Inverse of Laplace operator

Question

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 $

Answer 1

I believe this is equivalent to finding the Green's function for Laplace's equation in on $(0,1)$. Please do take my ideas with a grain of salt however as I have little experience with this.

You are in search of a function $G$ with $$ -G''(x)=\delta(x-x_0) $$ for some $x_0\in (0,1)$. That is, you want a $G$ $$ -\int_0^1G(x)\phi''(x)\mathrm dx=\phi(x_0) $$ for any $\phi\in H^2(0,1)\cap H_0^1(0,1)$.

Noting that this is really the "antiderivative" of the Heaviside function, which is the ramp function, we have $$ G(x)=-\max\{0,x-x_0\} $$ Then, for $\phi$ with the required adjectives, we have $$ -\int_0^1G(x)\phi''(x)\mathrm dx=x_0\phi'(x_0)-\phi(x_0)-x_0\phi'(x_0)=\phi(x_0) $$ as required.

Then, solving for $f$ in the equation $$ -\frac{d^2}{dx^2}f=g $$ is given by convolving $G$ with $g$, i.e. $$ f(x)=-\int_0^1G(x-y)g(y)\mathrm dy $$

edit: this answer is not correct and I am working to modify it. We need to use the boundary data given by $H_0^1(0,1)$ to find the parameter $x_0$ and insure that this inverse is actually well defined.

Answer 2

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 $c=-y$.