Linear Systems of Differential Equations Michael Taylor

Transcription

1 Linear Systems of Differential Equations Michael Taylor Contents. The matrix exponential 2. Exponentials and trigonometric functions 3. First order systems derived from higher order equations 4. Nonhomogeneous equations and Duhamel s formula 5. Simple electrical circuits 6. Second order systems 7. Curves in R 3 and the Frenet-Serret equations 8. Variable coefficient systems 9. Variation of parameters and Duhamel s formula. Power series expansions. Regular singular points A. Logarithms of matrices Introduction This chapter connects the linear algebra developed in Chapter 2 with Differential Equations. We define the matrix exponential in and show how it produces the solution to first order systems of differential equations with constant coefficients. We show how the use of eigenvectors and generalized eigenvectors helps to compute matrix exponentials. In 2 we look again at

2 2 Linear Systems of Differential Equations Michael Taylor connections between exponential and trigonometric functions, complementing results of Chapter,. In 3 we discuss how to reduce a higher order differential equation to a first order system, and show how the companion matrix of a polynomial arises in doing this. We show in 4 how the matrix exponential allows us to write down an integral formula (Duhamel s formula) for the solution to a non-homogeneous first order system, and illustrate how this in concert with the reduction process just mentioned, allows us to write down the solution to a non-homogeneous second order differential equation. Section 5 discusses how to derive first order systems describing the behavior of simple circuits, consisting of resistors, inductors, and capacitors. Here we treat a more general class of circuits than done in Chapter, 3. Section 6 deals with second order systems. While it is the case that second order n n systems can always be converted into first order (2n) (2n) systems, many such systems have special structure, worthy of separate study. Material on self adjoint transformations from Chapter 2 plays an important role in this section. In 7 we discuss the Frenet-Serret equations, for a curve in three-dimensional Euclidean space. These equations involve the curvature and torsion of a curve, and also a frame field along the curve, called the Frenet frame, which forms an orthonormal basis of R 3 at each point on the curve. Regarding these equations as a system of differential equations, we discuss the problem of finding a curve with given curvature and torsion. Doing this brings in a number of topics from the previous sections, and from Chapter 2, such as the use of properties of orthogonal matrices. Having introduced equations with variable coefficients in 7, we concentrate on their treatment in subsequent sections. In 8 we study the solution operator S(t, s) to a homogeneous system, show how it extends the notion of matrix exponential, and extend Duhamel s formula to the variable coefficient setting. In 9 we show how the method of variation of parameters, introduced in Chapter, ties in with and becomes a special case of Duhamel s formula. Section treats power series expansions for a first order linear system with analytic coefficients, and extends the study to equations with regular singular points. These sections provide a systematic treatment of material touched on in Chapter, 5. In these sections we use elementary power series techniques. Additional insight can be gleaned from the theory of functions of a complex variable. Readers who have seen some complex variable theory can consult [Ahl], pp , [Hart], pp. 7 83, or [T], Vol., pp. 3 3, for material on this.

3 . The matrix exponential 3 Appendix A treats logarithms of matrices, a construction inverse to the matrix exponential introduced in, establishing results that are of use in 8 and.. The matrix exponential Here we discuss a key concept in matrix analysis, the matrix exponential. Given A M(n, F), F = R or C, we define e A by the same power series used in Chapter to define e A for A R: (.) e A = k= k! Ak. Note that A can be a real or complex n n matrix. In either case, recall from of Chapter 2 that A k A k. Hence the standard ratio test implies (.) is absolutely convergent for each A M(n, F). Hence (.2) e ta = k= t k k! Ak is a convergent power series in t, for all t R (indeed for t C). As for all such convergent power series, we can differentiate term by term. We have (.3) d dt eta = = k= k= k tk A k k! t k (k )! Ak A. We can factor A out on either the left or the right, obtaining (.4) d dt eta = e ta A = Ae ta. Hence x(t) = e ta x solves the first-order system (.5) dx dt = Ax, x() = x. This is the unique solution to (.5). To see this, let x(t) be any solution to (.5), and consider (.6) u(t) = e ta x(t).

4 4 Linear Systems of Differential Equations Michael Taylor Then u() = x() = x and (.7) d dt u(t) = e ta Ax(t) + e ta x (t) =, so u(t) u() = x. The same argument yields (.8) Hence x(t) = e ta x, as asserted. d ( e ta e ta) =, hence e ta e ta I. dt Using a variant of the computation (.7), we show that the matrix exponential has the following property, which generalizes the identity e s+t = e s e t for real s, t, established in Chapter. Proposition.. Given A M(n, C), s, t R, (.9) e (s+t)a = e sa e ta. Proof. Using the Leibniz formula for the derivative of a product, plus (.4), we have (.) d ( e (s+t)a e ta) = e (s+t)a Ae ta e (s+t)a Ae ta =. dt Hence e (s+t)a e ta is independent of t, so (.) e (s+t)a e ta = e sa, s, t R. Taking s = yields e ta e ta = I (as we have already seen in (.8)) or e ta = (e ta ), so we can multiply both sides of (.) on the right by e ta and obtain (.9). Now, generally, for A, B M(n, F), (.2) e A e B e A+B. However, we do have the following. Proposition.2. Given A, B M(n, C), (.3) AB = BA = e A+B = e A e B.

5 . The matrix exponential 5 Proof. We compute (.4) d (e t(a+b) e tb e ta) dt = e t(a+b) (A + B)e tb e ta e t(a+b) Be tb e ta e t(a+b) e tb Ae ta. Now AB = BA AB k = B k A, hence (.5) e tb A = ( t) k B k A = Ae tb, k! k= so (.4) vanishes. Hence e t(a+b) e tb e ta is independent of t, so (.6) e t(a+b) e tb e ta = I, the value at t =. Multiplying through on the right by e ta and then by e tb gives (.7) e t(a+b) = e ta e tb. Setting t = gives (.3). We now look at examples of matrix exponentials. We start with some computations via the infinite series (.2). Take ( ) ( ) (.8) A =, B =. 2 Then (.9) A k = so (.2) e ta = ( ) 2 k, B 2 = B 3 = =, ( ) ( ) e t e 2t, e tb t =. Note that A and B do not commute, and neither do e ta and e tb, for general t. On the other hand, if we take ( ) (.2) C = = I + B, since I and B commute, we have without further effort that ( (.22) e tc = e ti e tb e t te t ) = e t.

6 6 Linear Systems of Differential Equations Michael Taylor We turn to constructions of matrix exponentials via use of eigenvalues and eigenvectors. Suppose v j is an eigenvector of A with eigenvalue λ j, (.23) Av j = λ j v j. Then A k v j = λ k j v j, and hence (.24) e ta v j = k= t k k! Ak v j = k= t k k! λk j v j = e tλ j v j. This enables us to construct e ta v for each v C n if A M(n, C) and C n has a basis of eigenvectors, {v j : j n}. In such a case, write v as a linear combination of the eigenvectors, (.25) v = c v + + c n v n, and then (.26) e ta v = c e ta v + + c n e ta v n = c e tλ v + + c n e tλ n v n. We illustrate this process with some examples. Example. Take (.27) A = ( ). One has det(λi A) = λ 2, hence eigenvalues (.28) λ =, λ 2 =, with corresponding eigenvectors ( ) (.29) v =, v 2 = Hence ( ). (.3) e ta v = e t v, e ta v 2 = e t v 2. To write out e ta as a 2 2 matrix, note that the first and second columns of this matrix are given respectively by ( ) ( ) (.3) e ta and e ta.

