Section 15.1 What Makes a System Nonlinear

In First-Order Linear Systems, every system we met could be written as \(\vec{X}' = A\vec{X}\) for one constant matrix \(A\text{.}\) That single line is what made the eigenvalue method work: one matrix, one characteristic equation, one formula covering every solution at once.

Most systems that come out of a real model are not that obliging. The moment two quantities interact through a product — predators eating prey, susceptible people meeting infected ones, two species drawing on the same food supply — the equations pick up a term like \(xy\text{,}\) and the matrix disappears. This section is about recognizing when that has happened, being precise about which tools we lose, and locating the points that organize everything else: the equilibria.

Subsection Recognizing a Nonlinear System

The test is the one we used for a single equation in Classification, now applied to both equations at once.

📙 Definition 316. Nonlinear System.

A first-order system

\begin{align*} \frac{dx}{dt} \amp = f(x,y)\\ \frac{dy}{dt} \amp = g(x,y) \end{align*}

is linear when \(f\) and \(g\) are sums of constant multiples of \(x\) and \(y\) (plus, possibly, a function of \(t\) alone). It is nonlinear when either right-hand side contains a product of unknowns, a power of an unknown, or an unknown inside a function such as \(\sin(x)\text{,}\) \(e^{y}\text{,}\) or \(\ln(x)\text{.}\)

Reading a system this way takes only a few seconds:

System	Verdict	Why
\(x' = 3x - 2y, \quad y' = x + y\)	linear	every term is a constant times \(x\) or \(y\)
\(x' = x - xy, \quad y' = xy - 2y\)	nonlinear	the product \(xy\) appears
\(x' = y, \quad y' = -\sin(x)\)	nonlinear	\(x\) sits inside \(\sin\)
\(x' = x(1 - x), \quad y' = -y\)	nonlinear	\(x(1-x) = x - x^2\) contains \(x^2\)
\(x' = -x + t^2, \quad y' = 4x - y\)	linear	\(t^2\) is a power of the independent variable, which is allowed

That last row is worth pausing on. Nonlinearity is about how the unknowns enter, never about how \(t\) enters. A coefficient may be as complicated a function of \(t\) as you like and the system stays linear.

📝: The Letters Will Change.

Subsection What Nonlinearity Costs Us

Three tools we have leaned on stop working. It is worth being precise about which three, because almost everything else survives.

Superposition is gone. If \(\vec{X}_1\) and \(\vec{X}_2\) both solve a linear system, so does \(C_1\vec{X}_1 + C_2\vec{X}_2\text{.}\) For a nonlinear system that fails: adding two solutions of \(x' = x - xy\) does not produce a third, because the \(xy\) term does not distribute across the sum.
There is no coefficient matrix. Without a matrix \(A\) there is no \(\det(A - rI) = 0\) to solve, so there are no eigenvalues for the system as a whole. Eigenvalues will come back in Linearization and the Jacobian, but attached to a single equilibrium point rather than to the system.
Closed-form solutions are the exception. The systems in this chapter have no formula for \(x(t)\) and \(y(t)\) in terms of elementary functions. That is not a gap in our technique; for most nonlinear systems no such formula exists.

What survives is everything geometric and everything numerical. The phase plane still works. Direction fields still work. Trajectories still cannot cross one another. Euler’s method still marches forward exactly as it did in Euler’s Method. So the plan for this chapter is to lean on the tools that survive:

Find the equilibrium points (this section).
Use nullclines to map the flow between them (Nullclines and the Phase Plane).
Zoom in on each equilibrium, where the system is nearly linear, and classify it with eigenvalues (Linearization and the Jacobian).

The payoff is a complete qualitative picture — where the system settles, what it runs away from, whether it oscillates — assembled without ever solving anything.

Subsection Finding Equilibrium Points

In Qualitative Methods, an equilibrium of \(y' = f(y)\) was a value \(c\) with \(f(c) = 0\text{:}\) start there and you never move. In the plane the idea is identical, except that both derivatives must vanish at once.

📙 Definition 317. Equilibrium Point of a System.

