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An $n$ by $n$ matrix A is said to be dissipative with respect to a norm $\|\cdot \|$ if for all $x$ and $t\geq 0$, we have $\|e^{At}x\|\leq\|x\|$. Two matrices $A$ and $B$ are said to be jointly dissipative if they are each dissipative with respect to the same norm.

Here $e^{At}$ is defined by

$$e^{At}:=\sum_{n=0}^{n=\infty}\frac{1}{n!}A^nt^n.$$

My question is the following: what are necessary and sufficient conditions (on $A$ and $B$) for the existence a norm for which $A$ and $B$ are jointly dissipative?

Of course depending on what we take $n$ to be, this question can be quite hard. If anyone has any particularly clever suggestions for an approach, or solutions to special cases ($n\geq 2$), I would love to hear from you.

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2 Answers 2

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The discrete-time analogue -- there exists a norm in which $|A_1^nx|\leq |x|$, $|A_2^nx|\leq |x|$ for all $n\geq 1$ and $x \in\mathbb{R}^d$ -- is equivalent to the property that the semigroup generated by $A_1$ and $A_2$ is bounded. To see this we just define $$|x|:=\sup_{n\geq 0} \|A_{i_n}\cdots A_{i_1}x\|$$ for all $x$. Determining whether or not this property holds is known to be an algorithmically undecidable problem by work of Blondel and Tsitsiklis.

There could easily be continuous-time analogues of these statements, but I'm not very familiar with that part of the literature.

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    $\begingroup$ I would say that the continuous analogue should be the same, but we need to involve the semigroup $S$ generated by all $\exp(At)$ and $\exp(Bt)$ with $t>0$. If this semigroup is unbounded, then the matrices are not jointly dissipative, otherwise one may define $\|x\|_S=\sup_{C\in S}\|Cx\|$. $\endgroup$ Dec 1, 2015 at 17:33
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The problem has been addressed in the infinite dimensional setting in the following paper:

Máté Matolcsi, On the relation of closed forms and Trotter’s product formula, J. Funct. Anal. 205 (2003), no. 2, 401--413.

Theorem 2 is of special interest here. If there are any practical consequences, I do not know.

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