7 . The matrix exponential 7 To compute this, we write (, ) t and (, ) t as linear combinations of the eigenvectors. We have ( ) (.32) = ( ) + ( ) ( ), = ( ) ( ) Hence ) ) ) (.33) e ta ( = ( 2 eta + ( 2 eta ( ) ( ) + e t = et 2 2 ( 2 = (et + e t ) ) 2 (et e t, ) and similarly ( ) ( (.34) e ta 2 = (et e t ) ). 2 (et + e t ) Recalling that (.35) cosh t = et + e t, sinh t = et e t, 2 2 we have ( ) (.36) e ta cosh t sinh t =. sinh t cosh t Example 2. Take (.37) A = ( ) 2. 2 One has det(λi A) = λ 2 2λ + 2, hence eigenvalues (.38) λ = + i, λ 2 = i, with corresponding eigenvectors ( ) 2 (.39) v =, v 2 = + i We have ( ) 2. i (.4) e ta v = e (+i)t v, e ta v 2 = e ( i)t v 2.

8 8 Linear Systems of Differential Equations Michael Taylor We can write (.4) to obtain (.42) and (.43) ( ) = i + ( ) 2 + i ( ) 2, 4 + i 4 i ( ) = i ( ) 2 + i ( ) 2, 2 + i 2 i ( ) e ta = i + ( ) 2 4 e(+i)t + i + i ( = et (2i + 2)e it + (2 2i)e it 4 2ie it + 2ie it 4 e( i)t ), ( ) e ta = i ( ) 2 2 e(+i)t + i ( ) 2 + i 2 e( i)t i ( = et 2ie it 2ie it ) 2 ( i)e it + ( + i)e it. ( ) 2 i We can write these in terms of trigonometric functions, using the fundamental Euler identities (.44) e it = cot t + i sin t, e it = cos t i sin t, established in of Chapter. (See 2 of this chapter for more on this.) These yield (.45) cos t = eit + e it, sin t = eit e it, 2 2i and an inspection of the formulas above gives ( ) ( ) ( ) cos t sin t (.46) e ta = e t, e ta sin t hence ( ) (.47) e ta = e t cos t sin t 2 sin t. sin t cos t + sin t ( ) 2 sin t = e t, cos t + sin t As was shown in Chapter 2, 6, if A M(n, C) has n distinct eigenvalues, then C n has a basis of eigenvectors. If A has multiple eigenvalues, C n might or might not have a basis of eigenvectors, though as shown in 7 of Chapter

9 . The matrix exponential 9 2, there will be a basis of generalized eigenvectors. If v is a generalized eienvector of A, say (.48) (A λi) m v =, then (.49) e t(a λi) v = k<m so (.5) e ta v = e tλ k<m t k k! (A λi)k v, t k k! (A λi)k v. Example 3. Consider the 3 3 matrix A used in (7.26) of Chapter 2: (.5) A = 2 3. Here 2 is a double eigenvalue and a simple eigenvalue. Calculations done in Chapter 2, 7, yield (.52) (A 2I) =, (A 2I) = 3, (A I) 6 3 =. Hence (.53) (.54) e ta e ta e 2t =, = e 2t t k (A 2I)k k! k= 3t = e 2t +, and 6 6 (.55) e ta 3 = e t 3.

10 Linear Systems of Differential Equations Michael Taylor Note that 6 (.56) e ta = e ta 3 6e ta + 3e ta. Putting these calculations together yields e 2t 3te 2t 6e t 6e 2t + 9te 2t (.57) e ta = e 2t 3e t + 3e 2t. e t Example 4. Consider the 3 3 matrix 2 (.58) A = A computation gives det(λi A) = (λ ) 3. Hence for N = A I we have Spec(N) = {}, so we know N is nilpotent (by Proposition 8. of Chapter 2). In fact, a calculation gives (.59) N = 3 3, N 2 =, N 3 = Hence (.6) e ta = e t[ I + tn + t2 2 N 2] + 3t 2 2t +3t 2 = e t 3t 3t. 3t 2 2t 3t 2 Exercises. Use the method of eigenvalues and eigenvectors given in (.23) (.26) to compute e ta for each of the following: ( ) ( ) ( ) ( ) i A =, A =, A =, A =. 2 i

11 . The matrix exponential 2. Use the method given in (.48) (.5) and illustrated in (.5) (.6) to compute e ta for each of the following: A = ( ) 2, A = 2 4 3, A = Show that e t(a+b) = e ta e tb, t = AB = BA. Hint. Set X(t) = e t(a+b), Y (t) = e ta e tb. Show that X Y = X (t) Y (t) = Be t(a+b) e ta Be tb, and hence that X Y = Be ta = e ta B, t. 4. Given A M(n, C), suppose Φ(t) is an n n matrix valued solution to Show that d Φ(t) = AΦ(t). dt Φ(t) = e ta B, where B = Φ(). Deduce that Φ(t) is invertible for all t R if and only if Φ() is invertible, and that in such a case (For a generalization, see (8.3).) e (t s)a = Φ(t)Φ(s). 5. Let A, B M(n, C) and assume B is invertible. Show that and use this to show that (B AB) k = B A k B, e tb AB = B e ta B.

12 2 Linear Systems of Differential Equations Michael Taylor 6. Show that if A is diagonal, i.e., a A =..., then e ta = e ta a nn.... e tann Exercises 7 bear on the identity (.6) det e ta = e t Tr A, given A M(n, C). 7. Show that if (.6) holds for A = A and if A 2 = B A B, then (.6) holds for A = A Show that (.6) holds whenever A is diagonalizable. Hint. Use Exercises Assume A M(n, C) is upper triangular: a a n (.62) A =..... a nn Show that e ta is upper triangular, of the form e (t) e n (t) e ta =...., e jj (t) = e ta jj. e nn (t). Deduce that (.6) holds when A has the form (.62). Then deduce that (.6) holds for all A M(n, C).. Let A(t) be a smooth function of t with values in M(n, C). Show that (.63) A() = = d dt ea(t) t= = A ().

13 . The matrix exponential 3 Hint. Take the power series expansion of e A(t), in powers of A(t). 2. Let A(t) be a smooth M(n, C)-valued function of t I and assume (.64) A(s)A(t) = A(t)A(s), s, t I. (.65) Show that Hint. Show that if (.64) holds, and apply Exercise. d dt ea(t) = A (t)e A(t) = e A(t) A (t). d dt ea(t) = d ds ea(s) A(t) e A(t) s=t, 3. Here is an alternative approach to Proposition.2. Assume (.66) A, B M(n, C), AB = BA. Show that (.67) (A + B) m = m j= ( ) m A j B m j, j ( ) m = j m! j!(m j)!. From here, show that (.68) e A+B = = m= m= j= (A + B)m m! m j!(m j)! Aj B m j. Then take n = m j and show this is (.69) so = = j= n= j= = e A e B, j! Aj j!n! Aj B n n= n! Bn (.7) e A+B = e A e B.

