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I am encountering difficulties while trying to numerically solve an equation in Mathematica that involves an unknown function and an integral. Specifically, the equation takes the form of \begin{equation} f(z)=g(z)+\int_{z}^1\mathrm{d}y\,g\left(1-\frac{z}{y}\right)\frac{1}{y}f(y),\quad z\in(0,1), \end{equation} where $g(z)$ is a known function, and $f(z)$ is the unknown function I am attempting to solve for. It requires for any $g(z)$, you should have a corresponding solution for $f(z)$.

I tried a specific $g(z)$:

g[z_] := z
equation[z_] := f[z] == g[z] + Integrate[g[1 - z/y]/y*f[y], {y, z, 1}]
solution = NSolve[{equation[z]}, f, {z, 0, 1}]
result[z_] := f[z] /. solution[[1]] 

But it didn't work.

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    $\begingroup$ You are looking for a numerical solution for unknown function g[z]? Isn't that a contradiction? $\endgroup$ Nov 13, 2023 at 8:13
  • $\begingroup$ No, I want to solve f[z], and g[z] is a known function, however I don’t know how g[z] looks like. So I just set it as an unknown function. I know it seems wired but this is the problem. $\endgroup$
    – dcmpsr
    Nov 13, 2023 at 8:45
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    $\begingroup$ g[z] is an undefined function and you're looking for a numerical solution? That's the contradiction! $\endgroup$ Nov 13, 2023 at 8:47
  • $\begingroup$ Yes, you are right. I reread the question and made an edit on my question. $\endgroup$
    – dcmpsr
    Nov 13, 2023 at 9:25
  • $\begingroup$ Ok, but be aware that the function domain of g[y] is -Infinity<y<0! $\endgroup$ Nov 13, 2023 at 9:28

1 Answer 1

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Although the general problem in the question probably requires the methods referenced by flinty in 285076, the specific problem contained in the question's code, g[z] = z can be solved as follows. Write the integral equation as

eq = f[z] - z - Integrate[(1 - z/y)/y*f[y], {y, z, 1}]; 

Now twice differentiate eq

deq = D[eq, {z, 2}]
(* -f[z]/z^2 + f''[z] *)

construct boundary conditions

bc0 = eq /. z -> 1
{* -1 + f[1] *}
bc1 = D[eq, z] /. z -> 1
(* -1 + f'[1] *)

and solve with DSolve

DSolve[{deq == 0, bc0 == 0, bc1 == 0}, f, z] // Flatten
(* {f -> Function[{z}, 1/10 z^(1/2 - Sqrt[5]/2) 
   (5 - Sqrt[5] + 5 z^Sqrt[5] + Sqrt[5] z^Sqrt[5])]} *)

This solution can be verified by back-substitution.

Simplify[eq /. %, 0 < z < 1]
(* 0 *)

Added note: The approach here should work for any g[z] that is a polynomial or can be approximated accurately by one. The order of the resulting ODE would be one greater than the order of the polynomial. Of course, obtaining a symbolic solution, as accomplished here, might not be possible. Numerical solutions of the resulting ODE would be possible except near z = 0, where the solutions typically would be singular. Even the symbolic solution obtained for g[z] = z is singular there.

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  • $\begingroup$ Thank you sir! It works well! I'm new to mma and I got invaluable experience from your answer. $\endgroup$
    – dcmpsr
    Nov 16, 2023 at 5:54

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