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# Diff Eq&Dynmc Sys I MATH 645

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Differential Equations and Dynamical Systems Classnotes for Math 645 University of Massachusetts v3 Fall 2008 Luc Rey Bellet Department of Mathematics and Statistics University of Massachusetts Amherst7 MA 01003 Contents H D 03 a Existence and Uniqueness 11 Introduction 12 Banach xed point theorem 13 Existence and uniqueness for the Cauchy problem 14 Peano Theorem 15 Continuation of solutions 16 Global existence 17 Wellposedness and dynamical systems 18 Exercises Linear Differential Equations 21 General theory 22 The exponential of a linear map A 23 Linear systems with constant coef cients 24 Stability of linear systems 25 Floquet theory 26 Linearization 27 Exercises Stability analysis 31 Stability of critical points of nonlinear systems 32 Stable and unstable manifold theorem 33 Center manifolds 34 Stability by Liapunov functions 35 Gradient and Hamiltonian systems 351 352 Gradient systems Hamiltonian systems Poincar Bendixson Theorem 41 Limit sets and attractors 42 Poincare maps and stability of periodic solutions 37 37 42 48 56 61 65 70 100 106 106 108 CONTENTS 3 43 Bendixson criterion 114 44 Poincare Bendixson Theorem 114 45 Examples 118 Chapter 1 Existence and Uniqueness 11 Introduction An ordinary dz erentz39al equation ODE is given by a relation of the form Ft777 7 397m 07 M where t E R7 Lx z E R and the function F is de ned on some open set of R gtlt R gtlt gtlt R A function x I a R 7 where I is an interval in R7 is a solution of 11 if t is of class C ie7 m times continuously differentiable and if F1tztztz t7 mt 0 for all t E I 12 We say that the ODE is of order m if the maximal order of the derivative occurring in 11 is m Example 111 Clairaut equation 1734 Let us consider the rst order equation xitxfz 07 13 where f is some given function It is given7 in implicit form7 by a nonlinear equation in x It is easy to verify that the lines t Of 7 fC are solutions of 13 for any 0 Consider for example fz 22 2 then one sees easily that given a point to7 x0 there exists either 0 or 2 such solutions passing by the point tome see Figure 11 As we see from this example7 it is in general very dif cult to obtain results on the uniqueness or existence of solutions for general equations of the form 11 We will therefore restrict ourselves to situations where 11 can be solved as a function of 0 xltmgt mg Hyman 14 CHAPTER 1 EXISTENCE AND UNIQUENESS 5 Figure 11 Some solutions for Clairault equation for f z2 z Such an equation is called an explicit ODE of order m One can always reduce an ODE of order m to a first order ODE for a vector in a space of larger dirnension For example we can introduce the new variables I l 1 123 mm 15 and rewrite 14 as the system 1 2 332 3 16 hz 1 771 m gt1y2y 39m39 This is an equation of order 1 for the supervector a 1 asm E Rmquot each 331 is in R and it has the form 23 ft Therefore in general7 it is sufficient to consider the case of first order equations rn1 If f does not depend explicitly on t7 ie7 fta fa the ODE 23 is called autonomous The function f U gt Rquot7 where U is an open set of R 7 defines a vector eld A solution of 23 is then a pararnetrized curve t which is tangent to the vector field at any point7 see figures 12 and 13 Note a non autonornous ODE 23 f ta with a E R can be written as an autonomous ODE in R 1 by setting 9 t y t 1 EFW 17 CHAPTER 1 EXISTENCE AND UNIQUENESS 6 Example 112 PredatorPrey equation Let us consider the equation a y dawn 6 18 Where oz y 6 are given positive constants Here t is the population of the preys and yt is the population of the predators If the population of predators y is below the threshold oz then a is increasing While if y is above oz then a is decreasing The opposite holds for the population y In order to study the solutions let us divide the rst equation by the second one and consider a as a function of y We obtain d9 1 MCI y or 391 6d a ydy d y T y Viv 5 y 19 Integrating gives yx loga alogy erConst 110 One can verify that the level curves 110 are closed bounded curves and each solution tyt stays on a level curve of 110 for any t E R This suggests that the solutions are periodic see Figure 12 l ldj1 l39I V 392 1 8 1 111111 H 1111111 41 tl P J Figure 12 The vector eld for the predator prey equation with oz 1 2 7 3 6 2 and the solutions passing through the point 1 1 and 0505 Example 113 van der Pol equation The van der Pol equation 3 1 2l 111 CHAPTER 1 EXISTENCE AND UNIQUENESS 7 It can written as a first order system by setting y 23 y y 1 2y 112 It is a perturbation of the harmonic oscillator e 0 3 a 0 whose solutions are the periodic solution t Acost gz and yt t Asint gz circles When 6 gt 0 one observes that one periodic solution survives which is the deformation of a circle of radius 2 and all other solution are attracted to this periodic solution limit cycle7 see Figure 13 Figure 13 The vector field for the van der Pol equation with 6 01 as well as two solutions passing through the points 1 2 and 2 3 We will discuss these examples in more details later For now we observe that7 in both cases7 the solutions curves never intersect This means that there are never two solutions passing by the same point Our first goal will be to find sufficient conditions for the problem l ft77 t0 0 to have a unique solution We say that to and 230 are the initial values and the problem 113 is called a Cauchy Problem or an initial value problem IVP 12 Banach xed point theorem We will need some simple tools of functional analysis Let E be a vector space with addition and multiplication by scalar A in R or C A norm on E is a map E gt R which satisfies the following three properties 0 N1 Z 0 and 0 if and only if a 0 CHAPTER 1 EXISTENCE AND UNIQUENESS 8 N2 llMll lAlll lli 0 N3 S triangle inequality A vector space E equipped with a norm is called a normed vector space In a normed vector space E we can de ne the convergence of sequence We say that the sequence converges to z E E7 if for any 6 gt 07 there exists N 2 1 such that7 for all n 2 N7 we have 7 S 6 We say that is a Cauchy sequence if for any 6 gt 07 there exists N 2 1 such that7 for all nm 2 N7 we have 7 S 6 De nition 121 A norrned vector space E is said to be complete if every Cauchy sequence in E converges to an element of E A complete norrned vector space E is called a Banach space Let and H denote two norrns on the vector space E We say that the norms H and H are equivalent if there exist positive constants c and C such that S lt for all z E E It is easy to check that the equivalence of norm de nes an equivalence relation Further more if a Cauchy sequence for a norm is also a Cauchy sequence for an equivalent norrn Example 122 The vector space E R or C with the euclidean norrn El12 is a Banach space Other examples of norms are El or 00 supl In any case R or C equipped with any norm is a Banach space7 since all norrn are equivalent in a nite dirnensional space see exercises The previous example shows that for nite dimensional vector spaces the choice of a norm does not matter much For in nite dirnensional vector spaces the situation is very different as the following example dernonstrate Proposition 123 Let COJD f 01 a R f continuous 114 With the norm Hflloo sup Wt 115 te01 C071 is a Banach space With either of the norms HfH101lftldt or HflzOliflttgti2dtl2 116 C071 is not complete CHAPTER 1 EXISTENCE AND UNIQUENESS 9 Proof We let the reader verify that Hle HfHZ and Hme are norms Let fn be a Cauchy sequence for the norm We have then lfnt 7 fmtl S Hf 7 meoo 3 E for all nm 2 N 117 This implies that7 for any t7 fnt is a Cauchy sequence in R which is complete Therefore fnt converges to an element of R which we call ft It remains to show that the function ft is continuous Taking the limit in 7 oo in 1177 we have lfnt 7 ftl S E for all n 2 N7 118 where N depends on E but not on t This means that fnt converges uniformly to ft and therefore ft is continuous Let us consider the sequence fn of piecewise linear continuous functions7 where fnt 0 on 012 7 171 and fnt 1 on 12 17171 and linearly interpolating in between One veri es easily that for any in 2 n we have Hf 7 fmlll S 171 and Hf 7 meZ S Therefore fn is a Cauchy sequence But the limit function is not continuous and therefore the sequence does not converge in CO7 I We have also Proposition 124 Let X be an arbitrary set and let us consider the space 8X f X 7 R f bounded 119 with the norm llflloo SUPlfW 120 16X Then 8X is a Banach space Proof The proof is almost identical to the rst part the previous proposition and is left to the reader In a Banach space E we can de ne basic topological concepts as in R o A open ball of radius r and center a is the set B5a x E E g 7 all lt r o A neighborhood ofa is a set V such that B5a C V for some 6 gt 0 o A set U C E is open if U is a neighborhood of each of its elernent7 ie7 for any x 6 U7 there exists 6 gt 0 such that B5x C U o A set V C E is closed if the limit of any convergent sequence is in V o A set K is compact if any sequence with xn E K has a subsequence which converges in K CHAPTER 1 EXISTENCE AND UNIQUENESS 10 0 Let E and F be two Banach spaces and U C E A function f U 7 F is continuous at 0 6 U if for all 6 gt 07 there exists 6 gt 0 such that z E U and Hz 7 zOH lt 6 implies that 7 fx0H lt 6 o The map z gt gt is a continuous function of E to R7 since 7 3 Hz 7 zOH by the triangle inequality Certain properties which are true in nite dimensional Banach spaces are not true in in nite dimensional Banach spaces such as the function spaces we have considered in Propositions 123 and 124 For example we show that o The closed ball x E E S 1 is not necessarily compact 0 Two norms on a Banach space are not always equivalent 0 The theorem of Bolzano Weierstrass which says each bounded sequence has a convergent subsequence is not necessarily true 0 The equivalence of K compact and K closed and bounded is not necessarily true The proposition 123 shows that H00 and H1 are not equivalent For7 if they were equivalent7 any Cauchy sequence for H1 would be a Cauchy sequence But we have constructed explicitly a Cauchy sequence for H1 which is not a Cauchy sequence for Let us consider the Banach space B01 and let fnt to be equal to 1 if 1n1 lt t S 171 and 0 otherwise We have llfnlloo 1 for all n and an7me00 1 for any n7m Therefore fn cannot have a convergent subsequence This shows at the same time7 that the unit ball is not compact7 that Bolzano Weierstrass fails7 and that closed bounded sets are not necessarily compact Let us suppose that we want to solve a nonlinear equation in a Banach space E Let f be a function from E to E then we might want to solve f z nd a xed point of f 121 The next theorem will provide a suf cient condition for the existence of a xed point Theorem 125 Banach Fixed Point Theorem 1922 Let E be a Banach space D C E closed and f D 7 E a map which satis es 1 fD C D 2 f is a contraction on D ie there eists 04 lt 1 such that Sallz7yll for all LyED 122 Then f has a unique cced point z in D f x CHAPTER 1 EXISTENCE AND UNIQUENESS 11 Proof We rst show uniqueness Let us suppose that there are two xed points z and y ie7 f z and fy y Since f is a contraction we have H iyll WW 7 MIN S all c yll 123 with 04 lt 17 this is possible only if z y To prove the existence we choose an arbitrary x0 6 D and we consider the iteration 1 f0xn1 fn7 