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

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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 $z-z_0$.

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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 '11 at 13:09
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 '11 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 '11 at 13:19

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 $(0,0)$.

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This function has a branch cut on the negative x axis. – Michael Renardy Jun 16 '11 at 14:48

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