APM346-2016F > APM346--Misc

Halloween Challenge 1


Victor Ivrii:

XinYu Zheng:
We are looking for traveling wave solutions. Let $u(x,t)=\phi(x-vt)$. Plugging this in, we find that
$$(v^2-c^2)\phi ''+\sin(\phi)=0$$
Note that this equation describes the motion of a pendulum, as the ODE for the pendulum is $\theta''+(g/L)\sin \theta=0$.
Now, if $v=\pm c$, the equation reduces to $\sin(\phi)=0$. In this case, the solution would be $\phi\equiv 0(\mod \pi)$, which are constant solutions. Let us try to find a solution for $v\neq \pm c$.
To find $\phi$, we ideally want some simplification when we apply $\sin$ to it. We also want the result to look like $\phi''$. A first guess would be to consider $\arctan(x)$. Note that the second derivative of $\arctan(x)$ contains $(x^2+1)^2$ in the denominator, but $\sin(\arctan(x))$ only has $\sqrt{x^2+1}$ in the denominator. However, we note
which has the same denominator as the second derivative of $\arctan(x)$. But the numerator of the second derivative of $\arctan(x)$ is $-2x$, and we want something that looks like $x(x^2-1)$ instead. We note that
$$\frac{d^2}{dx^2} \arctan(e^x)=\frac{e^x(1-e^{2x})}{(e^{2x}-1)^2} $$
Which looks like the form we want. So we take $\phi(x)=4\arctan(e^{ax})$ and plug in to the ODE we obtain:
To satisfy this equation, we want $a^2=1/(c^2-v^2)$, which implies $a=\pm 1/\sqrt{c^2-v^2}$. But this requires $c^2-v^2>0$. So for $|v|<c$, we have a solution
$$u(x,t)=4\arctan \left(e^{\pm\frac{x-vt}{\sqrt{c^2-v^2}}}\right)$$
Of course, $x+vt$ in place of $x-vt$ is also a solution.

Victor Ivrii:
When $v>c$ we indeed get a pendulum equation, studied in ODE course and the phase portrait is like on the picture. observe we have periodic oscillations and non-periodic movements. What to do as $v<c$? Simple: we plug $\phi:=\pi +\psi$ and then for $\psi$ we have the same equation albeit with $v^2-c^2$ replaced by $c^2-v^2$.


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