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.UNIT III: HIGHER-ORDER LINEAR EQUATIONS We give a comprehensive development of the theory of linear differential equations with con-stant coefficients. We use the operator calculus to deduce the existence and uniqueness. We presents techniques for finding a complete solution of the inhomogeneous equation from solu-tions of the homogeneous equation. We also give qualitative results on asymptotic stability. LECTURE 11. HIGHER-ORDER LINEAR EQUATIONS The n-th order linear differential equation with constant coefficient is (11.1) Ly = y(n) + p1y(n−1) + + pn−1y� + pny = f(t),··· where y(k) = dky is the k-th derivative of y with respect to t, and pj are real or complex constants, dtk f(t) is a continuous function on an interval I. The letter L stands for the (homogeneous) differen-tial operator. It is easy to see that L : Cn(I) C(I) is linear, where Ck(I) is the space of functions →differentiable k times on I. As for the second-order equations treated in Unit II, the principle of superposition and the principle of the complementary solution apply to (11.1). Principle of Superposition. If Lu = 0 and Lv = 0, where L is given in (11.1), then L(c1u+c2v) = 0 for any constants c1 and c2. Principle of the Complementary Solution. If u is a particular solution of Lu = f, where L is given in (11.1), and if v is any solution of Lv = 0, then L(u + v) = f and every solution of Ly = f can be obtained this way. Therefore, the general solution of (11.1) is given as y = yp + yh, where yp is a particular solution of (11.1) and yh is a solution of the corresponding homogeneous equation (11.2) Ly = y(n) + p1y(n−1) + + pn−1y� + pny = 0.··· The characteristic polynomial. We try y(t) = eλt , λ ∈ C, as a solution of the homogeneous equa-dk tion (11.2). Since (e λt) = λkeλt, the substitution yields dtk Leλt = (λn + p1λn−1 + + pn−1λ + pn)e λt = 0.··· Moreover, since eλt is never zero, Leλt = 0 if and only if λ is a root of the characteristic polynomial (11.3) pL(λ) = λn + p1λn−1 + ........... + pn−1λ + pn. of L. 1� � Example 11.1. We recall the results for the second-order equation (11.4) Ly = y�� + py� + qy, where p , q are constants. The roots of the characteristic polynomial are λ = −p ±√Δ , Δ = p 2 − 4q. 2 If p2 > 4q then y = eλt with the above λ are solutions of Ly = 0. We will show how the exponential substitution y = eλt applies to solve the general differential equations of all order (11.2). The operator calculus. The study of linear differential equations become easier if we introduce an abstract symbol D = d/dt for the operation of differentiation. A word of caution. The symbol d is used for differentials. e.g., D(t3) = 3t2, but d(t3) = 3t2dt. As an operator, D is linear. That is, D(u + v) = Du + Dv, D(cu) = cDu for all differentiable functions u, v and for any constant c. We now list some properties of D. By definition, dk D0 = id, Dk = k = 1, 2, . . . . dtk Moreover, (11.5) Dj Dk = Dj+k , (Dj )k = Djk, j, k = 1, 2, . . . . The proof is left as an exercise. With the notation D, we may write the differential operator in (11.2) as Ly = (Dn + p1Dn−1 + + pn−1D + pn)y. ··· Then, it can be recognized that the first factor of the right side is pL(D), the characteristic polyno-mial pL evaluated, formally, at D. In this sense, we say L = pL(D). Linear operators with constant coefficients are permutable, in the sense of the following. Lemma 11.2. If p(D) = aj Dj and q(D) = bkDk are two linear differential operator, where aj , bk are constants. then, � p(D)q(D) = q(D)p(D) = aj bkDj+k . The proof uses (11.5) and it is left as an exercise. Remark 11.3. The above lemma is not true of linear operators with variable coefficients. For exam-ple, D(tf) = (tf)� = tf� + f = (tD + id)f, where f is a differentiable function of t. In other words, Dt = tD + id. In many applications, one takes trial solutions of the form eλtu where λ ∈ C and u is a function with a certain degree of smoothness. Thus, it is useful to know how such a function works with the operator D. Lemma 11.4 (The Exponential Shift Laws). If p is a polynomial and λ is a constant, then p(D)(e λtf) = e λt p(D + λ)f. Lecture 11 2 18.034 Spring 2009� �Proof. By the rule for differentiation, D(e λtf) = e λtDf + λeλt = e λt(D + λ)f. Hence, (D − a)(e λtf) = e λt(D − a + λ)f for any constant a. Then, by induction, (11.6) (D − a)k(e λtf) = e λt(D − a + λ)kf. Finally, by the fundamental theorem of calculus, any polynomial p can be factorized as p(D) = (D − a1)k1 (D − a2)k2 (D − am)km , ··· where aj ∈ C are roots of the polynomial and kj � 1 are the corresponding multiplicity. The assertion then follows by Lemma 11.2 and (11.6). � As a consequence, moreover, we have (D − λ)(e λtf) = e λtDf, (D − λ)k(e λtf) = e λtDkf. Exercise. If a is not a root of the polynomial p , then show that ateb(t) = p(a) is a particular solution of the differential equation p(D)y = eat . We now present our main result. Theorem 11.5. If λ is a (complex) root of multiplicity k of the characteristic polynomial p(λ) = λn +∗ p1λn−1 + + pn−1λ + pn of the linear differential operator p(D) with constant coefficients, then the ··· functions treλ∗t, where r = 0, 1, . . . , k − 1, are solution of p(D)y = 0. Proof. By the exponential shift law, it follows that r r(D − λ )k(t e λ∗t) = e λ∗tDkt = 0 ∗for r = 0, 1, . . . , k − 1. On the other hand, p(λ) must contain the factor (λ − λ )k, and hence ∗p(D) = (D − λ )k q(D), q(D) = (D − λj )kj .∗λj =λ Finally, by Lemma 11.2 p(D)(tr e λ∗t) = q(D)(D − λ )k(tr e λ∗t) = 0.∗This completes the proof. � Corollary 11.6. If p(λ) = (λ − λ1)k1 (λ − λ2)k2 (λ − λm)km · then the functions treλj t , where r = 0, 1, 2, . . . , λj−1 and j = 1, 2, . . . , m, are solutions of the differential equation p(D)y = 0. Lecture 11 3 18.034 Spring 2009Complex solutions. An interesting feature of the analysis via the operator calculus is that many problems are best solved by the …
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