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I have to find conditions under which the solutions of $x'=Ax$ are bounded for $t\rightarrow \infty$ and $t\rightarrow \pm \infty$. We proved in our course that $\lim_{t \rightarrow \ \infty} x(t)=0$ and $\lim_{t \rightarrow - \infty} |x(t)|=\infty$, if all the eigenvalues of $A$ have negative real part (and for the last limit $x$ is not the zero solution); if $A$ has an eigenvalue greater that zero, then $\lim_{t \rightarrow \infty} |x(t)|=\infty$.

So I think that the conditions are: $x$ is bounded for $t\rightarrow \infty$ , if the real part of all the eigenvalues are negative (trivial, since it follows from my course) and bounded for $t\rightarrow \pm \infty$ if the real parts of all the eigenvalues vanishes. Do you think this is sufficient - or that my professor could want some other additional (finer) conditions ?

Side question: Since we didn't specify what $|x(t)|$ means, I took it means the norm of $x(t)$ and not the vector of the absolute values of the components of $x(t)$, because in the latter case $\lim_{t \rightarrow \infty} |x(t)|=\infty$ doesn't makes sense to me, because only a sequence of numbers can have the limit $\infty$. Am I right that it means the norm ?

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Yes, you are right that $|x(t)|$ means norm, although it is more common to denote it $\|x(t)\|$. Notice that all norms on $\mathbb{R}^n$ are equivalent so boundedness of $x(t)$ in any norm just means that the entries of $x(t)$ stay bounded. You can get boundedness of $\|x(t)\|$ on $[0, \infty)$ in essentially two different ways, you can read both of them off the formula for the exponential of $A$: $$x(t) = \exp(tA) = P\exp(tD+tN)P^{-1} = P\exp(tD)\exp(tN)P^{-1}$$ where $$A = P(D+N)P^{-1}$$ is the Jordan decomposition of $A$. This means that $D$ is diagonal whose entries are generalized eigenvalues of $A$ and the entries of $N$ consists of $0$s, except for the entries on the super-diagonal which may (or may not) be $1$. Notice that $N$ and $D$ always commute so it is valid to say $\exp(tD+tN) = \exp(tD)\exp(tN)$. $N$ is nilpotent so you can calculate its exponential to be a matrix whose entries are polynomial in $t$. Therefore, your typical entry of $x(t)$ is of the form $$\sum e^{\lambda t}p(t)$$ where $\lambda$ is an eigenvalue of $A$ and $p$ is a polynomial. If $\lambda$ has negative real part then this expression stays bounded. But it also stays bounded if $\lambda$ has real part equal to $0$ and $p$ is constant. This happens for example when $A$ is diagonalizable and the real parts of the eigenvalues of $A$ are less than or equal to $0$. To get boundedness on $(-\infty, \infty)$ you would need that $A$ is diagonalizable and all of its eigenvalues have real part $0$. So you were right in what you said for the diagonalizable case, but in general you have to look at the impact of the polynomials caused by any generalized eigenvectors of $A$.

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    But doesn't it still hold in the general case ( even if $A$ isn't diagnalizable, because every Jordan block of the Jordan decomposed matrix has the form $ e^{\lambda t} \left(\begin{array}{ccccc} 1 & t & t^{2} & \cdots & \frac{t^{n-1}}{(n-1)!}\\ & \ddots & \ddots & \ddots & \vdots\\ & & \ddots & \ddots & t^{2}\\ & 0 & & \ddots & t\\ & & & & 1 \end{array}\right)$ and thus my solution $x(t)=Q^{-1}y(t)$, where $y$ is the solution of the transformed system $y'=JQy$ where $Q$ is the change of base matrix and $J$ the JNF of $A$ [cont]2011-12-03
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    [cont] still contains in each component only expressions of the form $e^{\lambda t} p(t) $, so looking only at $e^{\lambda t} $ should suffice ?2011-12-03
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    if the eigenvalue has negative real part, the solutions blow up as $t \to -\infty$. So it must have 0 real part. But then any polynomial blows up as $t \to \pm \infty$ so the polynomials must be constant. Hence the JNF cannot have $1$s on the superdiagonal so the matrix must be diagonalizable to stay bounded on all of $\mathbb{R}$ is what I was trying to say. Or am I completely missing your point?2011-12-03
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    Ah, no, I think I was missing your point. I think I got it now. One last question though: Should a typical entry of $x(t)$ be $e^{\lambda t} p(t)$ instead of $\sum e^{\lambda t} p(t)$ ?2011-12-04
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    no, because the the $\lambda$s can vary over the sum also.2011-12-04
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Yes you are right in both remarks: 1) for boundedness of $x$ you need all the eigenvalues of $A$ having negative real part. In that case, $A$ is called "stable" or "Hurwitz". 2) $\left|x(t)\right|$ is a vector norm.