658
10 Variational Problems of Functionals with Vector, Tensor …
Putting ϕ =
∂ F
∂∇·(a×b/c·d)
, and using using the parallelepipedal product formula of
trivector, the integral of the last term in Eq. (2) can be written as
ϕδ∇ · (a × b/c · d) = ϕ∇ · δ(a × b/c · d) = ∇ · [ϕδ(a × b/c · d)] − (∇ϕ) · δ(a × b/c · d)
= ∇ · [ϕδ(a × b/c · d)] − (∇ϕ) ·
(c · d)(a × δb + δa × b) − (a × b)(c · δd + d · δc)
(c · d) 2
= ∇ · [ϕδ(a × b/c · d)] +
(a × ∇ϕ) · δb − (b × ∇ϕ) · δa
(c · d)
+
(∇ϕ) · (a × b)(c · δd + d · δc)
(c · d) 2
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂ a
−
b × ∇ϕ
(c · d)
· δa +
∂ F
∂ b
+
a × ∇ϕ
(c · d)
· δb +
∂ F
∂ c
+
∇ϕ · (a × b)
(c · d) 2 d
· δc
+
∂ F
∂ d
+
∇ϕ · (a × b)
(c · d) 2 c
· δd
dV +
V
∇ ·
ϕδ
a × b
c · d
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂ a
−
b
c · d
× ∇
∂ F
∂∇ · (a × b/c · d)
= 0
∂ F
∂ b
+
a
c · d
× ∇
∂ F
∂∇ · (a × b/c · d)
= 0
∂ F
∂ c
+
a × b
(c · d) 2 · ∇
∂ F
∂∇ · (a × b/c · d)
d = 0
∂ F
∂ d
+
a × b
(c · d) 2 · ∇
∂ F
∂∇ · (a × b/c · d)
c = 0
(5)
The natural boundary condition is
∂ F
∂∇ · (a × b/c · d)
S
= 0
( 6 )
Example 10.4.8 The functional with the divergence of product of the scalar, modulus
of vector and vector. Derive the Euler equations and natural boundary condition of
the functional
J =
V
F(u, a, ∇ · u|a|a)dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
δa +
∂ F
∂∇ · (u|a|a)
δ∇ · (u|a|a)
dV
(2)
Putting ϕ =
∂ F
∂∇·(u|a|a)
, the integral of the last term in Eq. (2) can be written as