Existence of unique global solution

Question

Let the function $f:[0,T] \times \mathbb R^d \to \mathbb R^d$ satisfy the Caratheodory conditions. Furthermore, let there be $l \in L^1(0,T)$ such that for all $t\in [0,T]$ and $v,w \in \mathbb R^d$ the inequalities $$\Vert f(t,0)\Vert \leq l(t)$$ and $$\vert f(t,v)-f(t,w) \Vert \leq l(t) \Vert v-w \Vert$$ hold.

How can I show that for all $(t_0,u_0) \in [0,T] \times \mathbb R^d$ there exists a unique global solution $u$ on $[0,T]$ of the initial value problem $$\begin{cases}u'(t)=f(t,u(t))\\u(t_0)=u_0\end{cases}$$?

Answer 1

For the existence of a solution you should look at any good book considering the Carathéodory conditions (the usual construction considers a certain sequences of functions and uses Arzelà-Ascoli).

To show that the solution is global and unique you can use similar arguments, both based on Gronwall's lemma. Indeed, $$ \|u(t)\|\le\|u_0\|+\int_0^t l(s)\|u(s)\|\,ds $$ for $t$ on some interval and so $$ \|u(t)\|\le\|u_0\|\exp\left(\int_0^t l(s)\,ds\right) $$ for $t$ in the same interval. Since $l$ is in $L^1$ the solution cannot explode and so it is global.

For the uniqueness, note that if $u$ and $v$ are solutions, then $$ \|u(t)-v(t)\|\le\int_0^t l(s)\|u(s)\|\,ds $$ now already for $t\in[0,T]$ and so $$ \|u(t)-v(t)\|\le0\exp\left(\int_0^T l(s)\,ds\right)=0 $$ for $t\in[0,T]$.