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Hi everyone, could someone help me give a characterization of $\Psi(z)$ such that

$$8\Psi''-4\frac{(\Psi')^2}{\Psi}-\theta(\Psi')^2\geq 0,\ \ \forall z\in\mathbb{R}$$


$$\Psi=\Psi(z),\ \ \Psi'=\Psi'(z),\ \ \Psi''=\Psi''(z)$$

and $\theta$ is some positive constant.

Many thanks!

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I am sorry to trouble you. I think that I have found a solution. Set $x=\Psi$, $y=\Psi'$ and $w=y^2$, then we have $$\Psi''=\frac{d\Psi'}{dz}=\frac{d\Psi'}{d\Psi}\times\frac{d\Psi}{dz}=\frac{dy}{‌​dx}\times y$$ which yields $$4\frac{dw}{dx}-4\frac{w}{x}-\theta w\geq 0$$ Notice that $w>0$ and $x=\Psi>0$, we have $$w(x)\geq Cxe^{\theta x/4}$$ for some constant $C$. – Higgs88 Aug 14 '12 at 9:04
But I have a new question about the following differential inequality: $$\left(1-\frac{z\Psi'(z)}{2\Psi(z)}\right)^2-4k\left(\frac{(\Psi') ^{2}(z)}{\Psi(z)}-2\Psi''(z)\right)-\theta k(\Psi')^{2}(z)\geq 0,\ \ z>0$$ where $k>0$ and $\theta>0$ are given constants. – Higgs88 Aug 14 '12 at 9:10
Could you give some conditions which are easier to describe for $\Psi$ such that the previous inequality holds. For example we can take such $\Psi$ satisfying: $$\left(\frac{(\Psi')^2}{\Psi}-2\Psi''\right)+\frac{\theta}{4}(\Psi')^2\leq 0,\ \ \forall z>0$$ which is just the previous question that I posed. But a problem is that we impose that $\Psi'(z)$ is bounded in $(0,+\infty)$. So we have to consider the original problem. – Higgs88 Aug 14 '12 at 9:16
So my question is to find the function $\Psi$ satifying the inequality and $\Psi'$ bounded. Thanks a lot for your help! – Higgs88 Aug 14 '12 at 9:17
I'd suggest to re-edit the question and state the new problem, including a few lines about its origin and motivation. As I understand, you want $\Psi:(0,+\infty)\rightarrow\mathbb{R}$ with bounded derivative $\Psi'$, and satisfying the new differential inequality. – Pietro Majer Aug 15 '12 at 7:41

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