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Lecture 3: Invariant Manifolds In this lecture we consider differential manifolds A manifold is a smooth surface that looks locally like Euclidean space We are interested in paths on these manifolds and how they characterise the solutions of differential equations; hence consider differential manifolds in phase space We will revisit hyperbolic equilibria and also consider non-hyperbolic equilibria and their associated manifolds We start with vector subspaces – generalisations of lines and planes. Manifolds are curved generalisations of vector subspaces. 1 Dynamical Systems: Lecture 3

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### Transcript of Lecture 3: Invariant Manifolds - GitHub PagesLecture 3: Invariant Manifolds •In this lecture we...

Lecture 3: Invariant Manifolds• In this lecture we consider differential manifolds

A manifold is a smooth surface that looks locally like Euclidean space

• We are interested in paths on these manifolds and how they characterise the solutions of differential equations;hence consider differential manifolds in phase space

• We will revisit hyperbolic equilibria and also consider non-hyperbolic equilibria and their associated manifolds

• We start with vector subspaces – generalisations of lines and planes. Manifolds are curved generalisations of vector subspaces.

1Dynamical Systems: Lecture 3

Differential Manifolds• Manifold M is locally made up of

patches copied from ℝ"

• e.g. M = surface of a sphere

• Definition of a differential manifold involves how local coordinate systems are placed on M and how neighbouring patches transform into each other

Dynamical Systems: Lecture 3 2

ℝ#

\$

Linear systems: Stable, Unstable and Centre subspaces

• We studied the linear system"̇ = \$"

• 3 dis:nct types of eigenvalue of interest:

– the stable set, with %& ' < 0, denoted '*+, '-+ ⋯'/+

– the unstable set, with %& ' > 0, denoted '*1, '-1 ⋯'21

– the centre set with, %& ' = 0, denoted '*3, '-3 ⋯'43

• There are 5 eigenvalues, so 6 + 8 + 9 = 5

Dynamical Systems: Lecture 3 3

Vector bases and the subspaces• If ! has real, distinct eigenvectors, the " eigenvectors span phase space

and any vector can be made up from a weighted sum of eigenvectors (denoted #\$ …#&)

Then, given the initial condition '(0):' 0 = ,\$#\$ + ,.#. + ⋯+ ,&#&

we can express the solution as' 0 = ,\$#\$1234 + ,.#.1254 + ⋯+ ,&#&1264

• If the eigenvalues are complex or repeated, other methods exist for defining #\$ …#&

Dynamical Systems: Lecture 3 4

• Split phase space into three subspaces spanned by eigenvector sets:!"# eigenvectors associated with stable (negative real part) eigenvalues

!\$% eigenvectors associated with unstable (positive real part) eigenvalues!&' eigenvectors associated with eigenvalues having zero real part

• Each eigenvector, multiplied by ()*, +()* or +,()* etc generates a component of the solution in -"# + ,-\$% + ,-&' + of the form

- + =0"12

34"-"# + +0

\$12

64\$-\$% + +0

&12

74&-&' +

These components are written as- + = -# + + -% + + -' +

e.g. -2# + = ()89*!2#.Dynamical Systems: Lecture 3 5

Effect of changing coordinates on A

• Last lecture we considered a change of coordinates ! = #\$%&.Let the matrix # be made up of the three set of vectors '(), '+,, '-. and call this matrix /.

• Let 0 = /\$%1 be the new coordinates:

0 = /\$%1 =0)0,0.

