Chapter 7.P3, Problem 1P

a) Show that there are no critical points when $c<\sqrt{0.5}$, one critical point for $c<\sqrt{0.5}$, and two criticalpoints when $c<\sqrt{0.5}$.

b) Find the critical point(s) and determine the eigenvalues of the associated Jacobian matrix when $c<\sqrt{0.5}$ and when $c=1$.

c) How do you think trajectories of the system will behavefor $c=1$? Plot the trajectory starting at the origin. Does itbehave the way that you expected?

d) Choose one or two other initial points and plot the corresponding trajectories. Do these plots agree with your expectations?

To determine

To prove: There are no critical points for the system $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$ where $a=0.25,b=0.5$ and $c>0$ when $c<\sqrt{0.5}$, one critical point for $c=\sqrt{0.5}$, and two critical points for $c>\sqrt{0.5}$.

Explanation of Solution

Given information:

The system of equations is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

Formula used:

Quadratic formula:

The roots of the equations $a{x}^2+bx+c=0$ are $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Proof:

Consider $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$

By plugging $a=0.25,b=0.5$,

The system becomes $x\text{'}=-y-z,y\text{'}=x+0.25y,z\text{'}=0.5+z\left(x-c\right)$

The points, if any, where $f(x)=0$ are called critical pointsof the autonomous system $x\text{'}=f(x)$.

The critical points of the system $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$ are found by solving the equations $-y-z=0,x+0.25y=0$ and $0.5+z\left(x-c\right)=0$.

Consider the first equation $-y-z=0$.

$\Rightarrow -y=z$

Consider the second equation $x+0.25y=0$.

$\Rightarrow x=-0.25y$

Now plug these value in the equation $0.5+z\left(x-c\right)=0$,

$\Rightarrow 0.5-y\left(-0.25y-c\right)=0$

$\Rightarrow 0.5+0.25y^2+cy=0$

$\Rightarrow 0.25y^2+cy+0.5=0$

$\Rightarrow y^2+4cy+2=0$

By using the quadratic formula,

$\Rightarrow y=\frac{-4c\pm \sqrt{16c^2-8}}{2}$

$\Rightarrow y=-2c\pm \sqrt{2\left(2c^2-1\right)}$

Thus the value of $y$ is not real when $2c^2-1<0$.

That is, $c<\sqrt{0.5}$.

Thus there is no critical point when $c<\sqrt{0.5}$.

Also, there are two distinct real values of when $2c^2-1>0$.

That is, $c>\sqrt{0.5}$

Thus there are two critical points when $c>\sqrt{0.5}$.

If $c=\sqrt{0.5}$ that is $2c^2-1=0$ then we get a unique value of $y$.

Thus there is only one critical point for the system.

To determine

The critical points of the system $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$ when $c=\sqrt{0.5}$ and $c=1$. Also, find the eigenvalues of the associated Jacobian matrix when $c=\sqrt{0.5}$ and $c=1$.

Answer to Problem 1P

Solution:

When $c=\sqrt{0.5}$ then thecritical point is $\left(\frac{\sqrt{2}}{4},-\sqrt{2},\sqrt{2}\right)$ and eigenvalues of the corresponding linear system are $0,-0.0517767\pm 1.52419i$.

When $c=1$ then the critical points are $\left(0.8536,-3.4142,3.4142\right)$ and $\left(0.1464,-0.5858,0.5858\right)$. The corresponding eigenvalues of the linear system are $-0.530252,-0.0366507\pm 1.154204i$ and $0.161186,-0.0288165\pm 2.094929i$ respectively.

Explanation of Solution

Given information:

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

Explanation:

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

By plugging $a=0.25,b=0.5$ and $c=\sqrt{0.5}$,

The system becomes, $x\text{'}=-y-z,y\text{'}=x+0.25y,z\text{'}=0.5+z\left(x-\sqrt{0.5}\right)$.

From part (a),

The critical point is $x=-0.25y,y=-2c\pm \sqrt{2\left(2c^2-1\right)}$ and $y=-z$.

Hence if $c=\sqrt{0.5}$ then $y=-2\sqrt{0.5}=-2\frac{\sqrt{2}}{2}=-\sqrt{2}$.

Therefore, the critical point is $\left(\frac{\sqrt{2}}{4},-\sqrt{2},\sqrt{2}\right)$.

