first order nonlinear equations

7/31/2019 First Order Nonlinear Equations

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First order nonlinear equations Separable equations

56

3 First order nonlinear equations

( ),dy

f x y

dx

=

wheref(x,y) is not (in general) a linear function ofy.

The equation is autonomous if there is no explicit dependence on the independent variable, i.e.

f(x,y)= f(y).

3.1 Separable equations

These are often straight forward to solve.

Suppose we can write ( )( )

( )

,g x

f x y

h y

=

Then the ode may be written as ( ) ( )dy

h y g xdx

=

( ) ( )h y dy g x dx=

H(y)= G(x)+ const h = dH/dy;g= dG/dx

2

sin

1

dy xy

dx y=

+

Rearrange:2

1 1 siny dy dyy xy dx y dx

+ = + =

Integrate 21 1 1

ln sin cos2 2

y dy y y x dx x Ay

+ = + = = +

Arbitrary constant of integrationA determined by boundary conditiony(0)=1

0 + = 1 + AA = 3

21 3

ln cos2 2

y y x+ =

lny = lny2 2 2ln 3 2 cosy y x+ = (*)

End of Lecture 7

Cannot give explicit solution fory, but clearly periodic with period 2.

For some quantity u, the value ofu +ln u is a monotonically increasing function ofu.

The minimum value of the right-hand-side of (*) is 32 = 1, atx = 2n, and the maximum valueis 3+2=5 atx = 2(n+1).

y2 + lny2=1 y =1 is the minimum value. The maximum value is aroundy =1.92.

Note: ify is a solution, then y is also a solution to this equation.


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57

3.1.1 FAMILIES OF SOLUTIONS

In the above example we considered a specific solution, selecting A =3 in order to satisfy theboundary conditiony(0)=1. This is one of a family of solutions that may be obtained using other

values ofA.

2.5 5 7.5 10 12.5 15 17.5

1

2

3

4

2 2ln 2cosy y A x+ = ; redA=3; greenA=4, blueA=5, blackA=6; magentaA=0.5, cyanA=0.1,

dark greyA=0.01.

[See S3_1_FamilyOfSolutionsForSeparableEquation.nb]

AsA ,y and dominated byy2~ A amplitude of oscillation becomes small.

AsA ,y 0 and dominated by lny2~ A amplitude of oscillation becomes small.

Isoclines: contours of constant dy/dx.

Here vertical lines atx = n(where dy/dx vanishes) are isoclines Can be helpful for sketching solution.Solution with greatest amplitude for 2 2ln 2cosy y A x+ = .

max(y) =yt2 2ln 2t ty y A+ = +

min(y) =yb2 2ln 2b by y A+ =

yt2yb

2+ 2 ln(yt/yb)=4

Let Y (yt+yb),B (ybyb) yt= Y+ B, yb = YB

4 2ln 4Y B

BY Y B

+

+ =

IncreasingA


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58

Look for extremum dB/dA =0

1 1 10

2

B BB Y B

Y B Y B

+ + + = +

( )

2 2 2 2 2 2

1 11 02

Y B Y B BB B BY B Y B Y B

+ + = = =

Y2 =B2+1 Y= (B2+1)

Substitute back2

2

2

1 11 ln 1 0

2 1

B BB B

B B

+ ++ + =

+

Which requires numerical solution, givingB =0.48, Y=1.11,yt=1.59,yb=0.63.

3.1.2 FLOW MAP

Consider ( )21dy t ydt = [Note: separable, but we wont separate]

Here we wish to map out the solution, without directly solving the ode.

Construct the flow, i.e. arrows showing the slope.

dy/dt= taty =0.

For |y| large, then dy/dtlarge and negative.

For |y|0.


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59

0.0 0.5 1.0 1.5 2.0 2.5 3.0

t

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

.

[Fields produced by Flow.dfc]

Arrows are vectors showing direction of solution; theflow. Arrows often referred to asflow vectors.

