Chapter 3
3.9 The Method of Eigenfunction Expansion
We have solved the linear homogeneous PDE by the method of separation of variables.
However this method cannot be used directly to solve nonhomogeneous PDE.
top surface
y
b
bottom surface
a
x
0
Figure 3.9-1 A thin rectangular plate with insulated top and bottom surfaces
The two-dimensional steady state heat equation for a thin rectangular plate with time
independent heat source shown in Figure 3.9-1 is the Poisson’s equation
 2u  2u
= f(x,y)

x 2 y 2
(3.9-1)
The heat equation for this case has the following boundary conditions
u(0,y) = g1(y),
u(a,y) = g2(y),
0<y<b
u(x,0) = f1(x),
u(x,b) = f2(x),
0<x<a
The original problem with function u is decomposed into two sub-problems with new
functions u1 and u2. The boundary conditions for the sub-problems are shown in Figure 3.9-2.
The function u1 is the solution of Poisson’s equation with all homogeneous boundary
conditions and the function u2 is the solution to Laplace’s equation with all nonhomogeneous boundary conditions. The original function u is related to the new functions by
u = u1 + u2
The function u2 is already evaluated in section 3.8 where

u2(x,y) =

n 1


n 1
 n 
 n

x  sinh 
(b  y )  +
Ansin 
 a 
 a

 n 
 n

y  sinh 
(a  x )  +
Cnsin 
 b 
 b

3-57


n 1


n 1
 n 
 n 
x  sinh 
y
Bnsin 
 a 
 a 
 n 
 n 
y  sinh 
x
Dnsin 
 b 
 b 
To complete the solution of the Poisson’s equation for the problem in Figure 3.9-1, we only
need to treat Poisson’s equation with zero boundary condition shown in Figure 3.9-2.
top surface
y
u=f2(x)
b
2
u=g1(y)
u=f(x,y)
u=f1(x)
0
u=g2(y)
bottom surface
a
x
u1=0
u2=f2(x)
2
u1=f(x,y)
u1=0
u1=0 u2=g1(y)
2
u2=0
u2=g2(y)
u1=0
u2=f1(x)
Figure 3.9-2 A thin rectangular plate with all non-homogeneous boundary conditions.
 2 u1  2 u1
 2 = f(x,y)
x 2
y
(3.9-2)
The Poisson’s equation for this case has the following boundary conditions
u(0,y) = 0,
u(a,y) = 0,
0<y<b
u(x,0) = 0,
u(x,b) = 0,
0<x<a
Since the solution in any direction x or y with homogenous solution is the sin function, we try
the following function that satisfies the zero boundary conditions
u1(x,y) =


n 1
m 1
 
 m 
 n 
x  sin 
y
Emnsin 
 b 
 a 
(3.9-3)
The constants Emn are to be determined by substituting (3.9-3) into the equation (3.9-2)
 2 u1
=
x 2
 2 u1
=
y 2


n 1
m 1


n 1
m 1
 
 
 m
 m 
Emn 
 sin 
 a 
 a
2

 n 
x  sin 
y
 b 

 n 
 m 
 n 
x  sin 
y
Emn 
 sin 
 b 
 b 
 a 
2
3-58



n 1
m 1
 
 m  2  n  2   m 
 n 
x  sin 
y  = f(x,y)
Emn 
 
  sin 
 b 
 a   b    a 
(3.9-4)
Equation (3.9-4) is a double Fourier sine series expansion of f(x,y), therefore
Emn = 
4
abmn
b
a
0
0

f ( x, y ) sin(
m
n
x) sin(
y)dxdy
a
b
 m  2  n  2 
In this equation mn = 
 
 
 a   b  
Example 3.9-1. ---------------------------------------------------------------------------------Solve the following equation in a 11 square (0 < x < 1, 0 < y < 1)
 2u  2u
=u+3

x 2 y 2
with the following boundary conditions
u(0,y) = 0,
u(1,y) = 0,
0<y<1
u(x,0) = 0,
u(x,1) = 0,
0<x<1
Solution -----------------------------------------------------------------------------------------We assume a trial function of the form
u(x,y) =


n 1
m 1
 
 m
Emnsin 
 a

 n 
x  sin 
y
 b 

Since a = 1 and b = 1, we have
u(x,y) =


n 1
m 1
 
Emnsin(mx) sin(ny)
Substituting the trial function and its second derivative into the original PDE yields



n 1
m 1
 
Emnmn sin(mx) sin(ny) =


n 1
m 1
 
Emnsin(mx) sin(ny) + 3
In this equation mn = 2(m2 + n2). Rearranging the equation yields
3-59


n 1
m 1
 
Emn( 1  mn)sin(mx) sin(ny) = 3
Therefore
1 1
Emn( 1  mn) = 4   3 sin(mx) sin(ny)dxdy
0 0
Evaluating the integral and then solving for Emn, we obtain
Emn = 
1
12
((1)m  1)((1)n  1)
2
1  mn mn
Emn = 0 if m = even or n = even, otherwise Emn = 
1
48
.
1  mn mn 2
A Matlab program is listed in Table 3.9-1 to plot u(x,y,) as shown in Figure 3.9-3.
__________ Table 3.9-1 Matlab program to plot u(x,
% Eigenfunction expansion method
%
[X,Y]=meshgrid(0:.02:1);
n1=length(X);
uxy=zeros(n1,n1);
t=1;pi2=pi*pi;
for n=0:4
for m=0:4
np=2*n+1;mp=2*m+1;
lamda=(np*np+mp*mp)*pi2;
Emn=1/(np*mp*(1+lamda));
uxy=sin(np*pi*X).*sin(mp*pi*Y)*Emn+uxy;
end
end
uxy=-48*uxy/(pi2);
mesh(X,Y,uxy)
3-60
y) ___________
0
-0.05
-0.1
-0.15
-0.2
-0.25
1
0.8
1
0.6
0.8
0.6
0.4
0.4
0.2
0.2
0
0
Figure 3.9-3 The solution surface for 2u = u + 3 over a unit square.
Equation (2u = u + 3) can also be solved using pdetool with the following PDE
specification.
The results from Matlab are shown in Figure 3.9-4.
3-61
Color: u Height: u
0
-0.02
0
-0.04
-0.05
-0.06
-0.08
-0.1
-0.1
-0.15
-0.12
-0.2
-0.14
-0.25
1
-0.16
0.8
1
0.6
0.8
-0.18
0.6
0.4
0.4
0.2
0.2
0
0
Figure 3.9-4 Matlab solution for 2u = u + 3 over a unit square.
3-62
-0.2