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) = $\lambda \int_{-1}^{1}(x+z)\,y(z)dz$.
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!
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– Louis Mar 06 '15 at 05:13