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Based on this question:Show that: a) $X^{-1}(t)$ is bounded in $[\beta,\infty)$. b)No system solution approaches zero solution when $t \rightarrow \infty.$

I am working on the following problem, which is the inverse statement of the cited question:

Let a system $x' = A(t)x$ and suppose there are values positives $k, \beta$ such that a positive fundamental matrix $X(t)$ satisfies $\|X(t)\| \leq k$, $t \geq 0$ and $X^{-1}(t)$ is bounded in $[0,\infty)$. Show that: $$ \liminf_{t \rightarrow \infty} \int^t_\beta \operatorname{tr}(A(s))\,ds > - \infty.$$

My attempt:

Note that for some constants $M>0$ so that $$ \|X^{-1}\|=\frac{\|adj(X(t))\|}{|det(X(t))|}\le M $$ Then $$ |det(X(t))|\ge\frac{\|adj(X(t))\|}{M} $$

If I apply Liouville's Formula: $$det(X(t))=det(X(0))e^{\int_{0}^t tr(A(s))ds}$$ Then $$ |det(X(0))|e^{\int_{0}^t tr(A(s))ds}\ge\frac{\|adj(X(t))\|}{M} $$

But I am stuck on there. Because we try to show that $e^{\int_{0}^t tr(A(s))ds}$ is bounded away from constant. I cannot take $t\to \infty$ since $\|adj(X(t))\|$ dependent on $t$. How to deal with this term? we just know that upper bound but not lower bound.

H.Y Duan
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1 Answers1

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Let me try...

I am unsure what exactly a positive fundamental matrix means, but I guess it is just a fundamental matrix considered for $t \ge 0$, is that correct? I.e., we have $X(0) = I$ and $\dot{X}(t) = A(t)X(t)$, $t \ge 0$. I also guess that we can take $\beta = 0$ since $\beta$ is not actually defined.

Coefficients of $X(t)$ are continuous in $t$, and so the determinant of $X(t)$ is a continuous function of time. At $t=0$, we have $X(0)=I$ and $\det X(0)=1$. Saying that the inverse exists and is bounded implies that the determinant of $X$ does not cross zero and remains strictly positive. So, for $d := \det X$, there exists $d_0>0$ such that $d(t)\ge d_0$ for all $t\ge 0$; note that $d(0)=1$ and $d$ is continuous and bounded.

Next, by Jacobi's formula, we have $$\dot{d}(t) = d(t)\operatorname{tr}\left(X^{-1}(t)\dot{X}(t)\right) =d(t)\operatorname{tr}\left(X^{-1}(t)A(t)X(t)\right) = d(t)\operatorname{tr}\left(A(t)\right).$$

So (we can also come here with Liouville's Formula), $$d(t) = e^{\int_0^\infty\operatorname{tr}\left(A(s)\right)ds}d(0)$$ and since $d(t)\ge d_0>0$, we conclude $\int_0^\infty\operatorname{tr}\left(A(s)\right)ds \ge \ln d_0 > -\infty$.

Arastas
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