Chapter 8

Stability theory

We discuss properties of solutions of a first order two dimensional , and for a special class of linear . We denote the independent variable by ‘t’ in place of ‘x’, and x,y denote dependent variables. Let I ⊆ R be an interval, and Ω ⊆ R2 be a domain. Let us consider the system

dx = F (t, x, y), dt (8.1) dy = G(t, x, y), dt where the functions are defined on I × Ω, and are locally Lipschitz w.r.t. variable (x, y) ∈ Ω.

Definition 8.1 (Autonomous system) A system of ODE having the form (8.1) is called an autonomous system if the functions F (t, x, y) and G(t, x, y) are constant w.r.t. variable t. That is,

dx = F (x, y), dt (8.2) dy = G(x, y), dt

Definition 8.2 A point (x0, y0) ∈ Ω is said to be a critical point of the autonomous system (8.2) if F (x0, y0) = G(x0, y0) = 0. (8.3)

A critical point is also called an equilibrium point, a rest point.

Definition 8.3 Let (x(t), y(t)) be a solution of a two-dimensional (planar) autonomous system (8.2). The trace of (x(t), y(t)) as t varies is a curve in the plane. This curve is called .

Remark 8.4 (On solutions of autonomous systems) (i) Two different solutions may represent the same trajectory. For,

(1) If (x1(t), y1(t)) defined on an interval J is a solution of the autonomous system (8.2), then the pair of functions (x2(t), y2(t)) defined by

(x2(t), y2(t)) := (x1(t − s), y1(t − s)), for t ∈ s + J (8.4)

is a solution on interval s + J, for every arbitrary but fixed s ∈ R.

(2) However, traces of both the solutions on the respective intervals is the same. If the independent variable t is interpreted as time, then we note that the two different solutions visit every point on the trajectory with a time lag s. Also see Example 8.5

67 68 §8.0

(ii) The do not cross. For,

(1) Suppose two trajectories γ1 and γ2 cross at a point (x0, y0) ∈ Ω. Let (x1(t), y1(t)) defined on interval I1 be a solution whose trace is γ1, and (x2(t), y2(t)) defined on interval I2 be a solution whose trace is γ2. By the assumption of crossing, there exist t1 ∈ I1, and t2 ∈ I2 such that

(x1(t1), y1(t1)) = (x0, y0) = (x2(t2), y2(t2)). (8.5)

Let us define a pair of functions (x3(t), y3(t)) on interval t1 − t2 + I2 by

(x3(t), y3(t)) := (x2(t − t1 + t2), y2(t − t1 + t2)) (8.6)

It can be easily checked, via Chain rule, that (x3(t), y3(t)) is a solution of (8.2) on the interval t1 − t2 + I2. Note that

(x3(t1), y3(t1)) = (x2(t1 − t1 + t2), y2(t1 − t1 + t2)) = (x2(t2), y2(t2)) = (x0, y0) (8.7) Thus, (x3(t), y3(t)) and (x1(t), y1(t)) are solutions of the same initial value prob- lem, contradicting the uniqueness of solutions to IVP. Therefore two trajectories do not cross each other.

(iii) The trajectories fill the domain Ω, since through every point a trajectory passes. This is a consequence of existence theorem. (iv) Through every point in the phase Ω, exactly one trajecory passes. This is a consequence of uniqueness of solutions to IVPs. (v) From the last two remarks, it follows that the trajectories partition the Ω. In fact, defining a relation on Ω by saying that two points (x1, y1), (x2, y2) ∈ Ω are related if (x1, y1) and (x2, y2) lie on the same trajectory, it is easy to verify that this relation is an equivalence relation and thereby giving rise to a partition of Ω in terms of equivalnece classes. Each equivalence class is a trajectory. (vi) Note that trajectories consisting of single point correspond to critical points. (vii) types of trajectories: For autonomous systems with two dimensional phase space, three types of trajectories are possible. A trajectory consisting of single point (corresponding to equilib- rium solutions), and if trajectory has more than one point then it could be a closed curve (corresponding to periodic solutions), or a curve without self-intersection. (viii) For linear autonomous systems, a special class of systems (8.2) for which F and G are linear in x, y, note that saturated solutions are global, i.e., saturated solutions are defined on the entire real line R. Hence for linear autonomous systems, we do not mention the interval on which a given solution is defined.

