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

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Sergio Manzetti 2017-12-1

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评论： Sergio Manzetti 2017-12-12

采纳的回答： Torsten

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 2017-12-1

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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

g(x)=1/(f(x)*h(x)*2*pi)

...

Best wishes

Torsten.

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Torsten 2017-12-12

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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 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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John D'Errico 2017-12-1

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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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Sergio Manzetti 2017-12-1

编辑：Sergio Manzetti 2017-12-1

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Thanks, I think the solution is not trivial, because the integral of h and f is:

W =

(150161193738355954586222597234585017169743443289080649840382803820382539880280999645654803891496677831582301321288235091356610775513114160448451324201927488546281633057*pi)/235872651551346334515097161311766805682307576276027526773512893765351510149509303255382920323603672971341773439422740306235661539370299539535089640714543502728087207936 - (857667986409676946755091410422420433765979555592970194729842043977566558535980733470902042660694127536712947665701541348018841649893734864216952836677423460461115319701245*sin((140170517671301*pi)/616503364953199))/27309427497697258096319858537523969165745773917815361139146936050648948950585748747769209155925329298757330388892969600751474273592254457836547309903960103405916641989001029353472 + (2731497041226781970742487761996229018221209713*pi*cos((140170517671301*pi)/616503364953199))/615656346818663737691860001564743965704370926101022604186692084441339402679643915803347910232576806887603562348544 + (13496744475957601245584225453924415705057921\\\n334313386459527769*pi^3)/6373655913801205184275058478629623527955117412724383078393329658404842718330226912869950027182751428026671953834958012184158952051832486872914239745199048223325975250024718628380737536

However, if I use "double(W) it is simply:

ans =

1.99999997942694e+000

and the reciprocal is:

500.000005143264e-003

and I have no information of the functional of g, except that it should be in the form e^(-i2*pi*x) and some constant. The absolute best way would be to solve g as an operator, and not as a function. That would solve this on a general level.

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.

Sergio Manzetti 2017-12-5

Thanks I will give it another look.

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Sergio Manzetti 2017-12-5

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编辑：Sergio Manzetti 2017-12-5

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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;