Lecture 2B Autonomous Equations

7/29/2019 Lecture 2B Autonomous Equations

1/7

1

AUTONOMOUS EQUATIONS

Consider now another simple version of ,; suppose that f is independent ofx, that is, suppose that We call such equationsautonomous first order differential equations. These equations have many applications in

the physical and biological sciences, demography, and other fields, and have very nice geometric properties, someof which will be illustrated in the examples that follow. In addition, autonomous equations may always be solved,

at least in principle, because they areseparable and therefore their integral curves obey

Example 1 Each of the following is an autonomous first order equation. 1 3

; 0 2.5

1 ; 0 10 , constantsAutonomous equations and systems, which we will study later, sometimes havecritical points:

Example 2 What are the critical points of 4 2 ?Solution 4 2 is zero when 0 and 2. Thus, the critical points are 0 and 2.

Example 3 Find the critical points of the autonomous equation 9

SolutionThe critical points are found by solving 9 0.1 9 0gives the roots 0 ; 13 ; 13

Notice that the isoclines of an autonomous equation (

3) are given by

These are horizontal lines on which the solution curves have derivative . Thus, critical points are really isoclinesalong which the solution curves have slope 0 and are, by definition, solutions of the equation. For example, theline 1 / 3 is the constant function 1/3 and it satisfies the equation 9

because both sides of the equation reduces to 0.

Critical Points is a critical point of the autonomous differential equation if 0.Definition


2/7

2

These solutions play a very important role; they divide the plane into regions in which the solution curves exhibit

very well-defined characteristics which may be used to explore the global behavior of the solutions of thedifferential equation. The following examples illustrate this idea.

Example 4 The equation 4 2 has critical points 0 and 2 and these divide the plane into threeregions:

0,

0 2, and

2.

By charting the sign of42 we then know what the sign of is and therefore whether is increasing,decreasing, or stationary.The following chart summarizes the information we want (This very useful diagram is called aphase diagram),

Phase Diagram

This diagram tells us that is negative on the interval 0, positive on the interval 0 2 andnegative again on 2 .Accordingly, is decreasing on 0, increasing on 0 2, and decreasing again on 2 .

Notice that 0 on the lines 0 and 2. These are the critical points and the integral curves go flat aswe approach them. To graph these integral curves, we place they axis vertically and draw the critical points (note

again that these are horizontal lines). Then we draw the curves so that they conform to the information provided by

the chart:

The three regions have been labeled as I, II, and III. In region I, the solutions curves decrease since 0. In region II, thesolution curves increase since

0. In region III, the solution curves decrease again.

The integral curves form a family of solutions, but the figure above shows only a representative of this family in

each of the regions. Such graph is called aphase portraitof the differential equation and the solutions 0and 2 are called equilibrium solutions.Equilibrium and stability are important themes in differential equations, but they belong to a more advancedcourse. Here we give only a quick brush with the idea of equilibrium via the following definition:

0 2

y

4 2y ++

+ +

4 2 + y dec. inc. dec.

I

II

III

y

x

y = 0

y = 20 2; solution curves areincreasing

2; solution curves are decreasing

0; solution curves are decreasing


3/7

3

Roughly speaking, solutions that start near a stable point stays near it for ever. In contrast, asymptotically stable

points have the property that solutions which start near them tend to them and can be made arbitrarily close to

them:

If a point is not stable then it may be unstable orsemi-stable. The distinction between these two is illustrated in

the examples below but we will not attempt to differentiate them via a formal definition. The terms attractor andrepeller are also used in connection with asymptotically stable and unstable points, respectively.

The determination of the nature of a critical point may be summarized in the following way: (1) determine the

behavior of as was done in the previous example. (2) in each subinterval that is thus created, draw an arrowthat indicates the direction in which solution curves move (left for decreasing and right for increasing).

A stable point is characterized by the fact that, regardless where we start, as long as is it close to the critical

point, the arrows will point toward that point. Thus, the phase diagram for example 8 looks like this:

The arrows in each region point in the direction in which solutions curves would move. At 0, the arrows point away.Therefore, this is an unstable point. On the other hand, the arrows always point toward 2 as long as we startsufficiently close to 2. This solution is therefore an attractor, or stable, solution.

Example 5 Identify its critical points and draw the phase portrait of 2 .Solution Setting 2 0 we get the critical points 0 , 2. We now proceed to make the phasediagram:

0 2y

42y ++ + +

y(42y) + y dec. inc. dec.

| | Stable Points

A critical point

is called stable if given any number

0, there exists a positive number

such that if

| | then

Definition

0 2

2 +++

+

+y

dec. dec. inc. 2

c

Asymptotically stable pointStable point

c


4/7

4

the phase portrait is shown in the figure below:

The point

2is unstable (also known as a repeller). As

, the curves diverge from it. On the other hand,

0is approached from above so it acts as an attractor, but repels the curve below so it behaves like a repeller (orunstable) point; we call it a semi-stable point.Example 6 The logistic equation of population dynamics given by

1 is an autonomous equation(why?). Identify its critical points and draw the phase portrait of its solution curves.

