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A B I L E N E C H R I S T I A N U N I V E R S I T Y

Department of Mathematics

Autonomous Equations and StabilitySections 2.4-2.5

Dr. John EhrkeDepartment of Mathematics

Fall 2012

A B I L E N E C H R I S T I A N U N I V E R S I T Y D E P A R T M E N T O F M A T H E M A T I C S

Autonomous Differential EquationsIn this lecture we will consider a special type of differential equation calledan autonomous differential equation. Autonomous differential equations arecharacterized by their lack of dependence on the independent variable. Thegeneral form of a first order autonomous equation is given by

dydt

= f (y). (1)

Keep in mind that y is still a function of an independent variable t, but itdoes not appear explicitly in the forcing term, f . These equations aresometimes called time independent. These equations are separable and sonormally are easily solved depending on the nature of f (y). Rather thandwell on solution techniques we will be concerned with how to get usefulinformation out of the ODE without actually solving it. In this section wewill investigate:

• Qualitative techniques for describing solutions of autonomous ODEs.• Models with harvesting, parameter studies.• Bifurcation analysis.

Slide 2/14 — Dr. John Ehrke — Lecture 5 — Fall 2012


Characteristics of Autonomous Slope FieldsThe nature of autonomous equationsmakes spotting constant solutions andinterpreting the general behavior ofsolutions fairly straightforward. A typicalslope field for an autonomous differentialequation is given below. What can weobserve?

Observations:1. Every horizontal line, y = y0 in the

direction field is an isocline withslope f (y0).

2. The integral curves (solutions) areinvariant under translation. (i.e. Ahorizontal shift of a solution isanother solution.)

3. The critical points of the ODE are thesolutions to the equation f (y) = 0where f (y) is the right hand side ofthe autonomous ODE. Saying that y0

is a critical point is akin to sayingf (y0) = 0, but notice this meansy = y0 is a solution and hence ahorizontal solution (barrier) throughwhich other solutions cannot pass.



Population Dynamic EquationsIn the next few examples, we will be layering a few basic principles of populationgrowth onto the uninhibited growth model,

dydt

= ry (2)

to obtain a more realistic model. In the case of equation, (2), r is called the growthrate if r > 0 and the decay rate if r < 0, and has solution y(t) = y0ert where y0 is theinitial size of the population. While we understand the dynamics of this model wellfrom pre-calculus and calculus, it paints a very limited picture of populationbehavior since it predicts uninhibited exponential growth. We will make this a littlemore interesting in the coming examples.

ExampleLet’s assume a constant amount of fish, ω, are pulled from a fresh fish farm withuninhibited growth rate r, and are modeled by the equation

dydt

= ry − ω (3)

Find the critical points and analyze the behavior of solutions as a function of ω.



Phase Lines and EquilibriaObservations:

• The figures to the left show the phaseportrait (above) and slope field (below)associated with the modeldy/dt = ry− ω. The value y0 = ω/rcorresponds to the zero of the right handside of the equation. This point is aconstant solution to the differentialequation, often called an equilibriumsolution.

• Looking at the phase portrait above thegreen arrow together with critical pointω/r denote the increasing/decreasingnature of the solutions, y(t). The arrowpointing left of ω/r indicates solutionswhich begin below ω/r decrease awayfrom ω/r while solutions that begin aboveω/r grow without bound.

• The y-axis of the phase portrait is oftencalled the phase line and is usuallyoriented vertically with the key pointslabeled.



Phase Analysis Example

ExampleGiven the first order autonomous equation dy/dt = y3 − y perform thefollowing quantitative analysis:

1 Find the critical points (if any exist) and sketch the phase line.2 Label each critical point according to its stability type.3 Sketch the direction field and solutions corresponding to each of the

following initial points: y(0) = 0, y(0) = 1/2, y(0) = −1/2, y(0) = 2, andy(0) = −2.

Solution: The critical points are the solutions of the equation y3 − y = 0which are y = 0,±1. The critical point at y = 0 is a stable sink, while thecritical points at y = ±1 are unstable sources.



Logistic EquationWhile the previous example is interesting it does not reflect the true conditionsunder which populations grow/decay. Limitations including spatial requirements,food/water, and other resources can effect the rate at which populations grow.Factoring this into our analysis requires we replace the constant r by a function h(y)and examine the autonomous equation

dydt

= h(y)y (4)

choosing h(y) in such a way that h(y) ≈ r when y is small (i.e. grows exponentiallyfor a while, but h(y) decreases as y grows larger, and eventually h(y) < 0 when thepopulation is too large to sustain.

