Ordinary Di↵erential Equations
Simon Brendle
Contents
Preface vii Chapter 1. Introduction 1 1.1. Linear ordinary di↵erential equations and the method of § integrating factors 1 1.2. The method of separation of variables 3 § 1.3. Problems 4 § Chapter 2. Systems of linear di↵erential equations 5 2.1. The exponential of a matrix 5 § 2.2. Calculating the matrix exponential of a diagonalizable matrix 7 § 2.3. Generalized eigenspaces and the L + N decomposition 10 § 2.4. Calculating the exponential of a general n n matrix 15 § ⇥ 2.5. Solving systems of linear di↵erential equations using matrix § exponentials 17 2.6. Asymptotic behavior of solutions 21 § 2.7. Problems 24 § Chapter 3. Nonlinear systems 27 3.1. Peano’s existence theorem 27 § 3.2. Existence theory via the method of Picard iterates 30 § 3.3. Uniqueness and the maximal time interval of existence 32 § 3.4. Continuous dependence on the initial data 34 § 3.5. Di↵erentiability of flows and the linearized equation 37 § 3.6. Liouville’s theorem 39 § v vi Contents
3.7. Problems 40 § Chapter 4. Analysis of equilibrium points 43 4.1. Stability of equilibrium points 43 § 4.2. The stable manifold theorem 44 § 4.3. Lyapunov’s theorems 53 § 4.4. Gradient and Hamiltonian systems 55 § 4.5. Problems 56 § Chapter 5. Limit sets of dynamical systems and the Poincar´e- Bendixson theorem 57 5.1. Positively invariant sets 57 § 5.2. The !-limit set of a trajectory 60 § 5.3. !-limit sets of planar dynamical systems 62 § 5.4. Stability of periodic solutions and the Poincar´emap 65 § 5.5. Problems 67 § Chapter 6. Ordinary di↵erential equations in geometry, physics, and biology 71 6.1. Delaunay’s surfaces in di↵erential geometry 71 § 6.2. The mathematical pendulum 75 § 6.3. Kepler’s problem 78 § 6.4. Predator-prey models 81 § 6.5. Mathematical models for the spread of infectious diseases 82 § 6.6. A mathematical model of glycolysis 83 § 6.7. Problems 87 § Chapter 7. Sturm-Liouville theory 89 7.1. Boundary value problems for linear di↵erential equations of § second order 89 7.2. The Sturm comparison theorem 92 § 7.3. Eigenvalues and eigenfunctions of Sturm-Liouville systems 95 § 7.4. The Liouville normal form 100 § 7.5. Asymptotic behavior of eigenvalues of a Sturm-Liouville § system 102 7.6. Asymptotic behavior of eigenfunctions 104 § 7.7. Orthogonality and completeness of eigenfunctions 108 § 7.8. Problems 112 § Bibliography 115 Preface
These notes grew out of courses taught by the author at Stanford University during the period of 2006 – 2009. The material is all classical. The author is grateful to Messrs. Chad Groft, Michael Eichmair, and Jesse Gell-Redman, who served as course assistants during that time.
vii
Chapter 1
Introduction
1.1. Linear ordinary di↵erential equations and the method of integrating factors
A di↵erential equation is an equation which relates the derivatives of an unknown function to the unknown function itself and known quantities. We distinguish two basic types of di↵erential equations: An ordinary di↵erential equation is a di↵erential equation for an unknown function which depends on a single variable (usually denoted by t and referred to as time). By contrast, if the unknown function depends on two or more variables, the equation is a partial di↵erential equation. In this text, we will restrict ourselves to ordinary di↵erential equations, as the theory of partial di↵erential equations is considerably more di cult. Perhaps the simplest example of an ordinary di↵erential equation is the equation
(1) x0(t)=ax(t), where x(t) is a real-valued function and a is a constant. This is an example of a linear di↵erential equation of first order. Its general solution is described in the following proposition:
Proposition 1.1. A function x(t) is a solution of (1) if and only if x(t)= ceat for some constant c.
Proof. Let x(t) be an arbitrary solution of (1). Then
d at at (e x(t)) = e (x0(t) ax(t)) = 0. dt
1 2 1. Introduction
at at Therefore, the function e x(t) is constant. Consequently, x(t)=ce for some constant c. Conversely, suppose that x(t) is a function of the form x(t)=ceat for at some constant c.Thenx0(t)=ca e = ax(t). Therefore, any function of at the form x(t)=ce is a solution of (1). ⇤
We now consider a more general situation. Specifically, we consider the di↵erential equation
(2) x0(t)=a(t) x(t)+f(t). Here, a(t) and f(t) are given continuous functions which are defined on some interval J R. Like (1), the equation (2) is a linear di↵erential equation ⇢ of first order. However, while the equation (1) has constant coe cients, coe cients in the equation (2) are allowed to depend on t. In the following proposition, we describe the general solution of (2):
t Proposition 1.2. Fix a time t0 J, and let '(t)= a(s) ds. Then a 2 t0 function x(t) is a solution of (2) if and only if R t '(t) '(s) x(t)=e e f(s) ds + c ✓ Zt0 ◆ for some constant c.
Proof. Let x(t) be an arbitrary solution of (2). Then
d '(t) '(t) (e x(t)) = e (x0(t) '0(t) x(t)) dt '(t) = e (x0(t) a(t) x(t)) '(t) = e f(t). Integrating this identity, we obtain t '(t) '(s) e x(t)= e f(s) ds + c Zt0 for some constant c.Thisimplies t '(t) '(s) x(t)=e e f(s) ds + c ✓ Zt0 ◆ for some constant c. Conversely, suppose that x(t) is of the form t '(t) '(s) x(t)=e e f(s) ds + c ✓ Zt0 ◆ 1.2. The method of separation of variables 3 for some constant c.Then t '(t) '(s) x0(t)='0(t) e e f(s) ds + c ✓ Zt0 ◆ t '(t) d '(s) + e e f(s) ds dt ✓ Zt0 ◆ = a(t) x(t)+f(t). Therefore, x(t) solves the di↵erential equation (2). This completes the proof. ⇤
1.2. The method of separation of variables
We next describe another class of di↵erential equations of first order that can be solved in closed form. We say that a di↵erential equation is separable if it can be written in the form
(3) x0(t)=f(x(t)) g(t). Here, f(x) and g(t) are continuous functions which we assume to be given. Moreover, we assume that U R is an open set such that f(x) = 0 for all ⇢ 6 x U. 2 In order to solve (3), we first choose a function ' : U R such that 1 ! '0(x)= f(x) . Suppose now that x(t) is a solution of (3) which takes values in U. Then we obtain d 1 '(x(t)) = '0(x(t)) x0(t)= x0(t)=g(t). dt f(x(t)) Integrating both sides of this equation gives
'(x(t)) = g(t) dt + c Z for some constant c. Thus, the solution x(t) can be written in the form
1 x(t)=' g(t) dt + c , ✓ Z ◆ 1 where ' denotes the inverse of '. Note that the general solution of the di↵erential equation involves an arbitrary constant c.Ifweprescribethe value of the function x(t) at some time t0,thenthisuniquelydetermines the integration constant c, and we obtain a unique solution of the given di↵erential equation with that initial condition. As an example, let us consider the di↵erential equation 2 x0(t)=t (1 + x(t) ). 4 1. Introduction
1 To solve this equation, we observe that 1+x2 dx = arctan(x). Hence, if x(t) is a solution of the given di↵erential equation, then R d 1 arctan(x(t)) = x0(t)=t. dt 1+x(t)2 Integrating this equation, we obtain t2 arctan(x(t)) = + c 2 for some constant c. Thus, we conclude that t2 x(t) = tan + c . 2 ⇣ ⌘ 1.3. Problems
Problem 1.1. Find the solution of the di↵erential equation x (t)= 2t x(t)+ 0 1+t2 1 with initial condition x(0) = 1. t Problem 1.2. Find the solution of the di↵erential equation x0(t)= t+1 y(t)+ 1 with initial condition x(0) = 0.
Problem 1.3. Find the general solution of the di↵erential equation x0(t)= x(t)(1 x(t)). This di↵erential is related to the logistic growth model. Problem 1.4. Find the general solution of the di↵erential equation x0(t)= 1 x(t) log x(t) . This equation describes the Gompertz growth model.
Problem 1.5. Let x(t) be the solution of the di↵erential equation x0(t)= cos x(t) with initial condition x(0) = 0. (i) Using separation of variables, show that log(1 + sin x(t)) log(1 sin x(t)) = 2t. 2 cos x cos x (Hint: Write cos x = 1+sin x + 1 sin x .) (ii) Show that t t e e t t x(t) = arcsin t t = arctan(e ) arctan(e ). e + e ⇣ ⌘ Chapter 2
Systems of linear di↵erential equations
2.1. The exponential of a matrix
n n Let A C ⇥ be an n n matrix. The operator norm of A is defined by 2 ⇥ A op =sup Ax . k k x Cn, x 1 k k 2 k k It is straightforward to verify that the operator norm is submultiplicative; that is, AB A B . k kop k kop k kop Iterating this estimate, we obtain Ak A k k kop k kop for every nonnegative integer k. This implies that the sequence m 1 Ak k! Xk=0 n n is a Cauchy sequence in C ⇥ .Itslimit m 1 exp(A):= lim Ak m !1 k! Xk=0 is referred to as the matrix exponential of A.
n n Proposition 2.1. Let A, B C ⇥ be two n n matrices satisfying AB = 2 ⇥ BA. Then exp(A + B)=exp(A)exp(B).
5 6 2. Systems of linear di↵erential equations
Proof. Since A and B commute, we obtain l l l j l j (A + B) = A B , j Xj=0 ✓ ◆ hence l 1 l 1 j l j (A + B) = A B . l! j!(l j)! Xj=0 Summation over l gives 2m 1 1 (A + B)l = Aj Bk. l! j! k! l=0 j,k 0,j+k 2m X X From this, we deduce that 2m m m 1 1 1 (A + B)k Aj Bk k! j! k! Xk=0 ✓ Xj=0 ◆✓Xk=0 ◆ 1 = Aj Bk. j! k! j,k 0,j+k 2m, max j,k >m X { } This gives 2m m m 1 1 1 (A + B)l Aj Bk l! j! k! l=0 ✓ j=0 ◆✓k=0 ◆ op X X X 1 A j B k j! k! k kop k kop j,k 0,j+k 2m, max j,k >m X { } 1 1 1 1 A j B k , j! k kop k! k kop ✓ j=Xm+1 ◆✓k=Xm+1 ◆ and the right hand side converges to 0 as m . From this, the assertion !1 follows. ⇤ n n Corollary 2.2. For any matrix A C ⇥ ,thematrixexp(A) is invertible, 2 and its inverse is given by exp( A). Corollary 2.3. We have exp((s+t)A)=exp(sA)exp(tA) for every matrix n n A C ⇥ and all s, t R. 2 2 In the remainder of this section, we derive an alternative formula for the exponential of a matrix. This formula is inspired by Euler’s formula for the exponential of a number: n n Proposition 2.4. For every matrix A C ⇥ , we have 2 1 m exp(A)= lim I + A . m m !1 ⇣ ⌘ 2.2. Calculating the matrix exponential of a diagonalizable matrix 7
Proof. We compute 1 m exp(A) I + A m ⇣1 m ⌘ 1 m = exp A I + A m m hm 1 ⇣ ⌘i ⇣ ⌘ 1 m l 1 1 1 1 l = exp A exp A I A I + A . m m m m Xl=0 h ⇣ ⌘i h ⇣ ⌘ i⇣ ⌘ This gives 1 m exp(A) I + A m op m 1 ⇣ ⌘ 1 m l 1 1 1 1 l exp A exp A I A I + A m op m m op m op Xl=0 ⇣ ⌘ ⇣ ⌘ m 1 m l 1 l A op 1 1 1 e m k k exp A I A 1+ A op m m op m k k Xl=0 ⇣ ⌘ ⇣ ⌘ m 1 m l 1 l A op 1 1 A op e m k k exp A I A e m k k m m op Xl=0 ⇣ ⌘ m 1 A op 1 1 = me m k k exp A I A . m m op ⇣ ⌘ On the other hand, 1 1 1 1 1 exp A I A = Ak, m m k! mk ⇣ ⌘ Xk=2 hence
1 1 1 1 1 1 k 1 2 A op exp A I A A op A op e m k k . m m op k! mk k k m2 k k ⇣ ⌘ k=2 X Putting these facts together, we conclude that m 1 1 2 A op exp(A) I + A A op ek k . m op m k k ⇣ ⌘ From this, the assertion follows easily. ⇤
2.2. Calculating the matrix exponential of a diagonalizable matrix
n n In this section, we consider a matrix A C ⇥ which is diagonalizable. 2 n n In other words, there exists an invertible matrix S C ⇥ and a diagonal 2 8 2. Systems of linear di↵erential equations matrix 1 0 ... 0 0 2 ... 0 D = 2 . . . . 3 . . .. . 6 7 6 0 ...... 7 6 n7 1 such that A = SDS . Equivalently,4 a matrix A5 is diagonalizable if there exists a basis of Cn which consists of eigenvectors of A. In order to compute the exponential of such a matrix we need two aux- iliary results. The first one relates the matrix exponentials of two matrices that are similar to each other. n n Proposition 2.5. Suppose that A, B C ⇥ are similar, so that A = 1 n2 n 1 SBS for some invertible matrix S C ⇥ . Then exp(tA)=S exp(tB)S 2 for all t R. 2 k k 1 Proof. Using induction on k, it is easy to show that A = SB S for all integers k 0. Consequently, k k 1 t k 1 t k 1 1 exp(tA)= A = SB S = S exp(tB)S . k! k! Xk=0 Xk=0 This completes the proof. ⇤ The second result gives a formula for the exponential of a diagonal ma- trix: Proposition 2.6. Suppose that
1 0 ... 0 0 2 ... 0 D = 2 . . . . 3 . . .. . 6 7 6 0 ...... 7 6 n7 is a diagonal matrix. Then 4 5 et 1 0 ... 0 0 et 2 ... 0 exp(tD)=2 . . . . 3 ...... 6 7 6 0 ...... et n 7 6 7 4 5 Proof. Using induction on k, we can show that k 1 0 ... 0 0 k ... 0 k 2 D = 2 . . . . 3 . . .. . 6 7 6 0 ...... k 7 6 n7 4 5 2.2. Calculating the matrix exponential of a diagonalizable matrix 9 for every integer k 0. Summation over k gives
tk k k1=0 k! 1 0 ... 0 k tk k 1 t 2 0 k1=0 k! 2 ... 0 3 exp(tD)= Dk = P . k! . . .. . k=0 6 . P . . . 7 X 6 tk k 7 6 0 ...... 1 7 6 k=0 k! n7 4 5 P From this, the assertion follows. ⇤
1 To summarize, if A = SDS is a diagonalizable matrix, then its matrix exponential is given by
et 1 0 ... 0 0 et 2 ... 0 1 1 exp(tA)=S exp(tD)S = S 2 . . . . 3 S ...... 6 7 6 0 ...... et n 7 6 7 4 5 As an example, let us consider the matrix
↵ A = , ↵ where ↵, R. The matrix A has eigenvalues 1 = ↵ + i and 2 = ↵ i . 2 Moreover, the vectors
1 v = 1 i and 1 v = 2 i are eigenvectors of A with eigenvalues 1 and 2. Consequently, A is diag- onalizable and ↵ + i 0 A = S S 1, 0 ↵ i where 11 S = . i i 10 2. Systems of linear di↵erential equations
Therefore, the matrix exponential of A is given by et(↵+i ) 0 exp(tA)=S S 1 0 et(↵ i ) 11 et(↵+i ) 0 1 i = 2 2 i i 0 et(↵ i ) 1 i 2 2 1 (et(↵+i ) + et(↵ i ) i (et(↵+i ) + et(↵ i )) = 2 2 i (et(↵+i ) + et(↵ i )) 1 (et(↵+i ) + et(↵ i ) 2 2 e↵t cos( t) e↵t sin( t) = . e↵t sin( t) e↵t cos( t) 2.3. Generalized eigenspaces and the L + N decomposition
In order to compute the exponential of a matrix that is not diagonaliz- able, it will be necessary to consider decompositions of Cn into generalized eigenspaces. We will need the following theorem due to Cayley and Hamil- ton: Theorem 2.7. Let A be a n n matrix, and let p ( )=det( I A) denote ⇥ A the characteristic polynomial of A. Then pA(A)=0.
Proof. The proof involves several steps. Step 1: Suppose first that A is a diagonal matrix with diagonal entries 1,..., n,i.e. 1 0 ... 0 0 2 ... 0 A = 2 . . . . 3 ...... 6 7 6 0 ...... 7 6 n7 Then 4 5 p( 1)0... 0 0 p( 2) ... 0 p(A)=2 . . . . 3 . . .. . 6 7 6 0 ...... p( )7 6 n 7 for every polynomial p. In particular,4 if p = pA is the characteristic5 polyno- mial of A,thenpA( j) = 0 for all j,hencepA(A) = 0. Step 2: Suppose next that A is an upper triangular matrix whose diag- onal entries are pairwise distinct. In this case, A has n distinct eigenvalues. In particular, A is diagonalizable. Hence, we can find a diagonal matrix B 1 and an invertible matrix S such that A = SBS . Clearly, A and B have 1 the same characteristic polynomial, so pA(A)=pB(A)=SpB(B)S =0 by Step 1. 2.3. Generalized eigenspaces and the L + N decomposition 11
Step 3: Suppose now that A is a arbitrary upper triangular matrix. We can find a sequence of matrices Ak such that limk Ak = A and each !1 matrix Ak is upper triangular with n distinct diagonal entries. This implies pA(A)=limk pA (Ak) = 0. !1 k Step 4: Finally, if A is a general n n matrix, we can find an upper ⇥1 triangular matrix B such that A = SBS . Again, A and B have the same 1 characteristic polynomial, so we obtain pA(A)=pB(A)=SpB(B)S =0 by Step 3. ⇤ We will also need the following tool from algebra: Proposition 2.8. Suppose that f( ) and g( ) are two polynomials that are relatively prime. (This means that any polynomial that divides both f( ) and g( ) must be constant, i.e. of degree 0.) Then we can find polynomials p( ) and q( ) such that p( ) f( )+q( ) g( )=1.
This is standard result in algebra. The polynomials p( ) and q( ) can be found using the Euclidean algorithm. A proof can be found in most algebra textbooks. Proposition 2.9. Let A be an n n matrix, and let f( ) and g( ) be two ⇥ polynomials that are relatively prime. Moreovr, let x be a vector satisfying f(A) g(A) x =0. Then there exists a unique pair of vectors y, z such that f(A) y =0, g(A) z =0,andy + z = x. In other words, ker(f(A) g(A)) = ker f(A) ker g(A). Proof. Since the polynomials f( ) and g( ) are relatively prime, we can find polynomials p( ) and q( ) such that p( ) f( )+q( ) g( )=1. This implies p(A) f(A)+q(A) g(A)=I. In order to prove the existence part, we define vectors y, z by y = q(A) g(A) x and z = p(A) f(A) x.Then f(A) y = f(A) q(A) g(A) x = q(A) f(A) g(A) x =0, g(A) z = g(A) p(A) f(A) x = p(A) f(A) g(A) x =0, and y + z =(p(A) f(A)+q(A) g(A)) x = x. Therefore, the vectors y, z have all the required properties. In order to prove the uniqueness part, it su ces to show that ker f(A) \ ker g(A)= 0 . Assume that x lies in the intersection of ker f(A) and { } ker g(A), so that f(A) x = 0 and g(A) x = 0. This implies p(A) f(A) x =0 and q(A) g(A) x = 0. Adding both equations, we obtain x =(p(A) f(A)+ 12 2. Systems of linear di↵erential equations q(A) g(A)) x = 0. This shows that show that ker f(A) ker g(A)= 0 , as \ { } claimed. ⇤ Corollary 2.10. Let A be a n n matrix, and denote by p ( )=det( I A) ⇥ A the characteristic polynomial of A. Let us write the polynomial pA( ) in the form p ( )=( )⌫1 ( )⌫m , A 1 ··· m where 1,..., m are the distinct eigenvalues of A and ⌫1,...,⌫m denote their respective algebraic multiplicities. Then we have the direct sum decom- position n ⌫1 ⌫m C =ker(A 1I) ... (A mI) . The spaces ker(A I)⌫j are referred to as the generalized eigenspaces of j A.
Proof. For abbreviation, let g ( )=( )⌫j , so that j j p ( )=g ( ) g ( ). A 1 ··· m For each k 1,...,m , the polynomials g1( ) gk 1( ) and gk( ) are 2{ } ··· relatively prime, as they have no roots in common. Consequently,
ker(g1(A) gk 1(A)gk(A)) = ker(g1(A) gk 1(A)) ker gk(A). ··· ··· Repeated application of this result yields the direct sum decomposition ker p (A)=kerg (A) ... ker g (A). A 1 m On the other hand, pA(A) = 0 by the Cayley-Hamilton theorem, so that n n ker pA(A)=C . As a result, we obtain the following decomposition of C into generalized eigenspaces: n C =kerg1(A) ... ker gm(A). ⇤ n n Theorem 2.11. Let A C ⇥ be given. Then we can find matrices L, N 2 2 Cn with the following properties: (i) L + N = A. (ii) LN = NL. (iii) L is diagonalizable. (iv) N is nilpotent, i.e. N n =0. Moreover, the matrices L and N are unique.
Proof. We first prove the existence statement. Consider the decomposition of Cn into generalized eigenspaces:
n ⌫1 ⌫m C =ker(A 1I) ... (A mI) . 2.3. Generalized eigenspaces and the L + N decomposition 13
Consider the linear transformation from Cn into itself that sends a vector x 2 ker(A I)⌫j to x (j =1,...,m). Let L be the n n matrix associated j j ⇥ with this linear transformation. This implies Lx = x for all x ker(A j 2 I)⌫j . Clearly, ker(L I)=ker(A I)⌫j for j =1,...,m. Therefore, j j j there exists a basis of Cn that consists of eigenvectors of L. Consequently, L is diagonalizable. We claim that A and L commute, i.e. LA = AL. Itsu ces to show that LAx = ALx for all vectors x ker(A I)⌫j and all j =1,...,m.Indeed, 2 j if x belongs to the generalized eigenspace ker(A I)⌫j ,thenAx lies in j the same generalized eigenspace. Therefore, Lx = jx and LAx = jAx. Putting these facts together, we obtain LAx = jAx = ALx, as claimed. Therefore, LA = AL. We now put N = A L. Clearly, L + N = A and LN = LA L2 = AL L2 = NL. Hence, it remains to show that N n = 0. As above, it is enough to show that N nx = 0 for all vectors x ker(A I)⌫j and all 2 j j =1,...,m. By definition of L and N,wehaveNx = Ax Lx =(A I)x j for all x ker(A I)⌫j . From this it is easy to see that N nx =(A I)nx. 2 j j However, (A I)nx =0sincex ker(A I)⌫j and ⌫ n.Thus,we j 2 j j conclude that N nx = 0 for all x ker(A I)⌫j . This completes the proof 2 j of the existence part.
