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Rayleigh's differential equation

by atrus_ovis
Tags: differential, equation, rayleigh
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atrus_ovis
#1
Dec29-12, 08:28 AM
P: 101
In Rayleigh's DE :
http://www.wolframalpha.com/input/?i...ntial+equation

What does mu stand for?
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dextercioby
#2
Dec29-12, 09:13 AM
Sci Advisor
HW Helper
P: 11,896
It's a mere real positive parameter, just like in other famous ODE's, for example the Bessel differential equation and <n>.

http://www.wolframalpha.com/input/?i...ntial+equation
atrus_ovis
#3
Dec29-12, 09:44 AM
P: 101
Well, i am asked to numerically solve it and produce a phase diagram.
Should its value be given to me?

dextercioby
#4
Dec29-12, 10:00 AM
Sci Advisor
HW Helper
P: 11,896
Rayleigh's differential equation

I guess it should, so you're free to choose any value you want: Take [itex] \mu =1 [/itex] and solve it numerically.
atrus_ovis
#5
Dec30-12, 08:28 AM
P: 101
You're right , it was supposed to be given.
Rayleigh's DE is [itex]y''-\mu y' + \frac{\mu (y')^3}{3} + y = 0[/itex]
By rearranging it to a system of DEs, you get
[tex]
y_1 = y , y_1' = y_2 \\
y_2' = \mu y_2 - \frac{\mu (y_2)^3}{3} - y_1
[/tex]

So i have only the derivative of y2 , i.e. the 2nd derivative of y1.
Since i don't have an analytical description of y2 , how do i compute it with specific parameters, according to the numerical method.
For example, for the classic Runge Kutta method,where f = y'
[tex]
k_1 = hf(x_n,y_n) = hy_2(n)\\
k_2 = hf(x_n + 0.5h,y_n + 0.5k_1) = ?
[/tex]
I should numerically approximate the intermmediate values as well?


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