Note that sincef(t,y) is single valued then solutions cannot cross.

dy/dt

y


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60

0.0 0.5 1.0 1.5 2.0 2.5 3.0

t

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

y

Can draw on solution easily using arrows. There is an analogy between the flow vectors andvelocity vectors in a fluid flow, and between the solutions and streamlines. Note: solutions drawnwith uniform spacing fory(0), but only up toy(0)=4.

0.0 0.5 1.0 1.5 2.0 2.5 3.0

t

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

y

The green lines are orthogonal to the red ones. These might represent, for example, height contoursabove a plane, or contours of constant potential (see IB Fluids next year). Note potential contours

drawn with uniform spacing fortaty =2.


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61

To construct this map of the solution we did not require the equation to be separable. However,since it is, we can determine the analytical solution.

( )21dy

t ydt

= 2

1 1 1 1

1 2 1 1

dy dyt

y dt y y dt

= + =

+

Integrating ( )21 1 1 1

ln 1 ln 1 ln2 2 1 2

yy y t c

y

+ + = = +

2 21

1

t t cyAe e

y

++ = =

2 2 2

2 2 2

1 1 1

1 1 1

t t t c

t t t c

Ae Ae ey

Ae Ae e

+

+

= = =

+

y = tanh (t2+c) or y = coth (t2+c)

Note: We could have approximated the behaviour for large |y| without solving the exact equation by

noting that 1 y2y2 so 2dy

tydt

2

1 dyt

y dt 2

1 1

2t const

y +

2 2

0

2y

t t

for

some arbitrary t0.

Exploring the behaviour

c >0 y(0)>0 y =tanh(t2+c) has 0< y 1;y 1 as t

c


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0.0 0.5 1.0 1.5 2.0 2.5 3.0

t

-2.0

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

y

Note: All curves asymptote to t=0 andy = 1.

This emphasises that approach to stable equilibrium (and departure from unstable equilibrium) isfaster at high t.

In the earlier example2

sin

1

dy xy

dx y=

+, we can set

2

sin

1 2

y xf

y

= =

+, say. The isoclines are therefore

given by

y2 (2/) y sin x +1=0

2

2

sin sin1

x xy

= for ||1

For solution to exist, require sin2x 2>0.

When sinx = , then y =1 and location of isoclines is single-valued so they must be vertical at

y =0.

As dy/dx =0 when sin x =0, thenx = nmust have vertical isoclines with =0.

Upper branch of isocline has2

2

sin sin sin1 2x x xy

= + as || becomes small.

c > 0

c < 0

c < 0


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First order nonlinear equations Stability

63

Lower branch of isolcine has

2 2

2 2

2 4

2 4

sin sin sin1 1 1

sin

sin 1 11 1

2 sin 8 sin

1cosec

2 sin 2

x x xy

x

x x

xx

= =

= + +

=

!

0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0

t

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

y

Yellow line: sin x Red line: cosec2x.

End of Lecture 8

3.2 StabilityThe technique we shall explore here is not really necessary for first order odes, but is much more

useful for higher order odes.

> 0 > 0

< 0


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64

3.2.1 PERTURBATION

Suppose thaty =y0 is the equilibrium solution, i.e.f(t,y0)=0.

Consider the behaviour ofy =y0 + u where 0 < $1 and u = u(t)= O(1)

For the example used in 3.1.2 ( )21dy t ydt

= ,y0=1 and considery = 1 + u.

Substitute into the ode: ( ) ( )( )20 01d y u t y udt

+ = +

( )( ) ( )2 2 20 0 01 2 2du

t y y u u t u y udt

= + + = +

and linearise (discard terms in 2

and higher)

02du

ty u

dt

Solve with u(0) = u02

0

0

y tu u e

y0 = 12

0

tu u e 0 as t

Solutions converge towardsy = 1y =1 is stable.

y0=12

0

tu u e as t

Solutions diverge fromy =1y =1 is unstable.

As we shall see, a solution is stable iff/y 0, but what iff/y =0?

