linear systems - operators

7/29/2019 Linear Systems - Operators

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Copyright Rene Barrientos Page 1

INTRODUCTION TO SYSTEMS OF DIFFERENTIAL EQUATIONS

Motivation

In our study of dynamical systems we encountered the equation and subsequently, we found that the same equation governs certain electric circuits. For example, theequation corresponding to anRLCcircuit is 1 Simple systems provide nice illustrations of physical phenomena, but reality is much more complicated.Very seldom will an electric circuit, for example, consist of a single loop. Similarly, harmonic oscillatorsare used to illustrate a much more complex system of spring and masses attached together, as in the figurebelow, for example:

In order to analyze the motion of such a system it is necessary to introduce two position variables, one foreach block, and derive the equations of motion by applying Newtons Laws. These equations formsystem ofdifferential equations, the subject of our present inquiry.

Systems do not only arise in connection with mathematical models of physical phenomena. They arisenaturally when we seek to find the geometric properties of higher order equations. For example, considerthe second order equation 0Solving for Let and . Then

We have a system of differential equations in the variables and: which we can also write as

We will soon see how much information about the geometry of the solutions is concealed by thesystemscoefficients, that is, the numbers in front of the variables and.

This procedure may be generalize to arbitrary second and higher differential equations. Let us consider theequation,, , 0 for example. If we could solve for , then

,,

Let and . Then and ,,

This gives rise to thefirst order system , ,


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We need not stop here. The nth order equation, , , , , , 0 may be reduced to asystem ofnfirst order equationsvia the substitution , , . ,

The resulting first order system is given by

... ,, , ,

or in an more symmetric way:

0 1 0 0 0 0 1 0 ... 0 0 0 0 , , , ,

where is the function that results when, , , , , , 0 is solved for . We will ofcourse omit the zero coefficients, but when we write the system in matrix form it will be essential to accountfor them.

Example 1 Write a first order system corresponding to the equation 2 4 .Solution

Solving for: 2 4 Let

,

and

. Then

2 4 We write this system like this:

4 2 Example 2 Write a first order system corresponding to the equation 4 8 c o s , 0 0,0 1, which corresponds to a driven vertical oscillator subjected to a retarding force.Solution

Solving for

:

4 8 c os Let and . Then, 4 8 cosor equivalently, 8 4 cos

The initial conditions are replaced by0 0 , 0 1.


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The foregoing discussion shows but a tiny part of the mathematical landscape of the subject of systems ofdifferential equations and we will limit ourselves even further for now, confining our study to first ordersystems of the type ,, ,, 1We study these equations because they afford us the luxury of exploring both their qualitative and

quantitative aspects in a relatively intuitive way. Let us first state some definitions and terminology.

Solution of a System

By a solution of (1) on an interval we mean two differentiable functions , defined on such that ,, , , for all . For example, the functions and are solutions of the system defined throughout the real line, as it can be verified by substitution.

Aut onomous sys tems

System (1) is autonomous ifand depend only on the dependent variables and. Thus, autonomoussystems have the form , ,

Thecritical pointsof an autonomous system are the points, such that, , 0Example 3 The system

is autonomous.

Example 4 The system 1 is not autonomous.

Example 5 Find the critical points of the system Solution

We find the critical points by solving the simultaneous equations 0 0

The first equation states that 1 0 0 or 1If 0, the second equation implies that 0 so one critical point is 0,0. If 1, then 1 so 1,1 are two additional critical points. Hence0,0, 1,1, 1,1is the set of critical points of this autonomous equation.


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Linear Systems

A system is linear if the functions and are linear in both and. When a system is linear it can alwaysbe expressed in the form

2

where andare arbitrary functions of. A linear system ishomogenous if the functions 0 (that is, they are identically 0).Example 6 The system 2

is homogeneous.

Example 7 The system 2 2 isnothomogeneous .

Example 8 The system 2 2 isnothomogeneous but it is linear.

Example 9 The system is homogeneous but it isnot linear.

Example 10 The system 2

l n is neither linear nor homogeneous.Example 11 The system 2 is both autonomous and linear.

