stability theory

Chapter 4 Stability Theory 4.1 Basic Concepts In this chapter we introduce the concepts of stability and asymptotic stability for solutions of a difierential equation and consider some methods that may be used to prove stability. To introduce the concepts, consider the simple scalar equation y0(t) = ay(t): (4.1.1) The solution is, of course, y(t) = y0e at, where y0 = y(0). In particular, y(t) · 0 is a solution. What happens if we start at some point other that 0? If a < 0, then every solution approaches 0 as t ! 1. We say that the zero solution is (globally) asymptotically stable. See Figure 4.1, which shows the graphs of a few solutions and the direction fleld of the equation, i.e., the arrows have the same slope as the solution that passes through the tail point. If we take a = 0 in (4.1.1), the solutions are all constant. This does have some relevance to stability: if we start near the zero solution, we stay near the zero solution. In this case, we say the zero solution is stable, (but not asymptotically stable). Finally, if a > 0 in (4.1.1), every nonzero solution goes to inflnity as t goes to inflnity. In this case, no matter how close to zero we start, the solution is eventually far away from zero. We say the zero solution is unstable. See Figure 4.2. 1 2 CHAPTER 4. STABILITY THEORY -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 -4 -3 -2 -1 0 1 2 3 4 t x x ' = - x Figure 4.1: The zero solution is globally asymptotically stable for y0 = ¡y. -1 -0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 -4 -3 -2 -1 0 1 2 3 4 t x x ' = x Figure 4.2: The zero solution is unstable for y0 = y. Lemma 4.1.1. Let fi > 0 be a real number an let m ‚ 0 be an integer. Then there is a constant C > 0 (depending on fi and m) such that tme¡fit • C; 8t ‚ 0: Proof. In the case m = 0, we can take C = 1. In the other cases, the function g(t) = tme¡fit satisfles g(0) = 0 and we know from Calculus that g(t)! 0 as t!1. Thus, g is bounded on [0;1). 4.1. BASIC CONCEPTS 3 Lemma 4.1.2. Let ‚ be a complex number and let m ‚ 0 be an integer. Suppose that Re(‚) < ¾. Then there is a constant C such thatflfltme‚tflfl • Ce¾t: Proof. Suppose flrst that fi is real and fi < ¾. Then, fi¡¾ is negative, so by the last lemma there is a constant C such that tme(fi¡¾)t • C; 8t ‚ 0: Multiplying this inequality by e¾t yields tmefit • Ce¾t; 8t ‚ 0: Suppose that ‚ is complex, say ‚ = fi + ifl, where fi and fl are real. If fi = Re(‚) < ¾, then flfltme‚tflfl = tmefit • Ce¾t; 8t ‚ 0: Lemma 4.1.3. Let P (‚) be a polynomial of degree n (with complex coe–cients). Let ‚1; : : : ; ‚k be the roots of P (‚) = 0 and suppose that Re(‚j) < ¾ for j = 1: : : : ; k. Then, if y is a solution of the difierential equation P (D)y = 0, there is a constant C ‚ 0 such that jy(t)j • Ce¾t; 8t ‚ 0: Proof. We can flnd a fundamental set of solutions y1; : : : ; yn, where each yj(t) is of the form tme‚‘t for some integer m and root ‚‘. By the last lemma, there is a constant Kj such that jyj(t)j • Kje¾t; 8t ‚ 0: If y is an arbitrary solution of P (D)y = 0, then y = c1y1 + ¢ ¢ ¢+ cnyn for some constants cj. Then, for t ‚ 0, jy(t)j = jc1y1(t) + ¢ ¢ ¢+ cnyn(t)j • jc1jjy1(t)j+ ¢ ¢ ¢+ jcnjjyn(t)j • jc1jK1e¾t + ¢ ¢ ¢+ jcnjKne¾t = µ jc1jK1 + ¢ ¢ ¢+ jcnjKn ¶ e¾t This completes the proof. 