650
10 Variational Problems of Functionals with Vector, Tensor …
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂u
− (a × b) · ∇ ·
∂ F
∂∇(ua × b)
= 0
∂ F
∂ a
− b × ∇ ·
∂ F
∂∇(ua × b)
= 0
∂ F
∂ b
+ a × ∇ ·
∂ F
∂∇(ua × b)
= 0
(5)
The natural boundary condition is
n ·
∂ F
∂∇(ua × b)
S
= 0
(6)
Example 10.3.14 The functional with the left, right gradients of a scalar multiplied
by a vector. Let the double inner product of tensors of second order conform to the
law of parallel operation, the functional is
J =
V
F(u, a, ∇ua, ua∇)dV
(1)
derive the Euler equations and natural boundary condition.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ua
: δ∇ua +
∂ F
∂ua∇
: δua∇
dV
(2)
Putting A =
∂ F
∂∇ua
, B =
∂ F
∂ua∇
, the integrals of the last two terms in Eq. (2) can be
written as
A : δ∇ua = ∇ · ( A · δua) − (∇ · A) · δua = ∇ · ( A · δua) − (∇ · A) · aδu − u(∇ · A) · δa
(3)
B : δua∇ = ∇ · (δua · B) − (B · ∇) · δua = ∇ · (δua · B) − (B · ∇) · aδu − u(B · ∇) · δa
(4)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− (∇ · A + B · ∇) · a
δu +
∂ F
∂ a
− u(∇ · A + B · ∇)
· δa
dV
+
V
[∇ · ( A · δua) + ∇ · (δua · B)]dV
(5)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
∂ F
∂u
−
∇ ·
∂ F
∂∇ua
+
∂ F
∂ua∇
· ∇
· a = 0
∂ F
∂ a
− u
∇ ·
∂ F
∂∇ua
+
∂ F
∂ua∇
· ∇
= 0
(6)
10 Variational Problems of Functionals with Vector, Tensor …
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂u
− (a × b) · ∇ ·
∂ F
∂∇(ua × b)
= 0
∂ F
∂ a
− b × ∇ ·
∂ F
∂∇(ua × b)
= 0
∂ F
∂ b
+ a × ∇ ·
∂ F
∂∇(ua × b)
= 0
(5)
The natural boundary condition is
n ·
∂ F
∂∇(ua × b)
S
= 0
(6)
Example 10.3.14 The functional with the left, right gradients of a scalar multiplied
by a vector. Let the double inner product of tensors of second order conform to the
law of parallel operation, the functional is
J =
V
F(u, a, ∇ua, ua∇)dV
(1)
derive the Euler equations and natural boundary condition.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ua
: δ∇ua +
∂ F
∂ua∇
: δua∇
dV
(2)
Putting A =
∂ F
∂∇ua
, B =
∂ F
∂ua∇
, the integrals of the last two terms in Eq. (2) can be
written as
A : δ∇ua = ∇ · ( A · δua) − (∇ · A) · δua = ∇ · ( A · δua) − (∇ · A) · aδu − u(∇ · A) · δa
(3)
B : δua∇ = ∇ · (δua · B) − (B · ∇) · δua = ∇ · (δua · B) − (B · ∇) · aδu − u(B · ∇) · δa
(4)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− (∇ · A + B · ∇) · a
δu +
∂ F
∂ a
− u(∇ · A + B · ∇)
· δa
dV
+
V
[∇ · ( A · δua) + ∇ · (δua · B)]dV
(5)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
∂ F
∂u
−
∇ ·
∂ F
∂∇ua
+
∂ F
∂ua∇
· ∇
· a = 0
∂ F
∂ a
− u
∇ ·
∂ F
∂∇ua
+
∂ F
∂ua∇
· ∇
= 0
(6)
