How do we prove existence of this Cauchy problem
$$
\begin{cases}
y'= f(x,y)= y \ln|y|\\
y(x_0)=y_0
\end{cases}
$$
using Peano theorem?
I try this:
First, we have that
$$
\dfrac{\partial f}{\partial y}
=
\begin{cases}
1+\ln(y) &: y>0\\
-1+\ln(y) &: y < 0\\
0 &: y=0.
\end{cases}
$$
We remark that $\dfrac{\partial f}{\partial y}$ isn't bounded in neiborhood of $y=0$, then this implies that $f$ isn't locally ipschitzian. So if $y_0 =0$ we haven't existance and unicity of an maximal solution. But if $y_0 \neq 0$ we have existance and unicity by Cauchy-Lipschitsz theorem.
If $y_0=0$ we consider the problem on $C=\{(x,y): x \in [x_0,+\infty[, |y| < +\infty\}$. We have that $f$ is countinuous and bounded on $C$, then by Peano theorem the problem admits at most one solution on $[x_0,+\infty[$.
Is my solution correct?
Thanks in advance
Best Answer
Your application of Peano's Existence Theorem is wrong, you're mostly seeking a Picard (Lipschitz) approach.
For Peano, let's see what we have. First of all, it is : $f(x,y) = y\ln|y|$.
Since the argument of $\ln$ is $|y|$, we have no issues with restrictions, thus $D_f = \mathbb R \times \mathbb R$ or in other words $f:\mathbb R \times \mathbb R \to \mathbb R$, since the variable $x \in \mathbb R$ doesn't come into explicit play but only implicit via $y \equiv y(x)$.
That obviously means that your function is continuous in $\mathbb R \times \mathbb R$, which by Peano means that every initial value problem $y(x_0) = y_0$ has a local solution.