CHAPTER 20
LAPLACE EQUATION
Team 6:
Bhanu Kuncharam
Tony Rochav
Wei Lu
20.1 Introduction
Remembering from chapter 16, the Laplacian operator (in Cartesian
coordinates):
2
2
2



2  2  2  2
 x  y  z
There are two types of Laplacian equations,
• Homogeneous
 2u  0
• Non-homogeneous (also known as the Poisson equation):
 2u  f
Where f is a “source” function that is prescribed over the region in question.
Introduction
A question to ask before we start going deeper into learning how to solve the Laplace
equation is: Does this equation come up naturally? The answer to this question is yes.
Let it be illustrated from the following example. Let’s consider the heat equation:
u
Q
 k 2u 
t
Cp
Under which considerations would the Laplace equation arise?
• Steady state (solution at equilibrium)
u
0
t
•
No heat sources
Q0
Under these conditions the Laplace equation arises from physical situations.
0  k 2 u   2 u  0
20.2 Separation of Variables: Cartesian
Coordinates
The method of separation of variables assumes that the solution for an
equation of multiple variable can be describes as the product of
functions of a single variable. One function for each variable. For
example.
T ( x, y )  F ( x)G ( y )
This method of solution can be applied to the Laplace equation when
we assume that the Laplacian applies to a 2 dimensional function.
How we solve Laplace’s equation will depend upon the geometry of the
2-D object we are solving it on.
Separation of Variables
Let’s start out by solving it on the rectangle given by 0 ≤ x ≤ L , 0 ≤ y ≤ H .
For the geometry shown in the right, Laplace’s equation along with the four boundary
conditions will be:
y
 2T  2T
 T  2  2 0
x
y
T (0, y )  g 1 ( y ),
2
T ( L, y )  g 2 ( y )
T=f2(x)
H
T=g1(y)
T=g2(y)
T ( x,0)  f 2 ( x)
T ( x, H )  f 1 ( x )
0
T=f1(x)
L
x
•First notices that we don’t need and initial condition for this heat problem since the
Laplace equation is homogeneous, meaning it is equal to zero, which is the equivalent of
assuming a steady state for the T variable.
•Next, let’s notice that while the partial differential equation is both linear and
homogeneous the boundary conditions are only linear and are not homogeneous. This
creates a problem because separation of variables requires homogeneous boundary
conditions.
Separation of Variables
To completely solve Laplace’s equation we are going to have to solve it four times. Each
time we solve it only one of the four boundary conditions can be nonhomogeneous while
the remaining three will be homogeneous.
Because we know that Laplace’s equation is linear and homogeneous and each of the
pieces is a solution to Laplace’s equation then the sum will also be a solution.
T2(x,H)=0
T1(x,H)=0
T1(L,y)=0
T1(0,y)=0
+
T2(L,y)=g2(y)
T2(0,y)=0
T2(x,H)=f2(x)
T4(L,y)=g2(y)
T3(0,y)=g1(y)
T1(x,0)=f1(x)
T2(x,0)=0
T3(x,H)= f2(x)
T4(x,H)=0
=
T1(x,0)=f1(x)
T3(L,y)=0
T3(0,y)=0
+
T4(L,y)=0
T4(0,y)= g1(y)
=g2(y)
T3(x,0)=f1(x)
T4(x,0)=0
+
Separation of Variables
Now, once solved all four of these problems the solution to our original system, (16), will
be,
T ( x, y)  T1 ( x, y)  T2 ( x, y)  T3 ( x, y)  T4 ( x, y)
Also, this will satisfy each of the four original boundary conditions. For example:
T ( x,0)  T1 ( x,0)  T2 ( x,0)  T3 ( x,0)  T4 ( x,0)  f 1 ( x)  0  0  0  f 1 ( x)
And so on for the remaining boundary conditions.
We will solve the problem in the following way:
• Apply separation of variables to the each problem and find a product solution that
will satisfy the differential equation and the three homogeneous boundary conditions.
•
Use the Principle of Superposition to find a solution to the problem and then apply
the final boundary condition to determine the value of the constant(s) that are left in
the problem.
Two of the cases will be shown here and the two can be done as an additional exercise.
Separation of Variables - Example 1
Find the solution to the following differential equation.
 2T3  2T3
 T3 

