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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