0
$\begingroup$

I'm having problem obtaining the solution of the homogeneous Fredholm integral equation of the 2nd kind with a separable kernel. I always get a zero if I use the normal method I was taught for the non homogeneous type.

I have an example: y(x) = $\lambda \int_{-1}^{1}(x+z)\,y(z)dz$.

$\endgroup$

migrated from mathematica.stackexchange.com Mar 6 '15 at 12:51

This question came from our site for users of Wolfram Mathematica.

  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Read the faq! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Louis Mar 6 '15 at 5:13
  • $\begingroup$ math.stackexchange.com/questions/1177932/… $\endgroup$ – JJacquelin Mar 6 '15 at 16:04
1
$\begingroup$

It is rather simple, as soon as one guesses that the solution is a linear function: $$y(x)= a_0 + a_1x$$ then placing all the terms of the equation into the left-hand side one finds there the following expression:

$$y(x)-\lambda x\int_{-1}^1 y(z)\mathrm dz -\lambda \int_{-1}^1 z y(z)\mathrm dz=a_0 + a_1 x - \frac{2}{3}a_1 \lambda - 2 a_0 x \lambda.$$

Which is, evidently, only equal to zero, if all coefficients are equal to zero, so $$ \left\{\begin{array}{ccccc}a_0&-&\frac{2}{3}a_1\lambda&=&0\\ -2a_0\lambda&+&a_1&=&0.\end{array}\right.$$ This system has only trivial solutions except when the determinant is equal to 0. The determinant is $\Delta=1\times1-(-2\lambda)(-\frac{2}{3}\lambda)=1-\frac{4}{3}\lambda^2$. The non trivial solutions are obtained for $\lambda=\pm\frac{\sqrt3}{2}$ and in this case are given by $a_1=a_0\sqrt{3}$ or $a_1=-a_0\sqrt3$.

That is the non-trivial solution you are looking for. The $\lambda$ values $\sqrt3/2$ and $\sqrt3/2$ are the corresponding eigenvalues.

Further, choosing

$$y(x)= a_0 + a_1x + a_2x^2 + a_3x^3 + a_4x^4 + a_5x^5$$ and using the same calculations one can make sure that the higher degrees do not interfere.

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.