0
2
2
x
y
2
y
Boundary Conditions:
T3=f2(x)
H
T3 (0, y )  0,
T3 ( L, y )  0
T3=0
T3 ( x,0)  0
T3=0
T3 ( x, H )  f 2 ( x)
0
T3=0
L
x
Notice that this example is a part of our original problem solution, it would be
the third component of our solution.
T4(x,H)=0
T2(x,H)=f2(x)
T1(x,H)=0
T4(L,y)=g2(y)
T3(0,y)=g1(y)
T1(x,0)=f1(x)
=
T2(x,H)=0
T1(L,y)=0
T1(0,y)=0
T1(x,0)=f1(x)
+
T3(x,H)= f2(x)
T2(L,y)=g2(y)
T2(0,y)=0
T2(x,0)=0
+
T3(L,y)=0
T3(0,y)=0
T3(x,0)=f1(x)
+
T4(L,y)=0
T4(0,y)= g1(y)
=g2(y)
T4(x,0)=0
Separation of Variables – Solution to Example 1
We assume the solution can be solved by separating the variable. In other words
the solution can be expressed in the form:
T3 ( x, y )  F ( x)G ( y )
Our homogeneous boundary conditions now become:
F (0)  0
F ( L)  0
G(0)  0
Next, we’ll plug the product solution into the differential equation.
2
2
F ( x)G( y)  2 F ( x)G( y)   0
2
x
y
2F
 2G
G ( y ) 2  F ( x) 2  0
x
y
1 2F
1  2G

2
F x
G x 2
y
H
T3=f2(x)
T3=0
0
T3=0
T3=0
L
x
Separation of Variables – Solution to Example 1
At this point we need to select our separation variable. We need
identify were the two homogeneous boundaries exist and select
the positive or negative variable that will give us a solution in the
form of sine and cosine.
After that you can find that how to build up the solution is
different from one book to another. While in the textbook tend
to find and combine all the solutions (using the superposition
principle), others notice when only the positive or negative
solution are non-trivial, thus reducing the need of working out
so many constants
In this section both approaches are exemplified, for the first two
examples we eliminated the possibility of  =0, integration
constant being zero, because we know the values of all four
boundary conditions.
Separation of Variables – Solution to Example 1
For the reasons mentioned before, for this case, we select our integration
constant as (-). So, after adding in the separation constant we get,
1 2F
1  2G

 
2
2
F x
G x
and two ordinary differential equations that we get from this case (along
with their boundary conditions) are,
2F
 F  0
2
x
F ( 0)  0
F ( L)  0
and
 2G
 G  0
2
x
G (0)  0
Separation of Variables – Solution to Example 1
Using our first ODE we find that the solution is, as intended, in the
form:
F ( x)  C1 cos(  x)  C 2 sin(  x)
Applying the first boundary condition gives
0  C1 (1)  C 2 (0)  C1  0
Applying the second boundary condition, and C1=0 to F(x) we get:
0  F ( L)  C 2 sin(  L)
Now, we after disregarding, trivial solutions, this means we must have:
0  sin(  L)
L   n
n  1,2,3,...
Separation of Variables – Solution to Example 1
The positive eigenvalues and their corresponding eigenfunctions of this
boundary value problem are then,
 n 
n  

L


 nx 
Fn ( x)  sin 

L


2
n  1,2,3,...
The second differential equation is then,
 2 G  n 

G  0
2
x
 L 
G ( 0)  0
Because the coefficient of G is positive we know that a solution to this is,
G ( y )  C1e
 ny 


 L 
 C2 e
  ny 


 L 
 ny 
 ny 
  C2 sinh 

 L 
 L 
or G( y )  C1 cosh
Separation of Variables – Solution to Example 1
Applying the boundary condition the solution becomes,
G (0)  C1 (1)  0
 ny 
G ( y )  C 2 sinh 

 L 
The product solution for this case becomes,
 ny   nx 
T3 ( x, y )  F ( x)G ( y )  Bn sinh 
 sin 

 L   L 
and solution to this partial differential equation is then,
 ny   nx 
T3 ( x, y )   Bn sinh 
 sin 

 L   L 
n 1

n  1,2,3
Separation of Variables – Solution to Example 1
Finally, let’s apply the nonhomogeneous boundary condition to get the
coefficients for this solution.
 nH   nx 
T3 ( x, H )  f 2 ( x)   Bn sinh 
 sin 

 L   L 
n 1

As expected, this is again a Fourier sine series and so using previously done
work instead of using the orthogonally of the sines to we see that,
 nH  2
 nx 
Bn sinh 
f 2 ( x) sin 

dx
 L  L0
 L 
L

2
 nx 
Bn 
f 2 ( x) sin 
dx
 nH  0
 L 
L sinh 

 L 
n  1,2,3,...
L

n  1,2,3,...
Separation of Variables – Solution to Example 1
Finally we see that the solution for this problem is:
 ny   nx 
T3 ( x, y )   Bn sinh 
 sin 

 L   L 
n 1

Where
2
 nx 
Bn 
f 2 ( x) sin 
dx
 nH  0
 L 
L sinh 

L


L

n  1,2,3,...
• Important remarks: For the Laplace equation, the sign of the separation
constant and the sequencing of the boundary conditions needs to be decided
on a case-by-case basis. However the following rule of thumb can be used as
guidance: “Anticipating the edge along which the eventual Fourier series will
take place, chose the  or - so as to obtain oscillatory functions along that
edge. Then apply the boundary conditions adjacent to that edge first”.
Separation of Variables – Visualization of Example 1
Consider f(x), to be a constant, say f(x)=100C and our rectangular body to have
dimensions L=200, H=100. In this case, we can see the boundary conditions
taking place in the eastern, western, and southern boundary were T=0.
http://dylanh.bol.ucla.edu/CA/temperature.html
Separation of Variables – Example 2
Find a solution to the following PDE
T4(x,H)=0
 T4  T4
 T4 

