Math Insight

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:
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   Hint

   You can use the stability theorem to determine the stability of the equilibria. To determine the direction of the vector field, you can test the sign of $\diff{ z }{t}$ at points between the equilibria to know if $z$ is increasing or decreasing on those intervals. (Alternatively, you can infer the direction from the stability of equilibria, knowing the flow must be toward stable equilibria and away from unstable equilibria.)

   Online, drag $n_e$ to specify the number of equilibria and drag the resulting red points to the locations of the equilibria. Click a point to change it between closed (stable) and open (unstable). Click the phase line between equilibria to show a vector field. Click again to change the direction of the vector field.

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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:
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   Hint

   Online, drag $n_e$ to specify the number of equilibria and drag the resulting red points to the locations of the equilibria. Click a point to change it between closed (stable) and open (unstable). Click the phase line between equilibria to show a vector field. Click again to change the direction of the vector field.

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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:
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   Hint

   Online, to specify the number of equilibria on each phase line that is now turned vertical, drag the $n_e$ attached to that line to the desired value. Then, move the points to the locations of the equilibria and click them to change between closed and open points. (We won't draw vector fields on these phase lines.)

   The next part will show you how to get credit in the applet for the bifurcation point and branches.

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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.
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   Hint

   If you aren't sure what the curves of equilibria should look like, you can calculate additional phase lines for other values of $\alpha$. Then, you can plot those points on the diagram to give you additional information about the shape of the curve.

   Online, click “show curve” to reveal a curve that you can move by dragging the three black points that appear. Move those points so that curve has the desired shape. Click each branch of the curve to change it between solid and dashed, to represent stable or unstable equilibria, respectively.

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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 decreases increases
   . For $\alpha < \alpha^*$, there are
   equilibria. For $\alpha > \alpha^*$, there are
   equilibria.

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.)

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   Hint

   At a bifurcation point, the number or stability of equilibria changes. At most values of $\alpha$, only the locations of the equilibria change. But, as you increase $\alpha$, at one particular value of $\alpha$, the left section of the graph of $f$ stops touching the horizontal axis, decreasing the number of equilibria.

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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 =
   .

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   Hint

   When the left part of the graph of $f$ stops touching the horizontal axis at $\alpha^*$, it removes a pair of equilibria. (OK, exactly at the moment where it is tangent to the horizontal axis, i.e., at $\alpha=\alpha^*$, there is only one equilibrium there, but if $\alpha \ne \alpha^*$, there are either zero or two places where the left half of the graph of $f$ touches the horizontal axis.)

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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:
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   Hint

   Online, you have three branches of equilibria that you can manipulate so they form the bifurcation diagram. Two branches are curves that come together at the bifurcation point (the red point). The third branch is a line.

   Move the red point to indicate the location of the bifurcation point. Move the two points on the adjoining curves to indicate how the equilibria change as a function of the parameter $\alpha$. Move two points on the line so that the line indicates how the third equilibrium changes as a function of the parameter $\alpha$. You can click each branch of equilibria to switch between a solid line (indicating a stable equilibrium) and a dashed line (indicating an unstable equilibrium).

   To put the bifurcation point in the right spot, look at the graph with $\alpha=\alpha^*$ and determine the value of $v$, let's call it $v^*$, where the graph of $f$ is just touching the horizontal axis. That value $v=v^*$ is the location of the equilibrium at the bifurcation point. Note that on the bifurcation diagram, the location of $v$ is flipped so that it is on the vertical axis. The horizontal axis is for the parameter $\alpha$. Therefore, move the red point indicating the bifurcation to the coordinates $(\alpha^*, v^*)$.

   Next, focus on the equilibria that come from the branch of the graph of $f$ where the bifurcation occurs. Those equilibria exist only for $\alpha$ either greater than or less than $\alpha^*$, so the two curves coming of the red point should either both go to the right or both go to the left. To pin down their exact locations, look at one of the graphs of $f$ where these extra equilibria exist and note the values of $v$ of those equilibria. Then, move the two points on curves so their vertical coordinates equal those values of $v$; their horizontal coordinates should be the value of $\alpha$ of the graph that you consulted.

   The third equilibrium that exists in all graphs of $f$ corresponds to the remaining line on the bifurcation diagram. To correctly position the line, look up the values of $v$ of that equilibrium on two of the graphs of $f$. Position the two points so their first coordinates are the values of $\alpha$ of the graphs you consulted and their second coordinates are the values of $v$ that you found from the graph.

   The last step is to get the stability of the branches of equilibria correct. Determine the stability of the equilibria from the graphs (either by thinking of the direction in which $v$ would move on either side of the equilibrium or by noting the sign of the slope of $f$ at each equilibrium and using the stability theorem). Then, click each curve in the bifurcation diagram so that stable equilibria are represented by solid curves and unstable equilibria are represented by dashed curves.

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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:
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   Hint

   If you take a vertical cross section of the bifurcation diagram at $a=-8$, the result is the phase line at that value of the parameter, only flipped vertically. You can read off directly from the diagram the values of the equilibria (the vertical coordinate of the curves at $a=-8$) and their stability.

   You cannot read off the direction field (or vector field) directly from the bifurcation diagram. But, once you have the equilibria and their stability, you can infer the directions of motions because the solutions should approach a certain class of equilibrium and move away from the other class.

   Online, for the phase line, drag $n_e$ to specify the number of equilibria and drag the resulting red points to the locations of the equilibria. Click a point to change it between closed (stable) and open (unstable). Click the phase line between equilibria to show a vector field. Click again to change the direction of the vector field.

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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:
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   Hint

   Exactly the same exercise as part a, only at a different value of $a$.

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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:
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   Hint

   Exactly the same exercise as part a, only at a different value of $a$.

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4. Identify any bifurcation points.

   Bifurcations points are at $a =$
   .

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   Hint

   Bifurcation points are where the number or stability of equilibria changes. Are there any special values of $a$ where something more interesting happens other than changes in the locations of the equilibria?

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