Version: Jan. 2004. MATH 2921 , ODE 2, Spring, 2004sph/ode_notes/2921-04-notes0.pdf · MATH 2921 , ODE 2, Spring, 2004 CLASS NOTES ... The versions will be dated so you will know

Version: Jan. 2004.

Changes starting on page 7 from version of Jan. 4, and a new section addedstarting on page 16 from the version of Jan. 8

MATH 2921 , ODE 2, Spring, 2004

CLASS NOTES

Topic: Ordinary di¤erential equations and dynamical systems

Text: No speci�c text. Material will be selected from the following books:

1. Nonlinear Oscillations, Dynamical Systems, and Bifurcations of Vector Fields,by Guckenheimer and Holmes.

2. Stability, Instability, and Chaos, by Glendinning

3. Nonlinear Ordinary Di¤erential Equations, by Grimshaw

4. Nonlinear dynamical systems and di¤erential equations, by Verhulst.

5. Ordinary Di¤erential Equations, by Coddington and Levinson

Some of these will be on reserve in the math library.

I expect to have class notes covering all material of the lectures on the web, at

www.math.pitt.edu/~sph/ode_notes

The versions will be dated so you will know if you have the latest one. I will alsonote if the changes in a new version are just trivial corrections.Initial versions of these notes are there now, but please note that they will

be continually revised up to the point where I discuss them in class.

O¢ ce Hours: M W F 1:15-1:45. M W 4:15-4:45, or by appointment.

Grades: Grades will be given on the basis of homework assignments to be madeevery couple of weeks, approximately.

1

Stable Manifold Theorem

Consider an autonomous system of n di¤erential equations in n unknown func-tions, written in vector form as

y0 = f (y) (1)

under the assumption that f (0) = 0 , and all of the second partial derivatives of fare continuous in some neighborhood of 0: Suppose that A is the Jacobian matrix�@fi@xj

�: We can then write

f (y) = Ay + g (y)

where

limjyj!0

g (y)

jyj = 0: (2)

An interesting situation is when some of the eigenvalues of A have negative realpart and the rest have positive real part. We start with a linear example:

y01 = �y1 (3)

y02 = y2:

The general solution is easily found:

y =c1e�t�10

�+ c2e

t

�01

�:

If c2 = 0 then limt!1 y (t) = 0: If c1 = 0 the limt!�1 y (t) = 0: (In this example,those solutions which approach zero as t tends to minus in�nity have the propertythat they tend to in�nity, in norm, as t tends to in�nity. However that is not truein all examples, as we shall see below.)

With this motivation we have two important de�nitions. For simplicity we willassume that f 2 C2 (Rn; Rn) ; that is, we choose = Rn. Consider once again (1) ;that is y0 = f (y) : Assume that f (y0) = 0: For c 2 ; let � (t; c) denote the valueat time t of the solution of (1) with initial condition y (0) = c:

De�nition 1 The stable manifold S for (1) at y0 is the set

S =nc j lim

t!1� (t; c) = y0

o:

The unstable manifold U is the set

U =�c j lim

t!�1� (t; c) = y0

�:

2

The stable manifold for (3) is the y1 axis. The unstable manifold is the y2 axis.They intersect only at (0; 0) : All points not on either axis are on neither the stablenor the unstable manifolds.

Example:

y01 = �y1y02 = y2 � 3y21:

Since we can solve for y1; the whole system is easy to solve. We �nd that the generalsolution is

y1 = ce�t

y2 = det + c2e�2t;

where c and d are arbitrary constants. If d = 0; then limt!1 y (t) = 0: If c = 0;then limt!�1 y (t) = 0: If c 6= 0 and d 6= 0; then y2 is unbounded both as t ! 1and as t! �1. Notice that y1(0) = c; y2(0) = d+ c2: Hence the stable manifold isthe curve y2 = y21; and the unstable manifold is the y2 axis. This is pictured in thephase diagram below.

