STABILITY. Phase portraits and local stability

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1 MAS271 Methods for differential equations Dr. R. Jain STABILITY Phase portraits and local stability We are interested in system of ordinary differential equations of the form ẋ = f(x, y), ẏ = g(x, y), (1) where f and g do not depend explicitly upon t i.e. the system is autonomous. Example 1. Predator-prey model (foxes and rabbits) ẋ = ax bx 2 cxy, ẏ = py + qxy, (2) where a, b, c, p, q are positive constants. x(t) and y(t) are rabbit and fox population respectively. Example 2. Simple Pendulum If θ is the angle of deviation of an undamped pendulum of length l whose bob has mass m, then equation of motion is: where ω 2 = g/l. Putting ml θ = mg sin θ θ + ω 2 sin θ = 0, (3) x = θ, y = θ, equation (3) may be written as ẋ = y, ẏ = ω 2 sin x. (4)

2 The xy- plane is called the phase plane and the solution curves of (1) are x = x(t), y = y(t) the orbits, trajectories or paths. The phase plane together with the orbits is the phase portrait. The point (x, y) such that f(x, y) = 0, g(x, y) = 0 (5) is called an equilibrium point, critical point or stationary point. An isocline is a curve in the phase plane connecting points at which the gradients of the orbits are the same. For (1) dy dx = ẏ ẋ = g(x, y) f(x, y) and so an isocline is a curve along which f(x, y)= constant g(x, y). In practice the two most important isoclines are f(x, y) = 0, g(x, y) = 0. The usual procedure to construct a phase portrait is: (1) find the equilibrium points (2) determine the nature of each one (3) draw the isoclines, especially ẋ = 0 dy dx infinite; ẏ = 0 dy dx = 0. (4) use artistic imagination along with any additional information! ( ) ẋ ẏ ( ) x = A ; A = y Find eigenvalues λ 1 and λ 2 : A λi = 0, I = Classification of equilibrium points ( ) ( ) a11 a 12 a 21 a 22

3 (1) λ 1 < 0, λ 2 < 0 (stable node)

4 (2) λ 1 > 0, λ 2 > 0 (unstable node)

5 (3) λ 1 = λ 2 = λ, A λi 0 (a) λ > 0 (unstable node)

6 (b) λ < 0 (stable node)

7 (4) λ 1 = λ 2 = λ, A λi = 0 (a) λ > 0, unstable improper node

8 (b) λ < 0, stable improper node

9 (5) λ 1 λ 2 < 0 saddle (always unstable)

10 (6) λ 1,2 = ±iω centre (stable)

11 (7) λ 1,2 = α ± iω focus or spiral (a) α > 0, unstable

12 (b) α < 0, stable

13 Here are some more examples of various types of equilibrium points.

14 Notes: (1). If the real parts of both λ 1 and λ 2 are negative, then lim t (x(t), y(t)) = (0, 0), for all choices of constants, and (0,0) is called an asymptotically stable equilibrium. (2). If λ 1 and λ 2 are conjugate and purely imaginary (λ 1 = iq, λ 2 = iq, q 0), then solutions are periodic and (0,0) is a stable equilibrium. (3). If either λ 1 or λ 2 has real part, then (0,0) is an unstable equilibrium. (4). If either or both of λ 1 and λ 2 is zero, all solution trajectories are confined to straight lines and may even reduce to single points.

15 Example Sketch the phase portrait of ẍ + ω 2 x = 0 ( ) where ω is a constant. Solution Putting y = ẋ the above equation ( ) is equivalent to ẋ = y, ẏ = ω 2 x or ( ) ẋ ẏ = ( ) ( ) 0 1 x ω 2 0 y ( ) x = A y (i) The equilibrium point is (0,0) (ii) A λi = λ 1 ω 2 λ = λ 2 + ω 2 λ = ±iω (purely imaginary) (0,0) is a centre (iii) dy dx = ẏ ẋ ẏ = 0 on x = 0 dy dx = 0 on y-axis (y 0). ẋ > 0 for y > 0 x increases as t increases ẋ < 0 for y < 0 x decreases as t increases ẋ = 0 on y = 0 dy dx infinite on x-axis ẏ > 0 for x < 0 y increases as t increases

