(a)

System is given as,

P˙  = P[(aR−S)−(a−1)(PR+RS+PS)]……………………………………… (1)

R˙ = R[(aS−P)−(a−1)(PR+RS+PS)]……………………………………… (2)

S˙   = S[(aP−R)−(a−1)(PR+RS+PS)]……………………………………… (3)

While E1 and E2 are,

E1(P,R,S) = P + R + SE2(P,R,S) = PRS

Differentiating E_1,

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[(aR−S)−(a−1)(PR+RS+PS)]+S[(aP−R)−(a−1)(PR+RS+PS)]+R[(aS−P)−(a−1)(PR+RS+PS)]}                     ={P[(aR−S)]+S[(aP−R)]+R[(aS−P)]−(a−1)(PR+RS+PS)(P+R+S)}

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

E˙1(P,R,S) ={P[(aR−S)]+S[(aP−R)]+R[(aS−P)]−E1(a−1)(PR+RS+PS)}                      ={[(aPR−SP)]+[(aSP−SR)]+[(aSR−PR)]−E1(a−1)(PR+RS+PS)}                      ={[(aPR+aSR+aSP−SP−PR−SR)]+[−E1(a−1)(PR+RS+PS)]}                     ={[(a(PR+SR+SP)−(SP+PR+SR))]+[−E1(a−1)(PR+RS+PS)]}

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

∴E˙1(P,R,S)  = (1−E1)(a−1)(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,

(1−E1)(a−1)(PR+SR+SP)=0                                           (1−E1)=0                                              (a−1)=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 1−P−R for l1 in the above equation,

Z1= 1−P−R……………………………… (7)

Differentiating,

Z˙1= 0−P˙−R˙     =−P˙−R˙

Substituting for P˙ and R˙,

Z˙1 =−{P[(aR−S)−(a−1)(PR+RS+PS)]+R[(aS−P)−(a−1)(PR+RS+PS)]}

To calculate fixed point, substitute Z˙1= 0

∴ Z˙ =−{P[(aR−S)−(a−1)(PR+RS+PS)]+R[(aS−P)−(a−1)(PR+RS+PS)]}∴  0=    P[(aR−S)−(a−1)(PR+RS+PS)]+R[(aS−P)−(a−1)(PR+RS+PS)] 

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

∴P[(aR−0)−(a−1)(PR+R(0)+P(0))]+R[(a(0)−P)−(a−1)(PR+R(0)+P(0))] =0∴P[aR−(a−1)(PR)]+R[(−P)−(a−1)(PR)]=0∴P R(a−(a−1)P−1−(a−1)R)=0∴P R(a−(a−1)P−1−(a−1)R)=0∴P R((a−1)−(a−1)P−(a−1)R)=0∴P R(a−1)(1−P−R)=0

Substituting 1−P−R= Z1 from (7),

∴P R(a−1)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[(aR−S)−(a−1)(PR+RS+PS)])+PS(R[(aS−P)−(a−1)(PR+RS+PS)])+PR(S[(aP−R)−(a−1)(PR+RS+PS)])}       =(a−1)(PR+RS+PS){RS(P[(aR−S)−1])+PS(R[(aS−P)−1])+PR(S[(aP−R)−1])}      =(a−1)(PR+RS+PS){RS([(aPR−PS)−P])+PS([(aSR−PR)−R])+PR([(aPS−RS)−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(a−1)(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(a−1)(PR+RS+PS)]       =E2(a−1)(1−3(PR+RS+PS))       =E2(a−1)((P+R+S)2−3PR−3RS−3PS)       =E2(a−1)(P2+2PR+2PS+R2+2RS+S2−3PR−3RS−3PS)

E˙2   =E2(a−1)(P2+2PR+2PS+R2+2RS+S2−3PR−3RS−3PS)         =E2(a−1)2(P2−2PR+R2+P2−2PS+S2+R2−2RS+S2)        =E2(a−1)2[(P−R)2+(P−S)2+(R−S)2]∴E˙2 =E2(a−1)2[(P−R)2+(P−S)2+(R−S)2]

Hence, it is proved.

(e)

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

∴E2(a−1)2[(P−R)2+(P−S)2+(R−S)2]=0∴E2(a−1)2=0 and (P−R)2+(P−S)2+(R−S)2=0

(P−R)2+(P−S)2+(R−S)2=0 Is possible only if P*= R*= S*

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

∴P[(aR−S)−(a−1)(PR+RS+PS)] =0∴(aR−S)=(a−1)(PR+RS+PS)

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

(aR−S)=(a−1)(PR+RS+PS)(aP*−P*)=(a−1)((P*)2+(P*)2+(P*)2) P*(a−1)=(a−1)(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).

(aR−S)=(a−1)(PR+RS+PS)

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

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