How to numerically evaluate this singular integral in integral2?

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Ryan
Ryan 2024 年 9 月 4 日
コメント済み: Ryan 2024 年 9 月 6 日
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!
  5 件のコメント
Matt J
Matt J 2024 年 9 月 5 日
編集済み: Matt J 2024 年 9 月 5 日
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
Ryan 2024 年 9 月 6 日
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 件)

Matt J
Matt J 2024 年 9 月 4 日
編集済み: Matt J 2024 年 9 月 5 日
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
  1 件のコメント
Ryan
Ryan 2024 年 9 月 6 日
Thanks for that, this is an interesting insight into the behaviour of the integral

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