
Introduction and Preliminary Concepts

Separation of Variables

Sturm-Liouville Theory

Fourier and Laplace Transforms
Linearity, Superposition and
Classifications:
A partial differential operator (PDO) L is considered linear if the
following is satisfied
L(u  cv)  Lu  cLv
where c is a scalar and u & v are functions
For example,


L(u  cv)    2 (u  cv)
 y

 (u y  2u )  c(v y  2v)
 Lu  cLv
therefore, PDO L is linear.
Linearity, Superposition and Classification (3)
It follows that for a linear PDO the equation
Lu  f
is homogeneous if f  0 and heterogeneous if f is any other function.
The order of a PDE is determined based on the highest order derivative
that is in it. For example, the inviscid Burgers’ equation below
ut  uu x  0
is first-order while the Korteweg-deVries (KdV) equation
ut  uu x  u xxx  0
is third-order due to the additional term.
If the linear homogeneous equation
Lv  0
is satisfied by u1 ,...un it also follows that u  c1u1  ...  cnun will satisfy it.
Linearity, Superposition and Classification (3)
In addition, PDEs can be classified based on the sign of the
discriminant B2-AC using the following criteria.
parabolic
hyperbolic
elliptic
if B 2  AC  0,
if B 2  AC  0,
if B 2  AC  0
The heat/diffusion equation, which we will be discussing, is
parabolic since,
 2u xx  ut
where  is a constant
B 2  AC  0 2  ( 2 )(0)  0
Examples of solutions of the equation are shown below.
Mathematical techniques for engineers and scientists (2)
Image modified from http://home.comcast.net/~sharov/PopEcol/lec12/diffus.html
Separation of
Variables
First order chemical reaction
takes place on the catalyst
wall of cylindrical container
X=R
Find out the concentration
change of A inside the
container
Set up PDE:
X=0
Catalyst
wall
r=kA
D*Axx=At, or D*∂2A/∂x2
=∂A/∂t
A-concentration of A
D-diffusivity
*from Chemical Engineering Kinetics (6)
First order chemical reaction
takes plane on the catalyst wall
Find out the concentration
change of A inside the
cylindrical contatiner
X=R
Boundary and initial conditions:
X=0
Catalyst
wall
r=kA
I.C t=0, A=A0
B.C1 x=0, ∂A/∂x │x=0 =0, by
symmetry
B.C2 x=R, D*∂A/∂x │x=R =k*A,
diffusion =reaction at wall
Run dimensionless:
u=A/A0, s=x/R, τ=t/tc
(=tD/R2)
X=R
Then we get:
tc*D/R2∂2u/∂s2 =∂u/∂τ,
let tc*D/R2=1, then tc=R2/D
X=0
Catalyst
wall
r=kA
BCs: ∂u/∂s =0 at s=0
∂u/∂s =kR/D*u=φu,
φ=kR/D
at s=1
Assume: u=F(s)*G(τ)
G(τ)*dF2/ds2=F(s)*dG(τ)/ds,
1/F(s)*dF2/ds2=1/G(τ)*dG(τ)/d
s=-λn2
X=R
Solve 1/G(τ)*dG(τ)/ds=-λn2
X=0
Catalyst
wall
r=kA
we get: dG(τ)=exp(-λn2τ)
For 1/F(s)*dF2/ds2=-λn2
assume
F(s)=∑(Ansinλns+ Bncosλns)
Use BC1: ∂u/∂s =0 at s=0
∂u/∂s=G(τ)*dF/ds=0
X=R
∑(Anλncosλn*0- Bnλnsinλn*0)=0
An=0
F(s)= ∑Bncosλns
X=0
Catalyst
wall
r=kA
Use BC2: ∂u/∂s =φu at s=1
-Bnλnsinλn=φBncosλn
-λntanλn=φ, λncan be solved
u(s, τ) =∑exp(-λn2 τ) Bncosλns
How to get Bn?
X=R
τ =0, u(s,0)=1= ∑1*Bncosλns
Use orthogonality principle
X=0
Catalyst
wall
r=kA
∫01 *cos(λns)ds
=Bn∫01cos2(λns)ds
Bn = ∫01
cos(λns)ds/∫01cos2(λns)ds
Bn =4sin(λn )/(2λn+sin2λn)
45
40
40
35
35
30
30
Concentration A
Concentration A
45
25
20
25
20
15
15
10
10
5
5
0
0
0
2
4
6
time(t)
8
10
0 12
2
4
6
time(t)
8
10


Second order, homogenous, linear differential
equation with boundary conditions.
Solutions to such problems contain an infinite
set of eigenvalues with corresponding
eigenfunction solutions

Equations from Advanced Engineering
Mathematics (9)

Problem from Introduction to Differential
Equations (7)

Problem from Introduction to Differential
Equations (7)

Problem from Introduction to Differential
Equations (7)

Problem from Introduction to Differential
Equations (7)

Problem from Introduction to Differential
Equations (7)

Problem from Introduction to Differential
Equations (7)


Many differential equations are in the form of
the Sturm-Liouville problem.
Sturm-Liouville analysis can help determine
integration constants for equations with an
infinite number of terms (eigenfunctions).

