dynamical systems assignmentntdoc

8/19/2019 Dynamical Systems assignmentntdoc

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Question 1a)

A fixed (or stationary) point is a point at which ˙ x= 0 . This system has two fixed

points at

x= 0and x= k 1 a / k 2 .

Graphically (fig 1) it can be demonstrated that x= 0 is an unstable fixed point and that x= k 1 a / k 2 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= k 1 a / k 2 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 x≥ 0 . This would imply anegative amount of chemical $ meaningless from the chemist's point of view$

b) Figure 2 shows how x tends to x= k 1 a / k 2 given various initial conditions$ unless theinitial value is %ero. &otice that below x= k 1 a / k 2 the curve is a logistic curve$ whereasabove this the curve that of exponential decay.

1

˙ x= k 1 ax − k 2 x2


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Figure 2

2

0 10 2 0 3 0 4 0 50 60 70 80 90 1 000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

t

x

0


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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= k 1 a − k 3 b /k 2 or x= c 1 /k 2 . ,f k 3 b k 1 a then c 1 is negative$ pointing to a mathematical fixed point below%ero (although clearly not relevant to the physical system)$ as indicated in (fig 3).

Also consider the derivative of ˙ x at %ero.d xdx

= k 1 a − k 3 b −2k 2 x 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 -.

3

A X k 1 = 2X

A X k 2

= 2X

B X k 3 = C

˙ x = k 1 ax − k 2 x2 − k 3 bx

˙ x = k 1 a − k 3 b x − k 2 x2

˙ x= c 1 x− c 2 x2

c 1= k 1 a − k 3 b c2= k 2

x= k 1 a − k 3 b /k 2

Figure 1

Figure 3


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Question 2a)

The system

,n matrix from the system is

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

Next 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 Δ 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 of h.

The red line on the graph (fig 4) denotes the points at which T 2= 4Δ $ in this casewhen h 2/ . ,f Δ is greater than one $(/$/) will be a spiral node$ otherwise it will be a stablesin3 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.

4

˙ x = − x− hy ̇y = x− y

˙ x˙ y

= − 1 − h1 − 1 x y

− 1 − h1 −1

T =− 2 Δ= h 1

Figure 4


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To summarise (Chart 1)

h h − 1 h =− 1 − 1 h 0 h = 0 h 0

&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 matrixis 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

5

˙ x = y hx x2 y2

˙ y = − x hy x2 y ?2

r 2= x2 y2

2r dr dt

= 2 ˙ x x 2 ˙ y y

2 r dr dt

= 2x y hx x2 y2 2y hx x2 y2 − x

2r ṙ = 2xy − 2yx 2h x2 y2 x2 y2 = 2hr 4

dr dt

= hr 3

˙ x˙ y

= 1 00 −1

x y

Chart 1

dr dt

= hr 3

∫ − r − 3 dr = ∫ 1 dt


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8here c and 9 are un3nown constants of integration.

As t ∞ the denominator will also tend towards infinity$ regardless of the magnitudeof 9$ therefore r will tend to /. Given that the radius of tra:ectories is decreasing to %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

6

y=± 1 y= x3

1 , 1−1 ,− 1

J = ∂2 x∂ x∂ t

∂2 x∂ y∂ t

∂2 y∂ x∂ t

∂2 y∂ y∂ t

= 0 2y3x 2 − 1

dxdt

= y2− 1

dydt

= x3− y

Figure 5

2r − 2 = t c

r = 1 2t K

limt ∞

r = 0


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and corresponding determinant and trace7

At point ( -1, -1)

oth the determinant and trace are positive$ and T 2− 4 Δ = − 23 0 . This is evidencethat (-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)

dxdt

= y2− 1 4/.5< /.#5 / /

dydt

= x3− y 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)

7

Δ = − x2

yT = − 1

Δ=T =− 1

Figure


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J has eigenvectors and respective eigenvalues1

1$ # and

− 2

3 $ 4? which

correspond to the manifolds of the saddle node. Given that the first eigenvalue is positive 1− 1 is the unstable manifold.

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

8

J = 0 23 −1

∂∂ x

G dxdt ∂∂ y G dydt

Figure 7


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is consistently positive or consistently negative ∀ x , y

+uch a function would be the positive constant C 2 which is clearly continuous anddifferentiable.

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

9

∂

∂ xC 2 dx

dt ∂

∂ yC 2 dy

dt = C 2× 0 C 2 × − 1 =− C 2


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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).

Howe er a chaotic system should not be confused with a random system (althoughit may appear so). 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 0niversity 5

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.

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

1 !" #arrison $ Physics Flash Ani ati!ns , %ni&ersity o' Toronto

10

Figure (

˙ x˙ y

= a bc d x y

Figure (


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,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 for the 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$ as

this would imply the existence of a point A at which ˙ x , ˙ y will have two separatedirections (Fig ").

6ue to this the !oincar@4 endixson theorem states that as tra:ectories cannotcross$ in any phase plane region that a tra:ectory enters but does not leave$ then it musteither 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$ ; # gives the example of two dimensional systems that have adiscontinuity in them

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

11

Figure )


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ibliography

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


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Appendix%%%% Code to see how the chemical system in Q1 tends to a fixed pointk1a = 0.5;k2 =0.25; for 0 = 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

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