# ENGI 9420 Lecture Notes 4 - Stability Analysis Page Stability Analysis for Non-linear Ordinary Differential Equations

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1 ENGI 940 Lecture Notes 4 - Stability Analysis Page Stability Analysis for Non-linear Ordinary Differential Equations A pair of simultaneous first order homogeneous linear ordinary differential equations for two functions x(t), y(t) of one independent variable t, x = = ax + by dy y = = cx + d y x a b x may be represented by the matrix equation = y c d y A single second order linear homogeneous ordinary differential equation for x(t) with constant coefficients, d x dy + p + qx = 0 + py + qx = 0 may be re-written as a linked pair of first order homogeneous ordinary differential equations, by introducing a second dependent variable: dy d d x = y = = dy = qx py x 0 1 x and may also be represented in matrix form = y q p y The general solution for (x, y) in either case can be displayed graphically as a set of contour curves (or level curves) in a phase space. Sections in this chapter: 4.01 Motion of a Pendulum 4.0 Stability of Stationary Points 4.03 Linear Approximation to a System of Non-Linear ODEs (1) 4.04 Reminder of Linear Ordinary Differential Equations 4.05 Stability Analysis for a Linear System 4.06 Linear Approximation to a System of Non-Linear ODEs () 4.07 Limit Cycles 4.08 Van der Pol s Equation 4.09 Theorem for Limit Cycles 4.10 Lyapunov Functions [for reference only - not examinable] 4.11 Duffing s Equation 4.1 More Examples 4.13 Liénard s Theorem

2 ENGI Pendulum Page Motion of a Pendulum Consider a pendulum, moving under its own weight, without friction. The pendulum bob has mass m, the shaft has length L and negligible mass, and the angle of the shaft with the vertical is x. The tension along the shaft is T. The acceleration due to gravity is g ( 9.81 m s ). Resolving forces radially (centripetal force) Resolving forces transverse to the pendulum: T + mg cos x = ml x mg sin x = ml x g (1) L x + k sin x = 0, where k = The Maclaurin series expansion of sin x is: 3 5 x x sin x = x + 3! 5! Provided the oscillations of the pendulum are small, (x << 1), sin x x and the ordinary differential equation governing the motion of the pendulum is, to a good approximation, d x + kx= 0 () (which is the ODE of simple harmonic motion)

3 ENGI Pendulum Page 4.03 Let the angular velocity of the pendulum be v Then, using the chain rule of differentiation, dv dv dv x = v = = = v = x =. The ODE () becomes dv v k x 0 v dv k x 0 + = + = v x c v dv + k x = 0 + k = v + k x = c (3) If at time t = 0 the pendulum is passing through its equilibrium position with angular speed v o, then the initial conditions are x v v ( 0) = 0, = ( 0) = o t = 0 Substituting the initial conditions into (), vo + k 0 = c, which leads to a complete solution for v as an implicit function of x, v + kx = v o (4) A plot of this solution for various choices of v o generates a family of concentric ellipses. Recall that this solution is valid only for small displacements x. The x-v plane is called the phase plane.

4 ENGI Pendulum Page 4.04 Returning to the more general case g x + k sin x = 0, where k = (1) L and again using dv dv dv x = v = = = v the ODE can be re-written as dv v k sin x 0 v dv k sin x 0 + = + = v c v dv + k sin x = 0 k cos x = 1 1 ml v + ( ml k cos x) = ml c (5) KE.. PE.. Total energy However, the kinetic energy is 1 ( ) m Lv. In the absence of friction, the sum of kinetic and potential energy is constant, so that the potential energy of the pendulum must be ml k cos x (= mg L cos x, which makes sense upon examining the diagram on page 4.0). Each value of total energy E = 1 ml c generates an orbit (or energy curve). The relationship between total energy and initial angular velocity is obtained from substituting the initial conditions (x = 0 and v = v o when t = 0) into (5): vo c k = c = vo k v k cos x = vo k (6) vo v = k k cos x = k ( 1 cos x) 0 vo v 0 vo v v vo The maximum angular speed of the pendulum, v o, occurs when x = 0. Also, using cos 1 sin sin 1 cos θ = θ θ = θ, v o v x = 1 cos x = sin k vo v x = sin 4k

5 ENGI Pendulum Page 4.05 Recall that the angular velocity is just v =. Differentiating the complete solutions (6): v k cosx = vo k implicitly with respect to time, we obtain dv dv v + k sin x = 0 = k sin x This expression can also be derived directly from the ODE x + k sin x = 0 (1). d x dv v 0 and k sinx 0 When 0 < x < π and v > 0, = > = = <. Therefore, in the phase plane, as x increases from the starting point (0, v o ), v decreases in the first quadrant, until the maximum value of x (label that maximum value of x as M). Tracking back into the second quadrant, before (0, v o ), d x dv π < x < 0 and v > 0 = v > 0 and = = k sinx > 0. The orbit increases from x = M to a maximum at (0, v o ), then decreases until x = +M. This tracks the motion of the pendulum on its complete swing from left to right. By symmetry, the swing in the opposite direction should generate a mirror image in the x axis of the phase plane, to complete the orbit.

