Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer

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 $F\leq G$

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.