Dixon Systems

8/21/2019 Dixon Systems

1/15

Question 1a)

A fixed (or stationary) point is a point at which x=0 . This system has two fixedpoints at x=0 and x=k1a / k2 .

Graphically (fig 1)it can be demonstrated that x=0 is an unstable fixed point and thatx=k1a / k2 is stable

From a chemist's point of view x=0 is a state in which there is no reaction due tothere being no chemical present to react with molecule A.

!oint x=k1a / k2 is the stable reaction where the system is in e"uilibrium with #being created at the same rate as it degrades into and A. The system will tend to thispoint.

!erhaps it should be noted that the x is only defined for x0 . This would imply anegative amount of chemical $ meaningless from the chemist's point of view$

b) Figure 2shows how x tends to x=k1a / k2 given various initial conditions$ unless

the initial value is %ero. &otice that below x=k1a / k2 the curve is a logistic curve$whereas above this the curve that of exponential decay.

Figure 2

1

x=k1axk2x2

0 10 20 30 40 50 6 0 70 80 90 1000

0. 5

1

1. 5

2

2. 5

3

3. 5

4

4. 5

5

t

x0

Figure1


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c)

Adding a third reaction to the system

y the aw of *ass Action the reaction is now

+o if written in the form

d)This chemical system has fixed points at x=0 and . x= k1 ak3 b/k2

or

x=c1/ k2 . ,f k3 bk1a then c1 is negative$ pointing to a mathematical fixed pointbelow %ero (although clearly not relevant to the physical system)$ as indicated in(fig 3).

Also consider the derivative of x at %ero.dx

dx=k1ak3b2k2x This is clearly

negative at $ indicating a stable point.

To the chemist this is indicating an experiment where all of chemical is used up tocreate chemical $ leading to the stable point of -.

2

AXk1=2X

AXk2=2X

BXk3=C

x=k1 axk2x2k3 bx

x=k1 ak3 bxk2x2

x=c1xc2x2

c1=k

1ak

3b c

2=k

2

x= k1 ak3 b/k2

Figure3


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Question 2a)

The system

,n matrix from the system is

learly the system has a fixed point at (/$/).

&ext consider the matrix

and the trace and determinant

0sing these facts it is possible to determine the nature of the fixed points (Fig 4)

,f is negative (/$/) will be a saddle$ if h is %ero there will be a non isolated fixed point$

that is a point on a manifold line of fixed points.

1owever if (/$/) is not a saddle point it will be a stable point as the trace is independent ofh.

The red line on the graph (fig 4) denotes the points at which T2=4 $ in this casewhen h 2/ . ,f is greater than one $(/$/) will be a spiral node$ otherwise it will be a stable

sin3 node. +hould h exactly e"ual one$ the fixed point will be a star node.

Given that the trace cannot e"ual %ero there is no possibility that the node is acentre of a fixed orbit.

3

x = xhy

y = xy

xy =1 h

1 1 xy

1 h1 1

T=2 =h1

Figure4


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To summarise (Chart 1)

h h1 h=1 1h0 h=0 h0

&ode +addle !oint &on isolated fixed point +table &ode +table +tar +table +piral

b) Ta3ing the system

( a ) ,f h is set to h=0 then the system becomes decoupled. ,t is clear frominspection that there exists a fixed point at the origin (/$/)

This matrix has a determinant of 45 so the point will be a saddle point. As the matrix

is diagonalised it is simple to see the eigenvectors and values are 5 01 and 45 10 $essentially showing that the x4axis is a stable manifold$ and that the y4axis is unstable.

( b ) +etting

6ifferentiating with respect to t gives

and then substituting the e"uations for x and y 7

which then rearranges to give

( c ) y separation of variables

4

x = yhx x2y2

y = xhy x2y ?

2

r2=x2y2

2rdr

dt=2 x x2 y y

2 rdr

dt=2x yhx x

2y

22y hx x

2y

2x

2rr=2xy2yx2h x2y2x2y2=2hr4

drdt

=hr3

xy =1 0

0 1 xy

Chart1

dr

dt=hr3

r3dr = 1dt


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8here c and 9 are un3nown constants of integration.