14 4 Linear Systems of Differential Equations Michael Taylor 4. As an alternative to the proof of (.4), given in (.3), which depends on term by term differentiation of power series, verify that, for A M(n, C), (.7) d ( dt eta = lim e (t+h)a e ta) h h = e ta ( lim e ha I ) h h = e ta A = Ae ta, the second identity in (.7) by (.7), the third by the definition (.2), and the fourth by commutativity. 2. Exponentials and trigonometric functions In Chapter we have seen how to use complex exponentials to give a selfcontained treatment of basic results on the trigonometric functions cos t and sin t. Here we present a variant, using matrix exponentials. We begin by looking at ( ) (2.) x(t) = e tj x, J =, which solves (2.2) x (t) = Jx(t), x() = x R 2. We first note that the planar curve x(t) moves about on a circle centered about the origin. Indeed, (2.3) d dt x(t) 2 = d ( ) x(t) x(t) = x (t) x(t) + x(t) x (t) dt = Jx(t) x(t) + x(t) Jx(t) =, since J t = J. Thus x(t) = x is constant. Furthermore the velocity v(t) = x (t) has constant magnitude; in fact (2.4) v(t) 2 = v(t) v(t) = Jx(t) Jx(t) = x(t) 2, since J t J = J 2 = I.

15 2. Exponentials and trigonometric functions 5 (2.5) For example, ( ) ( ) c(t) = e tj s(t) is a curve, moving on the unit circle x 2 + x2 2 =, at unit speed, with initial position x() = (, ) t and initial velocity v() = (, ) t. Now in trigonometry the functions cos t and sin t are defined to be the x and x 2 coordinates of such a parametrization of the unit circle, so we have ( ) ( ) cos t (2.6) = e tj. sin t The differential equation (2.2) then gives (2.7) Using d d cos t = sin t, dt ( ) e tj = e tj J sin t = cos t. dt ( ) ( ) = Je tj, we have a formula for e tj( ), which together with (2.6) yields ( ) (2.8) e tj cos t sin t = = (cos t)i + (sin t)j. sin t cos t Then the identity e (s+t)j = e sj e tj yields the following identities, when matrix multiplication is carried out: (2.9) cos(s + t) = (cos s)(cos t) (sin s)(sin t), sin(s + t) = (cos s)(sin t) + (sin s)(cos t). We now show how the treatment of sin t and cos t presented above is really quite close to that given in Chapter,. To start, we note that if C is regarded as a real vector space, with basis e =, e 2 = i, and hence identified with R 2, via ( ) x (2.) z = x + iy, y then the matrix representation for the linear transformation z iz is given by J: ( ) ( ) x y (2.) iz = y + ix, J =. y x

16 6 Linear Systems of Differential Equations Michael Taylor More generally, the linear transformation z (c + is)z has matrix representation ( ) c s (2.2). s c Taking this into account, we see that the identity (2.8) is equivalent to (2.3) e it = cos t + i sin t, which is Euler s formula, as in (.39) of Chapter. Here is another approach to the evaluation of e tj. eigenvalues and eigenvectors of J: (2.4) λ = i, λ 2 = i; v = ( ), v i 2 = Then, using the fact that e tj v k = e tλ kv k, we have ( ) (2.5) e tj = ( ) 2 eit + ( ) i 2 e it. i We compute the ( ). i Comparison with (2.6) gives (2.6) cos t = 2 (eit + e it ), sin t = 2i (eit e it ), again leading to (2.3). Exercises. Recall Skew(n) and SO(n), defined by (.7) and (2.4) of Chapter 2. Show that (2.7) A Skew(n) = e ta SO(n), t R. Note how this generalizes (2.3). 2. Given an n n matrix A, let us set (2.8) cos ta = 2 (eita + e ita ), sin ta = 2i (eita e ita ).

17 3. First-order systems derived from higher-order equations 7 Show that (2.9) d d cos ta = A sin ta, dt sin ta = A cos ta. dt 3. In the context of Exercise 2, show that (2.2) cos ta = k= ( ) k (2k)! (ta)2k, sin ta = k= ( ) k (2k + )! (ta)2k+. 4. Show that Av = λv = (cos ta)v = (cos tλ)v, (sin ta)v = (sin tλ)v. 5. Compute cos ta and sin ta in each of the following cases: A = ( ), A = ( ), A = ( ) ( ) i, A =. i 6. Suppose A M(n, C) and B = ( ) A M(2n, C). A Show that ( ) e tb cos ta sin ta =. sin ta cos ta 3. First-order systems derived from higher-order equations There is a standard process to convert an nth order differential equation (3.) to a first-order system. Set d n y dt n + a d n y n dt n + + a dy dt + a y = (3.2) x (t) = y(t), x (t) = y (t),..., x n (t) = y (n ) (t).

18 8 Linear Systems of Differential Equations Michael Taylor Then x = (x,..., x n ) t satisfies x = x (3.3). x n 2 = x n x n = a n x n a x, or equivalently (3.4) dx dt = Ax, with (3.5) A = a a a n 2 a n The matrix A given by (3.5) is called the companion matrix of the polynomial (3.6) p(λ) = λ n + a n λ n + + a λ + a. Note that a direct search of solutions to (3.) of the form e λt leads one to solve p(λ) =. Thus the following result is naturally suggested. Proposition 3.. If p(λ) is a polynomial of the form (3.6), with companion matrix A, given by (3.5), then (3.7) p(λ) = det(λi A). Proof. We look at λ λ (3.8) λi A = λ, a a a n 2 λ + a n and compute its determinant by expanding by minors down the first column. We see that (3.9) det(λi A) = λ det(λi Ã) + ( )n a det B,

19 3. First-order systems derived from higher-order equations 9 where (3.) Ã is the companion matrix of λ n + a n λ n a, B is lower triangular, with s on the diagonal. By induction on n, we have det(λi Ã) = λn + a n λ n a, while det B = ( ) n. Substituting this into (3.9) gives (3.7). Converse construction We next show that each solution to a first order n n system of the form (3.4) (for general A M(n, F)) also satisfies an nth order scalar ODE. Indeed, if (3.4) holds, then (3.) x (k) = Ax (k ) = = A k x. Now if p(λ) is given by (3.7), and say (3.2) p(λ) = λ n + a n λ n + + a λ + a, then, by the Cayley-Hamilton theorem (cf. (8.) of Chapter 2), (3.3) p(a) = A n + a n A n + + a A + a I =. Hence (3.4) x (n) = A n x = a n A n x a Ax a x = a n x (n ) a x a x, so we have the asserted nth order scalar equation: (3.5) x (n) + a n x (n ) + + a x + a x =. Remark. If the minimal polynomial q(λ) of A has degree m, less than n, we can replace p by q and derive analogues of (3.4) (3.5), giving a single differential equation of degree m for x.

20 2 Linear Systems of Differential Equations Michael Taylor Exercises. Using the method (3.2) (3.5), convert ( 2 dx dt = into a second order scalar equation. 2. Using the method (3.2) (3.3), convert into a 2 2 first order system. ) x y 3y + 2y = In Exercises 3 4, assume that λ is a root of multiplicity k 2 for the polynomial p(λ) given by (3.6). 3. Verify that e λ t, te λ t,..., t k e λ t are solutions to (3.). 4. Deduce that, for each j =,..., k, the system (3.3) has a solution of the form (3.6) x(t) = (t j + αt j + + β)e tλ v, (with v depending on j). 5. For given A M(n, C), suppose x = Ax has a solution of the form (3.6). Show that λ must be a root of multiplicity j + of the minimal polynomial of A. Hint. Take into account the remark below (3.5). 6. Using Exercises 3 5, show that the minimal polynomial of the companion matrix A in (3.5) must be the characteristic polynomial p(λ). 4. Non-homogeneous equations and Duhamel s formula In 3 we have focused on homogeneous equations, x Ax =. Here we consider the non-homogeneous equation (4.) dx dt Ax = f(t), x() = x C n.

