Return to Answer

Suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE with nondecreasing velocity must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.

Suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, suppose that the speed $|X'|$ is strictly increasing.

Then either $X'>0$ or $X'<0$ on $(0,\infty)$. (Otherwise, the continuous function $X'$ will take the value $0$ at some point of $(0,\infty)$, which will contradict the conditions that $X'(0)=0$ and $|X'|$ is strictly increasing.)

So, the continuous function $X$ is either strictly increasing or strictly decreasing on $(0,\infty)$. So, the function $X\colon[0,\infty)\to X([0,\infty))$ has an inverse $X^{-1}$. Letting now $v:=X'\circ X^{-1}$, we have $X'(t)=v(X(t))$ for all $t\in[0,\infty)$, whence $X''(t)=v'(X(t))X'(t)=v'(X(t))v(X(t))$. Now the ODE can be rewritten as $v'(x)v(x)+f'(x)=0$ for $x$ in the interval $X([0,\infty))$, or as \begin{equation*} \dfrac d{dx}\,(v(x)^2)=-2f'(x), \end{equation*} which implies $v(x)^2=2(f(0)-f(x))$, because $v(0)=v(X(0))=X'(0)=0$. So, \begin{equation*} X'(t)^2=v(X(t))^2=2(f(0)-f(X(t))) \tag{1} \end{equation*} for $t\in[0,\infty)$. 

Since $f$ is strongly convex, we have $f(x)\to\infty$ as $|x|\to\infty$. Noting that the left-hand side of (1) is $\ge0$, we conclude that the function $X$ must be bounded. Since $|X'|$ is strictly increasing, and either $X'>0$ or $X'<0$ on $(0,\infty)$, it follows that $X'$ is either increasing or decreasing (from $X'(0)=0$), and hence $X$ is either (i) convex and increasing or (ii) concave and decreasing, on $(0,\infty)$. So, the bounded function function $X$ must be constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE with strictly increasing speed must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.

Suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.

Suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE with nondecreasing velocity must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.

The answer is no.

Indeed, suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.

Suppose that $n=1$ and $f(x)=(x+1)^2$. Then $f$ is strongly convex and $$X(t)=\cos(t\sqrt2)-1.$$

So, $X(t)$ does not converge as $t\to\infty$.

On the other hand, still for $n=1$, if the velocity $X'$ is nondecreasing, then $X''\ge0$ and hence $X$ is convex. So, in view of the condition $X'(0)=0$, $X$ is nondecreasing. It also follows that $f'(X(t))=-X''(t)\le0$. Since $f$ is strongly convex, we have $f'(x)\to\infty$ as $x\to\infty$. So, $X$ is bounded from above. Since $X$ is convex and nondecreasing, it follows that $X$ is constant. In view of the condition $X(0)=0$, we conclude that $X=0$ identically.

Thus, any solution $X$ of the ODE must be the constant $0$. On the other hand, if the constant $0$ is indeed a solution of the ODE, then we must have $\nabla f(0)=f'(0)=0$.

Now, for any dimension $n$, if $\nabla f(0)=0$, then the constant $0$ is the only solution of the ODE.