3
$\begingroup$

Suppose we have an analytic function $f: U \to {\mathbb R}$, where $U\subset {\mathbb R}^n$ is an open subset and $0 \in U$ is a critical point of $f$. Thom conjectured that if a trajectory $x(t)$ of gradient of $f$ converges to $0$ as $t\to \infty$, then the unit tangent vector $$\frac{\dot{x}(t)}{\| \dot{x}(t)\|}$$ has a unique limit as $t\to \infty$. This conjecture was proved by Kurdyka, Mostowski & Parusinski.

I want to know the case for smooth functions. Is there any counterexample or not known yet? Or under what condition may one expect the same result as Thom conjectured?

$\endgroup$

2 Answers 2

12
$\begingroup$

It is easy to make a $C^\infty$ function on $\mathbb{R}^2$ for which the unit tangent vector of a gradient flow converging to an equilibrium has no limit. Instead of giving all detail, I'll give two hints.

1. Think of a function that has a unique critical point, a global minimum at the origin, and such that the gradient of $f$ at the upper half boundary of the circle of radius $r$ makes an angle $\theta(r)$ with the exterior normal, oscillating e.g. within $\pm\pi/4$. There is no difficulty to realize this for a function $C^\infty(\mathbb{R}^2)$ because the norm of the gradient may go to zero extremely fast (I'll add more details at request). Then, all gradient lines will converge to the origin, and all of them will present a oscillating unit tangent vector, till they remain in the upper half-plane. So the point is to arrange things so that at least one trajectory remains there for all times, which is a bit technical, yet feasible.

2. Even simpler: start with a simple curve $\Gamma$ with an end-point at the origin, such that $\Gamma\setminus\{0\}$ is parametrized by a $C^\infty$ injective regular curve $\gamma:\mathbb{R}\to\mathbb{R}^2\setminus\{0\}$ such that $\gamma(t)\to 0$, and $\|\dot\gamma(t)\|\le\exp(-1/\|\gamma(t)\|)$, but with a non converging unit tangent vector. Define a function $f_0$, and a field $f_1$, on $\Gamma$,setting $f_0(x):=\int_t^\infty\|\gamma(t)\|^2dt$ and $f_1(x)=-\dot\gamma(t)$ if $x=\gamma(t)$, and both vanishing at the origin. Use the Whitney extension theorem to extend the data $f_0$ and $f_1$ to a $C^\infty$ function $f$ and its gradient. Then $\gamma$ is a gradient-line trajectory of $f$ by construction, with the desired properties.

$\endgroup$
0
5
$\begingroup$

The result quoted by the author of the questions is not correct.

Suppose we have an analytic function $f: U \to {\mathbb R}$, where $U\subset {\mathbb R}^n$ is an open subset and $0 \in U$ is a critical point of $f$. It seems that, differently than in the claim of the author, the existence of the limit $$ \lim_{t\to\infty}\frac{\dot{x}(t)}{\| \dot{x}(t)\|} $$ for a trajectory $x(t)$ of $\nabla f$ that converges to $0$ as $t\to \infty$, is still an open problem. It is so called Generalized Thom’s Gradient Conjecture. I am not sure who is the author of this conjecture, but it is stated as Conjecture 2 in [1]. See also bottom of the page 768 in [2].

The original Thom's conjecture proved in the affirmative by Kurdyka, Parusiński and Mostowski [2] was about the existence of the limit of secants $$ \lim_{t\to\infty}\frac{{x}(t)-x_0}{\| {x}(t)-x_0\|}. $$

[1] P. Goldstein, Gradient flow of a harmonic function in $\mathbb{R}^3$. J. Differential Equations 247 (2009), 2517–2557.

[2] K. Kurdyka, Krzysztof T. Mostowski, A. Parusiński, Proof of the gradient conjecture of R. Thom. Ann. of Math. 152 (2000), 763–792.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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