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Repeat the steps from the previous problem set to graph the nullclines, equilibria, and direction vectors. (Complete steps 1-4 in the following phase plane applet.)
Feedback from applet
Step 1: nullclines:
Step 2: equilibria:
Step 2: equilibria classification:
Step 2: number of equilibria:
Step 2: saddle stable direction:
step 2: saddle unstable direction:
Step 3: vector directions in regions:
Step 3: vector locations in regions:
Step 4: vector directions on nullclines:
Step 4: vector locations on nullclines:
Step 5: initial condition:
Step 5: solution trajectory end point:
Step 5: solution trajectory follows vector field:
The system has four equilibria. Enter them in clockwise order, starting with the lower left.
$E_1=$
$E_2=$
$E_3=$
$E_4=$
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Calculate the Jacobian matrix of the system. Each component will be a function of $p_1$ and $p_2$.
Hint
Calculating these partial derivatives requires careful bookkeeping. Since the computer won't give you credit until you get each derivative completely correct, getting an answer accepted may be frustrating. Just calculate the derivative slowly and carefully.
You can try computing them one of two ways and see which one you find easier.
- You can use the product rule, where you treat the quantity in parentheses as one of the factors.
- You can multiply everything out (i.e., distribute so you can remove the parentheses) and then take the derivative.
Guidance for the product rule
We want to use the product rule to calculate the partial derivative of
$0.3 p_1 \left(1-\frac{p_1+2p_2}{ 2400 }\right)$ with respect to the first variable $p_1$. Let $0.3p_1$ be the first factor and let $1-\frac{p_1+2p_2}{ 2400 }$ be the second factor.
What is the derivative of the first factor with respect to $p_1$?
$\pdiff{}{p_1} \left( 0.3p_1 \right) = $
What is the derivative of the second factor with respect to $p_1$?
$\pdiff{}{p_1} \left( 1-\frac{p_1+2p_2}{ 2400 } \right) = $
$\pdiff{}{p_1} \left( 0.3p_1 \right) = 0.3$
$\pdiff{}{p_1} \left( 1-\frac{p_1+2p_2}{ 2400 } \right) = - \frac{1}{2400}$
(Hide)
By the product rule, the derivative of the product is the sum of (A) the first factor times the derivative of the second factor and (B) the second factor times the derivative of the first factor. This will be the upper left entry of the Jacobian matrix.
(Hide)
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Evaluate the Jacobian matrix at the first equilibrium, $E_1=＿$
Calculate the eigenvalues.
$\lambda_1 = $
$\lambda_2 = $
Classify the equilibrium $E_1=＿$:
On the above applet, click the equilibrium until it indicates its correct classification.
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Evaluate the Jacobian matrix at the second equilibrium, $E_2=＿$
Calculate the eigenvalues.
$\lambda_1 = $
$\lambda_2 = $
Classify the equilibrium $E_2=＿$:
On the above applet, click the equilibrium until it indicates its correct classification.
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Evaluate the Jacobian matrix at the third equilibrium, $E_3=＿$
Calculate the eigenvalues.
$\lambda_1 = $
$\lambda_2 = $
Classify the equilibrium $E_3=＿$:
On the above applet, click the equilibrium until it indicates its correct classification.
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Evaluate the Jacobian matrix at the fourth equilibrium, $E_4=＿$
Calculate the eigenvalues.
$\lambda_1 = $
$\lambda_2 = $
Classify the equilibrium $E_4=＿$:
On the above applet, click the equilibrium until it indicates its correct classification.
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Since both $E_2=＿$ and $E_4=＿$ are stable, we say that this system with strong competition exhibits bistability. These stable states correspond to the two outcomes we observed before. $E_2$ corresponds to population 2 winning and population 1 disappearing and $E_4$ corresponds to the reverse scenario.
Whether population 1 wins or population 2 wins depends on the initial condition $(p_1(0),p_2(0))$. The set of initial conditions that lead to population 1 eventually winning is the basin of attraction of the stable equilibrium $E_4$. The set of initial conditions that lead to population 2 eventually winning is the basin of attraction of the stable equilibrium $E_2$.
