APM346-2015F > Web Bonus = Sept

Web bonus problem : Week 3 (#2)

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Victor Ivrii:
http://www.math.toronto.edu/courses/apm346h1/20159/PDE-textbook/Chapter2/S2.3.P.html#problem-2.3.P.6 Problem 6

Bruce Wu:
a) By plugging $u=\phi(x-vt)$ into equations (16) and (17) on the problems page, we obtain:
\begin{equation}
v^{2}\phi''-c^{2}\phi''+m^{2}\phi=0
\end{equation}\begin{equation}
v^{2}\phi''-c^{2}\phi''-m^{2}\phi=0
\end{equation}
for (16) and (17), respectively.
Proceeding to solve (1) using methods of linear ODEs yields:
\begin{equation}
\phi''=-\frac{m^{2}}{v^{2}-c^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{v^{2}-c^{2}}}z\right)
\end{equation}
Where $a,b$ depend on initial conditions. This is assuming that $v^{2} > c^{2}$, which gives us a periodic solution that is bounded, as desired. Proceeding the same way with (2):
\begin{equation}
\phi''=\frac{m^{2}}{v^{2}-c^{2}}\phi
\end{equation}
If we assume again that $v^{2} > c^{2}$, we will get:
\begin{equation}\large
\phi(z)=ae^{\frac{m}{\sqrt{v^{2}-c^{2}}}z}+be^{-\frac{m}{\sqrt{v^{2}-c^{2}}}z}
\end{equation}
But this solution does not tend to $0$ as $|z|\rightarrow\infty$ unless $a=b=0$, and we do not want the trivial solution. Therefore $v^{2}$ must be less than $c^{2}$. Then:
\begin{equation}
\phi''=-\frac{m^{2}}{c^{2}-v^{2}}\phi\Rightarrow \phi(z)=a\sin \left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)+b\cos\left(\frac{m}{\sqrt{c^{2}-v^{2}}}z\right)
\end{equation}
Now $u=\phi(x-vt)$ using $\phi(z)$ in (3) and (6) solve the original problems (16) and (17), respectively.

b) Plugging $u=\phi(x-vt)$ into problem (18):
\begin{equation}
-v\phi'-K\phi'''=0
\end{equation}\begin{equation}
\phi'''+\frac{v}{K}\phi'=0
\end{equation}
If $v$ and $K$ have the same sign, then the characteristic equation will have imaginary roots and give a bounded periodic solution:
\begin{equation}
\phi(z)=a\sin \left(\sqrt{\frac{v}{K}}z\right)+b\cos \left(\sqrt{\frac{v}{K}}z\right)+c
\end{equation}
Same with problem (19):
\begin{equation}
-v\phi'-iK\phi''=0
\end{equation}\begin{equation}
\phi''-\frac{iv}{K}\phi'=0
\end{equation}
In this case, the solution will be suitable regardless of the value of $v$, and the solution will be complex:
\begin{equation}\large
\phi(z)=ae^{\frac{iv}{K}z}+b=a\left(\cos \left(\frac{vz}{K}\right)+i\sin \left(\frac{vz}{K}\right)\right)+b
\end{equation}
Same with problem (20):
\begin{equation}
v^{2}\phi''+K\phi''''=0\Rightarrow\phi''''+\frac{v^{2}}{K}\phi''=0
\end{equation}
Here, since $v$ only appears in terms of its square, we do not need to worry about its value. As long as $K>0$, solving the above ODE gives us a suitable solution:
\begin{equation}
\phi(z)=a\sin \left(\frac{vz}{\sqrt{K}}\right)+b\cos \left(\frac{vz}{\sqrt{K}}\right)+c+d
\end{equation}
Again, $u=\phi(x-vt)$ using $\phi(z)$ in (9), (12), and (14) solve the original problems (18), (19), and (20), respectively.

Victor Ivrii:
You do well but I just explain it simpler. Right--we are looking for solutions $u=\phi (x-vt)$ and get
\begin{equation}
(v^2-c^2)\phi'' + m^2 \phi=0.
\label{A}
\end{equation}
It is ODE and we are looking for its bounded non-trivial solutions. If $v^2-c^2>0$ then such solutions indeed exist and they are (up to the shift) $A\cos (\omega \xi)$ with $\omega= m/\sqrt{v^2-c^2}$ and they are periodic with period $2\pi /\omega = 2\pi \sqrt{v^2-c^2}/m$.

Conclusion: for each $v:|v|\ge c$ there exist required solutions which are periodic with period $2\pi \sqrt{v^2-c^2}/m$.

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