I 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 &gt; 0, if |x - y| &lt; \frac{\epsilon}{L} then <br /> |f(x) - f(y)| \leq L|x - y| &lt; \epsilon.

 

[psie]

A Lipschitz function is necessarily continuous: for any \epsilon &gt; 0, if |x - y| &lt; \frac{\epsilon}{L} then <br /> |f(x) - f(y)| \leq L|x - y| &lt; \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 $x'=f(t,x)$.