Chapter 8.1, Problem 12E

Interpretation Introduction

Interpretation:

For a bar magnet system ${\dot{\text{θ}}}_{\text{1}}{\text{ = K sin (θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}{\text{) - sin θ}}_{\text{1}}$ and ${\dot{\text{θ}}}_{\text{2}}{\text{ = K sin (θ}}_{\text{2}}{\text{ - θ}}_{\text{1}}{\text{) - sin θ}}_{\text{2}}$, determine and classify all the fixed points of the system. Show that a bifurcation occurs at $\text{K = }\frac{\text{1}}{\text{2}}$ and determine type ofbifurcation. Show that the system is a “gradient” system, ${\dot{\text{θ}}}_{\text{i}}\text{ = - }\partial \text{V/}\partial {\text{θ}}_{\text{i}}$ determined for some potential function ${\text{V(θ}}_{\text{1}}{\text{, θ}}_{\text{2}}\text{)}$. Prove that the system has no periodic orbits. Sketch the phase portrait for $\text{0 < K < }\frac{\text{1}}{\text{2}}$ and then for $\text{K > }\frac{\text{1}}{\text{2}}$.

Concept Introduction:

Determine the fixed point for the given system.

Determine the bifurcation using the Jacobian matrix and fixed point.

Sketch phase portrait using system equation.

Answer to Problem 12E

Solution:

The fixed point of the system is $\left({\text{θ}}_{\text{1}}{\text{,θ}}_{\text{2}}\right)\text{ = }\left(\frac{\text{π}}{\text{2}}\text{,}\frac{\text{-π}}{\text{2}}\right)$

It is shown that a bifurcation occurs at $\text{K = }\frac{\text{1}}{\text{2}}$, and at the fixed point, the bifurcation is saddle node.

It is shown that the system is a “gradient” system, ${\dot{\text{θ}}}_{\text{i}}\text{ = - }\partial \text{V/}\partial {\text{θ}}_{\text{i}}$.

It is proved that the system has no periodic orbits.

The phase portrait for $\text{0 < K < }\frac{\text{1}}{\text{2}}$ and then for $\text{K > }\frac{\text{1}}{\text{2}}$ is shown.

Explanation of Solution

Consider the Interacting bar magnets system with system equation ${\dot{\text{θ}}}_{\text{1}}{\text{ = K sin (θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}{\text{) - sin θ}}_{\text{1}}$ and ${\dot{\text{θ}}}_{\text{2}}{\text{ = K sin (θ}}_{\text{2}}{\text{ - θ}}_{\text{1}}{\text{) - sin θ}}_{\text{2}}$, where $\text{K }\ge \text{ 0}$. Let ${\text{θ}}_{\text{1}}$, ${\text{θ}}_{\text{2}}$ denote the angular orientations of the north poles of themagnets. Then the term ${\text{θ}}_{\text{2}}-{\text{θ}}_{\text{1}}$ represents a repulsiveforce that tries to keep the two north poles $\text{180}°$ apart. Thisrepulsion is opposed by the $\text{sin θ}$ terms, which model externalmagnets that pull the north poles of both bar magnets to theeast. If the inertia of the magnets is negligible compared toviscous damping, then the equations above are a decentapproximation of the true dynamics.

To calculate the fixed point of the system, equationsare as:

${\dot{\text{θ}}}_{\text{1}}\text{ = Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right){\text{- sinθ}}_{\text{1}}$

${\dot{\text{θ}}}_2\text{ = Ksin}\left({\text{θ}}_2{\text{- θ}}_1\right){\text{- sinθ}}_2$

For the calculation of the fixed point of the above system equation, first equating the first derivative of the above system equation to zero.

Substitute ${\dot{\text{θ}}}_1=0$ in the above system equation,

$\text{Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)\text{= 0}$

$\text{Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{= sin}\left({\text{θ}}_{\text{1}}\right)$

From the trigonometric identity, $\text{sin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{= sin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)$

Hence,

$\text{Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - Ksin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)\text{ = sin}\left({\text{θ}}_{\text{1}}\right)$

Substitute ${\dot{\text{θ}}}_2=0$ system equation,

$\text{Ksin}\left({\text{θ}}_2{\text{- θ}}_1\right)\text{- sin}\left({\text{θ}}_2\right)\text{= 0}$

