Question 8212d

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 $\dot{q} = \frac{\partial H}{\partial p} = \frac{p}{a} - b {e}^{- \alpha t} q$

solving for $p$ we have

$p = a {e}^{- t \alpha} \left(b q + {e}^{\alpha t} \dot{q}\right)$

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

$\frac{d}{\mathrm{dt}} \left(\frac{\partial L}{\partial \dot{q}}\right) - \frac{\partial L}{\partial q} = 0$

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 $\frac{1}{2} a {\dot{q}}^{2}$ plus potential $\frac{1}{2} {q}^{2}$ ) so the lagrangian is

${L}_{e q u} = \frac{1}{2} \left(a {\dot{q}}^{2} - k {q}^{2}\right)$

because $L$ and ${L}_{e q u}$ have the same movement equations.

and also

${H}_{e q u} = \frac{1}{2} \left({p}^{2} / a + k {q}^{2}\right)$