Given H(p,q,t)=p^2/(2 a) -
b q p e^(-alpha t) + (a b)/
2 q ^2 e^(-alpha t) (alpha + b e^(-alpha t)) + (k q^2)/2H(p,q,t)=p22a−bqpe−αt+ab2q2e−αt(α+be−αt)+kq22
by Legendre's transformation we have
H=dot q p -LH=.qp−L
now dotq = (partialH)/(partial p)=p/a - b e^(-alpha t) q.q=∂H∂p=pa−be−αtq
solving for pp we have
p=a e^(-t alpha) (b q + e^(alpha t) dot q)p=ae−tα(bq+eαt.q)
substituting into LL we have
L=a e^(-alpha t)
dot q (b q + e^(alpha t) dot q) -
1/2 a e^(-2 alpha t) (b q +
e^(alpha t) dot q)^2L=ae−αt.q(bq+eαt.q)−12ae−2αt(bq+eαt.q)2
The movement equation associated to that lagrangian is obtainable making
d/dt((partial L)/(partial dot q))-(partial L)/(partial q)=0ddt(∂L∂.q)−∂L∂q=0
so
a ddot q + k q=0 is the movement equation.
so an equivalent lagrangian is obtained as follows
a ddot q dot q+k dot q q= 0 integrating we obtain the movement total energy (kinetic 1/2a dot q^2 plus potential 1/2q^2 ) so the lagrangian is
L_(equ) = 1/2(a dot q^2-k q^2)
because L and L_(equ) have the same movement equations.
and also
H_(equ)=1/2(p^2/a+k q^2)