We next turn to the proof of the uniqueness statement. Suppose that n n L, N C ⇥ satsify (i) – (iv). We claim that Lx = jx for all vectors 2 x ker(A I)⌫j and all j =1,...,m. To this end, we use the formula 2 j L I =(A I) N.SinceN commutes with A I, it follows that j j j 2n 2n 2n l 2n l (L I) = ( N) (A I) . j l j Xl=0 ✓ ◆ Using the identity N n = 0, we obtain n 1 2n 2n l 2n l (L I) = ( N) (A I) . j l j Xl=0 ✓ ◆ Suppose that x ker(A I)⌫j .Since⌫ n,wehave(A I)2n lx = 2 j j j 0 for all l =0,...,n 1. This implies (L I)2nx = 0. Since L is j diagonalizable, we it follows that (L I)x = 0. Thus, we conclude that j Lx = x for all vectors x ker(A I)⌫j and all j =1,...,m. j 2 j Since n ⌫1 ⌫m C =ker(A 1I) ... (A mI) , there is exactly one matrix L such that Lx = x for x ker(A I)⌫j and j 2 j j =1,...,m. This completes the proof of the uniqueness statement. ⇤ 14 2. Systems of linear di↵erential equations
As an example, let us compute the L + N decomposition of the matrix
023 4 012 4 A = 2 3 . 1 2 46 6 0 1 237 6 7 4 5
We begin by computing the eigenvalues and eigenvectors of A. The charac- teristic polynomial of A is given by
2 34 0 1 24 det( I A)=det2 3 12 +4 6 601 2 37 6 7 4 1 245 2 34 = det 2 +4 6 +det 1 24 2 3 2 3 12 3 12 3 = ( 3 4 )+(3 2 + 1) 5 4 5 = 4 +2 2 +1 =( i)2 ( + i)2.
Thus, the eigenvalues of A are i and i, and they both have algebraic multiplicity 2. A straightforward calculation shows that the generalized eigenspaces are given by
4 4i 6i 2+8i 2 2 2i 2 4i 4+8i ker(A iI) =ker2 3 = span v1,v2 4+2i 2+4i 4+8i 6 12i { } 6 22i 4i 6i 7 6 7 4 5 and
44i 6i 2 8i 2 22i 2+4i 4 8i ker(A + iI) =ker2 3 = span v3,v4 , 4 2i 2 4i 4 8i 6 + 12i { } 6 2 2i 4i 6i 7 6 7 4 5 2.4. Calculating the exponential of a general n n matrix 15 ⇥
where v1,v2,v3,v4 are defined by 1 2+i i 0 v1 = 2 3 ,v2 = 2 3 , 0 2+i 607 6 1 7 6 7 6 7 4 15 4 2 i5 i 0 v3 = 2 3 ,v4 = 2 3 . 0 2 i 6 0 7 6 1 7 6 7 6 7 Thus, we conclude that Lv4 = 5iv , Lv = iv4, Lv =5 iv , Lv = iv . 1 1 2 2 3 3 4 4 Let S be the 4 4-matrix with column vectors v ,v ,v ,v : ⇥ 1 2 3 4 12+i 12i i 0 i 0 S = 2 3 . 0 2+i 0 2 i 60 10 1 7 6 7 The inverse of S is given by4 5 1 i 1 2 2 2 2 i 1 2i 1 00 2 2 S = 2 3 . 1 i 1 2 2 2 2 600 i 1+2i 7 6 2 2 7 Using the identities Lv = iv4 , Lv = iv , Lv =5 iv , Lv = iv ,we 1 1 2 2 3 3 4 4 obtain i 00 0 012 3 0 i 00 1 10 1 4 L = S 2 3 S = 2 3 . 00 i 0 00 25 600 0 i7 6 00 127 6 7 6 7 Consequently, 4 5 4 5 011 1 1110 N = A L = 2 3 . 1 2 21 6 0 1 117 6 7 It is easy to check that LN = NL4and N 2 = 0, as expected.5
2.4. Calculating the exponential of a general n n matrix ⇥ Let now A be an arbitrary n n matrix. By Theorem 2.11, we may write ⇥ A = L + N,whereL is diagonalizable, N is nilpotent, and LN = NL.This gives exp(tA)=exp(tL)exp(tN). 16 2. Systems of linear di↵erential equations
1 Since L is diagonalizable, we may write L = SDS ,whereD is a diagonal matrix. This gives 1 exp(tL)=S exp(tD)S . On the other hand,
n 1 1 tk tk exp(tN)= N k = N k k! k! Xk=0 Xk=0 since N n = 0. Putting these facts together, we obtain
n 1 k 1 t k exp(tA)=S exp(tD)S N . k! Xk=0 Since the exponential of a diagonal matrix is trivial to compute, we thus obtain an explicit formula for exp(tA). As an example, let us compute the matrix exponential of the matrix 023 4 012 4 A = 2 3 . 1 2 46 6 0 1 237 6 7 4 5 We have seen above that 012 3 10 1 4 L = 2 3 00 25 6 00 127 6 7 4 5 and 011 1 1110 N = 2 3 . 1 2 21 6 0 1 117 6 7 Moreover, 4 5 i 00 0 0 i 00 1 L = S 2 3 S , 00 i 0 600 0 i7 6 7 where 4 5 12+i 12i i 0 i 0 S = 2 3 . 0 2+i 0 2 i 60 10 1 7 6 7 4 5 2.5. Solving systems of linear di↵erential equations using matrix exponentials17
This gives eit 00 0 it 0 e 00 1 exp(tL)=S 2 it 3 S . 00e 0 6 00 0e it7 6 7 Furthermore, 4 5 1 tt t t 1+tt 0 exp(tN)=I + tN = 2 3 t 2t 1 2tt 6 0 t t 1+t7 6 7 4 5 since N 2 = 0. Putting these facts together, we obtain eit 00 0 1 tt t it 0 e 00 1 t 1+tt 0 exp(tA)=S 2 3 S 2 3 . 00e it 0 t 2t 1 2tt 6 00 0e it7 6 0 t t 1+t7 6 7 6 7 4 5 4 5 2.5. Solving systems of linear di↵erential equations using matrix exponentials
In this section, we consider a system of linear di↵erential equations of the form n
xi0 (t)= aij xj(t), Xj=1 where x1(t),...,xn(t) are the unkown functions and aij is a given set of real numbers. It is convenient to rewrite this system in matrix form. To that end, we write
a11 a12 ... a1n a21 a22 ... a2n A = 2 . . . . 3 . . .. . 6 7 6a a ... a 7 6 n1 n2 nn7 and 4 5 x1(t) x2(t) x(t)=2 . 3 . . 6 7 6x (t)7 6 n 7 With this understood, the system can4 be rewritten5 as
x0(t)=Ax(t). 18 2. Systems of linear di↵erential equations
Proposition 2.12. The function t exp(tA) is di↵erentiable, and 7! d exp(tA)=A exp(tA). dt Proof. Using the identity
1 1 hk 1 (exp(hA) I)= Ak, h k! Xk=1 we obtain 1 lim (exp(hA) I)=A. h 0 h ! This implies 1 1 lim (exp((t+h)A) exp(tA)) = lim (exp(hA) I)exp(tA)=A exp(tA). h 0 h h 0 h ! ! This proves the claim. ⇤ n n n Theorem 2.13. Given any vector x0 R and any matrix A R ⇥ , 2 2 there exists a unique solution of the di↵erential equation x0(t)=Ax(t) with initial condition x(0) = x0. Moreover, the solution can be expressed as x(t)=exp(tA)x0.
Proof. Let x(t) be a solution of the di↵erential equation x0(t)=Ax(t)with initial condition x(0) = x0.Then d (exp( tA) x(t)) = exp( tA) x0(t) exp( tA) Ax(t)=0. dt Consequently, the function t exp( tA) x(t) is constant. From this, we 7! deduce that exp( tA) x(t)=x ,hencex(t)=exp(tA)x .Thisprovesthe 0 0 uniqueness statement.
Conversely, if we define x(t) by the formula x(t)=exp(tA)x0,then x(0) = x0 and d x0(t)= exp(tA)x = A exp(tA)x = Ax(t). dt 0 0 Thus, x(t) solves the given initial value problem. ⇤
We next consider an inhomogeneous system of the form
x0(t)=Ax(t)+f(t), where f(t) is a given function which takes values in Rn. Theorem 2.14. Let A be an n n matrix, and let f : I Rn be a con- ⇥ ! tinuous function which is defined on an open interval I containing 0. Then 2.5. Solving systems of linear di↵erential equations using matrix exponentials19
there exists a unique solution of the di↵erential equation x0(t)=Ax(t)+f(t) with initial condition x(0) = x0. Moreover, the solution can be expressed as t x(t)=exp(tA) x + exp( sA) f(s) ds . 0 ✓ Z0 ◆
Proof. Let x(t) be a solution of the di↵erential equation x0(t)=Ax(t)+f(t) with initial condition x(0) = x0.Then d (exp( tA) x(t)) = exp( tA) x0(t) exp( tA) Ax(t)=exp( tA) f(t). dt Integrating this identity gives t exp( tA) x(t)=x + exp( sA) f(s) ds, 0 Z0 hence t x(t)=exp(tA) x + exp( sA) f(s) ds . 0 ✓ Z0 ◆ This proves the uniqueness statement. Conversely, if we define x(t) by the formula t x(t)=exp(tA) x + exp( sA) f(s) ds , 0 ✓ Z0 ◆ then x(0) = x0 and t x0(t)=A exp(tA) x + exp( sA) f(s) ds 0 ✓ Z0 ◆ d t +exp(tA) x + exp( sA) f(s) ds dt 0 ✓ Z0 ◆ = Ax(t)+f(t).
Therefore, x(t) is a solution of the given initial value problem. ⇤
As an example, let us find the solution of the inhomogeneous system
1 2 2t 2 x0(t)= x(t)+e 33 3 with initial condition 0 x(0) = . 0 The coe cient matrix 1 2 A = 33 20 2. Systems of linear di↵erential equations has eigenvalues =2+p5 i and =2 p5 i. The associated eigenvectors 1 2 are given by 2 2 v = ,v= . 1 1+p5 i 2 1 p5 i Hence, if we define 2 2 S = , 1+p5 i 1 p5 i then 2+p5 i 0 A = S S 1 02p5 i and (2+p5 i)t e 0 1 exp(tA)=S (2 p5 i)t S . " 0 e # Moreover, we compute
p5 it e 0 1 2 exp( tA) f(t)=S S 0 ep5 it 3 " # 1 e p5 it 0 2+p5 i = S 2p5 i 0 ep5 it 2+p5 i " # 1 e p5 it(2 + p5 i) = S . 2p5 i ep5 it( 2+p5 i) " # This implies
t p5 it 1 (e 1) (2 + p5 i) exp( sA) f(s) ds = S p5 it . 0 10 (e 1) (2 p5 i) Z " # Thus, the solution of the initial value problem is given by t x(t)=exp(tA) exp( sA) f(s) ds Z0 1 e(2+p5 i)t 0 (e p5 it 1) (2 + p5 i) = S 10 0 e(2 p5 i)t (ep5 it 1) (2 p5 i) " #" # 1 (e2t e(2+p5 i)t)(2+p5 i) = S 10 (e2t e(2 p5 i)t)(2 p5 i) " # 1 (e2t e(2+p5 i)t)(4+2p5 i) (e2t e(2 p5 i)t)(4 2p5 i) = 10 (e2t e(2+p5 i)t)(3 3p5 i) (e2t e(2 p5 i)t)(3+3p5 i) " # 1 4e2t +4e2t cos(p5 t) 2p5 e2t sin(p5 t) = . 5 3e2t +3e2t cos(p5 t)+3p5 e2t sin(p5 t) 2.6. Asymptotic behavior of solutions 21
2.6. Asymptotic behavior of solutions
Proposition 2.15. Assume that all eigenvalues of A have negative real part. Then exp(tA) 0 as t . In other words, every solution of the ! !1 di↵erential equation x (t)=Ax(t) converges to 0 as t . 0 !1
Proof. Let 1,..., m denote the eigenvalues of A, and let ⌫1,...,⌫m denote their algebraic multiplicities. Suppose that x ker(A I)⌫j , so that x is 2 j a generalized eigenvector of A.Then
n 1 tk exp(tA)x = et j exp(t(A I))x = et j (A I)kx. j k! j Xk=0 Since Re( j) < 0, it follows that
n 1 tk exp(tA)x = et Re( j ) (A I)kx 0 k k k! j ! k=0 X as t .Thus,limt exp(tA)x =0wheneverx is a generalized eigen- !1 !1 vector of A. Since every vector in Cn can be written as a sum of generalized n eigenvectors, we conclude that limt exp(tA)x = 0 for all x C . From !1 2 this, the assertion follows easily. ⇤ Proposition 2.16. Assume that all eigenvalues of A have negative real part. Moreover, suppose that f : R Rn is a continuous function which ! is periodic with period ⌧. Then there exists a unique solution x¯(t) of the di↵erential equation x0(t)=Ax(t)+f(t) which is periodic with period ⌧. Moreover, if x(t) is another solution of the di↵erential equation x0(t)= Ax(t)+f(t), then x(t) x¯(t) 0 as t . ! !1 Proof. We first observe that the matrix exp( ⌧A) I is invertible. In- deed, if 1 is an eigenvalue of the matrix exp( ⌧A), then 1 is an eigenvalue of the matrix exp( m⌧A) for every integer m. This is impossible since exp( m⌧A) 0 as m . Consequently, the matrix exp( ⌧A) I is ! !1 invertible.
We now define a vectorx ¯0 by ⌧ (exp( ⌧A) I)¯x = exp( sA) f(s) ds. 0 Z0 Moreover, we define
t x¯(t)=exp(tA) x¯ + exp( sA) f(s) ds . 0 ✓ Z0 ◆ 22 2. Systems of linear di↵erential equations
Since the function f(t)isperiodicwithperiod⌧,wehave t+⌧ x¯(t + ⌧)=exp((t + ⌧)A) x¯ + exp( sA) f(s) ds 0 ✓ Z0 ◆ t+⌧ =exp((t + ⌧)A) exp( ⌧A)¯x + exp( sA) f(s) ds 0 ✓ Z⌧ ◆ t =exp((t + ⌧)A) exp( ⌧A)¯x + exp( (s + ⌧)A) f(s + ⌧) ds 0 ✓ Z0 ◆ t =exp(tA) x¯ + exp( sA) f(s + ⌧) ds 0 ✓ Z0 ◆ =¯x(t) for all t. Therefore, the functionx ¯(t)isperiodicwithperiod⌧. Moreover, by Theorem 2.14, the functionx ¯(t) satisfiesx ¯0(t)=Ax¯(t)+f(t) for all t. This proves that there is at least one periodic solution. We next assume that x(t) is an arbitrary solution of the di↵erential equation x (t)=Ax(t)+f(t). Then the di↵erence x(t) x¯(t) satisfies the 0 di↵erential equation d (x(t) x¯(t)) = A(x(t) x¯(t)). dt This implies x(t) x¯(t)=exp(tA)(x(0) x¯(0)), and the right hand side converges to 0 as t . !1 Finally, we show thatx ¯(t) is the only solution which is periodic with period ⌧. To prove this, suppose that x(t) is another solution which is periodic with period ⌧. Then the function x(t) x¯(t)isperiodicwithperiod ⌧.Sincex(t) x¯(t) 0 as t , we conclude that x(t) x¯(t) = 0 for all ! !1 t. This completes the proof. ⇤ In the remainder of this section, we analyze what happens when A has some eigenvalues with negative real part and some eigenvalues with positive real part. As usual, we denote by 1,..., m the eigenvalues of A and by ⌫ ,...,⌫ their algebraic multiplicities. We assume that Re( ) =0 1 m j 6 for j =1,...,m. After rearranging the eigenvalues, we may assume that Re( j) < 0 for j =1,...,l and Re( j) > 0 for j = l +1,...,m. Let us define l ⌫j p ( )= ( j) jY=1 and m p ( )= ( )⌫j . + j j=Yl+1 2.6. Asymptotic behavior of solutions 23
We claim that p ( ) and p+( ) are relatively prime. Suppose by contra- diction that there is a polynomial q( ) of degree at least 1 which divides both p ( ) and p+( ). By the fundamental theorem of algebra, q( ) has at least one root in the complex plane, and this number must be a common root of p ( ) and p+( ). But this impossible since the roots of p ( ) all have positive real part and the roots of p+( ) all have negative real part. Thus, we conclude that p ( ) and p+( ) are relatively prime. Moreover, the product p ( ) p+( ) is the characteristic polynomial of A. Using the Cayley-Hamilton, we conclude that n C =ker(p (A) p+(A)) = ker p (A) p+(A). Let P and P+ denote the canonical projections associated with this direct sum decomposition. The subspaces ker p (A) and ker p+(A) are referred to as the stable and unstable subspaces, and the projections P and P+ are referred to as the spectral projections. We now assume that A has real entries. Since the eigenvalues of a real matrix occur in pairs of complex conjugate numbers, the polynomials p ( ) and p+( ) have real coe cients. Thus, the subspaces ker p (A), ker p+(A) n ⇢ C are invariant under complex conjugation, and the projections P and P+ are matrices with real entries.
Proposition 2.17. Assume that A has no eigenvalues on the imaginary axis. Let us choose ↵>0 small enough such that Re( ) >↵for all | j | eigenvalues of A.Ifx ker p (A), then e↵t exp(tA)x 0 as t . 2 ! !1 Similarly, if x ker p (A), then e ↵t exp(tA)x 0 as t . 2 + ! ! 1 Proof. Suppose first that x ker(A I)⌫j for some j 1,...,l .Then 2 j 2{ } n 1 tk exp(tA)x = et j exp(t(A I))x = et j (A I)kx. j k! j Xk=0 Since Re( ) < ↵, it follows that j n 1 tk e↵t exp(tA)x = et (Re( j )+↵) (A I)kx 0 k k k! j ! k=0 X ↵t ⌫ as t .Thus,limt e exp(tA)x =0wheneverx ker(A jI) j !1 !1 2 and j 1,...,l .Since 2{ } ⌫1 ⌫l ker p (A)=ker(A 1I) ... (A lI) , ↵t we conclude that limt e exp(tA)x = 0 for all x ker p (A). An !1 ↵t 2 analogous argument shows that limt e exp(tA)x = 0 for all x ! 1 2 ker p+(A). ⇤ 24 2. Systems of linear di↵erential equations
Corollary 2.18. Assume that A has no eigenvalues on the imaginary axis. Let us choose ↵>0 small enough such that Re( ) >↵for all eigenvalues | j | of A. Then there exist positive constants N and ↵ such that e↵t exp(tA) P 0 ! as t and !1 ↵t e exp(tA) P 0 + ! as t . ! 1 2.7. Problems
Problem 2.1. Consider the following 3 3 matrix: ⇥ 12 2 A = 01 3 . 2 3 00 2 Find a diagonalizable matrix L such4 that AL5= LA and A L is nilpotent. Problem 2.2. Consider the following 4 4 matrix: ⇥ 1 101 2 110 A = 2 3 . 00 12 600 117 6 7 Find a diagonalizable matrix L4such that AL =5LA and A L is nilpotent. Problem 2.3. Find the general solution of the inhomogeneous di↵erential equation 21 0 x (t)= x(t)+ . 0 4 1 5t n n Problem 2.4. Consider a matrix A C ⇥ . Show that 2 µ : µ is an eigenvalue of exp(tA) = et : is an eigenvalue of A . { } { } This result is called the spectral mapping theorem. (Hint: It su ces to prove this in the special case A is an upper triangular matrix.)
n n Problem 2.5. Consider a matrix A C ⇥ . As usual, let 1,..., m 2 denote the eigenvalues of A and by ⌫1,...,⌫m their algebraic multiplicities. We assume that Re( j) < 0 for j =1,...,k;Re( j) = 0 for j = k +1,...,l; and Re( j) > 0 for j = l +1,...,m. Show that k l n ⌫j x C :limsup exp(tA)x < = ker(A jI) ker(A jI). { 2 t k k 1} !1 Mj=1 j=Mk+1 2.7. Problems 25
n n Problem 2.6. Consider a matrix A R ⇥ . Suppose that all eigenvalues 2 of A have negative real part. Show that the limit ⌧ lim exp(tA) T exp(tA) dt =: S ⌧ !1 Z0 exists. Show that the matrix⇥ S is positive⇤ definite, and 1 Sx, Ax = x 2 h i 2 k k for all x Rn. 2 n n Problem 2.7. Consider a matrix A C ⇥ . Suppose that every eigenvalue 2 k of A satisfies < 1. Show that limk A = 0. | | !1
Chapter 3
Nonlinear systems
3.1. Peano’s existence theorem
The proof of Peano’s theorem relies on a compactness theorem due to Arzel`a and Ascoli:
n Arzel`a-Ascoli Theorem. Let J be a compact interval, and let xk : J R ! be a sequence of continuous functions. Assume that the functions xk(t) are uniformly bounded, so that
sup sup xk(t) < . k N t J k k 1 2 2 Suppose further that the sequence xk is equicontinuous. This means that, given any ">0, there exists a real number >0 so that
sup sup xk(s) xk(t) ". k N s,t J, s t | | 2 2 | | Then a subsequence of the original sequence xk(t) converges uniformly to some limit function x(t).
Proof. Let A = t ,t ,... be a countable dense subset of J.Sincethe { 1 2 } sequence xk(t1) is bounded, the Bolzano-Weierstrass theorem implies that we can find a sequence of integers k : l =1, 2,... going to infinity such { l,1 } that the limit liml xkl,1 (t1) exists. Similarly, since the sequence xk(t2) is bounded, there exists!1 a subsequence k : l =1, 2,... of the sequence { l,2 } kl,1 : l =1, 2,... with the property that the limit liml xk (t1)exists. { } !1 l,1 Proceeding inductively, we obtain for each m N a sequence of integers 2 kl,m : l =1, 2,... with the property that the limit liml xk (tm)exists. { } !1 l,m Moreover, the sequence k : l =1, 2,... is a subsequence of the previous { l,m } sequence kl,m 1 : l =1, 2,... . { } 27 28 3. Nonlinear systems
We now consider the diagonal sequence k : l =1, 2,... . Given any { l,l } integer m N,wehave kl,l : l = m, m+1,... kl,m : l =1, 2,... .Since 2 { }⇢{ } the limit liml xk (tm) exists, we conclude that the limit liml xk (tm) !1 l,m !1 l,l exists as well. To summarize, the limit liml xk (t) exists for each t A. !1 l,l 2 We want to show that the functions xkl,l (t) converge uniformly. Sup- pose that " is a given positive real number. Since the functions xk(t) are equicontinuous, we can find a real number >0 so that
sup sup xk(s) xk(t) ". k N s,t J, s t | | 2 2 | | Moreover, since the set A is dense, we can find a finite subset E A such m ⇢ that J (s , s + ). Finally, we can find a positive integer l such ⇢ s E 0 that 2 S max xk (s) xk (s) " s E | ˜l,˜l l,l | 2 for all ˜l l l . We claim that 0 xk (t) xk (t) 3" | ˜l,˜l l,l | for all t J and ˜l l l . Indeed, for each t J there exists a number 2 0 2 s E such that s t < . This gives 2 | | xk (t) xk (t) | ˜l,˜l l,l | xk (s) xk (s) + xk (s) xk (t) + xk (s) xk (t) 3", | ˜l,˜l l,l | | l,l l,l | | ˜l,˜l ˜l,˜l | as claimed. In particular, for each t J,thesequencex (t), l =1, 2,...,isa 2 kl,l Cauchy sequence. Consequently, the limit x(t)=liml xkl,l (t) exists for each t J. Moreover, !1 2 x(t) x (t) =lim x (t) x (t) 3" kl,l k˜l,˜l kl,l | | ˜l | | !1 for all t J and all l l .Since">0 is arbitrary, the sequence x (t) 2 0 k converges uniformly to x(t). This completes the proof. ⇤ Theorem 3.1. Let U Rn be an open set, and let F : U Rn be a ⇢ ! continuous mapping. Fix a point x U. Then the di↵erential equation 0 2 x0(t)=F (x(t)) with initial condition x(0) = x0 admits a solution x(t), which is defined on an interval ( , ). Here, is a positive real number which depends on x0.