0
2
2
x
y
2
2
2
Boundary conditions:
T4(L,y)=0
T4(0,y)= g1(y)
T4 (0, y )  g1 ( y )
T4 ( L, y )  0
T4(x,0)=0
T4 ( x,0)  0
T4 ( x, H )  0
Notice that this example is a part of our original problem solution, it would be
the fourth component of our solution.
T4(x,H)=0
T2(x,H)=f2(x)
T1(x,H)=0
T4(L,y)=g2(y)
T3(0,y)=g1(y)
T1(x,0)=f1(x)
=
T2(x,H)=0
T1(L,y)=0
T1(0,y)=0
T1(x,0)=f1(x)
+
T3(x,H)= f2(x)
T2(L,y)=g2(y)
T2(0,y)=0
T2(x,0)=0
+
T3(L,y)=0
T3(0,y)=0
T3(x,0)=f1(x)
+
T4(L,y)=0
T4(0,y)= g1(y)
=g2(y)
T4(x,0)=0
Separation of Variables – Solution to Example 2
We start from assuming the solution can be solved by separating the variable.
T4 ( x, y)  F ( x)G( y)
Boundary conditions:
F ( L)  0
G (0)  0
G( H )  0
Next, we plug the product solution into the differential equation.
2
2
F ( x)G( y)  2 F ( x)G( y)  0
2
x
y
2F
 2G
G ( y ) 2  F ( x) 2  0
x
y
1 2F
1  2G

2
F x
G y 2
Separation of Variables – Solution to Example 2
At this point we need to choose a separation constant, in this case (), Note
how the selection of this variable depends on where we have two
boundary conditions.
So, after adding in the separation constant we get,
1 2F
1  2G


2
2
F x
G y
and two ordinary differential equations that we get from this case (along with
their boundary conditions) are
2F
 F  0
2
x
F ( L)  0
and
 2G
 G  0
2
y
G ( 0)  0
G ( L)  0
Separation of Variables – Solution to Example 2
This time our second ODE has the sine, cosine, solution:
G( y)  C1 cos(  y)  C2 sin(  y)
Applying our boundary conditions we get to determine that
G (0)  C1  0
The positive eigenvalues and their corresponding eigenfunctions of this
boundary value problem are then,
 n 

 H 
n  
2
 ny 
Gn ( y )  sin 

 H 
n  1,2,3,...
Separation of Variables – Solution to Example 2
The second differential equation is then,
 2 F  n 

F  0
2
x
 L 
F ( L)  0
Because the coefficient of F is positive we know that a solution to this is,
 nx 
 nx 
F ( x)  C1 cosh 
  C 2 sinh 

 H 
 H 
However, this is not really suited for dealing with the h(L)= 0 boundary
condition. So, let’s also notice that the following is also a solution:
 n ( x  L) 
 n ( x  L) 
F ( x)  C1 cosh 

C
sinh



2
H
H




This is usually called a “shifted solution” and is very useful!
Separation of Variables – Solution to Example 2
Applying the lone boundary condition to this “shifted” solution gives,
F ( L)  C1 (1)  0
The solution to the first differential equation is now,
 n ( x  L) 
F ( x)  C 2 sinh 

H


And this is all the farther we can go with this because we only had a single
boundary condition. That is not really a problem however because we now
have enough information to form the product solution for this partial
differential equation. A product solution for this partial differential equation
is,
 n ( x  L)   nx 
Tn ( x, y)  F ( x)G( y)  Bn sinh 
 sin 

H
H

 

n  1,2,3
Separation of Variables – Solution to Example 2
The solution to this partial differential equation is then,
 n ( x  L)   nx 
T4 ( x, y )   Bn sinh 
 sin 

H

  H 
n 1

Finally, let us apply the non-homogeneous boundary condition to get the
coefficients for this solution.
 n ( x  L)   nx 
T4 (0, y )  g1 ( y )   Bn sinh 
 sin 

H

  H 
n 1

Now, in the previous problems we have done this has clearly been a Fourier
series of some kind and in fact it still is. The difference here is that the
coefficients of the Fourier sine series are now,
  nL 
Bn sinh 

 H 
Separation of Variables – Solution to Example 2
Remember that a Fourier sine series is a series of coefficients (depending on n)
times a sine. We still have that here, except the “coefficients” are a little
messier this time that what we saw when we first dealt with Fourier series. So,
the coefficients can be found using exactly the same formula from the
Fourier sine series section of a function on 0 < y < H we just need to be
careful with the coefficients.
  nL  2
Bn sinh 