16 ẏ < 0 for x > 0 y decreases as t increases (iv) dy = ω2 x dx y y dy = ω 2 x dx 1 2 y ω2 x 2 = constant ( ) y 2 + ω 2 x 2 = C (**) is a first integral of (*). The curves ω 2 x 2 = y 2 = C or x 2 C/ω 2 + y2 C = 1 are ellipses (ω 1) Notes 1. Each orbit corresponds to a different value of C. 2. Each orbit is closed periodic solution 3. For physical systems (**) is an energy equation. Linear approximation of non-linear system near the equilibrium point (x, y) To obtain the linear approximation near (x, y) set x = x + X(t), y = y + Y (t) (6) and ignore the non-linear terms in f and g. Example Find the linear approximation near (π, 0) of ẋ = y, ẏ = ω 2 sin x Solution Putting x = π + X, y = Y we get Ẋ = Y, Ẏ = ω 2 sin(π + X)

17 Now sin(π + X) = sin π cos X + cos π sin X = sin X X for small X. The linear approximation is therefore ( ẊẎ ) = ( ) ( ) 0 1 x ω 2. 0 y Here we used first principles to find the linear approximation. It can also be found using the Jacobian matrix ( f/ x f/ y J = g/ x g/ y ). The associated linear system near (x, y) is, using (6) ( ) ( X = A ẊẎ Y ), (7) where A = J evaluated at (x, y). In the example f(x, y) = y, g(x, y) = ω 2 sin x so that ( ) ( ) f/ x f/ y 0 1 J = = g/ x g/ y ω 2 cos x 0 and ( ) 0 1 A = J (π,0) = ω 2 = cos π 0 ( 0 ) 1 ω 2 0 as before. Why is this linear approximation useful? An important theorem, due to Poincaré, states that provided the eigenvalues of A are either non-zero, real and distinct or have non-zero real parts the equilibrium point of the non-linear system (1) and the linear approximation (7) are of the same type. This result enables us to deduce the nature of the equilibrium points of (1) by considering the linear approximation (7). If the equilibrium point is asymptotically stable for the linear system then it is asymptotically stable for the non-linear system. In other cases more work is necessary. λ 1 = λ 2 and real: system (1) can have either a node or a spiral

18 λ 1, λ 2, purely imaginary: system (1) can have either a centre or a spiral. Example Sketch the phase portrait of (A) ẋ = x(3 x 2y), ẏ = y(x 1) (predator-prey) (B) ẋ = y, ẏ = ω 2 sin x (simple pendulum) Solution A (i) the equilibrium points are given by x(3 x 2y) = 0 (a) y(x 1) = 0 (b) (b) is satisfied by y = 0 or x = 1. If y = 0 (a) x(3 x) = 0 x = 0 or x = 3 (0, 0) and (3, 0) If x = 1 (a) 2 2y = 0 y = 1 (1, 1) The equilibrium points are (0,0), (3,0) and (1,1). (ii) Here f(x, y) = 3x x 2 2xy, g(x, y) = xy y so that ( ) f/ x f/ y J = g/ x g/ y = ( ) 3 2x 2y 2x y x 1 At (0,0) A = J (0,0) = ( 3 0 ) 0 1 and the eigenvalues are 3 and -1 so that (0,0) is an unstable saddle for the linear approximation and hence for the non-linear system (by Poincaré). At (3,0) ( ) 3 6 A = J (3,0) = 0 2 and the eigenvalues are -3 and 2 so that (3,0) is an unstable saddle for linear approximation and hence for the non-linear system (by Poincaré). At (1,1) ( ) 1 2 A = J (1,1) = 1 0 and the eigenvalues are given by 1 λ 2 0 = A λi = = λ(1 + λ) + 2 = λ 1 λ 2 + λ + 2

