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Can a two variable Harmonic function f(x,y) be zero on a curve with a cusp?

flag	asked Jun 16 2011 at 11:16 mosen 175●4

Michael Renardy · Answer 1 · 2011-06-16 12:35:41Z UTC

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No. A two variable harmonic function is the real part of an analytic function. Near a zero, an analytic function behaves like a power of $z-z_0$.

answered Jun 16 2011 at 12:35

Michael Renardy
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It is the real part of an anlytic fuction, but there is no reason that it behave like that. May be the real part of an analytic function is zero "on a curve" but its imaginary not. Further, if you consider the function xy as a two variable harmonicreal function, it is zero on the two axis and at the origin but not always zero. – mosen Jun 16 2011 at 13:09

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Let u be your harmonic function, and suppose u(0,0)=0. Let v be the harmonic conjugate, and w.l.o.g. let v(0,0)=0. Then u+iv is an analytic function, which has a Taylor series $$u+iv=a_n z^n +O(z^{n+1}).$$ We can write this in the form $w(z)^n$, where $w$ is an analytic function of $z$ that is locally invertible. Now $u=0$ is equivalent to $Re(w^n)=0$. In the $w$ plane, this is given by $n$ straight lines intersecting at the origin. These lines map to smooth curves in the $z$ plane. – Michael Renardy Jun 16 2011 at 14:53

No, the lines might be mapped into parts of the curve with cusp and not cover it. If you are sure that my question has an answer NO, could you please explain it more? I really need it. – mosen Jun 20 2011 at 13:19

Gerald Edgar · Answer 2 · 2011-06-16 13:57:24Z UTC

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Perhaps study this one: $$ f(x,y) = \frac{2 \sqrt{x^{2} + y^{2}} - \sqrt{2 \sqrt{x^{2} + y^{2}} + 2 x}}{2 \sqrt{x^{2} + y^{2}} - 2 \sqrt{2 \sqrt{x^{2} + y^{2}} + 2 x} + 2} $$ which vanishes on a cardioid with a cusp at the origin. Harmonic except possibly at the one point $(0,0)$.

answered Jun 16 2011 at 13:57

Gerald Edgar
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This function has a branch cut on the negative x axis. – Michael Renardy Jun 16 2011 at 14:48

Zero sets of harmonic fucntions

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