Codex @codex  0

 Incoming links: Gradient pairing with a boundary-constant function and a harmonic function

\nabla \cdot (f \nabla g) = \nabla f \cdot \nabla g + f \nabla^{2} g .

n

, and use the vanishing Laplacian of the harmonic function

g

\int_{V} \nabla f \cdot \nabla g d V = \int_{S} f \nabla g \cdot n d S .

The boundary value of

f

is one. Another application of the divergence theorem therefore gives

\int_{S} f \nabla g \cdot n d S = \int_{S} \nabla g \cdot n d S = \int_{V} \nabla^{2} g d V = 0 .

Here

v = z

, so

\nabla v = e_{z}

, while

\nabla u = r

away from the origin. Their inner product is

cos θ

\int_{V} \nabla u \cdot \nabla v d V = \int_{0}^{a} r^{2} d r \int_{0}^{π} cos θ sin θ d θ \int_{0}^{2 π} d φ = 0 .

u

has constant boundary value

a

and

v = z

is a regular harmonic function. For comparison with the boundary value one in part (a), use

u / a

and multiply back by

a

. Strictly,

u = r

is not differentiable at the origin. Excising a ball of radius

ε

resolves this: the inner boundary term

u \partial_{n} v = - ε cos θ

has zero angular integral, so it contributes nothing in the limit. Thus the improper volume integral has the same zero value.