Let $\phi_+ (\phi_-)$ be a strictly increasing (decreasing) function defined on $R_+$ such that $\phi_+(\phi_-)\in\mathcal{C}^0(R_+)\cap\mathcal{C}^1(R_+^{\ast})$ and $\phi_+(0)=0(\phi_-(0)=0)$. Assume also

$$\int_0^1\Big(\frac{1}{\phi_+(l)}-\frac{1}{\phi_-(l)}\Big)dl<+\infty,~ \int_1^{+\infty}\Big(\frac{1}{\phi_+(l)}-\frac{1}{\phi_-(l)}\Big)dl=+\infty.$$



and I would like to solve the following system

\begin{eqnarray} 2\phi'_+(l)\phi_+(l)\mu\big(\phi_+(l)\big)&=&\exp\big(-\Delta(l)\big), \forall l>0 \\ 2\phi'_-(l)\phi_-(l)\mu\big(\phi_-(l)\big)&=&\exp\big(-\Delta(l)\big), \forall l>0 \end{eqnarray} where $\mu$ is a continuous density function of some probability distribution, i.e.

$$\mu(s)\ge 0,~ \forall s\in R \text{ and } \int_{R}\mu(s)ds=1.$$

I try to express $(\phi_+,\phi_-)$ by $\mu$, but I can not find an explicit expression. In fact, it is clear that


which yields by $\phi_+(0)=\phi_-(0)=0$


But I don't know what to do further. Does some have an idea? Thx for the reply.

  • $\begingroup$ For english speakers EDP = PDE $\endgroup$ – Anthony Quas Nov 5 '14 at 16:27
  • 2
    $\begingroup$ Regardless of language, is there a PDE here? All I see is two identical ODEs (typo?). $\endgroup$ – Michael Renardy Nov 6 '14 at 1:13
  • $\begingroup$ I edited the second equation in the system (and rendered Michael Renardy's comment obsolete), based on the assertion that "it is clear that $\phi_+'\mu\circ\phi_+ - \phi_-' \mu\circ\phi_- = \ldots$." $\endgroup$ – Willie Wong Nov 6 '14 at 8:39
  • $\begingroup$ Also, your last assertion seems incorrect. Integrating the difference should get you, writing $M(s) = \int_0^s \mu(s) \mathrm{d}s$ that $M(\phi_+) - M(\phi_-) = 1 - \exp (-\Delta)$. Note that the left hand side is equivalent to $\int_{\phi_-}^{\phi_+} \mu(s) \mathrm{d}s$. $\endgroup$ – Willie Wong Nov 6 '14 at 8:49
  • $\begingroup$ Lastly, observe that if $\mu(s) = 0$ on an interval $(-\epsilon,\epsilon)$, solutions verifying your assumptions cannot exist: for sufficiently small $l$ you must have $\int_{\phi_-}^{\phi_+} \mu(s) \mathrm{d}s = 0$ while $1 - \exp(-\Delta(l)) > 0$. $\endgroup$ – Willie Wong Nov 6 '14 at 9:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.