Formulae involving ∇ Vector Identities with Proofs: Nabla Formulae ...

Formulae involving ∇
Vector Identities with Proofs: Nabla Formulae for Vector Analysis
李国华 (Kok-Wah LEE) @ 08 May 2009 (Version 1.0)
No. 4, Jalan Bukit Beruang 5, Taman Bukit Beruang, 75450 Bukit Beruang, Melaka, Malaysia.
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Vector: A = A1 i + A2 j + A3 k B = B1 i + B2 j + B3 k C = C1 i + C2 j + C3 k
Scalar: φ = φ(x,y,z) ψ = ψ(x,y,z)
Nabla:
∇ =
∂ ∂ ∂
i + j + k
∂x
∂y
∂z
(1) (A x B).C ≡ (B x C).A ≡ (C x A).B
(2) A x (B x C) ≡ (A.C)B - (A.B)C
(3) Prove ∇(φ + ψ) = ∇φ + ∇ψ
( φ + ψ ) ∂(
φ + ψ ) ∂(
φ + )
i + j + k
⎛ ∂ ∂ ∂ ⎞ ∂
ψ
⎜ i + j + k ⎟(
φ + ψ ) =
⎝ ∂x
∂y
∂z
⎠
∂x
∂y
∂z
∂φ ∂ψ
∂φ
∂ψ
∂φ
∂ψ
= i + i + j + j + k + k
∂x
∂x
∂y
∂y
∂z
∂z
⎛ ∂φ
∂φ
∂φ
⎞ ⎛ ∂ψ
∂ψ
∂ψ
⎞
= ⎜ i + j + k ⎟ + ⎜ i + j + k ⎟
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
⎛ ∂ ∂ ∂ ⎞ ⎛ ∂ ∂ ∂ ⎞
= ⎜ i + j + k ⎟φ
+ ⎜ i + j + k ⎟ψ ⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
∴ ∇(φ + ψ) = ∇φ + ∇ψ
(4) Prove ∇(φ ψ) = φ ∇ψ + ψ ∇φ
( φψ ) ∂(
φψ ) ∂(
)
i + j + k
⎛ ∂ ∂ ∂ ⎞ ∂
φψ
⎜ i + j + k ⎟(
φψ
) =
⎝ ∂x
∂y
∂z
⎠ ∂x
∂y
∂z
1

∂ψ ∂φ
∂ψ
∂φ
∂ψ
∂φ
= φ i + ψ i + φ j + ψ j + φ k + ψ k
∂x
∂x
∂x
∂x
∂x
∂x
⎛ ∂ψ
∂ψ
∂ψ
⎞ ⎛ ∂φ
∂φ
∂φ
⎞
= φ ⎜ i + j + k ⎟ + ψ ⎜ i + j + k ⎟
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
⎛ ∂ ∂ ∂ ⎞ ⎛ ∂ ∂ ∂ ⎞
= φ⎜ i + j + k ⎟ψ
+ ψ ⎜ i + j + k ⎟φ
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
∴ ∇(φ ψ) = φ ∇ψ + ψ ∇φ
(5) Prove ∇.(A + B) = ∇.A + ∇.B
⎛ ∂ ∂ ∂ ⎞
∇ .( + B)
= ⎜ i + j + k ⎟ 1 1 2 2 3 +
⎝ ∂x
∂y
∂z
⎠
[ ( A + B ) i + ( A + B ) j + ( A B ) k]
A 3
=
( A + B ) ∂(
A + B ) ∂(
A + B )
∂ 1 1 2 2
3 3
∂x
+
∂y
+
∂z
2
= LHS
⎛ ∂ ∂ ∂ ⎞
⎛ ∂ ∂ ∂ ⎞
∇ . + ∇.
