Appendix: All Solutions to the Problems
943
10.14 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 a =
∂ F
∂∇×u∇u
, b = δu∇u = ∇uδu + u∇δu, 有
a · ∇ × b = −∇ · (a × b) + ∇ × a · (∇uδu + u∇δu)
= −∇ · (a × b) + (∇u · ∇ × a)δu + ∇ · (u∇ × aδu) − (∇ · u∇ × a)δu
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
+ ∇u · ∇ × a − ∇ · u∇ × a
δu
−∇ · (a × b) + ∇ · (u∇ × aδu)]dV
(3)
The Euler equation of the functional is
∂ F
∂u
+ ∇u · ∇ ×
∂ F
∂∇ × u∇u
− ∇ · u∇ ×
∂ F
∂∇ × u∇u
= 0
( 4 )
The natural boundary conditions are
n ×
∂ F
∂∇ × u∇u
S
= 0, u∇ ×
∂ F
∂∇ × u∇u
· n
S
= 0
( 5 )
10.15 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(u, a, ∇ × (∇u × a))dV .
Solution Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂∇ × (∇u × a)
· δ∇ × (∇u × a)
dV
(1)
Putting b =
∂ F
∂∇×(∇u×a)
, c = δ(∇u × a) = ∇δu × a + ∇u × δa, 有
b · δ∇ × (∇u × a) = b · ∇ × c = −∇ · (b × c) + (∇ × b) · c
= −∇ · (b × c) + ∇ × b · (∇δu × a + ∇u × δa)
= −∇ · (b × c) + (a × ∇ × b) · ∇δu + [(∇ × b) × ∇u] · δa
= −∇ · (b × c) + ∇ · [(a × ∇ × b)δu] − [∇ · (a × ∇ × b)]δu