Since $F\left(x,y,z\right)=-y-z,G\left(x,y,z\right)=x+0.25y$ and $H\left(x,y,z\right)=0.5+z\left(x-\sqrt{0.5}\right)$, the Jacobian matrix is given by,

$J\left(x,y,z\right)=\left(\begin{array}{ccc}{F}_x& F_y& F_z\\ G_x& G_y& G_z\\ H_x& H_y& H_z\end{array}\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ z& 0& x-c\end{array}\right)$

i) Near the critical point $\left(\frac{\sqrt{2}}{4},-\sqrt{2},\sqrt{2}\right)$, the Jacobian matrix becomes,

$J\left(\frac{\sqrt{2}}{4},-\sqrt{2},\sqrt{2}\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ \sqrt{2}& 0& \frac{\sqrt{2}}{4}-\frac{\sqrt{2}}{2}\end{array}\right)$

$J\left(\frac{\sqrt{2}}{4},-\sqrt{2},\sqrt{2}\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ \sqrt{2}& 0& \frac{-\sqrt{2}}{4}\end{array}\right)$

To find eigenvalues of the matrix $\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ \sqrt{2}& 0& \frac{-\sqrt{2}}{4}\end{array}\right)$:

Let $A=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ \sqrt{2}& 0& \frac{-\sqrt{2}}{4}\end{array}\right)$ and the eigenvalues are ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

The characteristics equation is $\left|A-\lambda I\right|=0$,

$\Rightarrow \left|\begin{array}{ccc}-\lambda & -1& -1\\ 1& 0.25-\lambda & 0\\ \sqrt{2}& 0& \frac{-\sqrt{2}}{4}-\lambda \end{array}\right|=0$

$\Rightarrow -\lambda \left(0.25-\lambda \right)\left(\left(-\sqrt{2}/4\right)-\lambda \right)+\left(\left(-\sqrt{2}/4\right)-\lambda \right)+\sqrt{2}\left(0.25-\lambda \right)=0$

$\Rightarrow -\lambda \left(\frac{1}{4}-\lambda \right)\left(\frac{-\sqrt{2}}{4}-\lambda \right)-\frac{\sqrt{2}}{4}-\lambda +\frac{\sqrt{2}}{4}-\sqrt{2}\lambda =0$

$\Rightarrow \frac{\sqrt{2}\lambda }{16}+\frac{-\sqrt{2}{\lambda }^2}{4}+\frac{{\lambda }^2}{4}-{\lambda }^3-\lambda -\sqrt{2}\lambda =0$.

$\Rightarrow -{\lambda }^3+\left(\frac{1-\sqrt{2}}{4}\right){\lambda }^2+\left(\frac{\sqrt{2}}{16}-1-\sqrt{2}\right)\lambda =0$

$\Rightarrow \lambda \left[\left(\frac{1-\sqrt{2}}{4}\right)\lambda -{\lambda }^2+\left(\frac{\sqrt{2}}{16}-1-\sqrt{2}\right)\right]=0$

$\Rightarrow \lambda =0,-0.0517767\pm 1.52419i$

Thus the eigenvalues when $c=\sqrt{0.5}$ are $0,-0.0517767\pm 1.52419i$.

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

By plugging $a=0.25,b=0.5$ and $c=1$,

The system becomes, $x\text{'}=-y-z,y\text{'}=x+0.25y,z\text{'}=0.5+z\left(x-1\right)$.

From part (a),

The critical point is $x=-0.25y,y=-2c\pm \sqrt{2\left(2c^2-1\right)}$ and $y=-z$.

Hence if $c=1$ then $y=-2\pm \sqrt{2}$.

That is, $y=-2+\sqrt{2}$ and $y=-2-\sqrt{2}$

Therefore, the critical points are $\left(0.8536,-3.4142,3.4142\right)$ and $\left(0.1464,-0.5858,0.5858\right)$.

Since $F\left(x,y,z\right)=-y-z,G\left(x,y,z\right)=x+0.25y$ and $H\left(x,y,z\right)=0.5+z\left(x-1\right)$, the Jacobian matrix is given by,

$J\left(x,y,z\right)=\left(\begin{array}{ccc}{F}_x& F_y& F_z\\ G_x& G_y& G_z\\ H_x& H_y& H_z\end{array}\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ z& 0& x-1\end{array}\right)$.

ii) Near the critical point $\left(0.8536,-3.4142,3.4142\right)$, the Jacobian matrix becomes,

$J\left(0.8536,-3.4142,3.4142\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 3.4142& 0& 0.8536-1\end{array}\right)$

$J\left(0.8536,-3.4142,3.4142\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 3.4142& 0& -0.1464\end{array}\right)$

To find eigenvalues of the matrix $\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 3.4142& 0& -0.1464\end{array}\right)$:

Let $A=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 3.4142& 0& -0.1464\end{array}\right)$ and the eigenvalues are ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

The characteristics equation is $\left|A-\lambda I\right|=0$,

$\Rightarrow \left|\begin{array}{ccc}-\lambda & -1& -1\\ 1& 0.25-\lambda & 0\\ 3.4142& 0& -0.1464-\lambda \end{array}\right|=0$

$\Rightarrow -\lambda \left(0.25-\lambda \right)\left(-0.1464-\lambda \right)+\left(-0.1464-\lambda \right)-3.4142\left(0.25-\lambda \right)=0$

$\Rightarrow 0.0366\lambda +0.25{\lambda }^2-0.1464{\lambda }^2-{\lambda }^3-0.1464-\lambda -0.8536+3.4142\lambda =0$.

