Find the Green’s function corresponding to the Sturm-Liouville problem

Question

My Sturm-Liouville problem was $y^{\prime\prime}+\lambda y=0$, $y(0)=0$, $y(1)=0$. I found the eigenvalues to be $(n\pi)^2$ and the associated eigenfunction to be $\sin(n\pi x)$. I now have to find the Green's Function corresponding to the Sturm-Liouville problem and I am stuck.

Answer 1

You are looking at the resolvent operator $(\lambda I-L)^{-1}$, where $L$ has the domain $\mathcal{D}(L)$ consisting of absolutely continuous functions $f : [0,1]\rightarrow\mathbb{C}$ such that $f(0)=0=f(1)$. For such $f$, the operator $L$ is $$ Lf=-f''. $$ The resolvent of this operator, which is $f=(L-\lambda I)^{-1}g$, is obtained by finding the unique solution $f$ of $$ (L-\lambda I)f=g,\;\; f(0)=0,\; f(1)=0. $$ (Here it is assumed that $g\in L^2[0,1]$, and that $(L-\lambda I)^{-1}$ exists for the given $\lambda$.) To do this, you can use variation of parameters on the solutions of $(L-\lambda I)f=0$. Start by solving $$ -\varphi''-\lambda\varphi = 0,\;\; \varphi(0)=0,\varphi'(0)=1 \\ -\psi''-\lambda\psi=0,\;\;\psi(1)=0,\;\psi'(1)=1 $$ These solutions are $$ \varphi_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}},\\ \psi_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}}. $$ The Wronskian of these solutions is independent of $x$: $$ W(\varphi_{\lambda},\psi_{\lambda})=\varphi_{\lambda}\psi_{\lambda}'-\varphi_{\lambda}'\psi_{\lambda} \\ = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}\cos(\sqrt{\lambda}(x-1))-\cos(\sqrt{\lambda}x)\frac{\sin(\sqrt{\lambda}(x-1))}{\sqrt{\lambda}} \\ = \frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}. $$ The integral representation of the resolvent $(\lambda I-L)^{-1}$ is $$ (\lambda I-L)^{-1}f = \\ =\frac{\sqrt{\lambda}}{\sin(\sqrt{\lambda})}\left[\int_0^x f(y)\psi_{\lambda}(y)dy\cdot\varphi_{\lambda}(x)+\int_x^1 f(y)\varphi_{\lambda}(y)dy\cdot\psi_{\lambda}(x)\right]. $$ If you rewrite this as one integral, then the Green function $G(x,y,\lambda)$ is the kernel.