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Hi everyone, I would like to find a function

$$\Psi\in\mathcal{C}^2: z\in\mathbb{R}\rightarrow\Psi(z)\in\mathbb{R_+}$$

satisfying the following conditions:

$$1-\frac{z\Psi'(z)}{\Psi(z)}+8s\Psi''(z)\geq 0$$

$$\Psi(z)^2+\frac{4}{\theta}\Psi(z)\leq\frac{z^2}{4\theta s}$$

where $\theta$, $s$ are given positive constants.

The second inequality yields

$$0\leq\Psi(z)\leq\sqrt{\frac{z^2}{4\theta s}+\frac{4}{\theta^2}}-\frac{2}{\theta}$$

Could someone have an idea to solve the first differential equation? Or give such a function $\Psi(z)$? Many thanks!

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  • $\begingroup$ Maybe such a function does not exist? However, if we allow $\Psi$ to map to the negative reals, then $\Psi(z)$ given by the negative solution to the quadratic equation above satisfies the other conditions. But I have not checked any further. It would be good to know a bit of your motivation for investigating these inequalities... $\endgroup$
    – Suvrit
    Commented Aug 27, 2012 at 14:20
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    $\begingroup$ The second consition implies that $\Psi(0)=0$. Shouldn't you exclude $z=0$ for the first condition? $\endgroup$
    – Ilya Bogdanov
    Commented Aug 27, 2012 at 14:48
  • $\begingroup$ Notice also that the substitution $z=4t\sqrt{s/\theta}$, $\psi(z)=2\phi(t)/\theta$ leads to cancellation of constants. The inequalities read as $1-t\phi'/\phi+\phi''\geq 0$, $\phi^2+2\phi\leq t^2$. $\endgroup$
    – Ilya Bogdanov
    Commented Aug 27, 2012 at 14:56

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