I have the equation,
$$(1+\lambda)\xi(r,\theta)=\gamma\int_{0}^{2\pi}w(r,\theta, R, \theta')\xi(R,\theta')d\theta'$$,
where $w=w(r,\theta, r', \theta')$
I know that,
$$\xi_{n}(R,\theta)=cos(n\theta)$$
$$\lambda_{n}=-1+\gamma\int_{0}^{2\pi}w(R,\theta,R,0)cos(n\theta)d\theta$$
are corresonding eigenfunctions when $r=R$. I need to now get the full eigenfunction with r-dependence, $\xi_{n}(r, \theta)$, by using the above two equations. I know that my final answer should look like,
$$\xi_{n}(r,\theta)=\int_{0}^{2\pi}w(r,\theta,R,\theta')cos(n\theta')d\theta'$$
Any help would be greatly appreciated, thanks!
Edit:I forgot to mention. The person in the paper says: the answer can be derived by exploiting the special form of the eigenvalue problem, if that helps at all.