In this phase diagram we have drawn some solution curves (the curves t !(y1 (t) ; y2 (t)) ), and we have also drawn the �nullclines�. These are the curves on

3

which either y01 = 0 or y02 = 0: At a point where a solution crosses a nullcline withy01 = 0; we have y02 6= 0; for otherwise the point is an equilibrium point, and thesolution is constant. In this case, the tangent vector (y01 (t) ; y

02 (t)) to the curve at

the point of crossing is vertical, pointing up or down depending on the sign of y02:Similarly, at a point where the solution curve crosses a nullcline where y02 = 0; thetangent vector is horizontal, pointing to the left or right depending on the sign ofy01: You should observe that the nullclines are usually not solution curves, but thestable and unstable manifolds are.

Notice that the stable manifold can be expressed in the form y2 = p (y1) ; andthe unstable manifold can be expressed in the form y1 = q (y2) (= 0).

This is the typical situation �locally�, that is, in some neighborhood of theequilibrium point. This is the content of the following very important theorem.We make the same smoothness assumptions of f as above, namely, that the partialderivatives

@fi@yj

are all continuous in some neighborhood of 0: As before, A denotes the Jacobianderivative

�@fi@yj

�jy=0 . We recall that we can transform A to Jordan canonical

form giving the form below. Here the matrices Ni are either zero, or have 1 on theupper o¤ diagonal. Thus they are �nilpotent�matrices of dimension equal to themultiplicity of the eigenvalue �i.

P�10 AP0 =

0BBBB@�1I1 +N1 0 0 0 0

0 �2I2 +N2 0 0 00 0 ::: 0 00 0 0 �k�1Ik�1 +Nk�1 00 0 0 0 �kIk +Nk

1CCCCA : (4)

Also, in the statement of the theorem, we are considering a system

y0 = Ay + g (y)

where y 2 Rn+m for some n � 1 and m � 1: The function g satis�es the samesmoothness conditions as above, and for each i; j;

@gi@yj

j0 = 0

4

In the theorem below c is not the initial condition. Instead, it is part of theinitial condition, being of dimension n: This is di¤erent from our previous use ofc; where it had the full dimension of the system under consideration. We use thenotation C to denote the initial point of a solution in Rn+m:

The statement of the theorem is fairly complicated for two reasons. First, thereis the need to allow for the transformation which puts A in Jordan canonical form,and second, there is the need to allow for the powers tj which appear in solutions ofthe linear equation y0 = Ay when Ni 6= 0: Let me say a few more words about thelatter complication. If A itself can be diagonalized, then there are no powers of tin the solution. For example, the system

y01 = y2

y02 = y1

has solutions

c1et

�11

�+ c2e

�t�

1�1

�:

Looking particularly at the second term, we have a bound����e�t� 1�1

����� � e�t:(Recall what norm we are using for vectors.)

But consider the system

y01 = y2

y02 = �2y2 � y1:

This is equivalent to the second order equation y00 + 2y0 + y = 0; and we know,because of repeated roots, that this has a solution y = te�t: This does not satisfythe bound jyj � Ke�t; for any K: But, if we choose some small " > 0; then therewill be a K such that jy (t)j � Ke�(1�")t: This is simply saying that te�t

e�(1�")t= te�"t is

bounded on [0;1): The constant K increases as " gets smaller. The solution decaysexponentially, but not quite at the rate suggested by the eigenvalues. The bounddepends on the particular solution, as seen by considering the solution 2te�t; whereobviously the K must be larger.

5

This is one reason for the appearance of �"� in the theorem below. We notethat in this theorem, when discussing the number of eigenvalues, we are includingmultiplicity, so that �i might equal �j for some i 6= j: In stating the theorem we useT to denote transpose, to turn some row vectors into column vectors. In the proofthere may be some inconsistency in the use of column and row vectors; think of allvectors as column vectors but as equivalent to points in Rn.