The above remark is illustrated by the following example.

Example 8.5 dx dy = y, = −4x. (8.8) dt dt

Note that (x1(t), y1(t)) := (cos 2t, −2 sin 2t) is a solution of (8.8). The trajectory passing through 2 2 y2 the point (1, 0) ∈ R is the ellipse x + 4 = 1 travelled counterclockwise. Consider the solution π π (x2(t), y2(t)) := (cos(t − π/2), −2 sin 2(t − π/2)). This solution satisfies (x2( 2 ), y2( 2 )) = (1, 0), and hence has the same trajectory as (x1(t), y1(t)). Draw figure

MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 8 : Stability theory 69

8.1 Solving linear planar systems with constant coefficients

Consider the system of ODE µ ¶ µ ¶ µ ¶ µ ¶ x0 a b x x = =: A . (8.9) y0 c d y y

8.1.1 Fundamental Definition 8.6 (Fundamental matrix) A matrix valued function Φ whose columns are solu- tions of the system of ODE (8.9) is called a solution matrix. A solution matrix Φ is called a fundamental matrix if the columns of Φ form a fundamental pair of solutions for the system (8.9). A fundamental matrix Φ is called the standard fundamental matrix if Φ(0) is the identity matrix.

Remark 8.7 Since the columns of a solution matrixΦ are solutions of (8.9), the matrix valued function Φ satisfies the system of ODE Φ0 = AΦ. (8.10)

Exercise 8.8 A solution matrix is a fundamental matrix if and only if its determinant is not zero.

Exercise 8.9 Prove that if Ψ is a fundamental matrix then ΨC is also a fundamental matrix for every constant invertible matrix C. Prove that all fundamental matrices occur this way.

Computation of fundamental matrix By definition of a fundamental matrix, finding a fundamental pair of solutions to system of ODE (8.9) is equivalent to finding a fundamental matrix. In view of Exercise 8.9, fundamental matrix is not unique but the standard fundamental matrix is unique. µ ¶ a (1) Observe that eλt is a non-trivial solution of (8.9) if and only if b µ ¶ µ ¶ µ ¶ µ ¶ a 0 a a 6= ,A = λ . (8.11) b 0 b b µ ¶ a That is, λ is an eigenvalue and is an eigenvector corresponding to λ. b

(2) Questionµ ¶ Is it possible to find a fundamental pair, both of which are of the form a eλt ? b µ ¶ µ ¶ a c Answer Supposing that φ (t) = eλ1t and φ (t) = eλ2t are two solutions of 1 b 2 d (8.9), φ1, φ2 form a fundamental pair if and only if ¯ ¯ ¯ a c ¯ ¯ ¯ 6= 0, (8.12) ¯ b d ¯

since the above determinant is the of φ1, φ2 at t = 0. That is, the matrix A should have two linearly independent eigenvectors. Note that this is equivalent to saying that A is diagonalisable.

Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 70 8.1. Solving linear planar systems with constant coefficients

(3) Question What if the matrix A does not have two linearly independent eigenvectors? This can happen when A has only one eigenvalue of multiplicity two. Inspired by a similar situation in the contextµ of¶ constant coefficientµ second¶ order linear ODE, we are a a tempted to try φ (t) = eλ1t and φ (t) = teλ1t as a fundamental pair. But 1 b 2 b note that φ1, φ2 does not form a fundamental pair since Wronskian at t = 0 will be zero, also note that φ2 is not even a solution of the linear system (8.9). Nevertheless, we can find a solution having the form of φ1. Therefore, we try a variant of above suggestion to find another solution that together φ1 constitutes a fundamental pair. Let µ ¶ µ ¶ a c φ (t) = teλ1t + eλ1t (8.13) 2 b d

Then φ2(t) solves the system (8.9) if and only if µ ¶ µ ¶ c a (A − λ I) = . (8.14) 1 d b µ ¶ µ ¶ a c One can easily verify that , are linearly independent. Thus, φ , φ defined b d 1 2 by µ ¶ µ ¶ µ ¶ a a c φ (t) = eλ1t , φ (t) = teλ1t + eλ1t (8.15) 1 b 2 b d µ ¶ µ ¶ a c is a fundamental pair, where , are related by the equation (8.14). b d

