July 2013
Chapter 27
Boundary-Value and Eigenvalue Problems
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Ordinary differential equation is accompanied by auxiliary conditions.
These conditions are used to evaluate the constants of integration that
result during the solution of the equation. For an nth order equation,
n conditions are required. If all the conditions are specified at the same
value of the independent variable, then we are dealing with an
initial-value problem (next figure a).
In contrast, there is another application for which the conditions are not
known at a single point, but rather, are known at different values of the
independent variable. Because these values are often specified at the
extreme points or boundaries of a system, they are customarily referred to
as boundary-value problems (previous figure b).
‫ يبقى‬،‫من يبحث عن صديق بال عيب‬
) ‫بال صديق( مثل تركي‬
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We will discuss two general approaches for obtaining a solution:
the shooting method and the finite-difference approach.
Additionally, we present techniques to approach a special type of
boundary-value problem: the determination of eigenvalues.
27.1 General Methods for Boundary-Value Problems
The conservation of heat can be used to develop a heat balance for a
long, thin rod. If the rod is not insulated along its length and the system is
at a steady state, the equation that result is
𝑑2 𝑇
𝑑𝑥 2
+ ℎ′ (𝑇𝑎 − 𝑇) = 0
(27.1)
where h’ is a heat transfer coefficient (m-2) that parameterizes the rate of
heat dissipation to the surrounding air and Ta is the temperature of the
surrounding air (⁰C).
To obtain a solution for the previous equation, there must be
appropriate boundary conditions. A simple case is where the temperatures
at the ends of the bar are held at fixed values. These can be expressed
mathematically as
T(0) = T1
T(L) = T2
‫يمكنك أن تكون أفضل من كل أصدقائك بأحد‬
‫ االجتهاد أو أن تصادق الحمقى‬:‫طريقين‬
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With these conditions, the previous equation can be solved analytically
using calculus. For a 10-m rod with Ta = 20, T1 = 40, T2 = 200 and h’ = 0.01,
the solution is
T = 73.4523 e0.1x – 53.4523 e-0.1x + 20
27.1.1 The Shooting Method
The shooting method is based on converting the boundary-value
problem into an equivalent initial-value problem. A trial-and-error
approach is then implemented to solve the initial-value version.
The approach can be illustrated by an example.
EXAMPLE 27.1 The Shooting Method
Problem Statement:
Use the shooting method to solve Eq.(27.1) for a 10-m rod with h’=0.01
m-2, Ta = 20, and the boundary conditions
T(0) = 40
T(10) = 200
Solution:
The second-order equation can be expressed as two first-order ODEs:
𝑑𝑇
𝑑𝑥
𝑑𝑧
𝑑𝑥
=𝑧
= ℎ′ (𝑇 − 𝑇𝑎 )
ُ ‫والر َما‬
....‫ح َن َوا ِه ٌل ِم ِنـي‬
ِ ‫ولَقَـ ْد ذَ َكرْ ت ُ ِك‬
ُ ‫اله ْنـ ِد ت َ ْق‬
ْ ‫ـر ِم‬
‫ـن د َِمـي‬
ُ ‫ط‬
ِ ُ‫َو ِبيْـض‬
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Continue:
To solve these equations, we require an initial value for z.
For the shooting method, we guess a value- say, z(0) = 10 and we can then
obtain the solution by integration. For example, using a fourth-order RK
method with a step size of 2, we obtain a value at the end of the interval
of T(10) = 168.3797 (next figure a), which differs from the boundary
condition of T(10) = 200. Therefore, we make another guess, z(0) = 20,
and perform the computation again. This time, the result of
T(10) = 285.8980 is obtained (next figure b).
Now, because the original ODE is linear, the values
z(0) = 10
T(10) = 168.3797
z(0) = 20
T(10) = 285.8980
and
are linearly related. As such, they can be used to compute the value of z(0)
that yields T(10) = 200. A linear interpolation formula can be employed for
this purpose:
𝑧(0) = 10 +
20 − 10
285.8980 − 168.3797
(200 – 168.3797) = 12.6907
This value can then be used to determine the correct solution, as shown in
the next figure (c).
،‫ عجبننن البنننن دم‬:‫منننن كنننالم بلنننيب‬
‫ ويبغضني ويطيعني‬،‫يحب هللا ويعصاه‬
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Nonlinear Two-Point Problems
For nonlinear boundary-value problems, linear interpolation or
extrapolation through two solution points will not necessarily result in an
accurate estimate of the required boundary condition to attain an exact
solution. An alternative is to perform three applications of the shooting
method and use a quadratic interpolating polynomial to estimate the
proper boundary condition.