$\Rightarrow 0.1036{\lambda }^2-{\lambda }^3-1+2.4508\lambda =0$

$\Rightarrow -{\lambda }^3+0.1036{\lambda }^2+2.4508\lambda -1=0$

$\Rightarrow {\lambda }^3-0.1036{\lambda }^2-2.4508\lambda +1=0$

$\Rightarrow \lambda =0.161186,-0.0288165\pm 2.094929i$

iii) Near the critical point $\left(0.1464,-0.5858,0.5858\right)$, the Jacobian matrix becomes,

$J\left(0.1464,-0.5858,0.5858\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 0.5858& 0& 0.1464-1\end{array}\right)$

$J\left(0.1464,-0.5858,0.5858\right)=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 0.5858& 0& -0.8536\end{array}\right)$

To find eigenvalues of the matrix $\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 0.5858& 0& -0.8536\end{array}\right)$:

Let $A=\left(\begin{array}{ccc}0& -1& -1\\ 1& 0.25& 0\\ 0.5858& 0& -0.8536\end{array}\right)$ and the eigenvalues are ${\lambda }_1,{\lambda }_2,{\lambda }_3$.

The characteristics equation is $\left|A-\lambda I\right|=0$,

$\Rightarrow \left|\begin{array}{ccc}-\lambda & -1& -1\\ 1& 0.25-\lambda & 0\\ 0.5858& 0& -0.8536-\lambda \end{array}\right|=0$

$\Rightarrow -\lambda \left(0.25-\lambda \right)\left(-0.8536-\lambda \right)+\left(-0.8536-\lambda \right)+0.5858\left(0.25-\lambda \right)=0$

$\Rightarrow -0.8536\left(-0.25\lambda +{\lambda }^2\right)-\lambda \left(-0.25\lambda +{\lambda }^2\right)-0.8536-\lambda +0.1464-0.5858\lambda =0$.

$\Rightarrow 0.2134\lambda -0.8536{\lambda }^2+0.25{\lambda }^2-{\lambda }^3-0.8536-\lambda +0.1464-0.5858\lambda =0$

$\Rightarrow -0.6036{\lambda }^2-{\lambda }^3-0.7072-1.3724\lambda =0$

$\Rightarrow {\lambda }^3+0.6036{\lambda }^2+1.3724\lambda +0.7072=0$

$\Rightarrow \lambda =-0.530252,-0.0366507\pm 1.154204i$

Thus the eigenvalues when $c=1$ are $-0.530252,-0.0366507\pm 1.154204i$ and $0.161186,-0.0288165\pm 2.094929i$.

To determine

How the trajectories of the system will behave for $c=1$. Also, plot the trajectory starting at the origin. Whether it behave the way that you expected or not.

Answer to Problem 1P

Solution:

As $t\to \infty$ all solutions converge to the point $\left(0.1464,-0.5858,0.5858\right)$.

The trajectory starting at origin is:

The graph is agreeing with our expectations.

Explanation of Solution

Given information:

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

Explanation:

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$.

From part (b),

The eigenvalues corresponding to the critical point $\left(0.8536,-3.4142,3.4142\right)$ are $0.161186,-0.0288165\pm 2.094929i$

Thus one of the eigenvalueshasa positive real part. Therefore, the critical point $\left(0.8536,-3.4142,3.4142\right)$ is unstable.

The eigenvalues corresponding to the critical point $\left(0.1464,-0.5858,0.5858\right)$ are $-0.530252,-0.0366507\pm 1.154204i$

Thus the eigenvalues have a negative real part. Therefore, the critical point $\left(0.1464,-0.5858,0.5858\right)$ is stable which is asymptotically stable.

Therefore, as $t\to \infty$ all solutions converge to the point $\left(0.1464,-0.5858,0.5858\right)$.

The trajectory starting at the origin is:

As $t\to \infty$ all solutions converge to the point $\left(0.1464,-0.5858,0.5858\right)$

Therefore, the graph is agreeing with our expectations.

To determine

To graph: The corresponding trajectories by choosing one or two initial points. Whether these plots agree with your expectations or not.

Explanation of Solution

Given information:

The system is $x\text{'}=-y-z,y\text{'}=x+ay,z\text{'}=b+z\left(x-c\right)$ and $c=1$.

Graph:

The trajectory with the initial condition $\left(0,0,0\right)$ is:

The trajectory with the initial condition $\left(1,1,1\right)$ is:

The trajectory with the initial point close to $\left(0.8536,-3.4142,3.4142\right)$ is:

The trajectory with the initial point close to $\left(0.1464,-0.5858,0.5858\right)$ is:

Interpretation:

From the above graph,

All trajectories converge to the point $\left(0.1464,-0.5858,0.5858\right)$.

From part (c),

As $t\to \infty$ all solutions converge to the point $\left(0.1464,-0.5858,0.5858\right)$

Therefore, the graph agrees with our expectations.