Bn sin nπx = 0, 0.5 < x < 1 n=1 183 Partial Differential Equations Then, following the discussion on Fourier series in Chapter 5, we get 2 0.5 1 ⋅ sin nπx d x 1 ∫0 nπ 2 = 1 − cos nπ 2 Bn = Finally, the solution to the model described by Equations 6.25 through 6.27 is ∞ 2 nπ 2 2 1 − cos e− n π t sin nπx 2 n=1 nπ u( x , t ) = ∑ It is important to note that the above models include the fixed homogeneous type boundary conditions and they both result in a Fourier sine series. Class 2: Bar with insulated boundaries As a second model, consider the previous model with one modification. Suppose we have the situation in which the ends of the rod are insulated instead of being at a fixed temperature of 0º. Then in this scenario, no heat flows out of, or into, either end. This is the so-called insulated boundary condition. This new model can be described using Equation 6.12 as l ∂u ∂2 u = , 0 < x < L, t > 0 k ∂t ∂ x 2 subject to Equations 6.31 and 6.14: ∂u(0, t ) =0 ∂x BC : for t > 0 (6.31) ∂u( L , t ) =0 ∂x IC : u( x , 0) = f ( x ), 0 < x < L To solve this new model by the separation of variables technique, the same procedure used above will be followed. Step 1: Substitute Equation 6.15 into 6.12 to get X ′′ = constant X (6.32) 184 Applied Mathematical Methods for Chemical Engineers and T′ = same constant kT (6.33) Also ∂( XT ) = X ′T ∂x Therefore, the interpretation of the boundary conditions given by Equation 6.31 results in ux (0, t ) = X ′(0)T (t ) = 0 and ux ( L , t ) = X ′( L )T (t ) = 0 such that X ′(0) = 0 and X ′( L ) = 0 (6.34) for nontrivial solutions. Step 2: In finding the solution to Equation 6.32, there will be three cases of the constant to consider. Case I: Suppose the constant is zero. Then Equation 6.32 becomes X ′′ = 0 with the general solution X ( x ) = c1 x + c2 The conditions given by Equation 6.34 when applied to this general solution result in X ( x ) = c2 , (6.35) a constant. This implies that the separation constant, or eigenvalue, can be zero. Case II: Suppose the constant is positive, say λ 2. Then Equation 6.32 becomes X ′′ − λ 2 X = 0 with general solution X ( x ) = c3 e λx + c3 e −λx Applying Equation 6.34 to this general solution results in 185 Partial Differential Equations c3 = c4 = 0 for λ 2 ≠ 0, and this case yields only the trivial solution. Therefore, the separation constant, λ, is not positive. Case III: Here we suppose the constant is negative, say −λ 2. Then Equation 6.32 results in X ′′ + λ 2 X = 0 with general solution X ( x ) = c5 cos λx + c6 sin λx Apply of Equation 6.34 to evaluate c5 and c6, results in c6 = 0 and c5 ≠ 0, and to obtain a nontrivial solution. That is, X ′( L ) = 0 = −c5 λ sin λL + c6 λ cos λL results in sin λL = 0 ⇒ λL = nπ nπ or λ = , n = 1, 2, 3, L and X n ( x ) = c5 cos nπx L (6.36) Step 3: Solve Equation 6.33 to get T (t ) = c7 for the case of the separation constant being zero (Step 2, Case I). Therefore, u(x, t) = XT produces a constant in the case of a zero separation constant, that is, u( x , t ) = XT = c2c7 = constant = A0 2 where A0/2 is used to reemphasize the relationship to a Fourier cosine series as discussed in Chapter 5. In the case of a negative separation constant, Equation 6.33 solves to Tn (t ) = c8 e − kλ nt 2 186 Applied Mathematical Methods for Chemical Engineers Then for each n, un ( x , t ) = X n Tn = An e− k ( nπ/L ) t cos 2 nπx L Step 4: By applying the superposition principle, we get the result nπx A0 ∞ 2 + ∑ An e− k ( nπ/ L ) t cos , 2 n=1 L u( x , t ) = which is the solution to Equations 6.12 and 6.31. In this case we note that a Fourier cosine series results when the boundary conditions are of the insulated type. Step 5: Here we determine the constants An (n ≥ 0) with the aid of the given IC, in this case Equation 6.14. Thus, u( x , 0) = f ( x ) = nπx A0 ∞ + ∑ An cos 2 n=1 L where the constants A0 and An are the coefficients of the Fourier cosine series for f(x) = u(x, 0). For example, suppose f(x) is given as 1, 0 < x < 0.5 f ( x ) = u( x , 0) = 0, 0.5 < x < 1 Then A0 = 2 0.5 dx = 1 1 ∫0 and for n ≥ 1 An = 0.5 ∫0 1 ⋅ cos nπx d x = 2 nπ sin nπ 2 Therefore, 1 ∞ 2 nπ 2 u( x , t ) = + ∑ sin e− k ( nπ ) t cos nπx 2 n=1 n 2 Class 3: Bar with periodic boundary conditions A third model can be created from the first one by considering the mixed or periodic boundary conditions instead of the fixed temperature at the ends. Here the previously considered straight rod is formed into a circular ring with perfectly joined ends such that the temperature is continuous across the joint. That is, u(− L , t ) = u( L , t ) 187 Partial Differential Equations Also, the derivative of the temperature function, u, with respect to length can be assumed continuous: ux (− L , t ) = ux ( L , t ) This new model can be described as 1 ∂u ∂2 u = , − L < x < L, t > 0 k ∂t ∂ x 2 (6.37) u(− L , t ) = u( L , t ) for t > 0 BC: (6.38) u x (− L , t ) = u x ( L , t ) IC : u( x , 0) = f ( x ), − L < x < L (6.39) Again, the solution procedure is the same as above. Step 1: Substitute u = XT into Equation 6.37 to get Equations 6.32 and 6.33 below: X ′′ = constant X T′ = same constant kT The boundary conditions in Equation 6.38 are interpreted in terms of u = XT to be X (− L ) = X ( L ) (6.40) X ′(− L ) = X ′( L ) (6.41) It should be observed here that Equations 6.32, 6.40, and 6.41 together form a periodic Sturm–Liouville problem on [−L, L] as discussed in Chapter 4. Step 2: Again there are three cases of the separation constant to consider. Case I: Separation constant is zero. This gives the solution to Equation 6.32 as X ( x ) = c1 x + c2 which is subject to the boundary conditions, Equations 6.40 and 6.41. That is, at −L X (− L ) = −c1 L + c2 188 Applied Mathematical Methods for Chemical Engineers and at L X ( L ) = c1 L + c2 such that X (− L ) = −c1 L + c2 = X ( L ) = c1 L + c2 ⇒ −c1 = c1 Therefore, c1 = 0. Further, since X′ = c1, no new information is obtained except that X ( x ) = c2 , a constant This constant will be conveniently denoted as A0/2 to correspond with the notation for the Fourier series coefficient discussed in Chapter 5. That is, X (x) = A0 2 This is taken to mean that zero is an eigenvalue with the corresponding eigenfunction a constant. Case II: Separation constant is positive, say λ 2. Then, Equation 6.32 gives the general solution X ( x ) = c3 e λx + c4 e −λx Application of Equation 6.40 results in X (− L ) = c3 e −λL + c4 e λL = c3 e λL + c4 e −λL = X ( L ) or (c3 − c4 )[e λL − e −λL ] = 0 ⇒ c3 = c4 since eλL ≠ e−λL . Substitution of Equation 6.41 into the general solution yields X ′(− L ) = λc3 e −λL − λc4 e λL = λc3 e λL − λc4 e −λL = X ′( L ) (6.42) But since c3 = c4, Equation 6.42 can be rearranged to give c3 (e −λL − e λL ) = −c3 (e −λL − e λL ) ⇒ c3 = −c3 Therefore, c3 = 0 = c4. From this, we can conclude that only the trivial solution is obtained for a positive separation constant. 189 Partial Differential Equations Case III: Separation constant is negative, say −λ 2. Equation 6.32 then solves to X ( x ) = c5 cos λx + c0 sin λx Then, applying Equation 6.40 to Equation 6.43 results in X (− L ) = c5 cos λL − c6 sin λL = c5 cos λL + c6 sin λL = X ( L ) or c6 sin λL = 0 and applying Equation 6.41 to 6.43 results in X ′(− L ) = λc5 sin λL + λc6 cos λL = −c5 λ sin λL + c6 λ cos λL = X ′( L ) which reduces to c5 sin λL = 0 Therefore, for nontrivial solution to exist sin λL = 0 ⇒ λL = nπ, n = 1,2,3, giving the eigenvalues, λ= nπ L and X ( x ) = c5 cos nπx nπx + c6 sin L L are the corresponding eigenfunctions. Step 3: As before, Equation 6.33 solves to T (t ) = c7 in the case of the zero eigenvalue and T (t ) = c8 e −λ kt = c8 e − k ( nπ /L ) 2 2t (6.43) 190 Applied Mathematical Methods for Chemical Engineers in the case of the negative eigenvalues −λ 2n. In terms of the assumption u(x, t) = XT we get nπx nπx − k ( nπ /L )2 t un ( x , t ) = An cos + sin e L L for each n. Step 4: By the principle of superposition, the temperature profile u(x, t) is given by u( x , t ) = A0 ∞ nπx nπx − k ( nπ/ L )2 t + ∑ An cos + Bn sin e 2 n=1 L L which is the solution to Equations 6.37 and 6.38 with the as-yet-undefined Fourier coefficients. Step 5: Application of Equation 6.39 defines the constants A0, An, and Bn. That is, u( x ,0) = f ( x ) = A0 ∞ nπx nπx + ∑ An cos + Bn sin 2 n=1 L L where nπx 1 L f ( x ) cos d x, n ≥ 0 ∫ − L L L nπx 1 L Bn = ∫ f ( x )sin d x, n ≥ 1 L −L L An = Class 4: Bar with fixed nonhomogeneous time-independent boundaries So far we have dealt with models that are strictly homogeneous. However, there are cases of nonhomogeneous problems that can be adjusted in such a way that the method of separation of variables can still be applied. One such case is that in which the boundary conditions are nonzero constants (fixed nonhomogeneous). It is important to also note that the initial condition is still being described by Equation 6.14. This case is illustrated below. A model can be constructed by maintaining the ends of the rod at fixed nonzero temperatures, that is, at x = 0, u = T1, and at x = L, u = T2. This model is the so-called fixed nonhomogeneous, time-independent boundary conditions case. However, the technique of separation of variables requires the boundary conditions and the differential equation to be homogeneous. Therefore, the model given by Equation 6.12, l ∂u ∂2 u = , 0 < x < L, t > 0 k ∂t ∂ x 2 191 Partial Differential Equations u(0, t ) = T1 (6.44) BC : u( L , t ) = T2 , t > 0 IC : u( x , 0) = f ( x ), 0 < x < L is a nonhomogeneous problem that must be reduced to a homogeneous one. The substitution of u( x , t ) = w( x , t ) + v ( x ) (6.45) where v(x) and w(x, t) represent the steady state and transient solution, respectively, reduces Equations 6.12 and 6.44 to l ∂ w ∂2 w = , 0 < x < L, t > 0 k ∂t ∂ x 2 (6.46) and w(0, t ) = 0 w( L , t ) = 0, t > 0 (6.47) with v(x) defined by v ′′( x ) = 0 v (0) = T1 v ( L ) = T2 (6.48) (6.49) Equations 6.46 and 6.47 now form a homogeneous boundary value problem, with the initial condition given by w( x ,0) = f ( x ) − v ( x ), 0 < x < L (6.50) These models (Equations 6.46 and 6.47) were solved previously with the fixed homogeneous boundary conditions. Equation 6.48 together with the boundary conditions given by Equation 6.49 form the steady-state model, and its solution is given as v ( x ) = T1 + T2 − T1 x L (6.51) The final solution of this nonhomogeneous model is given in Table 6.1. Also, the other discussed models are summarized in the table, according to boundary condition type. Examples demonstrating the use of this table are also given. 192 Applied Mathematical Methods for Chemical Engineers TABLE 6.1 Typical Boundary Conditions Associated with the Heat Equation Boundary Conditions Fixed, homogeneous: u(0, t) = 0 u(L, t) = 0 Insulated: ux(0, t) = 0 ux(L, t) = 0 Periodic: u(−L, t) = u(L, t) Eigenvalues λn = nπ L n = 1, 2, 3, … λn = nπ L n = 0, 1, 2, … λn = nπ L ux(−L, t) = ux(L, t) n = 0, 1, 2, … Fixed, nonhomogeneous: u(0, t) = T1 λn = u(L, t) = T2 n = 1, 2, 3, … nπ L Series Solution ∞ u( x , t ) = ∑ Bn e− kλn t sin λ n x 2 n=1 where Bn are the Fourier sine series coefficients of f(x) u( x , t ) = A0 ∞ 2 + ∑ An e− kλn t cos λ n x 2 n=1 where A0 and An are the Fourier cosine series coefficients of f(x) u( x , t ) = A0 ∞ 2 + ∑[ An cos λ n x + Bn sin λ n x ]e− kλn t 2 n=1 where A0, An, and Bn are the Fourier coefficients of f(x) u(x, t) = w(x, t) + v(x) T −T v ( x ) = T1 + 2 1 x L ∞ w( x , t ) = ∑ Bn e− kλ n t sin λ n x 2 n=1 where Bn are the Fourier sine series coefficients of f(x) − v(x) Other models may be developed from different combinations of the boundary conditions. In those cases, the five steps outlined above should yield the appropriate solutions, provided the conditions of linearity and homogeneity are satisfied. There are plenty of examples in this section to illustrate the method of separation of variables. It is hoped that the reader will explore the references for deeper discussions involving these methods. Example 6.1 Consider the conduction of heat in a copper rod 100 cm in length whose ends are maintained at 0ºC for all t > 0. Derive an expression for the temperature u(x, t) if the initial temperature distribution in the rod is u( x , 0) = 50, 0 ≤ x ≤ 100 and the thermal diffusivity of copper is 1.14 cm2/s. 193 Partial Differential Equations Solution This is a case of the model described by Equations 6.12 through 6.14. From Table 6.1, the fixed homogeneous boundary conditions correspond to eigenvalues λ= nπ n π = L 100 and series solution ∞ u( x , t ) = ∑ Bn e − k ( nπ /100) t sin 2 n =1 nπx 100 Then, the initial value ∞ u( x , 0) = 50 = ∑ Bn sin n =1 nπx 100 defines Bn, that is, for n ≥ 1 100 nπx d x 1 100 nπx 100 ∫ 50 sin 100 Bn = 0 100 = 50 sin dx = (1 − cos nπ ) π n x 2 nπ 50 ∫0 100 sin 100dx ∫0 = 200 (2n − 1)π Therefore ∞ 200 (2n − 1)πx ( 2 n −1)2 π 2 /10 4 t e −1.14 sin n − π (2 1) 100 n =1 u( x , t ) = ∑ Example 6.2 Consider an aluminum rod (k = 0.86 cm2/s) of length l, to be previously at the uniform temperature of 25ºC. Suppose that at time t = 0, the end x = 0 is cooled to 0ºC while the end x = l, is heated to 60ºC, and both thereafter are maintained at those temperatures. 1. Find the temperature distribution in the rod at any time t. 2.For, l = 20 cm, use the first term in the series for the temperature distribution to find the approximate temperature at x = 5 cm when t = 30 s; when t = 60 s. 3. Use the first two terms in the series for the temperature distribution to find an approximate value of u(5, 30). What is the percentage difference between the 194 Applied Mathematical Methods for Chemical Engineers one- and two-term approximations? Does the third term in the series have any appreciable effect for this value of t? 4. Use the first term in the series for the temperature distribution to estimate the time interval that must elapse before the temperature at x = 5 cm comes within 1% of its steady-state value. Solution The model for this phenomenon is 1 ∂u ∂ 2 u = , 0 < x < l, t > 0 k ∂t ∂ x 2 subject to u(0, t ) = 0 BC: u(l , t ) = 60 IC : ( x , 0) = 25 This is a problem with fixed nonhomogeneous boundary conditions and the eigenvalues and series solution are given in Table 6.1. That is, u( x , t ) = w( x , t ) + v ( x ) where T −T 60 x v ( x ) = T1 + 2 1 X = l l and ∞ w( x , t ) = ∑ Bn e − kλ t sin λx 2 n =1 λ= where nπ l 1. To define Bn, the initial condition is applied: w( x , 0) = u( x , 0) − v ( x ) = 25 − 60 x ∞ nπx = ∑ Bn sin l l n =1 Therefore, for n ≥ 1 Bn = nπx 2 l 60 x 10 dx = (7 cos nπ + 5) 25 − sin l ∫0 l l nπ and the temperature distribution in the rod at any time t is u( x , t ) = 60 x ∞ 10 nπx 2 + ∑ (7 cos nπ + 5)e −0.86( nπ / l ) t sin π l n l n =1 195 Partial Differential Equations 2.For, l = 20 cm, x = 5 cm, and t = 30 s, the first term of the series gives u(5, 30) ≅ 15 + 10 π (−2)(0.53)sin = 12.6°C π 4 When t = 60 s, u(5, 60) = 15 + 10 (−2)(0.28)(0.707) = 13.7°C π 3. Using the first two terms of the series, we get u2 (5, 30) = 15 + 10 10 (−2)(0.530)(0.707) + (12)(0.078) = 14.1°C π 2π Then the percentage difference between the one- and two-term approximations is u2 (5,30) − u(5,30) × 100 = 11% u2 (5,30) For t = 30 s, the third term of the series is −4.876 × 10 −3, which does not have any appreciable effect on either of the two approximations. 4. Time interval that must elapse before the temperature at x = 5 cm comes within 1% of its steady-state value with the first term of the series can be estimated as follows: u(5, τ) = 15 + 10 (−2)(0.707)e −0.0212 τ π where τ indicates the time interval. Then 0.15 = e −0.0212 τ ⇒ τ = 160 s −4.5 Therefore, a time interval of 160 s would have to elapse before the temperature at x = 5 cm comes within 1% of the steady-state value. Example 6.3 Consider a uniform rod of length, l, having an initial temperature distribution given by f(x), 0 < x < l. Assume that the temperature at the end x = 0 is held at 0ºC, while the end x = l is insulated, so that no heat passes through it. 1. Show that the fundamental solutions of the PDE and boundary conditions are un ( x , t ) = e − (2 n −1) 2 π 2 kt /4 l 2 sin (2n − 1)πx , n = 1,2,3, 2l 196 Applied Mathematical Methods for Chemical Engineers 2. Find a formal series expansion for the temperature u(x, t), ∞ un ( x , t ) = ∑ cnun ( x , t ), n =1 that also satisfies the initial condition u(x, 0) = f(x). Solution The model for this problem is 1 ∂u ∂ 2 u = , 0 < x < l,t > 0 k ∂t ∂ x 2 u(0, t ) = 0 t>0 ux (l , t ) = 0 u( x , 0) = f ( x ), 0 ≤ x ≤ l From Table 6.1, there is no model with this combination of boundary conditions. Therefore, the method must be directly applied; that is, let u( x , t ) = X ( x )T (t ) Then, after substitution into the differential equation, one gets 1 T ′(t ) X ′′( x ) = =λ k T (t ) X (x) or the boundary value problem X ′′ − λX = 0 X (0) = 0 X ′(l ) = 0 and T′ = λk ⇒ T (t ) = c1e λkt T We now consider the three cases of λ that must satisfy the above boundary value problem. That is, for λ = 0 we get the general solution X ( x ) = c2 x + c3 X (0) = 0 = c3 X ′(l ) = 0 = c2 , λ ≠ 0 197 Partial Differential Equations Suppose λ > 0, say λ = β 2, then X ( x ) = c4 eβx + c5e −βx and X (0) = c4 + c5 = 0 ⇒ c5 = − c4 Therefore X ( x ) = c4 (eβx − e −βx ) = 2c4 sinh βx then X ′(l ) = 2βc4 cosh βl = 0 ⇒ c4 = 0 and c5 = 0 since cosh βl ≠ 0 Therefore, λ > 0 yields the trivial solution. As the third case, suppose λ < 0, λ = −α2, then the general solution X ( x ) = c6 cos αx + c7 sin αx results; X (0) = 0 = c6 X ′(l ) = αc7 cos αl = 0 then for nontrivial solution to exist, c7 ≠ 0 and π cos αl = 0 ⇒ αl = (2n − 1) , n = 1,2, 2 (2n − 1)π α= 2l are the eigenvalues and X ( x ) = c7 sin (2n − 1)πx 2l are the eigenfunctions. Then, for each n un ( x , t ) = Tn (t ) X n ( x ) = cn e − (2 n −1) 2 π 2 kt /4 l 2 sin (2n − 1)πx 2l By the superposition principle, ∞ ∞ u( x , t ) = ∑ cnun ( x , t ) = ∑ cn e − (2 n −1) n =1 n =1 2 π 2 kt /4 l 2 sin (2n − 1)πx 2l 198 Applied Mathematical Methods for Chemical Engineers where ∞ f ( x ) = u( x ,0) = ∑ cn sin n =1 (2n − 1)πx 2l That is, for n ≥ 1 l ∫0 f ( x )sin l (2n − 1)πx (2n − 1)πx d x = cn ∫ sin 2 dx 0 2l 2l or cn = 2 l (2n − 1)πx f ( x )sin dx l ∫0 2l Example 6.4 Consider the model described by 1 ut = uxx , 0 < x < l , t > 0 k u(0, t ) = 0, ux (l , t ) + γu(l , t ) = 0, t > 0 u( x , 0) = f ( x ) 0 ≤ x ≤ l (6.52) where γ is a positive constant. 1.Let u(x, t) = X(x)T(t) and show that X ′′ − σX = 0, X (0) = 0, X ′(l ) + γX (l ) = 0 (6.53) and T ′ − σkT = 0 where σ is the separation constant. 2. Assume that σ is real, and show that Equation 6.53 has only trivial ­solutions if σ ≥ 0. 3.If σ < 0, let σ = −λ 2, λ > 0. Show that Equation 6.53 has nontrivial solutions only if λ is a solution of the equation λ cos λl + γ sin λl = 0 Solution 1. For u(x, t) = X(x)T(t), substitution into the PDE results in 1 T ′ X ′′ = =σ k T X (6.54) 199 Partial Differential Equations and u(0, t ) = X (0)T (t ) ⇒ X (0) = 0 Also ux ( x , t ) = X ′( x )T (t ) such that ux (l , t ) = X ′(l )T (t ) But ux (l , t ) + γu(l , t ) = 0 = X ′(l )T (t ) + γX (l )T (t ) = X ′(l ) + γX (l ) since T(t) is arbitrary. Therefore, we have X ′′ − σX = 0 subject to X (0) = 0, X ′(l ) + γX (l ) = 0 and T ′ − σkT = 0 2. Given that X ′′ − σX = 0 and X (0) = 0 X ′(l ) + γX (l ) = 0 then the general solution for the case σ = 0 is X ( x ) = c1 x + c2 and X ( x ) = c3e σx + c4 e − σx if σ > 0 200 Applied Mathematical Methods for Chemical Engineers But X (0) = 0 = c2 and X ′(l ) + γX (l ) = 0 = c1 + c2 γl ⇒ c1 = 0 Therefore, the case σ = 0 gives only the trivial solution. Also, for σ > 0, say σ = β 2, β > 0, X (0) = 0 = c3 + c4 ⇒ c4 = − c3 and X ′(l ) + γX (l ) = c3β(eβl − e −βl ) + γc3 (eβl + e −βl ) = 0 or c3{2β sinh βl + 2 γ cosh βl} = 0. Thus, either β tanh βl = −1 ⇒ βl < 0 or c3 = 0 but βl > 0, c3 = 0 λ Therefore, the case σ > 0, also results in the trivial solution. 3. For σ < 0, say σ = −λ 2, λ > 0 then X ′′ + λ 2 X = 0 X (0) = 0 X ′(l ) + γX (l ) = 0 give X ( x ) = c5 cos λx + c6 sin λx X ′( x ) = λc5 sin λx + λc6 cos λx then X (0) = 0 = c5 and X ′(l ) + γX (l ) = 0 = λc6 cos λl + γc6 sin λl Therefore, for nontrivial solution to exist we must choose c6 ≠ 0 and λ cos λl + γ sin λl = 0 Example 6.5 In this fifth example for this section, we will consider the temperature distribution in a circular cylinder of finite length with insulated ends. Also, the lateral surface ρ = c is kept at a temperature of 0ºK and the initial temperature distribution is a given function of ρ-only. 201 Partial Differential Equations z r r=c y f u=0 x For the cylinder shown, the model under consideration for a homogeneous material is ∂ 2 u 1 ∂u ∂u = k 2 + ∂t ∂ρ ρ ∂ρ (0 < ρ < c, t > 0) (6.55) u(c, t ) = 0 (t > 0) (6.56) u(ρ,0) = f (ρ) (0 < ρ < c) (6.57) Observe that there is an unstated boundary condition at ρ = 0. However, one can argue that the temperature must be finite along the axis of the cylinder and so the temperature distribution must be bounded there. Alternatively, the boundary value problem that will result from reducing the PDE is a singular Sturm–Liouville type. That is, the resulting second-order ordinary differential equation, when put in Sturm–Liouville form, will satisfy the requirement of ρ(0) = 0. One boundary condition is given by Equation 6.56 while the other requirement is the solution must be bounded at ρ = 0. This type was discussed as Class 3, type 1 in Chapter 4. Solution Assume a solution of the form u(ρ, t ) = R(ρ)T (t ) (6.58) 202 Applied Mathematical Methods for Chemical Engineers then follow the five steps previously given. Equation 6.55 gives 1 T′ 1 = R′′ + R′ = −λ 2 ρ kT R That is, ρR′′(ρ) + R′(ρ) + λ 2ρR(ρ) = 0 (6.59) subject to R(c) = 0 and T ′(t ) + λ 2 kT = 0 (6.60) Equation 6.59 is Bessel’s equation. Comparing Equation 6.59 with Equation 3.80, 2a − 1 a 2 − γ 2c 2 y′′ − y′ + b 2c 2 x 2 c − 2 + y = 0 x x2 gives 2a − 1 = −1 ⇒ a = 0, 2c − 2 = 0 ⇒ c = 1 b 2c 2 = λ 2 ⇒ b = λ , a 2 − γ 2c 2 = 0 ⇒ γ = 0 Also, Equation 3.80 has fundamental solutions y1 = x a J γ (bx c ) and y2 = x a J − γ (bx c ) where J γ (•) and J_γ (•) are Bessel Functions of the first kind of order γ as ­discussed in Chapter 3. Since γ = 0 in this case, then we have the zero-order Bessel Functions; that is, the general solution of Equation 6.59 is R(ρ) = c1 J 0 (λρ) + c2Y0 (λρ) where 2 Y0 (λρ) = {ln(λρ/2) + γ}J 0 (λρ) π 2 λ 2ρ2 λ 4ρ4 λ 6ρ6 1 1 1 + 2 − 2 2 1 + + 2 2 2 1 + + − π 2 2 ⋅4 2 2 ⋅4 ⋅6 2 3 (6.61) 203 Partial Differential Equations and J 0 (λρ) = 1 − λ 2ρ2 λ 4ρ4 λ 6ρ6 + 2 2 − 2 2 2 + 2 2 2 ⋅4 2 ⋅4 ⋅6 It is clear that the function Y0(λρ) is unbounded at ρ = 0. Therefore, to have a bounded solution, c2 must be chosen as zero. Application of the other boundary condition gives R(c) = 0 = c1 J 0 (λc) ⇒ J 0 (λc) = 0 for a nontrivial solution to exist. This means that the eigenvalues λ i are defined by J 0 (λ i c) = 0 for i = 1,2, (6.62) and Rk (ρ) = J 0 (λ i ρ) are the corresponding eigenfunctions. Furthermore, Equation 6.60 solves to T (t ) = c3e − λi kt 2 Therefore for each i, ui (ρ, t ) = Ai J 0 (λ i ρ)e − λi kt 2 and by the principle of superposition ∞ u(ρ, t ) = ∑ Ai J 0 (λ i ρ)e − λi kt 2 i =1 (6.63) satisfies the given differential equation and boundary conditions. The initial condition is incorporated as follows: ∞ f (ρ) = ∑ Ai J 0 (λ i ρ) (0 < ρ < c) i =1 Then, for i ≥ 1, c c ∫0 ρf (ρ) J0 (λ iρ) dρ = Ai ∫0 ρ[ J0 (λ iρ)]2 dρ That is, c Ai = 2 ∫ ρf (ρ) J 0 (λ i ρ) dρ 0 c 2 [ J1 (λ i c)]2 (6.64) Therefore, Equations 6.62 through 6.64 define the solution to the model given by Equations 6.55 through 6.57. 204 Applied Mathematical Methods for Chemical Engineers Example 6.6 Nonhomogenous Case Consider a circular plate that is composed of two different materials in the form of concentric circles. If the temperature in the plate is described by ∂ 2 u 1 ∂u ∂u + = , 0 < r < 2, t > 0 ∂r 2 r ∂r ∂t (6.65) u(2, t ) = 100, t > 0 (6.66) 200, 0 < r < 1 u(r , 0) = 100, 1 < r < 2 (6.67) Determine the temperature profile u ( r , t ) by making the substitution u(r , t ) = w(r , t ) + V (r ) (6.68) The substitution given in Equation 6.68 leads to ∂2 w 1 ∂ w ∂ w + = ∂r 2 r ∂r ∂t (6.69) d 2V 1 dV + =0 dr 2 r dr (6.70) u(2, t ) = w(2, t ) + V (2) = 100 (6.71) and with becoming w(2, t ) = 0, V (2) = 100 (6.72 and 6.73) Also, the initial condition becomes w(r , 0) = u(r , 0) − V (r ) ≡ f (r ) (6.74) Using separation of variables, w(r , t ) = R(r )T (t ) (6.75) Equation 6.69 becomes 1 R′′ + R′ − λR = 0 r T ′ =λ T (6.76 and 6.77) 205 Partial Differential Equations Upon comparing the R-equation with Equation 3.80, ( 2a − 1) a 2 − γ 2c 2 y′ + b 2c 2 x 2 c − 2 + y=0 x x2 having linearly independentsolutions y′′ − y1 = x a J γ (bx c ), y2 = x a J − γ (bx c ) we arrive at R(r ) = c1 J 0 (i λ r ) + c2 J −0 (i λ r ) (6.78) = c1 J 0 (i λ r ) + c2Y0 (i λ r ) Applying the boundary conditions: R(r ) is bounded at r = 0, then c2 must be chosen as zero because Y0 (βr ) → ∞ as r → 0 where β = i λ or λ = −β 2 (6.79) Recalling the discussion on singular Sturm-Liouville problems from Chapter 4, in this case r = 2 → R(2) = 0 = c1 J 0 (2β). However, for a nontrivial solution to exist, 0 = J 0 (2β) which defines the Eigenvalues. That is, J 0 (2β n ) = 0, n = 1,2,3, (6.80) One procedure to determine β n values: 1. Locate a table of values for the zeros of Bessel Functions. 2.Identify the case n = 1 (in this case) and locate the listed zero noting the order in which they are listed (most tables list smallest first); in this specific case 2β1 = 2.4048, 2β 2 = 5.5201, 2.4048 = 1.2024. 3.Calculate β1 from step 2 above .i.e. β1 = 2 Finally, Rn (r ) = c1n J 0 (β nr ) Is the eigenfunction and wn (r , t ) = Rn (r )Tn (t ) = c1n J 0 (β nr )c2 ne −βnt 2 Then applying the principle of superposition results in ∞ w(r , t ) = ∑ Cn J 0 (β nr )e −βn n =1 2 206 Applied Mathematical Methods for Chemical Engineers Also we need to solve (Steady State case) for V (r ) : d 2V 1 dV + =0 dr 2 r dr Subject to the conditions V (2) = 100 and V (0) is bounded. Solution for this part: V ′′ 1 a = − ⇒ ln(V ′) = − ln r + a1 ⇒ V ′ = e − ln r + a1 ≡ 2 V′ r r or V (r ) = a2 ln r + a3 Applying the boundednes condition requires that a2 be chosen as zero, since ln r → ∞, as r → 0. Applying the second boundary condition results in V (r ) = 100 where the initial condition for the unsteady problem can be described by Equation 6.74 w(r , 0) = u(r , 0) − V (r ) ≡ f (r ) Then we can find the Fourier Coefficients by employing Theorem 4.1 property 2, specifically: b ∫ r ( x )Φn ( x ) Φm ( x )dx = 0 if n ≠ m; where r ( x ) is the weighting function. a To identify the weighting function for the current problem, we need to rewrite Equation 6.76 in the Sturm-Liouville form: 1 d dR R′′ + R′ − λR ≡ r − λrR = 0 and compare with r dr dr ( p( x ) y′ )′ + [ q( x ) + λr ( x )] y = 0 noting that our weighting function is the lambda multiplier r. Therefore in our case, property 2 becomes 2 ∫ rJ0 (βn r )J0 (βm r )dr = 0 0 if n ≠ m; r is the weighting function. 207 Partial Differential Equations Returning to identify the Fourier Coefficients: 2 2 0 0 ∫ u(r , 0) − 100 r J0 (βm r )dr = Cn ∫ r J0 (βm r )J0 (βn r )dr , m ≥ 1 or 2 Cn = 2 ∫ u(r , 0) − 100 r J0 (βm r )d ∫ u(r , 0) − 100 r J0 (βm r )d 0 = 2 ∫ r J0 (βm r )J0 (βn r )dr 0 2 ∫ r J02 (βm r )dr 0 0 From the math tables (e.g., Schaums), x x n 2 2 ∫ xJn2 (αx )dx = 2 { Jn′ (αx )} + 2 1 − α 2 x 2 { Jn (αx )} 2 2 2 In our case, n = 0, therefore r2 ∫ r J02 (βm r )dr = 2 { J0′ (βm r )} 2 + 2 r2 (1) { J0 (β m r )} 2 But J 0′ (β m r ) ≡ d d J 0 (β m r ) = β m J 0 () = −β m J1 (β m r ) dr dr Since J 0 (β m r ) is a composite function and J 0′ ( x ) = − J1 ( x ) Finally, 2 { } r2 ∫ r J02 (βm r )dr = 2 −βm J1 (βm r ) 0 2 + { = 2β 2m J12 ( 2β m ) + 2 J 02 ( 2β m ) = 2β 2m J12 ( 2β m ) Therefore the Fourier Coefficients are given by 2 Cm = ∫ [u(r ,0) − 100 ]r J0 (βm r ) dr 0 2 β 2m J12 (2β m ) } 2 2 r2 J 0 (β m r ) 2 0 m ≥1 , m ≥1 208 Applied Mathematical Methods for Chemical Engineers however, ∫ x J0 ( x ) d x = xJ1 ( x ) such that 2 Cm = 1 ∫ [u(r ,0) − 100 ]r J0 (βm r ) dr ∫ [200 − 100 ]r J0 (βm r ) dr 0 2 β m2 J12 (2β m ) = 0 2 β 2m J12 (2β m ) 1 = = 50 ∫ r J 0 (β m r ) dr 0 β 2m J12 (2β m ) = 50 r J1 (β m r ) 1 0 β3m J12 (2β m ) 50 J1 (β m ) m ≥1 β3m J12 (2β m ) Summary ∞ u ( r , t ) = w ( r , t ) + V (r ) = 100 + ∑ Cm J 0 (β m r )e −βm 2 m =1 50 J (β m ) where the Cm = 3 21 m ≥ 1. β m J1 (2β m ) Until now, it has been presumed that the approach that was previously used to develop Fourier series is valid for use when Bessel Functions are involved. Although not a formal proof, in the subsequent development the Fourier Series–related approach will be justified as follows: Consider the problem given by ∂u ∂ 2 u 1 ∂u = k 2 + , u (1, 0 ) = 0, u ( r , 0 ) = f ( r ) and u ( r , t ) is bounded ∂r ∂t r ∂r Then assuming u (r , t ) = R (r ) T (t ) is a solution leads to T ′ + kλ 2T = 0, rR′′ + R′ + λ 2rR = 0 Then by comparison to Equation 3.80, a 2 − γ 2c 2 2a − 1 y′ + b 2c 2 x 2 c − 2 + y=0 x x 2 y′′ − we get a = 0, c = 1, ν = 0 and b = λ 209 Partial Differential Equations such that R(r ) = m1J0 ( λr ) + m2Y0 ( λr ) However, Y0 ( λr ) is unbounded at r = 0 and m2 must be chosen as zero. Therefore a possible solution is un (r , t ) = m3ne − k λ t m1n J0 ( λ nr ) = Ane − k λn t J0 ( λ nr ) 2 2 with J 0 ( λ n ) = 0, n = 1,2, 3, defining the eigenvalues by satisfying the condition u (1, 0 ) = 0. Applying the superposition principle results in ∞ u (r , t ) = ∑ An e − k n t J0 ( λ nr ) 2 n =1 (6.81) Then applying the given initial condition leads to ∞ f ( r ) = ∑ An J 0 ( λ nr ). (6.82) n =1 To determine the values of An, we can consider Rm = J 0 ( λ m r ) and Rn = J 0 ( λ nr ) as respective solutions of rRm′′ + Rm′ + λ 2m rRm = 0 and rRn′′ + Rn′ + λ 2nrRn = 0 Multiplying the first equation by Rn , the second by Rm and subtracting d ( Rm′ Rn − Rn′ Rm ) + ( Rm′ Rn − Rn′ Rm ) = ( λ 2n − λ 2m ) r Rm Rn dr or r d {r ( Rm′ Rn − Rn′ Rm )} = ( λ n2 − λ m2 ) r Rm Rn . dr Integrating results in using Rm = J 0 ( λ m r ) , Rn = J 0 ( λ nr ) , Rm′ = λ m J 0′ ( λ m r ) , Rn′ = λ n J 0′ ( λ nr ) ( λ 2n − λ 2m ) ∫ rJ0 ( λ mr ) J0 ( λ nr ) dr = r ( λ m J0′ ( λ mr ) J0 ( λ nr ) − λ n J0′ ( λ nr ) J0 ( λ mr )) (6.83) 210 Applied Mathematical Methods for Chemical Engineers Specifically, if we integrate between 0 and 1, we find that for λ m ≠ λ n , J 0 ( λ m ) = 0, J 0 ( λ n ) = 0 then 1 (6.84) ∫0 r J0 ( λ mr ) J0 ( λ nr ) dr = 0, m ≠ n. Therefore multiplying (I) by rJ 0 ( λ m r ) and integrating from 0 to 1 using (IV) results in 1 ∞ 1 1 0 n =1 0 0 ∫ rf (r ) J0 ( λ mr ) dr = ∑ An ∫ rJ0 ( λ mr ) J0 ( λ nr ) dr = Am ∫ rJ02 ( λ mr ) dr. Now, one can use the eigenvalues to calculate the Fourier Coefficients as needed to evaluate the solution for any or as many terms as needed. Example 6.7 In this example a model with a nonzero sink term is examined. That is, consider the problem described by ∂u 4 ∂ 2 u = − Au (0 < x < 9), t > 0 ∂t ∂x 2 (6.85) u(0, t ) = u(9, t ) = 0, t > 0 (6.86) u( x , 0) = 3 x (0 < x < 9) (6.87) where A is a positive constant. Then except for the term −Au, this would be similar to previous problems. One is therefore motivated to find a substitution that will recast the problem into a familiar homogeneous one. To accomplish this, consider the substitution u( x , t ) = w( x , t )e − At (6.88) Then Equations 6.85 through 6.87 become ∂ w 4 ∂2 w = ∂t ∂x2 (0 < x < 9), t > 0 (6.89) w(0, t ) = w(9, t ) = 0, t > 0 (6.90) w( x , 0) = 3 x (0 < x < 9) (6.91) a problem that can be solved using Table 6.1. From Table 6.1, the case of fixed, homogeneous boundary condition gives ∞ w( x , t ) = ∑ Bn e −4 λ nt sin λ n x n =1 2 211 Partial Differential Equations where λ n is defined by λn = nπ , n = 1,2,3, 9 and the Bn are the Fourier sine series coefficients of f(x) = 3x. That is, for n ≥ 1 Bn = 2 9 nπx 2 9 nπx 3 x sin d x = ∫ x sin dx ∫ 0 9 9 3 0 9 9 = 2 54 nπx 9 nπx 2 9x + sin − cos = − cos nπ 9 0 9 nπ nπ 3 nπ = 54 (−1)n +1 nπ Therefore, the solution to Equation 6.85 to 6.87 is ∞ 54 nπx 2 (−1)n +1 e −( 4 λ n + A)t sin n 9 n =1 π u( x , t ) = ∑ 6.3 NONHOMOGENEOUS PROBLEM AND EIGENFUNCTION EXPANSION In the previous section, the method of separation of variables was applied to some problems with special nonhomogeneous terms that could be recasted as homogeneous by using a suitable substitution. In this section, a method will be outlined that is applicable to those nonhomogeneous problems for which no simple substitution can be made to remove the nonhomogeneity. This method is called eigenfunction expansion [3, 4, 6]. Consider the flow of heat in a rod of length L that is uniformly constructed (Equation 6.11, Section 6.2). Further, the rod has temperature-independent heat sources distributed in some prescribed way throughout and is time-dependent. In addition, the temperature at the ends is allowed to be time-dependent. Then, for a prescribed initial temperature distribution, the following model is appropriate: ∂u ∂2 u = k 2 + Q( x , t ) ∂t ∂x (6.92) BC : u(0, t ) = A(t ) (6.93) u( L , t ) = B(t ) (6.94) IC : u( x , 0) = f ( x ) (6.95) Equations 6.92 through 6.94 describe a nonhomogeneous PDE with nonhomogeneous boundary conditions. 212 Applied Mathematical Methods for Chemical Engineers The associated homogeneous model is given by ∂v ∂2 v = k 2 + Q( x , t ) ∂t ∂x (6.96) v (0, t ) = 0 (6.97) v( L , t ) = 0 (6.98) v ( x , o) = f ( x ) (6.99) and was solved in the previous section. Also recall from the previous section that the associated homogeneous model produced the regular Sturm–Liouville boundary value problem: d 2 φn + λ n φn = 0 dx 2 (6.100) φn (0) = 0 (6.101) φn ( L ) = 0 (6.102) where λ n = (nπ/L)2 with corresponding eigenfunctions ɸ n(x) = sin(nπx/L). Any piecewise smooth function can be expanded in terms of these eigenfunctions. Again, as in Chapter 5, by piecewise smooth we mean a function, f(x) say, that is continuous on the closed interval [a, b] and whose first derivative, f′(x), is continuous on each of the subintervals xj < x < xj+1 and the limits f ( x +j ) and f ( x −j ) exist. Therefore, even though u(x, t) satisfies nonhomogeneous boundary conditions, it is valid that ∞ u( x , t ) = ∑ bn (t )φn ( x ) (6.103) n =1 except at x = 0 and x = L. This means that term-by-term differentiations with respect to x are not justified because u(x, t) and ɸ n(x) do not satisfy the same homogeneous boundary conditions. That is, ∂2 u ∞ d 2 φn ≠ b ( t ) ∑ n ∂ x 2 n =1 dx 2 However, term-by-term time derivatives are valid, such that ∂u ∞ dbn =∑ φn ( x ) ∂t n =1 dt (6.104) 213 Partial Differential Equations Therefore, Equation 6.92 becomes ∞ ∂2 u dbn + ( , ) = φn ( x ) Q x t ∑ ∂x 2 n =1 dt k (6.105) a generalized Fourier series from which L dbn = dt ∂2 u ∫0 k ∂ x 2 + Q( x , t ) φn ( x ) d x L ∫0 φn2 ( x ) d x , n ≥1 and can be rearranged to give ∂2 u L dbn = dt ∫0 k ∂ x 2 φn ( x ) d x ∫0 Q( x , t )φn ( x ) d x + L L ∫0 φn2 ( x ) d x ∫0 φn2 ( x ) d x L (6.106) The quantity L ∫0 Q( x , t )φn ( x ) d x L ∫0 φ2n ( x ) d x can be evaluated from known information; that is, we know the eigenfunctions ɸ n(x) from solving Equations 6.96 to 6.98 and Q(x, t) is given in Equation 6.92. The resulting quantity is expected to be a function of t only and is denoted as qn(t). That is, L ∫ Q( x , t )φn ( x ) d x qn (t ) = 0 L ∫0 φ2n ( x ) d x (6.107) Equation 6.106 may now be simplified to L dbn = qn (t ) + dt ∫0 ∂2 u φn ( x ) d x ∂x 2 L ∫ φn2 ( x ) d x k 0 To evaluate the quantity L ∫0 k ∂2 u φn ( x ) d x ∂x2 the Sturm–Liouville operator L≡ d d p +q dx dx (6.108) 214 Applied Mathematical Methods for Chemical Engineers is reintroduced. In this problem, p = 1, q = 0 and L= ∂2 ∂x 2 Further, recall that ∂ = d with t held constant. Then the formula ∂ x dx b du dv ∫a [uL(v) − vL(u)]d x = p u dx − v dx a b (6.109) known as Green’s formula, or the integral form of Lagrange’s identity can be used instead of integration by parts. Therefore, for this problem, we have L ∫0 L ∂2 u ∂u ∂2 v ∂v −v u 2 − v 2 d x = u ∂x ∂x ∂ ∂ x x 0 (6.110) The right-hand side of Equation 6.110 can be evaluated as ∂v ∂u ∂v ∂u u( L , t ) − v ( L , t ) − u(0, t ) + v(0, t ) ∂x x=L ∂x x=L ∂ x x =0 ∂ x x =0 nπ nπ nπ = B(t ) cos nπ − A(t ) = [ B(t )(−1)n − A(t )] L L L for v = φn ( x ) = sin nπx L dv ∂ v nπ nπx = = cos dx ∂ x L L v(0, t ) = φn (0) = 0 v ( L , t ) = φn ( L ) = 0 Therefore, Equation 6.110 becomes L ∂2 u nπ d 2 φn ( x ) − φn ( x ) 2 d x = [ B(t )(−1)n − A(t )] 2 dx L ∂x ∫0 u (6.111) 215 Partial Differential Equations But Equation 6.100 can be rearranged to give d 2 φn = −λ n φn dx 2 such that Equation 6.111 becomes ∂2 u L L ∫0 φn ( x ) ∂ x 2 d x = −λ n ∫0 uφn ( x ) d x − nπ [ B(t )(−1)n − A(t )] L (6.112) Then, substituting Equation 6.112 into 6.108 gives ∂2 u L ∫0 k ∂ x 2 φ n ( x ) d x dbn = qn (t ) + L dt ∫0 φn2 ( x ) d x nπk [ A(t ) − (−1)n B(t )] λ n k ∫ uφn ( x ) d x 0 L = qn (t ) − + L L ∫ φn2 ( x ) d x ∫ φn2 ( x ) d x L 0 (6.113) 0 However, Equation 6.103 is also a generalized Fourier series, which means that for n ≥ 1, L ∫ u( x , t )φn ( x ) d x bn (t ) = 0 L ∫0 φ2n ( x ) d x (6.114) Therefore, Equation 6.114 can be used to simplify Equation 6.113 to nπk [ A(t ) − (−1)n B(t )] dbn L + kλ n bn = qn (t ) + L dt φ2 ( x ) d x ∫0 (6.115) n a linear first-order ordinary differential equation. The required initial condition for bn(t) comes from Equation 6.95. That is, ∞ u( x ,0) = ∑ bn (0)φn ( x ) = f ( x ) n =1 such that for n ≥ 1 L ∫ b (0) = 0 n f ( x )φn ( x ) d x L ∫0 φ2n ( x ) d x As an illustration on the use of this method, an elementary example follows. (6.116) 216 Applied Mathematical Methods for Chemical Engineers Example 6.8 Given the model ∂u ∂ 2 u + sin(5 x )e −2t , 0 < x < π, t > 0 = ∂t ∂ x 2 u(0, t ) = 1, t > 0 u(π, t ) = 0 u( x , 0) = 0, 0 ≤ x ≤ π find the temperature distribution u(x, t). Solution The associated homogeneous model is given by Equations 6.96 through 6.98 and the boundary value problem satisfying the same homogeneous boundary conditions is given by Equations 6.100 through 6.102. Then nπ 2 λn = = n2 L since L = π and the corresponding eigenfunctions are φn ( x ) = sin (nx ) Also, from the given differential equation Q( x , t ) = sin (5 x )e −2t Then the quantity π L π ∫0 φ2n ( x ) d x = ∫0 sin 2 nx d x = 2 and L ∫ Q( x , t )φn ( x ) d x = 2 e−2t π sin 5x sin nx d x = 0, qn (t ) = 0 L ∫0 π ∫0 φn2 ( x ) d x = 2 −2t π 2 e ∫ sin 5 x d x = e −2t 0 π if n = 5 Further, the other needed quantities are u(0, t ) = 1, u(π, t ) = 0, v (0, t ) = φn (0) = 0 v (π, t ) = φn (π) = 0 since if n ≠ 5 (why?) 217 Partial Differential Equations φn = sin(nx ) = v dv = n cos (nx ) dx Therefore, the RHS of Equation 6.110 evaluates to −n, and Equation 6.115 becomes dbn 2n + n 2 bn = e −2t − dt π subject to bn (0) = 0 since f(x) = 0. The first-order linear differential equation solves to b5 (t ) = − 5π + 46 −25t 1 −2t 2 e + e − , 115π 23 5π using the integrating factor method described in Chapter 2. Finally, the temperature distribution is given by Equation 6.103 as ∞ 5π + 46 −25t 1 −2t 2 u( x , t ) = ∑ bn (t )φn ( x ) = − e + e − sin 5 x 23 5π 115π n =1 This method is suitable for nonhomogeneous problems when the nonhomogeneity occurs as a time-dependent source term or as time-dependent boundary conditions. If either the differential equation or the boundary conditions is nonhomogeneous, then direct substitution into Equation 6.103 may be applicable. For example, consider a plane wall of thickness L that is initially at a uniform temperature Ti. The ambient temperature on one side of the wall is suddenly changed to a new value of T∞ while the temperature on the other side of the wall is held at Ti. This can be mathematically modeled as ∂ 2 u 1 ∂u = , 0 < x < L, t > 0 ∂ x 2 α ∂t (6.117) u(0, t ) = Ti (6.118) ∂u − k = h[u( L , t ) − T∞ ] ∂x x=L (6.119) u( x , 0) = Ti (6.120) 218 Applied Mathematical Methods for Chemical Engineers The associated homogeneous problem was previously discussed as Example 6.4. Then since both u(x, t) and ɸ n(x) satisfy the same homogeneous boundary conditions, ∂2 u ∞ d 2φ n ( x ) = ∑ bn (t ) 2 dx 2 ∂x n =1 (6.121) is justifiable, and direct substitution can reduce the computational effort. Also, ∂u ∞ dbn =∑ φn ( x ) ∂t n =1 dt (6.122) Then from Example 6.4, h φn ( x ) = sin λ n x , λ n = − tan λ n L k d 2φ n = −λ 2nφn dx 2 Substitution of these quantities, along with Equations 6.121 and 6.122, into Equation 6.117 gives ∞ ∑ n =1 dbn + αλ n2 bn φn ( x ) = 0 dt (6.123) where Equation 6.103 has been used. Also, the required initial condition on the firstorder linear differential equation given by Equation 6.123 is ∞ u( x , 0) = Ti = ∑ bn (0)φn ( x ) n =1 which is a generalized Fourier series. Then, L ∫ Ti x1φn ( x ) d x for n ≥ 1 bn (0) = 0 L ∫0 φn2 ( x ) d x In this case, L L 1 L ∫0 φ2n ( x ) d x = ∫0 sin 2 λ n x d x = 2 − 4λ n sin 2λ n L and L L 1 1 ∫0 Tiφn ( x ) d x = ∫0 Ti sin λ n x d x = Ti λ n − λ n cos λ n L Therefore, for n ≥ 1 bn (0) = 2Ti (1 − cos λ n L ) sin 2λ n L λn L − 2λ n 219 Partial Differential Equations and dbn + αλ n2 bn = 0 dt which solves to bn (t ) = 2Ti (1 − cos λ n L ) −αλ 2nt e sin 2λ n L λn L − 2λ n Therefore, ∞ 2Ti (1 − cos λ n L ) −αλ 2nt e sin λ n x sin 2λ n L n =1 λn L − 2λ n u( x , t ) = ∑ So far, the methods that have been outlined are primarily applicable to problems involving finite space dimensions. In the next section, problems with semi-infinite domains will be discussed. 6.4 LAPLACE TRANSFORM METHODS In this section, the method of Laplace transform will be used. The properties of Laplace transforms, and especially Theorem 3.7 (Section 3.6.1), will be applicable. The Laplace transform was introduced earlier for use in solving ordinary differential equations. Now we will emphasize its use in solving PDEs. Assuming that the Laplace transform of the dependent variable exists, the usual procedure to solve a PDE is 1.Transform the PDE to an ordinary differential equation. 2. Transform the accompanying boundary conditions to those suitable for use with the ordinary differential equation. 3.Solve the resulting problem using the techniques discussed (Chapters 1–3). 4.Invert the results to recover the solution to the PDE. The inversion step can be relatively easy if the terms of step 3 can be located in a table of Laplace transforms. Without such a convenient table a more difficult technique involving the residue theorem has to be employed (see Example 6.11). Here, a semi-infinite rod with one end at x = 0 and extending to infinity along the positive x-axis will be a typical physical model. For example, consider the following model. Example 6.9 ∂u ∂2 u = α2 2 ∂t ∂x ( x > 0, t > 0) (6.124) 220 Applied Mathematical Methods for Chemical Engineers u( x , 0) = A (6.125) B, 0 < t < t0 u(0, t ) = 0, t > t0 (6.126) Then applying the Laplace transform to Equation 6.124 means ∂2 u ∂u L = α 2L 2 ∂t ∂x giving ∂u L = sv ( x , s) − u( x , 0) ∂t where ∞ v ( x , s) = L{u( x , t )} = ∫ e − st u( x , t ) dt 0 and ∂2 u ∂2 L 2 = 2 ∂x ∂x ∂2 ∞ ∫0 e− st u( x , t ) dt = ∂ x 2 [L{u( x , t )}] where differentiation is done treating s as a parameter, such that Equation 6.124 becomes sv ( x , s) − u( x , 0) = α 2 d 2 v( x , s) dx 2 or d2v s A − v=− 2 dx 2 α 2 α (6.127) using Equation 6.125. It should also be noted that the Laplace transform is carried out on the variable t while x is treated as a parameter. The general solution of Equation 6.127 is A v ( x , s) = c1 (s)e sx / α + c2 (s)e − sx/α + (6.128) s where the arbitrary constants c1(s) and c2(s) may depend on s. For a bounded solution, the constant c1(s) must be chosen as zero. Also, to determine c2(s), it is necessary to take the Laplace transform of the given boundary condition, Equation 6.126. That is, in terms of the unit step function u(0, t ) = B[1 − U (t − t0 )] 221 Partial Differential Equations where 0 if t < t0 U (t − t0 ) = 1 if t ≥ t0 is the unit step function. Then L{u(0, t )} = BL{[1 − U (t − t0 )]} = B B − st0 − e = v (0, s) s s Therefore v (0, s) = c2 (s) + A B B − st0 = − e s s s or c2 (s) = B − A B − st0 − e s s such that Equation 6.128 becomes B − A B − st0 − v ( x , s) = − e e s s s x α + A s Then taking the inverse transform of Equation 6.129 results in x x u( x , t ) = L−1{v ( x , s)} = ( B − A) erfc − B erf +A 2α t 2 α − t t 0 where the quantities erfc( x ) = 1 − erf( x ) = 2 π ∞ ∫x e − ξ 2 dξ and erf( x ) = 2 π x ∫0 e− ξ 2 dξ are the complementary error function and error function, respectively. As a second example illustrating the method, consider the following: Example 6.10 ∂u ∂2 u = α2 2 ∂t ∂x ( x > 0, t > 0) (6.129) 222 Applied Mathematical Methods for Chemical Engineers subject to u(0, t ) = 1 (t > 0) u( x ,0) = 0 ( x > 0) Then, as a procedure, the first step is to transform the PDE. Thus ∂2 v ∂2 u ∂u L = sv ( x , s) − u( x ,0) = α 2 L 2 = α 2 2 ∂x ∂t ∂x resulting in an ordinary differential equation α2 d2v = sv ( x , s) dx 2 since, in this case, the initial value is zero. The second step is to find the general solution of the ordinary differential equation, resulting in v ( x , s) = c1 (s)e − ( s /α) x + c2 e( s /α) x and observe that a bounded solution is expected, in which case the arbitrary constant, c2(s) in this case, must be chosen as zero. The remaining constant in the equation can be determined by taking the Laplace transform of the left boundary condition and comparing that result to v (0, s) = c1 (s) That is, L{u(0, t )} = L{1} = 1 s so that v (0, s) = c1 (s) = 1 s Therefore 1 − sx v ( x , s) = e α s The third step is to invert the Laplace transform or find the inverse transform of v(x, s) with the aid of a table [7] to get 1 − s x x u( x , t ) = L−1 e α = erfc 2α t s where erfc(•) was defined in the previous example. 223 Partial Differential Equations This third example demonstrates the flexibility of the Laplace transform method over the separation of variables method to efficiently solve some PDE problems when the domain is finite. Example 6.11 Consider ∂u ∂ 2 u = , 0 < x < 1, t > 0 ∂t ∂ x 2 subject to u(0, t ) = 1, t > 0 u(1, t ) = 0, t > 0 u( x ,0) = 0, 0 < x < 1 First, take Laplace transform of the given differential equation to get d 2 v( x , s) = sv ( x , s) − u( x ,0) dx 2 which reduces to d 2 v( x , s) = sv ( x , s) dx 2 (6.130) where use of the initial condition is made and ∞ v ( x , s) = L[{u( x , t )}] = ∫ e − st u( x , t ) dt 0 To find the constants of integration for the general solution to Equation 6.130, the Laplace transform of the boundary conditions are needed. That is, 1 s L{u(1, t )} = v (1, s) = 0 L{u(0, t )} = v (0, s) = Then, the general solution of Equation 6.130 can be represented as v ( x , s) = c1 (s)sinh x s + c2 (s) cosh x s Employing the transformed boundary conditions gives v (1, s) = 0 = c1 (s)sinh s + c2 (s) cosh s 224 Applied Mathematical Methods for Chemical Engineers which can be rearranged to give c1 (s) = − c2 (s) cosh s sinh s Then, use of the second boundary condition gives v (0, s) = 1 = c1 (s)sinh(0) + c2 (s) cosh(0) s which results in c2 (s) = 1 s Therefore c1 (s) = − 1 cosh s s sinh s The solution may now be represented as v ( x , s) = − 1 cosh s 1 sinh x s + cosh x s s sinh s s = 1 sinh s cosh x s − cosh s sinh x s s sinh s = 1 sinh[(1 − x ) s ] s sinh s Then sinh[(1 − x ) s ] u( x , t ) = L−1{v ( x , s)} = L−1 s sinh s 2 ∞ (−1)n − n2 π 2t e sin nπ(1 − x ) = 1− x + ∑ π n =1 n As a note, the quantity 1 – x can be represented in its Fourier sine series form to be 1− x = Another example follows. 2 ∞ 1 ∑ sin nπx π n =1 n 225 Partial Differential Equations Example 6.12 Solve: uxx = uy in y > 0, 0 < x < a, subject to: u(x, 0) = 1, u(0, y) = u(a, y) = 0 Solution Consider the Laplace transform of u with respect to y, that is, ∞ u ( x , s) = ∫ e − syu( x , y) d y 0 Then the differential equation and time-like condition transforms to uxx = su − 1 ⇔ d 2u = su − 1 dx 2 The general solution to the transformed ordinary differential equation is u ( x , s) = c1e sx + c2 e − sx + c3 ⇔ k1 cosh sx + k 2 sinh sx + c3 The particular constant c 3 can be evaluated by the method of undetermined coefficient, while the arbitrary constants k1 and k 2 must be evaluated from the transformed boundary conditions on x. Following this procedure, u(x, s) is given by 1 cosh s ( x − a2 ) u ( x , s) = − sa s s cosh 2 The first term can be located in a table of Laplace transforms. The second term requires a little more effort, say, the residue theorem. That is, let a sa P ( x , s) = − cosh s x − , Q(s) = s cosh 2 2 Then the residue at s = 0 is lim sP ( x , s)/Q(s) = −1. For s ≠ 0, the residue is given by s→ 0 ρn ( y) = P ( x , s n ) sn y e Q ′ ( sn ) Q′(s) = cosh sa a s sa sinh + 2 4 2 To determine the sn, Q(s) is set equal to zero, that is, cosh sa e = 2 sa /2 + e− 2 sa /2 =0 226 Applied Mathematical Methods for Chemical Engineers simplifies to e sa = −1 But the natural logarithm of a negative real number represents a multiple-­valued function [8–10]. That is, if z = ew, w = ln z = ln r + i(θ + 2nπ), n = 0, ± 1, ± 2, … where z = reiθ = rei(θ + 2nπ). However, in our case, θ = π, r = 1 and sa = i(π + 2nπ), n = 0, 1, 2, Therefore sn = −(1 + 2n)2 π 2 /a 2 Finally, 4 sin[(1 + 2n)πx / a] [ − (1+ 2 n )2 yπ 2 / a2 ] e (1 + 2n)π ρn ( y) = Therefore, the final result is 4 sin[(1 + 2n)πx /a]exp[− (1 + 2n)2 yπ 2 /a 2 ] (1 + 2n)π n=0 ∞ u ( x , y) = ∑ Additional examples involving the use of the residue theorem are discussed in Chapter 7. Example 6.13 Consider the problem given by ∂u ∂ 2 u = , 0 < x < 1, t > 0 ∂t ∂ x 2 Subject to the boundary conditions u ( 0, t ) = 0; u (1, t ) = 0 and the initial condition u ( x ,0 ) = 100. Then if we choose to use the methods of Laplace Transform to determine the profile, u ( x , t ) , we start with the definition { } ∞ L u ( x , t ) = Y ( x , s ) ≡ ∫ e − s t u ( x , t ) dt 0 227 Partial Differential Equations such that ∂2 u ∂u L = L 2 ∂t ∂x Then the left-hand side is ∂u L = sY ( x , s ) − 100 ∂t while the right-hand side is ∞ d 2Y ( x , s ) ∂2 u ∂2 L 2 = 2 ∫ e − s t u ( x , t ) dt = dt 2 ∂x ∂x 0 Therefore we get the transformed partial differential equation to become d 2Y − sY = −100, dx 2 which is now an ordinary differential equation. To complete the process of transforming the problem we need to transform the boundary conditions as well. Therefore L {u ( 0, t )} = Y ( 0, s ) = 0 and L {u (1, t )} = Y (1, s ) = 0 Finally the Laplace transformed partial differential equation and boundary conditions result in d 2Y − sY = −100, dx 2 Y ( 0, s ) = 0 and Y (1, s ) = 0 Using methods of solution of linear second order ordinary differential equations that were discussed in Chapter 3: Yg ( x , s ) = YP ( x , s ) + k 2 cosh sx + k3sinh sx In this case the particular solution is a constant and is determined as discussed in Chapter 3. YP ( x , s ) = k1 228 Applied Mathematical Methods for Chemical Engineers must satisfy d 2YP − sYP = −100, dx 2 resulting in k1 = Yg ( x , s ) = 100 s 100 + k 2 cosh sx + k3sinh sx s Applying the boundary conditions results in k2 = − ( ) ( ) 100 cosh s − 1 100 and k3 = s s sinh s Finally the solution in the Laplace domain is Y ( x, s) = 100 cosh s − 1 100 100 − cosh sx + sinh sx s s s sinh s To determine the solution in the time domain we need to invert the Laplace Transform. { } ( ) 100 100 cosh s − 1 100 − L−1 cosh sx + sinh sx s s sinh s s 100 sinh( x − 1) s 100sinh sx = 100 + L−1 − s sinh s s sinh s Y ( x , s ) = L−1 Make use of the residue theorem. Example 6.14 This example demonstrates one approach to inverting the Laplace transform when Bessel Functions are involved. ∂Ψ 0 ∂ 2 Ψ 0 1 ∂Ψ 0 = + ∂τ ∂ξ 2 ξ ∂ξ Ψ 0 = 1 at τ = 0 Ψ 0 is finite at ξ = 0; and − ∂Ψ 0 = w at ξ = 1 ∂ξ 229 Partial Differential Equations Let L {Ψ 0 ( ξ, τ )} = u ( ξ, s ) (6.131) Then the partial differential equation is transformed to the ordinary differential equation su ( ξ, s ) − 1 = d 2u 1 du + dξ 2 ξ dξ (6.132) subject to the transformed boundary conditions u ( ξ, s ) is finite at ξ = 0 − du w at ξ = 1. = dξ s (6.133) (6.134) This results in a general solution: 1 u ( ξ, s ) = a1 J 0 i s ξ + a2Y0 i s ξ + ; a1 and a2 are arbitraryconst . (6.135) s ( ) ( ) It helps to recall that Y0 ( x ) = 2 2 x2 x4 1 x6 1 1 ln ( x 2) + γ} J 0 ( x ) + 2 − 2 2 1 + + 2 2 2 1 + + − , { 2 3 2 246 π π 2 2 4 γ = 0.5772156… is Euler ’s const where J0 ( x ) = 1 − x2 x4 x6 + 2 2 − 2 2 2 + 2 2 2 i4 2 i4 i6 ( ) Then if we notice Y0 i s ξ is unbounded as ξ → 0 , hence a2 must be chosen as zero, in order for the solution to be bounded. This reduces Equation 6.135 to ( ) u ( ξ, s ) = a1 J 0 i s ξ + 1 s (6.136) Applying the boundary condition Equation 6.134, − du w w = ⇒ a1 = dξ s s i sJ1 i s ( ) (6.137) such that Equation 6.136 becomes u ( ξ, s ) = ( ) +1 ( ) s w J0 i s ξ s i sJ1 i s (6.138) 230 Applied Mathematical Methods for Chemical Engineers Inverting Equation 6.138, ( ) ( ) w J 0 i s ξ Ψ 0 ( ξ, τ ) = 1 + L−1 s i sJ1 i s (6.139) To complete the inversion, we first simplify the ratio and then apply the Residue theorem. Rewriting the Bessel Functions in series format, the fraction in Equation 6.139 results in s 3ξ 6 s 3ξ 6 sξ 2 s 2 ξ 4 sξ 2 s 2 ξ 4 w 1 + 2 + 2 2 + 2 2 2 + ... w 1 + 2 + 2 2 + 2 2 2 + ... wJ 0 (i sξ ) 2 24 246 2 24 246 = = (6.140) 2 s3 s s 1 si sJ1 (i s) s s2 − s + 2 + 2 2 + ... − s 2 + 2 + 2 2 + ... 2 2 4 2 4 6 2 2 4 2 4 6 Equation 6.140 without the quantity w can be further expressed as s 3ξ 6 sξ 2 s 2 ξ 4 1 + 22 + 22 4 2 + 22 4 262 + ... = A0 + A1s + A2 s 2 + ... s2 1 s − + 2 + 2 2 + ... 2 2 4 2 4 6 (6.141) or as s 3ξ 6 s2 sξ 2 s 2 ξ 4 1 s 2 1 + 22 + 22 4 2 + 22 4 262 + ... = − 2 − 22 4 − 22 4 26 − ... ( A0 + A1s + A2 s + ...) = s2 s2 1 s 1 s A 0 − − 2 − 2 2 − ... + A1s − − 2 − 2 2 − ... (6.142) 2 24 246 2 24 246 s2 1 s + A2 s 2 − − 2 − 2 2 − ... + ... 2 24 246 Equating the coefficients of both sides of Equation 6.142 results in the following: A0 = 1, 2 A0 = −2 (6.143) A0 A ξ2 − 1= , 2 •4 2 2 1 1 A1 = ( − ξ 2 ) 2 2 (6.144) − − 2 A0 A A ξ4 − 2 1 − 2 = 2 2, 2 2 2 •4 •6 2 •4 2 2 •4 1 2 1 ( 2 − ξ ) ξ4 A2 = 2 − − 224 22 4 4 6 (6.145) 231 Partial Differential Equations Therefore, the ratio can now be explicitly expressed in a form that exposes the singularities (Poles): s 1 1 1 ξ4 1 w −2 + − ξ 2 + s 2 2 − 2 − ξ 2 − 2 2 + 2 2 4 6 2 4 2 2 4 wJ 0 (i s ξ) P (s) = = −s2 si sJ1 (i s ) Q(s) The Residue theorem can now be applied to find the inverse Laplace transform. Recalling that the Inverse Laplace transform of a function F(s) can be written as Equation 3.97, ∞ f ( t ) = L−1 [ F ( s )] = ∑ ρn ( t ) 1 where Equation 3.98 defines the residues ρn ( t ) = P ( s n ) sn t e Q ' ( sn ) and Q ' ( sn ) is the value of dQ evaluated at the singular point of interest. Recall that ds P ( sn ) P (s) P (s) = lim ( s − sn ) = lim Q ' ( s n ) s → sn Q ( s ) − Q ( s n ) s → sn Q (s) s − sn When sn is a multiple pole of order m of F(s), then t2 β t m −1 ρn ( t ) = e snt β1 + β 2t + β3 + ... + m 2! ( m − 1)! (3.99) where 1 d m −i ( s − s n ) m F ( s ) s→ sn ( m − i )! ds m − i βi = lim Applying Equation 3.99 at the singularities (m = 2) , when s = 0 , ρ0 (t) = e 0 t β1 + t β2 β1 = lim s −> 0 d s 2 wJ 0 (i s ξ) = ds si sJ1 (i s ) 1 1 2 ξ4 s 1 2 2 1 w −2 + − ξ + s 2 − 2 − ξ − 2 2 + ... 2 2 46 24 2 24 d 2 = w 1 − ξ2 lim s 2 s −> 0 ds 22 s 232 Applied Mathematical Methods for Chemical Engineers and s 2 wJ 0 (i s ξ) β 2 = lim = s −> 0 si sJ (i s ) 1 s1 1 1 2 ξ4 2 2 1 w −2 + − ξ + s 2 − 2 − ξ − 2 2 + ... 