On sub and super solutions: Teschl and others

[psie]

TL;DR Summary

I'm confused about a comparison theorem related to ODEs and a definition of sub solution.

I'm reading Ordinary Differential Equations by Andersson and Böiers. There is a comparison theorem I have some questions about. I have also checked Teschl's Ordinary Differential Equations and Dynamical Systems, but there I have problems with his definition of a sub solution. I'll elaborate below. Here follows a theorem in the book I first stated:

Theorem. Assume that $f(t,x)$ is a continuous function in the strip $\{(t,x); t_0\leq t\leq t_1\}$ and satisfies a Lipschitz condition in a neighborhood of every point there. Furthermore, assume that $x(t)$ and $y(t)$ satisfy $$x'(t)=f(t,x)\quad\text{and}\quad y'(t)\geq f(t,y)$$ respectively, when $t_0\leq t\leq t_1$. Then $$x(t_0)=y(t_0)\implies x(t)\leq y(t)\quad\text{when }t_0\leq t\leq t_1.$$

This definition is not made in the book, but I guess $y(t)$ is called a super solution. What confuses me in this theorem are the inequalities and how the theorem is modified when we change some of the inequalities to strict inequalities.

1. First, I assume a corresponding result holds for a function $w(t)$ that satisfies $w'(t)\leq f(t,w)$, so that $x(t_0)=w(t_0)\implies x(t)\geq w(t)$ when $t_0\leq t\leq t_1$, right?
2. Second, I'm working a problem where a function $y(t)$ satisfies $y'(t)> f(t,y)$ on a half-open strip, i.e. $t_0\leq t<t_1$ (because it is undefined at $t_1$). So how is the conclusion of the theorem modified if we change the assumptions to $y'(t)> f(t,y)$ and a half-open strip?
3. Finally, in Teschl's book, he defines a sub solution $w(t)$ to be a function that satisfies $w'(t)< f(t,w)$ for $t_0\leq t<t_1$. However, in my problem, I have a function $w(t)$ that satisfies $w'(t)\leq f(t,w)$ for $t_0\leq t<t_1$ (in particular, $w'(t_0)=f(t_0,w(t_0))$. Is this not a sub solution then?

For completion, I post the proof of the theorem here. You can skip this of course. It uses the following lemma, stated without proof for the sake of brevity;

Lemma. Let $x(t)$ be a differentiable function such that $$x'(t)\leq Mx(t)+a,$$ where $M\neq 0$ and $a$ are fixed constants. Then $$x(t)\leq e^{M(t-t_0)}x(t_0)+\frac{a}{M}(e^{M(t-t_0)}-1),\quad t\geq t_0.$$

Proof (of theorem). Assume that there is some point $\tau$ in the interval $[t_0,t_1]$ where $x(\tau)>y(\tau)$. Then let $\bar t$ be the largest $t$ in $[t_0,\tau]$ with $x(t)\leq y(t)$. Put $z(t)=x(t)-y(t)$. Then $z(t)>0$ in $(\bar t,\tau]$ and $z(\bar t)=0$. Furthermore, for $t$ near $\bar t$, $$z'(t)=x'(t)-y'(t)\leq f(t,x(t))-f(t,y(t))\leq L(x(t)-y(t))=Lz(t).$$ The first inequality comes from the assumptions on $x(t)$ and $y(t)$, the second one makes use of the Lipschitz condition. [The] lemma (with $a=0$) now implies, for $t$ in a right neighborhood of $\bar t$, $$z(t)\leq e^{L(t-\bar t)}z(\bar t)=0.$$ We have arrived at a contradiction.

 

[pasmith]

1. First, I assume a corresponding result holds for a function $w(t)$ that satisfies $w'(t)\leq f(t,w)$, so that $x(t_0)=w(t_0)\implies x(t)\geq w(t)$ when $t_0\leq t\leq t_1$, right?
2. Second, I'm working a problem where a function $y(t)$ satisfies $y'(t)> f(t,y)$ on a half-open strip, i.e. $t_0\leq t<t_1$ (because it is undefined at $t_1$). So how is the conclusion of the theorem modified if we change the assumptions to $y'(t)> f(t,y)$ and a half-open strip?
3. Finally, in Teschl's book, he defines a sub solution $w(t)$ to be a function that satisfies $w'(t)< f(t,w)$ for $t_0\leq t<t_1$. However, in my problem, I have a function $w(t)$ that satisfies $w'(t)\leq f(t,w)$ for $t_0\leq t<t_1$ (in particular, $w'(t_0)=f(t_0,w(t_0))$. Is this not a sub solution then?

1) & 2) You have the proof; can you not step through it and check that it still holds in these modified cases? That will help you to understand the proof.
3) The exact solution satisfies $w(t) \leq f(t,w(t))$; would you call it a "sub solution"? In any event, the theorem itself is stated with a non-strict inequality.

 

[psie]

1) & 2) You have the proof; can you not step through it and check that it still holds in these modified cases? That will help you to understand the proof.

I can try.

My main concern is if the theorem still holds if we have a half-open strip $t_0\leq t<t_1$, but I can't see why it shouldn't. The point $t_1$ is not really used in the proof, except that $\tau$ could possibly be equal to $t_1$. The aim of the proof is to establish that $z(t)>0$ on $(\bar t,\tau]$. I don't see any issues with $\tau$ being equal to some number in $[t_0,t_1)$

That said, if we change $t_0\leq t\leq t_1$ to $t_0\leq t < t_1$ in the theorem, I think the proof goes through pretty much unchanged. All that is changed is that we assume $\tau$ is some number in $[t_0,t_1)$ instead.

Any thoughts on this? By the way, I don't see why $f$ needs to be continuous in the proof. Is this necessary?

 

[Office_Shredder]

You can also smash the open strip with the full theorem. It holds on any strip $t_0\leq t \leq t_2$ if $t_2<t_1$. Apply the theorem on this strip with $t_2$ picked arbitrarily close to $t_1$.

As far as your question about strict inequalities on the derivative, I suspect you don't get to win a strict inequality on x vs y, but I haven't constructed the counterexample yet

 

[pasmith]

By the way, I don't see why $f$ needs to be continuous in the proof. Is this necessary?

A Lipschitz function is necessarily continuous: for any $\epsilon > 0$, if $|x - y| < \frac{\epsilon}{L}$ then $$|f(x) - f(y)| \leq L|x - y| < \epsilon.$$

 

[psie]

A Lipschitz function is necessarily continuous: for any $\epsilon > 0$, if $|x - y| < \frac{\epsilon}{L}$ then $$|f(x) - f(y)| \leq L|x - y| < \epsilon.$$

Here, however, we have a function of two variables $f(t,x)$ that is only Lipschitz with respect to the second variable. Anyway, I assume they stipulated continuity of $f$ so that, according to the existence and uniqueness theorem (i.e. the Picard-Lindelöf theorem), we have a unique solution $x(t)$ to $x'=f(t,x)$.