3.2.2 SEMI-STABLE

Consider ( ) ( )2

1 0dy

y f ydt

=

Equilibrium solution requiresf(y)=0 y =1.

Sincefis independent oft, all solutions have the same shape.


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65

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

t

-1.0

-0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

y

Solutions fory 1 solutions diverge.

The equation is separable: ( )2

0

11

1y dy t t

y

= =

0

11y

t t=

Stability

Fort> t0,y ~11/t 1 from below as t.

Fort< t0,y ~ 1/(t0 t)+ as tt0 from below, soy rises from - as tincreases from t0.

Stability analysis: lety = 1 + u

( ) ( )2 2 21 1 1

d duu u u

dt dt + = = =

2

0du

udt =

t0


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66

Marginally stable.

3.2.3 GENERIC STABILITY PROBLEM

Consider ( ),dy

f y t

dt

=

1. Find fixed points (equilibrium solutions) by solvingf(y,t)=0 to find solutions withy = constt. [It is not sufficient fordy/dtto vanish for only some period of time.]

2. Expandfabout an equilibrium solutiony =y0, say. Note: Each equilibrium solution must beinvestigated separately.

a. y =y0+ u, say, and expandfin Taylor series abouty =y0:b. ( ) ( )

0 0

22

0 2

1, ,

2y y y y

f ff y t f y t u u

y y= =

= + + +

!

3. Substitute into equationa.

0 0

22

2

1

2y y y y

du f f u u

dt y y= =

= + +

!

Solutions:

If ( )0

0y y

ft

y

=

=

, then

duu

dt for sufficiently small u.

0dt

u u e

if>0 unstable 0 if t0, then u 0 as tincreases. Ift< t0, then u sign()

as tincreases towards t0 from below.


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67

[For previous example, ( )2

1dy

ydt

= ( )2 1f

yy

=

,

2

22

f

y

=

, so at y =1 equilibrium,

( )0

1

2u

t t

=

and semi-stable.

If=0, need to look at further terms of Taylor series in order to determine stability.3.2.4 PHASE PORTRAITS

Further insight may be gained by considering the phase portraitof the differential function. Thephase portrait is a plot ofdy/dtagainsty (effectivelyf(y,t) against t).

Consider

21

dyy

dt= .

This clearly has equilibrium solutions or fixed points aty =1.

Recall that f/y indicates the stability of the fixed points. When f/y 0 leads to dy/dtincreasing as we move away fromthe fixed point, giving an unstable solution.

A stable fixed point is an attractor, while an unstable fixed point is a repellor.

The semi-stable equation ( )2

1dy

ydt

= explored in 3.2.2 has the attractor and repellor merging:

y

dy/dt


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First order nonlinear equations Exact equations

68

End of Lecture 9

3.3 Exact equations

Recall that we can convert a linear equation of the form y+ qy = f into an exact equation of the

form ( )d

Iy Ifdx

= using the integrating factorq dx

I e= . Here we shall try to do a similar thing for

nonlinear equations.

Note: There is no simple guaranteed procedure for this it is not always possible. It will be

introduced by example.

Consider2 3

3 2 0dy

xy y xdx

+ + = .

This equation is neither linear nor separable. Can we solve it?

By inspection we can see ( )2 3 3 23 2 0dy d

xy y x xy xdx dx

+ + = + = , and the equation is exact.

Hence the solution is ( ) 3 2,x y xy x c + = (c = const)

12 3c x

yx

=

.

dis an exact differential.

is sometimes called apotential. It is a conserved quantity.

3.3.1 FINDING AN EXACT EQUATION

Consider ( ) ( ), , 0dy

f x y g x ydx

+ = , which we wish to write as 0d

dx

= for some (x,y).

Recalld dy

dx y dx x

= +

, so we need ( ),f x y

y

=

and ( ),g x y

x

=

.

In the previous example,

y

dy/dt


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First order nonlinear equations Exact equations

69

23xy

y

=

=xy3+ A(x),

3 2y xx

= +

=xy3 +x2+ B(y),

whereA(x) andB(y) are the constants of integration.

