Plotting Solutions to the Heat Equation Through a Truncated Series Evaluation

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Mark Fasano 2021-11-3

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评论： Yongjian Feng 2021-11-3

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

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) * ...

quadgk(h_FC_integrand,x,0,5));

Cn = quadgk(h_FC,0,.01);

sinVec = sin((pi*n*xVec)/5);

solutionVec1 = solutionVec1 + (Cn * exp(-((n*pi)/5)^2 * .01) ...

.*sinVec);

end

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).

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Answer 1

Yongjian Feng 2021-11-3

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1578355-plotting-solutions-to-the-heat-equation-through-a-truncated-series-evaluation#answer_823185

在 MATLAB Online 中打开

The error seems to be caused by this line:

quadgk(h_FC_integrand,x,0,5));

quadgk computes numerical integration. What do you expect from this please?

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Mark Fasano 2021-11-3

This significantly slowed my code. Changing h to be .1 and lowering n to only range from 1 to 10 now takes a lot of time but I don't really see why. Any insights? Ideally I'd want my space discretization to stay the same so the plot remains smooth

Yongjian Feng 2021-11-3

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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