Chapter 7.3, Problem 12E

Interpretation Introduction

Interpretation:

In special case of rock-paper–scissors cycle for $\text{a = 1}$, it was shown that there existtwo conserved quantities, ${\text{E}}_{\text{1}}$ and ${\text{E}}_{\text{2}}$, to show that ${\text{E}}_{\text{1}}$ is no longerconserved everywhere, but it isconserved if attention is restricted if to the set where ${\text{E}}_{\text{1}}\text{= 1}$. This set is defined by allordered triples of real numbers (P, R, S) such that $\text{P+R+S=1}$, 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 $\text{P,R, S}\ge 0$ and $\text{P+R+S=1}$, 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, ${\dot{\text{E}}}_{\text{2}}\text{= }\frac{{\text{(a-1)E}}_{\text{2}}}{\text{2}}{\text{[(P-R)}}^{\text{2}}{\text{+(R-S)}}^{\text{2}}{\text{+(S-P)}}^{\text{2}}\text{]}$

To show that ${\text{E}}_{\text{2}}$ vanishes at the boundary of T and at the interior fixed point $\text{(P*,R*,S*) = }\frac{1}{3}(1,1,1)$ using the previous results.

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

To show that for $\text{a<1}$, the heteroclinic cycle attractsall trajectories that start in the interior of T.

Concept Introduction:

Fixed point of a differential equation is a point where, $\text{f(x*) = 0 ; while substitution f(x*) = }\dot{\text{x}}\text{ is used and x\&*\#x00A0;is a fixed point}$

Nullclines are the curves where either $\dot{\text{x}}=0\text{ or }\dot{\text{y}}\text{ = 0}$ 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.

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 ${\text{E}}_{\text{1}}\text{= 1}$.

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.

${\dot{\text{E}}}_{\text{2}}\text{= }\frac{{\text{(a-1)E}}_{\text{2}}}{\text{2}}{\text{[(P-R)}}^{\text{2}}{\text{+(R-S)}}^{\text{2}}{\text{+(S-P)}}^{\text{2}}\text{]}$ Is proved.

It is shown that ${\text{E}}_{\text{2}}$ vanishes at the boundary of T and at the interior fixed point $\text{(P*,R*,S*) = }\frac{1}{3}(1,1,1)$.

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

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

Explanation of Solution

(a)

System is given as,

$\dot{\text{P}}\text{=}\text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]$……………………………………… (1)

$\dot{\text{R}}\text{=}\text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]$……………………………………… (2)

$\dot{\text{S}}\text{=}\text{S}\left[\left(\text{aP}-\text{R}\right)-\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\right]$……………………………………… (3)

While ${\text{E}}_{\text{1}}$ and ${\text{E}}_{\text{2}}$ are,

$\begin{array}{l}{\text{E}}_{\text{1}}\text{(P,R,S)}\text{=}\text{P}\text{+}\text{R}\text{+}\text{S}\\ {\text{E}}_{\text{2}}\text{(P,R,S)}\text{=}\text{PRS}\end{array}$

Differentiating E_1,

${\dot{\text{E}}}_{\text{1}}\text{(P,R,S)}\text{=}\dot{\text{P}}\text{+}\dot{\text{R}}\text{+}\dot{\text{S}}$

Now, Substituting $\left(\text{1}\right)\text{, }\left(\text{2}\right)\text{, }\left(\text{3}\right)$ in above expression, following can be obtained, $\begin{array}{l}{\dot{\text{E}}}_{\text{1}}\text{(P,R,S)}\text{=}\left\{\begin{array}{l}\text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]+\\ \text{S}\left[\left(\text{aP}-\text{R}\right)-\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\right]+\\ \text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\end{array}\right\}\\ \\ =\left\{\begin{array}{l}\text{P}\left[\left(\text{aR}-\text{S}\right)\right]+\text{S}\left[\left(\text{aP}-\text{R}\right)\right]+\\ \text{R}\left[\left(\text{aS}-\text{P}\right)\right]-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\left(\text{P+R+S}\right)\end{array}\right\}\end{array}$