$\text{Ksin}\left({\text{θ}}_2{\text{- θ}}_1\right)\text{ = sin}\left({\text{θ}}_2\right)$

From the trigonometric identity, $\text{sin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{= sin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)$

$\text{Ksin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)\text{ - Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ = sin}\left({\text{θ}}_{\text{2}}\right)$

To obtain $\text{sin}\left({\text{θ}}_1\right)\text{+ sin}\left({\text{θ}}_2\right)=0$,

$\text{sin}\left({\text{θ}}_{\text{1}}\right){\text{+ sinθ}}_{\text{2}}\text{ = Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - Ksin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right){\text{ + Ksinθ}}_{\text{2}}{\text{cosθ}}_{\text{1}}\text{ - Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ = 0 }$

Using trigonometric identity, $\text{sin}\left(\text{c}\right)\text{ + sin}\left(\text{d}\right)\text{ = 2sin}\left(\frac{\text{c+d}}{\text{2}}\right)\text{cos}\left(\frac{\text{c-d}}{\text{2}}\right)$

$\text{sin}\left({\text{θ}}_{\text{1}}\right){\text{+ sinθ}}_{\text{2}}\text{ = 0 }$

Apply trigonometric expression in the above expression of the fixed point,

$\text{sin}\left({\text{θ}}_{\text{1}}\right){\text{+ sinθ}}_{\text{2}}\text{ = 2sin}\left(\frac{{\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}}{\text{2}}\right)\text{cos}\left(\frac{{\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}}{\text{2}}\right)\text{ = 0}$

From the above, $\text{sin}\left(\frac{{\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}}{\text{2}}\right)\text{= 0}$ or $\text{cos}\left(\frac{{\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}}{\text{2}}\right)\text{= 0}$

By solving,

$\text{sin}\left(\frac{{\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}}{\text{2}}\right)\text{= 0}$

$\left(\frac{{\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}}{\text{2}}\right)\text{= 0}$

${\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}\text{ = 0}$

Similarly,

$\text{cos}\left(\frac{{\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}}{\text{2}}\right)\text{=0}$

$\frac{{\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}}{\text{2}}\text{ = }\frac{\text{π}}{\text{2}}$

${\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}\text{= π}$

Compare the above expression of ${\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}\text{ = 0}$ and ${\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}\text{= π}$,

${\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{2}}{\text{ + θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}\text{= π}$

${\text{θ}}_{\text{1}}{\text{ + θ}}_{\text{1}}\text{ = π}$

${\text{2θ}}_{\text{1}}\text{ = π}$

${\text{θ}}_{\text{1}}\text{ = }\frac{\text{π}}{\text{2}}$

Substitute $\frac{\text{π}}{\text{2}}$ for ${\text{θ}}_{\text{1}}$ in the above expression of ${\text{θ}}_{\text{1}}{\text{ - θ}}_{\text{2}}\text{= π}$,

$\frac{\text{π}}{\text{2}}{\text{ - θ}}_{\text{2}}\text{ = π}$

${\text{θ}}_{\text{2}}\text{ = }\frac{\text{π}}{\text{2}}\text{ - π}$

${\text{θ}}_{\text{2}}\text{ = -}\frac{\text{π}}{\text{2}}$

Therefore, the fixed point of the above system equation is $\left({\text{θ}}_{\text{1}}{\text{,θ}}_{\text{2}}\right)\text{ = }\left(\frac{\text{π}}{\text{2}}\text{,}\frac{\text{-π}}{\text{2}}\right)$.

Bifurcation is defined as the point or area at which something is divided into two parts and branches. The point in the system equation at which bifurcating occur.

For the calculation of the type of bifurcation, the Jacobian matrix is:

$\text{J = }\left(\begin{array}{cc}\frac{{\text{d}\dot{\text{θ}}}_{\text{1}}}{{\text{dθ}}_{\text{1}}}& \frac{{\text{d}\dot{\text{θ}}}_{\text{1}}}{{\text{dθ}}_{\text{2}}}\\ \frac{{\text{d}\dot{\text{θ}}}_{\text{2}}}{{\text{dθ}}_{\text{1}}}& \frac{{\text{d}\dot{\text{θ}}}_{\text{2}}}{{\text{dθ}}_{\text{2}}}\end{array}\right)$

By substituting ${\dot{\text{θ}}}_{\text{1}}\text{ = Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)$ and ${\dot{\text{θ}}}_2\text{ = Ksin}\left({\text{θ}}_2{\text{- θ}}_1\right)\text{- sin}\left({\text{θ}}_2\right)$,

