Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am interested in a system of differential equations that is non-linear, but it doesn't seem to be too crazy. I'm not very good at non-linear stuff, so I thought I'd throw it out there.

The actual equations I'm looking at have several parameters that'd I'd like to tweak eventually.

q' = k - m / r
r' = i - n r - j q

i, j, k, m and n are all real-valued constants. I'm guessing that this system would be cyclical in nature, but I'm not sure if it has any explicit solution, so I have produced a version of it with the constants removed to see if that can be solved:

q' = 1 - 1 / r
r' = 1 - r - q

Anyone know if either of these are solvable and what kind of techniques would be needed to solve them if so?

The first equation is based on a polar coordinate system where Q (or theta) is the angle and r is radius, and I've made a number of simplifications to make it somewhat tractable.

share|improve this question
    
If you don't mind, could you post the "polar form" you said you had? –  J. M. Nov 2 '10 at 10:26
add comment

2 Answers 2

up vote 5 down vote accepted

Taking that second question,

$r' = i - nr - jq$

and differentiating gives

$r'' = -nr' - jq' = -nr' - j(k-\frac{m}{r})$

or in other words

$r'' + ar' + \frac{b}{r} = c$

which is a much simpler differential equation only one variable. I think that you could probably solve this with power series or clever guessing, but it needs to be worked out.

share|improve this answer
    
In the resulting equation $t$ does not appear, so a standard trick can be used to reduce it to a first order equation. –  Mariano Suárez-Alvarez Nov 1 '10 at 22:43
1  
(I explained how to do this in an example here: math.stackexchange.com/questions/7341/…) –  Mariano Suárez-Alvarez Nov 1 '10 at 22:44
    
Thanks, neat trick. –  Tony Peterson Nov 2 '10 at 0:43
    
For $c=0$, this is a special case of Kummer's equation: en.wikipedia.org/wiki/…. How to deal with the case $c\neq 0$, I don't know. –  Hans Lundmark Nov 2 '10 at 6:52
    
I'm not sure I agree that it's a "much simpler" equation. In fact, a commonly used trick for analyzing second order ODEs is to rewrite them as a system of two first order ODEs so that one can use phase plane methods... –  Hans Lundmark Nov 2 '10 at 6:53
show 1 more comment

I don't think there's very much hope for an explicit solution. However, you can easily draw a phase portrait and see qualitatively how the solutions behave (in your case, the solutions in the upper half plane $r>0$ spiral towards the stable equilibrium $(q,r)=(0,1)$. Here's a Java applet by John Polking for drawing phase portraits (including nullclines); it can also find and analyze equilibria.

share|improve this answer
    
I agree, this system doesn't look like the sort that gives a neat closed form. @Tony, if you're going to go the numerical route and you want to (try to) find periodic solutions, try to look for a solver that has "event detection" capability. –  J. M. Nov 2 '10 at 0:33
    
@J. M.: For this particular system, it's clear from the phase plane analysis that there are no periodic solution. (By the way, that Java applet can also detect periodic (or nearly periodic) solutions.) –  Hans Lundmark Nov 2 '10 at 6:55
    
I haven't actually peered at the nullclines for the second system until your comment; you're right. However, the first system with more parameters might be a different story. –  J. M. Nov 2 '10 at 9:16
    
@J. M.: Good point. I thoughtlessly assumed that the second system was just the dimensionless form of the first one, but there are too many parameters. A dimensionless form would be for example $x'=1-\alpha/y$, $y'=1-x/\beta -y$. But this doesn't seem to really change anything. The Jacobian at the equilibrium $(x,y)=(\beta(1-\alpha),\alpha)$ has trace $-1$ and determinant $1/\alpha\beta$ (unless I made an error in my quick calculation). Hence it's a stable node or a stable spiral, so no periodic solutions in the general case either. –  Hans Lundmark Nov 2 '10 at 9:34
    
(Of course I'm being sloppy here about global questions, but computer plots indicate quite clearly that the fixpoint is globally attracting in the upper half plane. Although I haven't bothered proving that this is true, I would be quite surprised if it was possible to have limit cycles in this system.) –  Hans Lundmark Nov 2 '10 at 10:20
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.