how to show following estimate?

Question

Let $\phi_{+},\phi_{-} \in C^1(I_t)$ be such that $\phi_{+}^{'}(t)>f(t,\phi_{+}(t))$ and $\phi_{-}^{'}(t)<f(t,\phi_{-}(t))$, for all $t \in I_t$. If $\phi_{-}(t_0) \leq x_0 \leq \phi_{+}(t_0)$, for some $t_0 \in I_t$, then show that the solution $x=\phi(t)$, $t \in I_t$, of the IVP $x'=f(t,x), x(t_0)=x_0$ satisfies $\phi_{-}(t) < \phi (t)< \phi_{+}(t)$, $t \in (t_0,t_0+a)$.

Answer 1

Suppose that $\phi(t)\geq \phi_+(t)$ for some $\stackrel{\sim}{t}\, >t_0$. Consider $h(t)=\phi(t)-\phi_+(t)$. We have $h(t_0)\leq 0$ and $h(\stackrel{\sim}{t}) \geq 0$. By the intermediate value theorem, we must have some $t\in \left[t_0,\stackrel{\sim}{t}\right]$ with $h(t)=0$. Let $t^*$ be the smallest such $t$ (why can we guarantee there is a smallest $t$?).

Moreover, consider that

$$h'(t)=\phi'(t)-\phi_+'(t)=f(t,\phi(t))-\phi_+'(t)\tag{1}$$

Now, suppose that $t^*=t_0$. In this case we'd have:

$$\phi_+'(t_0)>f(t_0,\phi_+(t_0))=f(t_0,\phi(t_0)),$$

so that $h'(t_0)<0$. This implies that $h$ is decreasing at $t_0$, and hence must be negative on $(t_0,t_0+a)$ for some $a>0$, which concludes this case.

Finally, suppose that $t^*>t_0$. Then $h$ must be negative on $[t_0,t^*)$ so we may take $a=t^*-t_0$.

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The proof for $\phi_-$ is similar.