odechap3

Chapter 3 Linear Systems of Di®erential Equations Introduction This chapter connects the linear algebra developed in Chapter 2 with Dif- ferential Equations. We de¯ne the matrix exponential in x1 and show how it produces the solution to ¯rst order systems of di®erential equations with constant coe±cients. We show how the use of eigenvectors and generalized eigenvectors helps to compute matrix exponentials. In x2 we look again at connections between exponential and trigonometric functions, complement- ing results of Chapter 1, x1. Inx3 we discuss how to reduce a higher order di®erential equation to a ¯rst order system, and show how the \companion matrix" of a polynomial arises in doing this. We show in x4 how the matrix exponential allows us to write down an integral formula (Duhamel's formula) for the solution to a non-homogeneous ¯rst 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 di®erential equation. Section 5 discusses how to derive ¯rst order systems describing the be- havior of simple circuits, consisting of resistors, inductors, and capacitors. Here we treat a more general class of circuits than done in Chapter 1, x13. Section 6 deals with second order systems. While it is the case that second order n£nsystems can always be converted into ¯rst order (2 n)£(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. 149 150 3. Linear Systems of Di®erential Equations Inx7 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 ¯eld along the curve, called the Frenet frame, which forms an orthonormal basis of R3at each point on the curve. Regarding these equations as a system of di®erential equations, we discuss the problem of ¯nding 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 coe±cients in x7, we concen- trate on their treatment in subsequent sections. In x8 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 coe±cient setting. In x9 we show how the method of variation of parameters, intro- duced in Chapter 1, ties in with and becomes a special case of Duhamel's formula. Section 10 treats power series expansions for a ¯rst order linear sys- tem with analytic coe±cients, and x11 extends the study to equations with regular singular points. These sections provide a systematic treatment of material touched on in Chapter 1, x15. 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. 299{312, [ Hart ], pp. 70{83, or [ T], Vol. 1, pp. 30{31, for material on this. Appendix A treats logarithms of matrices, a construction inverse to the matrix exponential introduced in x1, establishing results that are of use in xx8 and 11. 1. The matrix exponential Here we discuss a key concept in matrix analysis, the matrix exponential. Given A2M(n;F);F=RorC, we de¯ne eAby the same power series used in Chapter 1 to de¯ne eAforA2R: (1.1) eA=1X k=11 k!Ak: Note that Acan be a real or complex n£nmatrix. In either case, recall fromx10 of Chapter 2 that kAkk · k Akk. Hence the standard ratio test implies (1.1) is absolutely convergent for each A2M(n;F). Hence (1.2) etA=1X k=0tk k!Ak 1. The matrix exponential 151 is a convergent power series in t, for all t2R(indeed for t2C). As for all such convergent power series, we can di®erentiate term by term. We have (1.3)d dtetA=1X k=1ktk¡1 k!Ak =1X k=1tk¡1 (k¡1)!Ak¡1A: We can factor Aout on either the left or the right, obtaining (1.4)d dtetA=etAA=AetA: Hence x(t) =etAx0solves the ¯rst-order system (1.5)dx dt=Ax; x (0) = x0: This is the unique solution to (1.5). To see this, let x(t) be any solution to (1.5), and consider (1.6) u(t) =e¡tAx(t): Then u(0) = x(0) = x0and (1.7)d dtu(t) =¡e¡tAAx(t) +e¡tAx0(t) = 0 ; sou(t)´u(0) = x0. The same argument yields (1.8)d dt³ etAe¡tA´ = 0;hence etAe¡tA´I: Hence x(t) =etAx0, as asserted. Using a variant of the computation (1.7), we show that the matrix expo- nential has the following property, which generalizes the identity es+t=eset for real s; t, established in Chapter 1. Proposition 1.1. Given A2M(n;C); s; t2R, (1.9) e(s+t)A=esAetA: 152 3. Linear Systems of Di®erential Equations Proof. Using the Leibniz formula for the derivative of a product, plus (1.4), we have (1.10)d dt³ e(s+t)Ae¡tA´ =e(s+t)AAe¡tA¡e(s+t)AAe¡tA= 0: Hence e(s+t)Ae¡tAis independent of t, so (1.11) e(s+t)Ae¡tA=esA;8s; t2R: Taking s= 0 yields etAe¡tA=I(as we have already seen in (1.8)) or e¡tA= (etA)¡1, so we can multiply both sides of (1.11) on the right by etA and obtain (1.9). Now, generally, for A; B2M(n;F), (1.12) eAeB6=eA+B: However, we do have the following. Proposition 1.2. Given A; B2M(n;C), (1.13) AB=BA=)eA+B=eAeB: Proof. We compute (1.14) d dt³ et(A+B)e¡tBe¡tA´ =et(A+B)(A+B)e¡tBe¡tA¡et(A+B)Be¡tBe¡tA¡et(A+B)e¡tBAe¡tA: Now AB=BA)ABk=BkA, hence (1.15) e¡tBA=1X k=0(¡t)k k!BkA=Ae¡tB; so (1.14) vanishes. Hence et(A+B)e¡tBe¡tAis independent of t, so (1.16) et(A+B)e¡tBe¡tA=I; the value at t= 0. Multiplying through on the right by etAand then by etB gives (1.17) et(A+B)=etAetB: Setting t= 1 gives (1.13). 1. The matrix exponential 153 We now look at examples of matrix exponentials. We start with some computations via the in¯nite series (1.2). Take (1.18) A=µ1 0 0 2¶ ; B =µ0 1 0 0¶ : Then (1.19) Ak=µ1 0 0 2k¶ ; B2=B3=¢¢¢= 0; so (1.20) etA=µet0 0e2t¶ ; etB=µ1t 0 1¶ : Note that AandBdo not commute, and neither do etAandetB, for general t6= 0. On the other hand, if we take (1.21) C=µ1 1 0 1¶ =I+B; since IandBcommute, we have without further e®ort that (1.22) etC=etIetB=µettet 0et¶ : We turn to constructions of matrix exponentials via use of eigenvalues and eigenvectors. Suppose vjis an eigenvector of Awith eigenvalue ¸j, (1.23) Avj=¸jvj: Then Akvj=¸k jvj, and hence (1.24) etAvj=1X k=0tk k!Akvj=1X k=0tk k!¸k jvj=et¸jvj: This enables us to construct etAvfor each v2CnifA2M(n;C) andCn has a basis of eigenvectors, fvj: 1·j·ng. In such a case, write vas a linear combination of the eigenvectors, (1.25) v=c1v1+¢¢¢+cnvn; and then (1.26)etAv=c1etAv1+¢¢¢+cnetAvn =c1et¸1v1+¢¢¢+cnet¸nvn: 154 3. Linear Systems of Di®erential Equations We illustrate this process with some examples. Example 1. Take (1.27) A=µ0 1 1 0¶ : One has det( ¸I¡A) =¸2¡1, hence eigenvalues (1.28) ¸1= 1; ¸ 2=¡1; with corresponding eigenvectors (1.29) v1=µ1 1¶ ; v 2=µ1 ¡1¶ : Hence (1.30) etAv1=etv1; etAv2=e¡tv2: To write out etAas a 2 £2 matrix, note that the ¯rst and second columns of this matrix are given respectively by (1.31) etAµ1 0¶ and etAµ0 1¶ : To compute this, we write (1 ;0)tand (0 ;1)tas linear combinations of the eigenvectors. We have (1.32)µ1 0¶ =1 2µ1 1¶ +1 2µ1 ¡1¶ ;µ0 1¶ =1 2µ1 1¶ ¡1 2µ1 ¡1¶ : Hence (1.33)etAµ1 0¶ =1 2etAµ1 1¶ +1 2etAµ1 ¡1¶ =et 2µ1 1¶ +e¡t 2µ1 ¡1¶ =µ1 2(et+e¡t) 1 2(et¡e¡t)¶ ; and similarly (1.34) etAµ1 1¶ =µ1 2(et¡e¡t) 1 2(et+e¡t)¶ : 1. The matrix exponential 155 Recalling that (1.35) cosh t=et+e¡t 2;sinht=et¡e¡t 2; we have (1.36) etA=µcoshtsinht sinhtcosht¶ : Example 2. Take (1.37) A=µ0¡2 1 2¶ : One has det( ¸I¡A) =¸2¡2¸¡2, hence eigenvalues (1.38) ¸1= 1 + i; ¸ 2= 1¡i; with corresponding eigenvectors (1.39) v1=µ¡2 1 +i¶ ; v 2=µ¡2 1¡i¶ : We have (1.40) etAv1=e(1+i)tv1; etAv2=e(1¡i)tv2: We can write (1.41)µ1 0¶ =¡i+ 1 4µ¡2 1 +i¶ +i¡1 4µ¡2 1¡i¶ ; µ0 1¶ =¡i 2µ¡2 1 +i¶ +i 2µ¡2 1¡i¶ ; to obtain (1.42)etAµ1 0¶ =¡i+ 1 4e(1+i)tµ¡2 1 +i¶ +i¡1 4e(1¡i)tµ¡2 1¡i¶ =et 4µ(2i+ 2)eit+ (2¡2i)e¡it ¡2ieit+ 2ie¡it¶ ; and (1.43)etAµ0 1¶ =¡i 2e(1+i)tµ¡2 1 +i¶ +i 2e(1¡i)tµ¡2 1¡i¶ =et 2µ2ieit¡2ie¡it (1¡i)eit+ (1 + i)e¡it¶ : 156 3. Linear Systems of Di®erential Equations We can write these in terms of trigonometric functions, using the fundamen- tal Euler identities (1.44) eit= cot t+isint; e¡it= cos t¡isint; established in x1 of Chapter 1. (See x2 of this chapter for more on this.) These yield (1.45) cos t=eit+e¡it 2;sint=eit¡e¡it 2i; and an inspection of the formulas above gives (1.46) etAµ1 0¶ =etµcost¡sint sint¶ ; etAµ0 1¶ =etµ¡2 sint cost+ sin t¶ ; hence (1.47) etA=etµcost¡sint¡2 sint sint cost+ sin t¶ : As was shown in Chapter 2, x6, ifA2M(n;C) has ndistinct eigenvalues, thenCnhas a basis of eigenvectors. If Ahas multiple eigenvalues, Cnmight or might not have a basis of eigenvectors, though as shown in x7 of Chapter 2, there will be a basis of generalized eigenvectors. If vis a generalized eienvector of A, say (1.48) ( A¡¸I)mv= 0; then (1.49) et(A¡¸I)v=X k<mtk k!(A¡¸I)kv; so (1.50) etAv=et¸X k<mtk k!(A¡¸I)kv: Example 3 . Consider the 3 £3 matrix Aused in (7.26) of Chapter 2: (1.51) A=0 @2 3 3 0 2 3 0 0 11 A: 1. The matrix exponential 157 Here 2 is a double eigenvalue and 1 a simple eigenvalue. Calculations done in Chapter 2, x7, yield (1.52) (A¡2I)0 @1 0 01 A= 0;(A¡2I)0 @0 1 01 A= 30 @1 0 01 A;(A¡I)0 @6 ¡3 11 A= 0: Hence etA0 @1 0 01 A=0 @e2t 0 01 A; (1.53) etA0 @0 1 01 A=e2t1X k=0tk k!(A¡2I)k0 @0 1 01 A (1.54) =e2t2 40 @0 1 01 A+0 @3t 0 01 A3 5; and (1.55) etA0 @6 ¡3 11 A=et0 @6 ¡3 11 A: Note that (1.56) etA0 @0 0 11 A=etA0 @6 ¡3 11 A¡6etA0 @1 0 01 A+ 3etA0 @0 1 01 A: Putting these calculations together yields (1.57) etA=0 @e2t3te2t6et¡6e2t+ 9te2t 0 e2t¡3et+ 3e2t 0 0 et1 A: Example 4 . Consider the 3 £3 matrix (1.58) A=0 @1 2 0 3 1 3 0¡2 11 A: 158 3. Linear Systems of Di®erential Equations A computation gives det( ¸I¡A) = (¸¡1)3. Hence for N=A¡Iwe have Spec( N) =f0g, so we know Nis nilpotent (by Proposition 8.1 of Chapter 2). In fact, a calculation gives (1.59) N=0 @0 2 0 3 0 3 0¡2 01 A; N2=0 @6 0 6 0 0 0 ¡6 0 ¡61 A; N3= 0: Hence (1.60)etA=eth I+tN+t2 2N2i =et0 @1 + 3 t22t1 + 3 t2 3t 1 3 t 1¡3t2¡2t1¡3t21 A: Exercises 1. Use the method of eigenvalues and eigenvectors given in (1.23){(1.26) to compute etAfor each of the following: A=µ1 1 1 0¶ ; A =µ1 1 0 2¶ ; A =µ1¡1 1 1¶ ; A =µ1i i1¶ : 2. Use the method given in (1.48){(1.50) and illustrated in (1.51){(1.60) to compute etAfor each of the following: A=µ1 2 0 1¶ ; A =0 @1 2 4 0 1 3 0 0 11 A; A =0 @1 0 1 0 2 0 ¡1 0 ¡11 A: 3. Show that et(A+B)=etAetB;8t=)AB=BA: Hint. SetX(t) =et(A+B); Y(t) =etAetB. Show that X´Y=)X0(t)¡Y0(t) =Bet(A+B)¡etABetB´0; and hence that X´Y=)BetA=etAB;8t: 1. The matrix exponential 159 4. Given A2M(n;C), suppose ©( t) is an n£nmatrix valued solution to d dt©(t) =A©(t): Show that ©(t) =etAB; where B= ©(0). Deduce that ©( t) is invertible for all t2Rif and only if ©(0) is invertible, and that in such a case e(t¡s)A= ©(t)©(s)¡1: (For a generalization, see (8.13).) 5. Let A; B2M(n;C) and assume Bis invertible. Show that (B¡1AB)k=B¡1AkB; and use this to show that etB¡1AB=B¡1etAB: 6. Show that if Ais diagonal, i.e., A=0 @a11 ... ann1 A; then etA=0 @eta11 ... etann1 A: Exercises 7{10 bear on the identity (1.61) det etA=etTrA; given A2M(n;C). 