Fourier transform & asymptotic expansion of Klein-Gordon equation

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 $X$: $$X(x) = sum_{n = 0}^infty left( a_n cos(npi x/L) + b_n sin(npi x / L) right).$$ 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 $$ C' e^{-x^2} = X(x) = sum_{n = 0}^infty left( a_n cos(npi x/L) + b_n sin(npi x / L) right).$$ 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 $$X(x) = sum_{n = 0}^infty a_n cos(npi x/L),$$ $$a_n = int_{-L}^L C' e^{-x^2} cos(npi x/L) dx,$$ 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.