Find a harmonic function on | z | &lt; 1 with value 1 on an arc of | z | = 1 and zero on the rest.Of course, if allow the arc with value 1 to be the whole | z | = 1 circle it would be trivial.Without that, how to construct such a function, I'm a bit lost. I suppose if I can find a holomorphic function f ( z ) whose real part full fill that it shall work, but playing around with the normal functions such as e x , sin ⁡ x etc, I can't find one with such property.

Question

Find a harmonic function on       |    z      |    &amp;lt;  1 with value 1 on an arc of       |    z      |    =  1 and zero on the rest.Of course, if allow the arc with value 1 to be the whole       |    z      |    =  1 circle it would be trivial.Without that, how to construct such a function, I&#039;m a bit lost. I suppose if I can find a holomorphic function   f  (  z  ) whose real part full fill that it shall work, but playing around with the normal functions such as       e    x  ,   sin  ⁡  x etc, I can&#039;t find one with such property.

Helena Howard · Accepted Answer

Let   0  ≤  α  ≤  2  π the length of the arc in question (where the function needs to go to 1) and for each       |    z      |    &amp;lt;  1 let   θ  (  z  ) the angle subtended by the arc at   z.Clearly,   θ is locally a difference of argument functions in   z (as the angle at   z with arc endpoints   a,  b counterclockwise is roughly   arg  ⁡  (  b  −  z  )  −  arg  ⁡  (  a  −  z  ) with appropriate choices which can be done the same near   z) so it is harmonic and by elementary geometry the function  u  (  z  )  =            2      θ      (      z      )      −      α              2      π       is harmonic,   0  ≤  u  ≤  1 and   u  →  1 on the arc (subtended angle tends to             2      π      +      α        2   as it is the exterior of the inscribed angle) and to 0 on the open arc exterior (where subtended angle tends to   α      /    2 of course).It is a consequence of Lindelof maximum principle for bounded harmonic functions that   u is the unique bounded real harmonic function satisfying the above(more generally if   u  ,  v are real bounded harmonic functions in a domain with boundary not a finite set and   u  −  v is 0 on the boundary except at finitely many points, then   u  =  v.Here one notes that   ℜ            z      +      1              1      −      z       is an unbounded harmonic function on the unit disc that tends to 0 on the boundary except at 1 so one can always add suitably modified multiples of such to get infinitely many solutions to the problem here that are however not bounded, so uniqueness indeed depends on the boundness condition)

Elsa Brewer · Accepted Answer

So we want a complex power series, with radius of convergence 1, and such that on the unit circle it is sometimes 0 and sometimes 1. There is really only one way to construct this, because the power series coefficients need to be precisely the Fourier coefficients of your function along the boundary. That is, take any interval   I  ⊂  [  0  ,  2  π  ] and take       a    n   to be the Fourier coefficients of the indicator function       χ    I   (extended to a   2  π-period function). Then      ∑          n      ∈              Z                  a    n        z    n  does what you want along the boundary. To obtain something convergent at 0 while keeping the function the same along the boundary, take the harmonic function      a    0    +      ∑          n      ≥      1        (      a    n        z    n    +      a          −      n                  z      n        ¯    )    .This power series has radius of convergence 1 because       a    n    =  O  (  1  ) (even   O  (  1      /    n  )). If you want a real-valued one, just take the real part of this function.Explicitly, if we choose   I  =  [  0  ,  π  ] we get      a    n    =      1          2      π            ∫          0              π            e          −      i      n      t        d  t  =      {                                        1            2                                    if                    n          =          0                                                                1              −                              e                                  −                  n                  π                  i                                                                    2              π              i              n                                                if                    n          ≠          0                    .                        Caveat: the function constructed in this way will be       1    2   at the endpoints of the arc.The functions constructed in this way can probably be made very explicit in terms of   log  ⁡  (  1  ±  z  ) and   log  ⁡  (  1  ±      z    ¯    ). The 0 vs 1 behavior should correspond to the different branch cuts.

Find a harmonic function on |z|<1 with value 1 on an arc of |z|=1 and zero on the rest.

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