(4) In case the matrix A does not have real eigenvalues, then eigenvalues are complex conjugates of each other. In this case,

(λ, v) is an eigen pair if and only if (λ, v) is also an eigen pair for A. (8.16) µ ¶ α + iβ Denoting λ = r + iq (note q 6= 0), v = , define γ + iδ µ ¶ µ ¶ α cos qt − β sin qt α sin qt + β cos qt φ (t) = ert , φ (t) = ert . (8.17) 1 γ cos qt − δ sin qt 2 γ sin qt + δ cos qt Verifying that the pair of functions defined above constitute a real-valued fundamental pair of solutions is left as an exercise.

8.1.2 Matrix exponentials Inspired by the formula for solutions to linear scalar equation y0 = ay, given by y(t) = ceat, where y(0) = c, we would like to say that the system (8.9) has solution given by µ ¶ µ ¶ µ ¶ µ ¶ x(t) a a x(0) = exp(tA) , where = . (8.18) y(t) b b y(0)

First of all, we must give a meaning for the symbol exp(A) as a matrix, which is called the exponental of matrix B. Then we must verify that the proposed formula for solution, in (8.18), is indeed a solution of (8.9). We do this next.

Definition 8.10 (Exponential of a matrix) If A is a 2 × 2 matrix, then exponential of A (de- noted by eA, or exp(A), is defined by

X∞ Ak eA = I + (8.19) k! k=1

MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 8 : Stability theory 71

We record below, without proof, some properties of matrix exponentials.

Lemma 8.11 Let A be a 2 × 2 matrix. (1) The series in (8.19) converges. (2) e0 = I. (3) (eA)−1 = e−A (4) exp(A + B) = exp(A) exp(B) if the matrices A and B satisfy AB = BA. (5) If J = P −1AP , then exp(J) = P −1 exp(A)P .

d tA tA (6) dt e = Ae . Theorem 8.12 If A is a 2 × 2 matrix, then Φ(t) = etA is a fundamental matrix of the system (8.9). If (x(t), y(t)) is a solution of the system (8.9) with (x(t0), y(t0)) = (x0, y0), then (x(t), y(t)) is given by µ ¶ µ ¶ x(t) x = e(t−t0)A 0 . (8.20) y(t) y0

Remark 8.13 Thanks to the above theorem, we do not need to struggle to find a fundamental matrix as we did earlier. We need to to take just the exponential of the matrix tA and that would give very easily a fundamental matrix. But it is not as simple as it seems to be. In fact, summing up the series for exponential of a matrix is not easy at all, even for relatively simpler matrices. To realise this, solve the exercise following this remark.

Exercise 8.14 Find exponential matrix for µ ¶ µ ¶ µ ¶ 2 0 1 1 1 2 A = ,A = ,A = . (8.21) 1 0 −1 2 0 1 3 2 3

Remark 8.15 After solving the above exercise, one would realise that it is easier to find funda- mental matrices by finding fundamental pairs of solutions instead of summing up a series! There is an alternate method to calculate exponential matrix for tA, via fundamental pairs, by observing that exponential matrix is nothing but the standard fundamental matrix. So, find a fundamental matrix Φ, then exponential matrix etA is given by

etA = [Φ(0)]−1Φ(t). (8.22)

8.2 Stability of an equilibrium point

Definition 8.16 (1) A critical point (x0, y0) of the system (8.2) is said to be stable if for each ² > 0 there exists a δ > 0 such that every solution (x(t), y(t)) for which there is an s such that k(x(s), y(s)) − (x0, y0)k < δ, exists for all t ≥ s and satisfies

k(x(t), y(t)) − (x0, y0)k < ², ∀ t ≥ s. (8.23)

(2) A critical point (x0, y0) of the system (8.2) is said to be unstable if it is not stable.

