Phase portraits

8/12/2019 Phase portraits

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Mathematical Models - DN2266

Homework 3 - Phase portraits

Teachers

Anna-Karin TORNBERG

Rikard OJALA

Authors

Pierre-Alexandre BEAUFORT

Hadrien VAN LIERDE

October 2013


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Contents

1 Scaling a differential equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Homogeneous differential equation. . . . . . . . . . . . . . . . . . . . . . 11.2 Non-homogeneous differential equation . . . . . . . . . . . . . . . . . . . 2

2 Stability of a first order ODE-system . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Manifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Asymptotical behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 Prey-predator model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

3.1 Setting the equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Behavior of the solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3 Critical points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.4 Scaling of the parameters . . . . . . . . . . . . . . . . . . . . . . . . . . 113.5 Analysis of the critical points . . . . . . . . . . . . . . . . . . . . . . . . 12

A Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A.1 Proposition1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17A.2 Proposition 1bis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18A.3 Proposition2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18A.4 Figures for the second set of datas. . . . . . . . . . . . . . . . . . . . . . 20

A.5 Phase_Portrait . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

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1 Scaling a differential equation

In this section, we show how to scale a homogeneous and a non-homogeneous differential equa-tion.

1.1 Homogeneous differential equation

The following differential equation describes damped free oscillations:

md2u

dt2 +c

du

dt +ku = 0

where u [m] is the deviation from a stationary position and t [s] is the time. Under thisform, the equation contains three parameters: the mass m [kg], the damping constant c [kg

s],

and the spring constant k [Nm

].

We first define the two following quantities:

0, the undamped frequency such that [0] =s1 (we will give the value of0 further),

=0t, a dimensionless time.

With this new variable, the equation becomes(i):

m20u+c0u

+ku = 0

Dividing the two members of the equality by m20 , we obtain:

u+ c

m0u+

k

m20u= 0

To simplify this expression, we choose 0 =

km

and we introduce the critical damping c0 =

2

km. We then obtain the following equation:

u+ 2c

c0u+u= 0

This equation only depends on one parameter, which we note = cc0

. The final equation is:

u+ 2u+u= 0

(i)Here, we note u = dud

(). As u is normally a function oft, the exact form would be dd

[u( 0

)].

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1.2 Non-homogeneous differential equation

In the case of forced oscillations, the equation is:

md2u

dt2 +c

du

dt +ku = Fcos(t)

We introduced two additional parameters. We follow the same steps as before but we introducea few more constants.We first define the four following quantities:

0, the undamped frequency such that [0] =s1 (we will give the value of0 further),

L [m], a reference length of displacement,

=0t, a dimensionless time.

u= uL

, a scaled displacement

With this new variable, the equation becomes(ii):

m20Lu+c0Lu

+kLu= Fcos(

0)

Dividing the two members of the equality by m20L, we obtain:

u+ c

m0 u+

k

m20 u= F

m20Lcos(

0 )

To simplify this expression, we choose 0 =

km

, L = Fm2

0

= Fk

and we introduce the critical

damping c0= 2

km. We then obtain the following equation:

u+ 2c

c0u+ u= cos(

0)

This equation only depends on two parameters, which we note = cc0

and = 0

. The finalequation is:

u+ 2u+ u= cos()

(ii)Here, we note u = dud

() = dd

[ 1L

u( 0

)].

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2 Stability of a first order ODE-system

We study here the stability of the linear system of differential equations :

u=

1 1P 1

Au (1)

where P . First, we determine the type of phase portrait according to the value ofPand -if it is possible - the slopes of the manifolds. Then, we compute lim

tu(t) and lim

tyx

(t) (iii),i.e.

the asymptotical behaviour of the integral curve(iv) .

2.1 Manifolds

The characteristic polynomial ofAis2 + 2+ (1 P), whose the roots are 1,2 = 1

P.Observe if1,2 0, the fast manifold is this with the smallest eigenvalue. According to thevalue ofP, the phase portrait is : P >0: eigenvectors are v1 = (1,

P) and v2= (1,

P).

