I'm trying to find the solution to the following IVP and prove its uniqueness by applying the Picard–Lindelöf theorem.
$$ y''+x^2y'+xy=0,\\ y(0)=y'(0)=0 $$
I've used the theorem in a few problems I've solved but the second order here confuses me. I know two things:
- To apply the theorem , my equation has to be of the following form : $g'(x)=f(x,g)$
- By setting $y'(x)=g(x)$ I can turn this into a first order system : $$y'(x)=g(x)\\g'(x)+x^2g(x)+xy=0$$
Now I can write the new equation like this:
$$g'(x)=-x^2g(x)-xy\ \implies g'(x)=f(x,g,y)$$
Because of the '$y$' term , this is not the form I'm looking for. Or is it? I know that $y$ is a function of $x$, so is it correct to write the following? $$g'(x)=f(x,g,y(x))=f(x,g)$$
Normally I would go on and prove that f is continuous and Lipschitz in $x,g$. What do I do in this case ? Do I have to show that f is continuous and Lipschitz in $y$ as well?
I think it should be like this \begin{align} y'(x)&=g(x)\\ g'(x)&=-x^2g(x)-xy(x) \end{align} With $v(x) = (y(x),g(x))^T$ as vector in $\mathbb R^2$ we get $$ v'(x) = f(x,v(x)) = f(x,g(x),y(x)) = A(x)v(x) $$ where $A(x)$ is the matrix $$ A(x) = \begin{pmatrix} 0 & 1 \\ -x & -x^2 \end{pmatrix} $$ The function $f(x,v(x))$ is Lipshitz in the second argument.