4 CHAPTER 4. STABILITY THEORY Theorem 4.1.4. Let A be an n £ n matrix and let ‚1; : : : ; ‚k be the (distinct) eigenvalues of A. Suppose that Re(‚j) < ¾ for j = 1; : : : ; k. Then there is a constant K such that keAtk • Ke¾t; 8t ‚ 0: Proof. Let P (‚) the characteristic polynomial of A. The roots of P (‚) are the same as the eigenvalues of A. By our algorithm for constructing eAt, we have eAt = n¡1X j=0 rj+1(t)A j; where each rj is a solution of P (D)r = 0. By the last lemma, there is a constant Cj such that jrj(t)j • Cje¾t for positive t. But then, for t ‚ 0, keAtk • •n¡1X j=0 Cj+1kAkj ‚ e¾t: Lemma 4.1.5. Let P (‚) be a polynomial of degree n. Suppose that every root of P (‚) has a nonpositive real part and that the roots with real part zero are simple. Then, if y is a solution of P (D)y = 0, there is a constant C such that jy(t)j • C; 8t ‚ 0: Proof. If ‚ is a root with negative real part, it contributes functions of the form tke‚t to the set of fundamental solutions. But, we know that these functions goes to zero as t goes to inflnity, and so is surely bounded on the right half axis. If ‚ = ifl is a simple pure imaginary root, it contributes one function eiflt to the funda- mental set of solutions, and this function is bounded. Thus, we get a fundamental set of solutions which are all bounded on [0;1). It follows easily that every solution is bounded on [0;1). (If we had a non-simple imaginary root, we would get a function like teiflt, which is not bounded, in our fundamental set of solutions.) Finally, we have the following theorem, which follows readily from the last lemma and an argument similar to the proof of Theorem 4.1.4. Theorem 4.1.6. Let A be an n £ n matrix and suppose that all of the eigenvalues A have real part less than or equal to zero, and that the eigenvalues with zero real part are simple (as roots of the characteristic polynomial). Then, there is a constant K such that keAtk • K; 8t ‚ 0: 4.1. BASIC CONCEPTS 5 Notation 4.1.7. If A is a square matrix, we will use ¾(A), the spectrum of A, to denote the collection of eigenvalues of A, counted with their multiplicities as roots of the characteristic polynomial. We write Re(¾(A)) < ¾ to say that all of the eigenvalues of A have real part less than ¾. For an nth order linear constant coe–cient equation (with real coe–cients) P (D)z = z(n) + a1z (n¡1) + a2z(n¡2) + ¢+ anz = 0; (4.1.2) the following classical theorem gives implicitly a test for the stability of solutions (i.e., the vanishing of solutions as t approaches inflnity) based on the coe–cients fajg of P (D). Theorem 4.1.8. If all the zeros of the characteristic polynomial P (‚) = ‚n + a1‚ (n¡1) + ¢ ¢ ¢+ an of (4.1.2) have negative real part, then given any solution z(t) there exists numbers a > 0 and M > 0 such that jz(t)j •Me¡at; t ‚ 0: Hence lim t!1 jz(t)j = 0. Theorem 4.1.9. (Routh-Hurwitz Criteria) Given the equation (4.1.2) with real coe–cients fajgnj=1. Let D1 = a1; D2 = det • a1 a3 1 a2 ‚ ; ¢ ¢ ¢ ; Dk = det 266666664 a1 a3 a5 ¢ ¢ ¢ a2k¡1 1 a2 a4 ¢ ¢ ¢ a2k¡2 0 a1 a3 ¢ ¢ ¢ a2k¡3 0 1 a2 ¢ ¢ ¢ a2k¡4 ... ... ... ... 