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.


1 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.

  • $\begingroup$ Thank you so much for your answer, you have been very helpful! I followed your suggestion for the image charge method and I ended up with the correct result. I also think I understood your argument for looking for a solution in terms of square roots. If G=ln z^alpha=ln (r^alpha cos(alphaphi)) has to be zero at phi=pi, then alpha must be (2n+1)/2. Why do I need a conformal map to the upper right quadrant to apply the image charge method? Why can't I do this computation directly on the upper half plane? $\endgroup$
    – samario28
    Mar 21, 2021 at 15:35
  • $\begingroup$ You can write any solution to Laplace's equation in terms of a charge and/or dipole density on a boundary outside the region of interest. So to that extent you can always use charges outside the region of interest to set the boundary conditions. The problem is with a small finite number of image charges only certain boundary conditions can be enforced - typically when there is a symmetry. You can do the calculation directly in the upper half plane, just not with a small finite number of image charges. The expansion in separated solutions would work, and could be summed to get the closed form. $\endgroup$
    – user200143
    Mar 22, 2021 at 19:03
  • $\begingroup$ Are you saying that by summing over a possibly infinite number of q_i ln |z-w_i|^2 (without square roots this time, since we are on the UHP), one should be able to impose the boundary conditions and to recover the correct solution? Do you have a reference for that? $\endgroup$
    – samario28
    Mar 23, 2021 at 13:04
  • $\begingroup$ J.D. Jackson, Classical Electrodynamics, third edition, section 1.10 has a discussion of how to calculate the Green's function in three dimensions, and how the boundary conditions can be set by a charge distribution outside the region. It also applies to two dimensions. $\endgroup$
    – user200143
    Mar 23, 2021 at 18:15

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