second order linear equations - a prelude to higher order linear equations a historical approach

7/29/2019 Second Order Linear Equations - A Prelude to Higher Order Linear Equations a Historical Approach

1/15Copyright Ren Barrientos Page 1

SECOND ORDER LINEAR EQUATIONS

AND

APPLICATIONS

,, Higher order equations have many applications in science and other fields of research. The intention of this set oflecture is to introduce you to the basic theory of linear equations and applications in the context of the familiarsecond order equation. This will serve as a springboard to higher dimensions since the results there arecompletely analogous.

We will proceed from the specific to the general, in much the same way as how the subject evolved. A historicalapproach is not the norm, but I believe that it will help you develop a firmer grasp of the theory.

Consider then the familiar problem of dynamics where a object of mass moves vertically under the influence ofthe force of gravity and a drag force which is proportional to its velocity: ; 0. NewtonsSecond Law tells us that the equation that governs the velocity function is

; 0 Rearranging terms,

; 0 1

This equation has the general form where is a depended variable (in the case of the moving object its velocity function) and the independentvariable (in this example time). This is how we begin with a concrete physical situation and develop an abstractmathematical model. By studying this model in general we are able to answer all questions of its special cases,such as the case that led to equations (1).But let us continue with the example; having obtained an equation for the velocity function, it is natural to askwhether we can obtain a differential equation for the position function and the answer is, of course, yes.

Introduce a vertical axis and call it the x-axis (the name is not too important), as shown below. Then we knowthat and are related by /.

Since /, / /. Hence equation (1) can be written in terms of the variable:

; 0

, 0

2and that is how a second order differential equation is born.Equation (2) has the general form

with 0, but it entirely possible that this term is present. So now, using more conventional notation, we turnour attention to the study of general second order linear equations with constant coefficients

3

1,0

x



where, , and are constants and is a function of. We assume, naturally, that 0 for otherwise we areback to a first order equation.

The only other additional abstraction that we will introduced at this time is that ofoperator notation.This help inwriting equation (3) more compactly, and later to work with systems of equations with more ease. We define thederivative operator by

With this in mind, we can write equation (3) as

The expression is an example of a differential operator. It is customary to denote operators byboldface capital letters. We will use to denote the operator corresponding to . Then equation (3)can be written very compactly and conveniently as

4Hence, hereinafter where,, and are constants. Later we will relax this requirement andallow these constants to be functions of the independent variable. If 0 for all (that is, it is identically0), then we call equation (4) homogeneous. Thus, homogeneous equations look like this:

0 5What might the solutions of such equation be? When this question was first posed, analysis (the branch ofmathematics to which differential equations belongs) was in its infancy and much of the work done by the greatmathematicians of the time was based on educated guessing and intuition. Following their footsteps, we look fora solution of (5) of the form Why? Because we know (as the masters did) that the exponential functions is very versatile. Pursuing thisline of thought, we assume that is a solution of (5) and substitute:

0Performing the required differentiations, 0

or 0Since is never0, 0. Thus, we have the following result:

Box1

The polynomial

is called the characteristic polynomial associated with equation (

5) and the equation

is called itsauxiliary equation. Compare and theoperator . Exceptfor the meaning of they are identical. This is a good way to remember how to correctly identify thecharacteristic polynomial of an equation.

What is more pertinent, however, is the fact that all the steps in the derivation of the result stated in box1 arereversible. In other words, the converse is also true:

0If is a solution of the equation

thenis a root of the polynomial



Box2

We have discovered a way to find solutions of equation (5): Just find the real roots of its characteristic polynomial.

Box3

Example 1 For what value(s) of is a solution of the equation 5 6 0?Solution

First, look at the equation in operator notation: 5 6 0. The characteristic polynomialin this example is 5 6. Compare and make sure that you can correctly identify thecharacteristic polynomial of an equation.

