1

I want to prove the following theorem:

If $u$ & $v$ are harmonic functions in unit disc $D$ and $uv$ is identically equal to $0$ in $D$, prove that either $u$ is identically zero or $v$ is identically zero in $D$.

I know a real valued function $H$ of two real variables $x$ and $y$ is said to be harmonic in a given domain of $xy$-plane if throughout that domain, it has continuous partial derivatives of the first and second order and satisfies the partial differential equation $H_{xx}(x,y)+H_{yy}(x,y)=0$.

The question is related to complex functions $u$ and $v$ which are harmonic and the definition above is for real valued function $H$. How should I prove this?

The region where functions are harmonic is a unit disc. So this question is different and not duplicate.

Vlad
  • 6,710
Kavita
  • 728
  • @Chappers isolated zeros ? $ Im(z)$ is harmonic – reuns Mar 22 '17 at 16:45
  • @user1952009 Closed zero-sets sounds better, with that in mind... – Chappers Mar 22 '17 at 16:46
  • 2
    Do you know the max/min principle for harmonic functions, or that a real harmonic function in $D$ is the real part of a holomorphic function there? – zhw. Mar 22 '17 at 16:47
  • 2
    @Chappers Well if $f(z)$ is real analytic then $f(x+iy) = \sum_{n=0}^\infty \sum_{m=0}^\infty c_{n,m} x^n y^m$ for $|x+iy|$ small enough, where $n! m! c_{n,m} = \frac{\partial f}{\partial x^n \partial y^m}$. But it is non-trivial to prove that harmonic $\implies$ real-analytic – reuns Mar 22 '17 at 16:47
  • @Chappers i don't understand your comment. – zhw. Mar 22 '17 at 16:53
  • @zhw sorry I don't know that max / min principle for harmonic functions. – Kavita Mar 22 '17 at 16:55
  • As I am bit new to this subject I am unaware of some results. – Kavita Mar 22 '17 at 16:58
  • 1
    what about being the real part of a holomorphic function? (you have the complex analysis tag for some reason). – zhw. Mar 22 '17 at 16:59
  • @zhw I am not getting any hint. How should I relate holomorphic function to harmonic function? Are you using min/max principle for that? – Kavita Mar 22 '17 at 17:04
  • 1
    How did you encounter this problem? what do you know about harmonic functions, what results have you convered? you're asking us a question in a vacuum at the moment – zhw. Mar 22 '17 at 18:18
  • 1
    I don't know about harmonic functions, but if $uv = 0$, then $u$ or $v$ have to be zero because the only way to get 0 by multiplying is to multiply by 0. – Travis Mar 23 '17 at 02:38

1 Answers1

1

Local maximum property for real harmonic functions: If $u$ is real and harmonic in a connected open set $U,$ and if $u$ has a local maximum at some $a\in U,$ then $u$ is constant in $U.$

Now in this problem we are given two complex valued harmonic functions $u,v$ in $D$ such that $uv=0$ in $D.$ Suppose $u\not \equiv 0$ in $D.$ Then there is $a\in D$ such that $u(a)\ne 0.$ By continuity, $u\ne 0$ in some $D(a,r)\subset D.$ It follows that $v\equiv0$ in $D(a,r).$ Writing $v= v_1 + iv_2,$ both $v_1,v_2$ are real and harmonic in $D$ and both $v_1,v_2$ are $\equiv 0$ in $D(a,r).$ But then both $v_1,v_2$ have a local maximum at $a.$ By the local maximum property, both $v_1,v_2$ are constant in $D.$ Since $v_1(a)=0=v_2(a),$ $v_1,v_2 \equiv 0$ in $D.$ Thus $v\equiv 0$ in $D.$

zhw.
  • 105,693