How can I solve an integral equation with an unknown kernel?

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Sergio Manzetti
Sergio Manzetti 2017-12-1
评论： Sergio Manzetti 2017-12-12
The equation I am trying to solve is:
where f(x) and h(x) are both complex and known, and g(x) is an unknown function. Presumably, the result should be a function g(x), however, it is not to be excluded that g(x) could actually be an operator instead. Can this be solved for either cases in MATLAB?
Thanks!

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Torsten
Torsten 2017-12-1
g is not unique - it can be of any function type you like (we already had this discussion).
g(x)=1/integral_{x=0}^{x=2*pi} f(x)*h(x)dx
or
g(x)=1/(f(x)*h(x)*2*pi)
or
...
Best wishes
Torsten.
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Torsten
Torsten 2017-12-12
Why don't you start from a solution that worked ?
syms L C x
assume (L>0);
h = 1;
g = 5;
y = C-exp(2*g*1i*x/h);
z = C-exp(-2*g*1i*x/h);
prod = y*z*(1+x^2);
Csol = solve(int(prod,x,0,L)-1==0,C);
Best wishes
Torsten.
Sergio Manzetti
Sergio Manzetti 2017-12-12
Dear Torsten, I did! However, you used a different solution, with "assume" in it I will try it now!
Thanks!

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更多回答（2 个）

John D'Errico
John D'Errico 2017-12-1
If g(x) is unknown, then if all you have is a single equation equal to a constant, then there is no simple solution. Or, you can look at it as if there are infinitely many solutions, one of them being a constant function.
Just compute the integral of h(x)*f(x). Take the reciprocal. That is the value of the constant g that will make int(h*f*g) equal 1. So as long as int(h*f) over [0,2*pi] is not identically 0, then A solution is trivial. Yes there may be infinitely many other solutions, but they cannot be found unless you have information as to the functional form of g(x).
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John D'Errico
John D'Errico 2017-12-4
I don't see why not. Integration is just a linear operator. If g is a constant, then it can be pulled outside the integral.

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Sergio Manzetti
Sergio Manzetti 2017-12-5
编辑：Sergio Manzetti 2017-12-5
Hi John, I found out that the non-hermitian nature of those functions makes them require a kernel, and not a constant, it appears to be (1+x^2). So I made this version to find the normalization constant when the integral is unity:
if true
% code
end
syms h g x C
h = 1
g = 5
y=@(x)(C - (exp(2.*g.*1i.*x./h));
z=@(x)(C - (exp(-2.*g.*1i.*x./h));
prod=@(x)y(x).*z(x)*(1+x^2);
W= integral(prod,0,2*pi)==1;
and I should get the result for "C", however that is simply 0. It looks incorrect, or is the W part correctly written?
Thanks!

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