Green function of the Sturm Liouville problem I am working on the following Sturm-Liouville problem
$Ly=-y^{''}$ with boundary conditions $y^{'}(0)=y(1)=0$
I have calculated the eigenvalues $(\frac{\pi}{2}+k\pi)^2$ but I don't know how to calculate the eigenfunctions and the green function.
Can someone help me? Thank you very much.
 A: Let $\varphi_{\lambda}(x)=\cos(\sqrt{\lambda}x)$ and $\psi_{\lambda}(x)=\sin(\sqrt{\lambda}(x-1))/\sqrt{\lambda}$. Then
$$
      \varphi_{\lambda}''=\lambda\varphi_{\lambda},\; \varphi_{\lambda}(0)=1,\; \varphi_{\lambda}'(0)=0 \\
      \psi_{\lambda}''=\lambda\psi_{\lambda},\;\psi_{\lambda}(1)=0,\;\psi_{\lambda}'(1)=1.
$$
The Green function solution of $(L-\lambda I)g=f$ subject to $g'(0)=0$, $g(1)=0$ is given by
$$
       g= \frac{\psi_{\lambda}(x)}{w(\lambda)}\int_0^x f(y)\varphi_{\lambda}(y)dy+\frac{\varphi_{\lambda}(x)}{w(\lambda)}\int_x^1f(y)\psi_{\lambda}(y)dy,
$$
where $\omega(\lambda)$ is the Wronskian of the solutions $\varphi_{\lambda}$ and $\psi_{\lambda}$ given by
$$
             \psi_{\lambda}'\varphi_{\lambda}-\varphi_{\lambda}'\varphi_{\lambda} \\=\cos(\sqrt{\lambda}(x-1))\cos(\sqrt{\lambda}x)+\sin(\sqrt{\lambda}x)\sin(\sqrt{\lambda}(x-1)) \\
 =\cos(\sqrt{\lambda}(x-1)-\sqrt{\lambda}x)=\cos(\sqrt{\lambda}).
$$
There are poles of order $1$ in the resolvent expression at $\sqrt{\lambda_n}=(n+1/2)\pi$ or $\lambda_n=(n+1/2)^2\pi^2$. These are the eigenvalues, and the eigenfunctions are $\psi_{\lambda_n}$.
