10.4 The Euler Equations of Divergence Type Functionals
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∂ F
∂ a
− (b × c) × ∇
∂ F
∂∇ · a × (b × c)
= 0
∂ F
∂ b
+ c ×
a × ∇
∂ F
∂∇ · a × (b × c)
= 0
∂ F
∂ c
− b ×
a × ∇
∂ F
∂∇ · a × (b × c)
= 0
(5)
The natural boundary condition is
∂ F
∂∇ · a × (b × c)
S
= 0
( 6 )
Example 10.4.13 The functional with the divergence of cross product of trivector,
where, a vector is the gradient of a scalar. Let the functional
J =
V
F(u, a, b, ∇ · ∇u × (a × b))dV
(1)
Derive its Euler equations and natural boundary conditions.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇ · ∇u × (a × b)
δ∇ · ∇u × (a × b)
dV (2)
Putting ϕ =
∂ F
∂∇·∇u×(a×b)
, c = δ∇u × (a × b) = ∇δu × (a × b) +∇u × (δa × b) +
∇u × (a × δb), and using the parallelepipedal product formula of trivector, there is
ϕδ∇ · [∇u × (a × b)] = ϕ∇ · δ[∇u × (a × b)] = ϕ∇ · c = ∇ · ϕc − c · ∇ϕ
= ∇ · ϕc − ∇ϕ · [∇δu × (a × b) + ∇u × (δa × b) + ∇u × (a × δb)]
= ∇ · ϕc − (a × b) × ∇ϕ · ∇δu − (∇ϕ × ∇u) · (δa × b) − (∇ϕ × ∇u) · (a × δb)
= ∇ · ϕc − ∇ · [(a × b) × ∇ϕδu] + ∇ · [(a × b) × ∇ϕ]δu
− [b × (∇ϕ × ∇u)] · δa − [(∇ϕ × ∇u) × a] · δb
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ ∇ · (a × b) × ∇ϕ
δu +
∂ F
∂ a
− b × (∇ϕ × ∇u)
· δa
+
∂ F
∂ b
+ a × (∇ϕ × ∇u)
· δb
dV +
V
{∇ · ϕc − ∇ · [(a × b) × ∇ϕδu]}dV (4)
The Euler equations of the functional are