$\text{ J = }\left(\begin{array}{cc}\frac{\text{d}\left(\text{Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)\right)}{{\text{dθ}}_{\text{1}}}& \frac{\text{d}\left(\text{Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)\right)}{{\text{dθ}}_{\text{2}}}\\ \frac{\text{d}\left(\text{Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)\right)}{{\text{dθ}}_{\text{1}}}& \frac{\text{d}\left(\text{Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)\right)}{{\text{dθ}}_{\text{2}}}\end{array}\right)$

$\text{J = }\left(\begin{array}{cc}\text{Kcos}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- cos}\left({\text{θ}}_{\text{1}}\right)& \text{- Kcos}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\\ \text{- Kcos}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right)& \text{Kcos}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right){\text{- cosθ}}_{\text{2}}\end{array}\right)$

The Jacobian matrix at the fixed point $\left({\text{θ}}_{\text{1}}{\text{,θ}}_{\text{2}}\right)\text{ = }\left(\frac{\text{π}}{\text{2}}\text{,}\frac{\text{-π}}{\text{2}}\right)$ is:

${\text{J}}_{\left(\frac{\text{π}}{\text{2}}\text{,}\frac{\text{-π}}{\text{2}}\right)}\text{ = }\left(\begin{array}{cc}\text{Kcos}\left(\frac{\text{π}}{\text{2}}\text{-}\left(\frac{\text{- π}}{\text{2}}\right)\right)\text{- cos}\left(\frac{\text{π}}{\text{2}}\right)& \text{- Kcos}\left(\frac{\text{π}}{\text{2}}\text{ - }\left(\frac{\text{- π}}{\text{2}}\right)\right)\\ \text{- Kcos}\left(\left(\frac{\text{- π}}{\text{2}}\right)\text{- }\frac{\text{π}}{\text{2}}\right)& \text{Kcos}\left(\left(\frac{\text{- π}}{\text{2}}\right)\text{-}\frac{\text{π}}{\text{2}}\right)\text{- cos}\left(\frac{\text{- π}}{\text{2}}\right)\end{array}\right)$

${\text{J}}_{\left(\frac{\text{π}}{\text{2}}\text{,}\frac{\text{-π}}{\text{2}}\right)}\text{=}\left(\begin{array}{cc}\text{-K}& \text{K}\\ \text{K}& \text{-K}\end{array}\right)$

The trace of the above Jacobian matrix is calculated as:

$\text{τ = -2K}$

The determinant of the above Jacobian matrix is calculated as:

${\text{Δ = K}}^{\text{2}}{\text{- K}}^{\text{2}}$

$\text{Δ= 0}$

Therefore, from the above calculation of the system equation, the determinant of the above system equation at the fixed point is zero, so the bifurcation is saddle node bifurcation.

The calculation of the potential gradient of the given systemis shown below.

The expression of the potential function is:

${\dot{\text{θ}}}_{\text{1}}\text{=}\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}$

Substitute ${\dot{\text{θ}}}_{\text{1}}\text{ = Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)$,

$\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}\text{= Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)$

Therefore, the potential function for the value of ${\text{θ}}_{\text{1}}$ is $\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}\text{= Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)$.

Similarly, for

${\dot{\text{θ}}}_2\text{=}\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}$

Substitute ${\dot{\text{θ}}}_2\text{ = Ksin}\left({\text{θ}}_2{\text{- θ}}_1\right)\text{- sin}\left({\text{θ}}_2\right)$,

$\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}=\text{Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)$

Therefore, the potential function for the value of ${\text{θ}}_{\text{2}}$ is $\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}=\text{Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right){\text{ - sinθ}}_{\text{2}}$.

For the calculation of the periodic orbit,

Add the potential function.

