Gross-Neveu model analytic solution 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.
 A: 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 $\omega=\sqrt{-k^2+1}$. 
A: 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 $x\rightarrow+\infty$. 