21 4. Non-homogeneous equations and Duhamel s formula 2 Here A M(n, C) and f(t) takes values in C n. The key to solving this is to recognize that the left side of (4.) is equal to (4.2) e ta d dt ( ) e ta x(t), as follows from the product formula for the derivative and the defining property of e ta, given in (.4). Thus (4.) is equivalent to (4.3) and integration yields d ( ) e ta x(t) = e ta f(t), x() = x, dt (4.4) e ta x(t) = x + t e sa f(s) ds. Applying e ta to both sides then gives the solution: (4.5) x(t) = e ta x + This is called Duhamel s formula. t e (t s)a f(s) ds. Example. We combine methods of this section and 3 (and also 2) to solve (4.6) y + y = f(t), y() = y, y () = y. As in 3, set x = (x, x ) = (y, y ), to obtain the system ( ) ( ) ( ) ( ) d x x (4.7) = +. dt f(t) x Recognizing the 2 2 matrix above as J, and recalling from 2 that ( ) (4.8) e (s t)j cos(s t) sin(s t) =, sin(s t) cos(s t) we obtain (4.9) ( ) ( ) ( x (t) cos t sin t y = x (t) sin t cos t and hence y ) + t (4.) y(t) = (cos t)y + (sin t)y + x ( ) ( ) cos(s t) sin(s t) ds, sin(s t) cos(s t) f(s) t sin(t s) f(s) ds.

22 22 Linear Systems of Differential Equations Michael Taylor Use of Duhamel s formula is a good replacement for the method of variation of parameters, discussed in 4 of Chapter. See 9 of this chapter for more on this. Here, we discuss a variant of the method of undetermined coefficients, introduced for single second-order equations in of Chapter. We consider the following special case, the first-order n n system (4.) dx dt Ax = (cos σt)v, given σ R, v R n, and A M(n, R) (or we could use complex coefficients). We assume (4.2) iσ, iσ / Spec A, and look for a solution to (4.) of the form (4.3) x p (t) = (cos σt)a + (sin σt)b, a, b R n. Substitution into (4.) leads to success with (4.4) a = A(A 2 + σi) v, b = σ(a 2 + σi) v. If (4.2) does not hold, (4.4) fails, and (4.) might not have a solution of the form (4.3). Of course, (4.5) will work; (4.) will have a solution of the form (4.5) x(t) = t (cos σs)e (t s)a v ds. When (4.2) holds and (4.3) works, the general solution to (4.) is (4.6) x(t) = e ta u + (cos σt)a + (sin σt)b, u R n, u related to x() by (4.7) x() = u + a. If all the eigenvalues of A have negative real part, e ta u will decay to as t +. Then e ta u is called the transient part of the solution. The other part, (cos σt)a + (sin σt)b, is called the steady state solution.

23 4. Non-homogeneous equations and Duhamel s formula 23 Exercises. Given A M(n, C), set E k (t) = k j= Verify that E k (t) = AE k (t) and that t j j! Aj. d ( Ek (t)e ta) = tk dt k! Ak+ e ta. 2. Verify that, if A is invertible, t s k e sa ds = k! A (k+)[ E k (t)e ta I ]. 3. Solve the initial value problem ( ) ( dx e dt = t x + e t ), x() = ( ). 4. Solve the initial value problem ( ) ( dx e t dt = x + e t ), x() = ( ). 5. Solve the initial value problem ( ) ( dx e t dt = x + e t ), x() = ( ). 6. Produce analogues of (4.8) (4.) for y 3y + 2y = f(t), y() = y, y () = y.

24 24 Linear Systems of Differential Equations Michael Taylor In Exercises 7 8, take X, Y M(n, C) and (4.8) U(t, s) = e t(x+sy ), U s (t, s) = U(t, s). s 7. Show that U s satisfies U s t = (X + sy )U s + Y U, U s (, s) =. 8. Use Duhamel s formula to show that (4.9) Deduce that U s (t, s) = t d ds ex+sy s= = e X e (t τ)(x+sy ) Y e τ(x+sy ) dτ. e τx Y e τx dτ. 9. Assume X(t) is a smooth function of t I with values in M(n, C). Show that, for t I, (4.2) d dt ex(t) = e X(t) e τx(t) X (t)e τx(t) dτ.. In the context of Exercise 9, assume t, t I = X(t)X(t ) = X(t )X(t). In such a case, simplify (4.2), and compare the result with that of Exercise in. 5. Simple electrical circuits Here we extend the scope of the treatment of electrical circuits in 3 of Chapter. Rules worked out by Kirchhoff and others in the 8s allow one to write down a system of linear differential equations describing the

25 5. Simple electrical circuits 25 voltages and currents running along a variety of electrical circuits, containing resistors, capacitors, and inductors. There are two types of basic laws. The first type consists of two rules known as Kirchhoff s laws: (A) (B) The sum of the voltage drops around any closed loop is zero. The sum of the currents at any node is zero. The second type of law specifies the voltage drop across each circuit element: (a) (b) (c) Resistor: Inductor: V = IR, V = L di dt, Capacitor: V = Q C. In each case, V is the voltage drop (in volts), I is the current (in amps), R is the resistance (in ohms), L is the inductance (in henrys), C is the capacitance (in farads), and Q is the charge (in coulombs). We refer to 3 of Chapter for basic information about these units. The rule (c) is supplemented by the following formula for the current across a capacitor: (c2) I = dq dt. In (b) and (c2), time is measured in seconds. Rules (A), (B), and (a) give algebraic relations among the various voltages and currents, while rules (b) and (c) (c2) give differential equations, namely (5.) (5.2) L di dt = V C dv dt = I (Inductor), (Capacitor). Note that (5.2) results from applying d/dt to (c) and then using (c2). If a circuit has k capacitors and l inductors, we get an m m system of first order differential equations, with m = k + l. We illustrate the formulation of such differential equations for circuits presented in Figure 5. and Figure 5.2. In each case, the circuit elements are numbered. We denote by V j the voltage drop across element j and by I j the current across element j.

26 26 Linear Systems of Differential Equations Michael Taylor Figure 5. Figure 5. depicts a classical RLC circuit, such as treated in 3 of Chapter. Rules (A), (B), and (a) give (5.3) V + V 2 + V 3 = E(t), I = I 2 = I 3, V = RI. Equations (5.) (5.3) yield a system of two ODEs, for I 3 and V 2 : (5.4) L di 3 dt = V 3, C dv 2 dt = I 2. We need to express V 3 and I 2 in terms of I 3, V 2, and E(t), using (5.3). In fact, we have (5.5) V 3 = E(t) V V 2 = E(t) RI V 2 = E(t) RI 3 V 2, I 2 = I 3, so we get the system (5.6) or, in matrix form, ( ) d I3 (5.7) = dt V 2 L di 3 dt = RI 3 V 2 + E(t), C dv 2 dt = I 3, ( ) ( ) R/L /L I3 + /C V 2 L ( ) E(t).

