932
Appendix: All Solutions to the Problems
H = y
F y − F = −
x 2 + y 2
1 + y 2
= −
x 2 + y 2 − p 2
The Euler equations of the Hamiltonian form are
dy
dx
=
p
x 2 + y 2 − p 2
,
d p
dx
= −
∂ H
∂ y
=
y
x 2 + y 2 − p 2
9.22 Find the Hamiltonian form of Euler equations of the functional J =
π
0 (y
2
1 + y
2
2 + y
2
2 )dx.
Solution Putting F = y
2
1 + y
2
2 + y
2
2 , then p 1 = F y 1
= 2y 1
, p 2 = F y 2
= 2y 2
,
y 1
=
p 1
2
, y 2
=
p 2
2
, the Hamiltonian function of the given functional is
H 1 = y 1
F y 1
− F = y
2
1 − y
2
2 − y
2
2 =
p
2
1
4
− y
2
2 −
p
2
2
4
H 2 = y 2
F y 2
− F = y
2
2 − y
2
2 − y
2
1 =
p
2
2
4
− y
2
2 −
p
2
1
4
The Euler equations of the Hamiltonian form are
dy 1
dx
=
p 1
2
,
d p 1
dx
= −
∂ H 1
∂ y 1
= 0
dy 2
dx
=
p 2
2
,
d p 2
dx
= −
∂ H 2
∂ y 2
= 2y 2
9.23 Find the Hamiltonian form of Euler equations of the functional J =
x 2
x 0
(x
2
+ y 1 y
2
1 + y 2 y
2
2 )dx.
Solution Putting F = x
2
+ y 1 y
2
1 + y 2 y
2
2 , then p 1 = F y 1
= 2y 1 y 1
, p 2 = F y 2
=
2y 2 y 2
, y 1
=
p 1
2y 1
, y 2
=
p 2
2y 2
, the Hamiltonian function of the given functional is
H 1 = y 1
F y 1
− F = y 1 y
2
1 − x
2
− y 2 y
2
2 =
p
2
1
4y 1
− x
2
−
p
2
2
4y 2
H 2 = y 2
F y 2
− F = y 2 y
2
2 − x
2
− y 1 y
2
1 =
p
2
2
4y 2
− x
2
−
p
2
1
4y 1
The Euler equations of the Hamiltonian form are
⎧
⎨
⎩
dy 1
dx
=
p 1
2y 1
,
d p 1
dx
= −
∂ H 1
∂ y 1
=
p
2
1
4y
2
1
dy 2
dx
=
p 2
2y 2
,
d p 2
dx
= −
∂ H 2
∂ y 2
=
p
2
2
4y
2
2
9.24 Find the Hamiltonian form of Euler equations of the functional J =
x 1
x 0
2x y 1 − y
2
1 +
y
3
2
3
dx.
Solution Putting F = 2x y 1 − y
2
1 +
y
3
2
3
, then p 1 = F y 1
= −2y 1
, p 2 = F y 2
= y
2
2 ,
y 1
= −
p 1
2
, y 2
= ±
√ p 2 , the Hamiltonian function of the given functional is