The Geomagnetic Field - Quantitative Plate Tectonics - page 125

Proof Let us consider the vector field A ( r ) D
¥( r ) r § and its flux through S ( R ). We remind that
the quantity r § d S is the directional derivative of
§ along the normal to S (see Appendix1 ) , which
is usually indicated by @§/@ n .
Then, by the divergence theorem we have:

domain where its harmonic potential is defined.
Alternatively, we can say that the flux of a conser-
vative vector field F through the closed boundary
S ( R ), where its harmonic potential is defined, is
zero.
Applying the divergence theorem to ( 4.63 )
yields the following alternative form of Gauss'
law:

Z Z

r ¥ r § C ¥ r

2 § dV

r .¥ r §/dV D

Z

R R

I I r F dV D 0 (4.64)

¥.r/ @ § ¥ r § dS D

@n dS

R

S. R / S. R /

Now let us assume that § is harmonic in R
and that § D 0on S ( R ). Setting ¥ D § in ( 4.61 )
yields:

Z

r ¥ r § C ¥ r

2 § dV

D

R

Z

The Green first identity ( 4.61 ) immediately fol-
lows. This completes the proof.

. r § / 2 dV D 0

R

A first interesting consequence of the Green
first identity follows when § is harmonic, so that
r

Therefore, r § D 0in R , thereby, § D const .
The hypothesis that § D 0on S ( R ) and the con-
tinuity of § then imply that § D 0in R .This
proves the important property that if § is har-
monic in R and § D 0on S ( R ), then § vanishes
in the whole region R . The following unique-
ness theorem is another corollary of Green's first
identity.

2 § D 0in R . In this instance, setting ¥ D 1in
( 4.61 )gives:

I

@§
@n dS D 0

(4.62)

S. R /

This equation states that the normal derivative
of a harmonic function averages to zero on the
frontier of the region where it is harmonic. It
also can be shown that if ( 4.62 ) holds, then § is
harmonic in R . Therefore, ( 4.62 ) is a necessary
and sufficient condition for § to be harmonic in
R . Now let us assume that F is the potential field
associated with §,sothat: F Dr §. Then, if n is
a versor normal to S ( R )wehave:

Corollary 4 (Stokes' Theorem)
A harmonic function is uniquely determined by its
boundary values .

Proof Let ¥ D ¥( r )and§ D §( r )betwo
harmonic functions in a closed region R .Let
us also assume that ¥ D § on the boundary
S ( R ). Clearly, ¥ § is harmonic in R and it
results ¥ § D 0on S ( R ). Therefore, according
to the previous corollary we must have ¥ § D 0
in R ,thatis,¥ D §. This proves that if two
functions coincide on the boundary of R then
they also coincide in R . Therefore, a harmonic
function is uniquely determined by the boundary
values.

@§
@n Dr § n D F n

Therefore, Eq. ( 4.62 ) can be rewritten as fol-
lows:

I

F dS D 0 (4.63)

S. R /

Stokes' theorem implies that the Dirichlet
boundary - value problem , that is, find a function
that solves Laplace's equation in the interior
of R given the values on the boundary of this

This equation is often indicated as Gauss ' law .
It says that the normal component of a potential
field averages to zero over the boundary of the