2 2 4 6 2 4 2 2 4 = − 2w lim s 2 s −> 0 s2 Such that the Laplace inverse at s = 0 is ρ0 ( τ ) = e 0 τ (β1 + τβ 2 ) = w 1 2 − ξ − 2 wτ 22 For the case s ≠ 0, ( ) = P ( ξ, s ) ( ) Q (s) w J0 i s ξ s i sJ1 i s with ( ) Q ( s ) = s i sJ1 i s = 0 resulting in ( ) J1 i s = 0. If we let i s = λ ⇒ s = −λ 2 then, J1 ( λ n ) = 0, n = 1,2,3, defines the eigenvalues. Also, d s i sJ1 i s ds 3 1 d = i sJ1 i s + is 3/2 s −1/2 J1 i s 2 2 ds And using a set of tables [7], the derivative of the Bessel function J1 ( x ) can be simplified: ( ) Q ( s ) = s i sJ1 i s ⇒ Q′ ( s ) = ( ) ( ) J1′( x ) = 1 [ J0 ( x ) − J2 ( x )] 2 ( ) 233 Partial Differential Equations J2 ( x ) = 2 J1 ( x ) − J 0 ( x ) ; such that J1′( x ) = J 0 ( x ) − 1 J1 ( x ) x x Therefore, the inverse ( ) ( ) ( ) ( ) ∞ wJ λ ξ w J0 i s ξ w 1 n 0 2 e −λn τ L−1 = − ξ 2 − 2 wτ + 2 λ 2 2 si sJ i s n n =1 1 J0 λ n ∑ 2 and the final solution is given in Example 6.15. Example 6.15 This example is another demonstration of inverting the Laplace Transform when Bessel Functions are involved. Given ( ( s2 ) P ( x, s) ≡ J0 i a s Q ( s ) J0 i x s L−1 ( ) ( ) where P ( x , s ) = J 0 i x s and Q ( s ) = s 2 J 0 i a s ( ) Q′ = 2 s J0 i a s + s 2 ) a d J0 i a s = 2 s J0 i a s + s 2 −i J0 i a s ds 2 s ( ) ( ) ( ( ) a J0 i a s 2 s = 2 s J 0 i a s − s 2i ( Setting Q ( s ) = 0 in order to determine the poles ( ) ( ) 0 = s 2 J 0 i a s ⇒ s = 0,0 or J 0 i a s = 0 For the case s = 0 the residues can be determined by Equation 3.99: m t m −1 Am t i −1 Ai t2 S t ρn ( t ) = e Sn t A1 + tA2 + A3 + + = e n ∑ Ai 2! ( m − 1)! ( i − 1)! i =1 where 1 d m −i [(s − sn )m F (s)] s→ sn (m − i )! ds m − i Ai = lim ) ) 234 Applied Mathematical Methods for Chemical Engineers In the present case, m = 2: 1 d 2−1 [(s − sn )2 F (s)] s→ s 0 (2 − 1)! ds 2−1 A1 = lim J0 (i x s ) 1 d 2−1 (s − 0)2 2 2 − 1 s→ s 0 (2 − 1)! ds s J0 (i a s ) = lim To further evaluate this limit, it helps to know that J0 ( x ) = 1− x2 x4 x6 + − + 22 22 i 4 2 22 i 4 2 i 62 J1 ( x ) = x5 x7 x x3 − 2 + 2 2 − 2 2 2 + 2 2 i4 2 i4 i6 2 i4 i6 i 8 J0′ ( x ) ≡ d J ( x ) = − J1 ( x ) dx 0 Therefore d 2 J0 ( i x s ) (s ) 2 s→s 0 ds s J0 ( i a s ) A1 = lim −J i a s i x 0 2 s = lim s→s 0 ( ) ( ) 3 i x s3 2 i x s − + J i x s ia + 0 2 22 i 4 2 s 2 J0 i a s ( ( ) ) 3 i a s3 2 i a s − + 2 22 i 4 ( ) 2 2 ix ia −1 ⋅ + 1 ⋅ 2 2 x 2 a2 = ⋅ 2 = 4 − 4 (1) and J0 (i x s ) 1 d1−1 =1 [(s − 0)2 F (s)] = lim (s − 0)2 2 1 1 − s→ 0 (1 − 1)! ds s→ 0 s J0 (i a s ) A2 = lim Therefore ρ0 (t ) = x 2 − a2 +t 4 235 Partial Differential Equations −λ 2n , n = 1,2,3, a2 Remark: J0 ( x ) = 0 means that each value of x is a zero (or root) of the Bessel function. For the case s ≠ 0, J0 (i a s ) = 0, letting λ n = i a sn ⇒ sn = Then Q ′ = − s 2i ( ) J0 i x s 2 s J0 i a s L−1 ( ) a 2 s J1 (i a s ) = −λ n −λ n2 J (λ ) 2 a 2 1 n xλ 2 ∞ ∞ J0 a n − λa2n t P ( x, s) 1 2 −1 2 2 e ≡ L = ρ0 + ∑ ρn ( t ) = ( x − a ) + t + 2a ∑ 3 4 Q (s) n =1 λ n J1 ( λ n ) n =1 Example 6.16 In this example on the use of Laplace transforms to solve the heat equation, we will use tabulated inversions as well as the residue theorem to invert the transforms. Consider the problem given by ∂2 y ∂ y = , 0 < x < 1, t > 0 ∂ x 2 ∂t y ( o, t ) = o, y (1, t ) = u0 , t > 0 y ( x , 0 ) = 0, 0 < x < 1 Applying the Laplace Transform results in d 2U − sU = 0 dx 2 (6.146) subject to the transformed boundary conditions U ( 0, s ) = 0 and U (1, s ) = u0 ; where L { y ( x , t )} = U ( x , s ) s The general solution to Equation 6.146 can be represented as U ( x , s ) = m1 cosh ( ) sx + m2 sinh ( sx ) Then following application of the boundary conditions leads to U ( x , s ) = u0 ) s sinh ( s ) sinh ( sx (6.147) 236 Applied Mathematical Methods for Chemical Engineers Equation 6.147 can then be restated as u0 ) =u s sinh ( s ) sinh ( sx 0 e sx − e − sx e( x −1) s − e −( x +1) = u0 s − s −2 s ) s ( e − e ) s (1 − e s However, 1 1 − e −2 s = ∞ ∑ e−2n s n=0 such that sinh ( sx ) ∞ e −(2 n+1− x ) = s sinh s n∑ s =0 s − e −(2 n+1+ x ) s s Therefore, one can locate the error functions in a table for each term: ( ) ( ) ∞ sinh sx e −( 2 n +1− x ) s −1 e −( 2 n +1+ x ) s y ( x , t ) = u0 L−1 = u0 ∑ L−1 − L s s n=0 s sinh s (6.148) ∞ 2n + 1 − x 2n + 1 + x = u0 ∑ erfc − erfc 2 t 2 t n=0 Alternatively, we could start with Equation 6.147 and apply the residue theorem: P ( x , s ) = sinh sx , Q ( s ) = s sinh s ; Q′ ( s ) = sinh s + s cosh s 2 Then the poles of Q ( s ) are located at s = 0 or sinh s = 0 The residues at those poles are s −1/2 x 2 cosh sx sinh sx ρ0 = lim (s − 0) = 1lim =x s→ 0 s sinh s s→0 s −1/2 cosh s 2 and for s ≠ 0, sinh s = 0 ⇒ s = i λ sin λ = 0 ⇒ λ = nπ , n = 1, 2, Therefore, 237 Partial Differential Equations sinh y ( x ,t ) = u0 L−1 ( s sinh sx ) ∞ ∑ = xu0 + 2u0 s n =1 ( ) sin ( nπx ) e − n2π2t n (6.149) It now remains to show that Equations 6.148 and 6.149 are equal, an exercise that is left to the reader. Example 6.17 This is a Bessel-related example in which the boundary conditions are somewhat more complicated. Given that a uniform solid sphere of radius 1 at an initial constant temperature u0 throughout is dropped into a large container of fluid that is kept at a constant temperature u1 ( u1 > u0 ) for all time. Since there will be heat transfer across the boundary r =1, the temperature profile u ( r , t ) in the sphere can be described by the model ∂ u ∂ 2 u 2 ∂u = + , ∂ t ∂ r 2 r ∂r 0 < r < 1, t > 0 (6.150) ∂u = − h u(1, t ) − u1 ,0 < h < 1 ∂r r =1 (6.151) where u1 is a constant fluid temperature u(r , 0) = u0 u(r , t ) 0 < r <1 is bounded as r → 0 (6.152) (6.153) Using Laplace Transform Methods, ∞ y (r , s) = L {u(r , t )} = ∫ e − st u(r , t )dt 0 The partial differential equation transforms to d 2 y 2 dy + dr 2 r dr (6.154) dy u = − h y(1, s) − 1 dr s (6.155) s y − u0 = subject to at r = 1: and y(r , s) is bounded as r → 0 (6.156) Then Equation 6.154 can be restated as d 2 y 2 dy + − sy = −u0 dr 2 r dr (6.157) 238 Applied Mathematical Methods for Chemical Engineers The associated homogeneous differential equation can be solved by comparing with Equation 3.80: d 2 y 2 dy + − sy = 0 dr 2 r dr 2a − 1 = −2 ⇒ a = − 1 2 2c − 2 = 0 ⇒ c = 1 a 2 − γ 2c 2 = 0 ⇒ γ 2 = 1 4 , γ = ± 1 2 b 2c 2 = − s ∴ b = ± i s 1 1 ∴ yc (r , s) = m, r − 2 J 1 (i s r ) + m2 r − 2 J 1 (i s r ) 2 2 k k = 1 cosh s r + 2 sinh s r r r assuming y p = k3 , y′p = y′′p = 0 . Substitution into Equation 6.154 gives uc = k3 s k u k ∴ y(r , s) = 1 cosh s r + 2 sinh s r + c r r s − sy p = − u0 ⇒ y p = (6.158) Using Equation 6.157, y(r , s) is bounded as r → 0, which implies that k1 must be ­chosen as zero, since cosh s r → ∞ as r → 0, such that u k2 sinh s r + 0 r s s cosh( sr ) 1 y′ = k 2 − 2 sinh( sr ) r r y(r , s) = at r = 1, u dy = − h y(1, s) − 1 s dr u u k2 s cosh s − sinh s = − h k2 sinh s + 0 − 1 s s h = − h k2 sinh s − (u0 − u)1 s or h k2 { s cosh s − sinh s + h sinh s } = − (u0 − u1 ) s or h − (u0 − u1 ) s k2 = s cosh s + (h − 1)sinh s h − (u0 − u1 )sinh( s r ) 1 u s ∴ y(r , s) = ⋅ + 0 s cosh s + (h − 1)sinh s r s 239 Partial Differential Equations Then u(r , t ) = L−1 { y(r , s)} = L−1 = u0 + { } (u1 − u0 ) h sinh( s r ) u0 1 + L−1 s r s( s cosh s + (h − 1)sinh s ) ∞ sinh s r h(u1 − u0 ) −1 P ( s n , r ) sn t L e ⇔∑ r s( s cosh s + (h − 1)sinh s ) n = 0 Q′(sn ) Letting P (sn , r ) = sinh s r and Q(s) = s( s cosh s + (h − 1)sinh s ) s( s cosh s + (h − 1)sinh s = 0 ⇒ s = 0 or s cosh s + (h − 1)sinh s = 0 for s ≠ 0 ⇒ s + (h − 1) tanh s = 0 s − s ,h > 1 = h −1 1− h i sin λ = iλ; s = −λ 2n , n = 1,2,.... s = iλ, then tanh(iλ) = iλ or cos λ tanh s = If −λ n h −1 in the case s = 0, tan λ n = s sinh s r 0 = s(cosh s + (h − 1)sinh s ) 0 r 12 s cosh s r r r 2 = lim = = 1 1 1 1 1 1 s→ 0 s 2 cosh s + s ⋅ s 2 sinh s + (h − 1) s 2 cosh s 1 + h − 1 h 2 2 2 ρ0 = lim s→ 0 for s ≠ 0, 1 1 Q′(s) = s cosh s + (h − 1)sinh s + s s 2 cosh s 2 1 −1 1 −1 1 + s 2 ⋅ s 2 sinh s + (h − 1) s 2 cosh s 2 2 s s 1 = cosh s + sinh s + (h − 1) s cosh s 2 2 2 1 = h s cosh s + s sinh s 2 i 1 = h iλ n cos λ n − λ 2ni sin λ n = (h λ n cos λ n − λ 2n sin λ n ) 2 2 240 Applied Mathematical Methods for Chemical Engineers Now h sinh s r r ∞ P ( s n , r ) e sn t h (u1 − u0 ) L−1 = (u1 − u0 ) + ∑ r h n = 1 Q ′ ( sn ) s s cosh s + (h − 1)sinh s r 2 i sin (λ n r ) e − λ n t h r ∞ = (u1 − u0 ) + ∑ i r h 2 = n 1 (hλ n cos λ n − λ n sin λ n ) 2 However, iλ n cos λ n + i(h − 1)sin λ n = 0 ⇒ λ n cos λ n = (1 − h)sin λ n = − (h − 1)sin λ n ∴ hλ n cos λ n = − h(h − 1)sin λ n resulting in ∞ r e − λ n t sin (λ n r ) h (u1 − u0 ) + 2∑ 2 r n =1 [ − h(h − 1) − λ n ] sin λ n h 2 ∞ r sin (λ n r )e − λ n t h (u1 − u0 ) + 2∑ 2 (1 ) sin − − λ λ r h h h n] n n =1 [ 2 Finally, 2 ∞ r h sin(λ nr )e − λ n t u(r , t ) = u0 + (u1 − u0 ) + 2∑ 2 r n [ h(1 − h) − λ n ] sin λ n h ∞ sin(λ nr )e − λ n t h = u0 + u1 − u0 + 2 (u1 − u0 )∑ 2 r n [ h(1 − h) − λ n ] sin λ n 2 ∞ sin(λ nr )e − λ n t h = u1 + 2 (u1 − u0 )∑ 2 r n [ h(1 − h) − λ n ] sin λ n 2 6.5 COMBINATION OF VARIABLES Another technique that is sometimes employed to reduce partial differential equations to ordinary differential equations is combination of variables or a similarity transformation. The process of normalization can be used to establish the applicability of combining the independent variables of the given PDE. Example 6.18 This example considers the flow of a fluid near a wall suddenly set in motion. Following [1], the problem statement is as follows: A semi-infinite body of liquid with constant density (ρ) and viscosity (μ) is bounded on one side by a flat surface (the xz-plane). Initially, the fluid and the solid surface are at 241 Partial Differential Equations rest, but at time t = 0, the solid surface is set in motion in the positive x-direction with a velocity V (shown below). y t < 0, fluid at rest V t = 0, wall, set in motion Vx( y, t)t > 0, fluid in unsteady flow One would like to know the velocity profile as a function of y and t. If there is no pressure gradient or gravity force in the x-direction and the flow is laminar, then this simplified problem can be solved in the following way. Solution In rectangular coordinates (x, y, z) ∂ρ ∂ ∂ ∂ + (ρv x ) + (ρv y ) + (ρvz ) = 0 ∂t ∂ x ∂y ∂z (6.159) is the equation of continuity [1]. The x-component equation of motion in terms of velocity gradients for a Newtonian fluid with constant ρ and μ is [1] ∂v ∂2 v ∂v ∂v ∂v ∂P ∂2 vx ∂2 vx + ρgx ρ x + v x x + v y x + vz x = − + µ 2x + + ∂t ∂x ∂x ∂y ∂z ∂x ∂ y2 ∂ z 2 However, v y = vz = 0 and v x = v x ( y, t ) such that Equation 6.159 reduces to ∂vx =0 ∂x (6.160) 242 Applied Mathematical Methods for Chemical Engineers and Equation 6.160 reduces to ∂vx ∂2 vx =γ ∂t ∂ y2 (6.161) where γ = μ/ρ. The initial and boundary conditions are at t ≤ 0, v x = 0 for all y (6.162) at y = 0, v x = V for all t > 0 (6.163) at y = ∞, v x = 0 for all t > 0 (6.164) vx = φ( η) V (6.165) y 4 γt (6.166) Now suppose where η= such that ∂( v x / V ) 1η =− φ′; 2t ∂t ∂ 2 ( v x / V ) η2 = 2 φ′′ ∂ y2 y then Equation 6.161 becomes ϕ′′ + 2 ηφ′ = 0 (6.167) φ = 1 at η = 0 (6.168) φ = 0 at η = ∞ (6.169) subject to Note that the initial and boundary conditions are combined to form Equation 6.169. Equation 6.167 solves to give the general solution η φ( η) = c1 ∫ e − η d η + c2 2 0 where the lower limit of the integral is a convenient selection. However, notice that an alternate choice of lower limit would only affect c 2, but c 2 is an arbitrary constant. 243 Partial Differential Equations With the aid of Equations 6.168 and 6.169, the solution to Equation 6.167 is φ( η) = 1 − 2 π η ∫0 e− η 2 y d η = 1 − erf 4 γt (6.170) where the error function erf( x ) = 2 π x ∫0 e− ξ 2 dξ was defined in the previous section. The combination of variables or similarity transformation approach can be a very useful technique when one is confronted with transport problems where one is interested in what happens at relatively short times or the system is very large and one boundary either does not exist or is indistinct. Further discussion of these and other types of problems that employs this technique can be found in [17]. Example 6.19 This example considers a semi-infinite solid, which is subjected to a step change in its surface temperature T [11]. Suppose the model can be described by 1 ∂T ∂ 2 T = α ∂t ∂ x 2 (6.171) T (0, t ) = Tc (6.172) lim T ( x , t ) = Ti (6.173) with and x →∞ as boundary conditions, and the initial condition given by T ( x ,0) = Ti (6.174) Further, suppose that the normalized variables are T= T − Ti x , x= T0 x0 and t = t t0 (6.175) Substitution into Equation 6.171 gives x 02 ∂T ∂ 2 T = αt0 ∂ t ∂ x 2 (6.176) 244 Applied Mathematical Methods for Chemical Engineers Equation 6.172 can be restated as T (0, t ) = Tc − Ti T0 then the choice of T0 = Tc − Ti gives T (0, t ) = 1 (6.177) Equation 6.173 and Equation 6.174 become lim T ( x , t ) = 0 (6.178) T ( x ,0) = 0, (6.179) x →∞ and respectively. Equation 6.176 can be further simplified if the quantity x 02 αt0 is taken as 1, since neither x0 nor t0 has been assigned any physical significance (see Chapter 8 on scaling). That is, if x 02 =1 αt0 then Equation 6.176 becomes ∂T ∂ 2 T = ∂t ∂x 2 (6.180) and a possible choice of a combined variable is ξ= x 2 ( x / x 0 )2 x 2 t0 x 2 = = 2 = t t / t0 t x 0 αt (6.181) Using Equation 6.181, the conditions described by Equation 6.178 and 6.179 can now be collapsed into one condition similar to Equation 6.169, namely, T ( ∞) = 0 (6.182) since ξ → ∞ as x → ∞ or t → 0. Therefore, Equation 6.171 or 6.180 can be reduced to a second-order ordinary differential equation following substitution of Equation 6.181. Given that a combined variable is available, its substitution into the PDE can be a delicate process and is demonstrated below. 245 Partial Differential Equations Example 6.20 This example considers the model described by Equation 6.171, subject to T = T0 at t = 0 for all x (6.183) T = Ts at x = 0 for all t (6.184) T → Ts , t → ∞, x > 0 (6.185) T → T0 , x → ∞, t > 0 (6.186) and further suppose that the combined variable is η= x 4αt (6.187) Then T ( x , t ) = f ( η) (6.188) dT ( x , t ) = df ( η) (6.189) which implies that The chain rule applied to Equation 6.187 gives ∂T ∂T df dx + dt = dη ∂x ∂t dη (6.190) also, the total derivative of η(x, t) gives dη = ∂η ∂η dx + dt ∂x ∂t (6.191) such that Equation 6.190 becomes ∂T ∂T ∂η ∂η dx + dt = f ′( η) dx + dt ∂x ∂t x ∂ ∂t Then equating like coefficients of both sides gives dx : ∂T ∂η = f ′( η) ∂x ∂x (6.192) dt : ∂T ∂η = f ′( η) ∂t ∂t (6.193) 246 Applied Mathematical Methods for Chemical Engineers To find the second derivative, the following device is useful: let H (x,t) = ∂T ∂x and φ( η, t ) = f ′( η) (6.194) ∂η ∂x (6.195) then dH ( x , t ) = dφ(ρ, t ) (6.196) such that ∂H ∂H ∂φ ∂φ dη + dt dx + dt = ∂x ∂t ∂η ∂t that is, ∂H ∂H ∂φ ∂η ∂η ∂φ dx + dt = dx + dt + dt ∂x ∂t ∂η ∂ x ∂t ∂t and equating like coefficients of both sides results in dx : ∂ H ∂φ ∂η = ∂ x ∂η ∂ x and dt : ∂ H ∂φ ∂η ∂φ = + ∂t ∂η ∂t ∂t Therefore ∂ 2 T ∂φ ∂η ∂η = = f ′′( η) ∂x ∂ x 2 ∂η ∂ x 2 since ∂η is independent of x. Finally, Equation 6.171 becomes ∂x f ′( η) ∂η ∂η = αf ′′( η) ∂x ∂t 2 or f ′′( η) + 2 ηf ′( η) = 0 using Equation 6.187. (6.197) 247 Partial Differential Equations 6.6 FOURIER INTEGRAL METHODS In chemical engineering, the use of Fourier integrals to solve problems is not as popular as separation of variables or Laplace transform. This is due to the fact that the incorporation of the boundary conditions associated with a particular application can usually be very challenging. For instance, consider a slab of finite thickness undergoing some heat transfer phenomena. Suppose the phenomena can be described by V ∂T ∂2 T ∂2 T = + ∂ x ∂ x 2 ∂ y2 ∂T | y = ±1 = ± qe x / δ , x < 0 ∂x ∂T | y = ±1 = ± qe − x / ∆ ∂y (6.198) (6.199) T (−∞, y) = 0 (6.200) T (0, ± 1) = 1 (6.201) and Then, using Fourier transforms, Equation 6.198 becomes ∂2 θ − ω 2 θ − iωVθ = 0 ∂ y2 (6.202) where ∞ θ(ω , y) = ∫ T ( x , y)e − iωx d x −∞ is defined by Equation 5.62. The general solution of Equation 6.202 includes two arbitrary constants, one of which can be eliminated based on the fact that the temperature field is symmetric. That is, the solution to Equation 6.202 reduces to θ(ω , y) = c1 cosh (ω 2 + iωV ) y Transforming the conditions given by Equation 6.199 results in 0 ∞ ∂θ x | y =1 = ∫ q exp − iωx d x + ∫ q exp(−( x / ∆) − iωx ) d x , −∞ 0 δ ∂y which integrates to ∂θ 1 1 | y =1 = q + ∂y (1 / δ) − iω iω + (1 / ∆) (6.203) 248 Applied Mathematical Methods for Chemical Engineers Therefore, Equation 6.203 becomes 1 1 1 θ(ω , y) = q + (ω 2 + iωV ) sinh (ω 2 + iωV ) 1 / δ − i ω i ω + (1 / ∆ ) × cosh (ω 2 + iωV ) y Application of Equation 5.63 gives the inverse transform T ( x , y) = cosh (ω 2 + iωV ) y exp(iωx ) dω q ∞ ∫ 2π −∞ [(1 / δ) − iω ] (ω 2 + iωV ) sinh (ω 2 + iωV ) + cosh (ω 2 + iωV ) y exp(iωx )dω q ∞ 2π ∫−∞ [iω + (1 / ∆)] (ω 2 + iωV ) sinh (ω 2 + iωV ) (6.204) There are some cases of practical value for which Fourier integrals can be helpful. Example 6.21 This example considers a semi-infinite thin slab whose surface is insulated. Suppose that the surface temperature of the bar is initially f(x), and a temperature of zero degrees is suddenly applied to the end x = 0 and is maintained. 1. Show that the solution to this boundary value problem can be represented as u( x , t ) ∞ ∞ 1 2 = f ( v )e − kλ t sin λv sin λx d λ d v π ∫0 ∫0 2.If f(x) is a constant, say u 0, show that x u( x , t ) = u0 erf kt 2 Solution The boundary value problem can be described by ∂u ∂2 u = k 2 x > 0, t > 0 ∂t ∂x u( x ,0) = f ( x ), u(0, t ) = 0, | u( x , t ) | < M (due to physical reasons) By separation of variables, the differential equation has a solution of the form u( x , t ) = e − kλ t ( A cos λx + B sin λx ) 2 249 Partial Differential Equations Using the condition at x = 0 reduces the solution to u( x , t ) = e − kλ t ( B sin λx ) 2 Since there are no restrictions on λ, B can be replaced by a function B(λ). Further, integration over λ from 0 to ∞ can be carried out analogous to the superposition principle that was applied to discrete values of λ. Therefore, a possible solution is ∞ u( x , t ) = ∫ B(λ )e − kλ t sin λx d λ 2 0 Application of the condition at t = 0 results in ∞ u( x ,0) = f ( x ) = ∫ B(λ )sin λx d λ 0 which is an integral equation for the determination of B(λ). This integral equation can be interpreted with the aid of Equation 5.64, suggesting that f(x) is an odd function. In which case, B( λ ) = 2 ∞ 2 ∞ f ( x ) sin λx d x = ∫ f ( v ) sin λv d v π ∫0 π 0 Therefore ∞ u( x , t ) = ∫ B(λ )e − kλ t sin λx d λ = 2 0 2 ∞ ∞ 2 f ( v )e − kλ t sin λv sin λx d λ d v π ∫0 ∫0 Since 1 sin λv sin λx = [cos λ ( v − x ) − cos λ ( v + x )] 2 then 1 ∞ ∞ 2 f ( v )e − kλ t [cos λ( v − x ) − cos λ( v + x )] d λ d v π ∫0 ∫0 ∞ ∞ 1 ∞ 2 2 = ∫ f ( v ) ∫ e − kλ t cos λ ( v − x ) d λ − ∫ e − kλ t cos λ ( v + x ) d λ dv 0 0 π 0 u( x , t ) = But ∞ ∫0 e−αλ 2 cos βλ d λ = 1 π −β2 /4 α e 2 α Therefore u( x , t ) = ∞ 1 ∞ 2/4 2/4 f ( v )e − ( v − x ) kt d v − ∫ f ( v )e − ( v + x ) kt 0 2 πkt ∫0 250 Applied Mathematical Methods for Chemical Engineers Letting ( v − x )/2 k t = p in the first integral and ( v + x )/2 k t = p in the second integral, u(x, t) reduces to u( x , t ) = 1 ∞ π ∫− x /2 f (2 p k t + x )e − p d p − ∫ 2 kt ∞ − x /2 kt 2 f (2 p k t − x )e − p d p but f(x) is a constant and can be taken outside of the integrals, thus giving ∞ u0 ∞ 2 2 e− p d p − ∫ e − p d p ∫ − x kt x /2 /2 kt π u0 x /2 kt − p2 2u0 x /2 kt − p2 = e dp = e d p = u0 erf( x / 2 kt ) π ∫− x /2 kt π ∫0 u( x , t ) = From the very brief introduction of Fourier integrals in Chapter 5 together with the above examples, the engineer should get the sense that these integrals are more useful for general problems than specific practical problems. When the various transforms are to be employed, one has to be careful in making sure that the boundary conditions can be utilized a priori. 6.7 REGULAR PERTURBATION APPROACHES Sometimes a problem is encountered in which the governing equation is almost identical to a simpler problem that one already knows how to solve. Hopefully, some modification of the new problem can resolve the difference and lead to useful results. Example 6.22 This example considers a model for a species A in a reacting system [12]: 1 ∂ r ∂CA ∂ ∂CA + =0 r ∂r ∂r ∂ z ∂ z (6.205) where r and z are the radial and axial coordinates, respectively, as shown in Figure 6.1. Suppose Equation 6.205 is subject to the following boundary conditions: ∂CA = 0 at z = 0 ∂z (6.206) ∂CA = 0 at r = 0 ∂r (6.207) CA ( RW, z ) = CABi ; 0 < z < δ (6.208) − DAB ∂ kC C CA = 1 A tot ∂z K + K ′CA at z = δ (6.209) 251 Partial Differential Equations 2d r Gas z L FIGURE 6.1 CVD Reactor. Then, except for Equation 6.209, the method of separation of variables is applicable. Observe that the terms of the denominator of the right-hand side of Equation 6.209 are all positive physical quantities. That is, the right-hand side of Equation 6.209 is expected to have a power series expansion. First, the variables are redefined in terms of dimensionless quantities by CA (r , z ) = F ( ξ , ζ) CABi where ξ= r z δ ; ζ= ; a= Rw δ Rw Then Equations 6.205 through 6.209 become −1 ∂ 2 F ∂ 2 F 1 ∂ F = + a 2 ∂ζ 2 ∂ξ 2 ξ ∂ξ (6.210) ∂ F (ξ,0) =0 ∂ζ (6.211) ∂ F (0, ζ) =0 ∂ξ (6.212) F (1, ζ) = 1 (6.213) − DABCABi ∂ F (ξ,1) = Reaction rate (heterogeneous) δ ∂ζ (6.214) Consider the heterogeneous reaction rate expression to be rate = k 0CABi (1 − εF + ε 2 F 2 − ε 3 F 3 + ) (6.215) F = F0 + εF1 + ε 2 F 2 + (6.216) where and ε= K ′CA0 <1 K (6.217) 252 Applied Mathematical Methods for Chemical Engineers The otherwise nonlinear problem can be reduced to a set of linear problems by substituting Equations 6.215 and 6.216 into Equations 6.210 through 6.214 to get ∂ 2 F0 ε ∂ 2 F1 −1 ∂ 2 F0 ∂2 F ∂ 2 F2 ∂ 2 F2 + ε 21 + ε 2 + = + + ε2 2 2 2 2 2 a ∂ζ ∂ζ ∂ζ ∂ξ ∂ξ 2 ∂ξ ∂F ∂F 1 ∂F + + 0 + ε 1 + ε 2 2 + ξ ∂ξ ∂ξ ∂ξ (6.218) ∂ F0 ∂F ∂F + ε 1 + ε 2 2 + = 0 at ζ = 0 ∂ζ ∂ζ ∂ζ (6.219) ∂ F0 ε ∂ F1 ∂F + + ε 2 2 + = 0 at ξ = 0 ∂ξ ∂ξ ∂ξ (6.220) F0 + εF1 + ε 2 F2 + = 1 at ξ = 1 (6.221) − DABCABi ∂ F0 ∂ F1 ∂ F2 2 ∂ζ + ε ∂ζ + ε ∂ζ + δ = k 0CABi [1 − ε( F0 + εF1 + ε 2 F2 + ) + ε 2 ( F0 + εF1 + ε 2 F2 + )2 + O(ε 3 ) ] (6.222) Then equate like powers of ε in both the differential equation and the given conditions as follows: ε0 : − 1 ∂ 2 F0 ∂ 2 F0 1 ∂ F0 = + a 2 ∂ζ 2 ∂ξ 2 ξ ∂ξ ∂ F0 (ξ,0) =0 ∂ζ ∂ F0 (0, ζ) =0 ∂ξ F0 (1, ζ) = 1 − ε: − DABCABi ∂ F0 (ξ,1) = k 0CABi δ ∂ζ 1 ∂ 2 F1 ∂ 2 F1 1 ∂ F1 = + a 2 ∂ζ 2 ∂ξ 2 ξ ∂ξ ∂ F1 (ξ,0) =0 ∂ζ ∂ F1 (0, ζ) =0 ∂ξ F1 (1, ζ) = 0 − DABCABi ∂ F1 (ξ,1) = k 0CABi F0 δ ∂ζ 253 Partial Differential Equations ε2 : − 1 ∂ 2 F2 ∂ 2 F2 1 ∂ F2 = + a 2 ∂ζ 2 ∂ξ 2 ξ ∂ξ ∂ F2 (ξ,0) =0 ∂ζ ∂ F2 (0, ζ) =0 ∂ξ F2 (1, ζ) = 0 − DABCABi ∂ F2 (ξ,1) = k 0CABi ( F02 + F1 ) δ ∂ζ Each set of linear problems (respective coefficient of ε) can now be solved by the method of separation of variables, and the above process can be continued up to any desired power of ε. Example 6.23 This example considers the cooling of a given system [13] in which the specific heat is not a constant. Further, suppose that the system has volume V, surface area A, density ρ, initial temperature Ti, and specific heat C. At an initial time, the system is exposed to a convective environment with heat transfer coefficient h and a temperature Ta. If the specific heat C is given by C = Ca [1 + β(T − Ta )] (6.223) where Ca is the specific heat at T = Ta and β is a given constant, what is the temperature profile for this cooling system? Solution The energy balance for this system results in pVC dT + hA(T − Ta ) = 0 dt (6.224) t = 0, T = T1 (6.225) subject to Then, for θ= T − T1 t , τ= , ε = β(T1 − Ta ) T1 − Ta ρVCa / (hA) Equations 6.224 and 6.225 become (1 + εθ) dθ +θ= 0 dτ (6.226) 254 Applied Mathematical Methods for Chemical Engineers τ = 0, θ = 1 (6.227) If ε is much smaller than 1, then the following infinite series solution can be adopted: ∞ θ = θ0 + εθ1 + ε 2θ2 + = ∑ ε nθn (6.228) n=0 Substitution into Equations 6.226 and 6.227 results in [1 + ε(θ0 + εθ1 + ε 2θ2 + )] dθ 0 dθ dθ + ε 1 + ε 2 2 + dτ dτ dτ + θ0 + εθ1 + ε 2θ2 + = 0 or dθ0 dθ dθ dθ dθ dθ + θ0 + ε 1 + θ1 + θ0 0 + ε 2 2 + θ2 + θ0 1 + θ1 0 + = 0 dτ dτ dτ dτ dτ dτ and θ0 + εθ1 + ε 2θ2 + = 1 at τ = 0 Then equating coefficients of like powers of ε gives dθ0 + θ0 = 0 dτ θ0 = 1 at τ = 0 ε0 : ε: dθ1 dθ + θ1 + θ0 0 = 0 dτ dτ θ1 = 0 at τ = 0 and41 dθ2 dθ dθ + θ2 + θ0 1 + θ1 0 = 0 dτ dτ dτ θ2 = 0 at τ = 0 ε2 : for coefficients of ε up to the second power. The resulting solutions for the respective set of new problems are 3 θ0 = e − τ ; θ1 = e − τ − e −2 τ ; θ2 = e − τ − 2e −2 τ + e −3τ 2 such that a three-term solution is 3 θ0 = e − τ + ε(e − τ − e −2 τ ) + ε 2 e − τ − 2e −2 τ + e −3τ 2 (6.229) 255 Partial Differential Equations Notice that the direct solution of Equations 6.226 and 6.227 results in ln θ + ε (θ − 1) = −τ, which compares well with the three-term result [13] for ε = 0, 0.2, and −0.2. Example 6.24 This example considers plane Couette flow with variable viscosity [13, 14]. Consider the steady flow of an incompressible Newtonian fluid between two infinite parallel plates separated by a distance, a, as shown in Figure 6.2. Each plate is maintained at a temperature T0. The upper plate is allowed to move with a uniform velocity V. The thermal conductivity, k, of the fluid is constant while its viscosity, μ, is allowed to vary according to µ = µ 0 e −α (T − T0 ) (6.230) where μ0 is the viscosity at T0 and α is a given constant. The respective momentum and energy equations [1] are d du µ =0 dy dy (6.231) and 2 d 2 T µ du + =0 dy 2 k dy (6.232) The boundary conditions are y = 0, u = 0, T = T0 y = a, u = V , T = T0 where u is axial velocity and T is the fluid temperature. Introducing the dimensionless quantities θ= T − T0 µ V2 y u , Y = , U = , β = αT0 , and ε = 0 T0 a V kT0 Equations 6.231 and 6.232 together with their boundary conditions become T0 y T0 V Moving plate Stationary plate FIGURE 6.2 Plane Couette flow. a 256 Applied Mathematical Methods for Chemical Engineers d −βθ dU e =0 dY dY (6.233) 2 d 2θ dU + εe −βθ =0 dY dY 2 Y = 0, U = 0, θ = 0 (6.234) Y = 1, U = 1, θ = 0 Notice that ε is the Brinkman number, which is the ratio of viscous to conduction heating. For small ε or negligible viscous heating effects, one may assume that U = U 0 + εU1 + ε 2U 2 + (6.235) θ = θ0 + εθ1 + ε 2θ2 + (6.236) Upon substituting Equations 6.235 and 6.236 into Equations 6.233 and 6.234, the term exp[−β(θ 0 + εθ 1 + ε 2θ 2 + …)] arises and needs to be expanded. That is, if we take three terms of the series given by Equation 6.236, then exp[−β(θ0 + εθ1 + ε 2θ2 + )] = exp(−βθ0 ) exp(−εβθ1 ) exp(−ε 2βθ2 ) (βθ1 )2 = e −βθ0 1 − ε(βθ1 ) + ε 2 (1 − ε 2βθ2 ) 2 (βθ )2 = e −βθ0 1 − ε(βθ1 ) + ε 2 1 − βθ2 2 2 ( ) βθ = e −βθ0 − εe −βθ0 βθ1 + ε 2 e −βθ0 1 − βθ2 2 where only terms up to ε 2 have been retained. Similarly, the term (dU/dY)2 is expanded as 2 dU = dU 0 + ε dU1 + ε 2 dU 2 dY dY dY dY 2 2 dU dU1 dU1 2 dU dU dU1 = 0 + ε 2 0 + ε2 2 0 + dY dY dY dY dY dY where only the first three terms of Equation 6.235 are retained. Employing these expansions together with Equations 6.233 and 6.234 and equating coefficients of like powers of ε result in the sets 257 Partial Differential Equations d −βθ0 dU 0 e =0 dY dY d 2θ 0 =0 dY 2 Y = 0, U 0 = 0, θ0 = 0 ε0 : Y = 1, U 0 = 1, θ0 = 0 d dY ε: dU 0 −βθ0 dU1 − βθ1 =0 e dY dY 2 d 2θ1 dU + e −βθ0 0 = 0 dY dY 2 Y = 0, U1 = 0, θ1 = 0 Y = 1, U1 = 0, θ1 = 0 ε2 : d −βθ0 dU 2 dU (βθ )2 dU − βθ1 1 + 1 − βθ2 0 = 0 e dY dY 2 dY dY 2 dU 0 dU1 dU − βθ1 0 = 0 2 Y Y Y d d d Y = 0, U 2 = 0, θ2 = 0 d 2θ 2 + e −βθ0 dY 2 Y = 1, U 2 = 0, θ2 = 0 for terms up to ε 2. The solution sequence should be θ 0, U0, θ 1, U1, θ 2, and U2 to get the following: θ0 = 0 U0 = Y 1 θ1 = Y (1 − Y ) 2 1 U1 = − β(Y − 3Y 2 + 2Y 3 ) 12 1 θ2 = − β(Y − 2Y 2 + 2Y 3 − Y 4 ) 24 1 2 U2 = β (Y − 5Y 2 + 10Y 3 − 10Y 4 + 4Y 5 ) 120 These results for three terms may be substituted into Equations 6.235 and 6.236 for comparison to the exact solution given by [14] 258 Applied Mathematical Methods for Chemical Engineers 1/2 εβ εβ eβθ = 1 + sech 2 (2Y − 1) sinh −1 8 8 U= 1 8 1+ 2 εβ 1/2 1/2 εβ tanh (2Y − 1) sinh −1 + 1 8 The above are three examples of regular perturbation. The method is applicable to both partial and ordinary differential equations. Example 6.25 This example emphasizes the meaning of the term regular. Consider now the first-order problem y′ + εy = 0; y(0) = 1, ε < 1 Then, as discussed in Chapter 2, the solution is y = e − εx which can be expanded in a Taylor series about ε = 0, to be y = 1 − εx + ε 2 x ε3 x 3 − + 2! 3! Suppose we did not know the analytical solution, but suspected that we could try an approximation of the form ∞ y = y0 ( x ) + εy1 ( x ) + ε 2 y2 ( x ) + ∑ ε n yn ( x ) n=0 Then substitution into the differential equation results in y0′ ( x ) + εy1′ ( x ) + ε 2 y2′ ( x ) + + ε [ y0 ( x ) + εy1 ( x ) + ε 2 y2 ( x ) + ] = 0 By equating coefficients of like powers of ε, the following differential equations result: ε 0 : y0′ ( x ) = 0 ε : y1′ ( x ) + y0 ( x ) = 0 ε 2 : y2′ ( x ) + y1 ( x ) = 0 Also, the condition y(0) = 1, becomes y(0) = 1 = y0 (0) + εy1 (0) + ε 2 y2 (0) + , 259 Partial Differential Equations which, on equating coefficients of like powers of ε, gives ε 0 : y0 (0) = 1 ε : y1 (0) = 0 ε 2 : y2 (0) = 0 such that y0 ( x ) = 1 y1 ( x ) = x y2 ( x ) = x2 2 Therefore y = 1 + εx + ε 2 x 2 /2 + One can observe that this final solution is identical to the Taylor series expansion of the analytical solution. Therefore, whenever the perturbed solution has a Taylor series expansion such as demonstrated, the perturbation is considered as regular. In general, given a linear or nonlinear differential equation L (u, ε) = 0 that depends on the small positive parameter ε and with appropriate boundary or initial conditions (data) that may also depend on ε, we say that such a problem is a regular perturbation problem if the reduced problem consists of L ( v , 0) = 0 and the reduced data has a unique solution [15]. Example 6.26 This example is a further demonstration of the regular perturbation method. Consider the two-dimensional Helmholtz equation L (u, ε) = uxx + u yy + ε 2u = 0 in the unit disk x2 + y2 < 1 subject to the boundary condition u( x , y) = 1; x 2 + y 2 = 1 Assume that ∞ u ( x , y ) = ∑ un ( x , y ) ε 2 n n=0 260 Applied Mathematical Methods for Chemical Engineers is a solution. Then, by definition of a solution to a differential equation, ∞ ∞ uxx u yy + ε 2u = ∇ 2u + ε 2u = ∇ 2 ∑ un ε 2 n + ∑ un ε 2 n + 2 n=0 n=0 ∞ = ∇ 2u0 + ∑ [∇ 2un + un −1 ]ε 2 n = 0 n =1 where summation and differentiation are assumed interchangeable. The boundary condition becomes ∞ u( x , y) = u0 ( x , y) + ∑ un ( x , y)ε 2 n = 1; x 2 + y 2 = 0 n =1 Therefore, equating like powers of ε gives ε 0 : ∇ 2u = 0 u0 ( x , y) = 1 ε 2 n : ∇ 2un = − un −1; n ≥ 1 un ( x , y) = 0; x 2 + y 2 = 1; x ≥ 1 Notice that the perturbation method has replaced the Helmholtz equation with a system of Laplace and Poisson equations. By observing that polar coordinates can be used and that there is no θ dependence, that is, u = u(r), we get ∇ 2u0 = 0 = ∂ 2 u0 1 ∂ u0 + = 0; u0 (1) = 1 ∂r 2 r ∂r whose bounded solution is u 0 = 1. Also, the equation for u1 is ∂ 2 u1 1 ∂u1 + = − u0 ; u1 (1) = 0 ∂r 2 r ∂r and results in u1 (r ) = 1− r2 4 Therefore, to leading orders the reduced problem solution is u = 1 + ε2 (1 − r 2 ) + O(ε 4 ) 4 where the symbol O(ε 4) means terms of order ε 4. It can be shown (see problem 9) that the polar coordinate form of the Helmholtz equation ∂ 2 u 1 ∂u + + ε 2u = 0; u0 (1) = 1, ∂r 2 r ∂r 261 Partial Differential Equations which is bounded at the origin, solves to u= J 0 (εr ) J 0 (ε) and J0(z) has the series expansion J0 (z ) = 1 − z2 + O( z 4 ) 4 Therefore u= J 0 (εr ) 1 − (εr )2 / 4 + O(ε 4 ) = 1 − (ε)2 / 4 + O(ε 4 ) J 0 (ε) = 1 + ε 2 (1 − r 2 ) / 4 + O(ε 4 ) It is interesting to note that ε must be smaller than the first zero of the Bessel function in order for the reduced problem to yield a useful result. Example 6.27 This example considers the freezing of a saturated liquid in a semi-infinite region [13]. A saturated liquid is initially at its freezing temperature Tf. At some time, the face is maintained at a constant subfreezing temperature T0(T0 < Tf ). As heat is removed from the liquid, it freezes. If the freezing front at any time t is at xf and if it can be assumed that the uniform liquid remains at Tf throughout the process, what is the temperature profile T(x, t) in the solid phase? Solution A model for this problem is 1 ∂T ∂ 2 T = α ∂t ∂ x 2 T (0, t ) = T0 , T ( x f , t ) = Tf k ∂T dx | x = xf = ρλ f ∂x dt where α is thermal diffusivity, k and ρ are thermal conductivity and density, respectively, of the solid phase and λ is the latent heat. Introducing the dimensionless quantities, θ= Tf − T x x kt C (Tf − T0 ) , X = , Xf = f , τ = , ε= ρCxs2 λ Tf − T0 xs xs where xs is a reference distance and C is the solid phase specific heat. Further, by changing the variables from (X, τ) to (X, Xf ), the dimensionless model becomes 262 Applied Mathematical Methods for Chemical Engineers ∂2 θ ∂θ ∂θ = −ε ∂X 2 ∂Xf ∂X X = Xf θ( X = 0, X f ) = 1, θ( X = X f , X f ) = 0 dX f ∂θ = −ε dτ ∂X X = Xf Here, the quantity ε is the Stefan number, which represents the ratio of sensible to latent heat during a phase change. For a process with small ε or a comparatively large latent heat, such as in this case, let θ = θ0 + εθ1 + ε 2θ2 + Then following substitution into the dimensionless model and equating ­coefficients of like powers of ε results in ∂2 θ0 =0 ∂X 2 θ0 ( X = 0, X f ) = 1 θ0 ( X = X x , X f ) = 0 ε0 : ε: ∂ 2 θ1 ∂θ ∂θ =− 0 0 ∂X 2 ∂Xf ∂X X = Xf θ1 ( X = 0, X f ) = 0, θ1 ( X = X f , X f ) = 0 and ∂θ ∂θ ∂2 θ2 ∂θ ∂θ = − 0 1 + 1 0 2 ∂X ∂ X f ∂ X X = Xf ∂ X f ∂ X X = Xf θ2 ( X = 0, X f ) = 0, θ2 ( X = X f , X f ) = 0 ε2 : for coefficients of terms up to ε 2. The three-term expansion becomes θ = 1− 2 2 4 X X X 1 X X 1 2 X ε − ε 1 − + 19 − 10 − 9 X f 6 X f X f 360 X f Xf X f To derive an expression for the progress of the freezing front with respect to time, recall that dX f ∂θ = −ε dτ ∂X = X = Xf 1 1 7 ε − ε 2 + ε 3 + higher-order terms, Xf 3 45 subject to the condition that X f = 0 at τ = 0 263 Partial Differential Equations Solution of this latter equation results in 1 7 X f2 = 2τ ε − ε 2 + ε 3 + higher-order terms, 3 45 or −1 1 2 −1 1 7 X f ε 1 − ε + ε 2 3 2 45 1 1 1 = ε −1 X f2 + X f2 − εX f2 + terms of order ε 2 . 2 6 45 τ= Hopefully, these examples (and more in Chapter 7) have demonstrated the importance of the regular perturbation method as a technique that can be used to reduce some nonlinear problems to sets of linear problems. And, as was stated earlier, we like to convert new problems to old ones that we know how to solve. 6.8 PROBLEMS 1. In each of the following problems, use the method of separation of variables to solve the one-dimensional heat equation. Also, define the steady-state temperature at the midpoint of the region. ∂T 1 ∂2 T = a. ∂t 2 ∂ x 2 , T(0, t) = T(3, t) = 0, T(x, 0) = sin πx, 0 < x < 3 ∂T ∂2 T ∂T (0, t ) ∂T (π, T ) = , = = 0, T(x, 0) = cos x, 0 < x < π ∂t ∂ x 2 ∂x ∂x ∂T (− π, t ) ∂T (π, t ) ∂T ∂2 T = , T (− π, t ) = T (π, t ), = c. ∂t ∂ x 2 ∂x ∂x T(x, 0) = x + π, – π < x < π ∂T 1 ∂2 T d.∂t = 2 ∂ x 2 , T(0, t) = 100, T(2, t) = 50, T(x, 0) = 100–13x, 0 < x < 2 2.Solve the heat equation b. ∂T ∂2 T = −T ∂t ∂ x 2 subject to 1, 0 < x < 0.5 T (0, t ) = T (1, t ) = 0, T ( x , 0) = 0, 0.5 < x < 1 264 Applied Mathematical Methods for Chemical Engineers 3.Use the method of eigenfunction expansions to solve, without reducing to homogeneous boundary conditions ∂w ∂2 w =k 2 ∂t ∂x w(0, t ) = A w( x , 0) = h( x ), constant w( L , t ) = B 4.Use the method of Laplace transforms to solve the following problem: ∂u ∂2 u =v 2 ∂t ∂y at t = 0, u = 0 y = 0, u = V (constant) y = ∞, u = 0 5.A fluid of constant density, ρ, and viscosity, μ, is contained in a very long horizontal pipe of length L and radius R. Initially, the fluid is at rest. At t = 0, a pressure gradient (p 0 − pL)/L is impressed on the system. Can the method of Laplace transform be used to determine how the velocity profiles change with time? Hint: Use cylindrical coordinates and assume that the z-component of velocity, vz, is a function of r and t, while the r- and θ-components of velocity are both zero. Then combine the equations of continuity and motion to get [1] ρ ∂ vz p0 − pL 1 ∂ ∂ vz = +µ r L r ∂r ∂r ∂t subject to the conditions: vz (r , 0) = 0 vz (0, t ) is finite v z ( R, t ) = 0 6.Find a bounded solution to Laplace’s equation ∇ 2w = 0 for the half plane y > 0 if w takes on the value f(x) on the x-axis. Hint: (a) Separate the variables and set each side equal to −λ2; (b) argue that the lack of restrictions on λ allow the replacement of the arbitrary constants 265 Partial Differential Equations with arbitrary functions of λ, thus leading to the Fourier integral defined by Equation 5.56. Answer : w( x , y) = 1 ∞ ∞ e −λy f (u) cos λ(u − x ) du d λ π ∫0 ∫u =−∞ 7. The surface of a cylinder of radius R is maintained at a constant temperature T0, whereas the initial temperature was zero throughout [16]. Use Laplace transform to derive the temperature distribution. Model: ∂T ∂2 T 1 ∂T = k 2 + ∂r ∂t r ∂r T (r ,0) = 0, T ( R, t ) = T0 lim T (r , t ) is finite r→ 0 2 ∞ J (α r ) exp(− kα 2n t ) Answer: T (r , t ) = T0 1 − ∑ 0 n R n =1 α n J1 (α n R) where J0(α nR) = 0 defines the α n. ∂u ∂2 u 8.Given the model =k 2 +q ∂t ∂x subject to u(0, t) = 0, u(x, 0) = 0; u (∞, t) is finite; derive u( x , t ) = t x −x2 x exp dt + qt − q ∫ erfc 3 0 4 kt kt 4 4 kπt where q is a constant. Hint: Use the formula L−1 { } t u (s ) = ∫ u(t ) dt o s where ū(s) is a known (tabulated) transform to carry out the inversion [16]. 9.Show that the solution to the Helmholtz equation ∂2 u 1 ∂u + + ε 2 u = 0; u(1) = 1 ∂r 2 r ∂r which is bounded at the origin, solves to u= J0 (εr ) J 0 (ε) 266 Applied Mathematical Methods for Chemical Engineers 10.In Example 6.5, Equation 6.55 was explicitly stated without a verbal description of the steps. Locate the equation of energy from an appropriate reference source and reduce it to Equation 6.55 by making appropriate assumption indicated in the example. 11.Consider a circular plate that is composed of two different materials in the form of concentric circles. If the temperature in the plate is described by ∂2 u 1 ∂u ∂u + = , 0 < r < 2, t > 0 ∂r 2 r ∂r ∂t (6.237) u(2, t ) = 100, t > 0 (6.238) 200, 0 < r < 1 u(r ,0) = 100, 1 < r < 2, (6.239) determine the temperature profile u(r , t ) by making the substitution u(r , t ) = w(r , t ) + V (r ) (6.240) 12.Solve the Heat Equation (Equation 6.12), subject to 1, 0 < x < L / 2 u(0, t ) = 0, u( L , t ) = 0; u( x ,0) = 0, L / 2 < x < L 13.Solve the Heat Equation (Equation 6.12), subject to u(0, t ) = 0, u( L , t ) = 0; u( x ,0) = x ( L − x ) 14.Suppose heat is lost from the lateral surface of a thin rod of length L into a surrounding medium at temperature zero. If the linear law of heat transfer applies, then the heat equation takes on the form ∂2 u ∂u − hu = , 0 < x < L, t > 0 2 ∂x ∂t h = aconst k Determine the temperature u ( x , t ) if the initial temperature is f ( x ) and the ends x = 0 and x = L are insulated. 15.Determine the concentration profile of α ∂2 C ∂C + 2C = , 0 < x < L, t > 0 ∂x 2 ∂t 267 Partial Differential Equations subject to the conditions C ( 0, t ) = 0, C ( L , t ) = 0 and C ( x ,0 ) = f ( x ) 16.Given the diffusion equation k ∂2 u ∂u = , 0 < x < L, t > 0 ∂ x 2 ∂t subject to the conditions u ( 0, t ) = 5, u ( L , t ) = 8 and u ( x ,0 ) = f ( x ) . Hints: a. Find an equilibrium distribution u0 ( x ) , satisfying d 2u0 = 0, u0 ( 0 ) = 5 and u0 ( L ) = 8. dx 2 b. Determine the problem (partial differential equation + conditions) does the difference between u ( x , t ) and u0 ( x ) must satisfy. c. Solve the problem in part (b) for w ( x , t ) then combine with u0 ( x ) to give the solution for u ( x , t ) . REFERENCES 1.Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena, John Wiley & Sons, New York, 1960. 2. Giordano, F.R. and Weir, M.D. Differential Equations: A Modeling Approach, AddisonWesley, Reading, MA, 1991. 3.Myint-U, T. and Debnath, L. Partial Differential Equations for Scientists and Engineers, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1987. 4. Haberman, R. Elementary Applied Partial Differential Equations with Fourier Series and Boundary Value Problems, Prentice Hall, Englewood Cliffs, NJ, 1983. 5. Churchill, R.V. and Brown, J.W. Fourier Series and Boundary Value Problems, 4th ed., McGraw-Hill, New York, 1987. 6.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 3rd ed., John Wiley & Sons, New York, 1977. 7. Spiegel, M.R. Mathematical Handbook, McGraw-Hill, New York, 1968. 268 Applied Mathematical Methods for Chemical Engineers 8. Churchill, R.V. and Brown, J.W. Complex Variables and Applications, McGraw-Hill, New York, 1984. 9. Spiegel, M.R. Complex Variables, McGraw-Hill, New York, 1964. 10. Williams, W.E. Partial Differential Equations, Oxford, 1980. 11.Myers, G.E. Analytical Methods in Conduction Heat Transfer, McGraw-Hill, New York, 1971. 12. Loney, N.W. On using a boundary perturbation technique to linearize a system of nonlinear pde, Chem. Eng. Educ., Winter, 58, 1996. 13. Aziz, A. and Na, T.Y. Perturbation Methods in Heat Transfer, Hemisphere, Washington, 1984. 14.Turian, R.M. and Bird, R.B. Viscous heating in the cone-and-plate viscometer II, Chem. Eng. Sci., 18, 689, 1963. 15.Zauderer, E., Partial Differential Equations of Applied Mathematics, John Wiley & Sons, New York, 1983. 16.Walas, S.M. Modeling with Differential Equations in Chemical Engineering, Butterworth-Heinemann, Boston, 1991. 17. Plawsky, J., Transport Phenomena Fundamentals, 2nd ed., Taylor & Francis, FL, 2010. 7 7.1 Applications of Partial Differential Equations in Chemical Engineering INTRODUCTION The primary objective of this chapter is to present several worked-out examples. These examples reflect typical chemical engineering science research results, and it is recommended that the reader dedicate some time to work through these examples. The examples also demonstrate the application of much of the mathematics discussed in the previous chapters. Examples are provided from heat transfer; mass transfer; simultaneous diffusion and convection; simultaneous diffusion and chemical reaction; simultaneous diffusion, convection, and chemical reaction; and viscous flow. Some examples may be worked out in more detail than others. Those worked-out examples with missing details provide an opportunity for the readers to check their knowledge of the mathematical techniques used. 7.2 HEAT TRANSFER Although we know from our experience that heat flows from regions of high temperature to regions of low temperature, one might want to know how the temperature in a solid changes with time and position. Assuming that the heat transfer is strictly by conduction, one can analyze this situation with reference to the sketch shown as follows. An energy balance on a small volume of the solid of cross-sectional area A and thickness Δ x is {Rate of accumulation of energy} = {Rate of heat conduction in} − {Rate of heat conduction out} (7.1) The rate of accumulation term is given by* A ∆xρCp ∂u ( J /s) ∂t 269 270 Applied Mathematical Methods for Chemical Engineers q x+Δ x L x+Δx x Δx qx where u(x, t) is the solid temperature (ºC), x is the position along the solid (cm), t is the time (s), ρ is the density (g/cm3), and Cp is the solid specific heat (J/g ºC). The rate of heat conduction into the solid volume is given by Aq|x where q|x is heat flux at position x (J/cm2 s); whereas Aq|x+Δx is the rate of heat conduction out. q|x+Δx is the heat flux at position x + Δ x. Substituting these terms into Equation 7.1 results in A ∆xρCp ∂u = Aq x − Aq x + ∆x ∂t Then, dividing through by A Δ xρ Cp gives ∂u 1 Aq x − Aq x + ∆x = ∂t A ρCp ∆x 1 q x + ∆x − q x = − ∆x ρCp Taking the limit as Δ x → 0, leads to 1 ∂q ∂u = − ∂t ρCp ∂ x Inserting Fourier’s first law q = −κ ∂u ∂x (κ is the thermal conductivity) results in the familiar form ∂u κ ∂2 u ∂2 u κ = k = ; k= ∂t ρCp ∂ x 2 ∂x2 ρCp (7.2) Applications of Partial Differential Equations in Chemical Engineering 271 Example 7.1 A sheet of polymer 1/2-in. thick is to be cured at 292ºF for 50 min [1]. If the sheet is initially at 70ºF, and heat is applied from both surfaces, find the time required for the temperature at the center of the sheet to reach 290ºF. It can be assumed that the surfaces are brought to 292ºF as soon as curing has begun and held at that temperature throughout the process. The thermal diffusivity k/ρ Cp of this polymer can be taken as 0.0028 ft2/h (see Figure 7.1). The differential equation for unsteady heat conduction in one direction (Equation 7.2) is ∂u k ∂2 u = ; 0 < x < 1 / 48 ∂t ρCp ∂ x 2 (7.3) The origin is chosen at the center of the sheet, so that x is the distance from the center, as shown in Figure 7.1, and is in feet. Also, x0 is the half thickness. The boundary conditions are t > 0 ∂u(0, t ) =0 ∂x u( x 0 , t ) = 292°F (7.4) and the initial condition is u( x ,0) = 70°F; 0 < x < 1 48 (7.5) Remark: It is good practice whenever possible to express the differential equation in dimensionless form so that the result can be portable. Let Y (ξ , τ) = 292 − u( x , t) x kt , ξ= , τ= 292 − 70 x0 Cpρ× 20 292˚F 292˚F x 1/ 4˝ x=0 x = x0 FIGURE 7.1 Unsteady heat conduction in polymer sheet. (7.6) 272 Applied Mathematical Methods for Chemical Engineers Then Equation 7.3 becomes subject to ∂Y ∂ 2 Y = ∂τ ∂ξ 2 (7.7) ∂Y (0, τ) =0 ∂ξ (7.8) Y (1, τ) = 0 (7.9) Y (ξ,0) = 1 (7.10) From our previous experience, we will try a solution of the form Y (ξ , τ) = f (ξ ) g(τ) (7.11) Then Equation 7.7 becomes g′ f ′′ = = −λ 2 g f where the negative sign is chosen based on our experience using separation of variables up to now (the reader should check this). Therefore f ′′ + λ 2 f = 0 (7.12) subject to f ′(0) = 0 and f (1) = 0 Also g(τ) = c1e − λ 2τ The general solution of Equation 7.12 is f (ξ) = c2 cos λξ + c3 sin λξ and f ′(0) = 0 = λc3 ⇒ c3 = 0 then for nontrivial solution to exist, c2 ≠ 0, so that f (1) = 0 = c2 cos λ ⇒ λ = (2n − 1)π , n = 1,2, 2 (7.13) Applications of Partial Differential Equations in Chemical Engineering 273 Therefore, for each n Yn (ξ, τ) = An e − λ n t cos λ n ξ 2 and by the principle of superposition ∞ Y (ξ, τ) = ∑ An e − λ n t cos λ n ξ 2 n =1 solves Equation 7.7 through Equation 7.9. Then Equation 7.10 gives ∞ Y (ξ,0) = 1 = ∑ An cos λ n ξ n =1 such that for n ≥ 1 1 1 ∫0 cos λ n ξ dξ = An ∫0 cos2 λ n ξ dξ or 4 (2n − 1)π sin (2n − 1)π 2 4 (−1)n +1 = (2n − 1)π An = That is, ∞ Y (ξ, τ) = ∑ (−1)n +1 n =1 2n − 1 2 4 2n − 1 exp − π τ cos πξ 2 (2n − 1)π 2 Then the time required for the temperature at the center of the sheet to reach 290ºF is estimated as follows: Y (0, τ) = 292 − u(0, t ) 292 − 290 = = 0.009 292 − 70 292 − 70 Therefore 0.009 = 4 −4π e π 2 τ 274 Applied Mathematical Methods for Chemical Engineers using only the first term of the series. That is, τ ≅ 2.01. Then, from Equation 7.6, τ= Cpρ 2 kt 2.01 , or t = = 0.313 h = 18.7 min x0 τ = 2 Cpρx 0 (0.0028)(48)2 k Remark: The reader should check the validity of the approximation, which employed only the first term of the series. Example 7.2 Consider a plane wall of thickness L [2] initially at temperature T0 when suddenly at t = 0, the surface temperature at x = 0 is changed to T∞, and the surface at x = L is suddenly exposed to a bath with an ambient temperature also of T∞ (see Figure 7.2). At the surface x = L qconducted = qconvected such that the boundary condition becomes −k ∂T = h(T − T∞) ∂x where h is the convective heat transfer coefficient and k is the thermal conductivity. Determine the temperature distribution in this wall. Solution The mathematical description of the problem is 1 ∂T ∂ 2 T = , 0 < x < L, t > 0 α ∂t ∂ x 2 Surface at x = L qconv qcond FIGURE 7.2 Plane wall. Applications of Partial Differential Equations in Chemical Engineering 275 subject to the boundary conditions T (0, t ) = T∞ −k ∂T ( L , t ) = h(T − T∞ ) ∂x and the initial condition T ( x ,0) = T0 To employ the method of separation of variables, the differential equation and boundary conditions are required to be linear and homogeneous. Presently, the linearity requirement is satisfied, but the boundary conditions are not homogeneous. In this problem, recasting the variables into dimensionless quantities reduces the problem to a homogeneous one. That is, let u( z , τ) = αt T − T∞ x , z = , and τ = 2 T0 − T∞ L L then the dimensionless problem becomes ∂u ∂ 2 u = ∂τ ∂ z 2 subject to u( z ,0) = 1 u(0, τ) = 0 uz (1, τ) = − N B u(1, τ) where NB is the Biot number defined by NB = hL L /kA = k 1/hA Notice that the Biot number compares the steady-state conduction resistance of a plane wall L-units thick to the convection resistance at the surface. Small or large Biot numbers can aid in deciding on the specific boundary condition to use in setting up a physical model. For example, a small Biot number implies that convection resistance is significant whereas a large Biot number implies that conduction resistance is significant. Using the substitution u ( z , τ ) = y( z ) f ( τ ) the dimensionless problem becomes f ′ y′′ = = −λ 2 f y 276 Applied Mathematical Methods for Chemical Engineers where the sign was chosen based on previous experience (Chapter 6). As a result, we get the two ordinary differential equations f ′ + λ 2 f = 0 or f (τ) = c1e − λ 2τ and y′′ + λ 2 y = 0 The conditions on the y-equation are derived as follows: u(0, τ) = 0 = y(0) f (τ) ⇒ y(0) = 0 since f(τ ) is arbitrary, and uz (1, τ) = y′(1) f (τ) = − N Bu(1, τ) = y(1) f (τ), ⇒ y′(1) = − N B y(1) Then the general solution of the y-equation is y( z ) = c2 cos λz + c3 sin λz and the condition at z = 0 gives y(0) = 0 = c2 whereas the condition at z = 1 results in y′(1) = λc3 cos λ = − N Bc3 sin λ Therefore, for a nontrivial solution, c3 ≠ 0 and the eigenvalues are given by λ cot λ = − N B The corresponding eigenfunctions are given by yn = c3, n sin λ n z , n = 1, 2, because there are infinitely many eigenvalues. Then, for each n un ( z , τ) = yn ( z ) fn (τ) = Bn e − λ n τ sin λ n z 2 Further, by the principle of superposition ∞ u( z , τ) = ∑ Bn e − λ n τ sin λ n z 2 n =1 is the solution to the dimensionless differential equation and the given boundary conditions. The initial condition will be satisfied if Applications of Partial Differential Equations in Chemical Engineering 277 ∞ u( z ,0) = 1 = ∑ Bn sin λ n z n =1 such that for n ≥ 1 1 Bn = ∫0 sin λ n z d z 1 1 2 − 4 λ sin 2λ n n 4(1 − cos λ n ) = 2λ n − sin 2λ n = 1 − cos λ n λ n sin 2λ n − 2 4 Finally, the dimensionless temperature distribution is ∞ u( z , τ) = ∑ n =1 4(1 − cos λ n ) − λ 2n τ e sin λ n z 2λ n − sin 2λ where λ n cot λ n = − N B , n = 1,2 defines the λ n. The results can now be reported in terms of the dimensioned variables. Example 7.3 Consider an infinitely long, solid cylinder of radius r0 whose initial temperature is T0 [2]. If the temperature of its outside surface is suddenly changed to T∞, determine the temperature distribution in the cylinder. Solution The governing equation is 1 ∂T ∂ 2 T 1 ∂T = + α ∂t ∂r 2 r ∂r and the boundary and initial conditions are T (r0 , t ) = T∞ and T (r , 0) = T0 respectively. It is important to point out that as we have a second-order differential equation in r, there should be two boundary conditions involving r. The second boundary condition is that the temperature is expected to be finite at r = 0. This conclusion is based on our “engineering understanding” of the problem. 278 Applied Mathematical Methods for Chemical Engineers An alternative boundary condition at r = 0 is ∂T =0 ∂r due to the symmetry of the temperature distribution in this problem. If, however, the temperature varies with angular position, then symmetry would not be valid. Further, to solve this problem using separation of variables, it is sufficient to express the temperature distribution as w(r , t ) = T − T∞ such that 1 ∂ w ∂2 w 1 ∂ w = 2 + α ∂t ∂r r ∂r subject to w(0, t ) < ∞ w(r0 , t ) = 0 w(r ,0) = T0 − T∞ is homogeneous. Then letting w(r , t ) = R(r ) g(t ) the partial differential equation reduces to 1 g′ 2 = −λ 2 , ⇒ g(t ) = c1e −αλ t α g and 1 R′′ + R′ + λ 2 R = 0 r subject to R(0) < ∞ and R(r0 , t ) = 0 The R-equation and the accompanying boundary conditions constitute a singular Sturm–Liouville boundary value problem. Also the R-equation is Bessel’s equation of order zero. Bessel’s equation can be solved indirectly by using the following procedure: Recall Equation 3.80 y′′ − 2a − 1 a 2 − ν 2c 2 y′ + b 2c 2 x 2 c − 2 + y = 0 x x2 with general solution: y = c1 xa Jv (bxc) + c2xa J−v(bxc) Applications of Partial Differential Equations in Chemical Engineering 279 By comparing the coefficients of the first and zeroth derivatives in both equations we get the coefficient of first derivative: 2a − 1 = −1, ⇒ a = 0 2c − 2 = 0, ⇒ c = 1 and the coefficient of the zeroth derivative: b 2c 2 = λ 2 , ⇒ b = λ a − ν2c 2 = 0, ⇒ ν = 0 2 resulting in a general solution R(r ) = c2 J 0 (λr ) + c3 J −0 (λr ) In customary notation, Y0(·) is used instead of J−0(·). Therefore, the general solution is R(r ) = c2 J 0 (λr ) + c3Y0 (λr ) where J0(·) is the zero-order Bessel function of the first kind and Y0(·) is the zero-order Bessel function of the second kind. Further, as R(r) must be finite at r = 0, then c3 must be chosen as zero due to the fact that Y0 is unbounded at r = 0. Also, for nontrivial solution c2 ≠ 0 such that R(r0 ) = 0 = c2 J 0 (r0 λ ), ⇒ J 0 (r0 λ ) = 0 The function J0(r0 λ ) oscillates about the r 0λ axis with infinitely many intersections, that is, r0 λ1 = 2.4048 r0 λ 2 = 5.5201 r0 λ 3 = 8.6537 and more. These points of intersection can be read from a table of zeros of the Bessel function [3]. Therefore, the eigenvalues are defined by J 0 (r0 λ n ) = 0, n = 1, 2, and R n = c2, n J 0 (λ n r ) are the corresponding eigenfunctions. Then for each n wn (r , t ) = Rn gn = An e −αλ n t J 0 (λ n r ) 2 By the principle of superposition ∞ w(r , t ) = ∑ An e −αλ n t J 0 (λ n r ) n =1 2 280 Applied Mathematical Methods for Chemical Engineers satisfies the differential equation and the boundary conditions. The initial condition gives ∞ w(r ,0) = ∑ An J 0 (λ n r ) = T0 − T∞ n =1 which is recognizable as a generalized Fourier series. The Fourier coefficient can be evaluated with the aid of the orthogonality property discussed in the context of Sturm– Liouville problems (Chapter 4). That is, r0 ∫0 rJ0 (λ mr ) J0 (λ nr ) dr = 0 for m ≠ n where r is the weighting function determined from the Sturm–Liouville form of the differential equation. Therefore, r0 ∫ (T0 − T∞ )rJ0 (λ nr ) dr , An = 0 r ∫0 r[ J0 (λ nr )]2 dr 0 n ≥1 The integrals can be evaluated from standard tables containing integrals of Bessel functions [3]. The overall solution to this problem is ∞ w(r , t ) = ∑ An e −αλ n t J 0 (λ n r ) 2 n =1 with the eigenvalues defined by J 0 (r0 λ n ) = 0, n = 1,2, and the An as given earlier. Example 7.4 This example considers unsteady-state heat conduction in one dimension (see Figure 7.3) [4]. Consider a section of a flat wall of thickness L ft, whose height and length are both large in comparison to L. Suppose the temperature distribution is uniform throughout the wall initially, and heat is supplied at a fixed rate per unit area to one surface. Determine the temperature profile as a function of position and time. Solution Temperature is constant in any plane parallel to the surface (as initial temperature is uniform and every part of each wall surface sees the same conditions). Then one space coordinate is sufficient (see sketch). Further, for a section of unit area through the wall, the thermal equilibrium of a slice of the wall between a plane at distance x from the heated surface and parallel to a plane at x + Δ x from the same surface gives the following: Rate of heat input at distance x and time t is −k(∂T/∂x) Applications of Partial Differential Equations in Chemical Engineering 281 ∆x x L FIGURE 7.3 One-dimensional heat transfer. Rate of heat input at distance x and time t + Δ t is −k ∂T ∂ ∂T + −k ∆t ∂ x ∂t ∂x which comes from the first two terms of the Taylor series expansion of T(x, t) while treating x as a constant. Similarly, Rate of heat output at distance x + Δ x and time t is −k ∂T ∂ ∂T + −k ∆x ∂x ∂x ∂x which is the two-term Taylor series expansion of T(x, t) holding t constant. Also, Rate of heat output at distance x + Δ x and t + Δ t is −k ∂ ∂T ∂ ∂T ∂T ∂ ∂T −k + −k + −k ∆t ∆x ∆t + ∂ x ∂ x ∂t ∂ x ∂ x ∂t ∂x The heat content of the element at time t is ρ Cp TΔ x, whereas that at t + Δ t is ∂T ρCp T + ∆t ∆x ∂t such that the accumulation of heat in time Δ t is ρCp ∂T ∆t∆x ∂t 282 Applied Mathematical Methods for Chemical Engineers Then by the conservation law, Input − Output = Accumulation, we get ∂T 1 ∂ ∂T − k ∂ x + 2 ∂t − k ∂ x ∆t ∆t − ∂T ∂ ∂ T 1 ∂ ∂T ∂ ∂T −k + + − k −k ∆x ∆t ∆t −k ∆x + ∂ x 2 ∂t ∂ x ∂ x ∂x ∂ x ∂ x ∂T = ρCp ∆t∆x ∂x where we use the arithmetic average input and output rates during the time interval Δ t. Following simplifications and taking the appropriate limits we get ∂T ∂2 T =α 2 ∂t ∂x for α = k ρCp where the physical properties are assumed constant. Further, notice that even though the differential equation was derived based on a finite thickness L, there is no thickness dependency, and the equation can be used to model semi-infinite domain (x > 0) problems. Therefore, if at t = 0, T = T0 and at x = 0, T = T1 ∂2 T ∂T =α 2 L ∂x ∂t gives α d 2 y( x , s ) = sy( x , s) − T0 , where y( x , s) = L{T ( x , t )} dx 2 subject to the boundedness of y(x, s) and L{T (0, t )} = y(0, s) = T1 s Therefore y( x , s ) = T1 − T0 − e s s x α + T0 s which inverts to x T ( x , t ) = (T1 − T0 ) erfc + T0 2 αt Example 7.5 A solid rectangular slab at a uniform temperature T0 has its four edges thermally insulated. The temperature of one exposed face is raised to and maintained at T1, Applications of Partial Differential Equations in Chemical Engineering 283 whereas the temperature of the other exposed face is held fixed at T0. Determine how the temperature distribution varies with time. Solution As the edges are insulated, heat is expected to flow in one direction, and the differential equation ∂T ∂2 T =α 2 ∂t ∂x can be used to model the temperature distribution subject to the conditions T ( x ,0) = T0 T (0, t ) = T0 T ( L , t ) = T1 Then using the method of Laplace transform we get sU ( x , s) − T ( x ,0) = α d 2U ( x , s) , U ( x , s) = L{T ( x , t )} dx 2 or sU ( x , s) − T0 = α d 2U ( x , s) dx 2 with general solution s U ( x , s) = c1 (s)e − α x + c2 (s)e − αs x + T0 s The transformed boundary conditions become T0 s T1 L{T ( L , t )} = U ( L , s) = s L{T (0, t )} = U (0, s) = Upon using the transformed boundary conditions, the constants of integration are determined such that U ( x , s) = T1 − T0 e s e (T − T ) = 1 0 s sx α − e− sL α s − e− α L s x T α + 0 s s L sin h α sin h sx α + T0 s 284 Applied Mathematical Methods for Chemical Engineers Two methods will be used to find T(x, t) = L−1 {U(x, s)}. Method I In a suitable table of Laplace transform [3], we can locate sinh x s x 2 ∞ (−1)n − n2 π 2 αt /L2 nπx L−1 e sin = + ∑ sinh π s a s a n a n =1 then, if a is replaced by L / α and x / α replaces x sinh x L−1 s sinh L s α s α x 2 ∞ (−1)n nπx 2 2 2 e − n π αt / L sin = + ∑ L L π n =1 n Therefore, x 2 ∞ (−1)n − n2 π 2 αt / L2 nπx T ( x , t ) = (T1 − T0 ) + ∑ e sin + T0 L π n L n =1 Method II Consider the quantity T1 − T0 s s α P (s ) = s Q (s ) sinh L α sinh x Then Q(s) = s sinh L s α and Q(s) = 0 at either s = 0 or s L = inπ, n = 1, 2, α because sinh in π = i sin(n π ). The singular values s = 0, sn = − n 2 π 2α L2 are poles and in particular, simple poles as opposed to multiple poles. Further, Q ′(s) = sinh L s L s s + cosh L α 2 α α Applications of Partial Differential Equations in Chemical Engineering 285 such that inπx (T1 − T0 )sinh P ( s ) sn t 2 2 2 L ρn (t ) = e = e− n π α / L inπ Q ′(s ) sinh inπ + cosh inπ 2 nπx − n2 π 2 αt / L2 (T1 − T0 )sin e L = nπ cos nπ 2 or ρn (t ) = (T1 − T0 ) nπx 2 2 2 2 (−1)n e − n π αt / L sin nπ L For the simple pole s = 0. ρ0 (t ) = (T1 − T0 ) lim s→ 0 sinh x s α sinh L s α = (T1 − T0 ) x L where the series expansion of the hyperbolic functions were used in the limiting process. Therefore, x 2 ∞ (−1)n − n2 π 2 αt / L2 P (s ) nπx f (t ) = L−1 e sin = (T1 − T0 ) + ∑ L Q (s ) L π n =1 n such that x 2 ∞ (−1)n − n2 π 2 αt / L2 nπx T ( x , t ) = (T1 − T0 ) + ∑ e sin + T0 L π n L 1 n = Recall Table 6.1. One could certainly reduce the labor by employing that table where appropriate. 