Now the two expressions formust be identical, soA(x) =x2+ c andB(y)= c,

thus =xy3

+x2+ c.

However, not all equations are exact, so this procedure is not always possible.

Since2 2

y y x

=

, then we must have

f g

y

=

; can therefore test to see if an exact differential

is possible before trying to find it!

2 33 2 0dy

xy ydx + =

Note: separable, so it can be written as1 2

3

dy

y dx x= , but we look to make it exact.

f= 3xy2f/x = 3y2

g= 2y3g/y = 6y2f/x equation is not exact.

But the equation can be made exact by multiplying the equation byx:

2 2 33 2 0dy

x y xy

dx

+ = f/x = 6xy2=g/y.

Integratingf= /y=x2y3+ A(x)

Integratingg= /x=x2y3+ B(x).

Equating =x2y3 + const, hencey = cx2/3.

Here we have converted the equation into an exact equation by multiplying it by an integrating

factor(herex).

3.3.2 INTEGRATING FACTOR FOR NONLINEAR EQUATIONS

The idea of an integrating factor was introduced in 2.3.2 for linear equations.The process of finding the integrating factor for a nonlinear equation is more complex.

Multiply the general equation 0dy

f gdx

+ = by the function (x,y), so

0dy

f gdx

+ = .

For this new equation to be exact, we must have

( ) ( )f g

x y

=

(*)


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First order nonlinear equations Examples

70

and fy

=

( ) ( ),y f dy A x = +

gx

=

( ) ( ),x y g dx B y = +

Unfortunately, determining by solving (*) may be as difficult (or impossible) as solving theoriginal differential equation. [Of course, equations on the Examples Sheet may be possible]

[Finding appropriateA(x) andB(y) is typically more straight forward.]

Sometimes the solution of (*) may be simplified by proceeding on the assumption that is a

function ofx ory alone.

2 33 2 0dy

xy ydx

+ = [Our previous example]

Multiply by integrating factor, assuming =(y) [We already know = x works!]

( ) ( )2 3 23 2 6f g

f y g g y yx x y y y y

= = = = + = +

1 3

2

d

dy y

= ~y3/2

So 1/ 23f xyy

= =

= 2xy3/2+ A(x)

and 3/ 22g yx

= =

= 2xy3/2+ B(y).

Hence = 2xy3/2+ c andy = Cx2/3 as before!

Note that the integrating factor is therefore not unique.

Since the solution hasy ~x2/3, then ~y

3/2~ x, the previous integrating factor.

Moreover, one would expect any combination ofx andy that is equal to the integrating factorx

would also be an integrating factor, e.g.=x1y

3.

3.4 Examples

3.4.1 CHEMICAL KINETICSConsider the chemical reactionX+ Y U+ V e.g.NaOH+HClH2O + NaCl

Suppose the concentration ofXisx(t), ofYisy(t) and Uis u(t) (the reaction gives equal quantities

ofUand V, so v(t)= u(t)).

Ifx(0) =x0,y(0) =y0 and u(0)=0, then the conservation relations givex + u =x0 andy + u =y0.

Suppose the reaction rate ris xy, where =(T) [Note that if 2X+ Y, then x2y, etc.]

Then ( )( ) ( )0 0du

r xy x u y u f udt

= = = .

Fixed points (equilibria) are where f(u)=0 u =x0 and u =y0 (i.e. all of one of the species isconverted).


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71

Need to keep in mind the physical meaning of x and y: clearly cannot have a negativeconcentration!

We shall assume x0


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72

0 0

0 0

0 0

y t x t

y t x t

Ae y eu

Ae e

=

.

Noting that u(0)=0 givesA =y0/x0 so

0 0

0 00 0

0 0

y t x t

y t x t

e eu x y y e x e

=

Note: Problems such as this including the derivation of the ode(s) are fair game for examiners.Maths is not only about solving the equations, but also about deriving them and drawing

conclusions from the solutions. In Part IA the examiners would help you with any derivation

required in an exam.

