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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?

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2  
$u(x,y)=x\,y$ is harmonic and vanishes on the lines $x=0$ and $y=0$. –  Julián Aguirre Feb 9 '13 at 16:54
1  
Your example for parallel lines does not work. Note that by reflection, we must have an infinite set of equally spaced parallel lines on which the function is zero. Think of periodic functions. –  user53153 Feb 9 '13 at 20:19
    
@5PM I corrected that. $e^x \sin y$ would do. Thank you. This helped me a lot! –  PeterM Feb 9 '13 at 21:49

1 Answer 1

up vote 9 down vote accepted

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 to reflections about other lines.

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 a dense set of lines on which the function vanishes. Therefore, it vanishes identically.

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 $z^n$.

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1  
Wow, you posted this while I was editing my question with a similar argument. :) –  PeterM Feb 9 '13 at 19:45
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A example could be: Take any divisor of $360$, lets say $n$. Consider the functions $f_k(x,y)=y-\tan(k\frac{2\pi}{n})x$ for $\big(k=1,...,\frac{n}{2}$ and $k\neq\frac{n}{4}\big)$ and $f_{\frac{n}{4}}(x,y)=x$. Then the function $f(x,y)=f_1(x,y)...f_k(x,y)$ is harmonic and identically zero on the lines. –  Tomás Feb 9 '13 at 20:03

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