Qualitative Analysis of Linear Autonomous Systems With Constant Coefficients

7/29/2019 Qualitative Analysis of Linear Autonomous Systems With Constant Coefficients

1/15Copyright Rene Barrientos Page 1

QUALITATIVE ANALYSIS OF LINEAR AUTNOMOUS EQUATIONS WITH CONSTANT COEFFICIENTS

The system

is called a linear autonomous system with constant coefficients. When 0, this system can always beconverted to a homogeneous system 1by a suitable transformation of the form

The goal of these notes is to develop a basic understanding of the qualitative aspects of the solutions of system(1). First some general terminology and basic concepts.

Consider the system

,, ,, 2 , is asolution of system (1) on an interval Iprovided that the functions and isare differentiable in and

,, ,, for all

.

4 4 2 1 4 42 1

2 1 2 22 1 2 4 2 4 4 4 4

Example 1 The functions , 2 1 are solutions of the system

on the interval ,. We verify that the second equation is satisfied and leave it as anexercise to verify that the first one is also satisfied. We need to show that

Differentiating

On the other hand,4 42 1 4 4. Therefore



It is interesting to write the solution of example1 in matrix notation: 2 1We can write this solution as a sum of two vectors:

12

01

which looks very much like a linear combination of the functions , . In fact, that is exactly whatthis identity states although the coefficients are not real numbers but vectors.Geometrically, the solutions of a system form a set of parametric equations of a plane curve (or in the moregeneral case in). The curve is oriented by the parameter as illustrated in the figure below.

The column vector traces the curve as ranges over the interval of existence.When we describe solutions geometrically by means of this vector function, we refer to the Cartesian plane asthephase planeand the solution curves astrajectories.

Initial Value Problems

An initial value problem consists of a system for which initial conditions , are specified.Initial value problems are stated like this:

,, ,, ; , The initial conditions are used to find a particular solution of the equation.

4 4 0 1, 0 0

0 1 10 0 2 0

24

Example 2 Solve theIVP

Given that

, 2

2 1

is a solution of the system.

Solution we obtain the particular solution by applying the initial conditions:

Solving the system gives us 1 and 2. Thus, the particular solution is

,



We may write the result in the previous example as the sum:

10 2 12Once again, this may look just like another way of writing the answer, but soon we will see the significance of

the vectors and.Critical points

Recall that an autonomous system is one of the type , , 3and itscritical pointsare the points, for which both, 0 and, 0.Example 3 Findthe critical points of the system 2

5

SolutionThe critical points are found by solving thesimultaneous equations. 2 0 5 0The first equation is satisfied only when . Substituting into the second equation, 2 5 0 The only critical point is, .

Example 4 Find the critical points of the system

2 Solution

We Solve the system 2 0 0The second equation implies that . Substituting into the first equation, 2 0 0, 1/2If

0,

0.

if 1/2, 1/2.Therefore, the critical points are , , .Trajectories and the Phase Plane

The solution 2 of the previous example is a compact way of writing the parametricequations , 4



What do these trajectories look like? In general, it is not easy to answer this question; however, one can obtain aqualitative idea by analyzing the original equations in the system. For the sake of reducing unnecessaryarithmetic, let us consider a simple system such as:

The general idea is this: the functions, and , tell us how each of the dependentvariables are changing at each instant. If we make the analogy with a point particle moving in the plane, wemay think of, and , as the velocity components of the particle, but keep in mind that the systemmight not even be remotely connected to physical motion in space.

Using the analogy of the point particle, these equations tell us that at the point1,2, for example, the particle ismoving in such a way that/ 2, / 3.Graphically:

The red arrow is the tangent vector to the trajectory at1,2. Imagine if we could do this for every point in thedomain of the system. Then we would obtain a field of vectors (thedirection field) associated with the systemwhich gives us a very clear idea of the qualitative behavior of its trajectories.