Since fD C D this implies that xn E D for any 71 Let us show that is a Cauchy sequence We have Han 7 7 fzn1 S allxn 7 mklll lterating this inequality we obtain HMH Ml S anllh oll 124 If m gt n this implies that H m 7 S 7 m71ll llm71 7 m72ll H n 7 S ltam71 am H961 oll an S llzl 7 0M 125 Therefore is a Cauchy sequence since a 7 0 Since E is a Banach space7 this sequence converges to z E E The limit z is in D since D is closed Since f is a contraction7 it is continuous and we have x 33130an nggoxn me 126 ie7 z is a xed point of f I The proof of the theorem is constructive and provides the following algorithm to construct a xed point Method of successive approximations To solve f z 0 Choose an arbitrary 0 0 lterate xn Even if the hypotheses of the theorem are dif cult to check7 one might apply this algorithm If the algorithm converges this gives a xed point7 although not necessarily a unique one Example 126 The function fx cosx has a xed point on D 01 By the mean value theorem there is g 6 my such that cosx 7 cosy sin y 7 z thus lcosz 7 cosyl S supt601lsintH 7 yl S sin1lz 7 yl and sin1 lt 1 One observes a quite rapid convergence to the solution 07390 For example we have 950 0 951 1 952 05403 952 08575 953 06542 954 07934 955 07013 956 07639 CHAPTER 1 EXISTENCE AND UNIQUENESS 12 Example 127 Consider the Banach space C071 with the norm Let f E C071 and let kt7 s be a function of 2 variables continuous on 01 gtlt 01 Consider the xed point problem 951 ft A01ktszsds 127 We assume that A is such that 04 E M supogtgl f0 kts d5 lt 1 Consider the map Txt ft Af01ktsys The map T C071 a C071 is well de ned and one has the bound 1 1 Tt TyW S W kt759 8 y8d5 S H cinoow sup kt78d5 0 egg 0 128 Taking the suprernurn over It on the left side gives Tz7Tyo0 lt 04x7ygt07 129 so that T is a contraction Hence the Banach xed point theorem with D C071 implies the existence of a unique solution for 127 The method of successive approx imation applies and the iteration is7 y0t f and 11111 11 A 01kltt sums ds 130 13 Existence and uniqueness for the Cauchy prob lem Let us consider the Cauchy problem Mt t 960 We 960 131 where f U a R U is an open set of R gtlt R is a continuous function In order to nd a solution we will rewrite 131 as a xed point equation We integrate the differential equation between to an t we obtain the integral equation t 951 x0 fs 955 d5 132 to Every solution of 131 is thus a solution of 132 The converse also holds If t is a continuous function which veri es 132 on some interval I7 then it is automatically of class C1 and it satis es 131 Let I be an interval and let us de ne the map T CI a CI given by we 950 fsxsds 133 CHAPTER 1 EXISTENCE AND UNIQUENESS 13 The integral equation 132 can then be written as the fixed point equation MW ie7 we have transformed the differential equation 131 into a fixed point problem The method of successive approximation for this problem is called W 134 PicardLindelof iteration 0t 0 or any other function n1t 0 tfsnsds 135 Example 131 Let us consider the Cauchy problem 23 2 330 1 136 The solution is t The Picard Lindelof iteration gives 0 17 1 1 t7 2 1 t t2 t337 and so on One sees from Figure 14 that it converges in a suitable interval around 0 but diverges for larger values of t Figure 14 The first four iterations for the Picard Lindelof iteration scheme for the Cauchy problem 23 2 330 1 The next result shows how to choose the interval I such that T maps a suitably chosen set D into itself We have Lemma 132 LetA t lt tol g a Has 0 g h f A gt R be a continuous function with M supltmgteA We set oz minabM The map T given by 133 is wellde ned on the set B as to at0 04 gt Rquot a continuous and 0 3 b 137 and it satis es TB C B CHAPTER 1 EXISTENCE AND UNIQUENESS 14 Proof The lemma follows from the estimate Wynn 7 xOH Mlt 7 tel Ma 3 b 138 A fsxs d5 We say that a function f A 7 R with A is in the previous lemma satis es a Lipschitz condition if WOW 7 ftyyll S Lll yll for a11t7967ty E A 139 The constant L is called the Lipschitz constant Remark 133 In order to illustrate the meaning of condition 1397 let us suppose that ft fx does not depend on t and that we have 7 S LHx 7 whenever z and y are in the closed ball Pb0 This clearly implies that f is continuous in Ebwo and f is called Lipschitz continuous The opposite does not hold7 for example the function f m is continuous but not Lipschitz continuous at 0 If f is of class Cl7 then f is Lipschitz continuous To see this consider the line 25 z 5y 7 n which interpolates between z and y We have Hfltygt7fltzgtl imam 01f 28y796d8 gup llf 2lllly7ll7 140 ZEBbWo and therefore f is Lipschitz continuous with L Sup Mkmoiigb On the other hand Lipschitz continuity does not imply differentiability as the function f demonstrates The condition 139 requires that ftx is Lipschitz continuous in z uniformly in t with lt 7 tel S a If ftx satisfy a Lipschitz condition we have7 for any t E I to 7 mtg 04 t llT96t TZWW S to Hft79 t ft72tll dt t LHW 7 2w dt to S ostuszt 72tH S ozLHz7zHoo 141 teI Taking the supremum over t on the left side shows that HTm 7 TzHc0 S ozLHn 7 lf 04L lt 1 we can apply the Banach xed point theorem to prove existence of a xed CHAPTER 1 EXISTENCE AND UNIQUENESS 15 point and show the existence and uniqueness for the solution of the Cauchy problem for t in some suf ciently small interval around to In fact one can omit the condition 04L lt 1 by applying the method of successive approximation directly without invoking the Banach Fixed Point Theorem This is the content of the following theorem which is the basic result on existence of local solutions for the Cauchy problem 131 Here local means that we show the existence only of zt is in some interval around to Theorem 134 Existence and uniqueness for the Cauchy problem Let A tz lt 7 tel S a 7 poll 3 b and let us suppose that f A 7 R o is continuous 0 satis es a Lipschitz condition Then the Cauchy problem x ftz zt0 p0 has a unique solution on I to 7 Olt0 04 where 04 minabM with M supltmgt6A Proof We prove directly that the Picard Lindelo39f iteration converge uniformly on I to a solution of the Cauchy problem In a rst step we show by induction that 95 t 7x m lt MLkW for lt7t lt a 142 k1 k k1 0 7 For k 0 we have let 7x0H fsxs dsH S Mlt7t0l Let us assume that 142 holds for h 7 1 Then we have t t s Mme 7fltszk4ltsgtlds L Hmltsgt7zk4ltsgtlds to to k t l5 tOlk 3 M l 11 HMHQ 7 mt 1710M kl d5 7 ML M 1 143 Using 142 we show that is a Cauchy sequence for the norm suptel We have llkmtktll S llkmt7km71tllllk1tktll M Lhm t 7 to hm Lh1 t 7 to thl S 7 L kml k1 M 00 Lay S 7 Z I 7 144 L17k1 3 and the right hand side is the reminder term of a convergent series and thus goes to 0 as h goes to 00 The right hand side is independent of t so is a Cauchy sequence which converges uniformly to a continuous function x I 7 R CHAPTER 1 EXISTENCE AND UNIQUENESS 16 To show that t is a solution of the Cauchy problem we take the limit 71 7 oo in 135 The left side converges uniformly to Since f is continuous and A is compact ftzkt converges uniformly to ftxt on A Thus one can exchange integral and the limit and t is a solution of the integral equation 132 It remains to prove uniqueness of the solution Let t and 2t be two solutions of 132 By recurrence we show that lt7t0lk1 lh7M0H 24 kDy 145 We have t 7 yt fszs 7 fsys ds and therefore 7 S 2Mlt 7 fol which 145 for k 0 If 145 holds for k 71 we have t t it k wo7wmsmmawewss2M Egim to to k lt tolk 2ML W 146 and this proves 145 Since this holds for all k this shows that zt I 14 Peano Theorem In the previous section we established a local existence result by assuming a Lipschitz condition Simple examples show that this condition is also necessary Example 141 Consider the ODE x 2 147 We nd that t t 7 c2 for t gt c and t 7c 7 t2 for t lt c is a solution for any constant 0 But t E 0 is also a solution The Cauchy problem with7 say7 0 0 has in nitely many solutions For t gt 07 t E 0 is one solution7 z t2 is another solution7 and more generally t 0 for 0 S t S c and then t t7c2 for t 2 c is also a solution for any 0 This phenomenon occur because W is not Lipschitz continuous at z 0 We are going to show that7 without Lipschitz condition7 we can still obtain existence of solutions7 but not uniqueness Instead of using the Picard Lindelo39f iteration we are using another approximation scheme It turns out to be the simplest algorithm used for numerical approximations of ODE7s CHAPTER 1 EXISTENCE AND UNIQUENESS 17 Euler polygon 1736 Fix some h 31 07 the idea is to approximate the solution locally by xt h 2 t hft7 Let us consider now the sequence tn7 n given recursively by tn1 tn h n1 in n We then denote by xht the piecewise linear function which passes through the points tmxn It is called the Euler polygon and is an approximation to the solution of the Cauchy problem Figure 15 Euler polygons for x 7z2 for h05 and h 025 Lemma 142 Let A t t 7 tel 3 a7 7 mil 3 b and f A 7 R be a continuous function with M supltmgt6A Hft7 We set 04 minabM If h iaN N an integer the Euler Polygon satis es t7 E A fort E t07047t004 and we have the bound Hx 7am g t7 l brute t07047t004y Proof Let us consider rst the interval t07t0 04 and choose h gt 0 We show rst7 by induction that tmzn E A for n 01N We have 7 zikl S hM and so Hzn 7 mil 3 nhM 3 04M 3 b if n S N Since xht is piecewise linear t7zht E A for any t E t07t0 04 The estimate 149 follows from the fact that the slope of xht is nowhere bigger than M On to 7 mtg the argument is similar I De nition 143 A family of functions f 11 7 R j 17 27 is equieontinuous if for any 6 gt 0 there exists 6 gt 0 such that for7 for all j it 7 tquot lt 6 implies that Hf fjt H S 6 Equicontinuity means that all the functions f are uniformly continuous and that7 moreover7 6 can be chosen to depend only on 6 but not on j The estimate 149 shows that the family zht7 with h ozN7 N 17 27 is equicontinuous CHAPTER 1 EXISTENCE AND UNIQUENESS 18 Theorem 144 Arzelz i Ascoli 1895 Let f ab 7 R be afarnily offunctions such that 0 f7 is equicontinuous o For any t 6 lab there erists Mt E R such that supj S Then the family fj has a convergent subsequence 9 which converges uniformly to a continuous function g on lab Remark 145 As we have seen a bounded closed set in Ca7 bl is not always compact The Arzela Ascoli theorem shows that a bounded set of equicontinuous function is a compact set in Ca7 bl and thus it can be seen a generalization of Bolzano Weierstrass to Ca7 Proof The subsequence is constructed proof via a trick which is referred to as 77diagonal subsequence The set of rational numbers in lab is countable and we write it as t17t27t37u Consider the sequence jgt17 by assumption it is bounded in R and7 by Bolzano Weierstrass7 it has a convergent subsequence which we denote by f1it1i21 and therefore f11tf12tf13t converges for t t1 150 Consider next the sequence f1it2i21 Again7 by Bolzano Weierstrass7 this sequence has a convergent subsequence denoted by f2it2i21 We have f21t7 f22t7 f23t converges for t t17 t2 151 After n steps we nd a sequence fmti21 of fjt such that fn1tfn2tfn3t converges for t t17t2tn 152 Next we consider the diagonal sequence gnt fmt This sequence converges for any t1 since gntl fmtln21 is a subsequence of fmtln21 which converges By equicontinuity7 given 6 gt 07 there exists 6 gt 0 such that for all n 2 17 lt7t l lt 6 implies that Hgnt 7 gnt lt 6 Let us choose rational points t17t2tq1 such that a to lt t1 lt lt tq1 lt tq b and 1241 7t