Plotting Solutions to the Heat Equation Through a Truncated Series Evaluation

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Mark Fasano
Mark Fasano on 3 Nov 2021
Commented: Yongjian Feng on 3 Nov 2021
I have a solution to the heat equation (a function of two variables $x$ and $t$) which is in the form of an infinite series. I want to know what the solution actually looks like at specific times. My solution is:
where the coefficients are given by
and the function is known and given at the bottom. I am having problems with my Matlab code when I try to actually calculate these things however. What I have right now is:
h = .01;
xVec = 0:h:5;
for n = 1:100
h_FC_integrand = @(x,s) (x.*(10-x).*sin((pi*n.*x)/5) ./ (1+ ...
(x-5 * (1 + sin(2*pi.*s)./(s+1))).^2));
h_FC = @(s) ((2/5) * exp(((n*pi)/5)^2 * s) * ...
Cn = quadgk(h_FC,0,.01);
sinVec = sin((pi*n*xVec)/5);
solutionVec1 = solutionVec1 + (Cn * exp(-((n*pi)/5)^2 * .01) ...
The problem is in the line where I try to evaluate for . Matlab is giving me the error "Not enough input arguments." but I don't see what the problem is. The function handle h_FC is of only one variable so I don't need to specify which variable I want to integrate in and I give both the endpoints of integration. Any help would be much appreciated!
For reference:
Also, this same basic idea worked quite easily when the coefficients had no time dependence (I tried essentially this exact same method for a different problem and had no issues).

Answers (1)

Yongjian Feng
Yongjian Feng on 3 Nov 2021
The error seems to be caused by this line:
quadgk computes numerical integration. What do you expect from this please?
Yongjian Feng
Yongjian Feng on 3 Nov 2021
If you are doing symbolic integration of h_FC_integrand, maybe you can take it out of the for loop? So it becomes a function of t and n, instead of a funciton of t. Then inside the for loop, you can substitute n.
It seems to be the int function can be called outside the for loop.

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