P (0;1) 2< 2 < 1< 1 < 0 Stable node with fast manifold along v2 and slow along v1

P = 1 1,2= 0,2 Marginallystable node(v)

P (1;+) < 2< 2< 0 < 1 < + Saddle point with stable manifold along v2 and unstable along v1

P= 0 1,2 = 1Matrix A is defective since there is a single eigenvector (; 0) (vi) according to

(A+I)z1= 0 =

0 10 0

xy

Then, we have to compute the solution in the defective case:

(A+I)z2 = z1

If we take = 1, we get (;

1) (vii). Let = 0. The solution of this system is given by:

u(t) = (az1+b(tz1+ z2))exp(t)

Observe the solution is asymptotically stable(viii).

P


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Now, we are going to compute the slopes of manifolds, if it is possible. Therefore, we have tocompute the nullspace of(A I):

1 + 1P 1 +

Bz= 0

As we know the case P = 0 gives a defective matrix A, we consider P= 0. The nullspacehas non-zero vector only if B is not full rank. We check thus if the second row is a linearcombination of the first and itself:

1 + 1

0 P(1+)2

P

It is the case ifP= (1 +)2.

Actually, it is always the case since is the solution of(1 + )2P = 0. Let z:= (x; y). Then,Kerf(B) y= (1 +)x. As 1,2 = 1P, the slopes of manifolds areP (ifP 0).

2.2 Asymptotical behaviour

We study the asymptotical behaviour owing to :

eigenvalues 1,2 = 1

P

associated eigenvectors v1,2 y=

P ; here, v1,2 = (1,

P) (ix)

solution u(t) =av1exp(1t) +bv2exp(2t)

initial conditions u0:= u(t= 0) =av1+bv2

Our study focuses on limt

u(t) and limt

yx

(t). Then, we notice these asymptotical behaviors

depends on the value ofPand the initial conditions(x). In all cases, we assume that(a, b) = (0, 0)(this case leads to the trivial solution u(t) = 0).

P (0;1) u= a

1

P

exp(1t) +b

1

P

exp(2t), with 2< 1


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P (1;) u= a

1

P

exp(1t) +b

1

P

exp(2t), with 2 < 0 < 1.

ifa = 0, both components ofu are unbounded ; ifa= 0,ut

0

y

x=

Pb exp(2t) a exp(1t)b exp(2t) +a exp(1t)

=

Pb exp((2 1)t) ab exp((2 1)t) +a

y

xt

P ifa = 0 ; ifa= 0 and b = 0, y

xt

P

P= 0 u=

a

10

+b

t

10

+

01

exp(t)

ut

0a, by

x=

ba+bt

yxt

0 if(a, b) = 0

P P0= 1, ifu0

= (;

P ),

C0. The following table1 gives the type of phase

portrait on either side.

P


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3 Prey-predator model

In this section, we consider the following two-species model:

dx

dt = a(x, y)x

dy

dt = b(x, y)y

(2)

where x and y denote the populations of two species and a and b denote the correspondinggrowth rate.

3.1 Setting the equation

Let us assume that y denotes the predator population and xdenotes the prey population. Weintroduce the following assumptions (both in english and in mathematics):

1. if there is not enough prey, the predator population declines:

for a certain 1 R+0, x < 1 b(x, y)< 0y R+

2. an increase in the prey population increases the growth rate of the predator:

b

x(x, y)> 0x, y R+

3. if no predators are present, a small prey population will increase: there exists >0 suchthat

a(x, 0)> 0x ]0, [4. if the prey population goes beyond a certain size, it must decrease:

for a certain 2 R+0, x > 2 a(x, y)< 0y R+

5. if the predator population increases, the growth rate of the prey population declines:

a

y(x, y)< 0x, y R+

Many types of functions could satisfy these assumptions. We thus consider the specific model

a(x, y) = y xb(x, y) = x y

with ,, ,, , >0.

Let us show that this model respects the five assumptions given above. First of all, we intro-duce an obvious assumption based on the definition of a prey-predator model: it is obvious

that, for a given population of prey, if the predator population increases a little bit, then thegrowth rate of the predator population should decrease which can be expressed as by

0. To justify the presence of all other constants, we use assumptions 1 to 5:

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2) bx

>0 >0;3) If we choose 0, then a(x, 0) < 0x > 0, we thus have to choose > 0, in which casecondition 3 is verified;5) a

y0

4) for a certain y 0, ifx > y , then a(x, y)< 0; in particular, ify = 0, this inequalitymust also be verified for x large enough; but, as > 0, if < 0, x > is never verified for

anyx >0; for this reason, we necessarily have >0;1) for a certainy >0, ifx < +y, we haveb(x, y)< 0; this must be true for0 < y < >0;as >0, we must have x < +y

