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« **on:** September 29, 2015, 09:49:35 PM »
I tried to solve this problem but I could not get all the way through. I will post what I was able to get.

We start with the equation \begin{equation}\label{eq:problem}

u_{tt} - c^2u_{xx} + \sin(u) = 0

\end{equation}

The equation can be solved by the use of the *characteristic coordinates*, as discussed in class. As a reminder, we define: \begin{equation} \begin{cases}

\zeta = x + ct \\

\eta = x - ct

\end{cases} \end{equation}

Following the logic used in lecture/the online textbook, we can see that \begin{equation} x = \frac{1}{2}(\zeta + \eta) \end{equation} and \begin{equation} t = \frac{1}{2c}(\zeta - \eta) \end{equation}

By the chain rule (omitting several steps, as they are given in the textbook), we can show that:

\begin{equation}

-4c^2u_{\zeta\eta} = -\frac{1}{4}(c\partial_x + \partial_t)(c\partial_c - \partial_t) = u_{tt} - c^2u_{xx} = -\sin(u)

\end{equation}

We can see that in the characteristic coordinates, the original equation \eqref{eq:problem} becomes: \begin{equation}

u_{\zeta\eta} = \frac{1}{4c^2}\sin(u)

\end{equation}

Now here I get a little stuck. I have tried to progress with the solution after some thought but I am not sure if it will make sense.

Rewriting the equation, we have: \begin{equation}

4c^2\frac{\partial u}{\sin(u)} = \partial_\zeta\partial_\eta

\end{equation}

So integrating both sides, we get: \begin{equation}

-4c^2\ln(\cot(u) + \csc(u)) = \phi(\zeta)\partial_\eta + C

\end{equation}

Now here is where I really get stuck. I guess I have to integrate again, but this will lead to a nasty solution. For now I will leave the post in case someone is able to add on to the solution or provide a hint. I will return to the problem after thinking about it for a while.