This defines three decoupled subspaces, each with their individual coordinates

Dynamical Systems: Lecture 3 6

New equations of motion

• Change of coordinates:"̇ = \$%&'̇ = \$%&(' = \$%&(\$"

• We chose the new coordinates so that

\$%&(\$ =() 0 00 (+ 00 0 (,

Where -.[0(())] < 0, -.[0((+)] > 0, -.[0(())] = 0New equations of motion: "̇6 = (67)

"̇+ = ()7+"̇8 = (,7,

Dynamical Systems: Lecture 3 7

Consequences of the change of coordinates

• By changing coordinates we have split the solution of the linear differential equation into three independent subspaces:

ES, EU, and EC

• If the solution starts in one of these subspaces it will stay in it and cannot cross into one of the other subspaces

• These subspaces are invariant with respect to the flow !"#

Dynamical Systems: Lecture 3 8

Example systemConsider the system

"̇ =−3 0 00 3 −20 1 1

"

The eigenvalues are −3 and 2 ± *with eigenvectors [1 0 0]- and [0 1 ± * 1]-The solution is

" . =/012 0 00 /32 cos. + sin. −2/32sin.0 /32sin. /32 cos. − sin.

" 0

Dynamical Systems: Lecture 3 9

Stable subspace and unstable spiral

Dynamical Systems: Lecture 3 10

Degenerate example

Consider the system

"̇ =\$ 1 00 \$ 00 0 '

"

Assume \$ < 0 and ' > 0. We have a repeated eigenvalue \$ and an eigenvalue '.

The upper left block is degenerate and thus there is only one eigenvector for the eigenvalue \$.

"* is an unstable subspace and (",, ".) spans a stable subspace.

Dynamical Systems: Lecture 3 11

Degenerate stable subspace and unstable subspace

Dynamical Systems: Lecture 3 12

-1-0.5

00.5

1

-1-0.5

00.5

1-1.5

-1

-0.5

0

0.5

1

1.5

x1x2

x 3

Centre exampleConsider the system

"̇ =0 −1 01 0 00 0 2

"

The eigenvalues ±) have eigenvectors [1 ± ) 0],and eigenvalue at 2 has eigenvector [0 0 1],

The "3-axis is thus an unstable subspace and the ("1 "2) plane is a centre subspace

Dynamical Systems: Lecture 3 13

Centre subspace and unstable subspace

Dynamical Systems: Lecture 3 14

Observations• For the stable subspace

lim\$→&'( ) = +

All points in the stable subspace end up at the origin.• For the unstable subspace

lim\$→,&'- ) = +

Tracing back points on the unstable subspace ends up at the origin.

Nothing can be said about the centre subspace without careful thought! Look at the example . = 0 0

1 0 in the notes.

Dynamical Systems: Lecture 3 15

The Hartman-Grobman Theorem Introduction

Hyperbolic equilibrium points are in some way special (nothing can be said yet about the centre subspace made up from points with eigenvalues with zero real part).

We start by considering the local linearizaBon of a nonlinear system about an equilibrium.

Dynamical Systems: Lecture 3 16

Nonlinear system local theory• Consider the autonomous nonlinear differential equation

"̇ = \$(")Recall that \$ is a vector of functions and " is a vector of unknowns.

• Linearize about an equilibrium "∗ by writing " = "∗ + ):)̇ = *\$ "∗ ) + + ) ,

• The phase space of the linearized system can be split into three subspaces by a coordinate transformation based on the eigenvectors of the Jacobian *\$ "∗ :

-̇. = /.-. + 0. --̇1 = /1-1 + 01 --̇2 = /2-2 + 02 -

Each / is a square real matrix (stable, unstable or centre). The 0’s are vector functions of the transformed errors that make up + ) ,.

Dynamical Systems: Lecture 3 17

Hartman-Grobman Theorem

Theorem: For each hyperbolic equilibrium point, there exists a bi-con;nuous func;on ! (a mapping that is con;nuous and whose inverse is also con;nuous) between an open set containing the equilibrium point and an open set containing the origin of the linearized model so that trajectories are mapped exactly and the parameterisa;on of ;me is preserved.

Dynamical Systems: Lecture 3 18

• Near the origin, the stable linear subspace is mapped to a stable manifold in a region surrounding the equilibrium point.

• Near the origin, the unstable linear subspace is mapped to an unstable manifold in a region surrounding the equilibrium point.

• Nothing is said about non-hyperbolic equilibria!

