Laplace equation with boundary conditions of solution value and normal derivative I want to solve the Laplace equation in 2 dimensions $(x,y)$ in the quadrant domain $x,y \geq 0$ $$\partial_{x^2} f + \partial_{y^2}f = 0 \; .$$ The boundary conditions I am trying to impose are $$f = 0 \; , \hspace{4mm} \partial_{y}f = g(x) \hspace{4mm} \text{at} \hspace{4mm} y = 0\\ f = 0 \; , \hspace{4mm} \partial_{x}f = h(y) \hspace{4mm} \text{at} \hspace{4mm} x = 0$$
The textbook on linear partial differential equations I am using (Handbook of linear partial differential equations for engineers and scientists) does not cover this case.
Does a solution to these boundary conditions exist? How do I express the solution in terms of the boundary conditions formally?

I am thinking that this problem is very similar to the problem of solving the Laplace equation in the upper half-plane $(-\infty \leq x \leq \infty, y\geq 0)$ with prescribed function value and normal derivative on the lower boundary. To this, I think a solution can be developed by the Fourier transform
$$ \hat{f}(k,y) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x,y) e^{-i k x} dx $$
and
$$ \hat{g}(k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} g(x) e^{-i k x} dx $$
where $g(x)$ is the normal derivative of $f$ on the lower boundary of the half-plane. The transformed Laplace equation is
$$ \nabla^2 f = -k^2 \hat{f} + \partial_{y^2} \hat{f} = 0$$ and the solution is
$$ \hat{f}(k,y) = A(k) e^{k y} + B(k) e^{-k y} \; . $$
Using the boundary conditions
$$ \left. \partial_{y} \hat{f} \right \vert_{y=0} = \hat{g} = A k - B k \\ \left. \hat{f} \right \vert_{y=0} = 0 = A + B $$
and applying the inverse transform gives the final result
$$ f(x,y) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \frac{\hat{g}(k)}{2 k} \left[ e^{i k (x-iy)} - e^{i k (x+i y)} \right] d k \; . $$
Is this result correct?
I still need to transform the upper half-plane to the quadrant. I guess I could use the Schwarz-Christoffel transform but I am not sure if this is possible and how it is done.
Do you have any ideas how to do that?

Update: I have obtained as the result in the half-plane
$$ f(x,y) = \frac{1}{2 i} \int_{x-iy}^{x+iy} g(\xi) \; d \xi \; . $$
Does this appear correct?
 A: I'm posting this as another answer, because the method is very different.
You can try to solve this in cylindrical coordinates. Laplace equation is solved by $\{\sin k\phi,\cos k\phi\} r^{\pm k}$, and in our case, the boundary conditions are satisfied by $\sin 2n\phi$ where $n$ is a positive integer.
$$f=\sum_{n=1}^\infty a_n  r^{2n}\sin 2n\phi +\sum_{n=1}^\infty b_n r^{-2n}\sin 2n \phi $$
Boundary conditions:
$$\partial_y f(y=0)=g(r)=\sum_{n=1}^\infty 2n  a_n  r^{2n-1} +\sum_{n=1}^\infty 2n b_n r^{-2n-1} $$
$$\partial_x f(x=0)=h(r)=-\sum_{n=1}^\infty (-1)^n 2n  a_n  r^{2n-1} -\sum_{n=1}^\infty (-1)^n 2n b_n r^{-2n-1} $$
Add and subtract, and rearrange:
$$r(g(r)-h(r))=\sum_{n=\text{even}}^\infty 4n  (a_n  (r^2)^{n} + b_n (r^2)^{-n})$$
$$r(g(r)+h(r))=\sum_{n=\text{odd}}^\infty 4n  (a_n  (r^2)^{n} + b_n (r^2)^{-n})$$
This looks very ill-posed (especially as it's not even a true power series, you are missing terms), but it's a useful form if your $g$ and $h$ are of power series form.
But the main reason for such a badly behaving result is, that the task itself is severely overdetermined anyway... normally, Laplace equation only allows setting one thing per boundary condition: either a value or a derivative. If you set both, you have to leave some other boundary condition completely open/free. In your case, this is the opening towards the infinity (and in half-plane, you see the consequence in existence of both decreasing and increasing exponential terms). But if you are enforcing very unusual conditions, things will behave very badly! For example, a natural solution $r^2 \sin(2\phi)=2xy$ implies a positive derivative on both sides. If you enforce $g(r)=r^2,\,h(r)=-r^2$, boundary conditions are "pulling the sheet" in opposite directions! The extra curvature must escape in the diagonal direction towards infinity by creating nasty folding. I might expect the sum to diverge in such cases.
A: Indeed as you suggested, you can transform one result to the other. Complex map $f(z)=\sqrt{z}$ takes the half-plane coordinates to the quadrant. Now consider what this map does to the boundary condition. Write
$$x+iy=\sqrt{x'+iy'}\iff x'+iy'=(x+iy)^2$$
You also have
$$F(x',y')=f(x,y)$$
And of course, the positive half-axis is mapped from the horizontal leg $(x,0)\mapsto (x^2,0)$ and the negative from the vertical leg, $(0,y)\mapsto (-y^2,0)$.
You need
$$\frac{dF(x',y')}{dy'}=\frac{df(x,y)}{dx}\frac{dx}{dy'}+\frac{df(x,y)}{dy}\frac{dy}{dy'}=
\partial_x f(x,y)\frac{1}{2y}+\partial_y f(x,y)\frac{1}{2x}$$
Evaluate at $y'=0$ and $y=0$ (horizontal leg):
$$\frac{dF}{dy'}(x'>0,y'=0)=\frac1{2x}\partial_y f(\sqrt{x'},0)=\frac{1}{2\sqrt{x'}}g(\sqrt{x'})$$
Evaluate at $y'=0$ and $x=0$ (vertical leg):
$$\frac{dF}{dy'}(x'<0,y'=0)=\frac1{2y}\partial_x f(0,\sqrt{-x'})=\frac{1}{2\sqrt{-x'}}h(\sqrt{-x'})$$
This tells you exactly how to transform the quadrant boundary condition to the half-plane boundary condition and then solve it there. The inverse mapping simply involves reading the function $f(x,y)=F(x^2-y^2,2xy)$.