19 λ = 1 2 ( 1 ± 7i) (1,1) is a stable focus for the linear approximation and hence for the non-linear system (by Poincaré) (iii) dy dx = ẏ ẋ = y(x 1) x(3 x 2y) ẏ = 0 either y = 0 ẋ = x(3 x) or x = 1, ẋ = 2(1 y) ẋ = 0 either x = 0, ẏ = y or x = 3 2y, ẏ = 2y(1 y)

20

21 (B) ẋ = y, ẏ = ω 2 sin x (i) Equilibrium points are given by y = 0, sin x = 0 x = nπ, n Z, (n = 0, ±1, ±2,...) (nπ, 0) (ii) Here and ( ) 0 1 J = ω 2 cos x 0 ( ) ( ) A = J (nπ,0) = ω 2 = cos nπ 0 ω 2 ( 1) n 0 so that the eigenvalues are given by 0 = A λi = λ 1 ω 2 ( 1) n λ = λ 2 + ( 1) n ω 2 n even: λ = ±iω i.e. purely imaginary centre (linear approximation but care needed for non-linear system) n odd: λ = ±ω unstable saddle (non-linear system) (iii) dy dx = ẏ ẋ ẏ = 0 on x = nπ dy dx = 0 on x = nπ, y 0 ẋ = 0 on y = 0 dy infinite on x-axis except at x = nπ. dx Also, since ω 2 sin x is bounded dy 0 as y for every x. dx (iv) Since replacing x by x + 2π gives the same equations the portrait repeats itself every interval of length 2π. Consider π x π. Also dy dx = ω2 sin x y which is separable and gives y dy = ω 2 sin x dx y 2 2 = ω 2 cos x + C ( )

22 Phase portrait is therefore symmetric about x- and y- axes (x x or y y does not alter the equation) and y = ± 2(ω 2 cos x + C) For real y, we require ω 2 cos x + C 0 C ω 2. Equation ( ) is the first integral of the equation of motion θ = ω 2 sin θ and expresses conservation of mechanical energy. [For the equation ẍ = f(x) similar approach gives 1 2 y2 = f(x)dx + C]

23 simple pendulum phase portrait - physical interpretation Each path corresponds to a different value of C e.g. the path passing through (π, 0) corresponds to C = ω 2 and crosses the y-axis when y = ±2ω. The other closed paths cross the x-axis at x nπ. When y = 0, ω 2 cos x + C = 0 and since cos x < 1 for this to be positive ω 2 < C < ω 2. For these solutions y = θ = 0 for some θ at which point the pendulum reverses direction and periodic oscillations occur (closed paths periodic solutions). These occur when y = θ < 2ω at θ = 0 and the initial speed l θ is not great enough for complete revolutions to occur. For values of C > ω 2, y = θ = 0 is impossible, the paths are not closed and the solutions are not periodic. When θ = 0, θ > 2ω and the initial speed is large enough for complete revolutions to occur in finite time or whirling motion. The path corresponding to C = ω 2 separates these two types of motion and is called the separatrix. The initial speed is just sufficient to give complete revolutions but they take an infinite time to complete. Equilibrium points: (0,0) mass hanging at rest vertically below P (stable). (π, 0) mass vertically above P (unstable).

24 For this range of C the paths also cross the y-axis since when x = 0 y = ± 2(ω 2 + C) and ω 2 + C > 0. Since the phase portrait is symmetric about x- and y- axes and for ω 2 C ω 2 the paths cross the x- and y-axes the points x = nπ for n even are centres for the non-linear system.

