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I'm curious about techniques for solving a nonautonomous* system in the case of a non-linear differential equation. There's a simple example in my textbook (Hirsch, Smale, Devaney) where we obtain the following nonautonomous equation (after linearizing about the origin)

$$ x'(t) = x + y_0^2 e^{-2 t}$$

In this case, we simply guess a particular solution (which is obvious from the given equation) and everything follows through in a straightforward manner. In a paper I'm working through I have an equation that looks something like

$$X'(t) = \frac{t}{6} - \frac{(a + X)^2}{t^2},$$

and I'm not sure what strategies I should have at my disposal here. The literature on non-autonomous non-linear systems seems to be rather scarce from the bit of googling I've done. Any insight would be much appreciated.

*By non-autonomous I simply mean there is a $t$ hanging around on the right-side of the equation.

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First of all, your equation is linear. Second, it is even one of the form $y' + p(x) y = q(x)$ for which there is a standard solution using a multiplier. I don't see what makes you think your equation is so special, unless you made some mistake in writing it down? –  Raskolnikov Aug 24 '11 at 11:39
    
@Raskolnikov Yes, I do apologise for the post. The $\frac{a + X}{t}$ term should be squared (corrected now). Though, I didn't intend for that equation to be the focus of the post. I'm mainly just curious about strategies for solving non-autonomous non-linear equations in general. cheers –  tentaclenorm Aug 24 '11 at 11:55
    
Thank you for editing it. It makes more sense now. –  Raskolnikov Aug 24 '11 at 12:11
    
If you look at the usual proof of existence in ODE textbooks for first-order nonlinear equations, there's an approach via fixed point of a mapping defined by taking an integral in the neighborhood of an initial point. To show the mapping is a contraction, and therefore has a fixed point, some strong continuity assumptions on the integrand are needed. In your case there's a singularity at t=0, so you would need to work "away" from that point. –  hardmath Aug 24 '11 at 13:10

1 Answer 1

Let $y=a+X$ ,

Then $y'=X'$

$\therefore y'=\dfrac{t}{6}-\dfrac{y^2}{t^2}$

Let $y=\dfrac{t^2u'}{u}$ ,

Then $y'=\dfrac{t^2u''}{u}+\dfrac{2tu'}{u}-\dfrac{t^2(u')^2}{u^2}$

$\therefore\dfrac{t^2u''}{u}+\dfrac{2tu'}{u}-\dfrac{t^2(u')^2}{u^2}=\dfrac{t}{6}-\dfrac{t^2(u')^2}{u^2}$

$\dfrac{t^2u''}{u}+\dfrac{2tu'}{u}-\dfrac{t}{6}=0$

$6tu''+12u'-u=0$

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