6 ENGI Pendulum Page 4.06 (6) v k cosx = v k o x v v k ( 1 cosx) v k sin = o = o x v = vo ksin Three cases arise: (7) v o < k : v will decrease to zero: x x vo o v= 0 0 = v ksin sin = ± k In the first quadrant of the phase plane, the orbit will move right and down to an intercept M vo on the x axis at (M, 0), where sin = + and 0 < M < π. Extending to the k other three quadrants, the orbits resemble ellipses, centred on the origin. v o = k : v will just barely decrease to zero: x x v= 0 0 = ( k) ksin = ( k) cos M = ± π The pendulum swings all the way to the upside-down position and comes to rest there, before either swinging back or continuing on in the same direction. v o > k : The pendulum will never come to rest, reaching a non-zero minimum speed as it passes through the upside-down position: x v sin o = ± has no real solution for x when v o > k. k If it is swinging anticlockwise, then the orbit stays in the first and second quadrants. If it is swinging clockwise, then the orbit stays in the third and fourth quadrants. These orbits are not closed and extend beyond the range π < x < π. π vmin = vo ksin = vo 4 k We can then generate the full set of orbits in the phase plane for the general pendulum problem.

7 ENGI Pendulum Page 4.07 As time progresses, one moves along an orbit to the right above the x axis, but to the left below the x axis v =. The relationship (7) between angular velocity v and angle x is itself a first order non-linear ordinary differential equation for x as a function of the time t: o sin x v k vo ksin x = = ± ± = t + C x vo 4k sin For the case of closed orbits ( v o < k), the time to complete one orbit (the period T of the pendulum) can be shown to be π / 4 dθ M vo g T =, where b= sin = and k = k 0 1 b sin θ k L This is a complete elliptic integral of the first kind, which has no analytic solution in terms of finite combinations of algebraic functions, (except for special choices of v o and k). As v o k, the period T diverges to infinity it takes forever for the zero energy pendulum to reach the upside-down position.

8 ENGI Stability Page Stability of Stationary Points Consider the (generally non-linear) system of simultaneous first order ordinary differential equations dy = x = P( x, y), = y = Q( x, y) (1) (, ) (, ) dy dy y Q x y Using the chain rule, = = x =. P x y This can be integrated with respect to x to obtain a solution for y as an implicit function of x, provided P(x, y) 0. At points where P(x, y) = 0 but Q(x, y) 0, one may P( x, y) integrate = instead. dy Q x, y ( ) Points on the phase plane where P(x, y) = Q(x, y) = 0 are singular points. A unique slope does not exist at such points. Alternative names for singular points are equilibrium points or stationary points (because both x and y do not [instantaneously] change with time there) or critical points or fixed points. A singular point is stable (and is called an "attractor") if the response to a small disturbance remains small for all time. Stable singular point: all paths starting inside the inner circle stay closer than the outer circle forever. Unstable singular point. (or source )

9 ENGI Stability Page 4.09 Consider the system x = Pxy,, y = Qxy,, x0 = x*, y0 = y*, P0,0 = Q0,0 = 0 () ( ) ( ) ( ) ( ) ( ) ( ) which has a stationary point at the origin. Let x(t; x*), y(t; y*) be the complete solution to this system. The stationary point at the origin is stable if and only if, for every ε > 0 (however small), there exists a δ(ε) such that whenever the point (x*, y*) = (x(0; x*), y(0; y*)) is closer than δ to the origin, the point (x(t; x*), y(t; y*)) remains closer than ε to the origin for all time, or ( x* ) ( y* ) δ x ( t; x* ) y ( t; y* ) + < + < ε t A stationary point is asymptotically stable (also known as a sink ) if it is stable and any disturbance ultimately vanishes: lim x ( t; x* ) + y ( t; y* ) = 0. t

10 ENGI Stability Page 4.10 Here are three types of stationary points with nearby orbits: Stable centre (but not asymptotically stable) Unstable saddle point [all saddle points are unstable] Asymptotically stable focus (or spiral sink)