160 3. Linear Systems of Di®erential Equations 7. Show that if (1.61) holds for A=A1and if A2=B¡1A1B, then (1.61) holds for A=A2. 8. Show that (1.61) holds whenever Ais diagonalizable. Hint. Use Exercises 5{6. 9. Assume A2M(n;C) is upper triangular: (1.62) A=0 @a11¢¢¢ a1n ...... ann1 A: Show that etAis upper triangular, of the form etA=0 @e11(t)¢¢¢ e1n(t) ...... enn(t)1 A; e jj(t) =etajj: 10. Deduce that (1.61) holds when Ahas the form (1.62). Then deduce that (1.61) holds for all A2M(n;C). 11. Let A(t) be a smooth function of twith values in M(n;C). Show that (1.63) A(0) = 0 = )d dteA(t)¯¯¯ t=0=A0(0): Hint. Take the power series expansion of eA(t), in powers of A(t). 12. Let A(t) be a smooth M(n;C)-valued function of t2Iand assume (1.64) A(s)A(t) =A(t)A(s);8s; t2I: Show that (1.65)d dteA(t)=A0(t)eA(t)=eA(t)A0(t): Hint. Show that if (1.64) holds, d dteA(t)=d dseA(s)¡A(t)eA(t)¯¯¯ s=t; and apply Exercise 11. 1. The matrix exponential 161 13. Here is an alternative approach to Proposition 1.2. Assume (1.66) A; B2M(n;C); AB =BA: Show that (1.67) ( A+B)m=mX j=0µm j¶ AjBm¡j;µm j¶ =m! j!(m¡j)!: From here, show that (1.68)eA+B=1X m=01 m!(A+B)m =1X m=0mX j=01 j!(m¡j)!AjBm¡j: Then take n=m¡jand show this is (1.69)=1X j=01X n=01 j!n!AjBn =1X j=01 j!Aj1X n=01 n!Bn =eAeB; so (1.70) eA+B=eAeB: 14. As an alternative to the proof of (1.4), given in (1.3), which depends on term by term di®erentiation of power series, verify that, for A2 M(n;C), (1.71)d dtetA= lim h!01 h¡ e(t+h)A¡etA¢ =etAlim h!01 h¡ ehA¡I¢ =etAA =AetA; the second identity in (1.71) by (1.70), the third by the de¯nition (1.2), and the fourth by commutativity. 162 3. Linear Systems of Di®erential Equations 2. Exponentials and trigonometric functions In Chapter 1 we have seen how to use complex exponentials to give a self- contained treatment of basic results on the trigonometric functions cos tand sint. Here we present a variant, using matrix exponentials. We begin by looking at (2.1) x(t) =etJx0; J =µ0¡1 1 0¶ ; which solves (2.2) x0(t) =Jx(t); x(0) = x02R2: We ¯rst note that the planar curve x(t) moves about on a circle centered about the origin. Indeed, (2.3)d dtkx(t)k2=d dt¡ x(t)¢x(t)¢ =x0(t)¢x(t) +x(t)¢x0(t) =Jx(t)¢x(t) +x(t)¢Jx(t) = 0; since Jt=¡J. Thus kx(t)k=kx0kis constant. Furthermore the velocity v(t) =x0(t) has constant magnitude; in fact (2.4) kv(t)k2=v(t)¢v(t) =Jx(t)¢Jx(t) =kx(t)k2; since JtJ=¡J2=I. For example, (2.5)µc(t) s(t)¶ =etJµ1 0¶ is a curve, moving on the unit circle x2 1+x2 2= 1, at unit speed, with initial position x(0) = (1 ;0)tand initial velocity v(0) = (0 ;1)t. Now in trigonometry the functions cos tand sin tare de¯ned to be the x1andx2 coordinates of such a parametrization of the unit circle, so we have (2.6)µcost sint¶ =etJµ1 0¶ : The di®erential equation (2.2) then gives (2.7)d dtcost=¡sint;d dtsint= cos t: 2. Exponentials and trigonometric functions 163 Using etJµ0 1¶ =etJJµ1 0¶ =JetJµ1 0¶ ; we have a formula for etJ¡0 1¢ , which together with (2.6) yields (2.8) etJ=µcost¡sint sintcost¶ = (cos t)I+ (sin t)J: Then the identity e(s+t)J=esJetJyields the following identities, when ma- trix multiplication is carried out: (2.9)cos(s+t) = (cos s)(cos t)¡(sins)(sint); sin(s+t) = (cos s)(sint) + (sin s)(cos t): We now show how the treatment of sin tand cos tpresented above is really quite close to that given in Chapter 1, x1. To start, we note that if Cis regarded as a real vector space, with basis e1= 1; e2=i, and hence identi¯ed with R2, via (2.10) z=x+iy$µx y¶ ; then the matrix representation for the linear transformation z7!izis given byJ: (2.11) iz=¡y+ix; Jµx y¶ =µ¡y x¶ : More generally, the linear transformation z7!(c+is)zhas matrix repre- sentation (2.12)µc¡s s c¶ : Taking this into account, we see that the identity (2.8) is equivalent to (2.13) eit= cos t+isint; which is Euler's formula, as in (1.39) of Chapter 1. Here is another approach to the evaluation of etJ. We compute the eigenvalues and eigenvectors of J: (2.14) ¸1=i; ¸ 2=¡i;v1=µ1 ¡i¶ ; v 2=µ1 i¶ : Then, using the fact that etJvk=et¸kvk, we have (2.15) etJµ1 0¶ =1 2eitµ1 ¡i¶ +1 2e¡itµ1 i¶ : Comparison with (2.6) gives (2.16) cos t=1 2(eit+e¡it);sint=1 2i(eit¡e¡it); again leading to (2.13). 164 3. Linear Systems of Di®erential Equations Exercises 1. Recall Skew( n) and SO(n), de¯ned by (11.7) and (12.4) of Chapter 2. Show that (2.17) A2Skew( n) =)etA2SO(n);8t2R: Note how this generalizes (2.3). 2. Given an n£nmatrix A, let us set (2.18) cos tA=1 2(eitA+e¡itA);sintA=1 2i(eitA¡e¡itA): Show that (2.19)d dtcostA=¡AsintA;d dtsintA=AcostA: 3. In the context of Exercise 2, show that (2.20) cos tA=1X k=0(¡1)k (2k)!(tA)2k;sintA=1X k=0(¡1)k (2k+ 1)!(tA)2k+1: 4. Show that Av=¸v=)(costA)v= (cos t¸)v; (sintA)v= (sin t¸)v: 5. Compute cos tAand sin tAin each of the following cases: A=µ0 1 1 0¶ ; A =µ0¡1 1 0¶ ; A =µ1 1 0 1¶ ; A =µ0i i0¶ : 6. Suppose A2M(n;C) and B=µ0¡A A 0¶ 2M(2n;C): Show that etB=µcostA¡sintA sintA costA¶ : 3. First-order systems derived from higher-order equations 165 3. First-order systems derived from higher-order equations There is a standard process to convert an nth order di®erential equation (3.1)dny dtn+an¡1dn¡1y dtn¡1+¢¢¢+a1dy dt+a0y= 0 to a ¯rst-order system. Set (3.2) x0(t) =y(t); x1(t) =y0(t); : : : ; x n¡1(t) =y(n¡1)(t): Then x= (x0; : : : ; x n¡1)tsatis¯es (3.3)x0 0=x1 ... x0 n¡2=xn¡1 x0 n¡1=¡an¡1xn¡1¡ ¢¢¢ ¡ a0x0; or equivalently (3.4)dx dt=Ax; with (3.5) A=0 BBBB@0 1 ¢¢¢ 0 0 0 0 ¢¢¢ 0 0............... 0 0 ¢¢¢ 0 1 ¡a0¡a1¢¢¢ ¡ an¡2¡an¡11 CCCCA: The matrix Agiven by (3.5) is called the companion matrix of the polynomial (3.6) p(¸) =¸n+an¡1¸n¡1+¢¢¢+a1¸+a0: Note that a direct search of solutions to (3.1) of the form e¸tleads one to solve p(¸) = 0. Thus the following result is naturally suggested. Proposition 3.1. Ifp(¸)is a polynomial of the form (3.6), with companion matrix A, given by (3.5), then (3.7) p(¸) = det( ¸I¡A): 166 3. Linear Systems of Di®erential Equations Proof. We look at (3.8) ¸I¡A=0 BBBB@¸¡1¢¢¢ 0 0 0¸¢¢¢ 0 0............... 0 0 ¢¢¢ ¸ ¡1 a0a1¢¢¢ an¡2¸+an¡11 CCCCA; and compute its determinant by expanding by minors down the ¯rst column. We see that (3.9) det( ¸I¡A) =¸det(¸I¡eA) + (¡1)n¡1a0detB; where (3.10)eAis the companion matrix of ¸n¡1+an¡1¸n¡2+¢¢¢+a1; Bis lower triangular, with ¡1's on the diagonal. By induction on n, we have det( ¸I¡eA) =¸n¡1+an¡1¸n¡2+¢¢¢+a1, while detB= (¡1)n¡1. Substituting this into (3.9) gives (3.7). Converse construction We next show that each solution to a ¯rst order n£nsystem of the form (3.4) (for general A2M(n;F)) also satis¯es an nth order scalar ODE. Indeed, if (3.4) holds, then (3.11) x(k)=Ax(k¡1)=¢¢¢=Akx: Now if p(¸) is given by (3.7), and say (3.12) p(¸) =¸n+an¡1¸n¡1+¢¢¢+a1¸+a0; then, by the Cayley-Hamilton theorem (cf. (8.10) of Chapter 2), (3.13) p(A) =An+an¡1An¡1+¢¢¢+a1A+a0I= 0: Hence (3.14)x(n)=Anx =¡an¡1An¡1x¡ ¢¢¢ ¡ a1Ax¡a0x =¡an¡1x(n¡1)¡ ¢¢¢ ¡ a1x0¡a0x; so we have the asserted nth order scalar equation: (3.15) x(n)+an¡1x(n¡1)+¢¢¢+a1x0+a0x= 0: Remark. If the minimal polynomial q(¸) ofAhas degree m, less than n, we can replace pbyqand derive analogues of (3.14){(3.15), giving a single di®erential equation of degree mforx. 4. Non-homogeneous equations and Duhamel's formula 167 Exercises 1. Using the method (3.12){(3.15), convert dx dt=µ1 1 0 2¶ x into a second order scalar equation. 2. Using the method (3.2){(3.3), convert y00¡3y0+ 2y= 0 into a 2 £2 ¯rst order system. In Exercises 3{4, assume that ¸1is a root of multiplicity k¸2 for the polynomial p(¸) given by (3.6). 3. Verify that e¸1t; te¸1t; : : : ; tk¡1e¸1tare solutions to (3.1). 4. Deduce that, for each j= 0; : : : ; k ¡1, the system (3.3) has a solution of the form (3.16) x(t) = (tj+®tj¡1+¢¢¢+¯)et¸1v; (with vdepending on j). 5. For given A2M(n;C), suppose x0=Axhas a solution of the form (3.16). Show that ¸1must be a root of multiplicity ¸j+ 1 of the minimal polynomial of A. Hint. Take into account the remark below (3.15). 6. Using Exercises 3{5, show that the minimal polynomial of the compan- ion matrix Ain (3.5) must be the characteristic polynomial p(¸). 4. Non-homogeneous equations and Duhamel's formula Inxx1{3 we have focused on homogeneous equations, x0¡Ax= 0. Here we consider the non-homogeneous equation (4.1)dx dt¡Ax=f(t); x(0) = x02Cn: 168 3. Linear Systems of Di®erential Equations Here A2M(n;C) and f(t) takes values in Cn. The key to solving this is to recognize that the left side of (4.1) is equal to (4.2) etAd dt³ e¡tAx(t)´ ; as follows from the product formula for the derivative and the de¯ning prop- erty of etA, given in (1.4). Thus (4.1) is equivalent to (4.3)d dt³ e¡tAx(t)´ =e¡tAf(t); x(0) = x0; and integration yields (4.4) e¡tAx(t) =x0+Zt 0e¡sAf(s)ds: Applying etAto both sides then gives the solution: (4.5) x(t) =etAx0+Zt 0e(t¡s)Af(s)ds: This is called Duhamel's formula. Example . We combine methods of this section and x3 (and also x2) to solve (4.6) y00+y=f(t); y(0) = y0; y0(0) = y1: As in x3, set x= (x0; x1) = (y; y0), to obtain the system (4.7)d dtµx0 x1¶ =µ0 1 ¡1 0¶µx0 x1¶ +µ0 f(t)¶ : Recognizing the 2 £2 matrix above as ¡J, and recalling from x2 that (4.8) e(s¡t)J=µcos(s¡t)¡sin(s¡t) sin(s¡t) cos( s¡t)¶ ; we obtain (4.9)µx0(t) x1(t)¶ =µcostsint ¡sintcost¶µy0 y1¶ +Zt 0µcos(s¡t)¡sin(s¡t) sin(s¡t) cos( s¡t)¶µ0 f(s)¶ ds; and hence (4.10) y(t) = (cos t)y0+ (sin t)y1+Zt 0sin(t¡s)f(s)ds: 4. Non-homogeneous equations and Duhamel's formula 169 Use of Duhamel's formula is a good replacement for the method of vari- ation of parameters, discussed in x14 of Chapter 1. See x9 of this chapter for more on this. Here, we discuss a variant of the method of undetermined coe±cients, introduced for single second-order equations in x10 of Chapter 1. We consider the following special case, the ¯rst-order n£nsystem (4.11)dx dt¡Ax= (cos ¾t)v; given ¾2R; v2Rn, and A2M(n;R) (or we could use complex coe±- cients). We assume (4.12) i¾;¡i¾ =2Spec A; and look for a solution to (4.11) of the form (4.13) xp(t) = (cos ¾t)a+ (sin ¾t)b; a; b 2Rn: Substitution into (4.11) leads to success with (4.14)a=¡A(A2+¾I)¡1v; b=¡¾(A2+¾I)¡1v: If (4.12) does not hold, (4.14) fails, and (4.11) might not have a solution of the form (4.13). Of course, (4.5) will work; (4.11) will have a solution of the form (4.15) x(t) =Zt 0(cos¾s)e(t¡s)Av ds: When (4.12) holds and (4.13) works, the general solution to (4.11) is (4.16) x(t) =etAu0+ (cos ¾t)a+ (sin ¾t)b; u 02Rn; u0related to x(0) by (4.17) x(0) = u0+a: If all the eigenvalues of Ahave negative real part, etAu0will decay to 0 as t!