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Consider a linear ODE: $\dot{x} = A x$ where $A$ is Hurwitz, i.e. all its eigenvalues have negative real parts. Thus the system is exponentially stable. We know that there exists positive numbers $\beta$ and $\alpha$ such that $\| e^{A t} \| \leq \beta e^{-\alpha t}$ for all $t$. I see this result being used in many analysis.

My question is how to (practically) compute these values? In particular, if I pick $\alpha$ so that $ -\alpha > \max_i \Re(\lambda_i)$, where $\lambda_i$ are the eigenvalues of $A$, then how to compute a tight value for $\beta$?

Truong
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2 Answers2

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Put $A$ in Jordan canonical form: $A = U J U^{-1}$. Then $e^{At} = U e^{Jt} U^{-1}$ so $\|e^{At}\| \le \|U\| \|U^{-1}\| \|e^{Jt}\|$. Of the eigenvalues with greatest real part (say $r = \max_i \Re(\lambda_i)$), take one with the largest Jordan block (say of size $m$). Then for $t \ge 0$, $\|e^{Jt}\| \le e^{rt} \sum_{k=0}^m t^k/k!$.

EDIT: This was not quite right. It is true for sufficiently large $t$. But in general, you have to say $$ \|e^{Jt}\| \le \max_i e^{r_i t} \sum_{k=0}^{m_i} t^k/k!$$ where the $i$'th Jordan block has size $m_i$ and eigenvalue $\lambda_i$ with $r_i = \Re(\lambda_i)$.

Robert Israel
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    But then we will have $p(t) e^{r t}$, where $p(t)$ is a polynomial in $t$ instead of a constant $\beta$. – Truong Nov 09 '11 at 22:12
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    If there is a Jordan block of size greater than 1, yes. For example, $A = \pmatrix{-1 & 1\cr 0 & -1\cr}$ has $e^{At} = e^{-t} \pmatrix{1 & t\cr 0 & 1\cr}$ and $|e^{At}| \approx t e^{-t}$. Note that $p(t) e^{rt} \le \beta e^{-\alpha t}$ for all $t \ge 0$ if $r < -\alpha$, where $\beta$ is the maximum of $p(t) e^{(\alpha+r)t}$. – Robert Israel Nov 09 '11 at 22:36
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I have the same problem here. What I can do is to obtain $\alpha $ and $\beta$, but I am not sure whether their values, especially the one for $\beta$, is tight or not.

At the moment my proof only concerns semi-simple matrices (but I think you can easily extend to the general case using the idea from this answer) and induced 2-norm of matrices (note the equivalence in norms). And it is easy to show that $$ \begin{align*} \left\Vert e^{At}\right\Vert _{2} & =\left\Vert e^{T^{-1}JtT}\right\Vert _{2}\\ & =\left\Vert T^{-1}e^{Jt}T\right\Vert _{2}\\ & \leq\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}\left\Vert e^{Jt}\right\Vert _{2}\\ & =\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}e^{\max_{i}\mathfrak{R}\left(\lambda_{i}\left(A\right)\right)t}\\ & \triangleq\beta e^{-\alpha t} \end{align*} $$ where $$ A=T^{-1}JT $$

It is easy to see that under this scenario $ \alpha=-\max_{i}\mathfrak{R}\left(\lambda_{i}\left(A\right)\right) $ is tight, but as I have checked $\beta=\left\Vert T^{-1}\right\Vert _{2}\left\Vert T\right\Vert _{2}$ can be way off.

It's been years since this question is posted, not sure if you have found a better answer @Truong.

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Alan Wang
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