### Author Topic: Web bonus problem : Week 3 (#2)  (Read 3143 times)

#### Victor Ivrii

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##### Web bonus problem : Week 3 (#2)
« on: September 26, 2015, 12:44:41 PM »

#### Bruce Wu

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##### Re: Web bonus problem : Week 3 (#2)
« Reply #1 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:

v^{2}\phi''-c^{2}\phi''+m^{2}\phi=0

v^{2}\phi''-c^{2}\phi''-m^{2}\phi=0

for (16) and (17), respectively.
Proceeding to solve (1) using methods of linear ODEs yields:

\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)

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):

\phi''=\frac{m^{2}}{v^{2}-c^{2}}\phi

If we assume again that $v^{2} > c^{2}$, we will get:
\large
\phi(z)=ae^{\frac{m}{\sqrt{v^{2}-c^{2}}}z}+be^{-\frac{m}{\sqrt{v^{2}-c^{2}}}z}

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:

\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)

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):

-v\phi'-K\phi'''=0

\phi'''+\frac{v}{K}\phi'=0

If $v$ and $K$ have the same sign, then the characteristic equation will have imaginary roots and give a bounded periodic solution:

\phi(z)=a\sin \left(\sqrt{\frac{v}{K}}z\right)+b\cos \left(\sqrt{\frac{v}{K}}z\right)+c

Same with problem (19):

-v\phi'-iK\phi''=0

\phi''-\frac{iv}{K}\phi'=0

In this case, the solution will be suitable regardless of the value of $v$, and the solution will be complex:
\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

Same with problem (20):

v^{2}\phi''+K\phi''''=0\Rightarrow\phi''''+\frac{v^{2}}{K}\phi''=0

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:

\phi(z)=a\sin \left(\frac{vz}{\sqrt{K}}\right)+b\cos \left(\frac{vz}{\sqrt{K}}\right)+c+d

Again, $u=\phi(x-vt)$ using $\phi(z)$ in (9), (12), and (14) solve the original problems (18), (19), and (20), respectively.
« Last Edit: September 27, 2015, 05:26:06 PM by Fei Fan Wu »

#### Victor Ivrii

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##### Re: Web bonus problem : Week 3 (#2)
« Reply #2 on: October 03, 2015, 04:41:17 AM »
You do well but I just explain it simpler. Right--we are looking for solutions $u=\phi (x-vt)$ and get

(v^2-c^2)\phi'' + m^2 \phi=0.
\label{A}

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 $2\pi \sqrt{v^2-c^2}/m$.