A point \((x^*, y^*)\) is an equilibrium point (also called a critical point) of the system \(x' = f(x,y)\text{,}\) \(y' = g(x,y)\) when

\begin{equation*} f(x^*, y^*) = 0 \quad \text{and} \quad g(x^*, y^*) = 0 . \end{equation*}

The constant functions \(x(t) = x^*\text{,}\) \(y(t) = y^*\) are then a solution of the system, and the point sits in the phase plane as a single motionless dot.

Here is the first real difference from the linear case. A linear system \(\vec{X}' = A\vec{X}\) with \(\det A \neq 0\) has exactly one equilibrium, the origin. A nonlinear system can have several — and where they sit, plus how they compete for nearby trajectories, is the whole story.

Finding them is an algebra problem, not a calculus problem: solve two equations in two unknowns. The reliable technique is to factor each equation and then take every combination of the resulting cases.

🌌 Example 318. Equilibria of a Predator-Prey Model.

Rabbits \(R(t)\) and foxes \(F(t)\) in a meadow are modeled by

\begin{align*} \frac{dR}{dt} \amp = 0.5R - 0.02RF\\ \frac{dF}{dt} \amp = -0.3F + 0.01RF \end{align*}

with \(R\) and \(F\) counted in animals and \(t\) in months. Find every equilibrium point.

Solution.

Factor each right-hand side:

\begin{align*} 0.5R - 0.02RF \amp = R(0.5 - 0.02F)\\ -0.3F + 0.01RF \amp = F(-0.3 + 0.01R) \end{align*}

The first expression is zero when \(R = 0\) or \(F = 0.5/0.02 = 25\text{.}\) The second is zero when \(F = 0\) or \(R = 0.3/0.01 = 30\text{.}\) Both must hold at once, so we test all four pairings:

\(R = 0\) with \(F = 0\text{:}\) the point \((0,0)\text{.}\) Valid.
\(R = 0\) with \(R = 30\text{:}\) contradictory, no point.
\(F = 25\) with \(F = 0\text{:}\) contradictory, no point.
\(F = 25\) with \(R = 30\text{:}\) the point \((30, 25)\text{.}\) Valid.

So there are two equilibria: \((R,F) = (0,0)\) and \((R,F) = (30,25)\text{.}\)

Both mean something. At \((0,0)\) the meadow is empty and stays empty. At \((30,25)\) the two populations balance exactly: rabbits are eaten as fast as they are born, and foxes die as fast as they reproduce, so \(30\) rabbits and \(25\) foxes coexist indefinitely. Whether a slightly-off population returns to that balance is a question we cannot answer yet — that is Linearization and the Jacobian.

Notice what the factoring bought us. Had we instead written \(0.5R = 0.02RF\) and divided through by \(R\text{,}\) we would have lost the equilibrium at the origin — exactly the lost-solutions trap from Separation of Variables. Factor, then case-split. Never divide by an unknown.

🌌 Example 319. Equilibria of Two Competing Species.

Two species share a single food supply. Measured in thousands of individuals,

\begin{align*} \frac{dx}{dt} \amp = x(3 - x - 2y)\\ \frac{dy}{dt} \amp = y(2 - x - y) \end{align*}

Each species would grow logistically on its own; each term carrying the other letter is the cost of competition. Find every equilibrium point.

Solution.

Both equations arrive already factored. The first vanishes when

\begin{equation*} x = 0 \quad \text{or} \quad 3 - x - 2y = 0, \end{equation*}

and the second vanishes when

\begin{equation*} y = 0 \quad \text{or} \quad 2 - x - y = 0 . \end{equation*}

Four pairings again:

\(x = 0\) with \(y = 0\text{:}\) the point \((0,0)\) — neither species present.
\(x = 0\) with \(2 - x - y = 0\text{:}\) gives \(y = 2\text{,}\) the point \((0,2)\) — only species \(y\text{.}\)
\(3 - x - 2y = 0\) with \(y = 0\text{:}\) gives \(x = 3\text{,}\) the point \((3,0)\) — only species \(x\text{.}\)
\(3 - x - 2y = 0\) with \(2 - x - y = 0\text{:}\) subtracting the second from the first gives \(1 - y = 0\text{,}\) so \(y = 1\text{,}\) and then \(x = 2 - y = 1\text{.}\) The point \((1,1)\) — coexistence.

