I need to find an analytic solution via asymptotic expansion for the following system of equations: \begin{align} & i(u_t+u_x) + v = 0 \\ & i(v_t-v_x) + u = 0 \end{align}

\begin{equation} u(x,0) = Ae^{-x^2} \hspace{0.1 in} v(x,0) = -Ae^{-x^2} \end{equation}

I uncoupled them \begin{align} & v_{tt}-v_{xx} + v = 0\\ & u_{tt}-u_{xx} + u = 0 \end{align}

Wrote the solutions in terms of fourier series

\begin{align} & u(x,t) = \int_{-\infty}^{\infty}U(k,t)e^{-ikx}dk\\ & v(x,t) = \int_{-\infty}^{\infty}V(k,t)e^{-ikx}dk \end{align}

Came to the following differential equation \begin{align} & V_{tt} + V(1+k^2) = 0\\ & U_{tt} + U(1+k^2) = 0 \end{align}

found initial conditions for the derivatives by using the original equations and initial conditions \begin{align} u_t(x,0) = Ae^{-x^2}(2x-i) \hspace{0.2 in} v_t(x,0) = Ae^{-x^2}(2x+i) \end{align} Now I need to solve

\begin{align} & u(x,t) = \int_{-\infty}^{\infty}\left[\left[\frac{-iAe^{-\frac{k^2}{4}}\sqrt{1+k^2}}{2k\sqrt{\pi}}\left[k -1\right]\right]\text{sin}\left(\sqrt{1+k^2}t\right) + \left[\frac{Ae^{-\frac{k^2}{4}}}{2\sqrt{\pi}}\right]\text{cos}\left(\sqrt{1+k^2}t\right)\right]e^{-ikx}dk \notag\\ & v(x,t) = \int_{-\infty}^{\infty}\left[\left[\frac{iAe^{-\frac{k^2}{4}}\sqrt{1+k^2}}{2k\sqrt{\pi}}\left[k +1\right]\right]\text{sin}\left(\sqrt{1+k^2}t\right) + \left[\frac{-Ae^{-\frac{k^2}{4}}}{2\sqrt{\pi}}\right]\text{cos}\left(\sqrt{1+k^2}t\right)\right]e^{-ikx}dk \notag \end{align}

I changed the sins and cosines to their exponential forms and tried to use the method of stationary phase to find a solution. However my solution only contributes to x = 0. Any idea how I would find the asymptotic expansion of this?

I need to ultimately find the large t behaviour of this integral:

\begin{equation} I = \int_{-\infty}^{\infty}F(k)e^{i\sqrt{1+k^2}t-ikx}dk \end{equation}

Except the only point of stationary phase is at k = 0 which eliminates the x dependence.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user Kevin Murray