10.5 The Euler Equations of Rotation Type Functionals
671
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂u
+ a · ∇ × ∇ ×
∂ F
∂∇ × ∇ × ua
= 0
∂ F
∂ a
+ u∇ × ∇ ×
∂ F
∂∇ × ∇ × ua
= 0
(4)
The natural boundary conditions are
n ×
∂ F
∂∇ × ∇ × ua
S
= 0, n × ∇ ×
∂ F
∂∇ × ∇ × ua
S
= 0
(5)
Corollary Let the functional
J =
V
F(u, a, (∇×)
n ua)dV
(1)
The Euler equations of the functional are
⎧
⎪ ⎨
⎪ ⎩
∂ F
∂u
+ a · (∇×)
n
∂ F
∂(∇×) n ua
= 0
∂ F
∂ a
+ u(∇×)
n
∂ F
∂(∇×) n ua
= 0
(2)
The natural boundary conditions are
n × (∇×)
i−1
∂ F
∂∇ n × ua
S
= 0 (i = 1, 2, · · · , n)
(3)
Example 10.5.3 The functional with the rotation of product of a scalar function and
a vector. Derive the Euler equations and natural boundary conditions of the functional
J =
V
F(u, a, ∇ × f (u)a)dV
(1)
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
dV
(2)
Let b =
∂ F
∂∇× f a
, c = δ( f a) = a f u δu + f δa, there is
b · δ∇ × f a = −∇ · (b × c) + c · ∇ × b = −∇ · (b × c) + ∇ × b · (a f u δu + f δa)
= −∇ · (b × c) + ( f u a · ∇ × b)δu + f ∇ × b · δa
(3)
Substituting Eq. (3) into Eq. (2), we get