# Asymptotic behaviour of soliton-antisoliton solution for the Sine Gordon equation

The question isn't about any actual homework, it's rather a (probably simple) intermediate step I've encountered on Rajaraman's Solitons and instantons : an introduction to solitons and instantons in quantum field theory, in the context of the Sine-Gordon equation. The full solution of the equation is a rather complicated problem, so we limit ourselves to particular solutions, one of which is the soliton-antisoliton scattering solution, which has the form: $$\phi(x,t)=4arctan\left( \frac{sinh(ut/\sqrt{1-v^2})}{u \ cosh(x/\sqrt{1-v^2})}\right)$$ He argues that in the limit that t goes to minus infinity, for example, this becomes $$\phi\rightarrow 4arctan\left[exp\left(\frac{x+v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right] - 4arctan\left[exp\left(\frac{x-v(t+\Delta/2)}{\sqrt{1-v^2}}\right) \right]$$ where $$\Delta\equiv \frac{1-v^2}{v}lnv$$ and a similar solution for the positive infinity case(Page 38 of Rajaraman's forementioned book). I tried to put the solution in a form in which I can use the arctangent addiction formula, but thus far no success. Closer I got was $$4arctan\left \{\left[\frac{exp\left(x + \gamma v(t+(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] - \left[\frac{exp\left(x - \gamma v(t-(lnv)/(v\gamma))\right)}{1+e^{2\gamma x}} \right] \right \}$$ Since we're studying an asymptotic case, I don't really understand why the first term on this last equation should even be there instead of going to zero.

I'm probably missing something pretty obvious, but whatever the reason is, I just don't get it.Thanks in advance for any help.

There is a typo in Rajaraman's book. The correct expression for the $\Delta$ time delay should be $$\frac{\Delta}{2}=\frac{\sqrt{1-u^2}}{u}\ln(u)$$ You can check this in Rajaraman's 1975 review article (R. Rajaraman; Physics Reports, Volume 21, Issue 5, p. 227-313. (1975)).
With this, the soliton-antisoliton solution can be written as $$\displaystyle\Phi_{+-}(x,t)=4\arctan\left(\frac{\sinh(\gamma ut)}{u\cosh(\gamma x)}\right)= \\ =4\arctan\left(\frac{\exp{\left[\gamma\left(x+u\left(t-\frac{\Delta}{2}\right)\right)\right]-\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]}}{1+\exp{(2\gamma x)}} \right)$$ with $$\gamma=\frac{1}{\sqrt{1-u^2}}$$ In the $t\rightarrow-\infty$ limit the second term in the bracket is large and the first term is exponentially small, so you can make the approximation $$\exp\left(u\left(t-\frac{\Delta}{2}\right)\right)\approx\exp\left(u\left(t+\frac{\Delta}{2}\right)\right)$$ since you only make an exponentially small error with this.
So in the $t\rightarrow-\infty$ limit the solution is $$\Phi_{+-}(x,t)\approx4\arctan\left(\frac{\exp{\left[\gamma\left(x+u\left(t+\frac{\Delta}{2}\right)\right)\right]-\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]}}{1+\exp{(2\gamma x)}} \right)= \\ =4\arctan\left(\exp\left[\gamma\left(x+u\left(t+\frac{\Delta}{2}\right)\right)\right]\right)-4\arctan\left(\exp\left[\gamma\left(x-u\left(t+\frac{\Delta}{2}\right)\right)\right]\right)= \\ =\Phi_+(\gamma(x+u(t+\Delta/2)))+\Phi_-(\gamma(x-u(t+\Delta/2)))$$ where we used the identity $$\arctan(x)-\arctan(y)=\arctan\left(\frac{x-y}{1+xy}\right)$$ and where the single soliton/antisoliton solution is $$\Phi_{\pm}(X)=\pm4\arctan(\exp(X))$$ The situation is the same for the $t\rightarrow+\infty$ limit: $$\exp\left(-u\left(t+\frac{\Delta}{2}\right)\right)\approx\exp\left(-u\left(t-\frac{\Delta}{2}\right)\right)$$ and $$\Phi_{+-}(x,t)\approx\Phi_+(\gamma(x+u(t-\Delta/2)))+\Phi_-(\gamma(x-u(t-\Delta/2)))$$