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I'm working on proving existence and uniqueness of a local solution to ODE's of the Lienard variety $$y'' + f(y)\,y' + g(y) = 0$$

I'm trying to put this into standard $y' = f(t,y)$ system

So I see that if I let $x = y'$ , then I can rearrange the equation so $x' = -f(y)\;x - g(y)$

That maybe gives us a first order ODE if we consider $$y'= x$$ $$x' = -f(y)\;x - g(y)$$ to be an acceptable 1st order ODE due to the Lienard equation being autonomous.

However, I'm not sure if this is good enough. Our professor was pretty adamant we use a transformation, not a substitution, and that when we are done we should have a system which is independent of $y(t)$.

Obviously looking up Lienard equations, I can see the transformation I need: $$x_1 = y$$ $$x_2 = y' + \int_0^y f(s)\;ds$$ and it follows directly from this that my proper system is $$d/dt\;x_1 = y' = x_2 - \int_0^y f(s)ds $$ $$d/dt\; x_2 = y'' + f(y)\;y' = -g(y)$$

I have 2 questions:

1) Why $x_1 = y$ and $x_2 = y' + \int_0^y f(s)\;ds$? What is the philosophical reasoning behind this choice?

AND

2) After setting up this form, can I then proceed directly to proving existence and uniqueness of a solution with a fixed point theorem (Picard-Lindelof or appropriate) or is there more I need to do first?

Phillip Hamilton
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2 Answers2

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If you look at $$ y'' +f(y)y' + g(y) = 0 $$ then we should notice that $$ \dfrac{d}{dx}\int_0^y f(s) ds = f(y)y' $$ This is a result of the fundamental theorem of Calculus. so we have $$ y'' + \dfrac{d}{dx}\int_0^y f(s) ds +g(y) = \dfrac{d}{dx}\left[y' + \int_0^y f(s) ds\right] +g(y) = 0 $$ so you can see the transformation that you would like to go for. Sidenote I spent an awful long time of my PhD working on these equations and found a few solutions to specific forms of $f(y)$ and $g(y)$

Chinny84
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  • Thanks! That makes perfect sense. So as a follow up question, once we have used the transformation, can we then proceed immediately to using local existence and uniqueness theorems? The definitions seem like they hold just fine in the presence of nonlinearities but I'm new to fixed point theorems. – Phillip Hamilton Feb 18 '16 at 16:56
  • I will take a look at the second question. – Chinny84 Feb 18 '16 at 22:14
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There is a way to transform a Lienard equation into an Abel-type first order ODE. Start with $$ y'' + f(y)y' + g(y) = 0. $$ Let $u(y(t)) = y'(t)$, then Lienard equation transforms into $$ u\frac{du}{dy} + f(y)u + g(y)=0. $$ If we now consider an independent variable $v=1/u$, we arrive at $$ \frac{dv}{dy} = v^2 f(y) + v^3 g(y). $$

mforets
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