# Harmonic functions with zeros on two lines

For which pairs of lines $L_1$, $L_2$ do there exist real functions, harmonic in the whole plane, that are $0$ at all points of $L_1 \cup L_2$ without vanishing identically?

Note: This is self-study — not homework.

My thoughts:

I tried to exploit certain configurations of $L_1$ and $L_2$ and apply the reflection principle to show that the function vanishes on a closed curve and hence everywhere, but I wasn’t successful.

OK, I think I have a working argument. Let $u$ be a real harmonic function that vanishes on two lines $L_1$ and $L_2$. Apply the necessary translation and/or rotation to make $L_1$ match $y = 0$ and the intersection point (if any) match (0, 0). This will translate/rotate the set of zeros of $u$, but it won’t remove or introduce new zeros. The resulting function will remain real and harmonic.

If $L_1$ and $L_2$ are parallel, then $L_2$ is of the form $y = a$ and, by a direct computation of the Laplacian, the following function satisfies the requirements:

$$u(x, y) = e^x \sin\left(\frac{2\pi y}{a}\right)$$

If $L_1$ and $L_2$ intersect at an angle $\theta = 2\pi\frac{p}{q}$ where $p, q \in \Bbb N, q \ne 0$, the following function satisfies the requirements:

$$u(x, y) = \prod_{k=0}^{q-1}\left(y-x \tan\left(2\pi k \frac{p}{q}\right)\right)$$

Otherwise, $L_1$ and $L_2$ intersect at an angle $\theta = 2 \pi s$ where $s \not \in \Bbb Q$. By the Schwarz reflection principle, the function also vanishes on $\overline L_2$. With repeated rotations and applications of the reflection principle, we find that a rotation of the function has zeros in the following set:

$$y = x \tan(2 \pi s n), \ n \in \Bbb N$$

But the set is dense because $s \not \in \Bbb Q$. Hence the function is identically zero.

My question: Any mistakes in my argument? Is there an easier way?

#### Solutions Collecting From Web of "Harmonic functions with zeros on two lines"

The idea to use the reflection principle is good. When a harmonic function vanishes on a line (say, the real axis),
redefining it in the lower half-plane as $\tilde (x,y)=-u(x,-y)$ also yields a harmonic function. Since $u=\tilde u$
in the upper half-plane, we conclude that $u\equiv \tilde u$, that is, $u(x,y)\equiv -u(x,-y)$. The same applies
As a corollary: if a harmonic function vanishes on two lines $L_1$ and $L_2$, then it also vanished on the line
$L_3$ that is obtained by reflecting $L_1$ about $L_2$. The process of reflection can be repeated.
If the angle between $L_1$ and $L_2$ is not a rational multiple of $\pi$, then repeated reflections produce
If the angle between $L_1$ and $L_2$ is a rational multiple of $\pi$, say $m\pi/n$, then such harmonic function exists, and can be obtained from
the holomorphic function $z^n$.