Phase Plane Analysis Phase plane analysis is one of the most important techniques for studying the behavior of nonlinear systems, since there is usually no analytical solution for a nonlinear system.
Background The response characteristics (relative speed of response) for unforced systems were dependent on the initial conditions Eigenvalue/eigenvector analysis allowed us to predict the fast and slow (or stable and unstable) initial conditions Another way of obtaining a feel for the effect of initial conditions is to use a phase-plane plot A phase-plane plot for a two-state variable system consists of curves of one state variable versus the other state variable (x 1 (t) vs. x 2 (t)), where each curve is based on a different initial condition
Phase-Plane Behavior of Linear Systems Example 1: A Stable Equilibrium Point (Node Sink) Consider the system = x 1 1 = 4x 2 2 x ( t) = x (0) e 1 1 x ( t) = x (0) e 2 2 Plot x 1 and x 2 as a function of time for a large number of initial conditions t 4t The solutions converge to (0,0) for all initial conditions The point (0,0) is a stable equilibrium point for the system stable node
Phase-Plane Behavior of Linear Systems Example 2: An Unstable Equilibrium Point (Saddle) Consider the system = x 1 1 = 4x 2 2 x ( t) = x (0) e 1 1 x ( t) = x (0) e 2 2 Plot x 1 and x 2 as a function of time for a large number of initial conditions t 4t If the initial condition for x 2 was 0, then the trajectory reached the origin. Otherwise, the solution will always leave the origin. The point (0,0) is an unstable equilibrium point for the system saddle point The x 1 axis represents a stable subspace and the x 2 axis represents an unstable subspace
Phase-Plane Behavior of Linear Systems Example 3: Another Saddle Point Problem Consider the system Eigenvalues Eigenvectors = 2x + x 1 1 2 = 2x x 2 1 2 x ɺ = Ax λ = 1.5616 λ = 2.5616 1 2 0.2703 0.8719 ξ1 = ξ2 = 0.9628 0.4896 (stable subspace) 2 1 A = 2 1 (unstable subspace)
Phase-Plane Behavior of Linear Systems Example 4: Unstable Focus (Spiral Source) Consider the system Eigenvalues = x + 2x 1 1 2 = 2x + x 2 1 2 x ɺ = Ax 1 2 A = 2 1 = 1± 2i (unstable system because the real part is positive) unstable spiral focus
Phase-Plane Behavior of Linear Systems Example 5: Center Consider the system Eigenvalues = x x 1 1 2 = 4x + x 2 1 2 = ±1.7321i x ɺ = Ax A 1 1 = 4 1 Since the real part of the eigenvaluesis zero, there is a periodic solution, resulting a phase-plane plot where the equilibrium point is a center
Generalization of Phase-Plane Behavior Consider second-order systems Eigenvaluesof A Phase-plane behavior resulting from and Sinks (stable nodes) : Saddles (unstable) : a11 a12 x ɺ = Ax Jacobian matrix A = a21 a 22 Sources (unstable nodes) : ( λi A) det = 0 2 ( ) ( a )( a ) a a det λi A = λ λ = λ tr( A) λ + det( A) = 0 λ = ( ) 2 11 22 12 21 tr( A) ± tr( A) 4det( A) λ 2 2 λ1 2 Spirals : 1 and are complex conjugate. If Re λ 1 < 0 then stable, if Re λ > 0 then unstable λ ( λ ) ( λ ) Re < 0 and Re < 0 1 2 ( λ ) ( λ ) Re < 0 and Re > 0 1 2 ( λ1 ) ( λ2 ) λ ( ) ( ) 1 Re > 0 and Re > 0
Dynamic behavior diagram for second-order linear systems The eigenvaluewill be complex if ( A ) 2 4det( A) > tr( )