Dynamical Systems: Lecture 3 19

Hartman-Grobman Theorem

Illustration

Dynamical Systems: Lecture 3 20

Unstable manifold (curve)

Stable manifold (Surface)

Unstable subspace (line)

Stable subspace (plane)

H

H-1

Example of Hartman-Grobman Theorem

Consider the non-linear autonomous system"̇# = −"#

"̇& = "& + "#&

Dynamical Systems: Lecture 3 21

-3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6

x1

x 2

-3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6

z1

z 2

H

H-1WU

WS

EU

ES

Non-Hyperbolic Equilibria

• We cannot say much about non-linear equilibria whose linearized model is a centre.

• Use some tricks…e.g. transform a 2D system into polar coordinates.

"̇# = %# "#, "'"̇' = %' "#, "'

Let "# = (cos, and "' = (sin, then

(̇ = "#"̇# + "'"̇'( , ,̇ = "#"̇' − "'"̇#

('Does ( grow, shrink or stay constant?

Dynamical Systems: Lecture 3 22

Example of polar transforma/on

Consider"̇ = −% − "%%̇ = " + "'

Then

(̇ = ""̇ + %%̇( = −" % + "% + % " + "'

( = 0

*̇ = "%̇ − %"̇(' = " " + "' + % % + "%

(' = 1 + "Near the origin (the equilibrium point) we have a nonlinear centre.

Dynamical Systems: Lecture 3 23

Symmetry

• A 2D nonlinear system (", \$) is symmetric with respect to e.g. the "-axis if it is invariant under the transformation

(&, \$) → (−&,−\$)

• If the system is symmetric with respect to either " or \$, and if the origin is an equilibrium point, then centres will map to centres between the non-linear and linear approximation

Dynamical Systems: Lecture 3 24

Example of symmetryConsider the system

"̇ = \$ − \$& ≝ ( ", \$\$̇ = −" − \$* ≝ + ", \$

The equilibrium point at the origin has Jacobian 0 1−1 0 and is thus a linear centre.

Now note:( ",−\$ = −( ", \$+ ",−\$ = + ", \$

Therefore."

. −/ = ( ",−\$. −\$. −/ = .\$

./ = + ",−\$… so the nonlinear system has a centre

Dynamical Systems: Lecture 3 25

Conservative Systems

• If there exists a non-constant function !(#) such that %!/%' = 0 along solutions of the nonlinear differential equation #̇ = + # , the equations are called conservative. !(#) does not change along the solution trajectories.

• If # = #∗ is an isolated equilibrium point and there is a !(#) that has a local min or max at #∗, then there is a region around that point that contains a closed orbit.

Dynamical Systems: Lecture 3 26

Conserva)ve exampleConsider the system

"̇ = \$\$̇ = % "

Integrate−% " "̇ + \$\$̇ = 0

to get

−)*+

*% , -, + \$

.

2 = constant

Potential energy + kinetic energy = constant (see the examples sheet)This is called a Newtonian system

Dynamical Systems: Lecture 3 27

Further example

Consider a system of the form"̇ = \$ " %& ''̇ = \$ ' %( "

Then%( "\$ " "̇ − %& '\$ ' '̇ = 0

Which can be integrated to yield +(", ')

Dynamical Systems: Lecture 3 28

Consider the system"̇ = " − "% = " 1 − %%̇ = −% + "% = % " − 1

The equilibrium points are (0,0) and (1,1). Jacobian at (0,0) is 1 0

0 −1 ⟹ nonlinear saddle point (by Hartman-Grobman Theorem).

Jacobian at (1,1) is 0 −11 0 ; is (1,1) a non-linear centre?

Here" − 1" "̇ − 1 − %% %̇ = 0

⟹ " − ln" + % − ln% = 0 ", % = constantAt (1,1), 60/6" = 0 and 60/6% = 0 and det(6:0/6"6%) > 0 so 0(1,1) is a minimum pointTherefore (1,1) is a nonlinear centre.

Dynamical Systems: Lecture 3 29

Further example