+1. Then etAu0is called the transient part of the solution. The other part, (cos ¾t)A+ (sin ¾t)b, is called the steady state solution. 170 3. Linear Systems of Di®erential Equations Exercises 1. Given A2M(n;C), set Ek(t) =kX j=0tj j!Aj: Verify that E0 k(t) =AEk¡1(t) and that d dt¡ Ek(t)e¡tA¢ =¡tk k!Ak+1e¡tA: 2. Verify that, if Ais invertible, Zt 0ske¡sAds=¡k!A¡(k+1)£ Ek(t)e¡tA¡I¤ : 3. Solve the initial value problem dx dt=µ0 1 1 0¶ x+µet e¡t¶ ; x(0) =µ0 0¶ : 4. Solve the initial value problem dx dt=µ0¡1 1 0¶ x+µet e¡t¶ ; x(0) =µ0 0¶ : 5. Solve the initial value problem dx dt=µ1 1 ¡1¡1¶ x+µet e¡t¶ ; x(0) =µ0 0¶ : 6. Produce analogues of (4.8){(4.10) for y00¡3y0+ 2y=f(t); y(0) = y0; y0(0) = y1: 5. Simple electrical circuits 171 In Exercises 7{8, take X; Y2M(n;C) and (4.18) U(t; s) =et(X+sY); U s(t; s) =@ @sU(t; s): 7. Show that Ussatis¯es @Us @t= (X+sY)Us+Y U; U s(0; s) = 0 : 8. Use Duhamel's formula to show that Us(t; s) =Zt 0e(t¡¿)(X+sY)Y e¿(X+sY)d¿: Deduce that (4.19)d dseX+sY¯¯¯ s=0=eXZ1 0e¡¿XY e¿Xd¿: 9. Assume X(t) is a smooth function of t2Iwith values in M(n;C). Show that, for t2I, (4.20)d dteX(t)=eX(t)Z1 0e¡¿X(t)X0(t)e¿X(t)d¿: 10. In the context of Exercise 9, assume t; t02I=)X(t)X(t0) =X(t0)X(t): In such a case, simplify (4.20), and compare the result with that of Exercise 10 in x1. 5. Simple electrical circuits Here we extend the scope of the treatment of electrical circuits in x13 of Chapter 1. Rules worked out by Kirchho® and others in the 1800s allow one to write down a system of linear di®erential equations describing the 172 3. Linear Systems of Di®erential Equations voltages and currents running along a variety of electrical circuits, containing resistors, capacitors, and inductors. There are two types of basic laws. The ¯rst type consists of two rules known as Kirchho®'s laws: (A) The sum of the voltage drops around any closed loop is zero. (B) The sum of the currents at any node is zero. The second type of law speci¯es the voltage drop across each circuit element: Resistor: V=IR; (a) Inductor: V=LdI dt; (b) Capacitor: V=Q C: (c) In each case, Vis the voltage drop (in volts), Iis the current (in amps), Ris the resistance (in ohms), Lis the inductance (in henrys), Cis the capacitance (in farads), and Qis the charge (in coulombs). We refer to x13 of Chapter 1 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 volt- ages and currents, while rules (b) and (c){(c2) give di®erential equations, namely LdI dt=V (Inductor) ; (5.1) CdV dt=I (Capacitor) : (5.2) Note that (5.2) results from applying d=dt to (c) and then using (c2). If a circuit has kcapacitors and `inductors, we get an m£msystem of ¯rst order di®erential equations, with m=k+`. We illustrate the formulation of such di®erential equations for circuits presented in Figure 5.1 and Figure 5.2. In each case, the circuit elements are numbered. We denote by Vjthe voltage drop across element jand by Ijthe current across element j. 5. Simple electrical circuits 173 Figure 5.1 Figure 5.1 depicts a classical RLC circuit, such as treated in x13 of Chapter 1. Rules (A), (B), and (a) give (5.3)V1+V2+V3=E(t); I1=I2=I3; V1=RI1: Equations (5.1){(5.3) yield a system of two ODEs, for I3andV2: (5.4) LdI3 dt=V3; CdV2 dt=I2: We need to express V3andI2in terms of I3; V2, and E(t), using (5.3). In fact, we have (5.5)V3=E(t)¡V1¡V2=E(t)¡RI1¡V2=E(t)¡RI3¡V2; I2=I3; so we get the system (5.6)LdI3 dt=¡RI3¡V2+E(t); CdV2 dt=I3; or, in matrix form, (5.7)d dtµI3 V2¶ =µ¡R=L ¡1=L 1=C 0¶µI3 V2¶ +1 LµE(t) 0¶ : 174 3. Linear Systems of Di®erential Equations Figure 5.2 Note that the characteristic polynomial of the matrix (5.8) A=µ¡R=L ¡1=L 1=C 0¶ is (5.9) ¸2+R L¸+1 LC; with roots (5.10) ¸=¡R 2L§1 2Lr R2¡4L C: Let us now look at the slightly more complicated circuit depicted in Figure 5.2. Again we get a 2 £2 system of di®erential equations. Rules (A), (B), and (a) give (5.11)V1+V2+V4=E(t); V 2=V3; I1=I2+I3=I4; V3=R3I3; V 4=R4I4: Equations (5.1){(5.2) yield di®erential equations for I1andV2: (5.12) CdV2 dt=I2; LdI1 dt=V1: 5. Simple electrical circuits 175 Figure 5.3 We need to express I2andV1in terms of V2; I1, and E(t), using (5.11). In fact, we have (5.13)I2=I1¡I3=I1¡1 R3V3=I1¡1 R3V2; V1=E(t)¡V2¡V4=E(t)¡V2¡R4I4=E(t)¡V2¡R4I1; so we get the system (5.14)CdV2 dt=¡1 R3V2+I1; LdI1 dt=¡V2¡R4I1+E(t); or, in matrix form, (5.15)d dtµV2 I1¶ =µ¡1=R3C 1=C ¡1=L ¡R4=L¶µV2 I1¶ +1 Lµ0 E(t)¶ : Exercises 1. Work out the 3 £3 system of di®erential equations describing the be- havior of the circuit depicted in Figure 5.3. Assume E(t) = 5 sin 12 tvolts: 176 3. Linear Systems of Di®erential Equations 2. Using methods developed in x4, solve the 2 £2 system (5.7) when R= 5 ohms ; L = 4 henrys ; C = 1 farad ; and E(t) = 5 cos 2 tvolts; with initial data I3(0) = 0 amps ; V 2(0) = 5 volts : 3. Solve the 2 £2 system (5.15) when R3= 1 ohm ; R 4= 4 ohms ; L = 4 henrys ; C = 2 farads ; and E(t) = 2 cos 2 tvolts: 4. Use the method of (4.11){(4.14) to ¯nd the steady state solution to (5.7), when E(t) =Acos¾t: Take A; ¾; R andL¯xed and allow Cto vary. Show that the amplitude of the steady state solution is maximal (we say resonance is achieved) when LC=1 ¾2; recovering calculations of (13.7){(13.13) in Chapter 1. 5. Work out the analogue of Exercise 4 with the system (5.7) replaced by (5.15). Is the condition for resonance the same as in Exercise 4? 6. Draw an electrical circuit that leads to a 4 £4 system of di®erential equations, and write down said system. 6. Second-order systems Interacting physical systems often give rise to second-order systems of dif- ferential equations. Consider for example a system of nobjects, of mass m1; : : : ; m n, connected to each other and to two walls by n+1 springs, with spring constants k1; : : : ; k n+1, as in Fig. 6.1. We assume the masses slide 6. Second-order systems 177 Figure 6.1 without friction. Denote by xjthe position of the jth mass and by yjthe degree to which the jth spring is stretched. The equations of motion are (6.1) mjx00 j=¡kjyj+kj+1yj+1;1·j·n; and for certain constants aj, (6.2)yj=xj¡xj+1+aj;2·j·n; y1=x1+a1; y n+1=¡xn+an+1: Substituting (6.2) into (6.1) yields an n£nsystem, which we can write in matrix form as (6.3) Mx00=¡Kx+b; where x= (x1; : : : ; x n)t; b= (¡k1a1+k2a2; : : : ;¡knan+kn+1an+1)t, (6.4) M=0 @m1 ... mn1 A; and (6.5) K=0 BBBBB@k1+k2¡k2 ¡k2k2+k3... ... ...kn¡1+kn ¡kn ¡kn kn+kn+11 CCCCCA: 178 3. Linear Systems of Di®erential Equations We assume mj>0 and kj>0 for each j. Then clearly Mis a positive de¯nite matrix and Kis a real symmetric matrix. Proposition 6.1. If each kj>0, then K, given by (6.5), is positive de¯- nite. Proof. We have (6.6) x¢Kx=nX j=1(kj+kj+1)x2 j¡2nX j=2kjxj¡1xj: Now (6.7) 2 xj¡1xj·x2 j¡1+x2 j; so (6.8)x¢Kx¸nX j=1kjx2 j+n+1X j=2kjx2 j¡1 ¡nX j=2kjx2 j¡nX j=2kjx2 j¡1 ¸k1x2 1+kn+1x2 n: Furthermore note that the inequality in (6.7) is strict unless xj¡1=xjso the inequality in (6.8) is strict unless xj¡1=xjforeach j2 f2; : : : ; n g, i.e., unless x1=¢¢¢=xn. This proves that x¢Kx > 0 whenever x2Rn; x6= 0. To be more precise, we can sharpen (6.7) to (6.9) 2 xj¡1xj=x2 j¡1+x2 j¡(xj¡xj¡1)2; and then (6.8) is sharpened to (6.10) x¢Kx=k1x2 1+kn+1x2 n+nX j=2kj(xj¡xj¡1)2: If we set (6.11) ·= min fkj: 1·j·n+ 1g; then (6.10) implies (6.12) x¢Kx¸·³ x2 1+x2 n+nX j=2(xj¡xj¡1)2´ : 6. Second-order systems 179 The system (6.3) is inhomogeneous, but it is readily converted into the homogeneous system (6.13) Mz00=¡Kz; z =x¡K¡1b: This in turn can be rewritten (6.14) z00=¡M¡1Kz: Note that (6.15) L=M¡1=2KM¡1=2=)M¡1K=M¡1=2LM1=2; where (6.16) M1=2=0 B@m1=2 1 ... m1=2 n1 CA: Proposition 6.2. The matrix Lis positive de¯nite. Proof. x¢Lx= (M¡1=2x)¢K(M¡1=2x)>0 whenever x6= 0. According to (6.15), M¡1KandLare similar, so we have: Corollary 6.3. ForMandKof the form (6.4){(6.5), with mj; kj>0, the matrix M¡1Kis diagonalizable, and all its eigenvalues are positive. It follows that Rnhas a basis fv1; : : : ; v ngsatisfying (6.17) M¡1Kvj=¸2 jvj; ¸ j>0: Then the initial value problem (6.18) Mz00=¡Kz; z (0) = z0; z0(0) = z1 has the solution (6.19) z(t) =nX j=1³ ®jcos¸jt+¯j ¸jsin¸jt´ vj; where the coe±cients ®jand¯jare given by (6.20) z0=X ®jvj; z 1=X ¯jvj: 180 3. Linear Systems of Di®erential Equations An alternative approach to the system (6.14) is to set (6.21) u=M1=2z; for which (6.14) becomes (6.22) u00=¡Lu; with Lgiven by (6.15). Then Rnhas an orthonormal basis fwj: 1·j·ng, satisfying (6.23) Lwj=¸2 jwj;namely wj=M1=2vj; with vjas in (6.17). Note that we can set (6.24) L=A2; Aw j=¸jwj; and (6.22) becomes (6.25) u00+A2u= 0: One way to convert (6.25) to a ¯rst order (2 n)£(2n) system is to set (6.26) v=Au; w =u0: Then (6.25) becomes (6.27)d dtµv w¶ =Xµv w¶ ; X =µ0A ¡A0¶ : It is useful to note that when Xis given by (6.26), then (6.28) etX=µcostA sintA ¡sintAcostA¶ ; where cos tAand sin tAare given in Exercises 6{7 in x2. One way to see this is to let ©( t) denote the right side of (6.28) and use (2.19) to see that (6.29)d dt©(t) =X©(t);©(0) =µI0 0I¶ : Then Exercise 4 of x1 implies etX´©(t). These calculations imply that the solution to (6.25), with initial data u(0) = u0; u0(0) = u1, is given by (6.30) u(t) = (cos tA)u0+A¡1(sintA)u1: 6. Second-order systems 181 Compare (6.18){(6.20). This works for each invertible A2M(n;C). We move to the inhomogeneous variant of (6.14), which as above we can transform to the following inhomogeneous variant of (6.25): (6.31) u00+A2u=f(t); u(0) = u0; u0(0) = u1: Using the substitution (6.26), we get (6.32)d dtµv w¶ =Xµv w¶ +µ0 f(t)¶ ;µv(0) w(0)¶ =µAu0 u1¶ : Duhamel's formula applies to give (6.33)µv(t) w(y)¶ =etXµAu0 u1¶ +Zt 0e(t¡s)Xµ0 f(s)¶ ds: Using the formula (6.28) for etX, we see that the resulting formula for v(t) in (6.33) is equivalent to (6.34) u(t) = (cos tA)u0+A¡1(sintA)u1+Zt 0A¡1sin(t¡s)A f(s)ds: This is the analogue of Duhamel's formula for the solution to (6.31). We now return to the coupled spring problem and modify (6.1){(6.2) to allow for friction. Thus we replace (6.1) by (6.35) mjx00 j=¡kjyj+kj+1yj¡djx0 j; where yjare as in (6.2) and dj>0 are friction coe±cients, Then (6.3) is replaced by (6.36) Mx00=¡Kx¡Dx0+b; with bas in (6.3), MandKas in (6.4){(6.5), and (6.37) D=0 @d1 ... dn1 A; d j>0: As in (6.13), we can convert (6.36) to the homogeneous system (6.38) Mz00=¡Kz¡Dz0; z =x¡K¡1b: If we set u=M1=2z, as in (6.21), then, parallel to (6.22){(6.24), we get (6.39) u00+Bu0+A2u= 0; 182 3. Linear Systems