0 0 0 ¢ ¢ ¢ ak 377777775 : (4.1.3) where aj = 0 if j > n. Then the roots of P (‚), the characteristic polynomial of (4.1.2), have negative real part if and only Dk > 0 for all k = 1; ¢ ¢ ¢ ; n. To formalize these notions, we make the following deflnitions. Deflnition 4.1.10. Consider a difierential equation x0(t) = f(t; x(t)), where x(t) 2 Rn. We assume, of course, that f is continuous and locally Lipschitz with respect to the second variable. Let t 7! x(t; t0; x0) denote the maximally deflned solution of the equation satisfying the initial condition x(t0) = x0. Let ’ : [t0;1)! Rn be a solution of the difierential equation. 6 CHAPTER 4. STABILITY THEORY 1. We say that the solution ’ is stable on [t0;1) if, for every " > 0, there is a – > 0 such that whenever j’(t0)¡x0j < –, the solution x(t; t0; x0) is deflned for all t 2 [t0;1) and j’(t)¡ x(t; t0; x0)j < "; 8 t ‚ t0: 2. We say that ’ is asymptotically stable (on [t0;1)) if it is stable and, given " as above, there is a –1 < – such that whenever j’(t0)¡ x0j < –1, we have lim t!1 j’(t)¡ x(t; t0; x0)j = 0: 3. If ’ is not stable, we say that it is unstable. This means that there is some " > 0 such that for every – > 0 there is some point x0 with j’(t0) ¡ x0j < – such that j’(t1)¡ x(t1; t0; x0)j ‚ " for some time t1 2 [t0;1). For autonomous systems x0(t) = f(x(t)), the initial time t0 does not play any essential role and we usually use the interval [0;1) when discussing stability (see the discussion below). Frequently we wish to examine the stability of an equilibrium point. A point xe is an equilibrium point of the difierential equation x0(t) = f(t; x(t)) if f(t; xe) = 0 for all t. This means that the solution with initial condition x(t0) = xe is x(t) · xe. In other words, if you start the system at xe, it stays there. Thus, in discussing the stability of an equilibrium point, we are considering the stability of the solution ’(t) · xe. One also sees the terms \flxed point" and sometimes \singular point" used for an equilibrium point. In analyzing what happens at flxed points, it is often useful to observe that one can assume that xe = 0, without loss of generality. To see this, suppose that xe is an equilibrium point of x0(t) = f(t; x(t)) and that x(t) = x(t; t0; x0) is some other solution. Let y(t) = x(t)¡ xe. Then, y0(t) = x0(t) = f(t; x(t)) = f(t; y(t) + xe) = g(t; y(t)) where we deflne g(t; y) = f(t; y + xe). Thus, g has an equilibrium point at 0 and studying the dynamics of y0(t) = g(t; y(t)) near zero is the same as studying the dynamics of x0(t) = f(t; x(t)) near xe. 4.2 Stability of Linear Systems 4.2.1 Constant Coe–cients Consider the linear homogeneous system x0(t) = Ax(t); (LH) 4.2. STABILITY OF LINEAR SYSTEMS 7 where A is an constant n £ n matrix. The system may be real or complex. We know, of course, that the solution is x(t) = eAtx0; x(0) = x0: Thus, the origin is an equilibrium point for this system. Using the results of the last section, we can characterize the stability of this equilibrium point. Theorem 4.2.1. Let A be an n£ n matrix and let the spectrum of A (i.e., the eigenvalues of A) be denoted by ¾(A) and consider the linear system of difierential equations (LH). 1. If Re(¾(A)) • 0 and all the eigenvalues of A with real part zero are simple, then 0 is a stable flxed point for (LH). 2. If Re(¾(A)) < 0, then 0 is a globally asymptotically stable solution of (LH). 