Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics V resit / Question 2 / a / iii

On the unit sphere, the outward unit normal vector is

n = r

and

u = r

. Hence the outward flux integral is

\iint_{r = 1} u \cdot n d S = \iint_{r = 1} 1 d S = 4 π .

(1)

Using spherical polar coordinates and the divergence from ii, the volume integral over the unit ball is

∭_{r \leq 1} \nabla \cdot u d V = 4 π (3 - α) \int_{0}^{1} r^{2 - α} d r = 4 π (3 - α) [\frac{r ^{3 - α}}{3 - α}]_{0}^{1} = 4 π (α < 3) .

(2)

This improper integral converges precisely because

2 - α > - 1

. When the vector field is singular or fails to be continuously differentiable at the origin, justify the equality by applying the divergence theorem on

ε < r < 1

. The outward flux through its inner sphere is

- 4 π ε^{3 - α}

, so

∭_{ε < r < 1} \nabla \cdot u d V = 4 π (1 - ε^{3 - α}) ⟶ 4 π .

(3)

Thus the two integrals agree for every $α < 3$ , with the limiting interpretation where needed; the classical smoothness hypothesis does not hold for every such