B = ⎜ i + j + k ⎟ 1 2 3
1 2 +
⎝ ∂x
∂y
∂z
⎠
⎝ ∂x
∂y
∂z
⎠
( Ai
+ A j + A k)
+ ⎜ i + j + k ⎟(
B i + B j B k)
A 3
=
=
∂ 3
A1
∂A2
∂A3
∂B1
∂B2
∂B
+ + + + +
∂x
∂y
∂z
∂x
∂y
∂z
( A + B ) ∂(
A + B ) ∂(
A + B )
∂ 1 1 2 2 3 3
∂x
LHS = RHS
+
∂y
∴ ∇.(A + B) = ∇.A + ∇.B
(6) Prove ∇x(A + B) = ∇xA + ∇xB
+
∂z
= RHS
⎛ ∂ ∂ ∂ ⎞
∇ x( A + B)
= ⎜ i + j + k ⎟x
1 1 2 2 3 + 3
⎝ ∂x
∂y
∂z
⎠
=
i
∂
∂x
A + B
1
1
j
∂
∂y
A + B
2
2
3
[ ( A + B ) i + ( A + B ) j + ( A B ) k]
k
∂
∂z
A + B
3
= ⎛ ∂( A3
+ B3
) ∂(
A2
+ B2
) ⎞ ⎛ ∂(
A3
+ B3
) ∂(
A1
+ B1
) ⎞ ⎛ ∂(
A2
+ B2
) ∂(
A1
+ B1
) ⎞
⎜ − ⎟i
−
−
j + ⎜ − ⎟k
⎝
∂y
∂z
⎠
⎜
⎝
∂x
∂z
⎟
⎠
⎝
∂x
∂y
⎠

=
⎡⎛
∂A
⎤ ⎡
⎤
3 ∂A2
⎞ ⎛ ∂A3
∂A1
⎞ ⎛ ∂A2
∂A1
⎞ ⎛ ∂B3
∂B2
⎞ ⎛ ∂B3
∂B1
⎞ ⎛ ∂B2
∂B1
⎞
⎢⎜
− ⎟i
− ⎜ − ⎟ j + ⎜ − ⎟k⎥
+ ⎢⎜
− ⎟i
− ⎜ − ⎟ j + ⎜ − ⎟k⎥
⎣⎝
∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎦ ⎣⎝
∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎦
=
i j k
∂ ∂ ∂ +
∂x
∂y
∂z
A A A
∴ ∇x(A + B) = ∇xA + ∇xB
(7) Prove ∇.(φA) = (∇φ).A + φ(∇.A)
⎛ ∂ ∂ ∂ ⎞
∇ .( φ ) = ⎜ i + j + k ⎟ 1 2 + φ
⎝ ∂x
∂y
∂z
⎠
1
2
3
i
∂
∂x
B
. ( φAi
+ φA
j A k)
A 3
= ∂( φ A1
) ∂(
φA2
) ∂(
φA3
) = LHS
∂x
+
∂y
+
∂z
⎛ ∂φ
∂φ
∂φ
⎞
⎡⎛
∂ ∂ ∂ ⎞
⎤
( ∇φ ). A + φ(
∇.
A)
= ⎜ i + j + k ⎟.
( A1
i + A2
j + A3k
) + φ ⎢⎜
i + j + k ⎟.
