Math 266: Phase Plane Portrait

Transcription

1 Math 266: Phase Plane Portrait Long Jin Purdue, Spring 2018

2 Review: Phase line for an autonomous equation For a single autonomous equation y = f (y) we used a phase line to illustrate the equilibrium solutions y(t) y 0 where f (y 0 ) = 0 as well as other solutions. Figure: Left: Graph of f ; Right: Phase line for y = f (y).

3 Phase plane portrait for autonomous planar system For an autonomous planar system x = f (x, y), y = g(x, y) we can use a similar two-dimensional picture called phase plane portrait to illustrate all the solutions.

4 Phase plane portrait for autonomous planar system For an autonomous planar system x = f (x, y), y = g(x, y) we can use a similar two-dimensional picture called phase plane portrait to illustrate all the solutions. Phase plane portrait We plot y against x instead of t. Every solution is represented by a curve which is a parametric curve with parameter t.

5 Phase plane portrait for autonomous planar system For an autonomous planar system x = f (x, y), y = g(x, y) we can use a similar two-dimensional picture called phase plane portrait to illustrate all the solutions. Phase plane portrait We plot y against x instead of t. Every solution is represented by a curve which is a parametric curve with parameter t. An equilibrium solution is represented by a point (x 0, y 0 ) (called critical point) satisfying f (x, y) = g(x, y) = 0.

6 Direction field We also have a direction field associated to the autonomous planar system x = f (x, y), y = g(x, y) in which we plot the vector (f (x, y), g(x, y)) at each point (x, y) in the plane.

7 Direction field We also have a direction field associated to the autonomous planar system x = f (x, y), y = g(x, y) in which we plot the vector (f (x, y), g(x, y)) at each point (x, y) in the plane. If we regard y as a function of x, then it satisfies the equation dy dx = dy/dt dx/dt = g(x, y) f (x, y). The direction field of the planar system and the one of this first order equation is different by a scaling of each vector.

8 Linear systems We have discussed the simplest situation: a homogeneous linear system of two equations with constant coefficients. ( ) ( ) x x(t) a b = Ax, x(t) =, A = y(t) c d or more explicitly x = ax + by, y = cx + dy.

9 Linear systems We have discussed the simplest situation: a homogeneous linear system of two equations with constant coefficients. ( ) ( ) x x(t) a b = Ax, x(t) =, A = y(t) c d or more explicitly x = ax + by, y = cx + dy. We have seen how to solve this system by finding eigenvalues and eigenvectors of A. Alternatively, the trajectories in the phase plane can also be obtained by solving the first order equation dy dx = cx + dy ax + by which is homogeneous: the right-hand side is a function of y/x.

10 Linear examples For a single autonomous equation, the equilibrium solutions (critical points) have a simple classification: asymptotically stable/unstable/semistable.

11 Linear examples For a single autonomous equation, the equilibrium solutions (critical points) have a simple classification: asymptotically stable/unstable/semistable. For planar systems, things become more complicated. We have seen some linear examples: ( ) 1 1 A =, 0 is a saddle point (always unstable). 4 1 ( 3 2 A = ), 0 is a node, which is asymptotically stable. 2 2 A = stable. A = ( ) , 0 is a spiral point, which is asymptotically 2 ( ), 0 is an improper node, which is unstable.

12 Linear examples For a single autonomous equation, the equilibrium solutions (critical points) have a simple classification: asymptotically stable/unstable/semistable. For planar systems, things become more complicated. We have seen some linear examples: ( ) 1 1 A =, 0 is a saddle point (always unstable). 4 1 ( 3 2 A = ), 0 is a node, which is asymptotically stable. 2 2 A = stable. A = ( ) , 0 is a spiral point, which is asymptotically 2 ( ), 0 is an improper node, which is unstable. We can use the software pplane to plot the direction fields and some solutions.

13 Nonlinear examples For nonlinear systems, the phase plane portrait may be much more complicated. Let us see some examples using pplane. Pendulum: lθ + γ m θ + g sin θ = 0. Predator-Prey model: x = x(a αx), y = y( c + γy). Competing species: x = x(a αx βy), y = y(b γx δy). Van der Pol equation: x = µ(x 1 3 x 3 y), y = 1 µ x. Duffing equation of nonlinear spring: mu + γu + ku + lu 3 = 0.

14 Equilibrium solutions in linear system Now we discuss in detail the different situation of the equilibrium solutions in linear systems: x = Ax. If A is invertible, i.e. det(a) 0, then 0 is not an eigenvalue of A and 0 is the only critical point. If A is not invertible, i.e. det(a) = 0, then there are infinity many critical points forming a line passing through 0, (if A 0) or the whole plane (if A = 0).

15 Equilibrium solutions in linear system Now we discuss in detail the different situation of the equilibrium solutions in linear systems: x = Ax. If A is invertible, i.e. det(a) 0, then 0 is not an eigenvalue of A and 0 is the only critical point. If A is not invertible, i.e. det(a) = 0, then there are infinity many critical points forming a line passing through 0, (if A 0) or the whole plane (if A = 0). ( ) a b We are mostly interested in the first case. Write A =, c d then the eigenvalues are the roots of the characteristic equation det(a ri ) = r 2 tr(a)r + det(a) = 0, tr(a) = a + d, det(a) = ad bc. We can classify the behavior of the equilibrium solution by the properties of eigenvalues r 1 and r 2.

16 Classification of equilibrium solutions: Real eigenvalues When the eigenvalues r 1 and r 2 are real (and nonzero), we have the following classification: If r 1 and r 2 are of opposite sign (one positive, one negative), then 0 is a saddle point. Saddle points are always unstable.

