8.6 Exercises

8.1: Linearization, critical points, and equilibria

Your Turn

Find the critical points and linearizations of the following systems.

$x'=\sin(\pi y)+(x-1)^2$ , $y'=y^2-y$ ,
$x'=x+y+y^2$ , $y'=x$ ,
$x'=(x-1)^2+y$ , $y'=x^2+y$ .

Answer

Critical points $(0, 0)$ and $(0, 1)$ . At $(0, 0)$ using $u=x$ , $v=y$ the linearization is $u'=-2u-\left(\frac{1}{\pi}\right)v$ , $v'=-v$ . At $(0, 1)$ using $u=x$ , $v=y-1$ the linearization is $u'=-2u+\left(\frac{1}{\pi}\right)v$ $v'=v$ .
Critical point $(0, 0)$ . Using $u=x$ , $v=y$ the linearization is $u'=u+v$ , $v'=u$ .
Critical point $\left(\frac{1}{2},-\frac{1}{4}\right)$ . Using $u=x-\frac{1}{2}$ , $v=y+\frac{1}{4}$ the linearization is $u'=-u+v$ , $v'=u+v$ .

Your Turn

The idea of critical points and linearization works in higher dimensions as well. You simply make the Jacobian matrix bigger by adding more functions and more variables. For the following system of 3 equations find the critical points and their linearizations:

$x' = x + z^2,\quad y' = z^2-y, \quad z' = z+x^2.$

Answer

Critical points are $(0, 0, 0)$ , and $(−1, 1, −1)$ . The linearization at the origin using variables $u=x$ , $v=y$ , $w=z$ is $u'=u$ , $v'=-v$ , $z'=w$ . The linearization at the point $(−1, 1, −1)$ using variables $u=x+1$ , $v=y-1$ , $w=z+1$ is $u'=u-2$ , $v'=-v-2w$ , $w'=w-2u$ .

8.2: Stability and classification of isolated critical points

Your Turn

Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.

$x''+ x+x^3 = 0$
$\theta''+\sin \theta = 0$
$z''+ (z-1)(z+1) = 0$
$x''+ x^2+1 = 0$

Your Turn

Find the implicit equations of the trajectories of the following conservative systems. Next find their critical points (if any) and classify them.

$x''+ x^2 = 4$
$x''+ e^x = 0$
$x''+ (x+1)e^x = 0$

Answer

$\frac{1}{2}y^{2}+\frac{1}{3}x^{3}-4x=C$ , critical points: $(-2,0)$ , an unstable saddle, and $(2,0)$ , a stable center.
$\frac{1}{2}y^{2}+e^{x}=C$ , no critical points.
$\frac{1}{2}y^{2}+xe^{x}=C$ , critical point at $(-1,0)$ is a stable center.

8.3: Applications of nonlinear systems

Your Turn

Suppose the hares do not grow exponentially, but logistically. In particular consider

$x' = (0.4-0.01y)x - \gamma x^2, ~~~~~ y' = (0.003x-0.3)y.$

For the following two values of $\gamma$ , find and classify all the critical points in the positive quadrant, that is, for $x \geq 0$ and $y \geq 0$ . Then sketch the phase diagram. Discuss the implication for the long term behavior of the population.

$\gamma=0.001$ ,
$\gamma=0.01$ .

Your Turn

Suppose we have the system predator-prey system where the foxes are also killed at a constant rate $h$ ( $h$ foxes killed per unit time): $x' = (a-by)x,$ $y' = (cx-d)y - h$ .

Find the critical points and the Jacobin matrices of the system.
Put in the constants $a=0.4$ , $b=0.01$ , $c=0.003$ , $d=0.3$ , $h=10$ . Analyze the critical points. What do you think it says about the forest?

Answer

Solving for the critical points we get $(0,-\frac{h}{d})$ and $\left(\frac{bh+ad}{ac},\frac{a}{b}\right)$ . The Jacobian matrix at $(0,-\frac{h}{d})$ is $\left[\begin{array}{cc}{a+\frac{bh}{d}}&{0}\\{-\frac{cd}{d}}&{-d}\end{array}\right]$ whose eigenvalues are $a+\frac{bh}{d}$ and $-d$ . The eigenvalues are real of opposite signs and we get a saddle. (In the application, however, we are only looking at the positive quadrant so this critical point is irrelevant.) At $\left(\frac{bh+ad}{ac},\frac{a}{b}\right)$ we get Jacobian matrix $\left[\begin{array}{cc}{0}&{-\frac{b(bh+ad)}{ac}}\\{\frac{ac}{b}}&{\frac{bh+ad}{a}-d}\end{array}\right]$ .
For the specific numbers given, the second critical point is $\left(\frac{550}{3},40\right)$ the matrix is $\left[\begin{array}{cc}{0}&{-\frac{11}{6}}\\{\frac{3}{25}}&{\frac{1}{4}}\end{array}\right]$ , which has eigenvalues $\frac{5\pm\ i\sqrt{327}}{40}$ . Therefore there is a spiral source; the solution spirals outwards. The solution eventually hits one of the axes, $x=0$ or $y=0$ , so something will die out in the forest.

8.4: Limit cycles

8.5: Chaos

Your Turn

(project)

A simple fractal attractor can be drawn via the following chaos game. Draw three points of a triangle (just the vertices) and number them, say $p_1$ , $p_2$ and $p_3$ . Start with some random point $p$ (does not have to be one of the three points above) and draw it. Roll a die, and use it to pick of the $p_1$ , $p_2$ , or $p_3$ randomly (for example 1 and 4 mean $p_1$ , 2 and 5 mean $p_2$ , and 3 and 6 mean $p_3$ ). Suppose we picked $p_2$ , then let $p_{\text{new}}$ be the point exactly halfway between $p$ and $p_2$ . Draw this point and let $p$ now refer to this new point $p_{\text{new}}$ . Rinse, repeat. Try to be precise and draw as many iterations as possible. Your points should be attracted to the so-called Sierpinski triangle. A computer was used to run the game for 10,000 iterations to obtain the picture in Figure 1.

Figure 1: $10,000$ iterations of the chaos game producing the Sierpinski triangle.

Your Turn

Find critical points of the Lorenz system and the associated linearizations.

Answer

Critical points: $(0,0,0)$ , $(3\sqrt{8},3\sqrt{8},27)$ , $(-3\sqrt{8}, -3\sqrt{8}, 27)$ . Linearization at $(0,0,0)$ using $u=x$ , $v=y$ , $w=z$ is $u'=-10u +10v$ , $v'=28u-v$ , $w'=-\left(\frac{8}{3}\right)w$ . Linearization at $(3\sqrt{8}, 3\sqrt{8}, 27)$ using $u=x-3\sqrt{8}$ , $v=y-3\sqrt{8}$ , $w=z-27$ is $u'=-10u+10v$ , $v'=u-v-3\sqrt{8}w$ , $w'=3\sqrt{8}u+3\sqrt{8}v-\left(\frac{8}{3}\right) w$ . Linearization at $(-3\sqrt{8}, -3\sqrt{8}, 27)$ using $u=x+3\sqrt{8}$ , $v=y+3\sqrt{8}$ , $w=z-27$ is $u'=-10u+10v$ , $v'=u-v+3\sqrt{8}w$ , $w'=-3\sqrt{8}u-3\sqrt{8}v-\left(\frac{8}{3}\right) w$ .

Adapted from Differential Equations for Engineers by Jiří Lebl (https://www.jirka.org/diffyqs/), © Jiří Lebl, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.