# Problem of the Week #28 - December 10th, 2012

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#### Chris L T521

##### Well-known member
Staff member
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:

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:

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

Remember to read the POTW submission guidelines to find out how to submit your answers!

#### Chris L T521

##### Well-known member
Staff member
No one took a bite at this week's question. Here's my work.

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.

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