 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-2/3/c/solution

Fix the observation point

x

and apply Green's second identity in the source variable

y

. Since

Δ u = 0

Δ_{y} G_{D} = δ_{x}

and

G_{D} = 0

on the boundary, it gives

u (x) = \int_{D} (u Δ G_{D} - G_{D} Δ u) d^{2} y = \int_{\partial D} f (y) \partial_{n_{y}} G_{D} (x, y) d s_{y} .

The identity can be justified by deleting a small circle about

x

and letting its radius tend to zero; the logarithmic flux contributes

u (x)

x = (r, ϕ)

and

y = (ρ, ϕ^{'})

, and write

c = cos (ϕ - ϕ^{'})

. The symmetric expression for the Green's function is

G_{D} = \frac{1}{4 π} lo g \frac{r ^{2} + ρ ^{2} - 2 r ρ c}{1 + r ^{2} ρ ^{2} - 2 r ρ c} .

At the boundary, the outward normal derivative is

\partial_{ρ}

ρ = 1

, and hence

\partial_{ρ} G_{D} ∣_{ρ = 1} = \frac{1}{4 π} \frac{( 2 - 2 rc ) - ( 2 r ^{2} - 2 rc )}{1 + r ^{2} - 2 rc} = \frac{1}{2 π} \frac{1 - r ^{2}}{1 - 2 rc + r ^{2}} .

Since

d s = d ϕ^{'}

P (r, ϑ) = \frac{1 - r ^{2}}{1 - 2 r cos ϑ + r ^{2}}, u (r, ϕ) = \frac{1}{2 π} \int_{0}^{2 π} f (ϕ^{'}) P (r, ϕ - ϕ^{'}) d ϕ^{'} .

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