This question is based on Zeidler II/B, Problem 30.2.

Consider the ODE: find $u:[0,T] \to \mathbb{R}^n$ s.t. $$u'(t) = F(t,u(t))$$ $$u(0) = u_0$$ given $F:[0,T]\times \mathbb{R}^n \to \mathbb{R}^n$ Caratheodory. We know that if it has a solution, it satisfies $|u(t)| \leq C$ for all $t$. Let us suppose that there is $f \in L^1(0,T)$ such that $|F(t,a)| \leq f(t)$ for all $t$ and all $a$ such that $|a| \leq 2C$.

Consider the ODE $$v'(t) = \hat F(t,v(t))$$ $$v(0) = u_0$$ where (*) $$\hat F(t,a) = \begin{cases} F(t,a) &: |a| \leq 2C\\ F(t, \frac{2Ca}{|a|}) &: |a| \geq 2C \end{cases}.$$ This ODE has a solution, since $\hat F(t,a)$ is dominated by $f$ on the whole of $\mathbb{R}^n$ by definition of $\hat F$.

How do I show that the solution $v$ of the second ODE is also a solution of the first ODE (Zeidler claims this)? Of course it's obvious if the solution $|v(t)| \leq 2C$, but how about if it's greater than $2C$?

(*) - in Zeidler, the second case is given as $F(t,2a/C|a|)$, which I think is wrong since it would not make $\hat F(t,\cdot)$ continuous.

I posted this on MSE some time ago and started a bounty a few days ago but no luck.



First, note that if $0<b<T$ and $u:[0,b] \to \mathbb{R}^n$ satisfies
$$u'(t) = F(t,u(t))$$ $$u(0) = u_0$$ then we also have that it satisfies $|u(t)| \leq C$ for all $t\in [0,b]$.

Now, we know that $|v(0)|=|u_0|=|u(0)| \leq C$. Suppose there is $t_1\in [0,T]$ such that $|v(t_1)| > 2C$. Then there is $t_2\in [0,t_1)$ such that, $|v(t_2)| = 2C$ and, for all $t\in [0,t_2]$, $|v(t)| \leq 2C$.

Then $v$ restricted to $[0,t_2]$ satisfies $$v'(t) = F(t,v(t))$$ $$v(0) = u_0$$

So, by our initial remark, we have that $|v(t)| \leq C$ for all $t\in [0,t_2]$. But $|v(t_2)| = 2C>C$. Contradiction. So there is no such $t_1\in [0,T]$ such that $|v(t_1)| > 2C$.


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