7.3 MASS TRANSFER Example 7.6 A slab of porous solid 1/2-in. thick is soaked in pure ethanol [1]. The void space in the solid occupies 50% of its volume. The pores are fine, so that molecular diffusion can take place through the liquid in the passages: there is no convective mixing. The effective diffusivity of the system ethanol–water in the pores is one-tenth that in the free liquid. If the slab is placed in a large well-agitated reservoir of pure water at 77ºF, how long will it take for the mass fraction of ethanol at the center of the slab to fall to 0.009? Assume that there is no resistance to mass transfer in the water phase and that the concentration of ethanol in the water, and thus at the surface of the slab, is constant at zero. 286 Applied Mathematical Methods for Chemical Engineers Solution Since the densities of alcohol and water differ by only 20%, then one may assume that the total density remains constant such that ∂CA ∂2 CA = DAB ∂t ∂ y2 is a reasonable mathematical description of the unsteady-state process under discussion. If the following substitutions τ= DAB t C y , Y = A , ξ= 2 y0 Ctot y0 are made, where the distance y is measured normal to the center of the slab, and y0 is the half thickness of the slab, then we get ∂Y ∂ 2 Y = ∂τ ∂ξ 2 subject to ∂Y (0, τ) =0 ∂ξ Y (1, τ) = 0 Y (ξ, 0) = 1 This system of equations was solved in the context of heat transfer previously (Example 7.1). There the solution was given as ∞ Y (ξ, τ ) = ∑ (−1)n n =1 2n − 1 2 2n − 1 4 πξ π τ cos exp − 2 (2n − 1)π 2 For ξ = 0 and Y = 0.009, and DAB = 1.0 cm2/s, it is determined that τ = 2.01. Therefore, t= τy02 (2.01)(1/48)2 = = 0.0217 h or 1.30 min DAB (1/10)(1.0)(3.87) Note that use is made of the solution from a previous problem that was solved in dimensionless form. Also, the analogy between heat conduction and molecular diffusion can best be exploited when dimensionless variables are utilized. Example 7.7 Diffusivities of gases in polymers [5] is a model of diffusion through a membrane, which separates two compartments of a continuous-flow permeation chamber. Essentially, at time t = 0, a penetrant is introduced into one compartment (the upstream compartment) and permeates through the membrane into a stream flowing through the other (downstream) compartment. Applications of Partial Differential Equations in Chemical Engineering 287 Assumptions: 1. Diffusion of the penetrant in the gas phase and absorption at the membrane surface are instantaneous processes. 2. Diffusion in the membrane is Fickian with a constant diffusivity D(cm2/s). 3. The concentration of dissolved gas at the downstream surface of the membrane is always sufficiently low compared to that at the upstream surface, such that the downstream surface concentration may be set equal to zero. Below are the models for a flat membrane of thickness h and a cylindrical membrane of inner radius a and outer radius b. The penetrant is introduced from the outside (cylindrical case). Determine the rate at which the gas permeates through the downstream membrane surface. Flat Membrane ∂C (t , x ) ∂ 2 C (t , x ) =D ∂t ∂x2 C (0, x ) = 0 C (t ,0) = C1 C (t , h) = 0 Cylinder ∂C (t , x ) D ∂ ∂C = r r ∂r ∂r ∂t C (0, r ) = 0 C (t , a) = 0 C (t , b) = C1 Solution For the flat membrane case: Let L{C (t , x )} = y(s, x ) then the differential equation transforms to d2 y s − y=0 dx 2 D whose general solution is s s y(s, x ) = k1 cosh x + k 2 sinh x D D subject to L{C (t ,0)} = and C1 s L{C (t , h)} = 0 288 Applied Mathematical Methods for Chemical Engineers Therefore s h −C1 cosh D C k1 = 1 , k 2 = s s s sinh h D The solution in the s-domain is s s s s h cosh h − cosh h sinh x sinh D D D D y(s, x ) = C1 s s sin h D and the inverse transform is derived using the residue theorem s sinh(h − x ) D −1 C (t , x ) = C1L s sinh s h D ∞ P (s1 , x ) exp(sn t ) = C1 ∑ n = 0 Q ′ ( sn ) s h− x D = For s0 = 0, the residue = lim s→ 0 h s s sinh h D s sinh(h − x ) using L’Hopital’s rule. For sn ≠ 0 h s s Q ′(sn ) = sinh h +s cosh h 2 sD D D Also, substituting s = iλ D simplifies the results and we get π λ = n , n = 1,2, h from sinh s / Dh = 0. Finally, P ( sn , x ) 2 nπx = − sin Q ′ ( sn ) nπ h Applications of Partial Differential Equations in Chemical Engineering 289 and the concentration profile is h − x 2 ∞ 1 n 2 π 2tD nπx C (t , x ) = C1 − ∑ exp − sin π n =1 n h2 h h The rate of penetrant across the surface x = h is given by ∞ DAC1 n 2 π 2 Dt ∂C = J (t ) = − DA 1 + 2∑ (−1)n exp − ∂x x =h h h2 n =1 for a flat membrane with a surface area A (cm2). Also, notice that the steady-state rate Jss, is given by Jss = D AC1 h Cylinder The differential equation is Laplace transformed to d 2u 1 du s + − u=0 dr 2 r dr D subject to u(s , a) = 0 u(s , b) = C1 s The general solution of the transformed differential equation is s s u(s, r ) = k1 J 0 i r + k 2Y0 i r D D where J0(·) are Y0(·) the zero-order Bessel functions of the first and second kind, respectively. Following the use of the transformed boundary conditions k1 = C1Y0 (αb) s[ J 0 (αa)Y0 (αb) − J 0 (αb)Y0 (αa)] and k2 = − C1 J 0 (αb) s[ J 0 (αa)Y0 (αb) − J 0 (αb)Y0 (αa)] where the substitution α=i s D 290 Applied Mathematical Methods for Chemical Engineers has been used. Therefore u(s, r ) = C1 [Y0 (αb) J 0 (αr ) − J 0 (αb)Y0 (αr )] s[ J 0 (αa)Y0 (αb) − J 0 (αb)Y0 (αa)] The inverse transform gives ∞ C (t , r ) = ρ0 (t , r ) + ∑ n =1 P ( sn , r ) exp(sn t ) Q ′ ( sn ) where ρ0 (t , r ) = lim s s→ 0 P (s , r ) ln(b / r ) ≡ lim s u(s, r ) = C1 0 s → Q (s ) ln(b / a) The results for small arguments of the Bessel functions were employed where appropriate. (Where?) Finally, the concentration profile for the cylindrical geometry is ∞ ln(b / r ) + π∑ J 0 (α n b0 (α n a) ln(b / a) n =1 C (t , r ) = C1 [ Y ( α b ) J ( α r ) − J ( α b ) Y ( α r )] 0 0 n n 2 × 0 n 0 n exp −α Dt ( ) n J 02 (α n a) − J 02 (α n b) where J 0 (α n a)Y0 (α n b) − J 0 (α n b)Y0 (α n a) = 0 defines the eigenvalues, α n ≠ 0, n = 1, 2, … The rate of penetrant across the surface at r = a, for a cylinder of length L is ∂C J (t ) = 2πDLa ∂r r = a = ∞ 2πDLaC1 J 0 (α n a) J 0 (α n b) b exp ( −α n2 Dt ) 1 + 2 ln ∑ 2 ln(b / a) a n =1 J 0 (α n a) − J 02 (α n b) Again, the steady-state permeation rate is Jss = 2πDL aC1 ln(b / a) for the cylinder. Example 7.8 This model describes the release of relatively small molecules, such as benzoic acid [6]. Initially, the solute to be released is present in solution in a reservoir. A microporous membrane (a ≤ x ≤ b) without nonporous coating bounds the r­ eservoir. The pores of the membrane are filled with a liquid that is immiscible with the r­ eservoir phase (0 ≤ x < a). 291 Applications of Partial Differential Equations in Chemical Engineering In addition, we assumed the following: 1. 2. 3. 4. 5. 6. 7. 8. Diffusion of the agent (benzoic acid) is Fickian. Interfacial boundary layers are not influential here. Diffusivities of the agent are independent of concentration. Aqueous–organic partition coefficient for the agent is independent of concentration. Reservoir and bath solutions are ideal. Uniform temperature exists throughout the system. Aqueous and organic phases are immiscible. Initially, the agent in the reservoir is at some concentration below its saturation value in the solvent. Following previous works in this area [7, 8], the governing equation for the agent concentration in the reservoir is and ∂C1 ∂ 2 C1 = D1 ∂t ∂x 2 (7.14) ∂C2 ∂ 2 C2 = D2 ∂t ∂x 2 (7.15) is the governing equation for the agent in the membrane. The subscripts refer to the respective regions [0, a] and [a, b] as shown in Figure 7.4. D2 is an effective diffusivity and is defined as D2 = Dε τ (7.16) where D is the agent diffusivity in the pore liquid, ε is the membrane porosity, and τ is the membrane tortuosity. Both the agent diffusivity in the pore liquid (D) and in the reservoir liquid (D1) are calculated (Wilke–Chang correlation [9]). The quantity τ was defined and measured for various systems[10]; here we adopt the value given x 0 a b FIGURE 7.4 Cross-section of membrane device. 292 Applied Mathematical Methods for Chemical Engineers for a hydrophobic membrane. The porosity value (0.38) is a manufacturer-supplied quantity. Equations 7.14 and 7.15 are subject to the following boundary conditions: (7.17) ∂C1 (0, t ) =0 ∂x (7.18) ∂C1 (a, t ) ∂C (a, t ) = D2 2 ∂x ∂x (7.19) ∂C2 (b, t ) ∂C (b, t ) = − D2α 2 m2, w 2 ∂t ∂x (7.20) D1 Vw C1 (a, t ) = m1,2C2 (a, t ) Further, as the entire agent is initially present in the reservoir phase, then C1 ( x ,0) = C10 (7.21) C2 ( x ,0) = 0 (7.22) Equation 7.17 is a statement of the equilibrium partitioning at the reservoir–pore liquid interface with m1,2 the partition coefficient. Equation 7.18 indicates that the solute concentration is expected to be finite at the bottom of the reservoir. Equation 7.19 displays the continuity of the agent flux across the reservoir–pore interface, whereas Equation 7.20 accounts for the material leaving the membrane and entering the surrounding water bath. The quantity α 2 is the membrane area at the outer wall (cm 2). In deriving the solution to the model, we recast the model in a dimensionless form by introducing the following quantities: u1 (ξ, θ) = C1 ( x , t ) C10 (7.23) u2 (ξ, θ) = C2 ( x , t ) C10 (7.24) x Dt , θ = 12 b b (7.25) where ξ= The dimensionless model now consists of the following eight equations: ∂u1 ∂ 2 u1 = ∂θ ∂ξ 2 (7.26) D1 ∂u2 ∂ 2 u2 = D2 ∂θ ∂ξ 2 (7.27) Applications of Partial Differential Equations in Chemical Engineering 293 u1 (a / b, θ) = m1,2u2 (a / b, θ) (7.28) ∂u1 (0, θ) =0 ∂ξ (7.29) D1 ∂u1 (a / b, θ) ∂u2 (a / b, θ) = D2 ∂ξ ∂ξ (7.30) ∂u2 (1, θ) ∂u (1, θ) =− 2 ∂θ ∂ξ (7.31) β u1 (ξ,0) = 1 (7.32) u2 (ξ,0) = 0 (7.33) where β= Vw D1 bm2, w D2α 2 (7.34) This type of coupled system of equations is very amenable to the technique of Laplace transform. As such, we let ∞ ∞ 0 0 u1 (ξ, s) = ∫ u1 (ξ, θ)e − sθ dθ and u2 (ξ, s) = ∫ u2 (ξ, θ)e − sθ dθ (7.35) such that Equations 7.26, 7.27, 7.32, and 7.33 transform to the second-order linear differential equations: d 2u1 dξ 2 (7.36) D1 d 2u2 su2 = D2 dξ 2 (7.37) su1 − 1 = and subject to the transformed boundary conditions: u1 (a / b, s) = m1,2u2 (a / b, s) (7.38) du1 (0, s) =0 dξ (7.39) D1 du1 (a / b, s) du2 (a / b, s) = D2 dξ dξ (7.40) 294 Applied Mathematical Methods for Chemical Engineers βsu2 (1, s) = − du2 (1, s) dξ (7.41) The solution to the dimensionless model, Equations 7.36 through 7.41, is D2 D2 a a D2 ξ λβ cos λ 1 − + sin λ 1 − cos λ b b D1 D1 D1 sq(λ ) 1 u1 = + s and u2 = [βλD2 / D1 sin λ (1 − ξ) − cos λ(1 − ξ)] a sin λ sq(λ) b (7.42) D2 D1 (7.43) where D a a a q(λ ) = m1,2 λβ 2 sin λ 1 − − cos λ 1 − sin λ b b b D1 − D2 D1 D2 D1 a D2 a a cos λ 1 − + sin λ 1 − cos λ λβ D b b b 1 (7.44) D2 D1 The substitution iλ = D1s D2 (7.45) where i is the imaginary unit, was used in Equations 7.42 through 7.44. Equations 7.42 and 7.43 are inverted by the residue theorem [11] to give u1 (ξ, θ) = m1,2 a / b D2 a 1+ β + (m1,2 − 1) D1 b ∞ +∑ n =1 D2 D1 D D2 − λ 2n D12 θ D2 a a λ nβ D cos λ n 1 − b + sin λ n 1 − b cos λ n D ξ e 1 1 λ dq 2 dλ λ = λ n (7.46) and u2 (ξ, θ) = a/b D2 a 1+ β + (m1,2 − 1) D1 b a D2 λ nβ D sin λ n (1 − ξ) − cos λ )n (1 − ξ) sin λ n b 1 +∑ λ dq n =1 2 dλ λ = λ n ∞ D D2 − λ 2n D12 θ e D1 (7.47) 295 Applications of Partial Differential Equations in Chemical Engineering 25 Benzoic acid released (mg) 20 15 Theory Expt. 10 5 0 0 20 40 60 80 100 120 Time (h) FIGURE 7.5 Release profile for benzoic acid. (From Ramraj, R., Farell, S., and Loney, N.W., Sep. Sci. Technol., 34, 225, 1999.) where the eigenvalues, λ n, are defined by tan λ n (1 − a / b) = m1,2 tan(λ n a / b D2 / D1 ) + D2 / D1 D2 / D1 βλ n λ n m1,2 D2 / D1β tan(λ n a / b D2 / D1 ) − D2 / D1 (7.48) Then using Equation 7.24, the concentration profile of the agent at x = b can be determined. Figure 7.5 shows the release profile for benzoic acid, which was calculated using the first eigenvalue (Equation 7.48) and the first two terms of Equation 7.47. Example 7.9 This example considers drug delivery. A mathematical model that is used to describe the sustained release of human immunoglobulin G from biodegradable polymers poly (l-lactide) and poly (d,l-lactide-co-glycolide) microspheres is given in [34]. In the development of this model, it was assumed that the mass transfer coefficient at the surface is finite. If the mass balance is given as ∂C ∂ 2 C 2 ∂C = D 2 + ; 0 <r < R ∂r r ∂r ∂t (7.49) subject to the initial condition C (r ,0) = C0 , 0 < r < R (7.50) 296 Applied Mathematical Methods for Chemical Engineers and boundary condition ∂C (0, t ) = 0, t > 0 ∂r (7.51) and C ( R, t ) = C ∞ , t > 0 (7.52) Determine the mass of drug released to total mass M t / M T . Solution Let ∞ L {C (r , t )} = y (r , s ) ≡ ∫ e − st C (r , t ) dt 0 then Equation 7.49 becomes d 2 y(r , s) s y (r , s ) − C (r ,0 ) = D dr 2 + 2 dy(r , s) r dr (7.53) and the associated boundary conditions transform to dy ( 0, s ) =0 dr (7.54) C∞ = y ( R, s ) s (7.55) Then the solution of the transformed system (Equations 7.49 through 7.55) is carried out in two parts: a. By comparison of the associated homogeneous differential equation to Equation 3.80 d 2 y(r , s) 2 dy(r , s) s + − y (r , s ) = 0 r dr D dr 2 we can see that 2a − 1 = −2 ⇒ a = −1 2; 2c − 2 = 0 ⇒ c = 1; a 2 − γ 2c 2 = 0 ⇒ γ = ±1 / 2; b 2c 2 = − s s ⇒ b = ±i D D Applications of Partial Differential Equations in Chemical Engineering leading to s s yc (r , s) = k1 r −1/ 2 J1/ 2 i r + k 2 r −1/ 2 J −1/ 2 i r D D s s r r sin i cos i D D ≡ K1 + K2 r r b. Determination of the particular solution, yP (in this case a constant) say y p = B and following substitution into Equation 7.53, we get s C C yP = in ⇒ yP = in D D s Such that the general solution is given by s s sin i r cos i r D D Cin yg (r , s) = K1 + K2 + r r s Applying the transformed boundary conditions: dy ( 0, s ) =0 dr implies that the general solution must be bounded at the origin since s cos i r D → ∞ as r → 0 r therefore K 2 must be chosen as zero to obtain a bounded solution. C∞ C At r = R, y = ∞ ⇒ = K1 s s s sin i R D Cin + r s resulting in K1 = (C∞ − Cin ) R s R Ssin i D 297 298 Applied Mathematical Methods for Chemical Engineers Finally, the dimensionless concentration profile (in the Laplace Domain) is s r sin i D Cin . + r s s s sin i D (C − Cin ) R y (r , s ) = ∞ The dimensional concentration profile becomes sin i R C ( r , t ) = L−1 { y ( r , s )} = Cin + (C∞ − Cin ) L−1 r s sin i s r D s R D employing the residue theorem to carry out the inversion sin i Ssin i s r D s R D ≡ s s P (r , s ) R = F ( s ) ,where P (r , s ) = sin i r , Q ( s ) = s sin i Q (s ) D D where L−1 { F ( s )} = ∞ ∑ ρ (t ), n n=0 ρn (t ) = P ( sn ) Q ′ ( sn ) e sn t and the poles of F (s) are the zeroes of Q(s). This means that s s 0 = s sin i R ⇒ s = 0 or sin i R =0 D D For the case sin i s = 0, ρn (t ) = ρ0 (t ) = lim s F (s) = lim s→ 0 s→ 0 sin i s 1 i r s −1/ 2 D −1/ 2 cos i r 2 D r = lim = s→ 0 s R 1 −1/ 2 −1/ 2 cos i iR s D R 2 D s r D s R D 299 Applications of Partial Differential Equations in Chemical Engineering and for the case s ≠ 0, sin i s s − n2π 2 D R = 0 ⇒i R = nπ, ⇒ sn = , n = 1,2,3 D D R2 and also, Q ′(s ) = s R nπ d n nπ R = cos nπ = ( −1) s sin i D R 2 ds 2 Consequently, r sin nπ n2 π2 D t R e − R2 ρn (t ) = n nπ ( −1) 2 Hence 2 R ∞ r −n π D t n C (r , t ) = Cin + (C∞ − Cin ) + (C∞ − Cin ) ∑ ( −1) sin nπ e R2 R π r n =1 2 2 R ∞ r −n π D t 2 n C∞ − Cin ) ∑ ( −1) sin nπ e R2 ( π r n =1 R 2 2 = C∞ + r − sin nπ e R ∂C 2 R n = − (C∞ − Cin ) 2 ∑ ( −1) ∂r π r n + (C∞ − Cin ) 2 ∞ r −n ( −1)n cos nπ e ∑ r n =1 R 2 n2 π2 D t R2 π2 D t R2 At r = R, ∞ − 2 ∂C ∂r = − R Cin − C∞ ∑ e n =1 ( ) n2 π 2 D t R2 Then n2 π 2 D τ t ∞ R3 − ∂C R2 dτ = −8 Cin − C∞ ∑ M (t ) = ∫ − D A e ∂r r = R n =1 n 2π 0 ( ) n2 π 2 D t − ∞ R3 R2 = 8 Cin − C∞ ∑ 1 − e 2 n =1 n π ( ) t 0 300 Applied Mathematical Methods for Chemical Engineers Finally, n − R3 − e 1 2 n =1 n π ∞ M (t ) = MT 8 (Cin − C∞ ) ∑ 2 π2 D t R2 4 3 πR (Cin − C∞ ) 3 = 6 ∞ 1 6 ∞ 1 −n − 2∑ 2e ∑ 2 2 π n =1 n π n =1 n 2 π2 D t R2 However, 1 1 1 1 + + + + 12 p 22 p 32 p 4 2 p = 2 2 p −1 π 2 p B p ( 2 p )! , Bk are the Bernoulli numbers therefore 1 2π 2 (1 6 ) π 2 ∑ 2 = 2! = 6 , B1 = 1 / 6 n =1 n ∞ resulting in M (t ) 6 ∞ 1 −n = 1− 2 ∑ 2 e MT π n =1 n 2 π2 D t R2 7.4 COMPARISON BETWEEN HEAT AND MASS TRANSFER RESULTS Quite frequently heat transfer problems have their analogs in mass transfer. This can be fortuitous if recognized, as time and labor can be saved when solving a given problem. One approach that can be very helpful in exposing the similarities between two problems requires the use of dimensionless quantities. That is, recast the given differential equation and its conditions into a dimensionless form. Then derive a dimensionless solution. At this juncture, the solution is not tied to either heat transfer or mass transfer and can be interpreted as needed. In two previously worked-out applications, Examples 7.1 and 7.6, both situations were reduced to the identical dimensionless differential equations ∂Y ∂2 Y = ∂τ ∂ξ 2 subject to ∂Y (0, τ) =0 ∂ξ Y (1, τ) = 0 and Y (ξ,0) = 1 This dimensionless model has the solution Y (ξ , τ) = ∞ ( −1)n ∑ n =1 2n − 1 2 4 2n − 1 exp − π τ cos πξ 2 (2n − 1)π 2 Applications of Partial Differential Equations in Chemical Engineering 301 This result may now be interpreted in terms of a heat transfer application (Example 7.1) or a mass transfer application (Example 7.6). As a second illustration, consider a solid body occupying the space from y = 0 to y = ∞ that is initially at a temperature T0. At time t = 0, the surface at y = 0 is suddenly raised to a temperature T1 and is maintained at that temperature for some t > 0. Determine the temperature profile T(y, t), if the mathematical statement of the problem is given by ∂T ∂2 T =α 2 ; y>0 ∂t ∂y T ( y,0) = T0 T (0, t ) = T1 T (∞, t ) = T0 Conveniently, the dimensionless temperature profile can be taken as θ( y, t ) = T − T0 T1 − T0 which recasts the differential equation into ∂θ ∂2 θ =α 2 ∂y ∂t subject to θ = 0 at t ≤ 0 ∀y θ = 1 at y = 0 ∀t > 0 θ = 0 at y = ∞ ∀t > 0 Then, if we let θ = f ( η), η = y 4αt (7.56) the dimensionless differential equation can be reduced to an ordinary differential equation: f ′′( η) + 2 ηf ′( η) = 0 This latter differential equation has the general solution η f ( η) = k1 ∫ e − η d η + k2 2 0 with arbitrary constants k1 and k2. By employing the conditions f (0) = 1 and f ( ∞) = 0 (7.57) 302 Applied Mathematical Methods for Chemical Engineers there results θ= T − T0 2 = 1− T1 − T0 π η ∫0 e−ξ 2 y d ξ = 1 − erf 4αt Comparing this dimensionless temperature profile with the dimensionless concentration profile (Example 7.11) results in identical form and substance: cA − cA∞ x = 1 − erf cA0 − cA∞ 4 DAB z / v0 The error functions were previously discussed and defined in Section 6.4 but with differing arguments. In the heat transfer interpretation, α represents thermal diffusivity whereas DAB represents mass diffusivity in the mass transfer case. Also, the quantity z/v0 represents time whereas x and y are coordinates in their respective systems. If one is able to anticipate analogies between conduction and diffusion in certain specific situations, then a given issue may be resolved in a fraction of the time it would otherwise take. A procedure on how to convert dimensional differential equations to their dimensionless forms is given in Chapter 8. 7.5 SIMULTANEOUS DIFFUSION AND CONVECTION By extending our definition of diffusion to include the process of heat transfer by conduction, the examples that follow demonstrate how some problems involving diffusion and flow can be treated effectively with the mathematical tools previously discussed. For example, consider the problem involving heat transfer to a flowing fluid. This problem was solved [12] with the use of the confluent hypergeometric function [13]. Example 7.10 An example of heat transfer to a laminar flow fluid in a circular tube, this, the so-called Graetz problem, involves the determination of the temperature profile in a fully developed laminar flow fluid inside a circular tube. The governing equation for the Graetz problem may be obtained from an energy balance in cylindrical coordinates. Alternatively, one can start with the equation of energy in terms of transport properties for Newtonian fluids of constant density ρ and thermal conductivity k [14]. In cylindrical coordinates, the steady-state equation of energy excluding the r and θ component of velocities and neglecting viscous dissipation is given by ρCp vz (r ) ∂T 1 ∂ ∂T 1 ∂ 2 T ∂ 2 T + 2 =k r + 2 2 ∂z ∂z r ∂r ∂r r ∂θ (7.58) Applications of Partial Differential Equations in Chemical Engineering 303 Equation 7.58 can be further simplified if both the terms ∂2 T ∂2 T and ∂θ2 ∂z 2 are neglected on the basis that the amount of heat conducted in these directions are negligible in comparison to that conducted in the r-direction as well as that conducted by convection. The dimensionless form of the reduced Equation 7.58 is (1 − ξ 2 ) ∂ u 1 ∂ ∂u = ξ ∂ζ ξ ∂ξ ∂ξ (7.59) subject to u(0, ζ) is finite (7.60) u(1, ζ) = 0 (7.61) u(ξ,0) = 1 (7.62) where the dimensionless quantities u ( ξ , ζ) = Tw − T r kz , ξ= , ζ= Tw − T0 r1 ρCp vz max r12 have been used. Here, the quantities T0, Tw, r1, and vzmax are entering fluid temperature, wall temperature, radius of tube, and maximum axial fluid velocity, respectively. Then, applying the method of separation of variables to Equation 7.59, by assuming that u(ξ, ζ) = f (ξ) g(ζ) (7.63) such that Equation 7.59 reduces to g′ + λ 2 g = 0, ⇒ g(ζ) = c1e − λ ζ (7.64) d 2 f df + + λ 2 ξ(1 − ξ 2 ) f = 0 dξ 2 dξ (7.65) 2 and ξ where λ is real. To solve Equation 7.65, the Frobenius series method is appropriate. However, the following substitutions will be made to recast the differential equation into a more recognizable form: (1) Let w = λ ξ 2 and (2) f(w) = e−w/2 y(w). Then Equation 7.65 becomes w d2 y dy 1 λ + (1 − w) − − y= 0 2 dw dw 2 4 (7.66) 304 Applied Mathematical Methods for Chemical Engineers following application of the chain rule. Equation 7.66 is a confluent hypergeometric equation for which there are Tabulated solutions [13]. That is, Equation 7.66 has two linearly independent solutions: y1 ( w) = 1F1[a; b; w] = 1 + a a(a + 1) w 2 a(a + 1)(a + 2) w 3 w+ + + b b(b + 1)2! b(b + 1)(b + 2)3! where, in this case b = 1, a = 1 λ − 2 4 and y2 ( w) = y1 log w + aw 1 2 − + 1!1! a 1 Therefore, the general solution to Equation 7.66 is y( w) = c2 y1 ( w) + c3 y2 ( w) The boundary condition given by Equation 7.60 requires that f(0) be finite, which means that y(w) must also be finite. However, y2(w) is unbounded at w = 0, thus c3 must be chosen as zero. For a nontrivial solution, c2 ≠ 0 such that f (ξ) = c2 e − λ n ξ 2 1 λ F ;1; λξ 2 2 4 /2 1 1 (7.67) gives 1 λ f (1) = c2 e − λ / 2 1F1 − ;1; λ = 0 2 4 following application of the boundary condition given by Equation 7.61 ­resulting in 1 λ F − n ;1; λ n = 0 for n ≥ 1 2 4 1 1 (7.68) Equation 7.68 defines the n eigenvalues whereas Equation 7.67 defines the corresponding eigenfunctions. Then, by the principle of superposition, ∞ 1 λ 2 n u(ξ, ζ) = ∑ Cn e−λ 2 ζ e−λ n ξ / 21F1 − n ;1; λ n ξ 2 2 4 n =1 (7.69) Applications of Partial Differential Equations in Chemical Engineering 305 is the solution to Equations 7.59 through 7.61. The condition given by Equation 7.62 reduces Equation 7.69 to ∞ u(ξ,0) = 1 = ∑ Cn e − λ n ξ n =1 2 /2 1 λ F − n ;1; λ n ξ 2 2 4 1 1 which is a generalized Fourier series. Then, by making use of the orthogonal properties of a Sturm–Liouville problem, Cn is given by Cn = 1 1 − λn / 2 3 λ n ; 2; λ n 1F1 − 2 − λ e 2 4 n 1 ∫0 ξ(1− ξ 2 2 )e − λnξ2 1 λn ;1; λ n ξ 2 dξ F − 1 1 2 4 (7.70) From the temperature profile given by Equation 7.69, the heat flux at the wall, the total rate of heat transfer, and the bulk temperature of the fluid at the exit can be evaluated. Also, using the arithmetic mean of terminal temperature differences together with the definition of heat transfer coefficient, the arithmetic mean Nusselt number can be expressed as a function of the Graetz number. Similarly, the logarithmic Nusselt number can also be obtained as a function of the Graetz number [12]. Example 7.11 This example considers diffusion into a falling film (see Figure 7.6) [8]. Consider the diffusion of a solute A into a moving liquid film B. The liquid is in laminar flow. Assuming that (1) the film moves with a flat velocity profile v0, (2) the film may be taken to be infinitely thick with respect to the penetration of the absorbed material, and (3) the concentration at the interface x = 0 is cA0, then the mathematical statement of this problem is v0 ∂c A ∂ 2 cA = DAB ∂z ∂x2 (7.71) subject to cA = cA0 at x = 0 (7.72) cA = cA∞ at x = ∞ (7.73) cA = cA∞ at z = 0 (7.74) a. Derive the concentration profile for the solute A. b. Derive a liquid-film mass transfer coefficient based on the driving force cA0 − cA∞. 306 Applied Mathematical Methods for Chemical Engineers x z v0 d FIGURE 7.6 Diffusion into a falling film. Solution Let cA − cA∞ x = f ( η), η = cA0 − cA∞ 4 DAB z / v0 Then ∂c A dη = (cA0 − cA∞ ) f ′( η) dz ∂z dη ∂c A = (cA0 − cA∞ ) f ′( η) dx ∂x 2 2 d2 η ∂ cA dη c c f ( ) ( ) = − η + f ′( η) 2 ′′ A0 A∞ 2 dx dx ∂x but dη 2 DAB =− dz v0 1 ; 4 DAB z / v0 4 DAB z / v0 x dη 1 = dx 4 DAB z / v0 Applications of Partial Differential Equations in Chemical Engineering 307 Therefore, the differential equation reduces to f ′′ = −2 η f′ and the boundary conditions become f ( η) = 1 at η = 0; f ( η) = 0 at η = ∞ Integration of this new ordinary differential equation results in η f ( η) = k1 ∫ e − ξ d ξ + k 2 2 0 where k1 and k 2 are arbitrary constants. Imposing the boundary conditions results in f ( η) = − 2 π η ∫0 e − ξ 2 x d ξ + 1 = 1 − erf 4 DAB z / v0 That is, cA − cA∞ x = 1 − erf cA0 − cA∞ D z v 4 / AB 0 c. For short contact times for absorption in wetted-wall towers, the following expression for the liquid-phase mass transfer coefficient was developed in the literature [15]. The total moles of A transferred per unit time per unit cross-sectional transfer area is NA = ν0 L ∞ ∫0 (cA − cA∞ )z = L d x = 4 DAB ν0 (cA0 − cA∞ ) = k L (cA0 − cA∞ ) πL such that kL = 4 DAB ν0 πL Note that the quantity L/v0 is the time required for the liquid film to traverse the length L. The next two examples are solutions to the models that were developed in Chapter 1. We apply some of the mathematical methods discussed in the last five chapters to address issues of some practical concerns. In both cases we were able to obtain favorable comparisons with available experimental data. It is understood that a mathematical model is useful so long as it reasonably describes the physics, chemistry, or biology it is attempting to capture. 308 Applied Mathematical Methods for Chemical Engineers Example 7.12 Recall the model developed in Chapter 1 describing the molecular contribution to the mass transfer coefficient of a turbulent flowing fluid in a smooth tube. Use Equations 1.29 through 1.31 to derive the concentration profile for species A. Substitute the derived concentration profile into Equation 1.14 and then develop a local average mass transfer coefficient, k AV = NAV/Cb,AV (NAV is the average molar flux and Cb,AV is the average bulk concentration). Vb dC b 2 =− N Ar dz Rb Rb (1.14) Vb ∂CA ∂2 CA x =D ∂x2 δ ∂z (1.28) CA = 0 at x = 0 (1.29) CA = Cb ( z ) at CA = C0 at x=δ z=0 (1.30) (1.31) Solution First, we recast Equations 1.28 through 1.31 by letting u = C A − C0 Then taking the Laplace transform with respect to z L{u( z , x )} = CA (s, x ) transforms the recast Equation 1.28 and 1.31 into Airy’s differential equation d 2CA − a 2 sxCA = 0 dx 2 where V a 2 = b Dδ The general solution to Airy’s equation can be given in either 1/3-Bessel functions or Airy functions [17], that is, 2 2 CA ( x , λ ) = x 1/ 2 k1 J1/ 3 aλx 3/ 2 + k 2 J −1/ 3 aλx 3/ 2 3 3 (7.80) Applications of Partial Differential Equations in Chemical Engineering 309 or C A ( y, λ ) = 3 1 (aλ )1/ 3 2 { × k1 3 Ai(− y) − Bi(− y) + k 2 3 Ai(− y) + Bi(− y) } (7.81) where 3 2 y = ⋅ aλx 3/ 2 2 3 2/3 = ( aλ ) 2 / 3 x and λ=i s Both forms are utilized to evaluate the constants k1 and k2. The quantities Ai(−y) and Bi(−y) are Airy functions [17]. The constants k1 and k2 turn out to be C (λ ) + C0 δ −1/ 2 − C0 (aλ )1/ 3 Γ 2 J 2 aλδ 3/ 2 b −1/ 3 λ 2 31/ 3 λ2 3 3 k1 = 2 J1/ 3 aλδ 3/ 2 3 and k2 = C0 (aλ )1/ 3 2 Γ 3 λ 2 31/ 3 where Cb (λ ) = L{Cb ( z )} symbolizes the Laplace transform and Γ (·) is the gamma function. The next step is to solve for the concentration of the transferring species in the turbulent core, Cb(z). As the mass fluxes of the turbulent core and the laminar sublayer are equal at the interface, Equation 7.75 may be written as dCb 2 D ∂CA =− | x =δ RVb ∂ x dz and Laplace transformed to sCb (s) − C0 = − 2 D dC A | x =δ RVb dx (7.82) 310 Applied Mathematical Methods for Chemical Engineers Then using Equation 7.80 and performing the necessary simplification, we get Dδ −1/ 2 (aλ )1/ 3 Γ (2 / 3)32 / 3 3 C b (λ ) 1 RVb π =− 2 − C0 λ 2 D 2 2 δ1/ 2 aλJ −1/ 3 aλδ 3/ 2 λ 2 λ 2 J1/ 3 aλδ 3/ 2 − 3 RV 3 b (7.83) Equation 7.83 is inverted using the residue theorem to give the dimensionless concentration profile at the intersection of the laminar sublayer and the turbulent core: Cb ( z ) Dδ −1/ 2 a1/ 3 37 / 6 2 = Γ × 3 C0 RVb π ∞ e− λn z 2 D 2 2 λ n2 RVb 2 n = 1 5/ 3 2 J1/ 3 aλ n δ 3/ 2 λn + δ a + δ 3 6 D 3 3 RVb 2 ∑ (7.84) The values for λ n, are defined by 2 2 D 1/ 2 2 λ n J1/ 3 aλ n δ 3/ 2 = δ aJ −2 / 3 aλ n δ 3/ 2 3 RVb 3 (7.85) Mass Transfer Coefficient To estimate a local mass transfer coefficient for comparison to available experimental data we take the approach that the mass transfer coefficient, k, can be defined as the flux at the wall divided by the concentration difference between the bulk and wall concentrations. In the case that the wall concentration is zero and the bulk concentration is the same as that at the intersection of the laminar sublayer and the turbulent core, Cb(z), we can define the flux at the wall, Nw, as N w = − N Ax x =δ since δ is small. Employing Equations 7.82 and 7.84 results in N Ax x =δ = R dCb Vb 2 dz Therefore, the mass transfer coefficient, k, is ∞ R k = Vb 2 ∑ λ 2n Bn exp ( −λ 2n z ) n =1 ∞ ∑ Bn exp ( −λ n2 z ) n =1 Applications of Partial Differential Equations in Chemical Engineering 311 where Bn = 1 2 D RV 2 2 2 λ λ 5/3 δ 2 a 2 + n δ b J1/3 aλ n δ 3/ 2 n + D 6 3 3 3 RVb In practice, an average value of the mass transfer coefficient, k AV, is used. Here, we consider the average flux at the wall, NAV, over the closed interval [0, z] to be N AV = = ∞ z RVbC0 Dδ −1/ 2 a1/3 37/6 Γ (2 / 3) ∫ ∑ λ 2n Bn exp ( −λ 2n z ) dz 0 2z RVb π n =1 ∞ ∞ DC0 δ1/ 2 a1/3 37/6 Γ (2 / 3) ∑ Bn − ∑ Bn exp ( −λ 2n z ) 2πz n =1 n =1 Also, the average bulk concentration is Cb ,AV = = ∞ z C0 Dδ1/ 2 a1/3 37/6 Γ (2 / 3) ∫ ∑ Bn exp ( −λ 2n z ) dz 0 π z RVb n =1 C0 Dδ1/ 2 a1/3 34 /6 ∞ Bn ∞ Bn exp ( −λ 2n z ) −∑ ∑ π n =1 λ 2n n =1 λ 2n z RVb Then the local average mass transfer coefficient, k AV, is defined to be k AV = N AV / Cb ,AV ∞ = ( R / 2)Vb ∑ Bn 1 − exp ( −λ 2n z ) ∞ n =1 ∑ ( Bn / λ n2 ) 1 − exp ( −λ n2 z ) n =1 Comparison with Experimental Data In the works cited, the experimentally derived mass transfer coefficients are referenced to the frictional velocity as f K ∞+ = K ∞ / V * ; V * = Vb ; 2 f= 0.0791 N Re where K∞ is the fully developed mass transfer coefficient (cm/s). Therefore, we also reference k AV to the frictional velocity to get * K AV = k AV / V * 312 Applied Mathematical Methods for Chemical Engineers TABLE 7.1 Model Predictions versus Experimental Data λ 1 × 103 5.9543 5.8209 5.6411 5.5643 5.3037 * K Avg × 104 K ∞+ × 10 4 K ∞+ × 10 4 (Model) (Expt [18]) (Correlation [19]) y+ δ = f(y+) NRe 3.46 3.47 3.47 3.47 3.47 3.52 3.50 3.46 3.46 3.50 3.71 3.71 3.71 3.71 3.71 1.2 1.2 1.2 1.2 1.2 7750 11200 18600 23200 50200 Nsc 2400 2400 2400 2400 2400 Source: Huang, Chem. Eng. Sci., Vol. 59, 1191–1197, 2004. With permission. Before making the comparisons, we needed to determine how many terms of the infinite series are required to give a reasonable estimate of kAV. Because the eigenvalues increase rapidly as n increases (λ2/λ1 ≈ 2 orders of magnitude), we discovered that for z ≥ 100 cm, kAV is influenced more by the first eigenvalue than by z. Also, the cited experimental mass transfer coefficients [18,19] were all measured in the fully developed flow region. Therefore, the criterion z ≥ 100 cm is satisfied in all cases, as the reported fully developed region is well beyond 100 cm. This fact (z ≥ 100 cm) is used to get R k AV = Vb λ12 2 such that * K AV = ( R2 )Vbλ12 V* The first eigenvalue λ 1 can be determined using the Mathematica routine FindRoot [20]. Then the local dimensionless fully developed mass transfer coefficient can be calculated. Results using the first eigenvalue, λ 1, are reported in Table 7.1. Example 7.13 This example considers the model of a subdivided mass exchanger that was developed in Example 1.2. Determine the local solute concentration profile, CA(z, r) on the tube side using the method of separation of variables discussed in Chapter 6. As part of your solution, you will need to define the relationship that will yield the eigenvalues. Also, use 2π R ∫ ∫ CA ( L, r )νz (r )r dr dθ CAb = 0 0 2π R ∫0 ∫0 νz (r )r dr dθ to develop an expression for the bulk concentration. Finally, develop an expression for CAb / CA0. 313 Applications of Partial Differential Equations in Chemical Engineering Model i. Tube-side: 1 ∂ ∂CA r 2 ∂C νmax 1 − 2 A = DA r R ∂z r ∂r ∂r (7.86) subject to the boundary conditions CA (0, r ) = CA0 (7.87) ∂CA ( z ,0) =0 ∂r (7.88) ∂CA ( z , R) = K (CA|r = R − CAS ) ∂r (7.89) dCAS = −2πRNK (CA|r = R − CAS ) dz (7.90) − DA ii. Shell-side: Q subject to CAS ( L ) = CASL (7.91) N is the number of fibers and Q is volumetric flow rate of the shell-side (sweep) stream. Total mass balance of species A between the two streams: Q(CAS0 − CASL ) = NπR 2 νmax 2 4 CA0 − R 2 r=R r2 ∫r = 0 C A| z = L 1 − R2 r dr (7.92) CAS0 is the outgoing sweep stream concentration of species A (Figure 1.3). Solution Equations 7.89 through 7.86 were put into dimensionless forms: (1 − ξ 2 ) ∂θ 1 ∂ ∂θ = ξ ∂ζ ξ ∂ξ ∂ξ θ = 1 at ζ = 0 (7.93) (7.94) 314 Applied Mathematical Methods for Chemical Engineers ∂θ = 0 at ξ = 0 ∂ξ − ∂θ KR = (θ + θ S − 1) ∂ξ DA (7.95) (7.96) = N Sh (θ + θ S − 1), at ξ = 1 The dimensionless variables were defined as follows: Dimensionless tube-side concentration θ(ζ, ξ) = CA − CASL CA0 − CASL Dimensionless shell-side concentration θ S (ζ) = CA0 − CAS CA0 − CASL where ξ = r/R and ζ = DA z/vmax R2 are the dimensionless radial and axial coordinates, respectively. Equation 7.90 for the dialysate side is transformed into dθ S 2πRNK νmax R 2 = (θ| ξ =1 + θS − 1) dζ Q DA = 4 N ShV (θ| ξ =1 + θ S − 1) (7.97) subject to the condition θS = 1 at ζ = DA L 1 L2 = 2 Pe R νmax R (7.98) The quantity V is the tube-side volumetric flow rate divided by the sweep-side (shellside) volumetric flow rate, Nsh is the Sherwood number and Pe is the length Peclet number. The method of separation of variables was used to solve the dimensionless system of Equations 7.96 through 7.93; however, one could have used the methods of Laplace transform. Notice that Equation 7.93 is identical to Equation 7.59 of Example 7.9. Then the boundary value problem will contain the identical second-order differential equation (Equation 7.65): ξ d 2 f df + + λ 2 ξ(1 − ξ 2 ) f = 0 dξ 2 dξ Applications of Partial Differential Equations in Chemical Engineering 315 In Example 7.10, this differential equation was shown to be the confluent hypergeometric equation (Equation 7.66): w d2 y dy 1 λ + (1 − w) − − y= 0 dw 2 dw 2 4 with linearly independent solutions a a(a + 1) w 2 a(a + 1)(a + 2) w 3 w+ + + b b(b + 1)2! b(b + 1)(b + 2)3! y1 ( w) = 1F1[a; b; w] = 1 + where in this case b = 1, a = 1 λ − 2 4 and y2 ( w) = y1 log w + aw 1 2 − + 1!1! a 1 The boundary condition given by Equation 7.95 requires that the solution be finite, which eliminates the second solution. The resulting dimensionless concentration profile is given by ∞ θ(ζ, ξ) = ∑ An exp ( −λ n2ζ ) exp(−λ n ξ 2 / 2) 1F1[1 / 2 − λ n / 4;1; λ n ξ 2 ] n =1 The substitution of this dimensionless concentration profile into Equation 7.97 and solving the subsequent linear first-order differential equation produces the shell-side dimensionless concentration profile, θ S (ζ ): θ S (ζ ) = (λ 4 N Sh V 2 n ∞ ∑ Ane− λ V) + 4 N Sh n =1 n 2 1 λ F − n ;1; λ n 2 4 1 1 λ + 4 N ShV + 4 N ShVζ − exp ( −λ n2ζ ) + 1 exp − 2 Pe ( R L ) 2 n The Fourier coefficient, An is defined (using Equation 7.94) by An = e− λn / 2 [1 F1[1 / 2 − λ n / 4;1; λ n ] − (1 − λ n / 2)1F1[3 / 2 − λ n / 4; 2; λ n ]] λn 1 ∫0 ξ(1 − ξ 2 )e − λ n ξ (1 F1[1 / 2 − λ n / 4;1; λ n ])2 d ξ 2 n ≥1 316 Applied Mathematical Methods for Chemical Engineers The eigenvalues, λ n are defined (Equation 7.96) by (λ n 1F1[1 / 2 − λ n /4;1; λ n ] − 2λ n (1 / 2 − λ n / 4)[1 F1 3/2 − λ n /4; 2; λ n ]) 1 L2 Pe R 1 L2 λ n2 λ 2 + 4 N V Pe R n Sh = N Sh 1F1[1 / 2 − λ n /4;1; λ n ] 4 N ShV 2 2 + 2 2 1 − exp − ( λ n + 4 N ShV / Pe( L / R) ( λ n + 4 N ShV ) ( ) Where and −λ n ξ exp(−λ n ξ 2 / 2) 1F1[1 / 2 − λ n / 4;1; λ n ξ 2 ] dθ(ζ, ξ) ∞ 2 = ∑ An exp ( −λ nζ ) d 2 2 exp( / 2) ( [1 / 2 / 4;1; + −λ ξ − λ λ ξ ]) F dξ n n n 1 n =1 dξ 1 λ d 1 ( F1[1 / 2 − λ n / 4;1; λ n ξ 2 ]) = 2λ n ξ − n 1 F1 [ 3 / 2 − λ n / 4; 2; λ n ξ 2 ] 2 4 dξ 1 Now the quantity, CAb / CA0 can be determined using the bulk concentration, CAb given by 2π R ∫ ∫ C A ( L, r )νz (r )r dr dθ CAb = 0 0 2 π R ∫0 ∫0 νz (r )r dr dθ such that ∞ A e − λ n / 2 1F1[1 / 2 − λ n / 4;1; λ n ] CAb C − CASL = 1 + A0 4 N Sh ∑ n CA0 CA0 λ 2n + 4 N ShV n =1 λ 2 + 4 N ShV × exp− n −1 Pe( R / L )2 Typically, in the design of a dialysis system, one would like to know how much material would be removed for a given set of flow characteristics and system geometry. To address that question, the eigenvalue equation was solved using the Mathematica routine FindRoot to determine the first eigenvalue. Then, working in the same Mathematica notebook, the Fourier coefficient was calculated for each case of interest. An application of the results from the above model (CASL = 0 and the first term of the infinite series) was compared with published experimental data for the dialysis of beer. That comparison is displayed in Figure 7.7. Applications of Partial Differential Equations in Chemical Engineering 317 1 Model—Alcohol Expt. —Alcohol Model—Extract Expt. —Extract Fractional removal 0.8 0.6 0.4 0.2 0 50 100 200 300 400 500 600 Beer flow rate (mL/min) FIGURE 7.7 7.6 Alcohol and extract removal as a function of beer flow rate. SIMULTANEOUS DIFFUSION AND CHEMICAL REACTION There are numerous practical situations in which both diffusion and chemical reaction may be occurring, for instance, during low-pressure chemical vapor deposition (LPCVD), solid catalytic processes, or decoking a pyrolysis furnace, to name a few. Example 7.14 This example involves the derivation of the differential equation and boundary conditions for a process step that is integral to microelectronics processing. Consider the process of LPCVD in a horizontal cylindrical reactor heated from the outside [16]. That is, the reactor is surrounded by a furnace that supplies heat to the reacting contents. This heating strategy is the so-called hot wall process, and it is usually assumed that heats of reaction are small in comparison to the supplied heat. In this assumed isothermal process, thin circular disks called wafers are supported in a special wafer holder consisting of prearranged slots of equal distance apart so that the disks are located axisymmetric with the cylindrical tube. Material flows from one end of the reactor in the annulus created by the wafers and the reactor wall, but diffusion is anticipated to be the dominant mass transfer mechanism in the region bounded by any two wafers. Derive the mathematical model using the following assumptions: 1. Diffusion is the mass transfer mechanism in the region between any two wafers. 2. Gap between two wafers (inter-wafer region) is 2δ long. 3. Surface reaction dominates over homogeneous reaction. 318 Applied Mathematical Methods for Chemical Engineers 4. Azimuthal flow effects can be neglected. 5. Pseudo-binary mixture prevails. 6. Homogeneous gas phase reactions are negligible. Also solve the derived system of equations for the concentration profile as a function of r (radial) and z (along the axis). Solution Conservation of mass applied to the reactant species A in the inter-wafer region: 1 ∂ ∂ (rN Ar ) + N Az = 0 r ∂r ∂z (7.99) where NA is the molar flux relative to stationary coordinates [14] and is defined by N A = −CDAB∇X A + X A ( N A + N B ) (7.100) Then the molar flux in the r-direction for a dilute system is N Ar = − DAB ∂CA + X A ( N Ar + N Br ) ∂r (7.101) which consists of a diffusion part and a bulk flow part. For a diffusion-dominated mechanism, Equation 7.101 becomes N Ar = − DAB ∂CA ∂r (7.102) N Az = − DAB ∂CA ∂z (7.103) and the z-directed flux becomes Therefore, substitution of Equation 7.102 and Equation 7.103 into Equation 7.99 gives 1 ∂ ∂CA ∂ 2 CA =0 r + r ∂r ∂r ∂ z 2 (7.104) ∂CA (0, z ) =0 ∂r (7.105) subject to Applications of Partial Differential Equations in Chemical Engineering 319 that is, no concentration gradient exists along the axis of the cylinder. By symmetry, ∂CA (r ,0) =0 ∂x (7.67) (7.106) Also CA ( RW , z ) = CAb 0 < z < δ) (7.107) where RW is the radius of the wafer and CAb is the bulk concentration of species A, which varies along the length of the reactor. In addition, only half the inter-wafer region need be considered. The surface reaction can be taken as a first-order reaction; that is, − DAB ∂CA = kCA ∂z at z = δ (7.108) Therefore, the mathematical model for the outlined process is 1 ∂ ∂CA ∂ 2 CA =0 r + r ∂r ∂r ∂ z 2 ∂CA (0, z ) =0 ∂r ∂CA (r ,0) =0 ∂z CA ( RW , z ) = CAb , 0 < z < δ − DAB ∂CA = kCA ∂z at z = δ To solve this system of equations, the method of separation of variables will be employed. Note that this is a homogeneous problem that requires special handling. At this point, we need to broaden the definition of homogeneous to include a linear condition or equation that when satisfied by a particular function F is also satisfied by cF, where c is an arbitrary constant. This means that Equation 7.108 is homogeneous, but cannot be treated in the usual way. In this problem, Equation 7.107 plays the role of the initial condition. Let CA(r, z) = R(r) Z(z), then substitute into the differential equation to get 1 R′′ + R′ − λR = 0 r (7.109) Z ′′ + λR = 0 (7.110) and 320 Applied Mathematical Methods for Chemical Engineers where λ is the separation constant (three cases to be considered). The homogeneous boundary conditions become R′(0) = 0 Z ′(0) = 0 The other two nonhomogeneous boundary conditions will be dealt with Section 7.6. Case 1: λ = 0 Equation 7.109 solves to R(r ) = c2 + c3 r and R′(0) → ∞ unless c2 is choosen as zero. Further, Equation 7.110 becomes Z ′′ = (0) ⇒ Z ( z ) = m1z + m2 and Z ′(0) = 0 = m1 Therefore, the implication is CA (r , z ) = a constant. However, Equation 7.108 would be violated; therefore, on physical grounds, the case λ is zero must be rejected. Suppose λ > 0, say λ = α 2; then we get from Equation 7.109 1 R′′ + R′ − α 2 R = 0 r a modified Bessel’s equation [3,21]. This equation can be solved by comparing with Equation 3.80 to give R(r ) = c4 I 0 (αr ) + c5 K 0 (αr ) where I0(α r) is the zero-order modified Bessel function of the first kind and K0(α r) is the zero-order modified Bessel function of the second kind, which contains a logarithmic term. The constant c5 must be chosen as zero, as K0(·) and its derivative becomes unbounded as r approaches zero. Therefore, R(r ) = c4 I 0 (αr ) Applications of Partial Differential Equations in Chemical Engineering Also, for this λ , Equation 7.110 gives Z(z) = m3 cos α z + m4 sin α z and Z ′(0) = 0 = αm4 ⇒ m4 = 0 such that Z ( z ) = m3 cos αz Then for each n, we expect that CA, n (r , z ) = Rn Z n = An I 0 (α n r ) cos α n z and by the principle of superposition ∞ CA (r , z ) = ∑ An I 0 (α n r ) cos α n z n =1 which must now satisfy − DAB ∂CA = kCA ∂z at z = δ from which we get ∞ ∞ n =1 n =1 DAB ∑ An α n I 0 (r , z )sin α n δ = k ∑ An I 0 (α n r ) cos α n δ Then, for n ≥ 1, the α n are defined by α n tan α n δ = k DAB Now, Equation 7.107 will be useful in defining the quantity An, that is, ∞ CA ( RW , z ) = CAb = ∑ An I 0 (α n RW ) cos α n z n =1 which is recognizable as a generalized Fourier series. Therefore, for n ≥ 1, δ δ ∫0 CAb cos α n z d z = An I 0 (α n RW ) ∫0 cos2 α n z d z δ 1 1 = An I 0 (α n RW ) ∫ + cos 2α n z dz 0 2 2 δ 1 1 = An I 0 (α n RW ) + sin 2α n δ 2 2 2 α n 321 322 Applied Mathematical Methods for Chemical Engineers or 1 An = δ sin 2α n δ I 0 (α n RW ) + 4α n 2 (∫ C δ 0 Ab cos 2 α n z d z ) For the case λ < 0, say, λ = −β 2, β > 0, Equation 7.109 becomes R′′ + 1 R′ + β 2 R = 0 2 a Bessel differential equation [3,21], which can be solved by the same process of comparison used earlier. That is, compare this new R-equation with a 2 − γ 2c 2 2a − 1 y′′ − y′ + b 2c 2 x 2 c − 2 + y = 0 x x2 whose general solution is y = x a J γ (bx c ) + x a J − γ (bx c ) Then, by comparison to the R-equation 2a − 1 = −1 ⇒ a = 0 2c − 2 = 0 ⇒ c = 1 a − γ 2c 2 = 0 ⇒ γ = 0 2 b 2c 2 = β 2 ⇒ b = β such that the R-equation has general solution R(r ) = c6 J 0 (βr ) + C7Y0 (βr ) where J0(β r) is the zero-order Bessel function of the first kind and Y0(β r) is the zeroorder Bessel function of the second kind, which contains a logarithmic term. As a result of the logarithmic term, Y0(β r) and its derivative become unbounded as their argument approaches zero. Therefore, the constant c7 must be chosen as zero, such that R(r ) = c6 J 0 (βr ) Also, Equation 7.110 gives the general solution Z ( z ) = m5eβz + m6 e −βz Applications of Partial Differential Equations in Chemical Engineering 323 and Z ′(0) = 0 = βm5 − βm6 ⇒ m5 = m6 such that Z ( z ) = m5 (eβz + e −βz ) = 2m5 cosh βz Then, for each n, all we can conclude at this point is CA, n = H n J 0 (β n r ) cosh β n z where the β n are to be defined. Then, by the principle of superposition, ∞ CA (r , z ) = ∑ H n J 0 (β n r ) cosh β n z n =1 and ∂CA (r , δ) ∞ = ∑ H nβ n J 0 (β n r ) cosh β n δ ∂z n =1 Therefore, Equation 7.108 becomes ∞ ∞ n =1 n =1 − DAB ∑ H nβ n J 0 (β n r ) cosh β n δ = k ∑ H n J 0 (β n r ) cosh β n δ or β n tanh β n δ = − k , n ≥1 DAB which defines β n. However, the argument of the hyperbolic tangent is positive, which means that the left-hand side is always positive, whereas the right-hand side is negative. This is impossible and therefore λ < 0 has to be rejected. Finally, the solution to this model is given as ∞ CA (r , z ) = ∑ An I 0 (α n r ) cos α n z n =1 where the Fourier coefficient is given by 1 An = δ sin 2α n δ I 0 (α n RW ) + 4α n 2 (∫ C δ 0 Ab cos α n z d z ) 324 Applied Mathematical Methods for Chemical Engineers and the eigenvalues are defined by α n tan α n δ = k DAB for n ≥ 1. Example 7.15 This example considers the steady-state absorption of a gas that is sparingly soluble in an agitated liquid [42] accompanied by an irreversible reaction. If the process is described by ∂C ∂2 C = D 2 − kC ∂t ∂x (7.111) C = Ci at x = 0, t > 0 (7.112) C = C0 at t = 0, x > 0 (7.113) C = C0 e − kt at x = ∞, t ≥ 0 (7.114) C ( x , t ) = u ( x , t ) e − kt (7.115) ∂u ∂2 u =D 2 ∂t ∂x (7.116) subject to the conditions determine C ( x , t ) . Solution Letting reduces Equation 7.111 to which can be solved by the Combination of Variables method. Substituting u ( x , t ) = f ( η) , η = x 4 Dt (7.117) into Equation 7.116 such that du ( x , t ) = df ( η) and applying the chain rule to Equation 7.118 results in ∂u ∂u df dx + dt = dη ∂x ∂t dη (7.118) 325 Applications of Partial Differential Equations in Chemical Engineering Recalling the total derivative of η ( x , t ) is dη = ∂η ∂η dx + dt ∂x ∂t results in ∂η ∂u ∂η ∂u dx + dt = f ′ ( η) dx + dt ∂x ∂x ∂t ∂t (7.119) Upon equating like coefficients of both sides, dx : ∂u ∂η = f ′ ( η) ∂x ∂x (7.120) dt : ∂u ∂η = f ′ ( η) ∂t ∂t (7.121) From Equation 7.120, we get ∂2 u ∂η = f ′′ ( η) ∂x ∂x2 since 2 (7.122) ∂η is independent of x . Substituting Equations 7.120 through 7.122 results in ∂x f ′′ ( η) + 2 ηf ′ ( η) = 0 subject to u ( 0, t ) = Ci = f ( 0 ) u ( x ,0 ) = C0 = f ( ∞ ) u ( ∞, t ) = C = f ( ∞ ) 0 The general solution to Equation 7.123 is f ( η) = m1 ∫ e − η d η+ m2 2 Applying the conditions results in f ( η) = ∞ 2 (C0 − Ci ) ∫ e − η2 d η+ Ci π 0 = C0 erf ( η) + Ci erfc ( η) x x C ( x , t ) = C0 e − kt erf + Ci e − kt erfc 4 Dt 4 Dt (7.123) 326 Applied Mathematical Methods for Chemical Engineers Example 7.16 The following system of equations have been used in the modeling of biodegradation processes: 1 ∂CA ∂ 2 CA 2 ∂CA 1 − ε s K = + − CA ∂r 2 Dse ∂t r ∂r ε s Dse (7.124) CA (0, t ) is finite (7.125) ∂CA k = (C A − Cb ) at r = R Dse ∂r (7.126) CA (r ,0) = CA0 (7.127) Determine the concentration profile for species A. Solution To simplify the problem let CA (r , t ) = u(r , t ) + F (r ) (7.128) where u(r, t) represents the transient solution and F(r) represents the steady-state solution. Further, we require that 2 F ′′ + F ′ − βF = 0 r (7.129) F is to be finite at r = 0 (7.130) dF ( R) k m = ( F ( R) − C b ) dr Dse (7.131) and Then Equations 7.124 through 7.126 reduce to 1 ∂u ∂ 2 u 2 ∂u = + − βu Dse ∂t ∂r 2 r ∂r (7.132) Applications of Partial Differential Equations in Chemical Engineering 327 subject to u(0, t ) is finite (7.133) ∂u k m u at r = R = ∂r Dse (7.134) where β= 1 − εs K ε s Dse As a further simplification to Equations 7.132 through 7.134, let u(r , t ) = e − Dse βt w(r , t ) (7.135) such that Equations 7.132 through 7.134 become 1 ∂ w ∂2 w 2 ∂w = 2 + Dse ∂t ∂r r ∂r (7.136) subject to w(0, t ) is finite (7.137) ∂ w km = w at r = R ∂r Dse (7.138) Then, by separation of variables, that is, for w = f(r)T(t), we get 1 T ′ f ′′ 2 f ′ = + =λ Dse T f r f For the case λ = 0, we get f(r) = a constant, which implies that zero is an eigenvalue for this problem. This solution must be rejected, however, since Equation 7.138 could be satisfied only for the particular case of w being identically zero. The case λ > 0, say, λ = α 2, produces T (t ) = c1e λDse t and f ′′ + 2 f ′ − α2 f = 0 r 328 Applied Mathematical Methods for Chemical Engineers which has f (r )c2r −1/ 2 J1/ 2 (iαr ) + c2r −1/ 2 J −1/ 2 (iαr ) as its general solution. However, w = f(r)T(t) is required to remain finite, even as t gets larger and larger. This condition cannot be met for λ > 0. For λ < 0, say, λ = −η 2, then we have f ′′ + 2 f ′ + η2 f = 0 r whose general solution is f (r ) = c4 J1/ 2 ( ηr ) J ( ηr ) + c5 −1/ 2 r r Again, for a bounded solution, c5 must be chosen as zero, as J −1/ 2 ( ηr ) r becomes unbounded as r approaches zero. However, for a nontrivial solution to exist, c4 ≠ 0. Recall now that J1/ 2 ( ηr ) = 2 sin ηr ηπr J1/ 2 ( ηr ) = r 2 sin ηr ηπ r such that Therefore, the solution up to now is ∞ w(r , t ) = ∑ Ai e − η Dse t i =1 2 2 sin ηr ηπ r (7.139) with the use of the superposition principle. That is, Equation 7.139 solves Equations 7.136 and 7.137. Equation 7.138 will define the eigenvalues for this system in the following way: ∂w ∞ 2 η cos ηr sin ηr 2 = ∑ Ai e − η Dse t − 2 r r ∂r n = 1 ηπ 329 Applications of Partial Differential Equations in Chemical Engineering then, at r = R η cos ηR sin ηR k m sin ηR − = , i ≥1 R R2 Dse R or k R ηi R = m + 1 tan ηi R; i ≥ 1 Dse (7.140) u(r ,0) = w(r ,0) = CA0 − F (r ) (7.141) The initial condition is where F(r) is the solution to Equations 7.129 through 7.131. That is, let F (r ) = q(r ) q′(r ) − q(r ) q′′(r ) 2 2q(r ) , then F ′(r ) = , and F ′′(r ) = − q′ + 3 r r r2 r r r Then these substitutions reduce Equation 7.129 to q′′(r ) − βq(r ) = 0 whose general solution is q(r ) = k1 cosh βr + k 2 sinh βr Therefore F (r ) = k1 cosh βr k 2 sinh βr + r r However, k1 must be chosen as zero, as F(0) is required to be finite. Also, Equation 7.131 gives k2 = R 2C b k m R k m Dse 1 + sinh β R − R β cosh β R Dse such that F (r ) = R 2Cb k m sinh βr R k m Dse 1 + sinh β R − R β cosh β R r Dse (7.142) 330 Applied Mathematical Methods for Chemical Engineers Finally, ∞ u(r , t ) = e − Dse βt w(r , t ) = ∑ Ai e − ( Dse β+ ηi Dse )t 2 i =1 2 sin ηi π ηi π r or ∞ CA (r , t ) = F (r ) + ∑ Ai e − ( Dse β+ ηi Dse )t 2 i =1 2 sin ηi π ηi π r then at t = 0, ∞ CA (r ,0) = F (r ) + ∑ Ai i =1 2 sin ηi π = CA0 ηi π r or ∞ CA0 − F (r ) = ∑ Ai i =1 2 sin ηi π ηi π r defines Ai. That is, for i ≥ 1, R ∫0 (CA0 − F (r ))r 2 2 sin ηi r 2 R 2 sin ηi r r dr = Ai dr r ηi π ∫0 r where r2 is the weighting function. 