Grimanijd

Answered

2022-07-03

Consider the folloing ODE system:

${x}^{\prime}(t)=A(t)x(t)+b(t)$

$x({t}_{0})={x}_{0}$

Show that the solutions of the system are globally defined.

${x}^{\prime}(t)=A(t)x(t)+b(t)$

$x({t}_{0})={x}_{0}$

Show that the solutions of the system are globally defined.

Answer & Explanation

soosenhc

Expert

2022-07-04Added 16 answers

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 $x(t)={\int}_{{t}_{0}}^{t}Ax+{\int}_{{t}_{0}}^{t}b+{x}_{0}$ and $||x(t)||\le {\int}_{{t}_{0}}^{t}||A||||x||+K+||{x}_{0}||$; according to Gronwall, $||x(t)||\le (K+||{x}_{0}||)\mathrm{exp}({\int}_{{t}_{0}}^{t}||A||)\le (K+||{x}_{0}||)\mathrm{exp}(L)$. 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, ${K}_{I}=\underset{t\in I}{sup}|{\int}_{{t}_{0}}^{t}b|,{L}_{I}=\underset{t\in I}{sup}|{\int}_{{t}_{0}}^{t}||A|||$. Then $\underset{t\in I}{sup}||x(t)||\le ({K}_{I}+||{x}_{0}||)\mathrm{exp}({L}_{I})$ and we are done.

Then $x(t)={\int}_{{t}_{0}}^{t}Ax+{\int}_{{t}_{0}}^{t}b+{x}_{0}$ and $||x(t)||\le {\int}_{{t}_{0}}^{t}||A||||x||+K+||{x}_{0}||$; according to Gronwall, $||x(t)||\le (K+||{x}_{0}||)\mathrm{exp}({\int}_{{t}_{0}}^{t}||A||)\le (K+||{x}_{0}||)\mathrm{exp}(L)$. 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, ${K}_{I}=\underset{t\in I}{sup}|{\int}_{{t}_{0}}^{t}b|,{L}_{I}=\underset{t\in I}{sup}|{\int}_{{t}_{0}}^{t}||A|||$. Then $\underset{t\in I}{sup}||x(t)||\le ({K}_{I}+||{x}_{0}||)\mathrm{exp}({L}_{I})$ and we are done.

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