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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.

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    If you don't mind, could you post the "polar form" you said you had?2010-11-02

2 Answers 2

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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.

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    (cont.) But I have to admit that your procedure has the advantage of making it possible to look for the equation in tables! I found it on p.183 of Zwillinger's "Handbook of Differential Equations", which referred to section 13.1.1 in Abramowitz & Stegun.2010-11-02
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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.

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    I suspect the "simplifications" the OP did may not have had the effect he intended. Maybe a look at the polar form he said he had might be more enlightening. (Eugh, I should be putting that in the comments to the question! :D )2010-11-02