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Hamilton’s Method Examples
Example 1
Finding the governing equation for the 2-dof pendulum (Fig. ↓) using Cartesian coordinates, x, y:
Kinetic energy of the system is
(1) KE = (1)/(2)m(x’(t)2 + y’(t)2)
The potential energy of the system is due to gravity and spring. The potential energy due to the gravity field is:
(2) PEm = mg(y0 − y(t))
where y0 is y coordinate at θ = 0.
The potential energy due to springs is given by:
(3) PEs = (1)/(2)k(√(x(t)2 + y(t)2) − y0)2
Using these with Hamilton’s formulation yields
(4) t2⌠⌡t1δKEdt  = t2⌠⌡t1δ(1)/(2)m(x’(t)2 + y’(t)2)dt      = t2⌠⌡t1(mx’(t)δx’ + my’(t)δy’)dt      = mx’(t)δx’|t2t1 + my’(t)δy’|t2t1 − t2⌠⌡t1(mx’’(t)δx + my’’(t)δy)dt
When deriving the governing equations, which are ordinary differential equations, one can neglect the first two terms, which are boundary terms. For a rigorous reasoning, refer to Meirovitch [\citenumMeirovitch1997]. Similarly neglecting the boundary terms in calculating the other terms of Hamilton’s formulation:
(5)  − t2⌠⌡t1δPEsdt  =  − t2⌠⌡t1δ(1)/(2)k(√(x(t)2 + y(t)2) − y0)2dt      = t2⌠⌡t1( − kx(t)Aδx − ky(t)Aδy)dt, 
where
(6) A = (√(x2(t) + y2(t)) − y0)/(√(x2(t) + y2(t))).
(7)  − t2⌠⌡t1δPEmdt  =  − t2⌠⌡t1δmg(y0 − y(t))dt      = t2⌠⌡t1mgδydt
The equations of motion using x, y as generalized coordinates can be found by substituting eq. 4↑, 5↑ and 7↑ into eq. ↓. There is no non-conservative force in this problem; therefore, δWnc = 0. Finally
(8) t2⌠⌡t1(δKE − δPE + δWnc)dt =      t2⌠⌡t1( − mx’’(t)δx − my’’(t)δy − kx(t)Aδx − ky(t)Aδy + mgδy)dt
The coefficients of δx and δy should be zero independently. So the governing equations beomce
(9) mx’’(t) + kx(t)A  = 0     my’’(t) + ky(t)A − mg  = 0
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