Partial Differential Equations in Two or More Dimensions



7.2 FFT for Parabolic Equations

Example 7.2-1. ----------------------------------------------------------------------------------

Solve the heat equation [pic]= [pic] with the following boundary conditions

u(0,t) = 1 and u(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) = [pic] (7.2-1)

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

(n(x) = [pic]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) = [pic]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

[pic][pic]dx = [pic][pic]dx (7.2-3)

The left hand side becomes

[pic][pic]dx = [pic][pic]u(x,t)dx = [pic]

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

d(wv) = wdv + vdw

[pic]= [pic]( [pic]( [pic]= [pic]( [pic]

Let w = (n(x) ( [pic] = [pic]

dv = [pic]dx ( v = [pic]

[pic][pic]dx = (n(x) [pic] ( [pic][pic]dx

Applying the integration by parts again yields

[pic][pic]dx = (n(x) [pic] ( [pic] + [pic]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) [pic] = 0 and [pic] = [pic]. From the basis functions

(n(x) = [pic]sin(n(x) ,

[pic] = n([pic]cos(n(x) ( [pic] = ( (n()2[pic] sin(n(x) = ( (n()2(n(x)

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

[pic][pic]dx = [pic]( [pic]udx

[pic][pic]dx = n([pic]cos(n(x)|x=0 ( (n()2[pic] u(x,y)dx

[pic][pic]dx = n([pic]cos(0) ( (n()2(n(t)

The original parabolic PDE becomes

[pic] = n([pic] ( (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

[pic] = [pic]

( [pic]ln[pic] = t

1 ( [pic]n((n = exp[((n()2t]

(n = [pic][pic]

The solution is then

u(x,t) = [pic]= [pic][pic][pic]sin(n(x)

u(x,t) = 2[pic][pic]

Since 1 ( x = 2[pic]

The final form of the solution is

u(x,t) = 1 ( x ( 2[pic][pic]

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 7.2-1 Matlab program to plot 1 ( x ( 2[pic][pic] ___________

% 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)')

-------------------------------------------------------------------------------------------------

[pic]

Figure 7.2-1 Temperature profile at various times.

Example 7.2-2. ----------------------------------------------------------------------------------

Solve the heat equation [pic]= [pic] with the following boundary conditions

[pic](0,t) = ( 1 and [pic](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) = [pic] (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) = [pic]cos(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) = [pic]u(x,t)dx (7.2-6)

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

[pic][pic]dx = [pic][pic]dx (7.2-7)

The left hand side becomes

[pic][pic]dx = [pic][pic]u(x,t)dx = [pic]

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

d(wv) = wdv + vdw

[pic]= [pic]( [pic]( [pic]= [pic]( [pic]

Let w = (n(x) ( [pic] = [pic]

dv = [pic]dx ( v = [pic]

[pic][pic]dx = (n(x) [pic] ( [pic][pic]dx

Applying the integration by parts again yields

[pic][pic]dx = (n(x) [pic] ( [pic] + [pic]udx

From the basis functions

(n(x) = [pic]cos(n(x) ,

[pic] = ( n([pic]sin(n(x) ( [pic] = (n()2[pic]cos(n(x) = ( (n()2(n(x)

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

[pic][pic]dx = ( (n(x) [pic] ( [pic]udx

[pic][pic]dx = (n(0) ( (n()2[pic] u(x,y)dx = (n(0) ( (n()2(n

For n = 0

[pic][pic]dx = 1

Equation (7.2-7) can be integrated

[pic][pic]dx = [pic][pic]dx (7.2-7)

[pic]= 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

[pic] = (n(0) ( (n()2(n = [pic]cos(n(x) ( (n()2(n (7.2-8)

[pic] = [pic] ( (n()2(n

Equation (7.2-8) can be integrated as follow

[pic] = [pic]

( [pic]ln[pic] = t

1 ( [pic](n()2(n = exp[((n()2t] ( (n = [pic][pic]

The solution is then

u(x,t) = [pic]= t + [pic][pic][pic]cos(n(x)

u(x,t) = t + [pic][pic]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 7.2-2 Matlab program to plot t + [pic][pic]cos(n(x) ___________

% Example 7.2-2

clf

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

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

npi=n*pi;

cosnx=cos(npi'*x);

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

nt=length(t);

hold on

for i=1:nt

ti=t(i);

expn=(1-exp(conn*ti))./npi.^2;

theta=ti+2*expn*cosnx;

plot(x,theta)

grid on

end

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

-------------------------------------------------------------------------------------------------

[pic]

Figure 7.2-2 Temperature profile at various times.

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