Next, box 3 tells us that must be a root of 3 2, that is, 3 or 2. Hence and are solution of the differential equation.Example 2 Find at least one solution of the equation 2 3 0.Solution

In operator form: 2 3 0. Hence, the characteristic polynomial is 2 3.Solve the auxiliary equation 2 3 0 to obtain the roots of the characteristic polynomial:

2 3 0 3 1 0

The roots are 3 and 1. Accordingly, two solutions are and

Example 3 Find at least one solution of the equation 2 3 2 0.Solution

In operator form: 2 3 2 0. Hence, 2 3 2.Solve2 3 2 0 to obtain the roots of the characteristic polynomial:

2 3 2 0 22 1 0

The roots are 2 and 1/2. Accordingly, two solutions are

and

/

Example 4 Find at least one solution of the equation 2 2 0.Solution

In operator form: 2 2 0. Hence the characteristic polynomial is 2 2.Solving 2 2 0requires the quadratic formula since the polynomial does not factor over the rational numbers:

2 4422

0If is a root of the polynomial , then is a solution of theequation

0Let . Then is a solution of the equation of the equationif, and only if, is a root of the polynomial .



or 1 3. The two roots are 1 3 and 1 3.Accordingly, two solutions are and

Example 5 Find at least one solution of the equation 4 8 0.Solution

In operator form: 4 8 0. Hence the characteristic polynomial is 4 8 and itsroots are2. Accordingly, two solutions are

and Example 6 Find at least one solution of the equation 5 4 0.Solution

In operator form: 5 4 0. Hence 5 4 5 4. The roots of thecharacteristic polynomial are 0 and 4/5. Accordingly, two solutions are

and Obviously, we instead write

and

.

Before moving on, we make the following observation: in every example above, any multiple of the solutionsobtained via the characteristic polynomial is also a solution. This is a fact which we now prove in general terms.

Theorem 1 If is a solution of 0, then so is for any constant.Pf.

Consider By the properties of the derivative,

But is a solution so 0. Therefore, 0.

Another important property of linear homogeneous equations in general is that they obey thesuperpositionprinciple:

Theorem 2 If and are solutions of 0, then so is Pf.

0

The last equality is true because we assumed that and are solutions so each term in parenthesis is0.These two theorems establish the fact that is a linear operator. You will encounter many operators in yourstudies (we will soon encounter the Laplace Transform, which is also a linear operator). Some are linear andsome are not.



With this knowledge, we can state the following important corollary:

The combination is calleda linear combination of the functions and. We will come back tothis term when we study linear dependence and independence.

This corollary allows us to write a two-parameter family of solutions corresponding to

0provided that we

can find two different solutions. The word different is stressed for a reason which we must address before movingon.

What do we mean by different? The functions 2 and 6 are different. But they are notfundamentally different. For one, they belong to the same family both are straight lines through the origin.Furthermore, notice that 3, in other words, is a constant multiple of and therefore, is also aconstant multiple of.On the other hand, the functions 2 and are also different. But no manipulation that involvesmultiplying by a constant will convert one into the other. These are really, fundamentally different functions.

When two functions are different at this fundamental level we call them linearly independent. We will definethis term more precisely at the end of these notes but for now, we will say that two functions are linearlyindependent if one is not a constant multiple of the other.

Clearly, when has two distinct real roots and , the functions and are linearly independent.Hence, the corollary tells us that equations whose characteristic polynomial has two distinct real roots always havea two parameter family of solutions.

Example 7 Find a two parameter family of solutions of the equation 2 3.Solution

First, this equation is really 2 3 0which we already solved and obtained and . Accordingly, is a two-parameter family of solutions.

Example 8 Find a two parameter family of solutions of the equation 10 4 0. Assume that is theindependent variable.

Solution 10 4 25 2whose roots are 0 and 2/5. Thus /

is a two-parameter family of solutions. Observe that 1 for all . hence we write /

So it would seem that we have reduced the problem of solving these equations to one of finding roots of theircharacteristic polynomial. But not so fast. There are two other the possibilities: repeated roots and complexroots. Should we just ignore them? after all, what good could possibly come out of complex roots when we areliving in a real world? We must explore.

has Repeated Roots:

If a polynomial of second degree has a repeated root, then it must be real (why?) and therefore it would appearthat the corresponding 0 has only one solution. Let us illustrate this with an example:Example 9 Find at least one solution of the equation 4 4 0.Solution

In operator form: 4 4 0. Hence, 4 4.Solving 4 4 2 0

If and are solutions of , then so is for any constants and .



gives us a double root 2. Therefore, one solution of the equation is . Are there others? Ifso, how do we find them?

