Initial value problem and Caratheodory

Question

Consider the initial value problem $\begin{cases} u'(t)=f(t,u(t)), \,\,\,\, t \in (0,T) \\u(0)=0 \end{cases}$

The function $$f:[0,T]\times\mathbb R \to \mathbb R$$ suffices a Caratheodory-condition and on the interval $[0,T]$ there is an integrable function $m=m(t)$ where $|f(t,u) \leq m(t)$ on $[0,T] \times\mathbb R$. Now let $l:(0,T) \to [0, \infty)$ be a nonnengative, integrable function and $\omega:(0,T) \to \mathbb R$ a continuous function where $\int_0^\delta\frac{1}{\omega(r)} \, dr =\infty $ for all $\delta>0$. For sufficiently small $t$ and $|v-w|$ the inequality $|f(t,v)-f(t,w)| \leq l(t)\omega(|v-w|)$ holds.How can I now show that in a neighborhood of $0$ the solution of the initial value problem is unique? Some hints are much appreciated.

Answer 1

Existence is a consequence of the Carathéodory conditions.

For the uniqueness, note that for two solutions $u$ and $v$ you have $$ |u'(t)-v'(t)|\le l(t)\omega(|u(t)-v(t)|). $$ Hint: Show that if $u\ne v$, then $$ +\infty=\int_0^\delta\frac{|u'(t)-v'(t)|}{\omega(|u(t)-v(t)|)}\,dt\le \int_0^\delta l(t)\,dt<+\infty. $$