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Exercises 15.5 Exercises
💡 Conceptual Quiz.
Answer the following questions to test your understanding of nonlinear systems.
1. Multiple-Choice.
(a) One Matrix or Many?
A linear system
\(\vec{X}' = A\vec{X}\) is described by a single coefficient matrix. How many matrices does classifying a nonlinear system with three equilibria require?
Three — one Jacobian evaluated at each equilibrium.
Correct. The linearization is local, so each equilibrium gets its own matrix and possibly its own type.
One — the Jacobian, evaluated once.
\(J(x,y)\) is computed once symbolically, but it must then be evaluated separately at each equilibrium.
None — nonlinear systems have no matrices.
They have no single coefficient matrix, but the Jacobian is very much a matrix.
Six — two per equilibrium, one for each variable.
Both variables are handled by the two rows of one \(2\times 2\) matrix.
(b) What Nullclines Do Not Tell You.
You sketch nullclines and find that the flow circles an equilibrium. Which question remains unanswered?
Whether nearby trajectories approach the equilibrium, leave it, or loop around it forever.
Correct — circulation is consistent with a spiral sink, a spiral source, and a center alike.
Where the equilibrium is.
The equilibrium is exactly the crossing you sketched.
Which direction the flow travels in each region.
That is precisely what the sign test in each region tells you.
Whether the system is nonlinear.
You knew that from the equations before drawing anything.
(c) The Borderline Case.
A Jacobian at an equilibrium has eigenvalues
\(r = \pm 2i\text{.}\) What is the correct conclusion about the nonlinear system there?
Nothing yet — the linearization is inconclusive, and the nonlinear terms must be examined.
Correct. Zero real part is exactly the case the linearization cannot settle.
It is a center, so trajectories are closed loops.
That is what the linearization predicts. The nonlinear system may spiral in or out instead.
It is stable, since the real part is not positive.
A zero real part gives no margin either way — cubic terms can push the trajectory outward.
It is a saddle.
Saddles come from real eigenvalues of opposite signs.
(d) Why Superposition Fails.
Why can you not add two solutions of
\(x' = x - xy\text{,}\) \(y' = xy - 2y\) to get a third solution?
The \(xy\) terms do not distribute across a sum: \((x_1+x_2)(y_1+y_2) \neq x_1y_1 + x_2y_2\text{.}\)
Correct — the cross terms are exactly what breaks superposition.
The system has two equilibria instead of one.
Multiple equilibria are a consequence of nonlinearity, not the reason superposition fails.
The solutions are not known in closed form.
Superposition would still be a valid principle even for solutions we could not write down.
The two equations are coupled.
Linear systems can be fully coupled and still obey superposition.
(e) Reading a Sign Pair.
In one region of the phase plane,
\(x' \lt 0\) and
\(y' \lt 0\text{.}\) Which arrow belongs there?
Down and to the left.
Correct — both coordinates are decreasing.
Down and to the right.
Rightward motion would need \(x' \gt 0\text{.}\)
Up and to the left.
Upward motion would need \(y' \gt 0\text{.}\)
Straight down.
Straight down requires \(x' = 0\) — that happens only on an \(x\) -nullcline.
(f) Equilibria on an Axis.
Why are Jacobians at equilibria lying on a coordinate axis often triangular?
One variable is zero there, which kills an off-diagonal partial derivative built from interaction terms.
Correct — and a triangular matrix hands you its eigenvalues on the diagonal.
Jacobians are always triangular.
The coexistence equilibria in this chapter give full matrices.
Because the equilibrium is stable.
Shape and stability are unrelated — triangular matrices come in both kinds.
Because partial derivatives with respect to \(y\) are always zero on the \(x\) -axis.
Not in general; \(f_y = -2x\) is nonzero at \((3,0)\) in the competing-species model.
🏋️♂️ Finding Equilibria.
Factor each right-hand side and test every pairing of cases, as in
🌌 Example 318 .
2.
Find every equilibrium point of
\begin{align*}
\frac{dx}{dt} \amp = x - xy\\
\frac{dy}{dt} \amp = xy - 2y
\end{align*}
Answer .
