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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?

I have posted the same question in classical physics forum a few days ago but not one replies me. I hope people here can show me where I am going wrong or why this result happens. Thanks so much!

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# Variable substitution in Langevin equation and Fokker-Planck equation

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