Proof. Since U is open, we can find a positive real number r>0 such that B (x ) U. Let 2r 0 ⇢ M =sup F (x) , x Br(x0) k k 2 r and suppose that is chosen such that 0 < < M . 3.1. Peano’s existence theorem 29
For every positive integer k, we define a continuous function x :[ , ] k ! B (x ) as follows. For t [ , ], we define r 0 2 k k xk(t)=x0 + tF(x0). Suppose now that j 1,...,k 1 , and we have defined x (t) for t 2{ } k 2 [ j , j ]. We then define k k j j j x (t)=x + t F x k k k k k k ⇣ ⌘ ⇣ ⌘ ⇣ ⇣ ⌘⌘ for t [ j , (j+1) ] and 2 k k j j j x (t)=x + t + F x k k k k k k ⇣ ⌘ ⇣ ⌘ ⇣ ⇣ ⌘⌘ for t [ (j+1) , j ]. Using induction on j, one can show that 2 k k x (t) M t It follows immediately from the definition of xk that x (s) x (t) M s t | k k | | | for all s, t [ , ]. Consequently, the functions x (t) are equicontinuous. 2 k By the Arzel`a-Ascoli theorem, there exists a sequence of positive integers k such that the functions x (t) converge uniformly to some limit l !1 kl function x(t) as l . It is clear that x is a continuous function which !1 maps the interval [ , ] into the closed ball of radius r centered at x . 0 We claim that x(t) is a solution of the given initial value problem. To see this, we write t xk(t)=x0 + F (yk(s)) ds, Z0 where the function yk(t)isdefinedby x0 if t [ k , k ] j 2 j (j+1) yk(t)= x if t ( , ] for some j 1,...,k 1 8 k k 2 k k 2{ } <>x j if t [ (j+1) , j ) for some j 1,...,k 1 . k k 2 k k 2{ } It is easy to:> see that M sup xk(t) yk(t) . t [ , ] k k k 2 This implies M sup x(t) ykl (t) sup x(t) xkl (t) + 0 t [ , ] k kt [ , ] k k kl ! 2 2 30 3. Nonlinear systems as l . In other words, the functions y (t) converge uniformly to x(t) as !1 kl l . Since continuous functions on compact sets are uniformly continu- !1 ous, we conclude that the functions F (ykl (t)) converge uniformly to F (x(t)) as l . In particular, !1 x(t)= limxk (t) l l !1 t =lim x0 + F (yk (s)) ds l l !1 ✓ Z0 ◆ t = x0 + F (x(s)) ds. Z0 Putting t = 0 gives x(0) = x0. Moreover, since x(t) is continuous, the func- tion F (x(t)) is continuous as well. Thus, x(t) is continuously di↵erentiable with derivative x0(t)=F (x(t)). This completes the proof. ⇤ It is instructive to consider the special case when x0(t)=Ax(t) for some n n matrix A R ⇥ . In this case, the approximating functions xk(t) satisfy 2 j j x = I + A x k k k 0 and ⇣ ⌘ ⇣ ⌘ j j x = I A x k k k 0 for j 1,...,k . Using⇣ Proposition⌘ ⇣ 2.4, it⌘ is not di cult to show that 2{ } x (t) exp(tA)x as k . k ! 0 !1 3.2. Existence theory via the method of Picard iterates Theorem 3.2. Let U Rn be an open set, and let F : U Rn be a ⇢ ! continuously di↵erentiable mapping. Fix a point x U. Then the di↵er- 0 2 ential equation x0(t)=F (x(t)) with initial condition x(0) = x0 admits a solution x(t), which is defined on an interval ( , ). Here, is a positive real number which depends on x0. Since U is open, we can find a positive real number r>0 such that B (x ) U. Moreover, we define 3r 0 ⇢ M =sup F (x) x B2r(x0) k k 2 and L =sup DF(x) op. x B2r(x0) k k 2 We now choose small enough so that r 1 0 < The strategy is to construct a family of functions xk(t) defined on the interval ( , ) such that x (t)=x and 0 0 t xk+1(t)=x0 + F (xk(s)) ds Z0 for all t ( , ). The functions x (t) are called the Picard iterates. 2 k Lemma 3.3. The function xk(t) is well-defined, and k+1 xk(t) xk 1(t) 2 r k k for all t ( , ). 2 Proof. We argue by induction on k. For k = 1, the assertion is trivial. We now assume that the assertion has been established for all integers less than or equal to k. In other words, x (t) is defined for all j k, and we have j j+1 xj(t) xj 1(t) 2 r k k for all t ( , ) and all j k. Summation over j gives 2 k k j+1 max xk(t) x0 , xk 1(t) x0 xj(t) xj 1(t) 2 r<2r {k k k k} k k Xj=1 Xj=1 for all t ( , ). Consequently, xk(t),xk 1(t) B2r(x0) U. Therefore, 2 2 ⇢ xk+1(t) can be defined. By definition of L,wehave F (⇠) F (⌘) L ⇠ ⌘ whenever ⇠,⌘ k k k k 2 B2r(x0). Since xk(t),xk 1(t) B2r(x0) for all t ( , ), we conclude that 2 2 k+1 F (xk(t)) F (xk 1(t)) L xk(t) xk 1(t) 2 rL k k k k for all t ( , ). Using the identities 2 t xk+1(t)=x0 + F (xk(s)) ds Z0 and t xk(t)=x0 + F (xk 1(s)) ds, Z0 we obtain t xk+1(t) xk(t)= (F (xk(s)) F (xk 1(s))) ds. Z0 This implies k+1 k+1 k x (t) x (t) 2 rL t 2 rL 2 r k k+1 k k | | for all t ( , ). 2 ⇤ 32 3. Nonlinear systems Using Lemma 3.3, we obtain k˜ k˜ j+1 k+1 x˜(t) xk(t) xj(t) xj 1(t) 2 r 2 r k k k k k j=Xk+1 j=Xk+1 for all k˜ k 1 and all t ( , ). Thus, for each t ( , ), the sequence 2 2 xk(t) is a Cauchy sequence. Let us define x(t):=lim xk(t). k !1 Then k+1 x(t) xk(t) =lim xk˜(t) xk(t) 2 r k k k˜ k k !1 for all t ( , ). Thus, the sequence x (t) converges uniformly to x(t). In 2 k particular, the limit function x(t) is continuous. Moreover, x(t)= lim xk+1(t) k !1 t =lim x0 + F (xk(s)) ds k !1 ✓ Z0 ◆ t = x0 + F (x(s)) ds Z0 for all t ( , ). Putting t = 0 gives x(0) = x . Moreover, since x(t)is 2 0 continuous, the function F (x(t)) is continuous as well. Thus, x(t) is con- tinuously di↵erentiable with derivative x0(t)=F (x(t)). This completes the proof of Theorem 3.2. As an example, let us consider the di↵erential equation x0(t)=Ax(t) where A is an n n matrix. In this case, the Picard iterates satisfy ⇥ t xk+1(t)=x0 + Axk(s) ds. Z0 Using induction on k, one can show that k tl x (t)= Alx . k l! 0 Xl=0 Thus, x (t) exp(tA)x as expected. k ! 0 3.3. Uniqueness and the maximal time interval of existence Theorem 3.4. Let U Rn be an open set, and let F : U Rn be a ⇢ ! continuously di↵erentiable mapping. Suppose that x(t) and y(t) are two solutions of the di↵erential equation x0(t)=F (x(t)). Moreover, suppose that x(t) is defined on some open interval I and y(t) is defined on some 3.3. Uniqueness and the maximal time interval of existence 33 open interval J.If0 I J and x(0) = y(0), then x(t)=y(t) for all 2 \ t I J. 2 \ Proof. We first show that x(t)=y(t) for all t I J (0, ). Suppose 2 \ \ 1 that this false. Let ⌧ =inft I J (0, ):x(t) = y(t) . Clearly, { 2 \ \ 1 6 } x(⌧)=y(⌧). For abbreviation, letx ¯ := x(⌧)=y(⌧). Let us fix a real number r>0 such that B (¯x) U. We can find a real number >0such 2r ⇢ that [⌧,⌧ + ] I J and x(t),y(t) B (¯x) for all t [⌧,⌧ + ]. Hence, if ⇢ \ 2 r 2 we put L =supx Br(¯x) DF(x) op, then we obtain 2 k k F (x(t)) F (y(t)) L x(t) y(t) k k k k for all t [⌧,⌧ + ]. This implies 2 d 2 x(t) y(t) =2 x(t) y(t),x0(t) y0(t) dtk k h i =2 x(t) y(t),F(x(t)) F (y(t)) h i 2L x(t) y(t) 2 k k for t [⌧,⌧ + ]. Consequently, the function t e 2Lt x(t) y(t) 2 is 2 7! k k monotone decreasing on the interval [⌧,⌧ + ]. Since x(⌧)=y(⌧), it follows that x(t)=y(t) for all t [⌧,⌧ + ]. This contradicts the definition of ⌧. 2 Thus, x(t)=y(t) for all t I J (0, ). An analogous argument shows 2 \ \ 1 that x(t)=y(t) for all t I J ( , 0). 2 \ \ 1 ⇤ In the following, we assume that U Rn is an open set, and F : U Rn ⇢ ! is continuously di↵erentiable. Theorem 3.4 guarantees that, given any point x U, there exists a unique maximal solution of the initial value problem 0 2 x (t)=F (x(t)) (4) 0 (x(0) = x0. To see this, we fix a point x U and define 0 2 ? J = ⌧ R : there exists a solution of (4) which is x0 { 2 defined on some open interval containing ⌧ . } ? In other words, Jx0 is the union of the domains of definition of all solutions ? of the initial value problem (4). Clearly, Jx0 is an open interval. Using The- ? orem 3.1 or Theorem 3.2, we conclude that Jx0 is non-empty. Moreover, by Theorem 3.4, two solutions of (4) agree on the intersection of their domains of definition. Hence, there exists a unique function x : J ? U which solves x0 ! (4). ? We next characterize the maximal interval Jx0 on which the solution is defined. 34 3. Nonlinear systems Theorem 3.5. Assume that U Rn is an open set, and F : U Rn is ⇢ ! continuously di↵erentiable. Fix a point x0 U, and let x(t) be the unique 2 ? maximal solution of the initial value problem (4), and let Jx0 =(↵, ) denote its domain of definition. If < , then 1 n 1 lim sup min dist(x(t), R U), =0. t \ x(t) % n k ko Proof. We argue by contradiction. Suppose that < and 1 n 1 inf min dist(x(tk), R U), > 0 k N \ x(tk) 2 n k ko for some sequence of times tk .Thenthesequence x(tk):k N is % { 2 } contained in a compact subset of U. Using either Theorem 3.1 or Theorem 3.2, we can find a uniform constant >0 with the property that ( , ) J ? for all k. It is important that ⇢ x(tk) the constant does not depend on k; this is possible since the sequence x(tk):k N is contained in a compact subset of U. Consequently, { 2 } [0,t + ) J ? for each k.Thisimpliest + for all k. This contradicts k ⇢ x0 k the fact that limk tk = . This completes the proof of Theorem 3.5. ⇤ !1 We now define n ? ⌦= (x0,t) U R : t J . { 2 ⇥ 2 x0 } Moreover, for each point (x ,t) ⌦, we define 0 2 (x0,t)='t(x0)=x(t), where x : J ? U denotes the unique maximal solution of (4). The mapping x0 ! :⌦ U is referred as the flow associated with the di↵erential equation ! x0(t)=F (x(t)). The map enjoys the following semigroup property: If (x ,s) ⌦and ( (x ,s),t) ⌦, then (x ,s+ t) ⌦and (x ,s+ t)= 0 2 0 2 0 2 0 ( (x0,s),t). This is an easy consequence of the uniqueness theorem above. 3.4. Continuous dependence on the initial data As above let U Rn R, and let F : U Rn be a continuous mapping ⇢ ⇥ ! which is continuously di↵erentiable in the spatial variables. We denote by the flow associated with this di↵erential equation, and by ⌦ U R the ⇢ ⇥ domain of definition of . Theorem 3.6. The set ⌦ is open, and :⌦ U is continuous. ! In order to prove this theorem, we consider an arbitrary point (x ,t ) 0 0 2 ⌦. Our goal is to show that a ⌦contains a ball of positive radius centered at (x0,t0) and that is continuous at (x0,t0). 3.4. Continuous dependence on the initial data 35 It su ces to consider the case t 0. (The case t < 0 can be handled 0 0 in an analogous fashion.) Let C = (x ,t):0 t t . { 0 0} The set C is closed and bounded subset of U.SinceU is open, the set C has positive distance from the boundary of U. Hence, we can find a positive real number r such that n 0 < 4r To summarize, we have shown that ⌦contains a neighborhood of the point (x0,t0), and is continuous at the point (x0,t0). 3.5. Di↵erentiability of flows and the linearized equation As usual, we consider an autonomous system of ordinary di↵erential equa- n tions of the form x0(t)=F (x(t)), where F : U R is a continuously ! di↵erentiable function defined on some open subset U of Rn. We denote by the flow associated with this di↵erential equation, and by ⌦ U R the ⇢ ⇥ domain of definition of . We have shown earlier that the set ⌦is open and the map :⌦ U is continuous. ! We next establish di↵erentiability of . To that end, we need the fol- lowing auxiliary result: n n Proposition 3.10. Let A :[0,T] R ⇥ be a continuous function which ! takes values values in the space of n n matrices. Then there exists a ⇥ n n continuously di↵erentiable function M :[0,T] R ⇥ such that M 0(t)= ! A(t)M(t) and M(0) = I. Proof. For abbreviation, let N =supt [0,T ] A(t) op. For each k N,we 2 nk n k 2 define a continuous function Mk :[0,T] R ⇥ by ! Mk(t)=I + tA(0) for t [0, 1 ] and 2 k jT jT jT M (t)= I + t A M k k k k k h ⇣ ⌘ ⇣ ⌘i ⇣ ⌘ for t [ jT , (j+1)T ]. Using induction on j, we can show that 2 k k M (t) eNt k k kop for all t [0, jT ]. By the Arzel`a-Ascoli theorem, a subsequence of the se- 2 k quence Mk(t) converges uniformly to some function M(t). The function n n M :[0,T] R ⇥ is clearly continuous. Moreover, by following the argu- ! ments in the proof of Theorem 3.1, one readily verifies that t M(t)=I + A(s)M(s) ds Z0 38 3. Nonlinear systems for all t [0,T]. Thus, M(t) is continuously di↵erentiable and M (t)= 2 0 A(t)M(t). ⇤ Theorem 3.11. Fix a point (x ,t ) ⌦. Let x(t)= (x ,t) denote the 0 0 2 0 unique maximal solution with initial vector x0, and let A(t)=DF(x(t)). n n Moreover, suppose that M :[0,t0] R ⇥ is a solution of the ODE M 0(t)= ! A(t)M(t) with initial condition M(0) = I. Then the map 't0 is di↵erentiable at x0, and its di↵erential is given by D't0 = M(t0). In the following, we describe the proof of Theorem 3.11. We need to show that 1 lim 't (x0) 't (y) M(t0)(x0 y) =0. y 0 x y k 0 0 k ! k 0 k To verify this, let ">0 be given. Since F is continuously di↵erentiable, we can find a real number r>0 such that for each t [0,t ]wehave 2 0 B2r(x(t)) U and supy Br(x(t)) DF(y) A(t) op ". Integrating this ⇢ 2 k k inequality over a line segment, we obtain F (x(t)) F (y) A(t)(x(t) y) " y x(t) k k k k for all t [0,t ] and all points y B (x(t)). 2 0 2 r Since is continuous, we can find a real number 0 < sup (x0,t) (y0,t) r t [0,t0] k k 2 for all points y satisfying x y . 0 k 0 0k We now consider a point y with 0 < x y . For abbreviation, 0 k 0 0k let K := supt [0,t0] A(t) and L =supt [0,t0] M(t) . The function y(t)= 2 k k 2 k k (y ,t) satisfies y (t)=F (y(t)). Hence, the function u(t):=x(t) y(t) 0 0 M(t)(x y ) satisfies 0 0 u0(t)=x0(t) y0(t) M 0(t)(x y ) 0 0 = F (x(t)) F (y(t)) A(t)M(t)(x y ) 0 0 = F (x(t)) F (y(t)) A(t)(x(t) y(t)) + A(t)u(t). This implies u0(t) F (x(t)) F (y(t)) A(t)(x(t) y(t)) + A(t) u(t) k kk k k kk k " x(t) y(t) + K u(t) k k k k " M(t)(x y ) +(K + ") u(t) k 0 0 k k k "L x y +(K + ") u(t) k 0 0k k k for all t [0,t ]. 2 0 3.6. Liouville’s theorem 39 Lemma 3.12. We have (K+")t0 e u(t ) t "L x y . k 0 k 0 k 0 0k Proof. Note that u(0) = 0 by definition of u(t). If u(t0) = 0, the assertion is trivial. Hence, we may assume that u(t ) = 0. Let ⌧ =sup t [0,t ]: 0 6 { 2 0 u(t)=0. Clearly, ⌧ [0,t ), u(⌧) = 0, and u(⌧) = 0 for all t (⌧,t ]. } 2 0 6 2 0 Note that d u(t) u0(t) "L x y +(K + ") u(t) dtk kk k k 0 0k k k for all t (⌧,t ]. This implies 2 0 d (K+")t (K+")t (e u(t) ) "L e x y "L x y dt k k k 0 0k k 0 0k for all t (⌧,t ]. Consequently, the function t e (K+")t u(t) t"L x 2 0 7! k k k 0 y is monotone decreasing on the interval (⌧,t ]. Since u(⌧) = 0, it follows 0k 0 that (K+")t0 (K+")t0 e u(t ) t "L x y e u(⌧) ⌧"L x y 0, k 0 k 0 k 0 0k k k k 0 0k which establishes the claim. ⇤ We now complete the proof of Theorem 3.11. Using Lemma 3.12, we obtain 1 1 ' (x ) ' (y ) M(t )(x y ) = u(t ) x y k t0 0 t0 0 0 0 0 k x y k 0 k k 0 0k k 0 0k t "L e(K+")t0 0 for all points y satisfying 0 < x y .Since">0 is arbitrary, we 0 k 0 0k conclude that 1 lim 't (x0) 't (y) M(t0)(x0 y) =0. y 0 x y k 0 0 k ! k 0 k This completes the proof. 3.6. Liouville’s theorem Proposition 3.13. Let M(t) be a continuously di↵erentiable function taking n n values in R ⇥ .IfM 0(t)=A(t)M(t), then t tr(A(s)) ds det M(t)=detM(0) e 0 . R Proof. It su ces to show that d det M(t)=tr(A(t)) det M(t) dt 1 for all t. In order to verify this, we fix a time t0. Let us write A(t0)=SBS , n n n n where S C ⇥ is invertible and B C ⇥ is an upper triangular matrix. 2 2 40 3. Nonlinear systems Moreover, let M˜ (t)=(I +(t t ) A(t ))M(t ). Then M˜ (t )=M(t ) and 0 0 0 0 0 M˜ 0(t0)=M 0(t0). This implies d d det M(t) = det M˜ (t) . dt t=t0 dt t=t0 Moreover, det M˜ (t)=det(I +(t t ) A(t )) det M(t ) 0 0 0 =det(I +(t t ) B)detM(t ) 0 0 n = (1 + (t t ) b )detM(t ). 0 kk 0 kY=1 This implies n d det M˜ (t) = bkk det M(t0) dt t=t0 k=1 X =tr(B)detM(t0) =tr(A(t0)) det M(t0). Putting these facts together, we conclude that d det M(t) =tr(A(t0)) det M(t0), dt t=t0 as claimed. ⇤ Theorem 3.14. Suppose that DF(x) is trace-free for all x U. Then ' 2 t is volume-preserving for each t;thatis,det D' (x)=1for all (x, t) ⌦. t 2 Proof. Fix a point (x ,t ) ⌦. For abbreviation, let x(t)= (x ,t) and 0 0 2 0 A(t)=DF(x(t)). Then D't0 (x0)=M(t0), where M(t) is a solution of the di↵erential equation M 0(t)=A(t)M(t) with initial condition M(0) = I. Since A(t) is tracefree, we conclude that det M(t) is constant in t.Thus, det D't0 (x0)=detM(t0)=detM(0) = 1. ⇤ 3.7. Problems Problem 3.1. Let F be a continuously di↵erentiable vector field on Rn. Moreover, suppose that F (x) ( x ), where :[0, ) (0, )isa k k1 k k 1 ! n 1 continuous function satisfying 1 dr = . Fix a point x0 R , and let 0 (r) 1 2 x(t) denote the unique solution of the di↵erential equation x (t)=F (x(t)) R 0 with initial condition x(0) = x0. Show that x(t) is defined for all t R. 2 Problem 3.2. Let us define a continuous function F : R R by F (x)= ! x . Show that the di↵erential equation x (t)=F (x(t)) with initial con- | | 0 dition x(0) = 0 has infinitely many solutions. p 3.7. Problems 41 Problem 3.3. Let U Rn Rm be an open set, and let F : U Rn ⇢ ⇥ ! be a continuously di↵erentiable mapping. Given any point (x , ) U,we 0 0 2 define (x0, 0,t)=x(t), where x(t) is the unique solution of the di↵erential equation x0(t)=F (x(t), 0) with initial condition x(0) = x0. Moreover, let ⌦be the set of all triplets (x0, 0,t) U R for which (x0, 0,t)isdefined. 2 ⇥ Show that the set ⌦is open and the map :⌦ Rn is continuous. (Hint: ! Apply Theorem 3.6 to the system x0(t)=F (x(t), (t)), 0(t) = 0.) Chapter 4 Analysis of equilibrium points 4.1. Stability of equilibrium points Definition 4.1. Suppose thatx ¯ is an equilibrium point of the autonomous system x0(t)=F (x(t)), so that F (¯x) = 0. We say thatx ¯ is stable if, given any real number ">0 we can find a real number >0 such that ' (x) t 2 B (¯x) for all x B (¯x) and all t 0. We say thatx ¯ is asymptotically " 2 stable ifx ¯ is stable and limt 't(x)=¯x for all points x in some open !1 neighorhood ofx ¯. We note that the condition that limt 't(x)=¯x for all points x in !1 some open neighorhood ofx ¯ does not imply stability (cf. Problem 4.3 below). The following result gives a su cient condition for an equilibrium point to be asymptotically stable: Theorem 4.2. Suppose that 0 is an equilibrium point of the system x0(t)= F (x(t)). Moreover, suppose that the eigenvalues of the matrix A = DF(0) all have negative real part. Then 0 is asymptotically stable. Proof. Suppose that ">0 is given. Since the eigenvalues of A have negative real part, we have limt exp(tA) = 0. Let us fix a real number T>0such !1 that exp(TA) < 1 . By Theorem 3.6, we can find a real number ⌘>0 k kop 2 such that supt [0,T ] 't(y) <"for all points y B⌘(0). Moreover, by 2 k k 2 Theorem 3.11, we have D' (0) = exp(TA). Since exp(TA) < 1 ,we T k kop 2 can find a real number (0,⌘) such that ' (y) 1 y for all points 2 k T k 2 k k y B (0). 