Solution of the following PDE How to solve the following pde: $$u_{xy}=u_{x}u_{y}$$ with $$\begin{cases}
u(0,0)=c\ :c\in \mathbb{R}\\
u(x,0)=f(x)\\
u(0,y)=g(y)
\end{cases}$$
Any hints are appreciated.
 A: I'm going to describe a solution for the special case where $c = f(0) = g(0)$. When this condition is not obeyed, the boundary conditions are inconsistent and discontinuous at $(x,y) = (0,0)$. I'm not sure what happens then.
We perform the following steps, starting with the original PDE as stated:
\begin{align*}
u_{xy} &\;=\; u_x \, u_y\\[0.1in]
\frac{u_{xy}}{u_y} &\;=\; u_x\\
\frac{\partial}{\partial x} \ln\left(u_y\right) &\;=\; \frac{\partial}{\partial x} u\\[0.1in]
\ln\left(u_y\right) &\;=\; u \;+\; w(y) \qquad\qquad (1)
\end{align*}
In the last line above, $w(y)$ is an as-yet-unknown function of $y$ alone. We apply one of the boundary conditions by taking $x\rightarrow 0$ in Eq. (1) to yield an expression for $w(y)$:
\begin{equation*}
\ln\left(g'(y)\right) \;=\; g(y) \;+\; w(y) \quad\rightarrow\quad
w(y) \;=\; \ln\left(g'(y)\right) \;-\; g(y) \qquad (2)
\end{equation*}
We exponentiate both sides of Eq. (1) and use Eq. (2) as follows:
\begin{align*}
\frac{\partial}{\partial y} u &\;=\; e^u \, e^{w(y)}\\
&\;=\; e^u \, \exp\Bigl[\ln\left(g'(y)\right) \;-\; g(y)\Bigr]\\
\left(\frac{\partial}{\partial y} u\right) \; e^{-u} &\;=\; g'(y) \, e^{-g(y)}\\
\frac{\partial}{\partial y}e^{-u} &\;=\; \frac{\partial}{\partial y} e^{-g(y)}\\
e^{-u} &\;=\; e^{-g(y)} \;+\; a(x)\\
u(x,y) &\;=\; -\ln \left[e^{-g(y)} \;+\; a(x)\right] \qquad (3)
\end{align*} 
Here $a(x)$ is an arbitrary function of $x$ alone.
By an entirely-analogous argument, we can also obtain
\begin{equation*}
u(x,y) \;=\; -\ln \left[e^{-f(x)} \;+\; b(y)\right] \qquad (4)
\end{equation*} 
for some unknown function $b(y)$. Combining Eqs. (3) and (4) then implies that
\begin{equation*}
u(x,y) \;=\; -\ln \left[e^{-f(x)} \;+\; e^{-g(y)} \;+\; K\right] \qquad (5)
\end{equation*}
Here $K$ is an unknown constant. By enforcing either boundary condition ($u(x,0) = f(x)$ or $u(0,y) = g(y)$) in Eq. (5) and using the fact that $c = f(0) = g(0)$, we find that $K = -e^{-c}$, in which case the final solution is:
\begin{equation*}
u(x,y) \;=\; -\ln \left[e^{-f(x)} \;+\; e^{-g(y)} \;-\; e^{-c}\right]
\end{equation*}
It is straightforward algebra to check that this solution obeys the original PDE as well as all of the boundary conditions.
