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I am looking for a time-dependent probability density $f(x,y,t)$ solving the equation $$-\frac{\partial f}{\partial t} = \alpha\cdot \big(y - F(x)\big)\frac{\partial f}{\partial x}+\beta\cdot \big(G(y)-x\big)\frac{\partial f}{\partial y},$$ where the coefficient functions $F$ and $G$ given by\begin{align}F(x) &= a_0 + a_1 x + o(x - x^{0}),\\ G(y) &=b_0 + b_1y + o(y-y^{0}).\end{align}

My question is: What general class of equations does this PDE belong to, and what would be your angle of attack on such problem (if it makes sense at all)?

Remark I: Under the assumption of existence of a solution, my initial angle of attack was to rewrite the equation as $$i \frac{\partial f}{\partial t} = L\,f$$ where $L$ is an operator acting on $f$ given by $L = -i\left[\alpha\big(y - F(x)\big)\frac{\partial}{\partial x}+\beta\big(G(y)-x\big)\frac{\partial}{\partial y}\right]$. Then - I suppose - we can formally express the density $f$ as $$f(x,y,t)=\sum_k c_ke^{-i\lambda_k t}\varphi_k(x,y),$$ where $\lambda_k$ and $\varphi_k$ are the eigenvalues and eigenfunctions corresponding to the operator $L$. I am thus led to considering the equations $$L(\varphi_k) = \lambda_k\,\varphi_k$$ and swap the solutions $\varphi_k$ and $\lambda_k$ back the into the formal solution. But I have not been able to find the eigenfunctions, perhaps my patience was too short.

Remark II: A necessary condition, I suppose, for the existence of a density solving the PDE is the existence of curves $x$ and $y$ solving the "predator-prey" equations $$\frac{dx}{dt} = \alpha\cdot (y - F(x)),\qquad \frac{dy}{dt} = \beta\cdot (G(y)-x).$$ If we neglect the little-o terms, by putting them equal to zero (is that completely illegal?), the curves $x$ and $y$ are given by \begin{align}x(t) &=-\frac{a^0}{a^1} + C_1 e^{-a^1\mu t} + \frac{1}{1-a^1}\left[\frac{b^0}{b^1}+\frac{a^0}{a^1b^1}-C_2 e^{b^1\lambda t}-\frac{C_4}{b^1}e^{-a^1\mu t}\right]\\ y(t) &= \frac{a^1}{1-a^1}\left[\frac{b^0}{b^1}+\frac{a^0}{a^1b^1}-C_5 e^{b^1\lambda t}-\frac{C_6}{b^1}e^{-a^1\mu t}\right] \end{align} where $C_1$, $C_2$, $C_3$, $C_4$, $C_5$, and $C_6$ are arbitrary constants.

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1 Answer 1

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The temporal part is easy. Since there is no explicit time dependence on the RHS f is seperable as far as f(x,y,t)= R(x,y)T(t). With a separation constant of k, T= $e^{-kt}$.

This leaves the spatial segment. I didn't work it out myself but I would try to use Laplace's transform to turn this into an ODE of one variable and algebraic in the other spatial coordinate.

Hope this helps.

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