Suppose that $b:\mathbb{R}\to\mathbb{R}$ is locally Lipschitz and of polynomial growth. Suppose further that there are constants $C_1,C_2>0$ such that $(x-y)(b(x)-b(y))\leq C_1-C_2(x-y)^2$ for all $x\in\mathbb{R}$. The ODE $$ \dot{x}(t)=b(x(t))+d(t)+u(t),\quad x(0)=x_0,\qquad t\in[0,1], $$ is then seen to have a unique solution for any integrable functions $d,u:[0,1]\to\mathbb{R}$.

Let us denote $\mathcal{C}_0^\infty([0,1])=\{f\in \mathcal{C}^\infty([0,1]):\, f(0)=0\}$.

Is under this set of conditions the following true?

Claim: Let $R>0$. Then there are $M,\delta>0$ such that given any initial condition $x_0\in\mathbb{R}$ and $d\in\mathcal{C}_0^\infty([0,1])$, we can find a function $u:[0,1]\to\mathbb{R}$ with $\|u\|_\infty\leq M$ and times $0\leq t_1<\cdots<t_n\leq 1$ such that $$ \sum_{i=1}^{n-1}t_{i+1}-t_i\geq\delta $$ and $|x(t)|\geq R$ for all $t\in [t_1,t_2]\cup\cdots\cup[t_{n-1},t_n]$.

I managed to prove this for $b$ globally Lipschitz, which is essentially enough to get it under the additional assumption $\inf_x b^\prime(x)$ (since then, in combination with the one-sided Lipschitz condition, the derivative is bounded).

The bounty is for proving that if $\inf_x b^\prime(x)=-\infty$, then it is not possible to find uniform $M,\delta$. For this, I'd content myself with an example, e.g. $b(x)=x-x^3$.

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    $\begingroup$ The property in question holds if $\inf_{\mathbb R}b′>−\infty$ and does not hold if, say, $\lim_{x\to\infty}b'(x)=-\infty$ (so $−x$ is OK, but $−x−x^3$ is not) $\endgroup$
    – fedja
    Feb 27, 2020 at 0:44
  • $\begingroup$ Thanks for the answer. Do you have an easy argument which shows that the property fails for $-x-x^3$? $\endgroup$
    – julian
    Feb 28, 2020 at 21:00

1 Answer 1


Do you have an easy argument...

Sorry for the late reply, but it is, actually, a rather simple story. Let $R=1$ and suppose that we have declared some $M$ and $\delta$. Then the adversary starts at some $x_0\in(-\frac 12,\frac 12)$ and he has some guaranteed time $T=T(b,M)>0$ during which the solution stays in $(-1,1)$ if he keeps $d=0$ on that interval of time. What he does after that is to choose some $L>0$ and make $d$ mimic a big multiple of the $\delta$-measure to achieve a nearly instantaneous shift of the solution by a fixed large constant, so that we find ourselves in, say, $2$-neighborhood of some point $x_1$ such that $b'(x)<-L$ whenever $|x-x_1|<2$. He then switches $d$ to $-b(x_1)$ so that $x_1$ becomes an equilibrium point without control and the convergence to that equilibrium from its $2$-neighborhood is exponential with factor $L$ in the exponent, so in time $C/L$ we reach the $1/4$-neighborhood of $x_1$. If $L$ is large enough (in terms of $M$), our control is next to useless in this area and we still find ourselves in the $1/3$-neighborhood of $x_1$ in time $C/L$ with fixed known $C>0$. The adversary then uses $d$ once more to inflict an almost instantaneous shift by almost $-x_1$ and the solution returns to the $1/2$-neighborhood of the origin, so the cycle can be repeated. All he needs to ensure is that the stabilization time $C/L$ is less than $\frac\delta 2T$, say (which can be done by choosing $L$ large enough) and that the "almost instantaneous shifts" are fast and precise enough.

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    $\begingroup$ I love the way you write your answers. $\endgroup$ Mar 5, 2020 at 21:12

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