Question 9.3: (Lemos, 1979) Use the Hamilton-Jacobi theory to solve the eq......

Equation (9.33) is generated by the Lagrangian (see Problem 1.16)

L = e^{\lambda t}\left(\frac{m\dot{q}^{2}}{2}-\frac{m\omega^{2}}{2}{q}^{2}\right),  (9.34)

whose corresponding Hamiltonian is

H =e^{-\lambda t}\frac{p^{2}}{2m}+\frac{m\omega^{2}}{2}{q}^{2}e^{\lambda t}.  (9.35)

The Hamilton-Jacobi equation associated with this Hamiltonian is

\frac{e^{-\lambda t}}{2m}\left(\frac{\partial S}{\partial q}\right)^{2}+\frac{m\omega^{2}}{2}{q}^{2}e^{\lambda t}+ \frac{\partial S}{\partial t}=0.  (9.36)

Owing to the explicit time dependence of H, this last equation does not admit separation of the time variable in the form of the two previous examples 9.1 , 9.2(check it). However, the form of H suggests that the transformation

Q =qe^{{\lambda t}/{2}},   P =pe^{{-\lambda t}/{2}} (9.37)

is likely to simplify the Hamiltonian. It turns out that this transformation is canonical with a generating function F_{2}(q, P, t) =e^{{\lambda t}/{2}}qP, and the transformed Hamiltonian

K(Q, P, t) = H(q, p, t) +\frac{\partial F_{2}}{\partial t}=\frac{P^{2}}{2m}+\frac{m\omega^{2}}{2}Q^{2}+\frac{\lambda }{2}QP (9.38)

does not explicitly depend on time. Thus, the same technique of the two previous examples 9.1 , 9.2 is applicable to the system described by the new canonical variables (Q, P) and the transformed Hamiltonian K. The Hamilton-Jacobi equation associated with K is written

\frac{1}{2m} \left(\frac{\partial S}{\partial Q}\right)^{2} + \frac{m\omega^{2}}{2}Q^{2}+\frac{\lambda}{2} Q \frac{\partial S}{\partial Q}+ \frac{\partial S}{\partial t}=0,   (9.39)

which has a solution of the form

S = W(Q) − αt , (9.40)

where

\frac{1}{2m} \left(\frac{dW}{dQ}\right)^{2}+ \frac{m\omega^{2}}{2}Q^{2}+\frac{\lambda}{2} Q\frac{dW}{dQ}=\alpha.  (9.41)

Note that α is equal to the constant value of the Hamiltonian K. Solving this seconddegree algebraic equation for dW/dQ we obtain

\frac{dW}{dx}= -ax\pm \left[b^{2}-\left(1-a^{2}\right)x^{2}\right]^{{1}/{2}}  (9.42)

with x =\left(m \omega\right)^{{1}/{2}}Q, a = λ/2ω, b =\left({2\alpha}/{\omega}\right)^{{1}/{2}}. As far as solving (9.33) is concerned, the choice of sign in (9.42) is irrelevant – any complete integral will do. So we take

W = -\frac{ax^{2}}{2}+\int{\left[b^{2}-\left(1-a^{2}\right)x^{2}\right]^{{1}/{2}}dx}.  (9.43)

Let us consider the case a < 1 – that is, λ < 2ω. Defining γ =(1 − a²)^{{1}/{2}} there results

S = -\alpha t – \frac{ax^{2}}{2}+\int{\left(\frac{2\alpha }{\omega }-\gamma ^{2} x^{2}\right)}^{{1}/{2}}dx,  (9.44)

whence

\beta =\frac{\partial S}{\partial \alpha } =-t+\frac{1}{\omega } \int{\frac{dx}{\left(b^{2}-\gamma ^{2} x^{2}\right)^{{1}/{2}} } }=-t+\frac{1}{\omega \gamma }\sin ^{-1}\left(\frac{\gamma x}{b} \right).  (9.45)

Solving for x, returning to the variable Q and, finally, to the variable q, we find

q(t) = A e^{{-\lambda t}/{2}} \sin \left(\Omega t + \delta\right), \Omega= \left(\omega^{2} – {\lambda^{2}}/{4}\right)^{{1}/{2}},   (9.46)

where A and δ are constants determined by the initial conditions. Equation (9.46) is the usual solution for the damped oscillator in the weak damping case.