How can I solve an integral equation with an unknown kernel?
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Sergio Manzetti
il 1 Dic 2017
Commentato: Sergio Manzetti
il 12 Dic 2017
The equation I am trying to solve is:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/170497/image.gif)
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
il 1 Dic 2017
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.
10 Commenti
Torsten
il 12 Dic 2017
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.
Più risposte (2)
John D'Errico
il 1 Dic 2017
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).
11 Commenti
John D'Errico
il 4 Dic 2017
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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