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« **on:** September 27, 2015, 05:21:51 PM »
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.