# Fourier transform & asymptotic expansion of Klein-Gordon equation

I am looking for an approximate analytical solution to the generalized Klein-Gordon equation $$\begin{equation} \frac{\partial^2{\phi}}{\partial{t^2}}+\frac{\partial^2{\phi}}{\partial{x^2}}+\phi=0 \end{equation}$$ using Fourier and subsequent asymptotic expansion. Initial condition is $$\phi(x,0)=C\mathrm{e}^{-x^2}$$, with periodic boundary conditions $$\phi(L,t)=\phi(-L,t)$$. Taking the Fourier transform $$\begin{equation} \Phi(\kappa, t)=\int_{-L}^L\phi(x,t)\mathrm{e}^{-2\pi i\kappa x}dx \end{equation}$$ I can now calculate the second spatial derivate and attain $$\begin{equation} \frac{\partial^2{\Phi}}{\partial{t^2}}-(\kappa^2+1)\Phi=0 \\ \rightarrow \Phi(\kappa, t)= a_+(\kappa)\mathrm{e}^{i\sqrt{\kappa^2+1}t}+a_-(\kappa)\mathrm{e}^{-\sqrt{\kappa^2+1}t} \end{equation}$$ I could transform this back to $$\phi(x,t)$$ and apply the principle of stationary phase, but I lack the initial temporal derivative, as used over here to determine the Fourier coefficients: Solving the Klein-Gordon equation via Fourier transform

How can I determine the coefficients $$a_+(\kappa)$$ and $$a_-(\kappa)$$ without knowing my initial temporal derivative?

Do you need to solve the equation in Fourier (i.e. wavenumber-frequency space? The equation with boundary conditions falls easily enough to standard separation-of-variables techniques.

Assume $\phi(x,t) = X(x)T(t)$. Then the equation can be written as $$T''/T + 1 = -X''/X.$$ This is the separated form of the equation. Assume a separation constant $\lambda^2$ and you get two ordinary differential equations $T'' + (1 - \lambda^2)T = 0$ and $X'' + \lambda^2 X = 0$.

Solutions to the equation for $T$ take the form $$T(t) = c \exp{(\sqrt{\lambda^2 - 1}t)} + d \exp{(-\sqrt{\lambda^2 - 1}t)}$$ and solutions to the equation for $X$ take the form $$X(x) = a \cos{\lambda x} + b \sin{\lambda x}.$$ I've chosen to write $X$ as cosines and sines rather than complex exponentials for convenience later.

We can look at the boundary and initial conditions. Let look at the boundary condition first. We have $\phi(-L,t) = \phi(L,t)$ which implies $X(-L)T(t) = X(L)T(t)$. This implies that either $X(-L) = X(L)$ or $T(t) \equiv 0$. If $T(t) \equiv 0$ then we have the trivial solution $\phi(x,t) \equiv 0$. Looking for other solutions, then, we set $X(-L) = X(L)$. Referring to our solution for $X$, we conclude that $\lambda = n \frac{\pi}{L}$ for $n = 0, 1, 2, \ldots$ As there is a different solution for each possible value of $\lambda$, we arrive at the general solution for $X$: $$X(x) = \sum_{n = 0}^\infty \left( a_n \cos(n\pi x/L) + b_n \sin(n\pi x / L) \right).$$ Next we look at the initial condition

We have that $\phi(x,0) = X(x) T(0) = Ce^{-x^2}$. If $T(0) \ne 0$ then we can rewrite this as $X(x) = C' e^{-x^2}$ where of course $C' = C/T(0)$. Thus we have $$C' e^{-x^2} = X(x) = \sum_{n = 0}^\infty \left( a_n \cos(n\pi x/L) + b_n \sin(n\pi x / L) \right).$$ We can exploit the orthogonality in the L$^2([-L,L])$ inner product of these sines and cosines on $[-L,L]$ to solve for the $a_n$ and $b_n$. Indeed, as $C' e^{-x^2}$ has even symmetry on $[-L,L]$ we quickly find that $b_n = 0$ for all $n$.

Finally, then we have the general solution $\phi(x,t) = X(x) T(t)$ with $$X(x) = \sum_{n = 0}^\infty a_n \cos(n\pi x/L),$$ $$a_n = \int_{-L}^L C' e^{-x^2} \cos(n\pi x/L) dx,$$ and $$T(t) = \sum_{n = 0}^\infty c_n \exp{(\sqrt{\lambda_n^2 - 1}t)} + d_n \exp{(-\sqrt{\lambda_n^2 - 1}t)}.$$

Without more boundary or initial conditions, we can't say more.