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Autonomous equations and the phase line

Equilibria, the sign of $f(y)$, and the stability of each — the long-run behaviour of every solution, read off one line without integrating.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to find the equilibria of an autonomous equation, draw its phase line from the sign of the right-hand side, classify each equilibrium as stable, unstable or semi-stable, say what any given initial value tends to, and name the set of starting values that share a limit.

2. What you already have

You can solve a separable equation, and every autonomous equation is separable. This lesson is about not doing that. The integral is often unpleasant and the questions actually asked of a model — where does this settle, what happens if it is disturbed — are answered from the sign of the right-hand side in a few lines.

3. Words this lesson uses

4. A whole family of solutions, read off one line

For $y' = f(y)$ the slope depends only on the height. Two consequences follow immediately, and they are the whole lesson.

Every root of $f$ is a constant solution. If $f(c) = 0$ then $y \equiv c$ has $y' = 0 = f(c)$, so it solves the equation exactly.

Between roots, the sign of $f$ never changes, so a solution there is monotonic — rising where $f > 0$, falling where $f < 0$ — and it cannot cross an equilibrium, because uniqueness forbids two solutions through one point and the constant solution is already there.

Put together: a solution starting between two equilibria rises or falls monotonically and approaches the one it is heading for, without ever reaching it in finite time. So the long-run behaviour of every solution is settled by the signs of $f$ on a handful of intervals.

Draw the $y$ axis vertically, mark the roots of $f$, and put an arrow up or down in each gap. That picture is the phase line, and reading stability off it is immediate: arrows pointing in from both sides means stable, arrows pointing away means unstable, one of each means semi-stable.

There is also a test that needs no picture. Near a simple root, $f'(c) < 0$ means stable and $f'(c) > 0$ means unstable — because $f'$ is the slope of $f$, and a negative slope is exactly $f$ crossing from positive to negative.

5. The logistic equation, read rather than solved

$P' = rP\left(1 - \dfrac{P}{K}\right)$ with $r, K > 0$ is the standard model of growth against a limit, and it is separable — the integral needs partial fractions and a page of algebra.

The phase line takes four lines. The roots are $P = 0$ and $P = K$. Between them the product is positive, so populations rise; above $K$ it is negative, so they fall. Therefore $0$ is unstable, $K$ is stable, and every positive starting population tends to $K$.

That is the answer the modeller wanted. The explicit solution adds one thing the phase line cannot give — how long it takes — and if nobody asked, it was a page of algebra spent on a question nobody asked.

The number $K$ even has a name from this reading: the carrying capacity is not a parameter someone chose to call that, it is the stable equilibrium of the model.

6. Where this goes wrong

Testing stability at the equilibrium. $f$ is zero at every equilibrium, stable or not. The information is in the sign on each side.

Expecting a solution to arrive. A solution approaching a stable equilibrium takes for ever to get there: reaching it in finite time would put two solutions through one point.

Forgetting a repeated root. $f(y) = y^{2}$ has a double root at zero, and $f$ is positive on both sides — the equilibrium is semi-stable, approached from below and left above. The derivative test is silent here, because $f'(0) = 0$, and only the signs settle it.

Using the phase line on a non-autonomous equation. If $x$ appears on the right, the arrows change with time and the whole picture is void.

7. The method, step by step, and how to check it

An autonomous equation $y' = f(y)$ has a right-hand side that does not mention the independent variable. That single fact is what makes the phase line work: the direction of travel depends only on where you are, never on when.

1. Confirm the equation is autonomous. If $t$ or $x$ appears on the right, stop; the arrows would change with time.
2. Find the equilibria by solving $f(y) = 0$. Each root is a constant solution.
3. Mark them on a vertical line, in order. They cut it into intervals.
4. Sample $f$ once in each interval. The sign is the direction: positive means up, negative means down.
5. Classify each equilibrium from the arrows on its two sides: both in is stable, both out is unstable, one of each is semi-stable.
6. Answer the question. A start inside an interval moves in that interval's direction and approaches the equilibrium it is heading for, or runs off to infinity if there is none.

Why one sample per interval is enough. $f$ is continuous, so it can change sign only by passing through zero, and every zero has already been marked. Why solutions cannot cross an equilibrium. The equilibrium is itself a solution, and the uniqueness theorem forbids two solutions through one point; the same argument says a solution never actually reaches a stable equilibrium in finite time.