\((0,0)\) and
\((2,1)\text{.}\)
3.
Find every equilibrium point of
\begin{align*}
\frac{dx}{dt} \amp = x(6 - x - 2y)\\
\frac{dy}{dt} \amp = y(4 - x - y)
\end{align*}
Answer .
\((0,0)\text{,}\) \((6,0)\text{,}\) \((0,4)\text{,}\) and
\((2,2)\text{.}\)
4.
Find the nullclines and every equilibrium point of
\begin{align*}
\frac{dx}{dt} \amp = y - x^2\\
\frac{dy}{dt} \amp = x - y
\end{align*}
Answer .
\(x\) -nullcline
\(y = x^2\text{;}\) \(y\) -nullcline
\(y = x\text{;}\) equilibria
\((0,0)\) and
\((1,1)\text{.}\)
🏋️♂️ Building and Reading Jacobians.
Compute partial derivatives, then evaluate and classify using the table in
The Jacobian Matrix .
5.
Compute the Jacobian matrix \(J(x,y)\) for
\begin{align*}
\frac{dx}{dt} \amp = x^2 y - 3x\\
\frac{dy}{dt} \amp = x + y^2
\end{align*}
Answer .
\(J(x,y) = \begin{bmatrix} 2xy - 3 \amp x^2 \\ 1 \amp 2y \end{bmatrix}\)
6.
Classify the equilibrium in each case, given the Jacobian evaluated there.
\begin{align*}
\text{(a)} \quad \amp \begin{bmatrix} -3 \amp 1 \\ 0 \amp -2 \end{bmatrix}
\qquad \text{(b)} \quad \begin{bmatrix} 1 \amp 2 \\ -2 \amp 1 \end{bmatrix}
\qquad \text{(c)} \quad \begin{bmatrix} 0 \amp 3 \\ -3 \amp 0 \end{bmatrix}
\end{align*}
Answer .
(a) stable node (
\(r = -3, -2\) ); (b) spiral source (
\(r = 1 \pm 2i\) ); (c) linearized center (
\(r = \pm 3i\) ), inconclusive.
7. For the system of
Exercise 15.5.4 ,
\(x' = y - x^2\) and
\(y' = x - y\text{,}\) classify both equilibria.
Answer .
\((0,0)\) is a saddle (
\(r = \frac{-1 \pm \sqrt{5}}{2}\) );
\((1,1)\) is a stable node (
\(r = \frac{-3 \pm \sqrt{5}}{2}\) ).
8. A damped pendulum of unit length satisfies
\(x' = y\text{,}\) \(y' = -\sin(x) - \tfrac{1}{2}y\text{,}\) where
\(x\) is the angle from straight down and
\(y\) the angular velocity. Find all equilibria and compute
\(J(x,y)\text{.}\)
Answer .
Equilibria
\((n\pi, 0)\text{;}\) \(J(x,y) = \begin{bmatrix} 0 \amp 1 \\ -\cos(x) \amp -\frac{1}{2} \end{bmatrix}\text{.}\)
✍🏻 Problems.
Work through the following problems.
9. Competitive Exclusion.
Two species compete for one resource, in thousands of individuals:
\begin{align*}
\frac{dx}{dt} \amp = x(6 - x - 2y)\\
\frac{dy}{dt} \amp = y(4 - x - y)
\end{align*}
Compute \(J(x,y)\text{.}\)
Classify each equilibrium.
Describe, in a sentence or two, what the classification predicts for the two species.
Answer .
\((0,0)\) unstable node (
\(r = 6, 4\) );
\((6,0)\) stable node (
\(r = -6, -2\) );
\((0,4)\) stable node (
\(r = -2, -4\) );
\((2,2)\) saddle (
\(r = -2 \pm 2\sqrt{2}\) ). One species excludes the other; which one depends on the initial populations.
10. The Damped Pendulum.
Continue with
\(x' = y\text{,}\) \(y' = -\sin(x) - \tfrac{1}{2}y\) from
Exercise 15.5.8 .
