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====== IAM 950 HW2 ======

You job for HW2 is to compute all periodic orbits of the Lorenz ODE up to period 4 and produce a plot for each, like these.

{{:gibson:teaching:spring-2016:iam950:period2orbit.png?direct&400|}}{{:gibson:teaching:spring-2016:iam950:period3orbit.png?direct&400|}}

Use the classic parameter values $\sigma = 10, \beta = 8/3, \rho = 28$. Use whatever software system you like (Matlab, Julia, Mathematica, C, Fortran, whatever). Compute the orbits with these steps. 

**Step 1:** Plot the three unstable equilibria with dots. Compute and plot the stable manifold of the origin and a portion its unstable manifold as shown (magenta for stable manifold, blue and green for stable).  

**Step 2:** Plot the pseudo-Poincare section given by the curve $(x,y,z)$

\begin{eqnarray*}
y &= x - 0.01 x (x-8) (x+8) \\
z &= 13 + 0.188 x^2
\begin{eqnarray*}

for $-8 \leq x \leq 8$. This 1d curve lies pretty nearly on the attracting set. (I got it by guessing the right form and setting the constants by trial and error). Parameterize this curve numerically by $\eta$ so that $\eta$ is $\pm 1$ at the end points and zero at the center (simplest strategy: let $\eta = x/8$). 

**Step 3:** Let $\eta_{n+1} = f(\eta_n)$ be the return map of $\eta$ onto itself (hand-wavey, because trajectories starting on the $\eta$ curve don't actually come back exactly to $\eta$). Create a table of 
data of $\eta_n, \eta_{n+1}$ pairs by numerical integration of $(x,y,z)$ points starting on the $\eta$ curve. Make a plot of $f(\eta)$ using this numerical data.

**Step 4:** Fit a function $f(\eta)$ to the the numerical data from above. A piecewise polynomial with three or four terms should do. Plot this $f(\eta)$ on top of your numerical values. 

**Step 5:** Plot $f^n(\eta)$ versus $\eta$ along with the diagonal to get an estimate of the $\eta$ value of the period-n orbit. Then use the corresponding $x,y,z$ values as an initial guess for a periodic orbit. 

**Step 6:** Compute the periodic orbit, either approximately by adjusting $\eta$ until the trajectory very nearly lands back on its starting point, or better, by setting up a function $(x(t+T), y(t+T), z(t+T)) = \phi(x(t),y(t),z(t),T)$ which maps $x,y,z$ forward time $T$ under the Lorenz dynamics, using numerical integration, and then applying a nonlinear numerical solution method to find a zero of the function $\phi(x,y,z,T) - (x,y,z)$.
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