1.3 Fundamentals of the Theory of Field
35
Example 1.3.11 In a rectangular coordinate system, let a = a x i + a y j + a z k, verify
a = i∇
2 a x + j ∇
2 a y + k∇
2 a z .
Proof From the definition of vector rotation, there is
∇ × a =
∂a z
∂ y
−
∂a y
∂z
i +
∂a x
∂z
−
∂a z
∂ x
j +
∂a y
∂ x
−
∂a x
∂ y
k
(1)
So there is
∇ × ∇ × a =
∂
∂ y
∂ay
∂ x
−
∂ax
∂ y
−
∂
∂z
∂ax
∂z
−
∂az
∂ x
i +
∂
∂z
∂az
∂ y
−
∂ay
∂z
−
∂
∂ x
∂ay
∂ x
−
∂ax
∂ y
j
+
∂
∂ x
∂ax
∂z
−
∂az
∂ x
−
∂
∂ y
∂az
∂ y
−
∂ay
∂z
k
(2)
The expansion in the first pair of square brackets of Eq. (2) is
∂
2 a x
∂ x 2 +
∂
2 a y
∂ y∂ x
+
∂
2 a z
∂z∂ x
−
∂
2 a x
∂ x 2 +
∂
2 a x
∂ y 2 +
∂
2 a x
∂z 2
=
∂
∂ x
∇ · a − a x (3)
Similarly, the expansions in the rear two pair of square brackets of Eq. (2) are
respectively
∂
2 a x
∂ x∂ y
+
∂
2 a y
∂ y 2 +
∂
2 a z
∂z∂ y
−
∂
2 a y
∂ x 2 +
∂
2 a y
∂ y 2 +
∂
2 a y
∂z 2
=
∂
∂ y
∇ · a − a y (4)
∂
2 a x
∂ x∂z
+
∂
2 a y
∂ y∂z
+
∂
2 a z
∂z 2
−
∂
2 a z
∂ x 2 +
∂
2 a z
∂ y 2 +
∂
2 a z
∂z 2
=
∂
∂z
∇ · a − a z
(5)
Substituting Eqs. (3)–(5) into Eq. (2), we obtain
∇ × ∇ × a =
i
∂
∂ x
∇ · a + j
∂
∂ y
∇ · a + k
∂
∂z
∇ · a
− (ia x + j a y + ka k )
= ∇(∇ · a) − a
(6)
In a rectangular coordinate system, there is
a = ia x + j a y + ka k
(7)
Quod erat demonstrandum.
Example 1.3.12 Given the vector a = a x i + a y j + a z k, verify
(∇ × a) · (∇ × a) =
∂a z
∂ y
−
∂a y
∂z
2
+
∂a x
∂z
−
∂a z
∂ x
2
+
∂a y
∂ x
−
∂a x
∂ y
2
= |∇ × a|
2
= (∇ × a)
2
(1.3.104)