; let us assume 0, find u= (x, y) C1([0,

[) such that

dxdt

= ( y x)x

dy

dt = (x y)y

x(0) = x0y(0) = y0

(3)

We want to show that the solution of this problem is strictly positive: x(t), y(t) > 0

t. To

prove this, we are going to use the following propositions:

Proposition 1

Let u0 R2. We define the following problem: findu C1([0,[) such thatdxudt

= ( yu xu)xu

dyudt

= (xu yu)yu

u(0) = u0

(4)

Then there exists a unique solution u to this problem andu C([0,[).

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Proposition 2

Let u0,v0 R2 such that u0= v0 andu= (xu, yu),v= (xv, yv) C1([0,[) verifying:

dxudt

= ( yu xu)xu

dyudt

= (xu yu)yu

u(0) = u0

(5)

dxvdt

= ( yv xv)xv

dyvdt

= (xv yv)yv

v(0) = v0

(6)

Then u(t) = v(t)t. Moreover, fort1 t2, u(t1) = v(t2) ifu(t) = v0t t1.

The proofs of propositions1and2are in appendicesA.1andA.3. Let u = (x, y)be the solutionof3. Let us assume that

t2>0 such that x(t2) 0 or y(t2) 0

then, by continuity of the solution, there exists t1

]0, t2[ such that exactly one of the three

following conditions is verified:

case 1: (x(t1), y(t1)) = (0, 0) and x(t), y(t)> 0t < t1 case 2: x(t1) = 0, y(t1)> 0 and x(t), y(t)> 0t < t1 case 3: x(t1)> 0, y(t1) = 0 and x(t), y(t)> 0t < t1

In each case, we will prove that there exists one other curve that verifies 3but with different

initial conditions and that verifies one of the three conditions above. This leads to a contradic-tion according to proposition2 which proves that x(t), y(t)> 0. Let us analyze the three casesone by one.

Case 1: (x(t1), y(t1)) = (0, 0) and x(t), y(t)> 0t < t1

Let u1= (x1,y1) be the solution of

dx1dt

= ( y1 x1)x1

dy1dt

= (x1 y1)y1

u1(0) = (0, 0)

(7)

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Then, by proposition1, the unique solution to this problem is u1(t) = (0, 0) t. We verify thusu(t1) = (0, 0) =u1(t1) which contradicts proposition2.

Case 2: x(t1) = 0, y(t1)> 0 and x(t), y(t)> 0t < t1

Let u2= (x2,y2) be the solution of

dx2dt

= ( y2 x2)x2

dy2dt

= (x2 y2)y2

u2(0) = (0, 2y(t1))

(8)

Then, one can prove the three following properties:

1. x2

C([0,

[) by proposition1,

2. dx2

dt(0) = 0 (obvious),

3. dnx2

dtn(0) = 0n: if d

n1x2dtn1

(0) = 0 then

dnx2dtn

(0) = dn1

dtn1

dx2dt

(0)

= ( y2(0) x2(0))dn1x2

dtn1(0) + ( d

n1y2dtn1

(0) dn1x2

dtn1 (0))x2(0) = 0

As dx2

dt(0) = 0,

dnx2dtn

(0) = 0n.

The first condition above allows us to write x2 in terms of its Taylor polynomial evaluated in0 and this expression is equal to 0 according to the third condition: x2(t) =

n=0

1n!

dnx2(0)dtn

tn = 0.

We may then reduce the problem, to determine the only nonzero component:

dy2dt

= ( y2)y2y2 = 2y(t1)

(9)

One can prove the two following properties of the solution of the equation above:

1.t, y2(t)> 0: if not then u2(t) = (0, 0) for some t. Then u2(t) = (0, 0) =u1(t) whichcontradicts proposition2.

2.t, dy2dt

(t)< 0: since y(t)< 0, dy2

dt = ( y2)y2 < 0 ; y2 is thus strictly decreasing.

Owing to these conditions, there exists K 0 such that limt

y2 = Kwhere (0, K) has to be

a critical point of the system. Obviously, the only positive value ofK such that (0, K) is acritical point is K = 0 and lim

ty2 = 0. Then by continuity of y2, there exists t3 such that

2y(t1)> y2(t3) =y(t1)> 0. Thus u2(t3) = u(t1) which contradicts proposition2.

