Let $f_1,f_2:\mathbb{R}^2\to\mathbb{R}$ be $C^{\infty}$ functions such that $f_1(x,y)\leq f_2(x,y)\; \forall(x,y)\in\mathbb{R}^2$. Suppose that $\psi_1:I_1\to \mathbb{R}$ and $\psi_2:I_2\to \mathbb{R}$ are solutions for $\dot{y}=f_1$ and $\dot{y}=f_2$.

Prove that if $\exists x_0\in I_0 \cap I_1$ such that $\psi_1(x_0)=\psi_2(x_0)$ then $\psi_1(x)\leq\psi_2(x)\;\forall(x,y)\in I_1\cap I_2$ if $x\geq x_0$. What happens if $x<x_0$.

This is my proof: $\psi_1$ is a solution for $\dot{y}=f_1$, which means that $\displaystyle\frac{d\psi_1(z)}{dx}=f_1(x,\psi_1(z))$; $\psi_2$ is a solution for $\dot{y}=f_2$ then $\displaystyle\frac{d\psi_2(z)}{dx}=f_2(x,\psi_2(z))$.

The relation $f_1\leq f_2$ implies $\displaystyle\frac{d\psi_1(z)}{dx}\leq \displaystyle\frac{d\psi_2(z)}{dx} \forall z\in I_1\cap I_2$ and a premise tell us that exists $x_0\in I_1 \cap I_2 $ such that $\psi_1(x_0)=\psi_2(x_0)$; if we put those things together we get that $\psi_1(x)\leq \psi_2(x)$ if $x\geq x_0$,and the direction reverses if $x<x_0$ (which es $\psi_1 (x) > \psi_2 (x)$).

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    $\begingroup$ You can't jump from $f_1 \le f_2$ to $\psi_1' \le \psi_2'$. The relationship $f_1 \le f_2$ only holds for the same parameters. $\endgroup$ – copper.hat Sep 26 '14 at 2:59
  • $\begingroup$ @copper.hat Hmm. Then I'm not sure if the statement is correct. According to what you said, I could say that since $f_1(x_0,\psi_1(x_0))\leq f_2(x_0,\psi(x_0))$ then $\frac{d\psi_1 (x_0)}{dx}\leq \frac{d\psi_2 (x_0)}{dx}$, but is it possible to make it general for all $x$ in $I_1 \cap I_2$?. $\endgroup$ – Cure Sep 26 '14 at 3:25
  • $\begingroup$ I think one approach is to use the Bellman-Gronwall lemma. $\endgroup$ – copper.hat Sep 26 '14 at 3:31
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    $\begingroup$ As an aside, these sort of things fall under the umbrella of 'differential inequalities'. I find them fascinating. $\endgroup$ – copper.hat Sep 26 '14 at 4:28

The reasoning is a little more delicate. Note that if $\psi_1(t_0) < \psi_2(t_0)$, then the result is immediate. Unfortunately this doesn't work here.

Let $\delta(t) = \psi_1(t) - \psi_2(t)$. We have $\delta(t_0) =0$. Note that the $f_k$ are lipschitz on some neighbourhood of $(t_0, \psi(t_0))$. Let $L$ be the Lipschitz constant. we have $\dot{\delta}(t) = \dot{\psi_1}(t) - \dot{\psi_2}(t) = f_1(t, \psi_1(t)) - f_2(t,\psi_2(t)) \le f_2(t, \psi_1(t)) - f_2(t,\psi_2(t)) \le L |\delta(t)|$.

Now we work by contradiction. Suppose $\delta(t_2) >0$, and let $t_1 = \sup\{ t \in [t_0,t_2] | \delta(t) \le 0 \}$. Hence $\delta(t) >0$ for $t \in (t_1,t_2]$, and $\delta(t_1) = 0$. Now we have $|\delta(t)| = \delta(t)$ for $t \in (t_1,t_2]$, and so $\dot{\delta}(t) \le L \delta(t)$. In particular, this gives $\delta(t) \le e^{L(t-t_1)}\delta(t_1) = 0$, which is a contradiction.

Applying this result to the reversed equations gives $\delta(t) \ge 0$ for appropriate $t \le t_0$.


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