 H  H
 ny 
g
(
y
)
sin

dx
0 1
 H 
H
2
 ny 
Bn 
g
(
y
)
sin

dx
1

  nL  0
 H 
H sinh 

 H 
n  1,2,3,...
H
n  1,2,3,...
Separation of Variables – Solution to Example 2
Finally we see that the solution for this problem is:
 n ( x  L)   nx 
T4 ( x, y )   Bn sinh 
 sin 

H

  H 
n 1

Where
2
 ny 
Bn 
g1 ( y ) sin 
dx

n

L
H

0


H sinh 

 H 
H

n  1,2,3,...
Separation of Variables – Closing of Examples 1&2
Until this point we have worked two of the four cases (T3 &T4) that would need
to be solved in order to completely solve the Laplace equation with nonhomogenous boundary conditions in all four sides.
T4(x,H)=0
T2(x,H)=f2(x)
T1(x,H)=0
T4(L,y)=g2(y)
T3(0,y)=g1(y)
T1(x,0)=f1(x)
=
T2(x,H)=0
T1(L,y)=0
T1(0,y)=0
T1(x,0)=f1(x)
+
T3(x,H)= f2(x)
T2(L,y)=g2(y)
T2(0,y)=0
+
T3(L,y)=0
T3(0,y)=0
T2(x,0)=0
T3(x,0)=f1(x)
+
T4(L,y)=0
T4(0,y)= g1(y)
=g2(y)
T4(x,0)=0
As we have seen each case was similar and yet also had some differences. We
saw the use of both separation constants and that sometimes we need to use
a “shifted” solution in order to deal with one of the boundary conditions.
The remaining cases can be easy to find after this explanation.
The figure on the right illustrates the solution of the
Laplace equation with non-homogeneous boundary
conditions for al four sides, in this case fixed
temperature in all four sides a square body such as the
one in the problem.
http://tigger.uic.edu/~rkodama/
Separation of Variables –Example 3
Consider the Dirichlet problem:
2
2

T

T
2
 T  2  2 0
x
y
With the Boundary Conditions:
y
T (0, y )  20,
T (5, y )  50
T ( x ,0 )  f ( x )
T ( x, y )
T=50
T=20
0 x5
bounded as
y
0
T=f(x)
5
x
Separation of Variables –Example 3
This example can very well represent a complicated three dimensional
domain. Take for example the following figure from Greenberg’s
book:
Fig. Cooling fin*
Suppose that our interest is finding the temperature of the object near
the end of the fin, within the rectangle ABCE, assume we don’t know
the boundary along AB and the boundaries T=20 along AE, T=50
along BC, and T=f(x) along EC. To a good approximation, we render
the problem tractable by letting D be the entire semi-infinite strips
shown for the Dirichlet problem.
*Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
Separation of Variables –Solution of Example 3
We assume the solution can be solved by separating the variable. In other words
the solution is in the form:
T ( x, y)  F ( x)G( y)
We apply this to the homogeneous boundary conditions first since we’ll need
those once we get reach the point of choosing the separation constant. The
boundary conditions become,
F (0)  20
F (5)  50
G (0)  f ( x)
Next, we plug the product solution into the differential equation.
2
2
F ( x)G( y)  2 F ( x)G( y)  0
2
x
y
2F
 2G
G ( y ) 2  F ( x) 2  0
x
y
1 2F
1  2G

F x 2
G y 2
Separation of Variables –Solution of Example 3
Now, at this point we need to choose a separation constant, in this case (-)
1 2F
1  2G

 
2
2
F x
G x
So,
F ( x)' 'F ( x)  0
G ( y )' 'G ( y )  0
The possible solutions to this ODEs are:
 C1  C 2 x
F ( x)  
C3 cos(  x)  C 4 sin(  x)
 0
 0
C 5  C 6 y
G( y)  
C 7 exp(  y )  C8 exp(   y )
 0
 0
Separation of Variables –Solution of Example 3
This ODE has three types of equations, lets superimpose all the types of
solutions. This will give:
T ( x, y)  C1  C 2 x C5  C6 y



 C3 cos(  x)  C 4 sin(  x) C7 exp(  y)  C8 exp(   y)

If we try to apply the boundness condition (y→∞) we note that the term y and
the positive exponential will grow to infinite. Since this is not possible it is
necessary that C6 =0and C7=0. This reduces our equation to:
T ( x, y)  C1  C 2 x C5



 C3 cos(  x)  C 4 sin(  x) C8 exp(   y)
Grouping the constants:



T ( x, y)  K1  K2 x  K3 cos(  x)  K4 sin(  x) exp(   y)
Separation of Variables –Solution of Example 3
Now, the boundary condition in the left side:
T (0, y)  20  K1  K 3 exp(   y)
Matching the coefficients of the K1 and exp( ) terms on both sides of our
previous equation gives K1= 20 and K3=0. Using this result, and updating
T(x,y) we get:
T ( x, y )  20  K 2 x  K 4 sin(  x) * exp(   y )
Next, we use the boundary condition at the right side:
T (5, y )  50  20  5K 2  K 4 sin( 5  ) * exp(   y )
So
50  20  5 K 2
0  sin( 5  )
K2  6
Give
n

