Can a two variable Harmonic function f(x,y) be zero on a curve with a cusp?
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3 Answers
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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
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1$\begingroup$ 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. $\endgroup$– mosenCommented Jun 16, 2011 at 13:09
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2$\begingroup$ 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. $\endgroup$ Commented Jun 16, 2011 at 14:53
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$\begingroup$ 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. $\endgroup$– mosenCommented Jun 20, 2011 at 13:19
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There exists an honor thesis dedicated to exactly this problem, and its generalizations: Problems in Harmonic Function Theory Ronald A. Walker, see http://scholarship.richmond.edu/cgi/viewcontent.cgi?article=1488&context=honors-theses
answered Feb 9, 2017 at 13:02
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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
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$\begingroup$ This function has a branch cut on the negative x axis. $\endgroup$ Commented Jun 16, 2011 at 14:48