3. If there is an eigenvalue of A with positive real part, then 0 is unstable. Remark 4.2.2. In the case where Re(¾(A)) • 0 but there is a multiple eigenvalue with zero real part, further analysis is required to determine the stability of 0. For example, consider x0 = Ajx where A1 = • 0 0 0 0 ‚ ; A2 = • 0 1 0 0 ‚ : In both cases we have a double eigenvalue with zero real part (namely ‚ = 0), but the origin is stable for x0 = A1x and unstable for x0 = A2x. Proof of Theorem. Suppose flrst that Re(¾(A)) • 0 and all imaginary eigenvalues are simple. By the results of the last chapter, we can flnd a constant K > 0 such that keAtk • K; t ‚ 0: Let " > 0 be given. Choose – = "=K. If x0 is an initial condition with j0¡x0j = jx0j < –, then j0¡ x(t; 0; x0)j = jeAtx0j • keAtk jx0j • Kjx0j < K("=K) = ": This shows that the zero solution is stable. Now suppose that Re(¾(A)) < 0. Then the zero solution is stable by the flrst part of the proof. We can choose a real number w < 0 such that Re(‚j) < w for all eigenvalues ‚j of A. By the results of the last chapter, there is a constant K such that keAtk • Kewt; 8t ‚ 0: 8 CHAPTER 4. STABILITY THEORY But then for any initial condition x0, jx(t; 0; x0)j = jeAtx0j • Kjx0jewt; 8t ‚ 0: Since w is negative, ewt ! 0 as t!1. Thus, x(t; 0; x0)! 0 for any initial condition x0. For the last part of the proof, consider flrst the complex case. Suppose that we have an eigenvalue ‚ = fi+ ifl with fi > 0. Let v be an eigenvector of A belonging to the eigenvalue ‚. The solution of the system with initial condition v is eAtv = e‚tv. Let " > 0 be given. If we let ‰ = "=(2jvj), then j‰vj = "=2 < ". On the other hand, the solution x(t) of the system with initial condition ‰v is x(t) = eAt‰v = ‰e‚tv. Thus, jx(t)j = ("=2)efit. Since fi > 0, we see that jx(t)j ! 1 as t!1. Thus, every neighborhood of 0 contains a point that escapes to inflnity under the dynamics of the system, so 0 is unstable. Consider the case where A has real entries. We would like to know what the dynamics of the system are on Rn. If there is a positive real eigenvalue, the argument above shows that there are real initial conditions arbitrarily close to zero that go to inflnity under the dynamics. What happens when we have a nonreal eigenvalue ‚ = fi + ifl, where fi > 0 and fl 6= 0? There is a complex eigenvector w for this eigenvalue. Since A has real entries, „‚ is also an eigenvalue with eigenvector „w. The vector w and „w are linearly independent in Cn, since they are eigenvectors for distinct eigenvalues. Write w = u + iv, where u and v have real entries. We claim that u and v are linearly independent in Rn. To see this, suppose that we have real numbers a; b such that au+ bv = 0. Then we have 0 = au+ bv = a 2 (w + „w)¡ bi 2 (w ¡ „w) = ( a 2 ¡ b 2 i)w + ( a 2 + b 2 i) „w: The coe–cients of w and „w must be zero, since these vectors are independent. But this implies that a and b are zero. We have, of course, e‚t = efit cos(flt) + iefit sin(flt). Since w is an eigenvector we have eAtw = e‚tw = (efit cos(flt) + iefit sin(flt)(u+ iv) = [efit cos(flt)u¡ efit sin(flt)v] + i[efit sin(flt)u+ efit cos(flt)v]: On the other hand, eAtw = eAtu+ ieAtv since A, and hence eAt, are real. Equating real and imaginary parts gives us eAtu = efit cos(flt)u¡ efit sin(flt)v (4.2.4) eAtv = efit sin(flt)u+ efit cos(flt)v: (4.2.5) 4.2. STABILITY OF LINEAR SYSTEMS 9 In particular, consider the solution x(t) of the difierential equation x0 = Ax with the initial condition ‰u, ‰ > 0. We have x(t) = ‰eAtu and so jx(t)j = ‰jeAtuj = ‰jefit cos(flt)u¡ efit sin(flt)vj = ‰efitjcos(flt)u¡ sin(flt)vj: (4.2.6) Consider the function h(t) = jcos(flt)u¡ sin(flt)vj: This function is never zero: If h(t) = 0, we would have to have cos(flt) = 0 and sin(flt) = 0 because u and v are linearly independent. But there is no point a which both sine and cosine vanish. On the other hand, h is clearly continuous and it is periodic of period 2…=fl. Thus, it assumes all of its values on the compact interval [0; 2…=fl], and so its minimum value M is strictly greater than zero. If we go back to (4.2.6), we see that jx(t)j ‚ ‰efitM . Since fi > 0, we see that jx(t)j ! 1 as t goes to inflnity. By choosing ‰ small we can make the initial condition ‰u as close to 0 as we want. Thus, the origin is unstable for the real system. 4.2.2 Autonomous Systems in the Plane Many important applications can be written as two-dimensional autonomous systems in the form x0 = P (x; y) y0 = Q(x; y): (4.2.7) The systems are called autonomous because P and Q do not depend explicitly on t. By deflning z = • x y ‚ and f(z) = • P (z) Q(z) ‚ we can the system in the for z0 = f(z). Note that 2nd order equations of the form x00 = g(x; x0) can be written in the form (4.2.7) as the system x0 = y; y0 = g(x; y): Following the presentation in [1], we describe some properties of planar autonomous systems. Lemma 4.2.3. If x = x(t), y = y(t), r1 < t < r2, is a solution of (4.2.7), then for any real number c the functions x1(t) = x(t+ c); y1(t) = y(t+ c) are solutions of (4.2.7). 10 CHAPTER 4. STABILITY THEORY Proof. Applying the chainrule we have x01 = x 0(t+ c), y01 = y 0(t+ c) we have x01 = x 0(t+ c) = P (x(t+ c); x(t+ c)) = P (x1; y1); y01 = y 0(t+ c) = P (x(t+ c); x(t+ c)) = P (x1; y1): So x1, y1 gives a solution of (4.2.7) which is deflned on r1 ¡ c < t < r2 ¡ c. This property does not hold in general for non-autonomous systems: Consider x0 = x; y0 = tx: A solution is x(t) = et, y(t) = (t¡ 1)et and we have y0(t+ c) = (t+ c)et+c 6= tx(t) unless c = 0. As t varies, a solution x = x(t) , y = y(t)4 of (4.2.7) describes parametrically a curve in the plane. This curve is called a trajectory (or orbit). Lemma 4.2.4. Through any point passes at most one trajectory of (4.2.7). Proof. Let C1 : with representation x = x1(t), y = y1(t) and C2 : with representation x = x2(t), y = y2(t) be two distinct trajectories with a common point (x0; y0). Then there exists times t1, t2 such that (x0; y0) = (x1(t1); y1(t1)) = (x2(t2); y2(t2)) Then t1 6= t2, since otherwise the uniqueness of solutions would be contradicted (i.e., the fundamental uniqueness and existence theorem). Now by Lemma 4.2.3, x(t) = x1(t+ t1 ¡ t2); y(t) = y1(t+ t1 ¡ t2) is a solution. Now (x(t2); y(t2)) = (x0; y0) implies that x(t) and y(t) must agree respectively with x2(t) and y2(t) by uniqueness. Thus C1 and C2 must coincide. Note the distinction: A trajectory is a curve that is represented parametrically by one or more solutions. Thus x(t), y(t) and x(t + c), y(t + c) for c 6= 0 represent distinct solutions but the same trajectory. In order to get some intuition about what goes on near the origin for the linear system x0 = Ax, we will study in