Computers can do these calculations very fast and the filed of this particular system is shown in the figurebelow:

By analyzing some key information we can get an idea of its behavior without the aid of a computer. This is the

is information we look for:1) where is/ 0 ?2) where is/ 0 ?3) where is/ 0, and where is it negative?4) where is/ 0, and where is it negative?5) what happens along the coordinate axes?

-3 - -1 1 3

-

-

-1

1

2

3

3

2 1,2



When these questions are answered, we have a pretty good idea of what the trajectories look like. Let us do itfor the system under consideration:

First, 0 when 0 (the x-axis). this means that at any point on the x-axis, points on the trajectories donot have a horizontal component of motion; trajectories intersect the x-axis at right angles. 0 when 0. Thus, below the x-axis, trajectories have a positive x-component of motion; they move tothe right. The opposite happens above the x-axis because there

0. 0when 0 or . thus, trajectories interest the line with no vertical component ofmotion. 0when 0 of . Thus, the trajectory points have a positive y-component above anda negative y-component below it.

On the x-axis,

0so

0and

. Thus, these points have positive y-component of motion on the

positive x-axis ( 0 and negative y-component on the negative x-axis.On the y-axis, 0 so is independent of this. However, which means that trajectory pointshave a positive x-component of motion when 0 and a negative one when 0.We can summarize this information by plotting a point in each of these special regions and assigning smalldirection arrow according to our analysis above.

When we plot a few representative trajectories in the plane, together with the equations critical points, theresulting figure is called aphase portrait.

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0



Trace a few curves that have the fields arrows as tangents and you will have the systems phase portrait. In thisfigure, only one representative curve has been traced. You need a few more before you call it a phase portrait.

Geometric Analysis of System (1)

We are ready to study the geometric properties if system (1) by first considering the analysis of its onedimensionalscalarcounterpart:

4whose general solution is (as we have seen many times and in many different way) Unfortunately, there is where the similarities end because whereas the latter has a simple solution, the former ismuch richer in structure. In order to appreciate just how rich, let us begin by solving the scalar equation by a

method that seems a bit ridiculous, but which will serve as a model. Let , being a number that isyet unknown, be a tentative solution. Then Since 0

0

or 0If we seek nontrivial solution, then 0 and therefore . Hence, the nontrivial solutions of the scalarequation are given by and we are not surprised.

Now consider system (1) which may be written in matrix notation as 5where

is the matrix of coefficients and

. Notice how similar (4) and (5) are in

appearance. Let us try to find its solution in the same way we proceeded in equation (4):

Let be a solution where is a constant vector and a real number. Then, or

Since 0, or

6Thus, in order for to be a solution, the equation above must be satisfied. This equation is called aneigenvalue problemand its nontrivial solutions are called its eigenvectors. In order to find them we must firstfind the values of for which there are nontrivial solutions.Unlike the illustration involving the scalar equation, we may not factor the vector because the operation is undefined. We may, however, introduce the2 2 identity matrix which will allow us to factor.Rewriting equation (6) as



or equivalently, The nontrivial solutions of this system correspond to the values of for which1det 0 7Hence, the solutions of system (1) are found by finding the eigenvalues and eigenvectors of the matrix ofcoefficients. Equation (5) is equivalent to

0Expanding this determinant results in the second degree equation called thecharacteristic polynomial associated withA. Since it is a second degree polynomial, it has two rootsand therefore we must consider some possibilities.

CASE PROFILE

Real eigenvalues

Case 1: 0 Unstable nodeCase 2:

0Stable node

Case 3: 0 Degenerate nodeCase 4: 0 Saddle pointComplex eigenvalues

Case 5: , CenterCase 6: , , 0 Spiral

Box 1

In a moment we will come to understand the meaning of the third column.

Once the roots of the characteristic polynomial are found, we find the corresponding eigenvectors by solving

for.Example 5 Find the eigenvalues and eigenvectors of the matrix0 11 0.Solution

The eigenvalues are found by solving A A 0or 0 1 0

Hence 1, 1 are the eigenvalues. To find the corresponding eigenvectors, we replace in or equivalently, in 0 11 0 00

If 1: 1 11 1 001See Linear Algebra Supplement at the end of these notes.