(3) A critical point (x0, y0) of the system (8.2) is said to be asymptotically stable if

(i) the critical point (x0, y0) is stable, and,

(ii) if there exists a δ0 > 0 (0 < δ0 < δ) such that

k(x(s), y(s))−(x0, y0)k < δ0 for some s =⇒ (x(t), y(t)) → (x0, y0) as t → ∞. (8.24)

Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 72 8.3. Phase space picture

Convention We are interested in the behaviour of solutions as t → ∞. Therefore, for us always t ≥ 0, and hence we indicate directions on trajectories only for increasing t ↑ ∞. Remark 8.17 In other words, a critical point is stable if every trajectory that comes within a δ distance from the critical point at some time, stays within an ² distance at all later times. Further if the trajectory approaches the critical point, then the critical point is asymptotically stable.

8.3 Phase space picture

Before we start discussing the nature of trajectories for linear systems, we need to understand the following questions. (1) What are all 2 × 2 real matrices? (2) Under a linear transformation, what will be the images of a point, a closed curve, a curve?

Let us analyse some special cases of linear autonomous system (8.9): Case 1: Invertible diagonal matrix µ ¶ λ 0 A = , with λ 6= 0 6= µ. (8.25) 1 0 µ

(1) If (x(t), y(t)) is a solution of the system (8.9) corresponding to A1, with (x(0), y(0)) = (x0, y0), then (x(t), y(t)) is given by µ ¶ µ ¶ µ ¶ µ ¶ λt x(t) tA1 x0 e 0 x0 = e . = µt (8.26) y(t) y0 0 e y0

(2) To draw trajectories in the plane, we have to draw the unique trajectory passing through each point. (i) The critical point (0, 0) is called non-degenerate since it is isolated (well-separated from other critical points, if any; in fact none!)

(ii) Let us fix (x0, y0) = (0, 0) The trajectory through origin consists of only one point, which is origin.

(iii) Let us consider the case of x0 = 0, but y0 6= 0. Then solution is given by µt (x(t), y(t)) = (0, e y0). Similarly, in the case where y0 = 0, but x0 6= 0, the λt solution is given by (x(t), y(t)) = (e x0, 0).

(iv) Let us fix (x0, y0) with x0 6= 0 6= y0. In this case, we can write µ ¶ µ ¶ x(t) µ y(t) λ = , (8.27) x0 y0

hence trajectory through (x0, y0) is given by µ ¶ µ ¶ x µ y λ x y = , with > 0, > 0. (8.28) x0 y0 x0 y0

(3) Now draw the trajectories and indicate the progress of the curve as t increases. The nature of the trajectories vary and depend on the nature of λ, µ. There are five distinct scenarios:

λ = µ < 0, λ < 0 < µ, λ < µ < 0, 0 < µ < λ, 0 < λ = µ. (8.29)

MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 8 : Stability theory 73

Case 1B: Non-invertible diagonal matrix µ ¶ λ 0 A = , with λµ = 0. (8.30) 2 0 µ

(1) The critical point (0, 0) is called degenerate since there are infinitely many critical points of the system in any neighbourhood of the origin. (critical points are not well-separated from each other) In fact, all points on x axis (respectively y axis) are critical points if λ = 0 (respectively, µ = 0).

(2) If (x(t), y(t)) is a solution of the system (8.9) corresponding to A2, with (x(0), y(0)) = (x0, y0), then (x(t), y(t)) is given by µ ¶ µ ¶ µ ¶ µ ¶ λt x(t) tA2 x0 e 0 x0 = e = µt (8.31) y(t) y0 0 e y0

(3) To draw trajectories in the plane, we have to draw the unique trajectory passing through each point.

(i) Let us fix (x0, y0) = (0, 0) The trajectory through origin consists of only one point, which is origin.

(ii) Let us fix (x0, y0) to be any other critical point. The trajectory through any critical point consists of only one point, which is the critical point itself.

