 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ib/paper-3/4d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

w = i \frac{1 - z}{1 + z}

maps the unit disc conformally onto the upper half-plane. On the upper semicircle of its boundary,

w

approaches the positive real axis; on the lower semicircle it approaches the negative real axis. If

0 < ar g w < π

ϕ (w) = ϕ_{0} (1 - \frac{2 a r g w}{π})

has the required boundary limits.

For

z = x + i y

w = \frac{2 y + i ( 1 - x ^{2} - y ^{2} )}{( 1 + x ) ^{2} + y ^{2}} .

Since the imaginary part is positive inside the disc,

\frac{π}{2} - ar g w = arctan \frac{Re w}{Im w} = arctan \frac{2 y}{1 - x ^{2} - y ^{2}} .

Therefore the solution of the Dirichlet problem is

ϕ (x, y) = \frac{2 ϕ _{0}}{π} arctan (\frac{2 y}{1 - x ^{2} - y ^{2}}), x^{2} + y^{2} < 1.

The exceptional boundary endpoints are precisely the jump points of the prescribed data.

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