27 5. Simple electrical circuits 27 Figure 5.2 Note that the characteristic polynomial of the matrix (5.8) A = is ( R/L ) /L /C (5.9) λ 2 + R L λ + LC, with roots (5.) λ = R 2L ± R 2L 2 4 L C. Let us now look at the slightly more complicated circuit depicted in Figure 5.2. Again we get a 2 2 system of differential equations. Rules (A), (B), and (a) give (5.) V + V 2 + V 4 = E(t), V 2 = V 3, I = I 2 + I 3 = I 4, V 3 = R 3 I 3, V 4 = R 4 I 4. Equations (5.) (5.2) yield differential equations for I and V 2 : (5.2) C dv 2 dt = I 2, L di dt = V.

28 28 Linear Systems of Differential Equations Michael Taylor Figure 5.3 We need to express I 2 and V in terms of V 2, I, and E(t), using (5.). In fact, we have (5.3) I 2 = I I 3 = I R 3 V 3 = I R 3 V 2, V = E(t) V 2 V 4 = E(t) V 2 R 4 I 4 = E(t) V 2 R 4 I, so we get the system (5.4) or, in matrix form, (5.5) d dt ( V2 I ) = C dv 2 dt = R 3 V 2 + I, L di dt = V 2 R 4 I + E(t), ( /R3 C /C /L R 4 /L ) ( V2 I ) + ( ). L E(t) Exercises. Work out the 3 3 system of differential equations describing the behavior of the circuit depicted in Figure 5.3. Assume E(t) = 5 sin 2t volts.

29 6. Second-order systems Using methods developed in 4, solve the 2 2 system (5.7) when R = 5 ohms, L = 4 henrys, C = farad, and with initial data E(t) = 5 cos 2t volts, I 3 () = amps, V 2 () = 5 volts. 3. Solve the 2 2 system (5.5) when R 3 = ohm, R 4 = 4 ohms, L = 4 henrys, C = 2 farads, and E(t) = 2 cos 2t volts. 4. Use the method of (4.) (4.4) to find the steady state solution to (5.7), when E(t) = A cos σt. Take A, σ, R and L fixed and allow C to vary. Show that the amplitude of the steady state solution is maximal (we say resonance is achieved) when LC = σ 2, recovering calculations of (3.7) (3.3) in Chapter. 5. Work out the analogue of Exercise 4 with the system (5.7) replaced by (5.5). Is the condition for resonance the same as in Exercise 4? 6. Draw an electrical circuit that leads to a 4 4 system of differential equations, and write down said system. 6. Second-order systems Interacting physical systems often give rise to second-order systems of differential equations. Consider for example a system of n objects, of mass m,..., m n, connected to each other and to two walls by n + springs, with spring constants k,..., k n+, as in Fig. 6.. We assume the masses slide

30 3 Linear Systems of Differential Equations Michael Taylor Figure 6. without friction. Denote by x j the position of the jth mass and by y j the degree to which the jth spring is stretched. The equations of motion are (6.) m j x j = k j y j + k j+ y j+, j n, and for certain constants a j, (6.2) y j = x j x j+ + a j, 2 j n, y = x + a, y n+ = x n + a n+. Substituting (6.2) into (6.) yields an n n system, which we can write in matrix form as (6.3) Mx = Kx + b, where x = (x,..., x n ) t, b = ( k a + k 2 a 2,..., k n a n + k n+ a n+ ) t, m (6.4) M =... and k + k 2 k 2. k 2 k 2 + k.. 3 (6.5) K = kn + k n k n k n k n + k n+ m n,

31 6. Second-order systems 3 We assume m j > and k j > for each j. Then clearly M is a positive definite matrix and K is a real symmetric matrix. Proposition 6.. If each k j >, then K, given by (6.5), is positive definite. Proof. We have (6.6) x Kx = Now n (k j + k j+ )x 2 j 2 j= (6.7) 2x j x j x 2 j + x 2 j, so n k j x j x j. j=2 (6.8) x Kx n j= n+ k j x 2 j + k j x 2 j j=2 n k j x 2 j j=2 k x 2 + k n+ x 2 n. n k j x 2 j j=2 Furthermore note that the inequality in (6.7) is strict unless x j = x j so the inequality in (6.8) is strict unless x j = x j for each j {2,..., n}, i.e., unless x = = x n. This proves that x Kx > whenever x R n, x. To be more precise, we can sharpen (6.7) to (6.9) 2x j x j = x 2 j + x 2 j (x j x j ) 2, and then (6.8) is sharpened to (6.) x Kx = k x 2 + k n+ x 2 n + If we set n k j (x j x j ) 2. j=2 (6.) κ = min{k j : j n + }, then (6.) implies ( (6.2) x Kx κ x 2 + x 2 n + n (x j x j ) 2). j=2

32 32 Linear Systems of Differential Equations Michael Taylor The system (6.3) is inhomogeneous, but it is readily converted into the homogeneous system (6.3) Mz = Kz, z = x K b. This in turn can be rewritten (6.4) z = M Kz. Note that (6.5) L = M /2 KM /2 = M K = M /2 LM /2, where (6.6) M /2 = m /2... m /2 n. Proposition 6.2. The matrix L is positive definite. Proof. x Lx = (M /2 x) K(M /2 x) > whenever x. According to (6.5), M K and L are similar, so we have: Corollary 6.3. For M and K of the form (6.4) (6.5), with m j, k j >, the matrix M K is diagonalizable, and all its eigenvalues are positive. It follows that R n has a basis {v,..., v n } satisfying (6.7) M Kv j = λ 2 jv j, λ j >. Then the initial value problem (6.8) Mz = Kz, z() = z, z () = z has the solution n ( (6.9) z(t) = α j cos λ j t + β ) j sin λ j t v j, λ j j= where the coefficients α j and β j are given by (6.2) z = α j v j, z = β j v j.

33 6. Second-order systems 33 An alternative approach to the system (6.4) is to set (6.2) u = M /2 z, for which (6.4) becomes (6.22) u = Lu, with L given by (6.5). Then R n has an orthonormal basis {w j : j n}, satisfying (6.23) Lw j = λ 2 jw j, namely w j = M /2 v j, with v j as in (6.7). Note that we can set (6.24) L = A 2, Aw j = λ j w j, and (6.22) becomes (6.25) u + A 2 u =. One way to convert (6.25) to a first order (2n) (2n) system is to set (6.26) v = Au, w = u. Then (6.25) becomes (6.27) d dt ( ) v = X w ( ) v, X = w ( ) A. A It is useful to note that when X is given by (6.26), then ( ) (6.28) e tx cos ta sin ta =, sin ta cos ta where cos ta and sin ta are given in Exercises 6 7 in 2. One way to see this is to let Φ(t) denote the right side of (6.28) and use (2.9) to see that (6.29) d Φ(t) = XΦ(t), Φ() = dt ( ) I. I Then Exercise 4 of implies e tx Φ(t). These calculations imply that the solution to (6.25), with initial data u() = u, u () = u, is given by (6.3) u(t) = (cos ta)u + A (sin ta)u.

