Solution

Hamilton’s formulation
Lagrange Hamilton

• L H
• L=L(t, 𝑞𝐾 , 𝑞𝐾ሶ ) H=H(t, 𝑞𝐾 , 𝑃𝐾 )
𝜕𝐿
where ( 𝑃𝐾 = )
𝜕𝑞𝐾ሶ

• L=T-V H=T+V (has physical meaning)

• Lagrange equations: Hamilton equations:
𝑑 𝜕𝐿 𝜕𝐿 𝜕𝐻 𝜕𝐻
− =0 𝑞𝐾ሶ = , 𝑃𝐾ሶ = −
𝑑𝑡 𝜕 𝑞𝐾ሶ 𝜕𝑞𝐾 𝜕𝑃𝐾 𝜕𝑞𝐾
K=1,2,………….n
How to solve any problem using Hamilton methode:

• Step 1: Determine generalized coordinates

• Step 2: find kinetic energy , potential energy and Lagrange (L)
• Step 3: formulate Hamilton (H)
• Step 4: Apply to Hamilton’s equations
Example:
In spring mass system find the equation of motion
using Hamilton procedure.
Solution
Step 1 : x
Step 2 :
1
𝑇= 𝑚 𝑣2 (𝑣 = 𝑥)ሶ
2

1
𝑇 = 𝑚 𝑥ሶ 2
2
• Step 3: Hamiltonian (H)
1 1
𝐻 = 𝑇 + 𝑉 = 𝑚𝑥ሶ + 𝐾 𝑥 2
2
2 2
𝜕𝐿
𝑃𝐾 =
𝜕 𝑞𝐾ሶ

𝜕𝐿
𝑃𝑥 = = 𝑚 𝑥ሶ
𝜕𝑥ሶ
𝑃𝑥
𝑥ሶ =
𝑚

1 𝑃𝑥 2 1 2 1 𝑃𝑥 2 1
𝐻 =𝑇+𝑉 = 𝑚 2 + 𝐾𝑥 = + 𝐾 𝑥2
2 𝑚 2 2 𝑚 2
• Step 4: Hamilton’s equations

𝜕𝐻 𝜕𝐻
𝑞𝐾ሶ = 𝑃𝐾ሶ = −
𝜕𝑃𝐾 𝜕𝑞𝐾

𝜕𝐻 𝜕𝐻
𝑥ሶ = 𝑃𝑥ሶ = −
𝜕𝑃𝑥 𝜕𝑥

𝑃𝑥
𝑥ሶ = 1 𝑃𝑥ሶ = −𝐾𝑥 (2)
𝑚
By diff. equ. (1) with respect to time
𝑃𝑥ሶ
𝑥ሷ =
𝑚
 ሶ
𝑃𝑥 𝐾𝑥
𝑥ሷ = = -
𝑚 𝑚

𝐾
𝑥=
ሷ - 𝑥
𝑚
Simple harmonic motion

𝑥ሷ = −𝑤 2 𝑥
It’s solution is
𝑥=A cos(wt+B)
• Example:

If lagrangian function for dynamic system is

1 1 2 2
𝐿= ( 𝑞ሶ 12 +𝑞12 𝑞ሶ 22 ) − 𝐾 𝑞1
2 2
Where K constant
Find
• A)Hamiltonion function
• B)Prove that 𝑞12 = 𝐴 𝑠𝑖𝑛 2𝐾𝑡 + 𝐵 + 𝐶
Where A, B , C constants
 Solution
𝐻 =𝑇+𝑉

1 2 1 2 2
𝐻= 𝑞ሶ 1 +𝑞12 𝑞ሶ 22 + 𝐾 𝑞1
2 2
α=1,2
𝜕𝐿
𝑃1 = = 𝑞ሶ 1
𝜕𝑞ሶ 1

𝜕𝐿 𝑃2
𝑃2 = = 𝑞ሶ 2 𝑞12 𝑞ሶ 2 = 2
𝜕 𝑞ሶ 2 𝑞1
 2
1 2 2
𝑃2 1 2 2
𝐻= 𝑃1 +𝑞1 4 + 𝐾 𝑞1
2 𝑞1 2

