On sub and super solutions; Teschl and others

Question

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. What follows is the comparison 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.

Answer 1

To handle the case of strict inequalities the following comparison theorem is often useful. Let $I$ be an interval of the form $[t_0,t_1]$, $[t_0,t_1)$ or $[t_0,\infty)$ and let $f:I \times \mathbb{R} \to \mathbb{R}$ be any function. If $x,y:I\to\mathbb{R}$ are differentiable on $I$ with

i) $x'(t)-f(t,x(t)) < y'(t)-f(t,y(t))$ $(t \in I\setminus \{t_0\})$, and

ii) $\exists \varepsilon > 0:$ $x(t)<y(t)$ $(t \in (t_0,t_0+\varepsilon])$

then $x(t)<y(t)$ $(t \in I\setminus \{t_0\})$.

Proof: Assume $x(t)<y(t)$ $(t \in I\setminus \{t_0\})$ is not true. Then there is a first point $\tau > t_0+\varepsilon$ where $x(\tau)=y(\tau)$. In particular $x(t)<y(t)$ $(t \in (t_0,\tau))$. With $d(t):=y(t)-x(t)$ we get $$ d'(\tau)=\lim_{t\to\tau-} \frac{d(t)-d(\tau)}{t-\tau} \le 0. $$ On the other hand by i) we have $$ d'(\tau) > f(\tau,y(\tau))- f(\tau,x(\tau))=0, $$ a contradiction. (End of proof.)

Concerning your second question: If you have $x,y:[t_0,t_1) \to \mathbb{R}$ with $$ x'(t)=f(t,x(t)), ~ y'(t) > f(t,y(t)) ~ (t \in [t_0,t_1)), \quad x(t_0)=y(t_0) $$ then clearly i) is satisfied and $y'(t_0)>x'(t_0)$. If you know in addition that $y'$ is continuous, then $y'(t) > x'(t)$ on a small interval $[t_0,t_0+\varepsilon]$ which yields ii). Thus $x(t)<y(t)$ $(t \in (t_0,t_1))$.

The definition of sub and super solutions (often also called lower and upper solutions) is far from uniform. Some authors use strict inequalities, others don't. You always have to check the definition. The book of W. Walter: Ordinary differential equations (where you can also find the comparison theorem above) uses strict inequalities.