Consider a second order gradient-like system with linear damping $$\ddot{x}+\dot{x}+\nabla f(x)=0, \quad x(0)=x_0,\quad\dot{x}(0)=0$$ Suppose $f\in C^2(\mathbb{R}^n)$ and $\inf_{x\in\mathbb{R}^n}f(x)>-\infty$. The solution $x:[0,\infty)\rightarrow \mathbb{R}^n$ is bounded, i.e., $\lVert x(t)\rVert\leq c$ for all $t\in [0,\infty)$, where $c$ is a constant.

If we change the so called "gravity constant" (see this paper) from $1$ to any positive constant $g$, i.e., $$\ddot{x}+\dot{x}+g\nabla f(x)=0, \quad x(0)=x_0,\quad\dot{x}(0)=0$$ Is the solution to this new equation still bounded?


1 Answer 1


$\newcommand\la\lambda$No. E.g., if $n=1$, $x_0\ne0$, and $f(u)=-u^2/2$ for all real $u$, then the solution $$x(t)=\frac{x_0}{2} \, \Big(\frac{e^{\la_+ t}-e^{\la_- t}}{\sqrt{4 g+1}}+e^{\la_- t}+e^{\la_+ t}\Big)$$ of the problem $$\ddot{x}+\dot{x}+g\nabla f(x)=0, \quad x(0)=x_0,\quad\dot{x}(0)=0 \tag{1}\label{1}$$ will be bounded in $t\ge0$ if $0<g\le2$ and unbounded in $t\ge0$ if $g>2$, where $$\la_\pm:=\frac{-1\pm\sqrt{4 g+1}}2.$$

Addendum: The OP later added the condition that $f$ be bounded below. This changes the problem dramatically.

First here, note that, if $x(t)$ is the solution to \eqref{1}, then for the "energy" $$E(t):=\frac{|\dot x(t)|^2}2+gf(x(t))$$ and all real $t\ge0$ we have $$E'(t) =(\ddot x(t)+g\,\nabla f(x(t))\cdot\dot x(t) =-\dot x(t)\cdot\dot x(t) =-|\dot x(t)|^2\le0,$$ so that the "energy" is nonincreasing; of course, here $|\cdot|$ is the Euclidean norm and $\cdot$ is the dot product. It follows that for all real $t\ge0$ $$f(x(t))\le c:=E(0)/g<\infty.$$

If now $f$ is coercive (that is, $f(x)\to\infty$ as $|x|\to\infty$), then the set $\{y\in\mathbb R^n\colon f(y)\le c\}$ is bounded, and hence the solution $x(t)$ is bounded (in real $t\ge0$).

In a comment, the OP said "we can assume $f$ to be polynomial".

If $n=1$, then any bounded from below non-constant polynomial is coercive, and the case of a constant polynomial is quite easy. So, this solves the case $n=1$ (still with a polynomial $f$).

Even if $n>1$, it seems possible to show that "almost all" bounded from below non-constant polynomials are coercive, and for such polynomials, the solution $x(t)$ will be of course bounded.

  • $\begingroup$ Thanks for your answer. I apologize that I missed one important assumption, that is, $f$ is lower bounded. Is it still possible to construct a counterexample like the above one? $\endgroup$ Commented Nov 11, 2022 at 1:10
  • $\begingroup$ @JeanLegall : I think this additional assumption will dramatically change the problem, and I think the answer will then be yes, but it seems very difficult to prove that, especially in the full generality. $\endgroup$ Commented Nov 11, 2022 at 2:10
  • $\begingroup$ Indeed, we can assume $f$ to be polynomial, but even in this case I don't have an idea. $\endgroup$ Commented Nov 11, 2022 at 2:40
  • $\begingroup$ Take $f(x,y)=(1-xy)^2$, this function is not coercive but solution is always bounded. Seems that it requires a proof without using coercive. $\endgroup$ Commented Nov 11, 2022 at 19:39
  • $\begingroup$ @JeanLegall : Yes, as I said, not all bounded from below polynomials for $n>1$ are coercive, but "almost all of them". Also, please remember that the addendum is a "free bonus", as your original question (without the boundedness of $f$ from below) was fully answered. $\endgroup$ Commented Nov 11, 2022 at 20:14

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