2 43 44 4. Analysis of equilibrium points We now consider an initial point y B (0). Using induction on k,we 0 2 can show that the solution ' (y ) is defined on [0,kT], and ' (y ) t 0 k kT 0 k 2 k .Since ' (y ) <⌘, we conclude that k kT 0 k sup 't(y0) =sup 't('kT (y0)) <" t [kT,(k+1)T ] k k t [0,T ] k k 2 2 for all k. Therefore, ' (y ) B (0) for all t 0. This shows that 0 is a t 0 2 " stable equilibrium point. Finally, since ' (y ) 0 as k , Theorem kT 0 ! !1 3.6 implies that sup 't(y0) =sup 't('kT (y0)) 0 t [kT,(k+1)T ] k k t [0,T ] k k! 2 2 as k . This completes the proof. !1 ⇤ 4.2. The stable manifold theorem In this section, we give a precise description of the qualitative behavior of a dynamical system near an equilibrium point. Our discussion loosely follows the one in [6]. We consider an open set U Rn and a map F : U Rn ⇢ ! of class C1. Let us assume that 0 is a hyperbolic equilibrium point for the di↵erential equation x (t)=F (x(t)). That is, we have 0 U, F (0) = 0, 0 2 and all eigenvalues of the matrix A = DF(0) have non-zero real part. For abbreviation, let G(x)=F (x) Ax. Clearly, DG(0) = 0. Let U =kerp (A) and U+ =kerp+(A) denote the stable and unstable subspaces of A, and let P and P+ denote the associated projections. Since U and U+ are invariant under A, the projections P and P+ commute with A. We can find positive constants ↵ and ⇤such that ↵t exp(tA) P op ⇤ e k k for all t 0 and exp(tA) P ⇤ e↵t k +kop for all t 0. Suppose that r>0 is su ciently small such that B (0) U 2Kr ⇢ and 3↵ sup DF(x) DF(0) op . x 2⇤r k k 16⇤ k k Since DG(x)=DF(x) A = DF(x) DF(0), it follows that 3↵ sup DG(x) op . x 2⇤r k k 16⇤ k k 4.2. The stable manifold theorem 45 Lemma 4.3. Given any vector x U with x r, there exists a 2↵t k k function x(t) such that x(t) 2⇤ e 2 x and k k k k t x(t)=exp(tA)x + exp((t s)A) P G(x(s)) ds Z0 (5) 1 exp( (s t)A) P G(x(s)) ds + Zt for all t 0. Proof. The strategy is to use an iterative method. We define a sequence of functions xk(t)by x0(t)=0 and t xk(t)=exp(tA)x + exp((t s)A) P G(xk 1(s)) ds Z0 1 exp( (s t)A) P+G(xk 1(s)) ds. Zt We claim that the function xk(t) is well-defined, and k+1 ↵t xk(t) xk 1(t) 2 ⇤ e 2 x k k k k for all t 0. Since x1(t)=exp(tA) x and x0(t) = 0, we have x1(t) x0(t) = ↵t k k exp(tA) x = exp(tA) P x ⇤ e x for all t 0. Therefore, k k k k k k the assertion holds for k = 1. We next assume that k 1 and the assertion holds for all integers less than or equal to k. In other words, the function xj(t) is well-defined and j+1 ↵t xj(t) xj 1(t) 2 ⇤ e 2 x k k k k for all j k. Summation over j gives k max xk(t) , xk 1(t) xj(t) xj 1(t) {k k k k} k k Xj=1 k j+1 ↵t 2 ⇤ e 2 x k k Xj=1 ↵t 2⇤ e 2 x 2⇤r k k for all t 0. In particular, the function xk+1(t) can be defined. Moreover, 3↵ G(⇠) G(⌘) ⇠ ⌘ whenever ⇠,⌘ B2⇤r(0). Since xk(t),xk 1(t) k k 16⇤ k k 2 2 B2⇤r(0), we obtain 3↵ 3↵ k ↵t G(xk(t)) G(xk 1(t)) xk(t) xk 1(t) 2 e 2 x k k16⇤ k k 8 k k 46 4. Analysis of equilibrium points for all t 0. By definition of x (t) and x (t), we have k+1 k t xk+1(t) xk(t)= exp((t s)A) P (G(xk(s)) G(xk 1(s))) ds Z0 1 exp((t s)A) P+ (G(xk(s)) G(xk 1(s))) ds, Zt hence t ↵(t s) xk+1(t) xk(t) ⇤ e G(xk(s)) G(xk 1(s)) ds k k k k Z0 1 ↵(s t) + ⇤ e G(xk(s)) G(xk 1(s)) ds k k Zt for all t 0. Using our estimate for G(xk(s)) G(xk 1(s)) , we obtain k k t 3↵ k ↵(t s) ↵s xk+1(t) xk(t) 2 ⇤ e e 2 x ds k k 8 k k Z0 1 3↵ k ↵(s t) ↵s + 2 ⇤ e e 2 x ds t 8 k k Z t ↵(t s) ↵t 3↵ k 2 ⇤ e 2 e 2 x ds 8 k k Z0 3↵(s t) ↵t 1 3↵ k + 2 ⇤ e 2 e 2 x ds 8 k k Zt k ↵t 2 ⇤ e 2 x k k for all t 0. This completes the proof of the claim. We now continue with the proof of Lemma 4.3. The functions xk(t) satisfy k ↵t xk+1(t) xk(t) 2 ⇤ e 2 x k k k k for all t 0 and all k 0. In particular, for every t 0thesequence xk(t):k N is a Cauchy sequence. Hence, we may define a function x(t) { 2 } by x(t)= lim xk(t). k !1 It is easy to see that 1 k+1 ↵t x(t) xk(t) xj+1(t) xj(t) 2 ⇤ e 2 x k k k k k k Xj=k 4.2. The stable manifold theorem 47 for all t 0 and k 0. Hence, the functions x (t) converge uniformly to k x(t). As a consequence, the function x(t) is continuous and satisfies t x(t)=exp(tA)x + exp((t s)A) P G(x(s)) ds Z0 1 exp( (s t)A) P G(x(s)) ds + Zt for all t 0. This completes the proof of Lemma 4.3. ⇤ In the next step, we show that the function x(t) is unique. In fact, we prove a stronger statement: Lemma 4.4. Consider two vectors x ,y U with x , y r. 2 k k k k Moreover, suppose that x(t) and y(t) are two functions satisfying x(t) , y(t) k k k k 2⇤r and t x(t)=exp(tA)x + exp((t s)A) P G(x(s)) ds Z0 1 exp( (s t)A) P G(x(s)) ds + Zt and t y(t)=exp(tA)x + exp((t s)A) P G(x(s)) ds Z0 1 exp( (s t)A) P G(x(s)) ds + Zt for all t 0. Then x(t) y(t) 2⇤ x y . k k k k Proof. We compute t x(t) y(t)=exp(tA)(x y )+ exp((t s)A) P (G(x(s)) G(y(s))) ds Z0 1 exp((t s)A) P (G(x(s)) G(y(s))) ds + Zt 48 4. Analysis of equilibrium points for all t 0. This implies t ↵t ↵(t s) x(t) y(t) ⇤ e x y + ⇤ e G(x(s)) G(y(s)) ds k k k k k k Z0 1 3↵ ↵(s t) + e G(x(s)) G(y(s)) ds t 16 k k Z t ↵t 3↵ ↵(t s) ⇤ e x y + e x(s) y(s) ds k k 16 k k Z0 1 3↵ ↵(s t) + e x(s) y(s) ds 16 k k Zt 3 ⇤ x y + sup x(s) y(s) k k 8 s 0 k k for all t 0. Thus, we conclude that 3 sup x(t) y(t) ⇤ x y + sup x(t) y(t) , t 0 k k k k 8 t 0 k k hence sup x(t) y(t) 2⇤ x y . t 0 k k k k This proves the assertion. ⇤ We define a function from U to U+ as follows: for every vector x U with x r,wedefine (x )=P+x(0) U+,wherex(t)isthe 2 k k 2 unique solution of the integral equation (5). It follows from Lemma 4.4 that (x ) is well-defined and continuous. We next show that is di↵erentiable: Lemma 4.5. Fix a point x U with x r, and let x(t) denote the 2 k k solution of the integral equation (5). Then there exists a bounded continuous n n function M :[0, ) R ⇥ such that 1 ! t M(t)=exp(tA)P + exp((t s)A) P DG(x(s)) M(s) ds Z0 (6) 1 exp( (s t)A) P DG(x(s)) M(s) ds. + Zt Moreover, the function is di↵erentiable at x , and the di↵erential D (x ): U U+ is given by D (x )=P+M(0) U . ! | Proof. The existence of a solution to the integral equation (6) follows from an iteration procedure similar to the one used in the proof of Lemma 4.3. We omit the details. We claim that is di↵erentiable at x with di↵erential D (x )= P+M(0) U .Toprovethis,let">0 be given. We can find a real number | >0 such that DG(⇠) DG(⌘) " for all points ⇠,⌘ B (0) k kop 2 2⇤r satisfying ⇠ ⌘ 2⇤ . Integrating this inequality over a line segment, k k 4.2. The stable manifold theorem 49 we conclude that G(⇠) G(⌘) DG(⇠)(⇠ ⌘) " ⇠ ⌘ whenever k k k k ⇠ ⌘ 2⇤ . k k We now consider a point y U with y r and x y . 2 k k k k We define u(t):=x(t) y(t) M(t)(x y ), where x(t) and y(t) denote the solutions of our integral equation associated with x and y .Then t x(t) y(t)=exp(tA)(x y )+ exp((t s)A) P (G(x(s)) G(y(s))) ds Z0 1 exp( (s t)A) P (G(x(s)) G(y(s))) ds, + Zt hence t u(t)= exp((t s)A) P [G(x(s)) G(y(s)) DG(x(s) M(s)(x y )] ds Z0 1 exp( (s t)A) P+ [G(x(s)) G(y(s)) DG(x(s)) M(s)(x y )] ds Zt for all t 0. We have shown above that x(t) y(t) 2⇤ x y 2⇤ k k k k for all t 0. This implies G(x(t)) G(y(t)) DG(x(t)) M(t)(x y ) k k G(x(t)) G(y(t)) DG(x(t)) (x(t) y(t)) + DG(x(t)) u(t) k k k k 3↵ " x(t) y(t) + u(t) k k 16⇤ k k 3↵ 2⇤" x y + u(t) k k 16⇤ k k for all t 0. Putting these facts together, we obtain t 3↵ u(t) exp((t s)A) P op 2⇤" x y + u(s) ds k k 0 k k k k 16⇤ k k Z ⇣ ⌘ 1 3↵ + exp( (s t)A) P op 2⇤" x y + u(s) ds t k k k k 16⇤ k k Z t ⇣ ⌘ ↵(t s) 3↵ ⇤ e 2⇤" x y + u(s) ds 0 k k 16⇤ k k Z ⇣ ⌘ 1 ↵(s t) 3↵ + ⇤ e 2⇤" x y + u(s) ds t k k 16⇤ k k Z ⇣ ⌘ 4⇤2" 3 x y + sup u(s) ↵ k k 8 s 0 k k for all t 0. Thus, we conclude that 4⇤2" 3 sup u(t) x y + sup u(t) , t 0 k k ↵ k k 8 t 0 k k 50 4. Analysis of equilibrium points hence 8⇤2" sup u(t) x y . t 0 k k ↵ k k In particular, 8⇤2" u(0) x y . k k ↵ k k This implies 8⇤3" (x ) (y ) P+M(0) (x y ) = P+u(0) x y . k k k k ↵ k k Since ">0 is arbitrary, we conclude that is di↵erentiable at x and D (x )=P+M(0) U . This completes the proof. ⇤ | Corollary 4.6. We have D (0) = 0. Proof. If x = 0, then x(t) = 0 for all t 0. This implies M(t)= exp(tA) P ,henceD (0) = P+M(0) U = 0. ⇤ | Lemma 4.7. The di↵erential D (x ) depends continuously on the point x . Proof. Fix a point x U with x r, and let ">0 be given. As 2 k k above, we can find a real number >0 such that DG(⇠) DG(⌘) " k kop for all points ⇠,⌘ B (0) satisfying ⇠ ⌘ 2⇤ . 2 2⇤r k k We now consider a point y U with y r and x y . n n 2 k n kn k k Let M :[0, ) R ⇥ and N :[0, ) R ⇥ be bounded continuous 1 ! 1 ! functions satisfying the integral equations t M(t)=exp(tA)P + exp((t s)A) P DG(x(s)) M(s) ds Z0 1 exp( (s t)A) P DG(x(s)) M(s) ds + Zt and t N(t)=exp(tA)P + exp((t s)A) P DG(y(s)) N(s) ds Z0 1 exp( (s t)A) P DG(y(s)) N(s) ds. + Zt Then t M(t) N(t)= exp((t s)A) P [DG(x(s)) M(s) DG(y(s)) N(s)] ds Z0 1 exp( (s t)A) P [DG(x(s)) M(s) DG(y(s)) N(s)] ds. + Zt Note that x(t) y(t) 2⇤ x y 2⇤ k k k k 4.2. The stable manifold theorem 51 for all t 0. This implies DG(x(t)) M(t) DG(y(t)) N(t) k kop (DG(x(t)) DG(y(t))) M(t) + DG(y(t)) (M(t) N(t)) k kop k kop 3↵ " M(t) + M(t) N(t) . k kop 16⇤ k kop This gives M(t) N(t) k kop t 3↵ exp((t s)A) P op " M(s) op + M(s) N(s) op ds 0 k k k k 16⇤ k k Z ⇣ ⌘ 1 3↵ + exp( (s t)A) P+ op " M(s) op + M(s) N(s) op ds t k k k k 16⇤ k k Z t ⇣ ⌘ ↵(t s) 3↵ ⇤ e " M(s) op + M(s) N(s) op ds 0 k k 16⇤ k k Z ⇣ ⌘ 1 ↵(s t) 3↵ + ⇤ e " M(s) + M(s) N(s) ds k kop 16⇤ k kop Zt 2⇤" ⇣ 3 ⌘ sup M(s) op + sup M(s) N(s) op ↵ s 0 k k 8 s 0 k k for all t 0. Thus, 2⇤" 3 sup M(t) N(t) op sup M(t) op + sup M(t) N(t) op, t 0 k k ↵ t 0 k k 8 s 0 k k hence 4⇤" sup M(t) N(t) op sup M(t) op. t 0 k k ↵ t 0 k k Putting t = 0, we obtain 4⇤2" P+M(0) P+N(0) op sup M(t) op. k k ↵ t 0 k k Since ">0 is arbitrary and supt 0 M(t) op < , the assertion follows. ⇤ k k 1 Let W s denote the graph of , so that W s = x + (x ):x U , x r . { 2 k k } We have shown that W s is a C1 manifold which is tangential to the stable subspace U at the origin. We next show that any solution starting in W s converges to the origin at an exponential rate: Lemma 4.8. Consider a vector x U with x r. Moreover, let 2 k k x0 = (x )+x . Then the unique solution of the initial value problem 52 4. Analysis of equilibrium points x (t)=F (x(t)), x(0) = x , is defined for all t 0 and satisfies 0 0 ↵t sup e 2 x(t) < . t 0 k k 1 In other words, the solution x(t) converges to the origin at an exponential rate. Proof. By Lemma 4.3 and Lemma 4.4, there exists a unique function x(t) ↵t such that x(t) 2⇤ e 2 x and k k k k t exp( tA) x(t)=x + exp( sA) P G(x(s)) ds Z0 1 exp( sA) P G(x(s)) ds + Zt for all t 0. This gives d (exp( tA) x(t)) = exp( tA) P G(x(t)) + exp( tA) P+G(x(t)) dt =exp( tA) G(x(t)) for all t 0. From this, we deduce that x0(t)=Ax(t)+G(x(t)) = F (x(t)) for all t 0. Therefore, x(t) is a solution of the di↵erential equation x (t)= 0 F (x(t)). Moreover, P x(0) = x and P+x(0) = (x ) by definition of . Thus x(0) = (x )+x = x0.Sincex(t) converges exponentially to 0, the assertion follows. ⇤ n Lemma 4.9. Consider a vector x0 R .Writex0 = x+ + x , where 2 x+ U+ and x U .Assumethat x+ r, x r,andx+ = (x ). 2 2 k k k k 6 Then the solution of the initial value problem x0(t)=F (x(t)), x(0) = x0, must leave the ball B2⇤r(0) at some point. Proof. Let x(t) be the solution of the di↵erential equation x0(t)=F (x(t)) with initial condition x(0) = x0. Suppose that x(t) remains in the ball B2⇤r(0) for all time. Using the identity x0(t)=Ax(t)+G(x(t)), we obtain t (7) exp( tA) x(t)=x + exp( sA) G(x(s)) ds 0 Z0 for all t 0. This implies t exp( tA) P x(t)=x + exp( sA) P G(x(s)) ds. + + + Z0 4.3. Lyapunov’s theorems 53 Since x(t) is bounded and exp( tA) P 0 as t , we obtain k k + ! !1 (8) 0 = x + 1 exp( sA) P G(x(s)) ds. + + Z0 Subtracting (8) from (7), we obtain t exp( tA) x(t)=x + exp( sA) G(x(s)) ds Z0 1 exp( sA) P G(x(s)) ds, + Z0 hence t exp( tA) x(t)=x + exp( sA) P G(x(s)) ds Z0 1 exp( sA) P G(x(s)) ds + Zt for all t 0. Therefore, x(t) is a solution of the integral equation (5). This implies (x )=P+x(0) = x+, contrary to our assumption. Thus, the solution of the initial value problem x0(t)=F (x(t)), x(0) = x0,must eventually leave the ball B2⇤r(0). ⇤ To summarize, we have proved the following theorem: Theorem 4.10. Suppose that F : U Rn is of class C1. Moreover, suppose ! that 0 U is a hyperbolic equilibrium point. Then there exists a submanifold 2 W s of class C1 with the following properties: (i) The origin lies on W s and the tangent space to W s at 0 is given by U . In particular, dim W s =dimU . (ii) If x W s is su ciently close to 0, then the unique solution of the 0 2 di↵erential equation x0(t)=F (x(t)) with initial condition x(0) = x0 converges exponentially to 0. (iii) If x / W s is su ciently close to 0, then the unique solution of the 0 2 di↵erential equation x0(t)=F (x(t)) with initial condition x(0) = x0 will leave the ball B2⇤r(0) at some time in the future. Note that, in part (iii), the solution may re-enter the ball B2⇤r(0) at some later time, and it may still converge to the equilibrium point as t ,but !1 it cannot do so without first leaving the ball B2⇤r(0). 4.3. Lyapunov’s theorems Theorem 4.11. Suppose that x¯ is an equilibrium point of the autonomous system x0(t)=F (x(t)). Moreover, suppose that L is a smooth function 54 4. Analysis of equilibrium points defined on some ball Br(¯x), which has a strict local minimum at x¯.Finally, we assume that L(x),F(x) 0 for all x B (¯x). Then x¯ is stable. hr i 2 r Proof. Let " (0,r) be given. Since L has a strict local minimum atx ¯,we 2 have L(¯x) < inf L. @B"(¯x) By continuity, we can find a real number (0,") such that 2 sup L< inf L. @B (¯x) B (¯x) " We claim that ' (x) B (¯x) for all x B (¯x) and all t 0. To prove t 2 " 2 this, we argue by contradiction. Fix a point x B (¯x), and suppose that 0 2 ' (x ) / B (¯x) for some t 0. Let t 0 2 " ⌧ =inf t 0:' (x ) / B (¯x) . { t 0 2 " } Clearly, ' (x ) @B (¯x). Moreover, for t (0,⌧), we have ' (x ) B (¯x). ⌧ 0 2 " 2 t 0 2 " This implies d L(' (x )) = L(' (x )),F(' (x )) 0 dt t 0 hr t 0 t 0 i for all t (0,⌧). Thus, we conclude that 2 inf L L('⌧ (x0)) L(x0) sup L. @B (¯x) " B (¯x) This contradicts our choice of . ⇤ Theorem 4.12. Suppose that x¯ is an equilibrium point of the autonomous system x0(t)=F (x(t)). Moreover, suppose that L is a smooth function defined on some ball Br(¯x), which has a strict local minimum at x¯.Finally, we assume that L(x),F(x) < 0 for all x B (¯x) x¯ . Then x¯ is hr i 2 r \{ } asymptotically stable. Proof. It follows from Theorem 4.11 thatx ¯ is stable. Suppose thatx ¯ is not asymptotically stable. Let us fix a real number >0 such that 't(x) B r (¯x) for all x B (¯x) and all t 0. Sincex ¯ is not asymptotically 2 2 2 stable, we can find a point x0 B (¯x) such that lim supt 't(x0) x¯ > 2 !1 k k 0. Consequently, we can find a sequence of times s such that k !1 lim inft 's (x0) x¯ > 0. After passing to a subsequence if neces- !1 k k k sary, we may assume that the sequence 'sk (x0) converges to some point y B (¯x) x¯. 2 r \ Since L(x),F(x) 0 for all x B (¯x), the function t L(' (x )) is hr i 2 r 7! t 0 monotone decreasing. Hence, the limit := limt L('t(x0)) exists. This !1 gives L('t(y)) = lim L('t('s (x0))) = lim L('s +t(x0)) = . k k k k !1 !1 4.4. Gradient and Hamiltonian systems 55 Therefore, the function L('t(y)) is constant. In particular, d L(y),F(y) = L('t(y)) =0, hr i dt t=0 which contradicts our assumption. ⇤ 4.4. Gradient and Hamiltonian systems Finding Lyapunov functions is a di cult task in general. However, there are certain classes of nonlinear system which always admit a monotone quantity. In the following, we assume that U Rn is an open set and F : U Rn is ⇢ ! continuously di↵erentiable mapping. Definition 4.13. We say that the system x0(t)=F (x(t)) is a gradient system if F (x)= V (x) for some real-valued function V . r There is a convenient criterion for deciding whether a given system is a gradient or Hamiltonian system. A necessary condition for the system x (t)=F (x(t)) to be a gradient system is that @Fj (x)= @Fi (x) for all 0 @xi @xj x U. Moreover, if the domain U is simply connected, then this condition 2 is a su cient condition for x0(t)=F (x(t)) to be a gradient system. Proposition 4.14. Consider a gradient system of the form x0(t)=F (x(t)), where F (x)= V (x). Moreover, suppose that the function V attains a r strict local minimum at x¯. Then x¯ is a stable equilibrium point. Proof. Since V (x),F(x) = V (x) 2 0, hr i kr k the function V (x) is a Lyapunov function. Consequently,x ¯ is a stable equi- librium point. ⇤ Definition 4.15. Assume that n is even, and let 01 10 2 3 01 6 10 7 J = 6 7 . 6 . 7 6 .. 7 6 7 6 017 6 7 6 107 6 7 We say that the system4x (t)=F (x(t)) is Hamiltonian5 if F (x)=J H(x) 0 r for some real-valued function H. Proposition 4.16. Consider a gradient system of the form x0(t)=F (x(t)), where F (x)= J H(x). Moreover, suppose that the function H attains a r strict local minimum at x¯. Then x¯ is a stable equilibrium point. 56 4. Analysis of equilibrium points Proof. Since J is anti-symmetric, we have H(x),F(x) = H(x),J H(x) =0. hr i hr r i Therefore, the function H(x) is a Lyapunov function. Consequently,x ¯ is a stable equilibrium point. ⇤ 4.5. Problems Problem 4.1. Let U be an open set in Rn containing 0, and let F : U Rn ! be a smooth vector field. Assume that 0 is an equilibrium point of the system x0(t)=F (x(t)), and let A = DF(0) be the di↵erential of F at 0. Suppose that all eigenvalues of A have negative real part. Show that there exists a quadratic function L(x)= Sx,x such that L(x) has a strict local minimum h i at 0 and L(x),F(x) < 0 hr i if x U 0 is su ciently close to 0. Conclude from this that 0 is asymp- 2 \{ } totically stable. (Hint: Use Problem 2.6.) Problem 4.2. Consider the system 2 x0 (t)=x (t) + x (t) x (t) 2 1 2 1 2 2 x0 (t)=x (t) + x (t) x (t) 2. 2 1 1 2 (i) Find all equilibrium points of this system. For each equilibrium point, de- cide whether or not it is hyperbolic. For each hyperbolic equilibrium point, decide whether it is a source, a sink, or a saddle. (ii) Sketch the phase portrait of this system. (iii) Find the stable and unstable curves for each saddle point. (Hint: Con- sider the functions x (t)+x (t) and x (t) x (t).) 1 2 1 2 Problem 4.3. This example is taken from [4]. Let us consider the system 2 2 x0 (t)=x (t)+x (t)x (t) x (t) + x (t) (x (t)+x (t)) 1 1 1 2 1 2 1 2 2 2 2 x0 (t)=x (t) x (t) + x p(t) + x (t) (x (t) x (t)). 2 2 1 1 2 1 2 (i) Let us write x1(t)=r(t) cos ✓p(t) and x2(t)=r(t)sin✓(t). Show that r(t) and ✓(t) satisfy the di↵erential equations r0(t)=r(t)(1 r(t)) ✓0(t)=r(t)(1 cos ✓). (ii) Show that any solution that originates in a neighborhood of (1, 0) will converge to (1, 0) as t . !1 (iii) Show that the equilibrium point (1, 0) is unstable. Chapter 5 Limit sets of dynamical systems and the Poincar´e-Bendixson theorem 5.1. Positively invariant sets Definition 5.1. Let us consider the ODE x (t)=F (x(t)), where F : U 0 ! Rn is continuously di↵erentiable. We say that a set A U is invariant under ⇢ the flow generated by this ODE if 't(x) A for all x A and all t R 2 2 2 for which ' (x) is defined. Moreover, we say that a set A U is positively t ⇢ invariant if ' (x) A for all x A and all t [0, ) for which ' (x)is t 2 2 2 1 t defined. In this section, we show that a set is positively invariant under the ODE x0(t)=F (x(t)) if the vector field F points inward along the boundary. Theorem 5.2. Let F : U Rn be continuously di↵erentiable, and let ! A U be relatively closed so that A¯ U = A. Then the following statements ⇢ \ are equivalent: (i) A is positively invariant under the ODE x0(t)=F (x(t)). (ii) F (y),y z 0 for all points y A, z Rn satisfying y z = h i 2 2 k k dist(z,A). Proof. We first show that (i) implies (ii). Assume that A is positively in- variant. Moreover, suppose that y A and z Rn are two points satisfying 2 2 57 585. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem y z =dist(z,A). Let x(t), t [0,T), denote the unique solution of the k k 2 ODE x0(t)=F (x(t)) with initial condition x(0) = y.SinceA is positively invariant, we have x(0) z = y z =dist(z,A) x(t) z k k k k k k for all t [0,T). This implies 2 d 2 0 ( x(t) z ) =2 x0(0),x(0) z dt k k t=0 h i =2 F (x(0)),x(0) z h i =2 F (y),y z . h i This shows that F (y),y z 0, as claimed. h i We now show that (ii) implies (i). Let us assume that condition (ii) holds, and let x(t), t [0,T), be a solution of the ODE x (t)=F (x(t)) with 2 0 x(0) A. We claim that x(t) A for all t [0,T). To prove this, we argue 2 2 2 by contradiction. Suppose that x(t ) / A for some t (0,T). Let 0 2 0 2 C = x(t):t [0,t ] . { 2 0 } Note that C is a compact subset of U. Let us fix positive real numbers r and L such that dist(C, Rn U) > 2r and \ sup DF(x) op L⌧ L⌧ Lt0 e x(⌧) y = e dist(x(t),A)=e k k and Lt Lt Lt0 e x(t) y e dist(x(t),A) e k k for all t [⌧,t ). From this, we deduce that 2 0 d 2Lt 2 (e x(t) y ) 0, dt k k t=⌧ hence F (x(⌧)),x(⌧) y L x(⌧) y 2. h i k k L(t0 ⌧) n Moreover, x(⌧) y = e Adding both inequalities, we conclude that F (x(⌧)) F (y),x(⌧) y L x(⌧) y 2. h i k k By definition of L,wehave DF (i) A is positively invariant under the ODE x0(t)=F (x(t)). (ii) F (y),⌫(y) 0 for all points y @A. h i 2 It is often useful to consider the domains with piecewise smooth bound- ary. For example, for planar domains we have the following result: Corollary 5.4. Let F : U R2 be continuously di↵erentiable, and let A be ! a closed subset of U.Weassumethattheboundary@A is piecewise smooth. Let E denote the set of corners of @A. For each point y @A E, we denote 2 \ by ⌫(y) the outward-pointing unit normal vector at y. Then the following statements are equivalent: (i) A is positively invariant under the ODE x0(t)=F (x(t)). (ii) F (y),⌫(y) 0 for all points y @A E. h i 2 \ Proof. It is easy to see that (i) implies (ii). We now show that (ii) implies (i). By Theorem 5.2, it su ces to show that F (y),y z 0 for all points h i y A, z Rn satisfying y z =dist(z,A). To verify this, we distinguish 2 2 k k two cases: Case 1: Suppose first that y/E. Clearly, y must lie on the boundary 2 @A. Since the point y has smallest distance from z among all points in A,it follows that the vector z y = ↵⌫(y) for some number ↵ 0. This implies F (y),y z = ↵ F (y),⌫(y) 0. h i h i Case 2: Suppose finally that y E. By assumption, the point y lies on 2 two boundary arcs, which we denote by 1 and 2. Let ⌫1 denote the out- ward pointing unit normal vector field along 1, and let ⌫2 be the outward- pointing unit normal vector field along 1.Sincey has smallest distance from z among all points in A, we conclude that z y = ↵ ⌫ (y)+↵ ⌫ (y ) 1 1 2 2 2 for some numbers ↵ ,↵ 0. Using our assumption (and the continuity of 1 2 F ), we conclude that F (y ),⌫(y ) 0 and F (y ),⌫(y ) 0. This implies h 1 1 i h 2 2 i 605. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem F (y),y z = ↵ F (y),⌫ (y) ↵ F (y),⌫ (y) 0. This completes the h i 1 h 1 i 2 h 2 i proof. ⇤ 5.2. The !-limit set of a trajectory Definition 5.5. Let F : U Rn be continuously di↵erentiable, and let ! x(t) be a solution of the ODE x (t)=F (x(t)) which is defined for all t 0. 0 A point y Rn is an !-limit point of the trajectory x(t)ifthereexistsa 2 sequence of times sk such that limk x(sk)=y. !1 !1 If we denote by ⌦denote the set of all !-limit points of x(t), then ⌦= x(s):s [t, ) , { 2 1 } t 0 \ where x(s):s [t, ) denotes the closure of the set x(s):s [t, ) . { 2 1 } { 2 1 } Since intersections of closed sets are always closed, we conclude that ⌦is a closed subset of Rn. Proposition 5.6. If ⌦ is bounded and non-empty, then limt dist(x(t), ⌦) = !1 0. Proof. Suppose this is false. Then there exists a sequence of times t k !1 such that dist(x(tk), ⌦) >"for all k. Let y be an arbitrary point in ⌦. Since y is an !-limit point of the trajectory x(t), we can find a sequence of numbers s such that s >t and x(s ) y <"for all k.Thisimplies k k k k k k dist(x(sk), ⌦) <"for all k. By the intermediate value theorem, we can find a sequence of times ⌧ (t ,s ) such that dist(x(⌧ ), ⌦) = ".Since⌦is k 2 k k k bounded, it follows that the sequence x(⌧k) is bounded. By the Bolzano- Weierstrass theorem, we can find a sequence of integers kl such that n !1 the sequence x(⌧k ) converges to some point z R as l . Consequently, l 2 !1 z is an !-limit point and dist(z,⌦) = liml dist(x(⌧k , ⌦) ".Thisisa !1 l contradiction. ⇤ Proposition 5.7. If ⌦ is bounded, then ⌦ is connected. Proof. Suppose that ⌦is not connected. Then there exist non-empty closed sets A and A such that A A =⌦and A A = . Since ⌦is bounded, 1 2 1 [ 2 1 \ 2 ; the sets A1 and A2 are compact. Since A1 and A2 are disjoint, it follows that dist(A1,A2) > 0. Let us fix two arbitrary points y A and y A .Sincey is an 1 2 1 2 2 2 1 !-limit point of x(t), we can find a sequence of times s such that k,1 !1 limk x(sk,1)=y1. Similarly, we can find a sequence of times sk,2 !1 !1 such that limk x(sk,2)=y2. In particular, limk dist(x(sk,1),A1)=0 !1 !1 5.2. The !-limit set of a trajectory 61 and limk dist(x(sk,2),A1)=dist(y2,A1) dist(A1,A2). By the interme- !1 diate value theorem, we can find a sequence of times ⌧ such that k !1 1 dist(x(⌧ ),A )= dist(A ,A ). k 1 2 1 2 Using the triangle inequality, we obtain 1 dist(x(⌧ ),A ) dist(A ,A ) dist(x(⌧ ),A ) dist(A ,A ). k 2 1 2 k 1 2 1 2 This implies 1 dist(x(⌧ ), ⌦) = min dist(x(⌧ ),A ), dist(x(⌧ ),A ) = dist(A ,A ). k { k 1 k 2 } 2 1 2 This contradicts Proposition 5.6. ⇤ Proposition 5.8. The set ⌦ U is an invariant set. \ Proof. Consider a y ⌦ U. Then there exists a sequence of times s 2 \ k !1 such that limk x(sk)=y. Hence, if t R is fixed, then !1 2 lim x(sk + t)= lim 't(x(sk)) = 't(y). k k !1 !1 Thus, 't(y) ⌦for all t R. This proves the assertion. 2 2 ⇤ Finally, we show that certain equilibrium points cannot arise as !-limit points. Proposition 5.9. Let x(t), t 0, be a solution of the di↵erential equation x0(t)=F (x(t)) and let x¯ be an !-limit point of x(t). Moreover, we assume that x¯ is a stable equilibrium point for the system x (t)= F (x(t)). Then 0 x(t)=¯x for all t 0. Proof. As usual, we denote by 't the flow generated by the di↵erential equation x0(t)=F (x(t)). Let us fix a real number ">0. By assumption, we can find a real number >0 such that ' t(x) B"(¯x) for all x B (¯x) 2 2 and all t 0. Moreover, sincex ¯ is an !-limit point, we can find a sequence of times sk such that limk x(sk)=¯x. In particular, x(sk) B (¯x) !1 !1 2 if k is su ciently large. This implies x(0) = ' s (x(sk)) B"(¯x). Since k 2 ">0 is arbitrary, we conclude that x(0) =x ¯. From this, the assertion follows. ⇤ As an application, we can prove a global version of Lyapunov’s second theorem: Proposition 5.10. Consider the di↵erential equation x0(t)=F (x(t)), where F : U Rn is continuously di↵erentiable. We assume that there exists a ! function L : U Rn with the following properties: ! 625. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem (i) For each µ R, the sublevel set x U : L(x) µ is a compact 2 { 2 } subset of U. (ii) We have L(x),F(x) < 0 for all points x U E, where E hr i 2 \ consists of isolated points. Then, given any point x U, the function ' (x ) is defined for all t 0 0 2 t 0 and converges to an equilibrium point. Proof. Fix a point x U, and let µ = L(x ). Since L(x),F(x) 0 0 2 0 hr i for all x U, we conclude that the function t L(' (x )) is monotone 2 7! t 0 decreasing. In particular, L(' (x )) µ for all t 0. Since the sublevel t 0 set x U : L(x) µ is a compact subset of U, we conclude that the { 2 } solution exists for all t 0. Since the function t L(' (x )) is monotone 7! t 0 decreasing, the limit := limt L('t(x0)) exists. !1 Let ⌦denote the !-limit set of the trajectory 't(x0). Given any point y ⌦, we can find a sequence of times sk such that limk 's (x0)= 2 !1 !1 k y.Thisimplies L('t(y)) = lim L('t('s (x0))) = lim L('s +t(x0)) = . k k k k !1 !1 Consequently, the function L('t(y)) is constant. This implies d L(y),F(y) = L('t(y)) =0. hr i dt t=0 Consequently, ⌦ x U : L(x),F(x) =0 E. ⇢{ 2 h i }⇢ In particular, ⌦consists of at most finitely many points. Since ⌦is con- nected, we conclude that ⌦consists of at most a single point. On the other hand, since the trajectory 't(x0) is contained in a compact set, the set ⌦ is non-empty by the Bolzano-Weierstrass theorem. Thus, ⌦consists of ex- actly one point. Moreover, this point must be an equilibrium point since ⌦ is a positively invariant set. Hence, the assertion follows from Proposition 5.6. ⇤ 5.3. !-limit sets of planar dynamical systems In this section, we will consider a planary dynamical system of the form 2 2 x0(t)=F (x(t)), where F : R R . Our goal is to analyze the !-limit set ⇥ of a given solution x(t) of the ODE x0(t)=F (x(t)). Definition 5.11. A line segment S = z +(1 )z : (0, 1) is said { 0 1 2 } to be transversal if, for each point x S¯, the vector F (x) is not parallel to 2 the vector z z . 1 0 5.3. !-limit sets of planar dynamical systems 63 Lemma 5.12. Let S be a transversal line segment, and let x¯ be an arbitrary point in S. Then there exists an open neighborhood U of x¯ and a smooth function h : U R such that h(¯x)=0and 'h(x)(x) S for all points ! 2 x U. 2 Proof. This follows immediately from the implicit function theorem. ⇤ Lemma 5.13. Let x(t) be a solution of the ODE x0(t)=F (x(t)) which is defined on some time interval J, and let S be a transversal line segment. Then the set E = t J : x(t) S is discrete. { 2 2 } Proof. Suppose this is false. Then there exists a number ⌧ R and a 2 sequence of numbers tk E ⌧ such that limk tk = ⌧. Clearly, x(tk) 2 \{ } !1 2 ¯ x(⌧) x(tk) S and x(⌧)=limk x(tk) S. Hence, the vector ⌧ t is tangential !1 2 k x(⌧) x(tk) to S. Consequently, the limit limk = x0(⌧)=F (x(⌧)) is also !1 ⌧ tk tangential to S. This contradicts the fact that S is a transversal line segment. ⇤ The following monotonicity property plays a fundamental role in the proof of the Poincar´e-Bendixson theorem: Lemma 5.14. Let x(t) be a solution of the ODE x0(t)=F (x(t)) which is not periodic, and let S = z +(1 )z : (0, 1) be a transversal { 0 1 2 } line segment. Let us consider three times t0 Proof. Since x(t) is not periodic, the numbers 0, 1, 2 are all distinct. Suppose now that the assertion is false. Then we either have > max , 1 { 0 2} or < min , . Without loss of generality, we may assume that 1 { 0 2} > max , . (Otherwise, we switch the roles of z and z .) More- 1 { 0 2} 0 1 over, we may assume that < . (Otherwise, we replace F by F and t 0 2 i by t2 i.) Consequently, 0 < 2 < 1. Let = x(t):t [t ,t ] z +(1 )z : ( , ) . { 2 0 1 }[{ 0 1 2 0 1 } In other words, is the union of the trajectory from x(t0)tox(t1)with a line segment from x(t )tox(t ). By assumption, x(t) / S for all t 0 1 2 2 (t0,t1), so is free of self-intersections. By the Jordan curve theorem, the complement R2 has exactly two connected components. Let us denote \ these components by D1 and D2. By assumption, the vector F ( z +(1 )z ) cannot be parallel to S 0 1 for any [ , ]. After switching the roles of D and D if necessary, 2 0 1 1 2 645. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem we may assume that the vector F ( z +(1 )z ) points into D for each 0 1 1 ( , ). Hence, at each point on the boundary @D = , the vector field 2 0 1 1 F points inward or is tangential to @D1. By Corollary 5.4, the closure D¯1 is a positively invariant set. Since x(t ) D¯ , we conclude that x(t) D¯ for 1 2 1 2 1 all t>t1. On the other hand, since the point x(t2) lies on the line segment z +(1 )z : ( , ) ,wehavex(t ") / D¯ if ">0issu ciently { 0 1 2 0 1 } 2 2 1 small. This is a contradiction. ⇤ Lemma 5.15. Let x(t) be a solution of the ODE x0(t)=F (x(t)) which is defined for all t 0 and is not periodic. Moreover, let ⌦ denote its !-limit set, and let S = z +(1 )z : s [0, 1] be a transversal line segment. { 0 1 2 } Then the intersection ⌦ S consists of at most one point. \ Proof. We argue by contradiction. Suppose that the intersection ⌦ S \ contains two distinct points y and y .Sincey ⌦, we can find a sequence 1 2 1 2 of real numbers sk,1 such that limk x(sk,1)=y1. Moreover, there !1 !1 exists a sequence of real numbers sk,2 such that limk x(sk,2)=y2. !1 !1 Using Lemma 5.12, we can find a sequence of real numberss ˜k,1 such that s˜ s 0 and x(˜s ) S for all k. Similarly, we obtain a sequence of k,1 k,1 ! k,1 2 real numberss ˜ such thats ˜ s 0 and x(˜s ) S for all k. k,2 k,2 k,2 ! k,2 2 By Lemma 5.13, the set E = t 0:x(t) S is discrete. Moreover, { 2 } sinces ˜ , s˜ E for all k, it follows that E is unbounded. Let us write k,1 k,2 2 E = tk : k N ,wheretk is an increasing sequence of times going to { 2 } infinity. Since x(t ) S,wemaywritex(t )= z +(1 )z for some k 2 k k 0 k 1 [0, 1]. By Lemma 5.14, the sequence is either monotone increasing k 2 k or monotone decreasing. In either case, the limit limk k exists. This !1 implies that the limit limk x(tk) exists. On the other hand, the sequences !1 x(˜sk,1):k N and x(˜sk,2 : k N are subsequences of the sequence { 2 } { 2 } x(tk):k N , and we have limk x(˜sk,1)=y1 and limk x(˜sk,2)=y2. { 2 } !1 !1 Thus, y1 = y2, contrary to our assumption. ⇤ Lemma 5.16. Let x(t) be a solution of the ODE x0(t)=F (x(t)) which is defined for all t 0 and is not periodic. Moreover, let ⌦ denote its !-limit set. Assume that ⌦ is bounded, non-empty, and contains no equilibrium points. Then every point in ⌦ lies on a periodic orbit. Proof. Let us consider an arbitrary point y ⌦, and let y(t)betheunique 0 2 maximal solution with y(0) = y . By Proposition 5.8, we have y(t) ⌦for 0 2 all t. Since ⌦is a compact set, we conclude that the solution y(t)isdefined for all t R. Let z be an !-limit point of the solution y(t). Since y(t) ⌦ 2 2 for all t R, we conclude that z ⌦. Consequently, z cannot be an 2 2 equilibrium point. Hence, we can find a transversal line segment S such that z S. By Lemma 5.15, the set ⌦ S consists of at most one point. 2 \ Therefore, ⌦ S = z . \ { } 5.4. Stability of periodic solutions and the Poincar´emap 65 Since z is an !-limit point of y(t), we can find a sequence s such k !1 that limk y(sk)=z. Using Lemma 5.12, we can find another sequence of timess ˜!1such thats ˜ s 0 and y(˜s ) S for all k.Thisimplies k k k ! k 2 y(˜s ) ⌦ S for all k. Consequently, y(˜s )=z for all k. From this, we k 2 \ k deduce that the solution y(t)isperiodic. ⇤ Poincar´e-Bendixson Theorem. Let x(t) be a solution of the ODE x0(t)= F (x(t)), which is defined for all t 0. Let ⌦ be the set of !-limit points of the trajectory x(t).Supposethat⌦ is bounded, non-empty, and contains no equilibrium points. Then there exists a periodic solution y(t) of the ODE y0(t)=F (y(t)) such that y(t):t R =⌦. { 2 } Proof. If x(t) is periodic, the assertion is trivial. In the following, we will assume that x(t) is not periodic. By Lemma 5.16, there exists a periodic solution y(t) of the ODE y0(t)=F (y(t)) such that A := y(t):t R ⌦. { 2 }⇢ We claim that the set ⌦ A is closed. To see this, we consider a sequence \ of points zk ⌦ A such that limk zk =¯z. Clearly,z ¯ ⌦since⌦is 2 \ !1 2 closed. In particular,z ¯ cannot be an equilibrium point. Consequently, we can find a transversal line segment S such thatz ¯ S. By Lemma 5.15, the 2 set ⌦ S consists of at most one point. Thus, ⌦ S = z¯ . \ \ { } By Lemma 5.12, there exists an open neighborhood U ofz ¯ and a smooth function h : U R such that h(¯z) = 0 and 'h(z)(z) S for all points ! 2 z U.Sincelimk zk =¯z,wehavezk U if k is su ciently large. 2 !1 2 Then ' (z ) S. Since ⌦is an invariant set and z ⌦, it follows h(zk) k 2 k 2 that ' (z ) ⌦ S if k is su ciently large. From this, we deduce that h(zk) k 2 \ ' (z )=¯z for k su ciently large. Since z / A, we conclude thatz/ ¯ A. h(zk) k k 2 2 This shows that the set ⌦ A is closed. \ To summarize, we have shown that ⌦can be written as a disjoint union of the closed sets A and ⌦ A.SinceA is connected, we must have ⌦ A = . \ \ ; This completes the proof. ⇤ 5.4. Stability of periodic solutions and the Poincar´emap We now return to the n-dimensional case. Suppose that x(t) is a non- constant periodic solution of the ODE x (t)=F (x(t)), and let = x(t): 0 { t R . Our goal in this section is to analyze whether this periodic orbit is 2 } stable; that is, whether a nearby solution y(t) will converge to as t . !1 To simplify the notation, we will assume that x(0) = 0. Since 0 is not an equilibrium point, we can find an (n 1)-dimensional subspace S Rn ⇢ such that F (0) / S. By the implicit function theorem, we can find an open 2 neighborhood U of 0 and a smooth function h : U R such that h(0) = T ! and ' (y) S for all points y U. We next define a map P : S U S h(y) 2 2 \ ! 665. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem by P (y)=' (y) S h(y) 2 for y S U.Since' (0) = 0 and h(0) = T , we obtain P (0) = 0. The 2 \ T map P is called the Poincar´emap. Theorem 5.17. Suppose that the eigenvalues of DP(0) all lie inside the unit circle in C.Ify0 is su ciently close to 0, then dist('t(y0), ) 0 as ! t . !1 Proof. Let A = DP(0) denote the di↵erential of P at the origin. By as- sumption, the eigenvalues of A all lie inside the unit circle in C. Using the k L + N decomposition, it is easy to see that limk A = 0 (cf. Problem 2.7 above). Let us fix a positive integer m such that!1Am < 1 . k kop 2 For abbreviation, let T˜ = mT . By the implicit function theorem, we can find an open neighborhood U˜ of 0 and a smooth function h˜ : U˜ R such ! that h˜(0) = T˜ and ' (y) S for all points y U˜. We now define a map h˜(y) 2 2 P˜ : S U˜ S by \ ! P˜(y)=' (y) S h˜(y) 2 for y S U˜. Note that 2 \ P˜(y)=P ... P (y) m times if y S is su ciently close to the origin. This implies 2 | {z } DP˜(0) = DP(0)m = Am. Consequently, DP˜(0) < 1 by our choice of m. Consequently, we can k kop 2 find a real number >0 such that B (0) U˜ and P˜(y) 1 y for all ⇢ k k 2 k k points y S B (0). 