Appendix: All Solutions to the Problems
933
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H 11 = (y 1
F y1 − F)
y2 =
√ p2
=
−2xy 1 − y 2
1 −
y 3
2
3
y2 =
√ p2
= −2xy 1 −
p 2
1
4 −
p 3
2
3
H 12 = (y 1
F y1 − F)
y2 =−
√ p2
=
−2xy 1 − y 2
1 −
y 3
2
3
y2 =−
√ p2
= −2xy 1 −
p 2
1
4 +
p 3
2
3
H 21 = (y 2
F y2 − F)
y2 =
√ p2
=
−2xy 1 + y 2
1 +
2y 3
2
3
y2 =
√ p2
= −2xy 1 +
p 2
1
4 +
2
p 3
2
3
H 22 = (y 2
F y2 − F)
y2 =−
√ p2
=
−2xy 1 + y 2
1 +
2y 3
2
3
y2 =−
√ p2
= −2xy 1 +
p 2
1
4 −
2
p 3
2
3
The Euler equations of the Hamiltonian form are
dy 1
dx
= −
p 1
2
,
d p 1
dx
= −
∂ H 11
∂ y 1
= 2x,
dy 1
dx
= −
p 1
2
,
d p 1
dx
= −
∂ H 12
∂ y 1
= 2x
dy 2
dx
=
√ p 2 ,
d p 2
dx
= −
∂ H 21
∂ y 2
= 0,
dy 2
dx
= −
√ p 2 ,
d p 2
dx
= −
∂ H 22
∂ y 2
= 0
Chapter 10: Solutions to the Problems in Variational
Problems of Functionals with Vector, Tensor
and Hamiltonian Operators
10.1 Find the Euler equations and natural boundary conditions of the functional
J =
V F(u, v, ∇ × (u∇v))dV .
Solution According to Eq. (1.3.98), there is ∇ × (u∇v) = ∇u × ∇v, the original
functional can be written as
J =
V
F(u, v, ∇u × ∇v)dV
(1)
Taking the variation to the functional, we get
δ J =
V
∂ F
∂u
δu +
∂ F
∂v
δv +
∂ F
∂(∇u × ∇v)
· δ(∇u × ∇v)
dV
(2)
Putting c =
∂ F
∂(∇u×∇v)
, the variation of the last term in Eq. (1) can be written as
c · δ(∇u × ∇v) = c · (∇δu × ∇v + ∇u × ∇δv) = (∇v × c) · ∇δu
+ (c × ∇u) · ∇δv = ∇ · [(∇v × c)δu]
− [∇ · (∇v × c)]δu + ∇ · [(c × ∇u)δv]
− [∇ · (c × ∇u)]δv
(3)
Substituting Eq. (3) into Eq. (2), we get