MATLAB Answers

Problem in 2 dimensional numerical integration

3 views (last 30 days)
Ashok Das
Ashok Das on 30 Jan 2020
I am trying to integrate a function of the form
, (where H is the heaviside function)
for the domain , where and are fixed constants, i.e.,
= ??
For this I have used the code:
y1 = 10; y2 = 20;
B = @(x1,x2) ( y1.*del(x1-y1).*heaviside(y2-y1) + y2.*del(x2-y2).*heaviside(y1-y2) )./(y1.*y2);
integral2(B,0,y1,0,y2)
where 'del' is the dirac delta function defined below:
%% definiiton of del
function retval = del(x)
if x==0
retval = 1;
else
retval = 0;
end
return
Now if I run the main code, it gives the following error:
Error using integral2Calc>integral2t/tensor (line 241)
Integrand output size does not match the input size.
Error in integral2Calc>integral2t (line 55)
[Qsub,esub] = tensor(thetaL,thetaR,phiB,phiT);
Error in integral2Calc (line 9)
[q,errbnd] = integral2t(fun,xmin,xmax,ymin,ymax,optionstruct);
Error in integral2 (line 106)
Q = integral2Calc(fun,xmin,xmax,yminfun,ymaxfun,opstruct);
Error in test (line 4)
integral2(B,0,y1,0,y2)
Any suggestions?

  0 Comments

Sign in to comment.

Answers (1)

David Goodmanson
David Goodmanson on 31 Jan 2020
Hi Ashok,
you can make a decent case that the answer is 1/2, at least if the variables are continuous. For the first term,
Int{0,y1} delta(x1-y1) dx1 = 1/2
Int{0,y2} dx2 = y2
so the first term is
(1/2) H(y2-y1)
Likewise, the second term is
(1/2) H(y1-y2)
and the whole thing is
(1/2)( H(y2-y1)+H(y1-y2)) = 1/2

  4 Comments

Show 1 older comment
David Goodmanson
David Goodmanson on 31 Jan 2020
Hi Ashok,
I was maintaining that the answer here is 1/2, not 1. However, I appreciate what you are saying about this just being a simple example of a more general problem that you want to solve. I will probably delete my answer but I believe that will also delete your comment, so in the meantime I would like to inquire, are the variables discrete by intent, or are discrete variables being used to approximate a continuous variable situation?
Ashok Das
Ashok Das on 1 Feb 2020
Hi David,
The variables are continuous by nature.
David Goodmanson
David Goodmanson on 1 Feb 2020
If that's the case, can't you basically do the delta function part of the integrals by hand first, or are there too many delta functions in your expressions for that to be feasible? (and I still like 1/2)

Sign in to comment.

Sign in to answer this question.

Tags