34 34 Linear Systems of Differential Equations Michael Taylor Compare (6.8) (6.2). This works for each invertible A M(n, C). We move to the inhomogeneous variant of (6.4), which as above we can transform to the following inhomogeneous variant of (6.25): (6.3) u + A 2 u = f(t), u() = u, u () = u. Using the substitution (6.26), we get ( ) ( ) ( ) d v v (6.32) = X +, dt w w f(t) ( ) v() = w() ( Au u ). Duhamel s formula applies to give ( ) ( ) v(t) (6.33) = e tx Au + w(y) u t ( ) e (t s)x ds. f(s) Using the formula (6.28) for e tx, we see that the resulting formula for v(t) in (6.33) is equivalent to (6.34) u(t) = (cos ta)u + A (sin ta)u + t A sin(t s)a f(s) ds. This is the analogue of Duhamel s formula for the solution to (6.3). We now return to the coupled spring problem and modify (6.) (6.2) to allow for friction. Thus we replace (6.) by (6.35) m j x j = k j y j + k j+ y j d j x j, where y j are as in (6.2) and d j > are friction coefficients, Then (6.3) is replaced by (6.36) Mx = Kx Dx + b, with b as in (6.3), M and K as in (6.4) (6.5), and d (6.37) D =... d n, d j >. As in (6.3), we can convert (6.36) to the homogeneous system (6.38) Mz = Kz Dz, z = x K b. If we set u = M /2 z, as in (6.2), then, parallel to (6.22) (6.24), we get (6.39) u + Bu + A 2 u =,

35 6. Second-order systems 35 where A 2 is as in (6.24), with L = M /2 KM /2, as in (6.22), and d /m (6.4) B = M /2 DM /2 =... d n /m n The substitution (6.26) converts the n n second order system (6.39) to the (2n) (2n) first order system (6.4) d dt ( ) ( v A = w A B ) ( ) v. w. We can write (6.4) as (6.42) ( ) ( ) d v v = (X + Y ), dt w w with X as in (6.27) and (6.43) Y = Note that (6.44) XY = ( ). B ( ) AB, Y X = ( ), BA so these matrices do not commute. Thus e t(x+y ) might be difficult to calculate even when A and B commute. Such commutativity would hold, for example, if m = = m n and d = = d n, in which case B is a scalar multiple of the identity matrix. When the positive, self adjoint operators A and B do commute, we can make the following direct attack on the system (6.39). We know (cf. Exercise 3 in, Chapter 2) that R n has an orthonormal basis {w,..., w n } for which (6.45) Aw j = λ j w j, Bw j = 2µ j w j, λ j, µ j >. Then we can write a solution to (6.39) as (6.46) u(t) = u j (t)w j, where the real-valued coefficients u j (t) satisfy the equations (6.47) u j + 2µ j u j + λ 2 ju j =,

36 36 Linear Systems of Differential Equations Michael Taylor with solutions that are linear combinations: ( ) e µ jt α j cos λ 2 µ 2 j t + β j sin λ 2 j µ2 t, λ j > µ j, (6.48) ( e µ jt µ α j e j λ2 j t + β j e µ 2 j λ2t) j, λ j < µ j, e µ jt (α j + β j t), λ j = µ j. These three cases correspond to modes that are said to be underdamped, overdamped, and critically damped, respectively. In cases where A and B do not commute, analysis of (6.39) is less explicit, but we can establish the following decay result. Proposition 6.4. ( If A, B M(n, ) C) are positive definite, then all of the A eigenvalues of Z = have negative real part. A B Proof. Let s say (v, w) t and Z(v, w) t = λ(v, w) t. Then (6.49) Aw = λv, Av + Bw = λw, and (6.5) (Z(v, w) t, (v, w) t ) = (Bw, w) + [(Aw, v) (Av, w)], while also (6.5) (Z(v, w) t, (v, w) t ) = λ( v 2 + w 2 ). The two terms on the right sie of (6.5) are real and purely imaginary, respectively, so we obtain (6.52) (Re λ)( v 2 + w 2 ) = (Bw, w). If (v, w) t, we deduce that either Re λ < or w =. If w =, then (6.49) gives Av =, hence v =. Hence w, and Re λ <, as asserted. Exercises. Find the eigenvalues and eigenvectors of.

37 6. Second-order systems Use the results of Exercise to find the eigenvalues and eigenvectors of K, given by (6.5), in case n = 3 and k = k 2 = k 3 = k 4 = k. 3. Find the general solution to u + Bu + A 2 u =, in case A 2 = K, with K as in Exercise 2, and B = I. 4. Find the general solution to ( u + ) u + ( ) u =. 5. Produce a second order system of differential equations for the spring system presented in Fig (Assume there is no friction.) 6. Generalizing the treatment of (6.25), consider (6.53) u + Lu =, L M(N, C). Assume C N has a basis of eigenvectors v j, such that Lv j = λ 2 j v j, λ j C, λ j. Show that the general solution to (6.53) has the form (6.54) u(t) = N (α j e λjt + β j e λjt )v j, α j, β j C. j= How is this modified if some λ j =? 7. Find the general solution to ( ) u 2 + u =, and to u + u =.

38 38 Linear Systems of Differential Equations Michael Taylor Figure Curves in R 3 and the Frenet-Serret equations Given a curve c(t) = (x(t), y(t), z(t)) in 3-space, we define its velocity and acceleration by (7.) v(t) = c (t), a(t) = v (t) = c (t). We also define its speed s (t) and arclength by (7.2) s (t) = v(t), s(t) = t t s (τ) dτ, assuming we start at t = t. We define the unit tangent vector to the curve as (7.3) T (t) = v(t) v(t). Henceforth we assume the curve is parametrized by arclength. We define the curvature κ(s) of the curve and the normal N(s) by (7.4) κ(s) = Note that dt, ds dt ds = κ(s)n(s). (7.5) T (s) T (s) = = T (s) T (s) =, so indeed N(s) is orthogonal to T (s). We then define the binormal B(s) by (7.6) B(s) = T (s) N(s).

39 7. Curves in R 3 and the Frenet-Serret equations 39 For each s, the vectors T (s), N(s) and B(s) are mutually orthogonal unit vectors, known as the Frenet frame for the curve c(s). Rules governing the cross product yield (7.7) T (s) = N(s) B(s), N(s) = B(s) T (s). (For material on the cross product, see the exercises at the end of 2 of Chapter 2.) The torsion of a curve measures the change in the plane generated by T (s) and N(s), or equivalently it measures the rate of change of B(s). Note that, parallel to (7.5), B(s) B(s) = = B (s) B(s) =. Also, differentiating (7.6) and using (7.4), we have (7.8) B (s) = T (s) N(s) + T (s) N (s) = T (s) N (s) = B (s) T (s) =. We deduce that B (s) is parallel to N(s). We define the torsion by (7.9) db ds = τ(s)n(s). We complement the formulas (7.4) and (7.9) for dt/ds and db/ds with one for dn/ds. Since N(s) = B(s) T (s), we have (7.) or (7.) dn ds = db ds T + B dt ds dn ds = τn T + κb N, = κ(s)t (s) + τ(s)b(s). Together, (7.4), (7.9) and (7.) are known as the Frenet-Serret formulas. Example. Pick a, b > and consider the helix (7.2) c(t) = (a cos t, a sin t, bt). Then v(t) = ( a sin t, a cos t, b) and v(t) = a 2 + b 2, so we can pick s = t a 2 + b 2 to parametrize by arc length. We have (7.3) T (s) = ( a sin t, a cos t, b), a 2 + b2

40 4 Linear Systems of Differential Equations Michael Taylor hence (7.4) By (7.4), this gives (7.5) κ(s) = Hence (7.6) B(s) = T (s) N(s) = dt ds = a 2 ( a cos t, a sin t, ). + b2 a a 2, N(s) = ( cos t, sin t, ). + b2 (b sin t, b cos t, a). a 2 + b2 Then (7.7) so, by (7.9), db ds = a 2 (b cos t, b sin t, ), + b2 (7.8) τ(s) = b a 2 + b 2. In particular, for the helix (7.2), we see that the curvature and torsion are constant. (7.9) Let us collect the Frenet-Serret equations dt ds = dn ds = κt db ds = κn τn + τb for a smooth curve c(s) in R 3, parametrized by arclength, with unit tangent T (s), normal N(s), and binormal B(s), given by (7.2) N(s) = κ(s) T (s), assuming κ(s) = T (s) >. B(s) = T (s) N(s), The basic existence and uniqueness theory, which will be presented in Chapter 4, applies to (7.9). If κ(s) and τ(s) are given smooth functions on an interval I = (a, b) and s I, then, given T, N, B R 3, (7.9) has a unique solution on s I satisfying (7.2) T (s ) = T, N(s ) = N, B(s ) = B.

