10.4 The Euler Equations of Divergence Type Functionals
661
∂ F
∂∇ · u∇v
S
= 0, u∇
∂ F
∂∇ · u∇v
· n
S
= 0
( 6 )
Example 10.4.11 The functional with the divergence of inner product of respective
gradients of two scalars multiplied by a vector. Derive the Euler equations and natural
boundary conditions of the functional
J =
V
F(u, v, a, ∇ · (∇u · ∇v)a)dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂ a
· δa +
∂ F
∂∇ · (∇u · ∇v)a
δ∇ · (∇u · ∇v)a
dV
(2)
Putting ϕ =
∂ F
∂∇·(∇u·∇v)a
, b = δ(∇u·∇v)a = a(∇v·∇δu+∇u·∇δv)+(∇u·∇v)δa,
the integral of the last term in Eq. (2) can be written as
ϕ∇ · b = ∇ · ϕb − ∇ϕ · [a(∇v · ∇δu + ∇u · ∇δv) + (∇u · ∇v)δa]
= ∇ · ϕb − (a · ∇ϕ)∇v · ∇δu − (a · ∇ϕ)∇u · ∇δv − (∇u · ∇v)∇ϕ · δa
= ∇ · ϕb − ∇ · [(a · ∇ϕ)∇vδu] + [∇ · (a · ∇ϕ∇v)]δu−
∇ · [(a · ∇ϕ)∇uδv] + [∇ · (a · ∇ϕ∇u)]δv − (∇u · ∇v)∇ϕ · δa
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
+ ∇ · (a · ∇ϕ∇v)
δu +
∂ F
∂v
+ ∇ · (a · ∇ϕ∇u)
δv
+
∂ F
∂ a
− (∇u · ∇v)∇ϕ
· δa
dV
+
V
{∇ · ϕb − ∇ · [(a · ∇ϕ∇v)δu] − ∇ · [(a · ∇ϕ∇u)δv]}dV
(4)
Using the Gauss formula, the second volume integral is transformed into the
surface integral, there is
V
{∇ · ϕb − ∇ · [(a · ∇ϕ∇v)δu] − ∇ · [(a · ∇ϕ∇u)δv]}dV
=
S
[ϕa(∇v · ∇δu) + ϕa(∇u · ∇δv) + ϕ(∇u · ∇v)δa − (a · ∇ϕ∇v)δu − (a · ∇ϕ∇u)δv] · d S (5)
The Euler equations of the functional are