lt 6 For t E tl7tl1 we have How 79mtll S Haw 79ntzll Howl 7m H ll9mt 79mtzll 153 By equicontinuity Hgnt 7 gntlH and Hgmt 7 gmtl l are smaller than 6 By the convergence of gntl there exists Nl such that Hgntl 7 gmtlH S E if nm 2 M If we choose N maxlNl we have that Hgnt 7 gmtl S 36 for all t 6 ab and n7n 2 N This shows that gnt converges uniformly to some gt which is then continuous I From this we obtain CHAPTER 1 EXISTENCE AND UNIQUENESS 19 Theorem 146 Peano 1890 Let A t7z lt 7 tol 3 a7 7 xOH S b f A 7 R a continuous function with M supltmgt6A Set 04 minabM The Cauchy problem 131 has a solution on to 7 mtg 04 Proof Let us consider the Euler polygons with h ozN7 N 17 27 The sequence is bounded since 7z0H S Mlt7t0l S Ma and equicontinuous by Lemma 142 By Arzela Ascoli Theorem7 the family xht has a subsequence which converges uniformly to a continuous function zt on to 7 mtg 04 It remains to show that t is a solution Let t E to7 to a and let tn7 z the approximation obtained by Euler method for If t E thtHl we have M0 0 hf o o hft171 hftl717l71 I tifti7l 154 Since ftzt is a continuous function of t it is Riemann integrable and7 using a Riemann sum with left end points have t x mm ds Wazoo mama O hftl717tl71 t tiftl7tl f 17 155 where limhno Hrh 0 By the uniform continuity of f on A and the uniform convergence of the subsequence of to t we have that Hft7 Lt7ft7 S E if h is suf ciently small and h is such that xh belongs to the convergent subsequence Using that htj 7 and subtracting 155 from 154 we nd that t MW 960 7 f87968d8ll S 1 0716 HWOH S a6 WW 156 0 which converges to 046 as h 7 0 Since 6 is arbitrary t is a solution of the Cauchy problem in integral form 132 I 15 Continuation of solutions So far we only considered local solutions7 ie7 solutions which are de ned in a neigh borhood of tone Simple examples shows that the solution t may not exist for all t7 for example the equation x 1 x2 has solution t tant 7 c and this solution does not exist beyond the interval c 77T27c1L7T27 and we have t 7 ioo as t 7 c i 7r2 To extend the solution we solve the Cauchy problem locally7 say from to to to 04 and then we can try to continue the solution by solving the Cauchy problem x ft7 u with new initial condition xt0 04 and nd a solution from to 04 to to 04 0 and so on In order do this we should be able to solve it locally everywhere and we will therefore assume that f satisfy a local Lipschitz condition CHAPTER 1 EXISTENCE AND UNIQUENESS 20 De nition 151 A function f U a R where U is an open set of Rgtlt R satis es a local Lipschitz condition if for any tone 6 U there exist a neighborhood V C U such that f satis es a Lipschitz condition on V7 see 139 Note that if the function f is of class C1 in U7 then it satis es a local Lipschitz condition Lemma 152 Let U C R gtlt R be an open set and let us assume that f U a R is continuous and satis es a local Lipschitz condition Then for any tone 6 U there epists an open interval Im u Wgt with foo S u lt to lt w 3 00 such that o The Cauchy problem x ftx xt0 0 has a unique solution on Im o Ify I a R is a solution ofp ftx yt0 0 thenI C Im andy plI Proof a Let x I a R and z J a R be two solutions of the Cauchy problem with to 6 I7 J Then t 2t on I J Suppose it is not true7 there is point t such that 31 Consider the rst point where the solutions separate The local existence theorem 134 shows that it is impossible b Let us de ne the interval Imax U I I open interval 7 to 6 I7 there exists a solution on I 157 This interval is open and we can de ne the solution on Imax as follows If t 6 Im then there exists I where the Cauchy problem has a solution and we can de ne The part a shows that t is uniquely de ned on Im I Theorem 153 Let U C RXR be an open set and let us assume that f U a R is continuous and satis es a local Lipschitz condition Then every solution ofp ft7 d has a continuation up to the boundary of U More precisely ifx Imax a R is the solution passing through tone 6 U then for any compact K C U there epists thbg 6 max b1 lt to lt b2 such b17b1 K b27b2 Remark 154 If U R gtlt R 7 Theorem 153 means that either 0 t exists for all t7 0 There exists t such that liming 007 The exists globally or the solution 77blows up77 at a certain point in time CHAPTER 1 EXISTENCE AND UNIQUENESS 21 Proof Let W u 7 an lf w oo7 clearly there exists a point t2 such that t2 gt to and t2t2 K because K is bounded lf w lt 007 let us assume that there exist a compact K such that tt E K for any t E tempt Since ftx is bounded on the compact set K7 we have7 for tt suf ciently close to w W W Ml This shows that limtnw t x exists and upr7 gt 6 K7 since K is closed Theorem 134 for the Cauchy problem with zw x implies that there exists a solution in a neighborhood of W This contradicts the maximality of the interval 1 For t1 the argument is similar I At fsxsdtH Mlt 7 w lt e 158 16 Global existence In this section we derive suf cient conditions for global eccistence of solutions7 ie7 absence of blow up for t gt to or for all t The following simple lemma and its variants will be very useful Lemma 161 Gronwall Lemma Suppose that gt is a continuous function with gt 2 0 and that there eccits constants ab gt 0 such that t 9a ab gsds t6 t0T 159 to Then we have go ae ltquot 0gt t6 M 160 Proof Set Gt a bfgg s ds Then Gt 2 gt7 Gt gt 07 for t E tT7 and G t bgt Therefore G t by t bGt 7 lt 7 b t t T 161 Ga Ga 7 G 7 Elm l7 or7 equivalently7 d amp logGt b t E t0T 162 or log Gt 7 logG0 S bt 7 to7 t E t07T7 163 or Gt G0ebltH0gt aebltH0gt t6 t0T 164 which implies that gt S ael oi m7 for t E t07T I The rst condition for global existence is rather restrictive7 but it has the advantage of being easy to check CHAPTER 1 EXISTENCE AND UNIQUENESS 22 De nition 162 We say that the function f R gtlt R a R is linearly bounded if there exists a constant C such that HftxH 3 C1 7 for all ta 6 R gtlt R 165 Obviously if ft7 a is bounded on Rgtlt R 7 then it is linearly bounded The functions xcosx27 or alog2 are examples of linearly bounded function The function ay x y7y2T is not linearly bounded Theorem 163 Let f Rgtlt R a R be continuous locally Lipschitz see De nition 151 and linearly bounded see De nition 162 Then the Cauchy problem x ftx zt0 zo has a unique solution for all t Proof Since f is locally Lipschitz7 there is a unique local solution We have the a priori bound on solutions if t lWUll S ll oll t llf87968lld8 S ll oll Ot 1 ll8lld87 166 0 0 Using Gronwall Lemma for gt 1 H we nd that 1 lWUll S 1 ll oll qtito OF ll ll S ll0ll60t t 60t t0 1 167 This shows that the norm of the solution grows at most exponentially fast in time From Remark 154 it follows that the solution does not blow up in nite time I We formulate additional suf cient conditions for global existence but7 for simplicity7 we restrict ourselves to autonomous equations we consider Cauchy problems of the form x x7 zt0 x0 168 where ft7 a does not depend explicitly on t Theorem 164 Liapunov functions Let f R a R be locally Lipschitz Sup pose that there eists afunction Vz R a R of class C1 such that o V 2 0 and limHmHH00 V oo o ltVla7 S a bVx Then the Cauchy problem x fx zt0 zo has a unique solution for to lt t lt 00 CHAPTER 1 EXISTENCE AND UNIQUENESS 23 Proof Since f is locally Lipschitz7 there is a unique local solution t for the Cauchy problem We have d r av d267 EVWU F1 67ch VVWUL f95tgt S a bV9 t 169 or7 by integrating7 14951 vane AabVzsds 170 Applying Gronwall lemma to gt a bVxt gives the bound 1 bVxt S a bV0 eW tO 171 Therefore Vt remains bounded for all t Since limHmHH00 V 007 the level sets of V7 V 1c are compact for all e and thus stays nite for all t gt to too I Remark 165 The function V in Theorem 164 is usually referred to as a Liapunov function We will also use similar function later to study the stability of solutions Note that there is no general method to construct Liapunov function7 it involves some trial and error and some a priori knowledge on the equation Example 166 Gradient systems Let V R a R be a function of class C2 A gradient systems is an ODE of the form x 7VVx 172 The negative sign is a traditional convention Note that in dimension n 17 any autonomous ODE x f is a gradient system since we can always write V so y dy Consider the level sets of the function V7 V 1c x Vz e If x E V 1c is a regular point7 ie7 if VV 31 07 then7 by the implicit function Theorem7 locally near 7 V 1c is a smooth hypersurface surface of dimension n 7 1 For example7 if n 27 the level sets are smooth curves Note that if z is a regular point of the level curve Vquotlc7 then the solution curve zt is perpendicular to the level surface V 1c lndeed let y be a vector which is tangent to the level surface V 1c at the point x For any curve yt in the level set V 1c with 70 z and y 0 y we have d 0 a and so VVz is perpendicular to any tangent vector to the level set V 1c at all regular points of V We have the following V Ytlt0 WW 7 9 173 CHAPTER 1 EXISTENCE AND UNIQUENESS 24 Lemma 167 Let V R 7 R be afunotion of class C2 with limHmHH00 V 00 Then any solution of the gradient system x 7VVz zt0 0 eaists for allt gt to Proof If t is a solution of 31077 then we have Vzt 7VVzt VVzt S 0 174 This shows that V is a Liapunov function I Example 168 Hamiltonian systems Let x E R 7 y E R 7 and H R gtlt R 7 R be a function of class C2 The function Hy is called a Hamiltonian function or energy function and the 2n dimensional ODE 96 VyHWw y 7ViHzy 175 is called the Hamiltonian equation for the Hamiltonian Since H is of class C27 the vector eld fy VyHxy7 7VyHzyT is locally Lipschitz so that we have local solutions Let xtyt be a solution of 175 We have then d ampHtyt VmH x VyH yt VmHVyH7 VyHVmH 0 176 This means that H is a integral of the motion7 for any solution Hptqt const and that any solution stays on a level set of the function H For Hamiltonian equations this usually referred to as conservation of energy Let us assume further that limewH00 Hy 00 This means that Hy is bounded below7 ie7 Hy 2 7e from some 0 E R and that the level sets c are closed and bounded hypersurfaces In this case Hx7 ye is a Liapunov function for the ODE 175 and the solution exist for all positive and negative times Example 169 van der Pol equationsThe second order equation x 61 7 z z 7 z is written as the rst order system y y 617 2y 7 z 177 and is a perturbation ofthe harmonic oscillator a x 0 which is an Hamiltonian sys tem with Hamiltonian Hz7 y 724723 harmonic oscillator Taking the Hamiltonian as the Liapunov function we have i 72 27 0 if 95221 ltVHyz fyzgt 7 61 95 y i S Egg if 9521 178 Therefore VH f S 26H and H is a Liapunov function and we obtain global existence of solutions CHAPTER 1 EXISTENCE AND UNIQUENESS 25 lLLL 411 AlaaN fJm 39I X ll 1 K H lllll ill lllll ill lllll ill lllll ill lllll ill lllll ill lllll ill JJLLL ill Lllll ill lllll ill LLJJJ JJJ LLLll ill LLLLl ill LLLLl ill LLLLL ill LLLLL ill LLLLL ill lLLLL ill lLLLL ill lLLLL ill lLLLL Lil l i H i ll 1 ll 1 H l H l H i ll L ll 1 Ll 1 LL 1 L L 1 LL 1 1L 1 1L 1 Ll J Li L Li L ll L 1L L 1L L Figure 16 The vector eld for the Hamiltonian z4i4x2y2 and two solutions between t 0 and t 3 with initial conditions 051 and 057 24 Another class of systems which have solutions for all times are given by dissipative systems Theorem 1610 Dissipative systems Let f R a R be locally Lipsehitz Suppose that there epists i E R and positive constants a and 1 such that 0001 S aibll llz 179 Then the Cauchy problem x fx zt0 zo has