Quantitative Analysis of Linear Systems (Elimination and Operator App roach)

In order to study the geometry of solutions of linear systems, it will be convenient (and almost essential) tointroducematrix notation. Let be a vector in

with components

and

. Geometrically, this vector traces a plane curve

whose parametric equations are .


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The system of coordinates thus introduces is called thephase planeand the curve is called atrajectory

We define the derivative of , , by1 Naturally, we also use the notation

/for the same purpose and these two will be used interchangeably

whenever one is more convenient than the other.We also introduce the matrices and

Then system (2) can be written in matrix form as 3We call thematrix of coefficientsof the system. The sum is called thetrace of, denotedby and the quantity is, of course, thedeterminantof the matrix denoted by . The curve traced by is called atrajectory.Example 12 Write the system

3 2 5 In matrix form. Identify the trace and determinant of the matrix of coefficients.Solution

For ease of identifying the coefficients, let us write the system as 3 1 2 0 5 Then the matrix of coefficients is 3 12 0. The determinant and trace are det 3 0 2 1 2 and tr 3 0 0. The system is not homogeneous because . In matrixform,

3 12 0 05Observe the similarity of the first order system (3) with the first order linear equation where , and and are real numbers. We will come back to this analogy when we study theeigenvalue method. For now, we wish to study the simple case of the linear system with constantcoefficients: 4where the coefficients

, , ,and

are constants. In this case, the system can be fairly easily by a very

mundane, yet illuminating method: elimination. We illustrate the general idea in the context of ahomogeneous system. Assume for example that 0. Solve the first equation in system (4) for: 1 then

1We will use the prime notation do denote differentiation with respect to time.


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1 (1) Substitute these in the second of system (4):1

or

Writing this equation in standard form gives asecond order homogeneous linear equation in the variable: 5 In a completely analogous way, we can show that obeys the same equation. However, once equation (5) issolved, we can obtain from the equation 6

(2) Notice that equation (5) can be written as

As we shall see, the trace and determinant of the matrix of coefficient will prove to be more than usefulnotation.

For non-homogeneous systems equation (5) takes the form So we have developed a method of solving (4) when 0. If 0, then the first equation in system (4)can easily be solved for and the result used in the second equation to find. If either 0 or 0, wesay that the system is partially coupled. If bothand are zero the system is said to be decoupled andsolving it amounts to solving two separate and independent first order equations.

Remark: in step (1), we could have solved the second equation for in terms of and and obtained asecond order equation in

identical to (

5). What this tells us is the it would seen that by solving

and So it would appear that we have solved system (4), but this is not quite right. Doing so according to theformulas above introduces four arbitrary constants. However, only two can be independent and thereforeone must find dependency relations that link them. We may avoid this complication by using equation in(6) to find if that was the variable originally eliminated.Example 13 Solve the system 2 2 by elimination.Solution

Solve the first equation for

in terms of

and

:

2 Hence 2 Substituting in the second equation,2 22 Writing this equation in standard form, 4 5 0


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Let us compare with with 2 , 1 , 1 and 2: 2 2 4 11 0which is exactly the same equation obtained earlier.

The characteristic polynomial is 4 5 and its roots are 2 and 2 .Accordingly, cos sin Since 2 We have 2 cos sin cos sin sin2cos2

The solution vector is therefore

cos sin sincos

Example 14 Solve the system by elimination.Solution

The matrix of coefficients is 0 11 0 and obeys the equation 0or 0 1 0

The equation

0has solutions

and

. Hence,

The first equation in the system tells us that Thus, The solution vector is

The Operator Approach

We can solve some systems by eliminating variables, but this procedure can become cumbersome whenworking with large systems or higher order equations. The operational approach which we now introduce an

is an efficient way to solve linear systems.

We introduce the operatorsD and such that if is a function of the variable, then for all where is differentiable for all t in the domain off. For obvious reasons we call id the identity operator. We also defineby , thekthderivative off provided it exists, and we use these to define thedifferential operator.


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such that Remark: We usually do not write the identity operator as. Instead, we simply use the numeral coefficient

as a multiplicative operator. Thus

2 3 is written simply as 2 3. If we apply this operator to a functiony, the result is 2 3 2 3 As we have seen before, is a linear operator.We define the product of operators as follows: Ifand are operators, then their product isdefined by If all the coefficients are constant, then these operators arecommutative, that is, ifand are linearoperators with constant coefficients, then

Commutativity is not a property that is shared by operators with non-constant coefficients.