The saddle $E_3=＿$ provides the key for determining the boundary between the two basins of attraction. This basin of attraction boundary will be a curve that comes out of the saddle in a direction determined by the eigenvectors of the Jacobian matrix. Although we can't compute the exact shape of the boundary curve, we can determine its general direction from the eigenvectors.
What is the negative eigenvalue of the Jacobian evaluated at the equilibrium $E_3=＿$?
$\lambda_{-} = $
What is its associated eigenvector?
$\vc{u}_{-} = $
What is the positive eigenvalue of the Jacobian?
$\lambda_{+} = $
What is its associated eigenvector?
$\vc{u}_{+} = $
Which of these two eigenvector is the direction along which solutions approach the equilibrium?
To draw this direction on the above phase plane, set step=2 and click the “show saddle vector” checkbox. (It will appear only when you have found $E_3$ correctly.) Change the pair of vectors pointing toward the equilibrium to line up in this direction.
Along the other eigenvector, solutions move directly away from the equilibrium $E_3$. Change the pair of vectors pointing away from the equilibrium to line up in this direction.
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To examine how the stable direction of the saddle determines the basins of attraction, execute the R script run_competition.R so that you can simulate the competition dynamical system with the run_competition command. (We assume that you already have the R package deSolve installed on your computer.) We'll try initial conditions that are along the stable direction of $E_3$.
At the equilibrium $E_3$, $p_1=$
and $p_2=$
. We want to find a nearby initial condition that is in the direction of the stable eigenvector $＿$. With such an initial condition, the solution trajectory should get very close to $E_3$, and small changes will switch between the two basins of attraction.
Rescale the stable direction $＿$ by dividing it by its first component. The result is a vector in the same direction but whose first component is 1:
Next multiply the vector by $100$ and subtract it from $E_3=＿$. The result should be a point whose first component is $1100$:
We'll use this point as an initial condition, since it is close to $E_3$ and in the direction of the stable eigenvector. In other words, we'll use the initial condition $p_1(0) =＿$ and $p_2(0) = ＿$.
So that we can have an integer starting population size, let's round $p_2(0)$ down to the nearest integer and use the initial condition $p_1(0) =＿$ and $p_2(0) = ＿$. Assuming you have executed the run_competition.R script, you can simulate with the rounded down initial condition for $200$ years using the command:
run_competition(r1=0.3, r2=0.2, K1=2400, K2=3000, a12=2, a21=2, p10=＿, p20=＿, tmax=200)
What happens with this initial condition?
The initial condition $p_1(0) =＿$ and $p_2(0) = ＿$ is in the basin of attraction of which stable equilibrium?
Let's change the initial condition every so slightly. Rather than rounding down the initial condition for species two, let's round it up to the nearest integer and use the initial condition $p_1(0) =＿$ and $p_2(0) = ＿$. Simulate with the rounded up initial condition for $200$ years using the command:
run_competition(r1=0.3, r2=0.2, K1=2400, K2=3000, a12=2, a21=2, p10=＿, p20=＿, tmax=200)
What happens with this initial condition?
The initial condition $p_1(0) =＿$ and $p_2(0) = ＿$ is in the basin of attraction of which stable equilibrium?
Recall that you calculated the starting point $＿$ by taking a small step from the saddle $＿$ in the direction of the stable eigenvector $＿$. You can think of this point as being on the edge of a knife. Just making tiny changes in the initial condition from $(＿,＿)$ to $(＿,＿)$ switched between the basins of attraction of the two stable equilibria. The stable direction of the saddle marks where this separation occurs.
The dividing line between the basins of attraction curves as you move away from the saddle. If you want to get a better picture of the curve in the phase line that divides the basins of attraction, try the initial conditions $(p_1(0),p_2(0))=(20,35)$ and versus $(p_1(0),p_2(0))=(20,34)$. For either initial condtion, the curve from the initial condition to the saddle is very close to the boundary between the basins of attraction.
On the above phase line applet, sketch the solution for the initial condition $(p_1(0),p_2(0))=(20,35)$. Set step to 5, move the green point to $(20,35)$ and increase nsegs to reveal more line segments with which to draw the trajectory. Since the applet can't compute the dividing line between the two basins of attraction, it will give you credit for going to either stable equilibrium. But, if you want to be accurate, follow the results you observe from the R simulation.