Can You Solve the Hamilton-Jacobi Equation for the Linear Harmonic Oscillator?

[Chris L T521]

Here's this week's problem (more along the lines of physics).

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Problem: Show that the function\[S(p,q,\alpha)=\frac{m\omega}{2}(q^2+\alpha^2)\cot(\omega t) - m\omega q\alpha\csc(\omega t)\]
is a solution of the Hamilton-Jacobi equation for Hamilton's principal function for the linear harmonic oscillator with Hamiltonian
\[H(p,q)=\frac{p^2+m^2\omega^2q^2}{2m}.\]
Show that this function generates a correct solution to the motion of the harmonic oscillator.

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Hints and Suggestions:

Spoiler

We first note that $S$ is a solution to the Hamilton-Jacobi equation if
\[H\left(q,\frac{\partial S}{\partial q}\right)+\frac{\partial S}{\partial t} = 0.\]

To show that $S$ generates a correct solution to the motion of the harmonic oscillator, you need to recover $q$ and $p$ from $S$. To do this, you need to use the transformation equations

\[\left\{\begin{aligned}Q &= \beta = \frac{\partial S}{\partial \alpha}\\ p &= \frac{\partial S}{\partial q}.\end{aligned}\right.\]

Last but not least, the following trigonometric identity may come in handy:

\[A\sin(\omega t) + B\cos(\omega t)= \sqrt{A^2+B^2}\sin(\omega t+\varphi)\text{ with }\varphi = \text{sgn}\,(B)\arccos\left(\frac{A}{\sqrt{A^2+B^2}}\right)\]

 

[Chris L T521]

No one took a bite at this week's question. Here's my work.

Spoiler

We first show that $S$ is a solution to the Hamilton-Jacobi equation\[H\left(q,\frac{\partial S}{\partial q}\right)+\frac{\partial S}{\partial t}=0.\]
Since $S$ is given by
\[S=\frac{m\omega}{2}(q^2+\alpha^2)\cot(\omega t) - m\omega q\alpha\csc(\omega t),\]
we see that
\[\frac{\partial S}{\partial q} = m\omega q\cot(\omega t) - m\omega\alpha\csc(\omega t)\]
and
\[\frac{\partial S}{\partial t} = -\frac{m\omega^2}{2}(q^2+\alpha^2)\csc^2(\omega t) + m\omega^2 q\alpha\csc(\omega t)\cot(\omega t).\]
It now follows that
\[\begin{aligned}H\left(q,\frac{\partial S}{\partial q}\right) &= \frac{1}{2m}\left[\left(\frac{\partial S}{\partial q}\right)^2 +m^2\omega^2q^2\right]\\ &= \frac{1}{2m}\left[\left(m\omega q\cot(\omega t) - m\omega\alpha\csc(\omega t)\right)^2 +m^2\omega^2q^2\right]\\ &= \frac{1}{2m}\left[m^2\omega^2q^2\left(\cot^2(\omega t)+1\right)-2m^2\omega^2q\alpha\csc(\omega t)\cot(\omega t)+m^2\omega^2\alpha^2\csc^2(\omega t)\right]\\ &= \frac{1}{2m}\left(m^2\omega^2(q^2+\alpha^2)\csc^2(\omega t)-2m^2\omega^2q\alpha\csc(\omega t)\cot(\omega t)\right)\\ &= \frac{m\omega^2}{2}(q^2+\alpha^2)\csc^2(\omega t) - m\omega^2q\alpha\csc(\omega t)\cot(\omega t)\\ &= -\frac{\partial S}{\partial t}.\end{aligned}\]
Therefore, $H\left(q,\dfrac{\partial S}{\partial q}\right) + \dfrac{\partial S}{\partial t}=0$.

To show that $S$ generates a correct solution solution to the motion of the harmonic oscillator, we need to recover $p$ and $q$ from $S$. To do that, we need to use the transformation equations
\[\left\{\begin{aligned}Q &= \frac{\partial S}{\partial \alpha}=\beta\\ p &= \frac{\partial S}{\partial q}.\end{aligned}\right.\]
With this, we see that
\[Q = \beta = \frac{\partial S}{\partial\alpha} = m\omega\alpha\cot(\omega t)-m\omega q\csc(\omega t).\]
This implies that
\[q=\alpha\cos(\omega t)-\frac{\beta}{m\omega}\sin(\omega t)=\frac{1}{m\omega}\left(m\omega\alpha\cos(\omega t)-\beta\sin(\omega t)\right).~~~~~(1)\]
Using the trigonometric identity
\[A\sin(\omega t) + B\cos(\omega t)= \sqrt{A^2+B^2}\sin(\omega t+\varphi)~~~~~(2)\]
with
\[\varphi = \text{sgn}\,(B)\arccos\left(\frac{A}{\sqrt{A^2+B^2}}\right)~~~~~(3)\]
we see that

\[\boxed{q=\dfrac{\sqrt{m^2\omega^2\alpha^2+\beta^2}}{m\omega} \sin(\omega t + \varphi_1) ;\qquad \varphi_1 = \text{sgn}\,(\alpha) \arccos\left(-\dfrac{\beta}{\sqrt{m^2 \omega^2 \alpha^2 + \beta^2}}\right)}. ~~~~~(4)\]

Now, we note that
\[p = \frac{\partial S}{\partial q} = m\omega q\cot(\omega t) - m\omega\alpha\csc(\omega t).~~~~~(5)\]
Substituting (1) into (5) yields
\[\begin{aligned}p &= \left(m\omega\alpha\cos(\omega t)-\beta\sin(\omega t)\right)\cot(\omega t) - m\omega a\csc(\omega t)\\ &= m\omega\alpha\cos^2(\omega t)\csc(\omega t) - \beta\cos(\omega t)-m\omega\alpha\csc(\omega t)\\ &= -m\omega\alpha(1-\cos^2(\omega t))\csc(\omega t)-\beta\cos(\omega t)\\ &= -m\omega\alpha\sin(\omega t)-\beta\cos(\omega t).\end{aligned}~~~~~(6)\]
Applying the trigonometric identity in (2) to (6), we see that
\[\boxed{p = \sqrt{m^2\omega^2\alpha^2+\beta^2} \sin(\omega t+\varphi_2);\qquad \varphi_2=\text{sgn}\,(-\beta)\arccos \left(-\dfrac{m\omega\alpha}{\sqrt{m^2\omega^2\alpha^2+ \beta^2}}\right)}. ~~~~~(7)\]
Thus, (4) and (7) together form a correct solution to the motion of the harmonic oscillator.