The derivative test is a shortcut for step 5: at an equilibrium $y^{}$, $f'(y^{}) < 0$ means stable and $f'(y^{}) > 0$ means unstable, and $|f'(y^{})|$ is the rate at which nearby solutions approach or leave. When $f'(y^{*}) = 0$ it says nothing, and the signs decide.

How to check the answer. Check that the arrows alternate sensibly: between two simple roots the sign always flips, so two neighbouring stable equilibria are impossible without an unstable one between them. When an explicit solution is available, check that it starts at the given value and tends to the equilibrium the phase line predicted.

8. In the world: a skydiver's terminal velocity

A skydiver falling with speed $v$ is pulled down by gravity and held back by air resistance that grows like the square of the speed: $v' = g - kv^{2}$, an autonomous equation. Its phase line answers the question every skydiver cares about without solving anything.

The right-hand side is zero at $v = \sqrt{g/k}$. Below that speed $g > kv^{2}$ and $v$ increases; above it $v$ decreases. So $v^{} = \sqrt{g/k}$ is a stable equilibrium, the terminal velocity, and every fall, from a standing start or from a plane moving faster, approaches it. With $g = 9.8$ and $k = 0.0035$ per metre for a spread-eagled body, $v^{} = \sqrt{9.8/0.0035} = \sqrt{2800} \approx 53$ metres a second, about $190$ kilometres an hour. Diving head first reduces the drag coefficient to about $0.0011$, and $v^{*} \approx 94$ metres a second.

The derivative test gives how quickly the speed settles: $f'(v) = -2kv$, so at $v^{}$ the rate is $-2kv^{} = -2 \times 0.0035 \times 53 \approx -0.37$ per second, and the gap from terminal speed shrinks by a factor of $e$ every $2.7$ seconds. A skydiver is within a few per cent of terminal velocity about ten seconds after leaving the aircraft, which is what jumpers report.

9. In the world: when an epidemic takes hold

A simple model of an infection that people can catch more than once tracks the number infected, $I$, in a population of $N$: new infections arrive at a rate $\beta I\left(1 - \frac{I}{N}\right)$ and people recover at a rate $\gamma I$. So $I' = (\beta - \gamma)I - \frac{\beta}{N}I^{2}$, which is autonomous.

Its equilibria are $I = 0$ and $I^{} = N\left(1 - \frac{\gamma}{\beta}\right)$. When $\beta > \gamma$ the second is positive and stable, and $I = 0$ is unstable: a single case grows until the infection is endemic at $I^{}$. When $\beta < \gamma$ the only equilibrium in range is $I = 0$, now stable, and the infection dies out. The threshold is the ratio $R_0 = \frac{\beta}{\gamma}$, the average number of people one case infects, and whether it exceeds $1$ is the phase line's verdict.

With $\beta = 0.3$ and $\gamma = 0.1$ per day in a town of $10\,000$, $R_0 = 3$ and $I^{} = 10\,000 \times \frac{2}{3} \approx 6667$. A vaccination campaign that halves contacts lowers $\beta$ to $0.15$, $R_0$ to $1.5$, and $I^{}$ to $3333$; one that lowers $\beta$ below $0.1$ moves the stable equilibrium to zero. Public health targets are set by exactly this sign change.

10. In the world: a harvest the stock can bear

Fisheries use the same line with a harvest term, $P' = rP\left(1 - \frac{P}{K}\right) - H$. Harvesting lowers the parabola, the two equilibria move together, and when the catch reaches $\frac{rK}{4}$ they merge into one semi-stable point and then vanish. The largest sustainable catch is therefore $\frac{rK}{4}$, taken from a stock held at $\frac{K}{2}$, and a catch above it collapses the fishery whatever the stock's size. The collapse of the Newfoundland cod fishery in 1992 is the standard warning of what happens past that point.

11. In the world: a chemical reaction settling

A reversible reaction that turns a substance into a product and back settles at a concentration set by the balance of the two rates: with $x$ the product, $x' = k_1(a - x) - k_2x$, an autonomous equation whose one equilibrium, $x^{*} = \frac{k_1a}{k_1 + k_2}$, is stable because $f'(x) = -(k_1 + k_2) < 0$. Chemists call it the equilibrium concentration, and the derivative gives how fast it is approached: the gap shrinks like $e^{-(k_1 + k_2)t}$.

