580
9 Variational Principles in Mechanics and Their Applications
At the moment, the variables t, q i , p i are called the Hamiltonian variable, where
p i is called the generalized momentum, ˙
q i is the generalized velocity, and every pair
of q i and p i are also collectively called the regular variable or conjugate variable.
The Euler equations corresponding to the functional (9.7.10) are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
˙
p i =
d p i
dt
= −
∂ H
∂q i
˙
q i =
dq i
dt
=
∂ H
∂ p i
(i = 1, 2, . . . , n)
(9.7.13)
Equations (9.7.13) are the differential equations of motion taking canonical variables q i and p i as variable, they are called the Hamilton(’s) canonical equations, Hamilton(’s) equations, Hamiltonian equations or canonical equations.
The Hamiltonian variables q i and p i sometimes are also called the coordinates of
Hamilton(’s) equation or coordinates of Hamiltonian equation.
It follows from Eqs. (9.7.13) that the canonical equations are the 2n first order
differential equations, and every pair of equations are symmetric, this makes it be
in form the simpler than the second order Euler equation or Lagrange equation.
Moreover because the canonical equation comes from the Lagrange equation after
transformation, they are equivalent equation.
When mechanical problems are solved by the canonical equations, the key step
is to write out the Hamiltonian function expressed by the Hamiltonian variables q i
and p i .
Example 9.7.1 Find the Hamiltonian form of the Euler equations of the functional
J [y 1 , y 2 ] =
π
0 (2y 1 y 2 − 2y
2
1 + y
2
1 − y
2
2 )dx.
Solution Putting F = 2y 1 y 2 − 2y
2
1 + y
2
1 − y
2
2 , then there are p 1 = F y
1
= 2y
1 ,
p 2 = F y
2
= −2y
2 , the determinant of partial derivatives is
F y
1 y
1
F y
1 y
2
F y
2 y
1
F y
2 y
2
=
2 0
0 −2
= −4 = 0
From the above equation to work out y
1 and y
2 , there are y
1 =
p 1
2
, y
2 = −
p 2
2
.
Thus the Hamiltonian function of the given functional
H = y
1 F y
1
+ y
2 F y
2
− F = y
2
1 − y
2
2 − 2y 1 y 2 + 2y
2
1 =
p
2
1
4
−
p
2
2
4
− 2y 1 y 2 + 2y
2
1
According to Eqs. (9.1.9), (9.1.7) and (9.1.8), the Hamiltonian form of Euler
equations is
⎧
⎪ ⎨
⎪ ⎩
dy 1
dx
=
p 1
2
,
dy 2
dx
=
p 2
2
d p 1
dx
= −4y 1 + 2y 2 ,
d p 2
dx
= 2y 1
9 Variational Principles in Mechanics and Their Applications
At the moment, the variables t, q i , p i are called the Hamiltonian variable, where
p i is called the generalized momentum, ˙
q i is the generalized velocity, and every pair
of q i and p i are also collectively called the regular variable or conjugate variable.
The Euler equations corresponding to the functional (9.7.10) are
⎧
⎪ ⎪ ⎨
⎪ ⎪ ⎩
˙
p i =
d p i
dt
= −
∂ H
∂q i
˙
q i =
dq i
dt
=
∂ H
∂ p i
(i = 1, 2, . . . , n)
(9.7.13)
Equations (9.7.13) are the differential equations of motion taking canonical variables q i and p i as variable, they are called the Hamilton(’s) canonical equations, Hamilton(’s) equations, Hamiltonian equations or canonical equations.
The Hamiltonian variables q i and p i sometimes are also called the coordinates of
Hamilton(’s) equation or coordinates of Hamiltonian equation.
It follows from Eqs. (9.7.13) that the canonical equations are the 2n first order
differential equations, and every pair of equations are symmetric, this makes it be
in form the simpler than the second order Euler equation or Lagrange equation.
Moreover because the canonical equation comes from the Lagrange equation after
transformation, they are equivalent equation.
When mechanical problems are solved by the canonical equations, the key step
is to write out the Hamiltonian function expressed by the Hamiltonian variables q i
and p i .
Example 9.7.1 Find the Hamiltonian form of the Euler equations of the functional
J [y 1 , y 2 ] =
π
0 (2y 1 y 2 − 2y
2
1 + y
2
1 − y
2
2 )dx.
Solution Putting F = 2y 1 y 2 − 2y
2
1 + y
2
1 − y
2
2 , then there are p 1 = F y
1
= 2y
1 ,
p 2 = F y
2
= −2y
2 , the determinant of partial derivatives is
F y
1 y
1
F y
1 y
2
F y
2 y
1
F y
2 y
2
=
2 0
0 −2
= −4 = 0
From the above equation to work out y
1 and y
2 , there are y
1 =
p 1
2
, y
2 = −
p 2
2
.
Thus the Hamiltonian function of the given functional
H = y
1 F y
1
+ y
2 F y
2
− F = y
2
1 − y
2
2 − 2y 1 y 2 + 2y
2
1 =
p
2
1
4
−
p
2
2
4
− 2y 1 y 2 + 2y
2
1
According to Eqs. (9.1.9), (9.1.7) and (9.1.8), the Hamiltonian form of Euler
equations is
⎧
⎪ ⎨
⎪ ⎩
dy 1
dx
=
p 1
2
,
dy 2
dx
=
p 2
2
d p 1
dx
= −4y 1 + 2y 2 ,
d p 2
dx
= 2y 1