a unique solution for to lt t lt 00 Proof Consider the balls B0 x E R S ab and the Liapunov function ll ivllzt 180 The condition 179 implies that for any solution Vxt S 0 outside of the ball B0 and therefore V is a Liapunov function I Remark 1611 The condition 179 means that for the balls B x E R 7 11112 S R with R 2 M is chosen so large that B0 is contained in the interior of E7 the vector eld f points toward the interior of B This implies that a solution which starts in B will stay in B forever There are many variants to Theorem 1610 see the exxercises The basic idea is to nd a family of sets large balls in Theorem 1610 but the set could have other shapes such that7 on the boundary of the sets the vector f points inward This implies that solutions starting on the boundary will move inward the set If one proves this for all suf ciently large sets7 then one obtains global existence for all initial data CHAPTER 1 werNcE AND UNIQUENESS ze Figure 17 I39hesnlulmmemmtzequsummth 7m 28endb end We mm 404015 Exampla 1 s 2 The meu equeme ere yvm by A4472 7 72m m 7 22 In m Despne 139 epme smuplmty the Imean equeums exhhm for suxtehle values 139 me Perememse e my complex hehevlnz A 501mm exe summed m e compel mv em set an whmh the men 5 saw Such en mvexxsm set w esueng eme eee gure 1 We shoal lhsl me mm 5 meme we take e 7 wow Chcmmgy bv end usmglhe mequel y m s f 23 we m1 047 7sz 72 7w ltv7vgtamme we 71 was eweaeeg7 e b m M bm39hem1791s seem wuh a 7 1 em 7 menu2 end we mum ex Lem meme em e e z gt 0 1397 Wellposedness and dynamical systems Rat me Cauchy problem 2 7 I my eon 7 aw we dam the when w mm whee we exphcnlv Indicate me depememem meme ume in end we ms 1mm e CHAPTER 1 EXISTENCE AND UNIQUENESS 27 De nition 171 The Cauchy problem x ftx7 zt0 0 is called locally well posed resp globally wellposed if there exists a unique local resp global solution pt7 t070 which depends continuously of tone Lemma 172 Let f U a R gtlt R U an open set ofR gtlt R be continuous and satisfy a local Lipschitz condition Then for any compact K C U there epists L 2 0 such that llfty 7 ft796ll S Lll ylh for 311t7967ty E K 183 Proof39 Let us assume the contrary Then there exists sequences tmzn and tmyn in K such that HfW cn ftmynll gt nlmiynll 184 Since f is bounded on K with M maxm6K HftxH7 it follows from 184 that Hzniynll S 2Mn 185 By Bolzano Weierstrass7 the sequence tmzn has an accumulation point ta7 and ftx satis es a Lipschitz condition in a neighborhood V of t7 The bound 185 implies that there are in nitely many indices n such that tn7 z E V and tmyn E V Then 184 contradicts the Lipschitz condition on V I Theorem 173 Let f U a R gtlt R U an open set of R gtlt R be continuous and satisfy a local Lipschitz condition Then the solution ztt0z0 of the Cauchy problem x ftx zt0 0 is a continuous function of tone Moreover the function zt7 t070 is a Lipschitz continuous function ofxo ie there epists a constant B Rt such that Hiltb7b070 7 b7b071H S 7 1M Proof39 We choose a closed subinterval a7b of the maximal interval of existence m with tt0 6 ab We choose 6 small enough such that the tubular neighborhood K around the solution ptt07 zo7 K W t 6 all H96 htowo l lt 6 187 is contained in the open set U By Lemma 1727 ft7 8 satis es a Lipschitz condition on K with a Lipschitz constant L The set V V alum t1 6 my H951 7 xt1t0x0ll we 188 is a neighborhood of tone which satis es V C K C U lf thzl E V we have H96tt1961 Elem ll l tiwi t7t17t17t070ll t H961 e zltt1tozogtl Lt Hzltst1z1gte zltst1zltt1tozogtgtl 189 1 CHAPTER 1 EXISTENCE AND UNIQUENESS 28 From Gronwall lemma we conclude that H9502 1317951 MRtOJONl S Cthitllll l 95t17t07950ll S 6 190 and this concludes the continuity of tt0z0 To prove the Lipschitz continuity in no one sets t1 to in 190 and this prove 186 with R eth tOl I This theorem shows that the Cauchy problem is locally wellposed provided f is continuous and satisfy a local Lipschitz condition If in addition the solutions exist for all times then the Cauchy problem is globally wellposed Let us consider the map two R 7 R given by quot Ozo 95021307960 191 The map onto maps the initial position no at time to to the position at time t zt to 0 By de nition the maps two satisfy the composition relations ltzgttsgtt0ltzogt sgt lt gt oltzogtgt 192 If the ODE is autonomous ie u does not depend explicitly on t we have Lemma 174 Translation property Suppose that zt is a solution of x fx then zt 7 to is also a solution Proof If x t fxt then zt 7 to t 7 to fxt 7 t0 I This implies that if t zt Ono is the solution of the Cauchy problem x fx 0 zo then zt 7 to is the solution of the Cauchy problem x fx zt0 0 In other words zt 7 to tt0z0 and so the solution depends only on t7 to For autonomous equations we can thus always assume that to 0 In this case we will denote then the map onto ot tmo simply by ot to The map oi has the following group properties 0 a Oz x 390 t s t995 C t t9 t t9 95 lf the solutions exists for all t E R the collection of maps 15 is called the flow of the differential equations x Note that Property c implies that the map 15 R 7 R is invertible More generally a continuous map if R gtlt R 7 R which satis es Properties a b c is called a continuous time dynamical system If the Cauchy problem is globally wellposed then the maps 15 are a continuous ow of homeomorphisms and we will say that the dynamical system is continuous CHAPTER 1 EXISTENCE AND UNIQUENESS 29 Remark 175 If the vector elds ftx are of class Ck then one would expect that t7 tome is also a function of class Ck We will discuss this in the next chapter Remark 176 Theorem 173 shows the following For xed t zt7 to 0 can be made arbitrarily close to t7 to z0 provided g is small enough depending on fl This does not mean however that the solutions which start close to each other will remain close to each other7 What we proved is a bound Hztt0 0 7 t7 tome S KH Heth tOl which show that two solutions can separate7 typically at an exponential rate Example 177 For the Cauchy problem z lt 012gt 0lt gt 193 the solution is 640 The solution with initial condition 1 T is 5455 If H S 0 both solutions stay a distance less than Kl for all time t 2 07 if H gt 0 the solutions diverge from each other exponentially with time For given t we can however make them arbitrarily close up to time t by choosing small enough7 hence the continuity 18 Exercises 1 Determine whether the following sequences of functions are Cauchy sequences with respect to the uniform norm 39 Hoe on the given interval I Determine the limit fnx if it exists a fnz sin27r7n7 I 01 x 7 1 b n I 711 lt gt f z x 1 l l n 77 I 071 c f z n2 1 7n d n I 01 MW Mm l 2 Show that HfHZ is a norm on CO7 3 Prove that all norms on R are equivalent by proving that any norm in R is equivalent to the euclidean norm Hint a Let 6 be the usual vector basis in R and write x 161 men and use the triangle inequality and Cauchy Schwartz to show that S b Using a prove that l i l S CHx yllz and thus the function on R with H2 is a continuous function c Consider now the function on the compact set K L 1 and deduce from this the equivalence of the two norms CHAPTER 1 EXISTENCE AND UNIQUENESS 4 CT I 30 a Let f U a R where U C R is an open set and suppose that f satis es a Lipschitz condition on U Show that f is uniformly continuous on U b Let f E a R where E C R is a compact set Suppose that f is locally Lipschitz on E7 show that f satis es a Lipschitz condition on E c Show that f 1z is locally Lipschitz but that it does not satisfy a Lipschitz condition on 01 d Show that f Mm is not locally Lipschitz e Does the Cauchy problem x 1x7 0 a gt 0 have a unique solu tion Solve it and determine the maximal interval of existence What is the behavior of the solution at the boundary of this interval a Derive the following error estimate for the method of successive approxima tions Let x be a xed point given by this method Show that Oz Hrimll S all crnillh 194 1 L where 04 is the contraction rate b Consider the function fx em4 on the interval 01 Show that f has a xed point on 01 Do some iterations and estimate the error rigorously using a Consider the function f R a R given by 7 ze m2 iszO we 7 em if 95 S 0 195 a Show that 7 lt lx iyl for z 31 y b Show that f does not have a xed point Explain why this does not contradict the Banach xed point theorem Consider the lVP x x3 0 a 196 a Apply the Picard Lindelo39f iteration to compute the rst three iterations z1tx2t3t b Find the exact solution and expand it in a Taylor series around It 0 Show that the rst few terms agrees with the Picard iterates c How does the number of correct terms grow with iteration CHAPTER 1 EXISTENCE AND UNIQUENESS 31 8 Apply the Picard Lindelo39f iteration to the Cauchy problem zaz12z2 z0 1 0 2 t2 x1 x20 0 Compute the rst ve terms in the taylor series of the solution Show that the assumption that 77D is closed77 cannot be omitted in general in the xed point theorem Find a set D which is not closed and a map f D a E such that fD C D7 f is a contraction7 but f does not have a xed point in D 10 a Let I to 7 mtg a and for a positive constant H de ne Mn suplltll6 quot0 tel Show that HC de nes a norm and that the space E x I a R 7 t continuous and HXHH lt 00 is a Banach space b Consider the lVP x ftx7 zt0 z Give a proof of Theorem 134 in the classnotes by applying the Banach xed point theorem in the Banach space E with norm HC for a well chosen H c Suppose that ftx satisfy a global Lipschitz condition ie7 there exists a positive L gt 0 such that Hft7ftyll S Lllxiyll for all my 6 R and for all t E R 199 Show that the Cauchy problem x ft7 z Mtg 0 has a unique solution for all t E R Hint Use the norm de ned in a H H Consider the map T given by Tfx sin27m Ajl dy a Show that if f E C7171R then so is Tf b Find a A0 such that T is a contraction if W lt A0 and T is not a contraction if W gt A0 Hint For the second part nd a pair g such that HTf 7 T9lloo gt Hf 79H CHAPTER 1 EXISTENCE AND UNIQUENESS 32 12 a Consider the norm of CO7 1 given by 7 42 we 7 WW 1100 Why is it a norm Let t Tft sfsds 1101 0 Show that llelloo S 3211me and llelle S llflle b Show that the integral equation 1 t 951 t2 wads t6 07a 1102 0 has exactly one solution Determine the solution by rewriting the equa tion as an initial value problem and solving it7 ii by using the methods of successive approximations starting with 0 E 0 13 Let us consider R2 with the norm maxlz1l7 Let f R2 a R2 be given by 2 2 7 1 2x2 5 cosx2 fz1z2 i lt 4z123 1103 Let K 1727 lxll lt 17 ml 3 2 Find an explicit Lipschitz constant L for f on K 14 Let f R2 a R be of class C1 and satisfy f00 0 Suppose that t is a solution of the ODE x ax7 1104 which is not identically 0 Show that t has simple zeros Examples the harmonic oscillator x z or the mathematical pendulum x sinx 0 15 Consider the initial value problem x ftx7 zt0 zo where ft is a continuous function Show that if the initial value problem has a unique solution then the Euler polygons xht converge to this solution 16 Consider the Cauchy problem x ftx7 0 07 where fax 4signm if 2 t2 1105 4signx lzl 4t 7 cos7r11 if lt t2 The function f is continuous on R2 Consider the Euler polygons xht with h 2 2 239 123 Show that xht does not converge as h a 07 compute its accumulation points7 and show that they are solution of the Cauchy problem Hint the solutions are i4t2 CHAPTER 1 EXISTENCE AND UNIQUENESS 17 Consider the the Cauchy problem x ftx7 H 00 20 33 0 0 where f is given by 0 if t 0 