Substituting $\text{P+R+S}$ for ${\text{E}}_{\text{1}}$ in the above expression,

$\begin{array}{l}{\dot{\text{E}}}_{\text{1}}\text{(P,R,S)}\text{=}\left\{\begin{array}{l}\text{P}\left[\left(\text{aR}-\text{S}\right)\right]+\text{S}\left[\left(\text{aP}-\text{R}\right)\right]+\\ \text{R}\left[\left(\text{aS}-\text{P}\right)\right]-{\text{E}}_{\text{1}}\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\end{array}\right\}\\ \\ \text{=}\left\{\begin{array}{l}\left[\left(\text{aPR}-\text{SP}\right)\right]+\left[\left(\text{aSP}-\text{SR}\right)\right]+\\ \left[\left(\text{aSR}-\text{PR}\right)\right]-{\text{E}}_{\text{1}}\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\end{array}\right\}\\ \\ \text{=}\left\{\begin{array}{l}\left[\left(\text{aPR}+\text{aSR}+\text{aSP}-\text{SP}-\text{PR}-\text{SR}\right)\right]+\\ \left[-{\text{E}}_{\text{1}}\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\end{array}\right\}\\ \\ =\left\{\begin{array}{l}\left[\left(\text{a}\left(\text{PR+SR+SP}\right)-\left(\text{SP}+\text{PR}+\text{SR}\right)\right)\right]+\\ \left[-{\text{E}}_{\text{1}}\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\end{array}\right\}\end{array}$

$\therefore {\dot{\text{E}}}_{\text{1}}\text{(P,R,S)}\text{=}\left[\left(\text{PR+SR+SP}\right)\left(\text{a}-\text{1}\right)\right]+\left[-{\text{E}}_{\text{1}}\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]$

$\therefore {\dot{\text{E}}}_{\text{1}}\text{(P,R,S)}=\left(1-{\text{E}}_{\text{1}}\right)\left(\text{a}-\text{1}\right)\left(\text{PR+SR+SP}\right)$.

Above expressions tells that ${\text{E}}_{\text{1}}$ is no longerconserved everywhere, but it isconserved if attention is restricted to the set where ${\text{E}}_{\text{1}}\text{= 1}$.

(b)

To find the subset of the system equation, set

${\text{E}}_{\text{1}}=1$

Differentiating it,

${\dot{\text{E}}}_{\text{1}}=0$

Substituting value of ${\dot{\text{E}}}_{\text{1}}$ from above expression,

$\begin{array}{l}\left(1-{\text{E}}_{\text{1}}\right)\left(\text{a}-\text{1}\right)\left(\text{PR+SR+SP}\right)=0\\ \left(1-{\text{E}}_{\text{1}}\right)=0\\ \left(\text{a}-\text{1}\right)=0\\ \left(\text{PR+SR+SP}\right)=0\end{array}$

It is only possible if $\text{P,R,S}\text{>}\text{0}$ and dynamics can be restricted to the plane $\text{P+R+S}\text{=1}$. 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,

${\text{l}}_1\text{:}\text{P}\text{+}\text{R}=1$……………………………… (4)

${\text{l}}_2\text{:}\text{P}\text{+ S}=1$……………………..……….. (5)

${\text{l}}_3\text{:}\text{R}\text{+}\text{S}=1$…………………………….… (6)

Let the function ${\text{Z}}_{\text{1}}$ equal to the ${\text{l}}_{\text{1}}$,

${\text{Z}}_{\text{1}}\text{=}{\text{l}}_{\text{1}}$

Substituting $\text{1}-\text{P}-\text{R}$ for ${\text{l}}_{\text{1}}$ in the above equation,

${\text{Z}}_{\text{1}}\text{=}\text{1}-\text{P}-\text{R}$……………………………… (7)

Differentiating,

$\begin{array}{l}{\dot{\text{Z}}}_{\text{1}}\text{=}0-\dot{\text{P}}-\dot{\text{R}}\\ \text{=}-\dot{\text{P}}-\dot{\text{R}}\end{array}$

Substituting for $\dot{\text{P}}$ and $\dot{\text{R}}$,

${\dot{\text{Z}}}_{\text{1}}\text{=}-\left\{\text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]+\text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\right\}$

To calculate fixed point, substitute ${\dot{\text{Z}}}_{\text{1}}=0$

$\begin{array}{l}\therefore \dot{\text{Z}}\text{ =}-\left\{\text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]+\text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\right\}\\ \therefore 0=\text{ P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]+\text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\end{array}$

As there is no S term in $\left(\text{4}\right)\text{,}$ we substitute $\text{S = 0}$ in above equation.