Example 15 Show that 1 and 3 commute.Solution

We need to show that 1 3 3 1.Applying the definition, 1 3 1 3 1 3 1 13

3

3

2 3On the other hand, 3 1 3 1 3 3 3 3 3 2 3Thus, 1 3 3 1Systems and Operators

Although we will use a

2 2system in this discussion, the results derived can be used just as well in the

general case. Consider the linear first order system We may write this system using operators as follows:


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Moving all terms that involve the variablesx andy to the left hand side,

Thus, Example 16 Write the system 4 2 3 in operator form.

SolutionWriting the system as

4

2 3 0we obtain 1 4 2 3 0Here 1 , 4, 2, and 3 . Two of the operators constant and thissystem is not homogeneous.

Example 17 Write the system

in operator form.

SolutionWriting the system as 0 0 0we obtain the homogeneous system 1 0 1 0

1 0Here 1 , 1, 1, 1, 1 , 1, 1, 1and 1 .Observe how double subscripts have been used to identify the row and column in which theoperator occurs.

If the coefficient functions are constant, then operators commute and systems may be solved by familiaralgebraic methods. Let us therefore assume that, , , and are constants. Then system(2) may be written as


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which has the form 7where

,

,

, and

We may eliminate either variable by adding a suitable multiple2 of one equation to another. Say, forexample, that we want to first obtain a single ordinary differential equation for . Then we eliminatey byoperating with on the second equation, with on the first equation, and subtracting the result: Since the operators involve constant coefficients, they commute. Therefore, . When wesubtract the second equation from the first, we obtain 8

The operator is very reminiscent of a determinant. In fact, because of the multiplicativeproperties of these constant coefficient operators, we can write

Similarly, Hence, equation (8) can be written compactly as

A similar elimination procedure gives us an equation for :

We call the determinant the operational determinant of the system and we will call theseequations theoperational equations.Observe how similar this is to Cramers Rule. But of course we must be mindful that these are operators,not real numbers, so unless we give a meaning to the reciprocal of an operator (which can be done) wecannot divide by the operational determinant.

There is one more point before we proceed. If we use the operational equations to solve for and separately, then we must introduce four arbitrary constants; two forx and two fory. However, not all fourare independent. The orderof the operational determinant tells us how may independent constants thereshould be, and that number is equal to its order. We will illustrate this point in in the examples that follow.

Example 18 Solve the system

by the operator method.

SolutionThis is example 14 so let us verify the answer we obtained through the method of elimination.Writing the equation using operators: 0 0

2 We are not multiplying the equations by a number. We are operating on them with an operator.


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The operational equations are

11 0 10 and

11 01 0

Solving the system, 1 0 1 0Hence, .Solving for , . Since the operational determinant is of order 2,only two arbitrary constants are independent. We may obtain the dependency relations bysubstituting these solutions in either equation of the system. Le us use the first equation: for all Hence, and . Therefore,

and

Example 19 Use the operator method to solve the system 2 2 Solution

Writing the system in operator form, 1 2 02 1 0Thus,

1 22 1 0 20 1and 1 22 1 1 02 0We may solve for and using these equations or we may solve for say for and thenuse it in the original system, whose first equation tells us that

12 to obtain. Thus, once we havex, we havey. Let us solve for using

1 22 1 0 20 1

Evaluating the determinants, 1 1 4 10 2 0Thus, 2 5 0This is the second order ordinary differential equation 2 5 0


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whose characteristic polynomial 2 5 has roots 1 2 . Accordingly, c os 2 sin2As mentioned before, we may obtain from the systems first equation:

12

However, in order to illustrate the point mentioned earlier, we will instead use the equation

1 22 1 1 02 0to find.First notice that it will always be the case that the determinant on the left hand side remains thesame, no matter how many variables the system has because after all, it is the systems operationaldeterminant. What will change, in general, is the determinant on the right hand side of the equation.

The operational determinant 1 22 1 is of order 2. Therefore, the solutions of system canonly have two arbitrary constants.

