CoCalc Public Files2020-11-30-171452.ipynbOpen with one click!
Authors: Andy Hon, Nyla Jafri, Sewon(Timothy) Jeong
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Compute Environment: Ubuntu 20.04 (Default)
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var("E,V,R") Vprime= 100*E - 2*V Rprime= 0.272 - 0.00136*R - 0.00027*R*V Eprime= 0.00027*R*V - 0.33*E list_plot(Vprime, plotjoined=True)+list_plot(Rprime, plotjoined=True)+list_plot(Eprime, plotjoined=True)
--------------------------------------------------------------------------- NameError Traceback (most recent call last) <ipython-input-5-0eb44a0c940a> in <module> ----> 1 var("E,V,R") 2 Vprime= 100*E - 2*V 3 Rprime= 0.272 - 0.00136*R - 0.00027*R*V 4 Eprime= 0.00027*R*V - 0.33*E 5 list_plot(Vprime, plotjoined=True)+list_plot(Rprime, plotjoined=True)+list_plot(Eprime, plotjoined=True) NameError: name 'var' is not defined
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#Hello World
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var("E") var("V") var("R") Vprime= 100*E - 2*V Rprime= 0.272 - 0.00136*R - 0.00027*R*V Eprime= 0.00027*R*V - 0.33*E t=srange(0,100,0.1) sol_6=desolve_odeint([Vprime, Rprime, Eprime], ics=[0,0,0], dvars=[V,R,E], times=t) list_plot(list([sol_6[:,0], sol_6[:,1], sol_6[:,2]]))
--------------------------------------------------------------------------- NameError Traceback (most recent call last) <ipython-input-3-496fd129b02b> in <module> ----> 1 var("E") 2 var("V") 3 var("R") 4 Vprime= 100*E - 2*V 5 Rprime= 0.272 - 0.00136*R - 0.00027*R*V NameError: name 'var' is not defined
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Vprime= 100*E - 2*V Rprime= 0.272 - 0.00136*R - 0.00027*R*V Eprime= 0.00027*R*V - 0.33*E t=srange(0,100,0.1) sol_6=desolve_odeint([Vprime, Rprime, Eprime], ics=[0,0,0], dvars=[E,V,R], times=t) list_plot(list([sol_6[:,0], sol_6[:,1], sol_6[:,2]]))
--------------------------------------------------------------------------- NameError Traceback (most recent call last) <ipython-input-13-5df68e0800d3> in <module> ----> 1 var(V) 2 var(R) 3 Vprime= 100*E - 2*V 4 Rprime= 0.272 - 0.00136*R - 0.00027*R*V 5 Eprime= 0.00027*R*V - 0.33*E NameError: name 'var' is not defined
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#3 # importing library sympy from sympy import symbols, Eq, solve # defining symbols used in equations # or unknown variables V, R, E = symbols('V,R,E') # defining equations eq1 = Eq((100*E - 2*V), 0) print("Equation 1:") print(eq1) eq2 = Eq((0.272*R - 0.00136*R - 0.00027*V), 0) print("Equation 2") print(eq2) eq3= Eq((0.00027*V - 0.33*E), 0) print("Equation 3") print(eq3) # solving the equation print("Values of 3 unknown variable are as follows:") print(solve((eq1, eq2, eq3), (V,R,E)))
Equation 1: Eq(100*E - 2*V, 0) Equation 2 Eq(0.27064*R - 0.00027*V, 0) Equation 3 Eq(-0.33*E + 0.00027*V, 0) Values of 3 unknown variable are as follows: {E: 0.0, R: 0.0, V: 0.0}
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#In order to find the stability, I used an eigenvector and eigenvalues calculator to compute the eigenvector and eigenvalue of this 3x3 matrix. The results are λ_1≈-2.240 λ_2≈-0.337 λ_3≈0.247 for eigenvalues. v_1≈({{58552.061}, {-7073.759}, {1}}) v_2≈({{30.581}, {-24.566}, {1}}) v_3=({{1946.186*x_3}, {2136.363*x_3}, {x_3}}) --> These are the eigenvectors for the 3x3 matrix representing our model. #Concluding stability on 3x3 matrix equilibrium points: http://utkstair.org/clausius/docs/che505/pdf/ODE_stability.pdf #Based on the website, We would thus conclude that the equilibrium points of (0,0,0) is thereby unstable. There is at least 1 positive eigenvalue, which satisfies the condition for unstable equilibrium points that just needs at least one positive real part.
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