STABLE MANIFOLD THEOREM. Suppose that the (m+ n) � (m+ n)matrix A has n � 1 eigenvalues �1; :::; �n; (not necessarily distinct) with real partsall less than or equal to ��1 < 0 and m � 1 eigenvalues �n+1; :::; �N with positive realparts. Choose " > 0 so that ��1+ " < 0: Then there is a neighborhood � of 0 inRn and a smooth function Q : �! Rm with Q (0) = 0; such that if c = (c1; ::::; cn) isa point in �; and (d1; :::; dm) = Q (c1; :::; cn) ; and if C = P0 (c1; :::; cn; d1; :::; dm)

T ;then limt!1 � (t;C) = 0: Further, there is a positive number � (") which alsodepends on jCj such that for all t � 0;

j� (t;C)j � � (") e�(�1�")t:

Also, for any � > 0 there is a � > 0 such that if jcj < �; then j� (t;C)j < � for allt � 0: Finally, there is a �0 > 0 such that if c 2 and (d1; :::; dm) 6= Q (c1; :::; cn),and C = P0 (c1; :::; cn; d1; :::; dm)

T ; then j� (t;C)j > �0 for some t � 0:

Remarks:

1. In this theorem, the �1; :::::; �n+m represent all of the eigenvalues of A: Itcould be that �i = �j for some i 6= j: However, the matrix P0 is still the matrixwhich puts A in Jordan form Its columns are eigenvectors or generalized eigenvectorsof A:

2. In this result we do not let "! 0: It is chosen, as small as we wish, and thena su¢ ciently small region is found.

We can perhaps understand this theorem better if we let x =P�10 y: Then

x0 = P�10 AP0x+P�10 g (P0x) :

6

From formula (4) we see that this can be broken into two systems, namely0@ x1:::xn

1A0

= A1

0@ x1:::xn

1A+ h1 (x1; ::::::; xm+n)0@ xn+1

:::xm+n

1A0

= A2

0@ xn+1:::xm+n

1A+ h2 (x1; :::::; xm+n) :Note that the two systems, of dimensions n and m; are coupled only throughthe higher order terms h1 and h2. The change of variables x =P0y removes anycoupling of the linear terms.

All the eigenvalues of A1 have real parts less than or equal to ��1: All theeigenvalues of A2 have positive real parts. In some neighborhood of 0; all pointsof the form 0@ xn+1

:::xN

1A = Q

0@ x1:::xn

1A (5)

lie on the stable manifold, and these solutions all tend to zero at an exponential rate.In other words, jx (t)j = O

�e�(�1�")t

�as t!1: Recall that �1 � " > 0:

It should be noted that there could be a point not of the form (5) which ison the stable manifold. But no matter how close this point is to the origin, thesolution starting at this point must leave some �xed neighborhood of the origin,before returning and tending to 0: Eventually (for large enough t), the points onthis solution would be of the form (5) : We will give an example of this below.The reason for this somewhat strange situation is that the de�nition we gave of the�stable manifold� is a global one. Any solution which eventually tends to zero ison the stable manifold, not just solutions which start, and remain, in some smallneighborhood of the origin.

For an example of a stable manifold where the transformation to x coordinatesis needed, consider

y01 = �y2 �3

4(y1 + y2)

2

y02 = �y1 +3

4(y1 + y2)

2 :

7

Here is the phase plane.

If we let

x1 =1

2(y1 + y2)

x2 =1

2(y1 � y2) ;

we get

x01 = �x1x02 = x2 � 3x21:

We analyzed this earlier and saw that the stable manifold is x2 = x21: This can easilyexpressed in terms of y; but it is not of the form y2 = Q (y1) : Look back at thephase plane for the system just above and compare it with the one for (y1; y2) :

Note that by the change of variables t! �t we change the sign of the eigenvalues.The stable manifold of the new system is the unstable manifold of the original system.Therefore the theorem gives the existence of both the stable and unstable manifolds.Solutions starting at points not on either manifold leave some given neighborhoodboth as t increases and as t decreases.