41 7. Curves in R 3 and the Frenet-Serret equations 4 In fact, the case when κ(s) and τ(s) are analytic will be subsumed in the material of of this chapter. We now establish the following. Proposition 7.. Assume κ and τ are given smooth functions on I, with κ > on I. Assume {T, N, B } is an orthonormal basis of R 3, such that B = T N. Then there exists a smooth, unit-speed curve c(s), s I, for which the solution to (7.9) and (7.2) is the Frenet frame. To construct the curve, take T (s), N(s), and B(s) to solve (7.9) and (7.2), pick p R 3 and set s (7.22) c(s) = p + T (σ) dσ, s so T (s) = c (s) is the velocity of this curve. To deduce that {T (s), N(s), B(s)} is the Frenet frame for c(s), for all s I, we need to know: (7.23) {T (s), N(s), B(s)} orthonormal, with B(s) = T (s) N(s), s I. In order to pursue the analysis further, it is convenient to form the 3 3 matrix-valued function (7.24) F (s) = (T (s), N(s), B(s)), whose columns consist respectively of T (s), N(s), and B(s). Then (7.23) is equivalent to (7.25) F (s) SO(3), s I, with SO(3) defined as in (2.4) of Chapter 2. The hypothesis on {T, N, B } stated in Proposition 7. is equivalent to F = (T, N, B ) SO(3). Now F (s) satisfies the differential equation (7.26) F (s) = F (s)a(s), F (s ) = F, where κ(s) (7.27) A(s) = κ(s) τ(s). τ(s) Note that (7.28) df ds = A(s) F (s) = A(s)F (s),

42 42 Linear Systems of Differential Equations Michael Taylor since A(s) in (7.27) is skew-adjoint. Hence (7.29) d ds F (s)f (s) = df ds F (s) + F (s) df ds = F (s)a(s)f (s) F (s)a(s)f (s) =. Thus, whenever (7.26) (7.27) hold, (7.3) F F = I = F (s)f (s) I, and we have (7.23). Let us specialize the system (7.9), or equivalently (7.26), to the case where κ and τ are constant, i.e., κ (7.3) F (s) = F (s)a, A = κ τ, τ with solution (7.32) F (s) = F e (s s )A. We have already seen in that a helix of the form (7.2) has curvature κ and torsion τ, with (7.33) κ = and hence (7.34) a = a a 2 + b 2, τ = b a 2 + b 2, κ κ 2 + τ 2, b = τ κ 2 + τ 2. In (7.2), s and t are related by t = s κ 2 + τ 2. We can also see such a helix arise via a direct calculation of e sa, which we now produce. First, a straightforward calculation gives, for A as in (7.3), (7.35) det(λi A) = λ(λ 2 + κ 2 + τ 2 ), hence (7.36) Spec(A) = {, ±i κ 2 + τ 2 }.

43 7. Curves in R 3 and the Frenet-Serret equations 43 An inspection shows that we can take τ (7.37) v =, v 2 =, v 3 = κ 2 + τ 2 κ κ 2 + τ 2 κ, τ and then (7.38) Av =, Av 2 = κ 2 + τ 2 v 3, Av 3 = κ 2 + τ 2 v 2. In particular, with respect to the basis {v 2, v 3 } of V = Span{v 2, v 3 }, A V has the matrix representation (7.39) B = ( ) κ 2 + τ 2. We see that (7.4) e sa v = v, while, in light of the calculations giving (2.8), (7.4) e sa v 2 = (cos s κ 2 + τ 2 )v 2 +(sin s κ 2 + τ 2 )v 3, e sa v 3 = (sin s κ 2 + τ 2 )v 2 +(cos s κ 2 + τ 2 )v 3. Exercises. Consider a curve c(t) in R 3, not necessarily parametrized by arclength. Show that the acceleration a(t) is given by (7.42) a(t) = d2 s ( ds ) 2N. dt 2 T + κ dt Hint. Differentiate v(t) = (ds/dt)t (t) and use the chain rule dt/dt = (ds/dt)(dt/ds), plus (7.4). 2. Show that (7.43) κb = v a v 3. Hint. Take the cross product of both sides of (7.42) with T, and use (7.6).

44 44 Linear Systems of Differential Equations Michael Taylor 3. In the setting of Exercises 2, show that (7.44) κ 2 τ v 6 = a (v a ). Deduce from (7.43) (7.44) that (7.45) τ = (v a) a v a 2. Hint. Proceed from (7.43) to d dt (κ v 3 ) B + κ v 3 db dt = d dt (v a) = v a, and use db/dt = τ(ds/dt)n, as a consequence of (7.9). Then dot with a, and use a N = κ v 2, from (7.42), to get (7.44). 4. Consider the curve c(t) in R 3 given by c(t) = (a cos t, b sin t, t), where a and b are given positive constants. Compute the curvature, torsion, and Frenet frame. Hint. Use (7.43) to compute κ and B. Then use N = B T. Use (7.45) to compute τ. 5. Suppose c and c are two curves, both parametrized by arc length over s L, and both having the same curvature κ(s) > and the same torsion τ(s). Show that there exit x R 3 and A O(3) such that c(s) = Ac(s) + x, s [, L]. Hint. To begin, show that if their Frenet frames coincide at s =, i.e., T () = T (), Ñ() = N(), B() = B(), then T T, Ñ N, B B. 6. Suppose c is a curve in R 3 with curvature κ >. Show that there exists a plane in which c(t) lies for all t if and only if τ. Hint. When τ, the plane should be parallel to the orthogonal complement of B. 8. Variable coefficient systems Here we consider a variable coefficient n n first order system (8.) dx dt = A(t)x, x(t ) = x C n,

45 8. Variable coefficient systems 45 and its inhomogeneous analogue. The general theory, which will be presented in Chapter 4, implies that if A(t) is a smooth function of t I = (a, b) and t I, then (8.) has a unique solution x(t) for t I, depending linearly on x, so (8.2) x(t) = S(t, t )x, S(t, t ) L(C n ). See of this chapter for power series methods of constructing S(t, t ), when A(t) is analytic. As we have seen, (8.3) A(t) A = S(t, t ) = e (t t )A. However, for variable coefficient equations there is not such a simple formula, and the matrix entries of S(t, s) can involve a multiplicity of new special functions, such as Bessel functions, Airy functions, Legendre functions, and many more. We will not dwell on this here, but we will note how S(t, t ) is related to a complete set of solutions to (8.). Suppose x (t),..., x n (t) are n solutions to (8.) (but with different initial conditions). Fix t I, and assume (8.4) x (t ),..., x n (t ) are linearly independent in C n, or equivalently these vectors form a basis of C n. Given such solutions x j (t), we form the n n matrix (8.5) M(t) = ( x (t),..., x n (t) ), whose jth column is x j (t). This matrix function solves (8.6) dm dt = A(t)M(t). The condition (8.4) is equivalent to the statement that M(t ) is invertible. We claim that if M solves (8.6) and M(t ) is invertible then M(t) is invertible for all t I. To see this, we use the fact that the invertibility of M(t) is equivalent to the non-vanishing of the quantity (8.7) W (t) = det M(t), called the Wronskian of {x (t),..., x n (t)}. It is also notable that W (t) solves a differential equation. In general we have (8.8) d dt det M(t) = ( det M(t) ) Tr ( M(t) M (t) ).

