How to numerically evaluate this singular integral in integral2?

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Ryan 2024-9-4，13:53

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评论： Ryan 2024-9-6，10:18

Hello,

I am trying to numerically evaluate a 2D integral with an integrand singular along an elliptical curve, please see the simple code below

a = 1;
b = 2;
c = 3;
k = @(x,y) 1./(sqrt(x.^2/a^2+y.^2/b^2)-c);
h = integral2(@(x,y) k(x,y) ,0,10,-10,10);

This yields warnings that "the result fails the global error test" which is evidently because the integrand is singular when

I am curious as to how this integral can be restated to make a numerical integration tractable?

Thanks!

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Matt J 2024-9-5，23:17

编辑：Matt J 2024-9-5，23:43

Let us make the change of variables,

Then the integral becomes,

Now, converting to polar coordinates,

which is obviously a non-convergent integral if and only if

for any

. Since the region of integration is a rectangle of width 10/b and height 10/a, that will happen if c<norm([10/a,10/b]), which it is in this case.

Ryan 2024-9-6，10:18

Thank you Walter and Matt, I was hoping there was a way to render it numerically integratable but as demonstrated it appears that it isn't convergent to begin with.

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回答（1 个）

Matt J 2024-9-4，18:08

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https://ww2.mathworks.cn/matlabcentral/answers/2150089-how-to-numerically-evaluate-this-singular-integral-in-integral2#answer_1510839

编辑：Matt J 2024-9-5，23:24

在 MATLAB Online 中打开

As demonstrated in the comment above, the integral is theoretically non-convergent unless c>=norm([10/a,10/b]). Here is a further, numerical test:

a = 1;
b = 2;
c0 = norm([10/a,10/b]); %critical threshold

c=c0*1.0001; %slightly greater than threshold
k = @(x,y) 1./(sqrt(x.^2/a^2+y.^2/b^2)-c);
h = integral2(@(x,y) k(x,y) ,0,10,-10,10)
h = -60.2647

c=c0*0.9999;%slightly less than threshold
k = @(x,y) 1./(sqrt(x.^2/a^2+y.^2/b^2)-c);
h = integral2(@(x,y) k(x,y) ,0,10,-10,10)
Warning: Reached the maximum number of function evaluations (10000). The result fails the global error test.
h = -60.4654