10.3 The Euler Equations of Gradient Type Functionals
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The natural boundary conditions are
n ·
∂ F
∂∇ua
S
= 0,
∂ F
∂ua∇
· n
S
= 0
(7)
Example 10.3.15 The functional with the left, right gradients of a scalar function
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, ∇ f (u)a, f (u)a∇)dV
(1)
derive the Euler equations and natural boundary condition.
Solution Putting f = f (u), Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ f a
· δ f a +
∂ F
∂ f a∇
: δ f a∇
dV
(2)
Putting A =
∂ F
∂∇ f a
, B =
∂ F
∂ f a∇
, the integrals of the last two terms in Eq. (2) can
be written as
A : δ∇ f a = ∇ · ( A · δ f a) − (∇ · A) · δ f a = ∇ · ( A · δ f a) − (∇ · A) · a f u δu − f (∇ · A) · δa (3)
B : δ f a∇ = ∇ · (δ f a · B) − (B · ∇) · δ f a = ∇ · (δ f a · B) − (B · ∇) · a f u δu − f (B · ∇) · δa (4)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− f u (∇ · A + B · ∇) · a
δu +
∂ F
∂ a
− f (∇ · A + B · ∇)
· δa
dV
+
V
[∇ · ( A · δ f a) + ∇ · (δ f a · B)]dV
(5)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
∂ F
∂u
− f u (u)
∇ ·
∂ F
∂∇ f (u)a
+
∂ F
∂ f (u)a∇
· ∇
· a = 0
∂ F
∂ a
− f (u)
∇ ·
∂ F
∂∇ f (u)a
+
∂ F
∂ f (u)a∇
· ∇
= 0
(6)
The natural boundary conditions are
n ·
∂ F
∂∇ f (u)a
S
= 0,
∂ F
∂ f (u)a∇
· n
S
= 0
(7)