Integral equation numerical solution with NDSolve

Question

I'm trying to solve something like: f[x] == Integrate[f[x]*g[x]] where g[x] is known and f[x] is the function to solve (numerically is fine).

I've seen some examples of integral equations but wonder if there is a direct way to solve it with Mathematica.

Answer 1

If you want to solve the Fredholm equation of the second kind which is an integral equation of the form

$$f(x) - \lambda\int_{a}^{b} K(x,y)f(y)dy = g(x), \quad \forall x \in [a, b]$$ you can use the following code. We follow the simplistic source. One can make it much better by considering further implementation aspects as well as injecting more state of the art algorithmic antibiotic!!

Code:

Options[FredholmKind2] = {Method -> Automatic};
FredholmKind2[{a_, b_, lambda_, k_, g_}, n_?IntegerQ,OptionsPattern[]] := 
Block[{step, SI, GI, KMatrix, W, DMatrix, f, deltaX, delta},
     step = (b - a)/n;
     SI = Range[a, b, step];
     GI = g /@ SI;
     KMatrix = Outer[k, SI, SI];
     W = {step/2}~Join~ConstantArray[step, n - 1]~Join~{step/2};
     DMatrix = DiagonalMatrix[W];
     f = If[OptionValue[Method] === NIntegrate,
            deltaX[x_?NumericQ] := W . (k[x, #] & /@ SI) - 
            NIntegrate[k[x, y], {y, a, b},AccuracyGoal -> 4];
            (*If the integral is expensive ParallelMap is an option here *)
            delta = deltaX /@ SI;
            Interpolation[
             Transpose@{SI, 
             LinearSolve[IdentityMatrix[n + 1] + lambda*(DiagonalMatrix[delta] -
                          KMatrix . DMatrix), GI]}]
             ,
            Interpolation[
             Transpose@{SI, 
             LinearSolve[IdentityMatrix[n + 1] - lambda*(KMatrix . DMatrix),GI]}]
          ];
      f
    ]

Testing:

Now lets test it for the following equation which has an exact solution Sin[x]!! $$f(x) - \frac{4}{\pi}\int_{0}^{\frac{\pi}{2}} \cos(x-y)f(y)dy = -\frac{2}{\pi}\cos(x), \quad \forall x \in \left[0, \frac{\pi}{2}\right]$$ There are two separate methods available called Automatic and NIntegrate. The second method is computationally more expensive but produces better result. The above function returns the solution f as an InterpolatingFunction which you can later use an ordinary function in MMA.

n=90;(*number of discretization*)
a = 0.;
b = 0.5*Pi;
lambda = 4./Pi;
Kpart[x_, y_] := Cos[x - y];
Gpart[x_] := -2. Cos[x]/Pi;
f1 = FredholmKind2[{a,b,lambda,Kpart, Gpart},n,Method -> Automatic];
f2 = FredholmKind2[{a,b,lambda,Kpart, Gpart},n,Method -> NIntegrate];
Needs["PlotLegends`"];
Plot[Evaluate@Sqrt@((Sin[x] - #)^2 & /@ {f1[x], f2[x]}), {x, a, b},Frame -> True,
Axes -> False,PlotLegend -> {"Automatic", "NIntegrate"},LegendPosition -> {1.1, -0.4}]

In the above plot one can see how accurate the numerical solutions perform w.r.t the exact solution.

To Do!

Now if you want to solve the Fredholm integral equation of the first kind which looks like

$$\int_{a}^{b} K(x,y) f(y) dy = g(x), \quad \forall x \in [a,b],$$

I suggest that you take look at this at page 213. After the above example it will be easy for you to implement the solution for the problem of first kind.

BR

Answer 2

I was not going to post this because in a previous edit (now gone) you showed a very difficult problem.

Anyway, the current question could be answered as follows:

You want to find f[x] satisfying

f[x] == Integrate[f[x] g[x], x] 

for a known g[x].

Differentiating:

f'[x] == g[x] f[x]  

So for example

g[x_] := Sin@x^2;
fs = DSolve[f'[x] == g[x] f[x] && f[0] == 1, f, x]  
(*
 ->{{f -> Function[{x}, E^(x/2 - 1/4 Sin[2 x])]}}
*) 

Testing it

Integrate[g[u] (f /. fs[[1]])[u], u]  
(*
-> E^(u/2 - 1/4 Sin[2 u])
*)