25 Liapunov s direct method for stability For systems in which linearisation fails to give useful information, an alternative approach is to consider the non-linear system ẋ = f(x, y), ẏ = g(x, y) (8) and use Liapunov s direct method. In essence this method seeks a scalar function of x and y (which can be regarded as a measure of the energy of system (1)) and then seeks to demonstrate that this function either decreases as t indicating stability or it increases indicating instability. Here we confine our attention to the case when the equilibrium point is the origin (0,0). Let V (x, y) be a scalar function defined and continuous with continuous partial derivatives in some neighbourhood (nhd) of the origin 0 and such that V (0, 0) = 0. Definition 1. V (x, y) is positive (negative) definite in a nhd N of O if (i)v (0, 0) = 0 (ii)v (x, y) > 0 (< 0) for all (x, y) (0, 0) in N 2. V (x, y) is positive (negative) semi-definite in N if (i) V (0, 0) = 0 (ii)v (x, y) 0 ( 0) for all (x, y) (0, 0) in N. Examples: x 2 + y 2 is positive definite (pd). x 2 2e x y 6 is negative definite (nd). x 2 is negative semi-definite (nsd) since it is 0 at points (0, y) (x y) 2 is positive semidefinite (psd) since it is zero at points where x = y. y 2 (x 2 1) is nsd on the strip x < 1 it is zero at points (x, 0). On a solution curve x = x(t), y = y(t) of system (1) V (x, y) = V (x(t), y(t)) = V (t) and the rate of change of V along this curve is ( V = dv ) = V dt x ẋ + V y ẏ = f V x + g V y If V is n.s.d (n.d) in N then (0,0) is stable (asymptoti- Theorem Let V (x, y) be p.d. cally stable). A p.d function V (x, y) such that V 0 ( V < 0) is called a weak (strong) Liapunov function. Example 1. ẋ = y, ẏ = ω 2 x (simple harmonic motion)

26 ẍ = ω 2 x Typical mass-spring system T = kx when spring has extension x mẍ = kx ẍ + ω 2 x = 0 (ω 2 = k/m) total energy = kinetic energy + elastic energy = 1 2 mẋ kx2 = m( 1 2 y ω2 x 2 ) Motivated by above working take V (x, y) = y 2 + ω 2 x 2 then V = V x ẋ + V y ẏ = 2ω2 x(y) + 2y( ω 2 x) = 0 Since V is p.d. it follows that V is a weak Liapunov function (0,0) is stable. Example 2: ẋ = y, ẏ = cy ω 2 x (c > 0) (damped oscillations) Take V (x, y) = y 2 + ω 2 x 2 then V = V x ẋ + V y ẏ = 2ω2 x(y) + 2y( cy ω 2 x) = 2cy 2 0 V is a weak Liapunov function. (0, 0) is stable. N.B. In fact (0,0) is asymptotically stable Ex. this. Show this using the linearisation method but this Liapunov function does not detect Exercise For the simple pendulum use v(x, y) = 1 2 y2 + ω 2 (1 cos x) to show that (0,0) is stable.

27 [y = ẋ ẏ = ω 2 sin x, V = ω 2 sin x(y) + y( ω 2 sin x) = 0 Since V is p.d. is a weak Liapunov function. (0,0) is stable. ] In the above examples V has been associated with the energy of the mechanical system. However finding the Liapunov function can be difficult. A quadratic expression can be helpful. Example Find necessary and sufficient conditions on the constants a, b, c for F (x, y) = ax 2 + 2bxy + cy 2 to be (a) p.d. ; (b) n.d. Solution First note that F (0, 0) = 0. Need F (x, 0) > 0. But F (x, 0) = ax 2 a > 0 a > 0 is necessary. Also { F (x, y) = a x 2 + 2b a xy + c } a y2 { = a (x + b ( ) } c a y)2 + a b2 y 2 a 2 ( = a x + b ) 2 ( ) ac b 2 a y + y 2 a Sufficient conditions for F (x, y) > 0 for every (x, y) (0, 0) are a > 0 and ac b 2 > 0. These are also necessary since a > 0 necessary from above and F ( ba y, y ) = ac b2 y 2 > 0 for every y 0 a ac b 2 > 0 Necessary and Sufficient Condition (NSC) for F (x, y) to be p.d. are a > 0 and ac b 2 > 0, (or c > 0 ac b 2 > 0). (b) Similar arguments shows that NSC for F (x, y) to be n.d. are

