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.

  • $\begingroup$ Maybe you should explain what asymptotic expansion you seek and give more details (although I cannot promise anything). $\endgroup$
    – akhmeteli
    Sep 5, 2014 at 13:27
  • $\begingroup$ I've been trying to use the method of stationary phase, so changing the sins and cosines to their exponential forms and finding the dominant contributions to the integral at the stationary points of phase. $\endgroup$ Sep 5, 2014 at 13:31
  • $\begingroup$ I only need to find the long term behaviour of the problem so I thought the method of stationary phase would be a suitable method. $\endgroup$ Sep 5, 2014 at 13:32
  • $\begingroup$ With all due respect, you don't give enough details. I, for one, am not going to do the calculation for you. $\endgroup$
    – akhmeteli
    Sep 5, 2014 at 13:47
  • $\begingroup$ I added a bit more if it helps, but I can't give much more detail than that. I'm basically asking if the method I'm trying to use is suitable. $\endgroup$ Sep 5, 2014 at 14:08

2 Answers 2


I wonder if there can be an error in your derivation. OK, you uncoupled the equations. Then you could consider a 2-dimensional Fourier expansion of $v$ into exponents, say, $\exp(i(\omega t +k x))$. Then the dispersion relation (what you get when you substitute the exponents into your linear differential equation for $v$) would be $-\omega^2+k^2+1=0$, so $\omega=\sqrt{k^2+1}$, whereas you get $\omega=\sqrt{-k^2+1}$.

  • $\begingroup$ Indeed, I had made a mistake its $\omega = \sqrt{k^2+1}$, the problem is still the same however, my coefficients may be little wrong but they are irrelevant. The method of stationary phase still yields nonsensical results. $\endgroup$ Sep 5, 2014 at 13:19

You are making your life harder than it needs to be. I will sketch the solution for you.

You already know that $$ U = U_0(k) e^{-\imath\sqrt{1+k^2}t} $$ and likewise for $V$. Then plug this into the expression for $u(x,t)$, and set $t=0$. You obtain $$ u(x,0) = e^{-x^2} = \int U_0(k) e^{-\imath k x } d\!x $$ which gives you immediately $U_0(k) = \alpha e^{-q k^2}$ for some value of $\alpha$ and $q$ which you will determine. At this point, you may get the general solution as a closed integral, $$ u(x,t) = \int \alpha e^{-q k^2} e^{-\imath\sqrt{1+k^2} t} e^{-\imath k x} d\!x $$ which you may compute with Mathematica, look it up in Gradshteyn and Rhyzik, or compute with the saddle-point method for $x\rightarrow+\infty$.


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