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\title{Continuous and Discrete Inverse of the 1D Laplacian}
\author{Jos van Kan}

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\maketitle

\begin{abstract}
We are interested in constructing a continuous and discrete inverse to the 1D Laplacian on the entire real axis.
\end{abstract}

\section{Formulation of the problem}
Consider the ODE 
\begin{equation} \label {eq:p1}
 u-\tau^2 u''=f
\end{equation}
on the entire real axis. ($\tau>0$) Boundary conditions only require that the solution remains bounded in $\pm \infty$.
\subsection{Continous solution}
We rewrite \eqref{eq:p1} as follows
\begin{equation}
 u - \tau u' + \tau u' - \tau^2 u'' = f
\end{equation}
and multiply on both sides of the equation by the integrating factor $e^{-x/\tau}$. We integrate from $x$ to $\infty$ and obtain
\begin{equation}
 \tau e^{-x/\tau}(u + \tau u') = \int_x^\infty e^{-\xi/\tau}f(\xi)\, d\xi
\end{equation}
as is easily verified by differentiating. We bring the factor $-\tau e^{-x/\tau}$  to the other side and redefine the integration variable to obtain
\begin{equation}
 u+\tau u' = \frac 1\tau \int_0^\infty e^{-\xi/\tau} f(x+\xi)\, d\xi
\end{equation}
We now repeat the procedure, this time with integrating factor $e^{x/\tau}$. We integrate from $-\infty$ to $x$ to obtain:
\begin{equation}
\tau e^{x/\tau}u=\frac{1}{\tau}\int_{-\infty}^x e^{\eta/\tau} \int_0^\infty e^ {-\xi/\tau} f(\eta+\xi)\, d\xi\,d\eta 
\end{equation}
or equivalently, redefining $\eta'=x-\eta$ and dropping the prime
\begin{equation}
 u = \frac 1{\tau^2} \int_0^\infty \int_0^\infty e^{-(\xi+\eta)/\tau} f(x+\xi-\eta)\,d\xi\,d\eta
\end{equation}
Introducing new variables $t=\xi+\eta$ and $s=\xi-\eta$ this transforms to
\begin{align} \label{eq:res}
 u&=\frac 1{2\tau^2} \int_{-\infty}^\infty f(x+s) \int_{|s|}^\infty e^{-t/\tau}\, dt\, ds\\
&= \frac 1{2\tau} \int_{-\infty}^\infty e^{-|s|/\tau} f(x+s)\, ds\\
&= \frac 1{2\tau} \int_0^\infty e^{-s/\tau} (f(x+s)+f(x-s))\, ds
\end{align}
Remark, that this result makes sense for distributions $f$ that increase exponentially slower than $e^{x/\tau}$. If $f$ is bounded over $\mathbb R$ and has a limit for $x$ to $\pm \infty$ so has $u$ and in fact
\begin{equation}
 \lim_{x\rightarrow\pm\infty} u(x) = \lim_{x\rightarrow\pm\infty} f(x)
\end{equation}
as is easily proved.

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