Four equilibria: \((0,0)\text{,}\) \((3,0)\text{,}\) \((0,2)\text{,}\) and \((1,1)\text{.}\)

Four equilibria in one small model — and no linear system of two equations can produce more than one. That difference is why nonlinear models describe outcomes linear ones cannot: an ecosystem that could settle into either of two very different states, a disease that either dies out or explodes, a switch that stays wherever you last flipped it.

📝: When the Algebra Fights Back.

Checkpoint 320. 🤔💭 What Makes a System Nonlinear Reading Questions.

(a) 📖❓ Spot the Nonlinear Systems.

Select every system below that is nonlinear.

\(x' = 2x - xy, \ y' = y + x\)
Yes — \(xy\) is a product of two unknowns.
\(x' = y, \ y' = -\sin(x) - y\)
Yes — the unknown \(x\) sits inside \(\sin\text{.}\)
\(x' = -3x + 7y, \ y' = x - y\)
Every term is a constant times an unknown, so this system is linear.
\(x' = x + e^{t}, \ y' = 2x - 5y\)
The \(e^t\) involves the independent variable, which is allowed. This system is linear.
\(x' = x(4 - x), \ y' = xy\)
Yes — expanding the first gives \(4x - x^2\text{,}\) and the second is a product.

(b) 📖❓ What Does an Equilibrium Require?

A point \((x^*, y^*)\) is an equilibrium point of \(x' = f(x,y)\text{,}\) \(y' = g(x,y)\) exactly when

\(f(x^*,y^*) = 0\) and \(g(x^*,y^*) = 0\text{.}\)
Correct — both derivatives must vanish, or the point would still be moving.
\(f(x^*,y^*) = 0\) or \(g(x^*,y^*) = 0\text{.}\)
If only one is zero, one variable is still changing and the point drifts.
\(f(x^*,y^*) = g(x^*,y^*)\text{.}\)
Equal derivatives are not zero derivatives — both could equal \(7\text{.}\)
\(x^* = y^* = 0\text{.}\)
That is the only option for most linear systems. Nonlinear equilibria can sit anywhere.

(c) 📖❓ Count the Equilibria.

How many equilibrium points does \(x' = x(2-y)\text{,}\) \(y' = y(x-1)\) have?

Two: \((0,0)\) and \((1,2)\text{.}\)
Correct. The first equation needs \(x=0\) or \(y=2\text{;}\) the second needs \(y=0\) or \(x=1\text{.}\) Only \((0,0)\) and \((1,2)\) satisfy both.
One: only \((0,0)\text{.}\)
Check the pairing \(y = 2\) with \(x = 1\) — it works too.
Four, one for each factor.
All four pairings must be tested, but two of them are contradictory: \(x=0\) with \(x=1\text{,}\) and \(y=2\) with \(y=0\text{.}\)
Infinitely many, along the lines \(x=0\) and \(y=0\text{.}\)
A point on the \(x\)-axis away from the origin still has \(x' = x(2-0) = 2x \neq 0\text{.}\)

(d) 📖❓ Which Tools Survive?

Select every statement that is true about a nonlinear system.

The phase plane and direction fields still describe its motion.
Euler’s method still approximates its solutions step by step.
It may have more than one equilibrium point.
Adding two solutions produces another solution.
Superposition is exactly what nonlinearity destroys.
Its behavior is set by the eigenvalues of one coefficient matrix.
There is no single coefficient matrix. Eigenvalues return in the next sections, but one equilibrium at a time.

(e) 📖❓ Why Factor Instead of Divide?

To find the equilibria of \(x' = 4x - x^2\text{,}\) why should you factor to \(x(4-x) = 0\) rather than write \(4x = x^2\) and divide both sides by \(x\text{?}\)

Dividing by \(x\) assumes \(x \neq 0\) and throws away the equilibrium at \(x = 0\text{.}\)
Exactly — the same lost-solutions trap you met with separation of variables.
Dividing produces the wrong value for the other equilibrium.
Dividing does give \(x = 4\) correctly. The problem is what it silently discards.
Factoring is required whenever a system is nonlinear.
Factoring is a reliable habit, not a rule — plenty of equilibria are found other ways.

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