I am trying to find a non-constant harmonic function $\phi$ on $Q = \{z \mid \mathfrak{R}(z), \mathfrak{I}(z) > 0\}$, (i.e. the first quadrant) such that $\phi$ extends continuously to $\overline{Q} \setminus \{0\}$ with constant values on each the components of $\partial Q – \{0\}$ ($\phi$ need not be continuous at the origin). My method for doing this is to solve Laplace's equation in the strip $S = \{0 < \mathfrak{I}(z) < 1\}$ subject to appropriate boundary conditions, then carrying the solution forward under the conformal map $f(z) = e^{\pi z/2}$ which is a conformal equivalence between $S$ and $Q$ (I'm doing it this way because in an earlier part of the problem they ask us to find $f$).
Solving $\Delta u = 0$ on $S$ in general seems straightforward enough. Assume $u(x,y) = \alpha(x)\beta(y)$ then we get $\alpha''(x) = \lambda \alpha(x)$ and $\beta''(y) = -\lambda \beta(y)$ by simple algebra. This gives $\alpha(x) = Ae^{\sqrt \lambda x} + Be^{-\sqrt \lambda x}$ and $\beta(y) = Ce^{-\sqrt \lambda y} + De^{-i\sqrt \lambda y}$. Then we allow $u$ in general to be any linear combination of these solutions. My issue here is that I'm not sure how to carry the boundary conditions in $Q$ to boundary conditions on $S$. Intuitively it seems that it should just that $u$ must be constant on $\mathfrak{I}(z) = 0$ and $\mathfrak{I}(z) = 1$, but this forces $\alpha(x) = 0$ which gives $u = 0$, clearly not the desired result.My question is, how can I carry the boundary conditions over carefully (in general or in this case – I can probably figure out the general case if someone can work out this example)?
Best Answer
Your approach is too complicated. The principal value of the argument $\arg(z)$ is harmonic in the first quadrant as the imaginary part of the complex logarithm, with constant values on both the positive real axis and positive imaginary axis.
Therefore an affine transform $$ \phi (z) = a \arg(z) + b $$ is the solution to your problem.
(The corresponding harmonic functions on the strip $S$ are of the form $u(z)= a' \operatorname{Im}(z) + b'$. )