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

∂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
⎠