How to numerically evaluate this singular integral in integral2?
이전 댓글 표시
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!
댓글 수: 5
Ryan
2024년 9월 5일
format long g
a = 1;
b = 2;
c = 3;
k = @(x,y) 1./(sqrt(x.^2/a^2+y.^2/b^2)-c);
N = 20000;
M = 10;
Areas = zeros(1,M);
for K = 1 : M
x = sort(rand(1,N) * (10-0) + 0) .';
y = sort(rand(1,N) * (10-(-10)) + (-10));
f = k(x, y);
Areas(K) = trapz(y, trapz(x, f));
end
Areas
Random sampling is showing a wide range of values. It is unlikely that the integral converges.
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
2024년 9월 6일
답변 (1개)
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)
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)
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