81
« 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.