7.7 SIMULTANEOUS DIFFUSION, CONVECTION, AND CHEMICAL REACTION This type of phenomenon is typified by flowing systems, in which both surface and homogeneous chemical reactions are occurring. However, sometimes the mathematics associated with these phenomena may be too nontractable and a simpler model may be sought. Simplifying assumptions that are used to reduce a model would still require some measure of justification, at least in theory. In the very next example, an approach is taken that can be used to determine the effect of diffusion and surface reaction on a chemical vapor deposition process. Example 7.17 In multiwafer LPCVD systems, the convection contribution to the transport of reactants to surface sites in the interwafer region is usually neglected [22]. Consider laminar flow with a velocity profile in the annulus, between the wafers and the reactor walls. This velocity profile is assumed to be constant and contains the ratio of wafer Applications of Partial Differential Equations in Chemical Engineering 331 radius to that of the reactor, denoted by κ . Then in this case, an average velocity, vavg can be described by [14] vavg = ( P0 − PL ) R 2 1 − κ 4 1− κ2 − 2 8 µL ln(1 / κ ) 1 − κ (7.143) where R is the reactor radius. A mathematical description of the process occurring in this annulus region is given as vavg 1 ∂ ∂CA ∂CA = DAB r − k bCA ∂z r ∂r ∂r (7.144) subject to at z = 0 (7.145) CA is finite at r = 0 (7.146) CA = CA0 and − DAB ∂CA = k sC A ∂r at r = R (7.147) The differential equation is the equation of continuity of species A for constant density ρ and binary diffusivity DAB [14]. Also, the r and θ components of velocity are neglected. Develop an equation that can be used to justify the assumption of transport by diffusion in the interwafer region of a multiwafer LPCVD reactor. Solution Equations 7.144 through 7.147 can be recast in dimensionless form as ∂ F ∂2 F 1 ∂ F = + − αF ∂ζ ∂ξ 2 ξ ∂ξ (7.148) F = 1 at ζ = 0 (7.149) F is finite at ξ = 0 (7.150) ∂F = βF at ξ = 1 ∂ξ (7.151) subject to − 332 Applied Mathematical Methods for Chemical Engineers where the dimensionless quantities F ( ξ , ζ) = CA r D z R2 kb kR ; ξ = ; ζ = AB 2 ; α = ; β= s CA0 R vavg R DAB DAB were used. Equation 7.148 is nonhomogeneous and can be made homogeneous by the substitution w(ξ, ζ) = Feαζ (7.152) Therefore, the system ∂ w ∂2 w 1 ∂ w = + ∂ζ ∂ξ 2 ξ ∂ξ w = 1 at ζ = 0 w is finite at ξ = 0 − ∂w = βw at ξ = 1 ∂ξ can now be solved using the method of separation of variables to give ∞ F = ∑ Bn J 0 (λ n ξ)e − ( λ n +α )ζ 2 n =1 (7.153) where the eigenvalues λ n are defined by β J 0 (λ n ) = λ n J1 (λ n ), n ≥ 1 (7.154) and Bn = 2 J1 (λ n ) λ n [( J1 (λ n ))2 + ( J 0 (λ n ))2 ] (7.155) The axial concentration profile can be predicted from ∞ ∂F 2 = − ∑ Bn λ 2n J 0 (λ n ξ) + αBn J 0 (λ n ξ)e − ( λ n +α )ζ ∂ζ n =1 = −λ n2 F − αF { (7.156) The left-hand side of the Equation 7.156 can now be written in dimensional form as dCA /dt. That is, D dC A = − λ n2 AB + k b CA R2 dt (7.157) D dpA = − λ n2 AB + k b pA R2 dt (7.158) or by the ideal gas law where t = z/vavg has been used. Applications of Partial Differential Equations in Chemical Engineering 333 Equation 7.157 or 7.158 can be used to argue the effect of diffusion and surface reaction on the overall deposition process. Notice that the eigenvalues, λ n, are defined in relation to the overall surface rate coefficient ks, and the term ( λ 2n DAB /R 2 ) grows as n increases, whereas kb tends to remain relatively constant. Example 7.18 This is an example of rapid chemical reaction in the laminar boundary layer on a flat plate. Consider the steady-state dissolution of a slightly soluble solid in a flowing dilute solution, and suppose the solid is acidic whereas the flowing solution is basic [23]. The geometry under consideration is a flat plate consisting of acid, located at zero incidence to the flow. The process may be assumed isothermal, and the fluid properties are to be treated as independent of position. Species A dissolves from the plate and diffuses into the reaction zone, whereas species B diffuses into the reaction zone from the main body of the solution. In the negligibly thin reaction region, the two substances undergo a fast irreversible reaction: aA + bB → products (7.159) In the region between the reaction zone and the surface, the diffusional process can be described by u ∂CA ∂C ∂2 CA + v A = DA ∂x ∂y ∂ y2 (7.160) u ∂CB ∂C ∂2 CB + v B = DB ∂x ∂y ∂ y2 (7.161) whereas describes the diffusional process in the region outside the reaction zone. DA and DB represent the diffusivities of species A and B in the fluid, whereas CA and CB are molar concentrations. The quantities u and v are respectively the x and y components of the boundary layer velocity and satisfy the continuity equation [14, 24] ∂u ∂ v + =0 ∂x ∂ y (7.162) and the x-component equation of motion [14, 24] u ∂u ∂u ∂2 u +v =γ 2 ∂x ∂y ∂y (7.163) Then, the velocity components [24] are given by u= ∂Ψ ∂Ψ ∂η = = U ∞ f ′( η) ∂ y ∂η ∂ y (7.164) 334 Applied Mathematical Methods for Chemical Engineers and v=− ∂Ψ 1 νU ∞ = ( ηf ′ − f ) ∂x 2 x (7.165) with U∞ being the free stream velocity (far away from the surface) and η the combined variable defined as η= y U∞ νx (7.166) where Y is the kinematic viscosity. The dependent variable Ψ is the stream function [14]. Upon substitution of Equations 7.164 through 7.166 into Equation 7.163, there results ff ′′ + 2 f ′′′ = 0 (7.167) the so-called Blasius equation [24]. Now, one would like to know the rate at which dissolution takes place as a function of distance from the leading edge of the plate and the concentration of reactant in the mainstream (far from the surface). Solution Let CA C , cB = B CAS CBM cA = (7.168) where CAS is the surface concentration (moles/L) of A, and CBM is the mainstream concentration (moles/L) of B. Then, for cA ( x , y) = f ( η) (7.169) ∂c A ∂c df dx + A dy = dη ∂x ∂y dη (7.170) η = η( x , y) (7.171) but Therefore, dη = ∂η ∂η dx + dy ∂x ∂y Substituting Equation 7.172 into Equation 7.170 gives df ∂η ∂c A ∂c ∂η dx + A dy = dx + dy dη ∂ x ∂x ∂y ∂ y (7.172) Applications of Partial Differential Equations in Chemical Engineering Equating coefficients of dx and dy, dx : ∂cA df ∂η ∂η = = f ′( η) ∂ x dη ∂ x ∂x dy : ∂cA df ∂η ∂η = = f ′( η) ∂ y dη ∂ y ∂y and By letting H ( x , y) = ∂cA ∂η and φ( η, y) = f ′( η) ∂y ∂y suggest that dH ( x , y) = dφ( η, y) such that ∂H ∂H ∂φ ∂φ dx + dy = dη + dy ∂x ∂y ∂η ∂y = ∂φ ∂η ∂η ∂φ dx + dy + dy ∂η ∂ x ∂ y ∂ y and equating coefficients give ∂ H ∂φ ∂η = ∂ x ∂η ∂ x ∂ H ∂φ ∂η ∂φ = + dy : ∂ y ∂η ∂ y ∂ y dx : that is, 2 ∂η ∂ ∂η ∂ H ∂ 2 cA ∂2 η + η + η = = f η f f ) ( ) 2 ( ) ( ′ ′ ′′ ∂y ∂η ∂ y ∂y ∂ y2 ∂ y2 But ∂η U∞ = , ∂y νx ∂2 η ∂ U∞ =0= 2 ∂y ∂η νx and ∂η η =− ∂x 2x 335 336 Applied Mathematical Methods for Chemical Engineers Therefore, Equation 7.160 becomes − uηf ′( η) U U + v ∞ f ′( η) = 2 DA ∞ f ′′( η) 2x νx νx Then, with Equations 7.164 and 7.165, − U ∞ η[ f ′( η)]2 1 U ∞ ν 1/ 2 U ∞ 1/ 2 U∞ f ′′( η) + f ′( η) ( ηf ′ − f ) = 2 DA 2x 2 x νx νx which reduces to − ff ′ D = 2 A f ′′ 2 ν That is, 2 d 2cA f dcA + =0 SA dη2 2 dη (7.173) Equation 7.161 can be similarly transformed to 2 d 2cB f dcB + =0 SB dη2 2 dη (7.174) The quantities SA and SB are Schmidt numbers (Y /D) for the respective species. Equations 7.173 and 7.174 are subject to the conditions cA = 1 at η = 0 cA = cB = 0 at η = ηR cB = 1 at η = ∞ and cB = 1 at η = ∞. Substituting Equation 7.167 into Equation 7.173 and integrating results in η ∫ ( f ′′)SA / 2 d η , cA = 1 − η0 ∫0 ( f ′′)SA / 2 d η R 0 < η < ηR (7.175) , ηR < η < ∞ (7.176) Similarly, Equation 7.174 gives η ∫η ( f ′′)S cB = 1 − ∞ ∫η ( f ′′)S B/2 dη B/2 dη R R 337 Applications of Partial Differential Equations in Chemical Engineering Example 7.19 This is an example of rapid chemical reaction in the laminar boundary layer on a flat plate. Reconsider the previous example with nonisothermal condition and constant fluid properties. In this case, assume that the surface temperature TS and the mainstream temperature TM are constant. This implies that the temperature of the reaction zone TR is also constant. Under these conditions, determine the temperature profile for species A. Solution The equation of energy [14] can be modified to u ∂TA ∂T ∂ 2 TA +v A =α ∂x ∂y ∂ y2 (7.177) for the problem under consideration. The temperature of species A can be made dimensionless by TR − TA TR − TS tA = (7.178) Equation 7.177 becomes ∂t A ∂t ∂2 t + v A = α 2A ∂x ∂y ∂y u (7.179) Then, following the procedure of Example 7.18, Equation 7.179 can be reduced to 2 d 2 t A f dt A + =0 Pr dη2 2 dη (7.180) subject to the conditions t A = 1 at η = 0 (7.181) t A = 0 at η = ηR where the quantity Pr is the Prandtl number (cp ρ Y / k). Equations 7.180 and 7.181 solve to T − TA tA = R = 1− TR − TS η ∫0 ( f ′′)P η ∫0 ( f ′′)P r/2 R r/2 dη dη , 0 < η < ηR 338 Applied Mathematical Methods for Chemical Engineers Example 7.20 In this example of carrier-facilitated transport in membrane separation, consider a bundle of parallel hollow fibers through which a fluid flows [25]. One can examine the concentration profile in a single fiber and then predict the performance of the separation device. Further, consider a fluid (Newtonian) from which the solute is to be extracted while entering the reactive section of the hollow fiber in fully developed, one-dimensional laminar flow. As shown in Figure 7.8, at z equal zero, the fluid contacts the reactive membrane. At such a location the solute concentration CA is uniform and has the value CA0. As the fluid flows further into the reactive section, the solute diffuses through the membrane by carrier-facilitated transport and emerges into the second fluid (shell side), which surrounds the hollow fiber. CA is negligible axially on the shell side, as the incoming shell-side fluid is devoid of the solute, and is at a much higher flow rate. An alternative to this condition is a constant solute concentration on the shell side. Following the development in the literature [26], a reversible equilibrium reaction of the form k1 A + B = AB (7.182) k2 occurs inside the membrane, where A is the solute, B is the carrier, and AB is the solute–carrier complex. The quantities k1 and k 2 are the forward and reverse rate coefficients. The equation of continuity for the solute (species A) in this system is 1 ∂ ∂CA r 2 ∂C 2 νavg 1 − 2 A = DA r R ∂z r ∂r ∂r (7.183) Shell side Reactive section R r Impermeable wall z Fluid side A + B = AB Shell side Inlet plane FIGURE 7.8 Hollow fiber with reactive walls containing the supported carrier. Applications of Partial Differential Equations in Chemical Engineering 339 subject to CA = CA0 − DA at z = 0 (7.184) ∂CA = 0 at r = 0 ∂r (7.185) ∂CA = k w Sf (CA ) at r = R ∂r (7.186) where DA is the solute diffusivity in the fluid, kw is the membrane mass transfer coefficient for the solute, and S is the shape factor based on the inside radius, R, of the hollow fiber. This shape factor [25,27] behaves as a correction factor for geometry. It permits the translation of experimental data derived in flat membranes to nonflat membrane configurations for the same experimental conditions. Equation 7.186 is the boundary condition that incorporates the relevant information given in Equation 7.182 as to the rate of disappearance of species A. Usually, f(CA) is a quotient of two polynomials, and is typically of the form [25] f (CA ) = DB′ CT K eq DA′ (1 + K eqCA H ) CA (7.187) where the prime indicates diffusivity of that substance in the membrane, Keq is k1/k2. The equilibrium (thermodynamic) distribution coefficient, H, is defined as a proportionality factor relating the concentration of species A in the two phases (membrane and fluid). The quantity CT accounts for the concentration of species B and AB. Determine the concentration profile for species A. Solution Introducing the dimensionless variables C= CA r zDA ; ξ= ; ζ= CA0 R νavg R 2 into Equations 7.183 through 7.185, we get 2(1 − ξ 2 ) ∂C ∂ 2 C 1 ∂C = + ∂ζ ∂ξ 2 ξ ∂ξ (7.188) C = 1 at ζ = 0 (7.189) ∂C = 0 at ξ = 0 ∂ξ (7.190) Then, recognizing that f(CA) can be recast as an infinite series in dimensionless form to be 340 Applied Mathematical Methods for Chemical Engineers ∞ (1 + α )C + α ∑ (−1)n ε nC n +1 n =1 Equation 7.186 becomes − ∂C = w[(1 + α )C − εαC 2 + ε 2αC 3 − ε 3αC 4 + ] at ξ = 1 ∂ξ (7.191) where w is defined as w= Rk w S DA (7.192) a dimensionless group called the wall Sherwood number. The dimensionless quantities α and β are defined in the literature [25]. In this work ε is less than one, and can be associated with the quantity β in two cases. In one case, let ε be β , whereas in another let ε be the reciprocal of β (for β > 1). The system consisting of Equations 7.188 through 7.189 and Equation 7.191 can be recast by using the following form of the dimensionless concentration: C = F0 + εF1 + ε 2 F2 + (7.193) such that ∂C ∂ F0 ∂F ∂F = + ε 1 + ε2 2 + ∂ζ ∂ζ ∂ζ ∂ζ ∂C ∂ F0 ∂ F1 ∂ F = +ε + ε2 2 + ∂ξ ∂ξ ∂ξ ∂ξ ∂ 2 C ∂ 2 F0 ∂2 F ∂ 2 F2 = + ε 21 + ε 2 + 2 2 ∂ξ ∂ξ ∂ξ ∂ξ 2 which then results in ∂F ∂F ∂F 2(1 − ξ 2 ) 0 + ε 1 + ε 2 2 + ∂ζ ∂ζ ∂ζ ∂ 2 F0 ∂2 F 1 ∂F ∂F + ε 21 + 0 + ε 1 + 2 ∂ξ ∂ξ ξ ∂ξ ∂ξ (7.194) F0 + εF1 + ε 2 F2 + = 1 at ζ = 0 (7.195) ∂ F0 ∂F ∂F + ε 1 + ε 2 2 + = 0 at ξ = 0 ∂ξ ∂ξ ∂ξ (7.196) ∂F ∂F − 0 + ε 1 + = w{(1 + α ) F0 + ε((1 + α ) F1 − αF02 ) + O(ε 2 )} ∂ξ ∂ξ (7.197) = Applications of Partial Differential Equations in Chemical Engineering 341 Equating coefficients of like powers of ε gives ε 0 : 2(1 − ξ 2 ) − (7.199) ∂ F0 = 0 at ξ = 0 ∂ξ (7.200) ∂ F0 = w(1 + α ) F0 ∂ξ at ξ = 1 (7.201) ∂ F1 ∂ 2 F1 1 ∂ F1 = + ∂ζ ∂ξ 2 ξ ∂ξ (7.202) F1 = 0 at ζ = 0 (7.203) ∂ F1 = 0 at ξ = 0 ∂ξ (7.204) ∂ F1 = w[(1 + α ) F1 − αF02 ] at ξ = 1 ∂ξ ε 2 : 2(1 − ξ 2 ) − (7.198) F0 = 1 at ζ = 0 ε : 2(1 − ξ 2 ) − ∂ F0 ∂ 2 F0 1 ∂ F0 = + ∂ζ ∂ξ 2 ξ ∂ξ ∂ F2 ∂ 2 F2 1 ∂ F2 = + ∂ζ ∂ξ 2 ξ ∂ξ (7.205) (7.206) F2 = 0 at ζ = 0 (7.207) ∂ F2 = 0 at ξ = 0 ∂ξ (7.208) ∂ F2 = w[(1 + α ) F2 − 2αF0 F1 + αF03 ] at ξ = 1 ∂ξ (7.209) The system of Equations 7.198 through 7.201 is linear, and can be solved using separation of variables to give ∞ λ F0 = ∑ Bn exp − n (λ nζ + ξ 2 ) 2 n=0 1 λ F − n ; 1; λ n ξ 2 2 4 1 1 (7.210) 342 Applied Mathematical Methods for Chemical Engineers where the λ n are defined by 1 λ {λ n − w(1 + α )} 1F1 − n ; 1; λ n 2 4 1 λ 3 λ −2λ n − n 1F1 − n ; 2; λ n = 0 2 4 2 4 (7.211) and Bn by { − λn 2 ξ 2 } 1 λ F − n ; 1; λ n ξ 2 d ξ 2 4 , n≥0 Bn = 1 λn 2 1 2 2 2 ∫0 ξ(1 − ξ ) exp{−λ n ξ } 1F1 2 − 4 ; 1; λ n ξ dξ 1 ∫0 ξ(1 − ξ) exp 1 1 (7.212) where 1 λ F − n ;1; λ n ξ 2 2 4 1 1 is one solution of the confluent hypergeometric equation or Kummer’s equation [13]. Values of both λ n and Bn can be derived as described in the literature [28], or by employing standard software packages, such as Mathematica [20,29]. The system described by Equations 7.202 through 7.205 is also linear, but separation of variables is not the appropriate solution technique. Here Laplace transform is more suitable. That is, by transforming the ζ variable while treating ξ as a parameter, one gets Kummer’s equation as follows: ∞ Let L{F1 (ζ, ξ)} ≡ ∫ e − sζ F1 (ζ, ξ) dζ = F1 (s, ξ) 0 (7.213) Then Equations 7.202 and 7.203 become d dF1 − 2sξ(1 − ξ 2 ) F1 = 0 ξ dξ dξ (7.214) where s is a complex number. Solving Equation 7.214 and applying the transformed Equation 7.204 results in 1 i s 2s 2 ; 1; i 2s ξ 2 F1 (s, ξ) = η(s) exp − i ξ 1F1 − 2 2 2 2 and one can see that replacing i 2s with λ results in 1 λ F1 − n ;1; λ n ξ 2 2 4 (7.215) 343 Applications of Partial Differential Equations in Chemical Engineering the identical solution of Kummer’s equation previously obtained, where η (s) is defined by η(s) = { } −αF02 exp i s 2 [i 2s − w(1 + α)] 1F1 12 − i2 22s ;1; i 2s 1 i s 3 i 2s ; 2; i 2s − 2i 2s − 1 F1 − 2 2 2 2 2 2 following the Laplace transformation of Equation 7.205. Here the quantity F02 represents the Laplace transform of F02 , where F0 is the solution of Equations 7.198 through 7.201. Equation 7.215 can now be expressed in terms of λ as F1 (s, ξ) = −αwF02 exp{λ /2(1 − ξ 2 )} 1F1 12 − λ4 ; 1; λξ 2 [λ − w(1 + α )] 1F1 12 − λ4 ; 1; λ − 2λ ( 12 − λ4 ) 1F1 32 − λ4 ; 2; λ (7.216) where i 2s is replaced by λ . Equation 7.216 can be inverted with the use of the residue theorem [30]. That is, ∞ p(sn ) exp{snζ} q n = 0 ′ ( sn ) L−1{F1 (s, ξ)} = ∑ (7.217) where p(sn) and q’(sn) are the numerator and denominator, respectively, of Equation 7.216. This latter example demonstrates that several mathematical techniques may be required to achieve an analytical solution to a complicated transport model. In addition, the regular perturbation technique is used to reduce the problem to a set of linear problems that we know how to solve. 7.8 VISCOUS FLOW Example 7.21 A fluid of constant density (ρ ) and viscosity (μ) is contained in a very long horizontal pipe of length L and radius R [14, 35]. The fluid is at rest initially. At t = 0, a pressure gradient p0 − pL L is impressed on the system. Determine how the velocity profiles change with time. 344 Applied Mathematical Methods for Chemical Engineers Solution Cylindrical coordinates are convenient. Assumptions: 1. Both the r and θ -component of velocity are zero. 2. vz = vz(r, t). Problem Setup: The equations of motion and continuity [14] are combined to give ρ 1 ∂ ∂ vz ∂ vz p0 − pL = +µ r L r ∂r ∂r ∂t subject to vz (r ,0) = 0, 0 ≤ r ≤ R vz (0, t ) is finite, t > 0 vz ( R, t ) = 0, t > 0 Then it is convenient to introduce the following dimensionless variables: φ(ξ, τ) = vz µt r ; ξ= ; τ= 2 ( p0 − pL ) R 2 /4µL ρR R Substitution of the dimensionless variables into the differential equation gives 1 ∂ ∂φ ∂φ = 4+ ξ ∂τ ξ ∂ξ ∂ξ and the conditions become φ(ξ,0) = 0 φ(1, τ) = 0 and φ(0, τ) is finite This is a case in which the differential equation is nonhomogeneous. However, the fact that the system is expected to reach a steady state as τ → ∞ can be used to reduce the differential equation to a homogeneous one. That is, assume a solution of the dimensionless system to be φ(ξ, τ) = φ∞ − w(ξ, τ) where ɸ ∞ is the steady-state solution satisfying 0= 4+ 1 d dφ ∞ ξ ξ dξ dξ φ∞ (1) = 0 and φ∞ (0) is finite Applications of Partial Differential Equations in Chemical Engineering 345 and w(ξ , τ ) is the transient solution satisfying ∂w 1 ∂ ∂w = ξ ∂τ ξ ∂ξ ∂ξ w(ξ,0) = φ∞ w(1, τ) = 0 w(0, τ) is finite Further, assuming a solution of the form w(ξ, τ) = Z (ξ)T (τ) gives T ′ Z ′′ 1 Z ′ = + = −α 2 T Z ξ Z or T′ 2 = −α 2 ⇒ T (τ) = c1e −α τ T and Z ′′ + 1 Z ′ + α2 Z = 0 ξ subject to Z (0) is finite and Z ( I ) = 0 which is a singular Sturm–Liouville problem involving Bessel’s differential equation. As in previous examples, the general solution to this differential equation is Z (ξ) = c2 J 0 (αξ) + c3Y0 (αξ) where J0(·) and Y0(·) are zero-order Bessel functions. Applying the boundary conditions, we see that the constant c3 must be chosen as zero, as Y0(α ξ ) becomes unbounded as ξ → 0 and Z(ξ ) is to be finite. The second boundary condition gives J 0 (α ) = 0 for nontrivial solutions to exist. Further, as J0(α ) crosses the α -axis infinitely many times, then for each n Z n = c2, n J 0 (α n ξ) 346 Applied Mathematical Methods for Chemical Engineers is the eigenfunction corresponding to the eigenvalue α n satisfying J 0 (α n ) = 0 Also for each n wn (ξ, τ) = Bn e −α n τ J 0 (α n ξ) 2 where the constants c1 and c2,n are combined as Bn. Then, by the principle of superposition ∞ wn (ξ, τ) = ∑ Bn e −α n τ J 0 (α n ξ) 2 n =1 satisfies the differential equation and the boundary conditions. Application of the initial condition gives ∞ (1 − ξ 2 ) = ∑ Bn J 0 (α n ξ) n =1 which is a generalized Fourier series. Then the Fourier coefficient can be determined by (Sturm–Liouville theory of Chapter 4) 1 1 ∫0 ξ(1 − ξ 2 ) J0 (α n ξ) dξ = Bn ∫0 ξ[ J0 (α n ξ)]2 dξ, n ≥1 Here, the weight function is ξ , for the case of a singular Sturm–Liouville problem as discussed in Chapter 4. Then with the aid of integral tables for Bessel functions [3,21], we get Bn = 8 α 3n J1 (α n ) for n ≥ 1 Therefore ∞ J 0 (α n ξ) −α 2n τ e 3 α n = 1 n J1 (α n ) φ(ξ, τ) = (1 − ξ 2 ) − 8∑ is the solution. An alternative to the steady-state hypothesis is to solve the dimensionless problem using Laplace transform. That is, reconsider 1 ∂ ∂φ ∂φ = 4+ ξ ∂τ ξ ∂ξ ∂ξ Applications of Partial Differential Equations in Chemical Engineering 347 and the conditions φ(ξ,0) = 0 φ(1, τ) = 0 φ(0, τ) is finite Then let ∞ L{φ(ξ, τ)} = u(ξ, s) = ∫ φ(ξ, τ)e − sτ d τ 0 The differential equation transforms to d 2u 1 du −4 + − su = 2 ds s ξ dξ subject to {L{φ(0, τ)} = u(0, s) L{φ(0, τ)} = u(1, s) = 0} The general solution to the nonhomogeneous differential equation is ( ) ( ) u(ξ, s) = c1 J 0 i s ξ + c2Y0 i s ξ + 4 s2 As ɸ (0, τ ) is finite, then its Laplace transform is also expected to be finite. This means that c2 must be chosen as zero, since Y0 (i s ξ) → ∞ as ξ → 0. The second boundary condition gives c1 = −4 s 2 J0 i s ( ) Therefore u(ξ , s ) = ( )+ 4 ( ) s −4 J 0 i s ξ s 2 J0 i s 2 348 Applied Mathematical Methods for Chemical Engineers The inverse transform for u(ξ ,s) can be located in a table of Laplace transforms, for example [3], ( ) ( ) 2 ∞ J 0 i s ξ 1 e − λ n τ J 0 (λ n ξ ) L−1 = (ξ 2 − 1) + τ + 2∑ 3 2 n = 1 λ n J1 (λ n ) s J 0 i s 4 where λ 1, λ 2, … are the positive roots of J0 (λ ) = 0. Therefore ∞ e − λ n τ J 0 (λ n ξ ) 3 n = 1 λ n J1 (λ n ) 2 φ(ξ, τ) = (1 − ξ 2 ) − 8∑ Example 7.22 Consider now the problem of tangential Newtonian flow in annuli. Suppose one is interested in studying the velocity profiles of an isothermal, incompressible viscous fluid in the annular space between two cylinders, with either one or both cylinders rotating (Figure 7.9). Then, following the literature [31], ∂ vθ ∂ 1 ∂ (rvθ ) =µ ∂t ∂r r ∂r vθ = κRΩi at r = κR ρ vθ = RΩo at r = R vθ = 0 at t = 0 Ω Ωi (r, q) r θ Fluid κR R FIGURE 7.9 Fluid in annular space between rotating cylinders. Applications of Partial Differential Equations in Chemical Engineering 349 models the phenomena of interest. If the following dimensionless variables are substituted, ξ = r /R : radical coordinate τ = µt / ρR 2 : dimensionless time vθ φ= : tangential velocity R(Ωo − Ωi ) Ωi −α = : angular velocity (Ωo − Ωi ) the model can be reduced to ∂φ ∂ 1 ∂ = (ξφ) ∂τ ∂ξ ξ ∂ξ subject to φ = −ακ at ξ = κ φ = 1 − α at ξ = 1 (7.218) φ = 0 at τ = 0 The condition given by Equation 7.218 motivates the use of the Laplace transform method to determine ɸ (τ , ξ ). If ∞ L{φ(τ, ξ)} = ∫ e − sτ φ(τ, ξ) d τ ≡ y(s, ξ) 0 (7.219) then the dimensionless differential equation and Equation 7.218 become sy(s, ξ) − φ(0, ξ) = d 2 y ( s , ξ ) 1 dy ( s , ξ ) y + − 2 dξ 2 ξ dξ ξ subject to the Laplace transformed boundary conditions: y(s, k ) = −κα / s y(s,1) = (1 − α ) / s Rewriting, the second-order ordinary differential equation is recognizable as a Bessel equation where s is a parameter and ξ is the independent variable: d 2 y(s, ξ) 1 dy(s, ξ) 1 + −s+ 2 y = 0 2 dξ ξ dξ ξ 350 Applied Mathematical Methods for Chemical Engineers Then, by comparison to a 2 − v 2c 2 2a − 1 y′′ − y′ + b 2c 2 x 2 c − 2 + y = 0 x x2 whose solution is y = c1 x a J ν (bx c ) + c2 x a J − ν (bx c ) where a = 0, c = 1, b = i s , and Y = 1, we get y(s, ξ) = c1 J1 (i s ξ) + c2Y1 (i s ξ) where the second solution, J−1(·), is denoted by Y1(·). Then at ξ = κ , c1 J1 (i sκ ) + c2Y1 (i sκ ) = −κα / s and at ξ = 1, c1 J1 (i s ) + c2Y1 (i s ) = (1 − α )/s Therefore ακ Y1 (i sκ ) s 1− α Y1 (i s ) s − ακ 1− α Y1 (i s ) − Y1 (i sκ ) s s c1 = = J1 (i sκ )Y1 (i s ) − J1 (i s )Y1 (i sκ ) J1 (i sκ ) Y1 (i sκ ) − J1 (i s ) Y1 (i s ) and 1− α ακ J1 (i sκ ) + J1 (i s ) s s c2 = J1 (i sκ )Y1 (i s ) − J1 (i s )Y1 (i sκ ) Therefore y( s , ξ ) = − Y1 (i s )ακ + (1 − α )Y1 (i sκ ) J1 (i s ξ) s( J1 (i s )Y1 (i sκ ) − J1 (i sκ )Y1 (i s )) (1 − α ) J1 (i sκ ) + ακJ1 (i s ) Y1 (i s ξ) s( J1 (i s )Y1 (i sκ ) − J1 (i sκ )Y1 (i s )) Applications of Partial Differential Equations in Chemical Engineering Let λ j = i s j ⇒ s j = −λ 2j and dλ j ds = −1 2λ j then in terms of λ , y(s, ξ ) becomes y( λ , ξ ) = [Y1 (λ )ακ + (1 − α )Y1 (λκ )]J1 (λξ) [(1 − α ) J1 (λκ ) + ακJ1 (λ )]Y1 (λξ) − −λ 2 ( J1 (λ )Y1 (λκ ) − J1 (λκ )Y1 (λ )) −λ 2 ( J1 (λ )Y1 (λκ ) − J1 (λκ )Y1 (λ )) Now, by applying the formula ∞ L−1{y(s, ξ)} = φ(τ, ξ) = ∑ ρn (τ, ξ) n=0 where ρn (τ, ξ) = P ( s n , ξ ) sn τ e Q ′ ( sn ) the inverse Laplace transform of y(s, ξ ) can be derived. First, recall that P ( sn ) P (s ) P (s) = lim (s − sn ) = lim Q ′ ( s n ) s → sn Q ( s ) − Q ( s n ) s → sn Q(s) s − sn such that for s0 = 0 ρ0 (τ) = P (0) P (s ) = lim s 0 s → Q ′(0) Q (s ) Then for y( λ , ξ ) = P (λ , ξ ) Q (λ ) where P (λ , ξ) = [ακY1 (λ ) + (1 − α )Y1 (λκ )]J1 (λξ) − [(1 − α ) J1 (λκ ) + ακJ1 (λ )]Y1 (λξ) and Q(λ ) = −λ 2 [ J1 (λ )Y1 (λκ ) − J1 (λκ )Y1 (λ )] ρ0 (τ) = lim s s→ 0 P (s ) P (λ , ξ ) = lim(−λ 2 ) Q(s) λ→ 0 Q (λ ) 351 352 Applied Mathematical Methods for Chemical Engineers But −λ 2 P (λ , ξ) [ακY1 (λ ) + (1 − α )Y1 (λκ )]J1 (λξ) − [(1 − α ) J1 (λκ ) + ακJ1 (λ )]Y1 (λξ) = Q(λ ) J1 (λ )Y1 (λκ ) − J1 (λκ )Y1 (λ ) and for small values of λ J1 (λκ ) = λκ /2 and Y1 (λκ ) = −2/πλκ such that [ακY1 (λ ) + (1 − α )Y1 (λκ )]J1 (λξ) − [(1 − α ) J1 (λκ ) + ακJ1 (λ )]Y1 (λξ) J1 (λ )Y1 (λκ ) − J1 (λκ )Y1 (λ ) k2 1 − α (1 − κ 2 ) = ξ− ξ −1 1− κ2 1− κ2 lim λ→ 0 That is, ρ0 (τ) = k2 1 − α (1 − κ 2 ) ξ− ξ −1 1− κ2 1− κ2 is the steady-state solution. Then for sn ≠ 0 ⇔ λ n ≠ 0, n = 1,2, J1 (λ n )Y1 (λ n κ ) − J1 (λ n κ )Y1 (λ n ) = 0 (7.220) With the use of Equation 7.220, P(λ , ξ ) becomes P (λ , ξ ) = 1 [(1 − α ) J1 (λκ ) + ακJ1 (λ )][Y1 (λκ ) J1 (λξ) J1 (λκ ) − J1 (λκ )Y1 (λξ) Differentiating Q(λ ) and using Equation 7.220 results in dQ = −λ 2 [ J1′(λ)Y1 (λκ ) + J1 (λ)Y1′(λκ ) − J1′(λκ )Y1 (λ) − J1 (λκ )Y1′(λ )] dλ But 1 J1′( z ) = J 0 ( z ) − J1 ( z ) z (7.221) Applications of Partial Differential Equations in Chemical Engineering 353 and 1 Y1′( z ) = Y0 ( z ) − Y1 ( z ) z therefore J (λκ ) J1 (λ )Y1 (λκ ) dQ = −λ 2 J 0 (λ )Y1 (λκ ) − J1 (λκ )Y0 (λ ) + κ J1 (λ )Y0 (λκ ) − 0 J1 (λκ ) dλ which reduces to −2λ dQ = [ J12 (λ) − J12 (λκ )] dλ πJ1 (λκ ) J1 (λ ) following substitution of J1 (λ )Y0 (λ ) − J 0 (λ )Y1 (λ ) = 2/πλ and J1 (λκ )Y0 (λκ ) − J 0 (λκ )Y1 (λκ ) = 2/πλκ Therefore Q ′ ( sn ) = −1 dQ 1 = [ J12 (λ n ) − J12 (λ nκ )] 2λ dλ λ = λ n πJ1 (λ n κ ) J1 (λ ) Finally ρn (τ, ξ) = = P (λ n , ξ) − λ n2 τ e Q ′ ( sn ) πJ1 (λ n )[(1 − α ) J1 (λ n κ ) + ακJ1 (λ n )][Y1 (λ n κ ) J1 (λ n ξ) − J1 (λ n κ )Y1 (λ n ξ)] − λ 2n τ e J12 (λ n ) − J12 (λ n κ ) and 1 − α (1 − κ 2 ) κ ξ− ξ −1 1− κ2 1− κ2 ∞ πJ (λ )[(1 − α ) J1 (λ n κ ) + ακJ1 (λ n )][Y1 (λ n κ ) J1 (λ n ξ) − J1 (λ n κ )Y1 (λ n ξ)] − λ 2n τ e +∑ 1 n J12 (λ n ) − J12 (λ n κ ) n =1 φ(τ, ξ) = Using these dimensionless results, now back-substitute to recover the dimensioned variables in which the problem was stated. 354 Applied Mathematical Methods for Chemical Engineers Example 7.23 This example considers Laminar Flow in a Flat Duct with Permeable Walls [38]. In this example a system is modeled, which consists of a dilute solution flowing in a fully developed laminar mode through a semi-infinite flat duct. The material concentration is uniform up to a point ( z = 0 ) , where the fluid contacts a permeable wall, outside of which the concentration is constant. One would like to determine the concentration profile and mass flux for the region z > 0. Assumptions include steady-state conditions, absence of convection through the wall, homogeneous fluid with no sources or sinks, a solute partition coefficient of one between the ducted fluid and outside the wall and neglecting axial diffusion. The dimensionless model is given as 3 ∂Ψ ∂ 2 Ψ (1 − 4 y 2 ) = 2 2 ∂x ∂y Ψ = 1, x ≤ 0, ∀y subject to ∂Ψ = 0, y = 0 , ∀x ∂x ∂Ψ 1 − = N Sh Ψ , y = ∂x 2 (7.222) (7.223) (7.224) (7.225) The dimensionless quantities are defined as Ψ= (c − c0 ) ; x = z D ci − c0 v h 2 ; y = r h; N Sh = local Sherwood number , k h D where c is concentration, D is diffusion coefficient, v is average velocity, h is channel height, z axial coordinate, k is local mass transfer coefficient, and r is transverse coordinate. Solution Employing the method of separation of variables leads to X′ 2 = − β2 X 3 (7.226) d 2Y + β 2 (1 − 4 y 2 ) Y = 0 dy 2 (7.227) dY = 0, at y = 0 dy (7.228) and subject to the conditions − dY = N ShY at y = 1 2 dy (7.229) Applications of Partial Differential Equations in Chemical Engineering 355 Here the assumed product solution was Ψ ( x , y ) = X ( x )Y ( y ) (7.230) At this point we deviate from the approach used in [38] to emphasize an approach that was used in Section 7.5 (i.e., Example 7.10). Here we seek to transform Equation 7.227 into a prototype, Kummers differential equation [13], which then allows us to write down the linearly independent solutions without having to determine the summation of the infinite series used in [39]. If we let (7.231) u = β y2 then du = 2βy dy and dY dY du dY d 2Y dY d 2Y 2 2 = = 2βy ; = 2 β + 4 β y dy du dy du d y 2 du du 2 2 dY dY = 2β + 4βu 2 du du Back-substitution into Equation 7.227, results in u d 2Y 1 dY β u + + 1− 4 Y = 0 du 2 2 du 4 β (7.232) Employing a second substitution Y ( u ) = e − uW ( u ) (7.233) into Equation 7.232 results in u d 2W 1 dW 1 β + − 2u − − W = 0. du 2 4 du 2 2 (7.234) Equation 7.233 is to be compared with Kummers differential equation [13] z d2 R dR + (b − z ) − aR = 0 dz 2 dz (7.235) whose linearly independent solutions are a z b a ( a + 1) z 2 a ( a + 1)( a + 2 ) z 3 + + + b ( b + 1) 2! b ( b + 1)( b + 2 ) 3! R1 ( z ) = 1 F1 [ a; b; z ] = 1 + (7.236) 356 Applied Mathematical Methods for Chemical Engineers R2 ( z ) = R1 ( z ) logz+ + az 1 2 − + 1!1! a 1 ( a )r 1 1 2 2 − − − + ++ r r !r ! a a + r −1 1 (7.237) such that R ( z ) = m1 R1 ( z ) + m2 R2 ( z ) (7.238) is the general solution to Kummers equation, where m1 and m2 are arbitrary constants. In our case, a = 1 / 2 − β 4 , b = 1 / 2, z = 2u Such that the solutions to Equation 7.234 are W1 ( u ) = 1 F1 1 / 2 − β 4;1 / 2; 2u (7.239) W2 ( u ) = W1 ( u ) log 2u + (7.240) 2 (1 / 2 − β 4 ) u 1 2 − + 1!1! (1 / 2 − β 4 ) 1 Therefore Y ( y ) = e −βy 1 F1 1 / 2 − β n 4;1 / 2; 2β n y 2 2 = 1+ (1 / 2 − β 4 ) 2βy + (1 / 2 − β 4 )(3 / 2 − β 4 ) ( 2β y ) + n 2 1/ 2 n n n 2 (7.241) 1 / 2 ( 3 / 2 ) 2! is the appropriate eigenfunction, following substitution into Equations 7.231 and 7.233 and noting that Equation 7.228 implies that the eigenfunction must be finite as y → 0. We note that R2 ( z ) and its derivative will approach infinity as y → 0, such that R ( z ) would become unbounded unless m2 is chosen as zero. Equation 7.229 implies that β 3 β 3 β 1 β −β n e −βn / 4 1 F1 − n ;1 / 2; n + 2e −βn / 4β n (1 − β n / 2 ) 1 F1 − n ; ; n 4 2 4 2 4 2 4 (7.242) 1 β β n n = N Sh 1 F1 − ;1 / 2; , n ≥ 1 4 2 4 and defines the eigenvalues β n , noting that d a { 1 F1 [a; b;z ]} = 1 F1 [a + 1; b + 1;z ] dy b (7.243) since the series 1 F1 [ a; b; z ] is absolutely convergent (see Example 7.13). The Fourier coefficient can now be determined as given in Colton and coworkers [38] (see Problem 18). 357 Applications of Partial Differential Equations in Chemical Engineering Example 7.24 This example considers a proposed analysis for the wall region velocity profile [40]. Assuming that the wall exchange process on the average can be represented by the motion of fluid masses with fixed velocity, uL and a constant wall contact time, tc . Additionally, the transport of momentum within such masses is proposed to be represented as ∂u µ ∂ 2 u = ∂t ρ ∂ y 2 subject to the conditions u = 0 at y = 0 u = uL at y = ∞ u = uL at t = 0. Using the method of Combination of Variables (Section 6.5) similar to Example 7.11, let φ ( η) = u y ; η= , γ =µ ρ 4 γt uL such that ∂ η dφ ( η) ∂ φ ( η) = − ; φ ( η) = ∂t 2t dη ∂y 1 dφ ( η) ∂2 1 d 2φ ( η) φ ( η) = and 2 4 γt dη ∂y 4 γt dη2 results in the ordinary differential equation φ′′ + 2 ηφ′ = 0 (7.244) φ ( 0 ) = 0 and φ ( ∞ ) = 1. (7.245) subject to The general solution to this resulting ordinary differential Equation 7.244 is φ ( η) = c1 ∫ e − η d η + c2 2 Applying the accompanying conditions, Equation 7.245 leads to y u ( t , y ) = uL erf ( η) = uL erf µ 2 t ρ (7.246) 358 Applied Mathematical Methods for Chemical Engineers In summary, this chapter attempts to capture varied examples adapted from published research in the chemical engineering literature. Primarily, an attempt is made to emphasize the analytical techniques that are discussed in the undergraduate prerequisite mathematics courses but are under-applied in undergraduate chemical engineering curricula. Additional methods such as regular perturbation and combination of variables are introduced as a path to advanced analysis techniques that are also useful in chemical engineering research. More on the combination of variables as applied to the analysis of semi-infinite systems can be found in [36]. 7.9 PROBLEMS 1. a. Use Equations 7.162, 7.163, and 7.166 to derive Equation 7.167. b. Solve Equation 7.167, subject to y = 0; u = v = 0 y = ∞; u = U ∞ 2. Show all the intermediate steps to derive Equations 7.175 and 7.176 using Equation 7.167 and the given conditions in Example 7.17. 3. Solve f (0) = 1 2 ff ′ f ′′ + = 0; subject to f ( ηR ) = 0 2 SA 4. For the region ηR < η < ∞ reduce the equation of energy to 2 d 2 TB f dTB + =0 ρ dη2 2 dη Solve the reduced equation subject to TB = 0 at η = ηR TB = 1 at η = ∞ 5. In a porous medium, the continuity equation can be written as ε ∂ρ = −∇ ⋅ ρu ∂t where ε is the porosity, and the equation of motion (Darcy’s law) can be written as κ u = − ∇p µ Applications of Partial Differential Equations in Chemical Engineering 359 where gravity is neglected, κ is the permeability and μ is the viscosity [14,4]. a. Show that for an incompressible fluid, ∇ 2p = 0. b. Show that for isothermal flow of a compressible gas, 2εµρ0 ∂ρ = ∇ 2 ρ2 κ ∂t 6. Two concentric spherical metallic shells of radii a and b cm (a < b) are separated by a solid of thermal diffusivity α (cm 2/s). The outer surface of the inner shell is maintained at T0 ºC and the inner surface of the outer shell at T1ºC. Derive the differential equation governing the unsteadystate temperature distribution in the solid as a function of time and radial coordinate. Show that the solution takes the form T(t,r) = T1 b − T0 a ab T0 − T1 ∞ Bn + + ∑ sin[β(r − a)]exp(−β 2 αt) b−a r b−a n =1 r where β = nπ/(b − a). Demonstrate how Bn can be determined from any initial temperature distribution. 7. If a liquid is flowing through a fixed bed at a rate v cm/s when a pulse of a tracer is added, show that the later distribution of the tracer is given by the solution of E ∂2 C ∂C ∂C −v = 2 ∂x ∂ x ∂t here E is the mixing coefficient which operates in the axial direction (x) w only. Use the substitution z = x − vt, and assume infinite bed length to find C. 8. The steady laminar flow of a liquid through a heated cylindrical pipe has a parabolic velocity profile if natural convection effects, and variation of physical properties with temperature are neglected [4]. If the fluid entering the heated section is at a uniform temperature (T1) and the wall is maintained at a constant temperature (Tw), develop Graetz’s solution by neglecting the thermal conductivity in the axial direction. 9. The sudden closure of a valve generates a pressure wave within the liquid flowing in the pipe leading to the valve. The passage of this wave causes compression of the liquid and expansion of the pipe. Show that the velocity of the liquid and the pressure are related by − ∂ p ρ ∂v = ∂ x g ∂t and − ∂2 p ∂2 p 2 = c ∂t 2 ∂x 2 360 Applied Mathematical Methods for Chemical Engineers where c is the velocity of propagation of the pressure wave. If a uniform pipe of length L connects a reservoir at x = 0 to the valve at x = L, show that p( x , t ) = p0 + 4c ρv0 πg ∞ (−1)n πx πct ∑ (2n + 1) sin(2n + 1) 2 L sin(2n + 1) 2 L n =1 10. Consider the case of plug flow with homogeneous chemical reaction and reaction at the wall of a tubular reactor [32]. For an average velocity, a firstorder homogeneous reaction and a first-order wall reaction, show that a reasonable model is vavg subject to and ∂CA 1 ∂ ∂CA = DAB r − kCA ∂z r ∂r ∂r ∂CA = 0 at r = 0 ∂r CA = CA0 at z = 0 ∂CA = k W CA at r = R ∂r Find the concentration profile and use it to show that − DAB dC A D λ2 βJ (λ ) k = −( kd + k )CA , kd = AB2 n , λ n = o n , β = wR dt R J1 (λ n ) DAB 11. Consider a tubular flow reactor modeled by r 2 ∂C 1 ∂ ∂CA ∂2 CA v0 1 − A = DAB + kCAn r + 2 R ∂ z r ∂ r ∂ r z ∂ where DAB is the binary diffusivity and v0 is the axial velocity at the center of the tube of radius R. The differential equation is subject to the following boundary conditions CA = CA0 at z = 0 ∂CA = 0 at r = 0 ∂r ∂C − DAB A = k W CA ∂r at r = R Applications of Partial Differential Equations in Chemical Engineering 361 Further, supposing that n = 1 and that axial diffusion is negligible in comparison to bulk flow, determine the “cup mixing” concentration 1 CA = 4 ∫ [1 − u 2 ]θu du 0 where u = r C , θ= A R CA0 12. Consider evaluating the diffusion coefficient, D, for the resin phase in an ion exchange system modeled by ∂qi ∂2 q 2 ∂qi = D 2i + ∂t r ∂r ∂r where qi represents point concentration (meq/g dry resin), and t is time (s). The differential equation is subject to qi = 0 for t < 0, 0 < r < b and qi = qs at r = b, t > 0 where qs = A + Bt + Ct2; A, B, and C are experimentally determined constants. Find the total amount of acid, w, adsorbed per particle up to time t where w is defined as t ∂q w = ∫ Dρ i 1 / 2(4πb 2 ) dt 0 ∂r r = b in which ρ is the bulk density of the resin (g dry resin/cm3). 13. Bounded equimolal counterdiffusion [8]. Consider a system containing A and B with a partition at z = 0. At the boundaries, z = ±L, ∂cA /∂z = 0 for all t and at t = 0, cA = cA− for −L < z < 0 and cA = cA+ for 0 < z < L. The differential equation describing this system is given by ∂cA ∂ ∂c = DAB A ∂t ∂z ∂z Derive the concentration profile for the case when the diffusion coefficient is independent of concentration. 362 Applied Mathematical Methods for Chemical Engineers Answer: cA − cA− 1 2 ∞ sin(n + 1 / 2)πz / L = 1 + ∑ exp− [(n + 1 / 2)π / L ]2 DABt + − cA − cA 2 π n = 0 (n + 1 / 2) 14. Diffusion in a two-phase system [8].Consider the system consisting of two immiscible phases, I (−∞ < z < 0) and II (0 < z < +∞), separated by a partition. Initially, the solute concentration in phase I is cI0 and cII0 in phase II. At time t = 0 the partition at z = 0 is removed and diffusion is allowed to take place. Assuming that the diffusion can be described by Fick’s second law in both phases, and there is equilibrium at the interface (cII = mcI), derive the concentration profile for each phase in terms of the respective initial concentration. The mathematical statement of the problem is Diffusion in phase I: ∂cI ∂2 c = DI 2I ∂t ∂z Diffusion in phase II: ∂cII ∂2 cII = DII ∂t ∂z 2 Boundary conditions: At z = 0 cII = mcI At z = 0 DI (∂cI / ∂ z ) = DII (∂cII / ∂ z ) At z = −∞ ∂cI / ∂ z = 0 At z = +∞ ∂cII / ∂ z = 0 Initial conditions : At t = 0 cI = cI0 cII = cII0 Answer: 1 + erf( z / 4 DI t ) cI − cI0 = cII0 − mcI0 m + DI /DII 1 − erf( z / 4 DII t ) cII − cII0 = c − (1/m)cII0 (1 / m) + DII /DI 0 I where m is the distribution coefficient, DI and DII are the diffusion coefficients in the respective phase. Applications of Partial Differential Equations in Chemical Engineering 363 15. If the flat velocity profile assumed in Example 7.10 is incorrect, use the Navier–Stokes equations of motion to show that 0=µ d 2 vz + ρg dx 2 For the boundary conditions vz = 0 at x = δ , and dvz/dx = 0 at x = 0, show that vz = ρgδ 2 2µ x2 x2 1 − = vmax 1 − δ δ where the maximum velocity occurs at the film surface. Also show that the film thickness is δ= 3 3µΓ = ρ2 g 3 3µ 2 Re 4ρ2 g where Re = 4Γ /µ and Γ = ρ vavgδ is the mass rate of flow in the z-direction per unit width of wetted wall in the y direction. Use the derived velocity profile together with the boundary conditions cA = cA0 at x = 0 ∂cA / ∂ x = 0 at x = δ cA = cA1 at z = 0 to derive the concentration profile for cA. 16. Reconsider the model discussed in Example 7.8, but with cylindrical ­geometry [8,33]. Following the development in the literature [7,8], the governing ­equation for the agent concentration in the reservoir is ∂C1 ∂2 C 1 ∂C1 = D1 21 + ∂r ∂t r ∂r (7.247) ∂C2 ∂2 C2 1 ∂C2 = D2 + ∂r 2 ∂t r ∂r (7.248) and is the governing equation for the agent in the membrane. The subscripts refer to the respective region as shown in Figure 7.4. D2 is an effective diffusivity and is defined as 364 Applied Mathematical Methods for Chemical Engineers D2 = Dε τ (7.249) where D is the agent diffusivity in the pore liquid, ε is the membrane porosity, and τ is the membrane tortuosity. Both the agent diffusivity in the pore liquid (D) and in the reservoir liquid (D1) are calculated with the Wilke– Chang correlations [9]. The quantity τ was defined and measured for various systems [10]; we use the values given for hydrophobic membrane. The porosity value is a manufacturer-supplied quantity. Equations 7.247 and 7.248 are subject to the following boundary conditions: C1 (a, t ) = m1,2 C2 (a, t ) (7.250) ∂C1 (0, t ) =0 ∂r (7.251) ∂C1 (a, t ) ∂C (a, t ) = D2 2 ∂r ∂r (7.252) ∂C2 (b, t ) ∂C (b, t ) = − D2 α 2 m2, w 2 ∂t ∂r (7.253) D1 Vw Further, since the entire agent is initially present in the reservoir phase, then C1 (r ,0) = C10 (7.254) C2 (r ,0) = 0 (7.255) Equation 7.250 is a statement of the equilibrium partitioning at the reservoir–pore liquid interface with m1,2 being the partition coefficient. Equation 7.251 indicates that the solute concentration is expected to be finite at the bottom of the reservoir. Equation 7.252 displays the continuity of the agent flux across the reservoir–pore interface, whereas Equation 7.253 accounts for the material leaving the membrane and entering the surrounding water bath. The quantity α 2 is the membrane area at the outer wall (cm2). Determine the concentration profile exiting the membrane (region 2). H int: In deriving the solution to the model, first recast the model in dimensionless form by introducing the following quantities: u1 (ξ, θ) = C1 (r , t ) C10 (7.256) u2 (ξ, θ) = C2 (r , t ) C10 (7.257) Applications of Partial Differential Equations in Chemical Engineering 365 where ξ= r Dt , θ = 12 b b (7.258) 17. Abdekhodaie [37] presented an exact solution for the diffusional release from theophylline microspheres coated with ethyl vinyl acetate copolymer into a finite external volume using the Laplace transform method. The solute was assumed to have very low solubility in the polymeric membrane and the diffusion coefficient was assumed to be independent of concentration. The diffusion of drug into the surroundings was described by Fick’s second law of diffusion as ∂C D ∂ 2 ∂C = r ∂t r 2 ∂r ∂r subject to the conditions C ( r ,0 ) = Cs , C ( a, t ) = Cs , C ( b, t ) = KCb ( t ) where Cs is the solubility limit of the drug and K is the equilibrium distribution coefficient between polymeric membrane and external bulk concenMt tration Cb ( t ) . Determine the cumulative amount of solute released. M∞ REFERENCES 1. Bennett, C.O. and Myers, J.E. Momentum, Heat and Mass Transfer, McGraw-Hill, New York, 1962. 2. Myers, G.E. Analytical Methods in Conduction Heat Transfer, McGraw-Hill, New York, 1971. 3. Spiegel, M.R. Mathematical Handbook, McGraw-Hill, New York, 1968. 4. Jenson, V.G. and Jeffreys, G.V. Mathematical Methods in Chemical Engineering, Academic Press, London, 1963. 5. Felder, R.M., Spence, R.D., and Ferrell, J.K. A method for the dynamic measurement of diffusivities of gases in polymers, J. Appl. Polym. Sci., 19, 3193, 1975. 6. Ramraj, R., Farrell, S., and Loney, N.W. Analytical solution to controlled release using microporous membrane, Sep. Sci. Technol., 34, 225, 1999. 7. Jost, W. Diffusion in Solids, Liquids and Gases, Academic Press, New York, 1960. 8. Bird, R.B. Theory of diffusion, in Advances in Chemical Engineering, Vol. I, Drew, T.B. and Hoopes, J.W., Eds., Academic Press, New York, 1956, p. 156. 9. Reid, R., Prausnitz, J., and Sherwood, T. Properties of Gases and Liquids, McGrawHill, New York, 1977. 10. Prasad, R. and Sirkar, K. Dispersion-free solvent extraction with microporous hollowfiber modules, AIChE J., 34, 177, 1988. 366 Applied Mathematical Methods for Chemical Engineers 11. Loney, N.W. Analytical solution to mass transfer in laminar flow in hollow fiber with heterogeneous chemical reaction, Chem. Eng. Sci., 51, 3995, 1996. 12. Huang, C.R., Matlosz, M., Pan, W.D., and Snyder, W. Heat transfer to a laminar flow fluid in a circular tube, AIChE J., 30, 833, 1984. 13. Slater, L.C.J. Confluent Hypergeometric Functions, Cambridge University Press, London, 1960. 14. Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena, 2nd ed., John Wiley & Sons, New York, 2002. 15. Higbie, R. The rate of absorption of a pure gas into a still liquid during short periods of exposure, Trans. Am. Inst. Chem. Eng., 31, 365, 1935. 16. Loney, N.W. and Huang, C.R. Analytical solution of an LPCVD reactor interwafer region, Thin Solid Films, 226, 15, 1993. 17. Abramowitz, M. and Stegun, I.A. Handbook of Mathematical Functions, Dover, New York, 1968. 18. Son, J.S. and Hanratty, T.J. Limiting relation for the Eddy diffusivity close to a wall, AIChE J., 13, 689, 1967. 19. Shaw, D.A. and Hanratty, T.J. Turbulent mass transfer rates to a wall for large Schmidt numbers, AIChE J., 23, 28, 1977. 20. Wolfram, S.M. Mathematica Book, Cambridge, University Press, 1996. 20a. Huang, C.R., Denny, A.F., and Loney, N.W. Molecular diffusion in the laminar sublayer during turbulent flow in a smooth tube, Chem. Eng. Sci., 59, 1191, 2004. 21. Watson, G.N. A Treatise on the Theory of Bessel Functions, 2nd ed., Cambridge University Press, London, 1966. 22. Loney, N.W. Justification of the assumption of transport by diffusion in the interwafer region of a multiwafer CVD reactor, J. Mater. Sci. Lett., 15, 1219, 1996. 23. Friedlander, S.K. and Litt, M. Diffusion controlled reaction in a laminar boundary layer, Chem. Eng. Sci., 7, 229, 1958. 24. Schlichting, H., Boundary-Layer Theory, 7th ed., McGraw-Hill, New York, 1979. 25. Kim, J.I. and Stroeve, P. Mass transfer in separation devices with reactive hollow fibers, Chem. Eng. Sci., 43, 247, 1988. 26. Noble, D.R. and Way, D.J. Facilitated transport, in Membrane Handbook, Ho, W.S.W. and Sirkar, K., Eds., Van Nostrand Reinhold, New York, 1992, chap. 44. 27. Noble, R.D. Shape factors in facilitated transport through membranes, Ind. Chem. Fundam., 22, 139, 1983. 28. Brown, G.M. Heat or mass transfer in a fluid in laminar flow in a circular or flat conduit, AIChE J., 6, 179, 1960. 29. Wolfram, S. Mathematica: A System for Doing Mathematics by Computer, AddisonWesley, Redwood City, 1991. 30. Mickley, H.S., Sherwood, T.K., and Reed, C.E. Applied Mathematics in Chemical Engineering, McGraw-Hill, New York, 1957. 31. Bird, R.B. and Curtiss, C.F. Tangential Newtonian flow in annuli-I, Chem. Eng. Sci., 11, 108, 1959. 32. Smith, G.K., Krieger, B.B., and Herzog, P.M. Experimental and analytical study of wall reaction and transport effects in fast reaction systems, AIChE J., 26, 567, 1980. 33. Ramraj, R., Farrell, S., and Loney, N.W. Mathematical modeling of controlled release from a hollow fiber, J. Membr. Sci., 162, 73, 1999. 34. Wong, H. M., Wang, J. J., and Wang, C.H. In vitro sustained release of human immunoglobulin G from biodegradable microspheres, IE&C Res. 40, 933, 2001. 35. Slattery, J. C. Advanced Transport Phenomena, Cambridge University Press, 1999. 36. Plawsky, J. L. Transport Phenomena Fundamentals, 2nd ed., CRC Press—Taylor & Francis, 2010. Applications of Partial Differential Equations in Chemical Engineering 367 37. Abdekhodaie, M.J. Diffusional release of a solute from a spherical reservoir into a finite volume, Journal of Pharmaceutical Sciences, 91(8), 1803, 2002. 38. Colton, C.K., Smith, K.A., Stroeve, P. and Merrill, E.W. Laminar flow mass transfer in a flat duct with permeable walls, AIChE J., 17, 773, 1971. 39. Van der Does De Bye, J.A.W. and Schenk, J. Applied Science Research, Sect. A., 3, 308, 1952. 40. Hanratty, T.J. Turbulent exchange of mass and momentum with a boundary, AIChE J., 2, 359, 1956. 41. Hughmark, G.A. Heat and mass transfer in the wall region of turbulent pipe flow, AIChE J., 17, 51, 1971. 42. Lightfoot, E.N. Steady state absorption of a sparingly soluble gas in an agitated tank with simultaneous irreversible reaction, AIChE J., 4, 499, 1958. This page intentionally left blank 8 Dimensional Analysis and Scaling of Boundary Value Problems 8.1 INTRODUCTION In the practice of chemical engineering, dimensional analysis and scaling are ­techniques that can be employed to emphasize similarities between a prototype and its model. Dimensionless quantities are developed, which can serve as new variables. Usually, the number of new variables is much smaller than the original number of physical variables of the system under consideration. For example, suppose one is interested in conducting an investigation to determine the power required to drive an ordinary house fan [1]. More specifically, suppose one wants to relate the size and shape of the fan to the rotational speed and torque, where Newton’s second law relates the torque (t) to the forces generated when the fan accelerates the air that passes through it. Suppose the torque is chosen as the dependent variable and the physical variables are the following: • Fan diameter (d) • Fan design or shape of fan (R) • Medium characteristics (air viscosity, density, sound velocity, and ratio of specific heats) • Rotative speed (n) Since the house fan is to be uncomplicated, one can neglect the effects of the air viscosity, sound velocity, and the ratio of specific heats. Then, using the mass (M), length (L), and time (T) system, Table 8.1 can be generated. If the torque is divided by the density, Table 8.2 results. If t/ρ is divided by the product D5n2, there results the quantity t ρD 5 n 2 = 1 Therefore, the final analysis yields t f ,R = 0 ρD 5 n 2 which means that the torque for a given design R is proportional to the dimensionless product t ρD 5 n 2 369 370 Applied Mathematical Methods for Chemical Engineers TABLE 8.1 Dimensional Analysis on a Domestic House Fan Dimensions Quantity Torque Fan diameter Fan design Air density Rotative speed Symbol M L T t D R ρ n 1 0 0 1 0 2 1 0 −3 0 −2 0 0 0 −1 Source: Taylor, E.S., Dimensional Analysis for Engineers, Oxford Uni­ versity Press, London, U.K., 1974. With permission. TABLE 8.2 Analysis to Develop Drag in a Pipe Symbol M L T t/ρ D N 0 0 0 5 1 0 −2 0 −1 Source: Taylor, E.S., Dimensional Analysis for Engineers, Oxford Uni­ versity Press, London, U.K., 1974. With permission. Therefore, instead of five variables, only one variable needs to be considered. Another example, more familiar to chemical engineers, involves fluid flow. Consider a very long cylindrical pipe having a uniform cross-section area (A) in which an incompressible fluid is flowing. Assume that the motion is steady and that viscosity (μ) and density (ρ ) are not negligible. One may want to characterize the motion of the fluid. In this case, the pressure drop along the pipe, the average velocity (v) of the fluid, or the fluid discharge per unit time through A is needed. Hence, the motion in the pipe can be determined by ρ , μ, v, and D, where D is the pipe inside diameter. That is, if one forms the dimensionless quantity (drag) φ= p1 − p2 ρv 2 D then the resistance of a section of pipe of length ℓ is ( p1 − p2 ) A = ∆pA = φ Aρv 2 D Dimensional Analysis and Scaling of Boundary Value Problems 371 Among the four physical variables, only one dimensionless quantity can be formed, namely, ρvD N Re = µ where NRe is the Reynolds number. Therefore φ = φ( N Re ) Again, instead of four variables, the Reynolds number can be used to characterize the phenomenon. 8.2 CLASSICAL APPROACH TO DIMENSIONAL ANALYSIS The pi theorem is a generalized method of dimensional analysis and detailed discussions can be found in [1–7]. Below is a brief review of the pi theorem. Consider the magnitude f1 of some physical quantity depending on other, independent magnitudes g1, g2, …, gn, then f1 = φ( g1 , g2 ,, gn ) or alternatively ψ ( f1 , g1 , g2 ,, gn ) = 0 (8.1) Equation 8.1 is required to be dimensionally homogeneous. The pi theorem says that if the number of distinct reference quantities required to express the dimensional formula of all n magnitudes is r, then the n magnitudes may be grouped into n − r independent dimensionless ∏ terms, resulting in the relation φ(Π1 , Π2 ,, Πn − r ) = 0 (8.2) An immediate advantage is demonstrated if Equation 8.1 is compared with Equation 8.2. That is, the number of independent variables to be studied is reduced to n − r. Usually, it is not difficult to determine the value of n; just add up the variables. However, determining r requires more effort. Above, r is stated as distinct reference quantities. The term distinct is significant. That is, in the linear algebra sense, r is the rank of the dimensional matrix. For example, suppose a problem involving force (F), velocity (v), density (ρ ), viscosity (μ), and length (L) is to be restated in terms of dimensionless groups. Then, the rank of the dimensional matrix must be determined. In solving such problems, a usual first step is the construction of a table, listing the variables and their associated dimensions. Such a table is shown in Table 8.3. TABLE 8.3 Classical Approach to Dimensional Analysis Variable Symbol Dimensions Force Velocity Density Viscosity Length F v ρ μ L M L/t2 L/t ML3 M/L t L 372 Applied Mathematical Methods for Chemical Engineers The next step is to define the dimensional matrix whose elements are the exponents of the fundamental dimensions M, L, and t appearing in Table 8.3. Dimensional Matrix F 1 1 −2 M L t ρ 1 −3 0 V 0 1 −1 L 0 1 0 μ 1 −1 −1 In order to determine r, recall that the definition of the rank, r, of a matrix is the number of rows (columns) in the largest nonzero determinant, which can be formed from the matrix. In our situation, we have a matrix that contains three rows and five columns. However, determinants are only defined for square matrices (see Appendix A). Therefore, the largest possible determinant would contain three rows and three columns (determinant of order 3). It turns out that, for this problem, the largest nonzero determinant is of order 3; therefore, the rank is 3. However, in other situations, one should verify that the determinant is, in fact, nonzero. As a second example of the determination of the rank, r, consider the dimensional matrix: M L T P Q R S 2 −1 1 1 6 20 3 −3 −3 4 0 8 Here, the largest determinant will be that which contains three rows and three columns. However, all such determinants are zero. That is, 2 −1 1 3 2 6 −3 = −1 1 20 −3 1 4 2 3 4 6 0 = −1 −3 0 = 1 20 8 1 −3 8 1 3 4 6 −3 0 = 0 20 −3 3 It turns out here that only second-order determinants have nonzero values. For example, the determinant 21 = 13 ≠ 0 −1 6 Therefore, the rank of this dimensional matrix is 2. 