Timeline for Zero sets of harmonic fucntions
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2011 at 13:19 | comment | added | mosen | 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. | |
| Jun 16, 2011 at 14:53 | comment | added | Michael Renardy | 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. | |
| Jun 16, 2011 at 13:09 | comment | added | mosen | 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. | |
| Jun 16, 2011 at 12:35 | history | answered | Michael Renardy | CC BY-SA 3.0 |