zen 2t if tgt0 950 WW 2 if tgt0 ogzltt2 L106 72t if tgt0 tzgz a Show that f is continuous What does that imply for the Cauchy problem b Show that f does not satisfy a Lipschitz condition in any neighborhood of the origin c Apply Picard Lindelo39f iteration with z0t E 0 Are the accumulation points solutions d Show that the Cauchy problem has a unique solution What is the solution This problem shows that existence and uniqueness of the solution does not imply that the Picard Lindelo39f iteration converges to the unique solution Consider the Cauchy problem x Ax 0 17 with A gt 0 and t E 01 Compute the Euler polygons xht with h 171 and show that A d h 7 lt 7 lt 1 Ahxhht 7 dt t 7 Ana 1 107 Deduce from this the classical inequality A n A n 1 7 g 5 3 1 7 1108 n n Hint Use Gronwall Lemma Let 17 b7 0 and d be positive constants Consider the Predator Prey equation x za 7 by7 y ycx 7 d with positive initial conditions zt0 gt 0 and yt0 gt 0 Show that the solutions exists for all t and that the solution curves xtyt are periodic Hint You can use the change of variables p logx and q 109M a Show that any second order ODE x f 0 can be written as a Hamil tonian system for the Hamiltonian function Hy 1122 lx7 where y x and Vz Om ftdt b Compute the Hamiltonian function7 and it level curves and draw the solu tions curves for the following ODE7s i z iwzs the harmonic oscillator CHAPTER 1 EXISTENCE AND UNIQUENESS 21 A C7 V A O A A CL V D 34 ii x 7a sinx the mathematical pendulum One end A of weightless rod of length l is attached to a pivot7 and a mass m is attached to the other end B The system moves in a plane under the in uence of the gravitational force of amplitude mg which acts vertically downward Here t is the angle between the vertical and the rod and a gl mr 77Mmr2 Vertical motion of a body of mass m in free fall due to the gravity of a body of mass Depending on the energy Hz0y0 of the initial condition discuss in details the different types of solutions which can occur Are the solutions bounded or unbounded Are there constant solutions or periodic solutions Do the solutions converge as t 7 i007 Consider the Hamiltonian function Hy 1122 Suppose that we have initial conditions 0 0 and z 0 yo gt 0 with initial energy E H0y0 Use the conservation of energy to show the solution t is given implicitly by the formula Mt 1 t ds 0 x2E V8 Assume that Vz l7z7 ie7 V is an even function Show that if t is a solution then so are xc 7 t and 7xt Furthermore show that if do 0 then xc t 7zc 7 t and that if z d 0 then d t d 7 t Assume that V V7x and consider periodic solutions We denote by R the largest swing7 ie7 the maximal positive value of t along the periodic solution Using a show that the period p of the periodic solution is given by R 1 Hint Consider the quarter oscillation starting at the point 0 0 and y0 yo gt 0 and ending at zT R gt 0 and yT 0 Use also the symmetry of V and Use c to show the period for the harmonic oscillator is independent of the energy E Use c to show that for the mathematical pendulum the period is given by 4 7r2 1 d p 7 u V5 0 17k2sin2u where k sin r2 This integral is an elliptic integral of the rst type Hint Use 1 7 cosa sinzg and the substitution sin 52 ksinu CHAPTER 1 EXISTENCE AND UNIQUENESS 35 22 Show that the following ODE7s have global solutions ie7 de ned for all t gt to a x 4y3 2x y 7423 7 2y 7 cosx b s x30 c z z zx3 0 7 sin2t2zm3 d 95 i 7 21 y 7 1m2y2 x 5x 7 2y 7 y2 y 2y6y7y339 23 Prove the following generalizations of Gronwall Lemma 0 Let a gt 0 be a positive constant and gt and ht be nonnegative continuous functions Suppose that for any t E 07 T 91 At15 d5 1109 Then7 for any t E 0T t 9t S agfo MS d5 0 Let ft gt 0 be a positive function and gt and ht be nonnegative con tinuous functions Suppose that for any t E 0T t 91 f10 hsgsds 1111 Then7 for any t E 0T th 11 91 flttgtefo lt9gt 5 1112 24 Consider the FitZHugh Nagurno equation 3 f1172 9W1 27 xZ f2z1z2 0217 yx2 1113 where 039 and y are positive constants and the function g is given by 92 7x 712x 71 a In the zl m plane draw the graph ofthe curves f11 2 0 and f21 x2 0 CHAPTER 1 EXISTENCE AND UNIQUENESS 36 b Consider the rectangles ABCD whose sides are parallel to the X1 and x2 axis with two opposite corners located on the f2z1x2 0 Show that if the rectangle is taken suf ciently large7 a solution which start inside the rect angle stays inside the rectangle forever Deduce from this that the equations for any initial conditions 0 have a unique solutions for all time t gt 0 25 Show that the solutions of 3 13 7 4x1 7 2x2 2lt4 7 21 7 327 7 2 7 have a unique solution for all t 2 07 for any initial conditions x10 7 x20 wich are nonnegative Hint A possibility is to use a similar procedure as in the previous exercise 26 Continuous dependence on parameters Consider the lVP x ftxu7 zt0 0 where f V 7 R V C R gtlt R gtlt Rk an open set We denote by 02 the solution of the lVP we have suppressed the dependence on 157 Let us assume 0 f is a continuous function on V o ftc satis es a local Lipschitz condition in the following sense Given 007t07z0 E V and positive constants abc such that A E t77L lt 7 fol 3 a7 7 xOH 3 b7 My 7 poll 3 c C V then there exists a constant L such that llfltt79 7W 7 fty7Mll S Lll t yll for a11t7967u7t7yu E A Show that 02 depends continuously on M for t in some interval J containing to Chapter 2 Linear Differential Equations We denote by Rn the set of linear maps A R a R which we identify with the set of n gtlt n matrices A aij with real entries We write Ax instead of A for the vector with coef cients 2291 aijxj In this chapter we consider linear di erential equations7 ie7 ODEYs of the form z Atz gt 21 where z E R 7 g I a R 7 and A I a Rn with I some interval The linear ODE is called homogeneous if gt E 07 and inhomogeneous otherwise If At A is independent oft and g E 07 the linear ODE x Ax is called a system with constant coe cients 21 General theory We discuss rst general properties of the differential equations x At gt Theorem 211 Existence and uniqueness Let I a7b be an interval and suppose that At and gt are continuous function on I Then the Cauchy problem x At gt zt0 0 with to E I x0 6 R has a unique solution on I Proof The function ftx Atzgt is continuous and satis es a Lipschitz condi tion on I gtlt R Therefore the solution is unique wherever it exists Moreover on I gtlt R we have the bound HftxH S alluH b where a suptd HAt and b suptd Therefore we have the bound S HxOH aHzs b ds for t0t E I Gronwall lemma implies that remains bounded if t E I I Remark 212 If At and gt are continuous on R then7 applying Theorem 211 to 7T7 T for arbitrary T shows that the solution exists for all t E R CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 38 Theorem 213 Superposition principle Let I be an interval and let At 91t 92t be continuous function on I If 1 I a R is a solution of x At 91t7 x2 I a R is a solution of x At 92t7 then t clz1t 02z2t I a R is a solution of x At 0191 02920 Proof This a simple exercise I This theorem has very important consequences Homogeneous equations Let us consider homogeneous Cauchy problems x At7 zt0 0 and let denote its solutions ztt0z0 to indicate explicitly the dependence on the initial data a The solution zt7 t070 depends linearly on the initial condition 0 ie7 021307 01 0290 01t7 towo 02t7 t07yo 22 This follows by noting that7 by linearity7 both sides are solutions of the ODE and have the same initial conditions The uniqueness of the solutions implies then the equality As a consequence there exists a linear map Rt7 to R a R such that R02 t00 t to 0 lt maps the initial condition 0 at time to to the position at time t The linear map Rt7 to is called the resolvent of the differential equation x At The i th column of Rtt0 is a solution x Atz with initial condition x0 00107 0T where 1 is in i th position b lf 0 07 then zt E 0 for all t E I The point 0 is called a critical point As a consequence if t is a solution and it vanishes at some point t7 then it is identically 0 c The set of solutions of x At form a vector space We call a set of solutions x1t7 7kt linearly dependent if there exists constants cl 70k with at least one 017 07 such that clz1t ckkt 0 24 Note that by b7 if 24 holds at one point t7 it holds at any point t Therefore if the initial condition ta7 zkt0 are linearly dependent7 then the corresponding solu tions are linearly dependent for any t The k solutions are called linearly independent if they are not linearly dependent7 ie7 clz1t ckxkt 0 implies that that elck0 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 39 d From c it follows that there exist exactly n linearly independent solutions x1xn Every such set of n linearly independent solutions is called a fundamental system of solutions Any solution x of x At can be written in a unique way as a linear combination t a1z1t annt 25 e A system of n linearly independent solutions can be arranged in a matrix ltIgtt x1t nt In this notation the t th column of ltIgtt is the column vector The matrix ltIgtt is called a fundamental matrics or a Wronshz39an for x At lt satis es the matrix differential equation d Eta Atlt1gtt 26 f If ltIgtt is a fundamental matrix then the resolvent is given by ROWE ltIgttlt1gtto 1 27 lndeed t ltIgttltIgtt0 1x0 satis es x Atz because of 26 and zt0 zo Theorem 214 Properties of the resolvent Let At be continuous on the m terual I Then the resoluent of x At satis es I swam AlttgtRltttogt 2 Rt0t0 I the identity matna 3 Rtt0 Rtt1Rt1tO 2 Rtt0 is invertible and Rtt0 1 Rt0t Proof We have Rtt0z0 ztt0z0 AtRtt00 and Rt0t0x0 x0 for any x0 6 R This proves 1 and 2 Item 3 simply says that tt0z0 tt1zt1t0x0 Item 4 follows from 2 and 3 by setting t to Example 215 The harmonic oscillator x nx 0 can be written with 1 z and 2 x as 1 7 0 1 1 HMMJ 98gt One can nd easily two linearly independent solutions namely lt s gitgt gt and lt S z 29 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 40 By de nition7 the resolvent is the fundamental solution x1tz2t with z1t 17 0T and 2t0 01T so that we have cosht 7 to sinEt 7 t0 7EsinEt 7 t0 cosEt 7 t0 39 Note that the relation Rtt0 Rt7 sRst0 is simply the addition formula for sine and cosine mm 210 Theorem 216 Liouville Let At be continuous on the interval I and let lt1gtt be afundamental matnw of x Atx Then t det lt1gtt det lt1gtt0 exp trace As d5 7 211 to where trace At a11t amt Proof Let lt1gtt a517 tZJ1 From linear algebra we know that detA is a multilinear function of the rows of A It follows that Z5110 mlttgt detlt1gt1idetni1 where Dit 51 m 212 The matrix Dit is obtained from lt1gtt by replacing the t th line by its derivative We have lt1gt t Atlt1gtt7 ie7 Mgt 21 aikt kjt Using the multilinearity of the determinant we nd Z5110 mlttgt detlt1gtt ifaiktdet 15111 We 727th line 45211 We lt2anlttgtdetltlgtt 213 This is a scalar differential which can be solved by separation of variables and gives 211 I CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 41 Remark 217 The Liouville theorem has the following useful interpretation If V o1 on is a matrix whose columns are the vectors 111 1 then ldetVl is the volume of the parallelepiped spanned by 111 1 Using det A l 1 det A Liouville Theorem is equivalent to det Rt t0 exp 1 traceAs d5 214 If at time to we start with a set of initial conditions B of volume say 1 eg a unit cube at time t the set E is mapped to a set a parallelepiped Rtt0B of volume exp traceAs d5 In particular if trAt E 0 then the