of Di®erential Equations where A2is as in (6.24), with L=M¡1=2KM¡1=2, as in (6.22), and (6.40) B=M¡1=2DM¡1=2=0 @d1=m1 ... dn=mn1 A: The substitution (6.26) converts the n£nsecond order system (6.39) to the (2n)£(2n) ¯rst order system (6.41)d dtµv w¶ =µ0 A ¡A¡B¶µv w¶ : We can write (6.41) as (6.42)d dtµv w¶ = (X+Y)µv w¶ ; with Xas in (6.27) and (6.43) Y=µ0 0 0¡B¶ : Note that (6.44) XY=µ0¡AB 0 0¶ ; Y X =µ0 0 BA 0¶ ; so these matrices do not commute. Thus et(X+Y)might be di±cult to cal- culate even when AandBcommute. Such commutativity would hold, for example, if m1=¢¢¢=mnandd1=¢¢¢=dn, in which case Bis a scalar multiple of the identity matrix. When the positive, self adjoint operators AandBdo commute, we can make the following direct attack on the system (6.39). We know (cf. Exercise 3 inx11, Chapter 2) that Rnhas an orthonormal basis fw1; : : : ; w ngfor which (6.45) Awj=¸jwj; Bw j= 2¹jwj; ¸ j; ¹j>0: Then we can write a solution to (6.39) as (6.46) u(t) =X uj(t)wj; where the real-valued coe±cients uj(t) satisfy the equations (6.47) u00 j+ 2¹ju0 j+¸2 juj= 0; 6. Second-order systems 183 with solutions that are linear combinations: (6.48)e¡¹jt³ ®jcosq ¸2¡¹2 jt+¯jsinq ¸2 j¡¹2t´ ; ¸ j> ¹j; e¡¹jt³ ®jeq ¹2 j¡¸2 jt+¯je¡q ¹2 j¡¸2 jt´ ; ¸ j< ¹j; e¡¹jt(®j+¯jt); ¸ j=¹j: These three cases correspond to modes that are said to be underdamped, overdamped, and critically damped, respectively. In cases where AandBdo not commute, analysis of (6.39) is less explicit, but we can establish the following decay result. Proposition 6.4. IfA; B2M(n;C)are positive de¯nite, then all of the eigenvalues of Z=µ0 A ¡A¡B¶ have negative real part. Proof. Let's say ( v; w)t6= 0 and Z(v; w)t=¸(v; w)t. Then (6.49) Aw=¸v; Av +Bw=¡¸w; and (6.50) ( Z(v; w)t;(v; w)t) =¡(Bw; w ) + [( Aw; v )¡(Av; w )]; while also (6.51) ( Z(v; w)t;(v; w)t) =¸(kvk2+kwk2): The two terms on the right sie of (6.50) are real and purely imaginary, respectively, so we obtain (6.52) (Re ¸)(kvk2+kwk2) =¡(Bw; w ): If (v; w)t6= 0, we deduce that either Re ¸ < 0 or w= 0. If w= 0, then (6.49) gives Av= 0, hence v= 0. Hence w6= 0, and Re ¸ <0, as asserted. Exercises 1. Find the eigenvalues and eigenvectors of 0 @0 1 0 1 0 1 0 1 01 A: 184 3. Linear Systems of Di®erential Equations 2. Use the results of Exercise 1 to ¯nd the eigenvalues and eigenvectors of K, given by (6.5), in case n= 3 and k1=k2=k3=k4=k: 3. Find the general solution to u00+Bu0+A2u= 0; in case A2=K, with Kas in Exercise 2, and B=I. 4. Find the general solution to u00+µ0 1 1 0¶ u0+µ1 0 0 1¶ u= 0: 5. Produce a second order system of di®erential equations for the spring system presented in Fig. 6.2. (Assume there is no friction.) 6. Generalizing the treatment of (6.25), consider (6.53) u00+Lu= 0; L2M(N;C): AssumeCNhas a basis of eigenvectors vj, such that Lvj=¸2 jvj; ¸j2 C; ¸j6= 0. Show that the general solution to (6.53) has the form (6.54) u(t) =NX j=1(®je¸jt+¯je¡¸jt)vj; ® j; ¯j2C: How is this modi¯ed if some ¸j= 0? 7. Find the general solution to u00+µ2 1 0¡1¶ u= 0; and to u00+0 B@¡1 1 1 11 CAu= 0: 7. Curves inR3and the Frenet-Serret equations 185 Figure 6.2 7. Curves in R3and the Frenet-Serret equations Given a curve c(t) = ( x(t); y(t); z(t)) in 3-space, we de¯ne its velocity and acceleration by (7.1) v(t) =c0(t); a(t) =v0(t) =c00(t): We also de¯ne its speed s0(t) and arclength by (7.2) s0(t) =kv(t)k; s(t) =Zt t0s0(¿)d¿; assuming we start at t=t0. We de¯ne the unit tangent vector to the curve as (7.3) T(t) =v(t) kv(t)k: Henceforth we assume the curve is parametrized by arclength. We de¯ne the curvature ·(s) of the curve and the normal N(s) by (7.4) ·(s) =°°°dT ds°°°;dT ds=·(s)N(s): Note that (7.5) T(s)¢T(s) = 1 = )T0(s)¢T(s) = 0 ; so indeed N(s) is orthogonal to T(s). We then de¯ne the binormal B(s) by (7.6) B(s) =T(s)£N(s): 186 3. Linear Systems of Di®erential Equations 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 x12 of Chapter 2.) Thetorsion 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) = 1 = )B0(s)¢B(s) = 0 : Also, di®erentiating (7.6) and using (7.4), we have (7.8) B0(s) =T0(s)£N(s) +T(s)£N0(s) =T(s)£N0(s) =)B0(s)¢T(s) = 0 : We deduce that B0(s) is parallel to N(s). We de¯ne the torsion by (7.9)dB ds=¡¿(s)N(s): We complement the formulas (7.4) and (7.9) for dT=ds anddB=ds with one for dN=ds . Since N(s) =B(s)£T(s), we have (7.10)dN ds=dB ds£T+B£dT ds=¿N£T+·B£N; or (7.11)dN ds=¡·(s)T(s) +¿(s)B(s): Together, (7.4), (7.9) and (7.11) are known as the Frenet-Serret formulas. Example . Pick a; b > 0 and consider the helix (7.12) c(t) = (acost; asint; bt): Then v(t) = ( ¡asint; acost; b) and kv(t)k=p a2+b2, so we can pick s=tp a2+b2to parametrize by arc length. We have (7.13) T(s) =1p a2+b2(¡asint; acost; b); 7. Curves inR3and the Frenet-Serret equations 187 hence (7.14)dT ds=1 a2+b2(¡acost;¡asint;0): By (7.4), this gives (7.15) ·(s) =a a2+b2; N (s) = (¡cost;¡sint;0): Hence (7.16) B(s) =T(s)£N(s) =1p a2+b2(bsint;¡bcost; a): Then (7.17)dB ds=1 a2+b2(bcost; bsint;0); so, by (7.9), (7.18) ¿(s) =b a2+b2: In particular, for the helix (7.12), we see that the curvature and torsion are constant . Let us collect the Frenet-Serret equations (7.19)dT ds= ·N dN ds=¡·T +¿B dB ds= ¡¿N for a smooth curve c(s) inR3, parametrized by arclength, with unit tangent T(s), normal N(s), and binormal B(s), given by (7.20) N(s) =1 ·(s)T0(s); B (s) =T(s)£N(s); assuming ·(s) =kT0(s)k>0. The basic existence and uniqueness theory, which will be presented in Chapter 4, applies to (7.19). If ·(s) and ¿(s) are given smooth functions on an interval I= (a; b) and s02I, then, given T0; N0; B02R3, (7.19) has a unique solution on s2Isatisfying (7.21) T(s0) =T0; N (s0) =N0; B (s0) =B0: 188 3. Linear Systems of Di®erential Equations In fact, the case when ·(s) and ¿(s) are analytic will be subsumed in the material of x10 of this chapter. We now establish the following. Proposition 7.1. Assume ·and¿are given smooth functions on I, with · >0onI. Assume fT0; N0; B0gis an orthonormal basis of R3, such that B0=T0£N0. Then there exists a smooth, unit-speed curve c(s); s2I, for which the solution to (7.19) and (7.21) is the Frenet frame. To construct the curve, take T(s); N(s), and B(s) to solve (7.19) and (7.21), pick p2R3and set (7.22) c(s) =p+Zs s0T(¾)d¾; soT(s) =c0(s) is the velocity of this curve. To deduce that fT(s); N(s); B(s)g is the Frenet frame for c(s), for all s2I, we need to know: (7.23) fT(s); N(s); B(s)gorthonormal, with B(s) =T(s)£N(s);8s2I: 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)2SO(3);8s2I; withSO(3) de¯ned as in (12.4) of Chapter 2. The hypothesis on fT0; N0; B0g stated in Proposition 7.1 is equivalent to F0= (T0; N0; B0)2SO(3). Now F(s) satis¯es the di®erential equation (7.26) F0(s) =F(s)A(s); F (s0) =F0; where (7.27) A(s) =0 @0¡·(s) 0 ·(s) 0 ¡¿(s) 0 ¿(s) 01 A: Note that (7.28)dF¤ ds=A(s)¤F(s)¤=¡A(s)F(s)¤; 7. Curves inR3and the Frenet-Serret equations 189 since A(s) in (7.27) is skew-adjoint. Hence (7.29)d dsF(s)F(s)¤=dF dsF(s)¤+F(s)dF¤ ds =F(s)A(s)F(s)¤¡F(s)A(s)F(s)¤ = 0: Thus, whenever (7.26){(7.27) hold, (7.30) F0F¤ 0=I=)F(s)F(s)¤´I; and we have (7.23). Let us specialize the system (7.19), or equivalently (7.26), to the case where ·and¿areconstant , i.e., (7.31) F0(s) =F(s)A; A =0 @0¡·0 ·0¡¿ 0¿ 01 A; with solution (7.32) F(s) =F0e(s¡s0)A: We have already seen in that a helix of the form (7.12) has curvature ·and torsion ¿, with (7.33) ·=a a2+b2; ¿ =b a2+b2; and hence (7.34) a=· ·2+¿2; b =¿ ·2+¿2: In (7.12), sandtare related by t=sp ·2+¿2. We can also see such a helix arise via a direct calculation of esA, which we now produce. First, a straightforward calculation gives, for Aas in (7.31), (7.35) det( ¸I¡A) =¸(¸2+·2+¿2); hence (7.36) Spec( A) =f0;§ip ·2+¿2g: 190 3. Linear Systems of Di®erential Equations An inspection shows that we can take (7.37) v1=1p ·2+¿20 @¿ 0 ·1 A; v 2=0 @0 1 01 A; v 3=1p ·2+¿20 @¡· 0 ¿1 A; and then (7.38) Av1= 0; Av 2=p ·2+¿2v3; Av 3=¡p ·2+¿2v2: In particular, with respect to the basis fv2; v3gofV= Span fv2; v3g,AjV has the matrix representation (7.39) B=p ·2+¿2µ0¡1 1 0¶ : We see that (7.40) esAv1=v1; while, in light of the calculations giving (2.8), (7.41)esAv2= (cos sp ·2+¿2)v2+(sin sp ·2+¿2)v3; esAv3=¡(sinsp ·2+¿2)v2+(cos sp ·2+¿2)v3: Exercises 1. Consider a curve c(t) inR3, not necessarily parametrized by arclength. Show that the acceleration a(t) is given by (7.42) a(t) =d2s dt2T+·³ds dt´2 N: Hint. Di®erentiate 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) ·B=v£a kvk3: Hint. Take the cross product of both sides of (7.42) with T, and use (7.6). 8. Variable coe±cient systems 191 3. In the setting of Exercises 1{2, show that (7.44) ·2¿kvk6=¡a¢(v£a0): Deduce from (7.43){(7.44) that (7.45) ¿=(v£a)¢a0 kv£ak2: Hint. Proceed from (7.43) to d dt(·kvk3)B+·kvk3dB dt=d dt(v£a) =v£a0; and use dB=dt =¡¿(ds=dt )N, as a consequence of (7.9). Then dot with a, and use a¢N=·kvk2, from (7.42), to get (7.44). 4. Consider the curve c(t) inR3given by c(t) = (acost; bsint; t); where aandbare given positive constants. Compute the curvature, torsion, and Frenet frame. Hint. Use (7.43) to compute ·andB. Then use N=B£T. Use (7.45) to compute ¿. 5. Suppose cand ~care two curves, both parametrized by arc length over 0·s·L, and both having the same curvature ·(s)>0 and the same torsion ¿(s). Show that there exit x02R3andA2O(3) such that ~c(s) =Ac(s) +x0;8s2[0; L]: Hint. To begin, show that if their Frenet frames coincide at s= 0, i.e., eT(0) = T(0);eN(0) = N(0);eB(0) = B(0), then eT´T;eN´N;eB´B. 6. Suppose cis a curve inR3with curvature · >0. Show that there exists a plane in which c(t) lies for all tif and only if ¿´0. Hint. When ¿´0, the plane should be parallel to the orthogonal complement of B. 8. Variable coe±cient systems Here we consider a variable coe±cient n£n¯rst order system (8.1)dx dt=A(t)x; x (t0) =x02Cn; 192 3. Linear Systems of Di®erential Equations and its inhomogeneous analogue. The general theory, which will be pre- sented in Chapter 4, implies that if A(t) is a smooth function of t2I= (a; b) andt02I, then (8.1) has a unique solution x(t) fort2I, depending linearly onx0, so (8.2) x(t) =S(t; t0)x0; S(t; t0)2 L(Cn): Seex10 of this chapter for power series methods of constructing S(t; t0), when A(t) is analytic. As we have seen, (8.3) A(t)´A=)S(t; t0) =e(t¡t0)A: However, for variable coe±cient 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; t0) is related to a \complete set" of solutions to (8.1). Suppose x1(t); : : : ; x n(t) arensolutions to (8.1) (but with di®erent initial conditions). Fix t02I, and assume (8.4) x1(t0); : : : ; x n(t0) are linearly independent in Cn; or equivalently these vectors form a basis of Cn. Given such solutions xj(t), we form the n£nmatrix (8.5) M(t) =¡ x1(t); : : : ; x n(t)¢ ; whose jth column is xj(t). This matrix function solves (8.6)dM dt=A(t)M(t): The condition (8.4) is equivalent to the statement that M(t0) is invertible. We claim that if Msolves (8.6) and M(t0) is invertible then M(t) is invertible for all t2I. 