8.3 FINDING THE ΠS In this section, two ways of deriving the ∏s are discussed. The first is based on the pi theorem, whereas the second is a more practical approach. Recall the example of fluid flow through a long, smooth pipe of circular cross section. There are five Dimensional Analysis and Scaling of Boundary Value Problems 373 variables representing the n magnitudes, namely Δ p/ℓ, ρ , μ, v, and D. The number of distinct reference quantities (r) is 3 (rank of the dimensional matrix). Therefore, there should be 5 − 3 = 2 independent dimensionless products (∏1 and ∏2). Further, the theorem requires that, from among the n original variables, r of them is to be selected to form a recurring set. This recurring set further requires that the r variables, together, must involve the r distinct reference quantities. Here v, μ, and ρ form a suitable recurring set. Each ∏ will be formed from the recurring set and one of the remaining variables. For example, ∏1 is constructed from v, ρ , μ, and Δ p/ℓ, whereas ∏2 from ρ , μ, v, and D. For the formulation of ∏1, let the dimensionless number be vaμ bρ cΔ p/ℓ, where a, b, and c are constants to be determined. Then, the dimensional formula of this product is [ LT −1 ]a [ ML−1T −1 ]b [ ML−3 ]c [ ML−2 T −2 ] = [ M 0 L0 T 0 ] Equating exponents of like magnitudes gives L : a − b − 3c − 2 = 0 M : b + c +1= 0 T : −a−b−2=0 resulting in a = −3, b = 1, and c = −2. Therefore, the dimensionless number ∏1 is µ∆p / v 3ρ2 Similarly, for the formation of ∏2, let vaμ bρcD be the dimensionless number. Then, the dimensional formula of this product is [ LT −1 ]a [ ML−1T −1 ]b [ ML−3 ]c [ L ] = [ M 0 L0 T 0 ] giving a = 1, b = −1, and c = 1. Therefore, Equation 8.2 for this example is φ(Π1 , Π2 ) = 0 That is, µ∆p / vDρ φ 3 2 =0 µ vρ or, in terms of Δ p/ℓ ∆p / = v 3ρ2 φ1 ( N Re ) µ It is important to note that the recurring set must provide the opportunity for canceling of any dimensional components that are involved in the other magnitudes. Therefore, its constituent magnitudes must involve each of the reference magnitudes at least once. However, it must not be possible to form a dimensionless group from the recurring set alone. 374 Applied Mathematical Methods for Chemical Engineers As a second illustration on finding the ∏s, consider a screw propeller operating in a fluid of constant density, such as seawater. Here, one is interested in the thrust produced by the propeller, and the goal is to determine how the magnitude of this force depends on other associated quantities. Following Massey [2], the fluid is assumed to be homogeneous (no air bubbles) and the propeller is remote from all other surfaces (this latter assumption excludes ships, etc.). Listed in Table 8.4 are the relevant quantities. The quantity, g, is included to account for work done against gravity in moving the fluid surface vertically. However, surface tension is being neglected, as the propeller is assumed to be large. From Table 8.4, it can be seen that the n magnitudes are 7 and the number of reference magnitudes (mass, length, and time interval) is 3. Therefore, 7 − 3 = 4 dimensionless quantities (products) are expected. That is, φ(Π1 , Π2 , Π3 , Π4 ) = 0 (8.3) is the appropriate form to expect. As the number of reference magnitudes is 3, then 3 magnitudes (variables) must be selected from among the 7 for use as a recurring set. The variables F, ρ , and D are suitable, whereas the set D, v, and ω is not suitable because the reference magnitude M is not included. In addition, the set consisting of F, ρ , and μ is an unsuitable choice, even though all the reference magnitudes are included. This latter set is unsuitable because a dimensionless group, Fρ /μ2, can be formed from these variables alone. Using F, ρ , and D as the recurring set, the dimensionless numbers (∏s) take the forms F a1ρb1 D c1 v; F a 2 ρb 2 D c 2 ω; F a 3ρb 3 D c 3µ; F a 4 ρb 4 D c 4 g For the first ∏, the dimensional formula is [ MLT −2 ]a1 [ ML−3 ]b1 [ L ]c [ LT −1 ] = [ M 0 L0 T 0 ] such that al = −1/2, b1 = 1/2, and c1 = 1. Therefore, the first ∏ may be written as F−1/2ρ 1/2 Dv. However, the fractional exponents may prove to be inconvenient and can be removed by squaring to get Π1 = ρD 2 v 2 F TABLE 8.4 Relevant Quantities of a Screw Propeller in a Constant Density Fluid Quantity Thrust (force) Fluid density Propeller diameter Speed of advance Angular speed Fluid viscosity Weight per unit mass Symbol Dimensional Formula F ρ D v ω μ g [MLT−2] [ML−3] [L] [LT−1] [T−1] [ML−1T−1] [LT−2] Dimensional Analysis and Scaling of Boundary Value Problems 375 Similarly, the other ∏s are derived: Π2 = µ2 ρD 3 g ρD 4 ω 2 , Π3 = , Π4 = F Fρ F Then, Equation 8.3 may be expressed as ρD 2 v 2 ρD 4 ω2 µ 2 ρD 3 g , , , φ =0 F Fρ F F (8.4) It is important to note that the recurring set F, ρ , and D is not unique. However, if analysis is performed with any other suitable recurring set, the same information, appearing in different forms, would result. For example, the recurring set consisting of ρ , D, v results in ρD 2 v 2 Dω ρDv Dg , , , φ =0 v µ v F (8.5) It appears that only the first of the arguments in Equation 8.4 corresponds directly with any in Equation 8.5. However, the first two arguments of Equation 8.4 may be divided to yield Π1 Dω = Π2 v −2 which is a power of the second ∏ in Equation 8.5. The other two products in Equation 8.5 may also be obtained from combinations of those in Equation 8.4. This type of transformation of ∏s may be useful, but none of the original magnitudes must be completely removed from the set of ∏s. Also, the number of independent ∏s must be that specified by the pi theorem, n – r. That is, given φ(Π1 , Π2 , Π3 , Π4 ) = 0 then φ(Π1 , Π2 , Π1 / Π2 , Π4 ) = 0 is not appropriate, whereas the following is: φ(Π1 / Π2 , Π3 , Π4 ) = 0 The following example illustrates the key steps in the application of the pi theorem [8]. Example 8.1 Determine the dimensionless groups formed from the variables involved in the flow of a fluid external to a solid body. The force exerted on the body is a function of v, ρ , μ, and L[8]. 376 Applied Mathematical Methods for Chemical Engineers Solution Typically, a table of the variables and their dimensions is constructed. Variable Force Velocity Density Viscosity Length Symbol F v ρ μ L Dimensions M L /t2 L /t M / L3 M/L t L Source: Welty, J.R. et al.: Fundamentals of Momentum, Heat and Mass Transfer, 3rd ed. 1984. Copyright Wiley-VCH Verlag GmbH & Co. KGaA. Reproduced with permission. Then, a dimensional matrix is formed whose elements are the exponents of the fundamental dimensions M, L, and t appearing in each of the variables. In this case, the dimensional matrix is M L t F v ρ μ L 1 1 −2 0 1 −1 1 −3 0 1 −1 −1 0 1 0 From the above matrix, the rank is determined. This was done in the previous section, and there, as well as in this case, the rank r is 3. To determine the number of independent dimensionless groups, the number of variables is determined. In this case, n = 5. Therefore, the number of ∏s is n – r = 2. Next, a recurring set of r variables is selected. This recurring set will be made up of those variables, which will appear in each pi group, and among them, contain all of the fundamental dimensions. One approach in choosing this set is to exclude from it those variables whose effect is under investigation. In this problem, one would like to isolate the drag force effect; therefore, it will not be a part of the recurring set. For the remaining exclusion, the viscosity is chosen. The remaining variables are v, ρ, and L, which include the fundamental dimensions M, L, and t among them. That is, the recurring set consists of v, ρ, and L. Each of the two ∏s will include the recurring set and one of the previously excluded variables. That is, Π1 = v aρb Lc F and Π 2 = v d ρe L f µ Notice that each ∏ is required to be dimensionless. As such, the variables in the recurring set are raised to certain exponents, which will satisfy this dimensionless condition. Each pi group is evaluated independently. Therefore Π1 = v aρb Lc F Dimensionally, L a M b ML [ M 0 L0t 0 ] = 3 [ L ]c 2 t t L Dimensional Analysis and Scaling of Boundary Value Problems 377 Equating exponents of M, L, and t, on both sides, results in M: 0 = b + 1 L: 0 = a −3b + c + 1 t: 0 = − a − 2 giving a = −2, b = −1, and c = −2. Therefore Π1 = F / L2 ρv 2 which is the Euler number. Similarly, evaluating the exponents for the second pi group results in Π 2 = µ / ρvL = 1 / N Re where NRe is the Reynolds number. Therefore, the application of dimensional analysis has reduced a 5-variable problem to a 2-parameter problem: Eu = φ1 ( N Re ) The second method for finding the ∏s in this section is demonstrated using the following example. Reconsider the illustration involving the flow of a fluid through a very long, smooth pipe. This method involves the construction of a series of tables containing the variables and their dimensions as follows. Following the construction of Table 8.5, the dependence of each variable on the reference magnitudes is systematically eliminated: eliminate the dependence on M by dividing each variable having dimensions in respect to M by ρ . This results in Table 8.6. Eliminate the dependence on T. Each variable with dimensions involving T is multiplied or divided by an appropriate power of v, so as to make the T exponent zero, as shown in Table 8.7. Finally, eliminate the dependence on L by using appropriate powers of D, as shown in Table 8.8. The resulting two dimensionless groups (∏s) are DΔ p/ℓρ v2 and μ/ρ vD. If a particular variable is to be used as the dependent variable, then that variable (or any power of it) must be excluded as a multiplier or divisor in the elimination steps (Δ p/ℓ). Also, the order in which reference magnitudes are eliminated is only important insofar as expediency is concerned. Even though this latter technique may involve more labor, it is more straightforward to apply. TABLE 8.5 Systematic Elimination of Dependence Variables on Reference Magnitudes Δ p/ℓ D v μ ρ [M] [L] 1 0 0 1 1 −2 1 1 −1 −3 [T] −2 0 −1 −1 0 378 Applied Mathematical Methods for Chemical Engineers TABLE 8.6 Result of Systematic Elimination of Dependence on Reference Magnitudes Δ p/ℓρ D v μ/ρ ρ /ρ = 1 [M] [L] [T] 0 0 0 0 0 1 1 1 2 0 −2 0 −1 −1 0 TABLE 8.7 Elimination of the Dependence on T [M] Δ p/ℓρ v D v/v = 1 μ/ρ v 2 0 0 0 0 [L] −1 1 0 1 [T] 0 0 0 0 TABLE 8.8 Elimination of the Dependence on L DΔ p/ℓρ v2 D/D = 1 μ/ρ vD [M] [L] 0 0 0 0 0 0 [T] 0 0 0 So far, the discussion has been centered on developing dimensionless groups from a set of given process variables. These dimensionless groups are available as design parameters instead of the individual process variables. This is especially helpful when relationships are to be established between a model and an as yet designed larger scale process. Scaling of the independent and dependent variables in given differential equations to establish dimensionless group is presented later. 8.4 SCALING BOUNDARY VALUE PROBLEMS In studying transport phenomena one is guaranteed to encounter situations involving fluid–solid or gas–liquid interfaces. These problems usually become more complex when fluid motion is involved. Traditionally, such complex problems are analyzed with the aid of the concept of a boundary layer. The boundary layer equations are well established for fluid flow, heat transfer, and mass transfer [8, 9–13]. This set of Dimensional Analysis and Scaling of Boundary Value Problems U∞ U∞ Free stream 379 d(x) U y d t t x FIGURE 8.1 Velocity boundary layer. equations is a convenient starting point for the scaling of the independent and dependent variables. However, before we begin it is helpful to briefly review descriptions of the three types of boundary layers. In fluid flow, one may consider a flat plate over which the fluid is flowing (Figure 8.1). Then fluid particles making contact with the surface are presumed to have zero velocity. These particles further act to retard the motion of particles in the adjacent fluid layers until, at some distance, y = δ from the solid surface, the fluid velocity is no longer influenced by the surface. The quantity δ is the boundary layer thickness. This retardation of fluid motion is associated with shear stresses, τ, acting in planes parallel to the fluid velocity (see Figure 8.1). In addition, as one moves away from the solid surface in the y-direction, the x component of the fluid velocity, u, increases until it approaches the free stream value u∞. This is the velocity boundary layer and is expected to develop whenever there is fluid flow over a surface. The velocity boundary layer is important in establishing surface frictional effects, such as local friction coefficient Cf: τ (8.6) Cf = 2 s ρu∞ / 2 where the subscript s refers to the surface. For Newtonian fluids, the surface shear stress is ∂u τs = µ ∂ y y= 0 (8.7) where μ is the fluid viscosity. Similar to a velocity boundary layer developing whenever there is fluid flow over a surface, a thermal boundary layer develops whenever there is a temperature gradient between the surface and the free stream fluid. That is, flow over an isothermal flat plate (see Figure 8.2) is expected to have a uniform temperature profile T(y) = T∞ at its leading edge. However, those fluid particles that come into contact with the plate achieve thermal equilibrium at the surface temperature. These equilibrated particles exchange energy with those in the adjacent layers, thus developing a temperature gradient in a region of the fluid. The region in which such temperature gradients develop is the thermal boundary layer whose thickness is δ t. 380 Applied Mathematical Methods for Chemical Engineers T U∞ Free stream d(x) T T y dt x FIGURE 8.2 Ts Thermal boundary layer. CA∞ U∞ Free stream dc(x) CA dc y C x FIGURE 8.3 CA,S Species concentration boundary layer. A relation between conditions in the thermal boundary layer and the convection heat transfer coefficient, h, is k (∂T / ∂ y) y = 0 (8.8) h=− Ts − T∞ since at the surface (y = 0), there is no fluid motion and energy transfer occurs by conduction. Analogous to velocity and thermal boundary layers is the concentration boundary layer. That is, it determines convection mass transfer, similar to the velocity boundary layer determining wall friction or the thermal boundary layer determining convection heat transfer. When a binary mixture of species A and B flows over a surface (see Figure 8.3) and the concentration of species A at the surface, CA,S, differs from CA,∞, the free stream concentration, a concentration boundary layer is expected to develop. That is, the region of the fluid in which concentration gradients exist is the concentration boundary layer. The thickness of the concentration boundary layer is δ c. Similar to Equation 8.8, a relationship between conditions in the concentration boundary layer and the mass transfer coefficient, hm, is given by hm = − DAB (∂CA / ∂ y) y = 0 CA,S − CA,∞ (8.9) Dimensional Analysis and Scaling of Boundary Value Problems 381 For steady, two-dimensional velocity, thermal, and concentration boundary layers, the equations of change apply. These equations are developed elsewhere [8–10,13]. However, it is unusual for all the terms in the equations of change to be needed to resolve a given situation. Therefore, it is customary to simplify these equations by assuming fluids with constant physical properties (ρ , k, and μ), negligible body forces, no chemical reaction, and no heat generation. In addition, the usual boundary layer approximations are incorporated to further reduce the equations of change [10] to Continuity : X-momentum : u ∂u ∂ v + =0 ∂x ∂ y (8.10) ∂u ∂u 1 ∂p ∂2 u +v =− +v 2 ρ ∂x ∂x ∂y ∂y ∂T ∂T ∂2 T v ∂u +v =α 2 + Energy : u ∂x ∂y ∂y Cp ∂ y Species continuity : u (8.11) 2 ∂CA ∂C ∂2 CA + v A = DAB ∂x ∂y ∂ y2 (8.12) (8.13) Equation 8.10 through 8.13 are the convection transfer equations. When appropriate boundary conditions are included, these equations can be solved to determine spatial variations of u, v, T, and CA in the boundary layers. In this section, these reduced equations will be scaled to establish important analogies between momentum, heat, and mass transfer, while identifying key design parameters. To achieve the stated purpose, we will define dimensionless independent and dependent variables in the following way: Let x * = x L and y* = y L (8.14) where L is a characteristic length for the surface of interest, such as the length of a flat plate. Also, define the dimensionless velocities v* and u* as u* = u u∞ and v* = v u∞ (8.15) and the dimensionless temperature, T*, and species concentration, as CA*, T* = T − Ts T∞ − Ts and CA* = CA − CA,S CA,∞ − CA,S (8.16) The substitution of these dimensionless quantities into the convection transfer equations results in ∂u* ∂ v* (8.17) Continuity : + =0 ∂ x * ∂ y* 382 Applied Mathematical Methods for Chemical Engineers Velocity : u* ∂u* ∂u* dp* 1 ∂ 2 u* + v* =− + ∂ x* ∂ y* dx * N ReL ∂ y* 2 (8.18) * 2 where p = p /ρu∞ and NReL = u∞L/ν is the Reynolds number, based on length as opposed to diameter. Thermal : u* ∂T * ∂T * 1 ∂2 T * + v* = ∂ x* ∂ y* N ReL Pr ∂ y*2 (8.19) where Pr = ν / α is the Prandtl number and the viscous dissipation term (∂u/∂y)2 neglected. Concentration : u* ∂CA* ∂C * 1 ∂2 C *A + v* A = ∂ x* ∂ y* N ReL Sc ∂ y*2 (8.20) where Sc = ν /DAB is the Schmidt number. Notice the similarity between Equations 8.19 and 8.20. Either one of these equations can now be used to model heat or mass transfer, and only the interpretation would change for heat or mass transfer. The dimensionless differential equations suggest how important results may be generalized. For example, in functional form, the dimensionless velocity is dp * u* = f1 x * , y* , N ReL , dx * (8.21) such that the friction coefficient may be expressed as Cf = 2 f2 ( x * , N ReL ) N ReL (8.22) for a given geometry. Similarly, the dimensionless temperature is dp* T * = f3 x * , y* , N ReL , Pr , dx * (8.23) and can be used to establish the Nusselt number: Nu = + ∂T * = f4 ( x * , N ReL , Pr ) ∂ y* (8.24) An average Nusselt number may be derived based on an average heat transfer coefficient, which is the result of integrating over the surface of the body under consideration. Such an average Nusselt number Nu would be more familiar, and is given in functional form as Nu = f5 ( N ReL , Pr ) (8.25) Dimensional Analysis and Scaling of Boundary Value Problems 383 Also, the Sherwood number, Sh, which is the mass transfer analog of the Nusselt number, is derivable from dp* CA* = f6 x * , y* , N ReL , Sc, dx * (8.26) That is, hm = + DAB ∂CA* L ∂ y* y* = 0 (8.27) but Sh = ∂C * hm L = f7 ( x * , N ReL , Sc) = + A ∂ y* y * = 0 DAB (8.28) or an average Sherwood number, Sh: Sh = f8 ( N ReL , Sc) (8.29) From the above discussion, we observe that scaling offers significant advantage in ­identifying important dimensionless groups. In fluid flow, the friction coefficient is important and for a given geometry, only the Reynolds number is needed (Equation 8.22). In the case of heat (or mass) transfer, the Nusselt (Sherwood) number can be ­correlated through the Reynolds and Prandtl (Schmidt) numbers (Equation 8.25 or 8.29). Application of Equations 8.22, 8.24, 8.25, 8.28, and 8.29 is not limited to the boundary layers. These are general results that are repeatedly used in other situations, such as for flow in conduits. Another advantage resulting from scaling is the ability to reduce the order of a given differential equation. For example, Equation 8.20 could be reduced to a ­first-order equation if the product of the Reynolds and Schmidt numbers is large enough. This reduction would not be as straightforward based on Equation 8.13 alone. To further emphasize this application, consider the steady-state laminar flow in a tube of radius R in which a solute A contained in the fluid undergoes a first-order reaction at the wall [14]. Suppose one wants to determine the conditions justifying two different approximations: 1.Assume that the reaction causes total depletion of A at the pipe wall. 2.Assume the “plug-flow reactor” approximation, for which the radial concentration gradient is ignored while the flow is taken to be equal to the average velocity. The mathematical statement of the problem is as follows (see Figure 8.4): vz ∂CA 1 ∂ ∂CA = DAB r r ∂r ∂r ∂z 384 Applied Mathematical Methods for Chemical Engineers Nonreactive wall Reactive wall r R z vz(r) CA(r, z) CA0 L FIGURE 8.4 Schematic of steady-state laminar flow in a tube with reactive walls. CA = CA0 at z = 0 ∂CA = 0 at r = 0 ∂r − DAB ∂CA = k1CA ∂r at r = R, 0 ≤ z ≤ L where CA0 is the initial concentration of the solute, DAB is the binary diffusivity, and k1 is the first-order reaction rate coefficient. The laminar flow velocity vz is given by vz = 3 r 2 v 1 − 2 R where v is the average velocity. Let CA* = r z CA , r * = , and z * = rs zs Cs Then substitute these scaled values into the differential equations and their boundary conditions. Following the substitution, divide through by the dimensional coefficient of one term. In this case it is convenient to divide by vCs /zs to obtain 3 rs 2 2 ∂CA* DAB zs 1 ∂ ∂CA* * r 1 − r* * = ∂r * 2 R v rs2 r* ∂r * ∂z CA* = CA0 Cs at z * = 0 ∂CA* = 0 at r* = 0 ∂r * ∂C * kr R L − A = 1 s CA* at r* = 0 ≤ z * ≤ ∂r* DAB rs zs Dimensional Analysis and Scaling of Boundary Value Problems 385 In this problem, the dimensionless groups suggest the following choices for the scale factors: CA0 = 1, Cs rs L = 1, and =1 R zs Therefore, the dimensionless equations become 3 ∂C * L 1 1 ∂ ∂C * r* A [1 − r* 2 ] A = ∂ z * R Pe r * ∂r * ∂r * 2 CA* = 1 at z * = 0 ∂CA* = 0 at r * = 0 ∂r * − ∂CA* = N ShCA* ∂r * at r * = 1, 0 ≤ z * ≤ L zs where the quantity Pe = vR/DAB is the Peclet number and is the ratio of convective mass transfer to molecular mass transfer, and N Sh = k1 R DAB is a Sherwood number or a Damkohler number. If, NSh ≫ 1, then CA* ≈ 0 to assure that ∂CA* / ∂r * is of order 1 at r * = 1. Therefore, for this limiting case (very fast reaction), the boundary condition can be replaced by CA* ≈ 0 at r * = 1 If NSh ≪ 1, then since CA* is of order 1, ∂CA* / ∂r * 1 at r * = 1. However, ∂CA* / ∂r * is largest at r * = 1. Therefore, ∂CA* / ∂r * 1 throughout the tube, and one can conclude that CA* ≈ C*A ( z * ). As the radial concentration gradient is negligible, the heterogeneous reaction term can be directly included into the species mass balance to obtain v dC A 2k = − 1 CA R dz for 0 ≤ z ≤ L subject to CA = CA0 at z = 0 which describes the classic plug-flow reaction assumption. 386 Applied Mathematical Methods for Chemical Engineers In summary, scaling analysis can be reduced to a stepwise procedure. Following Krantz and Sczechowski [14]: 1.Write down the dimensional differential equations and their initial and boundary conditions appropriate to the transport or reactor design process being considered. 2.Form dimensionless variables by introducing unspecified scale factors for each dependent and independent variable: this also may involve introducing unspecified reference factors for some variables whose values we seek to normalize to zero. 3.Introduce these dimensionless variables into the describing differential equations and their initial and boundary conditions. 4.Divide through by the dimensional coefficient of one of the terms (preferably, one which will be retained) in each of the describing equations and their initial and boundary conditions. 5.Determine the scale and reference factors by ensuring that the principal terms in the describing equations are of order 1; identifying the principal terms is dependent on the particular conditions for which the scaling is being done. (For example, a highly viscous flow, a conductive heat t­ ransfer process, etc. This step may require introducing a “region of influence,” wherein the dependent variable[s] goes through a characteristic change in value.) 6. Above steps result in the minimum number of dimensionless groups (parameters) for the problem. The describing equations can now be explored for very small or very large values of the dimensionless group. Finally, a second example is due to Krantz and Sczechowski [14]: consider the problem of steady-state, fully developed laminar flow between two infinitely wide parallel plates, shown in Figure 8.5. The lower plate is stationary, whereas the upper plate moves at a constant velocity Vp. The flow is also subjected to a constant axial pressure gradient such that Δ P > 0. Determine the conditions for which the effect of the upper plate velocity Vp can be neglected. Vp P x=0 H P x=L y x L FIGURE 8.5 plates. Schematic of steady-state, fully developed laminar flow between two parallel Dimensional Analysis and Scaling of Boundary Value Problems 387 Solution The equations of motion and their boundary conditions are as follows: 0=− ∂P d2v + µ 2x ∂x dy (8.30) ∂P + ρg ∂y (8.31) 0=− vx = 0 at y = 0 v x = VP (8.32) at y = H (8.33) Equation 8.31 can be integrated and combined with Equation 8.30 to obtain 0= ∆P d2v + µ 2x L dy (8.34) where Δ P = PX = 0 − PX = L . Define the following dimensionless variables vx* = vx Us and y* = y ys (8.35) Substitute Equation 8.35 into Equations 8.32 through 8.34 to yield 0= ∆P µU s d 2 vx* + 2 L ys dy*2 U s vx* = 0 at ys y* = 0 U s vx* = VP at ys y* = H (8.36) (8.37) (8.38) Since the viscous term in Equation 8.36 must be retained to satisfy the two no-slip conditions at the solid boundaries, divide through by its dimensional coefficient. Similarly, in the boundary conditions, divide through by the dimensional coefficient of the dimensionless dependent variable. This results in ∆Pys2 d 2 vx* + LµU s dy*2 (8.39) vx* = 0 at y* = 0 (8.40) 0= vx* = vP Us at y* = H ys (8.41) Since this problem is being scaled for conditions such that the flow is caused principally by the pressure gradient, the pressure force is to be balanced by the viscous term given in Equation 8.39. That is, ∆Pys2 =1 LµU s (8.42) 388 Applied Mathematical Methods for Chemical Engineers which ensures that the magnitude of the dimensionless derivative, d2vx/dy*2, is of order 1. Further, the dimensionless variable, y*, will be bounded of order 1 if we impose the condition that H =1 ys (8.43) Therefore, using Equations 8.42 and 8.43 gives Us = H 2 ∆P µL (8.44) The velocity scale given by Equation 8.44 is directly proportional to the maximum velocity for flow between two flat plates driven only by a pressure gradient. This scaling ensures that dimensionless velocity goes from its minimum value of 0 to its maximum value of 1. Finally, the dimensionless equations are described by Equations 8.45, 8.40, and 8.46. 0 = 1+ d 2 vx dy*2 (8.45) vx* = 0 at y* = 0 vx* = VP µL H 2 ∆P at y* = 1 (8.46) Therefore, to ignore the effect of the moving upper plate on the flow relative to that of the imposed pressure gradient, VP µL 1 H 2 ∆P (8.47) If one is interested in determining conditions such that flow is caused principally by the upper moving boundary, then the velocity scale would be U s = Vp according to Equation 8.41. In this case, the dimensionless equations would become 0= ∆PH 2 d 2 vx* + LµVP dy*2 Then, to neglect the effect of the pressure gradient on the flow, the criterion must be satisfied: H 2 ∆P 1 µVP L The analytical solution of this problem permits assessing the error incurred by neglecting the plate velocity. For example, if (Vp μ L/H2Δ P) ≤ 0.1, comparison with the exact Dimensional Analysis and Scaling of Boundary Value Problems 389 analytical solution shows that the error is 20% in the drag at the wall, whereas a 2% error in the drag results if (Vp μ L/H2Δ P) ≤ 0.01. From this discussion, one can see that scaling can provide the criteria for simplifying the equations of change. Scaling is also helpful in assessing the error incurred in making these simplifying assumptions. Generally, one can observe that the error incurred in using the discussed scaling technique will be of the same order as the dimensionless group, which must be small enough to be neglected. 8.5 PROBLEMS 1.Consider a fluid flowing in a closed conduit at some average velocity, v, with a temperature difference existing between the fluid and the tube wall. If the important variables, their symbols, and dimensional representations are tabulated below. a. Determine the number of pi groups. b. List the contents of the recurring set. c. What are the dimensionless groups? Variable Tube diameter Fluid density Fluid viscosity Fluid heat capacity Fluid thermal conductivity Velocity Heat transfer coefficient Symbol Dimensions D ρ μ cp K V H L M/L3 M/L t Q/M T Q/t L T L/t Q/t L2T Note: The quantity Q represents heat, whereas T is temperature. Answer: Nu = f ( N Re , Pr ) or N St = f ( N Re , Pr ) 2. In the study of natural convection heat transfer from a vertical plane wall to an adjacent fluid, the variables are as follows: Variable Significant length Fluid density Fluid viscosity Fluid heat capacity Fluid thermal conductivity Fluid thermal expansion Gravitational acceleration Temperature difference Heat transfer coefficient Symbol Dimensions L ρ μ cp K β G ΔT H L M/L3 M/L t Q/M T Q/t L T 1/T L/t2 T Q/t L2 T 390 Applied Mathematical Methods for Chemical Engineers a. Determine the number of pi groups. b. List the contents of the recurring set. c. What are the dimensionless groups? Answer: Nu = f (Gr , Pr ) 3.Consider the transfer of mass from the walls of a circular conduit to a fluid flowing through the conduit. The important variables are as follows: Variable Tube diameter Fluid density Fluid viscosity Fluid velocity Fluid diffusivity Mass transfer coefficient Symbol D ρ μ V DAB kc Dimensions L M/L3 M/L t L/t L2/t L/t a. Determine the number of pi groups. b. List the contents of the recurring set. c. What are the dimensionless groups? Answer: NuAB = f ( N Re , Sc) REFERENCES 1.Taylor, E.S. Dimensional Analysis for Engineers, Oxford University Press, London, 1974. 2. Massey, B.S. Measures in Science and Engineering, Ellis Horwood Limited, Chichester West, U.K., 1986. 3.Massey, B.S. Units, Dimensional Analysis and Physical Similarity, Van Nostrand, London, 1971. 4.Pankhurst, R.C. Dimensional Analysis and Scale Factors, Chapman and Hall, New York, 1964. 5.Hansen, A.G. Similarity Analysis of Boundary Value Problems in Engineering, Prentice Hall, Englewood Cliffs, NJ, 1964. 6. Murphy, G. Similitude in Engineering, Ronald Press Company, New York, 1950. 7. Isaacson, E. de St Q. and Isaacson, M. de St Q. Dimensional Methods in Engineering and Physics, John Wiley & Sons, New York, 1975. 8.Welty, J.R., Wicks, C.E., and Wilson, R.E. Fundamentals of Momentum, Heat and Mass Transfer, 3rd ed., John Wiley & Sons, New York, 1984. 9.Bird, R.B., Stewart, W.E., and Lightfoot, E.N. Transport Phenomena, John Wiley & Sons, New York, 1960. Dimensional Analysis and Scaling of Boundary Value Problems 391 10.Incropera, F.P. and DeWitt, D.P. Fundamentals of Heat and Mass Transfer, 4th ed., John Wiley & Sons, New York, 1996. 11. Kays, W.M. and Crawford, M.E. Convective Heat and Mass Transfer, 3rd ed., McGrawHill, New York, 1993. 12. White, F.M. Viscous Fluid Flow, 2nd ed., McGraw-Hill, New York, 1991. 13. Schlichting, H. Boundary Layer Theory, 7th ed., McGraw-Hill, New York, 1979. 14.Krantz, W.B. and Sczechowski, J.G. Scaling initial and boundary value problems, Chemical Engineering Education, 28, 236, 1994. This page intentionally left blank 9 Selected Numerical Methods and Available Software Packages 9.1 INTRODUCTION AND PHILOSOPHY Numerical methods will play a more significant role in engineering science as computer technology continues to improve. Methods that were previously thought to be too effort consuming are now commercially available in a number of software packages, such as Mathematica® (Wolfram Research, Inc.), Maplesoft (Waterloo Maple, Inc.), or MATLAB® (The Math Works, Inc.). For example, finding the roots of an equation, involving several confluent hypergeometric functions, can be done with a single Mathematica command. There are also websites, such as http://gams.nist.gov/ cgi-bin/serve.cgi/Packages or http://www.netlib.org/index.html, where math packages are available for downloading and use on local machines. In this chapter, the notion will be to alert the reader to some of the commercially available packages and identify which numerical methods are applied. It is not the intention of this chapter to produce an exposition on numerical analysis. However, it may be helpful to be aware of what kind of results a particular method might yield and be able to make informed decisions. To this end, the chapter will briefly review various methods and highlight their central ideas to allow a user to be mindful of the methods’ virtues as well as their limitations. There will be examples worked out to varying level of completion as needed to demonstrate critical steps. In the last few sections under the discussion of the method of lines approach, given examples will need to include computer programming. The computer programming details will be referenced and provided in the appendices as developed for the solution of that example given in that section. 9.2 SOLUTION OF NONLINEAR ALGEBRAIC EQUATIONS Nonlinear algebraic equations turn up quite frequently in chemical engineering and may appear in several different forms. For example, in thermodynamics, pressure– volume–temperature relationships of real gases are often described by equations of state, such as, PV 4 − RTV 3 − βV 2 − γV − δ = 0 (9.1) where P, V, and T are the pressure, specific volume, and temperature, respectively. R is the gas constant and β, γ, and δ are empirical functions of temperature, specific 393 394 Applied Mathematical Methods for Chemical Engineers to each gas. Equation 9.1 is the Beattie–Bridgeman equation of state and is a fourthdegree polynomial in the specific volume. In multicomponent distillation, it may be necessary to estimate the minimum reflux ratio using classical methods [1, 2]. This estimate usually requires the solution of a polynomial in ɸ of degree n, such as the equation n α z jF F ∑ αj j =1 j −φ − F (1 − q) = 0 (9.2) where F is the molar feed flow rate, n the number of components in the feed, zjF the mole fraction of each component in the feed, q the feed quality, α j the relative volatility of each component at some average column conditions, and ɸ the sought-after root of the equation. Fanning friction factor, f, for turbulent flow of an incompressible fluid in a smooth pipe is f 2 1 = ln N Re + B− A f k 8 (9.3) The quantities A, B, and k are constants. NRe is the Reynolds number. Equation 9.3 is not in polynomial form, but can be rearranged to have all nonzero terms on one side of the equation. In fact, all three of the equations mentioned can be represented by the general form f (x) = 0 (9.4) where x is a variable, which can have multiple values that satisfy the equation. It is important to remember that the roots of Equation 9.4 can be real and distinct or real and repeated, or have complex conjugates or some mixture of real and complex. Regardless of the type of roots, it may be helpful to have some sense as to which method one should use to attack the problem. That is, one would like to know if, by choosing a particular iterative procedure, there would be convergence. A standard approach rearranges Equation 9.4 to develop a convergence criterion [3]. Suppose x is a root of Equation 9.4; then f (x ) = 0 . and the equation is satisfied. However, the value x, which is not a root when substituted into Equation 9.4, would result in f ( x ) ≠ 0 This guessing could go on indefinitely. But suppose we perform some algebraic manipulations on Equation 9.4 to extract an x out of it and bring it to the other side of the equation to get Selected Numerical Methods and Available Software Packages x = g( x ) 395 (9.5) Then, if x is substituted into Equation 9.5, we would get the result x = g( x ) (9.6) whereas x ≠ g( x ) Notice that g(x) is a new function that is different from the original f(x). Because we usually have no idea where the roots of f(x) lie, it is typical to make a guess, and check that guess to determine which way to go next. However, it turns out that the criterion [3] g ′( x ) < 1 (9.7) can be used to increase the chance of finding the roots of Equation 9.4 in an efficient way. That is, the absolute value of the derivative of g(x) needs to be less than 1 in the region containing the root x to guarantee convergence. For example, if f ( x ) = e x − 3x = 0 then x= ex ≡ g( x ) 3 such that g ′( x ) = ex 3 and the range of x, for which the absolute value of the derivative is less than 1, is given by ex <1 3 Therefore x < ln 3 ≅ 1.1 Essentially, one should guess starting values of x smaller than 1.1 to expect convergence. While this rearranging of the independent variable seems straightforward, one should exercise caution in accepting what appears to be obvious. For suppose the model 396 Applied Mathematical Methods for Chemical Engineers dCA F0 F = CA0 − 0 CA − kCA2 dt V V is used to represent the material balance of species A in an isothermal, constant volume chemical reactor system with second-order reaction. If one is interested in the steady-state concentration, CA,s, then for an initial concentration, CA0 of 1 mol/L, a feed to volume ratio, F0/V of 1 min−1, and a rate coefficient, k = 1 L/(mol min), the model becomes 2 1 − CA,s − CA,s =0 In the form of Equation 9.4, substituting x = CA,s, we get 1 − x − x2 = 0 a quadratic whose direct solutions are x = −1.618 and 0.618. Of course, being a physical system, the negative result is ignored. For solving this problem by a substitution method, one would develop the form of Equation 9.5. In this example, there are two possibilities: 1. g1 ( x ) ≡ x = 1 − x 2. g2 ( x ) ≡ x = − x 2 + 1 The first case results in g ′( x ) = −1 2 1− x such that x < 3/4 satisfies the requirement of Equation 9.7. After several iterations using direct substitution, the result x = 0.618 is achieved. The second possibility results in g ′( x ) = −2 x such that x < 1/2 satisfies the requirement of Equation 9.7. However, initial guess values of x < 1/2 would be useless, as convergence to the correct value of 0.618 could never be achieved. If one were to solve either of these examples explicitly using an iterative procedure, the Mathematica routine “FindRoot” [4] could perform this task with a starting value, x0, a minimum value, xmin, and a maximum value, xmax, selected from the suggested range. FindRoot employs either of the two approaches, depending on the number of starting values of x. If only one starting value is given, FindRoot employs Newton’s method [3,5,6]. On the other hand, if one specifies two starting values, then a variant of the Secant method is employed [5]. Selected Numerical Methods and Available Software Packages 397 As a demonstration of the use of FindRoot, consider the following equation, which defines the eigenvalues in a model of controlled release from a hollow fiber [7]: a ax a ax [ kJ1 ( x ) + βxJ0 ( x )] Y1 x J0 − m kY0 x J1 b b b k b k (9.8) ax a ax a − + [ kY1 ( x ) + βxY0 ( x )] m k J1 = J x J J x 0 0 b k 1 b b k 0 b where a, b, β, k, and m are given constants and J0(.), Y0(.), J1(.), and Y1(.) are Bessel functions. With given values of the constants, the line ax a BesselY 1, x BesselJ 0, kBesselJ [1, x ] b b k + ax a +βxBesselJ [0, x −m k BesselY 0, x b BesselJ 1, b k FindRoot ax a kBesselY [1, x ] m k BesselJ 1, b k BesselJ 0, x b − = 0, ax a +βxBesselY [0, x ] BesselJ 0, BesselJ 1, x b b k {x ,{x , x }} 0 1 returns a numerical value for x. It is clear from this example that very complicated nonlinear functions can be dealt with in a straightforward manner by this routine. Previously, several lines of Fortran codes with careful attention to stability issues would be required to effect a solution. In addition to a single nonlinear equation, this routine can also be used to produce numerical solutions to systems of equations without regard to linearity. In addition to FindRoot, there is also “NSolve” [4], which is more suited to finding the roots of polynomials. Following the Mathematica system, polynomial root finding is based on the Jenkins–Traub algorithm [8]. As mentioned previously, FindRoot can give results for systems of equations without regard to linearity. Therefore, similar to Equation 9.7, a criterion for systems can be established [3]. For a system of two equations, we have ∂ g1 ∂g + 2 ≤ M <1 ∂x ∂x ∂ g1 ∂g + 2 ≤ M <1 ∂y ∂y (9.9) 398 Applied Mathematical Methods for Chemical Engineers for all x and y in the region containing all the values xi and yi, the correct values x and y. When the bound, M, on the partial derivatives are very small in the entire region, the iteration converges very quickly. The quantities xi and yi are the iterants, whereas g1 and g2 are formed exactly the way Equation 9.5 was developed. Two common methods for finding roots to nonlinear systems are (1) Newton–Raphson and (2) the modified Newton–Raphson. Both approaches are briefly discussed in the subsections below. 9.2.1 NEWtON–RAPhsON MEthOD Suppose one wishes to solve the simultaneous equations f1 ( x , y) = 0 f2 ( x , y ) = 0 If both functions are continuous and differentiable, the following system can be developed [3]: ∂ f1 ∂f h + 1 k = − f1 ( xi , yi ) ∂x ∂y ∂ f2 ∂f h + 2 k = − f2 ( xi , yi ) ∂x ∂y (9.10) This system is the heart of the Newton–Raphson procedure. Here the partial derivatives are evaluated at xi and yi. The quantities h and k are the unknowns, and are defined as x i +1 = x i + h (9.11) yi +1 = yi + k The system given by Equation 9.10 is very straightforward to solve, but requires the Jacobian determinant, J (see Appendix A), J= ∂ f1 ∂ f1 ∂x ∂ y ∂ f2 ∂ f2 ∂x ∂ y (9.12) to be nonsingular. For example, if one were to solve the system y = cos x; x = sin y using the Newton–Raphson procedure, the first step is to rewrite the system as f1 ( x , y) = cos x − y = 0 f2 ( x , y) = x − sin y = 0 Selected Numerical Methods and Available Software Packages 399 Then differentiate to get ∂ f1 ∂ f2 = − sin x; =1 ∂x ∂x ∂ f1 ∂ f2 = −1; = − cos y ∂y ∂y such that the Jacobian is J= − sin x −1 = sin x cos y + 1 1 − cos y Then, in order for the Jacobian to vanish, either sin x or cos y must be negative; that is, sin x cos y = −1. Therefore, as long as the iterants xi and yi are selected from the first quadrant, the Jacobian will not vanish. Then h and k can be determined for each iteration until some tolerance is met. The need to know if the Jacobian is nonzero a priori cannot be met without some difficulty. In most cases, the procedure has to be carried out before such knowledge is in hand. Also, many computations are required for the use of this procedure; that is, for n simultaneous equations in n unknowns, n2 partial derivatives must be determined. One must also solve for n increments h, k, … by solving n simultaneous equations for n unknowns. These two disadvantages make it attractive to select other approaches of solving systems of nonlinear equations. One such approach involves a modification to the Newton–Raphson procedure and is highlighted in the next section. 9.2.2 MODIFIED NEWtON–RAPhsON MEthOD This procedure is essentially applying the single variable Newton–Raphson method n times, once for each variable. Each time the other variables are held constant. The approach is, given f1 ( x , y) = 0 f2 ( x , y ) = 0 then x1 = x 0 − f1 ( x 0 , y0 ) ∂ f1 / ∂ x (9.13) where ∂f1/∂x is evaluated at x0 and y0. Next, f 2 and the most recent values of x and y, in this case x1 and y0, are used to calculate y1: y1 = y0 − f2 ( x1 , y0 ) ∂ f2 / ∂ y (9.14) 400 Applied Mathematical Methods for Chemical Engineers ∂f 2/∂y is evaluated at x1 and y0. With xi and yi, f1 is reused to calculate x2. Then f 2 and the most recent values of x and y are used to calculate y2 [3]. Notice that the choice of using f1 to calculate a new x and using f 2 to calculate a new y appears arbitrary. Sometimes, that is, indeed, the case. However, the choice is supposed to be based on choosing the function with the steeper slope at the solution point ( x , y) to determine the next x. Otherwise, if both functions have the same slope in a neighborhood of the solution, the iteration will oscillate about a true solution. The real disadvantage in the modified Newton–Raphson method is the need to know a priori which function has the steeper slope at the solution point. In practice, a graph of the functions can help to identify the curve with the steeper slope at the solution point. If a graph is not available, then one has to resort to an arbitrary choice of functions, and if divergence occurs, switch the roles of f 1 and f 2. In summary, whether we are faced with a single nonlinear equation or a system, it is very likely that only one starting value will be available. A systematic selection of that value can improve the likelihood of a convergent solution. Equations 9.7 and 9.9 provide such a systematic approach. At this point, it is important to point out that there are other software packages, such as MATLAB, MAPLE, and MATHCAD, that can also be employed to solve nonlinear algebraic equations. As a matter of fact, there are books providing a wide range of problems solved with MATLAB [9]. Mathematica has packages available for calling MATLAB from within Mathematica and vice versa. MATLAB also has toolboxes that incorporate some of the symbolic capabilities of MAPLE, as pointed out in the literature [10]. MATHCAD, on the other hand, is lacking in its ability to carry out symbolic computations when compared to Mathematica or MATLAB. However, it is intuitive and very popular with the beginner, because there is no need to memorize commands. MAPLE can produce program statements for a Fortran compiler in addition to its ability to perform symbolic algebra and numeric approximations. However, in a comparative study with Mathematica and MATLAB, it was ranked behind Mathematica and MATLAB, respectively [11]. 9.3 SOLUTION OF SIMULTANEOUS LINEAR ALGEBRAIC EQUATIONS When compared to other areas of numerical analysis, numerical linear algebra is well developed, because there is general agreement as to the best algorithms, and why they are best [5,13]. There is a large body of work on the implementation of algorithms. This has resulted in packages of very efficient and reliable subroutines for the solution of problems in linear algebra [5]. A primary goal of this section is to familiarize the reader with some of the background reasoning for the selection of algorithms. Familiarity with matrix–vector notation and results involving matrices will be assumed, but most of what is needed is provided in the Appendix A. Detailed derivations will be replaced by clarifying examples, most of which will be borrowed from Selected Numerical Methods and Available Software Packages 401 the literature. These borrowed examples are chosen, because they have already been tested in the context of the algorithm or method under discussion. Starting the discussion with the example [14], x1 + x 2 + 2 x3 = 9 : E1 3 x1 − 2 x 2 + 3 x3 = 8 : E 2 (9.15) 4 x1 + 2 x 2 − 2 x3 = 2 : E 3 The objective is to determine the values of x1, x2, and x3 that will simultaneously solve all three equations. In general, the System 9.15 is a subset of the n linear equations in n unknowns represented by Ax = b (9.16) where A is an n × n matrix and both x and b represent column vectors (n × 1). Specifically, Equation 9.16 can be expanded into a11 x1 + a12 x 2 + a1n x n = b1 : E1 a21 x1 + a22 x 2 + a2 n x n = b2 : E2 ak 1 x1 + ak 2 x 2 + akn kn = bk : Ek an1 x1 + an 2 x 2 + ann x n = bn (9.17) : En or condensed using summation notation to n ∑ aki xi = bk (k = 1,2,, n) (9.18) i =1 Regardless of the form, the objective will be the same. Linear equations result naturally when we conduct material and energy balances, but most applications occur when we implement other numerical methods. One of the most basic solution techniques for systems such as Equation 9.15 is Gaussian elimination [3,5,9,13,14], which is illustrated using the System of Equations 9.15. We start by using the coefficient of x1 in the first equation, designated as E1, to eliminate the coefficients of x1 in the other equations—designated as E2 and E3, respectively. That is, (−4) times E1 added to E3 produces 0 x1 − 2 x 2 − 10 x3 = −34 and (−3) times E1 added to E2 produces 0 x1 − 5 x 2 − 3 x3 = −19 402 Applied Mathematical Methods for Chemical Engineers The new equivalent system of equations resulting from the above elementary operation is x1 + x 2 + 2 x3 = 9 −2 x2 − 10 x3 = −34 (9.19) −5 x 2 − 3 x3 = −19 The next step is to repeat the procedure on this equivalent system, starting with the coefficient of the second row to eliminate the coefficient of x2 in the third row. That is, (−5/2) times the new row 2, add to the new row 3, gives 0 x 2 + 22 x3 = 66 The final equivalent system of equations is in triangular form: x1 + x 2 + 2 x3 = 9 −2 x2 − 10 x3 = −34 (9.20) 22 x3 = 66 It is not difficult to see how to derive the final results by solving for x3 then x2 and x1, respectively, such that x1 1 x = x2 = 2 x3 3 or x1 = 1, x2 = 2, and x3 = 3. The procedure just described can be summarized into two basic steps: 1. Reduce or transform the given problem to an equivalent one, which is more easily solved 2.Solve the reduced problem By equivalent problem we mean a problem that has the same solution as the original one. The procedure can be generalized and is available in algorithms in many texts on numerical analysis [5, 9, 13–15]. The first equation in the system of Equations 9.17 times (a21/a11) is subtracted from the second to eliminate the first term of the second equation; likewise, the first term of every equation thereafter, k > 2, is eliminated by subtracting the first equation times (ak1/a11). The result of this step should be Selected Numerical Methods and Available Software Packages 403 a11 x1 + a12 x 2 + a1n x n = b1 a22 ′ x 2 + a2′ n x n = b2′ ak′ 2 x 2 + akn ′ xn = bk′ an′ 2 x 2 + ann ′ xn = bn′ where akj′ = aij − mk 1aij ; mkl = (ak 1 /a11 ) Next, the second term of every equation in the third through the last equation, k > 2, is eliminated by subtracting the second equation times ak′ 2 /a22 ′ = mk′ 2. Following this step, the third terms of the fourth through the last equation are eliminated. This is continued until the forward elimination process is completed and the form a11 x1 + a12 x 2 + a13 x3 + + a1n x n = b1 a22 ′ x 2 + a23 ′ x3 + + a2′ n x n = b2′ a33 ′′ x3 + + a3′′n x n = b3′′ (9.21) ( n −1) nn n a x = bn′ ( n −1) results. The leading terms in each equation are called pivots. The backward substitution procedure starts with the last equation, giving the solution xn = bn( n −1) ( n −1) ann Subsequently, x n −1 = ( bn(n−−11) − an(n−−1,1)n xn ) an( n−−1,2)n −1 n x1 = b1 − ∑ aij x j a11 j=2 completing Gaussian elimination. (9.22) 404 Applied Mathematical Methods for Chemical Engineers In actual machine computation, the form given by Equation 9.16 is more useful for large systems. Here, we will rework the example using the above procedure, but with the alternate form given by Equation 9.16: x1 + x 2 + 2 x3 = 9 : E1 1 1 2 x1 9 3 x1 − 2 x 2 + 3 x3 = 8 : E 2 ⇔ 4 2 −2 x 2 = 2 4 x1 + 2 x 2 − 2 x3 = 2 : E 3 3 −2 3 x3 8 1 1 2 9 ⇒ 4 2 −2 2 3 −2 3 8 where the first three columns are the coefficients of the equations in the System 9.15 and the last column is the vector b as represented in Equation 9.16 or the nonhomogeneous term. This vector b is appended unto the array of coefficients to give what is known as an augmented array. Notice that the second and third equations in the System 9.15 are switched for convenience. To start the forward elimination process, the first row times 4 is subtracted from the second row. The first row times 3 is subtracted from the third row. The resulting augmented matrix is 1 1 2 9 0 −2 −10 −34 0 −5 −3 −19 Continuing the forward elimination, the second row times (5/2) is subtracted from the third row. The augmented matrix is now 1 1 2 9 0 −2 −10 −34 , 0 0 22 66 which ends the forward elimination part of the procedure. Notice the form of the array; that is, each element below the main diagonal is zero. This is an example of an upper triangular array (see Appendix A). The same triangular form resulted earlier. The backward substitution starts with the last row. Interpreting the last row as 22 x3 = 66 results in x3 = 3. Similarly, the second and first rows, respectively, result in 2 x 2 − 10 x3 = −34 ⇒ x 2 = 2 and x1 = 1 Selected Numerical Methods and Available Software Packages 405 Also demonstrated in this example is the fact that Gaussian elimination works when all the pivots are nonzero and in most cases are of the same order of magnitude as the other elements. The method fails if the pivots are zero. When the pivots are relatively small, round-off error can create erroneous results unless corrective measures are introduced early. Here is an example in which the pivots are of very different orders of magnitude. The system 0.0030 x1 + 59.14 x 2 = 59.17 : E1 5.291x1 − 6.130 x 2 = 46.78 : E 2 has exact solution x1 = 10.00 and x2 = 1.000. However, if we perform Gaussian elimination using four-digit arithmetic with rounding, the first pivot is a11 = 0.0030, and its associated multiplier, m21, is 5.291/0.0030 = 1764. Carrying out the operation (E2 − m21E1) is supposed to give the equivalent system: 0.0030 x1 + 59.14 x 2 = 59.17 −104300 x 2 = −104400 Then, backward substitution gives x2 = 1.001, and x1 = −10.000. However, notice that a small error of 0.001 in x2 leads to a very large error in x1 (factor of 20,000). If we revisit the system, but this time institute the corrective action by switching E1 with E2 to get 5.291x1 − 6.130 x 2 = 46.78 : E1 0.0030 x1 + 59.14 x 2 = 59.17 : E 2 then the multiplier for this system is m21 = (0.0030/5.291) and the operation (E2 − m21E1) results in the equivalent system 5.291x1 − 6.130 x 2 = 46.78 59.14 x 2 = 59.14 such that the four-digit results, using backward substitution, are x1 = 10.00 and x2 = 1.000. This illustrates that Gaussian elimination can run into difficulties in cases where (k ) is small relative to the entries aij( k ) for k ≦ i ≦ n and k ≦ j ≦ the pivot element akk n. Pivoting strategies are usually accomplished by selecting a new element for the pivot, a (pqk), and then interchanging the kth and pth rows followed by the interchange of the kth and qth columns, if necessary. The strategy used in the latter example is termed partial pivoting or maximal column pivoting. Detailed discussions on pivoting can be found elsewhere [5, 9, 13, 16–18]. 