Explicit solution for a particular linear second-order elliptic PDE with boundary conditions? I have come across an elliptic PDE of the form:
$$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\left(\dfrac{\partial f}{\partial x}+y\right)-H=0$$
where all of the constants $C,D,E,F,G,H\geq0$. We seek a function 
$$f(x,y):[0,B]\times\mathbb{R}\rightarrow\mathbb{R}$$ for $B>0$ satisfying the PDE on $(0,B)\times\mathbb{R}$ subject to the boundary conditions 
$$\dfrac{\partial f}{\partial x}(0,y)=-y, \quad  \dfrac{\partial f}{\partial x}(B,y)=0$$ for all $y\in\mathbb{R}$ .
Anyone know if there is any hope for obtaining a solution in any kind of explicit form? What would be the best approach to attack this?
Thanks.
 A: Since you specify this PDE is elliptic, this also implies that $C,D\neq0$ .
$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\left(\dfrac{\partial f}{\partial x}+y\right)-H=0$
$C\dfrac{\partial^2f}{\partial x^2}+D\dfrac{\partial^2f}{\partial y^2}-E(y-F)\dfrac{\partial f}{\partial y}+G\dfrac{\partial f}{\partial x}+Gy-H=0$
Let $f=g+\dfrac{x^2y}{2B}-xy$ ,
Then $\dfrac{\partial f}{\partial x}=\dfrac{\partial g}{\partial x}+\dfrac{xy}{B}-y$
$\dfrac{\partial^2f}{\partial x^2}=\dfrac{\partial^2g}{\partial x^2}+\dfrac{y}{B}$
$\dfrac{\partial f}{\partial y}=\dfrac{\partial g}{\partial y}+\dfrac{x^2}{2B}-x$
$\dfrac{\partial^2f}{\partial y^2}=\dfrac{\partial^2g}{\partial y^2}$
$\therefore C\biggl(\dfrac{\partial^2g}{\partial x^2}+\dfrac{y}{B}\biggr)+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\biggl(\dfrac{\partial g}{\partial y}+\dfrac{x^2}{2B}-x\biggr)+G\left(\dfrac{\partial g}{\partial x}+\dfrac{xy}{B}-y\right)+Gy-H=0$
$C\dfrac{\partial^2g}{\partial x^2}+\dfrac{Cy}{B}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}-\dfrac{Ex^2(y-F)}{2B}+Ex(y-F)+G\dfrac{\partial g}{\partial x}+\dfrac{Gxy}{B}-Gy+Gy-H=0$
$C\dfrac{\partial^2g}{\partial x^2}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}+G\dfrac{\partial g}{\partial x}=\dfrac{Ex^2(y-F)}{2B}-Ex(y-F)-\dfrac{Gxy}{B}-\dfrac{Cy}{B}+H$ with $\dfrac{\partial g}{\partial x}(0,y)=0$ and $\dfrac{\partial g}{\partial x}(B,y)=0$
When solving $C\dfrac{\partial^2g}{\partial x^2}+D\dfrac{\partial^2g}{\partial y^2}-E(y-F)\dfrac{\partial g}{\partial y}+G\dfrac{\partial g}{\partial x}=0$ by separation of variables:
Let $g(x,y)=X(x)Y(y)$ ,
Then $CX''(x)Y(y)+DX(x)Y''(y)-E(y-F)X(x)Y'(y)+GX'(x)Y(y)=0$
$(CX''(x)+GX'(x))Y(y)=-(DY''(y)-E(y-F)Y'(y))X(x)$
$\dfrac{CX''(x)+GX'(x)}{X(x)}=-\dfrac{DY''(y)-E(y-F)Y'(y)}{Y(y)}=-\dfrac{4C^2t^2+G^2}{4C}$
$\begin{cases}CX''(x)+GX'(x)+\dfrac{4C^2t^2+G^2}{4C}X(x)=0\\DY''(y)-E(y-F)Y'(y)-\dfrac{4C^2t^2+G^2}{4C}Y(y)=0\end{cases}$
$\therefore$ Let $g(x,y)=\sum\limits_{n=0}^\infty C(n,y)e^{-\frac{Gx}{2C}}\cos\dfrac{n\pi x}{B}$ so that it automatically satisfies $\dfrac{\partial g}{\partial x}(0,y)=0$ and $\dfrac{\partial g}{\partial x}(B,y)=0$ ,
A: For starting point
$$df(x,y)=g(x,y)dx+h(x,y)dy$$
where
$$g(0,y)=-y\qquad g(B,y)=0\Rightarrow g(x,y)=y\frac{x-B}{B}$$
which leads to $$\frac{\partial^2 f}{\partial x^2}=\frac{\partial g}{\partial x}=\frac{y}{B}$$