2 \ We now consider an initial point y S. We assume that y is su ciently 0 2 0 close to the origin such that y0 U˜ and '˜ (y0) < .Weinductively 2 k h(y0) k define a sequence of numbers tk such that t0 = h˜(y0) and tk+1 = tk + ˜ h('tk (y0)). Clearly, ˜ 'tk+1 (y0)='˜ ('tk (y0)) = P ('tk (y0)). h('tk (y0)) Using induction on k, we can show that ' (y ) < 2 k . Therefore, k tk 0 k ' (y ) 0 as k .Thisimplies tk 0 ! !1 t t = h˜(' (y )) h˜(0) = T˜ k+1 k tk 0 ! as k .Since' (y ) 0 as k , we conclude that !1 tk 0 ! !1 sup 't(y0) 't tk (0) =sup 't('tk (y0)) 't(0) 0 t [t ,t ] k k t [0,t t ] k k! 2 k k+1 2 k+1 k 5.5. Problems 67 as k . In particular, !1 sup dist('t(y0), ) 0 t [t ,t ] ! 2 k k+1 as k . From this, the assertion follows. !1 ⇤ Finally, we describe how the di↵erential of the Poincar´emap is related to the di↵erential of 'T . Proposition 5.18. Let us define a linear transformation Q : Rn S by ! Qy = y for all y S and QF(0) = 0. Then 2 DP(0) y = QD'T (0) y for all y S. 2 Proof. Fix a vector y S. Using the chain rule, we obtain 2 d d P (sy) = 'h(sy)(sy) = D'T (0) y + F(0), ds s=0 ds s=0 where = d h(sy) . On the other hand, since P (sy) S for all s,we ds s=0 2 conclude that d P (sy) S. Thus, we conclude that ds s=0 2 d P (sy) = Q (D'T (0) y + F(0)) = QD'T (0) y. ds s=0 From this, the assertion follows. ⇤ 5.5. Problems Problem 5.1. Let p(x) be a polynomial of odd degree whose leading coe - 2 cient is positive. Moreover, let (x0,y0) be an arbitrary point in R . Suppose that (x(t),y(t)) is the unique maximal solution of the system x0(t)=y(t) 3 y0(t)= y(t) p(x) with the initial condition (x(0),y(0)) = (x0,y0). 2 2 (i) Show that supt 0(x(t) + y(t) ) < . (Hint: Look for a monotone 1 quantity of the form f(x)+g(y).) (ii) Show that the solution (x(t),y(t)) is defined for all t 0. (iii) Suppose that (¯x, y¯) is an !-limit point of the trajectory (x(t),y(t)). Show that p(¯x)=¯y = 0. (iv) Show that the !-limit set of the trajectory (x(t),y(t)) consists of exactly one point. 685. Limit sets of dynamical systems and the Poincar´e-Bendixson theorem Problem 5.2. Let F : Rn Rn be a continuously di↵erentiable mapping. ! Assume that x(t) is a solution of the system x0(t)=F (x(t)) which is defined for all t 0, and let ⌦be its !-limit set. We assume that ⌦is bounded and x ⌦: x r = v : [0,r] { 2 k k } { 2 } for some positive real number r and some unit vector v. The goal of this problem is to show that every point on the line segment v : [0,r] is { 2 } an equilibrium point. (i) Fix a number (0,r). Consider a sequence of times s such that 2 k !1 x(s ) rv, and define ⌧ =sup t [0,s ]: x(t) r and x(t),v . k ! k { 2 k k k h i } Show that ⌧ , x(⌧ ) v, and F (x(⌧ )),v 0. Deduce from this k !1 k ! h k i that F ( v),v 0. h i (ii) Fix a number (0,r). Consider a sequence of times s such 2 k !1 that x(s ) 0, and define ⌧ =sup t [0,s ]: x(t) r or x(t),v . k ! k { 2 k k k h i } Show that ⌧ , x(⌧ ) v, and F (x(⌧ )),v 0. Deduce from this k !1 k ! h k i that F ( v),v 0. h i (iii) Show that F ( v) = 0 for all (0,r). In other words, every point on 2 this line segment is an equilibrium point. Problem 5.3. Let F be a vector field on R2, and let x(t) be a solution of the ODE x0(t)=F (x(t)) which is defined for all t R and is not periodic. 2 Suppose thatx ¯ R2 is both an ↵-limit point and an !-limit point of the 2 trajectory x(t). In other words, there exists a family of times sk : k Z { 2 } such that limk x(sk)=limk x(sk)=¯x. Show thatx ¯ is an equilib- rium point. (Hint:!1 Consider a! 1 transversal line segment passing throughx ¯ and use the monotonicity property.) Problem 5.4. Let us consider the unique maximal solution of the system 2 2 (1 x1(t) ) 2 x10 (t)= 2 (x1(t)+(1 x1(t) ) x2(t)) 1+x1(t) 2 x0 (t)= x (t)+(1 x (t) ) x (t) 2 1 1 2 with initial condition (x1(0),x2(0)) = (0, 1). x1(t) (i) Let us write y1(t)= 2 and y2(t)=x2(t). Show that 1 x1(t) 2 y0 (t)= (y1(t)+y2(t)) 1 2 1+ 1+4y1(t) 2 y0 (t)= p (y2(t) y1(t)). 2 2 1+ 1+4y1(t) p (ii) Show that the !-limit set of (x1(t),x2(t)) is a union of two parallel lines. 5.5. Problems 69 Problem 5.5. Consider the system 2 x0 (t)=(1 x (t) )(x (t)+2x (t)) 1 1 1 2 2 x0 (t)=(1 x (t) )(x (t) 2 x (t)). 2 2 2 1 (i) Show that the square Q =[ 1, 1] [ 1, 1] is an invariant set for this ⇥ system. (ii) Let (x1(t),x2(t)) denote the unique maximal solution of this system with 1 initial condition (x1(0),x2(0)) = ( 2 , 0), and let ⌦denote its !-limit set. Show that ⌦ @Q. (Hint: Consider the function (1 x (t)2)(1 x (t)2).) ⇢ 1 2 (iii) Show that ⌦cannot consist of a single point. (Hint: Suppose that (x1(t),x2(t)) converges to an equilibrium point, and use the stable manifold theorem to arrive at a contradiction.) (iv) Let L = 1 [ 1, 1], L =[ 1, 1] 1 , L = 1 [ 1, 1], and 1 { }⇥ 2 ⇥{ } 3 { }⇥ L =[ 1, 1] 1 . Show that ⌦= L for some non-empty set 4 ⇥{ } j J j J 1, 2, 3, 4 . 2 ⇢{ } S (v) Show that ⌦= @Q. (Hint: Use Problem 5.2.) Chapter 6 Ordinary di↵erential equations in geometry, physics, and biology 6.1. Delaunay’s surfaces in di↵erential geometry Let ⌃be a surface of revolution in R3 so that ⌃= (r(t) cos s, r(t)sins, t):t I { 2 } for some function r : I (0, ). At each point on ⌃, we have two curvature ! 1 radii. The reciprocals of the curvature radii are referred to as the prinicpal curvatures, and their sum is referred to as the mean curvature of ⌃. In 1841, C. Delaunay investigated surfaces of revolution which have con- stant mean curvature 2. The mean curvature of a surface of revolution is given by r(t)r (t) (1 + r (t)2) H = 00 0 . 2 3 r(t)(1+r0(t) ) 2 Hence, the condition H = 2 is equivalent to the di↵erential equation 2 2 3 r(t)r00(t)=(1+r0(t) ) 2r(t)(1+r0(t) ) 2 . If we put x(t)=r(t) and y(t)=r0(t), we obtain the system x0(t)=y(t) 2 1 2 y0(t)=(1+y(t) ) 2 1+y(t) x(t) ⇣ p ⌘ 71 72 6. Ordinary di↵erential equations in geometry, physics, and biology in the half-plane U = (x, y) R2 : x>0 . In order to analyze this system, { 2 } we consider the function x L(x, y)=x2 . 1+y2 p A straightforward calculation yields @L 1 (x, y)=2x @x 1+y2 p and @L xy (x, y)= . @y 1+y2 (1 + y2) p This implies @L 1 @L y (x, y)+(1+y2) 2 1+y2 (x, y) @x x @y y ⇣1 p ⌘ xy =2xy + 2 1+y2 1+y2 x 1+y2 ⇣ p ⌘ =0. p p Hence, if (x(t),y(t)) is a solution x0(t)=y(t) 2 1 2 y0(t)=(1+y(t) ) 2 1+y(t) , x(t) ⇣ p ⌘ d then dt L(x(t),y(t)) = 0. In other words, every solution curve is contained in a level curve of the function L(x, y). The level curves are shown in the figure below: 6.1. Delaunay’s surfaces in di↵erential geometry 73 ...... 2.0 ...... 1.5 ...... 1.0 ...... 0.5 ...... 0.0 ...... -0.5 ...... -1.0 ...... -1.5 ...... -2.0 ...... 0.0 0.5 1.0 1.5 2.0 1 The system above has only one equilibrium point, which is ( 2 , 0). The 1 coe cient matrix of the linearized system at ( 2 , 0) is given by 01 , 40 which has eigenvalues 2i and 2i. Nonetheless, the point ( 1 , 0) is a stable 2 equilibrium point. To see this, we apply Lyapunov’s theorem. We first 1 observe that the gradient of L at the point ( 2 , 0) vanishes, and the Hessian 1 of L at the point ( 2 , 0) is given by 20 . 0 1 2 Since this matrix is positive definite, we conclude that L has a strict local 1 1 minimum at ( 2 , 0). Therefore, the point ( 2 , 0) is stable by Lyapunov’s the- 1 orem. However, the point ( 2 , 0) is clearly not asymptotically stable, since every trajectory is contained in a level curve of L. Note that the surface of 1 revolution associated with the constant solution ( 2 , 0) is a cylinder of radius 1 2 . 1 We next fix a real number 2 Since L is constant along any solution curve, we must have L(x(t),y(t)) = L(a, 0) = a (1 a) for all t J.Thisimplies 2 x(t) x(t)(1 x(t)) x(t)2 = a(1 a) 1+y(t)2 for all t J. From this it follows that p 2 1 a x(t) a for all t J. Moreover, 2 x(t) 1+y(t)2 = a (1 a)+x(t)2 p 1 ( a(1 a) x(t))2 = 1 2 a(1 a) a(1 a)+x(t)2 ✓ p ◆ 1 p , 2 a(1 a) hence p 1 (2a 1)2 y(t)2 1= . 4a(1 a) 4a(1 a) From this, we deduce that the solution is defined for all times, i.e. J = R. (Otherwise, we could find a sequence of times tk J such that limk x(tk)= 2 2 2 !1 0 or limk x(tk) + y(tk) = . That would contradict the previous in- !1 1 equalities.) In the next step, we claim that the function (x(t),y(t)) is periodic. To prove this, let T =inf t>0:y(t)=0 . { } (Note that T might be infinite. We will later show that this is not the case.) Note that y(t) < 0 for t (0,T). Using the identity L(x(t),y(t)) = 2 a (1 a), we obtain x(t)2 (a2 x(t)2)(x(t)2 (1 a)2) y(t)2 = 1= . (a(1 a)+x(t)2)2 (a(1 a)+x(t)2)2 Therefore, 1 a 6.2. The mathematical pendulum Let us considered an idealized pendulum. Let ✓ denote the angle between the pendulum and the vertical axis. If we neglect friction, the angle ✓ satisfies the di↵erential equation g ✓00(t)= sin ✓(t), l where l is the length of the pendulum and g denotes the acceleration due to g gravity. By a suitable choice of units, we can arrange that l = 1, so ✓00(t)= sin ✓(t). This di↵erential equation is equivalent to the system x0(t)=y(t) y0(t)= sin x(t). 76 6. Ordinary di↵erential equations in geometry, physics, and biology This system is Hamiltonian, and the Hamiltonian function is 1 H(x, y)= y2 cos x. 2 In particular, 1 y(t)2 cos x(t) = constant 2 for every solution (x(t),y(t)). This reflects the law of conservation of energy. The level curves of the function H(x, y) are shown below: 3 ...... 2 ...... 1 ...... 0 ...... -1 ...... -2 ...... -3 -5 -4 -3 -2 -1 0 1 2 3 4 5 The system above has infinitely many equilibrium points, which are of the form (⇡k, 0) with k Z. The point (0, 0) is a strict local minimum of the 2 function H(x, y). Consequently, the point (0, 0) is a stable equilibrium by Lyapunov’s theorem. The point (0, 0) is not asymptotically stable, however. We next consider the equilibrium point (⇡, 0). The coe cient matrix of the linearized system at (⇡, 0) is given by 01 . 10 This matrix has eigenvalues 1 and 1, and the associated eigenvectors are 1 1 and . By the stable manifold theorem, we can find a curve W s 1 1 passing through the point (⇡, 0) such that limt 't(x, y)=(⇡, 0) for every !1 point (x, y) W s.SinceH(x, y) is constant along a solution of the ODE, 2 we have H(x, y)= limH('t(x, y)) = H(⇡, 0) = 1 t !1 6.2. The mathematical pendulum 77 for all points (x, y) W s. Consequently, 2 1 W s (x, y): y2 cos x =1 ⇢{ 2 } x x = (x, y):y = 2 cos (x, y):y = 2 cos . { 2 }[{ 2 } 1 Since W s is tangent to the vector at the point (⇡, 0), we must have 1 x W s (x, y):y = 2 cos . ⇢{ 2 } Similarly, we can find a curve W u passing through the point (⇡, 0) such that s limt 't(x, y)=(⇡, 0) for every point (x, y) W . A similar argument ! 1 2 as above gives x W u (x, y):y = 2 cos . ⇢{ 2 } We next fix a real number 0 T =inf t>0:y(t)=0 . { } (Note that T might be infinite. We will later show that this is not the case.) Clearly, y(t) = 0 and cos x(t) = cos a for t (0,T). Since the functions 6 6 2 x(t) and y(t) are continuous, we have a y(t)= 2 (cos x(t) cos a) p 78 6. Ordinary di↵erential equations in geometry, physics, and biology for t (0,T). Therefore, we obtain 2 ⌧ 1 ⌧ = x0(t) dt y(t) Z0 ⌧ 1 = x0(t) dt 0 2 (cos x(t) cos a) Z a 1 = p dx x(⌧) 2 (cos x cos a) Z a 1 p dx a 2 (cos x cos a) Z for all ⌧ (0,T). Since thep integral 2 a 1 dx a 2 (cos x cos a) Z is finite, we conclude that T Consider the following system of two coupled second order di↵erential equa- tions: x1(t) x100(t)= 3 2 2 (x1(t) + x2(t) ) 2 x2(t) x200(t)= 3 . 2 2 (x1(t) + x2(t) ) 2 This system describes the motion of a point mass in a central force field, 2 where the force is proportional to r . Proposition 6.1. Suppose that (x1(t),x2(t)) is a solution of the system of di↵erential equations given above. Then we have the following conserved 6.3. Kepler’s problem 79 quantities: L = x (t) x0 (t) x (t) x0 (t) = constant, 1 2 2 1 x1(t) A1 = (x1(t) x20 (t) x2(t) x10 (t)) x20 (t)+ 1 = constant, 2 2 (x1(t) + x2(t) ) 2 x2(t) A2 =(x1(t) x20 (t) x2(t) x10 (t)) x10 (t)+ 1 = constant. 2 2 (x1(t) + x2(t) ) 2 Proof. We compute d (x (t) x0 (t) x (t) x0 (t)) dt 1 2 2 1 = x (t) x00(t) x (t) x00(t) 1 2 2 1 x2(t) x1(t) = x1(t) + x2(t) 2 2 3 2 2 3 (x1(t) + x2(t) ) 2 (x1(t) + x2(t) ) 2 =0. This proves the first statement. Since the function x (t) x (t) x (t) x (t) 1 20 2 10 is constant, we obtain d x1(t) (x1(t) x20 (t) x2(t) x10 (t)) x20 (t)+ 1 dt (x (t)2 + x (t)2) 2 ⇣ 1 2 ⌘ = (x (t) x0 (t) x (t) x0 (t)) x00(t) 1 2 2 1 2 x (t) x (t)2 x (t)+x (t)x (t) x (t) + 10 1 10 1 2 20 2 2 1 2 2 3 (x1(t) + x2(t) ) 2 (x1(t) + x2(t) ) 2 = (x (t) x0 (t) x (t) x0 (t)) x00(t) 1 2 2 1 2 x2(t) (x1(t) x20 (t) x2(t) x10 (t)) 3 2 2 (x1(t) + x2(t) ) 2 =0. This proves the second statement. Finally, if we replace (x1(t),x2(t)) by (x2(t),x1(t)), we obtain d x2(t) (x1(t) x20 (t) x2(t) x10 (t)) x10 (t)+ 1 =0. dt (x (t)2 + x (t)2) 2 ⇣ 1 2 ⌘ This completes the proof. ⇤ We note that the first identity reflects the conservation of angular mo- mentum. These conservation laws hold for any central force field. The conservation laws for A1 and A2 are much more subtle, and are special to Kepler’s problem. The vector (A1,A2) is called the Runge-Lenz vector. 80 6. Ordinary di↵erential equations in geometry, physics, and biology Corollary 6.2. Suppose that (x1(t),x2(t)) is a solution of the system of di↵erential equations given above. Then 2 2 2 2 x1(t) + x2(t) =(A1x1(t)+A2x2(t)+L ) . In particular, the path t (x (t),x (t)) is contained in a conic section. 7! 1 2 Proof. We compute 2 2 2 1 A x (t)+A x (t)= (x (t) x0 (t) x (t) x0 (t)) +(x (t) + x (t) ) 2 1 1 2 2 1 2 2 1 1 2 2 2 2 1 = L +(x (t) + x (t) ) 2 . 1 2 This implies 2 2 2 2 x1(t) + x2(t) =(A1x1(t)+A2x2(t)+L ) . In other words, the path t (x (t),x (t)) is contained in the set 7! 1 2 2 2 2 2 2 = (x1,x2) R : x + x =(A1x1 + A2x2 + L ) . { 2 1 2 } 2 If A2 + A2 < 1, then is an ellipse with principle axes L and 1 2 p1 A2 A2 1 2 L2 2 2 2 2 1 A2 A2 .IfA1 + A2 = 1, then is a parabola. Finally, if A1 + A2 > 1, then 1 2 is a hyperbola. ⇤ Finally, let us derive a formula for the position (x1(t),x2(t)) as a function of t. For simplicity, we only consider the case when the orbit is an ellipse. Proposition 6.3. Suppose that (x1(t),x2(t)) is a solution of the system of di↵erential equations given above. Moreover, suppose that A1 = " and A =0, where 0 "<1 and A and A are defined as in Proposition 6.1. 2 1 2 Then we may write L2 x (t)= (cos ✓(t)+"), 1 1 "2 L2 x2(t)= sin ✓(t), p1 "2 where ✓(t) satisfies Kepler’s equation 2 3 L 2 ✓(t)+" sin ✓(t)= t + constant. 1 "2 ⇣ ⌘ 2 3 L 2 In particular, the solution (x1(t),x2(t)) is periodic with period 2⇡ 1 "2 . Proof. By Corollary 6.2, 2 2 2 2 x1 + x2 =("x1 + L ) . Rearranging terms gives L2 " 2 L4 (1 "2) x + x2 = . 1 1 "2 2 1 "2 ⇣ ⌘ 6.4. Predator-prey models 81 Consequently, we can find a function function ✓(t) such that L2 x (t)= (cos ✓(t)+") 1 1 "2 L2 x2(t)= sin ✓(t). p1 "2 Di↵erentiating these identities with respect to t gives L = x (t) x0 (t) x (t) x0 (t) 1 2 2 1 4 4 L 2 L 2 = 3 (cos ✓(t)+" cos ✓(t)) ✓0(t)+ 3 sin ✓(t) ✓0(t) (1 "2) 2 (1 "2) 2 L4 = 3 (1 + " cos ✓(t)) ✓0(t). (1 "2) 2 Thus, we conclude that 2 3 L 2 (1 + " cos ✓(t)) ✓0(t)= . 1 "2 Integrating this equation with respect to t⇣yields ⌘ 2 3 L 2 ✓(t)+" sin ✓(t)= t + constant. 1 "2 2 3⇣ ⌘ L 2 Hence, if we define T =2⇡ 1 "2 ,then✓(t + T )=✓(t)+2⇡.Thisimplies (x1(t + T ),x2(t + T )) = (x1(t),x2(t)). Thus, the solution (x1(t),x2(t)) is 2 3 L 2 periodic with period T =2⇡ 1 "2 . ⇤ 6.4. Predator-prey models In this section, we analyze a model for the growth of the populations of two species, one of which preys on the other. This model was proposed by Volterra and Lotka in the 1920s. Let x(t) denote the size of the prey population at time t, and let y(t) denote the size of the predator population at time t. The dynamics of x(t) and y(t) is modeled by the di↵erential equations x0(t)=ax(t) ↵x(t) y(t) y0(t)= cy(t)+ x(t) y(t), where a, ↵, c, are positive constants. c a This system has two equilibrium points, (0, 0) and ( , ↵ ). The coe cient matrix of the linearized system at the point (0, 0) is given by a 0 . 0 c 82 6. Ordinary di↵erential equations in geometry, physics, and biology Therefore, the point (0, 0) is a saddle point. The stable curve is given by x =0 , and the unstable curve is y =0 . { } { } Similarly, the coe cient matrix of the linearized system at the point c a ( , ↵ ) is given by 0 a ↵ . ↵c 0 The eigenvalues of this matrix are pac i and pac i, so we need additional c a arguments to decide whether the equilibrium point ( , ↵ ) is stable. We next consider the function L(x, y)= x c log x + ↵y a log y for x, y > 0. We claim that this function is a Lyapunov function. It is clear that the function x x c log x attains its global minimum at the point c 7! . Similarly, the function y ↵y a log y attains its global minimum at a 7! the point ↵ . Therefore, the function L attains its global minimum at the c a point ( , ↵ ). Moreover, this is a strict minimum. Suppose now that (x(t),y(t)) is a solution of the system of di↵erential equations considered above satisfying x(t),y(t) > 0. Then d L(x(t),y(t)) dt c a = x0(t)+ ↵ y0(t) x(t) y(t) ⇣ c ⌘ ⇣ ⌘ a = (ax(t) ↵x(t) y(t)) + ↵ ( cy(t)+ x(t) y(t)) x(t) y(t) =(⇣ x(t) c⌘)(a ↵y(t)) + (↵y(t) ⇣a)( c + ⌘x(t)) =0. Therefore, the function L(x, y) is a conserved quantity. By Lyapunov’s c a theorem, the equilibrium point ( , ↵ ) is stable. Finally, it is not di cult to show that each level set of the function L is smooth curve which lies in a compact subset of the quadrant (x, y) R2 : { 2 x, y > 0 . Hence, every solution that originates in this quadrant is either } constant or periodic. 6.5. Mathematical models for the spread of infectious diseases In this section, we discuss a model for the spread of infectious diseases which is due to Kermack and McKendrick. This model relies on several assump- tions. First, we assume that the incubation period can be neglected, so that any infected person can immediately infect others. Second, we assume that 6.6. A mathematical model of glycolysis 83 any person who has recovered from the disease gains permanent immunity. Let us denote by x(t) the number of persons who are susceptible to the dis- ease. Moreover, let y(t) be the number of persons who are currently infected. In other words, x(t) is the number of persons who have not contracted the disease prior to time t, and y(t) is the number of persons who have been infected but have not yet recovered. Finally, let z(t) denote the number of persons who have recovered from the disease. The dynamics of x(t),y(t),z(t) can be modeled by the following system of di↵erential equations: x0(t)= x(t) y(t) y0(t)= x(t) y(t) y(t) z0(t)= y(t). Here, and are positive constants. This system is an example of what is called an SIR-model, and was first proposed by Kermack and McKendrick. We note that the sum x(t)+y(t)+z(t) is constant. Hence, it is enough to solve the two-dimensional system x0(t)= x(t) y(t) y0(t)= x(t) y(t) y(t). To that end, we consider the quantity L(x, y)=x(t) log x(t)+y(t). A straightforward calculation gives d L(x(t),y(t)) = 1 x0(t)+y0(t) dt x(t) = ⇣ ( x(t) ⌘) y(t)+ x(t) y(t) y(t) =0. Thus, L(x, y) is a conserved quantity. If x(0) , then the function x(t) is monotone decreasing and converges to 0 as t . On the other hand, if x(0) > , then the number of !1 infected persons will increase at first. The epidemic will reach its peak when x(t)= . Afterwards, the number of infected persons will decrease, and will converge to 0 as t . For that reason, the ratio is referred to !1 as the epidemiological threshold. 