1 𝑃22 1 2 2
𝐻= 𝑃12 + + 𝐾 𝑞1 Hamiltonian function
2 𝑞12 2
Hamiltonian equations
𝜕𝐻 𝜕𝐻
𝑞ሶ 𝛼 = 𝑃𝛼ሶ = −
𝜕𝑃𝛼 𝜕𝑞𝛼
α=1,2

𝜕𝐻
𝑞ሶ 1 = = 𝑃1 1
𝜕𝑃1

𝜕𝐻 𝑃2
𝑞ሶ 2 = = 2
𝜕𝑃2 𝑞12
 𝜕𝐻 𝑃22
𝑃1ሶ = − = − − 3 + 𝐾 2 𝑞1 (3)
𝜕𝑞1 𝑞1

𝜕𝐻
𝑃2ሶ = − =0
𝜕𝑞2

𝑃2 = constant = 𝐶1 (4)
• By diff. eq. (1)

𝑞1ሷ = 𝑃1ሶ
From 𝑃1ሶ in (3)
𝑃22
𝑞1ሷ = 3 − 𝐾 2 𝑞1
𝑞1

𝐶12
𝑞1ሷ = 3 − 𝐾 2 𝑞1
𝑞1
Diff. eq. second order 𝑞1ሷ = 𝑓(𝑞1 )
‫ ونكامل‬2𝑞ሶ 1 ‫حلها نضرب الطرفين في‬
 2𝑞ሶ 1 𝐶12
2𝑞ሶ 1 𝑞1ሷ = − 2𝑞ሶ 1 𝐾 2 𝑞1
𝑞13
Integrate with respect to time
2
−𝐶 1
𝑞ሶ 12 = 2 − 𝐾 2 𝑞12 + 𝐶2
𝑞1
‫𝑞 ونكامل‬12 ‫نضرب الطرفين في‬

𝑞ሶ 12 𝑞12 = −𝐶12 − 𝐾 2 𝑞14 + 𝐶2 𝑞12

𝐶2 2 𝐶12
𝑞ሶ 12 𝑞12 = −𝐾 2 (𝑞14 − 2 𝑞1 + 2 )
𝐾 𝐾
𝑎 𝑎2
𝑥 2 + 𝑎𝑥 + 𝑏 = (𝑥 + )2 +b −
2 4
 𝐶2 2 𝐶12
𝑞ሶ 12 𝑞12 = −𝐾 2 (𝑞14 − 2 𝑞1 + 2 )
𝐾 𝐾
𝑎 𝑎2
𝑥 2 + 𝑎𝑥 + 𝑏 = (𝑥 + )2 +b −
2 4

2
𝐶2 𝐶12 𝐶22
𝑞ሶ 12 𝑞12 = 𝐾 2 [− 𝑞12 − − 2+ ]
2𝐾 2 𝐾 4𝐾 4

2
𝐶2
𝑞ሶ 12 𝑞12 2
= 𝐾 [− 𝑞12 − + 𝐴2 ]
2𝐾 2
2
𝐶2
𝑞ሶ 12 𝑞12 = 𝐾 2 [𝐴2 − 𝑞12 − ]
2𝐾 2

2
𝐶2
𝑞ሶ 1 𝑞1 = 𝐾 𝐴2 − 𝑞12 −
2𝐾 2
 𝑞1 𝑑𝑞1
න 2 = න 𝐾𝑑𝑡
𝐶2
𝐴2 − 𝑞12 −
2𝐾 2
𝑓′ (𝑥)𝑑𝑥 𝑓(𝑥)
‫𝐴 ׬‬2 −𝑓 2 = 𝑠𝑖𝑛−1 +𝑐
𝑥 𝐴

2𝑞1 𝑑𝑞1
න 2 = න 2𝐾𝑑𝑡
𝐶2
𝐴2 − 𝑞12 −
2𝐾 2
𝐶
𝑞12 − 22
𝑠𝑖𝑛−1 ( 2𝐾
)=2Kt + 𝐶3
𝐴
𝐶
𝑞12 − 22
2𝐾 = sin(2𝐾𝑡 + 𝐶 )
3
𝐴
 𝐶2
𝑞12 − 2
= 𝐴sin(2𝐾𝑡 + 𝐶3 )
2𝐾

𝐶2
𝑞12 = 𝐴 sin 2𝐾𝑡 + 𝐶3 +
2𝐾 2