However, it is unlikely that such approach would yield the exact
answer, and additional iterations would be necessary to obtain the
solution.
‫قد يرى الناس الجرح الذي في رأسك‬
‫لكنهم ال يشعرون باأللم الذي تعانيه‬
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Another approach for a nonlinear problem involves recasting it as a
roots problem. Recall that the general form of a root problem is to find the
value of x that makes the function f(x) = 0. Now, let us use Example 27.1
to understand how the shooting method can be recast in this form.
First, recognize that the solution of the pair of differential equations
is also a “function” in the sense that we guess a condition at the left-hand
end of the bar, z0, and the integration yields a prediction of the
temperature at the right-hand end, T10.
Thus, we can think of the integration as
T10 = f(z0)
That is, it represents a process whereby a guess of z 0 yields a
prediction of T10. Viewed in this way, we can see that what we desire is the
value of z0 that yields a specific value of T10. If, as in the example,
we desire T10 = 200, the problem can be posed as
200 = f(z0)
By bringing the goal of 200 over to the right-hand side of the
equation, we generate a new function, g(z0), that represents the
difference between what we have, f(z0), and what we want, 200.
g(z0) = f(z0) – 200
if we drive this new function to zero, we will obtain the solution.
The next example illustrates the approach.
‫امنح الناس دائما أكثر مما‬
.‫يتوقعون الحصول عليه‬
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EXAMPLE 27.2 The Shooting Method for Nonlinear Problems
Problem Statement:
Suppose that the following nonlinear ODE is used to simulate the
temperature of the heated bar:
𝑑2 𝑇
𝑑𝑥 2
+ ℎ′′ (𝑇𝑎 − 𝑇)4 = 0
where h’’ = 5 × 10-8. The reaming problem conditions are as specified in
Example 27.1
Solution:
The second-order equation can be expressed as two first-order ODEs:
𝑑𝑇
𝑑𝑥
𝑑𝑧
𝑑𝑥
=𝑧
= ℎ′′ (𝑇 − 𝑇𝑎 )4
‫ نننو كالغننذا تحتننا‬: ‫الننناس ثالثننة أنننوا‬
‫ وننننو كالننندوا تحتنننا لينننه‬،‫لينننه دائمنننا‬
‫ والثالث كالدا يضرك فقط‬،‫أحيانا‬
Now, these equations can be integrated using any of the methods
described in the previous chapter. We used the constant step-size version
of the fourth-order RK and we implemented this approach as an Excel
macro function written in Visual BASIC. The function integrated the
equations based on an initial guess for z(0) and returned the temperature
at x = 10. The difference between this value and the goal of 200 was then
placed in a spreadsheet cell. The Excel Solver was then invoked to adjust
the value of z(0) until the difference was driven to zero.
The result is shown in the next figure along with the original linear case.
As might be expected, the nonlinear case is curved more than the linear
model. This is due to the power of four term in heat transfer relationship.
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27.1.2 Finite-Difference Methods
The most common alternatives to the shooting method are
finite-difference
approaches.
In
these
techniques,
finite
divided
differences are substituted for the derivatives in the original equation.
Thus, a linear differential equation is transformed into a set of
simultaneous algebraic equations that can be solved using other methods.
The finite divided-difference approximation for the second derivative is
𝑑2 𝑇
𝑑𝑥 2
=
𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1
∆𝑥 2
This approximation can be substituted into Eq.(27.1) to give
𝑇𝑖+1 − 2𝑇𝑖 + 𝑇𝑖−1
∆𝑥 2
− ℎ′ (𝑇𝑖 − 𝑇𝑎 ) = 0
Collecting terms gives
−𝑇𝑖−1 + (2 + ℎ′ ∆𝑥 2 )𝑇𝑖 − 𝑇𝑖+1 = ℎ′ ∆𝑥 2 𝑇𝑎
‫أعرف أن مالبب وسيارتك‬
‫ لكن‬،‫وموبايلك قيمتها عالية‬
‫حاول أن تكون أن األغلى‬
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This equation applies for each of the interior nodes of the rod.
The first and last interior nodes, 𝑇𝑖−1 and 𝑇𝑖+1 , respectively, are specified
by the boundary conditions. Therefore, the resulting set of linear algebraic
equations will be tridiagonal.
