Existence of solution of singular ODE Suppose $f$ is a Lipschitz continuous function defined on $\mathbb{R}$. How can one prove that the following ODE admits at least one solution.
\begin{equation}
y'' + \frac{1}{x}y' + f(y) = 0
\end{equation}
with $y(0) = y_0, y'(0) = 0$.
Thanks a lot!
Jack
 A: Given $\phi\colon[0,\infty)\to\mathbb{R}$ continuous, consider the equation
$$
y''+\frac{1}{x}\,y'=\phi,\quad y(0)=y_0,\quad y'(0)=0.
$$
It is easy to see that a solution is
$$
\psi(x)=y_0+\int_0^x\frac{1}{t}\Bigl(\int_0^ts\,\phi(s)\,ds\Bigr)\,dt.
$$
Moreover
$$
|\psi(x)-y_0|\le\frac{\sup_{0\le t\le x}|\phi(t)|}{4}\,x,\quad x\ge0.
$$
The strategy to prove existence of solution will be the following. Fix an interval $[0,a]$. Given $z\in C([a,b])$ define
$$
(Tz)(x)=y_0+\int_0^x\frac{1}{t}\Bigl(\int_0^ts\,f(z(s))\,ds\Bigr)\,dt.
$$
Show that $T$ is a contraction and has a fixed point. I will carry out this program assuming $f$ is bounded.
Let $L$ and $M$ be the Lipschitz constant and a bound of $f$ respectively. Choose $a>0$ such that $a^2<4/L$. Let $X$ be the set of continuous functions defined on $[0,a]$ such that
$$
\sup_{0\le x\le a}|z(x)-y_0|\le\frac{M\,a^2}{4}.
$$
$X$ is a closed subspace of $C([a,b])$, so that it is a complete merit space for the uniform metric.
Let's see that $T$ takes $X$ into itself. If $z\in X$ then
$$
|(Tz)(x)-y_0|\le\int_0^a\frac{1}{t}\Bigl(\int_0^ts\,|f(z(s))|\,ds\Bigr)\,dt\le\frac{M\,a^2}{4}.
$$
Let's prove now that it is a contraction. If $z_1,z_2\in X$ then for all $x\in[0,a]$
$$\begin{align}
|(Tz_1)(x)-(Tz_2)(x)|&\le\int_0^a\frac{1}{t}\Bigl(\int_0^ts\,|f(z_1(s))-f(z_2(s))|\,ds\Bigr)\,dt\\
&\le\frac{L\,a^2}{4}\sup_{0\le x\le a}|z_1(x)-z_2(x)|
\end{align}$$
with
$$
\frac{L\,a^2}{4}<1.
$$