Solving for

,

c os 2 sin2Notice that we have introduced two more constants because we do not know ahead of time whetherthey have the same values as and. What we do know is that all these constants are connectedbecause the operational determinant has order 2. Therefore, only two arbitrary constants cansurvive.

To find thisdependency relation between the constants, we substitute our solution into either of theoriginal equations. Say we use the second one: 2

Then

c os 2 sin2 2 c os 2 sin2 c os 2 sin2Differentiating and collecting like terms: 2 cos2 2 sin2 2 cos2 2 sin2

This equation holds true for all if and only if 2 2 and 2 2

These are the dependency equations and they tells that Thus, the solution of the system is given by the equations c os 2 sin2 s in2 cos2

Exercise:Use the equation to verify this answer.


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Example 20 Solve the initial value problem 2 4 5 2 3 1 ; 0 1 , 0 1Solution

We first write the system in operational form:

4 2 5 0 2 3 3Observe that this is not a homogeneous system. We proceed to solve it:

4 2 52 3 0 2 53 and

4 2 52 3 4 02 3 3The systems operational determinant is

4 2 32 5 5

2 0 1 5

A second order determinant tells us that the system has only two arbitrary constants.

Solving for, 5 2 0 1 5 2 53Dividing by 5, 4 3 2 5 35

or 4 3 3This is a non-homogeneous equation whose characteristic polynomial 4 3 hasroots 3 and 1. Thus, its complementary solution is

Let be a particular solution. Then 1, which is obtained by substitution. Hence, 1Similarly, 5 2 0 1 5 43From which we obtain 4 3 125 Hence and we may obtain a particular solution by letting . Substitutinginto the equation above gives us

45Thus, 45

The dependency relations may be obtained from either equation in the original system.

Since2 3 1 seems to involve less work, let us use it.2 1 45 3 1 1

Performing the required operations,


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23 3 3 3Combining like terms:6 3 2 3 3

Therefore,

6 3 32 3Solving for two of the variables in terms of the others, and . Hence, 45

Finally, we apply the initial conditions0 1 , 0 1:0 1 1 10 1 45 1

Solving for

and

:

and

The solution of the initial value problem is

Example 21 (inconsistent system) Solve the system 6 2 3 0Solution

We have

3 2 2

3 2 0 The systems operational determinant is 3 2 3 2 O

This is thezero operator which transform all functions into the zero function. What we have then isO which is absurd because it states that 0 for all . Therefore, the system is inconsistent it hasno solutions.

Example 22 (infinitely many solutions) Solve the system 0 0

Solution 1 0 0 or 0

Once again we have the zero operator: O 0


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However, this equation is satisfied by all functions . For any choice of, we can obtain acorresponding from the original system. For example, if we let s in, then thecorresponding may be obtained from its second equation:

s i n c o s Notice for example that

s in c o s satisfies the system.Example 23 (arbitrary constants) How many arbitrary constants will the system below have? 2 1 0Solution

The operational determinant is

1

1

This is a 4

th

order operator. Therefore, there are four arbitrary constants in the system.Example 24 Solve the system 0 2 0Solution

The operational determinant is

2 2Therefore, the system has four arbitrary constants. We proceed to solve it:

2 0 0 and 2 0 2 0

This gives us 2 0Similarly, 2 0

The characteristic polynomial 2 2 has roots 0 and 1 byinspection. There must be two more roots, which may be real or a complex conjugate pair.

In order to obtain them, we divide 2 by 1 to obtain 2. Therefore, the othertwo roots are complex:

1 1 82 12 12 7Thus, / cos 72 sin 72 Similarly, / cos 72 sin 72


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Exercise: Of all these eight constants only four are arbitrary. Find the dependency relations andwrite the solution of the system.

Example 25 (application) Write the system of equations that govern the coupled spring-mass system shownbelow and solve it. Assume that both springs have the same spring constant.

SolutionLet us introduce a coordinate system

We introduce the variables and which represent the displacement of the two masses fromtheir equilibrium positions, respectively.Applying Newtons Second Law to the figure on the right,

Simplifying,

2

Observe that the first equation lost the variable. This is only because we assumed both springshave the same stiffness.