Phase-plane behavior as a function of eigenvalue location
Nonlinear Systems Nonlinear systems will often have the same general phaseplane behavior as the model linearizedabout the equilibrium (steady-state) point, when the system is close to that particular equilibrium point Nonlinear systems often have multiple steady-state solutions. Phase-plane analysis of nonlinear systems provides an understanding of which steady-state solution that a particular set of initial conditions will converge to The local behavior(close to one of the steady-state solutions) can be understood from a linear phase-plane analysis of the particular steady-state solution (equilibrium point)
Consider the system Nonlinear System Example dz dt dz dt 1 2 ( z 1) = z + 2 1 ( z 3) = z + 1 2 Two steady-state (equilibrium) solutions Equilibrium 1: trivial z = 0 z = 0 Equilibrium 2: nontrivial 1s 2s Linearizing the system, we have z = 1 z = 3 1s 2s A z z + 1 2s 1s = z2s + 3 z 1s
Equilibrium 1 (Trivial) (0,0) 0 1 A = 3 0 λ = 3 λ = 3 1 2 This equilibrium point is a saddle point 0.5 0.5 Stable eigenvector ξ Unstable eigenvector 1 = 0.866 ξ2 = 0.866 The phase-plane of the linearized model around equilibrium point 1 Linearized model : 0 1 = 3 0 x where x = z - z s
Equilibrium 2 (Nontrivial) (-1,-3) 3 0 A = 0 1 λ = 3 λ = 1 1 2 This equilibrium point is a stable node 1 Fast stable eigenvector ξ1 = Slow stable eigenvector 0 ξ2 The phase-plane of the linearized model around equilibrium point 2 0 = 1
The phase-plane diagram of the nonlinear model The linearizedmodels capture the behavior of the nonlinear model when close to one of the equilibrium points
Exercise: Interacting tanks Two interacting tank in series with outlet flowratebeing function of the square root of tank height Parameter values 2.5 2.5 ft 5 ft 2 2 1 = 2.5 2 = 1 = 5ft 2 = 10ft R min R 6 min A A Input variable F = 5 ft 3 /min Steady-state height values : h 1s = 10, h 2s = 6 Perform a phase-plane analysis and discuss your results Linearized model: >> [V, D] = eig(a) A 0.125 0.125 = 0.0625 0.1042 V = -0.8466-0.7827 0.5323-0.6224 D = -0.2036 0 0-0.0256
Program Reference % Nonlinear Model for h10 = 8 : 12 for h20 = 4 : 8 y0=[h10 h20]; [t,y]=ode45('phaseplane', [0 200], y0); plot(y(:,1),y(:,2), y(1,1),y(1,2),'o'); hold on; xlabel('h_1'); ylabel('h_2'); title('phase-plane Plot'); end end % Linearized Model for h10 = -2 : 2 for h20 = -2:2 y0=[h10 h20]; [t,y]=ode45('phaseplane_l', [0 200], y0); y(:,1) = y(:,1)+10; y(:,2) = y(:,2)+6; plot(y(:,1),y(:,2),'r', y(1,1),y(1,2),'o'); hold on; end end h 2 function dy = phaseplane(t,y) A1=5; A2=10; R1=2.5; R2=5/sqrt(6); F=5; dy(1) = F/A1-R1/A1*sqrt(y(1)-y(2)); dy(2) = R1/A2*sqrt(y(1)-y(2))-R2/A2*sqrt(y(2)); dy = dy'; function dy = phaseplane_l(t,y) A = [-0.125 0.125; 0.0625-0.1042]; dy = A*y; 8 7.5 7 6.5 6 5.5 5 4.5 Phase-Plane Plot 4 8 8.5 9 9.5 10 10.5 11 11.5 12 h 1
Exercise: Bioreactor A model for a bioreactor dx dt dx dt 1 ( µ 0.4) = ( x ) x 2 = 4 1 2 0.4 0.53x2 µ = 0.12 + x 2 1 µ x 0.4 (growth rate) x 1 : biomass concentration x 2 : substrate concentration Perform a phase-plane analysis.