stability

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Jan 25, 2021 at 18:14	history	edited	Denny	CC BY-SA 4.0	added 505 characters in body
Jan 25, 2021 at 17:38	comment	added	Denny		@FabianWirth I will consider the set $\mathcal{C} = \{v: \langle v, -\nabla V(x) \rangle \geq 0\}$. If $V$ is convex, then $\mathcal{C}$ is the relaxation of such constraint $f = -\nabla V(x)$. So I can choose anything in $\mathcal{C}$.
Jan 20, 2021 at 16:22	comment	added	Fabian Wirth		So far nothing in the problem makes this a particular matrix problem. If I understand correctly, we have a system $\dot x = f(x,u)$ on a vector space (the vectors happen to be matrices). Now find a smooth Lyapunov function $V$ such that you can choose $u$ in dependence of $x$ such that $\dot V$ becomes negative definite. First of all, already in the vector case this is not always easy (as your wording seems to suggest). What you are looking for is called a control Lyapunov function (clf) and maybe you first read up on those. BTW: the condition $f(x,u)= -\nabla V(x)$ is quite restrictive.
Jan 20, 2021 at 9:49	comment	added	Denny		@JonathanJ. Thanks for good catch. I modify my question a little bit.
Jan 20, 2021 at 9:48	history	edited	Denny	CC BY-SA 4.0	added 123 characters in body
Jan 19, 2021 at 20:22	comment	added	Jonathan J.		Is there some additional structure or restrictions to this problem? For example, if $X$ is a real $n \times n$ matrix, is there anything wrong with treating it as a vector in $ \mathbb{R}^{n\times n}$? Usually with Lyapunov functions, one starts with the vector field $f$ and then one looks for an appropriate Lyapunov function. However here it looks like you might have the freedom to adjust your function $f$?
Jan 17, 2021 at 16:57	review	First posts
Jan 17, 2021 at 21:01
Jan 17, 2021 at 16:55	history	asked	Denny	CC BY-SA 4.0