As t the denominator will also tend towards infinity$ regardless of themagnitude of 9$ therefore r will tend to /. Given that the radius of tra:ectories is decreasingto %ero$ this would suggest that (/$/) is now a stable spiral.c)

onsider the system

This system has the nullclines7

8hich intersect at the fixed points7

Graphically

To analyse the nature of these fixed points the system is linearised about the saddlepoints$ generating the ;acobian matrix

5

y=1 y=x3

1, 11,1

J = 2xx t

2x y t

2y

x t

2y

y t = 0 2y

3x21

dx

dt=y

21

dy

dt=x

3y

Figure 5

2r2 = tc

r=1

2tK

limt

r=0


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and corresponding determinant and trace7

At point ( -1, -1)

oth the determinant and trace are positive$ and T24 = 23 0 . This is evidence

that (-1 , -1) is a stable spiral. To determine the nature of this spiral it is necessary toconsider some trial points

!oint ( 45 $4/.< ) ( 45$ 45.5 ) (4 /.< $ 45) (45.5 $ 45)

dx

dt=y

21 4/.5< /.#5 / /

dy

dt=x

3y 4/.5 /.5 0.271 0.331

+3etching these points (Fig 6) and the associated changes in x and y shows that this inan anti clockwise sink spiral.

8hereas at point (1,1)$ the determinant is 4=$ which is evidence of a saddle node.

>ecall the ;acobian matrix$ at (1,1)

6

= 6x2

yT = 1

=6T=1

Figure6


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J has eigenvectors and respective eigenvalues

1

1

$ # and

2

3

$ 4? which

correspond to the manifolds of the saddle node. Given that the first eigenvalue is positive

11 is the unstablemanifold.

A s3etch of the phase portrait (Fig 7)7

losed orbitsTo exclude the possibility of closed orbits$ by 6ulac's criterion it is sufficient to find a

continuous and differentiable yapunov function G x , y such that

7

J = 0 23 1

x Gdx

dt

yGdy

dt

Figure 7


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is consistently positive or consistently negative x , y

+uch a function would be the positive constant C2 which is clearly continuous and

differentiable.

This is clearly negative$ for all values of x and y$ therefore there are no closed orbits inthis system.

8

x C

2dx

dt

y C

2dy

dt

=C

20C

21=C

2


9/15

Question 3

a)a chaotic dynamical system

A chaotic dynamical system is a system in which relatively minor differences is theinitial conditions of a system lead to vastly different behaviour of the system. The system isthus hard to predict over a long period of time. ( True a computerised system can performseveral iterations very fast$ but there is no way to predict the outcome without runningevery iteration $ e"ually by its nature a chaotic system is very susceptible to computationalrounding errors).

Howeer a chaotic system should not be confused with a random system (althoughit may appearso). These systems are deterministic$ in that if the initial conditions andnature of the system is 3nown then it is possible to 3now the system at any time t. ruciallyif an experiment with a chaotic system was repeated with exactly the same initialconditions it would reach the same state.

A classical example of a chaotic system that was studied by !oincar@ is the threemass problem. ,llustrated (Fig !)a diagram by 6* 1arrison of Toronto 0niversity5

This shows the motion of a planet$ being affected by the gravity of two similar suns$following standard &ewtonian physics. ,f this simulation does have a periodic path$ it could

not be predicted how long it is. These two simulations differ only in the initial position of theorbiting plant$ but are mar3edly different.

1 DM Harrison ,Physics Flash Aniati!ns , ni!"rsity o# $oronto

9

Figure%

Figure %


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b) Chaos can never occur in the phase plane

A phase plane is a system in which only two dimensions are considered$ such asthe problems considered earlier in this piece of the form

,nitially it is tempting to cite the diagram above (Fig!) as a counter example of achaotic system in # dimensions. This is a fallacy$ the three mass problem also accounts forthe velocity vectors of the planet changing through gravitational acceleration$ so it is nottwo4dimensional.

The tra:ectories of the planets cross$ impossible in a two4dimensional system$ asthis would imply the existence of a point A at which x ,y will have two separatedirections(Fig ").

6ue to this the !oincar@4endixson theorem states that as tra:ectories cannotcross$ in any phase plane region that a s#oothtra:ectory enters but does not leave$ then itmust either tend to a fixed point or a closed orbit$ as in such a system tra:ectories can bepredicted to tend towards an orbit or a fixed point$ is not chaotic.

