# Gross-Neveu model analytic solution

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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}

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

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:

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

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

edited Nov 11, 2014
Should I be using method of steepest descent? The reason why I haven't is because my phase has real roots.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user Kevin Murray
I suspect that you should substitute the exact point of stationary phase, which depends on $x$. $t$ may be large, but $x$ is not limited. Maybe your solution contains some parts moving with a pretty much constant velocity - at least the opposite is not obvious, and then effective $x$ can be as large as $t$.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user akhmeteli
I am not sure if you should use the method of steepest descent, but it is not obvious that your coefficients are real.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user akhmeteli
You are right, just came to that conclusion right now with a friend. The stationary point must depend on x and t for a sensible answer. I understand the method a bit better now. Well the i in front of the sin coefficient cancels with the 1/i encoded in the sin and k is real so I don't see a way in which they can be complex?

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user Kevin Murray
Now that you offer some arguments, it looks like the coefficients at the exponents are indeed real. Glad you've sorted out your problem.

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

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user Kevin Murray
With all due respect, I may have forgotten the stationary phase method, but I cannot believe you on your word that "the only points of stationary phase is at $k = 0$", and I don't quite understand why your coefficients are "irrelevant". It is not obvious that your coefficients are real. As far as I remember, in the method of stationary phase (or steepest descent - en.wikipedia.org/wiki/Method_of_steepest_descent ) you must allow $k$ to be complex - your contour of integration can pass pretty much anywhere in the complex plane.

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

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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 $\omega=\sqrt{-k^2+1}$.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user akhmeteli
answered Sep 5, 2014 by (40 points)
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.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user Kevin Murray
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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 $x\rightarrow+\infty$.

This post imported from StackExchange Physics at 2014-11-11 09:08 (UTC), posted by SE-user MariusMatutiae
answered Sep 5, 2014 by (0 points)

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