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. $b(x)=x-x^3$.