SolutionAlthough Prepresents population and must therefore be considered a non-negative quantity, we will treatit as just another real variable for illustrative purposes. We assume is a positive number since itrepresents a physical quantity.

The critical points: 1 0 0 , 1 /

Phase Diagram 0 is an unstable point (repeller) and 1 / is a stable point. What does this mean in terms of thepopulation? Let us consider its phase portrait, limiting ourselves to 0 since this variable represents apopulation.

xy =0

y = 2I

II

III

y

0

1 ++

+ +

+

P

dec. inc. dec.1 1/

If the initial condition is close to the critical point

0and above it, then the solution curve

approaches that critical point asymptotically.However, if the initial condition is below thiscritical point, then the solution tends away fromit. for this reason, 0 is called semi-stable.On the other hand, as , the solutioncurves that start just above or just below line 2 diverge. 2 is an unstable point.

t

P=0

P = 1/k

I

II

P

Phase portrait


5/7

5

This diagram indicates is that, regardless of what the initial population is at 0, it will evolve toward anequilibrium state of 1 / as long as 0 . If initially 0 1 / , then it will increase toward the value 1/. On the other hand, if 0 then it will decrease toward that same equilibrium state 1/. Thisvalue is called the populationscarrying capacity and it is a characteristic of the logistic curve.

As a final remark, the solution of this differential equation can be expressed by the function whereA is an arbitrary constant. Notice that it has the desired property thatlim 1/As you can see, much information can be obtained from a qualitative analysis of a differential equation.

Exercise 1: solve 1 and show its solution is

Exercise 2: if0 0, what is for 0?Example 7 What are the critical points of the equation

1 sin ? Suppose in addition that we knowthat 0 0.5. what can be concluded about lim ?Solution

The critical points are given by 1 s i n 0 1 , Thus, all critical points are located at the integers: 0,1,2,3,Since 0 0.5 puts us initially on the interval [0,1], let us chart the function 1 sin on theinterval 1,2 only.

By the Existence and Uniqueness Theorem we know that this initial value problem has exactly one

solution: the one that satisfies the initial condition and is therefore confined to the strip 0 1. Thefigure below shows the profile of that solution in dark blue.

Thus, .

0sin1

+

+

+

+

+

y

Dec. Inc. Inc.

/ 211

ty = 0

y = 1y

y = 20.5

y = 1


6/7

6

Optional

I. (Technology and Differential Equations)

Let us revisit the first order equation , but let us write it like this:

, 1 This tells us that at point , where , is defined, the tangent to the a solution curve has slope , andtherefore we can associate a small vector with the curve this point of tangency, manely. the vector , ,,

This new object , is an example of a vector fieldand it associates with each point , a vector , that is tangent to the solution curves of the differential equation.

In this regard, it is called the direction field(sometimes also called slope field) of the differential equation and it

gives us a very clear picture of how the solution curves behave.

Using the computer program MATHEMATICA we can generate fields with the command

VectorPlot[{u(x,y),v(x,y)},{x,a,b},{y,c,d}]with , 1 and , ,. Then, VectorPlot[{1,f(x,y)},{x,a,b},{y,c,d}]will plot the slope field of

,Example O1 Plot the fields associated with and . Compare the results.Solution

These equations look very similar, but as you will see they describe entirely different phenomena.

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

VectorPlot1,^2,,1,1,,1,1 VectorPlot1,^2,,1,1,,1,1

solution curve

Tangent

,

1

,


7/7

7

Example O2 The direction field associated with the equation 1 can be plotted using the commandVectorPlot1, 1 2 1,,2,2,,1,1.

This is, of course, an autonomous equation and, as you can see, 0 and 1 are the two critical points of theequation.

0is anasymptotically stable solutions.

1is unstable.

One final remark. As you can see, the field vectors change in length in general and this is not convenient when all

we want is direction. We can fix this: given a non-zero vector there is associated with it a unit vector whichhas the same direction but is of unit magnitude hence the name unit vector. We may find this unit vector bydividing the components of the original vector by its norm.

Example O3 Find the unit vector associated with 1,2.Solution

The norm of this vector is 1 2 5. Therefore, the unit vector in the direction of1,2 is 15 , 25

Thus, the unit vector associated with

, 1 ,, is

, , , , , When we carry out this process we say that we havenormalized the field.

Example O4 The normalized direction of with 4 ,for example, is , . Thefield is shown in the figure below.

-2 -1 0 1

-2

-1

0

1

2

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0

Documents

Lecture 2B Autonomous Equations