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 offx1(t); : : : ; x n(t)g. It is also notable that W(t) solves a di®erential equation. In general we have (8.8)d dtdetM(t) =¡ detM(t)¢ Tr¡ M(t)¡1M0(t)¢ : 8. Variable coe±cient systems 193 (See Exercises 1{3 below.) Let ~I½Ibe the maximal interval containing t0on which M(t) is invertible. Then (8.8) holds for t2~I. When (8.6) holds, we have Tr( M(t)¡1M0(t)) = Tr( M(t)¡1A(t)M(t)) = Tr A(t), so the Wronskian solves the di®erential equation (8.9)dW dt=¡ TrA(t)¢ W(t): Hence (8.10) W(t) =eb(t;s)W(s); b(t; s) =Zt sTrA(¿)d¿: This implies ~I=Iand hence gives the asserted invertibility. From here we obtain the following. Proposition 8.1. IfM(t)solves (8.6) for t2IandM(t0)is invertible, then (8.11) S(t; t0) =M(t)M(t0)¡1;8t2I: Proof. We have seen that M(t) is invertible for all t2I. Ifx(t) solves (8.1), set (8.12) y(t) =M(t0)M(t)¡1x(t); and apply d=dt. Using the identity (8.13)d dtM(t)¡1=¡M(t)¡1M0(t)M(t)¡1; (see Exercise 4 below), we have (8.14)dy dt=¡M(t0)M(t)¡1A(t)M(t)M(t)¡1x(t) +M(t0)M(t)¡1A(t)x(t) = 0; hence y(t) =y(t0) for t2I, i.e., (8.15) M(t0)M(t)¡1x(t)´x(t0): Applying M(t)M(t0)¡1to both sides of (8.15) gives (8.11). Note also that, for s; t2I, (8.16) S(t; s) =M(t)M(s)¡1 194 3. Linear Systems of Di®erential Equations gives S(t; s)x(s) =x(t) for each solution x(t) to (8.1). We also have (8.17) S(t; t0) =S(t; s)S(s; t0); S(t; s) =S(s; t)¡1: There is a more general version of the Duhamel formula (4.5) for the solution to an inhomogeneous di®erential equation (8.18)dx dt=A(t)x+f(t); x(t0) =x0: Namely, (8.19) x(t) =S(t; t0)x0+Zt t0S(t; s)f(s)ds: In fact, clearly (8.19) satis¯es x(t0) =x0and applying d=dt to (8.19) gives the equation (8.18). In view of (8.16), we have the alternative formula (8.20) x(t) =M(t)M(t0)¡1x0+M(t)Zt t0M(s)¡1f(s)ds; 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 (8.1) in case the following commutativity hypothesis holds: (CA) A(t)A(t0) =A(t0)A(t);8t; t02I: We claim that if (8.21) B(t; s) =¡Zt sA(¿)d¿; then (8.22) ( CA) =)d dt¡ eB(t;s)x(t)¢ =eB(t;s)¡ x0(t)¡A(t)x(t)¢ ; from which it follows that (8.23) ( CA) =)S(t; s) =eRt sA(¿)d¿: (This identity fails in the absence of the hypothesis (CA).) To establish (8.22), we note that (CA) implies (CB) B(t; s)B(t0; s) =B(t0; s)B(t; s);8s; t; t02I: 8. Variable coe±cient systems 195 Next, (8.24)(CB) =)lim h!01 h³ eB(t+h;s)¡eB(t;s)´ = lim h!01 heB(t;s)³ eB(t+h;s)¡B(t;s)¡I´ =¡eB(t;s)A(t) =)d dteB(t;s)=¡eB(t;s)A(t); from which (8.22) follows. Here is an application of (8.23). Let x(s) be a planar curve, on an interval about s= 0, parametrized by arc-length, with unit tangent T(s) =x0(s). Then the Frenet-Serret equations (7.1) simplify to T0=·N, with N=JT, i.e., to (8.25) T0(s) =·(s)JT(s); with Jas in (2.1). Clearly the commutativity hypothesis (CA) holds for A(s) =·(s)J, so we deduce that (8.26) T(s) =e¸(s)JT(0); ¸(s) =Zs 0·(¿)d¿: Recall that etJis given by (2.8), i.e., (8.27) etJ=µcost¡sint sintcost¶ : Exercises Exercises 1{3 lead to a proof of the formula (8.8) for the derivative of detM(t). 1. Let A2M(n;C). Show that, as s!0, det(I+sA) = (1 + sa11)¢¢¢(1 +sann) +O(s2) = 1 + sTrA+O(s2); henced dsdet(I+sA)¯¯ s=0= Tr A: 196 3. Linear Systems of Di®erential Equations 2. Let B(s) be a smooth matrix valued function of s, with B(0) = I. Use Exercise 1 to show that d dsdetB(s)¯¯ s=0= Tr B0(0): Hint. Write B(s) =I+sB0(0) + O(s2). 3. Let C(s) be a smooth matrix valued function, and assume C(0) is in- vertible. Use Exercise 2 plus detC(s) =¡ detC(0)¢ detB(s); B (s) =C(0)¡1C(s) to show that d dsdetC(s)¯¯ s=0=¡ detC(0)¢ TrC(0)¡1C0(0): Use this to prove (8.8). Hint. Fixtand set C(s) =M(t+s), so d dtdetM(t) =d dsdetC(s)¯¯ s=0: 4. Verify the identity (8.13). Hint. SetU(t) =M(t)¡1and di®erentiate the identity U(t)M(t) =I. Exercises 5{6 generalize (8.25){(8.27) from the case of zero torsion (cf. Exercise 6 of x7) to the case (8.28) ¿(t) =¯·(t); ¯ constant. 5. Assume x(t) is a curve in R3for which (8.28) holds. Show that x(t) = x(0) +Rt 0T(s)ds, with (8.29)¡ T(t); N(t); B(t)¢ =e¾(t)K¡ T(0); N(0); B(0)¢ ; where (8.30) K=0 @0¡1 0 1 0 ¡¯ 0¯ 01 A; ¾(t) =Zt 0·(s)ds: 8. Variable coe±cient systems 197 Hint. Use (7.26){(7.27) and (CA) )(8.21){(8.23). 6. Let e1; e2; e3denote the standard basis of R3, and let v1= (1 + ¯2)¡1=2(¯e1+e3); v 2=e2; v 3= (1 + ¯2)¡1=2(e1¡¯e3): Show that v1; v2; v3form an orthonormal basis of R3and, with Kas in (8.30), Kv1= 0; Kv 2=¡(1 +¯2)1=2v3; Kv 3= (1 + ¯2)1=2v2: Deduce that e¾Kv1=v1; e¾Kv2= (cos ´)v2¡(sin´)v3; e¾Kv3= (sin ´)v2+ (cos ´)v3; where ´= (1 + ¯2)1=2¾. 7. Given B2M(n;C), write down the solution to dx dt=etBx; x (0) = x0: Hint. Use (8.23). Exercises 8{9 deal with a linear equation with periodic coe±cients: (8.31)dx dt=A(t)x; A (t+ 1) = A(t): SayA(t)2M(n;C). 8. Assume M(t) solves (8.6), with A(t) as in (8.31), and M(0) is invertible. Show that (8.32) M(1) = C=)M(t+ 1) = M(t)C: 9. In the setting of Exercise 8, we know M(t) is invertible for all t, soCis invertible. Results of Appendix A yield X2M(n;C) such that (8.33) eX=C: Show that (8.34) P(t) =M(t)e¡tX=)P(t+ 1) = P(t): The representation (8.35) M(t) =P(t)etX is called the Floquet representation ofM(t). 198 3. Linear Systems of Di®erential Equations 9. Variation of parameters and Duhamel's formula An inhomogeneous equation (9.1) y00+a(t)y0+b(t)y=f(t) can be solved via the method of variation of parameters, if one is given a complete set u1(t); u2(t) of solutions to the homogeneous equation (9.2) u00 j+a(t)u0 j+b(t)uj= 0: The method (derived already in x12 of Chapter 1 when a(t) and b(t) are constant) consists of seeking a solution to (9.1) in the form (9.3) y(t) =v1(t)u1(t) +v2(t)u2(t); and ¯nding equations for vj(t) which can be solved and which work to yield a solution to (9.1). We have (9.4) y0=v1u0 1+v2u0 2+v0 1u1+v0 2u2: We impose the condition (9.5) v0 1u1+v0 2u2= 0: Then y00=v0 1u0 1+v0 2u0 2+v1u00 1+v2u00 2, and plugging in (9.2) gives (9.6) y00=v0 1u0 1+v0 2u0 2¡(au0 1+bu1)v1¡(au0 2+bu2)v2; hence (9.7) y00+ay0+by=v0 1u0 1+v0 2u0 2: Thus we have a solution to (9.1) in the form (9.3) provided v0 1andv0 2solve (9.8)v0 1u1+v0 2u2= 0; v0 1u0 1+v0 2u0 2=f: This linear system for v0 1; v0 2has the explicit solution (9.9) v0 1=¡u2 Wf; v0 2=u1 Wf; where W(t) is the Wronskian: (9.10) W=u1u0 2¡u2u0 1= detµu1u2 u0 1u0 2¶ : 9. Variation of parameters and Duhamel's formula 199 Then (9.11)v1(t) =¡Zt t0u2(s) W(s)f(s)ds+C1; v2(t) =Zt t0u1(s) W(s)f(s)ds+C2: So (9.12) y(t) =C1u1(t) +C2u2(t) +Zt t0£ u2(t)u1(s)¡u1(t)u2(s)¤f(s) W(s)ds: We can connect this formula with that produced in x8 as follows. If y(t) solves (9.1), then x(t) = (y(t); y0(t))tsolves the ¯rst order system (9.13)dx dt=A(t)x+µ0 f(t)¶ ; where (9.14) A(t) =µ0 1 ¡b(t)¡a(t)¶ ; and a complete set of solutions to the homogeneous version of (9.13) is given by (9.15) xj(t) =µuj(t) u0 j(t)¶ ; j = 1;2: Thus we can set (9.16) M(t) =µu1(t)u2(t) u0 1(t)u0 2(t)¶ ; and as in (8.20) we have (9.17)µy(t) y0(t)¶ =M(t)M(t0)¡1µy0 y1¶ +M(t)Zt t0M(s)¡1µ0 f(s)¶ ds; solving (9.13) with y(t0) =y0; y0(t0) =y1. Note that (9.18) M(s)¡1=1 W(s)µu0 2(s)¡u2(s) ¡u0 1(s)u1(s)¶ ; with W(s), the Wronskian, as in (9.10). Thus the last term on the right side of (9.17) is equal to (9.19)µu1(t)u2(t) u0 1(t)u0 2(t)¶Zt t01 W(s)µ¡u2(s)f(s) u1(s)f(s)¶ ds; and performing this matrix multiplication yields the integrand in (9.12). Thus we see that Duhamel's formula provides an alternative approach to the method of variation of parameters. 200 3. Linear Systems of Di®erential Equations Exercises 1. Use the method of variation of parameters to solve (9.20) y00+y= tan t: 2. Convert (9.20) to a 2 £2 ¯rst order system and use Duhamel's formula to solve it. Compare the result with your work on Exercise 1. Compare also with (4.6){(4.10). 3. Do analogues of Exercises 1{2 for each of the following equations. y00+y=et; (a) y00+y= sin t; (b) y00+y=t; (c) y00+y=t2: 4. Show that the Wronskian, de¯ned by (9.10), satis¯es the equation (9.21)dW dt=¡a(t)W; ifu1andu2solve (9.2). 5. Show that one solution to (9.22) u00+ 2tu0+ 2u= 0 is (9.23) u1(t) =e¡t2: Set up and solve the di®erential equation for W(t) =u1u0 2¡u2u0 1. Then solve the associated ¯rst order equation for u2, to produce a linearly independent solution u2to (9.22), in terms of an integral. 6. Do Exercise 5 with (9.22) replaced by u00+ 2u0+u= 0; one of whose solutions is u1(t) =e¡t: 10. Power series expansions 201 10. Power series expansions Here we produce solutions to initial value problems (10.1)dx dt=A(t)x+f(t); x(0) = x0; in terms of a power series expansion, (10.2) x(t) =x0+x1t+x2t2+¢¢¢=1X k=0xktk; under the hypothesis that the n£nmatrix-valued function A(t) and vector- valued function f(t) are given by power series, (10.3) A(t) =1X k=0Aktk; f(t) =1X k=0fktk; convergent for jtj< R 0. The coe±cients xkin (10.2) will be obtained recur- sively, as follows. Given x(t) of the form (10.2), we have (10.4)dx dt=1X k=1kxktk¡1=1X k=0(k+ 1)xk+1tk; and (10.5)A(t)x=1X j=0Ajtj1X `=0x`t` =1X k=0³kX j=0Ak¡jxj´ tk; so the power series on the left and right sides of (10.1) agree if and only if, for each k¸0, (10.6) ( k+ 1)xk+1=kX j=0Ak¡jxj+fk: In particular, the ¯rst three recursions are (10.7)x1=A0x0+f0; 2x2=A1x0+A0x1+f1; 3x3=A2x0+A1x1+A0x2+f2: To start the recursion, the initial condition in (10.1) speci¯es x0. We next address the issue of convergence of the power series thus pro- duced for x(t). We will establish the following Proposition 10.1. Under the hypotheses given above, the power series (10.2) converges to the solution x(t)to (10.1), for jtj< R 0. 