For that matter, we could have asked the same question about the previous examples in which we obtained twodistinct, real roots. Do those differential equations have other solutions besides the ones we found via thecharacteristic polynomial? If they havesingular solutions, how do we find them? If not, and there isa generalsolution, how do we find it? These are theoretical questions that we will answer at the end of this lecture.

Let us then return to example

9with its lone solution

and ask whether it is possible to obtain a second,

linearly independent solution. If there is one, it cannot be a multiple ofso let us try to find a solution of theform where is a yet undetermined function. If this is to be a solution, then it must satisfy the differential equation

4 4 0Let us find the derivatives of: 2

2 2 2 4 4

Substituting into the differential equation,

4 4 4 2 4 0Collecting like terms and cancelling, 0

or 0Hence, where and are constant. In other words, the second solution has the form

Observe the term is just a multiple of the solution we already found. However, the term, which is asolution for al values of, is not. Setting 1 we obtain a second, linearly independent solution:

Hence, a two-parameter family of solutions of the differential equation of example9 is The derivation of a second linearly independent solution illustrated above can be generalized and it tells us that if is a repeated root, then we can construct a second, linearly independent solution using by setting . One will find that invariably . Thus, we have another important result:

Box4

Exercise: Suppose that 0 is such that 4 0. Show that /2 is a root of multiplicitytwo (a double root), that / isasolutionofthedifferentialequation,andthatif isanothersolution, then where and are arbitrary numbers. Hence conclude that / and / aretwolinearlyindependentsolutions.Example 10 Find two linearly independent solutions corresponding to9 30 25 0.Solution

We have9 30 25 0. Hence, 9 30 25 3 5

Let be a repeated root of. Then and are linearly independent solutions of theequation 0 andIs a two-parameter family of solutions.



The roots of the auxiliary equation are 5/3 and 5/3. Accordingly, / is a solutionand box4 tells us that / is another linearly independent solution. Thus,

/ /is a two parameter family of solutions.

It therefore seems that even when the roots are repeated these differential equations have at least two linearlyindependent solutions and therefore a two-parameter family of solutions. However, in mathematics it is highly

desirable to make categorical statements such as all second order linear homogeneous differential equations withconstant coefficients have exactly two linearly independent solutions. We are not there yet (and we may neverwill) but its worth a try. Let us move on.

has Complex Roots:Consider the equation 0. The characteristic polynomial of this equation is 1 whose rootsare .Following the previous discussion, we are tempted to claim that and are solutions. Theproblem is, we have not idea of whatis, and neither did Leonhard Euler (1707-1783) who eventually showedus the way.

But just for the sake of argument, let us see whether these solutions involving complex numbers work. Wemust make the bold assumption that we can differentiate them in the same way that we differentiate exponential

functions with real exponents. Hence, and

Substituting in the equation 0 we obtain confirmation that, at least formally, is asolution. In the same way, we can confirm that is also a solution. But all this is conjecturebecause we dont even know what is. What could it be? The hint as to its nature comes from the observationthat, by inspections, sin and cos are also solutions of this equation [verify].Following Euler we argue that one can expend into an infinity series just as one would when is real:Recall the series expansion of

:

! ! ! ! ! Let . Then,

1 2!

3!

4!

5!

! 1 2!

3!

4!

5!

!

Combining the terms containing: !

!

!

!

We recognize the terms in parenthesis: they are just the cosine and sine expansions! Hence,

This is Eulers Formula (one of many, but perhaps the most important) and it tells us what is and how tocompute its values. For example, we now know that cos sin .Here are some other interesting identities that are consequence of this wonderful formula:

/



and finally, a classic: /Try to figure out the last one.

