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10 Variational Problems of Functionals with Vector, Tensor …
J =
V
F(u, ∇ f (u))dV
(1)
Write the Euler equation and natural boundary condition.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂∇ f (u)
· δ∇ f (u)
dV
(2)
Putting a =
∂ F
∂∇ f (u)
, the integral of the latter term in Eq. (2) can be written as
a · δ∇ f (u) = a · ∇δ f (u) = ∇ · [aδ f (u)] − (∇ · a)δ f (u) = ∇ · [aδ f (u)] − (∇ · a) f u (u)δu
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− f u (u)∇ · a
δu + ∇ · [aδ f (u)]
dV
(4)
The Euler equation of the functional is
∂ F
∂u
− f u (u)∇ ·
∂ F
∂∇ f (u)
= 0
( 5 )
The natural boundary condition is
∂ F
∂∇ f (u)
· n
S
= 0
( 6 )
Example 10.3.3 The functional with the gradient of function of the n scalars. Let
the functional be
J =
V
F(u 1 , u 2 , · · · , u n , ∇ f (u 1 , u 2 , · · · , u n ))dV
(1)
Find the Euler equation and natural boundary condition.
Solution Putting f = f (u 1 , u 2 , · · · , u n ), Taking the variation to the functional, we
get
δ J =
V
n
i=1
∂ F
∂u i
δu i +
∂ F
∂∇ f
· δ∇ f
dV
(2)
Putting a =
∂ F
∂∇ f
, the integral of the latter term in Eq. (2) can be written as
10 Variational Problems of Functionals with Vector, Tensor …
J =
V
F(u, ∇ f (u))dV
(1)
Write the Euler equation and natural boundary condition.
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂∇ f (u)
· δ∇ f (u)
dV
(2)
Putting a =
∂ F
∂∇ f (u)
, the integral of the latter term in Eq. (2) can be written as
a · δ∇ f (u) = a · ∇δ f (u) = ∇ · [aδ f (u)] − (∇ · a)δ f (u) = ∇ · [aδ f (u)] − (∇ · a) f u (u)δu
(3)
Substituting Eq. (3) into Eq. (2), we get
δ J =
V
∂ F
∂u
− f u (u)∇ · a
δu + ∇ · [aδ f (u)]
dV
(4)
The Euler equation of the functional is
∂ F
∂u
− f u (u)∇ ·
∂ F
∂∇ f (u)
= 0
( 5 )
The natural boundary condition is
∂ F
∂∇ f (u)
· n
S
= 0
( 6 )
Example 10.3.3 The functional with the gradient of function of the n scalars. Let
the functional be
J =
V
F(u 1 , u 2 , · · · , u n , ∇ f (u 1 , u 2 , · · · , u n ))dV
(1)
Find the Euler equation and natural boundary condition.
Solution Putting f = f (u 1 , u 2 , · · · , u n ), Taking the variation to the functional, we
get
δ J =
V
n
i=1
∂ F
∂u i
δu i +
∂ F
∂∇ f
· δ∇ f
dV
(2)
Putting a =
∂ F
∂∇ f
, the integral of the latter term in Eq. (2) can be written as