5
n  1,2,3,...
Separation of Variables –Solution of Example 3
Using this result to update T(x,y):
 nx 
 ny 
T ( x, y )  20  6 x   K n sin 
*
exp



 5 
 5 
n 1

This is the solution to our problem. Now we only need to find the value of Kn.
To do this we use the lower boundary condition:
 nx 
T ( x,0)  f ( x)  20  6 x   K n sin 

 5 
n 1

We find that
 nx 
f ( x)  20  6 x   K n sin 

 5 
n 1

or
 nx 
K
sin

  f ( x)  20  6 x

n
 5 
n 1

Separation of Variables –Solution of Example 3
Applying the Fourier sine series we can find Kn as:
2
 f ( x)  20  6 xsin  nx dx
Kn 
50
 5 
5

n  1,2,3,...
Which is the constant that was missing to obtain the result of the Temperature
in our semi-infinite slab equation:
 nx 
 ny 
T ( x, y )  20  6 x   K n sin 
 * exp 

 5 
 5 
n 1

20.3 Separation of variables; Non-Cartesian
Coodinates
20.3.1 Plane polar coordinates
y
r sinθ
From figure polar coordinates can be
written in terms of Cartesian coordinates
x  r cos 
θ
r cosθ
y  r sin 
x
The laplacian in terms of polar coordinates where r, θ are
independent variables
2
2
1

1

2  2 
 2 2
r r r 
r
which is same as equation (24) from section 16.7
Example 1: Direchlet Problem
Problem definition:
1 u 1  2 u
 u 2 
 2
0
2
r r r 
r
u (r,0)  u1
(a  r  b)
2
 2u
u (r, )  u 2
(a  r  b)
(a  r  b, 0    )
(1)
(2a)
(2b)
u (a , )  0
(0    )
(2c)
u (b, )  f ()
(0    )
(2d)
u=f(θ)
u=u2
u=0
α
r
θ
Physical Interpretation:
a
b
The variable ‘u’ used here can be
u=u1
physically interpreted as temperature (T) in heat transfer problem
Or concentration (C) in diffusion problem.
The problem at hand can be seen as a temperature or concentration distribution in a
cylindrical disk where the disk thickness is ‘z’, which is ignored (the temperature
gradient in ‘z’ direction).
Solution using separation of variables method
We seek solution in the form of
u(r, )  R (r)()
(3)
Using this as solution the laplacian equation transforms into:
1
1
R   R   2 R  0
(4)
r
r
Which can be simplified r and divide by RΘ to separate the
variables
r 2 R   rR 


 constant   2
(5)
R

Here +κ2 is chosen to obtain solution in the form of cosκθ and sinκθ instead of coshκθ
and sinhκθ since we anticipate that to satisfy the u(b,θ)=f(θ) boundary condition we
need to expand f(θ) in a Fourier series (discussed later sections)
The previous differential equation can be written in a more familiar form as
r 2 R   rR    2 R  0
and
   2   0
(6)
(7)
The solution can be written as:
0
A  B ln r
R (r )  
(8a)


0
 Cr  Dr
and
0
 E  F
()  
(8b)
0
 G cos   H sin 
Note:
A, B, C,D E,F, G,
H are arbitrary
integration
constants and
needs to be
evaluated.
Superposition of these two solutions (8a, 8b), we can write a general solution as
u(r, )  (A  B ln r)(E  F)  (Cr   Dr  )(G cos   H sin )
(9)
u(r, )  (A  B ln r)(E  F)  (Cr   Dr  )(G cos   H sin )
Apply boundary conditions
When θ=0
u(r,0)  u1  (A  B ln r)E  (Cr   Dr  )G
(10)
u=0
α
r
θ
a
(11)
u=u1
b
Boundary conditions
When θ=α

u(r, )  u1  u1  I  (Pr  Qr

)sin  (12)
So u1+Iα=u2 and sinkα=0 gives
I  (u 2  u1 )
and   n
u=f(θ)
u=u2
Set AE=u1 , B=0, and G=0 which gives
u(r, )  u1  I  (Pr  Qr  )sin 
(9)


(n  1,2,3...)
(13a)
(13a)
u (r,0)  u1
(a  r  b)
u (r, )  u 2
(a  r  b)
u (a , )  0
u (b, )  f ()
(0    )
(0    )
We can write the solution after substituting equation 13a, 13b


 
n
u (r, )  u1  (u 2  u1 )   Pn r n /   Q n r  n /  sin
 n 1

Using next boundary condition:
u(a, )  0
(14)
(0    ) (2c)


 
n
u (a , )  0  u1  (u 2  u1 )   Pn a n /   Q n a  n /  sin
 n 1

(15)
Which can be written as:




n
n / 
 n / 
 u1  (u 2  u1 )   Pn a
 Qna
sin
 n 1

(16)
Where
P a
n
n / 
 Qna
 n / 

2 

n
   u1  (u 2  u1 )  sin
d
 0


(17)


