Grimanijd

2022-07-03

Consider the folloing ODE system:
${x}^{\prime }\left(t\right)=A\left(t\right)x\left(t\right)+b\left(t\right)$
$x\left({t}_{0}\right)={x}_{0}$
Show that the solutions of the system are globally defined.

soosenhc

Expert

We assume that $b,A$ are continuous over $\mathbb{R}$ and that (1) $t\in \mathbb{R}\to |{\int }_{{t}_{0}}^{t}b|$ is bounded by $K$ and (2) $t\in \mathbb{R}\to |{\int }_{{t}_{0}}^{t}||A|||$ is bounded by $L$.
Then $x\left(t\right)={\int }_{{t}_{0}}^{t}Ax+{\int }_{{t}_{0}}^{t}b+{x}_{0}$ and $||x\left(t\right)||\le {\int }_{{t}_{0}}^{t}||A||||x||+K+||{x}_{0}||$; according to Gronwall, $||x\left(t\right)||\le \left(K+||{x}_{0}||\right)\mathrm{exp}\left({\int }_{{t}_{0}}^{t}||A||\right)\le \left(K+||{x}_{0}||\right)\mathrm{exp}\left(L\right)$. Thus the solution $x\left(t\right)$ 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, ${K}_{I}=\underset{t\in I}{sup}|{\int }_{{t}_{0}}^{t}b|,{L}_{I}=\underset{t\in I}{sup}|{\int }_{{t}_{0}}^{t}||A|||$. Then $\underset{t\in I}{sup}||x\left(t\right)||\le \left({K}_{I}+||{x}_{0}||\right)\mathrm{exp}\left({L}_{I}\right)$ and we are done.

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