Problem from Heat Conduction (8)

The boundary conditions are such
that
Figure from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)

Problem from Heat Conduction (8)
• Other types of problems
•
•
Wave equation, diffusion equation…
Other types of boundaries:
• Fixed, Periodic, Moving
• Handled same as in regular Sturm-Liouville Problem
Diffusion Equation:
The energy balance over the infinitesimal volume of a rod shown in the
figure is as follows
kAux
 kAux x  hsx(u  u )  vAcu x  vAcu x  x 
x  x
 (mcu)
t
where, m=AΔxσc. Dividing the full equation by AΔxσc and taking the limit
as x goes to 0 yields
 2u xx  ut  vux  H (u  u )
wher e,  2 
Variables: c  specific heat
k
hs
&H 
c
Ac
hsx(u  u )
  mass density
s  circumfere nce
kAux x
v  speed of rod
u  u  temperatur e difference
x
A  cross - sectional area
h  heat trans fer coefficien t
k  thermal conductivi ty
kAux
x  x
x  x
Heat Conduction and the Heat Equation (5)
Image modified from http://www.ceb.cam.ac.uk/pages/finite-element-simulations.html
Initial/Boundary Conditions:
For all subsequent calculations we assume that diffusion in the rod is
one-dimensional and thus set H and and v to zero.
 2u xx  ut
The initial condition is simply a set function for u(x,t) over the length of the
rod at t=0 and is written thus,
u ( x,0)  f ( x)
(0  x  L )
The boundary condition can fall into one of three categories.
Dirichlet boundary condition:
u (0, t )   (t )
Meaning that the left end of the bar is kept at a set temperature.
Neumann boundary condition:
q(0, t )  kux (0, t )   (t )
This boundary condition specifies the heat flow out of the rod. Particularly,
setting  (t )  0 means insulating the left end of the rod.
Fundamental Solutions (4)
Initial/Boundary Conditions (cont`d):
Robin boundary condition:
Starting with Newton’s Law of Cooling
q  hA(u s  u f )
where, us and uf are temperatures of the solid and fluid
respectively. It then follows that,
 ( L, t )  h[u f  u ( L, t )]
 kux ( L, t )  h[u f  u ( L, t )]
 kux ( L, t )  hu ( L, t )  hu f
Heat Conduction and the Heat Equation (5)
For non-periodic
functions [f(x)] the
heat equation can be
solved using Fourier
t
exponential transform
and a known
Gaussian function
x  
g(x,t) the steps
follow…..
u ( x ,0 )  f ( x )
f (x )
x
Mathematical techniques for engineers and scientists (2)
The Fourier exponential transform can be written

as
F u ( x, t ); x  s   e isx u ( x, t )dx  U ( s, t )

Using the above definition we can write
F u xx ( x, t ); x  s   s 2U ( s, t )
F ut ( x, t ); x  s  U t ( s, t )
The heat equation can then be rewritten as
ut   2 s 2U  0,
t 0
IC : U (s,0)  G( s)
where, G(s)  F  f ( x); s
We can solve for U using the initial condition to
yield
2 2
U ( s, t )  G ( s )e  s t ,
  s  
2 2
2
2
1
Using the inverse Fourier tr ansform (e   s 
e  x /( 2 ) )
2 

and the convolutio n theorem ( f ( s ) g ( s ) 

u( x, t ) 

 f ( x   ) g ( )d ) yields

f ( ) g ( x  t )d
where,
g ( x, t ) 

Thus,
u ( x, t ) 
1
2

 (t ) 

f ( )e ( x  )
2
/ 4 2 t
1
2 t
e x
2
/ 4 2 t
d ……(1)
Mathematical techniques for engineers and scientists (2)
Special Case 1:
The simplest solution to the heat equation is
for
f ( x)  constant  F
by setting a 
 x
equation (1) can be rewritten as
2 t

u ( x, t ) 
F
2 t
e
a 2
2 t da

Assuming that all terms except a are constant the function can be
integrated to yield
u ( x, t ) 
F 

F
Thus, for the limiting case of f(x) = constant the heat will also
equal that constant.
Advanced engineering mathematics (9)
Special Case 2:
In this case, assume the initial condition is as
follows
Using these initial conditions the heat equation can be solved
to yield

u ( x, t ) 
100
e

4t 0

( x  ) 2
4t
d  50 * (1  erf (
x
2 t
))
since the integral can then be rewritten as
erf ( z ) 
2

z
e
 2
d
0
Beny Neta Presentation (1)
From the chart we can see
the function “smoothing”
out at large values of t.
This is due to the behavior
of the error function,
erf(∞)=1 and erf(~0)=0.
Thus, for positive values of
x at small times u(x,t)
approaches F and at large
times u(x,t) approaches
F/2. For negative values of
x the negative sign can be
taken outside of the error
function so the value of the
error function is subtracted
from 1. Thus at small times
u(x,t) approaches 0 and at
large times u(x,t)
approaches F/2.
t=10
t=500
t=1000
t=10000
u(x,t)
100
90
80
100000
70
10000000
60
50
40
30
20
10
0
-400 -300 -200 -100
0
100
200
300
x
400
Typical Behavior
The following curves are of
temperature collected at
different times. From the plots
a “smoothing” effect can again
be seen as the temperature
along entire length of the rod
approaches an equilibrium
value.
Images taken from
http://en.wikipedia.org/wiki/File:Heatequation_exampleB.gif
Special Case 3:
Similar to the error function the integral of equation (1) can be
rewritten as the kernel
 ( x  ) 2 / 4 2 t
K (  x; t ) 
and the function becomes….
e
2 t

 f ( ) K (  x; t )d
it can be shown that u ( x, t )   K (  x; t )d  1
u ( x, t ) 
 x
by setting a 
2 t



