Prove that the solutions of ODE system are globally defined

Question

Consider the folloing ODE system:

$$x'(t)= A(t) x(t)+ b(t)$$ $$x(t_0)=x_0$$

Show that the solutions of the system are globally defined.

Hint: We have to use gronwall inequality lemma.

Could someone help with this? I can not get to the right answer. Thanks!!

Answer 1

We assume that $b,A$ are continuous over $\mathbb{R}$ and that (1) $t\in\mathbb{R}\rightarrow |\int_{t_0}^tb|$ is bounded by $K$ and (2) $t\in\mathbb{R}\rightarrow |\int_{t_0}^t||A|||$ is bounded by $L$.

Then $x(t)=\int_{t_0}^t Ax+\int_{t_0}^t b+x_0$ and $||x(t)||\leq \int_{t_0}^t||A||||x||+K+||x_0||$; according to Gronwall, $||x(t)||\leq (K+||x_0||)\exp(\int_{t_0}^t||A||)\leq (K+||x_0||)\exp(L)$. Thus the solution $x(t)$ is bounded over any segment and, consequently, the maximal solution is defined over whole $\mathbb{R}$.

EDIT. We can do better (although I feel that my post does not interest the OP). The conditions (1),(2) are so that the solution exists and is bounded over $\mathbb{R}$. If you want only the existence, then (1),(2) are useless.

Proof. Let $I$ be a segment, $K_I=\sup_{t\in I}|\int_{t_0}^tb|,L_I=\sup_{t\in I} |\int_{t_0}^t||A|||$. Then $\sup_{t\in I}||x(t)||\leq (K_I+||x_0||)\exp(L_I)$ and we are done.