# Steering an ODE out of a ball

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 $$0\leq t_1<\cdots such that $$\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. $$b(x)=x-x^3$$.

• 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) Feb 27, 2020 at 0:44
• Thanks for the answer. Do you have an easy argument which shows that the property fails for $-x-x^3$? Feb 28, 2020 at 21:00

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.