ow de ned by the equation y Aty preserves volume We have such a situation in Example 215 see 28 Inhomogeneous equations We consider the equation z Atz gt 215 Theorem 218 Let it be a coed solution of the inhomogeneous equation 215 If zt is a solution of the homogeneous equation then t it is a solution of the homogeneous equation and all solutions of the inhomogeneous equation are obtained in this way Proof This is an easy exercise I If we know how to solve the homogeneous problem ie if we know the resolvent Rt to our task is then to nd just one solution of the inhomogeneous equation The following theorem provides an explicit formula for such solution Theorem 219 Variation of constants or Duhamells formula Let At and gt be continuous on the interval I and let Rtt0 be the resolvent of the homogeneous equation x Atx Then the solution of the Cauchy problem x At gt is given by W Rtt0x0 Rtsgsds 216 Proof The general solution ofthe homogeneous equation has the form Rt t0c with c E R The idea is to 77 vary the constants77 and to look for a solution of the inhomogeneous problem of the form xt Rt t0ct 217 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 42 We must then have7 using 1 of Theorem 2147 6 Mt 5R ttoct Rttoc t AtRttoct Rttoc t AtRtt0ct gt 218 Thus 0 00 Rt7to 19t Rt07t9t7 219 and7 integrating7 this gives Ct 0 Rt0sgs d5 Inserting this formula in 217 gives the result I It should be noted that7 in general7 the computation of the resolvent for x Atz is not easy and can rarely be done explicitly if A depends on t Example 2110 Forced harmonic oscillator We consider the differential equation x z ft7 or equivalently the rst order system 2 y y 72 7 ft The resolvent is given by 210 The solution of the above system with initial conditions MOM0T 9507110T is 83 gt W l 2 gt 35832281 gt 220 zt costxo sinty0 1 fs sint 7 5 d5 221 For example if ft cos Et we nd so that i COS 7 COS K K zt costzosintyo gamma 2 k 222 The motion is quasi periodic if is irrational7 periodic if is rational and 31 17 and the solution grows as t a 00 if H 1 resonance 22 The exponential of a linear map A In this section we let K R or C We equip K with a norm7 for example7 n n 2 12 12112121 1196112 1sz lrgiglmh 223 then K is a Banach space All norms being equivalent on K the choice is a matter of convenience CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 43 A n gtlt 71 matrix A aij with aij E K de nes a linear map A K a K and we denote by Kn the set of all linear maps from K into K The set Kn is also a vector space7 of real or complex dimension n2 and is a Banach space if equipped with a norm In addition to being a vector space Kn is naturally equipped with multiplication composition of linear maps and it is natural and advantageous to equip Kn with a norm which is compatible with matrix multiplication De nition 221 For A E Kn we de ne A W sup HAzH mm 224 M31 0 M The number is called the operator norm of A This de nition means that is the smallest real number R such that S R 7 for all z E K 7 225 and we have the bound llA ll S llAllll ll 226 The properties N1 and N2 are easily veri ed For the triangle inequality7 we have for AB 6 Kn A B ll S llA ll llB ll S HAN HBH ll ll 227 Dividing by and taking the supremum over all z 31 0 one obtains the triangle inequality HA BH S Simple but important properties of are summarized in Lemma 222 Let I E Kn be the identity map Ix x and let AB 6 Then we have 1 HIH 1 2 HABH S HAM HBH 3 lWH S HAM Proof 1 is immediate7 3 is a consequence of 2 To estimate HABH7 we apply twice 225 HMBWH S HAN llB ll S llAllllBllll ll 228 To conclude we divide by and take the supremum over z 31 0 I CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 44 Example 223 Let us denote by HAMp7 p 01700 the operator norm of A acting on K with the norm HzHP p 07100 see 223 Then we have the formulas 71 Mill gigglaijb 7L HAHoo lrgggiaut HAHZ ibiggest eigenvalue of AfA 229 Proof For we have 11A96Hl Z i1 m E ai j j1 7L m 7L m m 221 mi 2w 21am 1113 21am Hth i1j1 j1 i1 4 i1 230 and therefore HAHl S maxjE1aij1 To prove the equality7 choose jg such that 221 aijO manE1 aijD and then set x 010T where the 1 is in position jg Then for such x we have equality in 230 This shows that HAHi Cannot be smaller than maxi221 aiji The formula for HAHO0 is proved similarly For the norm Hg we have ltz7x where lty 27192111 is the usual scalar product Note that the matrix AA is symmetric and positive semi de nite ltxAAx HAzHg 2 0 From linear algebra we know that AA can be diagonalized and there exists an unitary matrix U UU 1 such that UAAU diagL7 A 7 where AZ 2 0 With z Uy we obtain HAM lt967AV1Igt lty7UAAUygt ZMyiiZ S AmaleHS 1W 231 i1 This implies that HAHZ S xAmax To show equality choose z to be an eigenvector for In order to solve linear ODE7s we will need to construct the exponential of a n gtlt 71 matrix A which we will denote by EA We will de ne it using the series representation of the exponential function If Ck is a sequence with Ck E Kn we de ne in nite series as usual 0 2200 Ck if and only if the partial sums converge The convergence of C 220 Ck is equivalent to the convergence of the 712 series of the coef cients 2k 0 We say that the series converges absolutely if the real series 2 converges For any norm7 there exist positive constants a and A such that a217 S S A EU and therefore absolute convergence of the series is equivalent to the absolute i k convergence of the 712 series 2k cgj CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 45 Proposition 224 Let A E Then 1 For any T gt 0 the series WV 5 14 I 7 232 7 jl 770 converges absolutely and uniformly on 7T7T e is a continuous function oft and we have He AH WA 233 2 The map t a eM is everywhere di erentiable and gem AetA etAA 234 dt Proof Item 1 For t E 7T7T we have tiAi lt ltwmw j 7 j T llAlV j lt 235 Let us denote by Snt the partial sum 2290 Then7 for m gt n7 we have llSntSmtH i Tquot A i T lf lj 236 jn1 j l jn1 j l This implies that Snt is a Cauchy sequence in Kn7 uniformly in t E 7T7T7 since the right side 236 is the remainder term for the series eTllAll The function Snt are continuous function7 they converge uniformly on 7T7T and Kn is a Banach space so that the limit em7 exists and is continuous The bound 233 follows immediately from 235 Item 2 The partial sum Snt are differentiable function oft with The same argument as in 1 shows that Sgt converges uniformly on 7T7T Since both Snt and Sgt converge uniformly we can exchange limit and differentiation If we take the limit n a 00 in 237 we obtain 234 I We summarize some properties of the exponential in Proposition 225 Let ABC E Then 1 If AB BA then eAB eAeB 2 IfC is invertible then eCTIAC C leAC CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 46 3 IfA diag17 7A then 6 4 diagei7 7 6 Proof If AB BA then7 using the binomial theorem7 we obtain A B Z ltngtAan k 238 k0 k and therefore ABiooltABgtniOOTLAkB r LikiOOApOOquAB 5 i 7 7 7 7 7 i e e 7 239 7 nl gig101716 123101 ql and this proves 1 It is easy to see that C lAkC C lACk Dividing by 17 summing and taking the limit proves 2 If A diag17 7 A797 then Ak diaglf7 7 AZ and this proves 3 I As a consequence we obtain Corollary 226 Let A E Then 1 e 1 6 239 6tsA etAesA39 3 eAHA eAeA Example 227 Let us compute the exponential of some simple matrices 1 Let J 31 2 then J2 7 and thus by induction J2 71 1 and J2n1 71 7 We obtain it 70 impel 7 com sing eJ QTHQWA i isiw COW 240 7 0 b m 112 GOSH Sing 2 Let A i lt 7b 0 gt then using 1 we have 6 76 7 isinwt COSUJt a 3 LetBltib Z gt then we have A a 1 and I and J commute Thus at at tB m1 w 6 008031 e s1nbt 6 6 6 lt 76 sinbt eat cosbt 39 23941 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 4 LetOlt0 E 47 00 0 1 5 Let D be the n gtlt 71 matrix with entries 6 on the off diagonal and otherwise gt then 020 and thus 6 0 lt1 6t 0 E e D 242 0 E 0 We have 0 0 62 0 0 0 Ema 0 0 62 0 0 0 D2 HDnA 0 0 e 0 0 0 0 0 0 0 0 0 243 and Dl0forl2n Then we have tZDZ twianil tD 1 tD 6 1 1 2 1 7171 622 Enil H71 1 6t Ti 72112 J tquot 1 6t n72 3 244 622 1 6t 7t 1 6t 1 6 Let A E A E E 7 A E A then etE 6tAID eAtetD 245 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 48 2 2 71 71 EA ate a t 6A1 aquot tquot EA 2 ngl 2 At At an tquot At e ete n72 e 39 39 2 21 246 6A1 are 5 t 6A1 6A1 are 23 Linear systems with constant coef cients From Proposition 224 one obtains immediately Theorem 231 The resoluent of the lineaiquot equation with constant eoe eients x Ax is given by Rt to 5040 247 Proof From proposition 224 we have Law4 Meow 248 dt Thus the j th column of et t0A is the solution of the Cauchy problem x Ax7 xt0 0 70107 T where the 1 is in j th position I Solving x Ax is thus reduced to the problem of computing the exponential of a matrix A7 see Example 227 for some simple examples We will present here a general technique to compute such an exponential In the scalar case the ODE x Ax has the general solution Xt Ce With this intuition in mind let us try to nd solutions of the form xt e Vi where i is a nonzero vector lnserting into the equation we deduce that e Vi is a solution if and only if A1 A117 249 ie7 A is an eigenvalue of Aand i is an eigenvector for the eigenvalue A If A is real and A is a complex eigenvalue with eigenvector w u ii then we have Ail Ail ie the eigenvalues and eigenvectors occur in complex conjugate pairs Proposition 232 Let A be a real n gtlt n matiix and consider the di erential equation x Ax A Z The funetiont gt gt e to is a real solution if and only if A E R i E R and A1 A1 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 49 2 Ifw 31 0 is an eigehuectoiquot for A with eigenvalue A oz i6 with B 31 0 then the imaginary part ofw uio is not zero In this case there are two real solutions t gt gt em cos tu 7 sin to 7 250 t gt gt em sin tu cos to 251 Proof If A1 Ai then e Vi is a solution If A oz i 7 then since A is real an eigenvector w u ii has nonzero imaginary part The real and imaginary parts of the corresponding solution e tu iii eltai tu iii 7 e tcos t isin tu iii 7 em cos tu 7 sin to ieu t sin tu cos to 252 are real solutions In order to show that these real solutions are linearly independent7 let us suppose that some linear combinations of them vanishes identically Evaluating at t 0 and t 7r2 yields clu ego 0 CW 7 C11 07 253 This implies that c c w 0 and thus c1 c2 0 This proves item 2 The proof of item 1 is easy I The problem now is reduced to the question whether we can nd it linearly inde pendent eigenvectors of A As we know from linear algebra this is not always possible7 for example the matrix A10 A0A1 254 00A has 1 as its only eigenavalue and 17 070T as its only eigenvector De nition 233 Let A be an eigenvalue of A then we de ne i The eigenvalue A of A has an algebraic multiplicity equal to l if A is a zero of order l of the charateristic polynomial detA 7 AI ii The eigenvalue A of A has a geometric multiplicity equal to h if h is the di mension of the subspace spanned by the eigenvectors of A for the eigenvalue A7 ie h dimkerA 7 AI The algebraic multiplicity of A for the matrix A given by 254 is 3 but its algebraic multiplicity is 1 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 50 De nition 234 The matrix A is called semi simple or diagonalizable if for each eigen value A algebraic and geometric multiplicity coincide In this case it is7 in principle7 easy to compute em7 we have Proposition 235 Let A be a semi simple n gtlt n matriw real or complew with eigen values A1n