3.4.2 POPULATION DYNAMICS

Constant birth and death rates

Suppose birth rate is proportional to populationy,

birthrate = y

[For many species, the number of females matter, but not the number of males, so might not be

constant. Moreover, for some species, e.g. rabbits, overcrowding causes the birth rate to decline.]

Suppose also that the death rate is proportional to the population,

deathrate =y

[This may be true for death by natural causes and hunting by another species. It will typically not

be true for epidemics, combat between members, etc.]

( )dy y rydt

= =

0rty y e=

population grows or decreases exponentially, depending on sign of.

End of Lecture 10

Fighting for limited resources

If>then the population will grow too large for the available resources (e.g. food supply).Scarcity of food may lead to fighting (& death) between individuals who happen across the samefood at the same time. The probability of one individual being at a given location at a given time is

proportional to the population,y, so the probability of two individuals meeting at a given location isproportional toy2, hence

deathrateduetofighting=sy2

wheres is a constant. Thus

( )2 1dy y

ry sy r y f ydt Y

= =

with Y= r/s. Note that this should hold even if


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73

The phase portrait shows critical points (fixed points or equilibria) aty =0, Y.

The y =0 fixed point has f/dy >0 so is unstable, whereas at y = Yf/dy


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0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

t

0.0

0.5

1.0

1.5

2.0

2.5

3.0

y

Clearly,y = Yis a stable equilibrium, whereasy =0 is unstable.

Logistics equation is separable and can be solved explicitly:

1 1 1

1

dy dyry dt y Y y dt

yY

= + =

lny

rt cY y

= +

rty

AeY y

=

Supposey =y0 at t= t0, thenA =y0ert0

/(Yy0) and

negative

curvature

positive

curvature


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75

( )

( )

0

0

0

0 0

r t t

r t t

Yy ey

Y y y e

=

+

which givesy Yas t.

But can population really rise from just above zero? With careful management it is possible (e.g.Old Blue see later), but in general

Mating opportunities

Individuals need to find mates to procreate. This can become difficult when population densities arevery low. The probability of finding a mate is similar to the probability of fighting over food and is

again proportional toy2. Of course, fighting or procreating depends on who meets!

For low population densities (where fighting is not an issue) the birth rate is y2, while the death

rate remainsy, so

( )2 1dy yy y y f ydt X = + =

This equation has the same structure as previously, with critical points at y =0 andy = X.

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

t

0.0

0.5

1.0

1.5

2.0

2.5

3.0

y

Now,y = Xis unstable, so if population falls belowy = X, then it will fall to zero and extinction!


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76

Passenger pigeon

The North American passenger pigeon is one of the most famous recent examples of extinction.

In mid 19th century,y ~ 109, the most numerous bird on earth. Widely hunted for sport and food in late 19th century; as many as 30,000 shot at a time! By 1896 only 250,000 remaining Extinct by 1900.Extinction was ultimately caused not by hunters killing the remaining birds, but by their inability tofind mates.

Their extinction was one of the early triggers for conservation.

[Lots of information on the web.]

Coming back from the brink

Introduction of mammals (cats, rats, ferrets, opossum, etc.) to New Zealand has devastated thepopulations of many species of birds. Some have become extinct, but others have brushed with the

edge of extinction only to recover through careful management, with human intervention (divineintervention?) effectively bypassing the problem of finding a mate.


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77

http://www.kcc.org.nz/birds/blackrobin/oldblue.asp

The New Zealand black robin is an extreme example: in 1980 there was only a single breading

female and four males. Now there are around 250 and the population is growing.

Logistic growth with a threshold

Populations above the critical threshold y = X cannot really grow without limit, so some newlimiting factor must come into play wheny is large enough.

Could model this as

1 1dy y y

r y fdt Y Z

=

ForY< Zthe phase portrait is

Clearly equilibria arey =0, Y, Z, with two stable and one unstable fixed points.

y

f

Y Z


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First order nonlinear equations Comparison with discrete equations

78

0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

t

0.0

0.5

1.0

1.5

2.0

2.5

3.0

y

Stable equilibria aty =0, Z, and unstable aty = Y.

