CoCalc -- Lab09_slides.ipynb

¹⁶⁰⁷ views

Kernel: SageMath (stable)

Finding and classifying equilibrium points

Steps:

Find equilibrium point
Get the Jacobian
Test the eigenvalues

Solve Equilibrium Points for Continuous Time System

Example: Shark Tuna model:

$S' = 0.01ST-0.2S$

$T' = 0.05T-0.01ST$

In [3]:

_=var('S,T')
S_prime = 0.01*S*T-0.2*S
T_prime = 0.05*T-0.01*S*T

sol = solve([S_prime==0,T_prime==0],(S,T))
show(sol)

Out[3]:

How to retrieve the result?

In [4]:

sol

Out[4]:

[[S == 0, T == 0], [S == 5, T == 20]]

In [5]:

sol[0]

Out[5]:

[S == 0, T == 0]

In [7]:

sol[0][0]
sol[0][1]

Out[7]:

T == 0

In [9]:

sol[0][1].rhs()
sol[0][1].rhs()

Out[9]:

0

In [16]:

t = srange(0,100,0.1)
sol = desolve_odeint([S_prime,T_prime],ics =[4,18],times=t,dvars=[S,T])

fig1 = plot_vector_field((S_prime,T_prime),(S,3,7),(T,12,25),title='S=5,T=20')
fig2= point((5,20),color='red')
fig3 = list_plot(zip(sol[:,0],sol[:,1]),axes_labels=['S','R'])
show(fig1+fig2+fig3)

j = jacobian([S_prime,T_prime],[S,T])
j2 = j.subs({S:5,T:20})
show(j2.eigenvalues())

Out[16]:

In [17]:

t = srange(0,100,0.1)
sol = desolve_odeint([S_prime,T_prime],ics =[1,1],times=t,dvars=[S,T])

fig1 = plot_vector_field((S_prime,T_prime),(S,0,70),(T,0,100),title='S=0,T=0')
fig2= point((0,0),color='red')
fig3 = list_plot(zip(sol[:,0],sol[:,1]),axes_labels=['S','R'])
show(fig1+fig2+fig3)

j = jacobian([S_prime,T_prime],[S,T])
j1 = j.subs({S:0,T:0})
show(j1.eigenvalues())

Out[17]:

In [98]:

Out[98]:

Solve Does not Always Work

Model we are going to use today:

$H' = \frac{1}{1+G^n}-0.2H$

$P'=H-0.2P$

$G'=P-0.2G$

In [11]:

_ = var('H,P,G')
n=12
H_prime=1/(1+G^n)-0.2*H
P_prime=H-0.2*P
G_prime=P-0.2*G

sol = solve([H_prime,P_prime,G_prime],(H,P,G))
show(sol)

Out[11]:

None of them are real number!!! But there is at least one root!!!

What should we do when solve does not work?

Use: find_root!!!

find_root Can find "One" equilibrium point

For a "one" variable equation

Example: $x^7+cos(x)+2x-300$

In [90]:

_=var('x')

In [91]:

plot(x^7+cos(x)+2*x-300,(x,0,3.5))

Out[91]:

There is definitely a root between 2 and 3

find_root(fun,x_min,x_max,tolerlance)

In [100]:

find_root(x^7+cos(x)+2*x-300,2,3,rtol=0.001)

Out[100]:

2.2545915040920135

But there are 3 variables, not one!

$H' = \frac{1}{1+G^n}-0.2H$

$P'=H-0.2P$

$G'=P-0.2G$

However, if we only need to find equilibrium point, we only need to solve one variable!!!

$0 = \frac{1}{1+G^n}-0.2H$

$0=H-0.2P\implies 5H = P$

$0=P-0.2G\implies 5P=G$

$\implies 25H = G$

Finding equilibrium point is just solving a function of H!!!!

In [0]:

Finding and classifying equilibrium points

Solve Equilibrium Points for Continuous Time System

How to retrieve the result?

Solve Does not Always Work

Model we are going to use today:

None of them are real number!!! But there is at least one root!!!

What should we do when solve does not work?

Use: find_root!!!

find_root Can find "One" equilibrium point

For a "one" variable equation

Example: $x^7+cos(x)+2x-300$

But there are 3 variables, not one!

However, if we only need to find equilibrium point, we only need to solve one variable!!!

Product

Resources

Company

Finding and classifying equilibrium points

Solve Equilibrium Points for Continuous Time System

How to retrieve the result?

Solve Does not Always Work

Model we are going to use today:

None of them are real number!!! But there is at least one root!!!

What should we do when solve does not work?

Use: find_root!!!

find_root Can find "One" equilibrium point

For a "one" variable equation

Example: x7+cos(x)+2x−300x^7+cos(x)+2x-300x7+cos(x)+2x−300

But there are 3 variables, not one!

However, if we only need to find equilibrium point, we only need to solve one variable!!!

Example: $x^7+cos(x)+2x-300$