(iii) Let us fix (x0, y0) different from any critical point. Then solution is given by λt µt (x(t), y(t)) = (e x0, e y0). (4) Now draw the trajectories and indicate the progress of the curve as t increases. The nature of the trajectories vary and depend on the nature of λ, µ. There are three distinct scenarios: λ < µ = 0, 0 = µ < λ, 0 = µ = λ. (8.32)

µ ¶ 0 1 Case 2: Matrix having the form λI + 0 0 µ ¶ λ 1 A = , λ ∈ R (8.33) 3 0 λ

(1) The exponential matrix is given by µ µ ¶¶ µ ¶ µ ¶ µ ¶ 0 1 eλt 0 1 t eλt teλt etA3 = exp(λtI) exp t = = (8.34) 0 0 0 eλt 0 1 0 eλt

(2) If (x(t), y(t)) is a solution of the system (8.9) corresponding to A3, with (x(0), y(0)) = (x0, y0), then (x(t), y(t)) is given by µ ¶ µ ¶ µ ¶ µ ¶ λt λt x(t) tA3 x0 e te x0 = e = λt (8.35) y(t) y0 0 e y0

(3) To draw trajectories in the plane, we have to draw the unique trajectory passing through each point.

Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 74 8.3. Phase space picture

(i) If λ 6= 0, the critical point (0, 0) is called non-degenerate since it is isolated (well-separated from other critical points, if any; in fact none!). If λ = 0, then critical point (0, 0) is called degenerate since there are infinitely many critical points of the system in any neighbourhood of the origin. (critical points are not well-separated from each other); in this case, all points on x axis are critical points.

(ii) The trajectory through any critical point consists of only one point, which is the critical point itself.

(iii) Let us fix (x0, y0) different from a critical point. Let us assume that x0 = 0 and y0 6= 0. Then µ ¶ y(t) x(t) = ty(t), and λt = log . (8.36) y0 Thus trajectories are given by µ ¶ y y λx = y log , with > 0. (8.37) y0 y0

(iv) In the general case of both x0 6= 0 and y0 6= 0, solution (x(t), y(t)) satisfies y(t) x(t) = [x0 + ty0], (8.38) y0

and thus passes through y axis at time t = − x0 . and thus trajectory is already y0 described in the previous point. Recall that trajectory does not change by intro- ducing a delay in the time at which a solution visits a point, and even have the same behaviour as t → ∞.

(v) The other case is y0 = 0, but x0 6= 0. Then solution is given by (x(t), y(t)) = λt (e x0, 0). (4) Now draw the trajectories and indicate the progress of the curve as t → ∞. The nature of the trajectories vary and depend on the nature of λ. There are three distinct scenarios: 0 < λ, λ < 0, λ = 0. (8.39) µ ¶ Case 3: Matrix having the form a −b b a µ ¶ a −b A = , λ ∈ R (8.40) 4 b a

(1) The exponential matrix is given by µ µ ¶¶ 0 −b etA4 = exp(atI) exp t (8.41) b 0 µ ¶ 0 −1 Denoting J = , we must evaluate exp(tJ). By evaluating various powers of 1 0 tJ and substituting them in the definition of exponential, it can be seen that µ ¶ µ ¶ (bt)2 (bt)4 (bt)6 bt (bt)3 (bt)5 (bt)7 exp(tJ) = 1 − + − + ··· I+ − + − + ··· J 2! 4! 6! 1! 3! 5! 7! (8.42)

MA 417: Ordinary Differential Equations Sivaji Ganesh Sista Chapter 8 : Stability theory 75

On noting that the series expansions in the above equation correspond to well-known functions, the above equation reduces to µ ¶ cos bt − sin bt exp(tJ) = cos btI + sin btJ = (8.43) sin bt cos bt

Thus exponential matrix becomes µ µ ¶¶ µ ¶ 0 −b cos bt − sin bt etA4 = exp(atI) exp t = eat (8.44) b 0 sin bt cos bt

(2) If (x(t), y(t)) is a solution of the system (8.9) corresponding to A3, with (x(0), y(0)) = (x0, y0), then (x(t), y(t)) is given by µ ¶ µ ¶ µ ¶ µ ¶ x(t) x cos bt − sin bt x = etA4 0 = eat 0 (8.45) y(t) y0 sin bt cos bt y0

That is, µ ¶ µ ¶ at x(t) e (x0 cos bt − y0 sin bt) = at . (8.46) y(t) e (x0 sin bt + y0 cos bt) We can write the above equation in the form µ ¶ µ ¶ at x(t) k(x0, y0)k e cos(bt − α) = at , (8.47) y(t) k(x0, y0)k e sin(bt − α) ³ ´ where α = tan−1 −y0 . Also, note x0

y(t) = tan(bt − α) (8.48) x(t)

(3) To draw trajectories in the plane, we have to draw the unique trajectory passing through each point. (i) The critical point (0, 0) is called non-degenerate since it is isolated (well-separated from other critical points, if any; in fact none!). (ii) The trajectory through origin, being critical point, consists of only one point, which is the critical point itself.

