MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.LECTURE 14. STABILITY The notion of stability. Roughly speaking, a system is called stable if its long-term behavior does not depend on significantly the initial conditions. An important result of mechanics is that a system of masses attached in (damped or undamped) springs is stable. A similar result is in network theory. In these notes, we study the differential equation of the form (14.1) y�� + py� + qy = f (t), where p, q are constatnts and f(t) represents the external forces. We learned that the general solution of (14.1) has the form (14.2) y = c1y1 + c2y2 + yp, where c1, c2 are arbitrary constants and yp is a particular solution of (14.1); c1y1 + c2y2 is the complementary solution, that is, the general solutions of the homogeneous equation (14.1) with f(t) = 0. The initial conditions determine the values of c1 and c2. Thus, we say the system (14.1) is stable if c1y1 + c2y2 → 0 as t → ∞ for any coice of c1 and c2. If (14.1) is stable then yp is called the steady-state solution and c1y1 + c2y2 is called transient. Physically, in a stable system, the output is the sum of a transient term, which depends on the initial conditions, but whose effects die out over time, and a steady-state, which represents the response of the system to the input f(t) after a long time. Stability conditions. We study under what circumstances the differential equation Ly = f, where (14.3) L = Dn + p1Dn−1 + ......... + pn−1D + pn, where pj are constants, is stable. Definition 14.1. The differential equation Ly = f, where L is given in (14.3) is called: (i) asymptotically stable if every solution of Ly = 0 tends to zero as t → ∞; (ii) stable if every solution of Ly = 0 remains bounded a t → ∞; (iii) unstable if it is not stable. We note that stability concerns only the behavior of the solutions of the corresponding homo-geneous equation Ly = 0. When f (t) = 0, then a steady-state solution is y ≡ 0. In this case, the system is stable if small initial departures from the steady-state remain small with the lapse of time. By definition, Ly = f is asymptotically stable if every basis solution of Ly = 0 tends to zero as t → ∞ and it is stable if the basis solutions remain bounded. In view of the characteristic polynomial of L and the fundamental theorem of algebra, we write L = (D − λ1)k1 (D − λ2)k2 (D − λm)km ,··· where λj ∈ C are all distinct and k1 + k2 + ...... + km = n. Exercise. The general solution of the homogeneous equation Ly = 0 is given by y(t) = c1(t)e λ1t + c2(t)e λ2t ...... + cm(t)e λmt , 1��������� ��������� where cj (t) is an arbritary polynomial of degree kj − 1. Exercise. If r is a nonnegative integer and λ ∈ C, show that lim |tr e λt| = 0 if Reλ < 0. t→∞ Therefore, Ly = f is asymptotically stable if Reλj < 0 for all j, and it is stable if Reλj < 0 or Reλj = 0 and kj = 1. We summarize the result. Theorem 14.2. The differential equation Ly = f is asymptotically stable if every root of the characteristic polynomial of L has a negative real part, and it is stable if every multiple root has a negative real part and no simple root has a positive real part. Example 14.3. We consider the second-order differential equation (14.4) y�� + py� + qy = 0, p, q are constants. We recall that the discriminant Δ = p2 − 4q tells us about the nature of the solutions, and hence about the stability of (14.4) If q < 0 then Δ > 0 and the characteristic polynomial λ2 +pλ+q has two real roots with opposite signs. Therefore, (14.4) is unstable. If p < 0 then at least one root of the characteristic polynomial must have a positive real part. Hence, (14.4) is unstable. If p = 0 and q > 0, then (14.4) reduces to y�� + qy = 0 with q > 0. Hence, it is stable but asymptotically stable. Finally, let p > 0 and q > 0. If Δ � 0 then the roots of the characteristic polynomial have negative real parts, and (14.4) is asymptotically stable. If Δ > 0 then Δ = p2 − 4q < p2 and thus √Δ < p. Therefore, (14.4) is asymptotically stable. In summary, (14.4) is asymptotically stable if and only if p > 0 and q > 0, and stable if and only if p � 0 and q > 0. Stability of higher-order differential equations. The above example phrases the stability crite-rion for (14.4) in terms of the coefficients of the equation. This is convenient since it does not require one to calculate the roots of the characteristic polynomial. For higher-order equations, (14.5) y(n) + p1y(n−1) + + pn−1y� + pny = 0, pj are constants, ··· it is not too hard to show that if (14.5) is asymptotically stable then pj > 0 for all j (Exercise). But, the converse is not true (Exercise). For the implication of a criterion for coefficients of (14.5) for stability, the coefficients must satisfy a more complicated set of inequalities, which we state without proof in the following. Routh-Hurwitz Criterion for Stability. The differential equation (14.5) is asymptotically stable if and only if in the determinant p1 1 0 0 ... 0 p3 p2 p1 1 ... 0 . . . . . . . . . . . . . . . . . . p2n−1 p2n−2 pn··· ··· ··· , Lecture 14 2 18.034 Spring 2009���� ���� ������ ������ ������ ������ where pk = 0 if k > n, all of its n principal minors, that is, the subdeterminants in the upper left corner having sizes respectively 1, 2, . . . , n, p1 1 0 p1, p1 1 p3 p2 , p3 p2 p1 p5 p4 p3 , . . . are positive. Exercise. We consider (D4 + 2D3 + 6D2 + 5D + 2)y = 260 sin 2t. (a) Find a particular solution. (Answer. 11 cos 2t − 3 sin 2t.) (b) Show that the corresponding characteristic polynomial is factorized as p(λ) = (λ2 + 3λ + 2)(λ2 + λ + 1), and hence the zeros have negative real parts. (c) Show that the determinant 4 1 0 0 5 6 4 1 0 0 0 2 satisfies the Routh-Hurwitz criterion. Lecture 14 3 18.034 Spring
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