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1 Preliminaries
Example 1.3.8 Verify ∇(a · b) = (b · ∇)a + (a · ∇)b + b × ∇ × a + a × ∇ × b.
Proof It can be known from the double vector product formula that
a × (c × b) = c(b · a) − b(a · c) = c(a · b) − (a · c)b
b × (c × a) = c(a · b) − a(b · c) = c(a · b) − b(c · a)
Taking ∇ as c, there is
∇(a · b) = ∇(a c · b) + ∇(a · b c ) = a c × (∇ × b) + (a c · ∇)b + b c × (∇ × a) + (b c · ∇)a
= (b · ∇)a + (a · ∇)b + b × ∇ × +a × ∇ × b
Quod erat demonstrandum.
Example 1.3.9 Verify ∇ × (a × b) = (b · ∇)a + (∇ · b)a − (a · ∇)b − (∇ · a)b.
Proof It can be known from the double vector product formula that
c × (a × b) = a(b · c) − b(a · c) = a(c · b) − (a · c)b = (b · c)a − b(c · a)
Taking ∇ as c, there is
∇ × (a × b) = ∇ × (a c × b) + ∇ × (a × b b ) = a c (∇ · b) − (a c · ∇)b + (b c · ∇)a − b c (∇ · a)
= (b · ∇)a + (∇ · b)a − (a · ∇)b − (∇ · a)b
where
(b · ∇)a =
b x
∂
∂ x
+ b y
∂
∂ y
+ b z
∂
∂z
a = b x
∂ a
∂ x
+ b y
∂ a
∂ y
+ b z
∂ a
∂z
Example 1.3.10 Verify ∇ × (∇ × a) = ∇(∇ · a) − a.
Proof It can be known from the double vector product formula that
c × (a × b) = a(c · b) − (a · c)b
Taking ∇ as a and c, b as a, there is
∇ × (∇ × a) = ∇(∇ · a) − (∇ · ∇)a = ∇(∇ · a) − a
Quod erat demonstrandum.