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Appendix: All Solutions to the Problems
+ [∇ · (a∇ · b)]δu − (∇u∇ · b) · δa
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
+ ∇ · (a∇ · b)
δu +
∂ F
∂ a
− ∇u∇ · b
· δa
+∇ · (bϕ) − ∇ · [(a∇ · b)δu]}dV
(3)
The Euler equations of the functional are
∂ F
∂u
+ ∇ ·
a∇ ·
∂ F
∂∇(∇u·a)
= 0
∂ F
∂ a
− ∇u∇ ·
∂ F
∂∇(∇u·a)
= 0
(4)
The natural boundary conditions are
∂ F
∂∇(∇u · a)
· n
S
= 0, a · n∇ ·
∂ F
∂∇(∇u · a)
S
= 0
( 5 )
10.5 Derive the Euler equations and natural boundary conditions of the functional
J =
V F(u, a, (a · ∇)u)dV . Where, a and ∇ are the inner product relation, it is not
a to take the right divergence.
Solution Taking the variation to the functional, there is
δ J =
V
∂ F
∂u
δu +
∂ F
∂ a
· δa +
∂ F
∂(a · ∇)u
δ(a · ∇)u
dV
(1)
Putting ϕ =
∂ F
∂(a·∇)u
, the variation of the last term in Eq. (1) can be written as
ϕδ(a · ∇)u = ϕδ(a · ∇u) = ϕ(∇u · δa + a · ∇δu) = ϕ∇u · δa
+ ∇ · (ϕaδu) − (∇ · ϕa)δu
(2)
Substituting Eq. (2) into Eq. (1), we get
δ J =
V
∂ F
∂u
− ∇ · ϕa
δu +
∂ F
∂ a
+ ϕ∇u
· δa + ∇ · (ϕaδu)
dV
(3)
The Euler equations of the functional are
∂ F
∂u
− ∇ ·
∂ F
∂(a·∇)u
a = 0
∂ F
∂ a
+
∂ F
∂(a·∇)u
∇u = 0
(4)
The natural boundary conditions are