# Math Insight

### Introduction to a bifurcation diagram

Math 1241, Fall 2020
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Due date: Dec. 9, 2020, 11:59 p.m.
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Total points: 3
1. Consider the dynamical system \begin{align*} \diff{ z }{t} &= \alpha - 2 \left(z - 8\right)^{2} + 4, \end{align*} where $\alpha$ is a parameter.
1. Let $\alpha=4$ so that the dynamical system becomes \begin{align*} \diff{ z }{t} &= - 2 \left(z - 8\right)^{2} + 8. \end{align*} How many equilibria are there?

Sketch a phase line of the dynamical system showing equilibria and vector field, using solid points for any stable equilibria and unfilled points for any unstable equilibria.
Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
2. Let $\alpha=-2$ so that the dynamical system becomes \begin{align*} \diff{ z }{t} &= - 2 \left(z - 8\right)^{2} + 2. \end{align*} How many equilibria are there?

Sketch a phase line of the dynamical system showing equilibria and vector field, using solid points for any stable equilibria and unfilled points for any unstable equilibria.
Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
3. Let $\alpha=-4$ so that the dynamical system becomes \begin{align*} \diff{ z }{t} &= - 2 \left(z - 8\right)^{2}. \end{align*} How many equilibria are there?

Sketch a phase line of the dynamical system showing equilibria and vector field. For this value of the parameter, the stability theorem might not help you determine stability (remember, the stability theorem only works if the derivative is positive or is negative). Instead, determine the vector field on either side of each equilibrium (by checking the value of $\diff{ z }{t}$). If the vector field is pointing toward an equilbrium, it is stable; if it is pointing away from an equilibrium, it is unstable. If on the other hand, the vector field is pointing toward an equilibrium on one side and away from the equilibrium on the other, we might call the equilibrium semi-stable. However, we'll still just call it unstable and represent it with an open circle.
Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
4. Let $\alpha=-6$ so that the dynamical system becomes \begin{align*} \diff{ z }{t} &= - 2 \left(z - 8\right)^{2} - 2. \end{align*} How many equilibria are there?

Sketch a phase line of the dynamical system showing equilibria and vector field, using solid points for any stable equilibria and unfilled points for any unstable equilibria.
Feedback from applet
number of equilibria:
vector field:
5. For the above cases with $\alpha = 4, -2, -4, -6$, flip the phase lines to be vertical and draw the equilibria on a plot where $\alpha$ is the horizontal axis and $z$ is the vertical axis. The plot should contain up to four columns of points corresponding to the equilibria at each value of $\alpha$. Use closed points for stable equilibria and open points for unstable equilibria.
Feedback from applet
bifurcation point:
branches of equilibria:
Equilibria:
Number of equilibria:
Stability of equilibria:
stability of equilibrium branches:
6. We just calculated the phase lines for four different values of $\alpha$, as they are representative of the different phase lines we would get as we change $\alpha$. Except for one special value of $\alpha$, the number of equilibria doesn't change as you vary $\alpha$ and the locations of the equilibria change smoothly as you change $\alpha$. Therefore, rather than calculate the equilibria for zillions of phase lines, we can show what the equilibria would be for those phase lines by drawing smooth curves through the points we drew on the above diagram. Draw this “curve of equilibria” on the above diagram, using a solid line for stable equilibria and a dashed line for unstable equilibria. Each vertical cross section will correspond to a phase line at the given value of $\alpha$, so the diagram allows you to immediately see what the different phase lines will look like. The resulting plot is called a bifurcation diagram.
7. On the bifurcation diagram, there should be one special value of $\alpha$ where the number or stability of the equilibria in the phase line changes. Let's denote this particular value by $\alpha^*$, which is called a bifurcation point. What is the value of the bifurcation point $\alpha^*$? $\alpha^* =$

How does the number of equilibria of the dynamical system change as $\alpha$ changes from being less than $\alpha^*$ to being greater than $\alpha^*$? The number of equilibria
. For $\alpha < \alpha^*$, there are
equilibria. For $\alpha > \alpha^*$, there are
equilibria.

2. For the dynamical system $\diff{ v }{t} = g(v,\alpha),$ the function $f$ of $v$ depends on a parameter $\alpha$. Rather than show the formula for $f$, you can see its behavior through these graphs for $\alpha=-23, -11, -7, -3$.

{}

$\alpha=-7$

$\alpha=-11$

$\alpha=-3$

For values of $\alpha$ in between those shown, $f$ changes smoothly, so its graph will be somewhere in between the snapshots shown. Online, you can change the value of $\alpha$ in the upper left graph using the slider to see how $f$ changes with $\alpha$ in more detail.

1. There is a bifurcation that occurs for a particular value of $\alpha$, which we'll denote by $\alpha^*$. What is the value of $\alpha^*$? $\alpha^*=$
($\alpha^*$ is one of the original four values of $\alpha$ shown above.)

2. At the bifurcation point, the number of equilibria changes.

For $-23 \le \alpha < \alpha^*$, the number of equilibria =
.

For $\alpha^* < \alpha \le -3$, the number of equilibria =
.

3. Sketch a bifurcation diagram with respect to the parameter $\alpha$. The bifurcation diagram should represent how the number, location, and stability of the equilibria depend on the value of $\alpha$ for $-23 \le \alpha \le -3$. Draw curves to show the location of the equilibria as a function $\alpha$. Use a solid line to indicate stable equilibria and a dashed line to indicate unstable equilibria.

Feedback from applet
bifurcation point:
branches of equilibria:
stability of equilibrium branches:

3. For the dynamical system \begin{align*} \diff{ x }{t} = f(x, a), \end{align*} where the function $f$ of $x$ also depends on a parameter $a$, a bifurcation diagram with respect to the parameter $a$ is shown below. In this diagram, solid lines represent stable equilibria and dashed lines represent unstable equilibria.
1. When $a= -8$, how many equilibria are there?
Determine their values, rounded to the nearest integer, and their stability.

Equilibria:
(Enter rounded equilibria in increasing order, separated by commas.)
Stability of equilibria:
(Enter stable for stable and unstable for unstable. Enter in same order as equilibria, separated by commas.)

Sketch the phase line for when $a= -8$, including equilibria and direction field. Use a solid circle for stable equilibria and an open circle for unstable equilibria.

Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
2. When $a= -1$, how many equilibria are there?
Determine their values, rounded to the nearest integer, and their stability.

Equilibria:
(Enter rounded equilibria in increasing order, separated by commas.)
Stability of equilibria:
(Enter stable for stable and unstable for unstable. Enter in same order as equilibria, separated by commas.)

Sketch the phase line for when $a= -1$, including equilibria and direction field. Use a solid circle for stable equilibria and an open circle for unstable equilibria.

Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
3. When $a= 7$, how many equilibria are there?
Determine their values, rounded to the nearest integer, and their stability.

Equilibria:
(Enter rounded equilibria in increasing order, separated by commas.)
Stability of equilibria:
(Enter stable for stable and unstable for unstable. Enter in same order as equilibria, separated by commas.)

Sketch the phase line for when $a= 7$, including equilibria and direction field. Use a solid circle for stable equilibria and an open circle for unstable equilibria.

Feedback from applet
equilibria:
number of equilibria:
stability of equilibria:
vector field:
4. Identify any bifurcation points.

Bifurcations points are at $a =$
.