202 3. Linear Systems of Di®erential Equations Proof. The hypotheses on (10.3) imply that for each R < R 0, there exist a; b2(0;1) such that (10.8) kAkk ·aR¡k;kfkk ·bR¡k;8k2Z+: We will show that, given r2(0; R), there exists C2(0;1) such that (10.9) kxjk ·Cr¡j;8j2Z+: Such estimates imply that the power series (10.2) converges for jtj< r, for each r < R 0, hence for jtj< R 0. We will prove (10.9) by induction. The inductive step is to assume it holds for all j·kand to deduce that it holds for j=k+ 1. This deduction proceeds as follows. We have, by (10.6), (10.8), and (10.9) for j·k, (10.10)(k+ 1)kxk+1k ·kX j=0kAk¡jk ¢ kxjk+kfkk ·aCkX j=0Rj¡kr¡j+bR¡k =aCr¡kkX j=0³r R´k¡j +bR¡k: Now, given 0 < r < R , (10.11)kX j=0³r R´k¡j <1X j=0³r R´j =1 1¡r R=M(R; r)<1: Hence (10.12) ( k+ 1)kxk+1k ·aCM (R; r)r¡k+br¡k: We place on Cthe constraint that (10.13) C¸b; and obtain (10.14) kxk+1k ·aM(R; r) + 1 k+ 1r¢Cr¡k¡1: This gives the desired result (10.15) kxk+1k ·Cr¡k¡1; 10. Power series expansions 203 as long as (10.16)aM(R; r) + 1 k+ 1r·1: Thus, to ¯nish the argument, we pick K2Nsuch that (10.17) K+ 1¸£ aM(R; r) + 1¤ r: (Recall that we have a; R; r , and M(R; r).) Then we pick C2(0;1) large enough that (10.9) holds for all j2 f0;1; : : : ; K g, i.e., we take (in addition to (10.13)) (10.18) C¸max 0·j·Krjkxjk: Then for all k¸K, the inductive step yielding (10.15) from the validity of (10.9) for all j·kholds, and the inductive proof of (10.9) is complete. For notational simplicity, we have discussed power series expansions about t= 0 so far, but the same considerations apply to power series about a more general point t0. Thus we could replace (10.1) by (10.19)dx dt=A(t)x+f(t); x(t0) =x0; with A(t) and f(t) given by power series (10.20) A(t) =1X k=0Ak(t¡t0)k; f(t) =1X k=0fk(t¡t0)k; forjt¡t0j< R 0, and ¯nd x(t) in the form (10.21) x(t) =1X k=0xk(t¡t0)k: The recursive formula for the coe±cients xkis again given by (10.6), and (10.8){(10.18) apply without further change. It is worth noting that, in (10.20){(10.21), (10.22) Ak=1 k!A(k)(t0); f k=1 k!f(k)(t0); x k=1 k!x(k)(t0): These formulas, say for fk, arise as follows. Setting t=t0in (10.20) gives f0=f(t0). Generally, if the power series for f(t) converges for jt¡t0j< R 0, so does the power series (10.23) f0(t) =1X k=1kfk(t¡t0)k¡1; 204 3. Linear Systems of Di®erential Equations and more generally, (10.24) f(n)(t) =1X k=nk(k¡1)¢¢¢(k¡n+ 1)fk(t¡t0)k¡n; and setting t=t0in (10.24) gives f(n)(t0) =n!fn. As an aside, we mention that a convenient way to prove (10.23) is to de¯ne g(t) to be the power series on the right side of (10.23) and show that (10.25)Zt 0g(s)ds=1X k=1fk(t¡t0)k=f(t)¡f(t0): Compare the discussion regarding (1.45){(1.50) in Chapter 1. We next establish the following important fact about functions given by convergent power series. Proposition 10.2. Iff(t)is given by a power series as in (10.20), conver- gent for jt¡t0j< R 0, then fcan also be expanded in a power series in t¡t1, for each t12(t0¡R0; t0+R0), with radius of convergence R0¡ jt0¡t1j. Proof. As in (10.22), such a power series would necessarily have the form (10.26) f(t) =1X n=0f(n)(t1) n!(t¡t1)n: We claim this converges when jt¡t1j< R 0¡ jt1¡t0j. To see this, apply (10.24) with t=t1to estimate jf(n)(t1)j. We have (10.27)f(n)(t1) n!=1X k=nµk n¶ fk(t1¡t0)k¡n: The assertion that the power series (10.20) for f(t) converges for jt¡t0j< R 0 implies that, for each r < R 0, there exists C <1such that kfkk ·Cr¡k. Hence (10.27) gives (10.28)kf(n)(t1)k n!·C1X k=nµk n¶ r¡kjt1¡t0jk¡n =Cr¡n1X k=nµk n¶³jt1¡t0j r´k¡n : To evaluate this last quantity, note that (10.29)'(s) = (1 ¡s)¡(n+1))'(`)(0) = ( n+ 1)¢¢¢(n+`) )'(s) =1X `=0µn+` n¶ s`=1X k=nµk n¶ sk¡n; 10. Power series expansions 205 forjsj<1, so (10.28) gives (10.30)kf(n)(t1)k n!·Cr¡n³ 1¡jt1¡t0j r´¡(n+1) =C¡ r¡ jt1¡t0j¢¡n³ 1¡jt1¡t0j r´¡1 : This estimate implies the series (10.26) converges for jt¡t1j< r¡ jt1¡t0j, for each r < R 0, hence for jt¡t1j< R 0¡ jt1¡t0j. The fact that the sum of the series is equal to f(t) is a consequence of the formula (10.32) given below for the remainder Rk(t; t1) in the expansion (10.31) f(t) =kX n=0f(n)(t1) n!(t¡t1)n+Rk(t; t1); namely, (10.32) Rk(t; t1) =1 k!Zt t1(t¡s)kf(k+1)(s)ds: See Exercise 1 below for an approach to proving (10.32). The asserted convergence in (10.26) is equivalent to the statement that (10.33) kRk(t; t1)k ¡! 0;ask! 1 ; forjt¡t1j< R 0¡ jt1¡t0j, which can be obtained from (10.30), or rather the analogous estimates on kf(n)(s)k=n!, when sis between t1andt. See Exercise 3 below. A (vector-valued) function fde¯ned on an interval I= (a; b) is said to be an analytic function onIif and only if for each t12I, there is an r1>0 such that for jt¡t1j< r1,f(t) is given by a convergent power series in t¡t1. Again, such a power series is necessarily of the form (10.26). What has just been demonstrated implies that if fis given by a comvergent power series int¡t0forjt¡t0j< R 0, then fis analytic on the interval ( t0¡R0; t0+R0). The following is a useful fact about analytic functions. Lemma 10.3. Iffis analytic on (a; b)and0< " < jb¡aj=2, then there exists ± >0such that for each t12[a+"; b¡"], the power series (10.26) converges whenever jt¡t1j< ±. We sketch an approach to the proof of Lemma 10.3. By hypothesis, each point p2[a+"; b¡"] =J"is the center of an open interval Ipon which fis given by a convergent power series about p. Using results about compactness 206 3. Linear Systems of Di®erential Equations ofJ"presented in Appendix A of Chapter 4, it can be shown that there is a ¯nite setfp1; : : : ; p Kg ½J"such that the intervals Ipj;1·j·Kcover J". Given this, one can deduce the existence of ± >0 such that for each t12J", the interval ( t1¡±; t1+±) is contained in one of the intervals Ipj;1·j·K, and then the convergence of the power series of fabout t1on (t1¡±; t1+±) follows from Proposition 10.2. With these tools in hand, we have the following result. Proposition 10.4. Assume A(t)andf(t)are analytic on an interval (a; b) andt02(a; b). Then the initial value problem (10.19) has a unique solution x(t), analytic on (a; b). Proof. We know that A(t) and f(t) have power series (10.20), convergent forjt¡t0j< R 0, for some R0>0, and the calculations from the ¯rst part of this section construct the solution x(t) to (10.19) for t2(t0¡R0; t0+R0). If this interval is not all of ( a; b), we can extend x(t) as follows. Suppose t0+R0·b¡", and for such " >0 take ± >0 as in Lemma 10.1. Take t1=t0+R0¡±=2. Then the previous argument constructs ~ x(t) as an analytic function on ( t1¡±; t1+±), satisfying (10.19) with the initial condition given at t1: ~x(t1) =x(t1). It is readily veri¯ed that ~ x=xon (t1¡±; t0+R0), so the solution x(t) is extended past t0+R0. Iterating this argument gives the conclusions of the proposition. To conclude this section, we mention a connection with the study of functions of a complex variable, which the reader could pursue further, con- sulting texts on complex analysis, such as [ Ahl]. Here is the general set-up. Let ­ ½Cbe an open set, and f: ­!C. We say fiscomplex di®erentiable atz02­ provided (10.34) lim w!0f(z0+w)¡f(z0) w exists. Here, w!0 inC. If this limit exists, we call it (10.35) f0(z0): We say fis complex di®erentiable on ­ if is complex di®erentiable at each z02­. The relevance of this concept to the material of this section is the fol- lowing. If f(t) is given by the power series (10.20), absolutely convergent for real t2(t0¡R0; t0+R0), then (10.36) f(z) =1X k=0fk(z¡t0)k 10. Power series expansions 207 is absolutely convergent for z2Csatisfying jz¡t0j< R 0, i.e., on the disk (10.37) DR0(t0) =fz2C:jz¡t0j< R 0g; and it is complex di®erentiable on this disk. Furthermore, f0is complex di®erentiable on this disk, etc., including the kth order derivative f(k), and (10.38) fk=f(k)(t0) k!: More is true, namely the following converse. Theorem 10.5. Assume fis complex di®erentiable on the open set ­½C. Lett02­and assume DR0(t0)½­. Then fis given by a power series, of the form (10.36), absolutely convergent on DR0(t0). This is one of the central basic results of complex analysis. A proof can be found in Chapter 5 of [ Ahl]. Example. Consider (10.39) f(z) =1 z2+ 1: This is well de¯ned except at §i, where the denominator vanishes, and one can readily verify that fis complex di®erentiable on Cn fi;¡ig. It follows from Theorem 10.5 that if t02Cnfi;¡ig, this function is given by a power series expansion about t0, absolutely convergent on DR(t0), where (10.40) R= min fjt0¡ij;jt0+ijg: In particular, (10.41) t02R=)R=q t2 0+ 1 gives the radius of convergence of the power series expansion of 1 =(z2+ 1) about t0. This is easy to see directly for t0= 0: (10.42)1 z2+ 1=1X k=0(¡1)kz2k: However, for other t02R, it is not so easy to see directly that this function has a power series expansion about t0with radius of convergence given by (10.41). The reader might give this a try. 208 3. Linear Systems of Di®erential Equations To interface this example with Proposition 10.4, we note that, by this proposition, plus the results just derived on 1 =(z2+ 1), the equation (10.43)dx dt=µ1 (t2+ 1)¡1 0 ¡1¶ x+µet e¡t¶ ; x(0) =µ1 ¡1¶ has a solution that is analytic on ( ¡1;1). The power series expansion for the solution x(t) about t0converges for jtj<1 ift0= 0 (this is an easy consequence of Proposition 10.4 and (10.42)), and for other t02R, it converges for jt¡t0j<p t2 0+ 1 (as a consequence of Proposition 10.4 and Theorem 10.5). Exercises 1. Assume fis a smooth (vector valued) function of t2(a; b). As in (10.31), de¯ne Rk(t; y) for t; y2(a; b), by (10.44) f(t) =kX n=0f(n)(y) n!(t¡y)n+Rk(t; y): Apply @=@y to both sides of (10.47). Observe that the left side becomes 0 and there is considerable cancellation on the right side. Show that (10.45)@ @yRk(t; y) =¡f(k+1)(y) k!(t¡y)k; R k(t; t) = 0 : Integrate this to establish the remainder formula (10.46) Rk(t; y) =1 k!Zt y(t¡s)kf(k+1)(s)ds: 2. Given g, continuous on [ y; t], show that there exists s12[y; t] such that (10.47)Zt yg(s)ds= (t¡y)g(s1): 10. Power series expansions 209 3. To show that (10.13) holds when the estimate (10.30) holds, and com- plete the proof of Proposition 10.4, argue as follows. By (10.30), (10.48) kRk(t; t1)k ·C(k+ 1) max s2[t1;t]³ 1¡js¡t0j r´¡1Zt t1(t¡s)k(r¡ js¡t0j)¡k¡1ds ·C0(k+ 1)Zt t1(t¡s)k(r¡ jt0¡sj)¡k¡1ds; provided jt1¡t0j< r. By (10.47), we have, for some s12[t1; t], (10.49)kRk(t; t1)k ·C0(k+ 1)jt1¡tj ¢ jt¡s1jk(r¡ js1¡t0j)¡k¡1 ·C0(k+ 1)jt1¡tj r¡ js1¡t0j³jt¡s1j r¡ js1¡t0j´k : Show that, provided jt¡t1j< r¡ jt1¡t0jands12[t1; t], this last quantity tends to 0 as k! 1 . 