( A1i
+ A2
j + A3k
)⎥
⎝ ∂x
∂y
∂z
⎠
⎣⎝
∂x
∂y
∂z
⎠
⎦
=
1
j
∂
∂y
B
2
k
∂
∂z
B
⎛ ∂φ
∂φ
∂φ
⎞ ⎛ ∂A1
∂A2
∂A3
⎞
⎜ A1
+ A2
+ A3
⎟ + φ⎜
+ + ⎟
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
= ⎛ ∂φ
∂A1
⎞ ⎛ ∂φ
∂A2
⎞ ⎛ ∂φ
∂A3
⎞
⎜ A1
+ φ ⎟ + ⎜ A2
+ φ ⎟ + ⎜ A3
+ φ ⎟
⎝ ∂x
∂x
⎠ ⎝ ∂y
∂y
⎠ ⎝ ∂z
∂z
⎠
= ∂( φ A1
) ∂(
φA2
) ∂(
φA3
) = RHS
∂x
+
∂y
LHS = RHS
+
∂z
∴ ∇.(φA) = (∇φ).A + φ(∇.A)
(8) Prove ∇x(φA) = (∇φ)xA + φ(∇xA)
∇ x
( φA)
i
∂
=
∂x
φA
1
j
∂
∂y
φA
2
k
∂
∂z
φA
3
= ⎛ ∂( φ A3
) ∂(
φA2
) ⎞ ⎛ ∂(
φA3
) ∂(
φA1
) ⎞ ⎛ ∂(
φA2
) ∂(
φA1
) ⎞
⎜ − ⎟i
− − j + ⎜ − ⎟k
⎝
∂y
∂z
⎠
⎜
⎝
∂x
∂z
⎟
⎠
⎝
∂x
= ⎛ ∂A3 ∂φ
∂A2
∂φ
⎞ ⎛ ∂A3
∂φ
∂A1
∂φ
⎞ ⎛ ∂A2
∂φ
∂A1
∂φ
⎞
⎜ + A −φ
− A ⎟i
− φ + A −φ
− A j + ⎜φ
+ A −φ
− A ⎟k
=
φ 3
2
3
1
2
1
⎝
∂y
∂y
∂y
∂y
⎠
⎜
⎝
∂x
∂x
⎡⎛
∂φ
∂φ
⎞ ⎛ ∂φ
∂φ
⎞ ⎛ ∂φ
∂φ
⎞ ⎤
⎢⎜
A3
− A2
⎟i
− ⎜ A3
− A1
⎟ j + ⎜ A2
− A1
⎟k⎥
⎣⎝
∂y
∂y
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎦
3
3
∂y
∂z
⎠
⎟
∂z
⎠
⎝
∂x
∂x
∂y
∂y
⎠

=
1
⎡⎛
∂A
3 ∂A
2 ⎞ ⎛ ∂A3
∂A1
⎞ ⎛ ∂A
2 ∂A1
⎞ ⎤
+ ⎢⎜φ
−φ
⎟i
− ⎜φ
−φ
⎟ j + ⎜
⎟
⎜
φ −φ
⎟
k⎥
⎢
⎜
⎟ ⎜
⎟
⎣⎝
∂y
∂y
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎥⎦
i j k i
∂φ ∂x
∂φ
∂y
∂φ
+ ∂
φ
∂z
∂x
A A A A
2
3
1
j
∂
∂y
A
2
k
∂
∂z
A
∴ ∇x(φA) = (∇φ)xA + φ(∇xA)
(9) Prove ∇.(AxB) = B.(∇xA) - A.(∇xB)
i
.( AxB) i j k . A1
x y z
⎟
B
⎟
⎛ ∂ ∂ ∂ ⎞
∇ = ⎜ + +
⎝ ∂ ∂ ∂ ⎠
⎛ ∂
⎝ ∂x
∂
∂y
∂
∂z
⎞
1
3
j
A
B
2
2
k
A
B
3
3
= ⎜ i + j + k ⎟.
[ ( A B − A B ) i − ( A B − A B ) j + ( A B − A B ) k]
⎠
2
=
∂( A2
B3
− A3B2
) ∂(
A1B
3 − A3B
1)
∂(
A1B
2 − A2
B1
)
∂x
−
⎡⎛
∂A
⎤
3 ∂A2
⎞ ⎛ ∂A3
∂A1
⎞ ⎛ ∂A2
∂A1
⎞
B.(
∇xA) = ( B1
i + B2
j + B3k
) . ⎢⎜
− ⎟i
− ⎜ − ⎟ j + ⎜ − ⎟k⎥
⎣⎝
∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎦
=
3
∂y
3
2
+
4
1
3
∂z
⎛ ∂A3
∂A2
⎞ ⎛ ∂A3
∂A1
⎞ ⎛ ∂A2
∂A1
⎞
B1
⎜ − ⎟ − B2⎜
