lecture10

Phase Plane Analysis : Phase Plane Analysis Dongsup Kim Department of Biosystems, KAIST Fall, 2004


Phase-plane analysis, I : Phase-plane analysis, I Two differential equations and solutions, Eliminating the variable t, Phase plane Trajectories Phase portraits phase plane


Predator-Prey Model : Predator-Prey Model Canadian lynx and snowshoe hare pelt-trading records of the Hudson Bay Company over almost a century


Phase-plane analysis, II : Phase-plane analysis, II Suppose tow coupled differential equations, The steady state solution satisfy Theorem: trajectories can cross one another only at steady-state points. Proof: in phase space, Except for the case where f=g=0 (steady-state points), dy/dx (slope) is uniquely specified. If trajectories cross at (x0, y0), there would be more than one slope at that point. Therefore, no trajectory crossing except at steady-state points.


Phase-plane analysis, III : Phase-plane analysis, III Phase portraits near steady states: behavior at the immediate neighborhood of the steady-state points. New variables, Assuming x’ and y’ are small perturbations, linearize the eq. in matrix notation, Eliminating y’,


Phase-plane analysis, IV : Phase-plane analysis, IV 2> 4: real roots Case I: >0, <0  1>0, 2>0: unstable node Case II: >0, >0  1<0, 2<0: stable node Case III: <0  1>0, 2<0: saddle point 2< 4: complex roots Case IV: >0  p<0: stable spiral Case V: <0  p>0: unstable spiral Case VI: =0  p=0: center Unstable node Stable node Saddle point


Phase-plane analysis, V : Phase-plane analysis, V Proposition: The long-term behavior of a system of linear equations is such that each variable either goes to zero or becomes arbitrarily large with the exception of a few special cases that essentially never occur. If an actual nonclosed system exhibits robust stable behavior, then a model reflecting this behavior must be nonlinear. Predator-Prey model


Phase-plane analysis, VI : Phase-plane analysis, VI Example: Steady-state solutions: Linearization: Stability: (0,0): <0  saddle point (v,s): >0, >0  stable node


Phase-plane analysis, VII : Phase-plane analysis, VII Qualitative behavior of the phase plane Find the steady-state points, and examine their stability Plot horizontal nullclines where g(x,y)=0, and indicate by small line segments that trajectories have horizontal tangents along these curves Plot vertical nullclines where f(x,y)=0, and indicate by small line segments that trajectories have vertical tangents along these curves Indicate by arrows whether the horizontal tangents point to the left (dx/dt<0) or to the right (dx/dt>0), and whether the vertical tangents point upward (dy/dt>0) or downward (dy/dt<0) Place further small arrows to indicate the direction of trajectories along the axes x=0 and y=0 Try to combine all the foregoing information into a consistent picture


Phase-plane analysis, VIII : Phase-plane analysis, VIII S=v/(1+v) ds/dt<0 ds/dt>0 ds/dt=0 dv/dt=0 v s  ln(/) dv/dt=0


Predator-Prey Model, I : Predator-Prey Model, I Canadian lynx and snowshoe hare pelt-trading records of the Hudson Bay Company over almost a century


Predator-Prey Model, II : Predator-Prey Model, II Also called Lotka-Volterra Model There are two species interacting: a prey x and predator y In the absence of y, x exhibits exponential growth, In the absence of x, y dies out exponentially, When x and y are present, x decreases and y increases at a rate xy (law of mass action) Steady-state points: Linearization: Stability at steady-state points: (0,0): a=a, b=0, c=0, d=-d  =-a+d,  = -ad < 0: saddle point (d/c, a/b): a=0, b= -bd/c, c=ca/b, d=0  =0, = ad > 0: center


Predator-Prey Model, III : Predator-Prey Model, III New variables: u = x – d/c, v = y – a/b in phase plane,


Predator-Prey Model, IV : Predator-Prey Model, IV Lotka-Volterra Model in chemical kinetics R + X → 2X, υa = κ1·[R]·[X] X + Y → 2Y, υb = κ2·[X]·[Y] Y → P, υc = κ3·[Y] = υ(a) − υ(b) = κ1·[R]·[X] − κ2·[X]·[Y] = υ(b) − υ(c) = κ2·[X]·[Y] − κ3·[Y]


Systems of 1st-order homogeneous linear DE with constant coefficients, I : Systems of 1st-order homogeneous linear DE with constant coefficients, I Suppose a system of linear differential equations, In matrix notation, Try ; eigenvalue problem (i, vi): eigenvalue-eigenvector of matrix A one solution: General solution:


Systems of 1st-order homogeneous linear DE with constant coefficients,II : Systems of 1st-order homogeneous linear DE with constant coefficients,II Example: Eigenvalues: