Nonlinear Dynamics and Chaos
Nonlinear Dynamics and Chaos
2nd Edition
ISBN: 9780813349107
Author: Steven H. Strogatz
Publisher: PERSEUS D
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Chapter 7.3, Problem 12E
Interpretation Introduction

Interpretation:

In special case of rock-paper–scissors cycle for a = 1, it was shown that there existtwo conserved quantities, E1 and E2, to show that E1 is no longerconserved everywhere, but it isconserved if attention is restricted if to the set where E1= 1. This set is defined by allordered triples of real numbers (P, R, S) such that P+R+S=1, is invariant; any trajectory that starts on it stays on it forever. Also to describe this set geometrically and to tell it’s shape.

Byconsideringa subset of the set in (a), defined by P,R, S 0 and P+R+S=1, to show that thissubsetis also invariant and also to tell its shape. Call this subset T

To show that the boundary of Tconsists of three fixed points connected by threetrajectories, all oriented in the same directionand hence, it is a heteroclinic cycle. (Cyclic graph.)

To show that, E˙2(a-1)E22[(P-R)2+(R-S)2+(S-P)2]

To show that E2 vanishes at the boundary of T and at the interior fixed point (P*,R*,S*) = 13(1,1,1) using the previous results.

To explain why the previous results mean that, for a>1, theinterior fixed point attracts all trajectories interior to the T.

To show that for a<1, the heteroclinic cycle attractsall trajectories that start in the interior of T.

Concept Introduction:

Fixed point of a differential equation is a point where, f(x*) = 0 ; while substitution f(x*) = x˙ is used and x&*#x00A0;is a fixed point

Nullclines are the curves where either x˙=0 or y˙ = 0 they show whether the flow is completely vertical or horizontal.

Vector fields in this aspect represent the direction of flow and whether flow is going away from fixed point or coming towards it.

Phase portraits represent the trajectories of the system with respect to the parameters and give qualitative idea about evolution of the system, its fixed points, whether they will attract or repel the flow etc.

Expert Solution & Answer
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Answer to Problem 12E

Solution:

It is shown that E1 is no longerconserved everywhere, but it isconserved if attention is restricted if to the set where E1= 1.

It is shown that Tis also invariant and its shape is triangle.

It is shown that the boundary of Tconsists of three fixed points connected by threetrajectories, all oriented in the same direction, hence it is proved that it is a heteroclinic cycle.

E˙2(a-1)E22[(P-R)2+(R-S)2+(S-P)2] Is proved.

It is shown that E2 vanishes at the boundary of T and at the interior fixed point (P*,R*,S*) = 13(1,1,1).

It is explained why interior fixed point attracts all trajectories interior to the T.

It is shown that for a<1, the heteroclinic cycle attracts all trajectories that start in the interior of T.

Explanation of Solution

(a)

System is given as,

P˙=P[(aRS)(a1)(PR+RS+PS)]……………………………………… (1)

R˙=R[(aSP)(a1)(PR+RS+PS)]……………………………………… (2)

S˙=S[(aPR)(a1)(PR+RS+PS)]……………………………………… (3)

While E1 and E2 are,

E1(P,R,S)=P+R+SE2(P,R,S)=PRS

Differentiating E1,

E˙1(P,R,S)=P˙+R˙+S˙

Now, Substituting (1)(2)(3) in above expression, following can be obtained, E˙1(P,R,S)={P[(aRS)(a1)(PR+RS+PS)]+S[(aPR)(a1)(PR+RS+PS)]+R[(aSP)(a1)(PR+RS+PS)]}={P[(aRS)]+S[(aPR)]+R[(aSP)](a1)(PR+RS+PS)(P+R+S)}

Substituting P+R+S for E1 in the above expression,

E˙1(P,R,S)={P[(aRS)]+S[(aPR)]+R[(aSP)]E1(a1)(PR+RS+PS)}={[(aPRSP)]+[(aSPSR)]+[(aSRPR)]E1(a1)(PR+RS+PS)}={[(aPR+aSR+aSPSPPRSR)]+[E1(a1)(PR+RS+PS)]}={[(a(PR+SR+SP)(SP+PR+SR))]+[E1(a1)(PR+RS+PS)]}

E˙1(P,R,S)=[(PR+SR+SP)(a1)]+[E1(a1)(PR+RS+PS)]

E˙1(P,R,S) = (1E1)(a1)(PR+SR+SP).

Above expressions tells that E1 is no longerconserved everywhere, but it isconserved if attention is restricted to the set where E1= 1.

(b)

To find the subset of the system equation, set

E1=1

Differentiating it,

E˙1=0

Substituting value of E˙1 from above expression,

(1E1)(a1)(PR+SR+SP)=0(1E1)=0(a1)=0(PR+SR+SP)=0

It is only possible if P,R,S>0 and dynamics can be restricted to the plane P+R+S=1. Hence this subset is also invariant. As the plane cuts all thee axes, the shape of T is a triangle.