1owever +prott#references the example of two dimensional systems that have adiscontinuity in them. The example given is the 6ixon system$

This system does not brea3 the !oincar@4endixson Theorem as this system is notsmooth$ in that at the point (/$/) both x and y are indeterminate. Four such systemswere plotted (Fig 1$% &erleaf)% with an initial position of (/.$/.)$ "=0.1 and

=[0.65 0.6& 0.70 0.71] .

# +prott$ ;. (#/5/) 'legant Chaos .+ingapore 8orld +cientific pp 5/< 4 55

10

x

y =a b

c d xy

Figure &

x = xy

x2y

2"x

y = y

2

x2y

2#y#1


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,t should be pointed out that a graphical explanation is somewhat insufficient$ due tothe nature of these computer generated plots are not smooth functions$ but rather mapsgenerated at small intervals.

1owever the plots do seem to indicate a sensitivity to initial conditions$ and a hardto predict behaviour$ and by passing through the point (/$/)$ the tra:ectory is able to bothescapeand re-enter a region enclosed by a tra:ectory$ (Fig 11) demonstrated by changingthe plot colour midway. +omething which would be impossible without the indeterminatepoint.

11

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

x

y

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

- 1 . 5 - 1 - 0 . 5 0 0 . 5 1 1 . 5- 1 . 5

- 1

- 0 . 5

0

0 . 5

1

1 . 5

Figure10


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+o does this constitute a chaotic system in the two4dimensional phase planeB 8hilea graphical demonstration of this sort is not a conclusive proof$ there is some evidencethat a chaoticCli3e system can occur in the phase plane.

12

- 1 - 0 . 8 - 0 . 6 - 0 . 4 - 0 . 2 0 0 . 2 0 . 4 0 . 6 0 . 8 1- 0 . 3

- 0 . 2

- 0 . 1

0

0 . 1

0 . 2

0 . 3

0 . 4

0 . 5

0 . 6

0 . 7

Figure11


13/15

ibliography

Arrowsmith$ 6.9. D !lace$ .*. (5


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Appendicies

%%%% Code to see how the chemical system in Q1 tends to a fixed point

k1a = 0.5;

k2 =0.25;

for0 = 0!0.1!5

x"1# = 0;

t"1# = 0;

for$ = 2!1!100

t"$# = t"$1# & 0.1

x"$# = x"$1# & 0.1'"k1a'"x"$1## k2'"x"$1##(2#

end

hold on

plot"x)*r*#

end

%%%% Code to plot the +ixon system for different ,al-es of )/

h = 0.001

= 0.1

/ = 0.5

x"1# = 0.5

y"1# = 0.5

for$ = 2!1!100000

dotx = x"$1#'y"$1#"x"$1#(2 & y"$1#(2# 'x"$1#;

doty = "y"$1#(2#"x"$1#(2 & y"$1#(2# /'y"$1# & / 1;

x"$#= x"$1# & h'dotx;

y"$#= y"$1# & h'doty;

end

plot"x)y)*r*#

%%%% Code to plot the +ixon system for different ,al-es of )/

clear

h = 0.01

= 0.1

/ = 0.0

x"1# = 0.5

y"1# = 0.5

3im = 5000

for$ = 2!1!3im

dotx = x"$1#'y"$1#"x"$1#(2 & y"$1#(2# 'x"$1#;

doty = "y"$1#(2#"x"$1#(2 & y"$1#(2# /'y"$1# & / 1;

x"$#= x"$1# & h'dotx;

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y"$#= y"$1# & h'doty;

end

plot"x)y)*r*#

hold on

x1"1# = x"3im# ; y1"1# = y"3im#;

%%% Chan4e the plot colo-r at a halfway point )

fori = 2!1!3im

dotx = x1"i1#'y1"i1#"x1"i1#(2 & y1"i1#(2# 'x1"i1#;

doty = "y1"i1#(2#"x1"i1#(2 & y1"i1#(2# /'y1"i1# & / 1;

x1"i#= x1"i1# & h'dotx;

y1"i#= y1"i1# & h'doty;

end

plot"x1)y1)**#

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Dixon Systems