some detail what happens for a real two dimensional system. Thus, we study the system • x y ‚0 = • a b c d ‚ • x y ‚ ; (4.2.8) 4.2. STABILITY OF LINEAR SYSTEMS 11 where the entries of the matrix A = • a b c d ‚ are real numbers. The characteristic polynomial of A is easily computed to be P (‚) = ‚2 ¡ (a+ d)‚+ ad¡ bc = ‚2 ¡ tr(A)‚+ det(A) (4.2.9) If ‚1 and ‚2 are the eigenvalues of A (not necessarily distinct), we have P (‚) = (‚¡ ‚1)(‚¡ ‚2) = ‚2 ¡ (‚1 + ‚2)‚+ ‚1‚2 Thus, we have the identities ‚1 + ‚2 = tr(A) (4.2.10) ‚1‚2 = det(A): (4.2.11) For brevity, let T = tr(A) and D = det(A) in the following discussion. By the quadratic formula, the eigenvalues are given by ‚1; ‚2 = T §pT 2 ¡ 4D 2 : The nature of the roots is determined by the discriminant ¢ = T 2 ¡ 4D: The data T;D; and ¢ allow us to determine the stability of the origin in many cases. For example, if ¢ > 0, we have two distinct real eigenvalues. If D > 0, the product of the eigenvalues is positive, by (4.2.11), and so they must have the same sign. By (4.2.10), the common sign is the same as the sign of T . If D < 0, the eigenvalues are non-zero and have opposite sign. If D = 0 one of the eigenvalues is zero and the other is not. The nonzero eigenvalue is equal to T . If ¢ = 0, we have a real eigenvalue ‚ = T=2 of multiplicity two. If T < 0, the origin is stable and if T > 0 the origin is unstable. If T = 0, determining the stability of the origin requires additional data, as we saw in Remark 4.2.2. If ¢ < 0, we have two complex conjugate eigenvalues ‚1 = fi + ifl and ‚2 = fi¡ ifl with fl 6= 0. In this case T = 2fi. Thus, if T 6= 0, we can determine the stability of the origin form sign of T . If T = 0, we have two simple pure imaginary eigenvalues, so the origin is stable. We can summarize all of this in the following Proposition. 12 CHAPTER 4. STABILITY THEORY Proposition 4.2.5. Consider the system (4.2.8) and let D = det(A), T = tr(A) and ¢ = T 2 ¡ 4D. Then, we have the following cases. 1. The origin is globally asymptotically stable in the following cases. (a) ¢ > 0, T < 0. (b) ¢ = 0, T < 0. (c) ¢ < 0, T < 0. 2. The origin is stable, but not asymptotically stable, in the following cases. (a) ¢ > 0, D = 0, T < 0. (b) ¢ < 0, T = 0. 3. The origin is unstable in the following cases. (a) ¢ > 0, D ‚ 0, T > 0. (b) ¢ = 0, T > 0. (c) ¢ < 0, T > 0. 4. In the case where T = 0 and D = 0, further analysis is required to determine the stability of the origin. We next consider what the behavior of the system (4.2.8) is in the various cases for the eigenvalues. Case 1. Distinct real eigenvalues ‚1 6= ‚2. In this case, the matrix A can be diagonalized. So, by a change of coordinates, the system can be transformed to the system • x y ‚0 = • ‚1 0 0 ‚2 ‚ • x y ‚ which has the solution x(t) = x0e ‚1t; x0 = x(0) (4.2.12) y(t) = y0e ‚2t; y0 = y(0): (4.2.13) Let us consider the cases for the signs of ‚1 and ‚2. Subcase 1A. Both eigenvalues are negative. 4.2. STABILITY OF LINEAR SYSTEMS 13 Say, for deflniteness, that we have ‚2 < ‚1 < 0. In this case, (x(t); y(t))! 0 as t!1 for any initial conditions, so the origin is asymptotically stable. In this case, we have ‚2=‚1 > 1. If x0 = 0, the integral curve approaches the origin along the y-axis. If x0 6= 0, jx(t)=x0j = e‚1t. so y(t) = y0e ‚2t = y0(e ‚1t)‚