6.6. A mathematical model of glycolysis In this section, we discuss Sel’kov’s model for glycolysis. Our treatment closely follows [5], Section 7.3. The di↵erential equations governing Sel’kov’s 84 6. Ordinary di↵erential equations in geometry, physics, and biology model are as follows: 2 x0(t)= x(t)+ay(t)+x(t) y(t) 2 y0(t)=b ay(t) x(t) y(t). Here, the functions x(t) and y(t) describe the concentrations, at time t, of two chemicals (adenosine diphosphate and fructose-6-phosphate), and a and b are positive constants. We will focus on the case when the initial values x(0) = x0 and y(0) = y0 are positive. Proposition 6.4. Given any real number b , the set a 2 A = (x, y) R : x 0, 0 y ,andx + y + b { 2 } is positively invariant. Proof. It su ces to show that the vector field F (x, y)=( x + ay + x2y, b ay x2y) is inward-pointing along the boundary of A. In other words, we need to show that F (x, y),⌫ 0 h i for every point (x, y) @A,where⌫ denotes the outward-pointing unit 2 normal vector to @A. The boundary of A consists of four line segments: @A = L L L L , 1 [ 2 [ 3 [ 4 where 2 L1 = (x, y) R : x = 0 and 0 y { 2 } 2 L2 = (x, y) R : y = 0 and 0 x + b { 2 } 2 L3 = (x, y) R : x + y = + b and b x + b { 2 } 2 L4 = (x, y) R : y = and 0 x b . { 2 } Step 1: Consider a point (x, y) L . The outward-pointing unit normal 2 1 vector to @A at (x, y)is⌫ =( 1, 0). This implies F (x, y),⌫ = F (0,y),⌫ = (ay, b ay), ( 1, 0) = ay 0 h i h i h i since y 0. Step 2: We next consider a point (x, y) L . In this case, the outward- 2 2 pointing unit-normal vector is given by ⌫ =(0, 1). Therefore, F (x, y),⌫ = F (x, 0),⌫ = ( x, b), (0, 1) = b 0. h i h i h i 6.6. A mathematical model of glycolysis 85 Step 3: Consider now a point (x, y) L . In this case, the outward- 2 3 pointing unit-normal vector is ⌫ =( 1 , 1 ). Hence, p2 p2 F (x, y),⌫ h i 1 1 = ( x(t)+ay(t)+x(t)2 y(t),b ay(t) x(t)2 y(t)), , p2 p2 D1 ⇣ ⌘E = (b x) 0 p2 since x b. Step 4: Finally, let (x, y) be a point on the line segment L4.The outward-pointing unit normal vector to @A is ⌫ =(0, 1). From this it follows that F (x, y),⌫ = F (x, ),⌫ h i h i = ( x + a + x2 , b a x2 ), (0, 1) h i = b a x2 b a 0 since b . a Therefore, the region A is positively invariant as claimed. ⇤ We next discuss the asymptotic behavior of the solution as t . !1 Proposition 6.5. Let (x0,y0) be a pair of positive real numbers. More- over, let (x(t),y(t)) be the unique maximal solution of the system above with x(0) = x0 and y(0) = y0, and let ⌦ be the set of all !-limit points of (x(t),y(t)). Then ⌦ is bounded and non-empty. Proof. Consider the region 2 A = (x, y) R : x 0, 0 y , and x + y + b , { 2 } where = max x + y b, y , b . It follows from Proposition 6.4 that { 0 0 0 a } (x(t),y(t)) A for all t 0. Since A is a bounded region, it follows from 2 the Bolzano-Weierstrass theorem that the solution (x(t),y(t)) has at least one !-limit point. This shows that ⌦is non-empty. On the other hand, since (x(t),y(t)) A for all t 0, it follows that ⌦ A, which implies that 2 ⇢ ⌦is bounded. ⇤ Finally, we analyze the equilibrium points: b Proposition 6.6. The point (b, a+b2 ) is the only equilibrium point of the system. If b4 +(2a 1) b2 +(a2 + a) < 0, then both eigenvalues of the matrix b 4 2 2 DF(b, a+b2 ) have positive real part. Similarly, if b +(2a 1) b +(a +a) > 0, b then both eigenvalues of the matrix DF(b, a+b2 ) have negative real part. 86 6. Ordinary di↵erential equations in geometry, physics, and biology Proof. Suppose that (¯x, y¯) is an equilibrium point. Then x¯ + a y¯ +¯x2 y¯ =0 and b a y¯ x¯2 y¯ =0. Adding both identities, we obtain b x¯ =0, hencex ¯ = b. Substituting this into the first equation, we obtain (a + b2)¯y = b, hence b y¯ = . a + b2 The di↵erential of F is given by 1+2xy a + x2 DF(x, y)= . 2xy a x2 The trace and determinant of this matrix are 1+2xy a + x2 tr = 1+2xy x2 a 2xy a x2 and 1+2xy a + x2 det = a + x2. 2xy a x2 In particular, the determinant of DF(x, y) is always positive. This shows b that the matrix DF(b, a+b2 ) cannot have two real eigenvalues with opposite b signs. The sign of the trace of DF(b, a+b2 ) indicates whether the eigenvalues b of the matrix DF(b, a+b2 ) have positive or negative real part. ⇤ Theorem 6.7. Let (x0,y0) be a pair of positive real numbers such that (x ,y ) =(b, b ). Moreover, let (x(t),y(t)) be the unique maximal solution 0 0 6 a+b2 with x(0) = x0 and y(0) = y0. Then the following holds: (i) If b4 +(2a 1) b2 +(a2 + a) < 0, then the solution (x(t),y(t)) approaches a periodic solution as t . !1 4 2 2 (ii) If b +(2a 1) b +(a + a) > 0, then either limt (x(t),y(t)) = b !1 (b, a+b2 ) or the solution (x(t),y(t)) converges to a periodic solution as t . !1 Proof. As above, let ⌦be the set of all !-limit points of (x(t),y(t)). We first assume that b4 +(2a 1) b2 +(a2 + a) < 0. In this case, both eigenvalues b of the matrix DF(b, a+b2 ) have positive real part. By Theorem 4.2, the equilibrium point (b, b ) is asymptotically stable after a time reversal t a+b2 ! t. Using Proposition 5.9, we conclude that (b, b ) / ⌦. In particular, a+b2 2 6.7. Problems 87 the set ⌦contains no equilibrium points. Hence, the Poincar´e-Bendixson theorem implies that the solution (x(t),y(t)) approaches a periodic solution as t . !1 We now assume that b4 +(2a 1) b2 +(a2 + a) > 0. In this case, b both eigenvalues of the matrix DF(b, a+b2 ) have negative real part. By b Theorem 4.2, the equilibrium point (b, a+b2 ) is asymptotically stable. If (b, b ) / ⌦, it follows from the Poincar´e-Bendixson theorem that the so- a+b2 2 lution (x(t),y(t)) converges to a periodic solution as t . On the other b !1 b hand, if (b, 2 ) ⌦, then we have limt (x(t),y(t)) = (b, 2 )since a+b !1 a+b b 2 (b, a+b2 ) is asymptotically stable. ⇤ 6.7. Problems Problem 6.1. Let (x(t),y(t)) be the unique solution of the di↵erential equations x0(t)=y(t) y0(t)= sin x(t) with initial values x(0) = 0 and y(0) = 2. (i) Show that x(t) x0(t) = 2 cos . 2 (ii) Using Problem 1.5, conclude that x(t) t t = arctan(e ) arctan(e ) 2 and 4 y(t)= t t . e + e What can you say about the asymptotic behavior of (x(t),y(t)) as t ? !1 Problem 6.2. Let (x1(t),x2(t)) be a solution of the system x1(t) x100(t)= 3 2 2 (x1(t) + x2(t) ) 2 x2(t) x200(t)= 3 . 2 2 (x1(t) + x2(t) ) 2 (i) Show that 1 2 2 1 E = (x10 (t) + x20 (t) ) 1 = constant. 2 2 2 (x1(t) + x2(t) ) 2 2 2 2 (ii) Show that A1 + A2 =2EL + 1, where L, A1,A2 denote the conserved quantities from Proposition 6.1. Chapter 7 Sturm-Liouville theory 7.1. Boundary value problems for linear di↵erential equations of second order In this chapter, we will study linear di↵erential equations of second order with variable coe cients. Definition 7.1. Let p(t) and q(t) be two continuously di↵erentiable func- tions that are defined on some interval [a, b]. We then consider the following linear di↵erential equation of second order for the unknown function u(t): d d (9) p(t) u(t) + q(t) u(t)=0. dt dt The di↵erential equationh (9) is calledi a Sturm-Liouville equation. Moreover, we say that (9) is a regular Sturm-Liouville equation if p(t) > 0 for all t [a, b]. 2 In the following, we will only consider regular Sturm-Liouville equations. Proposition 7.2. Suppose that (9) is a regular Sturm-Liouville equation. Then the set of all functions u(t) satisfying (9) is a vector space of dimen- sion 2. In other words, there exist two linearly independent solutions w0(t) and w1(t) of (9), and any other solution of (9) can be written as a linear combination of w0(t) and w1(t) with constant coe cients. Proof. The di↵erential equation (9) is equivalent to the non-autonomous linear system 1 x0 (t)= x (t) 1 p(t) 2 x0 (t)= q(t) x (t). 2 1 89 90 7. Sturm-Liouville theory By Proposition 3.10, there exists a unique solution of (9) which satisfies the initial condition u(a) = 1 and u0(a) = 0. Similarly, there exists a unique solution of (9) which satisfies the initial conditions u(a) = 0 and u0(a) = 1. Let us denote these solutions by w0(t) and w1(t). It follows from Proposition 3.10 that the functions w0(t) and w1(t) are defined on entire interval [a, b]. It is clear that w0(t) and w1(t) are linearly indepedent (i.e. neither function is a constant multiple of the other). It remains to show that any solution of (9) can be written as a linear combination of w0(t) and w1(t). To see this, let u(t) be an arbitrary solution of (9). Then the function v(t)= u(t) u(a) w (t) u (a) w (t) is a solution of (9), and we have v(a)=u(a) 0 0 1 u(a) w (a) u (a) w (a) = 0 and v (a)=u (a) u(a) w (a) u (a) w (a) = 0. 0 0 1 0 0 0 0 10 Hence, the uniqueness theorem implies that v(t) = 0 for all t [a, b]. Thus, 2 we conclude that u(t)=u(a) w (t)+u (a) w (t) for all t [a, b]. This shows 0 0 1 2 that u(t) can be written as a linear combination of w0(t) and w1(t)with constant coe cients. ⇤ So far, we have focused on initial value problems for systems of ordinary di↵erential equations. However, in many situations, it is more natural to impose boundary conditions instead. For example, we can prescribe the values of the function u at the endpoints of the interval [a, b]. Alternatively, we may prescribe the values of u0(t) at the endpoints of the interval [a, b]. The following list shows the most common boundary conditions: Dirichlet boundary conditions: u(a)=A, u(b)=B. • Neumann boundary conditions: u (a)=A, u (b)=B. • 0 0 Mixed Dirichlet-Neumann boundary condition: u(a)=A, u (b)= • 0 B. Periodic boundary condition: u(a)=u(b), u (a)=u (b). • 0 0 Here, A and B are given real numbers. A Sturm-Liouville equation together with a set of boundary conditions is called a Sturm-Liouville system. While the initial value problem for a regular Sturm-Liouville system always has exactly one solution, a boundary value problem may have infinitely many solutions or no solutions at all. To see this, we first consider the di↵erential equation d2 u(t)+u(t)=0 dt2 with the Dirichlet boundary conditions u(0) = 0 and u(⇡) = 0. This equa- tion has infinitely many solutions. In fact, every constant multiple of the function sin t is a solution. 7.1. Boundary value problems for linear di↵erential equations of second order91 We next consider the di↵erential equation d2 u(t)+u(t)=0 dt2 with the Dirichlet boundary conditions u(0) = 0 and u(⇡) = 1. This equa- d2 tion has no solution. In fact, if u(t) is a solution of dt2 u(t)+u(t)=0with u(0) = 0, then u(t) must be a constant multiple of sin t.Butthenu(⇡)=0 since sin ⇡ = 0. Therefore, there is no solution of the di↵erential equation d2 dt2 u(t)+u(t)=0withu(0) = 0 and u(⇡) = 1. In the sequel, we will mostly focus on the Dirichlet boundary value problem. The following result gives a necessary and su cient condition for the Dirichlet boundary value problem to admit a unique solution: Proposition 7.3. Suppose that (9) is a regular Sturm-Liouville equation. Then the following statements are equivalent: (i) Given any pair of real numbers (A, B), there exists a unique solu- tion of (9) such that u(a)=A and u(b)=B. (ii) Every solution of (9) satisfying u(a)=u(b)=0vanishes identi- cally. Proof. It is clear that (i) implies (ii). To prove the reverse implication, we consider the set V of all solutions of the di↵erential equation (9). By Proposition 7.2, V is a vector space of dimension 2. We next consider the linear transformation L : V R2, which assigns to every function u V the ! 2 pair (u(a),u(b)) R2. If condition (ii) holds, then L has trivial nullspace. 2 Since V has the same dimension as R2, it follows that L is invertible. This implies that statement (i) holds. ⇤ Thus, in order to decide whether the Dirichlet boundary value problem has a unique solution, it su ces to study solutions of the di↵erential equation (9) which satisfy the boundary condition u(a)=u(b) = 0. We now develop an important method for studying the solutions of a regular Sturm-Liouville equation d d p(t) u(t) + q(t) u(t)=0. dt dt Suppose that u(t) is a solutionh of (9)i which is not identically zero. Our strategy is to rewrite (9) as a system of two di↵erential equations of first order, and pass to polar coordinates: d p(t) u(t)=r(t) cos ✓(t) dt u(t)=r(t)sin✓(t). 92 7. Sturm-Liouville theory This change of variables is known as the Pr¨ufer substitution. Clearly, d 1 0= (r(t)sin✓(t)) r(t) cos ✓(t) dt p(t) 1 = r0(t)sin✓(t)+r(t) ✓0(t) cos ✓(t) r(t) cos ✓(t). p(t) Moreover, since u(t) is a solution of (9), we have d 0= (r(t) cos ✓(t)) + q(t) r(t)sin✓(t) dt = r0(t) cos ✓(t) r(t) ✓0(t)sin✓(t)+q(t) r(t)sin✓(t). This gives a system of two linear equations for the two unknowns r0(t) and ✓0(t). Solving this system yields 1 (10) r0(t)= q(t) r(t) cos ✓(t)sin✓(t) p(t) ⇣ ⌘ and 2 1 2 (11) ✓0(t)=q(t)sin ✓(t)+ cos ✓(t). p(t) This is a system of two nonlinear di↵erential equations for r(t) and ✓(t). A key observation is that the di↵erential equation for the function ✓(t)does not make any reference to the function r(t). Hence, we can first look for a solution ✓(t) of (11). Once ✓(t) is known, the function r(t)isdeterminedby the formula t 1 r(t)=r(a)exp q(s) cos ✓(s)sin✓(s) ds . a p(s) ✓ Z ⇣ ⌘ ◆ Moreover, if u(t) satisfies a boundary condition, we obtain additional re- strictions on r(t) and ✓(t). For example, if u satisfies the Dirichlet bound- ary conditions u(a)=u(b) = 0, then sin ✓(a)=sin✓(b) = 0. Simi- larly, the Neumann boundary condition u0(a)=u0(b) = 0 is equivalent to cos ✓(a) = cos ✓(b) = 0. Moreover, the mixed Dirichlet-Neumann boundary u(a)=u0(b) = 0 leads to the equation sin ✓(a) = cos ✓(b) = 0. In all these cases, the boundary condition for u(t) leads to a set of endpoint conditions for the function ✓(t). This is true for many boundary conditions, with the notable exception of periodic boundary conditions. 7.2. The Sturm comparison theorem In this section, we describe a method for comparing the phase functions of two Sturm-Liouville systems. To that end, we need the following comparison principle: 7.2. The Sturm comparison theorem 93 Proposition 7.4. Suppose that ✓(t) and ✓˜(t) satisfy the di↵erential inequal- ities 2 1 2 ✓0(t) q(t)sin ✓(t)+ cos ✓(t) p(t) and 2 1 2 ✓˜0(t) q(t)sin ✓˜(t)+ cos ✓˜(t), p(t) where p(t) is a positive function. If ✓(a) ✓˜(a), then ✓(t) ✓˜(t) for all t [a, b]. Moreover, if ✓(b)=✓˜(b), then ✓(t)=✓˜(t) for all t [a, b]. 2 2 Proof. We can find a positive constant L>0 such that d 1 (✓(t) ✓˜(t)) q(t)(sin2 ✓(t) sin2 ✓˜(t)) + (cos2 ✓(t) cos2 ✓˜(t)) dt p(t) L ✓(t) ✓˜(t) | | for all t [a, b]. 2 Suppose now that there exists a real number t [a, b] such that ✓(t ) < 1 2 1 ✓˜(t ). By assumption, t (a, b]. We now define 1 1 2 t =sup t [a, t ):✓(t) ✓˜(t) . 0 { 2 1 } It is easy to see that ✓(t )=✓˜(t ) and ✓(t) < ✓˜(t) for all t (t ,t ]. This 0 0 2 0 1 implies d (✓(t) ✓˜(t)) L (✓(t) ✓˜(t)) dt for all t (t ,t ]. Consequently, the function e Lt (✓(t) ✓˜(t)) is monotone 2 0 1 increasing on the interval [t ,t ]. Since ✓(t ) ✓˜(t ) = 0, we conclude that 0 1 0 0 ✓(t) ✓˜(t) 0 for all t [t ,t ]. This contradicts our choice of t .Thus,we 2 0 1 1 conclude that ✓(t) ✓˜(t) for all t [a, b]. 2 Consequently, d (✓(t) ✓˜(t)) L (✓(t) ✓˜(t)) dt for all t [a, b]. Therefore, the function eLt (✓(t) ✓˜(t)) is nonnegative and 2 monotone increasing on the interval [a, b]. Hence, if ✓(b) ✓˜(b) = 0, then ✓(t) ✓˜(t) = 0 for all t [a, b]. This completes the proof. 2 ⇤ Corollary 7.5. Suppose that ✓(t) satisfies the di↵erential inequality 2 1 2 ✓0(t) q(t)sin ✓(t)+ cos ✓(t), p(t) where p(t) is a positive function. If ✓(a) ⇡k, then ✓(b) >⇡k. 94 7. Sturm-Liouville theory Proof. Let us define ✓˜(t)=⇡k.Since✓˜(t) is constant, we have 2 1 2 ✓˜0(t) < q˜(t)sin ✓˜(t)+ cos ✓˜(t). p˜(t) Hence, Proposition 7.4 implies that ✓(b) ⇡k. Moreover, the inequality is be strict. In fact, if ✓(b)=⇡k, then Proposition 7.4 implies that ✓(t)=⇡k for all t [a, b], which is impossible. 2 ⇤ Corollary 7.6. Suppose that 0 p˜(t)˜u0(t)=˜r(t) cos ✓˜(t) u˜(t)=˜r(t)sin✓˜(t). The functions ✓(t) and ✓˜(t) satisfy the di↵erential equations 2 1 2 ✓0(t)=q(t)sin ✓(t)+ cos ✓(t) p(t) and 2 1 2 ✓˜0(t)=˜q(t)sin ✓˜(t)+ cos ✓˜(t) p˜(t) 1 q(t)sin2 ✓˜(t)+ cos2 ✓˜(t). p(t) Hence, the assertion follows from Proposition 7.4. ⇤ As another consequence of the comparison principle, we obtain the fol- lowing result: 7.3. Eigenvalues and eigenfunctions of Sturm-Liouville systems 95 Sturm Comparison Theorem. Suppose that 0 Proof. As above, we write d p(t) u(t)=r(t) cos ✓(t) dt u(t)=r(t)sin✓(t) and p˜(t)˜u0(t)=˜r(t) cos ✓˜(t) u˜(t)=˜r(t)sin✓˜(t). We can arrange that ✓(a) [0,⇡). Moreover, sinceu ˜(a) = 0, we may 2 assume that ✓˜(a) = 0. Sinceu ˜(b) = 0, it follows that ✓˜(b) must be an integer multiple of ⇡. Moreover, ✓˜(b) > 0 by Corollary 7.5. Consequently, ✓˜(b) ⇡. On the other hand, since ✓(a) ✓˜(a), it follows from Corollary 7.6 that ✓(b) ✓˜(b) ⇡.Since✓(a) <⇡, the intermediate value theorem implies the existence of a real number ⌧ (a, b] such that ✓(⌧)=⇡.This 2 implies u(⌧) = 0, as claimed. ⇤ 7.3. Eigenvalues and eigenfunctions of Sturm-Liouville systems Suppose that p(t), q(t), and ⇢(t) are three functions defined on some time interval [a, b]. We assume that p(t) and q(t) are continuously di↵erentiable, and ⇢(t) is continuous. Moreover, we assume that p(t) and ⇢(t) are positive for all t [a, b]. In this section, we will consider the Sturm-Liouville system 2 of the form d d (12) p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=u(b)=0, dt dt h i where is a constant. 96 7. Sturm-Liouville theory Definition 7.7. We say that is said to be an eigenvalue of the Sturm- Liouville (12) if the system (12) admits a non-trivial solution. If u(t)is a non-trivial solution of (12), we say that u(t) is an eigenfunction with eigenvalue . Our goal in this section is to analyze the eigenvalues of (12). To that end, we consider an arbitrary number R. It follows from the basic 2 existence and uniqueness theorem that the initial value problem d d p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=0,u0(a)=1 dt dt h i has a unique solution. We will denote this solution by u (t). Proposition 7.8. A real number is an eigenvalue of (12) if and only if u (b)=0. In this case, the function u (t) is an eigenfunction, and any other eigenfunction is a constant multiple of the function u (t). Proof. Suppose that u(t) is an eigenfunction with eigenvalue .Thefunc- tions u(t) and u0(a) u (t) satisfy the same di↵erential equation with the same initial values. Hence, it follows from the existence and uniqueness theorem that u(t)=u (a) u (t) for all t [a, b]. Since u(b) = 0, it follows that 0 2 u0(a) u (b) = 0. On the other hand, since u(t) is a non-trivial solution, we must have u (a) = 0. Therefore, u (b) = 0, and u(t) is a constant multiple 0 6 of u(t). Conversely, if u (b) = 0, then the function u (t) is an eigenfunction with eigenvalue . ⇤ In order to study the zeroes of the function u (t), we use the Pr¨ufer substitution. Let us write d p(t) u (t)=r (t) cos ✓ (t) dt u (t)=r (t)sin✓ (t), where ✓ (a) = 0. It was shown above that the function ✓ (t) satisfies the di↵erential equation 2 1 2 ✓0 (t)=(q(t)+ ⇢(t)) sin ✓ (t)+ cos ✓ (t). p(t) Using Corollary 7.5, we obtain ✓ (t) > 0 for all t (a, b]. 