Author Topic: Web bonus problem : Week 3 (#3)  (Read 868 times)

Victor Ivrii

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Emily Deibert

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Re: Web bonus problem : Week 3 (#3)
« Reply #1 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.

Victor Ivrii

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Re: Web bonus problem : Week 3 (#3)
« Reply #2 on: October 03, 2015, 04:20:41 AM »
You are trying to find the general solution, which is impossible. However what is required? To find solutions which are travelling waves that means
\begin{equation}
u= \phi (x-vt)
\label{A}
\end{equation}
 with unknown function $\phi$ and constant $v$. Then $u_{xx}=phi ''$, $u_{tt}=v^2 \phi''$ and we get equation
\begin{equation}
(v^2-c^2) \phi '' + \sin (\phi) =0.
\label{B}
\end{equation}
This is ODE and we are looking for its  bounded solutions. When $v =\pm c$ we get constant solutions $\phi(\xi)=\pi n$ (which are definitely of no interest).

Let $v^2>c^2$.  Then we get equation describing mathematical pendulum $\phi '' + k\sin (\phi)=0$ https://en.wikipedia.org/wiki/Pendulum_(mathematics) which you definitely considered in the ODE class at Chapter 9 of Boyce and Di Prima, $k=g/L$ where $L$ is a distance from the anchor to the center of mass ($\phi$ is a deviation from the stable equilibrium $2\pi n$). While explicit solution is impossible we know that such equation has periodic (with respect to $\xi$) solutions which we are interested in (there are also non-periodic solutions corresponding pendulum having so large speed that it passes through the top point but we are not interested in those).

Let $v^2<c^2$. Then $k<0$ but replacing $\phi$ by $\phi+\pi$ we get the same equation albeit with $k$ replaced by $-k$.

So, conclusion which we are looking:

For each $v\ne \pm c$ equation $u_{tt}-c^2 u_{xx} + \sin (u)=0$ has solutions of the form (\ref{A}) with periodic function $\phi$ (satisfying (\ref{B}).

Question: which periods can have $\phi$? Since $\phi''+  k\sin(\phi)=0$ has solution for any period $T\in (2\pi/\sqrt{k},\infty)$, we have all periods  in the interval ${(2\pi |v^2-c^2|^{-1/2},\infty)}$.