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Suppose we have the following differential inequality

$F''(w)\le \frac{p-1}{p}\frac{(F'(w))^2}{F(w)}$ on $w\in(w_0,w_0+\varepsilon)$, $w_0>0$, $\varepsilon>0$, $p>1$. In addition, $F(w)>0$, $F'(w)>0$ for $w\in[w_0,w_0+\varepsilon)$.

Let's consider the solution of the Cauchy problem

$\begin{cases}G''(w)= \frac{p-1}{p}\frac{(G'(w))^2}{G(w)},& w\in(w_0,w_0+\varepsilon),\\G'(w_0)=F'(w_0),&\\G(w_0)=F(w_0),&\end{cases},$ which can be written explicitly:

$G(w) = F(w_0)\left(1+\frac{F'(w_0)}{F(w_0)}\frac{w-w_0}{p}\right)^p$.

Is it possible to compare the functions $F$ and $G$ in any meaningful way? This case fits neither the canonical version of the Gronwall's lemma nor its generalisation by Bihari, because it involves second derivatives.

I'd be glad to hear all suggestions!

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up vote 1 down vote accepted

You can calculate the seconde derivative of $F^{1/p}$. you'll find that it is negative, and the seconde derivative of $G^{1/p}$ is 0, so we have $(F^{1/p})^{''}\leq (G^{1/p})^{''} $ Then by integrating twice you'll have $F^{1/p}\leq G^{1/p} $ and then $F\leq G$

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Thank you! In fact, this problem itself came from hypothesis of concavity of the function $F^{1/p}$. – TZakrevskiy May 14 '13 at 19:21

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