Solving a differential system

Question

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!

Answer 1

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.