From the trigonometric identity, $\text{sin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{= sin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)$

$\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}=\text{Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)+\text{Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)$

$\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}=\text{Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_{\text{2}}\right)\text{ - Ksin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)\text{- sin}\left({\text{θ}}_{\text{1}}\right)+\text{Ksin}\left({\text{θ}}_{\text{2}}\right)\text{cos}\left({\text{θ}}_{\text{1}}\right)\text{ - Ksin}\left({\text{θ}}_{\text{1}}\right)\text{cos}\left({\text{θ}}_2\right)\text{ - sin}\left({\text{θ}}_{\text{2}}\right)$

$\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\text{-}\partial \text{V}}{\partial {\text{θ}}_2}=\text{ - sin}\left({\text{θ}}_{\text{1}}\right)\text{- sin}\left({\text{θ}}_{\text{2}}\right)$

Hence,

$\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}{\text{= sinθ}}_{\text{2}}{\text{+ sinθ}}_{\text{1}}$

For the periodicity of the above system equation, calculate the closed orbit of the above system equation with the help of the Greens theorem,

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}{\displaystyle \underset{\text{0}}{\overset{\frac{\text{-π}}{\text{2}}}{\int }}{\displaystyle \underset{\text{0}}{\overset{\frac{\text{π}}{\text{2}}}{\int }}\left({\text{sinθ}}_{\text{2}}{\text{+ sinθ}}_{\text{1}}\right)}}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}{{\displaystyle \underset{\text{0}}{\overset{\frac{\text{-π}}{\text{2}}}{\int }}\left({\text{sinθ}}_{\text{2}}{\text{. θ}}_{\text{1}}{\text{- cosθ}}_{\text{1}}\right)|}}_{\text{0}}^{\frac{\text{π}}{\text{2}}}{\text{dθ}}_{\text{2}}$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}{\displaystyle \underset{0}{\overset{\frac{\text{-π}}{\text{2}}}{\int }}\left[\left({\text{sinθ}}_{\text{2}}\text{. }\left(\frac{\text{π}}{\text{2}}\right)\text{- cos}\left(\frac{\text{π}}{\text{2}}\right)\right)-\left({\text{sinθ}}_{\text{2}}\text{. }\left(\text{0}\right)\text{- cos}\left(\text{0}\right)\right)\right]}{\text{ dθ}}_{\text{2}}$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}{\displaystyle \underset{0}{\overset{\frac{\text{-π}}{\text{2}}}{\int }}\left[\left({\text{sinθ}}_{\text{2}}\text{. }\left(\frac{\text{π}}{\text{2}}\right)\right)+1\right]}{\text{ dθ}}_{\text{2}}$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}{\left[\left(\text{- cos }\left({\text{θ}}_{\text{2}}\right)\text{. }\left(\frac{\text{π}}{\text{2}}\right)\right)+{\text{θ}}_{\text{2}}\right]|}_{\text{0}}^{\frac{\text{-π}}{\text{2}}}$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{=}\left[\left(\text{- cos }\left(\frac{\text{-π}}{\text{2}}\right)\text{. }\left(\frac{\text{π}}{\text{2}}\right)\right)+\left(\frac{\text{-π}}{\text{2}}\right)\right]-\left[\left(\text{- cos }\left(\text{0}\right)\text{. }\left(\frac{\text{π}}{\text{2}}\right)\right)+\left(\text{0}\right)\right]$

${\displaystyle \iint \left(\frac{\partial \text{V}}{\partial {\text{θ}}_1}+\frac{\partial \text{V}}{\partial {\text{θ}}_2}\right)}{\text{dθ}}_{\text{1}}{\text{dθ}}_{\text{2}}\text{= }0$

$\text{= 0}$

Therefore, from above calculation of the potential function, it is clear that the system has no periodic orbit.

A phase portrait is defined as the geometrical representation of the trajectories of the dynamical system in the phase plane of the system equation. Every set of the initial condition is represented by a different curve or point in the phase plane.

The phase portrait for the given system equation is shown. The system equations are given as:

${\dot{\text{θ}}}_{\text{1}}\text{ = Ksin}\left({\text{θ}}_{\text{1}}{\text{- θ}}_{\text{2}}\right){\text{ - sinθ}}_{\text{1}}$

${\dot{\text{θ}}}_{\text{2}}\text{ = Ksin}\left({\text{θ}}_{\text{2}}{\text{- θ}}_{\text{1}}\right){\text{ - sinθ}}_{\text{2}}$

The phase portrait plot for the above system equation for the value of K lying in the range of $\text{0 < K < }\frac{1}{2}$ is provided below.

The phase portrait of the above system equation at $\text{K = 0}\text{.5}$ is:

The phase portrait plot for the above system equation for the value of K lying in the range of $\text{K > }\frac{\text{1}}{\text{2}}$ is provided below.

The phase portrait of the above system equation at $\text{K = 1}$ is provided below: