Lorenz Equations

dx

dt=σ(y−x)

dydt

=rx−y−xz

dzdt

=xy−bz

3 state variables 3dimension system

3 parameters

seemingly simple equations

note only 2 nonlinear terms

but incredibly rich nonlinearbehavior in the system

σ > 0

r > 0

b > 0

fixed points

(x*,y*,z*)1 (0,0,0)

(x*,y*,z*)1 (0,0,0)

(x*,y*,z*)2

(x*,y*,z*)3

0 < r < 1

+ b(r −1),+ b(r −1),r −1( )

− b(r −1),− b(r −1),r −1( )

r ≥ 1

C+

C-

the origin is always a fixed point

The existence of C+ and C- depends only on r, not b or σ

stability of the origin

Δ =σ (1− r)

τ = −σ −1

τ 2 − 4Δ = (σ −1)2 + 4σ r

δ fδ x

δ f

δ y

δg

δ x

δg

δ y

⎛

⎝

⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟(0,0,0)

=−σ σ

r −1

⎛⎝⎜

⎞⎠⎟

when r >1⇒ Δ < 0 ⇒

τ < 0 when σ > −1 ⇒ always true

τ < 0 for all parameter values

τ 2 − 4Δ > 0 for all parameter values

when 0< r <1⇒ Δ > 0 ⇒ have to look at τ and τ 2 −4Δ

stable node

saddle node

y

x

z

r > 1

saddle nodeat the origin

z= -b, vz = (0,0,z)

1= 1, v1 = (1,2,0)

Example for σ = 1r = 4

2= -3, v2 = (1,-2,0)

unstable manifold

stable manifold

stable manifold

b does not affect the stabilty.b only affects the rate of decay in the z eigendirection

Summary of Bifurcation at r = 1

0< r < 1 r > 1

stable node saddle node

new fixed point, C+

new fixed point, C-The origin looses stability and 2 new symmetric fixed points emerge.

What type of bifurcation does this sound like?

What is the classification of the new fixed fixed points?

origin stable origin unstable

Stability of the symmetric fixed points?

x

r

example for b=1

other b values would lookqualitatively the same

Plotting the location of the fixed points as a function of r

Looking like a supercritical pitchfork

stability of C+ and C-

δ fδ x

δ f

δ y

δ f

δ z

δ g

δ x

δg

δ y

δg

δ z

δh

δ x

δh

δ y

δh

δ z

⎛

⎝

⎜⎜⎜⎜⎜⎜⎜

⎞

⎠

⎟⎟⎟⎟⎟⎟⎟C+

=

−σ σ 0

r − z −1 −x

y x −b

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟C+

=

−σ σ 0

1 −1 − b(r − 1)

b(r − 1) b(r − 1) −b

⎛

⎝

⎜⎜⎜

⎞

⎠

⎟⎟⎟

C+ =(x* ,y* ,z* ) =( b(r −1), b(r −1),r −1)

need to findeigenvalues to classify

eigenvalues of a 3x3 matrix

det(A−I ) =0

det(A−I ) =a11 − a12 a13

a21 a22 − a23

a31 a32 a33 −=0

in general …

eigenvalues are found by solving the characteristic equation

for a 3x3 matrix

result is the characteristic polynomial with 3 roots: 1, 2, 3

Remember for 2x22D systems (I.e. 2 state variables)

Tip: can use mathematica to find a characteristic polynomial of a matrix

A =a bc d

⎛⎝⎜

⎞⎠⎟

det(A−I ) =deta− b

c d−⎛⎝⎜

⎞⎠⎟=0Characteristic equation

Characteristic polynomial (a−)(d−)−bc=0

2 −(a+d) + ad−bd=0

2nd order polynomial for a 2x2 matrix

The eigenvalues are the roots of the characteristic polynomial

Therefore 2 eigenvalues for a 2x2 matrix of a 2 dimension system

eigenvalues of a 3x3 matrix

a11 − a12 a13

a21 a22 − a23

a31 a32 a33 −=0

det

a11 a12 a13

a21 a22 a23

a31 a32 a33

⎛

⎝

⎜⎜

⎞

⎠

⎟⎟=

a11 a12 a13

a21 a22 a23

a31 a32 a33

=a11

a22 a23

a32 a33

−a12

a21 a23

a31 a33

+ a13

a21 a22

a31 a32

In general: The determinent of a 3x3 matrix can be found by hand by :

(a11 −)a22 − a23

a32 a33 −−a12

a21 a23

a31 a33 −+ a13

a21 a22 −a31 a32

=0

So the characteristic equation becomes:

− 3 + (a11 + a22 + a33)λ 2 + (a12a21 − a11a22 + a13a31 + a23a32 − a11a33 − a22a33)λ

+(-a13 a22 a31 + a12 a23 a31 + a13 a21 a32 - a11 a23 a32 - a12 a21 a33 + a11 a22 a33) = 0

Characteristic Polynomial

− 3 + Tr(A)λ 2 + (a12a21 − a11a22 + a13a31 + a23a32 − a11a33 − a22a33)λ + det(A) = 0

Trace of ADet of A

Homework problem

Due Monday

Problem 9.2.1

Parameter value where the Hopf bifurcation occurs

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.

C+ and C- are stable for r > 1 but less than the next critical parameter value

1< r < rH

where rH =σ(σ +b+ 3)σ −b−1

and σ −b−1> 0unstable limit cycle

1D stable manifold

2D unstable manifold

C+ is locally stable because all trajectories near stay near and approach C+ as time goes to infinity

Supercritical pitchfork at r=1

x*

r