 
n
u (r, )  u1  (u 2  u1 )   Pn r n /   Q n r  n /  sin
 n 1

Using next boundary condition:
u(b, )  f ()
(14)
(0    )
(2d)
Which can be written as:


 
n
u (b, )  f ()  u1  (u 2  u1 )   Pn b n /   Q n b  n /  sin
(18)
 n 1





n
f ()  u1  (u 2  u1 )   Pn b n /   Q n b  n /  sin
 n 1

(19)
Similar to previous expansion
P b
n
n / 
 Qn b
 n / 

2 

n
  f ()  u1  (u 2  u1 )  sin
d (20)
 0


Integrals in Equations (17) and (19) can be evaluated when f(θ)
is specified which allows us to render them into algebraic
equations. They can be solved to obtain a unique solution
because the determinant matrix of coefficients is not zero.
a n /  a n /   a 
 
n / 
 n / 
b
b
b
n / 
a
 
b
 n / 
0
Numerical Example: Temperature distribution
1 T 1  2 T
 T

 2
0
2
2
r r r 
r
T(r,0)  0
(a  r  b)
2
 2T
T(r, )  0
T(a , )  0
T(b, )  100
(a  r  b)
(0    )
(0    )
(a  r  b, 0    )
Solution:
Using the general solution from previous derivation


 
n
T(r, )  T1  (T2  T1 )   Pn r n /   Q n r  n /  sin
 n 1

Apply boundary conditions in above equation


T(r, )   Pn r
n / 
n 1
 Qn r
 n / 

n
sin

Evaluate Pn and Qn by solving the algebraic equations below:
P a
n / 
n
 Qn a
 n / 

2
n



0
sin
d  0

0

and
P b
n
n / 
 Qn b
 n / 

2
n
200 
n
 400



100
sin
d


cos
1




0

 

n

P a
n / 
n
P b
n
n / 

 Qn a n /   0
 Q n b  n / 

400

n
Qn 
 b
 n / 
a
Solving above algebraic equations we get
Pn 
 b
a
 
 
n / 
n / 
 r
 a 
b
 a
n / 
n / 


n
 sin



 b
n / 
 a
a  n / 
 n / 
Final solution for temperature distribution in disk is
given by following equation
 r

400
1 a
T(r, ) 
 
 n 1n  b
 a
 a  n / 
 b
 a
n / 
20.3.2 Cylindrical coordinates
x  r cos 
z
r
y  r sin 
θ
zz
b
Z=0
Z=L
The laplacian in terms of polar coordinates where r, θ,z are
independent variables
2
2
2
1

1


2  2 
 2 2 2
r r r 
r
z
(1)
Applying operator to a potential function and set it to zero, ignoring the source term.
 2 
 2
r
2
1  1  2   2 

 2
 2 0
2
r r r 
z
(2)
Assuming solution of the form and applying method of separation of variables
  R (r)()Z(z)
(3)
Substituting equation (3) in equation (2) we get
   Z(z)()
2
 R (r )()
 2 R (r )
 2 Z(r )
z
2
r 2
0
Z(z)() R (r ) R (r ) Z(z)  2 ()


r
r
r2
 2
(4)
R (r )()Z(z) )
Simplify by dividing by
and use short hand notation for brevity
1  2 R 1 R
1  2 1  2 Z

 2

0
2
2
2
R r
Rr r r 
Z z
(5)
Or separating ‘Z’ term
1  2 R 1 R
1  2
1 2Z

 2

2
2
R r
Rr r r 
Z z 2
(6)
Equations for the three components of solution can be written as follows
For “Z” component
1 2Z
2

k
Z z 2
2Z
 2  k 2Z  0
z
(7.1)
For “θ” component
 2

For “R” component
2
 2R
r
2
 v 2  0
(7.2)
1 R  2 v 2 

 k  2 R0
r r 
r 
(7.3)
Equation (7.3) is also known as Bessele Equation
Solutions for equations (7.1), (7.2) is well known
Z(z)  e  kz
(8)
and
()  e
iv
(9)
Solution to equation (7.3) is little more complicated, we will it find case by case below
The solution to the equation that is regular at origin is
R  J v (kr )
Case 1: kr<<1
Case 1: kr>>1
(10)
The solution is in terms of bessel function… Jv(kr)
J v (kr)  (kr) v
J v (kr ) 
(11)
2
v  

cos kr 
 
kr
2 4

(12)
References
(1) http://planetmath.org/encyclopedia/LaplaceEquationInCylindricalCoordinates.html
(2) http://www.owlnet.rice.edu/~hill/phys532/L1b.pdf
(3) Arfken, George, Weber, Hans, Mathematical Physics. Academic Press, San Diego, 2001
Laplace equation solution for Axisymmetry problem in
cylindrical coordinates
 2u
1 u 1  2 u  2 u
 u 2 
 2
 2 0
2
r r r 
r
z
2
(1)
Assuming Axisymmetry we discard the term containing, “θ”
 2u
2
1

