Partial Differential Equations in Two Or More Dimensions

Chapter 7

7.2 FFT for Parabolic Equations

Example 7.2-1. ------

Solve the heat equation = with the following boundary conditions

u(0,t) = 1andu(1,t) = 0, for t > 0

The initial temperature distribution is u(x,0) = 0 for 0 < x < 1.

Solution ------

Assume that the solution can be written in a series expansion as

u(x,t) = (7.2-1)

In this equation, n(x) are the basis functions that can be obtained from Table 7-1, case I.

n(x) = sin(nx) ,n = 1, 2, …

In a one-dimensional transient problem the basis functions cannot involve time because the differential equation has only a first derivative in t; there is no way to formulate an eigenvalue problem in t. For this problem, the finite Fourier transform (FFT) of the temperature is defined as

n(t) = u(x,t)dx(7.2-2)

In this equation n(t) is defined as the transformed temperature. It is an integral transform in which the original function is multiplied by a kernel or basis function, and the result integrated over the finite interval 0 x  1. The limits of integration correspond to the range of x in the problem; these limits will be modified for different boundary conditions of x.

The function n(t) may be obtained by applying the FFT to both sides of the Laplace equation

dx = dx(7.2-3)

The left hand side becomes

dx = u(x,t)dx =

The right hand side of equation (7.2-3) is rearranged using integration by parts

d(wv) = wdv + vdw

= = 

Let w = n(x)  =

dv = dxv =

dx = n(x) dx

Applying the integration by parts again yields

dx = n(x)  + udx

From the boundary conditions

n(x = 0) = 0, n(x = 1) = 0 and u(0,t) = 1, u(1,t) = 0

Therefore n(x) = 0 and = . From the basis functions

n(x) = sin(nx) ,

= ncos(nx)  =  (n)2 sin(nx) =  (n)2n(x)

The right hand side of equation (7.2-3) becomes

dx = udx

dx = ncos(nx)|x=0 (n)2 u(x,y)dx

dx = ncos(0) (n)2n(t)

The original parabolic PDE becomes

= n (n)2n(7.2-4)

The initial temperature distribution is u(x,0) = 0, therefore n(0) = 0. Equation (7.2-4) can be integrated as follow

ln = t

1 nn = exp[(n)2t]

n =

The solution is then

u(x,t) = = sin(nx)

u(x,t) = 2

Since1 x = 2

The final form of the solution is

u(x,t) = 1 x  2

Table 7.2-1 lists the Matlab program used to plot the temperature u(x,t) at various time using ten partial sum. Figure 7.2-1 presents the temperature profile at t = 0.01, 0.02, 0.05, 0.1, and 0.2.

______Table 7.2-1Matlab program to plot 1 x  2______

% Example 7.2-1

clf

x=0:.05:1;

n=1:10;

conn=-(n*pi).^2;

npi=n*pi;

sinnx=sin(npi'*x);

t=[0.01 .02 .05 .1 .2];

nt=length(t);

hold on

for i=1:nt

ti=t(i);

expn=exp(conn*ti)./npi;

theta=1-x-2*expn*sinnx;

plot(x,theta)

grid on

end

xlabel('x');ylabel('u(x,t)')

------

Figure 7.2-1 Temperature profile at various times.

Example 7.2-2. ------

Solve the heat equation = with the following boundary conditions

(0,t) =  1and(1,t) = 0, for t > 0

The initial temperature distribution is u(x,0) = 0 for 0 < x < 1.

Solution ------

Assume that the solution can be written in a series expansion as

u(x,t) = (7.2-5)

In this equation, n(x) are the basis functions that can be obtained from Table 7-1, case IV.

n(0) = 1, n(x) = cos(nx) ,n = 1, 2, …

n(t) = u(x,t)dx(7.2-6)

The function n(t) may be obtained by applying the FFT to both sides of the Laplace equation

dx = dx(7.2-7)

The left hand side becomes

dx = u(x,t)dx =

The right hand side of equation (7.2-7) is rearranged using integration by parts

d(wv) = wdv + vdw

= = 

Let w = n(x)  =

dv = dxv =

dx = n(x) dx

Applying the integration by parts again yields

dx = n(x)  + udx

From the basis functions

n(x) = cos(nx) ,

= nsin(nx)  = (n)2cos(nx) =  (n)2n(x)

Therefore = = 0. From the boundary condition (1,t) = 0 and (0,t) =  1, the right hand side of equation (7.2-7) becomes

dx = n(x) udx

dx =n(0)  (n)2 u(x,y)dx = n(0)  (n)2n

For n = 0

dx = 1

Equation (7.2-7) can be integrated

dx = dx(7.2-7)

= 1

The initial temperature distribution is u(x,0) = 0, therefore n(0) = 0 and we have

0(t) = t

For n > 0, the original parabolic PDE becomes

= n(0)  (n)2n = cos(nx)  (n)2n(7.2-8)

=  (n)2n

Equation (7.2-8) can be integrated as follow

ln = t

1 (n)2n = exp[(n)2t] n =

The solution is then

u(x,t) = = t + cos(nx)

u(x,t) = t + cos(nx)

Table 7.2-2 lists the Matlab program used to plot the temperature u(x,t) at various time using ten partial sum. Figure 7.2-2 presents the temperature profile at t = 0.01, 0.02, 0.05, 0.1, and 0.2.

______Table 7.2-2Matlab program to plot t + cos(nx) ______

% Example 7.2-2

clf

x=0:.02:1; n=1:10;