repeated according to their algebraic multplicity then there ewists a basis v1 vn of C where v is an eigenvectors ofA for the eigenvalue Ai Let P 711 717 255 be the matricc whose ith column is given by the vector vi Then e m 0 5 P P l 256 0 6AM Proof We have A1 0 AP Av1Avn Alvlgtnvn vlvn 257 0 An ie7 D E P lAP is a diagonal matrix whose entries are the eigenvalues of A Then the resolvent em for x Ax is given by e m 0 eAt PP leA PP l PeP lAP UJ 1 P P l 258 Remark 236 The change of variable y P lz transform the system x Ax into a system of decoupled equations Indeed we have y P lw P lAw Dy where D is diagonal Thus we have n equations Ajyj whose solutions e Vtvj form a fundamental matrix for y Dy Example 237 z z172z2 1720 c2 2x1 7 33 A 2 0 71 259 zg4172z273 472 71 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 51 The eigenvalues are the root of detA 7 AI 2 7 A 7 A3 1 7 AA2 A 2 The eigenvalues are A12 712i239x 27 and A3 1 The eigenvectors are computed to be 013 32 i 2W27274T and 113 17 02T and B 111702113 Three real linearly independent solutions are given by at 0 260 32 2 542 cos t2 2 i sin t2 0 261 4 0 32 2 542 sin t2 2 cos t2 0 262 4 0 Suppose that A is not semi simple7 ie7 if7 for at least one eigenvalue7 the geometric multiplicity is smaller than the algebraic multiplicity One way to solve the system is to transform A into a simpler form7 for example in a triangular form or in Jordan normal form7 ie7 one nds an invertible T such that T lAT has such form see the exercises for a proof than any matrix can be transformed into triangular form With the transformation z Ty and x Ty the ODE x Ax becomes y Sy For example if S has a triangular form we have the system 91 51191 81292 39 Sinyn 11 52292 quot 82 y 2 7 n 263 y Snnyn One can then solve the system iteratively one solves rst the equation for y then the one for yn1 and so on up to the equation for yl see the example below Finally one obtains z Ty Example 238 Consider the system of equations 23 7321 2x2 5 73 2 5 2 2 2 7 3 7 A 0 1 71 264 mg 2x3 0 0 2 with initial conditions 10720730 10720730 The third equations has solution z3t 62t30 lnserting into the second equations gives the inhomogeneous equation 2 2 2 7 62t30 and is solved using Duhamel7s formula t z2t etzgo 70 Ctisezs go etzgo 6t 7 62txgo 265 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 52 Inserting the solutions z2t and z3t into the rst equations gives the equation 23 7321 26t20 26 362txgo Again with Duhamel7s formula one nds 21t e gtzlo 26 7 e gtz20 26 352 7 55 3t230 266 The resolvent is then 673t7to 26t7to 7 2673t7to 26t7to 362t7to 7 5673t7to Rtt0 0 5040 5040 i 62t t0 267 0 0 62t t0 The resolvent can be computed easily if S is in Jordan normal form Let us consider rst the complex Jordan normal form Then S is block diagonal J1 J2 S t 268 Jk where each Jordan block J has the form J 269 A7 Since S is block diagonal we have etJl 5 5 270 wk so that it is enough to compute 6 where J is a Jordan block This has been computed in Example 227 57 and 6 Example 239 The system of equations 23 7221 22 2 2 72 7 271 mg 2x3 is already in Jordan normal form and its resolvent is 672t t672t 0 5 5 0 572 0 272 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 53 7 There is also a real Jordan normal form if A is a real matrix If A 04 i and A Oz 7 i are a pair complex eigenvalues then S has the form 268 where the block corresponding to the pair A7 A is given by RI R0 OI RI R0 0 E T where I is a 2 gtlt 2 identity matrix and R has the form Rltg 273 cos t 7 sin t sin t cos t mute with M we have that e eTteMt and this can be computed easily We will not discuss in detail here the algorithm used to put the matrix in Jordan normal form7 since this is not necessary to compute the resolvent eAt We will use a slightly simpler algorithm to compute eAt It is based on a fundamental result of linear algebra which we quote here without proof The exponential of R is given by eRt e gt Noting that T com De nition 2310 Let A be an eigenvalue of A The generalized eigenspace ofA con sists of the subspace EA 11A 7 AIkZ 07 for some h 21 274 The elements of the generalized eigenspace are called generalized eigenvectors Note that if A is semi simple the generalized eigenspace are obtained by taking only h 1 and thus consist only of eigenvectors We will need the following simple result Lemma 2311 The generalized eigenspace EA is invariant under A Proof If i 6 EA then A 7 AIkZ 0 Then A i AIkAi A i AIkAi i AA 7 MW A i AA i AIW 0 275 and thus Ti 6 EA We have the fundamental result CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 54 Theorem 2312 Let A be a n gtlt n matrim Then there erists a basis of C which consists of generalized eigenvectors ie C EA 276 eigenvalues Using this we will show that any matrix A can be decomposed into a semi simple part and a nilpotent part An example of nilpotent matrix is given in Example 227 De nition 2313 A matrix N is said to be nilpotent with nilpotency h if Nk 0 but Nk71 7g 0 Proposition 2314 Let A be a n gtlt n matrim then there erists a decomposition A S N 277 where A is semi simple N is nilpotent and commute with A and with nilpotency no larger than the maximum of the algebraic multiplicities of the eigenvalues Proof Let 1171n be a basis consisting of generalized eigenvectors and set P o1on be the matrix whose ith column is 111 Let A diag1n be the diagonal matrix where AZ A if i 6 EA Then we de ne SEPAPA NEA S 278 This provides a decomposition S A N By construction S is semi simple and has the same eigenvalues as A with the same algebraic multiplicities Note that SN 7 NS SA 7 S 7 A 7 SS SA 7 AS and so it is enough to show that S commutes with A Let i 6 EA then So Ar Moreover AV 6 EA by Lemma 2311 and thus A1 is an eigenvector for S So we have SA 7 ASi SAi 7 AM S 7 AIAi 0 279 By Theorem 2312 any i E C can be written as a sum of generalized eigenvectors and thus SA 7 ASi 0 280 foranyoECnso SA7AS0andsoSN7NS0 Finally we show that N is nilpotent Let choose in to be larger than the largest algebraic multiplicity of the eigenvalues of A If i 6 EA we have So Ar and thus using that S commute with A we obtain NW1 A7Smo A7Sm71A7IZ A7AIA7SW 1 A7AImo 0 281 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 55 Since any i can be written as a sum of generalized eigenvectors we obtain Nmi 0 for any i E C and so Nm 0 This concludes the proof of Proposition 2314 I Note that Proposition 2314 provides a algorithm to compute etA Example 2315 Let x Ax with 42 41 42 A 42 42 42 282 2 1 2 We have detA 7 AI2 2 and so A 0 has algebraic multiplicity 2 and A 72 has algebraic multiplcity 1 The vector 17 27 71T is an eigenvector for 72 We have 2 2 2 47 012 42 4 4 4 283 42 42 42 and so we can choose 1071T and 0171T has generalized eigenvectors We obtain 1 1 0 1 1 1 1 P 2 0 1 1344 17171 284 717171 2 i2 0 i2 and 72 0 0 717171 A 0 0 0 SPAP 1 72 72 72 285 0 0 0 1 1 1 710 71 NA7S 0 0 0 286 10 1 Finally we compute the exponential by 1 e Zt 1 7 2t e Zt 7 1 e Zt 7 2t etA PetAP 1ItN i 2541 4 2 25 25 4 2 287 75 12t 75 2t1 75 2 32t A simple but important consequence of this decomposition is the following Proposition 2316 IfA is a realngtltn matrim then e is a matrico whose components are sums of terms of the form pt em sin t and pte t cos t where 04 are real numbers such that A 04 i6 is an eigenvalue ofA and pt is a polynomial of degree at most n 7 1 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 56 24 Stability of linear systems For the ODE7 x ft7 a we say that 0 is a critical point if ft7 x0 0 for all t This implies that the constant solution t 0 is a solution of the Cauchy problem with zt0 zo Critical points are also called equilibrium points For a linear system with constant coef cients7 ie7 f Ax x0 0 is always critical point and it is the only critical point if detA 31 0 If detA 07 then 0 is an eigenvalue and any point in the eigenspace of the eigenvalue 0 is a critical point We de ne next the concept of stability of a solution De nition 241 Let f R gtlt R 7 R be continuous and locally Lipschitz Let ztt0 be the solution of the Cauchy problem x ftx zt0 x0 which we assume to eccist for all times t gt to Z The solution tt0z0 is stable in the sense of Liapunou iffor any 6 gt 0 there ewists 6 gt 0 such that for all with S 6 we have Htt0z0 g 7 xtt0z0H S E for all t 2 to 288 2 The solution tt0z0 is asymptotically stable if it is stable and there ewists 6 gt 0 such that hat for all with S 6 we have Htt0z0 0 7 t7t0z0H 0 289 3 The solution tt0z0 is unstable if it is not stable It a is a critical point7 we will say the critical point a is stable or unstable if the solution zt E a is stable or unstable Example 242 The solution t 0 of x Ax is asymptotically stable of A lt 07 stable if A 07 unstable if A gt 0 Example 243 The solutions of the equation x z 0 are stable but not asymp totically stable The general solution is a cost bsint7 7a sint bcost is a periodic solution of period 27139 on the circle of radius a2 b2 Two solutions starting at nearby points zhyl and zmyo will remain close forever Example 244 The solution zt0z0 0 lt 0 but unstable for 0 2 0 i 2 130 of x i z is asymptotically stable for Example 245 For the ODE x x2 7 17 there are two critical points 0 and 1 The solution zt 0 is unstable and the solution t 1 is stable CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 57 For a linear homogeneous equation x At we have zt t0 n3 72t t00 Rtt0 xtt0 7 tt0 0 and so it suf ces to study the stability of the crit ical point 0 For a linear inhomogeneous equation x At ft the difference t t0z0 g 7 zt to 0 is again equal to Rtt0 where Rtt0 is the resolvent of the homogeneous equation x At and thus the stability properties of a solution zt of the inhomogeneous problem are the same as the stability of the trivial solution of the homogeneous problem Therefore in the case of linear differential equations all the solutions have the same stability properties and one can talk about the stability of the differential equation As we have seen in Section 23 the solutions of linear systems with constant co ef cients are determined by the eigenvalues and the generalized eigenvectors of the matrix A We de ne the stable unstable and center subspaces denoted respectively by ES E and E0 and de ned by E9 QB 290 AReltO E 291 ARegtO E0 QB 292 ReO 293 By Lemma 2311 amd Theorem 2312 the generalized eigenspaces span the whole space and are invariant under A and thus also under etA So we have R ESEBEueBEC 294 and eAtE E for all t E R suc 295 From the proposition 2316 we obtain Theorem 246 Let x Ax be a linear system with constant coe cients x Ax and let A1 Ak be the eigenvalues of A a The critical point 0 is asymptotically stable if and only if all the eigenvalues ofA have a negative real part Re A lt 0 for i 1 k ie if E9 R b The critical point 0 is stable if and only 1 All the eigenvalues have a nonpositive real part Re A S 0 fori 1k ie ESEBEC R 2 If Re 0 the Jordan blocks have dimension 1 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 58 Proof If E9 R then any solutions has components which are linear combina tions of terms of the form tkeu tsinbt and tkeu tcosbt with a lt 0 In particular since limtH00 tkeu t sinbt 0 we see that every solution goes to 0 This implies that limtH00 HetAH 0 and so 0 is asymptotically stable If If E9 8 EC R then any solutions tkeu tsinbt and tkeu tcosbt with a S 0 If E0 is non trivial there will be some terms with a 0 and those terms will remain bounded