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

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

Denote

$$\Delta(l):=\int_0^l\Big(\frac{1}{2\phi_+(m)}-\frac{1}{2\phi_-(m)}\Big)dm$$

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

$$\phi'_+\mu\big(\phi_+\big)-\phi'_-\mu\big(\phi_-\big)=\exp\big(-\Delta(l)\big)\Delta'(l),$$

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

$$\mu\big(\phi_+\big)-\mu\big(\phi_-\big)=1-\exp\big(\Delta(l)\big).$$

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