Love Dynamics

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Love Dynamics, Love Triangles, & Chaos: 

Love Dynamics, Love Triangles, andamp; Chaos DON’T TRY THIS AT HOME!

Overview: 

Overview The Numerical Method: Runge-Kutta-Fehlberg Modeling Stability Simulations Special Solutions

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Assumption: f(t, y) is smooth enough IVP: y’ = f(t, y) a ≤ t ≤ b y(a) = y0 Taylor Polynomial about ti y(ti+1) = y(ti) + h y’(ti) +…+ hn/n! y(n)(ti) + O(hn+i) y’ = f(t, y), y’’ = f’(t, y),…, y(k) = f(k-1)(t, y)

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Then, y(ti+1) = y(ti) + h f(ti, y(ti)) +… + hn/n! f(n-1)(ti, y(ti)) + O(hn+i) OR yi+1 = yi + h f(ti, yi) +… + hn/n! f(n-1)(ti, yi) + O(hn+i)

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Taylor’s Method, Order 2 w0 = y0 wi+1 = wi + h f(ti, wi) + h2/2 f’(ti, wi) Good: Truncation Error is O(hn) Bad: Computation of Derivatives (complicated and time consuming)

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Runge Kutta Methods: Truncation Error is O(hn) No Computation of Derivatives

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Illustration: Taylor’s Order 2 Needs (1) f(t, y) + h/2 f’(t, y) = f(t, y) + h/2 [ ft(t, y) + fy(t, y) y’] = f(t, y) + h/2 ft(t, y) + h/2 fy(t, y) f(t, y) Taylor’s (again!) c f(t+a, y+b) = c f(t,y) + a c ft(t,y) + b c fy(t,y) + c R(*)

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Matching Coefficients: c = 1 c a = h/2 c b = h/2 which gives c = 1 a = h/2 b = h/2 f(t, y)

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Then (1) can be written: f( t + h/2, y + h/2 f(t, y) ) RK Order 2: w0 = y0 k1 = h f(ti, wi) k2 = h f(ti + h/2, wi + k1/2) wi+1 = wi + k2

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg RK Order 4: w0 = y0 k1 = h f(ti, wi) k2 = h f(ti + h/2, wi + k1/2) k3 = h f(ti + h/2, wi + k2/2) k4 = h f(ti+1, wi + k3) wi+1 = wi + 1/6 ( k1 + 2 k2 + 2 k3 + k4 )

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Further Improvement: Control the Error (predefined tolerance) Minimize the Number of Mesh Points

Runge-Kutta-Fehlberg: 

Runge-Kutta-Fehlberg Runge-Kutta-Fehlberg Compute RK Order 4 approximation, wi+1 Compute RK Order 5 approximation, ŵi+1 τi+1 (q h) = q4/h (ŵi+1 - wi+1) ≤ TOL Take q ≤ ( h TOL / | ŵi+1 - wi+1 |) ¼ Result: ODE45 Command in MATLAB

Modeling: 

Modeling Linear Systems: ů = A*u Solution: u(t) = u(0)exp(At) Predetermined No Chaos Well Documented Non-Linear Systems: ů = f(u, λ)

Stability: 

Stability Stability Linear Re(λ) andlt; 0 implies Asymptotic Stability Non-Linear Linearize Local Stability

Some Models: 

Some Models R’ = aR + bJ J’ = cR + dJ Romantic Styles Eager Beaver: a andgt; 0, b andgt; 0 Narcissistic Nerd: a andgt; 0, b andlt; 0 Cautious Lover: a andlt; 0, b andgt; 0 Hermit: a andlt; 0, b andlt; 0

Some Models: 

Some Models R’ = aR + bJ J’ = cR + dJ Simple Linear Model Out of Touch with One’s Own Feelings: a = d = 0 Fire and Ice: c = -b, d = -a Peas in a Pod: c = b, a = d

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Some Models: Love Triangles: 

Some Models: Love Triangles Rj’ = aRj + b(J-G) J’ = cRj + dJ Rg’ = aRg + b(G-J) G’ = eRg+ fG

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Some Models: Nonlinear: 

Some Models: Nonlinear R’ = aR + bJ(1-|J|) J’ = cR(1-|R|) + dJ Simple Nonlinear Model Eager Beaver: c = d = 1 Hermit: a = b = -2

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Some Models: Nonlinear Love Triangles: 

Some Models: Nonlinear Love Triangles Rj’ = aRj + b(J - G)(1 - |J – G|) J’ = cRj(1 - |Rj|)+ dJ Rg’ = aRg + b(G - J )(1 - |G – J|) G’ = eRg(1 - |Rg|)+ fG Love Triangle Nonlinear Model Cautious Lovers: a = -3, b = 4; e = 2, f = -1 Narcissistic Nerd: c = -7, d = 2

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Special Solutions: 

Special Solutions Chaos Nonlinear Unpredictable Non-stable Periodic Orbits Out of Touch: Nerd plus Lover Fire and Ice: Nerd plus Lover (|a| andlt; |b|)

Special Solutions: 

Special Solutions Strange Attractors – Nonlinear Love Triangle: Romeo: Lover Juliet: Nerd Guinevere: Lover

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Stability (cont.): 

Stability (cont.) Hyperbolic Equilibrium Point: An equilibrium point is hyperbolic if the Jacobian has no eigenvalues with the real part equal to zero (stability is based on the real part)

Stability (cont.): 

Stability (cont.) Hartman-Grobman Theorem Let ů=A*u be the linearization of ů=f(u). If A is hyperbolic, then both systems are equivalent around the equilibrium point.

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