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Consider the sine-Gordon equation, $$ \varphi_{tt}-\varphi_{xx}+\sin \varphi=0. $$

It is said that, using light-cone coordinates $$ u=\frac{x+t}{2},\quad v=\frac{x-t}{2}, $$ it is transformed into $$ \varphi_{uv}=\sin\varphi. $$

How does that work?

If I consider $\varphi=\varphi(u(x,t),v(x,t))$, using the chain rule, $$ \varphi_t = \varphi_u u_t+\varphi_v v_t=\frac{1}{2}\varphi_u-\frac{1}{2}\varphi_v\implies\varphi_{tt}=\frac{1}{4}\varphi_{uu}+\frac{1}{4}\varphi_{vv} $$ and, similarly, $\varphi_{xx}=\frac{1}{4}(\varphi_{uu}+\varphi_{vv})$.

There must be a mistake...

Where is my misunderstanding?

selector
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    I always find it confusing to use the same name for two different functions. Given a solution $\varphi(x,t)$ to the original Sine-Gordon equation, you want to verify that the function $\psi(u,v) = \varphi(u+v, u-v)$ satisfies the second version of the equation. – Deane Jul 13 '21 at 17:27
  • Okay! With this, I get $$\Psi_{v}=\varphi_x-\varphi_t$$ and $$\Psi_{uv}=-\varphi_{tt}+\varphi_{xx}+\varphi_{xt}-\varphi_{tx}=\sin\varphi(u+v,u-v)=\sin\Psi$$ since $\varphi_{xt}=\varphi_{tx}$ and, by assumption, $\varphi_{xx}-\varphi_{tt}=\sin\varphi$. – selector Jul 13 '21 at 17:44
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    Your mistake is that $\varphi_{tt} \ne \frac{1}{4} \left( \varphi_{uu} + \varphi_{vv} \right)$, you didn't differentiate the cross terms. You have $$\varphi_{tt} = \frac{1}{2} \bigg( \partial_{t} (\varphi_{u}) - \partial_{t} (\varphi_{v}) \bigg) = \frac{1}{2} \bigg( \varphi_{uu} u_{t} + \color{red}{\varphi_{uv} v_{t}} - \color{blue}{\varphi_{vu} u_{t}} - \varphi_{vv} v_{t} \bigg) = \frac{1}{2} \left( \frac{1}{2} \varphi_{uu} - \varphi_{uv} + \frac{1}{2} \varphi_{vv} \right)$$ You have a similar issue for $\varphi_{xx}$, which is the same as $\varphi_{tt}$ but with a $+ \varphi_{uv}$. – Matthew Cassell Jul 13 '21 at 17:56

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