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 [itex]w(t) \leq f(t,w(t))[/itex]; 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 [itex]\epsilon > 0[/itex], if [itex]|x - y| < \frac{\epsilon}{L}[/itex] then [tex]
|f(x) - f(y)| \leq L|x - y| < \epsilon.[/tex]

 

[psie]

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

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)##.