

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Hi everyone, I would like to find a function


satisfying the following conditions:


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

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!

share|cite|improve this question
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... – Suvrit Aug 27 '12 at 14:20
The second consition implies that $\Psi(0)=0$. Shouldn't you exclude $z=0$ for the first condition? – Ilya Bogdanov Aug 27 '12 at 14:48
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 $1-t\phi'/\phi+\phi''\geq 0$, $\phi^2+2\phi\leq t^2$. – Ilya Bogdanov Aug 27 '12 at 14:56

Your Answer



Browse other questions tagged or ask your own question.