Proof of Stable Manifold Theorem in case m = n = 1:

8

We assume that �1 = �1 and �2 = +1; and we assume that the system hasalready been transformed to the x coordinates. It can thus be written as

x01 = �x1 + g1 (x1; x2) (6)

x02 = x2 + g2 (x1; x2) :

Further,@gi@xj

(0; 0) = 0 for i; j = 1; 2:

This implies that

limjxj!0

jg (x)jjxj = 0:

The technique, which is important, is to write this as an integral equation inthe correct way, using the variation of parameters formula in such a way that onlysolutions which tend to zero as t!1 are included. Using inde�nite integrals, thevariation of parameters formula applied to each equation separately gives somethingof the form

x1 (t) = ce�t +

Zes�tg1 (x1 (s) ; x2 (s)) ds:

x2 (t) = det +

Zet�sg2 (x1 (s) ; x2 (s)) ds

But we must choose appropriate limits in the two integrals.

For the �rst, since we have a term e�t in the integral, we can expect that ifResg1 (x1 (s) ; x2 (s)) ds is not too large, the term will tend to zero. But in the

second integral we have et in both terms, and some cancellation is necessary to getthe limit to be zero. Consider the following system of integral equations.

x1 (t) = ce�t + e�t

Z t

0

esg1 (x1 (s) ; x2 (s)) ds

x2 (t) = et

�d+

Z t

0

e�sg2 (x1 (s) ; x2 (s)) ds

�As t!1 we can expect that the integral in the second line will converge, becauseof the term e�s. But we need to know that the solution exists on [0;1): Assuming

9

this we would choose d = �R10e�sg2 (x1 (s) ; x2 (s)) ds in order to get cancellation.

We therefore consider the following system of equations:

x1 (t) = ce�t + e�t

Z t

0

esg1 (x1 (s) ; x2 (s)) ds (7)

x2 (t) = �et�Z 1

t

e�sg2 (x1 (s) ; x2 (s)) ds

�(8)

Our goal is to �nd a unique solution to the pair of integral equations (7) � (8) ;for each c in some neighborhood of 0: We need to use the properties of g:

It is not immediately obvious that an x1 and x2 satisfying these equations willtend to zero at in�nity. This will be part of the proof. But it is interesting to notethe following L�hôpital�s rule calculation: If h is continuous on some interval [T;1);and limt!1 h (t) = 0 then

limt!1

e�tZ t

0

esh (s) ds = limt!1

R t0esh (s) ds

et

= limt!1

eth (t)

et= 0;

andR1te�sh (s) ds converges, with

limt!1

etZ 1

t

e�sh (s) ds = limt!1

R1te�sh (s) ds

e�t

= limt!1

�e�th (t)�e�t = 0:

(You should check that the assumptions of L�hôpital�s rule are satis�ed.)

We recall that @gi@xjj(0;0) = 0; and that the partial derivatives @gi

@xjare continuous

in some region jxj < �0. Therefore, for any � 2 (0; �0) ; g must satisfy a Lipschitzcondition of the form

jg (x1)� g (x2)j � L (�) jx1 � x2j (9)

for all x1;x2 in the region jxj < � , and further, lim�!0 L (�) = 0: In particular,this inequality holds with x2 = 0;giving

jg (x)j � L (�) jxj

10

if jxj < �; for su¢ ciently small �:

Now suppose that for some " 2 (0; 1) and some � < �0; x (t) = (x1 (t) ; x2 (t)) isa continuous vector valued function on [0;1); and

jx (t)j � �e�(1�")t for all t � 0:

Thenjg (x (t))j � L (�) �e�(1�")t

for all t � 0: Recalling the de�nition of the norm of a vector, this inequality is alsosatis�ed by each gi individually. We then haveZ t

0

e�(t�s) jg1 (x1 (s) ; x2 (s))j ds �Z t

0

e�(t�s)e�s+"sL (�) �ds

=�L (�)

"e�t�e"t � 1

�� �L (�)

"e�(1�")t; (10)

andZ 1

t

e(t�s) jg2 (x1 (s) ; x2 (s))j ds �Z 1

t

e(t�s)e�(1�")sL (�) �ds =�L (�)

2� " e�(1�")t: (11)

We now construct a solution to (7)� (8) by successive approximations, with startingfunctions

x01 (t) = ce�t

x02 (t) = 0:

Assume that jcj � 12�0: With x = (x1; x2) ; we have��x0 (t)�� � �0e�(1�")t;

and so ��g �x0��� � �0L (�0) e�(1�")t:This implies, in particular, that

R1te�sg2 (x

0 (s)) ds converges. We set

x11 (t) = ce�t + e�t

Z t

0

esg1�x01 (s) ; x

02 (s)

�ds

x12 (t) = �etZ 1

t

e�sg2�x01 (s) ; x

02 (s)

�ds:

11

Then, using equations (10) and (11) ; we �nd that

��x11 (t)�� � ce�t + e�t Z t

0

es��g1 �x01 (s) ; x02 (s)��� ds

� ce�t + e�tZ t

0

es�0L (�0) e�(1�")sds � ce�t + �0L (�0)

"e�(1�")t

and ��x12 (t)�� � �0L (�0)

2� " e�(1�")t:

Recall that we are holding " �xed. Hence we can choose �0 so small that for0 < � � �0;

L (�)

"<1

2: (12)

Then L(�)2�" < 1 as well. If we assume that jcj < 1

2�0; then we obtain that���x11 (t) ; x12 (t)��� < �0e�(1�")t (13)

for all t � 0. We continue by induction, letting

xj+11 (t) = ce�t + e�tZ t

0

esg1�xj1 (s) ; x

j2 (s)

�ds

xj+12 (t) = �etZ 1

t

e�sg2�xj1 (s) ; x

j2 (s)

�ds:

Suppose that xj has been de�ned, and��xj (t)�� � �0e�(1�")t (14)

for t � 0: Then the steps leading to (13) can be repeated with xj substituted forx0 to show that xj+1 is de�ned and also satis�es (14) :

In particular, each xji (t) is bounded by �0 on [0;1) and tends to zero at anexponential rate as t!1:

Also, the Lipschitz condition continues to apply to the di¤erence between thesuccessive approximations. We therefore have

�� xj+11 (t)� xj1 (t)�� � e�t Z t

0

esL (�0)��xj (s)� xj�1 (s)�� ds

12

and �� xj+12 (t)� xj2 (t)�� � et

Z 1

t

e�sL (�0)��xj (s)� xj�1 (s)�� ds

Further, ��x1 (t)� x0 (t)�� = jcj e�t:

The inductive steps are then almost the same as those de�ning the xj: We �ndthat if ��xj (t)� xj�1 (t)�� � �1

2

�j�1jcj e�(1�")t

for t � 0; then

��xj+11 (t)� xj1 (t)�� � e�t Z t

0

esL (�0)

�1

2

�j�1jcj e�(1�")sds

� e�(1�")t�1

2

�j�1jcj L (�0)

"��1

2

�jjcj e�(1�")t

because of (12) : Also,

��xj+12 (t)� xj2 (t)�� � et Z 1

t

e�sL (�0)

�1

2

�j�1jcj e�(1�")sds

��1

2

�j�1jcj L (�0)2� " e

�(1�")t ��1

2

�jjcj e�(1�")t:

From this it follows as in the usual successive approximations procedure that x0 +�1j=1 (x

j � xj�1) converges uniformly on [0;1) to a unique solution of (7) � (8) :Furthermore this solution tends to zero at in�nity in such a way that jx (t)j ��0e

�(1�")t for all t:

We can now give the formula of the local stable manifold as follows. The region� in the statement of the stable manifold theorem is, in this two dimensional case,the region jxj < 1

2�0; where �0 was chosen above. For any c in � there is a solution

(x1 (t; c) ; x2 (t; c)) ; the solution with x1 (0) = c; which remains in jxj � �0 and tendsto zero with a certain exponential rate. Also, x2 (0; c) =

13

d = �Z 1

0

e�sg2 (x1 (s; c) ; x2 (s; c)) ds:

This is the formula for the local part of the stable manifold as in the statement ofthe theorem. That is,

Q (c) = �Z 1

0

e�sg2 (x1 (s; c) ; x2 (s; c)) ds:

I should remark that the stable manifold S is the graph of the function Q: Thus,in the n +m dimensional case, it is a smooth surface which is a graph in rotatedcoordinates.

