Appendix: All Solutions to the Problems
937
∂ F
∂(a · ∇)u
a · n
S
= 0
( 5 )
10.6 Write out the Euler equations and natural boundary conditions of the functional
J =
V F(u, f (|∇u|))dV .
Solution Taking the variation to the functional, we have
δ J =
V
∂ F
∂u
δu +
∂ F
∂ f (|∇u|)
δ f (|∇u|)
dV
(1)
Putting f = f (|∇u|), ϕ =
∂ F
∂ f (|∇u|)
, 有
ϕδ f (|∇u|) = ϕ
∂ f
∂|∇u|
δ|∇u| = ϕ
∂ f
∂|∇u|
∇u
|∇u|
· δ∇u = ∇ ·
ϕ
∂ f
∂|∇u|
∇u
|∇u|
δu
− ∇ ·
ϕ
∂ f
∂|∇u|
∇u
|∇u|
δu
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
− ∇ ·
ϕ
∂ f
∂|∇u|
∇u
|∇u|
δu + ∇ ·
ϕ
∂ f
∂|∇u|
∇u
|∇u|
δu
dV (3)
The Euler equation of the functional is
∂ F
∂u
− ∇ ·
∂ F
∂ f (|∇u|)
∂ f
∂|∇u|
∇u
|∇u|
= 0
( 4 )
The natural boundary conditions are
∂ F
∂ f (|∇u|)
∂ f
∂|∇u|
∇u
|∇u|
· n
S
= 0 or
∂ F
∂ f (|∇u|)
∂ f
∂|∇u|
S
= 0
( 5 )
10.7 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(a, b, ∇|a × b|)dV .
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂ a
· δa +
∂ F
∂ b
· δb +
∂ F
∂∇|a × b|
· δ∇|a × b|
dV
(1)
Putting c =
∂ F
∂∇|a×b|
, the variation of the last term in Eq. (1) can be written as
c · δ∇|a × b| = c · ∇δ|a × b| = ∇ · (cδ|a × b|) − (∇ · c)δ|a × b|