Lab10_slides.ipynb

Kernel: SageMath (stable)

Chaos

We have learned Two Stable System Behaviors

1. Converge to Equilibrium

2. Stable Limit Cycle

Is there any other type?

Chaos System$\begin{cases} \text{Bounded Output}\\ \text{Not Periodic} \end{cases}$

Romeo Juliet System (Continuous Time Example)

$R' = R + 0.1J$

$J' = -J-T$

$T' = 0.1-cT+RT$

Let $c = 14,R(0)=5, J(0)=5, T(0)=1$

_=var('J,R,T')
c = 14
jprime = R+0.1*J
rprime = -J-T
tprime = 0.1-c*T+R*T

time = srange(0,100,0.1)
sol = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [5,5,1])

list_plot(zip(time,sol[:,0]),plotjoined=True,axes_labels=['time','J'])

list_plot(zip(time,sol[:,1]),plotjoined=True,axes_labels=['time','R'])

list_plot(zip(time,sol[:,2]),plotjoined=True,axes_labels=['time','T'])

It do not converge to any equilibrium point, nor stable limit cycle

However, it is still bounded!

list_plot(sol,plotjoined=True)

Initial Values Matters a Lot

$R' = R + 0.1J$

$J' = -J-T$

$T' = 0.1-cT+RT$

time = srange(0,500,0.1)
sol1 = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [5,5,1])
sol2 = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [5.1,5.1,1.1])

fig1 = list_plot(zip(time[0:1000],sol1[0:1000,0]),plotjoined=True)+list_plot(zip(time[0:1000],sol2[0:1000,0]),plotjoined=True,color='red')
fig2 = list_plot(zip(time[4000:-1],sol1[4000:-1,0]),plotjoined=True)+list_plot(zip(time[4000:-1],sol2[4000:-1,0]),plotjoined=True,color='red')

show(fig1)
show(fig2)

list_plot(sol1,plotjoined=True,color='blue')+list_plot(sol2,plotjoined=True,color='red')

time = srange(0,100,0.1)
sol = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [5,8,2])
fig1 = list_plot(sol,plotjoined=True)
sol = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [1,4,3])
fig2 = list_plot(sol,plotjoined=True,color='red')

sol = desolve_odeint([jprime,rprime,tprime],dvars = [J,R,T],times = time,ics = [1.3,-1.6,50])
fig3 = list_plot(sol,plotjoined=True,color='green')
show(fig1+fig2+fig3)

Review: Jacobian

$\begin{bmatrix}d g(X,Y)\\dh(X,Y)\end{bmatrix} = \begin{bmatrix}\frac{\partial g(X,Y)}{\partial X}&\frac{\partial g(X,Y)}{\partial Y}\\\frac{\partial h(X,Y)}{\partial X}&\frac{\partial h(X,Y)}{\partial Y}\end{bmatrix}\begin{bmatrix}dx\\dy\end{bmatrix}$

If you are at equilibrium point $(x_0,y_0)$

$g(x_0,y_0)=0 ,h(x_0,y_0)=0$

For a point nearby $(x_0,y_0)$, e.g., $(x_1,y_1)$

$g(x_1,y_1) = dg(x,y)$

$h(x_1,y_1) = dh(x,y)$

$\begin{bmatrix}g(x_1,y_1)\\h(x_1,y_1)\end{bmatrix} = \begin{bmatrix}\frac{\partial g(X,Y)}{\partial X}&\frac{\partial g(X,Y)}{\partial Y}\\\frac{\partial h(X,Y)}{\partial X}&\frac{\partial h(X,Y)}{\partial Y}\end{bmatrix}_{X=x_0,Y=y_0}\begin{bmatrix}(x_1-x_0)$y_1-y_0)\end{bmatrix}$

Partial Derivative

$g(X,Y,Z)=(X^2-Y^2)(4X+2Z)-\frac{YZ^3}{X+Z^3}$

Find: $\frac{\partial g}{\partial Z}$

Two Rules:

Product Rule: $\frac{d(ab)}{dx} = \frac{da}{dx}b + a\frac{db}{dx}$

$\frac{dg}{dZ} = (X^2-Y^2)\frac{d(4X+2Z)}{dZ}-\frac{d(YZ^3(X+Z^3)^{-1})}{dZ}$

$=2(X^2-Y^2)-\frac{d(YZ^3)}{dZ}(X+Z^3)^{-1}-YZ^3\frac{d(X+Z^3)^{-1}}{dZ}$

But you can also do in this way:

$\frac{d\frac{num}{den}}{dx}$

$= \frac{\frac{d(num)}{dx}den - num\frac{d(den)}{dx}}{den^2}$

$\frac{dg}{dZ} = 2(X^2-Y^2)- \frac{3YZ^2(X+Z^3)-YZ^33Z^2}{(X+Z^3)^2}$

$=2(X^2-Y^2)-\frac{3XYZ^2}{(X+Z^3)^2}$

Optimization

Suppose the number of sparrows is defined by two parameters:

$f(x,y)=9x^2+6y^2-4x^3-2y^2-3x^2y^2$

At which point in the state space will the population be stable?

_= var('x,y,z')
f = 9*x^2+6*y^2-4*x^3-2*y^2-3*x^2*y^2
fx = diff(f,x)
fy = diff(f,y)

plot3d(-1*f+10^8,(x,-100,100),(y,-100,100))  #I flip the z, so maximum is the lowest point

plot_vector_field((fx,fy),(x,-100,100),(y,-100,100))

plot3d(f,(x,-100,100),(y,-100,100))

Highest point: Local maximum

$X' =0, Y' = 0$

Eigen values of Hessian Matrix <0