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 consttan 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 solutiion 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) differentiia 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 equadd 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 polynomiia 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 examplle 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 asp(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 λis 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 rr(D − λ)k(te 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 2009 Complex solutions. An interesting feature of the analysis via the operator calculus is that many problems are best solved by the use of complex-valued functions, even when the coefficients are real, e.g. (11.4) when p2 < 4q. Theorem 11.5 holds whether the coefficients pj of p(D)y =0 are real or complex. In fact, althooug t is interpreted as real (in particular in the discussion of stability) the operator calculus and the solutions constructed with it apply equally well to functions of the complex variable. But, when the coefficients pj are real, then we may obtain a stronger conclusion. We recall that the complex roots of a polynomial come in pair, µ ± iν, µ,ν ∈ R. We also recall the complex exponential eµ±iν = eµ(cos ν ± iν). Lemma 11.7 (Principle of Equating Real Parts). If a complex-valued function y(t)= u(t)+iv(t), where u and v are real-valued functions, satisfies the differential equation (11.2) with real coefficients, then u(t) and v(t), the real and imaginary parts of y, both satisfy (11.2). Proof. Let Ly = L(u + iv)=0. Since the coefficients of L are real, taking the complex conjugate we have Ly = L(u − iv)=0. Then, by linearity, u = y + y and v = y − y 22i both satisfy (11.2). � It holds true for differential equations with real-variable coefficients. Corollary 11.8. Each pair of complex roots µ ± iν, µ,ν ∈ R, of multiplicity k of the polynomial p gives real solutions treut cos νt ,treut sin νt, where r =0, 1,...,k − 1, of the differential equation p(D)y =0. Lecture 11 4 18.034 Spring 2009