How does one obtain the fundamental solution of the differential operator $(1-\partial_x^2)$? There does not seem to be any easily accesible literature specifically describing how this is done, except in the case of very simple or highly specific differential operators.
The attempted solution is as follows: \begin{align} &(1-\partial_x^2)G(x) = \delta(x)\\ \Rightarrow &G(x)-G''(x) = \delta(x)\\ \Rightarrow &\mathcal{L}\{G\}-\mathcal{L}\{G''\} = 1\\ \Rightarrow &\mathcal{L}\{G\} - s^2\mathcal{L}\{G\}+sG(0)+G'(0)=1\\ \Rightarrow &(1-s^2)\mathcal{L}\{G\} = 1-sG(0)-G'(0)\\ \Rightarrow &\mathcal{L}\{G\} = \frac{1-G'(0)}{1-s^2} - G(0)\frac{s}{1-s^2}\\ \Rightarrow &\mathcal{L}\{G\} = \frac{1-G'(0)}{1-s^2} + G(0)\frac{s}{s^2-1}\\ \end{align} Then apply the inverse Laplace transform to get $$ G(x) = \frac{1-G'(0)}{2}e^{-|x|} + G(0)\cosh(x) $$ We now attempt to determine the value of $G(0)$ and $G'(0)$: $$ G(0) = \frac{1-G'(0)}{2} + G(0) $$ Therefore, $G'(0)=1$. However, this then means that $G(x)=G(0)\cosh(x)$, which implies that $G'(x)=G(0)\sinh(x)$, which gives that $G'(0)=0$, so we end up getting $0=1$. Something, clearly, must be wrong.
Help would be appreciated!
Given that the objective is to find the fundamental solution of $1−\partial^2_x$, we can approach this similarly to how you began by solving the corresponding differential equation, which we will do via Fourier transforms. \begin{align} & \\(1−\partial^2_x )G(x) = \delta(x) & \\\Rightarrow G(x)-G''(x) = \delta(x) & \\\Rightarrow \mathcal{F}(G(x))-\mathcal{F}(G''(x)) = \mathcal{F}(\delta(x)) & \\\Rightarrow \mathcal{F}(G(x)) - (2\pi i \epsilon)^2\mathcal{F}(G(x)) = 1 & \\\Rightarrow (1 + 4\pi^2 \epsilon^2)\mathcal{F}(G(x)) = 1 & \\\Rightarrow \mathcal{F}(G(x)) = \frac{1}{(1 + 4\pi^2 \epsilon^2)} & \\\Rightarrow \mathcal{F^{-1}}\mathcal{F}(G(x)) = \mathcal{F^{-1}}\frac{1}{(1 + 4\pi^2 \epsilon^2)} & \\\Rightarrow G(x) = \frac{e^{-|x|}}{2} \end{align} This should be the solution Fundamental solution of the Differential operator