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Case 3: x(t1)> 0, y(t1) = 0 and x(t), y(t)> 0t < t1

For x0 > 0, let u3= (x3,y3) be the solution of

dx3dt

= ( y3 x3)x3

dy3dt

= (x3 y3)y3

u3(0) = (x0, 0)

(10)

As in case 2 for the componentx2, one can prove that y3(t) = 0 t. The only nonzero componentsatisfies then the following equation:

dx3dt

= ( x3)x3

x3(0) = x0

(11)

We then distinguish the three following cases:

case 3.1: x(t1) =

case 3.2: x(t1)>

case 3.3: x(t1)

With x0 = 2x(t1), one can show that by proposition 2 x3(t) >

and that x3(t) is strictlydecreasing. Then x3 tends to a critical point K with K . Obviously, the only possiblevalue ofK is

. Thus there existst3 such that 2x(t1)> x3(t3) =x(t1)>

which contradictsproposition 2as u3(t3) = u(t1).

Case 3.3: x(t1)< With x0 = x(t1)/2, one can show that by proposition 2 x3(t) 1.Then, u Cn([0,[):

dnxu

dtn =

dn1

dtn1[

dxu

dt ] = ( yu xu) d

n1xudtn1

+ ( dn1yu

dtn1 d

n1xudtn1

)xu

dnyu

dtn =

dn1

dtn1[

dyu

dt ] = (xu yu) d

n1xudtn1

+ (dn1xudtn1

dn1yu

dtn1 )yu

(16)

As u C1([0,[), we know by recursion that u C([0,[).

Let us prove the uniqueness of the solution. Assume that u and v both satisfy the differentialequation above. Let us note e= (xe, ye) = u v. Then e satisfies the following system:

dxe

dt = ( ye xe)xe

dye

dt = (xe ye)ye

e(0) = (0, 0)

(17)

One can prove the three following properties of e

:1. As u,v C([0,[), e C([0,[)2.

dne

dtn(0) = (0, 0)n. This is true for n = 1. For n > 1, let us assume it is true for n 1.

Then,

dnxe

dtn (0) =

dn1

dtn1[

dxe

dt (0)]

= ( ye(0) xe(0)) dn1xe

dtn1(0) + ( d

n1yedtn1

(0) dn1xe

dtn1 (0))xe(0)

= 0dnye

dtn (0) =

dn1

dtn1[

dye

dt (0)]

= (xe(0) ye(0)) dn1xe

dtn1(0) + (

dn1xedtn1

(0) dn1ye

dtn1 (0))ye(0)

= 0

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3. e(t) = 0t. Owing to point 1 above, e can be written in terms of its Taylor polynomialwhich is zero according to point 2: e(t) =

n=0

1

n!

dne

dtn(0)tn = 0.

The last properties of e shows that u(t) = v(t)t which proves the uniqueness of the solution.

A.2 Proposition 1bis

This proposition is very similar to the preceding proposition. We just change the signs of theright hand terms of the differential equation.

Proposition (1bis)

Let u0 R2. We define the following problem: findu= (xu, yu) C1([0,[) such thatdxu

dt

=

(

yu

xu)xu

dyudt

= (xu yu)yu

u(0) = u0

(18)

Then the solution u to this problem is unique andu C([0,[).

We will not prove this proposition as the proof is essentially the same as for proposition 1.

A.3 Proposition2Here is the proof of proposition2in section 2.

Proposition 2

Let u0,v0 R2 such that u0= v0 andu= (xu, yu),v= (xv, yv) C1([0,[) verifying:

dxudt

= ( yu xu)xu

dyudt

= (xu yu)yu

u(0) = u0

dxvdt

= ( yv xv)xv

dyvdt

= (xv yv)yvv(0) = v0

Then,

1. u(t) = v(t)t.

2. fort1 t2, u(t1) = v(t2) ifu(t) = v0t t1.

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Proof :

Let us prove the first conclusion. Let us assume there existst1 > 0 such that u(t1) = v(t1) = u1.Then, we consider the following problem: findw= (xw, yw) C1([0,[) such that

dxw

dt = ( yw xw)xw

dyw

dt = (xw yw)yw

w(0) = u1

According to the proposition 1 bis (in appendixA.2), the solution of this problem is unique. Onecan show that u(t) = u(t1 t) andv(t) = v(t1 t) are both solutions of this problem. But,u(t1) = u(0) = v(0) = v(t1). Which contradicts the fact the solution is unique.