406 Applied Mathematical Methods for Chemical Engineers A class of procedures that are not as sensitive to round-off errors is the so-called iterative procedures. One such procedure is the Gauss–Seidel method, briefly described below [3, 5, 9, 14]. Consider the System 9.17, in which the diagonal elements a11, a22, …, ann are all nonzero, if necessary, the equations must be rearranged so that all diagonal elements are nonzero. Solve the first equation for x1, solve the second equation for x2, and so on, to get 1 (b1 − a12 x 2 − a13 x3 − a1n x n ) a11 1 (b2 − a21 x1 − a23 x3 − a2 n x n ) x2 = a22 1 (b3 − a31 x1 − a32 x 2 − a3n x n ) x3 = a33 x1 = (9.23) xn = 1 (bn − an1 x1 − an 2 x 2 − an ,n −1 x n −1 ) ann Then start the iteration by choosing an initial set of guesses for all the xi and inserting these into the first equation to calculate a new x1. This value of x1 along with the previously guessed values for x3, x4, …, xn are inserted into the second equation to calculate x2. The procedure is continued down the list until a new value of xn is determined, and then return to the top of the list to repeat the procedure if necessary. As an illustrative example, consider the system 6 x1 − x 2 + x3 = 7 : E1 x1 + 4 x 2 − x3 = 6 : E 2 x1 − 2 x 2 + 4 x3 = 9 : E 3 Then, solving for x1 using the first equation, E1, x2 using E2 and x3 using E3 results in 7 1 1 + x 2 − x3 6 6 6 3 1 1 x 2 = − x1 + x3 2 4 4 9 1 1 x3 = − x1 + x 2 4 4 2 x1 = Next, we guess x = (0, 0, 0)t as the starting point, and successive application leads to Selected Numerical Methods and Available Software Packages 407 7 1 7 29 = 1.16, x 2 = 3 / 2 − = = 1.21, 6 4 6 24 1 7 1 29 x3 = 9/4 − + = 2.56 4 6 2 24 x1 = The vector x = (1.16, 1.21, 2.56)t constitutes the values for the next iteration. This process was carried out until the result x = (1, 2, 3)t was achieved. This required six iterations. Convergence is assumed when a change in each of the answers satisfies a preset criterion. Generally, the procedure is guaranteed to converge whenever the diagonal element is larger than the sum of the row elements or column elements. Another in the class of iterative procedure is the Jacobi method. This method is illustrated by solving the following system: 10 x1 − x 2 + 2 x3 = 6 : E1 − x1 + 11x2 − x3 + 3 x 4 = 25 : E 2 2 x1 − x2 + 10 x3 − x4 = −11: E 3 3 x2 − x3 + 8 x 4 = 15 : E 4 This system has exact solution x = (1, 2, −1, 1)t. The goal of the method is to convert Ax = b to the form x = Tx + c. This is accomplished by solving equation Ei for xi, for each i = 1, 2, 3, 4 to obtain 1 1 3 x 2 − x3 + 10 5 5 1 1 3 25 x 2 = x1 + x3 − x 4 + 11 11 11 11 1 1 1 11 x3 = − x1 + x 2 + x 4 − 5 10 10 10 3 1 15 x 4 = − x 2 + x3 + 8 8 8 x1 = Thus, the matrix T is given by 0 1 / 10 −1 / 5 0 1 / 11 0 1 / 11 −3 / 11 T= −1 / 5 1 / 10 0 1 / 10 0 −3 / 8 1 / 8 0 and the vector c, by c = (3/5, 25/11, −11/10, 15/8)t. 408 Applied Mathematical Methods for Chemical Engineers To start this procedure, we guess x(0) = (0, 0, 0, 0)t and generate xi(1) by 1 (0) 1 (0) 3 x 2 − x3 + = 0.6000 10 5 5 1 1 3 25 = x1(0) + x3(0) − x 4(0) + = 2.2727 11 11 11 11 1 1 1 11 = − x1(0) + x 2(0) + x 4(0) − = −1.1000 5 10 10 10 3 1 15 = − x 2(0) + x3(0) + = 1.8750 8 8 8 x1(1) = x 2(1) x3(1) x 4(1) Additional iterates, k, can be generated for x ( k ) = ( x1( k ) , x 2(k ) , x3(k ) , x 4(k ) ) . By continuing the procedure up to k = 10, we get t 1 (9) 1 (9) 3 x 2 − x3 + = 1.0001 10 5 5 1 1 3 25 = x1(9) + x3(9) − x 4(9) + = 1.9998 11 11 11 11 1 1 1 11 = − x1(9) + x 2(9) + x 4(9) − = −0.9998 5 10 10 10 3 1 15 = − x 2(9) + x3(9) + = 0.9998 8 8 8 x1(10) = x 2(10) x3(10) x 4(10) The decision to stop after 10 iterations is due to the criterion that the relative difference between the last two iterations is to be smaller than some appropriately chosen tolerance. Notice that one difference between the Gauss–Seidel and Jacobi methods is in their use of the newly calculated xi value. In Gauss–Seidel, the newly calculated xi value is used to determine the xi+1 value, but this is not so in the Jacobi method. In most cases, this leads to faster convergence of the Gauss–Seidel over the Jacobi method. Generally, iteration methods are most useful for large matrices with a substantial number of zero elements. In procedures such as the Jacobi and Gauss–Seidel methods, a residual vector r, r = b − Ax (9.24) where x is an approximation to the solution vector, is associated with each calculation of a component to the solution vector [5]. Then, if ri(k) = ( rli( k ) , r2(ik ) ,, rni( k ) ) t (9.25) is the residual vector for the Gauss–Seidel method corresponding to the approximate t (k ) solution vector ( x1( k ) , x 2(k ) ,, xi(−k1) , xi( k −1) ,, x n( k −1) ) , the mth component of ri is Selected Numerical Methods and Available Software Packages i −1 n j =1 j =1 i −1 n 409 rmi( k ) = bm − ∑ amj x (j k ) − ∑ amj x (j k −1) = bm − ∑ amj x (j k ) − j =1 ∑ (9.26) amj x (j k −1) − amj xi( k −1) j = i +1 for each m = 1, 2, …, n. Therefore, the ith component of ri( k ) is i −1 rii( k ) = bi − ∑ aij x (j k ) − j =1 n ∑ aij x (jk −1) − aii xi( k −1) (9.27) j = i +1 such that i −1 rii( kk ) + aii xi( k −1) = bi − ∑ aij x (j k ) − j =1 n ∑ aij x (jk −1) (9.28) j = i +1 However, in the Gauss–Seidel method, xi( k −1) = 1 aii i −1 (k ) b − i ∑ aij x j − j =1 n ∑ aij x (jk −1) j = i +1 Therefore rii( k ) + aii xi( k −1) = aii xi( k ) (9.29) or xi( k ) = xi( k −1) + rii( k ) aii (9.30) The result given by Equation 9.30 is obtained by requiring that the ith coordinate of the vector, ri(+k1) be zero at each stage. However, the reduction to zero of one coordinate of the residual vector is not the most effective way to reduce the norm of ri(+k1). A modification that can lead to a more significant improvement in convergence speed is xi( k ) = xi( k −1) + ω rii( k ) aii (9.31) for certain choices of the positive relaxation parameter, ω. Methods using Equation 9.30 are called relaxation methods [5]. For choices of 0 < ω < 1, we have under-relaxation methods, which are successful for some systems that are not convergent for Gauss–Seidel. Those methods 410 Applied Mathematical Methods for Chemical Engineers associated with ω > 1 are called over-relaxation methods and are useful in accelerating the convergence for systems that are already convergent by Gauss– Seidel. These over-relaxation methods are also named successive over-relaxation (SOR), and find application in the numerical solution of certain partial differential equations. In practice, Equations 9.28 and 9.31 are appropriately combined to give xi( k ) = (1 − ω ) xi( k −1) + ω aii i −1 (k ) bi − ∑ aij x j − j =1 n j = i +1 ∑ aij x (jk −1) (9.32) which is the central theme of the SOR methods. Below is an illustration comparing the SOR, Equation 9.32, and the Gauss–Seidel methods. The following system 4 x1 + 3 x 2 = 24 : E1 3 x1 + 4 x 2 − x3 = 30 : E 2 − x 2 + 4 x3 = −24 : E 3 has exact solution x = (3, 4, −5)t. For both methods the initial guess is x(0) = (1, 1, −1)t. The equations for Gauss–Seidel are x1( k ) = −0.75 x 2(k −1) + 6 x 2(k ) = −0.75 x1( k ) + 0.25 x3(k −1) + 7.5 x3(k ) = 0.25 x 2(k ) − 6 for each k = 1, 2, … and the equations for the SOR method with ω = 1.25 are x1( k ) = −0.25 x1( k −1) − 0.9375 x 2(k −1) + 7.5 x 2(k ) = −0.9375 x1( k ) − 0.25 x 2(k −1) + 0.3125 x3(k −1) + 9.375 x3(k ) = 1.3125 x 2(k ) − 0.25 x3(k −1) − 7.5 Table 9.1 summarizes the results of the comparison for up to seven iterations. From this table, we see that the SOR method is converging much faster than the Gauss– Seidel. Not shown in the table is the fact that the Gauss–Seidel method required 34 iterations, versus 14 for the SOR to obtain seven-decimal-place accuracy. Guidelines on how one may choose ω, including Kahan and Ostrowski-Reich conditions, can be found in the literature [5]. Selected Numerical Methods and Available Software Packages 411 TABLE 9.1 Comparison of Gauss–Seidel and SOR Methods Gauss–Seidel SOR with ω = 1.25 (1) 1 x 5.2500 6.3125 x 2(1) 3.8125 3.5195 x3(1) −5.0469 −6.6501 x1(5) 3.0343 3.0037 x 2(5) 3.9714 4.0029 x3(5) −5.0071 −5.0057 x1(7) 3.0134 3.0000 x 2(7) 3.9888 4.0002 x3(7) −5.0028 −5.0003 9.3.1 ERROR EstIMAtE In addition to what has been said above about achieving a good approximation to the solution vector, x in Ax = b, it is necessary to add a few more words regarding “very difficult to converge” systems and suggest how one may attack such problems. It is not unreasonable to expect that if x is an approximation to the solution x, then the residual r = A x − b should have the property that the norm, ∥r∥ is small such that the norm of the difference, x − x would be small. Quite often, this is exactly the case. However, certain systems occurring in practice fail to display this behavior. For example, the system given by 1 2 x1 3 1.0001 2 x = 3.0001 2 has the approximate solution, x = (3,0) t with the residual vector 3 1 2 3 0 r = b − Ax = − = 3.0001 1.0001 2 0 .0002 Using matrix property 8, given in the Appendix A (Equation A.20), we can evaluate the ℓ∞ norm of the vector r to be ∥r∥ℓ ∞ = .0002. Even though the norm of this residual vector appears to be quite small, it is clear that the approximation x = (3,0) t is poor; in fact, using Equation A.22, x − x ∞ = 2. 412 Applied Mathematical Methods for Chemical Engineers This example demonstrates what can happen when we approximate the solution to a system representing two almost parallel lines. The point (3, 0) lies on one line: x1 + 2x2 = 3 and is very close to the other line given by 1.0001x1 + 2x2 = 3.0001, but is very different from the actual intersection point (1, 1). Generally, the system geometry may not be available to help us determine a priori when problems might occur; therefore, we must find an alternative. Such help can be obtained from the norms of the matrix A and its inverse [5,16,19,20]. In this section, we will use the ℓ∞ norm as opposed to the ℓ2 norm for matrices (as opposed to vectors), where A ∞ = max Ax x∞ =1 (9.33) is the ℓ∞ norm and the ℓ2 norm is A 2 = max Ax 2 x2 =1 (9.34) Further, if A = [aij] is an n × n matrix, then n A ∞ = max ∑ | aij | 1≤ i ≤ n j =1 To see how Equation 9.35 works, suppose 1 2 −1 A = 0 3 −1 5 −1 1 then n ∑ | a1 j | = | 1 | + | 2 | + | −1 | = 4 j =1 n ∑ | a2 j | = | 0 | + | 3 | + | −1 | = 4 j =1 n ∑ | a3 j | = | 5 | + | −1 | + | 1 | = 7 j =1 therefore n A ∞ = max ∑ | aij | = max{4,4,7} = 7 1≤ i ≤ n j =1 (9.35) 413 Selected Numerical Methods and Available Software Packages To continue the discussion on how one might use the norms of A and its inverse, A−1, to anticipate accuracy based on the residual of an approximation, we need the following [5]: x − x ≤ r A −1 (9.36) x − x r ; x ≤ A A −1 x b (9.37) and A quantity called the condition number, K(A), is defined relative to a norm to be K ( A) = A A −1 (9.38) r b (9.39) x − x r ≤ K ( A) r b (9.40) such that x − x ≤ K ( A) and Then, since 1 = I = A ⋅ A −1 ≤ A A −1 = K ( A) it is expected that the matrix A will be a well-conditioned matrix if K(A) is close to 1 and an ill-conditioned matrix if K(A) is much greater than 1. Returning to the first example of this section 1 2 A= 1.0001 2 has ∥A∥ ∞ = 3.0001. This quantity does not appear to be too large; however, −10000 10000 A −1 = 5000.05 −5000 such that ∥A−1∥ ∞ = 20000 and K(A) = 60002—indeed a very significant difference from one. We have omitted the computation details of the inverse here, and refer the reader to the Appendix A for a method on how to calculate it. 414 Applied Mathematical Methods for Chemical Engineers In practice, the calculation of the inverse is subject to round-off error, and will depend on the accuracy with which the calculation is performed. Based on this reasoning, it is desirable to estimate the condition number by an alternate procedure. Such a procedure can be found in the literature [5]. Essentially, using t-digit arithmetic and Gaussian elimination to approximate the solution to Ax = b, it has been shown [19] that the residual vector r for the approximation x has the property r ≈ 10 − t x x (9.41) From Equation 9.41, an estimate for the effective condition number in t-digit arithmetic can be obtained without the need to invert the matrix A. Also, Equation 9.41 assumes that all the arithmetic operations in the Gaussian elimination are performed using t-digit arithmetic, but that the operations that are needed to determine the residual are done in 2t-digit arithmetic. To approximate the t-digit (effective) condition number, consider the system Ay = r. (9.42) The approximate solution to Equation 9.42 is readily available, since the multipliers for the Gaussian elimination method have been calculated (and retained). Now, then, the approximate solution of Ay = r, satisfies y ≈ A −1r = A −1 ( b − Ax) = A −1 Ax − A −1 An = nn − n (9.43) such that is an estimate of the error in approximating the solution to the original system. Then, using Equation 9.41, y ≈ x − x = A −1r ≤ A −1 r ≈ A −1 (10 − t A x ) = 10 − t x K ( A) resulting in K ( A) ≈ y t 10 x (9.44) an estimate for the condition number involved with solving Ax = b using Gaussian elimination. Below is an illustrative example summarizing some of the main points of this section [5]. Consider the system 3.3330 15920 −10.333 x1 15913 2.2220 16.710 9.6120 x = 28.544 2 1.5611 5.1791 1.6852 x3 8.4254 415 Selected Numerical Methods and Available Software Packages Using Gaussian elimination and five-digit arithmetic, this system results in the augmented matrix 3.3330 15920 −10.333 15913 0 −10596 16.501 −10580 0 0 −5.0790 −4.7000 whose approximate solution is x = (1.2001, .99991, .92538)t. The residual vector corresponding to x is computed in double precision to be 15913 3.3330 15920 −10.333 1.2001 r = b − Ax = 28.544 − 2.2220 16.710 9.6120 .99991 8.4254 1.5611 5.1791 1.6852 .92538 −.00518 = .27413 −.18616 such that ∥r∥ ∞ = .27413. Using five-digit arithmetic for the calculations gives the approximation for the inverse of A to be −1.1701 × 10 −4 −1.4983 × 10 −1 8.5416 × 10 −1 A −1 = 6.2782 × 10 −5 1.2124 × 10 −4 −3.0662 × 10 −4 −8.6631 × 10 −5 1.3846 × 10 −1 −1.9689 × 10 −1 and Equation 9.35 gives ∥A−1 ∥ ∞ = 1.0041 and ∥A∥ ∞ = 15934, such that this estimate of the condition number K(A) is 15,999 using Equation 9.38. The estimate for the condition number using the t-digit approach, Equations 9.41 through 9.44, involves solving the system 3.3330 15920 −10.333 y1 .00518 2.2220 16.710 9.6120 y = .27413 2 1.5611 5.1791 1.6852 y3 −.18616 resulting in y = (−.20008, 8.9987 × 10 −5, .07461)t. Then, using Equation 9.44 we get K ( A) ≈ y ∞ t (.20008) 10 = 10 5 = 16672, x ∞ 1.2001 which gives the same quality information without involving the computation of A−1. 416 Applied Mathematical Methods for Chemical Engineers In the development of Equation 9.44, the estimate y = x − x was used. A reasonable expectation is that x + y is a more accurate approximation to the solution This notion is developed into a method called iterative of Ax = b compared to x. refinement [5]. To see how this improvement works, reconsider the latter example. The approximation to the given problem using five-digit arithmetic and Gaussian elimination is x (1) = (1.2001, .99991, .92538) t and the solution to Ay = r(1) is y(1) = (−.20008, 8.9987 × 10 −5, .07461)t. Then, using the iterative refinement approach gives x (2) = x (1) + y (1) = (1.0000, 1.0000, .99999) t where the actual error in this approximation is x − x (2) ∞ = 1 × 10 −5 as compared to x − x ∞ = .2001 or x − x ∞ = .2001 x ∞ resulting from five-digit arithmetic with Gaussian Elimination. One more iteration involves computing r (2) = b − A x (2) and solving Ay(2) = r(2), resulting in y(2) = (1.5002 × 10 −9, 2.0951 × 10 −10, 1.0000 × 10 −5). Then x (3) = x + y (2) = y (2) = (1.0000, 1.0000, 1.0000) t which agrees with the actual solution of the given system. 9.3.2 SPEcIAL TYPEs OF MAtRIcEs In this section some additional discussion on matrices is presented to facilitate subsequent numerical approaches that arise during the application of mathematics in chemical engineering. In the previous section we applied Gaussian Elimination to reduce a given system to an equivalent system. We then used backward substitution to extract the final results for the given variables. For example, if we start with the linear system x1 + x 2 + 3 x 4 = 4, 2 x1 + x 2 − x3 + x 4 = 1, 3x1 − x 2 − x3 + 2 x 4 = − 3, − x1 + 2 x2 + 3 x3 − x4 = 4. and use Gaussian elimination, we produce the equivalent system 1 0 0 0 1 0 −1 −1 0 3 0 0 4 −5 −7 13 13 − 13 − 13 3 Selected Numerical Methods and Available Software Packages 417 and then apply backward substitution to identify the final solutions for the given variables. An equivalent matrix 1 0 U = 0 0 1 0 −1 −1 0 3 0 0 3 −5 13 − 13 is produced whose form is upper-triangular. In general, an upper-triangular n × n matrix U has for each j entries uij = 0 for i = j + 1, j + 2,, n and a lower-triangular matrix L whose j entries are given by lij = 0 for each i = 1,2,, j − 1, i ≠ j. It is important to note that to derive both an upper- and a lower-triangular matrix, the Gaussian elimination procedure must be performed without row or column interchanges. To find a lower-triangular corresponding to U , we consider i = 1,2,3, and define m ji for each j = i + 1, i + 2,,4, to be the number used in the elimination step E j − m ji Ei → E j; that is, m21 = 2, m31 = 3, m41 = −1, m32 = 4, m42 = − 3 and m43 = 0. Then if L is defined to be the 4 × 4 lower-triangular matrix with entries lij given by ( ) 0, lij = 1, m ij when i = 1,2,, j − 1, when i = j, when i = j + 1, j + 2,, n, then 1 2 L= 3 −1 0 0 0 1 0 0 4 1 0 − 3 0 1 Note! The student should check that A = LU . In addition to illustrating what is meant by upper- and lower-triangular matrices, we have also factored the given matrix. However, more direct approaches to the factorization of matrices are available [5, 25, 36, 37] for example, if we consider the strictly diagonally dominant 4 × 4 matrix: 6 2 A = 1 −1 2 4 1 0 −1 1 0 4 −1 −1 3 1 418 Applied Mathematical Methods for Chemical Engineers Then A can be factored in the form A = LU , where L= l11 2 l21 l22 l41 l42 U= u11 u12 l31 0 0 l32 l33 l43 0 0 0 l44 and 0 u13 u22 u23 0 0 u33 0 0 0 u 14 u24 u34 u44 In this example we will employ the Doolittle’s method, which arbitrarily requires that the elements l11 = l22 = l33 = l44 = 1. To arrive at the product LU that satisfies the first row of the given matrix A, u11 = 16, u12 = 2, u13 = 1, u14 = −1. The portion of the multiplication that determines the remaining entries in the first column of the given matrix A, provides equations l21u11 = 1 3 1 l31 = 6 1 l 41 = − 6 l21 = 2 l31u11 = 1 and l41u11 = −1 At this stage of the proceedings the matrices L and U assume the form 1 1 3 L= 1 6 1 −6 1 0 1 0 l32 1 l42 l43 0 0 0 1 and U= 6 2 1 0 u22 u23 0 0 u33 0 0 0 −1 u24 u34 u44 Selected Numerical Methods and Available Software Packages 419 The part of the multiplication that satisfies the remaining entries in the second row of the given matrix A are 2 10 + u22 = 4 u22 = 3 3 1 2 + u23 = 1 such that u23 = 3 3 1 1 − + u24 = 0 u24 = 3 3 and the remaining entries in the second column of the given matrix A are 2 10 + l32 = 1 6 3 l32 = 1 5 l31 = 1 6 such that 2 10 − + l42 = 0 6 3 Continuing the process, alternate columns and rows between L and U to finally arrive at 1 1 3 L= 1 6 1 −6 1 0 1 0 1 5 1 10 1 − 9 37 0 0 0 1 and U= 6 2 1 0 10 3 0 0 2 3 37 10 0 0 0 −1 1 3 9 − 10 191 74 It is very important to keep in mind the specific aim is to solve the linear system Ax = b. One path to achieving the solution of a given system is to use the Gaussian elimination with backward substitution. Furthermore, the matrix A can be factored 420 Applied Mathematical Methods for Chemical Engineers into a lower-triangular matrix L and an upper-triangular matrix U whenever the given system Ax = b can be solved uniquely by Gaussian elimination without row (or column) interchanges. The system LUx = Ax = b can then be transformed into the system Ux = L−1b and since U is upper-triangular, backward substitution can be applied. However to be able to solve many systems using the matrix A it is desirable to have available more direct methods to determine L and U so that only a forward and backward substitution need to be performed. In addition to Doolittle’s method [5], there is Choleski’s Method, which requires that lii = uii for each value of i and Crout’s Method which requires that all the diagonal elements of U be one. To illustrate the key steps involved in using Choleski’s method [5, 38], we consider the matrix (positive definite) 4 A = −1 1 −1 4.25 2.75 2.75 3.5 1 Step 1: l11 = a11 = 4 = 2 Step 2: l21 = a21 −1 a 1 = = 0.5, l31 = 31 = = 0.5 l11 2 l11 2 Step 3: i=2 Step 4: 2 l22 = ( a22 − l21 ) = ( 4.25 − (0.5)2 ) 12 12 =2 Step 5: l32 = 1 1 ( a32 − l31l21 ) = ( 2.75 − 0.5(−0.5)) = 1.5 l22 2 Step 6: 2 2 l33 = ( a33 − l31 − l32 ) = (3.5 − (0.5)2 − (1.5)2 ) 12 12 =1 Selected Numerical Methods and Available Software Packages 421 In this procedure it is required that U = Lt , t ≡ transpose, that is, l12 = l13 = l23 = 0, u11 = l11 = 2, u12 = l21 = − 0.5, u13 = l31 = 0.5, u21 = l12 = 0, u22 = l22 = 2, u23 = l32 = 1.5, u31 = l13 = 0, u32 = l23 = 0, u33 = l33 = 1, Such that 2 L = − 0.5 0.5 0 0 2 0 1.5 1 2 − 0.5 U= 0 2 0 0 0.5 1.5 1 The key steps of Crout’s method [5] are illustrated through the given tridiagonal system of equations Ax = b: 2 x1 − x2 =1 − x1 + 2 x 2 − − x2 =0 x3 + 2 x3 − x4 = 0 − x3 + 2 x 4 = 1 whose augmented matrix is 2 −1 0 0 −1 0 1 −1 0 0 2 −1 0 2 1 −1 0 2 −1 0 where 2 −1 A= 0 0 −1 2 0 −1 −1 2 0 −1 0 0 −1 2 422 Applied Mathematical Methods for Chemical Engineers Step 1: l11 = 2, u12 = a12 1 =− l11 2 Step 2: i=2 1 3 l21 = a21 = − 1, l22 = a22 − l21u12 = 2 − ( −1) − = 2 2 1 2 a =− u23 = 23 = − 3 l22 3 2 i=3 2 4 l32 = a32 = − 1, l33 = a33 − l32 u23 = 2 − ( −1) − = 3 3 3 a34 =− u34 = 4 l33 Step 3: 3 5 l43 = a43 = − 1, l44 = a44 − l43 u 34 = 2 − ( −1) − = 4 4 Step 4: z1 = a15 1 = l11 2 Step 5: 1 z2 = a25 − l21z1 = l22 1 0 − ( −1) 2 1 = 3 3 2 1 0 − ( −1) 3 1 = 4 4 3 1 1 − ( −1) 4 1 =1 z4 = a45 − l43 z3 = l44 5 4 1 z3 = a35 − l32 z2 = l33 Selected Numerical Methods and Available Software Packages 423 Step 6: x4 = 1 Step 7: 1 3 x3 = z3 − u34 x 4 = − − = 1 4 4 1 2 x 2 = z2 − u23 x3 = − − = 1 3 3 1 1 x1 = z1 − u12 x 2 = − − = 1 2 2 The matrix A is now factored into 2 −1 0 0 2 −1 = 0 0 0 0 3 2 0 −1 4 3 0 −1 −1 0 2 −1 −1 2 0 −1 0 0 0 5 4 1 0 0 0 0 0 −1 2 − 1 2 1 0 0 0 2 0 − 3 3 1 − 4 0 1 0 such that x1 = x 2 = x3 = x 4 = 1. Note! Crout’s method requires that lii ≠ 0, i = 1,2,, n. These matrices are recommended for use when the linear system is small enough to be effectively accommodated in the main memory of a computer. Generally, it is most effective to use a direct technique that minimizes rounding error effects. 9.4 SOLUTION OF ORDINARY DIFFERENTIAL EQUATIONS In keeping with the philosophy of this chapter, we will highlight certain numerical strategies for approximating the solutions of ordinary differential equations. In doing so, we aim to point out the advantages and disadvantages of a given method or type of method, while leaving the actual algorithms and their derivations to the numerical analysis literature. Also, where appropriate, we will suggest available software packages that employ the numerical algorithm under discussion. 424 Applied Mathematical Methods for Chemical Engineers Following the traditional approach, this section will be discussed under two subheadings: (1) initial value problems and (2) boundary value problems. Supporting information, such as the algebra or calculus of finite differences, can be found in the literature [5, 9]. 9.4.1 INItIAL VALUE PROBLEMs For chemical engineers, the unsteady-state problems are of great interest, since they can help us to decide on process control strategies that can manage process upsets. As it turns out, very few of these will yield analytical results without a substantial number of simplifying assumptions. When such assumptions are unwarranted or unjustifiable, we have to turn to a numerical approach. Once this decision is made, we are then faced with choosing a method that will give a reliable approximation of the solution to the given problem. In deciding on a method, one must be mindful of the various errors that can occur and, in particular, propagation error, that can lead to instability in the method. Even though error analysis is an important aspect of numerical analysis, we will leave the in-depth discussion to the numerical analysis texts. The discussion that follows assumes that the problem under consideration for a numerical approach will be well posed [5,21]. Given a differential equation of the form dy = f (t , y) dt (9.45) and the condition that y = b when t = a, the analytically derived solution, F(t), passing through (a, b) would be an equation of the form y = F (t ) (9.46) The issue now is, how does Equation 9.45 describe the solution given by Equation 9.46, when all we know are Equation 9.45 and the initial value? Generally, the procedures will rely on the fact that given any point (t, y) on the solution curve, we can obtain the direction of the curve through that point. In principle, we have a starting point and a set of directions given by Equation 9.45, and, without any other information, we must follow the directions to the desired final destination, the approximate solution. The most straightforward method, the Euler method, employs the strategy of approximating the solution by constructing a tangent through the given initial point. Then, use the given differential equation to determine the slope of this tangent. Once this information is in hand, a small step in the right direction is taken at whose end point another tangent is constructed and its slope determined using Equation 9.45. By iterating this process, one can go as far as one wishes; however, round-off error usually limits the use of this technique in practice. Formally, the procedure is as follows: we wish to generate an approximation y corresponding to the point t such that y = F(t ) (9.47) Selected Numerical Methods and Available Software Packages 425 The first step is to subdivide the interval a ≦ t ≦ t into n-subintervals, each of width h= t − t0 n (9.48) and the resulting boundaries between subintervals called mesh points are denoted by t1, t2, …, tn−1, tn, such that t = tn and t0 = a. Then, starting from (t0, b), draw a tangent with slope dy = f (t0 , b) dt (t0 ,b ) such that f (t0 , b) = y1 − b t1 − t0 (9.49) Solving for y1, the y-coordinate of the tangent at the location t = t1, we get y1 = f (t0 , b)(t1 − t0 ) + b = hf (t0 , b) + b (9.50) In terms of Equation 9.46, F (t1 ) ≈ F (t0 ) + F ′(t0 )(t1 − t0 ) (9.51) since b = F(t0) and f(t0, b) = F′(t0). The approximation (as opposed to equality) comes from the fact that Equation 9.51 neglects second- and higher-order derivatives of F(t) by comparison to the Taylor series expansion of F(t). Whatever the size of the error created in a step, we would like to minimize it by shrinking h so that the tangent will closely approximate the actual solution, while forcing the remaining terms of the Taylor series to vanish. However, shrinking h requires increasing the number of steps to reach t . This then leads to a larger roundoff error and more than likely, an unacceptable result. This brief discussion of Euler’s method raises two important issues in all methods for solving the initial value problem: to improve the approximate solution to the initial value problem, we need to pay close attention to how the solution curve is modeled (straight line vs. polynomials, etc.) and the size of the step to use (h). How well these issues are resolved by a particular method is measured by a comparison with the Taylor series for the expected solution F(t). Many methods are based on the direct use of the Taylor series to develop the approximate solution curve. Such methods are called Taylor methods. Euler’s method is a Taylor method with a local truncation error, τ i, for the ith step τi = yi − yi −1 = f (ti −1 , yi −1 ), i = 1,2, n h (9.52) 426 Applied Mathematical Methods for Chemical Engineers where yi is the exact value of the solution at ti. The local truncation error has been defined as a measure of the amount by which the exact solution to the differential equation fails to satisfy the difference equation used for the approximation [3, 5, 18, 21]. Euler’s method as well as the other Taylor methods are considered as singlestep methods. These are methods for which the approximation for the mesh point, ti, involves information from only one of the previous mesh points ti − 1. In addition, Euler’s method is first order, because the second- and higher-order derivatives are neglected in the comparable Taylor series. Taylor methods are developed for nth-order accuracy; that is, the (n+1)th and higher derivatives are neglected. However, the need to determine the higher-order derivatives of f(t, y) introduces a major difficulty, and consequently, these methods are not very popular in practice. Alternatives to the Taylor methods are the Runge–Kutta methods. These methods were developed to utilize the high-order accuracy of the local truncation error of the Taylor methods, while avoiding computation and evaluation of the derivatives of f(t,y). To appreciate some of the similarities between the Taylor and Runge–Kutta methods we will state the associated difference equations without the detailed derivations. Full derivations and algorithms can be found in many texts on numerical analysis [5, 9, 18, 22–24]. All methods seek to obtain an approximation, w, to the well-posed initial value problem: dy = f (t , y), a ≤ t ≤ b, dt y( a ) = α (9.53) In the case of Euler’s method, w0 = α wi +1 = wi + hf (ti , wi ) (9.54) where wi = y(ti) for each i = 1, 2, …, n. For Taylor method of order n, we have w0 = α wi +1 = wi + hT ( n ) (ti , wi ), i = 0,1,, n − 1 (9.55) where T ( n ) (ti , wi ) = f (ti , wi ) + h h n −1 ( n −1) f ′(ti , wi ) + + f (ti , wi ) 2 n! (9.56) In the case of the Runge–Kutta methods, we have the so-called midpoint method, w0 = α h h wi +1 = wi + hf ti + , wi + f (ti , wi ) , i = 0,1,, n − 1 2 2 (9.57) Selected Numerical Methods and Available Software Packages 427 the modified Euler method, w0 = α h wi +1 = wi + [ f (ti , wi ) + f (ti +1 , wi + hf (ti , wi ))], i = 0,1,, n − 1 2 (9.58) the Heun method, w0 = α wi +1 = wi + 2 2 h f (ti , wi ) + 3 f ti + h, wi + hf (ti , wi ) , i = 0,1,, n − 1 (9.59) 4 3 3 and the very popular Runge–Kutta order-four method, given by w0 = α k1 = hf (ti , wi ) h k k2 = hf ti + , wi + 1 2 2 h k k3 = hf ti + , wi + 2 2 2 k4 = hf (ti +1 , wi + k3 ) (9.60) 1 wi +1 = wi + ( k1 + 2 k2 + 2 k3 + k4 ), i = 0,1,, n − 1 6 Equation 9.57 through 9.59 are second-order or Runge–Kutta order-two methods. The approach used to develop Equation 9.57 is typical, in that a number of parameters are evaluated by comparing their coefficients with the appropriate derivatives in the expansion of T(n) (ti, wi) given by Equation 9.56. For a secondorder Runge–Kutta, such as the result given by Equation 9.57, three parameters, a1, a 2, and a3 are used. These parameters have the property that a1f(t + a 2, y + a3) approximates T (2) (t , y) = f (t , y) + h f ′(t , y) 2 Since ∂ ∂ dy f (t , y) + f (t , y) ∂t ∂y dt ∂ ∂ = f (t , y) + f (t , y) ⋅ f (t , y) ∂t ∂y f ′(t , y) = (9.61) 428 Applied Mathematical Methods for Chemical Engineers then comparison of T(2) (t, y) with the truncated Taylor expansion of f(t + a2, y + a3) about (t, y) gives ∂ f (t , y) ∂ f (t , y) + a1a3 ∂t ∂y h ∂ f (t , y) h ∂ f (t , y) = f (t , y) + + ⋅ f (t , y) 2 ∂t 2 ∂y a1 f (t + a2 , y + a3 ) = a1 f (t , y) + a1a2 which result in a1 = 1, a2 = h/2, and a3 = h/2 · f(t, y) following the equating of coefficients. Notice that the truncated term in the Taylor expansion of f(t + a2, y + a3) about (t, y) involves the second-order derivative, which implies the order of the local truncation error. Example 9.1 This example demonstrates an alternative approach to arrive at Equation 9.54. Consider the differential equation dy = f ( x , y), y ( x 0 ) = y0 dx Then if y ( x k ) = yk , y ′ ( x k ) = yk′ = f ( x k , yk ) The differential equation can be approximated by the difference equation yk +1 = yk + h yk′ = yk + h f ( x k , yk ) This is accomplished by yk +1 ∫ dy = yk x k +1 ∫ xk f ( x , y ) dx or yk +1 = yk + x k +1 ∫ f ( x , y) dx xk if f ( x , y ) is considered to be relatively constant between the points x k and x k +1 then, Equation 9.54 describes the k + 1 value of y yk +1 = yk + x k +1 x k +1 xk k ∫ f ( xk , yk ) dx = yk + f ( xk , yk ) x∫ dx = yk + f ( x k , yk ) ( x k +1 − x k ) = yk + h f ( x k , yk ) ; k = 0,1, 2, , n Selected Numerical Methods and Available Software Packages 429 Example 9.2 This example demonstrates how one may use Euler’s Method Equation (9.54) in practice [5, 34, 35]. Suppose the initial values of x and y are given by x 0 = 0 and y0 = 1 , and the differential equation is dy = 2x + y dx Determine the value of y at x = 0.5 with step size h = 0.1. Solution results are provided in Table 9.2: f ( x k , yk ) = 2 x k + yk ; k = 0,1, 2,3, 4,5 Then in the context of Equation 9.54, we have ( ) y2 = y1 + h f ( x1, , y1 ) = 1.1000 + (0.1)[2(0.1) + 1.1000] = 1.2300 y3 = y2 + h f ( x2, , y2 ) = 1.2300 + (0.1)[2(0.2) + 1.2300] = 1.3930 y1 = y0 + h f x0, , y0 = 1.0000 + (0.1)(1.0000) = 1.1000 Upon solving the given problem using the method for linear first order differential equation described in Chapter 2, we get y ( x ) = 3e x − 2 x − 2 then y ( 0.5) = 3e 0.5 − 2(0.5) − 2 = 1.9462 TABLE 9.2 Euler Method k xk 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 yk 1.0000 1.1000 1.2300 1.3930 1.5923 1.8315 y ′ = f ( x k , yk ) y k + 1 = y k + h f ( x k , yk ) 1.0000 1.3000 1.6300 1.9930 2.3923 1.1000 1.2300 1.3930 1.5923 1.8315 430 Applied Mathematical Methods for Chemical Engineers By comparison, the actual (analytical) result is better by 5.89% when compared to the Euler approximation. Example 9.3 This example demonstrates the use of the Modified Euler Method Equation (9.58) in practice [5, 34, 35]: w0 = α h wi +1 = wi + [ f (ti , wi ) + f (ti +1 , wi + hf (ti , wi ))], i = 0,1, , n − 1 9.55 2 h = wi + ( wk′ + wk′ +1 ) 2 to resolve Example 9.2. Solution results are presented in Table 9.3. Solution From the given differential equation, we have f ( xi , yi ) = 2 xi + yi ; i = 0,1, 2,3, 4,5 Then, at i = 0, w1 ≡ y1 = f ( x 0 , y0 ) = w0 + h w0′ ≡ y(0.1) = 1.0000 + 0.1(1.0000 ) = 1.1000, which uses the Euler Method to approximate the value of y at h = 0.1. Returning to the given differential equation, we determine that (at i = 1 ): w1′ = f ( x1 , w1 ) = 2 x1 + y1 = 2(0.1) + 1.1000 = 1.3000. TABLE 9.3 Modified Euler Method Approximation I Approximation II Approximation III i xi wi wi ′ wi wi ′ wi wi ′ wi 0 1 2 3 4 5 0 0.1 0.2 0.3 0.4 0.5 1.0000 1.1000 1.2474 1.4313 1.6557 1.9247 1.0000 1.3000 1.6474 2.0313 2.4557 2.9247 1.1150 1.2640 1.4496 1.6759 1.9471 1.3150 1.6640 2.0496 2.4759 2.9471 1.1158 1.2648 1.4505 1.6769 1.9482 1.3158 1.6648 2.0505 2.4769 2.9482 1.1158 1.2648 1.4506 1.6770 1.9483 Selected Numerical Methods and Available Software Packages 431 Therefore, the Modified Euler Method yields w0 = 1 h w1 = w0 + [ f (t0 , w0 ) + f (t1 , w0 + hf (t1 , w1 ))] 2 h = w0 + ( w0′ + w1′ ) 2 = 1.0000 + 0.1 [1.0000 + 1.3000 ] = 1.1150 2 This is an improved approximation (Approximation II) of y1 to that given by the Euler Method used in Example 9.3. The process is now repeated using Approximation II to determine the value of the derivative, that is, w1′ = f ( x1 , w1 ) = 2 x1 + y1 = 2(0.1) + 1.1150 = 1.3150 determining a third (more improved) approximation of w1 (Approximation III): 1 1 w1 = y0 + h ( y0′ + y1′ ) ≡ w0 + h ( w0′ + w1′ ) 2 2 0.1 = 1.0000 + [1.0000 + 1.3158] = 1.1158 2 This newer value of y1 is now used to re-estimate the derivative w1′ = f ( x1 , w1 ) = 2 x1 + y1 = 2(0.1) + 1.1158 = 1.3158 which yields the approximation of w1 as h ( w0′ + w1′ ) 2 0.1 = 1.0000 + [1.0000 + 1.3158] = 1.1158 2 w1 = w0 + Since the last two approximations of w1 are the same, we now proceed to estimate w2 ( which is the value of y(0.2) ) by following the same procedure but starting with the value y1 ≡ w1 = 1.1158, which corresponds to x1 = 0.1. w2 = w1 + hf ( x1 , w1 ) = w1 + h w1′ = 1.1158 + 0.1(1.3158 ) = 1.2474 and w2′ = 2 x 2 + w2 = 2(0.2) + 1.2474 = 1.6474 432 Applied Mathematical Methods for Chemical Engineers Continuing the process we can obtain the required value of w5 ≡ y ( 0.5) = 1.9483 as indicated in Table 9.3 below. The major computational effort in applying the Runge–Kutta methods occurs in the evaluation of f. For second-order methods, two functional evaluations per step are required, and for the fourth-order method, four evaluations per step are required. As a consequence, the lower-order methods with smaller step size may be less costly than the higher-order methods using a larger step size. Relationships between evaluations per step and the order of local truncation error can be found in the literature [24]. Also available in the numerical analysis literature [15] are developments that combine the best features of the Euler and a second-order Runge–Kutta method. Example 9.4 This example illustrates the use of the Runge–Kutta order-four method. Estimate w2 using the Runge–Kutta order-four method with an h = 0.1 for the given initial value problem: dy = − y + t + 1; 0 ≤ t ≤ 1, y(0) = 1. dt Solution w0 = α ≡ 1.0000, f ( ti , wi ) ≡ − wi + ti + 1 Case i = 0 : k1 = hf (ti , wi ) ≡ 0.1( −1 + 1) = 0 h k 0.1 k 2 = hf ti + , wi + 1 ≡ 0.1 −1 + 1 + = 0.005 2 2 2 h k 0.005 0.1 + + 1 = 0.00475 k3 = hf ti + , wi + 2 ≡ 0.1 −1 − 2 2 2 2 k 4 = hf (ti +1 , wi + k3 ) ≡ 0.1( 0.1 + 1 − 1 − 0.00475) = 0.009525 1 1 wi +1 = wi + ( k1 + 2 k 2 + 2 k3 + k 4 ) ⇒ w1 = w0 + ( k1 + 2 k 2 + 2 k3 + k 4 ) 6 6 1 = 1 + ( 0 + 2(0.005) + 2(0.00475) + 0.009525) 6 = 1.0048375 Case i = 1: k1 = hf (ti , wi ) ≡ 0.1( −1.0048375 + 0.1 + 1) = 0.00951625 h k 0.00951625 0.1 k 2 = hf ti + , wi + 1 ≡ 0.1 (−1.0048375 − ) + 1 + (0.1 + ) 2 2 2 2 = 0.0140404375 h k 0.0140404375 0.1 k3 = hf ti + , wi + 2 ≡ 0.1 −1.0048375 − + 1 + 0.1 + 2 2 2 2 = 0.0138142281 k 4 = hf (ti +1 , wi + k3 ) ≡ 0.1( −1.0048375 − 0.0138142281 + 1 + .2) = 0.0181348272 Selected Numerical Methods and Available Software Packages 433 1 1 wi +1 = wi + ( k1 + 2 k 2 + 2 k3 + k 4 ) ⇒ w2 = w1 + ( k1 + 2 k 2 + 2 k3 + k 4 ) 6 6 1 0.00951625 + 2(0.0140404375) = 1.0048375 + 6 + 2(0.0138142281) + 0.0181348272 = 1.0187309014 All of the above methods choose the step size prior to carrying out the approximation step. Further, the step size remains fixed throughout the iterations. This feature can become a limitation when irregular regions are encountered, or when the solution to the differential equation behaves nonuniformly in parts of the interval. An adaptive strategy that allows for changes in the step size during iterations is the Runge–Kutta–Fehlberg. This method also reduces the number of required evaluations of the function f per step size. Various versions of this adaptive strategy can be found in the literature [5,9,13,25,33]. Algorithms that suggest mechanisms for adjusting the step size are also available [13]. The optimum step size should be such that the local errors are within a given tolerance, to ensure accuracy at the same time not adversely affecting efficiency. To further clarify this idea, we use the following result, which is typical in the numerical analysis literature [5,12,18,20,22,23]: y(t ) − yn ≤ kh p (9.62) where k is a constant that accounts for the smoothness of the solution y and the interval of approximation, p is related to the order of the difference scheme, and h is the step size. Even though Equation 9.62 is written for the global (entire interval) error, the essential idea remains the same. That is, reducing h by a factor of 10 reduces the error given by Equation 9.62 by a factor 10p. However, the number of algebraic operations is increased by about the same factor, which leads to an increase in the round-off error. Therefore, to get a sense of the optimal h, one that will not be too costly, a trial-anderror strategy is the usual approach taken. Many software packages have incorporated adaptive difference schemes for ordinary differential equations as the core of their designs. NDSolve of Mathematica and ode45 of MATLAB are two applications that are based on an adaptive extension of the fourth-order Runge–Kutta method. In addition to the single-step methods, there are methods that utilize the approximation at more than one previous mesh point to determine the approximation at the next point. These are the so-called multistep methods, and there are two types. Generally, the multistep method for solving the initial value problem described by Equation 9.53 is one whose difference equation for wi+1 at ti+1 is given by wi +1 = am −1 wi + am − 2 wi −1 + + a0 wi +1− m + h[bm f (ti +1 , wi +1 ) + bm −1 f (ti , wi ) + + b0 f (ti +1− m , wi +1− m )]; i = m − 1, m,… , n − 1 for the given starting values w0 = α 0 , w1 = α1 , w2 = α 2 , , wm −1 = α m −1 and an integer m > 1. (9.63) 434 Applied Mathematical Methods for Chemical Engineers If bm = 0, we have an explicit or open method, in which wi + 1 is explicitly given in terms of previously determined values. The explicit method constitutes one of the two types of multistep methods. For example, w0 = α 0 , w1 = α1 , w2 = α 2 , w3 = α 3 h [55 f (ti , wi ) − 59 f (ti −1 , wi −1 ) + 37 f (ti − 2 , wi − 2 ) − 9 f (ti − 3 , wi − 3 )], (9.64) 24 i = 3, 4, , n − 1 wi +1 = wi + defines the explicit four-step method known as the fourth-order Adams–Bashforth technique. If bm ≠ 0, we have an implicit or closed method, since wi + 1 occurs on both sides of Equation 9.63. The implicit method constitutes the second of the two types of multistep methods. For example, w0 = α 0 , w1 = α1 , w2 = α 2 h [9 f (ti +1 , wi +1 ) + 19 f (ti , wi ) − 5 f (ti −1 , wi −1 ) + f (ti − 2 , wi − 2 )], 24 i = 2,3, , n − 1 wi +1 = wi + (9.65) defines the implicit three-step method known as the fourth-order Adams–Moulton technique. Derivations of these and other multistep methods can be found in the numerical analysis literature [5,20]. The required starting values in either Equation 9.64 or Equation 9.65 are derived by assuming that the initial value, α, is the same as α 0. The remaining values, α 1, α 2, or α 3, are generated by either a Runge–Kutta method or some other single-step procedure. The following example illustrates the potential use of both types of multistep methods. Example 9.5 Consider dy = − y + t + 1; 0 ≤ t ≤ 1, y(0) = 1 dt Both Equations 9.64 and 9.65 will be used to approximate the solution. Also, we will use a step size of h = 0.1 together with the values from the exact solution, y(t) = e−t + t, as starting values. Therefore, for h = 0.1, ti = 0.1i, Equation 9.64 becomes (see Problem 11) 1 [18.5wi + 5.9 wi −1 − 3.7 wi − 2 + 0.9 wi − 3 + 0.24i + 2.52], 24 i = 3, 4, ,9 wi +1 = and Equation 9.65 becomes (see Problem 12) (9.66) 435 Selected Numerical Methods and Available Software Packages 1 [−0.9 wi +1 + 22.1wi + 0.5wi −1 − 0.1wi − 2 + 0.24i + 2.52] 24 i = 2,3, ,9 wi +1 = (9.67) which can be rearranged to give 1 [22.1wi + 0.5wi −1 − 0.1wi − 2 + 0.24i + 2.52], 24.9 i = 2,3, ,9 wi +1 = (9.68) Results for this example are given in Table 9.4. The table shows that the implicit method gives better results than the explicit method of the same order. Generally this is true. However, implicit methods have an inherent disadvantage in requiring that an algebraic conversion to an explicit representation be made; for example, the conversion from Equation 9.67 to 9.68. This type of algebraic conversion can be very difficult or even impossible to accomplish and is very dependent on the differential equation to be solved. For example, if we needed to solve the differential equation given by dy = e y ; 0 ≤ t ≤ 0.25, y(0) = 1 dt using Equation 9.65, we would get wi +1 = wi + h [9e wi+1 + 19e wi − 5e wi−1 + e wi−2 ] 24 as its difference equation. There is no clear way to solve this equation algebraically, to isolate the wi+1 to one side. In practice, implicit multistep methods are used to improve upon approximations obtained by explicit methods. This combination is the so-called predictor–corrector method. Predictor–corrector methods employ a single-step method, such as the Runge–Kutta of order 4, to generate the starting values to an explicit method, such as an Adams–Bashforth. Then the approximation from the explicit method is improved upon by use of an implicit method, such as an Adams–Moulton method. Also, there are variable step size algorithms associated with the predictor–corrector strategy in the literature [5,25]. TABLE 9.4 Comparison of Implicit and Explicit Methods ti .3 .4 .5 .6 .7 Adams–Bashforth wi Starting value 1.0703229200 1.1065354755 1.1488184077 1.1965933934 Error — 2.874E-6 4.816E-6 6.772E-6 8.090E-6 Adams–Moulton wi 1.0408180061 1.0703196614 1.1065301384 1.1488110076 1.1965845932 Error 2.146E-07 3.846E-07 5.213E-07 6.285E-07 7.106E-07 436 Applied Mathematical Methods for Chemical Engineers Example 9.6 This example demonstrates the predictor–corrector procedure using the four-step Adams–Bashforth technique as the predictor and the three–step Adams–Moulton technique as the corrector. Consider the initial value problem dy = 1 − x − 4 y; 0 ≤ x ≤ 1, y(0) = 1 dt Determine the approximate value of y at x = 0.4 with h = 0.1 using the ­predictor– corrector method. Solution Starting values are determined, using the Runge–Kutta order-four method, to be y0 = 1.000, y1 = 1.6089333, y2 = 2.5050062 and y3 = 3.8294145 Additional data needed are the values of the corresponding derivatives for each corresponding estimate, using dy = 1 − x − 4 y ≡ f ( x, y) dx That is, y0′ = 1 − x 0 + 4 y0 = 1 − 0 + 4(1) = 5, y1′ = 1 − x1 + 4 y1 = 1 − 0.1 + 4(1.6089333) = 7.3357332 y2′ = 1 − x 2 + 4 y2 = 1 − 0.2 + 4 (2.5050062) = 10.820025, y3′ = 1 − x3 + 4 y3 = 16.017658 Next we employ the Adams–Bashforth as the predictor, Equation 9.64: yi +1 = yi + h [55 y i′− 59 y i′−1 + 37 y i − 2 − 9 y i′− 3 ]. 24. y4 = 3.8294145 + 0.1 ( 469.011843) = 5.783631. 24 We now apply the corrector—Adams–Moulton, Equation 9.65: yi +1 = yi + h [y i′+1 + 19 y i′− 5 y i −1 + y i′− 2 ]. 24. Noticing that we need the value of yi′+1 , we use dy = 1 − x − 4 y ≡ f ( x, y) dx Selected Numerical Methods and Available Software Packages 437 again, but with the recently determined estimate of y4 = 5.783631, such that y4′ = 1 − 0.4 + 4 (5.783631) = 23.734524 Now the use of Equation 9.65 results in y4 = 3.8294145 + 0.1 [471.181828] = 5.792673 24. The actual value, correct to eight digits, is y(0.4) = 5.7942260. 9.4.2 BOUNDARY VALUE PROBLEMs Boundary value problems appear in many different areas of chemical engineering. Numerical approaches to solving them involve either a conversion to initial value problems (Case I) or, more directly, by finite difference methods (Case II). Because there are theorems that guarantee the existence of and the uniqueness of the solutions to initial value problems, one may favor that approach. However, we must not forget that quite often the physical situation is too complicated for us to ascertain that all the conditions of such theorems are satisfied. A very brief discussion of each approach follows. Case I: Conversion to initial value problems. The general approach of converting a given boundary value problem to a system of initial value problems is called the shooting method. In the numerical analysis literature [5,9,25,26], the shooting method is discussed separately for the linear and nonlinear problems. Burden et al. [5] present algorithms for both the linear and nonlinear problems. Following Burden et al. [5] to solve the linear boundary value problem y ′′ = p( x ) y ′ + q( x ) y + r ( x ), a ≤ x ≤ b, y(a) = α , y(b) = β (9.69) one would consider the two initial value problems given by y ′′ = p( x ) y ′ + q( x ) y + r ( x ), a ≤ x ≤ b, y(a) = α , y ′(a) = 0 (9.70) y ′′ = p( x ) y ′ + q( x ) y + r ( x ), a ≤ x ≤ b, y(a) = 0, y ′(a) = 1 (9.71) and Then, if y1(x) is the solution to Equation 9.70 and y2(x) is the solution to Equation 9.71, the solution to Equation 9.69 is y( x ) = y1 ( x ) + β − y1 (b) y2 ( x ), y2 (b) y2 (b) ≠ 0 (9.72) 438 Applied Mathematical Methods for Chemical Engineers The solution methods discussed in the section on initial value problems can be used to approximate y1(x) and y2(x). For the nonlinear case, the technique remains the same as that used to obtain a solution to Equation 9.69, except that a sequence of initial value problems of the form y ′′ = f ( x , y, y ′) a ≤ x ≤ b, y( a ) = α , y ′(a) = t k (9.73) are now required. The parameter, tk, must be chosen such that lim y(b, t k ) = y(b) = β (9.74) k →+∞ where y(b, tk) is the solution to Equation 9.73. Choosing the parameter tk to satisfy Equation 9.74 is not easy, and can be complicated by the fact that y( b , t k ) − β = 0 (9.75) is a nonlinear equation. In principle, the methodology in Section 9.2 can be employed here to solve t k = t k −1 − ( y ( b , t k −1 ) − β ) (dy /dt b ,tk −1 ) (9.76) However, we cannot evaluate dy dt b , tk −1 directly because we do not have an explicit representation for y(b, tk). All we know are the values y(b, t0), y(b, t1), …, y(b, tk −1). Burden et al. [5] show that we can rewrite Equation 9.73 to reflect its dependence on tk, as well as on x: y ′′( x , t ) = f ( x , y( x , t ), y ′( x , t )), a ≤ x ≤ b, y( a , t ) = α , y ′(a, t ) = t k (9.77) The subscript k is dropped inside the functional notation for convenience. Differentiating Equation 9.77 with respect to t and assuming that the order of differentiation of x and t is reversible gives ∂ ∂ ∂ ∂ ∂ y ′′( x , t ) = f ( x , y( x , t ), y ′( x , t )) y( x , t ) + y ′f ( x , y( x , t ), y ′( x , t )) y ′( x , t ) ∂t ∂y ∂t ∂y ∂t ∂ ∂ (a, t ) = 0, y ′(a, t ) = 1 ∂t ∂t Selected Numerical Methods and Available Software Packages 439 Simplification, using z(x, t) to represent ∂y(x, t)/∂t, results in z ′′ = ∂ ∂ f ( x , y, y ′ ) z + ( x , y, y ′) z ′, a ≤ x ≤ b, z (a) = 0, z ′(a) = 1 ∂y ∂y′ Then, Newton’s method becomes t k = t k −1 − ( y ( b , t k −1 ) − β ) z ( b , t k −1 ) (9.78) Equation 9.78 requires that two initial value problems be solved at each iteration. In practice, some version of the methods discussed under initial value problems (Section 9.4.1) is used to approximate the solution required by Newton’s method. In closing this section on the shooting methods, it is important to remember that round-off error may become very significant when we use these methods. For instance, β may be small enough such that the term (β − y1(b) )/y2(b) is dominated by −y1(b)/y2(b). When β is too small, an alternate method must be employed. Case II: Finite difference methods. Finite difference formulations may occur as any one of three types, namely forward, central, or backward finite difference [5,9,25]. Generally, these formulations lead to nonlinear systems of equations. The methods and approaches discussed in Section 9.2 can be employed. However, if the resulting system of equations is linear, then the methods of Section 9.3 apply. Next, we will briefly discuss a linear central difference and a nonlinear central difference formulation. In the linear case, we may reconsider the boundary value problem given by Equation 9.69: y ′′ = p( x ) y ′ + q( x ) y + r ( x ), a ≤ x ≤ b, y( a ) = α , y( b ) = β Then the interval [a, b] is divided into n + 1 equal subintervals having mesh points xi = a + ih, i = 0, 1, 2,, n + 1 and h = (b − a)/(n + 1). At the interior mesh points, xi, i = 1, 2, …, n, the derivatives y″ and y′ are approximated by centered difference formula for y″(xi) and y′(xi) as y ′′( xi ) ≈ 1 [ y( xi +1 ) − 2 y( xi ) + y( xi −1 )] h2 (9.79) 1 [ y( xi +1 ) − y( xi −1 )] 2h (9.80) and y ′( xi ) ≈ 440 Applied Mathematical Methods for Chemical Engineers Substitution of Equations 9.79 and 9.80 into the differential equation results in 1 1 [ y( xi +1 ) − 2 y( xi ) + y( xi −1 )] = p( xi ) [ y( xi +1 ) − y( xi −1 )] + q( xi ) y( xi ) + r ( xi ) h2 2h which can be restated as 2 wi − wi +1 − wi −1 wi +1 − wi −1 + p( xi ) 2 h 2h + q( xi ) wi = − r ( xi ), i = 1,2,, n (9.81) w0 = α , wn +1 = β where w is the approximation to the given boundary value problem. A further convenient rearrangement gives h h − 1 + p( xi ) wi −1 + (2 + h 2 q( xi )) wi − 1 − p( xi ) wi +1 = − h 2 r ( xi ) (9.82) 2 2 Then, if we let h ai = − 1 + p( xi ) 2 d i = 2 + h 2 q( xi ) h ci = − 1 − p( xi ) 2 bi = − h 2 r ( xi ) Equation 9.82 becomes ai wi −1 + di wi + ci wi +1 = bi Equation 9.83 can now be represented as a system of linear equations: d1 w1 + c1 w2 = b1 − a1α ai wi −1 + di wi + ci wi +1 = bi , 2 ≤ i ≤ n − 1 an wn −1 + d n wn = bn − cnβ This system may be solved by Gaussian elimination. (9.83) Selected Numerical Methods and Available Software Packages 441 Example 9.7 This example demonstrates the use of the central finite difference approximation to solve a second-order boundary value problem given by y ′′ − 4 y = 0; y(0) = 0, y(1) = 5 Use the finite difference Equations 9.82 and 9.83 given above to derive an approximate solution for three steps, n = 3. Solution h= b − a 1− 0 ≡ =1/ 4 n +1 4 We can identify that the quantities, p( x ) = 0, q( x ) = 4 and r ( x ) = 0 by comparing the given differential equation with Equation 9.69. Also the end points α = 0 and β = 5 are given. Employing Equation 9.83 leads to the derived system of equations: 2.25 w1 − w2 = 0 where a2 = −1, d2 = 2.25, c2 = −1, b2 = 0 − w1 + 2.25 w2 − w3 = 0; − w2 + 2.25 w3 = 5; Derived System Notice that in this case a1 = a2 = a3 , b1 = b2 = b3 , c1 = c2 = c3 and d1 = d 2 = d3 The solution of the derived system of equations results in the following estimates: w1 = 0.7256, w2 = 1.6327 and w3 = 2.9479. Comparison of these estimates with the closed form solution y( x ) = 5 sinh(2 x ) sinh(2) at the specified x-values is shown in Table 9.5. It is important to notice that the error is reduced as the number of steps increases, which means smaller increments (more interior points), and we can anticipate more accurate results with more steps. 442 Applied Mathematical Methods for Chemical Engineers TABLE 9.5 Comparison of Analytic Solution to Finite Difference Estimates xi yi(Num) y(Analytical) % error 0.25 0.50 0.75 0.7256 1.6327 2.9479 0.7184 1.6201 2.9354 1.000 0.778 0.426 When the differential equation is nonlinear, an n × n nonlinear system results. For example, the boundary value problem given by y ′′( x ) = f ( x , y, y ′), a ≤ x ≤ b, y( a ) = α , y( b ) = β (9.84) would result in the approximation 1 y ( x i +1 ) − y ( x i ) [ y( xi +1 ) − 2 y( xi ) + y( xi −1 )] = f xi , y( xi ), , h2 2h i = 1, 2, , n (9.85) following the same steps as in the linear case. Then, in terms of w, Equation 9.85 becomes w0 = α , wn +1 = β wi +1 − wi −1 2 wi − wi +1 − wi −1 = 0, i = 1, 2, , n + f xi wi , 2h h2 Finally, the n × n nonlinear system representation is w − α 2 w1 − w2 + h 2 f x1 , w1 , 2 −α = 0 2h w − w1 − w1 + 2 w2 − w3 + h 2 f x 2 , w2 , 3 =0 2h w − wn − 2 − wn − 2 + 2 wn −1 − wn + h 2 f x n −1 , wn −1 , n = 0 2h β − wn − 1 − w n − 1 + 2 w n + h 2 f x n , wn , −β = 0 2h This system may be solved by the Newton–Raphson method. Although finite difference formulations are usually more stable than the shooting methods, they are still approximations. The added level of approximation for the derivatives can generate errors, and care must be exercised when applying these methods. Selected Numerical Methods and Available Software Packages 9.4.3 443 SYstEMs OF ORDINARY DIFFERENtIAL EQUAtIONs In Section 9.4.1, selected numerical methods are examined for solving the initial value problems associated with first-order differential equations. Those methods are also applicable to higher-order differential equations following the reduction to a system of first-order equations. For example, the second-order differential equation d2 y = f (t , y, y ′) dt 2 (9.86) can be reduced to a system of two first-order differential equations [5,17,21]: dy =v dt (9.87) dv = f (t , y, v ) dt (9.88) In general, an nth-order differential equation such as y( n ) = f (t , y, y ′,, y( n −1) ) (9.89) can be reduced to a system of n first-order differential equations of the form x1′ = f1 (t , x1 ,…, x n ) x 2′ = f1 (t , x1 ,…, x n ) (9.90) x n′ = f1 (t , x1 ,…, x n ) In the particular case of the second-order equation, a procedure to approximate the solution to a system of two first-order equations x ′ = f (t , x , y) (9.91) y ′ = g(t , x , y) (9.92) x (t0 ) = x 0 , y(t0 ) = y0 (9.93) subject to the initial conditions is an extension of those methods discussed in Section 9.4.1. 