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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?

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    @anon: *Native language* should not be very difficult to determine, starting from dissonance's user page... :-)2011-08-01

2 Answers 2

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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.

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    This is just to let you know of the following answer. Looks like the @name functionality is disabled outside comments.2011-09-10
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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$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 $s$.