Find solutions of $$u(x)-\lambda\int^{2\pi}_0\sum_{j=1}^n\frac{1}{j}cos(jy)cos(jx)u(y)dy=sin^2x$$ for all values of $\lambda$. Find the resolvent kernel for this equation. (Find the least squares solution if necessary.)
I have to solve it using the methods of compact operator. I searched for few days, but I can't seem to get a good resource on how to use this method. Any help on solving this problem will be appreciated.
This is set up to use Fourier series. First, $$ \cos(2x) = \cos^{2}(x)-\sin^{2}(x)=1-2\sin^{2}(x) \\ \sin^{2}(x) = \frac{1-\cos(2x)}{2}. $$ So, for $n \ge 2$, it is consistent to assume $$ u(x) = A + B\cos(2x). $$ Assuming $n \ge 2$, $$ A + B\cos(2x) - \lambda B \frac{1}{2}\pi\cos(2x) = \frac{1}{2}-\frac{1}{2}\cos(2x) \\ A = \frac{1}{2},\;\; B(1-\lambda\pi/2) = -\frac{1}{2} \\ A=\frac{1}{2},\;\; B = \frac{1}{\lambda\pi-2} $$ The solution is $$ u(x) = \frac{1}{2}+\frac{1}{\lambda\pi-2}\cos(2x). $$ That should get you started.