444 Applied Mathematical Methods for Chemical Engineers Following Boyce and DiPrima [21], the Euler method would be extended to xi +1 = xi + hf (ti , xi , yi ), yi +1 = yi hg(ti , xi , yi ), i = 0,1,, n = xi + hxi′ = yi + hyi′ (9.94) for the mesh points ti = t0 + ih. Similarly, the fourth-order Runge–Kutta method applied to Equations 9.91 and 9.92 would become 1 xi +1 = xi + h[ ki1 + 2 ki 2 + 2 ki 3 + ki 4 ], 6 1 yi +1 = yi + h[ i1 + 2 i 2 + 2 i 3 + i 4 ], 6 (9.95) where ki1 = f (ti , xi , yi ), i1 = g(ti , xi , yi ) h k 1 h k 1 ki 2 = f ti + , xi + i1 , yi + i1 , i 2 = f ti + , xi + i1 , yi + i1 2 2 2 2 2 2 h k 1 h k 1 ki 3 = f ti + , xi + i 2 , yi + i 2 , i 3 = f ti + , xi + i 2 , yi + i 2 2 2 2 2 2 2 ki 4 = f (ti + h, xi + ki 3 , yi + i 3 ), i 4 = f (ti + h, xi + ki 3 , yi + i 3 ) The predictor–corrector method described in Section 9.4.1, which involves the use of Equations 9.64 and 9.65, would become 1 h[55 xi′1 − 59 xi′−1 + 37 xi′− 2 − 9 xi′− 3 ], 24 1 yi +1 = yi + h[55 yi′1 − 59 yi′−1 + 37 yi′− 2 − 9 yi′− 3 ] 24 (9.96) 1 h[9 xi′+1 + 19 xi′ − 5 xi′−1 + xi′− 2 ], 24 1 yi +1 = yi + h[9 yi′+1 + 19 yi′ − 5 yi′−1 + yi′− 2 ] 24 (9.97) x i +1 = x i + and x i +1 = x i + In the above discussion, it is assumed that the functions f and g satisfy the unique solution conditions. From Equations 9.95 through 9.97, it is evident that third- and higher-order equations will become cumbersome and are more manageable using matrix notations [16,17,19,22,27]. Selected Numerical Methods and Available Software Packages 445 Both Mathmatica and MATLAB, as well as other available software packages, have the capabilities to solve linear or nonlinear systems of algebraic and differential equations. Example 9.8 This example demonstrates the use of the Euler method for systems [21, 34, 35]. Given the second-order differential equation y ′′ = y; y ( 0 ) = 1, y ′ ( 0 ) = 2 reduce the differential equation to a system of first-order equations and use Euler’s method to estimate y (1) for h = 0.5. What is the absolute error in comparison to the exact solution? Solution Let y ′ = x; then x ′ = y or y′ = x ≡ f (t , x , y ) x ′ = y ≡ g (t , x , y ) subject to t0 = 0, x 0 = 2.0000, y0 = 1.000; then we have x1 = x 0 + h g ( t0 , x 0 , y0 ) = 2.0000 + 0.5(1.0000) = 2.5000 y1 = y0 + h f ( t0 , x 0 , y0 ) = 1.0000 + 0.5(2.0000) = 2.0000 Now t1 = t0 + 0.5 = 0.5; g ( t1 , x1 , y1 ) = y1 f ( t1 , x1 , y1 ) = x1 such that x 2 = x1 + h g ( t1 , x1 , y1 ) = 2.5000 + 0.5(2.0000) = 3.5000 y2 = y1 + h f ( t1 , x1 , y1 ) = 2.0000 + 0.5(2.5000) = 3.2500 The exact solution is y ( t ) = 3 t 1 −t e − e ⇒ y (1) = 3.8935. 2 2 Resulting in an absolute error of 3.8935 − 3.2500 × 100 = 16.53% 3.8935 446 Applied Mathematical Methods for Chemical Engineers 9.5 NUMERICAL SOLUTION OF PARTIAL DIFFERENTIAL EQUATIONS One of the most frequently occurring partial differential equation in chemical engineering is the so-called parabolic type. This equation is used to describe time-dependent diffusion processes and fluid flow. Therefore, the numerical solution methods for this type of partial differential equation are important in heat transfer, molecular diffusion, and fluid flow. There are many numerical approaches one can use to approximate the solution to the initial and boundary value problem presented by a parabolic partial differential equation. However, our discussion will focus on three approaches: an explicit finite difference method, an implicit finite difference method, and the so-called numerical method of lines. These approaches, as well as other numerical methods for all types of partial differential equations, can be found in the literature [5,9,18,22,25,28–33]. 9.5.1 EXPLIcIt AND IMPLIcIt FINItE DIFFERENcE MEthODs Following Burden et al. [5], we will consider the parabolic partial differential equation in one space dimension: ∂ ∂2 u( x , t ) = α 2 2 u( x , t ) + S ( x , t ), 0 < x < L , t > 0 ∂t ∂x (9.98) subject to the initial condition u( x ,0) = f ( x ), 0 ≤ x ≤ L (9.99) u(0, t ) = φ L , u( L , t ) = φ R , t > 0 (9.100) and the boundary conditions As part of the development of a finite difference scheme to approximate the solution to Equations 9.98 through 9.100, we must select two mesh constants, h and k, such that m = L/h and N = T k are an integers. The grid points are (xi, tj), where xi = ih, i = 0, 1,, m and t j = jk , j = 0, 1,, N Then, if we approximate the derivatives by u( xi , t j + k ) − u( xi , t j ) ∂ u( xi , t j ) ≈ ∂t k and u( xi + h, t j ) − 2u( xi , t j ) + u( xi − h, t j ) ∂2 u( xi , t j ) ≈ 2 ∂x h2 Selected Numerical Methods and Available Software Packages 447 and substitute into Equation 9.98 to get wi , j +1 − wi , j k − α2 wi +1, j − 2 wi , j + wi −1, j h2 = Si , j (9.101) Equation 9.101 can be rearranged to α2k α2k α2k w 1 2 w wi +1, j + Si , j , + − + i − 1, j i , j h2 h2 h2 i = 1,2,, m − 1, j = 1,2,, N wi , j +1 = (9.102) where wi,j approximates u(xi, tj), the source term, Si,j is S(xi, tj) and T is a convenient maximum time value. The initial and boundary conditions become respectively wi ,0 = u( xi ,0) = f ( xi ), i = 0,1,2,, m (9.103) w0, j = u(0, t j ) = φ L ; wm , j = u( x m , t j ) = φ R , j = 1,2,, N (9.104) and In the important case of no source term, and when both end conditions are zero, Equation 9.102 reduces to α2k α2k α2k wi −1, j + 1 − 2 2 wi , j + 2 wi +1, j , 2 h h h i = 1,2,, m − 1, j = 1,2,, N wi , j +1 = (9.105) with w0, j = wm , j = 0 (9.106) The method described by Equations 9.102 through 9.106 is an explicit method and is commonly called the forward difference method. It is explicit because knowledge of wi,j for tj at all the grid points means that wi,j + 1 can be calculated for the new time tj + 1 without solving simultaneous equations. The forward difference method is considered conditionally stable. Further, it can be shown that the method converges with a rate of convergence on the order of (k + h2) if the condition α2k ≤1/ 2 h2 (9.107) is satisfied [18,22]. This restriction means that a small value for h requires an even smaller k-value, which can easily lead to round-off errors. 448 Applied Mathematical Methods for Chemical Engineers Contrasting the forward difference method is the implicit method of Crank– Nicolson [5,9,18,22,25]. The Crank–Nicolson difference formulation for the homogeneous case of Equation 9.98 with zero end conditions ( φ L = φ R = 0 ) is wi , j +1 − wi , j k − α2 [( wi +1, j − 2 wi , j + wi −1, j ) + ( wi +1, j +1 − 2 wi , j +1 2h 2 + wi −1, j +1 )] = 0 (9.108) Burden et al. [5] give an algorithm involving the Crank–Nicolson method for solving Equation 9.108, while Constantinides et al. [9] have MATLAB examples involving the nonhomogeneous case of Equation 9.108. To facilitate the use of matrix methods discussed in previous sections (i.e., Section 9.3.2), Equation 9.108 can be restated as wi , j +1 − wi , j = α2k [( wi +1, j − 2 wi , j + wi −1, j ) + ( wi +1, j +1 − 2 wi , j +1 + wi −1, j +1 )] 2h 2 λ = [( wi +1, j − 2 wi , j + wi −1, j ) + ( wi +1, j +1 − 2 wi , j +1 + wi −1, j +1 )] 2 (9.109) α2k . h2 (9.110) where λ= upon expansion of wi , j +1 − wi , j = λ [( wi +1, j − 2 wi , j + wi −1, j ) + ( wi +1, j +1 − 2 wi , j +1 + wi −1, j +1 )] 2 and collection of terms involving subscript j + 1 to the left of the equal sign and terms involving subscript j to the right of the equal sign, as shown in the cases below for i = 1 and 2. i = 1: w1, j +1 − w1, j = w1, j +1 − λ [( w2, j − 2 w1, j + w0, j ) + ( w2, j +1 − 2 w1, j +1 + w0, j +1 )] 2 or λ λ λ λ w2, j +1 + λw1, j +1 − w0, j +1 = w1, j + w2, j − λw1, j + w0, j 2 2 2 2 i=2: w2, j +1 − λ λ λ λ w3, j +1 + λw2, j +1 − w1, j +1 = w2, j + w3, j + λw2, j + w1, j 2 2 2 2 or (1 + λ ) w2, j +1 − λ λ λ λ w3, j +1 − w1, j +1 = (1 − λ ) w2, j + w3, j + w1, j 2 2 2 2 Selected Numerical Methods and Available Software Packages 449 which leads to a matrix representation of the Crank–Nicolson as Aw( j +1) = Bw( j ) for each j = 0,1,2,, where ( w( j ) = w1, j , w2, j ,, wm −1 λ 0 − (1 + λ ) 2 λ − λ (1 + λ ) − 2 2 λ A = 0 − (1 + λ ) 2 λ − 0 0 2 ) t 0 0 λ − 0 2 (1 + λ ) λ 0 0 (1 − λ ) 2 λ λ 1 − λ) 0 ( 2 2 λ λ B= 0 0 (1 − λ ) 2 2 λ (1 − λ ) 0 0 2 and both matrices A and B are tridiagonal, with A being positive definite and strictly diagonal dominant. The Crank–Nicolson method is unconditionally stable and is more accurate than the forward difference method. 9.5.2 MEthOD OF LINEs As an alternative to the approaches described above for solving partial differential equations, we could take advantage of the methods described in Section 9.4.1. The numerical method of lines does exactly that [15, 33]. 450 Applied Mathematical Methods for Chemical Engineers Example 9.9 This example considers the problem described by ∂u ∂ 2 u = ∂t ∂ x 2 (9.111) subject to the typical initial condition u( x , 0) = f ( x ) and the boundary conditions u(0, t ) = 0 u( L , t ) = 0 Following Lee and Schiesser [33], the derivative ∂2 u/∂x2 is replaced with a finite difference approximation ∂ 2 u ui +1 − 2ui + ui −1 , i = 1, 2, , n ∂x 2 ( ∆x )2 Notice that ∆x is the same as h, used in earlier sections. Typically, a spatial grid is defined over the interval 0 ≤ x ≤ L. Along this grid, the value of a particular x is designated with the Index i. Therefore, i = 1 corresponds to x = 0 and i = n corresponds to x = L. The spacing between adjacent grid points is ∆x (or h). Upon combining the finite difference approximation with the left-hand side of Equation 9.111, we get dui ui +1 − 2ui + ui −1 , i = 1,2,, n dt ( ∆x )2 (9.112) Equation 9.112 has only one independent variable, namely t, therefore we have an ordinary differential equation, and specifically a system of ordinary differential equations (discretized) for i = 1, 2, …, n. Furthermore, this system is a set of initial value ordinary differential equations that are subject to the initial conditions u( xi ,0) = f ( xi ) ≡ ui (t ), i = 1, 2, , n For example, if the initial condition was given as f ( x ) = sin (πx ) Then we would restate the condition in terms of the discretized variables as u( xi , 0) = ui (t ) = sin(πxi ), i = 1, 2, , n 451 Selected Numerical Methods and Available Software Packages Finally all that remains is to step along the spatial grid and integrate the corresponding ordinary differential equations given by Equation 9.112. In carrying out this step we will include the boundary conditions in the following way: u(0, t ) ⇒ u( x1 , t ) = u1 (t ) = 0, du1 =0 dt Similarly, for i = n, u( L , t ) ⇒ u( x n , t ) = un (t ) = 0, dun =0 dt An example of a MATLAB program [33] that was used to solve the problem described above is given in Appendix B. The program is composed of the following modules containing functions as defined in the programming sense: • Main program (pdelin). • Initial parameters defined in a function (intpar); here parameters such as the number of first-order equations, n (=21) to be used in Equation 9.112. Also, contained in this function are the number of spatial steps to be taken in the specified time interval and the error tolerances for this problem. • Initial and boundary conditions are set in the function (inital); notice that there are 21 initial values and x is defined on the interval 0 ≤ x ≤ 1. • Derivative vector in a function called derv, which is the vectorized form of Equation 9.112. • Functions rkc4a and ssrkc4 combined as the programmed form of the Runge–Kutta order four method, which was given in Equation 9.60. • Formatted results are displayed using the function fprint. 9.5.3 SELEctED APPLIcAtIONs UsING thE MEthOD OF LINEs In addition to the types of applications discussed in Chapter 7, the method of lines can be employed in developing solutions to problems that are even more complex and nontraditional. Example 9.10 This example considers a simplified version of the transport and binding kinetics of an analyte, such as an antigen on an antibody surface of a fiber-optic biosensor. A 1-D version of the model is given [39] for a distance z, measured from the antibody interface as ∂C A ∂ 2 C A = , 0 < z < h, t > 0 ∂t ∂z 2 (9.113) subject to the conditions DA ∂C A = k f C A Cb , Sat − Cb − kr Cb ∂z ( ) at z = 0 (9.114) 452 Applied Mathematical Methods for Chemical Engineers C A = Cbulk at z = h C A = C0 at t = 0 (9.115) (9.116) where C A ( z , t ) represent the anylate concentration in the diffusion phase, Cb , Sat is the saturated value of the binding concentration, and D A is the anylate diffusivity. The concentration of the anylate bound to the antibody, Cb ( t ) , is described by ( ) dCb = k f C A ( 0, t ) Cb , sat − Cb − kr Cb dt (9.117) which is subject to the condition Cb ( 0 ) = Cb 0 (9.118) where k f and kr represent rate coefficients for the forward and reverse binding. The data for this example are D A = 1 × 10 −10 m 2 s , k f = 1 × 10 5 ( mole m 3 ) s −1 , kr = 1 × 10 −2 s −1 , Cb , Sat = 2.66 × 10 −8 mole m 2 , Cbulk = 4.48 × 10 −5 m, C0 = 0 mole m 3 Cb = 0 mole m 2 Determine the anylate concentration profile C A ( z , t ) mole m 3 and the bound anylate concentration profile Cb ( t ) mole m 2 using the method of lines. Solution Using the central finite difference approximations given by dyi 1 ( yi +1 − yi −1 ) and dx 2h 1 d 2 yi 2 ( yi +1 − 2 yi + yi −1 ) 2 dx h the spatial derivatives in Equations 9.113 and 9.114 can be approximated by the central differences, resulting in ∂2 C A ( z , t ) ∂z 2 C A ( z + ∆z ,t ) − 2C A ( z , t ) + C A ( z − ∆z , t ) ( ∆z )2 and ∂C A ( z , t ) ∂z C A ( z + ∆z , t ) − C A ( z − ∆z , t ) 2 ∆z with uniformly spaced grids: z (1) = 0, z ( 2 ) = ∆z , , z ( i ) = ( i − 1) ∆z , , z ( n ) = ( n − 1) ∆z = h 453 Selected Numerical Methods and Available Software Packages It is important to note that z ( i ) = ( i − 1) ∆z , i = 1, 2, , n is employed to accommodate software limitations that subscripts must be positive integers (0 cannot be used as a subscript). Substitution of the finite difference approximation into the left boundary condition results in DA C A ( z = ∆z , t ) − C A ( z = −∆z , t ) 2 ∆z ( ) = k f C A ( z = 0, t ) Cb , Sat − Cb − kr Cb (9.119) Observing that z = −∆z lies outside the given domain in z renders C A ( −∆z , t ) as a fictitious point and is addressed in the subsequent program as c f = C A ( −∆z , t ) , such that c f = C A ( z = ∆z , t ) − ( ) 2 ∆z k f C A ( z = 0, t ) Cb , Sat − Cb − kr Cb , DA ( ) which is programmed as cf + c(2) − ( 2 * dz / D ) * ( kf * c(1) * ( cbsat − cb ) − kr * cb ) . An example of the MATLAB program [39] that was used to solve this problem is given in Appendix C. The program is composed of the following key steps: pde_1 in which the parameters and variables are declared globally so that they can be shared with other routines along with the main program. • The derivative ∂C A ∂t is computed as ct over the grid of n points. • At grid point i = 1 corresponding to z = 0 the fictitious value cf is used to calculate ct(1). • ode15s is a MATLAB routine that is designed to facilitate systems that have convergence difficulties—typical of systems developed using finite differences such as in this example. As a third example using the method of lines, we consider a problem containing two partial differential equations, as follows: Example 9.11 This example considers a model of a drug delivery system in which the drug is initially embedded in a polymer matrix [39]. In this model two components are considered: the concentration profile for water (component 1) and the embedded drug (component 2) are described as (see Figure D.1) ∂C1 ∂ 2 C 1 ∂C1 ∂ 2 C1 = D1 21 + + ∂r ∂t r ∂r ∂z 2 (9.120) 454 Applied Mathematical Methods for Chemical Engineers and ∂C2 ∂ 2 C2 1 ∂C2 ∂ 2 C2 = D2 + + ∂r 2 ∂t r ∂r ∂z 2 (9.121) where D1 ( cm 2 s ) and D2 ( cm 2 s ) are the respective diffusivities of water and drug. The initial conditions for this system are C1 ( r , z , t = 0 ) = C10 and C2 ( r , z , t = 0 ) = C20 (9.122) while the boundary conditions are ∂C1 ( r , z , t ) ∂C2 ( r , z , t ) =0 = ∂r ∂r at r = 0 (9.123) and D1 ∂C1 = k1 (C1e − C1 ) at r = R ∂r (9.124) D2 ∂C2 = k 2 (C2e − C2 ) at r = R ∂r (9.125) D1 ∂C1 = k1 (C1e − C1 ) at z = z L ∂z (9.126) D2 ∂C2 = k 2 (C1e − C2 ) at z = z L ∂z (9.127) The quantities C1e and C2e are respectively exterior values of the water and drug concentrations, R ( cm ) is the radius of the polymer matrix and z L ( cm ) is the length. Also, k1 and k 2 are mass transfer coefficients. We attempt to determine the concentration profiles for water and the drug using the method of lines. Solution A MATLAB program [39] will be used to solve the system of equations and help develop the concentration profiles. To be consistent the variables used to describe the given model will be translated to the variables used in Figure D.1 (Appendix D) and the program listed in Appendix D: C1 ≡ u1, C2 = u 2, k1 = ku1 , k 2 = ku 2 , C1e = u1e , C2e = u2e , R = r0 D1 = Du1 and D2 = Du 2 Selected Numerical Methods and Available Software Packages 455 Finite difference approximations are ∂ 2 u1 u1( i + 1) − 2u1( i ) + u1( i − 1) ∂z 2 ( ∆z )2 ∂ 2 u1 u1( j + 1) − 2u1( j ) + u1( j − 1) ∂r 2 ( ∆r )2 and ∂ 2 u 2 u 2 ( i + 1) − 2 u 2 ( i ) + u 2 ( i − 1) ∂z 2 ( ∆z )2 ∂ 2 u 2 u 2 ( j + 1) − 2 u 2 ( j ) + u 2 ( j − 1) ∂r 2 ( ∆r )2 and are programmed for the axial direction as u1zz ( i , j ) = ( u1( i + 1, j ) − 2 * u1( i , j ) + u1( i − 1, j )) / dzs u 2 zz ( i , j ) = ( u 2 ( i + 1, j ) − 2 * u 2 ( i , j ) + u 2 ( i − 1, j )) / dzs and for the radial direction as u1rr ( i , j ) = ( u1( i , j + 1) − 2 * u1( i , j ) + u1(i , j − 1) ) u 2rr ( i , j ) = ( u 2 ( i , j + 1) − 2 * u 2 ( i , j ) + u 2(i , j − 1) ) Equations 9.120 and 9.121 are programmed as u1t ( i , j ) = Du1* ( u1rr ( i , j ) + u1r ( i , j ) + u1zz (i , j ) ) u 2t ( i , j ) = Du 2 * ( u 2rr ( i , j ) + u 2r ( i , j ) + u 2 zz (i, j) ) 1 ∂C requires special attention as r → 0 ; however, one needs only to recall r ∂r l’Hôpital’s rule, which is applicable in this case: The term lim r→0 such that 1 ∂C ∂ 2 C = r ∂r ∂r 2 456 Applied Mathematical Methods for Chemical Engineers ∂ 2 C1 1 ∂C1 ∂2 C + = 2 21 2 ∂r ∂r r ∂r (9.128) which gives (C1 ≡ ul ) ∂ 2 u1 u1( 2 ) − 2u1(1) + u1( 0 ) at j = 1 ∂r 2 ( ∆r )2 However, u1( 0 ) is a fictitious value corresponding to r = −∆r on 0 ≤ r ≤ r 0. The finite difference approximation for Equation 9.123 results in ∂u1 u1( 2 ) − u1( 0 ) = 0 ⇒ u1( 0 ) = u1( 2 ) ∂r 2 ∆r Substituting into Equation 9.128 gives u1( 2 ) − u1(1) ∂ 2 u1 2 ∂r 2 ( ∆r )2 and is programmed as u1r ( i, j ) = 2 * ( u1( i, j + 1) − u1( i, j )) / drs u 2r ( i , j ) = 2 * ( u 2 ( i, j + 1) − u 2 ( i , j )) / drs When r ≠ 0, the term 1 ∂u1 is approximated as r ∂r 1 ∂u1 1 u1( j + 1) − u1( j − 1) ≡ u1r ( i , j ) r ∂r r ( j ) 2 ∆r Then the interior points in r , i.e. ( j ≠ 1, nr ) , are programmed as ( ) ( ) u1r ( i , j ) = 1 r ( j ) * ( u1( i , j + 1) − u1(i , j − 1) / ( 2 * dr ) u 2r ( i , j ) = 1 r ( j ) * ( u 2 ( i , j + 1) − u 2(i , j − 1) / ( 2 * dr ) The point r = r 0 i.e ( j = nr ), involves the use of a fictitious point. Since Du1 u1( j + 1) − u1( j − 1) u1f − u1( j = nr − 1) ∂u1 Du1 = Du1 2 ∆r 2 ∆r ∂r 457 Selected Numerical Methods and Available Software Packages where u1f is a fictitious value at r = r 0 + ∆r , such that Equation 9.124 becomes u1f − u1( j = nr − 1) Du1 = k1( u1e − u1( j = nr )) 2 ∆r or, solving for u1f, u1f = u1( j = nr − 1) + 2 ∆r ( k1 Du1 ) ( u1e − u1( j = nr )) (9.129) and is programmed as u1f = u1( i , j − 1) + 2 * dr * ku1 Du1* ( u1e − u1(i , j ) ) The second derivative ∂ 2 u1 ∂r 2 is also approximated at r = r 0 + ∆r, resulting in ∂ 2 u1 u1( j + 1) − u1( j − 1) ∂r 2 ( ∆r )2 which is combined with Equation 9.129 and programmed as u1rr ( i, j ) = ( u1f − 2 * u1( i, j ) + u1( i , j − 1))/drs Similarly the second derivative in u 2 is programmed as u 2f = u 2 ( i, j − 1) + 2 * dr * ku 2 Du 2 * ( u 2e − u 2(i , j ) ) and u 2rr ( i , j ) = ( u 2f − 2 * u 2 ( i, j ) + u 2 ( i , j − 1))/drs The MATLAB program given in Appendix D is composed of the following modules: • pde_1(t, u), which contains the discretized equations discussed above. • pde_1m is the main program that calls pde_1(t, u). • Ode15s is a MATLAB routine that is designed to address slow convergent systems of equations. Such as those typically generated by finite difference discretizations. 458 Applied Mathematical Methods for Chemical Engineers An alternative set of programming (in MATLAB) which can be used to develop comparable solutions is also given in Appendix D. This approach employs library routines dss004.m, based on the derivations given in [40], and dSS044.m to d­ iscretize the derivatives and a main program pde_2(t,u) which calls each routine as needed. More detailed explanation of the use of dss004.m and dSS044.m is given in Appendix D. 9.6 SUMMARY Admittedly, the scope of the chapter is limited to a few of the methods that are used in chemical engineering problem solving. They certainly are not expected to produce desired results in all possible instances. However, they should provide the reader at least a cursory view of what can be expected when a problem is to be solved numerically. A notable exclusion is a discussion on solving the linear system Ax = 0. However, this important case of a linear system deserves more buildup than the allowable scope of this chapter. Hopefully, the curious reader will be stimulated enough to further explore the selected references. The chapter does highlight areas of concern to the practicing chemical engineer, namely, solution of nonlinear and linear algebraic equations, as well as solution of differential equations. Indeed, attempted numerical solution of differential equations usually leads to the solution of nonlinear or linear algebraic equations. Commercially available software packages that can aid the practicing chemical engineer are also mentioned, and many downloadable packages can be found at http://gams.nist.gov/cgi-bin/serve.cgi/Packages or http://www.netlib.org/index.html. Principally, the chapter’s thrust is to expose readers to numerical analysis methods without overwhelming them with premature details. 9.7 PROBLEMS 1.The equilibrium reactions A ↔ B + D (1) and A ↔ B + 2D (2) occur at a given temperature and pressure. At the given condition of temperature and pressure, the following relationships are observed among the mole fractions: yB yD = 3.75, yA yC yD2 = 0.135 yA a.On a basis of 100 mol of A, conduct the material balance to show that the relationships can be recasted in terms of the extent, ξi, of each reaction. ξ1 ( ξ1 + 2 ξ 2 ) = 3.75 (100 − ξ1 − ξ 2 )(100 + ξ1 + 2ξ 2 ) ξ 2 ( ξ1 + 2 ξ 2 ) 2 = 0.135 (100 − ξ1 − ξ 2 )(100 + ξ1 + 2ξ 2 ) Selected Numerical Methods and Available Software Packages 459 b. Determine ξ 1 and ξ 2, using the Newton–Raphson method. c. Determine ξ 1 and ξ 2, using a computer algebraic system. d. Determine ξ 1 and ξ 2, using the modified Newton–Raphson method. 2. Given the two functions f1 ( x ) = x 2 , f2 ( x ) = x +1 x determine the point of intersection a. Using the modified Newton–Raphson method with f1 to find x and f 2 to find y b. Switching the roles of f1 and f 2 c. Using the Newton–Raphson method 3. Use the Gauss–Seidel method to solve the following system: 2 x1 + x 2 + x3 − x 4 = −3 x1 + 9 x 2 + 8 x3 + 4 x 4 = 15 − x1 + 3 x 2 + 5 x3 + 2 x 4 = 10 x2 + x4 = 2 4. a.What happens if you attempt to use the Gauss–Seidel method to solve the rearranged system? − x1 + 3 x 2 + 5 x3 + 2 x 4 = 10 x1 + 9 x2 + 8 x3 + 4 x4 = 15 x2 + x4 = 2 2 x1 + x 2 + x3 − x 4 = −3 b. Solve this system using Gaussian elimination with backward substitution. c. Solve by inverting the coefficient matrix. d. Solve using a computer algebraic system. 5. a. U se Choleski’s Method to find a factorization of the form A = LLT for the following matrices: 2 −1 A = −1 2 0 −1 0 − 1 2 460 Applied Mathematical Methods for Chemical Engineers 4 1 B= 1 1 1 1 3 −1 −1 2 1 0 1 1 0 2 b. Use Crout’s Method to solve the following linear systems: i. x1 − x 2 = 0 −2 x1 + 4 x 2 − 2 x3 = − 1 − x 2 + 2 x3 = 1.5 ii. −2 x1 − x 2 = 3 − x1 + 2 x 2 − x3 = − 3 − x 2 + 2 x3 = 1 6.Use the Runge–Kutta order four method with step sizes of 0.02 and 0.01 to approximate the solution to y ′ = 1 − t + 3 y; y(0) = 1 at t = 1.0. Also, compare your result to the exact answer. 7.Use the Adams–Moulton predictor–corrector method to approximate the ­solution to dx = x − 4 y, x (0) = 1 dt dy = − x + y, y(0) = 0 dt with h = 0.1 at t = 0.4. Correct the predicted value twice. Hint: See Example 9.5. a. Use the exact solution values (five digits) to start the problem. b. Use the Euler method to obtain the appropriate starting values (x1, x2, x3, and y1, y2, y3) to the problem. c. Use the midpoint method to obtain the starting values. Selected Numerical Methods and Available Software Packages 461 8.Using the Runge–Kutta order four method for starting, the Adams– Bashforth method as the predictor, and the Adams–Moulton method as a corrector, approximate the solution to dx = x; x (0) = 1 dt at t = 1.0 with a step size of 0.01. 9.Repeat problem 7 using the Runge–Kutta order four method. 10.Repeat problem 7 using the Adams–Bashforth method and the Runge– Kutta order four method to obtain the appropriate starting values. 11. Repeat problem 7 using the Adams–Moulton method and the Runge–Kutta order four method to obtain the appropriate starting values. 12.Given the initial value problem dy = − y + t + 1; 0 ≤ t ≤ 1, y(0) = 1 dt use h = 0.1 and ti = 0.1i to verify Equation 9.66. 13.Verify Equation 9.67 using the same information given in Problem 11. 14. Given a slab of material 1.00 m thick at a uniform temperature of 100ºC. If the front surface is suddenly exposed to a constant temperature of 0ºC, and the back surface is insulated, calculate the temperature profile for a time of 5000 s. Data: ∂T ∂2 T = (2 × 10 −5 m 2 /s) 2 ; 0 ≤ x ≤ 1.00 m ∂t ∂x 15.Use the method of lines approach to solve problem 1 and problem 2 of Chapter 6. 16.Solve the problem consisting of Equations 9.130–9.134 below using the method of lines. ∂C A 1 ∂ ∂C A ∂ 2 C A 1 ∂C A = D AB ) + r = D AB ( ∂t ∂r r ∂r ∂r 2 r ∂r (9.130) Initial and boundary conditions: C A (r,0) = C A0 (9.131) 462 Applied Mathematical Methods for Chemical Engineers C A (0, t) is finite or in this case − D AB ∂C A (0, t ) =0 ∂r ∂C A kC ≅ 1 A (mol/cm2s) ∂r 1 + k 2 C A (9.132) (9.133) Hint: Numerical Solution letting ui (t ) ≡ C A (r , t ); i = 1,..., n (9.134) Then Equation 9.131 becomes ui (0) = C A 0 ; i = 1,..., n (9.135) Method of Lines: yi +1 − 2 yi + yi −1 y −y ; y′ ≅ i +1 i −1 Using the finite difference approximations y′′ ≅ 2 ( ∆x ) 2 ∆x the spatial component of the partial differential equation can be transformed so that the entire partial differential equation becomes dui ui +1 (t ) − 2ui (t ) + ui −1 (t ) 1 ui +1 (t ) − ui −1 (t ) + ; i = 2,..., n − 1 = dt ( ∆r )2 i ∆r 2 ∆r (9.136) for the interior region. The left boundary condition ( i = 1 case), using Equation 9.132. C A (0, t) is finite or in this case ∂C A (0, t ) =0 ∂r can be represented as ui +1 (t ) − ui −1 (t ) u2 (t ) − u0 (t ) ≡ = 0 ⇒ u2 (t ) = u0 (t ); i = 1. 2 ∆r 2 ∆r (9.137) such that du1 u2 (t ) − 2u1 (t ) + u0 (t ) 1 u2 (t ) − u0 (t ) = + ( ∆r )2 2 ∆r dt i ∆r (9.138) Selected Numerical Methods and Available Software Packages 463 1 ∂C A, However, the term in the wing bracket is derived from which is r ∂r undefined as r → 0. To resolve this issue we use the well-known l’Hôpital’s rule: 1 ∂C A ∂2 C A ui +1 (t ) − 2ui (t ) + ui −1 (t ) = ≅ r → 0 r ∂r ∂r 2 ( ∆r )2 lim Therefore du1 u2 (t ) − 2u1 (t ) + u0 (t ) 1 u2 (t ) − u0 (t ) = + dt ( ∆r )2 2 ∆r i ∆r u (t ) − u1 (t ) u (t ) − 2u1 (t ) + u0 (t ) = 4 2 = 2 2 2 ( ∆r ) ( ∆r )2 (9.139) The right boundary condition is described by Equation 9.136: − D AB kC ∂C A ≅ 1 A ∂r 1 + k 2 C A However, the nth boundary is given in the discretized format by Equation 9.138 for the radial coordinate as dun un +1 (t ) − 2un (t ) + un −1 (t ) 1 un +1 (t ) − un −1 (t ) + ;i=n = 2 dt ( ∆r ) n ∆r 2 ∆r But the boundary condition is discretized as − DAB [ k un (t ) un +1 (t ) − un −1 (t ) ]= 1 1 + k2 un (t ) 2 ∆r (9.140) or un +1 (t ) = un −1 (t ) − 2 ∆r k1 un (t ) DAB 1 + k2 un (t ) Therefore, for the case i = n, k1 un (t ) dun 2 ∆r k1un (t ) 1 [un −1 (t ) − − un (t )] − = dt ( ∆r )2 DAB 1 + k2un (t ) n ∆rDAB 1 + k2 un (t ) (9.141) 464 Applied Mathematical Methods for Chemical Engineers 9.7.1 SUMMARY The system of equations to be solved is dui ui +1 (t ) − 2ui (t ) + ui −1 (t ) 1 ui +1 (t ) − ui −1 (t ) + ; i = 2,..., n − 1 = dt ( ∆r )2 i ∆r 2 ∆r du1 u (t ) − u1 (t ) = 4 2 dt ( ∆r )2 k1 un (t ) dun 2 ∆r k1un (t ) 1 = [un−1 (t ) − − un (t )] − 2 dt ( ∆r ) DAB 1 + k2un (t ) n ∆rDAB 1 + k2 un (t ) REFERENCES 1.Underwood, A.J.V. Fractional distillation of multicomponent mixtures, Chem. Eng. Prog., 44, 603, 1948. 2. Treybal, R.E. Mass Transfer Operations, 3rd ed., McGraw-Hill, New York, 1980. 3. Stark, P.A. Introduction to Numerical Methods, Macmillan, New York, 1970. 4.Wolfram, S. The Mathematica Book, 3rd ed., Wolfram Media, Champaign, Illinois, and Cambridge University Press, London, 1996. 5.Burden, R.L., Faires, J.D., and Reynolds, A.C. Numerical Analysis, 2nd ed., Prindle, Weber and Schmidt, Boston, 1978. 6.Constantinides, A. Applied Numerical Methods with Personal Computers, McGrawHill, New York, 1987. 7. Ramraj, R., Farrell, S., and Loney, N.W. Mathematical modeling of controlled release from a hollow fiber, J. Membr. Sci., 162, 73, 1999. 8. Jenkins, M.A. and Traub, J.F. A three-stage algorithm for real polynomials using quadratic iteration, SIAM J. Numer. Anal. 7, 545, 1970. 9. Constantinides, A. and Mostoufi, N. Numerical Methods for Chemical Engineers with Matlab Applications, Prentice Hall, Upper Saddle River, NJ, 1999. 10. Pattee, H.A. Selecting computer mathematics, Mech. Eng., 117, 82, 1995. 11. Mackenzie, J. and Allen, M. Mathematical power tools, Chem. Eng. Educ., Spring, 156, 1998. 12.Malek-Madani, R. Advanced Engineering Mathematics with Mathematica and Matlab, Addison-Wesley Longman, Reading, MA, 1998. 13.Johnston, R.L. Numerical Methods: A Software Approach, John Wiley & Sons, New York, 1982. 14. Riggs, J.B. An Introduction to Numerical Methods for Chemical Engineers, Texas Tech University Press, Lubbock, TX, 1988. 15.Hanna, O.T. and Sandall, O.C. Computational Methods in Chemical Engineering, Prentice Hall, Upper Saddle River, NJ, 1995. 16. Strang, G., Linear Algebra and Its Applications, 2nd ed., Academic Press, New York, 1980. 17.Amundson, N.R. Mathematical Methods in Chemical Engineering: Matrices and Their Applications, Prentice Hall, Englewood Cliffs, NJ, 1966. 18. Nakamura, S. Applied Numerical Methods, Prentice Hall, Englewood Cliffs, NJ, 1991. Selected Numerical Methods and Available Software Packages 465 19.Forsythe, G.E. and Moler, C.B. Computer Solution of Linear Algebraic Systems, Prentice Hall, Englewood Cliffs, NJ, 1967. 20. Forsythe, G.E., Malcolm, M.A., and Moler, C.B. Computer Methods for Mathematical Computations, Prentice Hall, Englewood Cliffs, NJ, 1977. 21.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley & Sons, New York, 2005. 22.Isaacson, E. and Keller, H.B. Analysis of Numerical Methods, John Wiley & Sons, New York, 1966. 23.Higham, N.J. Accuracy and Stability of Numerical Algorithms, SIAM, Philadelphia, 1996. 24. Butcher, J.C. On the attainable order of Runge–Kutta methods, Math. Comp., 19, 408, 1965. 25. Cheney, W. and Kincaid, D. Numerical Mathematics and Computing, 2nd ed., Brooks and Cole Publishing Company, Monterey, CA, 1985. 26. Na, T.Y. Computational Methods in Engineering Boundary Value Problems, Academic Press, New York, 1979. 27.Gear, C.W. Numerical Initial Value Problems in Ordinary Differential Equations, Prentice Hall, Englewood Cliffs, NJ, 1971. 28.Myint, U.T. and Debnath, L. Partial Differential Equations for Scientists and Engineers, 3rd ed., Prentice Hall, Englewood Cliffs, NJ, 1987. 29.Kythe, P.K., Puri, P., and Schaferkotter, M.R. Partial Differential Equations and Mathematica, CRC Press, Boca Raton, FL, 1997. 30.Kreyszig, E. Advanced Engineering Mathematics, 7th ed., John Wiley & Sons, New York, 1993. 31.O’Neil, P.V. Advanced Engineering Mathematics, 4th ed., PWS-Kent Publishing Company, Boston, MA, 1995. 32. Cooper, J.M. Introduction to Partial Differential Equations with MATLAB, Birkhauser, Boston, MA, 1998. 33. Lee, H.J. and Schiesser, W.E. Ordinary and Partial Differential Equation Routines in C, C++, Fortran, Java®, Maple® and MATLAB®, Chapman & Hall and CRC Press, Boca Raton, FL, 2004. 34.Spiegel, M.R. Calculus of Finite Differences and Difference Equations, Schaum’s Outlines, 10th printing, McGraw-Hill, Inc., New York, 1971 35. Butenko, S. and Pardalos, P.M. Numerical methods and Optimization an Introduction, CRC Press (Taylor & Francis), Boca Raton, FL, 2014. 36.Sewell, G., Computational Methods of Linear Algebra, 2nd ed., Wiley-Interscience, John Wiley & Sons Inc., Hoboken, New Jersey, 2005. 37.Bronson, R. and Costa, G.B. Matrix Methods: Applied Linear Algebra, 3rd ed., Academic Press (Elsevier), Burlington, MA, 2009. 38.Sewell, G. The Numerical Solution of Ordinary and Partial Differential Equations, 2nd ed., Wiley-Interscience, John Wiley & Sons Inc., Hoboken, New Jersey, 2005. 39.Schiesser, W.E. Partial Differential Equation Analysis in Biomedical Engineering: Case Studies with MATLAB, Cambridge University Press, New York, 2013. 40. Bickley, W.G. Formulae for numerical differentiation, Math. Gaz., Vol 25, 1941. This page intentionally left blank Appendix A: Elementary Properties of Determinants and Matrices A.1 DETERMINANTS The reader may be familiar with determinants, but a brief review is useful to the discussion on the solution of linear equations. It turns out that defining the determinant is a more difficult task than listing its properties; however, we will make an attempt to bring out somewhat of a working definition [1]. A determinant of the nth order is usually written in one of the forms det A = A = a11 a12 a1a a21 a22 a2 a aa1 aa 2 aaa = dot a11 a12 a1a a21 a22 a2 a aa1 aa 2 aaa = aij (A.1) and is an array (matrix) of n2 (n × n) elements, where the elements aij may be real or complex numbers or functions. The determinant itself is a function, and if the aij are considered variables, then |A| is a function of n2 variables with a particular form. Before we express the functional form of |A|, a list of its useful properties are given and illustrated with a 2 × 2 case [1,2]. 1.The determinant is a linear function of the first row: ab a + a′ b + b′ a′ b′ = (a + a′)d − (b + b′)c = + cd c d c d ta tb ab = tad − tbc = t c d cd 2.The determinant changes sign when two rows are exchanged: a11 a12 a21 a22 = a11a22 − a12 a21 = − a21 a22 a11 a12 467 468 Appendix A 3.The determinant of the identity matrix (see discussion on matrices) is 1: 10 =1 01 4.If two rows of the array A are equal, then |A| = 0: ab = ab − ab = 0 ab 5.The operation of subtracting a multiple of one row from another leaves the determinant unchanged: a11 − ta21 a12 − ta22 a11 a12 = a21 a22 a21 a22 6.If the array A has a zero row, then |A| = 0: 0 0 =0 a21 a22 7. If the array is triangular, then |A| is the product a11 a22…ann of the entries on the main diagonal: a11 a12 a11 0 = a11a22 = 0 a22 a21 a22 8.If the |A| is zero, then the array A is singular: a11 a12 a21 a22 is singular if and only if a11 a12 = a11a22 − a12 a21 = 0 a21 a22 9.For any two n by n arrays (matrices), the determinant of the product is the product of the determinants; that is, given two arrays A and B, then |AB| = (|A|)(|B|). a11 a12 a21 a22 b11 b12 a11b11 + a12 b21 a11b12 + a12 b22 = b21 b22 a21b11 + a22 b12 a21b21 + a22 b22 469 Appendix A 10.The transpose of the array A has the same determinants as the array A: t A =A Now the functional form of the determinant may be given in terms of cofactor expansion: det A = ai1 Ai1 + ai 2 Ai 2 + + ain Ain (A.2) where Aij is called the cofactor and itself is given as Aij = (−1)1+ j det Mij (A.3) where Mij is called the minor and is formed by deleting row i and column j of the array A. An example of this expansion: 3 −2 3 A = 4 2 −2 1 1 2 Then 2 − 2 A11 = (−1)1+1 det = 6 1 2 4 − 2 A12 = (−1)1+2 det = 10 1 2 4 2 A13 = (−1)1+3 det = 2 1 1 Therefore, the |A| = a11 A11 + a12 A12 + a13A13 = 44. An alternative and useful formula for the determinant is det A = det P −1 det L det D det U = ±(product of the pivots) (A.4) where ±1 is the determinant of P−1 or of P and depends on whether the number of row exchanges is even or odd. L and U are lower and upper triangular arrays, respectively, and their determinants are 1. Triangular arrays are those whose elements are zeros above or below the main diagonal. That is, a lower triangular array is one in which all elements above the main diagonal are zeros. The main diagonal, D, is a diagonal array whose nonzero elements appear on the main diagonal. The array, P, is a permutation array (matrix) that reorders the rows of the array A, so that the product PA admits a factorization with nonzero pivots [2]. While determinants themselves may have limited uses, they do facilitate the helpful applications of matrices. 470 A.2 Appendix A MATRICES A matrix is a rectangular array of elements arranged in rows and columns as a11 a12 a1n a a a2 n A = 21 22 = an1 an 2 ann a11 a12 a1n a21 a22 a2 n = [aij ] an1 an 2 ann (A.5) where the square or round brackets are meant to be different from the straight lines used to represent determinants. The elements can be numbers (real or complex) or functions. The matrix given in Equation A.5 is an m by n or, more frequently, stated as m × n, which exploits the fact that there are m rows and n columns. If m = n, the matrix is said to be square or of nth order. Matrices will be denoted with bold capital letters. A special matrix containing only a single column is called a column vector, and will be denoted by a small boldface letter, such as x 1 x x = 2 x n = x1 x2 x n (A.6) The element positioned in the ith row and jth column is designated by aij, the first subscript identifying its row and the second its column. Associated with each matrix A is the matrix At, known as the transpose of A. This transpose is obtained by interchanging the rows and columns of A. Therefore, if A = [aij], then At = [aji]. The transpose xt of a column vector (n × 1 matrix) is a row vector (1 × n matrix). A square matrix in which all elements except those on the main diagonal are zero is called a diagonal matrix, and the case in which the diagonal elements are all unity is called the identity matrix I. The algebraic properties of matrices can be found in the literature on linear algebra [2–5]. Here we list those that facilitate the discussions in Chapter 9. Proofs are left to the linear algebra literature: 1. Equality: Two m × n matrices, A and B, are said to be equal if corresponding elements are equal; that is, if aij = bij for each i and j. 2. Zero: A matrix whose elements are all zero is called a null matrix and is denoted by 0. 3. Addition: The sum of two m × n matrices is defined as the matrix obtained by adding corresponding elements: A + B = [aij ] + [bij ] = [aij + bij ] (A.7) 4. Multiplication: The product AB of two matrices is defined whenever the number of columns of the first matrix is the same as the number of rows of the second matrix. When this condition exists, the matrices are said to be 471 Appendix A conformable. If A is n × m and B is n × r, then the product AB = C is an m × r matrix. The element in the ith row and jth column of C is obtained by multiplying each element of the ith row of A by the corresponding element of the jth column of B, and then adding the resulting products. That is, n cij = ∑ aik bkj k =1 (A.8) Also, matrix multiplication satisfies the associative law ( AB)C = A(BC) (A.9) A(B + C) = AB + AC (A.10) and the distributive law For this reason, it is necessary to use terminology which specifies the order of multiplication. For example, if the matrices A and B are n × n, then C = AB is read as B premultiplied by A, whereas D = BA is read as B postmultiplied by A. Further, the results AB and BA are not necessarily equal. This is a very different phenomenon from the usual symbolic algebra and may even result in the product of two matrices being the null matrix without either matrix being null. For instance, 1 3 5 2 0 6 A = −2 4 0 and B = 1 0 3 −1 0 −3 0 2 2 1x 2 + 3 x1 + 5 x (−1) 1x 0 + 3 x 0 + 5 x 0 1x 6 + 3 x 3 + 5 x (−3) AB = −2 x 2 + 4 x1 + 0 x (−1) −2 x 0 + 4 x 0 + 0 x 0 −2 x 6 + 4 x 3 + 0 x (−3) 0 x 2 + 2 x1 + 2 x (−1) 0 x 0 + 2 x 0 + 2 x 0 0 x 6 + 2 x 3 + 2 x (−3) 0 0 0 = 0 0 0 0 0 0 whereas 2 x1 + 0 x (−2) + 6 x 0 2x3 + 0 x 4 + 6x 2 2x5 + 0 x 0 + 6x 2 BA = 1x1 + 0 x (−2) + 3 x 0 1x 3 + 0 x 4 + 3 x 2 1x 5 + 0 x 0 + 3 x 2 −1x1 + 0 x (−2) + (−3) x 0 −1x 3 + 0 x 4 + −3 x 2 −1x 5 + 0 x 0 + −3 x 2 2 18 22 = 1 9 11 −1 −9 −11 472 Appendix A This example illustrates Equation A.8, as well as the fact that the product of two conformable matrices is not necessarily commutative. Matrix multiplication applies to the special cases of 1 × n and n × 1, row and column vectors, respectively. If we denote the 1 × n as xt and the n × 1 as y, then a x t y = ∑ xi yi (A.11) x t y = yx t , x t (y + z) = x t y + x t z, (αx t )y = αx t y = x t (αy) (A.12) i =1 Also A special product called the scalar or inner product is defined by n ( x, y) = ∑ xi yi = x t y (A.13) i =1 where y is the conjugate of y. A particular useful form of Equation A.13 is n n i =1 i =1 ( x, x) = ∑ x1 x1 = ∑ x1 2 (A.14) If (x,y) = 0, then the two vectors are said to be orthogonal. These vector properties can be illustrated with the following example. Let 2−i i x = −2 , y = i 1+ i 3 then x t y = (i)(2 − i) + (−2)(i) + (1 + i)(3) = 4 + 3i ( x, y) = (i)(2 + i) + (−2)(−i) + (1 + i)(3) = 2 + 7i x t x = (i)2 + (−2)2 + (1 + i)2 = 3 + 2i ( x, x) = (i)(−i) + (−2)(−2) + (1 + i)(1 − i) = 7 5. Multiplication by a number: The product of a matrix A by a number α (real or complex) is defined by αA = α[aij ] = [αaij ] (A.15) That is, each element of the matrix is multiplied by the number α. This differs from the multiplication of a determinant by a number where only the first row is multiplied. 473 Appendix A 6. Subtraction: The difference A – B of two conformable matrices is de­fined by A − B = A + (−B) = [aij ] − [bij ] = [aij − bij ] (A.16) 7. Inverse: This is the operation for square matrices that is analogous to division for numbers. That is, for a given square matrix A, we need to determine another matrix B such that AB = I, the identity matrix. If the matrix B exists, it is called the inverse of A and we write B = A−1. Multiplication is commutative between any matrix and its inverse. AA−1 = A−1 A = I (A.17) Also, when A has an inverse, it is said to be nonsingular, otherwise A is said to be singular. Property 8 for determinants is a necessary and sufficient condition for a matrix to be singular. Direct use of the determinant to find the inverse is possible. A procedure known as Cramer’s rule may be used on small systems of equations. That is, given Ax = b, then x = A−1b (premultiplication by A−1) can be determined as illustrated below: x1 + 3 x 2 = 0 2 x1 + 4 x 2 = 6 then x1 = 03 64 13 24 = 10 26 6 −18 = 9, x 2 = = = −3 2 −2 − 13 24 In general, the jth component of x = A−1b is given by a a b a 11 12 1 1n xj = ; Bj = det A an1 an 2 bn ann det B j (A.18) The vector b replaces the jth column of A. Another way to arrive at Equation A.18 is to start with the definition given in Equation A.3. That is, if aij is any element of a square matrix, and the cofactor of aij in |A| is Aij, the transpose of the matrix whose elements are made up of all the Aijs is called the adjoint of A and is denoted adj A = [Aji]. It can then be shown [2,5] that a typical element of A−1 is given by Aji/|A|. A more useful way to compute A−1 is by means of elementary operations: 1.Interchange of two rows 2.Multiplication of a row by a nonzero scalar 3.Addition of any multiple of one row to another row 474 Appendix A Generally, any nonsingular matrix A can be transformed into the identity I by a systematic sequence of the elementary operations. It can be shown that the same sequence of operations performed on I will yield A−1 [1,2,4]. An example illustrating the process is as follows: 1 −1 −1 A = 3 −1 2 2 2 3 Step 1 Obtain zeros in the off-diagonal position in the first column by adding (−3) times the first row to the second row and adding (−2) times the first row to the third row. 1 −1 −1 0 2 5 0 4 5 Step 2 Obtain a one in the diagonal position in the second column by multiplying the second row by 1/2. 1 −1 −1 0 1 5/2 5 0 4 Step 3 Obtain zeros in the off-diagonal positions in the second column by adding the second row to the first row and adding (−4) times the second row to the third row. 1 0 3/2 0 1 5/2 0 0 −5 Step 4 Obtain a one in the diagonal position in the third column by multiplying the third row by (−1/5). 1 0 3/2 0 1 5/2 1 0 0 Step 5 Obtain zeros in the off-diagonal positions in the third column by adding (–3/2) times the third row to the first row, and adding (–5/2) times the third row to the first row. 1 0 0 0 1 0 0 0 1 475 Appendix A If the same sequence of operations in the same order is now performed on I, the sequence of matrices as follows: 1 0 0 0 1 0 , 0 0 1 1 0 0 −3 1 0 , −2 0 1 1 0 0 −1 / 2 1 / 2 0 −3 / 2 1 / 2 0 , −3 / 2 1 / 2 0 0 1 4 −2 1 −2 −1 / 2 1 / 2 0 −3 / 2 1 / 2 0 −4 / 2 2 / 5 −1 / 5 , 7 / 10 −1 / 10 3 / 10 1 / 2 −1 / 2 1 / 2 −4 / 5 2 / 5 −1 / 5 The last matrix is A−1. 8. Distances: To define a distance in Rn, we will use the idea of the norm of a vector. The, ℓ2 and, ℓ∞ norms for the vector x = (x1, x2,…, xn)t are defined by 1/ 2 2 : x n = ∑ xi2 i =1 2 (A.19) which is the usual Euclidean norm of the vector x and ∞ : x ∞ = max | xi | (A.20) 1≤ i ≤ n Then if x = ( x1, x2,…, xn )t and y = ( y1, y2,…, yn )t are vectors in Rn, the, ℓ2 and, ℓ∞ distances between x and y are defined to be 2 : x − y 2 ∞ : x − y ∞ a = ∑ xi − yi i =1 1/ 2 (A.21) and = max | xi − yi | 1≤ i ≤ n (A.22) Our major concern with matrices has to do with solving linear algebraic equations as discussed in Chapter 9. There the method of Gaussian elimination was emphasized. Below is an alternate approach to solving the system given by Ax = b in which the matrix A−1 is exhibited. That is, the solution will be of the form x = A−1b, where b is premultiplied by A−1. 476 Appendix A Consider the set of linear equations given by 2 x1 −3 x 2 −2 x 4 3 x 2 +2 x3 + x 4 x1 −2 x 2 −4 x3 +2 x 4 2 x1 + x2 −3 x3 − x 4 =8 =5 =2 =6 They can be recasted in the form 2 −3 0 0 3 2 1 −2 −4 2 1 −3 −2 1 2 −1 x1 x2 x3 x4 8 =5 2 6 where A is the 4 × 4 matrix. We now seek the solution x = A−1b through the use of elementary operations. As a first step, we form the augmented matrix by appending the unit matrix of order 4 to the 4 × 5 matrix formed from A and the vector b: 2 −3 0 −2 8 1 0 0 0 0 3 2 1 5 0 1 0 0 1 −2 −4 2 2 0 0 1 0 2 1 −3 −1 6 0 0 0 1 Second, we carry out the three operations in the order given: 1.Divide the first row by 2. 2.Add (−1) times the new first row to the third row. 3.Subtract 2 times the new first row from the fourth row. These elementary operations result in 1 1.5 0 3 1 −0.5 2 1 0 −1 4 0.5 2 1 5 0 −4 3 −2 −0.5 −3 −1 2 −1 0 1 0 0 0 0 1 0 0 0 0 1 The third step requires the following elementary operations on the newest augmented matrix: 1.Add the fourth row to the first row. 2.Subtract the fourth row from the second row. 3. Subtract 3 times the fourth row from the third row. 477 Appendix A 1 2.5 0 −1 0 − 12.5 0 4 − 3 0 2 −0.5 0 0 1 5 0 7 1 1 0 −1 5 0 4 2.5 0 1 − 3 − 3 1 −2 −1 0 0 1 The fourth step requires the operations: 1.Add two times the second row to the first row. 2.Subtract 12.5 times the second row from the third row. 3. Add four times the second row to the fourth row. 1 0 0 0 0 1.5 1 0 −1 −1 5 0 7 1 0 −57.5 0 − 83.5 − 10.0 − 12.5 1 9.5 0 17 1 26 − 1 0 0 1 0 9.5 0 19.5 2 2.5 The fifth step involves the following elementary operations on the newest augmented matrix: 1.Divide the third row by (−57.5). 2.Subtract 9.5 times the new third row from the first row. 3.Subtract 5 times the new third row from the second row. 4.Subtract 17 times the new third row from the fourth row. 1 0 0 0 0.069 − 1 0 0 −0.261 0.130 − 0.087 0.087 − 0.174 0 1 0 1.452 0.174 0.217 − 0.017 − 0.165 0 0 1 1.313 0.043 0.304 0.296 − 0.191 0 0 0 5.704 0.348 0.438 0.165 Finally, multiplying the second row by (−1) results in 1 0 0 0 0.069 1 0 0 0.261 −0.130 0.087 −0.087 0.174 0 1 0 1.452 0.174 0.217 −0.017 − 0.165 0 0 1 1.313 0.043 0.304 0.296 − 0.191 0 0 0 5.704 0.348 0.438 0.165 where the 4 × 4 identity matrix has moved completely to the left, taking the place of A. This final augmented matrix represents the form Ix = A−1b, where the inverse matrix is 478 Appendix A A −1 0.348 −0.130 = 0.174 0.043 0.069 0.087 − 0.087 0.174 0.217 − 0.017 − 0.165 0.304 0.296 − 0.191 0.438 0.165 and the solution vector Ix = (5.704, 0.261, 1.452, 1.313)t. In summary, if we use Ei to denote any matrix representing the elementary operations performed above, then Ax = b can be adjusted to a succession of equivalent forms E n E n−1 E 2 E1 Ax = E n E n−1 E 2 E1 b (A.23) where the elementary matrices are selected to convert the square matrix A to the unit matrix I. This means that E n E n−1 E 2 E1 A = I (A.24) Since the same operations are being performed on the vector b, the column vector b is appended to the n × n matrix A to form an augmented matrix Ab. Then, premultiplying Ab with the operators Ei produces the desired result x = A−1b, and, if needed, the inverse of A is E n E n−1 E 2 E1 = A−1 (A.25) Another example: Find the inverse of 1 2 3 A = 2 5 7 1 1 1 Solution Form the 3 × 6 matrix [A⋮I3] and transform it by elementary row operations to the form [I3⋮A]. The pivot element at each stage is emboldened. 1 2 3 1 0 0 2 5 7 0 1 0 Matrix A augmented by I3. 1 1 1 0 0 1 1 2 3 1 0 0 0 1 1 −2 1 0 To make a21 = 0, replaced row 2 by the sum of itself and −2 0 −1 −2 −1 0 1 times row 1. To make a31 = 0, replaced row 3 by the sum of itself and −1 times row 1. 479 Appendix A 1 0 1 5 −2 0 0 1 1 −2 1 0 To make a12 = 0, replaced row one by the sum of itself and 0 0 1 3 −1 −1 −2 times row 2. To make a32 = 0, replace row 3 by the sum of itself and row 2. 1 0 1 5 −2 0 0 1 1 −2 1 0 Multiplied row 3 by −1. 0 0 1 3 −1 −1 1 0 0 2 −1 1 0 1 0 −5 2 1 To make a13 = 0, replaced row one by the sum of itself and 0 0 1 3 −1 −1 −1 times row 3. To make a23 = 0, replaced row 2 by the sum of itself and −1 times row 3. The final matrix is of the form [I3 ⋮ A−1], that is, 2 −1 1 A = −5 2 1 3 −1 −1 −1 A.3 ADDITIONAL PROPERTIES OF MATRICES Square matrices can appear in series (finite or infinite) and may exhibit behavior quite similar to those of scalar series. For instance, we could have a0 Y n + a1 Y n−1 + a2 Y n−2 + + an−1 Y + an I = 0 (A.26) where Y is a square matrix. Matrix polynomials may be factored in ways similar to that of scalar polynomials. For example: A3 − 9A2 + 26A − 24I can be factored into (A − 2I)(A − 3I)(A − 4I). It is also possible to use square matrices as exponents of scalar functions, such as eA = I + A A 2 A3 + + + 1! 2! 3! (A.27) Then, if B is a square matrix of the same order as A, e A e B = e B e A = e ( A+ B ) (A.28) e A e− A = I (A.29) and For a given square matrix A of order n whose elements are constants and a column vector x, the equation Ax = λx (A.30) 480 Appendix A or ( A − λI) x = 0 (A.31) has nontrivial solution if, and only if ( A − λI) = 0 (A.32) Equation A.32 is called the characteristic equation. A.4 CALCULUS OF MATRICES If the elements of a matrix are functions of some appropriate independent variable(s), the matrix can be differentiated or integrated with respect to the independent variable(s). Similar to the calculus of functions, we can differentiate matrices by differentiating the elements of the matrix in each case. The usual designation for the derivative of a square matrix Y is dY/dx if the matrix elements are functions of the scalar x. The differential coefficient of a product is also similar to that of a product of scalar functions; however, the order of the matrices in the product must be maintained. Therefore d dA dB dC BC + A C + AB ( ABC) = dx dx dx dx (A.33) where A, B, and C are conformable square matrices whose elements are functions of x. Integration of a matrix whose elements are derivatives or functions of an independent variable is accomplished by integrating each element with respect to the independent variable. Therefore, for a typical element, yij(x), of the matrix Y, the result ∫ Yd x = ∫ yij ( x ) d x (A.34) REFERENCES 1.Amundson, N.R. Mathematical Methods in Chemical Engineering, Prentice Hall, Englewood Cliffs, 1966. 2. Strang, G. Linear Algebra and Its Applications, 2nd ed., Academic Press, New York, 1980. 3.Boyce, W.E. and DiPrima, R.C. Elementary Differential Equations and Boundary Value Problems, 8th ed., John Wiley, New York, 2005. 4. Schneider, H. and Barker, G.P. Matrices and Linear Algebra, Holt, Rinehart & Winston, New York, 1968. 