CDS 140b Spring 2014 Homework 2
| R. Murray, D. MacMartin | Issued: 9 Feb 2014 (Wed) |
| CDS 140b, Spring 2014 | Due: 16 Feb 2014 (Wed) |
__MATHJAX__
Note: In the upper left hand corner of the second page of your homework set, please put the number of hours that you spent on this homework set (including reading).
- A planar pendulum (in the $x$-$z$ plane) of mass $m$ and length $\ell$ hangs from a support point that moves according to $x=a\cos (\omega t)$. Find the Lagrangian, the Hamiltonian, and write the first-order equations of motion for the pendulum.
- Perko, Section 3.3, problem 5.
Show that
<amsmath>\aligned \dot x &=y+y(x^2+y^2)\\ \dot y &=x-x(x^2+y^2)
\endaligned</amsmath>is a Hamiltonian system with $4H(x,y)=(x^2+y^2)^2-2(x^2-y^2)$. Show that $dH/dt=0$ along solution curves of this system and therefore that solution curves of this system are given by
<amsmath> (x^2+y^2)^2-2(x^2-y^2)=C
</amsmath>Show that the origin is a saddle for this system and that $(\pm 1,0)$ are centers for this system. (Note the symmetry with respect to the $x$-axis.) Sketch the two homoclinic orbits corresponding to $C=0$ and sketch the phase portrait for this system, noting the occurrence of a compound separatrix cycle.
- Perko, Section 3.6, problem 4.
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Consider the stability of the Lagrange points (with some simplifying steps).
With the mass of the sun as $1-\mu$ and planet as $\mu$, then in the rotating coordinate system the Hamiltonian is given by
<amsmath> H=\frac{(p_x+y)^2+(p_y-x)^2}{2}+\Omega(x,y)
</amsmath>where the potential function
<amsmath> \Omega(x,y)=\frac{x^2+y^2}{2}+\frac{1-\mu}{r_1}+\frac{\mu}{r_2}+\frac{\mu(1-\mu)}{2}
</amsmath>with $r_1=\sqrt{(x+\mu)^2+y^2}$ and $r_2=\sqrt{(x-1+\mu)^2+y^2}$. Show that the equilibrium points are given by the critical points of the (messy) function $\Omega$. (This leads to 5 solutions, for L1 through L5, with L1, L2, and L3 collinear, i.e., y=0. If you are interested, the collinear solutions are not too difficult to solve for numerically for some $\mu$.) To explore the linearized dynamics it is sufficient to retain only quadratic terms in $H$ (why?). For the collinear Lagrange points this leads to
<amsmath> H=\frac{(p_x+y)^2+(p_y-x)^2}{2}-ax^2+by^2
</amsmath>for $a>0$ and $b>0$. E.g., for the L1 point in the Earth-Jupiter system then a=9.892, b=3.446. Describe the linearized dynamics about this Lagrange point; are periodic orbits stable?