Of course we can compute for any real number ; just apply the formula and use radian mode in yourcalculator: cos1 sin1

0.54030 0.85147Although the idea of a complex number had surfaced by the mid 1500s and was used by Leonhard Euler, JohannBernoulli, and Joseph Louis Lagrange, as well as other mathematicians of the 17th and 18th centuries, it did notreally take hold until William Rowan Hamilton (1805-1865) developed an algebraic system which associated theordered pair , with the complex number . The story and history if is long and extremely interesting,but we will be content to accept that is well defined and may be used freely without fear of running into mathematical inconsistencies orcontradictions. Hence, when solving differential equations whose characteristic polynomial has complex roots,we will proceed as if they were real roots.

Example 11 Find a two-parameter family of solutions of the equation 9 0.Solution

9 has roots 3. Hence, and are two linearly independentsolutions and Rest assured that if you substitute this solution back in the equation it will satisfy it.

But we are still puzzled. What is the connection between the solutions sin and cos of 0. and the complex exponential solutions and ? The first give us the twoparameter of solutions cos sinwhereas the second Can these be reconciled? Yes, thought the use of Eulers Formula. Observe:

cos sin cos s in cos sin cos sin cos

Hence, both solutions represent the same thing:a linear combination of the sine and cosine functions.

In general, whenhas complex roots, they will be of the form where and are real numbers. Thecorresponding solutions are therefore and Using the properties of exponents and Eulers Formula,

cos sin Similarly, corresponding to the conjugate ,

cos sin

If we form a linear combination of these, we obtain



cos sin cos sin cos sin

This last expression has the form Where and . Hence, we have the following practical result:

Box5

Example 12 Find a two-parameter family of solutions of the equation

4

5 0. Assume

is the

independent variable.

Solution 4 5 has roots 2 . Hence, 2, 1 and a two parameter family ofsolutions is given by the complex exponential function

Box5 tell us that we can also write

Example 13 Find a two-parameter family of solutions of the equation2 3 0. Express the answer intrigonometric form.

Solution

2 3 has roots . Hence, 1/4, 23/4 and a two parameter familyof solutions is given by the complex exponential function

However, we should have directly written

/ 234 234 since that is what the instructions were.

We have reached the end of this exploration and we can now make an almost categorical statement:

The fact that these equations can have at most two linearly independent solutions, and therefore exactly twolinearly independent solutions whose linear combination forms the general solution is the topic of our nextdiscussion.

The Equation 0 always has at least two linearly independent solutions whose linear combinationforms a two-parameter family of solutions.

Let complex roots of. Then and are linearly independentsolutions of the equation 0 andIs a two-parameter family of solutions. Further more, this family can also be written as



The General Theory of Second Order Linear Equations

We relax the conditions imposed on the coefficients of the operator and allow them to be functions of theindependent variable. Let where we introduce the identify operator (sometimes written as whose property is to leave functionsunchanged: for all in the domain of. This operator is usually omitted unless its presence it isabsolutely necessary and we simply write

Hence,

The equation 6is called asecond order linear equation in general form. As with first order equations, the standard form of (6) isgiven by 7where /, /, and / and where we assume an interval in which

0. We call the zeros of

thesingularities or singular pointsof equation (6).

Example 14 The following are second order linear differential equations:

Equation Standard form

2 4 2 4 sin 2 1 4 24 csc csc ; ln 4 / / ln 4; 0

4 4On the other hand, the equation 2 0 is not linear because the term is not linear.Linear Operators

In mathematics, an operator is a mathematical object we use to operate on functions. It is a generalization of theconcept of function which we use to describe relations between numbers. You have already met two operators:the derivativeand the indefinite integral. the symbol / is meaningless unless I tell you what to use it on. Itis the derivative operator and it operates on differentiable functions. Thus, for example,

2 1 6Similarly, the symbol ___ represents the indefinite integral and it is meaningless unless I tell you what to useit on. Both the derivative and the integral enjoy a very important property: they are linearoperatorsThis meansthat if and are functions defined on some domain, and and are real numbers, then

and

where it is assumed that andare differentiable on in the first case and integrable on in the second.

