Ordinary Differential Equation Introduction and Preliminaries

Transcription

1 Ordinary Differential Equation Introduction and Preliminaries There are many branches of science and engineering where differential equations arise naturally. Now days, it finds applications in many areas including medicine, economics and social sciences. In this context, the study of differential equations assumes importance. In addition, in the study of differential equations, we also see the applications of many tools from analysis and linear algebra. Without spending more time on motivation, (which will be clear as we go along) let us start with the following notations. Let be an independent variable and let be a dependent variable of. The derivatives of (with respect to ) are denoted by The independent variable will be defined for an interval where is either or an interval with these notations, we are ready to define the term ``differential equation". A differential equation is a relationship between the independent variable and the unknown dependent function along with its derivatives. More precisely, we have the following definition. DEFINITION (Ordinary Differential Equation, ODE) An equation of the form (7.1.1) is called an ORDINARY DIFFERENTIAL EQUATION; where is a known function from to Remark The aim of studying the ODE (7.1.1) is to determine the unknown function which satisfies the differential equation under suitable conditions. 2. Usually ( 7.1.1) is written as and the interval is not mentioned in most of the examples. Some examples of differential equations are 1

2 DEFINITION (Order of a Differential Equation) The ORDER of a differential equation is the order of the highest derivative occurring in the equation. In Example 7.1, the order of Equations 1, 3, 4, 5 are one, that of Equations 2, 6 and 8 are two and the Equation 7 has order three. DEFINITION (Solution) A function equation ( 7.1.1) on if is called a SOLUTION of the differential 1. is differentiable (as many times as the order of the equation) on and 2. satisfies the differential equation for all. That is, for all. If is a solution of an ODE (7.1.1) on, we also say that satisfies (7.1.1). Sometimes a solution is also called an INTEGRAL. EXAMPLE Consider the differential equation on. We see that if we take, then is differentiable, and therefore 2

3 Hence, is a solution of the given differential equation for all. 2. It can be easily verified that for any constant is a solution of the differential equation on any interval that does not contain the point as the function is not defined at. Furthere it can be shown that is the only solution for this equation whenever the interval contains the point. 3. Consider the differential equation on. It can be easily verified that a solution of this differential equation satisfies the relation. DEFINITION (Explicit/Implicit Solution) A solution of the form is called an EXPLICIT SOLUTION (e.g., see Examples and ). If is given by an implicit relation and satisfies the differential equation, then is called an IMPLICIT SOLUTION (e.g., see Example ). Remark Since the solution is obtained by integration, we may expect a constant of integration (for each integration) to appear in a solution of a differential equation. If the order of the ODE is we expect arbitrary constants. To start with, let us try to understand the structure of a first order differential equation of the form (7.1.2) and move to higher orders later. 3

4 DEFINITION (General Solution) A function is called a general solution of (7.1.2) on an interval if is a solution of (7.1.2) for each for an arbitrary constant. Remark The family of functions is called a one parameter family of functions and is called a parameter. In other words, a general solution of (7.1.2) is nothing but a one parameter family of solutions of (7.1.2). EXAMPLE Determine a differential equation for which a family of circles with center at and arbitrary radius, is an implicit solution. Solution: This family is represented by the implicit relation (7.1.3) where is a real constant. Then is a solution of the differential equation (7.1.4) The function satisfying (7.1.3) is a one parameter family of solutions or a general solution of (7.1.4). 6. Consider the one parameter family of circles with center at and unit radius. The family is represented by the implicit relation (7.1.5) where is a real constant. Show that satisfies Solution: We note that, differentiation of the given equation, leads to 4

5 Now, eliminating from the two equations, we get 11. In Example , we see that is not defined explicitly as a function of but implicitly defined by (7.1.3). On the other hand is an explicit solution in Example Let us now look at some geometrical interpretations of the differential Equation ( 7.1.2). The Equation (7.1.2) is a relation between and the slope of the function at the point For instance, let us find the equation of the curve passing through and whose slope at each point is If is the required curve, then satisfies It is easy to verify that EXERCISE satisfies the equation 1. Find the order of the following differential equations: Show that for each is a solution of 3. Find a differential equation satisfied by the given family of curves: 1. real (family of lines). 2. real (family of parabolas). 3. is a parameter of the curve and is a real number (family of circles in parametric representation). 5

