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Suppose that for each $t\in\mathbf{R}$ we have an $n\times n$ matrix $A(t)$ and that there exists an $m\geq 0$ in such a way that for each $|t|\geq m$ the matrix $A(t)$ is positive definite. The family of matrices $A$ depends continuously on the parameter $t$. Moreover, the matrix norm of $A(t)$ is bounded by some constant. Now consider the following ODE $$\dot{x}(t)=A(t)x(t).$$ Then from the Cauchy-Lipschitz theorem, it follows that for each $y_0\in\mathbf{R}^n$ there exists a unique solution $x(t)$ of this ODE with $x(0)=y_0$. My question is: can I conclude that $x=0$ is the only bounded solution?

For $A$ independent of $t$ this is true, as we can compute the solution exactly and just see that any non-trivial solution is unbounded. For $n=1$, it is also true: for some fixed $t>m$ we can look at any solution $x$ with $x(t)>0$ (if $x(t)<0$ just look at $-x$), then the equation and the positive definiteness implies $x'(t)>0$. Hence $x$ is increasing at that point, which means that it grows to infinity as $t\rightarrow\infty$. Since any non-trivial solution to this ODE is non-zero at some $t>m$ it follows that $x=0$ is the only bounded solution. However, in the general non-autonomous $n\times n$ case I do not see how this works precisely. I do not necessarily need a proof, a reference to a book or paper is fine as well.

Update: Thanks to humanStampedist I discovered I at least need to assume that $A(t)\rightarrow A^{\pm}$ as $t\rightarrow\pm\infty$ for some positive definite matrices $A^{\pm}$. However, assuming this, I still do not know the solution, so more help is needed.

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The norm of every the non-trivial solutions tend to infinity, as $t\to\infty$, provided that the least eigenvalues of $A(t)$ is bounded from below by a $\lambda_0>0$, i.e., if $$ \big(x(t),A(t)x(t)\big)\ge \lambda_0 |x(t)|^2 $$

Indeed, is $x(t)$ is a solution, then $$ \frac{d}{dt}|x(t)|^2=2\big(x(t),x'(t)\big)=\big(x(t),A(t)x(t)\big)\ge \lambda_0 |x(t)|^2, $$ and hence $$ \left(\mathrm{e}^{-\lambda_0 t}|x(t)|^2\right)'\ge 0 $$ and thus $$ \mathrm{e}^{-\lambda_0 t}|x(t)|^2\ge |x(0)|^2 $$ or $$ |x(t)|^2\ge \mathrm{e}^{\lambda_0 t}|x(0)|^2 $$

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  • $\begingroup$ Thanks, this really seems to work. You do not even need the eigenvalues of $A(t)$ to be bounded from below by $\lambda_0$ for all $t\in\mathbf{R}$, just the ones with $t\geq m$ are fine. $\endgroup$ – WillemMSchouten Feb 27 '18 at 11:38
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I doubt that you proposition is true even for $n=1$. What might happen is that the positive definitiness of $A(t)$ might disappear if $t\rightarrow\infty$. Here is the example I have in mind: $$x'(t)=\frac{1}{\cosh(t)}x(t),\ x(0)=x_0>0$$ The solution is (see e.g. http://www.wolframalpha.com/input/?i=integrate+1%2Fcosh(x)) $$x(t)=x_0\exp(2\arctan(\tanh(\frac{t}{2})),$$ which is bounded, since the $\arctan$ is bounded.

You might be able to save you proposition if you additionally assume, that the eigenvalues of $A(t)$, let's call them $\lambda_i(t)$, satisfy $$\lambda_i(t)\geq C>0\ \forall i$$ for a fixed constant $C$ independent of $t$, but I do not have a proof or a reference for this at the moment.

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  • $\begingroup$ Thanks. Luckily for me, in the context in which I want to use it, the positive definiteness does not disappear in the limits. I will edit the OP accordingly. $\endgroup$ – WillemMSchouten Feb 27 '18 at 10:44

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