How to Solve a 3rd Order Differential Equation for Alfven Waves?

[anton]

Hi!

I need to solve a differential equation for my bachelor's project. Rather i need to find a asymptotic solution, when x goes to infinity.
That eq. describes Alfven Waves in rotating inhomogenious plasmas, accounting effects of finite larmor radius.
So the equation is as follows:

$$f'''(x)+xf'(x)+\frac{3}{2}f(x)=0$$

I made some research and I found that such equations are actually studied, but unfortunaly I can find the articles, which concern such an equation.

First article is:
Pfeiffer, G. W. Asymptotic solutions of y"'+qy'+ry=0. J. Differential Equations 11 (1972), 145-155.

Where q and r are some functions of x.

Second:

Hershenov, J. Solutions of the Differential equation y"'+ay'+by=0. Stud.
Appl. Math. 55 (1976), 301-314.

Where a and b are some constants.

The second article more likely matches my needs.

Also i know that equaion
$$f'''-4xf'-2f=0$$
which differs from mine only with constants, is Generalized Airy's Equation, and whose solution is well known and is expressed through products of Airy's functions.

If anyone can help me with this eq., or if you could find one of those article it would help my diploma very much!

thx for your concideration.

 

[Clausius2]

Hi!

I need to solve a differential equation for my bachelor's project. Rather i need to find a asymptotic solution, when x goes to infinity.
That eq. describes Alfven Waves in rotating inhomogenious plasmas, accounting effects of finite larmor radius.
So the equation is as follows:

$$f'''(x)+xf'(x)+\frac{3}{2}f(x)=0$$

The standard method for obtaining the asymptotic behavior as x goes to infinity when the coefficients are linear is to work out the leading behavior of the solution. Assume $$f=e^{S(x)}$$. This gives you a differential equation for S(x). There should be some terms which can be neglected for large x. Assume some of them as negligible and check your assumption afterwards. That'll be the leading behavior.

If I were you I would take $$xS'\sim -3/2$$ as the dominant balance. Thus, $$S(x)\sim -3/2lnx+c(x)$$ being $$c(x)<<lnx$$. Substitute again the expression for S(x) into the differential equation calculated before and work out c(x) assuming another dominant balance.

The final form (at first order) should be like:

$$f(x)\sim Ax^{-3/2}e^{c(x)}$$

 

[anton]

how come that I can take $$xS'\approx-3/2$$?
I've calculated c(x) as you proposed: $$c(x)=3/2lnx+C1x^2+C2x+C3$$. So the final form is like $$f(x)\approx Ae^{c(x)}$$
Daymn... i don't understand that WKB method...

 

[Clausius2]

how come that I can take $$xS'\approx-3/2$$?
I've calculated c(x) as you proposed: $$c(x)=3/2lnx+C1x^2+C2x+C3$$. So the final form is like $$f(x)\approx Ae^{c(x)}$$
Daymn... i don't understand that WKB method...

First you assume a dual dominant balance, and then you must check that every other term is negligible. Your c(x) is wrong, because x^2>>lnx as x goes to infinity, and that cannot be.

 

[Tide]

Here's another approach that you may find helpful.

Laplace transform your original ODE. This will give you a first order DE in the transform variable. Solve that DE. You can find the asymptotic behavior of the desired solution by applying the phase integral method to the inverse Laplace transform.

The phase integral method means finding points of "stationary phase" (you'll have three of them for your system). Typically, the integrand in the inverse transform will be something like

$$e^{\phi (\kappa x)}$$

where $\kappa$ is your transform variable. You want to find points in the complex $\kappa$ plane for which $d\phi/d\kappa = 0$ and, for large x, you'll expand the phase function $\phi$ about those points through second order terms.

These points of stationary phase thus become saddle points and you can integrate along the paths of steepest descent through those points. The basic idea is that the most significant contribution to the integrals comes from $\kappa$ close to those points when x is sufficiently large.

I recommend trying out the procedure on Airy's equation to get the hang of it. Once you've done that you can move on to dealing with Alfven waves.