<s>[INST] Inverse of Laplace operator Let $\ L=\frac{-d^2}{dx^2}$ defined on $\ H^2(]0,1[) ∩ H^1_0(]0,1[)$ \ $\ L$ is the laplacian operator in one dimension \ how can we express the inverse of $\ L $
 [/INST] The Green's function might need a slight modification. 
\begin{align*}
G\left(x,y\right) & =\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\sin\left(n\pi x\right)\sin\left(n\pi y\right)\\
 & =\begin{cases}
\left(1-y\right)x & 0\leq x\leq y\\
y\left(1-x\right) & y\leq x\leq1
\end{cases}.
\end{align*}
In details: 
Let $L=-\frac{d^{2}}{dx^{2}}$, with domain as specified. Then $L$
is selfadjoint, and it has eigenfunctions (normalized) $\left\{ \varphi_{n}\left(x\right):=\sqrt{2}\sin\left(n\pi x\right)\mathrel{;}n\in\mathbb{N}\right\} $,
so that 
\begin{align*}
L^{-1} & =\sum_{n=1}^{\infty}\lambda_{n}^{-1}\varphi_{n}\left(x\right)\varphi_{n}\left(y\right)=\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}\sin\left(n\pi x\right)\sin\left(n\pi y\right).
\end{align*}
For the Green's function, assume $LG\left(\cdot,y\right)=\delta\left(\cdot-y\right)$,
$y\in\left(0,1\right)$, subject to the condition $G\left(0,y\right)=G\left(0,1\right)=0$.
So 
\begin{align*}
G\left(x,y\right) & =\begin{cases}
ax & x\leq y\\
c\left(x-1\right) & x\geq y
\end{cases}
\end{align*}
where $a=\frac{c\left(y-1\right)}{y}$, since $G\left(\cdot,y\right)$
is continuous at $x=y$. But 
\begin{align*}
\lim_{x\rightarrow y^{-}}G_{x}\left(x,y\right)-\lim_{x\rightarrow y^{+}}G_{x}\left(x,y\right) & =1,
\end{align*}
so $c=-y$. 
</s>