Appendix: All Solutions to the Problems
939
The natural boundary conditions are
∂ F
∂∇ · u∇u
S
= 0, u∇
∂ F
∂∇ · u∇u
· n
S
= 0
( 5 )
10.9 Find the Euler equations and natural boundary conditions of the functional
J =
V F(u, v, ∇ · v∇u)dV .
Solution The variation of the functional is
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂∇ · v∇u
δ∇ · v∇u
dV
(1)
Putting ϕ =
∂ F
∂∇·v∇u
, a = δv∇u = ∇uδv + v∇δu, 有
φ∇ · a = ∇ · ϕa − ∇ϕ · a = ∇ · ϕa − ∇ϕ · (∇uδv + v∇δu)
= ∇ · ϕa − ∇ϕ · ∇uδv − v∇ϕ · ∇δu
= ∇ · ϕa − ∇ϕ · ∇uδv − ∇ · (v∇ϕϕδu) + (∇ · v∇ϕ)∇ · ∇δu
= ∇ · ϕa − ∇ϕ · ∇uδv − ∇ · (v∇ϕϕδu) + ∇ · [(∇ · v∇ϕ)∇δu]
= −[∇(∇ · v∇ϕ)] · ∇δu
= ∇ · ϕa − ∇ϕ · ∇uδv − ∇ · (v∇ϕϕδu) + ∇ · [(∇ · v∇ϕ)∇δu]
− ∇ · {[∇(∇ · v∇ϕ)]δu} + [∇ · v∇ϕ)]δu
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
+ ∇ · v∇ϕ
δu +
∂ F
∂v
− ∇ϕ · ∇u
δv
dV
+
V
{∇ · ϕa − ∇ · (v∇ϕϕδu) + ∇ · [(∇ · v∇ϕ)∇δu] − ∇ · [∇(∇ · v∇ϕ)δu]}dV
(3)
The Euler equations of the functional are
∂ F
∂u
+ ∇ · v∇
∂ F
∂∇·v∇u
= 0
∂ F
∂v
− ∇u · ∇
∂ F
∂∇·v∇u
= 0
(4)
The natural boundary conditions are
ϕ| S = 0, v∇ϕ · n| S = 0, ∇ · v∇ϕ| S = 0, ∇(∇ · v∇ϕ) · n| S = 0,
(5)
10.10 Find the Euler equations and natural boundary conditions of the functional
J =
V F(u, v, ∇ · u∇v)dV .