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Bifurcations

Summary of last class

Super critical pitchfork bifurcationsspin-off ofmultiple stable fixed points from one

Sub critical pitchfork bifurcationsspin-off of multiple unstable fixed points from one

4 3 2 1 0 1 2 3 43

2

1

0

1

2

3

parameter r

fixed

point(s)

Bifurcation diagram

4 3 2 1 0 1 2 3 43

2

1

0

1

2

3

parameter r

fixed

point(s)

Bifurcation diagram

Class 4 (TUE) Dynamical Systems 2013 Siep W eila nd 5 / 29

Example 1: Gravitation and rotation

Outline

1 Bifurcations

2 Example 1: Gravitation and rotation

3 Example 2: Population dynamics

4 Example 3: Catalyst reaction kinetics



Example 1: gravitation and rotation



Example 1: rotating ball in bowl

Physical parameters:

r: bowl radius, m: mass, : rotationvelocity, b: friction

(t): angle origin-vertical andorigin-mass

Forces acting on mass:

Ffriction= b: tangential friction force

Fgrav= mg: downward gravity

Fcentr = mrsin()2: sideways

centrifugal force

Projection along tangent on circle gives model

mr= b mgsin() +mrsin()2 cos()


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rotating ball in bowl

Nasty assumption:

frictionb>> m is so large that 0.

Yields the simplified 1st order model

b mgsin() +mrsin()2 cos() = 0

or

=mr2

b sin()cos()

mg

b sin() = f()

Need to determine:

Fixed pointsSet f() = m

b sin()

r2 cos() g

= 0

Stability of fixed pointsCalculate sign off() at fixed points.



determine fixed points

Fixed points:

Setf() =

m

b sin()

r2 cos() g

= 0

yields

sin() = 0 for = k with k= 0, 1, . . . r2 cos() g= 0 for cos() = g

r2.

This has solutionsnone if g

r2 >1

= cos1(g/r2) if g

r2 1



determine stability of fixed points

Stability of fixed points:Calculate

f() =m

b cos()

r2 cos() g

mr2

b sin2()

and evaluate at fixed points: f(0) = m

b(r2 g) yields

= 0stableif

gr

f() = mb

(r2 +g)> 0 yields= is unstable

if >

gr

then

f( cos1(g/r2)) = mr2

b

1 cos2(cos1(g/r2)

< 0

and hencetwo stable fixed points at = cos1(g/r2)



bifurcation diagram

Bifurcation diagram is of supercritical pitchfork type

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

3

2

1

0

1

2

3

parameter r*omega2/g

fixed

point(s)

Bifurcation diagram

Physical intuition: rotation speed high enough implies that down positionis no longer stable, ball searches new equilibria


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time simulations

Suppose m = 0.01, r= 0.1, b= 1, g= 10. Then =

gr

= 10.

Trajectories (t) for different initial conditions

0

10 20 30 40 50 600.2

0.15

0.1

0.05

0

0.05

0.1

0.15

0.2

time t

solution

phi(t)

trajectories of angle phi(t)

0

10 20 30 40 50 601.5

1

0.5

0

0.5

1

1.5

time t

solution

phi(t)

trajectoriesof angle phi(t)

rotation speed = 8< rotation speed = 15>

(simulations in IODE or using script inclass 2 pp. 10)



conclusions on ball in rotating bowl

Bifurcation value at =

(g/r).

Flow diagrams on bowl surface follow from bifurcation diagram

Solutions (t) can be simulated using iode

For > we have that || < /2 Taylor expansion off(, ) around (, ) = (0,

(g/r)) results in

equivalence between= f(, )

and thenormal form X= RX X3.


Example 2: Population dynamics

Outline

1 Bifurcations






Example 2: population dynamics

Extension of logistic equation for population dynamics

N= r0N

1

N

p(N)

withgrowth rate r0,carrying capacity and death ratep(N).Consider death rate due topredator function:

p(N) = N2

2 +N2 , >0, >0

Case = 1 and = 30 0 10 20 30 40 50 60 70 80 90 10000.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

http://dynsv2.pdf/http://dynsv2.pdf/

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fixed points two parameter model

Exercise:draw these four diagrams and determine stability of fixed points!