46 46 Linear Systems of Differential Equations Michael Taylor (See Exercises 3 below.) Let Ĩ I be the maximal interval containing t on which M(t) is invertible. Then (8.8) holds for t Ĩ. When (8.6) holds, we have Tr(M(t) M (t)) = Tr(M(t) A(t)M(t)) = Tr A(t), so the Wronskian solves the differential equation (8.9) Hence dw dt = ( Tr A(t) ) W (t). (8.) W (t) = e b(t,s) W (s), b(t, s) = t s Tr A(τ) dτ. This implies Ĩ = I and hence gives the asserted invertibility. From here we obtain the following. Proposition 8.. If M(t) solves (8.6) for t I and M(t ) is invertible, then (8.) S(t, t ) = M(t)M(t ), t I. Proof. We have seen that M(t) is invertible for all t I. If x(t) solves (8.), set (8.2) y(t) = M(t) x(t), and apply d/dt to x(t) = M(t)y(t), obtaining (8.3) dx dt = M (t)y(t) + M(t)y (t) = A(t)M(t)y(t) + M(t)y (t) = A(t)x + M(t)y (t). If x(t) solves (8.), this yields (8.4) hence y(t) = y(t ) for all t I, i.e., dy dt =, (8.6) M(t) x(t) M(t ) x(t ). Applying M(t) to both sides gives (8.). Note also that, for s, t I, (8.6) S(t, s) = M(t)M(s)

47 8. Variable coefficient systems 47 gives S(t, s)x(s) = x(t) for each solution x(t) to (8.). We also have (8.7) S(t, t ) = S(t, s)s(s, t ), S(t, s) = S(s, t). There is a more general version of the Duhamel formula (4.5) for the solution to an inhomogeneous differential equation (8.8) dx dt = A(t)x + f(t), x(t ) = x. Namely, t (8.9) x(t) = S(t, t )x + S(t, s)f(s) ds. t In fact, clearly (8.9) satisfies x(t ) = x and applying d/dt to (8.9) gives the equation (8.8). In view of (8.6), we have the alternative formula t (8.2) x(t) = M(t)M(t ) x + M(t) M(s) f(s) ds, t for invertible M(t) as in (8.5) (8.6). We note that there is a simple formula for the solution operator S(t, s) to (8.) in case the following commutativity hypothesis holds: (CA) A(t)A(t ) = A(t )A(t), t, t I. We claim that if (8.2) B(t, s) = then t s A(τ) dτ, (8.22) (CA) = d dt( e B(t,s) x(t) ) = e B(t,s)( x (t) A(t)x(t) ), from which it follows that (8.23) (CA) = S(t, s) = e t s A(τ) dτ. (This identity fails in the absence of the hypothesis (CA).) To establish (8.22), we note that (CA) implies (CB) B(t, s)b(t, s) = B(t, s)b(t, s), s, t, t I.

48 48 Linear Systems of Differential Equations Michael Taylor Next, (8.24) (CB) = lim (e B(t+h,s) e B(t,s)) h h ) = lim h h eb(t,s)( e B(t+h,s) B(t,s) I = e B(t,s) A(t) from which (8.22) follows. = d dt eb(t,s) = e B(t,s) A(t), Here is an application of (8.23). Let x(s) be a planar curve, on an interval about s =, parametrized by arc-length, with unit tangent T (s) = x (s). Then the Frenet-Serret equations (7.) simplify to T = κn, with N = JT, i.e., to (8.25) T (s) = κ(s)jt (s), with J as in (2.). Clearly the commutativity hypothesis (CA) holds for A(s) = κ(s)j, so we deduce that (8.26) T (s) = e λ(s)j T (), λ(s) = s Recall that e tj is given by (2.8), i.e., ( ) (8.27) e tj cos t sin t =. sin t cos t κ(τ) dτ. Exercises Exercises 3 lead to a proof of the formula (8.8) for the derivative of det M(t).. Let A M(n, C). Show that, as s, det(i + sa) = ( + sa ) ( + sa nn ) + O(s 2 ) = + s Tr A + O(s 2 ), hence d ds det(i + sa) s= = Tr A.

49 8. Variable coefficient systems Let B(s) be a smooth matrix valued function of s, with B() = I. Use Exercise to show that d ds det B(s) s= = Tr B (). Hint. Write B(s) = I + sb () + O(s 2 ). 3. Let C(s) be a smooth matrix valued function, and assume C() is invertible. Use Exercise 2 plus det C(s) = ( det C() ) det B(s), B(s) = C() C(s) to show that d ds det C(s) s= = ( det C() ) Tr C() C (). Use this to prove (8.8). Hint. Fix t and set C(s) = M(t + s), so d dt det M(t) = d ds det C(s) s=. 4. Show that, if M(t) is smooth and invertible on an interval I, then d dt M(t) = M(t) M (t)m(t). Then apply d/dt to (8.2), to obtain an alternative derivation of Proposition 8.. Hint. Set U(t) = M(t) and differentiate the identity U(t)M(t) = I. 5. Using the relation x(t) = M(t)y(t), show that the equation (8.8) is equivalent to dy dt = M(t) f(t). Use this to obtain a derivation of (8.2). Exercises 6 7 generalize (8.25) (8.27) from the case of zero torsion (cf. Exercise 6 of 7) to the case (8.28) τ(t) = βκ(t), β constant.

50 5 Linear Systems of Differential Equations Michael Taylor 6. Assume x(t) is a curve in R 3 for which (8.28) holds. Show that x(t) = x() + t T (s) ds, with (8.29) ( T (t), N(t), B(t) ) = ( T (), N(), B() ) e σ(t)k, where (8.3) K = β, σ(t) = β t Hint. Use (7.26) (7.27) and (CA) (8.2) (8.23). κ(s) ds. 7. Let e, e 2, e 3 denote the standard basis of R 3, and let v = ( + β 2 ) /2 (βe + e 3 ), v 2 = e 2, v 3 = ( + β 2 ) /2 (e βe 3 ). Show that v, v 2, v 3 form an orthonormal basis of R 3 and, with K as in (8.3), Kv =, Kv 2 = ( + β 2 ) /2 v 3, Kv 3 = ( + β 2 ) /2 v 2. Deduce that e σk v = v, e σk v 2 = (cos η)v 2 (sin η)v 3, e σk v 3 = (sin η)v 2 + (cos η)v 3, where η = ( + β 2 ) /2 σ. 8. Given B M(n, C), write down the solution to Hint. Use (8.23). dx dt = etb x, x() = x. Exercises 9 deal with a linear equation with periodic coefficients: (8.3) Say A(t) M(n, C). dx dt = A(t)x, A(t + ) = A(t).