11 ENGI Linear Approximation (1) Page Linear Approximation to a System of Non-Linear ODEs (1) The Taylor series of any function f (x, y) about the point (x 0, y 0 ) is f f f ( x, y) = f ( x0, y0) + ( x x0) + ( y y0) + x ( x, y ) y ( x, y ) ( ) ( )( ) ( ) f x x0 f x x0 y y0 f y y + + x ( 0, 0)! x y ( 0, 0)! y ( 0, 0)! x y x y x y provided that the series converges to f (x, y). 0 + (1) This allows us to create a linear approximation to the non-linear system x = Pxy,, y = Qxy,, x0 = x*, y0 = y*, P0,0 = Q0,0 = 0. () ( ) ( ) ( ) ( ) ( ) ( ) P P P( x, y) = P( 0,0) + x + y + P1 ( x, y) (3) x y (, ) ( 0,0) ( 0,0) P1 x y where lim = 0, (because P 1 (x, y) is at least second order in x, y) ( xy, ) ( 0,0) x + y and similarly for Q(x, y), so that the system becomes x = ax + by + P x, y 1 ( ) (, ) y = cx + d y + Q x y where a, b, c, d are all constants. In the neighbourhood of the singular point (0, 0), this system can be modelled by the linear system x = ax + by (5) y = cx + d y P P Q Q where a =, b =, c =, d =. x y x y ( 0,0) ( 0,0) ( 0,0) ( 0,0) 1 (4)

12 ENGI ODEs Page Reminder of Linear Ordinary Differential Equations To find the general solution of the homogeneous second order linear ODE d y dy + p + q y = 0, with constant real coefficients p and q, form the auxiliary equation or characteristic equation λ + pλ + q = 0 and evaluate the discriminant D = p ± D p 4q and the roots λ1, λ =. Three cases arise. D > 0 : The characteristic equation has a pair of distinct real roots λ 1, λ. The general solution is y = λ1 Ae x + λx Be. D = 0: The characteristic equation has a pair of equal real roots λ. x The general solution is y = ( Ax + B) e λ. D < 0 : The characteristic equation has a complex conjugate pair of roots λ 1, λ = a ± bj. The general solution is y ax = e ( Acosbx + Bsin bx), where A, B are arbitrary constants of integration.

13 ENGI ODEs Page 4.13 To find the general solution of the system of simultaneous first order linear ODEs = ax + by dy = cx + d y λ, t λt x t y t = αe, βe into the ODE, to obtain substitute the trial solution () () ( ) ( ) e λt αλ = aα e λt + bβ e λt βλe λt = cα e λt + dβ e λt or, in matrix form, a λ b α 0 = c d λ β 0 α = β = 0 is a solution (the trivial solution) for any choice of a, b, c, d and λ. Non-trivial solutions exist when the determinant of the matrix of coefficients is zero: a λ b = 0 ( a λ)( d λ) bc = 0 c d λ ( a d ) λ ( ad bc) λ + + = which is the characteristic equation of the system. The solutions to the characteristic equation are the eigenvalues of the coefficient matrix a b A = c d and, for each eigenvalue λ, a non-zero vector α that satisfies the equation β a λ b α 0 = c d λ β 0 is an eigenvector for that eigenvalue. The general solution to the system of ODEs is a linear combination of the solutions arising from each eigenvalue: ( x() t, y() t ) ( c1 1e λ 1 1 ) 1t c e λt, c e λ1t λ = α + α β + c β e t unless the eigenvalues are equal, in which case the general solution is t t x t, y t c c t e λ λ = α + α, cβ + c β t e ( ) ( () ()) ( 1 1 ) ( 1 1 ) (where, in this case, (α 1, β 1 ) is not necessarily an eigenvector). 0

14 ENGI Stability Analysis (Linear) Page Stability Analysis for a Linear System In the case where (0, 0) is the only critical point of the system = ax + by dy = cx + d y it follows that the characteristic equation λ ( a + d ) λ + ( ad bc) = has only non-zero roots and that det A = ad bc 0. 0 Proof: If λ = 0 then at least one eigenvalue of the coefficient matrix A is zero, from which it follows immediately that a 0 b = det A = ad bc = 0 c d 0 Both roots non-zero ad bc 0. If (0, 0) is the only critical point of the system, then no other choice of (x, y) satisfies both equations ax + by = 0 ( ad bc) x = 0 cx + d y = 0 ( ad bc) y = 0 from which it follows immediately that ad bc = det A 0. If the roots are both non-zero and (x, y) is a critical point of the system, then ax + by = 0 ( ad bc) x = 0 cx + d y = 0 ( ad bc) y = 0 But λ1, λ 0 ad bc 0 (0, 0) is the only solution to this pair of simultaneous linear equations. Therefore (0, 0) is the only critical point of the system if and only if both roots of the characteristic equation are non-zero.

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