$\begin{array}{l}\therefore \text{P}\left[\left(\text{aR}-\text{0}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+R(0)+P(0)}\right)\right]+\text{R}\left[\left(\text{a(0)}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+R(0)+P(0)}\right)\right]=0\\ \therefore \text{P}\left[\text{aR}-\left(\text{a}-\text{1}\right)\left(\text{PR}\right)\right]+\text{R}\left[\left(-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR}\right)\right]=0\\ \therefore \text{P}\text{R}\left(\text{a}-\left(\text{a}-\text{1}\right)\text{P}-1-\left(\text{a}-\text{1}\right)\text{R}\right)=0\\ \therefore \text{P}\text{R}\left(\text{a}-\left(\text{a}-\text{1}\right)\text{P}-1-\left(\text{a}-\text{1}\right)\text{R}\right)=0\\ \therefore \text{P}\text{R}\left(\left(\text{a}-\text{1}\right)-\left(\text{a}-\text{1}\right)\text{P}-\left(\text{a}-\text{1}\right)\text{R}\right)=0\\ \therefore \text{P}\text{R}\left(\text{a}-\text{1}\right)\left(1-\text{P}-\text{R}\right)=0\end{array}$

Substituting $1-\text{P}-\text{R}=$ ${\text{Z}}_{\text{1}}$ from $\left(\text{7}\right)\text{,}$

$\therefore \text{P}\text{R}\left(\text{a}-\text{1}\right){\text{Z}}_{\text{1}}=0{\text{ = }\dot{\text{Z}}}_1$

Therefore, the system has one fixed point for the lineofequation ${\text{l}}_{\text{1}}$, similarly for ${\text{l}}_{\text{2}}$ and ${\text{l}}_{\text{3}}$ there are two fixed point exists. Therefore, there exist three fixed point the given system of equations.

(d)

To prove the relation ${\dot{\text{E}}}_{\text{2}}\text{= }\frac{{\text{(a-1)E}}_{\text{2}}}{\text{2}}{\text{[(P-R)}}^{\text{2}}{\text{+(R-S)}}^{\text{2}}{\text{+(S-P)}}^{\text{2}}\text{]}$

It is given that,

${\text{E}}_{\text{2}}\text{(P,R,S)}\text{=}\text{PRS}$

Differentiating

${\dot{\text{E}}}_{\text{2}}\text{=}\dot{\text{P}}\text{RS}\text{+}\text{P}\dot{\text{R}}\text{S}\text{+}\text{PR}\dot{\text{S}}$

Substituting $\left(\text{1}\right)\text{, }\left(\text{2}\right)\text{ and }\left(\text{3}\right)$ in above equation,

$\begin{array}{l}{\dot{\text{E}}}_{\text{2}}=\left\{\begin{array}{l}\text{RS}\left(\text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\right)+\\ \text{PS}\left(\text{R}\left[\left(\text{aS}-\text{P}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]\right)+\\ \text{PR}\left(\text{S}\left[\left(\text{aP}-\text{R}\right)-\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\right]\right)\end{array}\right\}\\ =\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\left\{\begin{array}{l}\text{RS}\left(\text{P}\left[\left(\text{aR}-\text{S}\right)-1\right]\right)+\\ \text{PS}\left(\text{R}\left[\left(\text{aS}-\text{P}\right)-1\right]\right)+\\ \text{PR}\left(\text{S}\left[\left(\text{aP}-\text{R}\right)-1\right]\right)\end{array}\right\}\\ =\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\left\{\begin{array}{l}\text{RS}\left(\left[\left(\text{aPR}-\text{PS}\right)-\text{P}\right]\right)+\\ \text{PS}\left(\left[\left(\text{aSR}-\text{PR}\right)-\text{R}\right]\right)+\\ \text{PR}\left(\left[\left(\text{aPS}-\text{RS}\right)-\text{S}\right]\right)\end{array}\right\}\end{array}$

Again substituting $\text{PRS}$ for ${\text{E}}_{\text{2}}\text{(P,R,S)}$ in the above expression of the ${\dot{\text{E}}}_{\text{2}}$,

${\dot{\text{E}}}_{\text{2}}={\text{E}}_{\text{2}}\left[\text{a}\left(\text{R+S+P}\right)-\left(\text{S+P+R}\right)-3\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\right]$

Therefore, for the boundary of $\text{T}$ the above equation is written in the form of

$\begin{array}{l}{\dot{\text{E}}}_{\text{2}}={\text{E}}_{\text{2}}\left[\text{a}\left(\text{R+S+P}\right)-\left(\text{S+P+R}\right)-3\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)\right]\\ ={\text{E}}_{\text{2}}\left(\text{a}-1\right)\left(1-3\left(\text{PR+RS+PS}\right)\right)\\ ={\text{E}}_{\text{2}}\left(\text{a}-1\right)\left({\left(\text{P+R+S}\right)}^2-3\text{PR}-3\text{RS}-3\text{PS}\right)\\ ={\text{E}}_{\text{2}}\left(\text{a}-1\right)\left({\text{P}}^2+2\text{PR}+2\text{PS}+{\text{R}}^2+2\text{RS}+{\text{S}}^2-3\text{PR}-3\text{RS}-3\text{PS}\right)\end{array}$

