Math Insight

We'll imagine that we care only about non-negative values of the state variables $x$ and $y$. Perhaps $x$ and $y$ represent concentrations or populations sizes for which negative values don't make sense. Hence, in the below phase plane, we'll plot just the first quadrant.

1. We'll begin by determining where the variables are not moving, i.e., the nullclines and the equilibria.

   The $x$-nullcline is the set of points $(x,y)$ where $x$ is not changing, i.e., where $\diff{x}{t}=0$. (The $x$-nullcline will typically be composed of one or more curves in the phase plane.) Write the equation for the $x$-nullcline in terms of $x$ and $y$.
   
   
   This curve is a parabola circle cubic line
   with slope
   and $y$-intercept
   . Plot this curve on the above phase plane and label it as either “$x$-nullcline” or “$\diff{x}{t}=0$”. (Pencil is recommended unless you draw quite accurately. The curve needs to be sufficiently accurate so that it intersects the $y$-nullcline in the correct places, as discussed below.)

2. The $y$-nullcline is the set of points $(x,y)$ where $y$ is not changing, i.e., where $\diff{y}{t}=0$. Write the equation for the $y$-nullcline in terms of $x$ and $y$.
   
   
   This curve is a parabola cubic line circle
   with vertex at the point
   that opens leftward downward upward rightward
   .

   Plot this curve on the above phase plane and label it as either “$y$-nullcline” or “$\diff{y}{t}=0$”. (Pencil is recommended, or wait until the next step so that you know exactly where the curve should intersect the $x$-nullcline.)

3. The equilibria are the points where both $x$ and $y$ are unchanging, i.e., we need both $\diff{x}{t}=0$ and $\diff{y}{t}=0$. In other words, the equilibria are the points where the $x$-nullcline and the $y$-nullcline are perpendicular are parallel intersect
   

   To determine for the equilibria, solve the $x$-nullcline for $y$ (if you haven't already) and substitute this value for $y$ into the $y$-nullcline. The result is a quadratic equation for $x$:
   

   Solve this equation (e.g., by factoring or plugging into the quadratic formula) to determine the values of $x$ at the two different equilibria: $x_e =$
   and $x_e =$
   .

   Plug each of these values of $x$ into either nullcline equation (probably the $x$-nullcline is easier) to determine the values of $y$ at the equilibria. The final result is that the equilibria are $(x_e,y_e) =$
   and $(x_e,y_e) =$
   .

   Plot the equilibria on the above phase plane. At this point, you can finalize your sketches of the nullclines, if you haven't already, making sure they intersect at the equilibria.

4. At each point $(x,y)$, the rate of change of $x$, is determined by the first equation $\diff{x}{t} = - 2 x + 3 y$. To summarize how this rate of change depends on $x$ and $y$, let's define the function $f(x,y) = - 2 x + 3 y$ so that the dynamics of $x$ are given by $\diff{x}{t}=f(x,y)$. Similarly defined $g(x,y)=4 x - y^{2}$ so that the dynamics of $y$ are given by $\diff{y}{t}=g(x,y)$. Rewrite the equations for the $x$-nullcline and $y$-nullcline in terms of $f(x,y)$ and $g(x,y)$. The $x$-nullcline is the equation
   and the $y$-nullcline is the equation
   . The equilibria are the simultaneous solution to both equations \begin{gather*} ＿\\ ＿. \end{gather*}

   With those definitions of $f(x,y)$ and $g(x,y)$, we can compactly express the direction at which the solution must move at each point $(x,y)$. The solution must move through the point $(x,y)$ in the direction $(dx/dt, dy/dy) =$
   . (Your answer should be a vector in terms of $f(x,y)$ and/or $g(x,y)$.)

   Let's calculate the direction of movement at a few points in the phase plane. For each point, calculate $(dx/dt, dy/dt)=＿$. Then sketch the point on the phase plane along with a vector coming out of that point and whose direction is given by the direction of motion of the solution through that point. The length of the vector is arbitrary so make it a convenient length -- long enough to see, but not too long to cross a nullcline into another region. The exact direction of the vector is less important than making sure you have the signs of $dx/dt$ and $dy/dt$ correct. (In other words, your vector should point in the correct compass direction, such as southwest, north, or northeast.)