u

u
2u  2 
 2 0
r r z
r
 u0
2
Z=0
0
u=g(r)
u=f(r)
u=h(z)
(2)
u=g(r)
u=f(r)
0
=
 u1  0
2
Z=L
u=h(z)
2u 2  0
+
0
By superposition of solution as shown in figure we find solution in two
parts one for u1 and u2
(1) Solving the second part of the solution, u1
u1 (r, z)  R (r)Z(z)
(3)
Separating variables by substituting equation (3) in equation (2) gives
1
1
R Z  R Z  2 RZ  0
r
r
1
R   R 
Z
r

 constant   2
R
Z
These equation can be written as
1


R  R  2R  0
r
and
Z   2 Z  0
(4)
(5)
(6)
(7)
Solutions for the previous differentiation equations
 A  B ln r
R
CI 0 (kr )  DK 0 (kr )
k0
k0
(8)
k0
 E  Fz
Z
k0
G cos kz  H sin kz
(9)
I0 and K0 are modified bessel functions of the first and second kind,
respectively of zero order
The general solution is
u (r, z)  (A  B ln r )( E  FZ) 
(CI 0 (kr )  DK 0 (kr ))( G cos z  H sin z) (10)
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
Applying boundary conditions and simplifying the final solution is

u1 (b, z)  h (z)   Tn I 0 (
n 1
nb
nz
) sin
L
L
nz
 nb  2 L
Tn I 0 
dz

 h ( z) sin
L
L
L


0
Or
L
2
nz
Tn 
dz
 h ( z) sin
n

b
L

0
LI 0 

L


Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
(0  z  L)
(2) Solving the second part of the solution, u2
u 2 (r, z)  R (r) Z(z)
(1)
Separating variables by substituting equation and then writing the solution as done
previously…
 A  B ln r
R
CJ 0 (kr )  DY0 (kr )
k0
k0
(2)
E  Fz
k0

Z
k0
G cosh kz  H sinh kz
(3)
J0 and Y0 are modified bessel functions of the first and second kind,
respectively of zero order
u(r, z)  (A  B ln r)( E  FZ) 
(CJ 0 (kr )  DY0 (kr ))( G cosh z  H sinh z) (4)
Applying boundary conditions and simplifying the final solution is
r 
r 
r


u 2 (r, z)   J 0 (z n ) Sn cosh z n    Tn sinh z n 
b 
n 1
 b 
 b

Sn 
r

f
(
r
)
J
z

rdr
0
n
2
2 
b
b [J1 (z n )] 0

2
(5)
b
(6)
b
r
r
2
r

Sn cosh(z n )  Tn sinh(z n )  2
g
(
r
)
J
z

rdr
0
n
2 
b
b
b
b [J1 (z n )] 0

Final solution is obtained by solving equation (6) and equation (7) and substituting
equation (5)
The complete solution to the problem on ‘u’ is as below
u(r, z)  u1 (r, z)  u 2 (r, z)
(7)
20.3.3 Spherical coordinates
Laplace equation in spherical coordinates can be written as follows:
1    2  
1  
 
1  2 
 
 sin 
 
   2   
0
2
2 
  sin   
  sin   
   
2
r, , z   R (r )P()Q()
(1)
(2)
Substituting equation (2) and simplifying
  2 R 
1
 
P 
1
 2Q
 
  2
 sin    2
0
2
2
2
   Q sin  
 R      P sin   
1
(3)
Multiply with r2 sin2 (φ)
sin 2    2 R  sin   
P  1  2 Q
 
 
 sin   
0
2
R    
P  
  Q 
(4)
For “θ” component
 2Q

2
 m 2Q  0
(5)
This has a solution that is already known
Q()  e im
m  0,1,2,3.....
http://users.aber.ac.uk/ruw/teach/260/laplace.php
(6)
For “r” component
  2 R 
 
  L(L  1)R  0
   
(6)
For “φ” component
1  
P  
m2 
 sin    L(L  1)  2  P  0
sin   
  
sin  
(7)
Solutions to these are pretty difficult and we will limit discussion only to axisymmetric about the
polar axis z, that is where the variable is independent of θ
http://users.aber.ac.uk/ruw/teach/260/laplace.php
Assuming Axisymmetry for Polar axis
(c, )  f ()
Boundary condition
(0    )
(1)
This is analogous problem to the Direchlet problem in polar coordinates
The solution is in the form of
r, , z   R ()()
(2)
 2 R   2R 
  cot 

 constant  k 2
R

The ODE’s from equation (3)
Change of variables
2 R  2R  k 2 R  0
(4)
  cot   k 2  0
(5)
x  cos 
(6)
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
(3)
Change of variables to x gives us
(1  x )
2
d 2
dx
2
 2x
d
 L(L  1)  0
dx
(7)
Note that if x=-1, +1 are excluded from the problem “L” may be a non-integer.
The solution to equation (7) is the Legendre polynomial discussed in chapter 4, section 4.3
We will thus write the general solution to Laplace’s equation in spherical coordinates
for this case as follows:
1 

