Homework 6 Solutions
May 13, 2021
Problem 1
p.17 on pg. 487
a
To find the equilibrium points, we need to find the points (x, y) for which:
x(4 − x − y) = 0 y(2 + 2α − y − αy) = 0
From the first equation, we can see that in order for the first equation to hold, we must have either x = 0 or 4 − x − y = 0. Similarly, for the second equation to hold we must have that y = 0 or 2 + 2α − y − αx = 0. So, we get:
• Case 1: x = 0 and y = 0. This automatically gives the solution (x, y) = (0, 0).
• Case 2: x = 0 and 2 + 2α − y − αx = 0. This yields y = 2 + 2α, giving the solution (x, y) = (0, 2 + 2α).
• Case 3: 4 − x − y = 0 and y = 0. This yields x = 4 and we get the solution (x, y) = (4, 0).
• Case 4: 4 − x − y = 0 and 2 + 2α − y − αx = 0. Solving this pair of equations gives the solution (x, y) = (2, 2).
So, the equilibrium points at (x, y) = (0, 0), (0, 2 + 2α), (4, 0), (2, 2).
b, c
Since we need to compute the linearization of the system near (2, 2) for both of these parts, I think it will be easier to just do that here. For f (x, y) = x(4 − x − y) and g(x, y) = y(2 + 2α − y − αy), we have our system in the form:
(
x0(t) = f (x(t), y(t)) y0(t) = g(x(t), y(t))
Hence, we compute: ∂f ∂x = 4 − 2x − y ∂f ∂y = −x ∂g ∂x = −αy ∂g ∂y = 2 + 2α − 2y − αx And therefore the linearization near (2, 2) is given by:
Mα = "∂f ∂x(2, 2) ∂f ∂y(2, 2) ∂g ∂x(2, 2) ∂g ∂y(2, 2) # = −2 −2 −2α −2
Thus, for α = .75, we get the matrix:
M.75 =
−2 −2 −1.5 −2
The eigenvalues of which are, numerically, −3.73 and −0.26. As they are both negative, this is a stable node. On the other hand, for α = 1.25, we get the matrix:
M.75 =
−2 −2 −2.5 −2
This had eigenvalues: −4.23 and 0.23. Since these have different signs, this is a saddle point. The major change that happens in between is that we go from having a stable node to a saddle point.
d
We compute the characteristic polynomial of the matrix Mα above as:
chMα(z) = z
2
+ 4z − 4(1 − α) The roots, hence, are:
z± = 1 2 −4 ±p16 − 16(1 − α) = 1 2 −4 ±√16α = −2 ± 2√α
Since we transition from a stable node to a saddle point, we must go from having two negative eigenvalues to having one negative eigenvalue and one positive eigenvalue. The
transition point will happen when the larger of the two eigenvalues is equal to zero. That is, when:
−2 + 2√α = 0
and this clearly happens when α = 1, so we find that α0 = 1.
e
We produce the following phase portraits:
Figure 3: α = 1.25
Project 1 on pg.514 - 516
1
What we really need to show is that S(t) + I(t) + R(t) constant in time. For this, we simply take its derivative and replace in terms from the ODE. For the SIR model we have:
(S(t) + I(t) + R(t))0 = S0(t) + I0(t) + R0(t)
= (−βI(t)S(t)) + (βI(t)S(t) − νI(t)) + (νI(t)) = 0
and for the SIRS model we have
(S(t) + I(t) + R(t))0 = S0(t) + I0(t) + R0(t)
= (−βI(t)S(t) + γR(t)) + (βI(t)S(t) − νI(t)) + (νI(t) − γR(t)) = 0
Hence, the sum S(t) + I(t) + R(t) is constant in time, and therefore S(t) + I(t) + R(t) = S(0) + I(0) + R(0) = N
2
The vector field defining our ODE system is given by V (S, I) = (−βIS, βIS − νI), i.e. our ODE system is given as (S0(t), I0(t)) = V (S(t), I(t)). If we show that the vector field never points to the outside of the triangle at any point on the boundary of the triangle, then we can be certain that no trajectory leaves the triangle from the inside of it. The boundary of the triangle is given by the line segments:
L1 = { (S, I) | S = 0, 0 ≤ I ≤ N }
L2 = { (S, I) | I = 0, 0 ≤ S ≤ N }
L3 = { (S, I) | S + I = N, S ≥ 0, I ≥ 0 }
Along the segment L1, we have:
V (S, I) = V (0, I) = (0, −νI)
Since this runs parallel to the I-axis, we know that does not point to the outside of the triangle. So, we have this segment of the boundary covered.
Along the segment L2, we have:
V (S, I) = V (S, 0) = (0, 0)
So, every point along this boundary is an equilibrium point. Consequently, we find that the vector field does not point outward. So, this segment of the boundary is handled.
Along the segment L3, to show that the vector field does not point to the outside of the
triangle we will show that V (S, I)·n ≤ 0 along this segment, where n is the outward pointing normal to L3. This implies that the component of V (S, I) that is perpendicular to this line
is either vanishing or points inward. The outward normal is n = (1, 1). Thus, V (S, I) · n = βIS + βIS − νI = −νI ≤ 0
where −νI ≤ 0, since ν > 0 and I ≥ 0. Indeed, we have strict inequality except at the single point (S, I) = (N, 0) along the boundary. This means that trajectories are crossing into the triangle through this segment of the boundary, but they cannot leave the triangle through this segment of the boundary.
Since trajectories cannot leave the triangle through any segment of the boundary of the triangle Γ, trajectories cannot leave the triangle Γ.
3
Consider the I-nullcline, i.e. the set where the vector field has vanishing I-component. Since the I-component is βIS − νI, we find that this happens along the lines I = 0 and S = ρ = νβ. In particular, in the region I > 0, we know that the βIS −νI > 0 for S > ρ and βIS −νI < 0 for S < ρ. Considering the S component of the vector field −βIS, we see that for I, S > 0 this always points to the left.
Therefore, along any trajectory, we have that S(t) is decreasing in t. Thus, if a solution has S(0) < ρ, then we know that S(t) < ρ for all future times. Consequently, since the I-component of the vector field is negative in this region, we see that I(t) is always decreasing in time and this is not an epidemic.
On the other hand, if S(0) > ρ, then I(t) increases during the period of time that S(t) > ρ and decreases after S(t) < ρ. Thus, this trajectory is an epidemic.
Below is a plot of the phase diagram for β = 1, ν = 2 in the region S + I ≤ 4. The vertical line at S = 2 represents the value of ρ. Following the arrows should demonstrate the reasoning above. All trajectories that begin to the left of the vertical line decrease in I for all future times, where as those that begind to the right of the vertical line at first increase and then decrease.
The duration of infection is 1ν, and β is the contact rate. Rewriting ρ = βν = 11 νβ
we can see that ρ gets smaller as the duration of infection and contact rate increase. For a fixed N , this means that the epidemic region takes up more of the triangle, and hence the epidemic region is larger. 4 Consider I0(t) S0(t) = βI(t)S(t) − νI(t) −βI(t)S(t) = −1 + ν β 1 S(t)
From this we get the equation:
I0(t) = −S0(t) + ν β
S0(t) S(t) integrating in time, we obtain:
I(t) = −S(t) + ν
β log(S(t)) + C where C is the constant of integration. Thus, we should have:
I(t) + S(t) − ν
β log(S(t)) = C
is constant in time. This motivates us to choose H(S, I) = I + S − νβlog(S), and indeed following the above in reverse shows us that dtdH(S(t), I(t)) = 0 along solutions. Hence, this is a conserved quantity.
To get trajectories, we solve for I in the equation H(S, I) = C. Of course we find that:
I(t) = C − S(t) + ν
β log(S(t))
Thus, we may plot I as a function of S to see the shape of the trajectories. This lets us produce the desired phase portrait, which we plot below for β = 1, ν = 2 in the triangle for which N = 4. Notice that the trajectories can enter the triangle, but they do not leave the triangle.
5
From the above, we know that along any trajectory, S(t) is always positive and is decreasing. As S(t) passes ρ = νβ we find that I(t) is decreasing in time. From the phase portrait, we can see that I(t) tends toward 0. But, since the quantity S(t) + I(t) − νβ log(S(t)) = C a constant, we observe that S(t) cannot tend toward 0, since then the −βν log(S(t)) would go to ∞. Therefore, it must tend toward some positive value S0 as we observe above. Last, since
the quantity S(t) + I(t) + R(t) = N is preserved, we see that R(t) tends toward R0 = N − S0.
Now, we first notice that not all individuals become infected, since the limiting behavior has a positive susceptible population. Thus, we see that the reason that the epidemic stops is that the infected population I(t) tends toward 0, and this ends the transfer of the population S(t) → I(t) → R(t). Thus, indeed the epidemic dies out due to a lack of infected individuals, rather than a lack of susceptible individuals.
6
Vaccinating the population reduces the size of the total population participating in the disease process. Once the size of the susceptible population is below ρ, no matter the size to of the infected population, the size of the infected population will always be decreasing, and the trajectory does not represent an epidemic.