Only one equation results out of this computation: 0 Therefore, 11We may take

11to be the eigenvector associated the

1although as you can see, there

are infinitely many. Similarly,

If 1: 1 11 1 00And again only one equation results out of this computation: 0

Therefore, 11We may take

11to be the eigenvector associated the

1.

Now let us suppose we were solving the system whose matrix is 0 11 0. Having found the eigenvalues and their corresponding eigenvectors of thismatrix, we know that is a solution of the system. Thus,

11 and 11 are two solutions. Are there more that were not captured in this process? No. We have the following importantresult about homogeneous linear systems:

Observe that the vector solutions correspond to parametric equations of curves in the plane. For example,

11 Really represents ; Similarly, the parametric equations corresponding to

Can be obtained by expanding: 11 11 Thus, Trying to describe this curve is not easy, but if we consider the vector for

Fact: If and are two linearly independent solutions of (1), then so it their linear combinationwhere s and are arbitrary scalars. Furthermore,is thegeneral solutionof system (1).



11 11 we see that the solution curves are linear combinations of the eigenvectors 11 and 11. This isthe line of thought that we would like to explore.

What does 11 1

1 tell us? It tells us that the solution vector is composed in some way

of the eigenvectors and . For example, if , then looks mostly like 11 , which is just amultiple of11. Similarly, if , then looks mostly like 11. For intermediate values of, thesolution vector looks like a combination of the two. The constants and divide the plane into four sectors,as shown below:

Now let us trace a typical trajectory in the region 0, 0. Starting at , we see that the solutionvector looks mostly like11, that is, it is parallel to it. At 0, the solution vector is0 11 11Finally, as

, the solution vector looks mostly like

11. Let us translate this into a graph:

A similar analysis reveals that the trajectories follow the same general curve in the other sectors. The figureillustrates the profile of the solution curves in the plane.

11

11 0, 0 0, 0

0, 0

0, 0

11

110



11 11 As you can see, this method of analysis proves to be very efficient and gives us a clear picture of the systemsphase portrait.

Observe that in the previous illustration if we let

1

and

1

, then

0

. This corresponds toCase 4 in box

1. The Origin, which is the only critical point of the system, looks like asaddle point. The names

of the other cases similarly reflect the geometric nature of the corresponding phase portraits.

Example 6 (The Spiral) Determine the type of trajectory associated with the system 2 3 Solution

The matrix of this system is 2 31 1. This matrix has 1, 1. Thus,

1The roots of this polynomial are, 1 1 42 12 32 which corresponds tocase 6; the spiral. Look at the direction field. It is clear that the trajectories willfollow spirals as indicated by the representative red curve.

What is the general solution in the previous example? It is

-0.4 -0.2 0.0 0.2 0.4

-0.4

-0.2

0.0

0.2

0.4

0



Where and are the eigenvectors corresponding to the eigenvalues and , respectively.We may factor to obtain

/

which clearly indicates that the origin is an unstable node, that is, the trajectories more away from it if theinitial conditions are even slightly off. Case 6 really is composed of two sub-cases: if are thecomplex conjugate eigenvalues, then (a) the origin is an unstable node if 0 and (b) a stable node if 0.Example 7 (The Spiral) Determine the type of trajectory associated with the system 3 2Solution

The matrix of this system is 1 31 2. Hence, 1, 1. Thus, 1The roots are , 1 1 42 12 52 We expect a spiral because has complex roots. However, the real part of these roots is negative.

Thus, the spiral will circulate toward the origin.

Example 8 (The Saddle again) Determine the type of trajectory associated with the system

3 3 Solution

The matrix for this system is 1 33 1. We have 2, 8. Thus, 2 8The roots of this polynomial are

3 2The field corresponding to

forms a spiral with limit at the origin.



, 4 4322 2 3The roots straddle 0: 1, 5 which corresponds to Case 4: 0 . The origin acts asa saddle point. The tangent field is illustrated in the figure below.

Saddles are characterized by the appearance of asymptotes in the phase plane.