EXAMPLE 27.3 Finite Difference Approximation of Boundary-Value
Problems
Problem Statement:
Use the finite-difference approach to solve the same problem as in
Example 27.1
Solution:
Employing the parameters in Example 27.1, we can write the above
equation for the rod.
Using four interior nodes with a segment length of ∆𝑥 = 2m results in the
following equations:
𝑇1
2.04 −1 0 0
40.8
𝑇
[ −1 2.04 −1 0 ] { 2 } = { 0.8 }
𝑇3
0 −1 2.04 −1
0.8
0 0 −1 2.04 𝑇4
200.8
which can be solved for
{𝑇}𝑇 = ⌊65.9698 93.7785 124.5382 159.4795⌋
‫ال يمكنك أن تصنع ما هو أكثر حماقة‬
‫من أن تجلب على جانب الطريق حتى‬
‫يأتيك أحدهم ويحاول مساعدتك‬
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The next table provides a comparison between the analytical solution and
the numerical solutions obtained in Examples 27.1 and 27.3
For both numerical methods, these errors can be omitted by decreasing
their respective step sizes. Although both techniques perform well for the
present case, the finite-difference approach is preferred because of the
ease with which it can be extended to more complex cases.
27.2 Eigenvalue Problems
Eigenvalue, or characteristic-value, problems are a special class of
boundary-value problems that are common in engineering problem
contexts involving vibrations, elasticity, and other oscillating systems.
27.2.1 Mathematical Background
We dealt with methods for solving sets of linear algebraic equations
of the general form
[𝐴]{𝑋} = {𝐵}
‫صــداقـننننـات ال تـنـتـهننننـي (( الــمننننـرأة‬
.. ‫ الـقننننـارا بـكـتـابننننـة‬.. ‫بــمـر تـهننننـا‬
))‫الـمـنـافـق بـحـذا مـن يـنـافـقـة‬
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If the equations comprising such a system are linearly independent
(that is, have a nonzero determinant), they will have a unique solution.
In contrast, a homogenous linear algebraic system has the general
form
[A]{X} = 0
Although nontrivial solutions (that is, solutions other than all x’s = 0)
of such systems are possible, they are generally not unique.
Rather, the simultaneous equations establish relationships among the x’s
that can be satisfied by various combinations of values.
Eigenvalue problems associated with engineering are typically of the
general form
(𝑎11 − 𝜆) 𝑥1 +
𝑎12 𝑥2 + … + 𝑎1𝑛 𝑥𝑛 = 0
𝑎21 𝑥1 + (𝑎22 − 𝜆) 𝑥2 + … + 𝑎2𝑛 𝑥𝑛 = 0
.
.
.
.
.
.
𝑎𝑛1 𝑥1 +
𝑎𝑛2 𝑥2 + … + (𝑎𝑛𝑛 − 𝜆) 𝑥𝑛 = 0
where 𝜆 is an unknown parameter called the eigenvalue, or characteristic
value. A solution {X} for such a system is referred to as an eigenvector.
‫ تلك هي‬،‫روح واحدة حل في جسدين‬
‫ أرسطو‬......‫الصداقة ن سألتني عنها‬
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The above set of equations may also be expressed concisely as
[[𝐴] − 𝜆[1]] {𝑋} = 0
The solution of the previous equation depends on determining 𝜆.
One way to accomplish this is based on the fact that the determinant of
the matrix [[𝐴] − 𝜆[1]] must be equal to zero for non-trivial solutions to
be possible. Expanding the determinant yields a polynomial in 𝜆.
The roots of this polynomial are the solutions for the eigenvalues.
27.2.2 Physical Background
The mass-spring system in the next figure (a) is a simple context to
illustrate how eigenvalues occur in physical problem settings.
‫ال تأمن من كذب لك أن يكذب عليك وال‬
‫من اغتاب عندك أن يغتابك عند غيرك‬
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Assume that each mass has no external or damping forces acting on
it. In addition, assume that each spring has the same natural length l and
the same spring constant k.
Finally, assume that the displacement of each spring is measured
relative to its own local coordinate system (previous figure a).
Under these assumptions, Newton’s second law can be employed to
develop a force balance for each mass
𝑚1
𝑑 2 𝑥1
𝑑𝑡 2
= −𝑘𝑥1 + 𝑘(𝑥2 − 𝑥1 )
and
𝑚2
𝑑 2 𝑥2
𝑑𝑡 2
= −𝑘(𝑥2 − 𝑥1 ) + 𝑘𝑥2
where 𝑥𝑖 is the displacement of mass i away from its equilibrium position
(previous figure b).