− ⎟ + B3
⎜ − ⎟
⎝ ∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠
Similarly, by interchanging the variable of A and B, we have
⎡⎛
∂B
⎤
3 ∂B2
⎞ ⎛ ∂B3
∂B1
⎞ ⎛ ∂B2
∂B1
⎞
A.(
∇xB) = ( A1
i + A2
j + A3k
) . ⎢ ⎜ − ⎟
⎟i
− ⎜ − ⎟ j + ⎜ − ⎟
⎟k⎥
⎣⎝
∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠ ⎦
=
⎛ ∂B3
∂B2
⎞ ⎛ ∂B3
∂B1
⎞ ⎛ ∂B2
∂B1
⎞
A1
⎜ − ⎟ − A2
⎜ − ⎟ + A3
⎜ − ⎟
⎝ ∂y
∂z
⎠ ⎝ ∂x
∂z
⎠ ⎝ ∂x
∂y
⎠
B.(∇xA) - A.(∇xB) = ⎛ ∂A3
∂B1
⎞ ⎛ ∂A2
∂B1
⎞ ⎛ ∂A3
∂B2
⎞
⎜ B1
+ A3
⎟ − ⎜ B1
+ A2
⎟ − ⎜ B2
+ A3
⎟
⎝ ∂y
∂y
⎠ ⎝ ∂z
∂z
⎠ ⎝ ∂x
∂x
⎠
⎛ ∂A1
∂B2
⎞ ⎛ ∂A2
∂B3
⎞ ⎛ ∂A1
∂B3
⎞
+ ⎜ B2
+ A1
⎟ + ⎜ B3
+ A2
⎟ − ⎜ B3
+ A1
⎟
⎝ ∂z
∂z
⎠ ⎝ ∂x
∂x
⎠ ⎝ ∂y
∂y
⎠
= ∂( A3
B1
) ∂(
A2B1
) ∂(
A3
B2
) ∂(
A1B
2 ) ∂(
A2
B3
) ∂(
A1B3
)
∂y
−
∂z
−
∂x
+
∂z
+
3
1
∂x
= ∂( A2
B3
− A3B2
) ∂(
A1B3
− A3B1
) ∂(
A1B2
− A2B1
)
∂x
−
∂y
∴ ∇.(AxB) = B.(∇xA) - A.(∇xB)
+
∂z
−
1
2
∂y
2
1

(10) Prove ∇x(AxB) = (B.∇)A - B(∇.A) - (A.∇)B + A(∇.B)
∇
=
i
( AxB)
= ∇x
A A A = ∇x[
( A B − A B ) i − ( A B − A B ) j + ( A B − A B ) k]
B
j
B
k
x 1 2 3
2 3 3 2 1 3 3 1 1 2 2 1
i
∂
∂x
A B − A B
2
3
3
2
1
3
1
2
B
1
3
j
∂
∂y
A B − A B
3
k
∂
∂z
A B − A B
1
2
2
1
= ⎛ ∂( A1
B2
− A2
B1
) ∂(
A3B1
− A1B
3 ) ⎞ ⎛ ∂(
A1
B2
− A2
B1
) ∂(
A2
B3
− A3B
2 ) ⎞ ⎛ ∂(
A3B1
− A1B
3 ) ∂(
A2B3
− A3B2
) ⎞
⎜
−
⎟i
−
−
j + ⎜
−
⎟k
⎝
= LHS
∂y
∂z
⎠
⎜
⎝
(B.∇)A - B(∇.A) = ⎛ ∂ ∂ ∂ ⎞
⎛ ∂A1
∂A2
∂A3
⎞
⎜ B1 + B2
+ B3
⎟(
A1i
+ A2
j + A3k
) −⎜
+ + ⎟(
B1i
+ B2
j + B3k
)
⎝ ∂x
∂y
∂z
⎠
⎝ ∂x
∂y
∂z
⎠
∂x
= ⎛ ∂A1 ∂A1
∂A2
∂A3
⎞ ⎛ ∂A2
∂A2
∂A1
∂A3
⎞ ⎛ ∂A3
∂A3
∂A1
∂A2
⎞
⎜ B B B B ⎟i
B B B B j + ⎜ B + B − B − B ⎟k
⎝
2
+
∂y
3
−
∂z
1
−
∂y
1
+ ⎜
∂z
⎠ ⎝
1
+
∂x
3
−
∂z
2
−
∂x
Similarly, by interchanging the variable of A and B, we have
(A.∇)B - A(∇.B) = ⎛ ∂ ∂ ∂ ⎞
⎛ ∂B1
∂B2
∂B3
⎞
⎜ A1 + A2
