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The Gronwall's inequality can be stated as:

Suppose $u,v$ be continuous function on $[a,b]$ with $u\geq 0$, if $$v(t)\leq C+\int_a^t u(s)v(s)ds,\quad \forall t\in[a,b]$$ where $C$ is a constant, then $$ v(t)\leq C\exp\left(\int_a^t u(s)ds\right). $$

I only find a proof in WiKi, and in fact I don't know the idea of that proof there, so any guide of ideas of proof or another proof would be helpful to me.

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up vote 3 down vote accepted

Set $\displaystyle \alpha(t)= C+ \int_a^t u(s)\cdot v(s)\,ds\;$, $\quad \displaystyle \beta(t)=C \cdot \exp \left( \int_a^t u(s)\,ds \right)$ and $\displaystyle \gamma(t)= \frac{\alpha(t)}{\beta(t)}$ for $C>0$.

Then $\gamma(a)=1$ and $$ \gamma'(t)=u(t) \frac{v(t)- C- \int_a^t u(s)\cdot v(s)\,ds}{C \exp \left( \int_a^t u(s)\,ds \right)} \leq 0 $$

So $\gamma$ is nonincreasing, hence $\gamma(t) \leq 1$ that is $\displaystyle C+ \int_a^t u(s)\cdot v(s)\,ds \leq C \exp \left( \int_a^t u(s)\,ds \right)$.

Thus the inequality follows.

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Assuming for the moment $u,v$ are differentiable, then we have $v'(t)<u(t)v(t)$ for all $t$. So in particular $u(t)\ge \frac{v'(t)}{v(t)}=(\ln(v(t))'$. Therefore $\int_{a}^{x} u(t)\ge \ln(v(x))+K$ with $K=u(a)-\ln(v(a))=u(a)-\ln(C)$.

Now we need to prove $$u(t)v(t)\le Ce^{\int^{t}_{a} u}*u(t)\leftrightarrow v(t)\le Ce^{\int^{t}_{a}u(t)}$$By above assumption we have $$Ce^{\int^{t}_{a}u(t)}\ge Ce^{In(v(t))+u(a)-\ln(C)}=v(t)e^{u(a)}$$and it is trivial that $v(t)\le v(t)e^{u(a)}$ since by assumption $u(a)\ge 0$. This is not a proof as we assumed that two functions are differentiable(and $v\ge 0$, though this can be fixed), but I think this justifies the heuristics behind the proof.

For a formal short, nice proof wikipedia has one at here.

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How to get $v'(t)<u(t)v(t)$ (with $u,v\in C^1$)? –  van abel Jan 12 '13 at 13:59
    
differentiate on both sides. –  Bombyx mori Jan 12 '13 at 14:06
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