### Author Topic: Web bonus problem : Week 3 (#4)  (Read 2726 times)

#### Victor Ivrii

• Elder Member
• Posts: 2511
• Karma: 0
##### Web bonus problem : Week 3 (#4)
« on: September 26, 2015, 12:46:10 PM »

#### Jeremy Li 2

• Full Member
• Posts: 15
• Karma: 0
##### Re: Web bonus problem : Week 3 (#4)
« Reply #1 on: November 30, 2015, 05:49:35 PM »
I didn't really make it that far on this problem, but I thought I ought to post as far as I got in case anyone has any ideas.

Plugging $u(x,t)=\phi(x-vt)$ into the given PDEs:

v^2\phi''-\phi''+\phi-2\phi^3=0\\
v^2\phi''-\phi''-\phi+2\phi^3=0

And so we get

(v^2-1)\phi''=2\phi^3-\phi\\
(v^2-1)\phi''=\phi-2\phi^3

This ODE looks very difficult - any ideas?

#### Chi Ma

• Full Member
• Posts: 16
• Karma: 0
##### Re: Web bonus problem : Week 3 (#4)
« Reply #2 on: December 06, 2015, 12:46:20 AM »
We look for solutions such that $u_{tt}-u_{xx}=u(1-2u^2)=0$.
$u_{tt}-u_{xx}=0$ implies that $u(x,t) = f(x \pm t)$ for some function $f$.
$u(1-2u^2)=0$ implies that either $u = 0$ or $u = \pm \frac{1}{\sqrt 2}$.

A kink may be described by

u(x,t) = \left\{\begin{array}{21}
&\pm \frac{1}{\sqrt 2} \qquad & x \ge t\\
& 0 & x < t \end{array} \right.

A soliton may be described by

u(x,t) = \left\{\begin{array}{21}
&\pm \frac{1}{\sqrt 2} \qquad & x = t\\
& 0 & x \neq t \end{array} \right.

« Last Edit: December 10, 2015, 07:10:23 PM by Chi Ma »

#### Victor Ivrii

• Elder Member
• Posts: 2511
• Karma: 0
##### Re: Web bonus problem : Week 3 (#4)
« Reply #3 on: December 22, 2015, 02:58:59 AM »
We have 2nd order ODE with no explicit $x$. Could be reduced to 1st order by the standard $\phi'=\psi$, $\phi''=\psi'=\frac{d\psi}{d\phi}\phi'$.