Logistic growth and phase lines
For an equation , the steady states are the zeros of , and the sign of between them says which way every solution moves. Drawn as arrows on the -axis, this state-space diagram (phase line) tells you what a population does in the long run without solving anything. The logistic equation is the standard case: growth that slows as the population nears its carrying capacity.
In this topic
- Taught
- Before the final, flavour B
- Questions
- 11
- Total marks
- 32
- Suggested time
- 57 min
- Difficulty
Introductory to Capstone
Show the 3 hardest
Key results
- Steady states
For an equation whose right side depends on alone.
- Stability
For a steady state of , with the signs of taken on intervals just below and just above .
- The logistic equation
For constants and : the steady states are 0 (unstable) and (stable), and grows fastest at , at the rate .
- Solution of the logistic equation
For and .
- Harvesting
For constants , and ; is the maximum sustainable yield.
Read the theoryUBC Math 100 §12.1 Linear and nonlinear differential equationsUBC Math 100 §13.1 State-space diagramsUBC Math 100 §13.2 Logistic growth
Worked example
A fish stock of tonnes grows logistically and is harvested at a constant 32 tonnes per year, so that the equation below holds. Find the steady states. If the stock is 150 tonnes when harvesting begins, what does it approach in the long run?
Give the long-run stock in tonnes.
Strategy
Find the steady states as the zeros of the right-hand side, then use its sign between them to see which way a stock of 150 tonnes moves and where it stops.
Working
Multiply through by −1250 to clear the decimals.
Harvesting moves the steady states of the unfished stock, 0 and 500, to 100 and 400.
Negative below 100, positive between 100 and 400, negative above 400: 400 is stable and 100 is unstable.
The stock starts between the steady states, so it rises and approaches 400.
Answer
The answer is steady states at 100 and 400 tonnes; the stock approaches 400 tonnes.
Check
At 400 tonnes growth 0.4 · 400 · 0.2 = 32 a year replaces the catch, and the stock gains 10 tonnes a year at 350 but loses 14 at 450.
Where this goes wrong
Choosing the nearer steady state, 100. It is unstable: just above 100 the stock grows, so from 150 tonnes it climbs away from 100 to the stable steady state at 400.
On the paper
Find the equilibria by setting the rate to zero, draw the phase line with the sign of the rate between them, and classify each equilibrium from the arrows.
Leans on
Solving a quadratic equation, from Functions, graphs and inverses
Your turn
A fishery works with a fixed effort , so its catch is proportional to the stock: tonnes per year. The stock , in tonnes, then satisfies the equation below, with . Find the positive steady state in terms of , and the catch taken each year once the stock is there. Which effort makes that sustained catch as large as possible?
Give the effort E, per year.
Practice questions
5 of the 11, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.
- Introductory1 mark · 2 min
How many steady states does this differential equation have?
Answer this questionStrategy
The steady states are the zeros of the right-hand side. Factor it fully and count the distinct values of y that make it zero.
Working
Factor out y² rather than dividing by it, so the root y = 0 is kept.
A product is zero exactly when one of its factors is.
The squared factor still gives only one value of y, so there are three.
Answer
The answer is three steady states, at y = −2, 0 and 2.
Check
As a quadratic in y², y⁴ − 4y² is zero only for y² = 0 or y² = 4, and those give the three values 0, 2 and −2.
Where this goes wrong
Dividing both sides by y² to simplify, which throws away the steady state y = 0. Factor instead, and count each distinct zero of the right-hand side once.
Leans on
Factoring a polynomial, from Functions, graphs and inverses
- Introductory2 marks · 3 min
The differential equation below has steady states and . Which statement describes their stability?
Answer this questionStrategy
The sign of the right-hand side between the steady states says which way y moves there. A steady state is stable when the arrows on both sides point at it.
Working
The sign can change only at a steady state, so one test value per interval is enough.
Arrows point right where y increases and left where it decreases: both point at 1, and both point away from 4.
Answer
The answer is y = 1 is stable and y = 4 is unstable.
Check
The derivative of (y − 1)(y − 4) with respect to y is 2y − 5: −3 at y = 1, stable, and 3 at y = 4, unstable.
Where this goes wrong
Assuming the larger steady state is stable, as it is for the logistic equation. Here dy/dt > 0 above 4, so a solution starting just above 4 grows away from it: read the arrows from the sign of the right-hand side every time.
Leans on
Signs of a factored function, from Curve sketching
- Standard2 marks · 4 min
A patient receives a drug through an intravenous drip at 10 mg per hour. The liver removes 4 % of the drug in the body each hour, and the kidneys remove a further 1 % each hour. Write a differential equation for the amount of drug in the body, in mg, and find the amount it approaches in the long run.