   • $(x,y)=\left ( 8, \quad 4\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 4, \quad 1\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 7, \quad 10\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 11, \quad 7\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 10.5, \quad 7\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 10.24, \quad 6.4\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 0, \quad 8\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 6, \quad 4\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 4, \quad 4\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 3, \quad 3\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 0, \quad 0\right )$,    $(dx/dt,dy/dt)=$
     
   • $(x,y)=\left ( 9, \quad 6\right )$,    $(dx/dt,dy/dt)=$
     
5. Just considering the first quadrant (where $x \ge 0$ and $y \ge 0$), the nullclines cut the phase plane into
   pieces. Away from a nullcline $\diff{x}{t}$ cannot can
   be zero and $\diff{y}{t}$ can cannot
   be zero. Since we $f(x,y)$ and $g(x,y)$ are continuous functions, they cannot can
   change signs without passing through
   . Therefore, within each of the those four pieces, the basic compass direction (northwest, east, southeast, etc.) cannot can
   change. To get the basic direction everywhere in the phase plane, it suffices to draw just one direction arrow in each of those pieces, which you have not have
   already done.
6. The $y$-nullcline cuts the $x$-nullcline into
   pieces. Along each piece of the $x$-nullcline, $\diff{x}{t}=$
   and $\diff{y}{t}$ can cannot
   be zero. Any direction arrow crossing through one of those pieces of the $x$-nullcline must point down up or down left or right left up right
   . Along one piece of the $x$-nullcline, the sign of $\diff{y}{t}$ cannot can
   change so the direction arrows can cannot
   change direction on a single piece of the $x$-nullcline. To determine in which direction the solution crosses the $x$-nullcline, it suffices to draw just one direction arrow in each of those pieces, which you have have not
   already done.
7. Similarly, the $x$-nullcline cuts the $y$-nullcline into
   pieces. Along each piece of the $y$-nullcline, $\diff{y}{t}=$
   and $\diff{x}{t}$ can cannot
   be zero. Any direction arrow crossing through one of those pieces of the $y$-nullcline must point up or down down right left up left or right
   . Along one piece of the $y$-nullcline, the sign of $\diff{x}{t}$ can cannot
   change so the direction arrows can cannot
   change direction on a single piece of the $y$-nullcline. To determine in which direction the solution crosses the $y$-nullcline, it suffices to draw just one direction arrow in each of those pieces, which you have not have
   already done.
8. What must the solution look like when starting at the initial condition $(x(0),y(0)) = \left ( 8, \quad 1\right )$? To determine the shape of the solution, plot the point $\left ( 8, \quad 1\right )$ on the above phase plane. Starting at that point, in which direction must the solution move? straight up downward and to the left straight right straight left upward and to the right upward and to the left downward and to the right straight down
   (You don't actually have to compute $(dx/dt,dy/dt)$ for $(x,y) = \left ( 8, \quad 1\right )$; you can just determine the general direction from the arrow(s) drawn in its piece of the phase plane. Draw the point in $(x,y) = \left ( 8, \quad 1\right )$ in the above phase plane and draw a curve moving from that point in the correct general direction.

   As the curve keeps moving in that direction, what must it hit first? the y-nullcline an equilibrium the y-axis the x-axis the x-nullcline
   When it hits ＿, in which direction must it move? straight right upward and to the left downward and to the left straight down straight up downward and to the right straight left upward and to the right
   Once the solution trajectory crosses ＿, in which direction must it move? downward and to the left straight up straight right upward and to the right upward and to the left straight left straight down downward and to the right
   

   As the trajectory continues moving ＿, can it cross ＿ again? No. Given the direction arrows, trajectories can only cross from below.
   Can it cross the $y$-nullcline? No. Given the direction arrows, trajectories can only cross from the left.
   Since the trajectory is hemmed in and most continue moving ＿, where must the trajectory end up as time increases to infinity? It must end up at the non-equilibrium equilibrium
   $(x,y)=$
   .

9. Sketch the solution you determined on the below axes. Plot $x(t)$ versus time $t$ on the top axes and $y(t)$ versus time on the bottom axes. Since time is not represented on the phase plane, the values of $t$ are arbitrary (hence the $t$-axes have no scale). Although the shape will be approximate, certain values, such as the initial condition and the long term behavior must correctly match your solution. For intermediate values of time, just make sure that $x(t)$ and $y(t)$ move in the proper directions to match the sequence of directions that you determined in the phase plane.
   
   

   Since the graphs must be graded by hand, you can achieve at most 70% correct via computer grading.

1. Calculate the nullclines and plot them on the phase plane. To help you, we show a phase plane plot below. Dotted lines are included as a hint, but they are not intended to be complete. Be sure to label the nullclines.
2. Identify all equilibria. Give their values and show them in the phase plane.
3. Sketch a direction arrow in each of the regions of the phase plane divided by the nullclines. You only need to consider the part of the phase plane where $s(t) \ge 0 $ and $b(t) \ge 0$.
4. Sketch a direction arrow on each segment of each nullcline.
5. Sketch a plausible solution in the phase plane starting at the initial condition of 120 and 120 , i.e., $(s(0),b(0))=(120,120)$. The solution should be consistent with your direction arrows.
   
   (We need more information to determine exactly what the solution should look like. Whatever you choose for the behavior, it need only be consistent with the direction arrows.)