Remarks:

• The simplest function that has all the necessary properties is h(y) = a − bywhere a, b ≥ 0.

• Under these conditions, equation (4) becomes the logistic equation

dydt

= (r − ay)y (5)

• Note, the logistic equation has two critical points (and hence two equilibriumsolutions) y = r/a and y = 0.



Solution of the logistic equation

ExampleThe logistic equation is often written in the form

dydt

= r(

1 − yk

)y (6)

where k = r/a. This form is used to make the analytic solution more intuitive. Usingthis form, find the analytic solution of the logistic equation, and verify the solutionbehavior qualitatively.

Solution:The analytic solution to the logistic equation can be found in detail on pg. 82 of thetext.

• The logistic equation has two critical points y = 0 and y = k. The y = 0equilibrium corresponds to population extinction. The y = k equilibriumdenotes the population saturation level, often called the carrying capacity of thepopulation’s environment.

• Above the saturation level, dy/dt < 0 and so populations decrease becausetheir environment cannot sustain them. Similarly, below the saturation level,dy/dt > 0 and so populations have the required resources to thrive.



Critical Points and StabilityAs we saw in the previous example, there are different types of solutionbehaviors depending on the nature of the critical points. There are threetypes of critical point behavior possible.

Stable Critical Point: The solution y = k in the previous example was asolution that all other solutions wanted to be near. Suchcritical point solutions are called stable solutions.

Unstable Critical Point: On the other hand, the solution y = 0 is thesolution which in some sense repels the other solutions. Suchcritical point solutions are called unstable.

Semistable Critical Point: Functions, f (y) which have critical points withmultiplicity often exhibit critical point behavior wheresolutions on one side are attracted and solutions on the otherside are repelled. These critical points are called semistable.

In the examples which follow, we will look at how a change in parameterscan effect critical point behavior.



Logistic model with harvestingIn this example, we will consider the logistic model that includesharvesting. Think about a fresh fish farm in which a constant amount of fishare pulled out for processing. The ODE under this case looks like

dydt

= ay− by2 − h (7)

where h is the amount of fish harvested. If h = 0, this is the model exploredin the previous example. Analyze the behavior of solutions depending onthe value of the parameter. Assume h > 0.

Solution:• hm = a/2b is the maximum harvesting rate. Harvesting rates higher

than this quantity lead to solutions which are everywhere decreasingand in the context of a population would lead to extinction.

• Harvesting rates above hm exhibit two critical points, one stable andone unstable, while harvesting rates less than hm have no critical points.The point hm is the point at which this change occurs, and is a specialtype of point called a bifurcation.



Autonomous Equations and Parameters

Definition (Bifurcation)A phase line is a single line containing the information about theequilibrium points and the nature of solutions on each side of them. When aseries of phase lines are plotted on a graph of the parameter vs. thedependent variable (i.e. µ vs. y in this example) we obtain a bifurcationplot. A bifurcation occurs for a value of the parameter which changes thebasic nature of solutions.

ExampleConsider the first order autonomous ODE,

dydt

= y(1− y)2 + µ. (8)

Analyze the equilibria and behavior of solutions. Include a description ofthe bifurcation points and bifurcation diagram.



Analysis of ParametersWe consider the first order autonomous ODE dy/dt = y(1− y)2 + µ. To findthe points of bifurcation, we look for the critical points of

f (y) = y(1− y)2 + µ. (9)

Taking derivatives, yields

(y(1− y)2 + µ)′ = 0 =⇒ 3y2 − 4y + 1 = 0.

Solving for y gives y = 1 and y = 1/3. Plugging these critical values of yback into (9) and solving for µ we obtain µ = 0 and µ = −4/27. This createsthree regions of interest:

1 For µ < −4/27, y has a single equilibrium point.2 For −4/27 < µ < 0, y has three equilibrium points.3 For µ > 0, y has a single equilibrium point.4 For µ = 0 and µ = −4/27, there are two critical points, y = 0 and

y = 1/3.



Bifurcation Plot



Stability Theorem

Stability TheoremLet f (y) and f ′(y) be continuous. The equation

dydt

= f (y) (10)

is stable at y = y0 provided f (y0) = 0 and f ′(y0) < 0. The equation (10) isunstable at the point y = y1 provided that f (y1) = 0 and f ′(y1) > 0. If bothf (y2) = 0 and f ′(y2) = 0, then the point y2 is a point of bifurcation for (10).


autonomous equations and stability - acu blogsblogs.acu.edu/jee99a/files/2012/08/autonomous... ·...

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