Page 263 - Intro to Tensor Calculus

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The vector field ~u (1) can be eliminated from equation (2.4.37) by taking the divergence of both sides of the
equation. This produces

2
∂ ∇· ~u (2) 2 (2) 2 (2)
% =(λ + µ)∇ (∇· ~u )+ µ∇· ∇ ~u .
∂t 2
2
The displacement field is assumed to be continuous and so we can interchange the order of the operators ∇
and ∇ and write
2 (2)
∂ ~u 2 (2)
∇· % − (λ +2µ)∇ ~u =0.
∂t 2
This last equation implies that
2
∂ ~u (2) 2 (2)
% =(λ +2µ)∇ ~u
∂t 2
and consequently, ~u (2) is a vector wave which moves with the speed p (λ +2µ)/%. Similarly, when the vector
field ~u (2) is eliminated from the equation (2.4.37), by taking the curl of both sides, we find the vector ~u (1)
also satisfies a wave equation having the form

2
∂ ~u (1) 2 (1)
% = µ∇ ~u .
∂t 2
p (2) (1)
This later wave moves with the speed µ/%. The vector ~u is a compressive wave, while the wave u is
a shearing wave.
The exercises 30 through 38 enable us to write the Navier’s equations in Cartesian, cylindrical or
spherical coordinates. In particular, we have for cartesian coordinates

2
2
2
2
2
2
2
∂ u ∂ v ∂ w ∂ u ∂ u ∂ u ∂ u
(λ + µ)( + + )+ µ( + + )+ %b x =%
∂x 2 ∂x∂y ∂x∂z ∂x 2 ∂y 2 ∂z 2 ∂t 2
2
2
2
2
2
2
2
∂ u ∂ v ∂ w ∂ v ∂ v ∂ v ∂ v
(λ + µ)( + + )+ µ( + + )+ %b y =%
∂x∂y ∂y 2 ∂y∂z ∂x 2 ∂y 2 ∂z 2 ∂t 2
2
2
2
2
2
2
2
∂ u ∂ v ∂ w ∂ w ∂ w ∂ w ∂ w
(λ + µ)( + + 2 )+ µ( 2 + 2 + 2 )+ %b z =% 2
∂x∂z ∂y∂z ∂z ∂x ∂y ∂z ∂t
and in cylindrical coordinates

∂ 1 ∂ 1 ∂u θ ∂u z
(λ + µ) (ru r )+ + +
∂r r ∂r r ∂θ ∂z
2 2 2 2
∂ u r 1 ∂u r 1 ∂ u r ∂ u r u r 2 ∂u θ ∂ u r
µ( + + + − − )+ %b r =%
2
2
∂r 2 r ∂r r ∂θ 2 ∂z 2 r 2 r ∂θ ∂t 2
1 ∂ 1 ∂ 1 ∂u θ ∂u z
(λ + µ) (ru r )+ + +
r ∂θ r ∂r r ∂θ ∂z
2 2 2 2
∂ u θ 1 ∂u θ 1 ∂ u θ ∂ u θ 2 ∂u r u θ ∂ u θ
µ( + + + + − )+ %b θ =%
2
2
∂r 2 r ∂r r ∂θ 2 ∂z 2 r ∂θ r 2 ∂t 2
∂ 1 ∂ 1 ∂u θ ∂u z
(λ + µ) (ru r )+ + +
∂z r ∂r r ∂θ ∂z
2 2 2 2
∂ u z 1 ∂u z 1 ∂ u z ∂ u z ∂ u z
µ( + + + )+ %b z =%
2
∂r 2 r ∂r r ∂θ 2 ∂z 2 ∂t 2