Can we bound from above sub-solutions of Volterra integral equations? (Nonlinear Gronwall's Lemma)

Gronwall's lemma says the following. Assume that $v\in C^0([t_0, T])$ is a nonnegative function. If $u \in C^0([t_0, T])$ satisfies the integral inequality

$$u(t) \le c + \int_{t_0}^t u(s)v(s)\, ds,\qquad t \in [t_0, T]$$

where $c\in\mathbb{R}$, then

$$u(t) \le c \exp\left(\int_{t_0}^t v(s)\, ds\right), \qquad t \in [t_0, T].$$

In other words, sub-solutions of the linear integral equation

$$w(t)=c+ \int_{t_0}^t v(s)w(s)\, ds, \qquad t \in [t_0, T]$$

are dominated by solutions of the same equation, provided that the coefficient $v$ is nonnegative.

Question What can we say about the general Volterra equation $$\tag{1}\ w(t) = c + \int_0^t F(s, w(s))\, ds,\qquad t \in [0, T]?$$ Under what conditions on $F$ is a sub-solution of (1) dominated by a solution?

This is not an answer but a (long) comment to Did's answer. In a nutshell: I think that we need to add a Lipschitz condition on $F$ (assumption 2 below). Did probably assumed it implicitly.

Here's a possible way to argue. Assume that:

1. for all $s\in[0, T]$, one has that $x\le y\Rightarrow F(s, x)\le F(s, y)$;
2. there exists a constant $L>0$ such that $\lvert F(s, x)-F(s, y)|\le L\lvert x-y\rvert$.$^{[1]}$

Let

$$u(t)\le c+\int_0^tF(s, u(s))\, ds,\qquad w(t)=c+\int_0^tF(s, w(s))\, ds.$$

We claim that $$\tag{1}u(t)\le w(t).$$ Proof. Consider the set $$ X=\left\{t\in [0, T]\ :\ u(t)-w(t)>0\right\}.$$ Assume by contradiction that $X$ is nonempty and set $$ t_0=\inf X.$$ Since $u$ is a subsolution we have that $t_0>0$. By assumption 1 we have that $$ F(s, u(s))-F(s, w(s))\le 0\qquad \forall s\in[0, t_0]. $$ So for $t\in X$ we have by assumption 2 \begin{equation} \begin{split} 0<u(t)-w(t)&\le \int_0^tF(s, u(s))-F(s, w(s))\,ds \\ &\le \int_{t_0}^t F(s, u(s))-F(s, w(s)\, ds \\ &\le L\int_{t_0}^tu(s)-w(s)\, ds, \end{split} \end{equation} and Gronwall's inequality gives the contradiction $0<u(t)-w(t)\le 0$. $\square$

I don't know if condition 2 can be weakened, but surely it cannot be dropped altogether. One must have at least a uniqueness result for the integral equation, and condition 2 gives such a result (that's the standard Picard's existence and uniqueness theorem).

For example, consider this problem:

$$w(t)=\int_0^t \big(w(s)\big)^{1/3}\, ds.$$

We know that there exist more than one solution to this equation, and of course every solution is a subsolution. If our claim were true in this case we would have for a pair of distinct solutions $w_1, w_2$ the inequalities $w_1(t) \le w_2(t)$ and $w_2(t)\le w_1(t)$, that is, $w_1(t)=w_2(t)$, a contradiction.

An open question remains, and it it necessity of the monotonicity condition on $F(s, \cdot)$. This is intuitively reasonable, but should be proved.

$^{[1]}$ It is enough to assume those conditions for almost all values of $s$.

A sufficient condition is that every function $F(s,\cdot)$ is nondecreasing. That is:

For every $s$ in $[\tau,T]$, $v\le v'$ implies $F(s,v)\le F(s,v')$.

In full generality, this condition is probably also necessary.