3
$\begingroup$

I have often heard about the asymptotics of gradient flows converging to some "equilibrium point" as $t \to \infty$. This concept has come to my ear by word of mouth multiple times and is often verified by direct calculations e.g. as for the 1-dimensional heat equation. It also has come to my attention that minimizing movements as described in e.g. Braides Book on $\Gamma$-convergence try to use this concept. I would like to learn more about it, but can not find a good point to start.
As for the notation in this question, $\partial $ denotes the subdifferential and $'$ the derivative with respect to the time.

Let's consider a a gradient flow in the euclidean space, for simplicity and let $F:\mathbb{R}^n \to \mathbb{R}$ be a $\lambda$-convex function (for $\lambda > 0$ the function $F(x)+\frac{\lambda}{2} |x|^2$ is convex) and lets for simplicity assume it has a gradient $\nabla F$. Then consider the smooth solution of the IVP: $$ u'(t)=-\nabla F(u(t)) \\ u(0)=u_0 $$ You can estimate the difference of two solutions by their initial conditions; i.e. $$ |u_1(t)-u_2(t)|\leq e^{-\lambda t }|u_1(0)-u_2(0)| $$ So taking the limit $t\to \infty$ yields that both solutions seem to converge to the same point, that is: $\lim_{t \to \infty} u_1(t)=\lim_{t \to \infty} u_2(t)$. Now let say $F$ has 1 critical point $c \in \mathbb{R}^n$ such that $\nabla F(c)=0$. Now we can consider the problem with inital condition $u_3(0)=c$. Now the function $u_3(t)=c$ solves the IVP and so we get $$ \lim_{t \to \infty} u(t)=c $$ for all solutions of the problem. So we found some sort of equilibrium point and described the asymptotic behaviour of the function $u$ using $F$.

The question that I have now is if this concept still holds if I replace $\mathbb{R}^n$ with a Hilbert space (as for metric spaces, I might explore them later). For example, the heat equation in a suitable domain, suitable initial/boundary conditions with the Dirichlet energy $E(u)=\int \frac{1}{2}|\nabla u|^2$ statsifies $$ u'=-\nabla_{L^2}E(u) \\ u(0,x)=g(x) \\ u(t,x)|_{\partial \Omega}=f(x) $$ and we have that $u$ converges as $t \to \infty$ to the solution of the Laplace equation, a critical point of $E(u)$: $$ \Delta u=0 \\ u|_{\partial \Omega}(t,x)=f(x) $$

Now consider the setting for gradient flows as in Evans PDE; if you have a similar or more general setting, feel free to use it.
Let $H$ be a real Hilbert space, $I:H \to (-\infty,+\infty]$ be convex, proper and lower semicontinouos and the domain of the subdifferential statisifies $\overline{D(\partial I)}=H$.Then for each $g \in {D(\partial I)}$ there exists a unique function $$ u \in C([0,\infty);H) \; u'\in L^\infty(0,\infty;H) $$
such that $u(0)=g$, $u(t) \in D(\partial I)$ for each $t>0$ as well as $u'(t)\in-\partial I(u(t))$.
So is there any way or estimate to describe the existence (and maybe uniqueness) of $$ \lim_{t \to \infty}u(t)=h \in H $$ where the limit is taken with respect to the Hilbert space $H$? Is $h$ a critical value of $I$? Is there some sort of "exponential decay" estimate like $||u(t)-h||_H\leq Ce^{-t}||u(0)-h||_H$?
I am thankful for every reference, hint or answer covering any of the aspects of my question. If you need to modify the assumptions, feel free to do so. Any textbook suggestions are appreciated, as I want to learn more about this topic in a more rigorous way!

$\endgroup$
  • 1
    $\begingroup$ This question might be related: link $\endgroup$ – Schlubbidubbi Jan 11 at 10:55
2
$\begingroup$

It's basically very similar to the finite-dimensional case. If $I$ is $\lambda$-convex for some $\lambda\in\mathbb{R}$ (i.e. $I-\frac{\lambda}{2}\|\cdot\|^2$ is convex), then we have the contraction estimate $$ \|u_1(t)-u_2(t)\|\leq e^{-\lambda t}\|u_1(0)-u_2(0)\|. $$ In particular, if $\lambda>0$, then there is at most one equilibrium point. In the general case, uniqueness may fail, as is illustrated by the example $I=\mathrm{const}$.

Maybe not the easiest way to prove this, but the one that lends itself best to generalizations uses the following characterization: A curve $u$ is a gradient flow of the $\lambda$-convex functional $I$ if and only if it satisfies the evolution variational inequality (EVI) $$ \frac 1 2\frac{d}{dt}\|u(t)-v\|^2+\frac{\lambda}2\|u(t)-v\|^2+I(u(t))\leq I(v) $$ for $v\in H$, $t>0$.

Adding up the EVIs for $u_1$ and $u_2$ (with $v=u_2(t)$ resp. $v=u_1(t)$) and applying Grönwall's inequality, one gets the exponential contraction estimate from above.

Every critical point of $I$ (or, which is the same for a convex functional, a global minimizer) is an equilibirum point by the same argument as in the finite-dimensional case. Conversely, the energy dissipation equality (EDE) $$ I(u(0))=I(u(t))+\frac 1 2\int_0^t \|\dot u(s)\|^2\,ds+\frac 1 2\int_0^t \|\partial^\circ I(u(s))\|^2\,ds, $$ which is another equivalent characterization of gradient flows, implies that $I$ decreases along gradient flow curves. Thus, if there is a unique equilibrium point, it must be global minimizer of $I$. In fact, with your definition of gradient flow, this argument only yields $I(h)\leq I(g)$ for $g\in D(\partial^\circ I)$. However, if one relaxes the condition on the gradient flow to $$ u\in C([0,\infty);H)\cap\mathrm{AC}_{\mathrm{loc}}((0,\infty);H), $$ there exists a gradient flow curve for every starting point $g\in H$.

Both the EVI and the EDE are starting points to extend the theory of gradient flows to metric spaces. The standard reference for this topic is the book by Ambrosio, Gigli, and Savaré. It can be a little hard to read for beginners, a somewhat more gentle introduction is given in Philippe Clément's An Introduction to Gradient Flows in Metric Spaces, which should also fill all the gaps I left in this post.

$\endgroup$
  • $\begingroup$ Thank you very much for your response. It was exactly what I was looking for! $\endgroup$ – F. Conrad Jan 11 at 14:47

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.