Poisson equation on the complex plane with mixed boundary conditions

Question

I would like to compute the Green function for the Laplacian on the complex plane, with mixed Neumann and Dirichlet boundary conditions (see equation 4.150 of Blumenhagen, Lust, Theisen "Basic Concepts of String Theory".)

Let us start from a simpler case, to set the notation: consider Neumann conditions on both end points of the open string.
Using the method of the image charge, I know that the solution should be in the form \begin{align} G(z,w)=-\frac12 {\rm ln}|z-w|^2+a\ {\rm ln}|z-z'|^2,\quad (1) \end{align} where $a$ and $z'$ are to be determined by imposing the boundary conditions.

Neumann conditions on both end points correspond to $z\partial/\partial z=\bar z\partial/\partial \bar z$ when $z=\bar z$: this implies that \begin{align} -\frac12\frac{1}{z-w}+a\frac{1}{z-z'}=-\frac12\frac{1}{z-\bar w}+a\frac{1}{z-\bar z'}. \end{align} Since in general $w\neq\bar w$ and since the choice $a=+\frac12, \ z'=w$ leads to the trivial solution $G=0$, the only interesting solution is $a=-\frac12,\ z'=\bar w.$ This means that \begin{align} G(z,w)=-\frac12 \left({\rm ln}|z-w|^2+ {\rm ln}|z-\bar w|^2\right), \end{align} which is the correct solution (see formula 4.149.)

How do I proceed for mixed boundary conditions?

This time I want to impose Neumann for $\sqrt z=\sqrt{\bar z}$ (positive real line) and Dirichlet for $\sqrt z=-\sqrt{\bar z}$ (negative real line).
However, the ansatz $(1)$ does not have square roots or fractions in the argument of the logarithm, which instead appear in the final solution.

Answer 1

I do not have a copy of Blumenhagen, so I am making some assumptions about what you want to do. However, I think I can answer at least some of your questions.

The square roots occur because you need to use a conformal mapping from $z$ to $u = \sqrt{z}$ in order to use the image charge method. You can see that you need to have square roots by separating variables in Laplace's equation. The separated solutions are $z^{\pm\alpha}$ that is $r^{\pm\alpha}$ multiplying $\cos(\alpha\phi)$ and/or $\sin(\alpha\phi)$. To match your boundary conditions that the $\phi$ derivative is zero at $\phi=0$ and the function is zero at $\phi=\pi$, you need the cosine solution with $\alpha = (2n+1)/2$ with $n$ integer. This is the real part of $z^{\pm (2n+1)/2}$, so you would expect an expansion in the square root of $z$.

To use the image charge method, you can realize that if you want the solution in the upper half plane for the source at position $w$, then $u=\sqrt{z}$ with the branch cut chosen on the negative real axis so that with $z=re^{i\phi}$, $u=\sqrt{r}e^{i\phi/2}$, and $-\pi < \phi < \pi$, will map the upper half plane to the upper right quadrant. You can now use the image charge method. You have the mapped source at $\sqrt{w}$, so the particular solution is ${\rm Re}~ q\ln(u-\sqrt{w})$, the three image charges will be $q$ at $\sqrt{\bar w}$, and $-q$ at both $-\sqrt{w}$, and $-\sqrt{\bar w}$. Change $u$ to $\sqrt{z}$ and fix the magnitude of $q$ to match your source to give your desired solution.