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Fall 2015 - Problem 12

conformal mappings, harmonic functions

Find a function u(x,y)u\p{x, y} harmonic in the region between the circles ∣z∣=2\abs{z} = 2 and ∣z−1∣=1\abs{z - 1} = 1 which equals 11 on the outer circle and 00 on the inner circle (except at the point where the circles are tangent to each other).
Solution.

Our strategy will be to apply a conformal mapping to simplify our domain, and then undo the change of variables.

The main issue is that circles are hard to work with, so we wish to apply a Möbius transformation transform them into lines. Hence, a Möbius transform which sends 2↦∞2 \mapsto \infty suffices, and φ(z)=1z−2\phi\p{z} = \frac{1}{z - 2} works.

φ\phi sends the circle ∣z∣=2\abs{z} = 2 to the line containing

φ(−2)=−14andφ(2i)=−14−i4,\phi\p{-2} = -\frac{1}{4} \quad\text{and}\quad \phi\p{2i} = -\frac{1}{4} - \frac{i}{4},

i.e., the line Re⁡z=−14\Re{z} = -\frac{1}{4}, and it sends the circle ∣z−1∣=1\abs{z - 1} = 1 to the line containing

φ(0)=−12andφ(1+i)=−12−i2,\phi\p{0} = -\frac{1}{2} \quad\text{and}\quad \phi\p{1 + i} = -\frac{1}{2} - \frac{i}{2},

the line Re⁡z=−12\Re{z} = -\frac{1}{2}. Thus, we need a harmonic function which is 00 when Re⁡z=−12\Re{z} = -\frac{1}{2} and 11 when Re⁡z=−14\Re{z} = -\frac{1}{4}. For example, v(x,y)=4(x+12)=4x+2v\p{x, y} = 4\p{x + \frac{1}{2}} = 4x + 2 works, which gives

u(x,y)=v(φ(x+iy))=4Re⁡(1z−2)+2.u\p{x, y} = v\p{\phi\p{x + iy}} = 4\Re\p{\frac{1}{z - 2}} + 2.