Let us prove the second conclusion. Let t1 t2 and u(t1) = v(t2). Then, we formulate a newproblem: findw C1([0,[) such that

dxw

dt = ( yw xw)xw

dyw

dt = (xw yw)yw

w(0) = u(t1 t2)

Then, according to proposition 1, the solution of this problem is unique. One can show thatw(t) = u(t + (t1 t2))is the unique solution of this problem. Comparingwandv, we see that theinitial conditions are different: w(0) = u(t1 t2) = v(0) as0< t1 t2 < t1 (see the assumptionsof the second conculsion). Thus, according to the first conclusion, we havew(t)= v(t)t. Thiscontradicts the fact that w(t2) = u(t1) = v(t2).

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A.4 Figures for the second set of datas

On the figures here, the phase portrait shows that the manifolds describe well the behavior ofthe solution in the case of the linear approximation. But, in the non linear model, the curvedoes not follow the pattern suggested by the manifolds obtained in the linear model. To geta better analysis of the behaviour of the solution around the critical point, we could plot the

curves for negative initial conditions but we did not do it as it does not make sense for apopulation model.

0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

3phase portrait in the nonlinear case (vectors)

u component

vcomponent

Figure 5: Phase portrait with vectors showing the direction and magnitude of the derivativesvectors in various points (20 times 20 points between 0.1 and 3).

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0.5 0 0.5 1 1.5 2 2.5 30.5

0

0.5

1

1.5

2

2.5

3phase portrait in the nonlinear case (curves)

u coordinate

v

coordinate

Figure 6: Phase portrait with the curve representing the solution for several initial conditions(5 times 5 points between 0.1 and 3).

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3 2 1 0 1 2 3 41

0.5

0

0.5

1

1.5

2

2.5

3phase portrait with linear approximation (curves)

u coordinate

v

coordinate

Figure 7: Phase portrait with the curve with linear approximation representing the solution forseveral initial conditions (5 times 5 points between 0.1 and 3). The manifolds are in red

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A.5 Phase_Portrait

1 f u n c t i o n Phase_Portrait()2 %PHASE_PORTRAIT

3 % P lo ts t h e p h as e p o rt ra it s f o r t he t w o s e ts o f p ar am et er s ( A ,B a n d C )

4 % b ot h w i th a rr ow s ( g i vi ng t he n or m a n d d i re ct io n o f t h e d e ri va ti ve i n

5 % e ac h p oi nt ) a nd w i th t he c u rv e i ts el f.

6 c l o s e a l l7

8 % % c a s e 1 n o nl i ne a r

9 u1 =l i n s p a c e(0.1,3,20);10 v1 =l i n s p a c e(0.1,3,20);11 [u,v]= me s hgrid (u1,v1);12 du = z e r o s (s i z e( u ) ) ;

13 dv = z e r o s (s i z e( v ) ) ;14 f o r i=1:numel(u)15 dU=f1(0,[u(i),v(i)]);

16 du(i)=dU(1);

17 dv(i)=dU(2);

18 en d19 f i g u r e20 q u i v e r ( u , v , d u , d v ,r )

21 t i t l e( p ha s e p o rt r ai t i n t he n o nl i ne a r c as e ( v e ct o rs ) )22 x l a b e l ( u c o m p o ne n t )23 y l a b e l ( v c o m p o ne n t )

24

25 u1 =l i n s p a c e(0.1,3,5);26 v1 =l i n s p a c e(0.1,3,5);27 [u,v]= me s hgrid (u1,v1);

28 f i g u r e

29 f o r i=1:numel(u)30 [~,Yout]= ode45 ( @ f1 , [ 0 1 0 0] , [ u ( i ) ; v ( i ) ] ) ;31 uout=Yout(:,1);

32 vout=Yout(:,2);

33 p l o t ( u o u t , v o u t )34 hold on

35 en d36 t i t l e( p ha s e p o rt r ai t i n t he n o nl i ne a r c as e ( c u rv e s ) )

37 x l a b e l ( u c o o r d in a t e )38 y l a b e l ( v c o o r d in a t e )39

40 % % C a se 1 l in ea r


44 du = z e r o s (s i z e( u ) ) ;

