Solving a differential system Let $\mu$ be a probability measure on $\mathbb R$ with Lebesgue density, i.e. $\mu(dx)=\mu(x)dx$. We aime to find increasing and decreasing functions $\phi_{+}: \mathbb R_+\to \mathbb R_{+}$ and $\phi_{-}: \mathbb R_+\to \mathbb R_{-}$ s.t. $\phi_{\pm}(0)=0$ and
$$\phi_{\pm}'(x)~~=~~\frac{1~-~F_{\mu}\big(\phi_+(x)\big)~+~F_{\mu}\big(\phi_-(x)\big)}{2\phi_{\pm}(x)\mu\big(\phi_{\pm}(x)\big)} \mbox{ for all } x>0,$$
where $F_{\mu}$ denotes the cumulative distribution function of $\mu$. My question is how to specify $(\phi_+,\phi_-)$ in terms of $\mu$? If there is some numerical shcema, I'm equally glad to know about it.
I find a solution if $\mu$ is symmetric, i.e. $\mu(x)=\mu(-x)$ for all $x\in\mathbb R$. Then it is easy to guess that $\phi_{\pm}=\pm\phi$ with
$$\phi^{-1}(x)~~=~~\int_0^x\frac{y\mu(y)}{1~-~F_{\mu}(y)}dy \mbox{ for all } x\in\mathbb R_+.$$
If someone knows how to treat the general case, please let me know. Thanks a lot!
 A: For simplicity, let's assume initially  that the support of $\mu$ is the whole real line, so that $F$ is a homeomorphism $ \mathbb{R} \to (0,1) $. Let's denote $b:=F(0)=\mu(-\infty,0]=1-\mu[0,+\infty)\in(0,1)$.  
The relevant function is the strictly convex function $\Phi :(0,1)\to\mathbb{R}$ defined as
$$\Phi(t):=\int_{b}^t F^{-1}(s)\, ds  $$
which is the Legendre transformation of $\int_0^x F(t)dt$. It has minimum value at $t=b$, and $\Phi(t)\to+\infty$, both for $t\to0$ and  $t\to1$ (indeed 
$\Phi(0)=\int_{b}^0 F^{-1}(s)\, ds=\int_ {-\infty}^0  F(t) dt$: writing them in terms of the density function $\mu(s)$ as double integrals, and using Tonelli's theorem, one finds that both integrals diverge).
Let's denote $\Psi_{+}:=(\Phi_{|[b,1)})^{-1}: \mathbb{R}_+ \to[b,1)$ and $\Psi_-:=(\Phi_{|(0,b]})^{-1}:\mathbb{R}_+\to(0,b]$ . Note that $\Psi(t):=1-\Psi_+(t)+\Psi_-(t)=\big|\{\Phi\ge t\}\big|$ is the inverse function of the monotone decreasing rearrangement of $\Psi$.
Consider the solution $u:[0,+\infty)\to \mathbb{R}$ of the first order autonomous ODE 
$$u'(t) = \frac{1}{2} \Psi (u ),\quad t\ge0$$
with $u(0)=0$, given  explicitly by
$$u^{-1}(x)=2\int_0^x\frac{dy}{ \Psi (y) }.$$
This integral is finite for all $x\ge0$ and diverges for $x\to+\infty$, so $u$ is a homeo $\mathbb{R}_+\to\mathbb{R}_+$ .
Then it is easy to check that $\phi_{\pm}:=F^{-1}\circ\Psi_\pm\circ u $ are defined  $\mathbb{R}_+$, diverges to $\pm\infty$ for $x\to+\infty$, and solve the initial problem written as
$$ 2\phi_+ F'(\phi_+)\phi_+'=2\phi_- F'(\phi_-)\phi_-'=1-F(\phi_+)+F(\phi_-)$$
that we can write equivalently, because $\phi_\pm=F^{-1}\circ F\circ \phi_\pm=\Phi'\circ F\circ \phi_\pm$, in the form
$$ 2(\Phi\circ F\circ \phi_+)'=2(\Phi\circ F\circ \phi_-)'=1-\Psi_+\circ\Phi\circ F\circ \phi_+  +\Psi_-\circ\Phi\circ F\circ \phi_- .$$ 
Conversely, if $\phi_\pm$ solve the initial system, the latter equation shows that $\Phi\circ F\circ \phi_+=\Phi\circ F\circ \phi_-$ because they coincide at $0$ and have the same derivative, and in fact solve the above equation for $u$.  
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Rmk. The assumption on the support of $\mu$ is not crucial; if it is not the whole real line, $F$ is constant on some open set, and $F^{-1}$ has jumps and it is only defined as a left inverse of $F$; the integral for $\Phi$ is however well defined. 