(iii) Observe that 2 2 2at 2 2 (x(t)) + (y(t)) = e (x0 + y0) (8.49) There are two points to be noted from here: (i). If a = 0, then solution lies on circle of radius k(x0, y0)k. (ii). If a 6= 0, depending on the sign of a, we have k(x(t), y(t))k goes to zero or infinity. p In fact by switching to polar coordinates, x = r cos θ, y = r sin θ, r = x2 + y2,

θ = bt − α. (8.50)

Trajectories are given by q q p θ+α 2 2 at 2 2 a( b ) 2 2 r = (x(t)) + (y(t)) = e (x0 + y0) = e (x0 + y0), (8.51)

which is nothing but ( a )θ r = ce b (8.52) p ( aα ) 2 2 where c = e b (x0 + y0). These curves are spirals.

Sivaji Ganesh Sista MA 417: Ordinary Differential Equations 76 8.3. Phase space picture

(4) Now draw the trajectories and indicate the progress of the curve as t → ∞. The nature of the trajectories vary and depend on the nature of a, b. There are four distinct scenarios:

a < 0 < b, b < 0 < a, a = 0&b > 0, a = 0&b < 0. (8.53)

Exercise 8.18 Discuss the phase space picture of linear autonomous systems corresponding to each of the matrices µ ¶ µ ¶ µ ¶ µ ¶ −1 0 −2 0 −2 0 2 0 A = ,A = ,A = ,A = , 1 0 2 2 0 −1 3 0 −2 4 0 1 µ ¶ µ ¶ µ ¶ µ ¶ 2 0 −2 0 2 0 0 0 A = ,A = ,A = ,A = , 5 0 2 6 0 0 7 0 0 8 0 0 µ ¶ µ ¶ µ ¶ µ ¶ −2 1 2 1 0 1 −2 −1 A = ,A = ,A = ,A = , 9 0 −2 10 0 2 11 0 0 12 1 −2 µ ¶ µ ¶ 1 2 0 −1 A = ,A = . 13 −2 1 14 1 0

Exercise 8.19 Comment on the stability of origin in each of the above systems.

Phase space picture for a linear autonomous systems corresponding to a general matrix So far, we discussed phase space portrait for systems described by a matrix of various special forms. The discussion for a general matrix case can be done once we have answers to the two questions posed at the beginning of this Section 8.3. The answers are (1) The special matrices we considered are all the 2 × 2 real matrices. It means that any 2 × 2 real matrix is similar to one of the three types of special matrices we considered. (2) Under a linear transformation, images of a point, a closed curve, and a curve are a point, a closed curve, and a curve respectively! Linear transformation stands for “line goes to line”. Note that a circle will become an ellipse under a linear transformation, in general.

Let P −1AP = J. Introducing a new set of dependent variables (v, w) defined by µ ¶ µ ¶ v x = P −1 . (8.54) w y

It is easy to verify that µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ µ ¶ x x0 x v v0 v solves = A if and only if solves = J (8.55) y y0 y w w0 w

We discussed the phase space portrait of (v, w) since J is one of the special matrices for which we analysed the phase space picture. Now returning to the original dependent variables (x, y), starting from a solution (v(t), w(t)) corresponding to matrix J, solution (x(t), y(t)) corresponding to matrix A is given by µ ¶ µ ¶ x(t) v(t) = P . (8.56) y(t) w(t) The trajectories in xy plane are nothing but images of trajectories in vw plane corresponding to special matrix J, under a linear transformation defined by the matrix P . Note that the qualitative behaviour of a critical point remains the same for all systems that correspond to matrices which are similar. This finishes the discussion of trajectories for a linear planar autonomous system.

MA 417: Ordinary Differential Equations Sivaji Ganesh Sista