12. Stability is about the neighbours, not about the point

The instinct is to look for something special at the equilibrium itself — a value, a curvature, a sign — and there is nothing there: $f$ is zero at all of them, which is what makes them equilibria. Stability is entirely a statement about the behaviour nearby, and the only way to get it is to sample on both sides or to differentiate, which is the same thing done symbolically. The second habit worth correcting is reading an arrow as a speed. An arrow says only up or down; the solution slows as it nears an equilibrium, because $f$ is heading for zero, which is exactly why it approaches without arriving.

13. Three equilibria on one line

1. For $y' = y(y - 2)(y - 5)$, find the roots of the right-hand side.

   $y(y - 2)(y - 5) = 0 \quad\Rightarrow\quad y = 0,\ 2,\ 5$

   Roots first: they cut the line into four intervals, and each root is a constant solution.

2. Take one sample in each of the two lower intervals.

   $y = -1: (-1)(-3)(-6) = -18 < 0 \ \downarrow; \qquad y = 1: (1)(-1)(-4) = 4 > 0 \ \uparrow$

   The right side cannot change sign inside an interval without a root, so one sample decides it.

3. Take one sample in each of the two upper intervals.

   $y = 3: (3)(1)(-2) = -6 < 0 \ \downarrow; \qquad y = 6: (6)(4)(1) = 24 > 0 \ \uparrow$

   Count the negative factors: an odd number gives a negative product.

4. Read stability from the arrows either side of each root.

   $0: \text{away} \Rightarrow \text{unstable}; \quad 2: \text{in} \Rightarrow \text{stable}; \quad 5: \text{away} \Rightarrow \text{unstable}$

   Arrows pointing in on both sides mean nearby solutions return.

5. Answer every initial condition at once.

   $y(0) < 0: \ y \to -\infty; \quad 0 < y(0) < 5: \ y \to 2; \quad y(0) > 5: \ y \to \infty$

   Solutions cannot cross an equilibrium, so each stays in its interval and follows its arrow.

14. A semi-stable equilibrium

1. For $y' = (y - 3)^{2}$, find the root.

   $(y - 3)^{2} = 0 \quad\Rightarrow\quad y = 3 \text{ (repeated)}$

   A repeated root is the case to watch.

2. Sample below the root, at $y = 2$.

   $(2 - 3)^{2} = 1 > 0 \ \uparrow$

   Below $3$, solutions rise towards it.

3. Sample above the root, at $y = 4$.

   $(4 - 3)^{2} = 1 > 0 \ \uparrow$

   A square is positive on both sides, so both arrows point the same way: above $3$, solutions rise away.

4. Try the derivative test.

   $f'(y) = 2(y - 3), \qquad f'(3) = 0$

   The test decides only when $f'$ is non-zero at the equilibrium, so here it is silent.

5. Classify from the picture.

   $3: \ \text{in from below, away above} \Rightarrow \text{semi-stable}$

   The picture fills the test's gap: a start at $2.9$ creeps up to $3$, while a start at $3.1$ leaves.

15. Stability by the derivative test

1. For $y' = y(4 - y)$, find the equilibria.

   $y(4 - y) = 0 \quad\Rightarrow\quad y = 0 \ \text{or}\ y = 4$

   A product is zero when a factor is.

2. Differentiate the right-hand side with respect to $y$.

   $f(y) = 4y - y^{2} \quad\Rightarrow\quad f'(y) = 4 - 2y$

   Expand first, then use the power rule.

3. Evaluate $f'$ at the lower equilibrium.

   $f'(0) = 4 > 0 \quad\Rightarrow\quad \text{unstable}$

   Near $y = 0$, $f(y) \approx f'(0)y = 4y$, so a small positive $y$ grows like $e^{4t}$.

4. Evaluate $f'$ at the upper equilibrium.

   $f'(4) = 4 - 8 = -4 < 0 \quad\Rightarrow\quad \text{stable}$

   Near $y = 4$, the gap $u = y - 4$ obeys $u' \approx -4u$, so it shrinks like $e^{-4t}$.