+ A3
⎟(
B1i
+ B2
j + B3k
) −⎜
+ + ⎟(
A1i
+ A2
j + A3k
)
⎝ ∂x
∂y
∂z
⎠
⎝ ∂x
∂y
∂z
⎠
= ⎛ ∂B1 ∂B1
∂B2
∂B3
⎞ ⎛ ∂B2
∂B2
∂B1
∂B3
⎞ ⎛ ∂B3
∂B3
∂B1
∂B2
⎞
⎜ A A A A ⎟i
A A A A j + ⎜ A + A − A − A ⎟k
⎝
2
+
∂y
3
−
∂z
1
−
∂y
1
+ ⎜
∂z
⎠ ⎝
(B.∇)A - B(∇.A) - (A.∇)B + A(∇.B)
1
+
∂x
3
−
∂z
2
−
∂x
5
2
2
∂z
⎟
∂z
⎠
⎟
∂z
⎠
= ⎡⎛
∂A B A B A B A B ⎤
1 ∂ 2 ⎞ ⎛ ∂ 1 ∂ 3 ⎞ ⎛ ∂ 2 ∂ 1 ⎞ ⎛ ∂ 3 ∂ 1 ⎞
⎜ A ⎟ B A ⎜ B A ⎟ B A i
⎢ B2 + 1 + ⎜ 3 + 1 ⎟ − 1 + 2 − ⎜ 1 + 3 ⎟⎥
⎣⎝
∂y
∂y
⎠ ⎝ ∂z
∂z
⎠ ⎝ ∂y
∂y
⎠ ⎝ ∂z
∂z
⎠⎦
⎡⎛
∂A2
∂B2
⎞ ⎛ ∂A2
∂B3
⎞ ⎛ ∂A1
∂B2
⎞ ⎛ ∂A3
∂B2
⎞⎤
+ ⎢⎜
B1 + A1
⎟ + ⎜ B3
+ A2
⎟ −⎜
B2
+ A1
⎟ −⎜
B2
+ A3
⎟ j
x x z z x x z z
⎥
⎣⎝
∂ ∂ ⎠ ⎝ ∂ ∂ ⎠ ⎝ ∂ ∂ ⎠ ⎝ ∂ ∂ ⎠⎦
⎡⎛
∂A
B A B A B A B ⎤
3 ∂ 1 ⎞ ⎛ ∂ 3 ∂ 2 ⎞ ⎛ ∂ 1 ∂ 3 ⎞ ⎛ ∂ 2 ∂ 3 ⎞
+ ⎢⎜
B1 + A3
⎟ + ⎜ B2
+ A3
⎟ −⎜
B3
+ A1
⎟ −⎜
B3
+ A2
⎟⎥k
⎣⎝
∂x
∂x
⎠ ⎝ ∂y
∂y
⎠ ⎝ ∂x
∂x
⎠ ⎝ ∂y
∂y
⎠⎦
= ⎛ ∂( A1
B2
− A2
B1
) ∂(
A1
B3
− A3B1
) ⎞ ⎛ ∂(
A1B
2 − A2B1
) ∂(
A3B
2 − A2
B3
) ⎞ ⎛ ∂(
A3B1
− A1B
3 ) ∂(
A3B2
− A2
B3
) ⎞
⎜
+
⎟i
−
+
j + ⎜
+
⎟k
⎝
∂y
∂z
⎠
⎜
⎝
∂x
= ⎛ ∂( A1
B2
− A2
B1
) ∂(
A3B1
− A1B
3 ) ⎞ ⎛ ∂(
A1
B2
− A2
B1
) ∂(
A2
B3
− A3B
2 ) ⎞ ⎛ ∂(
A3B1
− A1B
3 ) ∂(
A2B3
− A3B2
) ⎞
⎜
−
⎟i
−
−
j + ⎜
−
⎟k
⎝
= RHS
∂y
RHS = LHS
∂z
⎠
⎜
⎝
∂x
∂z
∂z
⎟
⎠
⎟
⎠
⎝
⎝
⎝
⎝
⎟
⎠
1
1
⎝
∂x
∂x
∂x
∂x
2
2
∂x
∂y
∂y
3
3
∂x
∂x
∂y
∂y
3
∂y
3
∂y
⎠
∂y
⎠
⎠
⎠
⎠

6
∴ ∇x(AxB) = (B.∇)A - B(∇.A) - (A.∇)B + A(∇.B)
(11) Prove ∇(A.B) = (B.∇)A + (A.∇)B + Bx(∇xA) + Ax(∇xB)
∇(A.B) = ( )
3
3
2
2
1
1
B
A
B
A
B
A
k
z
j
y
i
x
+
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂ = LHS
(B.∇)A = ( )
k
A
j
A
i
A
z
B
y
B
x
B 3
2
1
3
2
1
+
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
( ) ⎥ ⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
+
=
∇ k
y
A
x
A
j
z
A
x
A
i
z
A
y
A
x
k
B
j
B
i
B
xA
Bx