Classify the equilibrium at \((0,0)\text{.}\)
Classify the equilibrium at \((\pi,0)\text{.}\)
Explain how each classification matches the physical behavior of a real pendulum.
Answer .
(a) Spiral sink,
\(r = -\frac14 \pm \frac{\sqrt{15}}{4}i\text{.}\) (b) Saddle,
\(r = -\frac14 \pm \frac{\sqrt{17}}{4}\text{.}\) (c) Hanging down is stable with damped oscillation; balanced upright is unstable.
11. What the Linearization Missed.
For a constant \(a\text{,}\) consider
\begin{align*}
\frac{dx}{dt} \amp = -y + a\,x(x^2+y^2)\\
\frac{dy}{dt} \amp = x + a\,y(x^2+y^2)
\end{align*}
Show that \(J(0,0)\) is the same matrix for every value of \(a\text{,}\) and find its eigenvalues.
Let \(\rho = x^2 + y^2\text{.}\) Compute \(d\rho/dt\) and simplify.
Use part (b) to classify the origin when \(a \lt 0\text{,}\) when \(a \gt 0\text{,}\) and when \(a = 0\text{.}\)
What does this show about the reliability of the linearization?
Hint .
For part (b), differentiate
\(\rho = x^2+y^2\) to get
\(d\rho/dt = 2x\,x' + 2y\,y'\text{,}\) then substitute the system.
Answer .
(a)
\(J(0,0) = \begin{bmatrix} 0 \amp -1 \\ 1 \amp 0 \end{bmatrix}\text{,}\) \(r = \pm i\text{,}\) for every
\(a\text{.}\) (b)
\(d\rho/dt = 2a\rho^2\text{.}\) (c) Spiral sink for
\(a \lt 0\text{,}\) spiral source for
\(a \gt 0\text{,}\) true center for
\(a = 0\text{.}\) (d) A purely imaginary pair makes the linearization inconclusive.
12. The Epidemic Threshold.
\begin{align*}
\frac{dS}{dt} \amp = -\frac{\beta S I}{N}\\
\frac{dI}{dt} \amp = \frac{\beta S I}{N} - \gamma I
\end{align*}
with \(\beta \gt 0\) the transmission rate, \(\gamma \gt 0\) the recovery rate, and \(R_0 = \beta/\gamma\text{.}\)
Show that every point with \(I = 0\) is an equilibrium, so the model has a whole line of them.
Compute \(J(S,I)\) and evaluate it at a point \((S,0)\) on that line.
Find the eigenvalues at \((S,0)\) and explain why one of them is zero.
Show that an outbreak grows exactly when \(S \gt N/R_0\text{,}\) and interpret that threshold.
Answer .
(a) Both equations carry a factor
\(I\text{,}\) so
\(I = 0\) makes both vanish for any
\(S\text{.}\) (b)-(c)
\(J(S,0)\) is triangular with eigenvalues
\(r_1 = 0\) and
\(r_2 = \beta S/N - \gamma\text{;}\) the zero comes from the line of equilibria. (d)
\(I' \gt 0\) exactly when
\(S \gt N/R_0\text{;}\) vaccinating a fraction
\(p \gt 1 - 1/R_0\) prevents an outbreak.
13. Stepping Through a Nonlinear System.
No formula exists for the solutions of
\begin{align*}
\frac{dx}{dt} \amp = x - xy\\
\frac{dy}{dt} \amp = xy - 2y
\end{align*}
so we approximate. Starting from
\(x(0) = 3\text{,}\) \(y(0) = 1\text{,}\) take two steps of Euler’s method with
\(h = 0.1\text{,}\) using the update from
Euler’s Method . Then explain how the result compares with the equilibrium you found in
Exercise 15.5.2 .
Answer .
\((x_1,y_1) = (3, 1.1)\) and
\((x_2,y_2) = (2.97, 1.21)\text{.}\) The steps move up and to the left, circling the equilibrium
\((2,1)\) counterclockwise rather than approaching it.
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