6 4. Find the equation of the curve which passes through and whose slope at each point is Separable Equations In general, it may not be possible to find solutions of, where is an arbitrary continuous function. But there are special cases of the function can be solved. One such set of equations is for which the above equation (7.2.1) The Equation (7.2.1) is called a SEPARABLE EQUATION and is equivalent to Integrating with respect to we get where is a constant. Hence, its implicit solution is EXAMPLE Solve: Solution: Here, and Then By using partial fractions and integrating, we get 6

7 where is a constant of integration. 2. Solve Solution: It is easy to deduce that solution., where is a constant; is the required Observe that the solution is defined, only if for any For example, if we let then exists as long as 7

8 Equations Reducible to Separable Form There are many equations which are not of the form 7.2.1, but by a suitable substitution, they can be reduced to the separable form. One such class of equation is where and are homogeneous functions of the same degree in and and is a continuous function. In this case, we use the substitution, to get Thus, the above equation after substitution becomes which is a separable equation in For illustration, we consider some examples. EXAMPLE Find the general solution of Solution: Let be any interval not containing Then Letting we have On integration, we get 8

9 or The general solution can be re-written in the form This represents a family of circles with center and radius 2. Find the equation of the curve passing through and whose slope at each point is Solution: If is such a curve then we have Notice that it is a separable equation and it is easy to verify that satisfies 3. The equations of the type can also be solved by the above method by replacing by and by where and are to be chosen such that This condition changes the given differential equation into Thus, if then the equation reduces to the form. 9

10 EXERCISE Find the general solutions of the following: Find the solution of Obtain the general solutions of the following: Solve and use it to determine This equation occurs in a model of population. 10

11 Exact Equations As remarked, there are no general methods to find a solution of (7.1.2). The EXACT EQUATIONS is yet another class of equations that can be easily solved. In this section, we introduce this concept. Let be a region in -plane and let and be real valued functions defined on Consider an equation (7.3.1) In most of the books on Differential Equations, this equation is also written as (7.3.2) DEFINITION (Exact Equation) The Equation (7.3.1) is called Exact if there exists a real valued twice continuously differentiable function subset of ) such that (or the domain is an open (7.3.3) Remark If (7.3.1) is exact, then This implies that (where is a constant) is an implicit solution of (7.3.1). In other words, the left side of (7.3.1) is an exact differential. EXAMPLE The equation is an exact equation. Observe that in this example, The proof of the next theorem is given in Appendix

12 THEOREM Let and be twice continuously differentiable function in a region The Equation (7.3.1) is exact if and only if (7.3.4) Note: If (7.3.1) or (7.3.2) is exact, then there is a function satisfying for some constant such that Let us consider some examples, where Theorem can be used to easily find the general solution. EXAMPLE Solve Solution: With the above notations, we have Therefore, the given equation is exact. Hence, there exists a function such that The first partial differentiation when integrated with respect to (assuming to be a constant) gives, 12

13 But then implies or where is an arbitrary constant. Thus, the general solution of the given equation is The solution in this case is in implicit form. 2. Find values of and such that the equation is exact. Also, find its general solution. Solution: In this example, we have Hence for the given equation to be exact, With this condition on and the equation reduces to This equation is not meaningful if Thus, the above equation reduces to whose solution is 13

14 for some arbitrary constant 3. Solve the equation Solution: Here Hence, Thus the given equation is exact. Therefore, (keeping as constant). To determine we partially differentiate with respect to and compare with to get Hence is the required implicit solution. 14

15 Integrating Factors On many occasions, may not be exact. But the above equation may become exact, if we multiply it by a proper factor. For example, the equation is not exact. But, if we multiply it with then the equation reduces to an exact equation. Such a factor (in this case, equation. Formally ) is called an INTEGRATING FACTOR for the given DEFINITION (Integrating Factor) A function is called an integrating factor for the (7.3.1), if the equation is exact. EXAMPLE Solve the equation Solution: It can be easily verified that the given equation is not exact. Multiplying by equation reduces to the 15

