1
$\begingroup$

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

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?

$\endgroup$
2
  • $\begingroup$ Have you seen How to solve a non-linear integral equation? $\endgroup$ – user9660 Feb 18 '16 at 7:41
  • $\begingroup$ Greetings! Make the most of Mma.SE and take the tour now. Help us to help you, write an excellent question. Edit if improvable, show due diligence, give brief context, include minimal working examples of code and data in formatted form. As you receive give back, vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned. $\endgroup$ – rhermans Feb 18 '16 at 9:25
4
$\begingroup$

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:

Solve[eq,A]

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.