KdV Equation - Modelling Soliton


by quid
Tags: equation, modelling, soliton
quid
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#1
Jan19-12, 05:24 PM
P: 19
Hi all,

I am attempting to model soliton formation numerically. The solitons will be formed by moving a body of some sort through a shallow channel of water with the free surface subject to atmospheric pressure.

My goal would be to numerically predict wave amplitudes, wavelengths, velocities etc.

I have read a bit about the Korteweg-de Vries equation however I have no idea where to start in terms of solving it to find the desired unknowns.

Could someone please help shed some light on where to start?

Thanks
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Studiot
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#2
Jan19-12, 06:43 PM
P: 5,462
google 'the Hirota Method'

go well
quid
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#3
Jan19-12, 07:57 PM
P: 19
Quote Quote by Studiot View Post
google 'the Hirota Method'

go well
Thanks.

What I am really struggling with is the implementation of the KdV solutions to real life applications. What I mean by that is I have no idea what the variables are/mean and how it applies to my situation.

Surely the equation relates somehow to the topography, the pressure source, velocity etc. but HOW??

Studiot
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#4
Jan20-12, 04:03 AM
P: 5,462

KdV Equation - Modelling Soliton


What form do you have the KDV in?

If you think back to the ordinary wave equation, it is written as spatial displacement in terms of amplitude, time and a constant that has the dimensions of velocity squared.

In the derivation of the ordinary (linear) wave equation additional properties of the medium are needed. These might be thermodynamic equations of state, elastic equations, continuity etc.
In order to recover other physical properties such as pressure you have to return to these equations.

The same is true with non linear equations and their solutions. KDV is not the only NL equation leading to soliton solutions, but at least it is couched in terms of some physical properties ( mean depth, displacement, and time etc).
quid
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#5
Jan22-12, 04:32 PM
P: 19
I have KdV in the form:

uxxx + 6uux + ut = 0;

I'm finding that there is many solutions to this equation. Alot of them I barely understand the derivation. Do I need to derive solutions myself or can I use solutions of others.

In my case I require a wave speed c of 5m/s and all other variables can be changed. What solution should I use? The hirota method?

Sorry but I'm really struggling with this topic.

Thanks so much Studiot
Studiot
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#6
Jan22-12, 05:11 PM
P: 5,462
Should your equation not be

Ut - 6UUxx + Uxxx = 0

ie -6, not +6

Incidentally even without TEX you can use the very convenient subscript and superscript functions directly from the icon on the full reply box.
quid
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#7
Jan22-12, 05:13 PM
P: 19
Quote Quote by Studiot View Post
Should your equation not be

Ut - 6Uxx + Uxxx = 0

ie -6, not +6
Sorry yes you are right,

Ut - 6UUxx + Uxxx = 0
Studiot
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#8
Jan22-12, 05:30 PM
P: 5,462
OK take this equation and use the method of characteristics to assume a solution

[tex]u(x,t) = f(\eta )[/tex]

where [tex]\eta [/tex] is parameter and c is a constant.

substitute

[tex] - cf' - 6ff + f''' = 0[/tex]

integrate once A is a constant

[tex] - cf - 3{f^2} + f'' = A[/tex]

Integrate again, B is another constant

[tex]\frac{1}{2}{\left( {f'} \right)^2} = {f^3} + \frac{1}{2}c{f^2} + Af + B[/tex]

for a single wave we need the solution to die away to zero in both directions so imposing boundary conditions

[tex]f,f',f' \to 0\;as\;\eta \to \pm \infty [/tex]

the equation becomes

[tex]{\left( {f'} \right)^2} = {f^2}(2f + c)[/tex]

rearrange and integrate

[tex]\int {\frac{{df}}{{f\sqrt {(2f + c)} }}} = \int {d\eta } [/tex]

Use substitution

[tex]f = \frac{1}{2}c\sec {h^2}\theta [/tex]

to end up with the standard solution for a water wave

[tex]f(x - ct) = - \frac{1}{2}c\sec {h^2}\{ \frac{{\sqrt c }}{2}\left( {x - ct - {x_0}} \right)[/tex]

does this help?
quid
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#9
Jan22-12, 05:35 PM
P: 19
Sorry mate, still pretty lost. Can you elaborate?
Studiot
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#10
Jan22-12, 05:37 PM
P: 5,462
I'd better draw a diagram.

Are you studying fluid mechanics or maths or computing?
quid
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#11
Jan22-12, 05:48 PM
P: 19
Fluid mechanics. My maths obviously isnt up to scratch.

I can use MatLab too if that helps.
Studiot
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#12
Jan22-12, 06:03 PM
P: 5,462
So do you need the derivation or can we work in a more 'fluid mechanicsy' format?

Why are you going for partial diff shorthand?
Studiot
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#13
Jan22-12, 06:10 PM
P: 5,462
Hirota's method works for this, but is a linearisation mthod.

You mentioned trying to work a numerical method. Are you trying to develop a numerical calculation 'molecule' or just to follow Hirota?
quid
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#14
Jan22-12, 06:11 PM
P: 19
No the derivation is not too important.

The fluid mechanics is what I need to get my head around.

I plan to run and experiment by running a pressure source through an open channel of water to generate solitons. What I need to predict (to certain accuracy) is the soliton height, velocity, profile, wavelength etc for certain froude depth numbers.

I have been advised that solving KdV was the way to go about it as any linear analysis cannot predict solitons accurately?
quid
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#15
Jan22-12, 06:12 PM
P: 19
Any derivation is fine. I just am getting confused as there is so many all involving different parameters. I'm hoping I can just use the solutions already discovered by other methods?
Studiot
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#16
Jan22-12, 06:44 PM
P: 5,462
Here is the wave in terms of some real world quantities.
a is the amplitude eta is the (wave) function which describes the action (shape), c0 is the velocity.
h0 is the undisturbed depth.
Note the usual wave (x-Vt) on the horiz axis ie a fucntion of x and time.

The soliton is travelling left to right and we ignore the left hand (negative) half.

The KDV for water solitons is

[tex]\frac{{\partial \eta }}{{\partial t}} + {c_0}\frac{{\partial \eta }}{{\partial x}} + \frac{{3{c_0}}}{{2{h_0}}}\eta \frac{{\partial \eta }}{{\partial x}} + \frac{{1{c_0}h_0^2}}{6}\frac{{{\partial ^3}\eta }}{{\partial {x^3}}} = 0[/tex]

where

[tex]{c_0} = \sqrt {gh_0^2} [/tex]

c0 = √gh0

edit see post#20


is the velocity of gravity waves.

A solution is

[tex]\eta = a\sec {h^2}\left\{ {\sqrt {\frac{{3a}}{{4h_0^3}}} } \right.\left. {\frac{{\left( {x - Vt} \right)}}{1}} \right\}[/tex]

Note I said 'A solution is'. As you note there are many, but we want one that decays away to infinity on either side as I said in my previous derivation.

Would you like the derivation in this format, rather than as previous?

Does this help?
Attached Thumbnails
soliton1.jpg  
Studiot
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#17
Jan22-12, 07:20 PM
P: 5,462
I should perhaps mention that the solution presented depends upon an initial disturbance

[tex]{\eta _0}[/tex]

If you are going to be investigating the relation between the initial disturbance and the dispersive terms in the equation then you may need to rework the derivation.
quid
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#18
Jan22-12, 09:11 PM
P: 19
Quote Quote by Studiot View Post
[tex]{c_0} = \sqrt {gh_0^2} [/tex]
is the velocity of gravity waves.
Is is not:

[tex]{c_0} = \sqrt {gh_0} [/tex]??


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