45 dv = z e r o s (s i z e( v ) ) ;46 f o r i=1:numel(u)47 dU=f1lin(0,[u(i),v(i)]);

48 du(i)=dU(1);

49 dv(i)=dU(2);


53 t i t l e( p h a se p o r t ra i t w i th l i n e ar a p p r o xi m a t i o n ( v e c t o r s ) )54 x l a b e l ( u c o m p o ne n t )55 y l a b e l ( v c o m p o ne n t )

56

57 u1 =l i n s p a c e (0.1-3/4,3-3/4,5);58 v1 =l i n s p a c e (0.1-1/4,3-1/4,5);59 [u,v]= me s hgrid (u1,v1);

60 f i g u r e61 f o r i=1:numel(u)62 [~,Yout]= ode45 (@f1lin ,[0 100],[u(i);v(i) ]);63 uout=3/4+Yout(:,1);

64 vout=1/4+Yout(:,2);

65 p l o t ( u o u t , v o u t )66 hold on67 en d

68 t i t l e( p h a se p o r t ra i t w i th l i n e ar a p p r o xi m a t i o n ( c u r v e s ) )

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69 x l a b e l ( u c o o r d in a t e )

70 y l a b e l ( v c o o r d in a t e )71

72 % % C a s e 2 n o nl i ne a r

73 u1 =l i n s p a c e(0.1,3,20);

74 v1 =l i n s p a c e(0.1,3,20);75 [u,v]= me s hgrid (u1,v1);76 du = z e r o s (s i z e( u ) ) ;

77 dv = z e r o s (s i z e( v ) ) ;78 f o r i=1:numel(u)79 dU=f2(0,[u(i),v(i)]);

80 du(i)=dU(1);

81 dv(i)=dU(2);


85 t i t l e( p ha s e p o rt r ai t i n t he n o nl i ne a r c as e ( v e ct o rs ) )

86 x l a b e l ( u c o m p o ne n t )87 y l a b e l ( v c o m p o ne n t )88

89 u1 =l i n s p a c e(0.1,3,5);

90 v1 =l i n s p a c e(0.1,3,5);91 [u,v]= me s hgrid (u1,v1);92 f i g u r e93 f o r i=1:numel(u)94 [~,Yout]= ode45 ( @ f2 , [ 0 1 0 0] , [ u ( i ) ; v ( i ) ] ) ;95 uout=Yout(:,1);

96 vout=Yout(:,2);

97 p l o t ( u o u t , v o u t )98 hold on99 en d100

101 t i t l e( p ha s e p o rt r ai t i n t he n o nl i ne a r c as e ( c u rv e s ) )102 x l a b e l ( u c o o r d in a t e )103 y l a b e l ( v c o o r d in a t e )104

105 % % C a se 2 l in ea r


109 du = z e r o s (s i z e( u ) ) ;110 dv = z e r o s (s i z e( v ) ) ;111 f o r i=1:numel(u)112 dU=f2lin(0,[u(i),v(i)]);

113 du(i)=dU(1);

114 dv(i)=dU(2);

115 en d116 f i g u r e

117 q u i v e r ( u , v , d u , d v ,r )118 t i t l e( p h a se p o r t ra i t w i th l i n e ar a p p r o xi m a t i o n ( v e c t o r s ) )119 x l a b e l ( u c o m p o ne n t )120 y l a b e l ( v c o m p o ne n t )

121

122 u1 =l i n s p a c e (1-17/18,3.9-17/18,5);123 v1 = l i n s p a c e (0.1-1/18,3-1/18,5);124 [u,v]= me s hgrid (u1,v1);

125 f i g u r e126 f o r i=1:numel(u)127 [~,Yout]= ode45 (@f2lin ,[0 100],[u(i);v(i) ]);128 uout=17/18+Yout(:,1);

129 vout=1/18+Yout(:,2);

130 p l o t ( u o u t , v o u t )131 hold on132 en d

133 t i t l e( p h a se p o r t ra i t w i th l i n e ar a p p r o xi m a t i o n ( c u r v e s ) )134 x l a b e l ( u c o o r d in a t e )135 y l a b e l ( v c o o r d in a t e )136 en d137 f u n c t i o n F = f 1 ( t , u )138 F= z e r o s(2,1);139 F(1)=(1-2*u(1)-u(2))*u(2);

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Phase portraits

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Transcript of Phase portraits