Existence of solution of singular ODE

Question

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 $y(0) = y_0, y'(0) = 0$.

Answer 1

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 $$ \frac{L\,a^2}{4}<1. $$