CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

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Partial Differential Equations in Two or More Dimensions

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.

CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

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

= f(x,y) (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 non-homogeneous 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) = A_nsin sinh + B_nsin sinh

C_nsin sinh + D_nsin sinh

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.

CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

Figure 3.9-2 A thin rectangular plate with all non-homogeneous boundary conditions.

= f(x,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) = E_mnsin sin (3.9-3)

The constants E_mnare to be determined by substituting (3.9-3) into the equation (3.9-2)

=  E_mn sin sin

 E_mn sin sin = f(x,y) (3.9-4)

Equation (3.9-4) is a double Fourier sine series expansion of f(x,y), therefore

E_mn =  sin( x) sin( y)dxdy

In this equation _mn =

Example 3.9-1. ----------------------------------------------------------------------------------

Solve the following equation in a 11 square (0 < x < 1, 0 < y < 1)

= u + 3

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) = E_mnsin sin

Since a = 1 and b = 1, we have

u(x,y) = E_mnsin(mx) sin(ny)

Substituting the trial function and its second derivative into the original PDE yields

 E_mn_mn sin(mx) sin(ny) = E_mnsin(mx) sin(ny) + 3

In this equation _mn = ²(m² + n²). Rearranging the equation yields

E_mn( 1  _mn)sin(mx) sin(ny) = 3

Therefore

E_mn( 1  _mn) = 4 sin(mx) sin(ny)dxdy

Evaluating the integral and then solving for E_mn, we obtain

E_mn =  ((1)^m  1)((1)ⁿ  1)

E_mn = 0 if m = even or n = even, otherwiseE_mn =  .

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, y) ___________

% 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

uxy=-48*uxy/(pi2);

mesh(X,Y,uxy)

CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

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.

CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

The results from Matlab are shown in Figure 3.9-4.

CHAPTER 3 39 THE METHOD OF EIGENFUNCTION EXPANSION

Figure 3.9-4 Matlab solution for ²u = u + 3 over a unit square.

CONFIGURING USER STATE MANAGEMENT FEATURES 73 CHAPTER 7 IMPLEMENTING
INTERPOLATION 41 CHAPTER 5 INTERPOLATION THIS CHAPTER SUMMARIZES POLYNOMIAL
PREPARING FOR PRODUCTION DEPLOYMENT 219 CHAPTER 4 DESIGNING A

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