A certain solution for Sine-Gordon Equation

Question

I'm stuck at a seemingly easy problem but I don't know how to approach it (partially due to the shape of the sine-Gordon equation). Let's say that $\omega(u,v)$ is a solution of the sine-Gordon equation $\omega_{uv} = \sin{\omega}$ such that it satisfies the additional condition of

$$\left(\log\left(\frac{\omega_u}{\omega_v}\right)\right)_{uv} = 0$$

The above equation results in

$$\frac{\omega_u}{\omega_v} = U(u)\,V(v)$$

How can I identify the nature of the functions $U(u)$ and $V(v)$? I actually expect them to be both constants. I would appreciate any suggestion for that.

Please let me know if you need additional info!

Answer 1

Let me start by writing $1/U$ instead of $U$. The case when $U$ or $1/U$ vanishes (or $V$ vanishes) would require special handling of course. Then your constraint is $$ \frac{\omega_u}{\omega_v} = \frac{V}{U} \quad \text{or} \quad (U\partial_u - V\partial_v) \omega = 0 . $$ That is, $\omega$ is constant along the flow lines of the corresponding vector field. The vector field $(U\partial_v + V\partial_v)$ is orthogonal to it, with respect to the obvious Lorentzian metric.

Rewriting the sine-Gordon equation using these vector fields gives $$ \frac{1}{4UV} [(U\partial_u+V\partial_v)^2\omega - (U\partial_u-V\partial_v)^2\omega] = \sin \omega . $$ The constraint kills the second term on the lhs, which gives after simple rewriting $$ \frac{(U\partial_u+V\partial_v)^2\omega}{4\sin\omega} = UV . $$ The two vector fields actually commute, so applying $(U\partial_u-V\partial_v)$ kills the lhs, while simplifying its action on the rhs gives $$ \partial_u U = \partial_v V , $$ which together with the underlying hypothesis $U=U(u)$, $V=V(v)$ implies that both functions can only be constants. Edit: I was a bit too hasty! Both sides of the last equation, $\partial_u U$ and $\partial_v V$, of course have to be constants. But integrating, $U(u)$ and $V(v)$ are actually allowed to be 1st order polynomials.