MHB Solve \(u_{tt} - u_{xx} = \sin(u)\): Let \(u(\xi)\)

[Dustinsfl]

Consider \(u_{tt} - u_{xx} = \sin(u)\). Let \(u(\xi) = u(x - ct)\).

How do we get \((1 - c^2)u_{\xi\xi} = \sin(u)\)?
\[
u_{\xi\xi} = u_{xx} - c^2u_{tt}
\]
Correct?

 

[Opalg]

Consider \(u_{tt} - u_{xx} = \sin(u)\). Let \(u(\xi) = u(x - ct)\).

How do we get \((1 - c^2)u_{\xi\xi} = \sin(u)\)?
\[
u_{\xi\xi} = u_{xx} - c^2u_{tt}
\]
Correct?

If $u$ is a function of $\xi$, where $\xi = x-ct$, then the chain rule says that $u_t = \tfrac{\partial u}{\partial t} = \tfrac{du}{d\xi}\tfrac{\partial \xi}{\partial t} = -c\tfrac{du}{d\xi} = -cu_{\xi}.$ The same calculation, repeated, says that $u_{tt} = c^2u_{\xi\xi}.$ In the same way, $u_{xx} = u_{\xi\xi}.$ So the equation $u_{xx} - u_{tt} = \sin u$ becomes $(1-c^2)u_{\xi\xi} = \sin u.$ That is the standard form of the sine Gordon equation, but you seem to have changed a sign somewhere, writing $u_{tt} - u_{xx}$ rather than $u_{xx} - u_{tt}.$