only if there are no polynomial factors thin those terms7 ie7 only if the restriction of A to E0 is semisimple In this case we have then HeAtH S K and thus 0 is stable If the restriction of A to E0 has a non trivial nilpotent there will be terms which diverge as ta 00 and 0 is not stable If some eigenvalue A has a positive real part7 then there exists solutions zt with a 00 as t a 00 In this case 0 is unstable I The qualitative behavior of solutions of linear systems with constant coef cients is as follows o If x E E9 then zt eAtx satis es limtaoox t 0 and limtn m oo o If x E E then t 6quot satis es limtH00 00 and limtn m zt 0 o If x E E0 then t either stays bounded for all t E R or limtaioo 00 We illustrate the behavior of solutions for linear 2 dimensional systems with con stant coef cients in the following gures In gure 21 we show the 2 stable linear systems ES R2 with distinct eigenvalues and Jordan normal forms 04 3 A1 0 B a gtalt0 lt0 A2gtA1lt0A2lt0 and in gure 22 the 2 stable linear systems ES R2 with one eigenvalue and Jordan normalforms A 0 A 1 lt0 AgtAlt0 lt0 AgtAlt0 In gure 23 we show 2 unstable linear systems Eu R2 with corresponding Jordan normal forms 04 3 A1 0 B a gtagt0 lt0 A2gtA1gt0A2gt0 and in gure 24 the 2 unstable linear systems ES R2 with one eigenvalue and Jordan normal forms A 0 A 1 lt0 AgtAgt0 lt0 AgtAgt0 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 59 In gure 25 we show a center stable but not asymptotically stable and an hyper bolic linear system in R2 unstable with corresponding Jordan normal forms 0 in A1 0 agt0 A1lt0ltA2 n 0 7 7 0 A2 7 llllllll lllliJ xxxxxxx viiiiiiii llllll111rxr 44 llllllll xxxxxxxv llLJIJIIIrrr 444 x xiiiiiiii lllll VV39 4 j xxiilliii xxx illllzgrr V ffI xp 33LL 1111 Jill1rrrryyu If I Ax XLil Lizlrlrk PPgt A llllllllll l llllK xxk LJJJJJJJll JIV kkkkkkkkkk lxxxxxlxltII r mmmm 2 m f7frfff17lli k quot quot 4244quot xrr rrrrrrrrrxwxxxxx kquot aa 444 rw RP I aaaaaazlx xxx rrrrrrrrtlixxxxx KNPK aha xxxffflrl Trf117iilli xknkl 4 flflll xxxx llllllllli xxkkkyyr A nxflfrfllil VVTXTXKVVK xxxxpk IrflllzKK VVrYYTKKi x xkkkr xxflfrfritl iiiiiilx xkxk rrrrrrr11rx Figure 21 Stable linear systems with E9 R2 stable spiral and stable focus with two distinct eigenvalues7 stable focus with a nontrivial Jordan block xxxxxLlL LJ11 xx lll LJJIIzzr xxx xxluL11zrzr xl Jillzr l 111 fry y xxxxxxxxx xxw 111 xixryyyr xxxxxxxxxxx llzxzrrrryr xxxxxxxx xxx xx 11 rVV E En xxxxx xxxxx Krrrrrquotquotgt r aaaxxxxxx xx rrrrr 24444444 44 xxxxkPLkJPkL 44 44 lvlVf I xxkk kxkxkkk 444A x ff xxx xkk k 4Av ff 39Kkk allxzzzxz7h xxxxxxxxx rrlfr xxxx l zlxzzzrrri ixxxxxxxx Ifflffh llK xzxxzfllrffr Y rffrft V7xx Figure 22 Stable linear systems with E9 metric multiplicity two and one JJJJJJJlllllll ll zrrr 14444444111111111111 Izr xxxxxxxxxxxxxxzzzlxziz 11111444444111111111 lllIzrrr V 11111144444411111111 lllzrrr K 11111111444441111111 xxx y 1111111111444411111 lllr 4411111111144411111 IIrry JJJJJJJAAAKIIIJJIIL x 11111444444111114411 1 ljf fj ji f KYllllfffffffif u i A hiiidiim ihiiiiiii lllllllllllllliLLgtr rrrvivvrrrrrrrrrr iiiillllliiiiill rrrrrr111vvvrrrrr l zw rrrrrrrr1111vvvvr 22rr rrrrrrrrr1111vv x 4 xxzrrr zrrrrrrrrrrrr111 xxx AAA xrzr1rr lrrrrrrrrrrrrrr N 4441f Iff IlfoVVVfffflflf V 4a393939ll39lff IlffffVVVffffff AxI l flitll f 77fVVVVffffff xA393939I3939ff IllffffVVVVVfff A393939 39fr 777fVVVVVff R2 with two identical eigenvalues geo lf At depends on t in general it is not enough to look at the eigenvalues of A One can construct examples of matrices At whose eigenvalues are negative but for which 0 is unstable see homework One needs stronger condition As an example we prove CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 60 x xzxx 2azx u 44na nn kkkkxxxxxXTiff rr 393939 vv 44444En nx bgk 11f 39fxxxxxle Al 4EAAAAE N Nk Ppkkkx V r r 39 rrrzx 45 1141 aa xxxukknkgtk kxK1F xxxxx I r AA4AAAa auua kkkykkkkxxxif fflxl flrxx4aaaa aa rkpkkk V 1 fffllrlffl L444 Nytu v r xK l ff I rrvrrrrIIrzxx a an kknkkxr xv1wr11rrrrrxxx a kyykkykykkkl wwx xnxxwxxx xxxxkitrtrr7zx4 xxxrrwrryryyyyyyykx w 1 1 21 l 2 2 3 1 1 l 2 1 Ei i3mmxmmwrrny V wquot KKKKK Fwy1llllxixxyyyyyyIzlrrI x a KKkk k k 111141rrrrr 1 i aa Prrrrr 11111111 I I K 1 4 4 k KkkkkrylllllIIi 1 ltquot kk kkr rrfllIIlquot r I JLX Aaa xkxkkkk krkyi I 1 r 1 Jll kkk quotquotquotquot 39Lyrrrrr1 zrz 1i i L e a K P H 39kvrrrr 1 11ll quot quot kkkekkekkvr ivy 1 11 l e Figure 23 Unstable linear systems with E R2 unstable spiral and unstable focus with two distinct eigenvalues V 1x1111aaaa Figure 24 Unstable linear systems with E9 R2 with two identical eigenvalues geornetric rnultiplicity two and one VVIIrzzzlzl 1J1lllllllll lllllllll 44quot rkrrrrr xszlz1LlllXX 4 m VV iiiiiixrIfri 4a altax rk yrr1 iiilllamprl N Peer rr 1i J1Ll1rrrfffAW gtPkrkrkr 1 JLixY r 11frf a quot quot 1 lllx Irrr xxx xx Kkquotquotk quotquot I l 1 Vr7r Kk kkkk l w ltltltK waaag a a lxnvxxltxk h 2 2 3 ii and a xxgnjjl K7V agagn vrfjlllllllll KT1 ff K AEAEAA kjjlllllll xxw1p ffIlxx xaaagaa rknkj14lil Klltrr 1 44 k kkkj14 1 xxxtrrrr z k kukgk yi jjjj rxxxvrn rxrxx axxxKxxk kkkrk1 Ki liifi flrf kg knkk V xxlwrrr1r Irrrrrzxxll Kkkkkk rk Figure 25 A linear hyperbolic system with EueBES i R2 and a center with E0 7 R2 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 61 Theorem 247 Let At be symmetiie ie At At and continuous on t0oo If the eigenvalues Ait of At satisfy Ait S 04 for t 6 to 00 then the solution zt of x At satisfy lltll2 S 6 t t0lltoll27 tgt 1to 296 In paitieulaiquot ifa S 0 then 0 is stable and ifa lt 0 then 0 is asymptotically stable Proof Since At is symmetric its eigenvalues are real and it is diagonalizable with an orthogonal matrix there exists a matrix Qt with QTt Q 1t such that QTtAtQt diag1t Ant We show that for all i and all t gt to we have lt1 Aogt S alto ogt 297 We set i Qw and then we have lt1 ogt ltw wgt and lt11 Aogt ltw QTAQwgt S altw wgt alto ogt 298 For a solution t of x Atz we obtain lltll 2amp07 At9 tgt S Qall t lz 299 Integrating gives WW3 S ll lttoll 2at ll8ll d87 2100 and therefore by Gronwall Lemma WW3 S lltoll 62 t to 2101 I 25 Floquet theory In this section we consider periodic At ie there exists p gt 0 such that At 10 At for all t E R Such equation can be reduced at least in principle to the case of constants coef cients and is this reduction goes under the name of Floquet theory As a preliminary we need to de ne the logarithm of matrix For this it is necessary to consider complex matrix since the logarithm of a real matrix will be in general complex Proposition 251 Logarithm of a matrix Let C be an invertible matiim then there exists a complecs matrics R such that O 5R 2102 CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 62 Proof We will use the decomposition of C S N into a semi simple ie7 diag onalizable matrix S and a nilpotent matrix N ie7 Nk 0 for some k 2 1 with SN NS 21 Let us rst consider the case 0 S7 ie7 C is semi simple Since S is invertible and semi simple there exists P such that S PAPA with A diag1n and Ak 31 0 for all k We set T13L13quotl7 Ldiaglog1logn 2103 Then we have 1 5T ePLP PeLP l PAP 1 S 2104 b To treat the general case 0 S N we note that S is invertible if and only if C is invertible they have the same eigenvalues and if SN NS then S lN NS l Then we have OSN SIS 1N 2105 and S lN is nilpotent and commute with S Recall the power series t2 t3 0 39 l 1 tt77 77 f tlt1 2106 ogltgt 23 j om lt gt and that the formal rearrangement of power series 71 NH 1 Z 2 j1t 2107 n0 j1 j n39 is valid for ltl S 1 this is simply a complicated way to write the identity 610g1t 1t Let us de ne now R T Q 2108 where T is de ned as in a and 71 1 S lN j Q EM 2109 11 j Since S lN is nilpotent the series de ning Q is actually a nite sum and we do need to worry about convergence From the formal rearrangement 2107 we conclude that 59 A 1N 2110 Finally since T and Q commute we obtain 5ReTeQ SIA 1NO 2111 and this concludes the proof of Proposition 251 I CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 63 Theorem 252 Floquet Let At be a continuous periodic function of period p Then any fundamental matricc ltlgtt for x At has a representation of the form ltlgtt PteRt Ptp Pt 2112 where R is a constant matrim Remark 253 The theorem 252 provieds us the form of the solutions If no is an eigenvector of R for the eigenvalue A then the solution t has the form 2t e 7 where 2t Pt0 is periodic with period p More generally7 by the discussion in Section 237 a general solution will have components which is a linear combination of terms of the form ozttke t7 where at is a vector periodic in t Note7 in particular7 that there exists periodic solutions of period p whenever 0 is an eigenvalue of R Proof of Theorem 252 a We note rst that if ltlgt1t and ltlgt2t are two fundamental matrices7 then there exists an invertible matrix C such that ltlgt1t ltlgt2tO 2113 This follows from the fact that Rtt0 ltIgt1tlt1gt1t0 ltIgtZtlt1gtg1t0 2114 ie7 ltlgt1t lt1gt2ilt1gt1iolt1gt1io 2115 b If t is a solution of x At7 then one veri es easily that yt xtp is also a solution Therefore if ltlgtt is a fundamental matrix7 then lt ltlgtt 10 is also a fundamental matrix By a and Proposition 2517 there exists a matrix R such that ltlgttp can 2116 We now de ne Pt E wasquot 2117 and Pt is periodic of period p since Ptp ltlgttpe tpR ltlgttepRe tpR Pt 2118 This concludes the proof I The matrix C epR is called the transition matrix and the eigenvalues eigenvalues7 M of C epR are called the Floquet multipliers The matrix 0 depends on the choice of the fundamental matrix Mt7 however the eigenvalues do not see exercises The eigenvalues of R7 u are given by Al em are called the characteristic erpo nents They are unique7 up to a multiple of 27Tip CHAPTER 2 LINEAR DIFFERENTIAL EQUATIONS 64 Remark 254 For the equation x Atx7 let us consider the transformation xt Ptyt 2119 where Pt is the periodic matrix given by Floquet theorem We obtain Mt P tyt POWt AtPtyt 2120 On the other hand Pt tk Rt so that P t may i lt1gtte RtR AtPt i PtR 2121 Thus we nd yt Ryt 2122 The transformation z Pty reduces the linear equation with periodic coef cients x At to the system with constant coef cients y By Nevertheless there are7 in general7 no methods available to compute Pt or the Floquet multipliers Each equation has to be studied for itself and entire books are devoted to quite simple looking equations The Floquet theory is however very useful to study the stability of periodic solutions7 as we shall see later Example 255 Let us consider the equation x bt at 07 where at and bt are periodic functions For the fundamental solution t Rt0 we have ltIgt0 I and so by Floquet Theorem up o W 9622 2123 The Floquet multipliers are given by the solutions of A2aA 0 2124 where a 7x1p 7221 5 detO detRp0 efoptrAlt9gtd9 WWW 2125 In the special case where bs E 07 then the equation is A2 04A 1 0 We have then i If 72 lt 04 lt 2 then the Floquet multipliers A and A are complex conjugate with modulus 1 and therefore the solutions are bounded for all t gt 0 ii If 04 gt 2 or 04 lt 72 at least one eigenvalue of C has modulus greater than 1 and there exists solutions such that lxtlz 12 goes to in nity as It goes to in nity iii If 04 72 then A 1 is the eigenvalue of C and therefore there exists a periodic solution of period p If 04 2 then A 71 is the eigenvalue of C and therefore there exists a periodic solution of period 2p

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