Thus as t  0, K   (  x)
K
to
t1
t2
x
From the graph you can see that similar
“smoothing” process occurs for the Kernel
as for the temperature profile. It is also
important to note that the Kernel is a
response function to an initial
temperature function described by the
delta function
figure modified from:
http://mobjectivist.blogspot.com/2010_06_01_archive.html
Heat Conduction of a Semi-Infinite Rod
(Laplace Transforms)
For non-periodic
functions [f(x)]
t
the heat equation
u ( x ,0 )  f ( x )
can be solved in
f (x )
u (0, t )
the semi-infinite
x
 g (t )
domain using
Laplace
 u u ,
( 0  x   , t  0)
exponential
BC : u(0, t )  f (t ),
u( x, t )  0 as x  , t  0
transform the
IC : u( x,0)  0,
(0  x  )
steps follow…..
2
xx
t
Mathematical techniques for engineers and scientists (2)

Using the inverse Laplace transform
U(x,p) becomes,
x2
2   x  p e 4 t
e
 3/ 2
x
t
2
xe x /( 4 t )
u ( x, t )  g (t )
2  (t 3 / 2 )
2
Then using the convolution theorem
2
t
( f * g )(t )   f ( ) g (t   )d
0
u(x,t) becomes,
u ( x, t ) 
x
2 
t
 g (t   )
0
e x

2
/ 4 2
3/ 2
d
………(2)
Mathematical techniques for engineers and scientists (2)
The Laplace exponential transform can be written as

L u( x, t ); t  p   e  pt u( x, t )dt  U ( x, p)

Using the above definition we can write
L u xx ( x, t ); t  p  U xx ( x, p)
L ut ( x, t ); t  p  pU ( x, p)
The heat equation can then be rewritten as
U xx 
p

2
U  0,
x0
BC : U (0, p )  G ( p )
where G ( p )  L  f (t )
U ( x, p)  0 as x  
We can solve for U using the boundary condition to yield
U ( x, p)  G( p)e
x p

,
  p  
Mathematical techniques for engineers and scientists (2)
Special Case :
The simplest solution to the heat equation is for
g ( x)  cons tan t  F
By setting
a2 
x2
4 
we can solve for
2
2
x2

2
x
  2 2 and d  2 3 da
4 a
4 a
Thus equation (2) becomes
x
2F
or otherwise

2 t
e
a 2
da
0
u ( x, t )  F * erfc(
x
2 t
)
Advanced engineering mathematics (9)
For F=50 and α=1 u(x,t) is plotted below
From the chart
we can see a
similar
“smoothing”
process as in
the case of the
infinite domain
at large values
of t. Again,
erf(∞)=1 and
erf(~0)=0.
Thus at small
times u(x,t)
approaches F
and at large
times u(x,t)
approaches 0.
u(x,t)
50
40
30
20
10
0
0
x
100 200 300 400 500 600 700 800 900
t=10
t=500
t=1000
t=10000
100000
10000000
References
1.Neta, Beny. "PARTIAL DIFFERENTIAL EQUATIONS MA 3132 LECTURE
NOTES." math.nps.navy. Department of Mathematics Naval Postgraduate
School, n.d. Web. 20 Oct. 2011.
<www.math.nps.navy.mil/~bneta/pde.pdf>.
2.Andrews, Larry C., and Ronald L. Phillips. Mathematical techniques for
engineers and scientists. Bellingham, Wash.: SPIE Press, 2003. Print.
3."Linearity, Superposition and Classification." math.unl. N.p., n.d. Web. 21
Oct. 2011. <www.math.unl.edu/~scohn1/8423/class.pdf>.
4."CHAPTER 2: The Diffusion Equation." mth.pdx. N.p., n.d. Web. 21 Oct.
2011. <www.mth.pdx.edu/~marek/mth510pde/notes%202.pdf>.
5."Heat Conduction and the Heat Equation." ncsu. N.p., n.d. Web. 21 Oct.
2011. <www4.ncsu.edu/~rsmith/MA573_F09/Heat_Equation.pdf>.
6.Platnaik. Introduction to Differential Equations. PHI Learning Pvt. Ltd.
7.Jiji, Latif M. Heat Conduction. 3rd Ed. Springer Publishing. 2009.
8.Greenberg, Michael D.. Advanced engineering mathematics. 2nd ed.
Upper Saddle River, N.J.: Prentice Hall, 1998. Print.