We now wish to give another important property of the map Q. I will state thisin the n+m dimensional case, but in the situation where the matrix A is in Jordancanonical form so that the matrix P0 in (4) and in the statement of the theorem isthe identity.

Theorem: Under the hypotheses of the stable manifold theorem as stated earlierin the notes, the function Q:�! Rm is di¤erentiable at c = 0 and DQ (0) is them� n matrix of zeros.

Geometrically, this means that the stable manifold is tangent at 0 to the invariantsubspace (eigenspace) of A corresponding to the n eigenvalues with negative realpart.

I will not prove this result, only give a brief outline of the proof in the two-dimensional case. First, though, I will give an example. We had earlier theexample

y01 = �y1y02 = y2 + 3y

21:

We saw that the stable manifold was y2 = y21: In our new notation, Q (c) = c2:We see that indeed, Q0 (0) = 0: This means that the stable manifold is tangent atthe origin to the y1 axis. The y1 axis is itself the stable manifold for the linearizedproblem,

14

y01 = y1

y02 = �y2:

To prove that Q is di¤erentiable, we note that g2 is di¤erentiable with respect tox1 and x2. Therefore we have to prove the x1 (t; c) and x2 (t; c) are di¤erentiablewith respect to c: We recall that they are de�ned by the integral equations

x1 (t; c) = ce�t + e�t

Z t

0

esg1 (x1 (s; c) ; x2 (s; c)) ds

x2 (t; c) = �et�Z 1

t

e�sg2 (x1 (s; c) ; x2 (s; c)) ds

�:

If we assume that @x1@c

and @x2@c

are exist everywhere, we can di¤erentiate theseequations to get two more integral equations. These equations are

@x1 (t; c)

@c= e�t+e�t

Z t

0

es�@g1 (x1 (s; c) ; x2 (s; c))

@x1

@x1 (s; c)

@c+@g1 (x1 (s; c) ; x2 (s; c))

@x2

@x2 (s; c)

@c

�ds

and

@x2 (t; c)

@c= �et

Z 1

t

e�ss�@g2 (x1 (s; c) ; x2 (s; c))

@x1

@x1 (s; c)

@c+@g2 (x1 (s; c) ; x2 (s; c))

@x2

@x2 (s; c)

@c

�We then need to show that these integral equations have a solution, using successiveapproximations. To do this we need to use that fact that not only are g1 (x1; x2) andg2 (x1; x2) small near x = 0; but so are their partial derivatives. This follows fromthe condition (2) and the related condition that L (�) ! 0 as � ! 0: The proof issimilar to the proof in chapter 2 of di¤erentiability of solutions with respect to initialconditions. I will not give the details here.

Assuming, then, that the necessary functions are di¤erentiable with respect to c;we consider the derivative Q0 (0) : Recalling that

Q (c) = �Z 1

0

e�sg2 (x1 (s; c) ; x2 (s; c)) ds;

we have

Q0 (c) = �Z 1

0

e�s�@g2 (x1 (s; c) ; x2 (s; c))

@x1

@x1 (s; c)

@c+@g2 (x1 (s; c) ; x2 (s; c))

@x2

@x2 (s; c)

@c

�ds:

15

However, because (0; 0) is an equilibrium point of the original system, x (s; 0) = 0:Also, the condition (9) on g; plus the accompanying statement that lim�!0 L (�) = 0;

implies that the @g2(x1)@x1

j(0;0) = @g2(x1)@x2

= 0: Therefore, Q0 (0) = 0 as claimed.

In n-dimensions without the assumption that the linearized system is alreadyin Jordan canonical form, for a system

y0 = f (y)

at an equilibrium point y0; this statement becomes the following:

Theorem: The stable manifold at y0 is tangent to the stable manifold for thelinear system y0 = Ay where A is the Jacobian matrix @fi

@yjjy0.