Example 9 (The unstable nod) Determine the type of trajectory associated with the system 3 2Solution 3 10 2

Thus, 5, det 6 and 5 6. The roots of are 2 and 3. Thesecorrespond to :

0where we let

3and

2.

The trajectories associated with an unstable node are illustrated in the figure below.

Unstable node

Example 10 (The center) Determine the type of trajectory associated with the system

-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.

-0.5

0.0

0.5

1.



2 Solution

1 21 1Thus, 0, det 1 and 1. The roots of are and . These correspondeigenvalues of the form to , .The trajectories associated with center are illustrated in the figure below.

Exercises

1) Is the system

2Linear? If not, which term or terms make it not linear?2) Find all critical points of the system 3 4 2 53) Find all critical points of the system

3

14) Write the system 2 4 2

in matrix form. What is its matrix of coefficients? What profile will its trajectories follow?

-1.0 -0.5 0.0 0.5 1.0

-1.0

-0.5

0.0

0.5

1.0



5) Write the system 2 2in matrix form. What is its matrix of coefficients? What profile will its trajectories follow?

6) Solve the following nonlinear system using itsphase plane equation (see steps below). 2 The phase plane equation is found by:

a) Writing the system is an equivalent scalar equation by eliminating the parameter : //:

2

b) This is a linear equation. Its solution curves correspond to the trajectories of the original system. Canyou determine their nature?

c) Since this is not a linear autonomous system (although it is autonomous) we cannot use the eigenvalueapproach. Show that 0,0 is a critical point of the system and read up on how to linearize it near theorigin. Write the corresponding linear system which can be used to analyze the solutions at thatcritical point. The figure below illustrates the direction field for this system.

Can you see the nature of the trajectories near the origin?

Supplement: Matrices and Matrix Notation

In the study of systems of differential equations it is useful, if not essential, to use matrices. Recall that amatrix is an array of numbers of the form

We use boldface capital letters such asA,B, etc., to name matrices and thedimension of a matrix, denoted by

-4 -2 0 4

-

-

2

4


15/15

dim , refers to the number ofrows and columns it contains. For example, the matrix 1 4 2 0 0

is a2 3matrix because it has2 rows and3 columns. Itsentriesare the numbers 1, 4, 2, 9, 0, and 0. Entry is9because it occurs inrow 2, column 1.Column matrices are matrices of dimension

and we refer to them acolumn vectors. Thus,

is a

column vector. Similarly, arow vector is a matrix of dimension .Matrix ArithmeticMatrices are added by adding their corresponding entries. Therefore, addition is defined only for matrices ofidentical dimensions. We denote addition by the usual symbol .Scalar multiplication is the operation of multiplying a matrix by a number. This operation consists ofmultiplying every entry by the number:

Matrix multiplication is defined only when the number of columns of the first matrix equals the number ofcolumns of the second matrix. We define the

,entry of the product of two matrices as thedot product of

the ith row vector of the first matrix with thejth column vector of the second matrix. LetA andBbe matrices ofdimensions and , respectively, and let be the, entry of the product , Then

Example M1 If 1 1 24 0 3 and 0 0 15 4 16 7 9, then the product is defined and the entry inposition 1,2, for example, is found by taking the dot product of the first row of matrix A and the secondcolumn of matrixB:

1 1 2 047 1 0 1 4 2 7

The other entries are found in the same way and the product is a matrix of dimension2 3 as well. Notice thatwhereas is defined, is not.Example M2 If 0 13 2 and 57, find the product Solution 0 13 2 57 0 5 1 73 5 2 7 722

0 0

Let

be any matrix. The traceofA, denoted by

,is defined to be the sum of the elements of its

main diagonal:

Thedeterminant ofA, denoted bydet or , is the number defined by .The matrix isnonsingularif 0. When a matrix is nonsingular, the homogeneous systemhas only thetrivial solution 0. Conversely, if 0, then the system has nontrivial solutions.

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Qualitative Analysis of Linear Autonomous Systems With Constant Coefficients