Thedifferential operator continues to be linear.Example 15 Write the equation 1 using operator notation.Solution

The operator is 1 and we may write the equation as



The coefficient functions are continuous throughout the real line and 1 are its onlysingularities.Properties of L

The operatorLis linear. Let and be differentiable functions. Then for any real numbers and,

Homogeneous and Non-homogeneous Equations

When 0 we say that (6) is a homogeneousequation. Otherwise, we say it is a non-homogeneousequation.Example 16 The equation 3 0 is homogeneous. On the other hand, 3 2 is nothomogeneous.

With every non-homogeneous linear equation we associate a homogeneous equation 0called the associated homogeneous equation. Thus, the associated homogeneous equation corresponding to 1 is 0.As we shall see, the first step in obtaining the general solution of a non-homogeneous equation will be to solve itsassociated homogeneous equation.

The Superposition PrincipleTheorem1 If and are solutions of 0 then so is for any arbitraryconstants and.Pf.

By linearity, . Since and are solutions, 0 and 0. Hence,

0Therefore, is also a solution.

The Existence and Uniqueness Theorem

Next we state theExistence and UniquenessTheoremfor second order linear initial value problems. We will not

attempt to prove this result, but you may consult several excellent textbooks on the subject1 if you are interested ina more in-depth study.

1 See for example W. G. Kelley and A. PetersonThe Theory of Differential Equations, 2nd Ed.

, ,

The Existence and Uniqueness TheoremLet

be an second order linear differential equation with continuous coefficients and forcing functionf defined on aninterval where 0 and suppose that , , where . Then thereexists an interval centered at and a twice differentiable function defined on such that

and

Furthermore, if is any other solution of the differential equation such that , then forall .



Example 17 The equation 1 ; 0 1, 0 5has 1 , , , and . All these function s are continuous andfurthermore 0 on any interval that contains0 and excludes1.

Therefore, there is an interval centered at the origin and a unique function defined in which satisfies theequation and its initial conditions. Naturally, must exclude the points1.

The points 1 are the onlysingular points of the equation.Example 18 Give an example of an interval on which the IVP 1 0; 3 2, 3 0 isguaranteed to have a unique solution.

SolutionThe singular point of this equation is 0 since the leading coefficient function is zerothere. Therefore, any interval centered at 3 and which excludes the origin will do. For example 2,4 works. Similarly, 0,5 works as does 0,5. Furthermore, while it is nice to have asymmetric interval centered at 3 it is not necessary. Therefore, 0,10 or for that matter 0, are also intervals where the IVP is guaranteed to have a unique solution.

As with first order differential equations, the largest interval on which an IVP has a unique solution is called theequations intervalofexistence. In the previous that interval would be 0,.Existence of a Fundamental Set of Solutions and The General Solution

The next result is central to all the work that follows in our study of higher order equations. It states that under thesame continuity conditions of the Existence and Uniqueness Theorem the homogeneous equation

0has exactly two linearly independent solutionsand defined on an interval in which 0 whose linearcombination forms the equations general solution. Here is the important and central result of this lecture:

Example 19 We know that both and are solutions of the equation 5 6 0.Furthermore, the set, is linearly independent. Therefore,

is the equations general solution and, is its fundamental set.

Linear Dependence and Independence

We stated an intuitive version of what linear independence of two function means: Two functions are linearlyindependent if one is not a constant multiple of the other. Hence, we define to functions to belinearly dependent ifone is a constant multiple of the other.

Soon we will have to extend this concept to more functions and these simple definitions will not do. We need tobe more precise about it. suppose then that and are linearly dependent on the interval in which both aredefined, that is, we can find some number such that for some constant. Then 0

0 8

Existence of a Fundamental Set of Solut ions

The equation

has exactly two linearly independent solutions anddefined on an interval in which 0 and its generalsolution is given by the linear combinationwhere and are arbitrary constants. We call the set, the fundamental set of solutions of (8).



