Convert the form of two-dimensional PDE (Klein-Gordon equation)

Question

Convert the form of two-dimensional PDE (Klein-Gordon equation) from
$u_{ξζ}=\sin⁡ u$ to $u_{tt}-u_{xx}=\sin⁡ u$.

Using the assumption \begin{align} ξ&=\frac{1}{2} (x+t), \\ ζ&=\frac{1}{2} (x-t). \end{align}

I tried to use the general solution for the wave equation \begin{align} u(ξ,ζ)=u(x+ct)+u(x-ct) \end{align}

From the assumption $x=ξ+ζ ,t=ξ-ζ$.

Now, \begin{align} u(ξ,ζ)=u(x)+u(t) \end{align} \begin{align} u_ξ&=u_x+u_t\\ u_{ξζ}&=u_{xx}-u_{tt}. \end{align} My problem is when I replaced it, I got this equation: \begin{align} u_{tt}-u_{xx}=-\sin⁡u \end{align} I don't know if I made a mistake or if there is another method that can be used.

I would like to point out that the same assumption should be used.

Answer 1

You can slightly change your substitution and set $$\xi = \frac{x + t}{2} \quad \zeta = \frac{t-x}{2}.$$ For $u(\xi, \zeta) = u\left( \frac{t+x}{2}, \frac{t-x}{2}\right),$ we obtain $$u_{tt} = \frac{1}{4}(u_{\zeta \zeta} + 2u_{\zeta \xi} + u_{\xi \xi})$$ and $$u_{xx} = \frac{1}{4}(u_{\zeta \zeta} - 2u_{\zeta \xi} + u_{\xi xi}).$$ Then $$u_{tt} - u_{xx} = u_{\xi \zeta}=\sin(u).$$

Answer 2

The sine Gordon should reduce to the pendulum equation

$$\partial_{tt} u(t) =-\sin(u(t))$$

for u independent of $x$.

$$\partial_{tt} u-\partial_{xx} u=(\partial_t-\partial_x)(\partial_t+\partial_x)u=\partial_{\xi\eta}u = -\sin u$$