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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, but as suggested in the comments this should hold if $\inf_x b^\prime(x)>-\infty$. The bounty is for a proof of this and counterexample showing that this condition is sharp.

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

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 $$ \dot{x}(t)=b(x(t))+d(t)+u(t),\quad x(0)=0,\qquad t\in[0,1], $$ it 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 $b(x)=-x$ but my argument relied on the explicit formula for the solution in this case. It is also straight-forward to see that $$ |x(t)-\bar{x}(t)|^2\leq\int_0^t e^{-C_2(t-s)}(2C_1+\frac{1}{C_2}u(s)^2)\,ds, $$ where $\bar{x}$ denotes the solution to the ODE $$ \dot{\bar{x}}(t)=b(\bar{x}(t))+d(t). $$

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, but as suggested in the comments this should hold if $\inf_x b^\prime(x)>-\infty$. The bounty is for a proof of this and counterexample showing that this condition is sharp.

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 $$ \dot{x}(t)=b(x(t))+d(t)+u(t),\quad x(0)=0,\qquad t\in[0,1], $$ it 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 $b(x)=-x$ but my argument relied on the explicit formula for the solution in this case. It is also straight-forward to see that $$ |x(t)-\bar{x}(t)|^2\leq\int_0^t e^{-C_2(t-s)}(2C_1+\frac{1}{C_2}u(s)^2)\,ds. $$

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 $$ \dot{x}(t)=b(x(t))+d(t)+u(t),\quad x(0)=0,\qquad t\in[0,1], $$ it 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 $b(x)=-x$ but my argument relied on the explicit formula for the solution in this case. It is also straight-forward to see that $$ |x(t)-\bar{x}(t)|^2\leq\int_0^t e^{-C_2(t-s)}(2C_1+\frac{1}{C_2}u(s)^2)\,ds, $$ where $\bar{x}$ denotes the solution to the ODE $$ \dot{\bar{x}}(t)=b(\bar{x}(t))+d(t). $$