1
2
1
3
2
3
3
2
1
)
(
=
y
A
x
A
x
A
z
A
z
A
y
A
B
B
B
k
j
i
∂
∂
−
∂
∂
∂
∂
−
∂
∂
∂
∂
−
∂
∂ 1
2
3
1
2
3
3
2
1
= k
B
z
A
y
A
B
x
A
z
A
j
B
z
A
y
A
B
y
A
x
A
i
B
x
A
z
A
B
y
A
x
A
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
2
2
3
1
3
1
3
2
3
1
1
2
3
3
1
2
1
2
Similarly, by interchanging the variable of A and B, we have
(A.∇)B = ( )
k
B
j
B
i
B
z
A
y
A
x
A 3
2
1
3
2
1
+
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
( ) ⎥ ⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
+
=
∇ k
y
B
x
B
j
z
B
x
B
i
z
B
y
B
x
k
A
j
A
i
A
xB
Ax
1
2
1
3
2
3
3
2
1
)
(
=
y
B
x
B
x
B
z
B
z
B
y
B
A
A
A
k
j
i
∂
∂
−
∂
∂
∂
∂
−
∂
∂
∂
∂
−
∂
∂ 1
2
3
1
2
3
3
2
1
= k
A
z
B
y
B
A
x
B
z
B
j
A
z
B
y
B
A
y
B
x
B
i
A
x
B
z
B
A
y
B
x
B
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
2
2
3
1
3
1
3
2
3
1
1
2
3
3
1
2
1
2
Hence
(B.∇)A + Bx(∇xA)
= k
z
A
B
z
A
B
z
A
B
j
y
A
B
y
A
B
y
A
B
i
x
A
B
x
A
B
x
A
B ⎟ ⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂ 2
2
1
1
3
3
3
3
1
1
2
2
3
3
2
2
1
1
(A.∇)B + Ax(∇xB)
= k
z
B
A
z
B
A
z
B
A
j
y
B
A
y
B
A
y
B
A
i
x
B
A
x
B
A
x
B
A ⎟ ⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂ 2
2
1
1
3
3
3
3
1
1
2
2
3
3
2
2
1
1

(B.∇)A + (A.∇)B + Bx(∇xA) + Ax(∇xB)
= ⎛ ∂( A1
B1
) ∂(
A2B
2 ) ∂(
A3B3
) ⎞ ⎛ ∂(
A2B
2 ) ∂(
A1B1
) ∂(
A3B3
) ⎞ ⎛ ∂(
A3B
3 ) ∂(
A1B
1)
∂(
A2B
2 ) ⎞
⎜ + + ⎟i
+ ⎜ + + ⎟ j + ⎜ + + ⎟k
⎝ ∂x
∂x
∂x
⎠ ⎝ ∂y
∂y
∂y
⎠ ⎝ ∂z
∂z
∂z
⎠
=
( A1
B1
+ A2B
2 + A3B
3 ) ∂(
A1B
1 + A2B
2 + A3B
3)
∂(
A1B
1 + A2B2
+ A3B
)
i +
j +
k
∂ 3
⎛
⎝ ∂x
∂x
∂y
= ∂ ∂ ∂
⎜ i + j + k ⎟(
A B + A B + A B ) = RHS
∂y
LHS = RHS
∂z
⎞
⎠
1
1
2
2
∴ ∇(A.B) = (B.∇)A + (A.∇)B + Bx(∇xA) + Ax(∇xB)
(12) Prove ∇.(∇φ) = ∇ 2 φ
∇.(∇φ) = ⎛ ∂ ∂ ∂ ⎞ ⎛ ∂φ
∂φ
∂φ
⎞
⎜i
+ j + k ⎟.
