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I'm trying to show:

Let $A:[0,\infty[\to \mathcal{M}(n,\mathbb{R})$ a function and suppose that all solutions of the system of differential equations: $$\vec{x'}(t)=A(t)\vec{x}(t) \ \ \ (\star)$$ are bounded when $t\geq 0$. If $X(t)$ is a fundamental matrix (a matrix of fundamental solutions) of the system $(\star)$ show that $X^{-1}(t)$ is bounded for $t\geq 0$ if and only if the function $t\to \int^t_0 \operatorname{tr}A(s)ds$ bounded below.

I know that for the system $(\star)$, a fundamental matrix is $e^{tA}=e^{tPJP^{-1}}=Pe^{tJ}P^{-1}$ where $J$ is the Jordan form of $A$, but I have no idea how to prove the exercise.

Thanks for your help.

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1 Answer 1

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The determinant $f(t)=\mathrm{det}\,X$ of the fundamental matrix satisfies the equation $$(\ln f)'=(\ln\mathrm{det}\,X)'=\left(\mathrm{Tr}\ln X\right)'=\mathrm{Tr}\left(X'X^{-1}\right)=\mathrm{Tr}\,A,$$ so that $\displaystyle f(t)=\mathrm{const}\cdot\exp\int^t\mathrm{Tr}\,A(t)\,dt$. The statement for $X^{-1}$ follows from this relation and boundedness of $X$.

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