940
Appendix: All Solutions to the Problems
Solution The variation of the functional is
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂∇ · u∇v
δ∇ · u∇v
dV
(1)
Putting ϕ =
∂ F
∂∇·u∇v
, a = δδu∇v = ∇vδu + u∇δv, 有
ϕ∇ · a = ∇ · ϕa − ∇ϕ · a = ∇ · ϕa − ∇ϕ · (∇vδu + u∇δv)
= ∇ · ϕa − ∇ϕ · ∇v∇ · ∇δu − u∇ϕ · ∇δv
= ∇ · ϕa − ∇ · (∇ϕ · ∇v∇δu) + ∇(∇ϕ · ∇v) · ∇δu − ∇ · [((u∇ϕ)δv]
+ [∇ · ((u∇ϕ)]δv
= ∇ · ϕa − ∇ · (∇ϕ · ∇v∇δu) + ∇ · [∇(∇ϕ · ∇v)δu] − (∇ϕ · ∇v)δu
− ∇ · [((u∇ϕ)δv] + (∇ · u∇ϕ)δv
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
− ∇v · ∇ϕ
δu +
∂ F
∂v
+ ∇ · u∇ϕ
δv
dV
+
V
{∇ · ϕa − ∇ · (∇ϕ · ∇v∇δu) + ∇ · [∇(∇ϕ · ∇v)δu] − ∇ · [((u∇ϕ)δv]}dV
(3)
The Euler equations of the functional are
∂ F
∂u
− ∇v · ∇
∂ F
∂∇·u∇v
= 0
∂ F
∂v
+ ∇ · u∇
∂ F
∂∇·u∇v
= 0
(4)
The natural boundary conditions are
ϕ| S = 0, ∇v · ∇ ϕ| S = 0, ∇(∇v · ∇ ϕ) · n| S = 0, ,u∇ ϕ · n| S = 0
(5)
10.11 Find the Euler equations and natural boundary conditions of the functional
J =
V F(u, v, ,(vu))dV .
Solution The variation of the functional is
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂∂(vu)
δδ(vu)
dV
(1)
Putting ϕ =
∂ F
∂∂(vu)
, ψ = δ(vu) = vδu + uδv, 有
ϕ∇ · ∇ψ = ∇ · ϕ∇ψ − ∇ϕ · ∇ψ = ∇ · ϕ∇ψ − ∇ · ψ∇ϕ + ψψϕ
= ∇ · ϕ∇ψ − ∇ · ψ∇ϕ + ϕ(vδu + uδv)