Let $\gamma_+$, $\gamma_-:\mathbb{R}_+\to\mathbb{R}$ be two given functions. Assume that $\gamma_+$ ($\gamma_-$) is smooth, strictly increasing (decreasing) and $\gamma_{+}(+\infty)=+\infty$ ($\gamma_{-}(+\infty)=-\infty$).

Now consider the following system:

\begin{eqnarray} \lambda''(\gamma_+)\gamma_+'\gamma_+&=&\lambda''(\gamma_-)\gamma_-'\gamma_-,~ \forall l>0 \\ \lambda'(\gamma_+)-\lambda'(\gamma_-)&=&2g'(l)+2\lambda''(\gamma_-)\gamma_-'\gamma_-,~ \forall l>0, \end{eqnarray}

wherer $\gamma_+=\gamma_+(l)$, $\gamma_-=\gamma_-(l)$ and $g=g(l)$ is a given function regular enough. Now my question is how to obtain explicitly the solution $\lambda: \mathbb{R}\to\mathbb{R}$ and under which conditions $\lambda$ is convex. Thx a lot for the reply!

  • $\begingroup$ Can you also assume $\gamma_+ \ge \gamma_- $ ? I guess this would simplify the problem, as it could be seen as a genuine 1st order differential system in two independent functions on $\mathbb{R}_+$, $u:=\lambda'(\gamma_+)$ and $v:=\lambda'(\gamma_-)$, perhaps with initial conditions at $l=0$ if $\gamma_+(0)= \gamma_-(0)$ . $\endgroup$ – Pietro Majer Sep 22 '14 at 23:13
  • $\begingroup$ Sure, exactly that is i forget writting here. $\endgroup$ – CodeGolf Sep 23 '14 at 5:38
  • $\begingroup$ And do you know whether $\gamma_+(0)=\gamma_-(0)$ or $\gamma_+(0) > \gamma_-(0)$? Thank you. $\endgroup$ – Pietro Majer Sep 23 '14 at 6:04
  • $\begingroup$ Yes, there are probably two cases. $\gamma_+(0)=\gamma_-(0)$ or $\gamma_+(0)=\gamma_-(0)$, but for the first case there will be singular (since the formal solution one gets is contains an integral like $\int_0^{+\infty}(\frac{1}{\gamma_-(0)}+\frac{1}{\gamma_+(0)})dx$) $\endgroup$ – CodeGolf Sep 23 '14 at 6:25

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