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I'm looking for the solution of the following partial differential equation:

\begin{equation} \frac{\partial}{\partial x} \Big (- a x f(x,y) + \frac{\partial f}{\partial x} (x,y) \Big ) + \frac {\partial}{\partial y} \Big (- b y f(x,y) + \frac{\partial f}{\partial y} (x,y) \Big ) = 0 \end{equation}

Hereby, $a$ and $b$ are constants.

A possible approach could be to try to separate the equation. Let's assume $f(x, y) = f_1(x) f_2(y)$. Then, it holds

\begin{align} & \frac{\partial}{\partial x} \Big (- a x f(x,y) + \frac{\partial f}{\partial x} (x,y) \Big ) + \frac {\partial}{\partial y} \Big (- b y f(x,y) + \frac{\partial f}{\partial y} (x,y) \Big ) = 0 \\ \Leftrightarrow & - a f(x, y) - a x \frac{\partial f}{\partial x} (x, y) + \frac{\partial^2 f}{\partial x^2} (x,y) - b f(x, y) - b y \frac{\partial f}{\partial y} (x, y) + \frac{\partial^2 f}{\partial y^2} (x,y) = 0 \\ \Leftrightarrow & - a - a x \frac{\partial f_1}{\partial x} (x) \frac{1}{f_1(x)} + \frac{\partial^2 f_1}{\partial x^2} (x) \frac{1}{f_1(x)} - b - b y \frac{\partial f_2}{\partial y} (y)\frac{1}{f_2(y)} + \frac{\partial^2 f_2}{\partial y^2} (y) \frac{1}{f_2(y)} = 0 \end{align}

Hereby, we assumed $f_1(x) \neq 0$ and $f_2(y) \neq 0$. We get

\begin{align} - a - a x \frac{\partial f_1}{\partial x} (x) \frac{1}{f_1(x)} + \frac{\partial^2 f_1}{\partial x^2} (x) \frac{1}{f_1(x)} = C \\ - b - b y \frac{\partial f_2}{\partial y} (y) \frac{1}{f_2(y)} + \frac{\partial^2 f_2}{\partial y^2} (y) \frac{1}{f_2(y)} = - C \end{align}

Thus, we transfomed the partial differential equation to two ordinary differential equations. However, is there a way to solve the partial differential equation with a different approach and without needing to introduce the restrictions $f_1(x) \neq 0$ and $f_2(y) \neq 0$?

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  • $\begingroup$ Whats your domain? Do you have specific BCs? $\endgroup$
    – maxmilgram
    Aug 6, 2020 at 20:22

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