Behavior o¤of the stable manifold. We need to prove the last statement ofthe Stable Manifold Theorem. As before we will only consider the two-dimensionalcase, when the system can be put in the form

x01 = �x1 + g1 (x1; x2)x02 = x2 + g2 (x1; x2) :

We have shown the existence, in some small neighborhood of 0; of a stable manifoldgiven as a function x2 = Q (x1) : This is a smooth curve S, with Q0 (0) = 0; suchthat if x (0) lies on this curve, then x (t) is on the curve for all t � 0: Therefore, ifx (0) 2 S; then

x2 (t) = Q (x1 (t))

for all t � 0: We di¤erentiate this to get

x02 (t) = Q0 (x1 (t))x

01 (t)

orx2 + g2 (x1; x2) = Q

0 (x1) (�x1 + g1 (x1; x2)) :But x2 = Q (x1) ; so

Q (x1) + g2 (x1; Q (x1)) = Q0 (x1) (�x1 + g1 (x1; Q (x1))) :

This last equation involves only x1: By taking all possible solutions on the stablemanifold, we see that it holds for all x1: It is an identity involving the function Q;and can be written as

Q (z) + g2 (z;Q (z)) = Q0 (z) (�z + g1 (z;Q (z))) (15)

16

with no reference to the di¤erential equation, once Q has been de�ned. z is simplythe independent variable in this equation, which is not a di¤erential equation.

Now suppose that (x1 (t) ; x2 (t)) is a solution with jx (0)j small, but which doesnot start on the stable manifold. Let

� = x2 (t)�Q (x1 (t)) :

Then

� 0 (t) = x2 (t) + g2 (x1 (t) ; x2 (t))�Q0 (x1 (t)) (�x1 (t) + g1 (x1 (t) ; x2 (t))) :

We use the identity (15) to substitute for the term Q0 (x1 (t))x1 (t). This gives us

� 0 (t) = x2 (t) + g2 (x1 (t) ; x2 (t))�Q (x1 (t))� g2 (x1 (t) ; Q (x1 (t))) +Q0 (x1 (t)) g1 (x1 (t) ; Q (x1 (t)))�Q0 (x1 (t)) g1 (x1 (t) ; x2 (t)) :

We see the term x2 (t)�Q (x1 (t)) on the right side, which is �. Hence we get

� 0 = � + g2 (x1; x2)� g2 (x1; Q (x1)) +Q0 (x1) (g1 (x1; Q (x1))� g1 (x1; x2)) :

Recall that Q0 (0) = 0; and therefore for small x1; Q0 (x1) is small. Consider theterm g2 (x1; x2) � g2 (x1; Q (x1)) : Since � = x2 � Q (x1) we use the mean valuetheorem to write

g2 (x1; x2)� g2 (x1; Q (x1)) =@g2@x2

(x1; x�2) �;

where x�2 is some point between x2 and Q (x1). Recall that @g2@x2(0; 0) = 0: Hence

the term on the right just above is a small multiple of �: Similarly the second termis a small multiple of �: We then obtain an inquality that as long as (x1 (t) ; x2 (t))is small, we have

� 0 � 1

2�:

Therefore, � grows exponentially until the solution (x1; x2) leaves some �xed neigh-borhood of (0; 0) . This proves the last sentence of the stable manifold theorem inthis special case. For the general case see Coddington and Levinson, section 13.5.

Finally, if we make the change of variables t! �t; letting u (t) = y (�t) ; thenu satis�es

u0 = �Au� g (u) :

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The signs of the real parts of the eigenvalues all change, and the stable manifoldconsists of solutions such that u! 0 as t ! 1: In other words, y! 0 ast! �1: This is called the �unstable manifold�for the original system.

We now study the stable and unstable manifold for one further example.

Example:

y00 = y � y3: (16)

Stable and unstable manifolds are subsets of the phase space for the system. Tosee the phase space for this example we follow the usual procedure of letting y1 = y;y2 = y

0: We then have

y01 = y2 (17)

y02 = y1 � y31:

The phase space is the y1 � y2 plane, or it can be called the y � y0 plane.