In other words, there are constant and not both zero such that 0

for all . Clearly here those constants are and 1. This is then the way we formally define lineardependence.

A set that is not lineally dependent is said to be linearly independent and the functions that comprise it are said to be linearlyindependent functions. Hence, when a set is linearly independent, the equation

0can be satisfied for all only if 0.Example 20 The set

4,6is linearly dependent over the real line because we can find two constants, not both

zero, such that 4 6 0for all real numbers. For example, pick 3 and 2.Example 21 The set , is linearly independent over the real line. Let us try to find two constants, not bothzero, such that 0for all real numbers. since at least one constant must be nonzero, let us assume that it is. Then we can solvefor and write

for all

which is of course absurd. Similarly, if we assume

0then we conclude that a parabola is really

a constant multiple of a straight line which equally absurd. Thus, the functions and arelinearly independent on the real line.Example 22 Show that the functions and are linearly independent over the real line.Solution

We need to show that 0 for all only if 0. Suppose not. Suppose that at leastone of these constants is not0, say it is . Then

for all . what is worse, since 0,

1 for all . Hence, must be0. But then 0 for all or equivalently, 0 for all . This canonly happen if 0 as well. Hence 0.

Example 23 Show that any set in which one of the functions is the zero function is linearly dependent.

SolutionWe need to show that 0 for all only if 0. Suppose not. Suppose that at leastone of these constants is not0, say it is . Then

for all . What is worse, since 0,

0

Linear Dependence

Let and be defined on an interval. The functions are said to be linearly dependent on that interval ifthere are constants and not both zero such thatfor all and we call the set, a linearly dependent set.



1 for all . Hence, must be0. But then 0 for all or equivalently, 0 for all . This canonly happen if 0 as well. Hence 0.

These examples show that establishing linear independence is not as straightforward as it may seem, and of coursethings get much more complicated when we consider larger sets of functions. For that reason we wish to developa more mechanical way of testing for linear independence. Linear algebra can give us a helping hand.

Suppose that and are two non-zero differentiable functions defined in an interval and are such that 0for all in. Then differentiating gives us 0For all .Consider the system

0 0 9Fix in the interval . Then the system (9) is a homogeneous linear system with variablesand. If thefunctions are linearly dependent, then by definition this system has nontrivial solutions and from linear algebra weknow that therefore

0Since is an arbitrary point in, the determinant must be0 for all if the functions are linearly dependent.

This determinant, called theWronskian2of and, is very important and it is denote it by, :,

Hence, we have the following result:

Example 24 We showed that the set4,6 is linearly dependent over the real line. Compute their Wronskian.Solution, 4 64 6

24 24

The converse of this result is not true.

Example 25 Show that the functions and || have0 Wronskian but are linearly independent.Solution

First observe that both functions are defined throughout the real line and are differentiable at each pointin this domain [verify].

That they are linearly independent can be immediately seen from their graphs (see below):

2Named after the Polish Mathematician Josef Hoene Wronski (1776 1853)

If and are differentiable and linearly dependent on the interval, then, 0 for all .


15/15

However, let us compute their Wronskian:

If 0, then and therefore certainly, 0. If 0, and . Hence

,

3 3

3 3

And certainly the Wronskian is0 if 0.So showing that a Wronskian is0 does not guarantee that two functions are linearly dependent. However, thelogical equivalent of the result stated earlier is true (of course):

Example 26 Show that the functions and are linearly independent over the real line.Solution

0

Thus, the functions are linearly independent over the reals.

Example 27 Show that the functions 1 and 1 2 are linearly independent over the realline.

Solution

1 122 4 4 4 2 4 6 0

Thus, the functions are linearly independent over the reals.

A final remark: To apply the Wronskian criterion you must be sure that the functions are differentiable.

- -2

-1 1 2 3

-20

-10

10

20

Suppose and are differentiable on the interval. Then if , 0 for at least one , thefunctions are linearly independent on that interval.

second order linear equations - a prelude to higher order linear equations a historical approach

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