10.3 The Euler Equations of Gradient Type Functionals
643
The natural boundary condition is
∂ F
∂∇(u/v)
· n
S
= 0
( 6 )
Example 10.3.6 The functional with the gradient of a scalar multiplied by the inner
product of two vectors. Derive the Euler equations and natural boundary condition
of the functional
J =
V
F(u, a, b, ∇(ua · b))dV
(1)
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇(ua · b)
· δ∇(ua · b)
dV
(2)
Putting c =
∂ F
∂∇(ua·b)
, ϕ = δ(ua · b) = a · bδu + ub · δa + ua · δb, the integral of
the last term in Eq. (2) can be written as
c · ∇ϕ = ∇ · ϕc − (∇ · c)(a · bδu + ub · δa + ua · δb)
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− a · b∇ · c
δu +
∂ F
∂ a
− ub∇ · c
· δa
+
∂ F
∂ b
− ua∇ · c
· δb + ∇ · (ϕc)
dV
(4)
The Euler equations of the functional are
⎧
⎪ ⎪ ⎪ ⎪ ⎪ ⎨
⎪ ⎪ ⎪ ⎪ ⎪ ⎩
∂ F
∂u
− a · b∇ ·
∂ F
∂∇(ua · b)
= 0
∂ F
∂ a
− ub∇ ·
∂ F
∂∇(ua · b)
= 0
∂ F
∂ b
− ua∇ ·
∂ F
∂∇(ua · b)
= 0
(5)
The natural boundary condition is
∂ F
∂∇(ua · b)
· n
S
= 0
( 6 )
Example 10.3.7 The functional with gradient of quotient of the inner product of
two vectors and the inner product of other two vectors. Derive the Euler equations
and natural boundary condition of the functional