$\begin{array}{l}{\dot{\text{E}}}_{\text{2}}={\text{E}}_{\text{2}}\left(\text{a}-1\right)\left({\text{P}}^2+2\text{PR}+2\text{PS}+{\text{R}}^2+2\text{RS}+{\text{S}}^2-3\text{PR}-3\text{RS}-3\text{PS}\right)\\ =\frac{{\text{E}}_{\text{2}}\left(\text{a}-1\right)}{2}\left({\text{P}}^2-2{\text{PR+R}}^2+{\text{P}}^2-2\text{PS}+{\text{S}}^2+{\text{R}}^2-2\text{RS}+{\text{S}}^2\right)\\ =\frac{{\text{E}}_{\text{2}}\left(\text{a}-1\right)}{2}\left[{\left(\text{P}-\text{R}\right)}^2+{\left(\text{P}-\text{S}\right)}^2+{\left(\text{R}-\text{S}\right)}^2\right]\\ \therefore {\dot{\text{E}}}_{\text{2}}=\frac{{\text{E}}_{\text{2}}\left(\text{a}-1\right)}{2}\left[{\left(\text{P}-\text{R}\right)}^2+{\left(\text{P}-\text{S}\right)}^2+{\left(\text{R}-\text{S}\right)}^2\right]\end{array}$

Hence, it is proved.

(e)

To calculate the fixed point, ${\dot{\text{E}}}_{\text{2}}=0$ can be substituted.

$\begin{array}{l}\therefore \frac{{\text{E}}_{\text{2}}\left(\text{a}-1\right)}{2}\left[{\left(\text{P}-\text{R}\right)}^2+{\left(\text{P}-\text{S}\right)}^2+{\left(\text{R}-\text{S}\right)}^2\right]=0\\ \therefore \frac{{\text{E}}_{\text{2}}\left(\text{a}-1\right)}{2}=0\text{ and }{\left(\text{P}-\text{R}\right)}^2+{\left(\text{P}-\text{S}\right)}^2+{\left(\text{R}-\text{S}\right)}^2=0\end{array}$

${\left(\text{P}-\text{R}\right)}^2+{\left(\text{P}-\text{S}\right)}^2+{\left(\text{R}-\text{S}\right)}^2=0$ Is possible only if ${\text{P}}^{\text{*}}\text{=}{\text{R}}^{\text{*}}\text{=}{\text{S}}^{\text{*}}$

Now, consider $\dot{\text{P}}$ for fixed point, $\dot{\text{P}}=0$

$\begin{array}{l}\therefore \text{P}\left[\left(\text{aR}-\text{S}\right)-\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\right]=0\\ \therefore \left(\text{aR}-\text{S}\right)=\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\end{array}$

Substitute ${\text{P}}^{\text{*}}$ for $\text{P,R}$ and $\text{S}$ in the above equation

$\begin{array}{l}\left(\text{aR}-\text{S}\right)=\left(\text{a}-\text{1}\right)\left(\text{PR+RS+PS}\right)\\ \left({\text{aP}}^{*}-{\text{P}}^{*}\right)=\left(\text{a}-\text{1}\right)\left({\left({\text{P}}^{*}\right)}^2+{\left({\text{P}}^{*}\right)}^2+{\left({\text{P}}^{*}\right)}^2\right)\\ {\text{ P}}^{*}\left(\text{a}-\text{1}\right)=\left(\text{a}-\text{1}\right){\left({\text{P}}^{*}\right)}^2(1+1+1)\\ {\text{ P}}^{*}=\frac{1}{3}\end{array}$

Therefore, from the above calculation, it is clear that the function ${\text{E}}_{\text{2}}$ is maximized if the point

${\text{P}}^{\text{*}}{\text{,R}}^{\text{*}},{\text{S}}^{\text{*}}=\frac{1}{3}(1,1,1)$

(f)

From (e), the fixed point for the above system of equations for $\text{a < 1}$ is $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$, all the solution of the above system of equations go through the point $\left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$

Therefore, for that fixed point the interior fixed point attracts all the trajectories on the interior of $\text{T}$.

(g)

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

$\left(\text{aR}-\text{S}\right)=\left(\text{a}-1\right)\left(\text{PR+RS+PS}\right)$

In above expression, if $\text{a < 1}$ than the solution of the above system of equation goes through the boundary of ${\text{T}}_{\text{1}}$.

Conclusion

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