3.5 Comparison with discrete equations

Consider the logistics equation 1dy y

r ydt Y

=

r>0; Y>0.

[y = Ystable,y =0 unstable]

The Euler finite difference approximation (see 2.2.3) to this may be written as

yn+1yn = rdt(1 yn/Y)yn

( )

11

1 11

1

nn n n

nn

nn

yy y y

Yy

yY

yy

+

= +

= + +

=

where= r dt, = (1 +) and = Y(1+)/. Let un =yn/

un+1 = (1 un) un

This normalised nonlinear first order difference equation has only one important parameter, ,

whereas in the original we had rand Y.


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79

3.5.1 EQUILIBRIUM SOLUTIONS

Put un+1 = un: un = (1 un) un

( 1 un)un=0

un=0 orun = (

1)/

U. [c.f.y =0, Y]

3.5.2 STABILITY

Similar to what we did for the differential equations.

Nearun=0, have un2$ |un| un+1un

unnu0

Hence unstable if ||>1. Now since = 1 +, and= r dt>0, then un=0 unstable equilibrium.

Nearun = U= ( 1)/, suppose un = ( 1)/+ vn |vn|$ U

( )( )

( )

1

2

2

1 1 1 1 11

11 1

12

n n n n n

n n

n n

v v v v v

v v

v v

+

+ = + = +

= +

= +

( )1 2n nv v+

( ) 02n

nv v

Stable if |2 |


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80

so we require ( )1 1nn

vf U

v+ <

for stability (vn0 as n ).

For the current equation, we therefore require

( )( ) ( )1 1 2f d

u u uu du

= =

is less than unity when u =U= ( 1)/(the equilibrium solution). Substituting

21 2 2 1

+ =


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81

0.0 2.0 4.0 6.0 8.0 10.0

n

0.0

0.2

0.4

0.6

0.8

1.0

u

Keylambda = 0.80U = -0.25y0 = 0.20y0 = 0.40y0 = 0.60y0 = 0.80y0 = 1.00

0.0 2.0 4.0 6.0 8.0 10.0

n

0.0

0.2

0.4

0.6

0.8

1.0

u

Keylambda = 1.50U = 0.3 3y0 = 0.20y0 = 0.40y0 = 0.60y0 = 0.80y0 = 1.00

0.0 2.0 4.0 6.0 8.0 10.0

n

0.0

0.2

0.4

0.6

0.8

1.0

u

Keylambda = 2.80U = 0.64y0 = 0.20y0 = 0.40y0 = 0.60y0 = 0.80y0 = 1.00

0.0 2.0 4.0 6.0 8.0 10.0

n

0.0

0.2

0.4

0.6

0.8

1.0

u

Keylambda = 3.20U = 0.6 9y0 = 0.20y0 = 0.40y0 = 0.60y0 = 0.80y0 = 1.00

The =1.5 and =2.8 solutions are stable and approach the equilibrium solution u = U. The

convergence can be monotonic, or oscillatory. Why?

End of Lecture 11

3.5.4 GRAPHICAL APPROACH TO DIFFERENCE EQUATION

Plotting un+1 against un provides a straight forward and instructive method of graphically solving the

difference equation. Ifun+1= f(un) then begin by plottingf(un) and the line un+1 = un:

For=0.8:


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82

Clearly un0 as n regardless of starting point.

For=1.5:

Convergence is monotonic close to equilibrium solution.

For=2.8:

un

un+1

un

un+1


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Initial monotonic convergence changes to oscillatory convergence.

For=3.2:

Initial convergence changes to a steady oscillation with period two. The appearance of this new

oscillatory solution is called a bifurcation. In this case the period two oscillation is stable; the

original solution still exists but it is now unstable. The phenomenon is calledperiod-doubling.