An �energy� function is useful. Multiplying (16) by y0 and integrating showsthat if E (t) = y02 � y2 + 1

2y4; then

E 0 (t) = y0�y00 � y + y3

�= 0;

so E is constant. In the variables of (17), E (t) = y22 � y21 + 12y41: We can see the

trajectories in phase space by plotting various curves of the form y22 � y21 + 12y41 = C;

a constant in the y1 � y2 plane. First suppose C = 0: Notice that for very smally1; the term y21 is much larger than the term 1

2y41; so we have something close to

y22 � y21 = 0: This is two straight lines with slopes �1. But further away from theorigin, the y41 becomes important, and we see that if y1 is too large, we could not getE = 0: Therefore the curve must be bounded.

In fact, we can ask Maple to plot the curve. This is what we get:

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1.2510.750.50.250-0.25-0.5-0.75-1-1.25

0.625

0.5

0.375

0.25

0.1250

-0.125

-0.25

-0.375

-0.5

-0.625

x

y

x

y

In order to interpret this we look at the equations again and try to determine inwhich directions t is increasing, along various parts of the curve. Since y01 = y2; wesee that the solution moves to the right in the upper half plane and to the left inthe lower half plane. From this, we see that each of the two loops tends to (0; 0)in both directions, that is, as t ! 1 and as t ! �1: In this example, the stableand unstable manifolds coincide, in a global sense.

But if we draw a small circle around (0; 0) ; and only consider these curves withinthis circle, then we see that the stable manifold appears to consist of a curve which istangent to y1 = �y2; and the unstable manifold appears to consist of a curve whichis tangent to y1 = y2:

The de�nitions of the stable and unstable manifold were global. The stablemanifold consists of all initial points, anywhere in the plane, such that the solutiontends to zero at +1. We cannot determine this using our theorem about the relationof the stable manifold to eigenvalues. All we can do is �nd a part of the stable, andunstable manifolds, in this way. Nevertheless, this theorem is very important, forit at least gives the local picture of the solutions near an equilibrium point. Thechallenge of getting a global picture is often insurmountable. This example is anexception, because of the existence of the energy function E:

In this case, in fact, we see that S=U. The orbit of one branch of S (or U) iscalled a �homoclinic orbit�.

Note that if the intersection of S and U contains a non-equilibrium point , thenthe solution starting on this point tends to y0 as t!1 and also as t! �1, andso its trajectory is a homoclinic orbit. This can happen without S and U beingidentical, or one being a subset of the other.

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One further example to illustrate the di¢ culty of studying S and U in a higherdimensional system when there is no conserved energy function. We consider thesystem known as the Lorenz equations:

x0 = � (y � x)y0 = rx� y � xzz0 = xy � bz

Here �; r;and b are positive parameters. The linearized system at (0; 0; 0) is

U0 = AU

where

A =

0@ �� � 0r �1 00 0 �b

1A :We see that one eigenvalue is �b with eigenvector v1 =

24 001

35 : The trace of Ais negative and its determinant is �b� (1� r). We consider the case r > 1; in whichcase the determinant is positive so the product of the eigenvalues is positive Hencethere must be two real negative eigenvalues and one real positive eigenvalue. Thusthe stable manifold is two dimensional and the unstable manifold is one dimensional.

By considering the equationAv =�v

where � > 0 we can easily show that the eigenvector v3 corresponding to the positive

eigenvalue points into the positive quadrant of the x; y plane, and has z componentzero. We also see that the z axis is part of the stable manifold, while the eigenvectorv2 for the second negative eigenvalue has z component zero.

We therefore have a local picture as follows: The unstable manifold is acurve tangent to v1 and therefore pointing into the region x > 0; y > 0: Theunstable manifold is a surface which is transverse (not tangent) to v1 and containsan interval of the z axis. Now I will give a numerical computation of the globalunstable manifold. We see that near 0 it is a straightforward curve passing through

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the origin, but globally it is more complicated. This system is by no means wellunderstood, and in particular, little is proved about the unstable manifold at theparameter range where this computation was done. It appears to lie or be close tosome sort of two dimensional surfact, but this is misleading.

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