28 a < 0 and ac > b 2, (or c < 0 and ac b 2 > 0) Example For the system ẋ = 2xy 2 x 3, ẏ = y + x 2 y show that (0,0) is asymptotically stable. Solution For a Liapunov function try V (x, y) = αx 2 + y 2 for suitable α. By previous example for V to be positive definite α > 0. Now V = V x ẋ + V y ẏ = 2αx( 2xy 2 x 3 ) + 2y( y + x 2 y) = 2αx 4 + 2(1 2α)x 2 y 2 2y 2 Taking α = 1/2 gives V = x 4 2y 2 < 0 for all (x, y) (0, 0) Since V (0, 0) = 0 it now follows that V = 1 2 x2 + y 2 is a strong Liapunov function as V is positive definite and V is negative definite (0,0) is asymptotically stable. Note 1. Linear approximation is ( ) ( ) ( ) ẋ 0 0 x = ẏ ) 1 y giving det A = 0 and previous theory does not apply. 2. (0, 0) is the only equilibrium point. Since V as x all orbits whatever their starting point tend to (0,0) as t (0, 0) is globally asymptotically stable. Example: ẋ = 3x 3 y, ẏ = x 5 2y 3 V (x, y) = αx 2m + βy 2n

29 Solution V (0, 0) = 0 and provided m, n are positive integers and α > 0, β > 0 V (x, y) > 0 for all (x, y) (0, 0) and so V (x, y) is positive definite. Now V = V x ẋ + V y ẏ = 2mαx 2m 1 ( 3x 3 y) + 2nβy 2n 1 (x 5 2y 3 ) = 6mαx 2m+2 + (2nβx 5 y 2n 1 2mαx 2m 1 y) 4nβy 2n+2 The choice m = 3, n = 1 gives V = 18αx 8 + (2β 6α)x 5 y 4βy 4 Taking α = 1, β = 3 gives V = 18x 8 12y 4 < 0 Since V (0, 0) = 0 V is negative definite. V is a strong Liapunov function. is asymptotically stable. Theorem If, in some region Σ R n that contains the origin, there exists a scalar function V (x, y) such that V (0, 0) = 0 and V (x, y) is either positive definite or negative definite, and if, in every neighbourhood N of the origin with N Σ, there exists at least one point (x a, y a ) such that V (x a, y a ) has the same sign as V (x a, y a ) then the origin is unstable. Example Consider the system ẋ = x 2 y 2 ẏ = 2xy then (0,0) is the equilibrium point. ( ) 2x 2y J = J 2y 2x (0,0) = ( ) 0 0 = A 0 0 A λi = λ 0 0 λ = 0 λ2 = 0 i.e.λ = 0, 0 So, both eigenvalues are zero. Linearisation fails. Let us take the Liapunov function

30 V (x, y) = αxy 2 x 3 V (0, 0) = 0 V = dv dt = ( αy 2 3x 2) (x 2 y 2 ) 4αx 2 y 2 = 3(1 α)x 2 y 2 αy 4 3x 4 We can choose α = 1, so that V = y 4 3x 4 which is negative definite. For α = 1, V = x(y 2 x 2 ) = 0 for x = 0 or y = ±x so changes sign six times on the unit circle around the origin. Now consider V = y 4 3x 4 This is negative everywhere. So, in every neighbourhood of the origin there is at least one point where V has the same sign as V and hence the origin is unstable. Bifurcation Theory Consider the equation ẋ = µ x 2 where µ is a parameter. For µ > 0, the equilibrium points are x = ± µ. For µ < 0, there are no real equilibrium points. For µ = 0, ẋ = x 2 and x = 0 which is unstable.

