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Appendix: All Solutions to the Problems
= ∇ · (cδ|a × b|) + (∇ · c)
a × b
|a × b|
· b × δa − (∇ · c)
a × b
|a × b|
· a × δb
= ∇ · (cδ|a × b|) + (∇ · c)
a × b
|a × b|
× b · δa − (∇ · c)
a × b
|a × b|
× a · δb (2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂ a
+ (∇ · c)
a × b × b
|a × b|
· δa
+
∂ F
∂ b
+
a × a × b
|a × b|
· δb + ∇ · (cδ|a × b|)
dV
(3)
The Euler equations of the functional are
∂ F
∂ a
+
a×b×b
|a×b| ∇ ·
∂ F
∂∇|a×b|
= 0
∂ F
∂ a
+
a×a×b
|a×b| ∇ ·
∂ F
∂∇|a×b|
= 0
(4)
The natural boundary condition is
∂ F
∂∇|a × b|
· n
S
= 0
( 5 )
10.8 Write out the Euler equations and natural boundary conditions of the functional
J =
V F(u, ∇ · u∇u)dV .
Solution The variation of the functional is
δ J =
V
∂ F
∂u
δu +
∂ F
∂∇ · u∇u
δ∇ · u∇u
dV
(1)
Putting ϕ =
∂ F
∂∇·u∇u
, a = δu∇u = ∇uδu + u∇δu, 有
ϕ∇ · a = ∇ · ϕa − ∇ϕ · (∇uδu + u∇δu) = ∇ · ϕa − ∇u · ∇ϕδu
− ∇ · (u∇ϕδu) + (∇ · u∇ϕ)δu
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
− ∇u · ∇ϕ + ∇ · u∇ϕ
δu + ∇ · ϕa − ∇ · (u∇ϕδu)
dV (3)
The Euler equation of the functional is
∂ F
∂u
− ∇u · ∇
∂ F
∂∇ · u∇u
+ ∇ · u∇
∂ F
∂∇ · u∇u
= 0
( 4 )