 ( r , )   A L r L  B L L 1 PL cos  
n 0 
r


(8)
The Legendre polynomials can be obtained from the following equation
(for details refer to section 4.3 in textbook)
PL ( x ) 
1
dL
2 L L! dx L
x
2

1
L
(9)
Rodrigues’ formula
The Legendre polynomials can also obtained from generating function given below:
F( x , ) 

1
(1  2 x   )
2 1/ 2
   L PL ( x )
(10)
L 0
Note: The polynomials form a complete, orthogonal set of function in the
domain of -1<=x<=1 or 0<=θ<=π

f ( x )   A L PL ( x )
L 0
(11.1)
Which gives us below equation (11.2) using Fourier Legendre
expansion of the function in equation (11.1)
AL
2L  1 1

 f ( x )PL ( x )dx
2 1
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
(11.2)
Dirichlet Problem in Spherical Coordinates:
Now using the given boundary conditions in equation (2) for dirichlet problem
1 

 ( r , )   A L r L  B L L 1 PL cos  
n 0 
r


(8)
In this equation we have BL =0 because of stipulated boundaries given.
Therefore from equation (8) we have the following final equation



(r, )   A L r L PL cos 
n 0

(11)
Use next boundary condition to solve for the constant AL



(r, )  f ()   A L r L PL cos 
n 0

(12)
We already obtained equations for constants in general solution in previous slides



(r, )  f ()   A L c L PL cos 
n 0

(12)
From equation 11.2 we can write AL as follows
2L  1 1
A Lc 
 f ( x )PL ( x )dx
2 1
L
AL 
2L  1
2c
L
(13)
1
 f ()PL (cos ) sin d
(14)
1
Equation (14) and Equation (14) form the final solution for the dirichlet problem for
spherical coordinates
Transform:
In mathematics, a function that results when a given function is
multiplied by a so-called kernel function, and the product is integrated
between suitable limits. (Britannica)

F(s)  Lf s    e st f ( t )dt.
0
Application condition
We can use a Fourier transform on x if the x
domain is    x  
Or a Fourier cosine or sine transform on x if
the domain is 0  x  
1. http://ase.tufts.edu/chemistry/sykes
2.Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
Fourier Transform
Example1
Fourier Transform of a Cosine f (t )  cos( 2st )

F (u ) 


f (t )e
i 2ut

dt   cos( 2st )e i 2ut dt

  cos( 2st )cos( 2ut )  i sin( 2ut )dt






  cos( 2st ) cos( 2ut )dt  i  cos( 2st ) sin( 2ut )dt




  cos( 2st ) cos( 2ut )dt  i  cos( 2st ) sin( 2ut )dt
1
1
  (u  s )   (u  s )
2
2
Fourier Transform
Example1:Dirichlet problem for Half Plane
 u ( x, y )  0 for
2
(  x  , 0  y  )
u ( x,0)  f ( x) for (  x  )
Solution by Fourier Transform
y
1.Apply the Fourier transform in x to the Laplace’s equation
(note: y is independent of x )
2. Use Operational Formula
3.Find the general solution of the resulting equation
www.faculty.kfupm.edu.sa/MATH/ffairag/math470_091/notes/6p9b.ppt
x
Fourier Transform
Solution
We assume that
F u xx  u yy   F 0
u ( x, y )  0asy  
F u xx   F u yy   0
get
lim uˆ ( , y )  lim
 2 u  i x
(i ) uˆ  
e dx  0
2
  y
2
2
d
  2u  2
dy


u ( x, y )e ix dx


y   
y 


u ( x, y )e ix dx
   lim u ( x, y )e ix dx  0

 
 y 

0
d uˆ
or 2   2 uˆ  0
dy
2
 y
 y
uˆ ( , y )  Ae  Be
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.):
Prentice Hall.
We need A=0,so
 y
uˆ ( , y )  Be
 uˆ
 fˆ ( )
y 0
 uˆ
y 0
 fˆ ( )  B

 uˆ ( , y )  fˆ ( )e
y
Fourier Transform
Solution
We obtain the final result

f ( )
u ( x, y )  
d   P(  x, y ) f ( )d
2
2




(x   )  y
y

Poisson kernel
This is the Poisson integral formula for the half plane
Figure: Poisson kernel P
Greenberg, M. D. (1998). Advanced Engineering Mathematics (2nd ed.): Prentice Hall.
Helpful References
• Greenberg, M. D. (1998). Advanced Engineering
Mathematics (2nd ed.): Prentice Hall.
• Kreyszig, E. (1998). Advanced Engineering
Mathematics (8th ed.): Wiley.
• http://tutorial.math.lamar.edu/Classes/DE/LaplacesEq
n.aspx
• http://users.aber.ac.uk/ruw/teach/260/laplace.php
End of chapter 20