31 If we plot a graph of x versus µ, we can see that there is a region of stable equilibrium and there is a region of unstable equilibrium which meet at x = 0, µ = 0. the diagram is called bifurcation diagram and the point x = 0, µ = 0 is the bifurcation point. The bifurcation associated with ẋ = µ x 2 is called a saddle-node bifurcation. Consider another example ẏ = λ 2λy y 2 This has equilibrium points at y = λ ± λ 2 + λ. So, there are no real equilibrium points for 1 < λ < 0 and two equilibrium points otherwise. Therefore, we expect saddle-node bifurcation at λ = 0, y = 0 and λ = 1, y = 1. Now, at the equilibrium points dẏ dy = 2λ 2y = 2 λ 2 + λ The equilibrium point with the larger value of y is stable, whilst the other one is unstable. The bifurcation diagram is shown below.

32 The bifurcation diagram for ẏ = λ 2λy y 2. Note there are two saddle-node bifurcation. Transcritical Bifurcation Consider ẋ = µx x 2 We have equilibrium points at x = 0 and x = µ. If µ 0, there are two real equilibrium points. f x = dẋ dx = µ 2x = { µ at x = µ µ at x = 0 At x = µ, J = f = A = µ which is stable for µ > 0 and unstable for µ < 0. x At x = 0, A = J = f = µ which is unstable for µ > 0 and stable for µ < 0. x

33 This is transcritical bifurcation. At the bifurcation point, the two equilibrium solutions pass through each other and exchange stabilities, so this bifurcation is referred to as an exchange of stabilities. Pitchfork Bifurcation Consider ẋ = µx x 3 f(x) = x(µ x 2 ) = 0 x = 0, x = ± µ J = f x = µ 3x2 At x = 0, A = J 0 = µ. Therefore, the eigenvalue is µ (stable if µ < 0 and unstable if µ > 0). At x = µ, A = J µ = 2µ (stable if µ > 0 and unstable if µ < 0). At x = µ, A = J µ = 2µ (stable if µ > 0 and unstable if µ < 0). Similarly, ẋ = µx + x 3.

34 Two new equilibrium points created at the bifurcation point are unstable. Structurally unstable If ẋ = f(x) and x 1 the equilibrium point, we can write d(x x 1 ) dt = dx dt = ẋ = f(x 1) + (x x 1 )f (x 1 ). If x 1 is the equilibrium solution, f(x 1 ) = 0 and assuming that f (x 1 ) 0, we can write x = ke f (x 1 )t for some constant k. ẋ = (x x 1 )f (x 1 ) for x << 1 or ẋ = xf (x 1 ) If f (x 1 < 0, x 0 as t and the equilibrium point is stable. If f (x 1 ) > 0, x as t and the equilibrium point is unstable. However, if f (x 1 ) = 0, we need more terms in the Taylor expansion. ẋ (x x 1)f (x 1 ) 2 x << 1 Such systems where we have an equilibrium point (or points) for which f (x) = 0, we have structurally unstable system. Second order ordinary differential equation: Bifurcation If the equilibrium point with real part of an eigenvalue is zero, then just like first order equations, second order systems are structurally unstable. A slight perturbation (change) in the governing equations can change the qualitative nature of the phase portrait. Consider the second order system ẋ = µ x 2 ẏ = y If µ > 0, equilibrium points are ( µ, 0) and ( µ, 0). If µ < 0, there are no real equilibrium point. J = ( 2x 0 ) 0 1 J ( µ,0) = ( ) 2 µ ( ) ( ) λ 0 2 µ λ 0 = 0 λ 0 1 λ J ( µ,0) = ( 2 µ 0 ) 0 1 saddle point If µ = 0, we have ẋ = x 2, ẏ = y and (0,0) is the equilibrium point. J (0,0) = ( 0 0 ) 0 1

35 ( ) λ 0 A λi = 0 1 λ λ + λ 2 = 0 λ(1 + λ) = 0 λ = 0, 1 For µ = 0, (0, 0) is a saddle-node point. At such point, a saddle and a node collide and disappear. Note that since ẏ = y, we will have y = ke t. All integral paths asymptote to the x-axis as t where the dynamics are controlled by ẋ = µ x 2. In effect, we can ignore the dynamics in y-direction and concentrate on the dynamics on the x-axis. This shows why it was important to study bifurcation in first order systems, since the important dynamics of higher order system occur on a lower order.

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