With three fixed points (case kis fixed large or r is fixed small):

One stable low level fixed point: refuge

One stable high level fixed point: outbreak

One unstable fixed point in between refuge and outbreak

Conclusion

The complete bifurcation diagram should plot fixed points x as functionoftwo parametersk and r. (Three-dimensional plot).We will look atprojectionsof this3D plot instead.



projection bifurcation diagram: x as function of k

Here is one: (click to animate)

2 4 6 8 10 12 14 160

5

10

15

k

fixed

points

x*

r=0.4

fixed points as function ofk

all points are approximations, no stability indication.

Note thebifurcation points k when r varies. What type are these?Class 4 (TUE) Dynamical Systems 2013 Siep W eila nd 22 / 29


projection bifurcation diagram: x as function of r

Here is an other: (click to animate)

0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

r

fixed

points

x*

k=14

fixed points as function ofr

all points are approximations, no stability indication.

Note thebifurcation points r when kvaries. What type are these?Class 4 (TUE) Dynamical Systems 2013 Siep W eila nd 23 / 29


stability of fixed points

Example

Projection of 3D bifurcation diagram in the k-rplane.

As follows: Let f(x) = r(1 xk

) x1+x2

. Thenbifurcation points(k, r, x) satisfy:

f(x

) = 0 and f

(x

) = 0This yields, after some simplification,

k= 2x3

x2 1

and r= 2x3

(1 +x2

)2

Thus, bifurcation points k and r can be viewed asfunctionsof fixedpoints x. This gives two curves k(x) and r(x), parametrized by x, inthek-rplane.


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projected bifurcation points in k-r plane

Here is how this looks:

0 5 10 15 20 25 30 35 400

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

k

r

1 fixed point at small x: refuge

3 fixed points

1 fixed point at large x: outbreak

1 fixed point at small x: refuge

3 fixed points

1 fixed point at large x: outbreak

plot of (k, r) = (k(x), r(x)) as x varies



questions on this population model

With this analysis, can you now answer the following questions?

Supposex(t) represents the number of people with an infectiousdisease at timet and you can influencek through, e.g., medicine,pharmacy, hygiene, quarantaine. How would you set kifr 0.55 and

x(0) is large (near outbreak)? x(0) is between refuge and outbreak? x(0) is below refuge?

What are critical values k for k ifr= 0.55?

What happens with x(t) ifkvaries around k?

What if you set kjust below k and rhappens to change?

Can you prevent from outbreaks/catastrophes? If so, how?


Example 3: Catalyst reaction kinetics

Outline

1 Bifurcations






Example 3: catalyst reaction

Consider model of chemical reaction

A+Xk1 2X; A+Xk2 2X; B+ Xk3 C

Definition

Law of mass action: rate of an elementary reaction is proportional to theproduct of the concentrations of the reactants.

Denote concentrations x= [X], a = [A], b= [B] and c= [C]. Thenreaction kinetics is described by

x= k1ax k2x2 k3bx= (k1a k3b)x k2x

2

with positivereaction rate constants k1, k2, k3.


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catalyst reaction - questions

Suppose surplus of catalystA and reactant B, i.e., a and bare constant.

Questions:

1 Find all fixed points of this equation and classify their stability

2 Determine the bifurcation value c for the constant c= k1a k3band draw a bifurcation diagram for the complete reaction dynamics.What kind of bifurcation is this?

3 Sketch (or simulate) the graphs of the concentrationx(t) as functionof timet from different initial conditions.

See solutions of Homework set 2.

to previous class to next class

http://dynsv3.pdf/http://dynsv5.pdf/http://dynsv5.pdf/http://dynsv3.pdf/

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