17 Classification of equilibrium solutions: Real eigenvalues When the eigenvalues r 1 and r 2 are real (and nonzero), we have the following classification: If r 1 and r 2 are of opposite sign (one positive, one negative), then 0 is a saddle point. Saddle points are always unstable. If r 1 and r 2 are of the same sign, then 0 is a node, which is proper if r 1 r 2 and improper or star if r 1 = r 2.

18 Classification of equilibrium solutions: Real eigenvalues When the eigenvalues r 1 and r 2 are real (and nonzero), we have the following classification: If r 1 and r 2 are of opposite sign (one positive, one negative), then 0 is a saddle point. Saddle points are always unstable. If r 1 and r 2 are of the same sign, then 0 is a node, which is proper if r 1 r 2 and improper or star if r 1 = r 2. Moreover, the node is asymptotically stable if both r 1 and r 2 are negative; unstable if both r 1 and r 2 are positive.

19 Classification of equilibrium solutions: Real eigenvalues When the eigenvalues r 1 and r 2 are real (and nonzero), we have the following classification: If r 1 and r 2 are of opposite sign (one positive, one negative), then 0 is a saddle point. Saddle points are always unstable. If r 1 and r 2 are of the same sign, then 0 is a node, which is proper if r 1 r 2 and improper or star if r 1 = r 2. Moreover, the node is asymptotically stable if both r 1 and r 2 are negative; unstable if both r 1 and r 2 are positive. Stability 0 is stable if nearby solutions stay nearby. 0 is asymptotically stable if nearby solutions converge to 0. 0 is unstable if (at least some) nearby solutions escapes away from 0.

20 Classification of equilibrium solutions: Complex eigenvalues When the eigenvalues r 1 and r 2 are complex: r 1,2 = λ ± iµ, µ 0, we have the following classification: If λ = 0, then 0 is a center. A center is always stable, but not asymptotically stable.

21 Classification of equilibrium solutions: Complex eigenvalues When the eigenvalues r 1 and r 2 are complex: r 1,2 = λ ± iµ, µ 0, we have the following classification: If λ = 0, then 0 is a center. A center is always stable, but not asymptotically stable. If λ 0, then 0 is a spiral point.

22 Classification of equilibrium solutions: Complex eigenvalues When the eigenvalues r 1 and r 2 are complex: r 1,2 = λ ± iµ, µ 0, we have the following classification: If λ = 0, then 0 is a center. A center is always stable, but not asymptotically stable. If λ 0, then 0 is a spiral point. Moreover, the spiral point is asymptotically stable if λ < 0; unstable if λ > 0.

23 Classification of equilibrium solutions: Complex eigenvalues When the eigenvalues r 1 and r 2 are complex: r 1,2 = λ ± iµ, µ 0, we have the following classification: If λ = 0, then 0 is a center. A center is always stable, but not asymptotically stable. If λ 0, then 0 is a spiral point. Moreover, the spiral point is asymptotically stable if λ < 0; unstable if λ > 0. We can not determine the direction of the spirals (clockwise or counterclockwise) by looking at the eigenvalues. Instead, we can look at the direction field at some point. For example, at (x, y) = (1, 0), the vector in the direction field is (a, c). So if c > 0, then the spirals are counterclockwise; if c < 0, then the spirals are clockwise.

24 Trace-Determinant picture Alternatively, we can use trace and determinant of A to classify.

25 Mass-spring system As an example, we consider the mass-spring system mu + γu + ku = 0 where u = u(t) is the displacement from the equilibrium position; m > 0 is the mass; γ 0 is the damping coefficient; k > 0 is the spring constant.

26 Mass-spring system As an example, we consider the mass-spring system where mu + γu + ku = 0 u = u(t) is the displacement from the equilibrium position; m > 0 is the mass; γ 0 is the damping coefficient; k > 0 is the spring constant. Taking x = u, y = u, then we can rewrite this equation as a linear system ( ) ( ) x x 0 1 = Ax, x =, A = y k m γ m

27 Mass-spring system x = Ax, x = The characteristic equation is ( ) ( ) x 0 1, A = y k m γ m det(a ri ) = r 2 + γ m r + k m = 1 m (mr 2 + γr + k) = 0. Undamped: γ = 0, r = ±i k/m, so 0 is a center. Underdamped: 0 < γ < 2 km, r = λ ± iµ, λ = γ/2m < 0, so 0 is an asymptotically stable spiral point. Critically damped: γ = 2 km, r 1 = r 2 = γ/2m < 0, so 0 is an improper node, which is asymptotically stable. Overdamped: γ > 2 km, then r 1 < r 2 < 0, so 0 is an asymptotically stable node.

28 Another example Consider the following example with parameter α ( ) x α 2 = Ax, A =. 2 0 The characteristic equation is ( ) α r 2 det(a ri ) = det = r 2 αr + 4 = 0 2 r and the eigenvalues are r = α ± α

29 Another example: Continue Consider the following example with parameter α ( ) x α 2 = Ax, A =. 2 0 α < 4: Two negative eigenvalues, asymptotically stable node. 4 < α < 0: Complex eigenvalues with negative real parts, asymptotically stable spiral point. (Clockwise) 0 < α < 4: Complex eigenvalues with positive real parts, unstable spiral point. (Clockwise) α > 4: Two positive eigenvalues, unstable node. Critical values for α α = 4: asymptotically stable improper node. α = 0: center. α = 4: unstable improper node.