⎜i
+ j + k ⎟
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
∂z
⎠
2 2 2
=
∂ φ ∂ φ ∂ φ 2
+ + = ∇ φ
2 2 2
∂x
∂y
∂z
∴ ∇.(∇φ) = ∇ 2 φ
(13) Prove ∇x(∇φ) = 0
∇x(∇φ) =
3
⎛ ∂ ∂ ∂ ⎞ ⎛ ∂φ
∂φ
⎜i
+ j + k ⎟x⎜
i + j + k
⎝ ∂x
∂y
∂z
⎠ ⎝ ∂x
∂y
=
i
∂
∂x
∂φ
∂x
j
∂
∂y
∂φ
∂y
3
k
∂
∂z
∂φ
∂z
7
∂φ
⎞
⎟
∂z
⎠
∂z
= ( φ − φ ) − ( φ −φ
) j + ( φ −φ
)k
zy
yz i zx xz yx xy
Since φ has continuous second order partial derivatives, we have
∴ ∇x(∇φ) = 0
φxy = φyx φyz = φzy φzx = φxz

8
(14) Prove ∇.(∇xA) = 0
∇.(∇xA) =
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
k
y
A
x
A
j
z
A
x
A
i
z
A
y
A
z
k
y
j
x
i
1
2
1
3
2
3
.
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
−
∂
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
−
∂
∂
∂
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
−
∂
∂
∂
z
y
A
z
x
A
y
z
A
y
x
A
x
z
A
x
y
A 1
2
2
2
1
2
3
2
2
2
3
2
= 0
∴ ∇.(∇xA) = 0
(15) Prove ∇x(∇xA) = ∇(∇.A) - ∇ 2 A
∇x(∇xA) =
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
+
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
−
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
k
y
A
x
A
j
z
A
x
A
i
z
A
y
A
x
z
k
y
j
x
i
1
2
1
3
2
3
=
y
A
x
A
x
A
z
A
z
A
y
A
z
y
x
k
j
i
∂
∂
−
∂
∂
∂
∂
−
∂
∂
∂
∂
−
∂
∂
∂
∂
∂
∂
∂
∂
1
2
3
1
2
3
= k
y
z
A
y
A
x
A
x
z
A
j
z
A
z
y
A
x
y
A
x
A
i
x
z
A
z
A
y
A
y
x
A
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
+
∂
∂
−
∂
∂
−
∂
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
∂
−
∂
∂
∂
−
∂
∂
−
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
∂
+
∂
∂
−
∂
∂
−
∂
∂
∂ 2
2
2
3
2
2
3
2
1
2
2
2
2
3
2
1
2
2
2
2
3
2
2
1
2
2
1
2
2
2
= LHS
∇(∇.A) - ∇ 2 A
= ( )
k
A
j
A
i
A
z
y
x
z
A
y
A
x
A
z
k
y
j
x
i 3
2
1
2
2
2
2
2
2
3
2
1 +
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
−
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
⎟
⎠
⎞
⎜
⎝
⎛
∂
∂
+
∂
∂
+
∂
∂
= k
y
A
x
A
z
y
A
z
x
A
j
z
A
x
A
y
z
A
y
x
A
i
z
A
y
A
x
z
A
x
y
A
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
−
∂
∂
∂
+
∂
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
−
∂
∂
∂
+
∂
∂
∂
+
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
∂
∂
−
∂
∂
−
∂
∂
∂
+
∂
∂
∂
2
3
2
2
3
2
2
2
1
2
2
2
2
2
2
2
3
2
1
2
2
1
2
2
1
2
3
2
2
2
= RHS
LHS = RHS
∴ ∇x(∇xA) = ∇(∇.A) - ∇ 2 A