# Does gravity constant affect boundedness of solution?

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?

$$\newcommand\la\lambda$$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 $$0 and unbounded in $$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.

• 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? Commented Nov 11, 2022 at 1:10
• @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. Commented Nov 11, 2022 at 2:10
• Indeed, we can assume $f$ to be polynomial, but even in this case I don't have an idea. Commented Nov 11, 2022 at 2:40
• 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. Commented Nov 11, 2022 at 19:39
• @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. Commented Nov 11, 2022 at 20:14