Problem 2
a
For f (x, y) = x − y2 and g(x, y) = y − x2, the ODE is of the form: x0(t) y0(t) =f (x(t), y(t)) g(x(t), y(t))
Thus, to find the equilibrium points, we solve the equations f (x, y) = 0 and g(x, y) = 0. From the first equation, we find x = y2 and form the second we get y = x2. Substituting the second into the first, we find that x = y2 = (x2)2 = x4. Hence, x(x3− 1) = 0. So, either
x = 0, in which case y = 0, or x3 = 1 in which case x = 1 and therefore y = 1. Thus, the
equilibrium points are:
b,c
I find it easiest to do both b and c at the same time. To compute the linearization, we first need to compute the partials of f and g. They are below:
∂f ∂x = 1 ∂f ∂y = −2y ∂g ∂x = −2x ∂g ∂y = 1
Linearziation around (0, 0): From the above, we get the Jacobian matrix:
M =1 0 0 1
which is certainly easy to analyze, since it is the identity matrix! The eigenvalues are 1 and 1, and a basis of eigenvecotrs is given by
{1 0 ,0 1 }
This is an unstable node, and below we give a plot of the linearized vector field near this point (with the vectors normalized to have norm one for ease of viewing), with the eigenlines indicated.
Linearization around (1, 1): Here, we have the Jacobian matrix:
M = 1 −2 −2 1
The eigenvalues are 3 and −1, so this is a saddle point. A basis of eigenvectors is given as
{1 1 , 1 −1 }
and the actual vector field in a neighborhood nearby the actual point:
d
First, notice that f (x, y) > 0 in the region x > y2, and f (x, y) < 0 in the region x < y2.
Similarly, g(x, y) > 0 in the region y > x2, and g(x, y) < 0 in the region y < x2. The point
(4, 2) sits on the x nullcine, and in the region where g(x, y) < 0, so y0(0) < 0. Hence, the trajectory immediately enters the region where f (x, y) > 0, and part of the trajectory in this region is marked in red. As f (x, y) > 0, we know that x0(t) is positive and therefore x(t) increases. Hence, the trajectory moves to the right, away from the equilibrium points and away from the region where g(x, y) > 0. Thus, for the entire red segment, we have that y0(t) < 0, and y(t) is decreasing. x(t) continues to increase until the trajectory crosses the curve x = y2 again, and into the region where f (x, y) < 0. In this region, now, we find that x0(t) < 0 and y0(t) < 0. But, now we observe that as x(t) and y(t) decrease, the trajectory remains in the region where both f (x, y) < 0 and g(x, y) < 0. Hence, x(t) and y(t) will continue to decrease for all times. Therefore, the trajectory could never return to be near the equilibrium points. Nor could it enter a periodic orbit. Therefore, the trajectory is not contained in any bounded region and it escapes to infinity.
3
c
Just i is true. All non-constant solutions converge either to the equilibrium point at (0, 0) or at (π, 0). For ii it is not true because some solutions converge to the unstable equilibrium. And iii is not true because there are only two constant trajectories, since the equilibrium points (π, 0) and (−π, 0) are identified with one another.
d
The points are indicated below. These are either the equilibrium points or the trajectories where the pendulum has exactly the energy to stop where the pendulum is upside down. For the other trajectories, the pendulum either swings without going all the way around, or swings all the way around continuously. In either case, the orbit is periodic.
e
Because the prey population is so small, the predator population is decreasing.
4
I chose the initial condition (.5, .5) to be the starting point for comparing the different behaviors of trajectories. First, we consider x-dominance. Here I chose Kx = 2, Ex = 1,
Figure 4: Trajectory for the initial condition (.5, .5) in the x dominant regime
Figure 5: Trajectories for various initial conditions in the x dominant regime
Observe that, independent of the initials conditions x(0), y(0) ≥ 0, the trajectories converge to (Kx, 0), as predicted.
Next, we consider y dominance. I chose Kx = 1, Ex = 2, Ky = 2 and Ey = 1. The results
Figure 6: Trajectory for the initial condition (.5, .5) in the y dominant regime
Figure 7: Trajectories for various initial conditions in the y dominant regime
Observe that, independent of the initials conditions x(0), y(0) ≥ 0, the trajectories converge to (0, Ky), as predicted.
Next, we look at the coexistent regime. Here I chose Kx = 1, Ex = 2, Ky = 1, Ey = 2.
Figure 8: Trajectory for the initial condition (.5, .5) in the coexistent regime
Figure 9: Trajectories for various initial conditions in the coexistent regime
Observe that independent of the initial conditions x(0), y(0) ≥ 0, the trajectories converge to the equilibrium point in the middle.
Last we consider the initial condition dependent regime. For this I chose Kx = 2, Ex = 1,
Figure 10: Trajectory for the initial condition (.5, .5) in the initial condition dependent regime
Figure 11: Trajectories for various initial conditions that converge to (0, Ky) in the initial
Figure 12: Trajectories for various initial conditions that converge to (Kx, 0) in the initial
condition dependent regime
