Question #8212d

1 Answer
Dec 6, 2016

See below.

Explanation:

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)/2

by Legendre's transformation we have

H=dot q p -L

now dotq = (partialH)/(partial p)=p/a - b e^(-alpha t) q

solving for p we have

p=a e^(-t alpha) (b q + e^(alpha t) dot q)

substituting into L 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)^2

The movement equation associated to that lagrangian is obtainable making

d/dt((partial L)/(partial dot q))-(partial L)/(partial 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)