These equations can be expressed as
𝑚1
𝑚2
𝑑 2 𝑥1
𝑑𝑡 2
𝑑 2 𝑥2
𝑑𝑡 2
− 𝑘(𝑥2 − 2𝑥1 ) = 0
− 𝑘 (𝑥1 − 2𝑥2 ) = 0
‫تستغرق مناقشة المسائل التافهة وقتا‬
‫طويال ألن بعضنا يعرف عنها أكثر‬
‫مما يعرف عن المسائل الهامة‬
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From vibration theory, it is known that solutions to the previous equations
can take the form
𝑥𝑖 = 𝐴𝑖 sin(𝜔𝑡)
where 𝐴𝑖 = the amplitude of the vibration of mass 𝑖 and 𝜔 = the frequency
of the vibration, which is equal to
2𝜋
𝜔=
𝑇𝑝
where 𝑇𝑝 is the period. It follows that
𝑥𝑖′′ = − 𝐴𝑖 𝜔2 sin(𝜔𝑡)
The previous two equations can be transformed to
(
2𝑘
𝑚1
−
− 𝜔2 ) 𝐴1 −
𝑘
𝑚2
𝐴1 + (
2𝑘
𝑚2
𝑘
𝑚1
𝐴2
= 0
− 𝜔2 ) 𝐴2 = 0
at this point, the solution has been reduced to an eigenvalue problem.
‫من الحكمة أن تعتذر لرجل ذا كن‬
‫ وأن تعتذر المرأة حتى‬....... ‫مخطئا‬
‫ولو كن على صواب‬
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July 2013
EXAMPLE 27.4 Eigenvalues and Eigenvectors for a Mass-Spring System
Problem Statement:
Evaluate the eigenvalues and the eigenvectors of the previous equations
for the case where 𝑚1 = 𝑚2 = 40 kg and 𝑘 = 200 N/m.
Solution:
Substituting the parameter values into the equations yields
(10 − 𝜔2 )𝐴1 − 5𝐴2
=0
− 5𝐴1 + (10 − 𝜔2 )𝐴2 = 0
The determinant of this system is
(𝜔2 )2 − 20𝜔2 + 75 = 0
which can be solved by the quadratic formula for 𝜔2 = 15 and 5 s-2.
Therefore, the frequencies for the vibrations of the masses are
𝜔 = 3.873 𝑠 −1 and 2.236 𝑠 −1 , respectively. These values can be used to
determine the periods for the vibrations. For the first mode, T p = 1.62 s,
and for the second, Tp = 2.81 s.
As stated before, a unique set of value cannot be obtained for the
unknowns. However, their ratios can be specified by substituting the
eigenvalues back into the equations. For example, for the first mode
(𝜔2 = 15 𝑠 −2 ), 𝐴1 = −𝐴2 . For the second mode (𝜔2 = 5 𝑠 −2 ), 𝐴1 = 𝐴2 .
‫الطفنننل يلهنننو بالحيننناة صنننغيرا دون أن‬
..‫يعلم أن الحياة سوف تلعب به كبيرا‬
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July 2013
Continue:
This example provides valuable information regarding the behavior of
the system in the previous figure. Aside from its period, we know that if
the system is vibrating in the first mode, the amplitude of the second mass
will be equal but of opposite sign to the amplitude of the first.
As in the next figure (a), the masses vibrate apart and then together
indefinitely.
In the second mode, the two masses have equal amplitudes at all
times. Thus, as in the figure (b), they vibrate back and forth in unison.
It should be noted that the configuration of the amplitudes provides
guidance on how to set their initial values to attain pure motion in either
of the two modes. Any other configurations will lead to superposition of
the modes.
‫ أو بالبريد اإللكتروني‬9 4444 062 ‫النوتات مجانية للنفع العام فيرجى المساهمة باإلبالغ عن أي خطأ أو مالحظات تراها ضرورية برسالة نصية‬
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July 2013
Problem 27.4
Use the shooting method to solve
7
𝑑2 𝑦
𝑑𝑦
𝑑𝑥
𝑑𝑥
2 − 2
− 𝑦+𝑥 =0
with the boundary conditions y(0) = 5 and y(20) = 8
Solution
The second-order ODE can be expressed as the following pair of first-order
ODEs,
𝑑𝑦
𝑑𝑥
𝑑𝑧
𝑑𝑥
=𝑧
=
2𝑧 + 𝑦 − 𝑥
7
These can be solved for two guesses for the initial condition of z.
