Computational Physics II
• Assignment 6 - Korteweg-deVries Equation for Solitons

Due: April 14

Code

$U_t+\epsilon UU_x+\mu U_{xxx}=0$
$u(x-ct)=\frac{3c}{\epsilon}\mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right]$

First I will solve for the various partial derivatives needed:
$u_t = \frac{3c^2}{\epsilon} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \sqrt{\frac{c}{\mu}}$
$u_x = \frac{-3c}{\epsilon} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \sqrt{\frac{c}{\mu}}$
$u_{xxx} = \frac{6c^2}{\epsilon \mu} \sqrt{\frac{c}{\mu }} \mathrm{sech}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^4 \mathrm{tanh}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right] - \frac{3 c^2}{\epsilon \mu} \sqrt{\frac{c}{\mu }} \mathrm{sech}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^2 \mathrm{tanh}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^3$

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$U_t+\epsilon UU_x+\mu U_{xxx}$

Now substituting these in...
$= \frac{3c^2}{\epsilon} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \sqrt{\frac{c}{\mu}} $
$+ \epsilon \frac{3c}{\epsilon}\mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \frac{-3c}{\epsilon} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \sqrt{\frac{c}{\mu}} $
$+ \mu \left( \frac{6c^2}{\epsilon \mu} \sqrt{\frac{c}{\mu }} \mathrm{sech}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^4 \mathrm{tanh}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]
- \frac{3 c^2}{\epsilon \mu} \sqrt{\frac{c}{\mu }} \mathrm{sech}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^2 \mathrm{tanh}\left[\frac{1}{2} (-c t+x) \sqrt{\frac{c}{\mu }}\right]^3 \right)$
$= \frac{3c^2}{\epsilon} \sqrt{\frac{c}{\mu}} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] $
$- \frac{9c^2}{\epsilon} \sqrt{\frac{c}{\mu}} \mathrm{sech}^4\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right]$
$+ \frac{6c^2}{\epsilon} \sqrt{\frac{c}{\mu }} \mathrm{sech}^4\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] $
$- \frac{3 c^2}{\epsilon} \sqrt{\frac{c}{\mu }} \mathrm{sech}^2\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] \mathrm{tanh}^3\left[\frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)\right] $

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Let $\alpha = \frac{1}{2}\sqrt{\frac{c}{\mu}}(x-ct)$, then we can write the equation as:

$\frac{3c^2}{\epsilon} \sqrt{\frac{c}{\mu}} \left[ \mathrm{sech}^2(\alpha) \mathrm{tanh}(\alpha) - \mathrm{sech}^4(\alpha) \mathrm{tanh}(\alpha) - \mathrm{sech}^2(\alpha) \mathrm{tanh}^3(\alpha) \right]$
$= \frac{3c^2}{\epsilon} \sqrt{\frac{c}{\mu}} \mathrm{sech}^2(\alpha) \mathrm{tanh}(\alpha) \left[ 1 - \mathrm{sech}^2(\alpha) - \mathrm{tanh}^2(\alpha) \right]$
$= \beta \left[ 1 - \frac{1}{\mathrm{cosh}^2(\alpha)} - \frac{\mathrm{sinh}^2(\alpha)}{\mathrm{cosh}^2(\alpha)} \right]$
$= \beta \left[ 1- \frac{1+ \mathrm{sinh}^2}{\mathrm{cosh}^2} \right] = 0$

• I will now use a numerical solver to show that the $c$ parameter of the KdV equation controls the speed of the soliton. For this I modified the given soliton.c file given to us in lecture. I initialized the wave to the soliton solution from Section 1. I then propogate the wave by updating the solution at each time step. Here is a series of pictures showing the first and final wave functions:
  
  
  
• The top images are the first and last time slices of the soliton for an initial position of 1.0 and speed 1.0. The next ones are for a position of 1.0 and speed 1.5. The last set is for a position of 2.0 and speed 2.0. For all tests, 8000 time steps were taken with a $\Delta t$ of 0.001, giving a time lapse of 8. Notice that the difference in position form the first and last time steps agrees with this value with the speeds noted.

• I modified the intitial wave to be a square wave with width b and height a. I then ran the code and observed the results. Here is a table that shows the interesting results in changing the parameters:
  
  

  	Initial Wave	Final State
  (a,b) = (3,5)
  (a,b) = (5,1)
  (a,b) = (5,10)

  
  
• From these plots it seems that the number of emerging solitons is related to the height of the square wave. The width of the wave seemed only to effect the amplitude of the solitons produced. Notice that in all of the plots (including the ones from Section 2) there is an oscillating background signal on all of the waves.

• In order to see the effects of the so-called shockwaves of the KdV equation's solutions I modified the code to alter the value of $\mu$ after the soliton has been initialized to a command line parameter. I found the relation between this $\mu$ parameter and the resulting stable width of the shockwave. Here is a graph of the relationship with a linear fit:
  
  
  
• This shows that as the damping parameter $\mu$ decreases the width increases, and consequently the amplitude increases. When $\mu$ goes down to 0 an interesting phenomenon occurs. There is a leading strong wave (with the narrow width and high amplitude from before), but there are also trailing waves that are sequentially smaller. Here is a picture of the final slice of the KdV equation for the case where $\mu=0$