4. Consider the function g:Cn f0g !Cgiven by (10.50) g(z) =e¡1=z2: Show that gis complex di®erentiable on Cn f0g. Use Theorem 10.5 to deduce that h:R!R, given by (10.51)h(t) =e¡1=t2; t6= 0; 0; t = 0; is analytic on Rn f0g. Show that his not analytic on any interval containing 0. Compute h(k)(0): 5. Consider the Airy equation (10.52) y00=ty; y (0) = y0; y0(0) = y1; Introduced in (15.9) of Chapter 1. Show that this yields the ¯rst order system (10.53)dx dt= (A0+A1t)x; x (0) = x0; with (10.54) A0=µ0 1 0 0¶ ; A 1=µ0 0 1 0¶ ; x 0=µy0 y1¶ : 210 3. Linear Systems of Di®erential Equations Note that (10.55) A2 0=A2 1= 0; and (10.56) A0A1=µ1 0 0 0¶ ; A 1A0=µ0 0 0 1¶ : 6. For a system of the form (10.53), whose solution has a power series of the form (10.2), the recursion (10.6) becomes (10.57) ( k+ 1)xk+1=A0xk+A1xk¡1; with the convention that x¡1= 0. Assume (10.55) holds. Show that (10.58) xk+3=1 k+ 3³1 k+ 2A0A1+1 k+ 1A1A0´ xk: Note that when A0andA1are given by (10.54), this becomes (10.59) xk+3=1 k+ 3µ1 k+2 1 k+1¶ xk: Establish directly from (10.58) that the seriesPxktkis absolutely con- vergent for all t. 11. Regular singular points Here we consider equations of the form (11.1) tdx dt=A(t)x; where xtakes values in Cn, and A(t), with values in M(n;C), has a power series convergent for tin some interval ( ¡T0; T0), (11.2) A(t) =A0+A1t+A2t2+¢¢¢: The system (1.1) is said to have a regular singular point at t= 0. One source of such systems is the following class of second order equations: (11.3) t2u00(t) +tb(t)u0(t) +c(t)u(t) = 0 ; 11. Regular singular points 211 where b(t) and c(t) have convergent power series for jtj< T0. In such a case, one can set (11.4) x(t) =µu(t) v(t)¶ ; v(t) =tu0(t); obtaining (11.1) with (11.5) A(t) =µ0 1 ¡c(t) 1¡b(t)¶ : A paradigm example, studied in Exercises 6{13 of x15, Chapter 1, is the Bessel equation (11.6)d2u dt2+1 tdu dt+³ 1¡º2 t2´ u= 0; which via (11.4) takes the form (11.1), with (11.7) A(t) =A0+A2t2; A 0=µ0 1 º20¶ ; A 1=µ0 0 1 0¶ : It follows from Proposition 10.4 that, given t02(0; T0), the equation (11.1), with initial condition x(t0) =v, has a unique solution analytic on (0 ; T0). Our goal here is to analyze the behavior of x(t) ast&0. A starting point for the analysis of (11.1) is the case A(t)´A0, i.e., (11.8) tdx dt=A0x: The change of variable z(s) =x(es) yields (11.9)dz ds=A0z(s); with solution (11.10) z(s) =esA0v; v =z(0); hence (11.11) x(t) =e(logt)A0=tA0v; t > 0; the latter identity de¯ning tA0, for t >0. Compare results on the \Euler equations" in Exercises 1{3, x15, Chapter 1. 212 3. Linear Systems of Di®erential Equations Note that if v2 E(A0; ¸), then tA0v=t¸v, which either blows up or vanishes as t&0, if Re ¸ <0 or Re ¸ >0, respectively, or oscillates rapidly ast&0, if¸is purely imaginary but not zero. On the other hand, (11.12) v2 N(A0) =)tA0v´v: It is useful to have the following extension of this result to the setting of (11.1). Lemma 11.1. Ifv2 N(A0), then (11.1) has a solution given by a conver- gent power series on some interval about the origin, (11.13) x(t) =x0+x1t+x2t2+¢¢¢; x 0=v; as long as the eigenvalues of A0satisfy a mild condition, given in (11.18) below. Proof. We produce a recursive formula for the coe±cients xkin (11.13), in the spirit of the calculations of x10. We have (11.14) tdx dt=X k¸1kxktk; and (11.15)A(t)x=X j¸0AjtjX `¸0x`t` =A0x0+X k¸1kX `=0Ak¡`x`tk: Equating the power series in (11.14) and (11.15) would be impossible without our hypothesis that A0x0= 0, but having that, we obtain the recursive formulas, for k¸1, (11.16) kxk=A0xk+k¡1X `=0Ak¡`x`; i.e., (11.17) ( kI¡A0)xk=k¡1X `=0Ak¡`x`: Clearly we can solve uniquely for xkprovided (11.18) 8k2N=f1;2;3; : : :g; k =2Spec A0: 11. Regular singular points 213 This is the condition on Spec A0mentioned in the lemma. As long as this holds, we can solve for the coe±cients xkfor all k2N, obtaining (11.13). Estimates on these coe±cients implying that (11.13) has a positive radius of convergence are quite similar to those made in x10, and will not be repeated here. Our next goal is to extend this analysis to solutions to (11.1), for general A(t), of the form (11.2), without such an hypothesis of membership in N(A0) as in Lemma 11.1. We will seek a matrix-valued power series (11.19) U(t) =I+U1t+U2t2+¢¢¢ such that under the change of variable (11.20) x(t) =U(t)y(t); (11.1) becomes (11.21) tdy dt=A0y: This will work as long as A0does not have two eigenvalues that di®er by a nonzero integer, in which case a more elaborate construction will be needed. To implement (11.20) and achieve (11.21), we have from (11.20) and (11.1) that (11.22) A(t)U(t)y=tdx dt=tU(t)dy dt+tU0(t)y; which gives (11.21) provided U(t) satis¯es (11.23) tdU dt=A(t)U(t)¡U(t)A0: Now (11.23) has the same form as (11.1), i.e., (11.24) tdU dt=A(t)U(t); where Utakes values in M(n;C) and A(t) takes values in L(M(n;C)); (11.25)A(t)U=A(t)U(t)¡U(t)A0 = (A0+A1t+A2t2+¢¢¢)U: In particular, (11.26) A0U=A0U¡UA0= [A0; U] =CA0U; 214 3. Linear Systems of Di®erential Equations the last identity de¯ning CA02 L(M(n;C)). Note that (11.27) U(0) = I2 N(CA0); so Lemma 11.1 applies to (11.24), i.e., to (11.23). In this setting, the recur- sion for Uk; k¸1, analogous to (11.16){(11.17), takes the form (11.28) kUk= [A0; Uk] +k¡1X j=0Ak¡jUj; i.e., (11.29) ( kI¡CA0)Uk=k¡1X j=0Ak¡jUj: Recall U0=I. The condition for solvability of (11.29) for all k2N= f1;2;3; : : :gis that no positive integer belong to Spec CA0. Results of Chap- ter 2, x7 (cf. Exercise 9) yield the following: (11.30) Spec A0=f¸jg=)Spec CA0=f¸j¡¸kg: Thus the condition that Spec CA0contain no positive integer is equivalent to the condition that A0have no two eigenvalues that di®er by a nonzero integer. We thus have the following result. Proposition 11.2. Assume A0has no two eigenvalues that di®er by a nonzero integer. Then there exists T0>0andU(t)as in (11.19) with power series convergent for jtj< T0, such that the general solution to (11.1) fort2(0; T0)has the form (11.31) x(t) =U(t)tA0v; v 2Cn: Let us see how Proposition 11.2 applies to the Bessel equation (11.6), which we have recast in the form (11.1) with A(t) =A0+A2t2, as in (11.7). Note that (11.32) A0=µ0 1 º20¶ =)Spec A0=fº;¡ºg: Thus Spec CA0=f2º;0;¡2ºg, and Proposition 11.2 applies whenever ºis not an integer or half-integer. Now as indicated in the exercises in x15 of Chapter 1, there is not an obstruction to series expansions consistent with (11.31) when ºis a half-integer. This is due to the special structure of (11.7), 11. Regular singular points 215 and suggests a more general result, of the following sort. Suppose only even powers of tappear in the series for A(t): (11.33) A(t) =A0+A2t2+A4t4+¢¢¢: Then we look for U(t), solving (11.23), in the form (11.34) U(t) =I+U2t2+U4t4+¢¢¢: In such a case, only even powers of toccur in the power series for (11.23), and in place of (11.28){(11.29), one gets the following recursion formulas for U2k; k¸1: (11.35) 2 kU2k= [A0; U2k] +k¡1X j=0A2k¡2jU2j; i.e., (11.36) (2 kI¡CA0)U2k=k¡1X j=0A2k¡2jU2j: This is solvable for U2kas long as 2 k =2Spec CA0, and we have the following. Proposition 11.3. Assume A(t)satis¯es (11.33), and A0has no two eigen- values that di®er by a nonzero even integer. Then there exists T0>0and U(t)as in (11.34), with power series convergent for jtj< T 0, such that the general solution to (11.1) for t2(0; T0)has the form (11.31). We return to the Bessel equation (11.6) and consider the case º= 0. That is, we consider (11.1) with A(t) as in (11.7), and (11.37) A0=µ0 1 0 0¶ : Proposition 11.3 applies to this case (so does Proposition 11.2), and the general solution to (11.6) is the ¯rst entry in (11.31), where U(t) has the form (11.34). Note that in case (11.37), A2 0= 0, so for t >0, (11.28) tA0=µ1 log t 0 1¶ : Thus there are two linearly independent solutions to (11.39)d2u dt2+1 tdu dt+u= 0; 216 3. Linear Systems of Di®erential Equations fort >0, one having the form (11.40)X k¸0akt2k; with coe±cients akgiven recursively, and another having the form (11.41)X k¸0(bk+cklogt)t2k; again with coe±cients bkandckgiven recursively. The solution of the form (11.40) is as in (15.12){(15.15) of Chapter 1 (with º= 0), while the solution of the form (11.41) is of a di®erent sort than those constructed in that exercise set. Proceeding beyond the purview of Propositions 11.2 and 11.3, we now treat the case when A0satis¯es the following conditions. First, (11.42) Spec CA0contains exactly one positive integer ; `; and second (11.43) A0is diagonalizable, which implies (11.44) CA0is diagonalizable; cf. Chapter 2, x7, Exercise 8. Later we discuss weakening these conditions. As in Proposition 11.2, we use a transformation of the form (11.20), i.e., x(t) =U(t)y(t), with U(t) as in (11.19), but this time our goal is to obtain fory, not the equation (11.21), but rather one of the form (11.45) tdy dt= (A0+B`t`)y; with the additional property of a special structure on the commutator [ A0; B`], given in (11.55) below. To get (11.45), we use (11.22) to obtain for U(t) the equation (11.46) tdU dt=A(t)U(t)¡U(t)(A0+B`t`); in place of (11.23). Taking A(t) as in (11.2) and U(t) as in (11.19), we have (11.47) tdU dt=X k¸1kUktk; 11. Regular singular points 217 (11.48) A(t)U(t)¡U(t)A0=X k¸1[A0; Uk]tk+X k¸1³k¡1X j=0Ak¡jUj´ tk; and hence solving (11.46) requires for Uk; k¸1, that (11.49) kUk= [A0; Uk] +k¡1X j=0Ak¡jUj¡¡k; where (11.50)¡k= 0; k < `; B`; k =`; Uk¡`B`; k > `: Equivalently, (11.51) ( kI¡CA0)Uk=k¡1X j=0Ak¡jUj¡¡k: As before, (11.51) has a unique solution for each k < ` , since CA0¡kIis invertible on M(n;C). For k=`, the equation is (11.52) ( `I¡CA0)U`=`¡1X j=0A`¡jUj¡B`: This time CA0¡`Iis not invertible. However, if (11.44) holds, (11.53) M(n;C) =N(CA0¡`I)© R(CA0¡`I): Consequently, givenP`¡1 j=0A`¡jUj2M(n;C), we can take (11.54) B`2 N(CA0¡`I) so that the right side of (11.52) belongs to R(CA0¡`I), and then we can ¯nd a solution U`. We can uniquely specify U`by requiring U`2 R(CA0¡`I), though that is of no great consequence. Having such B`andU`, we can proceed to solve (11.51) for each k > ` . Estimates on the coe±cients Uk guaranteeing a positive radius of convergence for the power series (11.19) again follow by techniques of x10. We have reduced the problem of repre- senting the general solution to (11.1) for t2(0; T0) to that of representing the general solution to (11.45), given that (11.54) holds. The following result accomplishes this latter task. Note that (11.54) is equivalent to (11.55) [ A0; B`] =`B`;i.e., A0B`=B`(A0+`I): 218 3. Linear Systems of Di®erential Equations Lemma 11.4. Given A0; B`2M(n;C)satisfying (11.55), the general so- lution to (11.45) on t >0is given by (11.56) y(t) =tA0tB`v; v 2Cn: Proof. As mentioned earlier in this section, results of x10 imply that for each v2Cn, there is a unique solution to (11.45) on t > 0 satisfying y(1) = v. It remains to show that the right side of (11.56) satis¯es (11.45). Indeed, if y(t) is given by (11.56), then, for t >0, (11.57) tdy dt=A0tA0tB`v+tA0B`tB`v: Now (11.55) implies, for each m2N, (11.58)Am 0B`=Am¡1 0B`(A0+`I) =¢¢¢ =B`(A0+`I)m; which in turn implies (11.59) esA0B`=B`es(A0+`I)=B`es`esA0; hence (11.60) tA0B`=B`t`tA0; so (11.57) gives (11.61) tdy dt= (A0+B`t`)tA0tB`v; as desired. The construction involving (11.45){(11.55) plus Lemma 11.4 yields the following. Proposition 11.5. Assume A02M(n;C)has the property (11.42) and is diagonalizable. Then there exist T0>0; U(t)as in (11.19,) and B`2 M(n;C), satisfying (11.55), such that the general solution to (11.1) on t2 (0; T0)is (11.62) x(t) =U(t)tA0tB`v; v 2Cn: The following is an important property of B`. 