Mathematical Physics Vol 1

4.6 Examples

131

∂ 2 ( x 2 + y 2 + z 2 ) − 1 2 ∂ z 2

2 z 2 − x 2 − y 2 ( x 2 + y 2 + z 2 )

(4.155)

=

.

5 2

By adding equations (4.153), (4.154) and (4.155) we now obtain ∂ 2 ∂ x 2 + ∂ 2 ∂ y 2 + ∂ 2 ∂ z 2 1 p x 2 + y 2 + z 2 ! = 0 . Recall that the equation ∇ 2 φ = 0 is called the Laplace equation, and its solution is called the harmonic function. Therefore, the solution of equation (4.156) φ = 1 / r is a harmonic function 18 . The problem in this Example can be solved in yet another way. Given that ∇ 2 1 r = ∇ · ∇ 1 r , and using the result from Example 33, on p. 119, we obtain ∇ 1 r = − r r 3 . According to Example 53, on p. 131, it follows that ∇ · r r 3 = 0 , which proves that ϕ = 1 / r is a harmonic function. (4.156)

Problem53 Prove that ∇ ·

r r 3

= 0.

Proof If we introduce the notation φ = r − 3 and A = r the problem is reduced to the application of the previously proved property of divergence, where Example 34 on p.120 is used, for n = − 3,: ∇ · ( r − 3 r )= ∇ r − 3 · r +( r − 3 ∇ · r ) = − 3 r − 5 r · r + 3 r − 3 = 0 .

Problem54 Prove that ∇ · ( U ∇ V − V ∇ U )= U ∇ 2 V − V ∇ 2 U .

18 Note that φ = 1 / r is a harmonic function in space E

3 , whereas y = ln r is a harmonic function in space E 2 .