Appendix: All Solutions to the Problems
941
= ∇ · ϕ∇ψ − ∇ · ψ∇ϕ + uϕδv + ∇ · vϕ∇δu − ∇(vϕ) · ∇δu
= ∇ · ϕ∇ψ − ∇ · ψ∇ϕ + uϕδv + ∇ · vϕ∇δu
− ∇ · [∇(vϕ)δu] + (v
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
+ (vϕ)
δu +
∂ F
∂v
+ ((u)((ϕ)
δv
dV
+
V
{∇ · ϕ∇ψ − ∇ · ψ∇ϕ + ∇ · vϕ∇δu − ∇ · [∇(vϕ)δu]}dV (3)
The Euler equations of the functional are
∂ F
∂u
+
v
∂ F
∂∂(vu)
= 0
∂ F
∂v
+ ((u))
∂ F
∂∂(vu)
= 0
(4)
The natural boundary conditions are
ϕ| S = 0, ∇ϕ · n| S = 0, vϕ| S = 0, ∇(vϕ) · n| S = 0
( 5 )
10.12 Derive the Euler equations and natural boundary conditions of the functional
J =
V F
u, v, a, ∇ ·
u
v
a
dV .
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂ a
· δa +
∂ F
∂∇ · (ua/v)
δ∇ · (ua/v)
dV
(1)
Putting ϕ =
∂ F
∂∇·(ua/v)
, b = δ
ua
v
, the integral of the last term in Eq. (1) can be
written as
ϕδ∇ ·
ua
v
= ϕ∇ · b = ∇ · ϕb − ∇ϕ · b = ∇ · ϕb − ∇ϕ ·
a
v
δu +
u
v
δa −
ua
v 2 δv
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
− ∇ϕ ·
a
v
δu +
∂ F
∂v
+ ∇ϕ ·
ua
v 2
δv
+
∂ F
∂ a
−
u∇ϕ
v
· δa + ∇ · ϕb
dV
(3)
The Euler equations of the functional are