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Let us review Gronwall's lemma. If $v \in C^0([\tau, T])$ is such that

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

where $c$ is a real constant and $u \in C^0([\tau, T])$ is nonnegative, then

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

In other words, sub-solutions of the integral equation

$$w(t)=c+ \int_\tau^t u(s)w(s)\, ds, \qquad t \in [\tau, T]$$

are dominated by solutions of the same equation.

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

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I am not sure taht your equation (1) is the general Volterra equation. Your equation is equivalent to the initial value problem for the differential equation $w'=F(t,w)$, $w(\tau)=c$. The Volterra equation of the first kind is the linear integral equation $w(t)=\int_a^b k(s,t)w(s)ds$, where $k(s,t)$ is the kernel. Of course, there is a second kind. –  Julián Aguirre Jun 6 '11 at 21:50
@Juliàn: Unfortunately I'm not an expert of those things. I'm used to call Fredholm integral equation what you have called Volterra equation of the first kind and Volterra equation (without specifying any kind) the one I refer to in the question above. My only reference on this is Debnath-Mikusinski's book on Hilbert spaces. –  Giuseppe Negro Jun 7 '11 at 0:32
@dissonance: In your question title I think you mean "bound above" not "estimate." –  anon Jul 31 '11 at 15:04
@anon: You found a flaw in my English. What is the difference between those terms in current language? I guess that "estimate" is reserved to bounds in absolute value, am I right? –  Giuseppe Negro Aug 1 '11 at 12:39
@anon: Native language should not be very difficult to determine, starting from dissonance's user page... :-) –  Did Aug 1 '11 at 14:52

2 Answers 2

up vote 1 down vote accepted

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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I'm a bit busy right now and so I will think about your answer as soon as possible. In all cases, thank you very much for your interest. –  Giuseppe Negro Aug 1 '11 at 12:42
This is just to let you know of the following answer. Looks like the @name functionality is disabled outside comments. –  Giuseppe Negro Sep 10 '11 at 17:21

This is not an answer but a (long) comment on @Didier Piau answer to the original question.

I think that we need to add an extra hypothesis (probably Didier took it for implicitly granted): that $F$ satisfies a Lipschitz condition with respect to the second variable. This is what I tried.

Suppose that $L$ is such a Lipschitz constant for $F$ and let

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

We claim that $u(t)\le w(t)$. To see this notice that, because of monotonicity of $F(s, \cdot)$, we have

$$v(t)\le c+\int_0^tF(s, v(s))\, ds.$$

If, by contradiction, $v(t)-w(t)>0$ at some point, then by continuity there would be an interval $I=(t_0, t_1)$ s.t. $v(t)-w(t)>0$ on $I$ and $v(t_0)-w(t_0)=0$. On this interval we would have

$$v(t)-w(t) \le \int_{t_0}^t \Big(F(s, v(s))-F(s, w(s))\Big)\, ds \le L\int_{t_0}^t\big( v(s)-w(s)\big)\, ds$$

and thus, by Gronwall's lemma, the contradiction $v(t)-w(t)\le 0$ on $I$.

Then $v(t)\le w(t)$ and so $u(t)\le w(t)$, which was our claim.

I believe also that a Lipschitz condition (or at least a uniqueness theorem for solutions) is really needed. Let us take for example

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

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 I have not crafted a rigorous proof. I will think about it.

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