5. Jenson, V.G. and Jeffreys, G.V. Mathematical Methods in Chemical Engineering, 2nd ed., Academic Press, London, 1992. Appendix B: Numerical Method of Lines Example Using MATLAB® EXAMPLE 9.9 % % Main program pdelin computes the numerical % Solution to a linear PDE by six integrators % % Declare global variables global nsteps; % % Step through six integrators % for int = 1:6 int = 3 %Selecting the clasical Runge-Kutta order four method % % Integration parameters [neqn,nout,nsteps,t0,tf,abserr,relerr] = intpar; % % Initial condition vector [u0] = inital(neqn,t0); % % Output interval dt = tf-t0; % % Compute solution at nout output points for j = 1:nout % % Print current solution [out] = fprint(int,neqn,t0,u0); % % Fixed step modified Euler integrator if int = = 1 [u] = euler2a(neqn,t0,tf,u0,nsteps); end % % Variable step modified Euler integrator if int = = 2 [u] = euler2b(neqn,t0,tf,u0,nsteps,abserr,relerr); end 481 482 Appendix B % % Fixed step classical fourth order RK integrator if int = = 3 [u] = rkc4a(neqn,t0,tf,u0,nsteps); end % % Variable step classical fourth order RK integrator if int = = 4 [u] = rkc4b(neqn,t0,tf,u0,nsteps,abserr,relerr); end % % Fixed step RK Fehlberg (RKF45) integrator if int = = 5 [u] = rkf45a(neqn,t0,tf,u0,nsteps); end % % Variable step RK Fehlberg (RKF45) integrator if int = = 6 [u] = rkf45b(neqn,t0,tf,u0,nsteps,abserr,relerr); end % % Advance solution t0 = tf; tf = tf+dt; u0 = u; % % Next output end % % Next integrator end % % End of pdelin function[neqn,nout,nsteps,t0,tf,abserr, relerr] = intpar % % Function intpar sets the parameters to control the % integration of the linear PDE % % Number of first order ODEs neqn = 21; % % Number of output points nout = 6; % % Maximum number of steps in the interval t0 to tf nsteps = 250; % % Initial, final values of independent variable t0 = 0.0; tf = 0.2; % Appendix B % Error tolerances abserr = 1.0e-05; relerr = 1.0e-05; function [u] = inital(neqn,t) % % Function inital sets the initial condition vector % for the linear PDE % % Problem parameters xl = 0.0; xu = 1.0; % % Initial condition for i = 1:neqn x = xl+(i-1)/(neqn-1)*(xu-xl); u(i) = sin(pi*x); end function [ut] = derv(neqn,t,u) % % Function derv computes the derivative vector % for the linear PDE % % Problem parameters xl = 0.0; xu = 1.0; % % BC at x = 0 ut(1) = 0.0; % % BC at x = 1 ut(neqn) = 0.0; % % Interior points dx = (xu-xl)/(neqn-1); dxs = dx*dx; for i = 2:neqn-1 ut(i) = (u(i+1)-2.0*u(i)+u(i-1))/dxs; end function [u] = rkc4a(neqn,t0,tf,u0,nsteps) % % Function rkc4a computes an ODE solution by a fixed step % classical fourth order RK method for a series of points % along the solution by repeatedly calling function ssrkc4 % for a single classical fourth order RK step. % % Argument list % % neqn number of first order ODEs % % t0 initial value of independent variable % 483 484 Appendix B % tf final value of independent variable % % u0 initial condition vector of length neqn % % nsteps number of rkc4 steps % % u ODE solution vector of length neqn after % nsteps steps % % Integration step h = (tf-t0)/nsteps; % % nsteps rkc4 steps for i = 1:nsteps % % rkc4 step [t,u,e] = ssrkc4(neqn,t0,u0,h); % % Reset base point values for next rkc4 step u0 = u; t0 = t; % % Next rkc4 step end % % End of rkc4a function [t,u,e] = ssrkc4(neqn,t0,u0,h) % % Function ssrkc4 computes an ODE solution by the classical fourth % order RK method for one step along the solution (by calls to derv % to define the ODE derivative vector). It also estimates the % truncation error of the solution, and applies this estimate as % a correction to the solution vector. % % Argument list % % neqn number of first order ODEs % % t0 initial value of independent variable % % u0 initial condition vector of length neqn % % h integration step % % t independent variable % % u ODE solution vector of length neqn after % one rkc4 step Appendix B % % e estimate of truncation error of the solution vector % % Derivative vector at initial (base) point [ut0] = derv(neqn,t0,u0); % % k1, advance of dependent variable vector and % independent variable for calculation of k2 k1 = h*ut0; u = u0+0.5*k1; t = t0+0.5*h; % % Derivative vector at new u, t [ut] = derv(neqn,t,u); % % k2, advance of dependent variable vector and % independent variable for calculation of k3 k2 = h*ut; u = u0+0.5*k2; t = t0+0.5*h; % % Derivative vector at new u, t [ut] = derv(neqn,t,u); % % k3, advance of dependent variable vector and % independent variable for calculation of k4 k3 = h*ut; u = u0+k3; t = t0+h; % % Derivative vector at new u, t [ut] = derv(neqn,t,u); % % k4 k4 = h*ut; % % Second order step sum2 = u0+k2; % % Fourth order step sum4 = u0+(1.0/6.0)*(k1+2.0*k2+2.0*k3+k4); t = t0+h; % % Truncation error estimate e = sum4-sum2; % % Fourth order solution vector (from 2,4 RK pair); % two ways to the same result are listed % u = sum2+e; u = sum4; % 485 486 % End of ssrkc4 function [out] = fprint(ncase,neqn,t,u) % % Function fprint displays the numerical and exact % solutions to the linear PDE % % Declare global variables global nsteps; % % Return current value of independent variable % (MATLAB requires at least one return argument) out = t; % % Problem parameters xl = 0.0; xu = 1.0; % % Print a heading for the solution at t = 0 if(t< = 0.0) % Label for ODE integrator % % Fixed step modified Euler if(ncase = = 1) fprintf(‘\n\n euler2a integrator\n\n’); % % Variable step modified Euler elseif(ncase = = 2) fprintf(‘\n\n euler2b integrator\n\n’); % % Fixed step classical fourth order RK elseif(ncase = = 3) fprintf(‘\n\n rkc4a integrator\n\n’); % % Variable step classical fourth order RK elseif(ncase = = 4) fprintf(‘\n\n rkc4b integrator\n\n’); % % Fixed step RK Fehlberg 45 elseif(ncase = = 5) fprintf(‘\n\n rkc45a integrator\n\n’); % % Variable step RK Fehlberg 45 elseif(ncase = = 6) fprintf(‘\n\n rkc45b integrator\n\n’); end % % Heading fprintf(‘ ncase = %2d neqn = %2d nsteps = %3d \n\n’,ncase,neqn,nsteps); fprintf(‘t u(num) Appendix B 487 Appendix B u(exact) diff\n’); % % End of t = 0 heading end % % Numerical and analytical solution output % % Midpoint value of x x = (xu-xl)/2.0; % % Analytical solution at midpoint ue = exp(-pi*pi*t)*sin(pi*x); % % Grid index of midpoint im = round((neqn+1)/2); % % Display the numerical and exact solutions, and their difference fprintf(‘%5.2f %11.6f %11.6f %13.4e\n’,t,u(im), ue,u(im)-ue); Below are the results for the case int = 3 (Runge–Kutta order four). The table is a display of the formatting set in fprint. rkc4a integrator ncase = 3 neqn = 21 nsteps = 250 t u(num) u(exact) diff 0.00 0.20 0.40 0.60 0.80 1.00 1.000000 0.139476 0.019453 0.002713 0.000378 0.000053 1.000000 0.138911 0.019296 0.002680 0.000372 0.000052 0.0000e + 000 5.6448e − 004 1.5714e − 004 3.2810e − 005 6.0893e − 006 1.0595e − 006 As is indicated in the main program, other integrators (such as the Euler or Runge– Kutta–Fehlberg) could have been used to solve this problem. That approach would provide a means to compare the results from each method. However, in the present example the classical Runge–Kutta method was used and the results are compared with those from the analytical solution under the column labeled u(exact). The other columns, u(num) and diff, are the numerical results and the difference between the exact and the numerical. This page intentionally left blank Appendix C: Program for a Transport and Binding Kinetics Model of an Analyte EXAMPLE 9.10 % % Clear previous files clear all clc % % Parameters shared with the ODE routine global zl zu z dz dz2 D kf cbsat kr n cbulk ndss ncall % % Parameter numerical values D=1.0e-10; kf=1.0e+05; cbulk=4.48e-05; h=5.0e-05; c0=0; cb0=0; % % Variation in interface binding saturation cbsat=1.66e-08; cbsat=1.66e-09; % % Variation in interface unbinding rate kr=1.0e-01; kr=1.0e+01; % % Spatial grid zl=0; zu=5.0e-05; n=21; dz=(zu-zl)/(n-1); dz2=dz^2; % % Initial condition for i=1:n u0(i)=c0; end u0(n+1)=cb0; % % Independent variable for ODE integration t0=0.0; tf=100; tout=(t0:2:tf); nout=51; ncall=0; 489 490 Appendix C % % ODE itegration % % Variation in error tolerances reltol=1.0e-06; abstol=1.0e-06; reltol=1.0e-07; abstol=1.0e-07; options=odeset('RelTol',reltol,'AbsTol',abstol); mf=1; if(mf==1) % explicit FDs [t,u]=ode15s(@pde_1,tout,u0,options); end if(mf==2) ndss=4; % ndss = 2, 4, 6, 8 or 10 required [t,u]=ode15s(@pde_2,tout,u0,options); end if(mf==3) ndss=44; % ndss = 42, 44, 46, 48 or 50 required [t,u]=ode15s(@pde_3,tout,u0,options); end % % Store numerical solutions for it=1:nout c_plot(it)=u(it,1); cb_plot(it)=u(it,n+1); theta_plot(it)=cb_plot(it)/cbsat; rate_plot(it)=kf*c_plot(it)*(cbsat-cb_plot(it))-kr*cb_ plot(it); end % % Display selected output fprintf('\n mf = %2d abstol = %8.1e reltol = %8.1e\n',... mf,abstol,reltol); fprintf('\n t c(0,t) cb(t) theta rate\n'); for it=1:nout fprintf('%6.0f%12.3e%12.3e%12.3e%12.3e\n',... t(it),c_plot(it),cb_plot(it),theta_plot(it),rate_plot(it)); end fprintf('\n ncall = %4d\n',ncall); % % Plot numerical solution figure(1); subplot(2,2,1) plot(t,c_plot); axis tight title('c(0,t) vs t'); xlabel('t'); ylabel('c(0,t)') subplot(2,2,2) plot(t,cb_plot); axis tight title('cb(t) vs t'); xlabel('t'); ylabel('cb(t)') subplot(2,2,3) plot(t,theta_plot); axis tight title('theta(t) vs t'); xlabel('t'); ylabel('theta(t)') subplot(2,2,4) plot(t,rate_plot); axis tight title('rate(t) vs t'); xlabel('t'); ylabel('rate(t)') Appendix C 491 % % Store numerical solution for 3D plot for it=1:nout for i=1:n c_3D(it,i)=u(it,i); end end z=[zl:dz:zu]; figure(2) surf(z,t,c_3D) xlabel('z (m)'); ylabel('t (sec)'); zlabel('c(z,t) (moles/ m^3)'); title('c(z,t) (moles/m^3), z=0,2.5\times10^{-6},..., 5\ times10^{-5} (m), t=0,2,...,100 (sec)') % print -deps pde.eps; print -dps pde.ps The routine below converts Equation 9.113 in the given example into a set of ODEs that are then combined with the given ODE, Equation 9.117. function ut=pde_1(t,u) % % Problem parameters global zl zu z dz dz2 D kf cbsat kr n cbulk ndss ncall % % ODE and PDE for i=1:n c(i)=u(i); end cb=u(n+1); % % BCs cf=c(2)-(2*dz/D)*(kf*c(1)*(cbsat-cb)-kr*cb); c(n)=cbulk; % % PDE for i=1:n if(i==1) ct(1)=D*(c(2)-2*c(1)+cf)/dz2; elseif(i==n) ct(n)=0; else ct(i)=D*(c(i+1)-2*c(i)+c(i-1))/dz2; end end % % ODE cbt=kf*c(1)*(cbsat-cb)-kr*cb; % % Derivative vector for i=1:n ut(i)=ct(i); end ut(n+1)=cbt; % 492 Appendix C % Transpose for ODE integrator ut=ut'; % % Increment calls to pde_1 ncall=ncall+1; This file is called in the main program as ndss==4 for use with ode15s Supporting routines being called (www.cambridge.org/9781107022805): both dss004 and dss044 are used to compute numerical derivatives of 1-dimension (1D) arrays. More details on the process used in dss004 and dss044 can be found in Partial Differential Equation Analysis in Biomedical Engineering Case Studies with MATLAB® Shiesser, W. E. pp. 363 –366. % % File: dss004.m function [ux]=dss004(xl,xu,n,u) % % % % % % % % % % % % % % % % Function dss004 computes the first derivative, u , of a x variable u over the spatial domain xl le x le xu from classical five-point, fourth-order finite difference approximations Argument list xl Lower boundary value of x (input) xu Upper boundary value of x (input) n N umber of grid points in the x domain including the boundary points (input) u O ne-dimensional array containing the values of u at t he n grid point points for which the derivative is to be computed (input) % % % % % % % % % % % % % % ux One-dimensional array containing the numerical v alues of the derivatives of u at the n grid points (output) The mathematical details of the following Taylor series (or polynomials) are given in routine dss002. Five-point formulas (1) Left end, point i = 1 493 Appendix C % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % 2 3 ( dx) + u1 ( dx) + u1 ( dx) x 1f 2x 2f 3x 3f a(u2 = u1 + u1 5 + u1 ( dx) 5x 5f 6 + u1 ( dx) 6x 6f 7 + u1 ( dx) 7x 7f + ...) 2 3 (2dx) + u1 (2dx) + u1 (2dx) x 1f 2x 2f 3x 3f b(u3 = u1 + u1 5 + u1 (2dx) 5x 5f 6 + u1 (2dx) 6x 6f (2dx) 7x 7f + ...) 2 3 (3dx) + u1 (3dx) + u1 (3dx) x 1f 2x 2f 3x 3f 5 (3dx) 5x 5f 6 + u1 (3dx) 6x 6f (3dx) 7x 7f + ...) 2 3 (4dx) + u1 (4dx) + u1 (4dx) x 1f 2x 2f 3x 3f 5 (4dx) 5x 5f 6 + u1 (4dx) 6x 6f 4 (3dx) 4x 4f + u1 7 + u1 d(u5 = u1 + u1 + u1 4 (2dx) 4x 4f + u1 7 + u1 c(u4 = u1 + u1 + u1 4 ( dx) 4x 4f + u1 4 (4dx) 4x 4f + u1 7 + u1 (4dx) 7x 7f + ...) C onstants a, b, c, and d are selected so that the coefficients of the u1 terms sum to one and the coefficients of the u1 , x u1 and u1 terms sum to zero 3x 4x % 2x % % % % a + 2b + 3c + 4d = 1 % % a + 4b + 9c + 16d = 0 % % a + 8b + 27c + 64d = 0 % % a + 16b + 81c + 256d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x 494 % Appendix C terms, for u1 gives the following five-point approximation x % % 4 % u1 = (1/12dx)(-25u1 + 48u2 - 36u3 + 16u4 - 3u5) + O(dx (1) % x % % (2) Interior point, i = 2 % % 2 3 4 % a(u1 = u2 + u2 (-dx) + u2 (-dx) + u2 (-dx) + u2 (-dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (-dx) + u2 (-dx) + u2 (-dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % b(u3 = u2 + u2 ( dx) + u2 ( dx) + u2 ( dx) + u2 ( dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 ( dx) + u2 ( dx) + u2 ( dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % c(u4 = u2 + u2 (2dx) + u2 (2dx) + u2 (2dx) + u2 (2dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (2dx) + u2 (2dx) + u2 (2dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % d(u5 = u2 + u2 (3dx) + u2 (3dx) + u2 (3dx) + u2 (3dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (3dx) + u2 (3dx) + u2 (3dx) + ...) % 5x 5f 6x 6f 7x 7f % % -a + b + 2c + 3d = 1 % % a + b + 4c + 9d = 0 % % -a + b + 8c + 27d = 0 % % a + b + 16c + 81d = 0 % % Simultaneous solution for a, b, c, and d followed by the solution of the preceding Taylor series, truncated after Appendix C 495 the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 % u2 = (1/12dx)(-3u1 - 10u2 + 18u3 - 6u4 + u5) + O(dx )(2) % x % % (3) Interior point i, i ne 2, n-1 % % 2 3 % a(ui-2 = ui + ui (-2dx) + ui (-2dx) + ui (-2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui (-2dx) + ui (-2dx) + ui (-2dx)+ ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % b(ui-1 = ui + ui ( -dx) + ui ( -dx) + ui ( -dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( -dx) + ui ( -dx) + ui ( -dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % c(ui+1 = ui + ui ( dx) + ui ( dx) + ui ( dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( dx) + ui ( dx) + ui ( dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % d(ui+2 = ui + ui ( 2dx) + ui ( 2dx) + ui ( 2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( 2dx) + ui ( 2dx) + ui ( 2dx) + ...) % 4x 4f 5x 5f 6x 6f % % -2a b + c + 2d = 1 % % 4a + b + c + 4d = 0 % % -8a b + c + 8d = 0 % % 16a + b + c + 16d = 0 % 496 Appendix C % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 % ui = (1/12dx)(ui-2 - 8ui-1 + 0ui + 8ui+1 - ui+2) + O(dx )(3) % x % % (4) Interior point, i = n-1 % % 2 3 % a(un-4 = un-1 + un-1 (-3dx) + un-1 (-3dx) + un-1 (-3dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 (-3dx) + un-1 (-3dx) + un-1 (-3dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % b(un-3 = un-1 + un-1 (-2dx) + un-1 (-2dx) + un-1 (-2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 (-2dx) + un-1 (-2dx) + un-1 (-2dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % c(un-2 = un-1 + un-1 ( -dx) + un-1 (- -x) + un-1 (-dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 ( -dx) + un-1 ( -dx) + un-1 ( -dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % d(un = un-1 + un-1 ( dx) + un-1 ( dx) + un-1 ( dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 ( dx) + un-1 ( dx) + un-1 ( dx) + ... % 4x 4f 5x 5f 6x 6f % % -3a - 2b c + d = 1 % % 9a + 4b + c + d = 0 % % -27a - 8b c + d = 0 % % 81a + 16b + c + d = 0 % Appendix C 497 % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 % u n-1 = (1/12dx)(-un-4 + 6un-3 - 18un-2 + 10un-1 + 3un) + O(dx ) % x % (4) % % (5) Right end, point i = n % % 2 3 % a(un-4 = un + un (-4dx) + un (-4dx) + un (-4dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un (-4dx) + un (-4dx) + un (-4dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % b(un-3 = un + un (-3dx) + un (-3dx) + un (-3dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un (-3dx) + un (-3dx) + un (-3dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % c(un-2 = un + un (-2dx) + un (-2dx) + un (-2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un (-2dx) + un (-2dx) + un (-2dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % d(un-1 = un + un ( -dx) + un ( -dx) + un ( -dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un ( -dx) + un ( -dx) + un ( -dx) + ...) % 4x 4f 5x 5f 6x 6f % % -4a - 3b - 2c d = 1 % % 16a + 9b + 4c + d = 0 % % -64a - 27b - 8c d = 0 498 Appendix C % % 256a + 81b + 16c + d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 % un = (1/12dx)(3un-4 - 16un-3 + 36un-2 - 48un-1 + 25un) + O(dx ) % x % (5) % % The weighting coefficients for equations (1) to (5) can be % summarized as % % -25 48 -36 16 -3 % % -3 -10 18 -6 1 % % 1/12 1 -8 0 8 -1 % % -1 6 -18 10 3 % % 3 -16 36 -48 25 % % which are the coefficients reported by Bickley for n = 4,m = % 1, p = 0, 1, 2, 3, 4 (Bickley, W. G., Formulae for Numerical % Differentiation, Math. Gaz., vol. 25, 1941. Note - the Bickley % coefficients have been divided by a common factor of two). % % Equations (1) to (5) can now be programmed to generate the % derivative u (x) of function u(x). % x % % Compute the spatial increment dx=(xu-xl)/(n-1); r4fdx=1./(12.*dx); nm2=n-2; % % Equation (1) (note - the rhs of equations (1), (2), (3),(4) % and (5) have been formatted so that the numerical weighting % coefficients can be more easily associated with the Bickley % matrix above) 499 Appendix C ux( 1)=r4fdx*... ( -25.*u( 1) +48.*u( 2) -3.*u( 5)); % % Equation (2) ux( 2)=r4fdx*... ( -3.*u( 1) -10.*u( 2) +1.*u( 5)); % % Equation (3) for i=3:nm2 ux( i)=r4fdx*... ( +1.*u(i-2) -8.*u(i-1) -1.*u(i+2)); end % % Equation (4) ux(n-1)=r4fdx*... ( -1.*u(n-4) +6.*u(n-3) +3.*u( n)); % % Equation (5) ux( n)=r4fdx*... ( 3.*u(n-4) -16.*u(n-3) +25.*u( n)); -36.*u( 3) +16.*u( 4) +18.*u( 3) 4) +0.*u( i) -6.*u( +8.*u(i+1) -18.*u(n-2) +10.*u(n-1) +36.*u(n-2) -48.*u(n-1) This file is called in the main program as ndss==44 for use with ode15s. % % File: ds044.m function [uxx]=dss044(xl,xu,n,u,ux,nl,nu) % % % % % % % % % % % % % % % % Function dss044 computes a fourth-order approximation of a second-order derivative, with or without the normal derivative at the boundary. Argument list xl L eft value of the spatial independent variable (input) xu R ight value of the spatial independent variable (input) n Number of spatial grid points, including the end points (input) u O ne-dimensional array of the dependent variable to be differentiated (input) 500 % % Appendix C ux O ne-dimensional array of the first derivative of u. % The end values of ux, ux(1), and ux(n), are used in % N eumann boundary conditions at x = xl and x = xu, % depending on the arguments nl and nu (see the de% scription of nl and nu below) % % uxx O ne-dimensional array of the second derivative of u % (output) % % nl I nteger index for the type of boundary condition at % x = xl (input). The allowable values are % % 1 - Dirichlet boundary condition at x = xl % (ux(1) is not used) % % 2 - Neumann boundary condition at x = xl % (ux(1) is used) % % nu I nteger index for the type of boundary condition at % x = xu (input). The allowable values are % % 1 - Dirichlet boundary condition at x = xu % (ux(n) is not used) % % 2 - Neumann boundary condition at x = xu % (ux(n) is used) % % The following derivation was completed by W. E. Schiesser, Depts % of CHE and Math, Lehigh University, Bethlehem, PA 18015, USA, on % December 15, 1986. Additional details are given in function % dss042. % % *********************************************************** % % (1) uxx at the interior points 3, 4,..., n-2 % % To develop a set of fourth-order correct differentiation formulas % for the second derivative uxx, we consider first the interior % grid points at which a symmetric formula can be used. % 501 Appendix C % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % If we consider a formula of the form a*u(i-2) + b*u(i-1) + e*u(i) + c*u(i+1) + d*u(i+2) T aylor series expansions of u(i-2), u(i-1), u(i+1), and u(i+2) can be substituted into this formula. We then consider the l inear albegraic equations relating a, b, c, and d which will retain certain terms, i.e., uxx, and drop others, e.g., uxxx, uxxxx and uxxxxx. Thus, for grid points 3, 4,..., n-2 To retain uxx 4*a + b + c + 4*d = 2 (1) 8*d = 0 (2) 16*a + b + c + 16*d = 0 (3) To drop uxxx -8*a - b + c + To drop uxxxx To drop uxxxxx -32*a - b + c + 32*d = 0 (4) E quations (1) to (4) can be solved for a, b, c, and d. If equation (1) is added to equation (2) -4*a + 2*c + 12*d = 2 (5) If equation (1) is subtracted from equation (3) 12*a + 12*d = -2 (6) If equation (1) is added to equation (4) -28*a + 2*c + 36*d = 2 E quations (5) to (7) can be solved for a, c, and d. equation (5) is subtracted from equation (7), and the result combined with equation (6) 12*a + 12*d = -2 (7) If (6) 502 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix C -24*a + 24*d = 0 (8) Equations (6) and (8) can be solved for a and d. From (8), a = d. From equation (6), a = -1/12 and d = -1/12. Then, from equation (5), c = 4/3, and from equation (1), b = 4/3. The final differentiation formula is then obtained as (-1/12)*u(i-2) + (4/3)*u(i-1) + (4/3)*u(i+1) + (-1/12)*u(i+2) ( -1/12 + 4/3 - 1/12 + 4/3)*u(i) + uxx(i)*(dx**2) + O(dx**6) or uxx(i) = (1/(12*dx**2))*(-1*u(i-2) + 16*u(i-1) - 30*u(i) + 16*u(i+1) - 1*u(i+2) (9) + O(dx**4) Note that the ux term drops out, i.e., the basic equation is -2*a - b + c + 2*d = -2*(-1/12) - (4/3) + (4/3) + 2*(-1/12) = 0 quation (9) was obtained by dropping all terms in the E underlying Taylor series up to and including the fifth derivative, uxxxxx. Thus, equation (9) is exact for polynomials up to and including fifth order. This can be checked by substituting the functions 1, x, x**2, x**3, x**4 and x**5 in equation (9) and computing the corresponding derivatives for comparison with the known second derivatives. This is done for 1 merely by summing the weighting coefficients in equation (9), which should sum to zero, i.e., -1 + 16 - 30 + 16 -1 = 0. or the remaining functions, the algebra is rather F involved, but Appendix C % 503 t hese functions can be checked numerically, i.e., numerical values % o f x**2, x**3, x**4, and x**5 can be substituted in equation (9) % and the computed derivatives can be compared with the know numeri% c al second derivatives. This is not a proof of correctness of % e quation (9), but would likely detect any errors in equation (9). % % *********************************************************** ******* % % (2) uxx at the interior points i = 2 and n-1 % % For grid point 2, we consider a formula of the form % % a*u(i-1) + f*u(i) + b*u(i+1) + c*u(i+2) + d*u(i+3) + e*u(i+4) % % T aylor series expansions of u(i-1), u(i+1), u(i+2), u(i+3), and % u (i+4) when substituted into this formula give linear algebraic % equations relating a, b, c, d, and e. % % To drop ux % % -a + b + 2*c + 3*d + 4*e = 0 (10) % % To retain uxx % % a + b + 4*c + 9*d + 16*e = 2 (11) % % To drop uxxx % % -a + b + 8*c + 27*d + 64*e = 0 (12) % % To drop uxxxx % % a + b + 16*c + 81*d + 256*e = 0 (13) % % To drop uxxxxx % % -a + b + 32*c + 243*d + 1024*e = 0 (14) % % E quations (11), (12), (13), and (14) can be solved for a, b, c, d, and e. If equation (10) is added to equation (11) % % 2*b + 6*c + 12*d +20*e = 2 (15) 504 % % % % % % % % % % % % % % % % % % % % % % % % % Appendix C If equation (10) is subtracted from equation (12) 6*c + 24*d + 60*e = 0 (16) If equation (10) is added to equation (13) 2*b + 18*c + 84*d + 260*e = 0 (17) If equation (10) is subtracted from equation (14) 30*c + 240*d + 1020*e = 0 (18) Equations (15), (16), (17), and (18) can be solved for b, c, d and e. 6*c + 24*d + 60*e = 0 (16) If equation (15) is subtracted from equation (17) 12*c + 72*d + 240*e = -2 (19) 30*c + 240*d + 1020*e = 0 (18) quations (16), (18), and (19) can be solved for c, d, and E e. If two times equation (16) is subtracted from equation (19), % % % 24*d + 120*e = -2 (20) % % If five times equation (16) is subtracted from equation (18), % % 120*d + 720*e = 0 (21) % % Equations (20) and (21) can be solved for d and e. From (21), % e = (-1/6)*d. Substitution in equation (20) gives d = -1/2. % thus, e = 1/12. From equation (16), c = 7/6. From equation % (15), b = -1/3. From equation (10), a = 5/6. % % The final differentiation formula is then obtained as % % (5/6)*u(i-1) + (-1/3)*u(i+1) + (7/6)*u(i+2) + (-1/2)*u(i+3) % % + (1/12)*u(i+4) = (5/6 - 1/3 + 7/6 - 1/2 + 1/12)*u(i) % % + uxx*(dx**2) + O(dx**6) % % or Appendix C 505 % % uxx(i) = (1/12*dx**2)*(10*u(i-1) - 15*u(i) - 4*u(i+1) % (22) % + 14*u(i+2) - 6*u(i+3) + 1*u(i+4)) + O(dx**4) % % Equation (22) will be applied at i = 2 and n-1. thus % % uxx(2) = (1/12*dx**2)*(10*u(1) - 15*u(2) - 4*u(3) % (23) % + 14*u(4) - 6*u(5) + 1*u(6)) + O(dx**4) % % uxx(n-1) = (1/12*dx**2)*(10*u(n) - 15*u(n-1) - 4*u(n-2) % (24) % + 14*u(n-3) - 6*u(n-4) + 1*u(n-5)) + O(dx**4) % % *********************************************************** ******* % % (3) uxx at the boundary points 1 and n % % Finally, for grid point 1, an approximation with a Neumann boundary condition of the form % % a*u(i+1) + b*u(i+2) + c*u(i+3) + d*u(i+4) + e*ux(i) + f*u(i) % % Will be used. the corresponding algebraic equations are % % To drop ux % % a + 2*b + 3*c + 4*d + e = 0 (25) % % To retain uxx % % a + 4*b + 9*c + 16*d = 2 (26) % % To drop uxxx % % a + 8*b + 27*c + 64*d = 0 (27) % % To drop uxxxx % % a + 16*b + 81*c + 256*d = 0 (28) % % To drop uxxxxx % % a + 32*b + 243*c + 1024*d = 0 (29) % % Equations (25) to (29) can be solved for a, b, c, d, and e. If % 506 % % % % % % % % % % % % % % % % % % Appendix C Equation (26) is subtracted from Equations (27), (28), and (29), 4*b + 18*c + 48*d = -2 (30) 12*b + 72*c + 240*d = -2 (31) 28*b + 234*c + 1008*d = -2 (32) Equations (30), (31), and (32) can be solved for b, c, and d 18*c + 96*d = 4 (33) 108*c + 672*d = 12 (34) Equations (3) and (34) can be solved for c and d, c = 8/9, d = -1/8. From equation (30), b = -3. a = 8. From equation (25), e = -25/6. From equation (26), % % % The final differentiation formula is then obtained as % % 8*u(i+1) - 3*u(i+2) + (8/9)*u(i+3) - (1/8)*u(i+4) % % - (25/6)*ux(i)*dx % % = (8 - 3 + (8/9) - (1/8))*u(i) + uxx*(dx**2) + O(dx**6) % % or % % uxx(i) = (1/12*dx**2)*((-415/6)*u(i) + 96*u(i+1) - 36*u(i+2) % (35) % + (32/3)*u(i+3) - (3/2)*u(i+4) - 50*ux(i)*dx) + O(dx**4) % % Equation (35) will be applied at i = 1 and i = n % % uxx(1) = (1/12*dx**2)*((-415/6)*u(1) + 96*u(2) - 36*u(3) % (36) % + (32/3)*u(4) - (3/2)*u(5) - 50*ux(1)*dx) + O(dx**4) % % uxx(n) = (1/12*dx**2)*((-415/6)*u(n) + 96*u(n-1) - 36*u(n-2) % (37) % + (32/3)*u(n-3) - (3/2)*u(n-4) + 50*ux(n)*dx) + O(dx**4) % % Alternatively, for grid point 1, an approximation with a Dirichlet % boundary condition of the form % Appendix C 507 % a*u(i+1) + b*u(i+2) + c*u(i+3) + d*u(i+4) + e*u(i+5) + f*u(i) % % can be used. The corresponding algebraic equations are % % To drop ux % % a + 2*b + 3*c + 4*d + 5*e = 0 (38) % % To retain uxx % % a + 4*b + 9*c + 16*d + 25*e = 2 (39) % % To drop uxxx % % a + 8*b + 27*c + 64*d + 125*e = 0 (40) % % To drop uxxxx % % a + 16*b + 81*c + 256*d + 625*e = 0 (41) % % To drop uxxxxx % % a + 32*b + 243*c + 1024*d + 3125*e = 0 (42) % % E quations (38), (39), (40), (41), and (42) can be solved for a, b, c, d, and e. % % 2*b + 6*c + 12*d + 20*e = 2 (43) % % 6*b + 24*c + 60*d + 120*e = 0 (44) % % 14*b + 78*c + 252*d + 620*e = 0 (45) % % 30*b + 240*c + 1020*d + 3120*e = 0 (46) % % E quations (43), (44), (45), and (46) can be solved for b, c, d and e % % 6*c + 24*d + 60*e = -6 (47) % % 36*c + 168*d + 480*e = -14 (48) % % 150*c + 840*d + 2820*e = -30 (49) % % ­ Equations (47), (48), and (49) can be solved for c, d, and e % % 24*d + 120*e = 22 (50) % % 240*d + 1320*e = 120 (51) 508 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix C rom equations (50) and (51), d = 61/12, e = -5/6. From F equation (47), c = -13. From equation (43), b = 107/6. From equation (38), a = -77/6. The final differentiation formula is then obtained as -77/6)*u(i+1) + (107/6)*u(i+2) - 13*u(i+3) + ( (61/12)*u(i+4) - (5/6)*u(i+5) = (-77/6 + 107/6 - 13 + 61/12 - 5/6)*u(i) + uxx(i)*(dx**2) + O(dx**6) or uxx(i) = (1/12*dx**2)*(45*u(i) - 154*u(i+1) + 214*u(i+2) (52) - 156*u(i+3) + 61*u(i+4) - 10*u(i+5)) + O(dx**4) Equation (52) will be applied at i = 1 and i = n uxx(1) = (1/12*dx**2)*(45*u(1) - 154*u(2) + 214*u(3) (53) - 156*u(4) + 61*u(5) - 10*u(6)) + O(dx**4) uxx(n) = (1/12*dx**2)*(45*u(n) - 154*u(n-1) + 214*u(n-2) (54) -156*u(n-3) + 61*u(n-4) - 10*u(n-5)) + O(dx**4) *********************************************************** Grid spacing dx=(xu-xl)/(n-1); 1/(12*dx**2) for subsequent use r12dxs=1./(12.0*dx^2); uxx at the left boundary Without ux (equation (53)) if nl==1 uxx(1)=r12dxs*... ( 45.0*u(1)... -154.0*u(2)... +214.0*u(3)... -156.0*u(4)... 509 Appendix C +61.0*u(5)... -10.0*u(6)); % % % % % % % % % % % % % % With ux (equation (36)) elseif nl==2 uxx(1)=r12dxs*... (-415.0/6.0*u(1)... +96.0*u(2)... -36.0*u(3)... +32.0/3.0*u(4)... -3.0/2.0*u(5)... -50.0*ux(1)*dx); end uxx at the right boundary Without ux (equation (54)) if nu==1 uxx(n)=r12dxs*... ( 45.0*u(n )... -154.0*u(n-1)... +214.0*u(n-2)... -156.0*u(n-3)... +61.0*u(n-4)... -10.0*u(n-5)); With ux (equation (37)) elseif nu==2 uxx(n)=r12dxs*... (-415.0/6.0*u(n )... +96.0*u(n-1)... -36.0*u(n-2)... +32.0/3.0*u(n-3)... -3.0/2.0*u(n-4)... +50.0*ux(n )*dx); end uxx at the interior grid points i = 2 (equation (23)) uxx(2)=r12dxs*... ( 10.0*u(1)... -15.0*u(2)... -4.0*u(3)... +14.0*u(4)... -6.0*u(5)... +1.0*u(6)); i = n-1 (equation (24)) uxx(n-1)=r12dxs*... 510 Appendix C ( % % 10.0*u(n )... -15.0*u(n-1)... -4.0*u(n-2)... +14.0*u(n-3)... -6.0*u(n-4)... +1.0*u(n-5)); i = 3, 4,..., n-2 (equation (9)) for i=3:n-2 uxx(i)=r12dxs*... ( -1.0*u(i-2)... +16.0*u(i-1)... -30.0*u(i )... +16.0*u(i+1)... -1.0*u(i+2)); end Appendix D: Programmed Model of a Drug Delivery System EXAMPLE 9.11 This example shows the drug delivery system as modeled by Equations 9.120 through 9.127. The following is a translation between the variables used in Example 9.9 and Figure D.1: C1 ≡ u1,C 2 = u2, k1 = ku1,k2 = ku2, C1e = u1e ,C 2e = u2e ,R = r0 D1 = Du1 and D2 = Du2. % % % % % % % % % % % % % % % % % % % % % % % % % % Dynamic analysis of drug delivery The following equations model a polymer matrix in cylindrical coordinates u1 material balance u1t = Du1*(u1rr + (1/r)u1r) (1) u2 material balance u2t = Du2*(u2rr + (1/r)*u2r) (2) he variables and parameters for this model are (in cgs T units) Water concentration u1 (eq. (1)) Drug concentration u2 (eq. (2)) Time t Radial position r Axial position z 511 512 Appendix D Du1 ∂u1(r = z0, z, t) ∂r = ku1(u1e – u1(r = r0, z, t)) Du1 ∂u1(r, z = zL, t) ∂z = ku1(u1e – u1(r, z = zL, t)) u1(r, z, t) u2(r, z, t) r0 r z ∂u1(r, z = zL/2, t) =0 ∂z ∂u1(r = 0, z/t) ∂r =0 zL FIGURE D.1 % % % % % % % % % % % % % % % % % % % % % % % % % Model of a drug delivery system Polymer matrix radius r0 1 Polymer matrix length zL 2 Initial u1 u10 0.5 Initial u2 u20 1 External u1 u1e 1 External u2 u2e 0 u1 diffusivity Du1 1.0e-06 u2 diffusivity Du2 1.0e-06 u1 mass transfer coefficient ku1 0.1 u2 mass transfer coefficient ku2 0.1 The solution to this system, u1(r,z,t) (from eq. (1)) and u2(r,z,t) (from eq. (2)) is computed by ode15s for the integration in t. Clear previous files clear all clc % % Global area % % Global area 513 Appendix D global nr r u1 ncall nz r0 u2 dr z u1e dz zL u2e drs Du1 ku1 dzs... Du2... ku2... % % Model parameters u10=0.5; u20=1; u1e=1; u2e=0; r0=1; zL=2; Du1=1.0e-06; Du2=1.0e-06; ku1=1.0e-01; ku2=1.0e-01; % % Grid in axial direction nz=11; dz=zL/(2*(nz-1)); for i=1:nz z(i)=zL/2+(i-1)*dz; end dzs=dz^2; % % Grid in radial direction nr=11; dr=r0/(nr-1); for j=1:nr r(j)=(j-1)*dr; end drs=dr^2; % % Independent variable for ODE integration tf=2*3600*24; tout=[0.0:2*900*24:tf]'; nout=5; ncall=0; % % Initial condition for i=1:nz for j=1:nr u1(i,j)=u10; u2(i,j)=u20; u0((i-1)*nr+j)=u1(i,j); u0((i-1)*nr+j+nz*nr)=u2(i,j); end end % % ODE integration reltol=1.0e-04; abstol=1.0e-04; options=odeset('RelTol',reltol,'AbsTol',abstol); mf=1; if(mf==1)[t,u]=ode15s(@pde_1,tout,u0,options);end if(mf==2)[t,u]=ode15s(@pde_2,tout,u0,options);end if(mf==3)[t,u]=ode15s(@pde_3,tout,u0,options);end 514 Appendix D if(mf==4)[t,u]=ode15s(@pde_4,tout,u0,options);end % % 1D to 2D matrices for it=1:nout for i=1:nz for j=1:nr u1(it,i,j)=u(it,(i-1)*nr+j); u2(it,i,j)=u(it,(i-1)*nr+j+nz*nr); end end end % % Display a heading and radial profiles % % z=zL/2 fprintf('\n nr = %2d nz = %2d\n',nr,nz); for it=1:nout fprintf('\n t = %4.1f z = %3.1f\n',t(it)/3600,z(1)); for j=1:nr fprintf(... ' r = %4.2f u1(r,z,t) = %6.3f u2(r,z,t) = %6.3f\n',... r(j),u1(it,1,j),u2(it,1,j)); end fprintf('\n ku2*(u2(r=r0,z=zL/2,t)-u2e) = %8.3e\n\n',... 3600*ku2*(u2(it,1,nr)-u2e)); end fprintf('\n ncall = %5d\n',ncall); % % Parametric plots, radial profiles, z=zL/2 % % 2D plots figure(1); subplot(2,2,1) for j=1:nr u1plot(j)=u1(2,1,j); u2plot(j)=u2(2,1,j); end plot(r,u1plot,'-o',r,u2plot,'-x'); axis([0 r0 0 1]); title('u1,u2(r,z=zL/2,t=12hr),o-u1,x-u2'); xlabel('r'); ylabel('u1,u2(r,z=zL/2,t=12hr)') subplot(2,2,2) for j=1:nr u1plot(j)=u1(3,1,j); u2plot(j)=u2(3,1,j); end plot(r,u1plot,'-o',r,u2plot,'-x'); axis([0 r0 0 1]); title('u1,u2(r,z=zL/2,t=24hr),o-u1,x-u2'); xlabel('r'); ylabel('u1,u2(r,z=zL/2,t=24hr)') subplot(2,2,3) for j=1:nr u1plot(j)=u1(4,1,j); u2plot(j)=u2(4,1,j); end plot(r,u1plot,'-o',r,u2plot,'-x'); axis([0 r0 0 1]); title('u1,u2(r,z=zL/2,t=36hr),o-u1,x-u2'); xlabel('r'); ylabel('u1,u2(r,z=zL/2,t=36hr)') subplot(2,2,4) for j=1:nr u1plot(j)=u1(5,1,j); u2plot(j)=u2(5,1,j); end plot(r,u1plot,'-o',r,u2plot,'-x'); axis([0 r0 0 1]); Appendix D title('u1,u2(r,z=zL/2,t=48hr),o-u1,x-u2'); xlabel('r'); ylabel('u1,u2(r,z=zL/2,t=48hr)') % % 3D plots for it=1:nout for j=1:nr u1surf(it,j)=u1(it,1,j); end for j=1:nr u2surf(it,j)=u2(it,1,j); end end figure(2) surf(r,t/(3600*24),u1surf); xlabel('r (cm)'); ylabel('t (day)'); zlabel('u1(r,z=zL/2, t)'); figure(3) surf(r,t/(3600*24),u2surf); view(-20,60) xlabel('r (cm)'); ylabel('t (day)'); zlabel('u2(r,z=zL/2, t)'); The following program is called by the main program listed above. function ut=pde_1(t,u) % % Global area global nr nz dr dz drs dzs... r r0 z zL Du1 Du2... u1 u2 u1e u2e ku1 ku2... ncall % % 1D to 2D matrices for i=1:nz for j=1:nr ij=(i-1)*nr+j; u1(i,j)=u(ij); u2(i,j)=u(ij+nr*nz); end end % % Step through the grid points in r and z for i=1:nz for j=1:nr % % (1/r)*u1r, (1/r)*u2r if(j==1) u1r(i,j)=2*(u1(i,j+1)-u1(i,j))/drs; u2r(i,j)=2*(u2(i,j+1)-u2(i,j))/drs; elseif(j==nr) u1r(i,j)=(1/r(j))*(ku1/Du1)*(u1e-u1(i,j)); u2r(i,j)=(1/r(j))*(ku2/Du2)*(u2e-u2(i,j)); else u1r(i,j)=(1/r(j))*(u1(i,j+1)-u1(i,j-1))/(2*dr); u2r(i,j)=(1/r(j))*(u2(i,j+1)-u2(i,j-1))/(2*dr); end 515 516 % % Appendix D u1rr, u2rr if(j==1) u1rr(i,j)=2*(u1(i,j+1)-u1(i,j))/drs; u2rr(i,j)=2*(u2(i,j+1)-u2(i,j))/drs; elseif(j==nr) u1f=u1(i,j-1)+2*dr*ku1/Du1*(u1e-u1(i,j)); u1rr(i,j)=(u1f-2*u1(i,j)+u1(i,j-1))/drs; u2f=u2(i,j-1)+2*dr*ku2/Du2*(u2e-u2(i,j)); u2rr(i,j)=(u2f-2*u2(i,j)+u2(i,j-1))/drs; else u1rr(i,j)=(u1(i,j+1)-2*u1(i,j)+u1(i,j-1))/drs; u2rr(i,j)=(u2(i,j+1)-2*u2(i,j)+u2(i,j-1))/drs; end % % u1zz, u2zz if(i==1) u1zz(i,j)=2*(u1(i+1,j)-u1(i,j))/dzs; u2zz(i,j)=2*(u2(i+1,j)-u2(i,j))/dzs; elseif(i==nz) u1f=u1(i-1,j)+2*dz*ku1/Du1*(u1e-u1(i,j)); u1zz(i,j)=(u1f-2*u1(i,j)+u1(i-1,j))/dzs; u2f=u2(i-1,j)+2*dz*ku2/Du2*(u2e-u2(i,j)); u2zz(i,j)=(u2f-2*u2(i,j)+u2(i-1,j))/dzs; else u1zz(i,j)=(u1(i+1,j)-2*u1(i,j)+u1(i-1,j))/dzs; u2zz(i,j)=(u2(i+1,j)-2*u2(i,j)+u2(i-1,j))/dzs; end % % PDEs u1t(i,j)=Du1*(u1rr(i,j)+u1r(i,j)+u1zz(i,j)); u2t(i,j)=Du2*(u2rr(i,j)+u2r(i,j)+u2zz(i,j)); end end % % 2D to 1D matrices for i=1:nz for j=1:nr ij=(i-1)*nr+j; ut(ij)=u1t(i,j); ut(ij+nr*nz)=u2t(i,j); end end % % Transpose and count ut=ut'; ncall=ncall+1; Appendix D 517 AN ALTERNATIVE PROGRAMMING APPROACH In this alternative approach library routines are employed to implicitly generate the finite differences as opposed to the explicit spatial derivatives generated in the approach used above. function ut=pde_2(t,u) % % Global area global nr nz dr dz drs dzs... r r0 z zL Du1 Du2... u1 u2 u1e u2e ku1 ku2... ncall % % 1D to 2D matrices for i=1:nz for j=1:nr ij=(i-1)*nr+j; u1(i,j)=u(ij); u2(i,j)=u(ij+nr*nz); end end % % Step through the grid points in r for i=1:nz u1_1d=u1(i,:); u2_1d=u2(i,:); % % u1r, u2r u1r_1d=dss004(0,r0,nr,u1_1d); u1r(i,:)=u1r_1d; u1r(i,1)= 0.0; u1r(i,nr)=(ku1/Du1)*(u1e-u1(i,nr)); u2r_1d=dss004(0,r0,nr,u2_1d); u2r(i,:)=u2r_1d; u2r(i,1)= 0; u2r(i,nr)=(ku2/Du2)*(u2e-u2(i,nr)); % % u1rr, u2rr u1r_1d( 1)=0; u1r_1d(nr)=(ku1/Du1)*(u1e-u1_1d(nr)); nl=2; nu=2; u1rr_1d=dss044(0,r0,nr,u1_1d,u1r_1d,nl,nu); u1rr(i,:)=u1rr_1d; u2r_1d( 1)=0; u2r_1d(nr)=(ku2/Du2)*(u2e-u2_1d(nr)); nl=2; nu=2; u2rr_1d=dss044(0,r0,nr,u2_1d,u2r_1d,nl,nu); u2rr(i,:)=u2rr_1d; % % (1/r)*u1r, (1/r)*u2r 518 Appendix D for j=1:nr if(j~=1) u1r(i,j)=(1.0/r(j))*u1r(i,j); u2r(i,j)=(1.0/r(j))*u2r(i,j); elseif(j==1) u1rr(i,j)=2.0*u1rr(i,j); u2rr(i,j)=2.0*u2rr(i,j); end % % Next j end % % Next i end % % Step through the grid points in z for j=1:nr u1_1d=u1(:,j); u2_1d=u2(:,j); % % u1zz, u2zz u1z_1d( 1)=0.0; u1z_1d(nz)=(ku1/Du1)*(u1e-u1_1d(nz)); nl=2; nu=2; u1zz_1d=dss044(zL/2,zL,nz,u1_1d,u1z_1d,nl,nu); u1zz(:,j)=u1zz_1d; u2z_1d( 1)=0.0; u2z_1d(nz)=(ku2/Du2)*(u2e-u2_1d(nz)); nl=2; nu=2; u2zz_1d=dss044(zL/2,zL,nz,u2_1d,u2z_1d,nl,nu); u2zz(:,j)=u2zz_1d; % % Next j end % % PDEs for i=1:nz for j=1:nr u1t(i,j)=Du1*(u1rr(i,j)+u1r(i,j)+u1zz(i,j)); u2t(i,j)=Du2*(u2rr(i,j)+u2r(i,j)+u2zz(i,j)); end end % % 2D to 1D matrices for i=1:nz for j=1:nr ij=(i-1)*nr+j; ut(ij)=u1t(i,j); ut(ij+nr*nz)=u2t(i,j); end end Appendix D 519 % % Transpose and count ut=ut'; ncall=ncall+1; Supporting routines being called (www.cambridge.org/9781107022805): both dss004 and dss044 are used to compute numerical derivatives of 1-dimension (1D) arrays. However, in this problem u1, u 2 are 2-dimension (2D) arrays, which requires a 2D to 1D mapping and the inverse in order to employ these library routines. This is the essential difference between pde_1 and pde_2. More details on the process used in dss004 and dss044 can be found in Partial Differential Equation Analysis in Biomedical Engineering Case Studies with MATLAB® Shiesser, W. E. pp. 363–366. % % File: dss004.m function [ux]=dss004(xl,xu,n,u) % % Function dss004 computes the first derivative, u , of a % x % v ariable u over the spatial domain xl le x le xu from classical % five-point, fourth-order finite difference approximations % % Argument list % % xl Lower boundary value of x (input) % % xu Upper boundary value of x (input) % % n N umber of grid points in the x domain including the boundary points (input) % % u O ne-dimensional array containing the values of u at the n grid point points for which the derivative is to be computed (input) % % ux One-dimensional array containing the numerical % v alues of the derivatives of u at the n grid points (output) % % The mathematical details of the following Taylor series (or % polynomials) are given in routine dss002. % % Five-point formulas % % (1) Left end, point i = 1 % % 2 3 4 % a(u2 = u1 + u1 ( dx) + u1 ( dx) + u1 ( dx) + u1 ( dx) % x 1f 2x 2f 3x 3f 4x 4f % 520 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix D 5 + u1 ( dx) 5x 5f 6 + u1 ( dx) 6x 6f 7 + u1 ( dx) 7x 7f + ...) 2 3 (2dx) + u1 (2dx) + u1 (2dx) x 1f 2x 2f 3x 3f b(u3 = u1 + u1 5 + u1 (2dx) 5x 5f 6 + u1 (2dx) 6x 6f 7 + u1 (2dx) 7x 7f + ...) 2 3 (3dx) + u1 (3dx) + u1 (3dx) x 1f 2x 2f 3x 3f c(u4 = u1 + u1 5 + u1 (3dx) 5x 5f 6 + u1 (3dx) 6x 6f (3dx) 7x 7f + ...) 2 3 (4dx) + u1 (4dx) + u1 (4dx) x 1f 2x 2f 3x 3f 5 (4dx) 5x 5f 6 + u1 (4dx) 6x 6f 4 (3dx) 4x 4f + u1 7 + u1 d(u5 = u1 + u1 + u1 4 (2dx) 4x 4f + u1 4 (4dx) 4x 4f + u1 7 + u1 (4dx) 7x 7f + ...) onstants a, b, c, and d are selected so that C the coefficients of the u1 terms sum to one and the coefficients of the u1 , x 2x u1 and u1 terms sum to zero 3x 4x % % % % % % a + 2b + 3c + 4d = 1 % % a + 4b + 9c + 16d = 0 % % a + 8b + 27c + 64d = 0 % % a + 16b + 81c + 256d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 Appendix D % u 1 = (1/12dx)(-25u1 + 48u2 - 36u3 + 16u4 - 3u5) + O(dx ) (1) x 521 % % % (2) Interior point, i = 2 % % 2 3 4 % a(u1 = u2 + u2 (-dx) + u2 (-dx) + u2 (-dx) + u2 (-dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (-dx) + u2 (-dx) + u2 (-dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % b(u3 = u2 + u2 ( dx) + u2 ( dx) + u2 ( dx) + u2 ( dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 ( dx) + u2 ( dx) + u2 ( dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % c(u4 = u2 + u2 (2dx) + u2 (2dx) + u2 (2dx) + u2 (2dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (2dx) + u2 (2dx) + u2 (2dx) + ...) % 5x 5f 6x 6f 7x 7f % % 2 3 4 % d(u5 = u2 + u2 (3dx) + u2 (3dx) + u2 (3dx) + u2 (3dx) % x 1f 2x 2f 3x 3f 4x 4f % % 5 6 7 % + u2 (3dx) + u2 (3dx) + u2 (3dx) + ...) % 5x 5f 6x 6f 7x 7f % % -a + b + 2c + 3d = 1 % % a + b + 4c + 9d = 0 % % -a + b + 8c + 27d = 0 % % a + b + 16c + 81d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % t erms, for u1 gives the following five-point approximation % x 522 % % Appendix D 4 2 = (1/12dx)(-3u1 - 10u2 + 18u3 u O(dx ) (2) x 6u4 + u5) + % % % (3) Interior point i, i ne 2, n-1 % % 2 3 % a(ui-2 = ui + ui (-2dx) + ui (-2dx) + ui (-2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui (-2dx) + ui (-2dx) + ui (-2dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % b(ui-1 = ui + ui ( -dx) + ui ( -dx) + ui ( -dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( -dx) + ui ( -dx) + ui ( -dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % c(ui+1 = ui + ui ( dx) + ui ( dx) + ui ( dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( dx) + ui ( dx) + ui ( dx) + ...) % 4x 4f 5x 5f 6x 6f % % 2 3 % d(ui+2 = ui + ui ( 2dx) + ui ( 2dx) + ui ( 2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + ui ( 2dx) + ui ( 2dx) + ui ( 2dx) + ...) % 4x 4f 5x 5f 6x 6f % % -2a b + c + 2d = 1 % % 4a + b + c + 4d = 0 % % -8a b + c + 8d = 0 % % 16a + b + c + 16d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation 523 Appendix D % % % % % % % % % x 4 u i = (1/12dx)(ui-2 - 8ui-1 + 0ui + 8ui+1 - ui+2) + O(dx ) (3) x (4) Interior point, i = n-1 a (un-4 = un-1 + un-1 un-1 (-3dx) (-3dx) + un-1 2 (-3dx) 3 + % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 (-3dx) + un-1 (-3dx) + un-1 (-3dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % b(un-3 = un-1 + un-1 (-2dx) + un-1 (-2dx) + un-1 (-2dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 (-2dx) + un-1 (-2dx) + un-1 (-2dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % c(un-2 = un-1 + un-1 ( -dx) + un-1 (- -x) + un-1 ( -dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 ( -dx) + un-1 ( -dx) + un-1 ( -dx) + ... % 4x 4f 5x 5f 6x 6f % % 2 3 % d(un = un-1 + un-1 ( dx) + un-1 ( dx) + un-1 ( dx) % x 1f 2x 2f 3x 3f % % 4 5 6 % + un-1 ( dx) + un-1 ( dx) + un-1 ( dx) + ... % 4x 4f 5x 5f 6x 6f % % -3a - 2b c + d = 1 % % 9a + 4b + c + d = 0 % % -27a - 8b c + d = 0 % % 81a + 16b + c + d = 0 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u 524 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix D 4x erms, for u1 gives the following five-point t approximation x 4 n-1 = (1/12dx)(-un-4 + 6un-3 - 18un-2 + 10un-1 + 3un) + u O(dx ) x (4) (5) Right end, point i = n 2 a(un-4 = un + un (-4dx) x 1f + un (-4dx) 2x 2f (-4dx) 3x 3f 4 + un 3 + un 5 (-4dx) 4x 4f + un (-4dx) 5x 5f b(un-3 = un + un (-3dx) x 1f + un 6 + un (-4dx) 6x 6f 2 + ...) (-3dx) 2x 2f (-3dx) 3x 3f 4 + un 3 + un 5 (-3dx) 4x 4f + un (-3dx) 5x 5f c(un-2 = un + un (-2dx) x 1f + un 6 + un (-3dx) 6x 6f 2 + ...) (-2dx) 2x 2f (-2dx) 3x 3f 4 + un 3 + un 5 (-2dx) 4x 4f + un (-2dx) 5x 5f d(un-1 = un + un ( -dx) x 1f + un 6 + un (-2dx) 6x 6f 2 + ...) ( -dx) 2x 2f ( -dx) 3x 3f 4 + un ( -dx) 4x 4f 3 + un 5 + un -4a - 3b - 2c - d = 1 16a + 9b + 4c + d = 0 -64a - 27b - 8c - d = 0 256a + 81b + 16c + d = 0 ( -dx) 5x 5f 6 + un ( -dx) 6x 6f + ...) Appendix D 525 % % Simultaneous solution for a, b, c, and d followed by the solu% tion of the preceding Taylor series, truncated after the u % 4x % terms, for u1 gives the following five-point approximation % x % 4 % u n = (1/12dx)(3un-4 - 16un-3 + 36un-2 - 48un-1 + 25un) + O(dx ) % x % (5) % % The weighting coefficients for equations (1) to (5) can be % summarized as % % -25 48 -36 16 -3 % % -3 -10 18 -6 1 % % 1/12 1 -8 0 8 -1 % % -1 6 -18 10 3 % % 3 -16 36 -48 25 % % which are the coefficients reported by Bickley for n = 4, m = 1, % p = 0, 1, 2, 3, 4 (Bickley, W. G., Formulae for Numerical % D ifferentiation, Math. Gaz., vol. 25, 1941. Note - the Bickley % coefficients have been divided by a common factor of two). % % E quations (1) to (5) can now be programmed to generate the derivative u (x) of function u(x). % x % % Compute the spatial increment dx=(xu-xl)/(n-1); r4fdx=1./(12.*dx); % nm2=n-2; % Equation (1) (note - the rhs of equations (1), (2), (3), (4) % a nd (5) have been formatted so that the numerical weighting % c oefficients can be more easily associated with the Bickley % matrix above) ux( 1)=r4fdx*... ( -25.*u( 1) +48.*u( 2) -36.*u( 3) +16.*u( 4) -3.*u( 5)); % % Equation (2) 526 Appendix D ux( 2)=r4fdx*... ( -3.*u( 1) -10.*u( +1.*u( 5)); % % % % % % 2) +18.*u( Equation (3) for i=3:nm2 ux( i)=r4fdx*... ( +1.*u(i-2) -8.*u(i-1) -1.*u(i+2)); end +0.*u( 3) i) -6.*u( 4) +8.*u(i+1) Equation (4) ux(n-1)=r4fdx*... ( -1.*u(n-4) +6.*u(n-3) -18.*u(n-2) +10.*u(n-1) +3.*u( n)); Equation (5) ux( n)=r4fdx*... ( 3.*u(n-4) -16.*u(n-3) +36.*u(n-2) -48.*u(n-1) +25.*u( n)); And % % File: dss044.m function [uxx]=dss044(xl,xu,n,u,ux,nl,nu) % % % % % % % % % % % % % % % % % % % Function dss044 computes a fourth-order approximation of a second-order derivative, with or without the normal derivative at the boundary. Argument list xl L eft value of the spatial independent variable (input) xu R ight value of the spatial independent variable (input) n Number of spatial grid points, including the end points (input) u O ne-dimensional array of the dependent variable to be differentiated (input) ux O ne-dimensional array of the first derivative of u. T he end values of ux, ux(1), and ux(n), are used in N eumann boundary conditions at x = xl and x = xu, 527 Appendix D % % % % uxx O ne-dimensional array of the second derivative of u (output) % % nl I nteger index for the type of boundary condition at x = xl (input). The allowable values are % % % % % % % % % % % % % % % % % % depending on the arguments nl and nu (see the description of nl and nu below) 1 - Dirichlet boundary condition at x = xl (ux(1) is not used) 2 - Neumann boundary condition at x = xl (ux(1) is used) nu I nteger index for the type of boundary condition at x = xu (input). The allowable values are 1 - Dirichlet boundary condition at x = xu (ux(n) is not used) 2 - Neumann boundary condition at x = xu (ux(n) is used) T he following derivation was completed by W. E. Schiesser, Depts of CHE and Math, Lehigh University, Bethlehem, PA 18015, USA, on December 15, 1986. % % *********************************************************** ******* % % (1) uxx at the interior points 3, 4,..., n-2 % % T o develop a set of fourth-order correct differentiation formulas % f or the second derivative uxx, we consider first the interior % grid points at which a symmetric formula can be used. % % If we consider a formula of the form % % a*u(i-2) + b*u(i-1) + e*u(i) + c*u(i+1) + d*u(i+2) % % T aylor series expansions of u(i-2), u(i-1), u(i+1), and u(i+2) % c an be substituted into this formula. We then consider the % l inear albegraic equations relating a, b, c, and d which will 528 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix D etain certain terms, i.e., uxx, and drop others, e.g., r uxxx, uxxxx and uxxxxx. Thus, for grid points 3, 4,..., n-2 To retain uxx 4*a + b + c + 4*d = 2 (1) 8*d = 0 (2) 16*a + b + c + 16*d = 0 (3) To drop uxxx -8*a - b + c + To drop uxxxx To drop uxxxxx -32*a - b + c + 32*d = 0 (4) Equations (1) to (4) can be solved for a, b, c, and d. equation (1) is added to equation (2) ­ -4*a + 2*c + 12*d = 2 If (5) If equation (1) is subtracted from equation (3) 12*a + 12*d = -2 (6) If equation (1) is added to equation (4) -28*a + 2*c + 36*d = 2 quations (5) to (7) can be solved for a, c, and d. E equation (5) is subtracted from equation (7), and the result combined with equation (6) (7) If 12*a + 12*d = -2 -24*a + 24*d = 0 quations (6) and (8) can be solved for a and d. E From (8), a = d. From equation (6), a = -1/12 and d = -1/12. Then, from equation (5), c = 4/3, and from equation (1), b = 4/3. (6) (8) 529 Appendix D % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % The final differentiation formula is then obtained as (-1/12)*u(i-2) + (4/3)*u(i-1) + (4/3)*u(i+1) + (-1/12)*u(i+2) (-1/12 + 4/3 - 1/12 + 4/3)*u(i) + uxx(i)*(dx**2) + O(dx**6) or uxx(i) = (1/(12*dx**2))*(-1*u(i-2) + 16*u(i-1) - 30*u(i) + 16*u(i+1) - 1*u(i+2) (9) + O(dx**4) Note that the ux term drops out, i.e., the basic equation is -2*a - b + c + 2*d = -2*(-1/12) - (4/3) + (4/3) + 2*(-1/12) = 0 E quation (9) was obtained by dropping all terms in the underlying T aylor series up to and including the fifth derivative, uxxxxx. T hus, equation (9) is exact for polynomials up to and including f ifth order. This can be checked by substituting the functions 1 , x, x**2, x**3, x**4 and x**5 in equation (9) and computing the c orresponding derivatives for comparison with the known second d erivatives. This is done for 1 merely by summing the weighting c oefficients in equation (9), which should sum to zero, i.e., -1 + 16 - 30 + 16 -1 = 0. F or the remaining functions, the algebra is rather involved, but t hese functions can be checked numerically, i.e., numerical values o f x**2, x**3, x**4 and x**5 can be substituted in equation (9) and the computed derivatives can be compared with the know n umerical second derivatives. This is not a proof of correctness of 530 % Appendix D quation (9), but would likely detect any errors in e equation (9). % % *********************************************************** ******* % % (2) uxx at the interior points i = 2 and n-1 % % For grid point 2, we consider a formula of the form % % a *u(i-1) + f*u(i) + b*u(i+1) + c*u(i+2) + d*u(i+3) + e*u(i+4) % % Taylor series expansions of u(i-1), u(i+1), u(i+2), u(i+3) and % u(i+4) when substituted into this formula give linear algebraic % equations relating a, b, c, d, and e. % % To drop ux % % -a + b + 2*c + 3*d + 4*e = 0 (10) % % To retain uxx % % a + b + 4*c + 9*d + 16*e = 2 (11) % % To drop uxxx % % -a + b + 8*c + 27*d + 64*e = 0 (12) % % To drop uxxxx % % a + b + 16*c + 81*d + 256*e = 0 (13) % % To drop uxxxxx % % -a + b + 32*c + 243*d + 1024*e = 0 (14) % % Equations (11), (12), (13), and (14) can be solved for a, b, c, d, and e. If equation (10) is added to equation (11) % % 2*b + 6*c + 12*d +20*e = 2 (15) % % If equation (10) is subtracted from equation (12) % % 6*c + 24*d + 60*e = 0 (16) % % If equation (10) is added to equation (13) % Appendix D 531 % 2*b + 18*c + 84*d + 260*e = 0 (17) % % If equation (10) is subtracted from equation (14) % % 30*c + 240*d + 1020*e = 0 (18) % % Equations (15), (16), (17), and (18) can be solved for b, c, d and e. % % 6*c + 24*d + 60*e = 0 (16) % % If equation (15) is subtracted from equation (17) % % 12*c + 72*d + 240*e = -2 (19) % % 30*c + 240*d + 1020*e = 0 (18) % % Equations (16), (18), and (19) can be solved for c, d, and e. If % two times equation (16) is subtracted from equation (19), % % 24*d + 120*e = -2 (20) % % If five times equation (16) is subtracted from equation (18), % % 120*d + 720*e = 0 (21) % % E quations (20) and (21) can be solved for d and e. From (21), % e = (-1/6)*d. Substitution in equation (20) gives d = -1/2. % t hus, e = 1/12. From equation (16), c = 7/6. From equation % (15), b = -1/3. From equation (10), a = 5/6. % % The final differentiation formula is then obtained as % % (5/6)*u(i-1) + (-1/3)*u(i+1) + (7/6)*u(i+2) + (-1/2)*u(i+3) % % + (1/12)*u(i+4) = (5/6 - 1/3 + 7/6 - 1/2 + 1/12)*u(i) % % + uxx*(dx**2) + O(dx**6) % % or % % uxx(i) = (1/12*dx**2)*(10*u(i-1) - 15*u(i) - 4*u(i+1) % (22) % + 14*u(i+2) - 6*u(i+3) + 1*u(i+4)) + O(dx**4) % % Equation (22) will be applied at i = 2 and n-1. thus 532 Appendix D % % uxx(2) = (1/12*dx**2)*(10*u(1) - 15*u(2) - 4*u(3) % (23) % + 14*u(4) - 6*u(5) + 1*u(6)) + O(dx**4) % % uxx(n-1) = (1/12*dx**2)*(10*u(n) - 15*u(n-1) - 4*u(n-2) % (24) % + 14*u(n-3) - 6*u(n-4) + 1*u(n-5)) + O(dx**4) % % *********************************************************** ******* % % (3) uxx at the boundary points 1 and n % % Finally, for grid point 1, an approximation with a Neumann boundary condition of the form % % a *u(i+1) + b*u(i+2) + c*u(i+3) + d*u(i+4) + e*ux(i) + f*u(i) % % Will be used. the corresponding algebraic equations are % % To drop ux % % a + 2*b + 3*c + 4*d + e = 0 (25) % % To retain uxx % % a + 4*b + 9*c + 16*d = 2 (26) % % To drop uxxx % % a + 8*b + 27*c + 64*d = 0 (27) % % To drop uxxxx % % a + 16*b + 81*c + 256*d = 0 (28) % % To drop uxxxxx % % a + 32*b + 243*c + 1024*d = 0 (29) % % Equations (25) to (29) can be solved for a, b, c, d, and e. If % % Equation (26) is subtracted from equations (27), (28), and (29), % % 4*b + 18*c + 48*d = -2 (30) % % 12*b + 72*c + 240*d = -2 (31) 533 Appendix D % % % % d % % % % % % % % % % % % % % % % % % % % 28*b + 234*c + 1008*d = -2 (32) Equations (30), (31), and (32) can be solved for b, c, and 18*c + 96*d = 4 (33) 108*c + 672*d = 12 (34) E quations (3) and (34) can be solved for c and d, c = 8/9, d = -1/8. F rom equation (30), b = -3. From equation (26), a = 8. From equation (25), e = -25/6. The final differentiation formula is then obtained as 8*u(i+1) - 3*u(i+2) + (8/9)*u(i+3) - (1/8)*u(i+4) - (25/6)*ux(i)*dx = (8 - 3 + (8/9) - (1/8))*u(i) + uxx*(dx**2) + O(dx**6) or u xx(i) = (1/12*dx**2)*((-415/6)*u(i) + 96*u(i+1) - 36*u(i+2) % (35) % + (32/3)*u(i+3) - (3/2)*u(i+4) - 50*ux(i)*dx) + O(dx**4) % % Equation (35) will be applied at i = 1 and i = n % % uxx(1) = (1/12*dx**2)*((-415/6)*u(1) + 96*u(2) - 36*u(3) % (36) % + (32/3)*u(4) - (3/2)*u(5) - 50*ux(1)*dx) + O(dx**4) % % uxx(n) = (1/12*dx**2)*((-415/6)*u(n) + 96*u(n-1) - 36*u(n-2) % (37) % + (32/3)*u(n-3) - (3/2)*u(n-4) + 50*ux(n)*dx) + O(dx**4) % % A lternatively, for grid point 1, an approximation with a Dirichlet % boundary condition of the form % % a *u(i+1) + b*u(i+2) + c*u(i+3) + d*u(i+4) + e*u(i+5) + f*u(i) % % can be used. The corresponding algebraic equations are % 534 % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % Appendix D To drop ux a + 2*b + 3*c + 4*d + 5*e = 0 (38) To retain uxx a + 4*b + 9*c + 16*d + 25*e = 2 (39) To drop uxxx a + 8*b + 27*c + 64*d + 125*e = 0 (40) To drop uxxxx a + 16*b + 81*c + 256*d + 625*e = 0 (41) To drop uxxxxx a + 32*b + 243*c + 1024*d + 3125*e = 0 (42) quations (38), (39), (40), (41), and (42) can be solved E for a, b, c, d, and e. 2*b + 6*c + 12*d + 20*e = 2 (43) 6*b + 24*c + 60*d + 120*e = 0 (44) 14*b + 78*c + 252*d + 620*e = 0 (45) 30*b + 240*c + 1020*d + 3120*e = 0 (46) quations (43), (44), (45), and (46) can be solved for b, E c, d, and e 6*c + 24*d + 60*e = -6 (47) 36*c + 168*d + 480*e = -14 (48) 150*c + 840*d + 2820*e = -30 (49) quations (47), (48), and (49) can be solved for c, d, E and e 24*d + 120*e = 22 (50) 240*d + 1320*e = 120 (51) rom equations (50) and (51), d = 61/12, e = -5/6. From F equation (47), c = -13. From equation (43), b = 107/6. From equation (38), Appendix D % % % % % 535 a = -77/6. The final differentiation formula is then obtained as ( -77/6)*u(i+1) + (107/6)*u(i+2) - 13*u(i+3) + (61/12)*u(i+4) % % - (5/6)*u(i+5) = (-77/6 + 107/6 - 13 + 61/12 - 5/6)*u(i) + % % uxx(i)*(dx**2) + O(dx**6) % % or % % uxx(i) = (1/12*dx**2)*(45*u(i) - 154*u(i+1) + 214*u(i+2) % (52) % - 156*u(i+3) + 61*u(i+4) - 10*u(i+5)) + O(dx**4) % % Equation (52) will be applied at i = 1 and i = n % % uxx(1) = (1/12*dx**2)*(45*u(1) - 154*u(2) + 214*u(3) % (53) % - 156*u(4) + 61*u(5) - 10*u(6)) + O(dx**4) % % uxx(n) = (1/12*dx**2)*(45*u(n) - 154*u(n-1) + 214*u(n-2) % (54) % -156*u(n-3) + 61*u(n-4) - 10*u(n-5)) + O(dx**4) % % *********************************************************** ******* % % Grid spacing dx=(xu-xl)/(n-1); % % 1/(12*dx**2) for subsequent use r12dxs=1./(12.0*dx^2); % % uxx at the left boundary % % Without ux (equation (53)) if nl==1 uxx(1)=r12dxs*... ( 45.0*u(1)... -154.0*u(2)... +214.0*u(3)... -156.0*u(4)... +61.0*u(5)... -10.0*u(6)); % % With ux (equation (36)) elseif nl==2 uxx(1)=r12dxs*... 536 Appendix D (-415.0/6.0*u(1)... +96.0*u(2)... -36.0*u(3)... +32.0/3.0*u(4)... -3.0/2.0*u(5)... -50.0*ux(1)*dx); end % % % % % % % % % % % % % uxx at the right boundary Without ux (equation (54)) if nu==1 uxx(n)=r12dxs*... ( 45.0*u(n )... -154.0*u(n-1)... +214.0*u(n-2)... -156.0*u(n-3)... +61.0*u(n-4)... -10.0*u(n-5)); With ux (equation (37)) elseif nu==2 uxx(n)=r12dxs*... (-415.0/6.0*u(n )... +96.0*u(n-1)... -36.0*u(n-2)... +32.0/3.0*u(n-3)... -3.0/2.0*u(n-4)... +50.0*ux(n )*dx); end uxx at the interior grid points i = 2 (equation (23)) uxx(2)=r12dxs*... ( 10.0*u(1)... -15.0*u(2)... -4.0*u(3)... +14.0*u(4)... -6.0*u(5)... +1.0*u(6)); i = n-1 (equation (24)) uxx(n-1)=r12dxs*... ( 10.0*u(n )... -15.0*u(n-1)... -4.0*u(n-2)... +14.0*u(n-3)... -6.0*u(n-4)... +1.0*u(n-5)); Appendix D % i = 3, 4,..., n-2 (equation (9)) for i=3:n-2 uxx(i)=r12dxs*... ( -1.0*u(i-2)... +16.0*u(i-1)... -30.0*u(i )... +16.0*u(i+1)... -1.0*u(i+2)); end 537 This page intentionally left blank This page intentionally left blank ENGINEERING – CHEMICAL Focusing on the application of mathematics to chemical engineering, Applied Mathematical Methods for Chemical Engineers addresses the setup and verification of mathematical models using experimental or other independently derived data. The book provides an introduction to differential equations common to chemical engineering, followed by examples of first-order and linear second-order ordinary differential equations. Later chapters examine Sturm–Liouville problems, Fourier series, integrals, linear partial differential equations, regular perturbation, combination of variables, and numerical methods emphasizing the method of lines with MATLAB® programming examples. Fully revised and updated, this Third Edition: • Includes additional examples related to process control, Bessel Functions, and contemporary areas such as drug delivery • Introduces examples of variable coefficient Sturm–Liouville problems both in the regular and singular types • Demonstrates the use of Euler and modified Euler methods alongside the Runge–Kutta order-four method • Inserts more depth on specific applications such as nonhomogeneous cases of separation of variables • Adds a section on special types of matrices such as upper- and lower-triangular matrices • Presents a justification for Fourier–Bessel series in preference to a complicated proof • Incorporates examples related to biomedical engineering applications • Illustrates the use of the predictor-corrector method • Expands the problem sets of numerous chapters APPLIED MATHEMATICAL METHODS for CHEMICAL ENGINEERS Third Edition Applied Mathematical Methods for Chemical Engineers, Third Edition uses worked examples to expose several mathematical methods that are essential to solving real-world process engineering problems. K15249 ISBN: 978-1-4665-5299-9 Norman W. Loney 90000 9 781466 552999 K15249_Cover_PubGr.indd All Pages 9/1/15 1:52 PM