Vector_Identity

# Vector_Identity - 1 Formulae involving ∇ Vector...

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Unformatted text preview: 1 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. Email: [email protected]; [email protected] Tel.: +6013-6134998; +606-2312594; +605-4664998 www.xpreeli.com All rights reserved. Vector: A = A 1 i + A 2 j + A 3 k B = B 1 i + B 2 j + B 3 k C = C 1 i + C 2 j + C 3 k Scalar: φ = φ ( x , y , z ) ψ = ψ ( x , y , z ) Nabla: z k y j x i ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ (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 ( ) ( ) ( ) ( ) k z j y i x k z j y i x ∂ + ∂ + ∂ + ∂ + ∂ + ∂ = +         ∂ ∂ + ∂ ∂ + ∂ ∂ ψ φ ψ φ ψ φ ψ φ = k z k z j y j y i x i x ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ ψ φ ψ φ ψ φ =         ∂ ∂ + ∂ ∂ + ∂ ∂ +         ∂ ∂ + ∂ ∂ + ∂ ∂ k z j y i x k z j y i x ψ ψ ψ φ φ φ = ψ φ         ∂ ∂ + ∂ ∂ + ∂ ∂ +         ∂ ∂ + ∂ ∂ + ∂ ∂ k z j y i x k z j y i x ∴ ∇ ( φ + ψ ) = ∇ φ + ∇ ψ (4) Prove ∇ ( φ ψ ) = φ ∇ ψ + ψ ∇ φ ( ) ( ) ( ) ( ) k z j y i x k z j y i x ∂ ∂ + ∂ ∂ + ∂ ∂ =         ∂ ∂ + ∂ ∂ + ∂ ∂ φψ φψ φψ φψ 2 = k x k x j x j x i x i x ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ φ ψ ψ φ φ ψ ψ φ φ ψ ψ φ =         ∂ ∂ + ∂ ∂ + ∂ ∂ +         ∂ ∂ + ∂ ∂ + ∂ ∂ k z j y i x k z j y i x φ φ φ ψ ψ ψ ψ φ = φ ψ ψ φ         ∂ ∂ + ∂ ∂ + ∂ ∂ +         ∂ ∂ + ∂ ∂ + ∂ ∂ k z j y i x k z j y i x ∴ ∇ ( φ ψ ) = φ ∇ ψ + ψ ∇ φ (5) Prove ∇ .( A + B ) = ∇ . A + ∇ . B ( ) ( ) ( ) [ ] k B A j B A i B A k z j y i x B A 3 3 2 2 1 1 ) .( + + + + +         ∂ ∂ + ∂ ∂ + ∂ ∂ = + ∇ = ( ) ( ) ( ) z B A y B A x B A ∂ + ∂ + ∂ + ∂ + ∂ + ∂ 3 3 2 2 1 1 = LHS ( ) ( ) k B j B i B k z j y i x k A j A i A k z j y i x B A 3 2 1 3 2 1 . . + +         ∂ ∂ + ∂ ∂ + ∂ ∂ + + +         ∂ ∂ + ∂ ∂ + ∂ ∂ = ∇ + ∇ = z B y B x B z A y A x A ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ 3 2 1 3 2 1 = ( ) ( ) ( ) z B A y B A x B A ∂ + ∂ + ∂ + ∂ + ∂ + ∂ 3 3 2 2 1 1 = RHS LHS = RHS ∴ ∇ .( A + B ) = ∇ . A + ∇ . B (6) Prove ∇ x( A + B ) = ∇ x A + ∇ x B ( ) ( ) ( ) [ ] k B A j B A i B A x k z j y i x B A x 3 3 2 2 1 1 ) ( + + + + +         ∂ ∂ + ∂ ∂ + ∂ ∂ = + ∇ = 3 3 2 2 1 1 B A B A B A z y x k j i + + + ∂ ∂ ∂ ∂ ∂ ∂...
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## This note was uploaded on 10/09/2011 for the course FLUID MECH 4020 taught by Professor Chandra during the Spring '11 term at Université de Liège.

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Vector_Identity - 1 Formulae involving ∇ Vector...

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