At =3.449 there is a second period-doubling bifurcation in which the period two solution

becomes unstable and a stable period four oscillation appears.

un

un+1

un

un+1


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84

0.0 2.0 4.0 6.0 8.0 10.0

n

0.0

0.2

0.4

0.6

0.8

1.0

u

Keylambda = 3.50U = 0.71y0 = 0.20y0 = 0.40y0 = 0.60y0 = 0.80y0 = 1.00

Forslightly larger get another period doubling to an oscillation with period eight, then to sixteen,

etc., with values ofconverging to c=3.5699

3.5.5 LOGISTIC MAP

Valuable insight into the behaviour of the difference equation may be gained by plotting its

asymptotic solutions (as n ) as a function of the control parameter(s).

For our logistic difference equation un+1 = (1 un) un, this means plotting un against for largevalues ofn.

un

un+1


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85

1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5

lambda

0.0

0.2

0.4

0.6

0.8

1.0

1.2

u

When


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86

3.5.6 BANDS

3.5 3.52 3.54 3.56 3.58 3.6

lambda

0.3

0.4

0.5

0.6

0.7

0.8

0.9

u

For > c there are intervals of values of in which stable periodic orbits exist. These attractalmost all initial conditions and so show up as clear bands in the diagram. If we look hard enough it

turns out that there is an interval containing a stable periodic orbit of every integer period.

P4

P8P16


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87

3.568 3.57 3.572 3.574 3.576 3.578

lambda

0.46

0.47

0.48

0.49

0.5

0.51

0.52

0.53

0.54

u

For > c there are intervals of values of in which stable periodic orbits exist. These attract

almost all initial conditions and so show up as clear bands in the diagram. If we look hard enough it

turns out that there is an interval containing a stable periodic orbit of every integer period.

For>4, typical initial conditions are mapped out of the interval [0,1] and of to .

3.5.7 LOCATION OF BIFURCATIONS

Recall that when we were looking for equilibrium solutions we set un+1 = un and solving

un = (1 un) un

For the period two orbit, we can do a similar thing, but set un+2 = un. Now

un= un+2 = (1 un+1) un+1 = (1 (1 un) un) (1 un) un (*)

which yields a quartic in un. However, we already know two of the roots: since un+1=0, Usatisfy

both un+2 = un+1 and un+1 = un, then they are also solutions to (*). Factorising,

un (1 + un)[1 + (1+)un + 2un

2]=0.


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The term in square brackets provides additional roots

21 2 3

2n

u

+ = (**)

provided 2 23>0 < 1 or>3.

The new roots exist when 3 and correspond to the new period two orbit. From a perturbationanalysis we can show that they are stable for close to 3; in this case the period-doubling

bifurcation is termedsupercritical.

Note that at =3, the new root is a double root with un=2/3, whereas the equilibrium root also has

un=2/3.

In 3.5.2 we saw that |f/u| < 1 for stability of the equilibrium solution. We can use the same ideashere for the oscillatory solution, where we have un =f(f(un)). The stability boundary therefore

requires

( )( ) ( )( )2 21 2 1 2 2 1f f u u u uu

= + =

Substituting in either of the oscillatory solutions (**) and simplifying gives

4 + 22 = 1,

with roots = 1 6, = 1, =3 and = 1+6. Only the last two of these are relevant, and showthat the period two solution is stable for 3 <


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0.0 0.2 0.4 0.6 0.8 1.0

u(n)

0.0

0.2

0.4

0.6

0.8

1.0

u(n+?)

lambda= 2.90u(n+1)u(n+2)u(n+3)u(n+4)u(n+5)u(n+6)u(n+7)u(n+8)

0.0 0.2 0.4 0.6 0.8 1.0

u(n)

0.0

0.2

0.4

0.6

0.8

1.0

u(n+?)


0.0 0.2 0.4 0.6 0.8 1.0

u(n)

0.0

0.2

0.4

0.6

0.8

1.0

u(n+?)


0.0 0.2 0.4 0.6 0.8 1.0

u(n)

0.0

0.2

0.4

0.6

0.8

1.0

u(n+?)


first order nonlinear equations

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