2 Proposition 7.9. A real number is an eigenvalue of (12) if and only if ✓ (b)=⇡n for some positive integer n. Proof. By Proposition 7.8, is an eigenvalue of (12) if and only if u (b) = 0. This is equivalent to saying that ✓ (b)=⇡n for some integer n. Finally, n must be a positive integer since ✓ (b) > 0. ⇤ 7.3. Eigenvalues and eigenfunctions of Sturm-Liouville systems 97 Proposition 7.10. If ✓ (b)=⇡n, then the function u (t) has exactly n 1 zeroes in the interval (a, b). Proof. For each k 0, 1,...,n ,wedefine 2{ } ⌧ =inf t [a, b]:✓ (t) ⇡k . k { 2 } It is easy to see that a = ⌧0 <⌧1 <...<⌧n 1 <⌧n = b and ✓ (t) <⇡k for all t [a, ⌧ ). Moreover, ✓ (⌧ )=⇡k for each k 0, 1,...,n . Using 2 k k 2{ } Corollary 7.5, we conclude that ✓ (t) >⇡kfor all t (⌧ ,b]. Putting these 2 k facts together, we conclude that ⇡k<✓ (t) <⇡(k + 1) for all t (⌧ ,⌧ ). 2 k k+1 Consequently, u (⌧ ) = 0 and u (t) = 0 for all t (⌧ ,⌧ ). Therefore, the k 6 2 k k+1 function u (t) has exactly n 1 zeroes in the interval (a, b). ⇤ As an example, let us consider the eigenvalue problem d2 u(t)+ u(t)=0,u(a)=u(b)=0. dt2 In this case, the function u (t) is given by 1 sin(p (t a)) if >0 p u (t)=8t a if =0 > > 1 sinh(p (t a)) if <0. < p In particular, u (b) = 0> if and only if is positive and p (t a)=⇡n for : some positive integer n. Therefore, the n-th eigenvalue is given by ⇡2n2 = , n (b a)2 and the associated eigenfunction is given by t a constant sin ⇡n . · b a 2 ⇣ ⌘ Note that n = O(n ). We now return to the general case. In view of Proposition 7.9, the problem of finding the eigenvalues of (12) comes down to finding all numbers for which ✓ (b)=⇡n,wheren is a positive integer. This is a nonlinear equation in . We will show that this equation has exactly one solution for every positive integer n. This is a consequence of the following theorem: Theorem 7.11. The function ✓ (b) is continuous and strictly increas- 7! ing. Moreover, lim ✓ (b)=0 ! 1 and lim ✓ (b)= . 1 !1 98 7. Sturm-Liouville theory Proof. The proof involves several steps: Step 1: Using Theorem 3.6, it is easy to show that the function ( , t) 7! ✓ (t) is continuous (see also Problem 3.3). In particular, the function 7! ✓ (b) is continuous. Step 2: We next show that the function ✓ (b) is strictly increasing. 7! Suppose that and µ are two real numbers such that >µ. The function u (t) satisfies the di↵erential equation d d p(t) u (t) +(q(t)+ ⇢(t)) u (t)=0. dt dt h i Similarly, the function uµ(t) satisfies the di↵erential equation d d p(t) u (t) +(q(t)+µ⇢(t)) u (t)=0. dt dt µ µ Since >µand ⇢(ht) is positive,i we have q(t)+ ⇢(t) >q(t)+µ⇢(t) for all t [a, b]. Moreover, ✓ (a)=✓ (a) = 0. By Corollary 7.6, we have 2 µ ✓ (b) ✓ (b). We claim that ✓ (b) >✓(b). Indeed, if ✓ (b)=✓ (b), µ µ µ then Corollary 7.6 implies that ✓ (t)=✓ (t) for all t [a, b]. Using the µ 2 di↵erential equations 2 1 2 ✓0 (t)=(q(t)+ ⇢(t)) sin ✓ (t)+ cos ✓ (t) p(t) and 2 1 2 ✓0 (t)=(q(t)+µ⇢(t)) sin ✓ (t)+ cos ✓ (t), µ µ p(t) µ we conclude that = µ, contrary to our assumption. Consequently, ✓ (b) > ✓ (b). Therefore, the function ✓ (b) is strictly increasing. µ 7! Step 3: We now show that lim ✓ (b)= . Suppose that a positive !1 1 integer n is given. The function t a uˆ(t)=sin ⇡n b a ⇣ ⌘ satisfies the di↵erential equation d d ⇡2n2 p uˆ(t) + p uˆ(t)=0. dt max dt (b a)2 max h i Let us write d p uˆ(t)=ˆr(t) cos ✓ˆ(t) max dt uˆ(t)=ˆr(t)sin✓ˆ(t), where ✓ˆ(a) = 0. 7.3. Eigenvalues and eigenfunctions of Sturm-Liouville systems 99 Suppose that is chosen such that >0 and ⇡2n2 ⇢ p q . min (b a)2 max min Then ⇡2n2 q(t)+ ⇢(t) p (b a)2 max for all t [a, b]. Moreover, 2 p(t) p max for all t [a, b]. Hence, it follows from Corollary 7.6 that ✓ (b) ✓ˆ(b). 2 Sinceu ˆ(b) = 0, the number ✓ˆ(b) must be an integer multiple of ⇡. Moreover, since the functionu ˆ(t) has exactly n 1zeroesbetweena and b,wehave ✓ˆ(b)=⇡n by Proposition 7.10. Putting these facts together, we obtain ✓ (b) ✓ˆ(b)=⇡n.Sincen is arbitrary, we conclude that lim ✓ (b)= . !1 1 Step 4: It remains to show that lim ✓ (b) = 0. Suppose that a number " (0,⇡) is given. Let us choose! 1 a real number such that <0 2 and 2 1 2 (qmax + ⇢min)sin " + cos "<0. pmin We claim that ✓ (b) ". Suppose this is false. In this case, we define ⌧ =inf t (a, b]:✓ (b) >" . { 2 } Clearly, ✓ (⌧)=" and ✓ (⌧) 0. On the other hand, 0 2 1 2 ✓0 (⌧)=(q(⌧)+ ⇢(⌧)) sin ✓ (⌧)+ cos ✓ (⌧) p(⌧) 1 =(q(⌧)+ ⇢(⌧)) sin2 " + cos2 " p(⌧) 2 1 2 (qmax + ⇢min)sin " + cos " pmin < 0 by our choice of . This is a contradiction. Consequently, ✓ (b) ".Since ">0 can be chosen arbitrarily small, we conclude that lim ✓ (b) = 0. ! 1 This completes the proof of Theorem 7.11. ⇤ Theorem 7.12. There exists an increasing sequence of real numbers 1 < 2 <...with the following properties: (i) The real number n is the unique solution of the equation ✓ (b)= ⇡n. (ii) as n . n !1 !1 (iii) A real number is an eigenvalue if and only if there exists a positive integer n such that = n. 100 7. Sturm-Liouville theory (iv) For every positive integer n, the function u n (t) is an eigenfunction with eigenvalue . Moreover, the function u (t) has exactly n 1 n n zeroes in the open interval (a, b). Proof. Let n be a positive integer. It follows from Theorem 7.11 that lim ✓ (b) = 0 and lim ✓ (b)= . Hence, by the intermediate ! 1 !1 1 value theorem, there exists a real number n such that ✓ n (b)=⇡n.Since the function ✓ (b) is strictly increasing, we have ✓ (b) <⇡nfor < 7! n and ✓ (b) >⇡nfor > n. Therefore, n is the only solution of the equation ✓ (b)=⇡n.Thisproves(i). In order to prove (ii), we observe that ✓ n (b)=⇡n by definition of n.Thisimplieslimn ✓ (b)= . Since the function ✓ (b)is !1 n 1 7! n continuous, this can only happen if limn n = . !1 1 Finally, (iii) and (iv) follow immediately from Proposition 7.9 and Propo- sition 7.10. This completes the proof of Theorem 7.12. ⇤ 7.4. The Liouville normal form Consider the eigenvalue problem d d (13) p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=u(b)=0. dt dt h i We continue to assume that p(t),⇢(t) > 0. We say that a system is in Liouville normal form if p(t)=⇢(t) = 1 for all t [a, b]. 2 We claim that every system is equivalent to one that is in Liouville normal form. To explain this, let b ⇢(s) T = ds. p(s) Za s We then consider the system d2 (14) w(⌧)+(Q(⌧)+ ) w(⌧)=0,w(0) = w(T )=0, d⌧ 2 where Q :[0,T] R is defined by ! t ⇢(s) 1 1 d d 1 Q ds = ⇢(t) q(t)+(p(t) ⇢(t)) 4 p(t) (p(t) ⇢(t)) 4 . a sp(s) dt dt ✓ Z ◆ ⇢ h i The systems (13) and (14) are equivalent in the following sense: 7.4. The Liouville normal form 101 Theorem 7.13. Suppose that u :[a, b] R and w :[0,T] R are related ! ! by t 1 ⇢(s) u(t)=(p(t)⇢(t)) 4 w ds . p(s) ✓ Za s ◆ Then u is a solution of (13) if and only if w is a solution of (14). Proof. Di↵erentiating the identity t 1 ⇢(s) u(t)=(p(t)⇢(t)) 4 w ds p(s) ✓ Za s ◆ with respect to t gives t d 1 d ⇢(s) p(t) u(t)=(p(t)⇢(t)) 4 w ds dt d⌧ p(s) ✓ Za s ◆ t d 1 ⇢(s) + p(t) (p(t)⇢(t)) 4 w ds . dt p(s) ✓ Za s ◆ This implies 2 t d d 1 d ⇢(s) 4 p(t) u(t) = ⇢(t)(p(t)⇢(t)) 2 w ds dt dt d⌧ a sp(s) h i ✓ Z ◆ t d 1 d ⇢(s) + (p(t)⇢(t)) 4 w ds dt d⌧ p(s) ✓ Za s ◆ t 1 d 1 d ⇢(s) +(p(t)⇢(t)) 2 (p(t)⇢(t)) 4 w ds dt d⌧ p(s) ✓ Za s ◆ t d d 1 ⇢(s) + p(t) (p(t)⇢(t)) 4 w ds dt dt a sp(s) h i ✓ Z ◆ 2 t 1 d ⇢(s) = ⇢(t)(p(t)⇢(t)) 4 w ds d⌧ 2 p(s) ✓ Za s ◆ t d d 1 ⇢(s) + p(t) (p(t)⇢(t)) 4 w ds . dt dt a sp(s) h i ✓ Z ◆ 102 7. Sturm-Liouville theory Therefore, we obtain d d p(t) u(t) + q(t) u(t) dt dt h i 2 t 1 d ⇢(s) = ⇢(t)(p(t)⇢(t)) 4 w ds d⌧ 2 p(s) ✓ Za s ◆ t t 1 ⇢(s) ⇢(s) + ⇢(t)(p(t)⇢(t)) 4 Q ds w ds . p(s) p(s) ✓ Za s ◆ ✓ Za s ◆ From this, the assertion follows easily. ⇤ In other words, the two systems (13) and (14) have the same eigenvalues, and there is a one-to-one correspondence between eigenfunctions of (13) and eigenfunctions of (14). 7.5. Asymptotic behavior of eigenvalues of a Sturm-Liouville system In this section, we analyze the asymptotic behavior of the eigenvalues of a Sturm-Liouville system. We first consider a system which is in Liouville normal form: Proposition 7.14. Consider the system d2 u(t)+(q(t)+ ) u(t)=0,u(a)=u(b)=0. dt2 The n-th eigenvalue of this system can be bounded by ⇡2n2 ⇡2n2 q q . (b a)2 max n (b a)2 min Proof. As before, we denote by u (t) the unique solution of the initial value problem d2 u(t)+(q(t)+ ) u(t)=0,u(a)=0,u0(a)=1. dt2 As usual, we write d u (t)=r (t) cos ✓ (t) dt u (t)=r (t)sin✓ (t), where ✓ (a) = 0. Moreover, we consider the function t a uˆ(t)=sin ⇡n . b a ⇣ ⌘ 7.5. Asymptotic behavior of eigenvalues of a Sturm-Liouville system 103 The functionu ˆ(t) is a solution of the Sturm-Liouville equation d2 ⇡2n2 uˆ(t)+ uˆ(t)=0. dt2 (b a)2 We may write d uˆ(t)=ˆr(t) cos ✓ˆ(t) dt uˆ(t)=ˆr(t)sin✓ˆ(t), where ✓ˆ(a) = 0. We first define ⇡2n2 = q 0. (b a)2 max Then ⇡2n2 q(t)+ (b a)2 for all t [a, b]. Hence, Corollary 7.6 implies that ✓ (t) ✓ˆ(t) for all 2 t [a, b]. In particular, 2 ✓ (b) ✓ˆ(b)=⇡n = ✓ (b). n Since the function ✓ (b) is strictly monotone increasing, we conclude 7! that ⇡2n2 = q . n (b a)2 max In the next step, we define ⇡2n2 = q . (b a)2 min Then ⇡2n2 q(t)+ (b a)2 for all t [a, b]. By Corollary 7.6, we have ✓ (t) ✓ˆ(t) for all t [a, b]. 2 2 Hence, we obtain ✓ (b) ✓ˆ(b)=⇡n = ✓ (b). n Since the function ✓ (b) is strictly monotone increasing, we conclude 7! that ⇡2n2 = q . n (b a)2 min This completes the proof of Proposition 7.14. ⇤ 104 7. Sturm-Liouville theory Corollary 7.15. Consider the system d2 u(t)+(q(t)+ ) u(t)=0,u(a)=u(b)=0. dt2 If n denotes the n-th eigenvalue of this system, then ⇡2n2 = + O(1). n (b a)2 Combining Corollary 7.15 with Theorem 7.13, we can draw the following conclusion: Theorem 7.16. Consider the system d d p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=u(b)=0. dt dt h i Let n be the n-th eigenvalue of this system. Then ⇡2n2 = + O(1), n T 2 where b ⇢(t) T = dt. p(t) Za s 7.6. Asymptotic behavior of eigenfunctions We now analyze the asymptotic behavior of the n-th eigenfunction of a Sturm-Liouville system. In view of the discussion above, it su ces to con- sider systems in Liouville normal form. Proposition 7.17. Suppose that u :[a, b] R is a solution of the equation ! d2 u(t)+(q(t)+ ) u(t)=0 dt2 with u(a)=0. Moreover, suppose that u is normalized such that b u(t)2 dt =1. Za If is large, then 2 1 u(t)= sin(p (t a)) + O . ± b a p r ⇣ ⌘ Proof. Let us fix a large number >0. The di↵erential equation d2 u(t)+(q(t)+ ) u(t)=0 dt2 7.6. Asymptotic behavior of eigenfunctions 105 can be rewritten as d 1 d 1 1 (15) 2 u(t) +( 2 q(t)+ 2 ) u(t)=0. dt dt h i We now perform a Pr¨ufer substitution for (15). In other words, we write 1 d 2 u(t)=r(t) cos ✓(t) dt u(t)=r(t)sin✓(t), where ✓(a) = 0. Without loss of generality, we may assume that r(a) > 0. (Otherwise, we replace u(t)by u(t).) The functions r(t) and ✓(t) satisfy the di↵erential equations 1 r0(t)= 2 q(t) r(t) cos ✓(t)sin✓(t) and 1 1 2 1 2 ✓0(t)=( 2 q(t)+ 2 )sin ✓(t)+ 2 cos ✓(t). This implies d 1 log r(t)= 2 q(t) cos ✓(t)sin✓(t) dt and d 1 1 (✓(t) 2 t)= 2 q(t). dt Consequently, d 1 log r(t) C 2 dt 1 and d 1 1 (✓(t) 2 t) C 2 , dt 1 where C1 =supt [a,b] q(t ) . Integrating these inequalities gives 2 | | r(t) 1 log C 2 r(a) 2 and 1 1 ✓(t) 2 (t a) C 2 | | 2 for all t [a, b]. This implies 2 r(t) 1 1 C 2 r(a) 3 and 1 1 sin ✓(t) sin( 2 (t a)) C 2 | | 3 106 7. Sturm-Liouville theory for all t [a, b]. Putting these facts together, we obtain 2 1 1 r(t) 1 u(t) sin( 2 (t a)) = sin ✓(t) sin( 2 (t a)) r(a) r(a) r(t) 1 (16) 1 + sin ✓(t) sin( 2 (t a)) r(a) | | 1 2 C 2 . 3 We now estimate the term r(a). To that end, we observe that 1 2 2 1 1 u(t) sin ( 2 (t a)) 8 C 2 r(a)2 3 for all t [a, b]. We now integrate this inequality over the interval [a, b]. 2 Using the identities b u(t)2 dt =1 Za and b 2 1 b a 1 1 1 sin ( 2 (t a)) dt = cos( 2 (b a)) sin( 2 (b a)), 2 1 Za 2 2 we obtain 1 b a 1 C 2 . r(a)2 2 4 Consequently, 2 1 r(a) C 2 . b a 5 r In particular, r(a) C 6,whereC6 is a uniform constant that does not depend on . Using (16), we obtain 1 1 u(t) r(a)sin( 2 (t a)) C 2 , | | 7 hence 2 1 1 u(t) sin( 2 (t a)) C 2 . b a 8 r This completes the proof. ⇤ Corollary 7.18. Consider the system d2 u(t)+(q(t)+ ) u(t)=0,u(a)=u(b)=0. dt2 Let n be the n-th eigenvalue of this system. Moreover, suppose that un is the corresponding eigenfunction, normalized such that b 2 un(t) dt =1. Za 7.6. Asymptotic behavior of eigenfunctions 107 For n large, we have the asymptotic expansion 2 t a 1 u (t)= sin ⇡n + O . n ± b a b a n r ⇣ ⌘ ⇣ ⌘ Proof. By Proposition 7.17, 2 1 u (t)= sin( (t a)) + O . n n p ±rb a n p ⇣ ⌘ Moreover, it follows from Corollary 7.15 that ⇡n 1 = + O . n b a n p ⇣ ⌘ Putting these facts together, the assertion follows. ⇤ Combining Corollary 7.18 with Theorem 7.13, we can draw the following conclusion: Theorem 7.19. Consider the system d d p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=u(b)=0. dt dt h i Let n be the n-th eigenvalue of this system. Moreover, suppose that un is the corresponding eigenfunction, normalized such that b 2 ⇢(t) un(t) dt =1. Za For n large, we have t 2 1 ⇡n ⇢(s) 1 un(t)= (p(t)⇢(t)) 4 sin ds + O , ± T T a sp(s) n r ✓ Z ◆ ⇣ ⌘ where b ⇢(s) T = ds. p(s) Za s Proof. We may write t 1 ⇢(s) u (t)=(p(t)⇢(t)) 4 w ds n n p(s) ✓ Za s ◆ for some function wn :[0,T] R. The function wn satisfies ! d2 w (⌧)+(Q(⌧)+ ) w (⌧)=0,w(0) = w (T )=0. d⌧ 2 n n n n n 108 7. Sturm-Liouville theory Using the substitution rule, we obtain T b ⇢(t) t ⇢(s) 2 w (⌧)2 d⌧ = w ds dt n p(t) n p(s) Z0 Za s ✓ Za s ◆ b 2 = ⇢(t) un(t) dt Za =1. By Corollary 7.18, the function wn satisfies 2 ⇡n 1 w (⌧)= sin ⌧ + O n ± T T n r ⇣ ⌘ ⇣ ⌘ for n large. Thus, t 2 1 ⇡n ⇢(s) 1 un(t)= (p(t)⇢(t)) 4 sin ds + O ± T T a sp(s) n r ✓ Z ◆ ⇣ ⌘ for n large. ⇤ 7.7. Orthogonality and completeness of eigenfunctions Let us consider a Sturm-Liouville system of the form d2 u(t)+(q(t)+ ) u(t)=0,u(a)=u(b)=0. dt2 Let us denote by 1 < 2 <...the eigenvalues of this Sturm-Liouville system, and let u1(t),u2(t),... denote the associated eigenfunctions. We assume that the eigenfunctions are normalized such that b 2 un(t) dt =1 Za for all n. Proposition 7.20. We have b um(t) un(t) dt =0 Za for m = n. 6 7.7. Orthogonality and completeness of eigenfunctions 109 Proof. We compute b d d d 0= um(t) un(t) un(t) um(t) dt a dt dt dt Z ⇣ ⌘ b d2 d2 = u (t) u (t) u (t) u (t) dt m dt2 n n dt2 m Za b ⇣ ⌘ = ( ) u (t) u (t) dt. m n m n Za Since = , the assertion follows. m 6 n ⇤ Corollary 7.21. For every continuous function v, we have 2 1 b b u (t) v(t) dt v(t)2 dt. n n=1 a a X ✓ Z ◆ Z Proof. Let m b w = v u (t) v(t) dt u . n n n=1 a X ✓ Z ◆ Then b b m b 2 0 w(t)2 dt = v(t)2 dt u (t) v(t) dt . n a a n=1 a Z Z X ✓ Z ◆ Since m is arbitrary, the assertion follows. ⇤ In the remainder of this section, we establish a completeness property of the set of eigenfunctions. More precisely, we will show that b b 2 2 1 v(t) dt = un(t) v(t) dt a n=m+1 a Z X ✓ Z ◆ for every continuous function v. In order to prove this, we will follow the arguments in Birkho↵and Rota’s book [2]. Let 2 t a uˆ (t)= sin ⇡n . n b a b a r ⇣ ⌘ It is a well known fact that every continuous function on the interval [a, b] can be represented by a Fourier series. Moreover, for every continuous function v we have b b 2 2 1 v(t) dt = uˆn(t) v(t) dt . a n=m+1 a Z X ✓ Z ◆ This relation is known as Parseval’s identity. Moreover, Corollary 7.18 im- plies that b 2 1 (un(t) uˆn(t)) dt = O 2 , a n Z ⇣ ⌘ 110 7. Sturm-Liouville theory hence 1 b (u (t) uˆ (t))2 dt < . n n 1 n=1 a X Z Proposition 7.22. Fix an integer m such that 1 b 1 (u (t) uˆ (t))2 dt . n n 4 n=m+1 a X Z Moreover, suppose that v is a function satisfying b uˆn(t) v(t) dt =0 Za for n =1,...,m. Then 2 b 1 b v(t)2 dt 4 u (t) v(t) dt . n a n=m+1 a Z X ✓ Z ◆ Proof. Using Parseval’s identity for Fourier series, we obtain b b 2 2 1 v(t) dt = uˆn(t) v(t) dt a n=m+1 a Z X ✓ Z ◆ 2 1 b 2 u (t) v(t) dt n n=m+1 a X ✓ Z ◆ 2 1 b +2 (u (t) uˆ (t)) v(t) dt n n n=m+1 a X ✓ Z ◆ 2 1 b 2 u (t) v(t) dt n n=m+1 a X ✓ Z ◆ 1 b b +2 (u (t) uˆ (t))2 dt v(t)2 dt n n n=m+1 a a X ✓ Z ◆✓Z ◆ 2 1 b 1 b 2 u (t) v(t) dt + v(t)2 dt. n 2 n=m+1 a a X ✓ Z ◆ Z From this, the assertion follows. ⇤ Corollary 7.23. Fix an integer m such that 1 b 1 (u (t) uˆ (t))2 dt . n n 4 n=m+1 a X Z Moreover, suppose that v span u ,...,u is a function satisfying 2 { 1 m} b uˆn(t) v(t) dt =0 Za 7.7. Orthogonality and completeness of eigenfunctions 111 for n =1,...,m. Then v =0. Proposition 7.24. Fix a continuous function v and a real number ">0. Then there exists an integer m>0 and a function w span u ,...,u 2 { 1 m} such that b (v(t) w(t))2 dt ". Za Proof. Let us fix an integer m such that 1 b 1 (u (t) uˆ (t))2 dt n n 4 n=m+1 a X Z and 2 1 b " u (t) v(t) dt . n 4 n=m+1 a X ✓ Z ◆ By Corollary 7.23, we can find a function w span u ,...,u such that 2 { 1 m} b uˆ (t)(v(t) w(t)) dt =0 n Za for n =1,...,m. Using Proposition 7.22, we obtain 2 b 1 b (v(t) w(t))2 dt 4 u (t)(v(t) w(t)) dt n a n=m+1 a Z X ✓ Z ◆ 2 1 b =4 un(t) v(t) dt n=m+1 a X ✓ Z ◆ ", as claimed. ⇤ Corollary 7.25. For every continuous function v, we have b 2 b 1 2 un(t) v(t) dt = v(t) dt. n=1 a a X ✓ Z ◆ Z Proof. Given any ">0, we can find an integer m>0 and a function w span u ,...,u such that 2 { 1 m} m b 2 b u (t)(v(t) w(t)) dt (v(t) w(t))2 dt ". n n=1 a a X ✓ Z ◆ Z 112 7. Sturm-Liouville theory This gives b b v(t)2 dt w(t)2 dt + p" s s Za Za m b 2 = u (t) w(t) dt + p" v n un=1 ✓ a ◆ uX Z t m b 2 u (t) v(t) dt +2p". v n un=1 ✓ a ◆ uX Z t Since ">0 is arbitrary, the asserion follows. ⇤ Combining Corollary 7.18 with Theorem 7.13, we obtain the following result: Theorem 7.26. Consider the system d d p(t) u(t) +(q(t)+ ⇢(t)) u(t)=0,u(a)=u(b)=0. dt dt h i Let n be the n-th eigenvalue of this system. Moreover, suppose that un is the corresponding eigenfunction, normalized such that b 2 ⇢(t) un(t) dt =1. Za Then b 2 b 1 2 ⇢(t) un(t) v(t) dt = ⇢(t) v(t) dt n=1 a a X ✓ Z ◆ Z for every continuous function v. 7.8. Problems Problem 7.1. Suppose that a and b are real numbers such that a Bibliography [1] H. 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