Give the amount in milligrams.
Answer this questionStrategy
Input minus removal gives an equation of the form dA/dt = a − bA. Its single steady state, a/b, is stable, so every solution ends up there.
Working
Each organ removes a fixed fraction of what is present, so the two removal terms add.
At the steady state the drip exactly replaces what is removed.
Both arrows point at 200, so it is stable and A(t) approaches it from any starting amount.
Answer
The answer is dA/dt = 10 − 0.05A, and the amount approaches 200 mg.
Check
At 200 mg the liver removes 8 mg/h and the kidneys 2 mg/h, together exactly the 10 mg/h dripping in.
Where this goes wrong
Using one removal route only, which gives 10/0.04 = 250 mg (liver) or 10/0.01 = 1000 mg (kidneys). Both organs clear the drug at once, so the fractions add and the removal rate is 0.05A.
Leans on
Equations of the form y′ = a − by, from Exponential growth and decay
- Challenging4 marks · 7 min
Left alone, a fish stock of tonnes would grow logistically with per year and carrying capacity 4000 tonnes. A fishery removes a constant tonnes per year, so the stock satisfies the equation below. Find the maximum sustainable yield: the largest for which the stock has a steady state. For any larger the stock collapses, whatever its starting size.
Give h in tonnes per year.
Answer this questionStrategy
A steady state is a stock whose natural growth exactly replaces the catch. That is possible only while the catch is no more than the largest growth the stock can produce, so the critical harvest is the maximum of the growth function.
Working
At a steady state the catch equals the natural growth.
The peak of the parabola is at half the carrying capacity.
The largest natural growth the stock can produce: rK/4.
Above 300 the stock falls whatever its size, so it collapses.
Answer
The answer is 300 tonnes per year, taken from a stock held at 2000 tonnes.
Check
The steady-state equation 0.000075N² − 0.3N + h = 0 has discriminant 0.09 − 0.0003h, which is negative exactly when h > 300.
Where this goes wrong
Giving 2000, the stock at which the yield is largest, instead of the yield. The catch that stock can sustain is its natural growth there, G(2000) = 300 tonnes per year.
Leans on
Finding a maximum with the derivative, from Optimization
- Capstone6 marks · 12 min
A protein switches on its own production. Its concentration satisfies the equation below, where the first term is production and is decay.
(a) Find every steady state. (b) Draw the state-space diagram (phase line) for c ≥ 0 and classify each steady state as stable or unstable. (c) Describe what happens in the long run if c(0) = 0.5, and if c(0) = 3.
Answer this questionStrategy
Combine the right side into one fraction and factor it. Its zeros are the steady states, and its sign between them gives the arrows on the state-space diagram, which settle both stability and the long run from each start.
Working
One factored fraction shows every zero and every sign change.
The denominator 9 + c² is never zero, so these are all the steady states.
The sign can change only at a steady state, so one test value per interval fixes it.
The arrows point at 0 from above and at 9 from both sides, and away from 1 on both sides.
Below the threshold 1 the protein dies away; above it, production wins and the switch turns on.
Answer
The answer is steady states 0 and 9 are stable and 1 is unstable; from 0.5 the concentration decays to 0, and from 3 it rises to 9.
What earns the marks
- Sets the right-hand side to zero without dividing by c, and finds c = 0 together with the roots of c² − 10c + 9 = 0: c = 1 and c = 9.
- Writes the right side as −c(c − 1)(c − 9)/(9 + c²), or tests one value in each interval, and finds dc/dt < 0 on (0, 1), dc/dt > 0 on (1, 9) and dc/dt < 0 for c > 9.
- Draws arrows on the c-axis from those signs, and classifies c = 0 and c = 9 as stable and c = 1 as unstable.
- Concludes that from c(0) = 0.5 the concentration decreases towards 0, and from c(0) = 3 it increases towards 9, neither crossing a steady state.
Check
Production 10c²/(9 + c²) never reaches 10, so decay c wins for every c ≥ 10, and at c = 0.5 production is only about 0.27, below the decay 0.5.
Where this goes wrong
Dividing 10c²/(9 + c²) = c by c and losing the steady state c = 0, or assuming every solution ends at the largest steady state. The unstable steady state c = 1 is a threshold: below it the protein dies away, above it the concentration rises to 9.
Leans on
Signs of a factored function, from Curve sketching
Related topics
Needs
Slope fields and Euler’s method
Checking that a function solves a differential equation, reading a slope field, and stepping a solution forward with Euler’s method.
14 questions