The ``mean value theorem'' for solutions to the Laplace equation
Consider an electrostatic field F(r) in a
charge-free region so that it satisfies the Laplace equation:
Ñ2 F(r) = 0.
(1)
The ``mean value theorem'' value theorem states that the value of
F(r) at the arbitrary (charge-free) point
r is equal to the average of
F(r¢) over the surface of any sphere
centered on the point r (see Jackson problem #1.10).
One way to prove this theorem is the following. Consider a point
r¢ = r + u, where u will
describe a sphere of radius R about the fixed point
r. We can make a Taylor series expansion of the
electrostatic potential F(r¢) about the
fixed point r:
F(r + u ) = F(r) + u·Ñ F(r) +
1 2!
(u·Ñ)2 F(r) +
1 3!
(u·Ñ)3 F(r) +
1 4!
(u·Ñ)4 F(r) + ¼.
(2)
According to the premise of the theorem, we want to integrate both
sides of the equation 2 over a sphere of radius R in
the variable u:
ó õ
sphere
dSu = R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu).
(3)
We note that
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) 1 = 4 pR2,
(4)
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) u ·Ñ = 0,
(5)
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) (u ·Ñ)2 =
4 pR4 3
Ñ2,
(6)
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) (u ·Ñ)3 = 0,
(7)
and
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) (u ·Ñ)4 =
4 pR4 5
Ñ4.
(8)
Since Ñ2 F(r) = 0, the only non-zero term of
the average it thus the first term:
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) F(r + u) = 4 pR2 F(r),
(9)
or
F(r) =
1 4 pR2
R2
ó õ
2p
0
dfu
ó õ
+1
-1
dcos(qu) F(r + u).
(10)
Since this result is independent of the radius R, we see that
we have proven the theorem.
File translated from TEX by TTH, version 2.20. On 22 Jan 2001, 08:56.