Let Ω={z∈C∣∣z∣>1 and Rez>−2}. Suppose u:Ω→R is bounded, continuous, and harmonic on Ω and also that u(z)=1 when ∣z∣=1 and that u(z)=0 when Rez=−2. Determine u(2).
Solution.
First, consider the conformal map z↦z1. This fixes the unit circle, and it maps the line Rez=−2 to the line containing
−21,−2+i1=5−2−i,and∞1=0.
This is the circle centered at z=−41 and with radius 41. Thus, we have an annular region, but the circles are not concentric. To fix this, we will apply a Blaschke factor, which is a Möbius transformation fixing the unit disk of the form
φ(z)=1−αzz−α.
By symmetry, we need to map −21↦−r and 0↦r, where r>0 will be the radius of the resulting inner circle. We also expect α∈R, so we get the system
φ(−21)=1+21α−21−α=2+α−1−2α=−r,φ(0)=−α=r.
Substituting, we get
−1−2α=α(2+α)⟹4α+α2=−1⟹α∈{−2−3,−2+3}.
Since α∈D, we see that α=−2+3 is the correct choice, and so φ maps ∣∣z+41∣∣=41 to the circle of radius 2−3 centered at the origin. From here, we see
v(z)=log1/rlog∣z∣/r
satisfies f(z)=0 on the inner circle ∣z∣=r and that f(z)=1 on the unit circle. Thus, v(z)=u(φ−1(z)1) on ∣z∣=r and ∣z∣=1, so because these are both harmonic functions, we see that