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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.