Setting Lineard Equation ODE into standard form, proving existence of unique solutions

Question

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?

Answer 1

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 $g(y)$

Answer 2

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 $$ \frac{dv}{dy} = v^2 f(y) + v^3 g(y). $$