Variable substitution in Langevin equation and Fokker-Planck equation

Tianwu Zang

Dear all,
I have a question about the variable substitution in Langevin equation and Fokker-Planck equation and this has bothered me a lot. The general Langevin equation is:
$$\frac{dx}{dt}=u(x)+\sqrt{2 D(x)}\eta(t)$$
and the corresponding Fokker-Planck equation is thus:
$$\frac{\partial \rho(x)}{\partial t}=-\frac{\partial}{\partial x}\left[u(x)\rho(x)\right]+\frac{\partial^2}{\partial x^2}\left[D(x)\rho(x)\right]$$
which means the stationary distribution of x should satisfy
$$u(x)\rho(x)=\frac{\partial}{\partial x}\left[D(x)\rho(x)\right]$$
However, problem emerges when I want to use a variable substitution y(x), since the Langevin equation becomes
$$\frac{dy}{dt}=u(x)y'(x)+\sqrt{2 D(x)}y'(x)\eta(t)$$
which the corresponding F-P equation
$$\frac{\partial \rho(y)}{\partial t}=-\frac{\partial}{\partial y}\left[u(x)y'\rho(y)\right]+\frac{\partial^2}{\partial y^2}\left[D(x)y'^2\rho(y)\right]$$
and the stationary distribution of y is thus
$$u(x)y'\rho(y)=\frac{\partial}{\partial y}\left[D(x)y'^2\rho(y)\right]$$
Considering
$$\rho(x)dx=\rho(y)dy \Rightarrow \rho(x)=\rho(y)y'$$
we can rewrite the stationary ρ(y) equation before as
$$u(x)\rho(x)=\frac{\partial}{\partial y}\left[D(x)y'\rho(x)\right]=\frac{\partial}{\partial x}\left[D(x)y'\rho(x)\right]x'(y)=\frac{1}{y'}\frac{\partial}{\partial x}\left[D(x)y'\rho(x)\right]$$
which is not equal to the stationary ρ(x) derived before. Is there anything wrong with my derivation? Can anyone help me to figure this out?

Tianwu Zang

Nobody knows?