(c)

From (b), considered the lines connecting three fixed points in the system of equations are,

l1:P+R=1……………………………… (4)

l2:P+ S=1……………………..……….. (5)

l3:R+S=1…………………………….… (6)

Let the function Z1 equal to the l1,

Z1=l1

Substituting 1PR for l1 in the above equation,

Z1=1PR……………………………… (7)

Differentiating,

Z˙1=0P˙R˙=P˙R˙

Substituting for P˙ and R˙,

Z˙1={P[(aRS)(a1)(PR+RS+PS)]+R[(aSP)(a1)(PR+RS+PS)]}

To calculate fixed point, substitute Z˙1= 0

 Z˙ ={P[(aRS)(a1)(PR+RS+PS)]+R[(aSP)(a1)(PR+RS+PS)]}  0=    P[(aRS)(a1)(PR+RS+PS)]+R[(aSP)(a1)(PR+RS+PS)]

As there is no S term in (4), we substitute S = 0 in above equation.

P[(aR0)(a1)(PR+R(0)+P(0))]+R[(a(0)P)(a1)(PR+R(0)+P(0))]=0P[aR(a1)(PR)]+R[(P)(a1)(PR)]=0PR(a(a1)P1(a1)R)=0PR(a(a1)P1(a1)R)=0PR((a1)(a1)P(a1)R)=0PR(a1)(1PR)=0

Substituting 1PR= Z1 from (7),

PR(a1)Z1=0 = Z˙1

Therefore, the system has one fixed point for the lineofequation l1, similarly for l2 and l3 there are two fixed point exists. Therefore, there exist three fixed point the given system of equations.

(d)

To prove the relation E˙2(a-1)E22[(P-R)2+(R-S)2+(S-P)2]

It is given that,

E2(P,R,S)=PRS

Differentiating

E˙2=P˙RS+PR˙S+PRS˙

Substituting (1)(2) and (3) in above equation,

E˙2={RS(P[(aRS)(a1)(PR+RS+PS)])+PS(R[(aSP)(a1)(PR+RS+PS)])+PR(S[(aPR)(a1)(PR+RS+PS)])}=(a1)(PR+RS+PS){RS(P[(aRS)1])+PS(R[(aSP)1])+PR(S[(aPR)1])}=(a1)(PR+RS+PS){RS([(aPRPS)P])+PS([(aSRPR)R])+PR([(aPSRS)S])}

Again substituting PRS for E2(P,R,S) in the above expression of the E˙2,

E˙2=E2[a(R+S+P)(S+P+R)3(a1)(PR+RS+PS)]

Therefore, for the boundary of T the above equation is written in the form of

E˙2=E2[a(R+S+P)(S+P+R)3(a1)(PR+RS+PS)]=E2(a1)(13(PR+RS+PS))=E2(a1)((P+R+S)23PR3RS3PS)=E2(a1)(P2+2PR+2PS+R2+2RS+S23PR3RS3PS)

E˙2 =E2(a1)(P2+2PR+2PS+R2+2RS+S23PR3RS3PS)  =E2(a1)2(P22PR+R2+P22PS+S2+R22RS+S2) =E2(a1)2[(PR)2+(PS)2+(RS)2]E˙2=E2(a1)2[(PR)2+(PS)2+(RS)2]

Hence, it is proved.

(e)

To calculate the fixed point, E˙2=0 can be substituted.

E2(a1)2[(PR)2+(PS)2+(RS)2]=0E2(a1)2=0 and (PR)2+(PS)2+(RS)2=0

(PR)2+(PS)2+(RS)2=0 Is possible only if P*=R*=S*

Now, consider P˙ for fixed point, P˙= 0

P[(aRS)(a1)(PR+RS+PS)]=0(aRS)=(a1)(PR+RS+PS)

Substitute P* for P,R and S in the above equation

(aRS)=(a1)(PR+RS+PS)(aP*P*)=(a1)((P*)2+(P*)2+(P*)2) P*(a1)=(a1)(P*)2(1+1+1) P*=13

Therefore, from the above calculation, it is clear that the function E2 is maximized if the point

P*,R*,S*=13(1,1,1)

(f)

From (e), the fixed point for the above system of equations for a < 1 is (13,13,13), all the solution of the above system of equations go through the point (13,13,13)

Therefore, for that fixed point the interior fixed point attracts all the trajectories on the interior of T.

(g)

For the nature of the trajectories for the above system of equationsat a < 1, consider the expression of the fixed point from (e).

(aRS)=(a1)(PR+RS+PS)

In above expression, if a < 1 than the solution of the above system of equation goes through the boundary of T1.

Conclusion

Fixed points are found and phase portraits, and vector fields are drawn.

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