For our cases we used –1 and
. We solved the ODEs with the Heun
method without iteration using a step size of 0.125. The results are
z(0)
1
y(20) 11,837.64486
0.5
22,712.34615
Clearly, the solution is quite sensitive to the initial conditions.
These values can then be used to derive the correct initial condition,
𝑧(0) = −1 +
−0.5 + 1
22712.34615 − ( −11837.64486 )
= −0.82857239
( 8 − ( −11837.64486 ))
‫أفضل طبيب في العالم هو البيطري‬
‫فهو ال يستطيع أن يسأل مريضه عن‬
‫شكواه بل يكتشف ذلك بنفسه‬
‫ أو بالبريد اإللكتروني‬9 4444 062 ‫النوتات مجانية للنفع العام فيرجى المساهمة باإلبالغ عن أي خطأ أو مالحظات تراها ضرورية برسالة نصية‬
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‫‪July 2013‬‬
‫‪Continue:‬‬
‫‪The resulting fit is displayed below:‬‬
‫‪x‬‬
‫‪y‬‬
‫‪5‬‬
‫‪0‬‬
‫‪4.151601‬‬
‫‪2‬‬
‫‪4.461229‬‬
‫‪4‬‬
‫‪5.456047‬‬
‫‪6‬‬
‫‪6.852243‬‬
‫‪8‬‬
‫‪10 8.471474‬‬
‫‪12 10.17813‬‬
‫‪14 11.80277‬‬
‫‪16 12.97942‬‬
‫‪18 12.69896‬‬
‫‪20 8‬‬
‫أجمنننل األنهنننار لنننم نرهنننا بعننند‪ ..‬أجمنننل‬
‫الكتب لم نقرأها بعد‪ ..‬أجمل أيام حياتنا‬
‫لم تأ ٍ‬
‫ت بعد‪ ........‬ناظم حكم‬
‫النوتات مجانية للنفع العام فيرجى المساهمة باإلبالغ عن أي خطأ أو مالحظات تراها ضرورية برسالة نصية ‪ 9 4444 062‬أو بالبريد اإللكتروني‬
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July 2013
Problem 27.3
Use the finite-difference approach with ∆𝑥 = 1 to solve
𝑑2 𝑇
𝑑𝑥 2
− 0.15𝑇 = 0
Solution
A centered finite difference can be substituted for the second derivative to
give,
𝑇𝑖+1 − 2𝑇 + 𝑇𝑖−1
ℎ2
− 0.15 𝑇𝑖 = 0
or for h = 1,
−𝑇𝑖−1 + 2.15 𝑇𝑖 – 𝑇𝑖+1 = 0
The first node would be
2.15 𝑇1 – 𝑇2 = 240
and the last node would be
−𝑇9 + 2.15 𝑇10 = 150
The tridiagonal system can be solved with the Thomas algorithm or
Gauss-Seidel for (the analytical solution is also included)
‫الخبرة هي المشط الذي تقدمه الطبيعة‬
‫لإلنسان عندما يتساقط شعر رأسه‬
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‫‪July 2013‬‬
‫‪Continue:‬‬
‫‪Analytical‬‬
‫‪x‬‬
‫‪T‬‬
‫‪240‬‬
‫‪240‬‬
‫‪0‬‬
‫‪165.3290‬‬
‫‪165.7573‬‬
‫‪1‬‬
‫‪115.7689‬‬
‫‪116.3782‬‬
‫‪2‬‬
‫‪83.7924‬‬
‫‪84.4558‬‬
‫‪3‬‬
‫‪64.5425‬‬
‫‪65.2018‬‬
‫‪4‬‬
‫‪55.0957‬‬
‫‪55.7281‬‬
‫‪5‬‬
‫‪54.0171‬‬
‫‪54.6136‬‬
‫‪6‬‬
‫‪61.1428‬‬
‫‪61.6911‬‬
‫‪7‬‬
‫‪77.5552‬‬
‫‪78.0223‬‬
‫‪8‬‬
‫‪105.7469‬‬
‫‪106.0569‬‬
‫‪9‬‬
‫‪150‬‬
‫‪10 150‬‬
‫ذا انتقم لنفسك من المسي ليك‬
‫ساوي نفسك به‪ ..‬و ذا صفح عنه‬
‫استعبدته‪( .‬فرنسيب بيكون)‬
‫النوتات مجانية للنفع العام فيرجى المساهمة باإلبالغ عن أي خطأ أو مالحظات تراها ضرورية برسالة نصية ‪ 9 4444 062‬أو بالبريد اإللكتروني‬
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