Writing the system in operator form, 2 0 0The operational equations are

2

0

0

and

2 0 0If we solve for , then the first equation of the system

2 0 1 2may be used to obtain. Thus, solving for :

0 0

At some arbitrary time

0 0


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2 0This 4th order homogeneous linear differential equation has auxiliary polynomial 2 3This is a quadratic equation in the variable

and has roots

2 2 122 An interesting case is the one in which for then

32 122 3 32 Since the characteristic polynomial is of degree4, is the sum of a four linearly independent solutionsas is.As you can see, even this very simple problem leads to rather complicated solutions. For this and othermore profound reasons, the mathematicians of the late 19th and early 20th centuries developed qualitativemethods of analysis based on the geometric properties of systems. In the next lecture, we will explore someof these geometric techniques can come to appreciate their power.

Example 26 (application) Two tanks, tankA and tank B, are connected as shown in the figure below. TankA, which originally contained 1L of a 10% mixture receives additional mixture (same concentration) at arate of 2L/min. The well stirred mixture is then evacuated into tank B, which originally had 5L of purewater, at a rate of3 L/min. Let and be the amounts of chemical in each tank. Find expressions for and.

Solution

TANK A IN OUT

RATE OF MIXTURE 2 L/min 3 L/min

CONCENTRATION 10% /RATE OF CHEMICAL

2 0 . 1

3/

Therefore, 2 0 . 1 3

mixture of concentration10% in at a rate of 2 L/min

Mixture transferred arate of 3 L/min

Initially 1 liter,0.1 L of chemicalNo mixture leavestankB

Initially 5 liters ofpure waterTankA TankB


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TANK B IN OUT

RATE OF MIXTURE 3 L/min 0 L/minCONCENTRATION x/ N/A

RATE OF CHEMICAL

3/

N/A

Thus, 3 0Tank A has1 liter initially and we add mixture at a rate2 L/min while simultaneously transfer it totank B at a rate of3 L/min. Hence, 1 2 3 We must therefore take .Tank Bhas5 liters initially to these we add solution at a rate3 L/min from tankA. Thus,

Therefore, the system is . ; . ; The first equation is independent of and is a first order linear differential equation: 31 0.2 ; 0 0.1It has an integrating factor . Thus,

11 0 . 2 11 Integrating, 11 0 0 . 2 11 11 0 . 1 0.22 11 111 110 110 11 110 1

10 1

We obtain

from

31 31 110 1 310Hence, 0 310

or


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310 The solution vector is

110 110

310 1/10

0 1

101

3 ; 0 1

Writing the solution is this form is revealing because it tells us that the trajectory in phase space is astraight line.

Example 27 (application) Newtons Cooling Law states that when a body is placed in thermal contact withanother body of very large heat capacity whose temperature is, the first bodys temperature, which wewill denote by obeys the equation ; 0We may generalize this law to two (or more) bodies whose temperatures change until an equilibrium state isachieved. Let be the temperature of one of the bodies and be the temperature of the other body. Thentheir thermal interaction can be modeled by

where is a constant whose value depends on the material that separates the two bodies. If three bodies areplaced in thermal contact, the system that arises has three equations.

Let, and be the temperatures of the three regions in the figure above. Then applying NewtonsCooling Law,

For the sake of simplicity, assume 1. Then the system is


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2 2

2 Or in matrix form: 2 1 11 2 11 1 2

Let us solve this system using operators: 2 1 1 0 1 2 1 0 1 1 2 0

2 1 11 2 11 1 2 0 1 10 2 10 1 2 2 1 11 2 11 1 2 2 0 11 0 11 0 2 2 1 11 2 11 1 2

2 1 01 2 01 1 0Let us solve for since by the symmetry the expressions for and will be the same. Evaluatingthe determinants, 6 9 0

The roots of the characteristic polynomial are

0 , 3 , 3. Accordingly,

Similarly, and

Since the operational determinant is of order 3, there can only be three independent arbitrary constants andthey may be found by substituting the solutions in the system. This will result in a linear system of algebraicequations and it is one way of handling this problem, but not a very efficient one. later we will learnanother, more elegant, approach.

linear systems - operators

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