11. Regular singular points 219 Proposition 11.6. In the setting of Proposition 11.5, B`is nilpotent. Proof. This follows readily from (11.55), which implies that for each ¸j2 Spec A0, (11.63) B`:GE(A0; ¸j)¡! GE (A0; ¸j+`): Remark. Note that if Bm+1 `= 0, then, for t >0, (11.64) tB`=mX k=01 k!(logt)kBk `: Let us apply these results to the Bessel equation (11.6) in case º=nis a positive integer. We are hence looking at (11.1) when (11.65) A(t) =A0+A2t2; A 0=µ0 1 n20¶ ; A 2=µ0 0 1 0¶ : We have (11.66) Spec A0=fn;¡ng;Spec CA0=f2n;0;¡2ng: Clearly A0is diagonalizable. The recursion (11.51) for Uktakes the form (11.67) ( kI¡CA0)Uk= § k+ ¡k; where (11.68)§k= 0; k < 2; A2; k = 2; A2Uk¡2; k > 2; and (11.69)¡k= 0; k < 2n; B2n; k = 2n; Uk¡2nB2n; k > 2n: In particular, the critical equation (11.52) is (11.70) (2 nI¡CA0)U2n=A2U2n¡2+B2n; 220 3. Linear Systems of Di®erential Equations and we solve this after picking (11.71) B2n2 N(CA0¡2nI); such that the right side of (11.69) belongs to R(CA0¡2nI). We have from Chapter 2, x7, Exercise 9, that (since A0is diagonalizable) (11.72) N(CA0¡2nI) = Span fvwt:v2 E(A0; n); w2 E(At 0;¡n)g: ForA0in (11.64), vis a multiple of (1 ; n)tandwtis a multiple of ( ¡n;1), so (11.73) B2n=¯nB# 2n; ¯ n2C; B# 2n=µ¡n1 ¡n2n¶ : Note that B2 2n= 0. Consequently the general solution to (11.1) in this case takes the form (11.74) x(t) =U(t)tA0tB2nv; with (11.75) tB2n=I+ (log t)B2n: Note that (11.76) N(B2n) = Spanµ1 n¶ =E(A0; n); so (11.77) U(t)tA0tB2nµ1 n¶ =U(t)tA0µ1 n¶ =U(t)tnµ1 n¶ is a regular solution to (11.1). Its ¯rst component is, up to a constant multiple, the solution Jn(t) (in case º=n) studied in Chapter 1, x15, Exercises 6{13. The recursion gives results similar to (15.13){(15.15) of Chapter 1, and U(t) has in¯nite radius of convergence. Note also that (11.78) A0µ1 ¡n¶ =¡nµ1 ¡n¶ ; B 2nµ1 ¡n¶ =¡2n¯nµ1 n¶ ; which in concert with (11.75){(11.76) gives (11.79)U(t)tA0tB2nµ1 ¡n¶ =U(t)tA0µ1 ¡n¶ ¡2n¯n(logt)U(t)tA0µ1 n¶ =U(t)t¡nµ1 ¡n¶ ¡2n¯n(logt)U(t)tnµ1 n¶ : 11. Regular singular points 221 The ¯rst component gives a solution to (11.6), with º=n, complementary toJn(t), for t >0. Remark. When (11.1) is a 2 £2 system, either Proposition 11.2 or Proposi- tion 11.5 will be applicable. Indeed, if A02M(n;C) and its two eigenvalues di®er by a nonzero integer `, then A0is diagonalizable, and Spec CA0=f`;0;¡`g; so (11.42) holds. To extend the scope of Proposition 11.5, let us ¯rst note that the hy- pothesis (11.43) that A0is diagonalizable was used only to pass to (11.53), so we can replace this hypothesis by (11.80) `2N\Spec CA0=)M(n;C) =N(CA0¡`I)© R(CA0¡`I): We now show that we can drop hypothesis (11.42). In general, if Spec CA0\ N6=;, we have a ¯nite set (11.81) Spec CA0\N=f`j: 1·j·mg; say`1<¢¢¢< `m. In this more general setting, we use a transformation of the form (11.20), i.e., x(t) =U(t)y(t), with U(t) as in (11.19), to obtain for yan equation of the form (11.82) tdy dt= (A0+B`1t`1+¢¢¢+B`mt`m)y; with commutator properties on [ A0; B`j] analogous to (11.55) (see (11.85) below). In this setting, in place of (11.46), we aim for (11.83) tdU dt=A(t)U(t)¡U(t)(A0+B`1t`1+¢¢¢+B`mt`m): We continue to have (11.47){(11.51), with a natural replacement for ¡ k, which the reader can supply. The equation (11.51) for k =2 f`1; : : : ; ` mgis uniquely solvable for Ukbecause kI¡CA0is invertible. For k=`j;1·j· m, one can pick (11.84) B`j2 N(CA0¡`jI); and solve the appropriate variant of (11.52), using (11.80). Note that (11.84) is equivalent to (11.85) [ A0; B`j] =`jB`j;i.e., A0B`j=B`j(A0+`jI): 222 3. Linear Systems of Di®erential Equations Proposition 11.7. Assume A02M(n;C)has the property (11.80). For each `jas in (11.81), take B`jas indicated above, and set (11.86) B=B`1+¢¢¢+B`m: Then there exist T0>0andU(t)as in (11.19) such that the general solution to (11.1) on t2(0; T0)is (11.87) x(t) =U(t)tA0tBv; v 2Cn: Proof. It su±ces to show that the general solution to (11.82) is (11.88) y(t) =tA0tBv; v 2Cn; given that B`jsatisfy (11.85). In turn, it su±ces to show that if y(t) is given by (11.88), then (11.82) holds. To verify this, write (11.89) tdy dt=A0tA0tBv+tA0(B`1+¢¢¢+B`m)tBv: Now (11.85) yields (11.90) Ak 0B`j=B`j(A0+`jI)k;hence tA0B`j=B`jt`jtA0; which together with (11.89) yields (11.82), as needed. Parallel to Proposition 11.6, we have the following. Proposition 11.8. In the setting of Proposition 11.7, Bis nilpotent. Proof. By (11.85), we have, for each ¸j2Spec A0, (11.91) B:GE(A0; ¸j)¡! GE (A0; ¸j+`1)© ¢¢¢ © GE (A0; ¸j+`m); which readily implies nilpotence. Exercises 1. In place of (11.3), consider second order equations of the form (11.92) tu00(t) +b(t)u0(t) +c(t)u(t) = 0 ; 11. Regular singular points 223 where b(t) and c(t) have convergent power series in tforjtj< T 0. In such a case, show that setting (11.93) x(t) =µu(t) u0(t)¶ yields a sysyem of the form (11.1) with (11.94) A(t) =µ0 t ¡c(t)¡b(t)¶ : Contrast this with what you would get by multiplying (11.92) by tand using the formula (11.4) for x(t). 2. Make note of how to extend the study of (11.1) to (11.95) ( t¡t0)dx dt=A(t)x; when A(t) =P k¸0Ak(t¡t0)kforjt¡t0j< T0. We say t0is a regular singular point for (11.95). 3. The following is known as the hypergeometric equation: (11.96) t(1¡t)u00(t) + [°¡(®+¯+ 1)t]u0(t)¡®¯u(t) = 0 : Show that t0= 0 and t0= 1 are regular singular points and construct solutions near these points, given ®; ¯, and °. 4. The following is known as the con°uent hypergeometric equation: (11.97) tu00(t) + (°¡t)u0(t)¡®u(t) = 0 : Show that t0= 0 is a regular singular point and construct solutions near this point, given ®and°. 5. Let B(t) be analytic in tforjtj> a. We say that the equation (11.98)dy dt=B(t)y has a regular singular point at in¯nity provided that the change of vari- able (11.99) x(t) =y³1 t´ 224 3. Linear Systems of Di®erential Equations transforms (11.98) to an equation with a regular singular point at t= 0. Specify for which B(t) this happens. 6. Show that the hypergeometric equation (11.96) has a regular singular point at in¯nity. 7. What can you say about the behavior as t&0 of solutions to (11.1) when A(t) =A0+A1t; A 0=0 @2 1 1 01 A; A 1=0 @0 0 0 1 01 A? 8. What can you say about the behavior as t&0 of solutions to (11.1) when A(t) =A0+A1t; A 0=0 @1 1 1 01 A; A 1=0 @0 0 0 1 01 A? 9. In the context of Lemma 11.4, i.e., with A0andB`satisfying (11.55), show that B`and e2¼iA0commute. More generally, in the context of Proposition 11.7, with A0andBsat- isfying (11.85){(11.86), show that Band e2¼iA0commute. Deduce that for all t >0 (11.100) tA0e2¼iA0tBe2¼iB=tA0tBC; C =e2¼iA0e2¼iB: 10. In the setting of Exercise 9, pick E2M(n;C) such that (11.101) e2¼iE=C: (Cf. Appendix A.) Set (11.102) Q(t) =tA0tBt¡E; t > 0: A. Logarithms of matrices 225 Show that there exists m2Z+such that tmQ(t) is a polynomial in t, with coe±cients in M(n;C). Deduce that, in the setting of Proposition 11.7, the general solution to (11.1) on t2(0; T0) is (11.103) x(t) =U(t)Q(t)tEv; c 2Cn; with U(t) as in (11.19), Eas in (11.101), and Q(t) as in (11.102), so thattmQ(t) is a polynomial in t. A. Logarithms of matrices Given C2M(n;C), we say X2M(n;C) is a logarithm of Cprovided (A.1) eX=C: In this appendix, we aim to prove the following: Proposition A.1. IfC2M(n;C)is invertible, there exists X2M(n;C) satisfying (A.1). Let us start with the case n= 1, i.e., C2C. In case Cis a positive real number, we can take X= log C, de¯ned as in Chapter 1, x1; cf. (1.21){ (1.27). More generally, for C2Cn0, we can write (A.2) C=jCjeiµ; X = logjCj+iµ: Note that the logarithm XofCis not uniquely de¯ned. If X2Csolves (A.1), so does X+ 2¼ikfor each k2Z. As is customary, for C2Cn0, we will denote any such solution by log C. Let us now take an invertible C2M(n;C) with n >1, and look for a logarithm, i.e., a solution to (A.1). Such a logarithm is easy to produce if Cis diagonalizable, i.e., if for some invertible B2M(n;C), (A.3) B¡1CB=D=0 @¸1 ... ¸n1 A: Then (A.4) Y=0 @¹1 ... ¹n1 A; ¹ k= log ¸k=)eY=D; and so (A.5) eBY B¡1=BDB¡1=C: 226 3. Linear Systems of Di®erential Equations Similar arguments, in concert with results of Chapter 2, xx7{8, show that to prove Proposition A.1 it su±ces to construct a logarithm of (A.6) C=¸(I+N); ¸2Cn0; Nn= 0: In turn, if we can solve for Ythe equation (A.7) eY=I+N; given Nnilpotent, then (A.8) ¹= log ¸=)e¹I+Y=¸(I+N); so it su±ces to solve (A.7) for Y2M(n;C). We will produce a solution Yin the form of a power series in N. To prepare for this, we ¯rst strike o® on a slight tangent and produce a series solution to (A.9) eX(t)=I+tA; A 2M(n;C);ktAk<1: Taking a cue from the power series for log(1 + t) given in Chapter 1, (1.56), we establish the following. Proposition A.2. In case ktAk<1, (A.9) is solved by (A.10)X(t) =1X k=1(¡1)k¡1 ktkAk =tA¡t2 2A2+t3 3A3¡ ¢¢¢ : Proof. IfX(t) is given by (A.10), we have (A.12)dX dt=A¡tA2+t2A3¡ ¢¢¢ =A(I¡tA+t2A2¡ ¢¢¢ ) =A(I+tA)¡1: Hence (A.12)d dte¡X(t)=¡e¡X(t)A(I+tA)¡1; forjtj<1=kAk; cf.x1, Exercise 10. It follows that (A.13)d dt³ e¡X(t)(I+tA)´ =e¡X(t)³ ¡A(I+tA)¡1(I+tA) +A´ = 0; so (A.14) e¡X(t)(I+tA)´e¡X(0)=I; which implies (A.9). The task of solving (A.7) and hence completing the proof of Proposition A.1 is accomplished by the following result. A. Logarithms of matrices 227 Proposition A.3. IfN2M(n;C)is nilpotent, then for all t2R, (A.15) eY(t)=I+tN is solved by (A.16) Y(t) =n¡1X k=1(¡1)k¡1 ktkNk: Proof. IfY(t) is given by (A.16), we see that Y(t) is nilpotent and that eY(t)is a polynomial in t. Thus both sides of (A.15) are polynomials in t, and Proposition A.2 implies they are equal for jtj<1=kNk, so (A.15) holds for all t2R.