5. Confirm with a sample on each side of $4$.

   $y = 3: \ 3 \cdot 1 = 3 > 0 \ \uparrow; \qquad y = 5: \ 5 \cdot (-1) = -5 < 0 \ \downarrow$

   Both arrows point towards $4$, agreeing with the test.

6. Read what the size of $f'$ adds.

   $|f'(4)| = 4: \quad \text{the gap from } 4 \text{ falls by a factor of } e \text{ every } \tfrac{1}{4} \text{ unit of time}$

   The sign gives stability; the size gives the speed of return, which the phase line alone cannot.

16. Your turn: classify the equilibria of $y' = 4y - y^{3}$

1. Factor the right-hand side and find the equilibria.

   $4y - y^{3} = y(4 - y^{2}) = y(2 - y)(2 + y) = 0 \quad\Rightarrow\quad y = -2,\ 0,\ 2$

   Factor before sampling.

2. Your turn: work this step out. Its working is at the end of the packet.

   Sample one value in each of the four intervals.

3. Your turn: work this step out. Its working is at the end of the packet.

   Read stability from the arrows.

17. Guided practice

Put the steps of drawing the phase line of $y' = (y - 3)(5 - y)$ into the order they must be done.

Number the steps in order (write the number in the box):

• Solve $f(y) = 0$ for the equilibria
• Mark them on the $y$ axis
• Evaluate $f$ at one sample point per interval
• Draw an arrow in each interval
• Read stability off the arrows

18. Guided practice

Complete the worked solution: draw the phase line of $y' = (y - 1)(7 - y)$.

1. Set the right-hand side to zero to find the equilibria.

   $(y - 1)(7 - y) = 0 \quad\Rightarrow\quad y = 1 \ \text{or}\ y = 7$

   A product is zero exactly when one of its factors is.

2. Substitute a sample below the lower equilibrium, $y = 0$.

   $(0 - 1)(7 - 0) = (-1)(7) =$ _p

   Subtract inside each bracket, then multiply; the sign gives the arrow.

3. Substitute the midpoint, $y = 4$.

   $(4 - 1)(7 - 4) = 3 \cdot 3 =$ _q

   Between the equilibria both brackets are positive.

4. Substitute a sample above the upper equilibrium, $y = 8$.

   $(8 - 1)(7 - 8) = 7 \cdot (-1) =$ _r

   Above the upper equilibrium the second bracket is negative.

5. Draw the arrows from the signs and read stability.

   $\downarrow \ 1 \ \uparrow \ 7 \ \downarrow$

   The upper equilibrium has arrows pointing in (stable); the lower has arrows pointing away (unstable).

19. Guided practice

For $y' = (y - 3)(y - 7)$, is the equilibrium at $y = 3$ stable?

• Stable — nearby solutions come back to it
• Unstable — nearby solutions move away from it

20. Practice

For $y' = (y - 1)(3 - y)$, fill in the value of the right-hand side at each of these four values of $y$.

21. Practice

A quantity obeys $y' = (y - 2)(8 - y)$ and starts at $y(0) = 5$. What does $y$ approach as time goes on?

Answer:

22. Practice

A population of fish, in thousands, is introduced into a lake that can support at most $4$ thousand; it grows logistically, with $t$ in years. Solve $y' = y\left(1 - \dfrac{y}{4}\right)$ with $y(0) = 1$. Write $y$ as a formula in $t$ (type the exponential as e^(...)).

Answer:

23. Somewhere new

For $y' = (y - 2)(4 - y)$, which starting values $y(0)$ lead to $y \to 4$?

This task has no paper form; do it on a device.

24. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

25. Test question

A population of fish, in thousands, is introduced into a lake that can support at most $3$ thousand; it grows logistically, with $t$ in years. Solve $y' = y\left(1 - \dfrac{y}{3}\right)$ with $y(0) = 1$. Write $y$ as a formula in $t$ (type the exponential as e^(...)).

Answer:

26. What you can do now

You can draw a phase line, classify every equilibrium on it, and read the long-run behaviour of any initial value off the picture. Say in your own words why a solution never crosses an equilibrium.

Working for the steps left to you

16. Your turn: classify the equilibria of $y' = 4y - y^{3}$, step 2

16. Your turn: classify the equilibria of $y' = 4y - y^{3}$, step 3