$u(x)= 1/3+\int_{0}^{1}x\,t\sqrt{u(t)}\,dt$

u[x] == 1/3 + Integrate[x t Sqrt[u[t]], {t, 0, 1}]

Any ideas on how to treat such a problem with Mathematica functions?

  • $\begingroup$ Have you seen How to solve a non-linear integral equation? $\endgroup$
    – user9660
    Commented Feb 18, 2016 at 7:41
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    – rhermans
    Commented Feb 18, 2016 at 9:25

1 Answer 1


This is not a Fredholm equation. Nevertheless it seems to have a simple solution. Indeed: one evidently finds the solution in the form

u[x_] := 1/3 + A*x;

where A is a constant to be determined later on. Substituting into the integral one finds:

Integrate[t Sqrt[u[t]], {t, 0, 1}]

(*  (2 (2 + (1 + 3 A)^(3/2) (-2 + 9 A)))/(135 Sqrt[3] A^2)  *)

This brings one to the equation imposed on A:

eq = (2 (2 + (1 + 3 A)^(3/2) (-2 + 9 A)))/(135 Sqrt[3] A^2) == A

This equation can be solved exactly:


But the result is so cumbersome, that (if you have no special reasons) it is better to solve it numerically:

sl = NSolve[(2 (2 + (1 + 3 A)^(3/2) (-2 + 9 A)))/(135 Sqrt[3] A^2) == 
   A, A]

{{A -> 0.382266}}

Thus, your result is u(x)=1/3+0.382x

Have fun!


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