16 Thus, by definition, is an integrating factor. Hence, a general solution of the given equation is for some constant That is, 2. Find a general solution of the differential equation Solution: It can be easily verified that the given equation is not exact. METHOD 1: Here the terms and are homogeneous functions of degree It may be checked that an integrating factor for the given differential equation is Hence, we need to solve the partial differential equations (7.3.5) (7.3.6) Integrating (keeping constant) (7.3.5), we have (7.3.7) 16

17 and integrating (keeping constant) (7.3.6), we get (7.3.8) Comparing (7.3.7) and (7.3.8), the required solution is for some real constant Or equivalently, the solution is METHOD 2: Here the terms and are polynomial in and Therefore, we suppose that is an integrating factor for some We try to find this and Multiplying the terms and with we get For the new equation to be exact, we need That is, the terms and 17

18 must be equal. Solving for and we get and That is, the expression is also an integrating factor for the given differential equation. This integrating factor leads to and Thus, we need for some constant Hence, the required solution by this method is Remark If (7.3.1) has a general solution, then it can be shown that (7.3.1) admits an integrating factor. 2. If (7.3.1) has an integrating factor, then it has many (in fact infinitely many) integrating factors. 3. Given (7.3.1), whether or not it has an integrating factor, is a tough question to settle. 4. In some cases, we use the following rules to find the integrating factors. 1. Consider a homogeneous equation If is an Integrating Factor. 2. If the functions and are polynomial functions in then works as an integrating factor for some appropriate values of and 18

19 3. The equation has as an integrating factor, if is a function of alone. 4. The equation has as an integrating factor, if 5. For the equation is a function of alone. with the function is an integrating factor. EXERCISE Show that the following equations are exact and hence solve them Find conditions on the function so that the equation is exact. 3. What are the conditions on and so that the equation is exact. 19

20 4. Verify that the following equations are not exact. Further find suitable integrating factors to solve them Find the solution of 1. with 2. with 20

21 Linear Equations Sometimes we might think of a subset or subclass of differential equations which admit explicit solutions. This question is pertinent when we say that there are no means to find the explicit solution of where is an arbitrary continuous function in in suitable domain of definition. In this context, we have a class of equations, called Linear Equations (to be defined shortly) which admit explicit solutions. DEFINITION (Linear/Nonlinear Equations) Let and be real-valued piecewise continuous functions defined on interval The equation (7.4.1) is called a linear equation, where stands for The Equation (7.4.1) is called Linear nonhomogeneous if and is called Linear homogeneous if on A first order equation is called a non-linear equation (in the independent variable) if it is neither a linear homogeneous nor a non-homogeneous linear equation. EXAMPLE The equation is a non-linear equation. 2. The equation is a linear non-homogeneous equation. 3. The equation is a linear homogeneous equation. Define the indefinite integral ( or ). Multiplying (7.4.1) by we get On integration, we get 21

22 In other words, (7.4.2) where is an arbitrary constant is the general solution of (7.4.1). Remark If we let in the above discussion, ( 7.4.2) also represents (7.4.3) As a simple consequence, we have the following proposition. PROPOSITION (where is any constant) is the general solution of the linear homogeneous equation (7.4.4) In particular, when is a constant, the general solution is with an arbitrary constant. EXAMPLE Comparing the equation with (7.4.1), we have Hence, Substituting for in (7.4.2), we get as the required general solution. We can just use the second part of the above proposition to get the above result, as 22

23 2. The general solution of is where is an arbitrary constant. Notice that no non-zero solution exists if we insist on the condition A class of nonlinear Equations (7.4.1) (named after Bernoulli linear equation. These equations are of the type ) can be reduced to (7.4.5) If or then (7.4.5) is a linear equation. Suppose that We then define and therefore or equivalently (7.4.6) a linear equation. For illustration, consider the following example. EXAMPLE For constants and solve Solution: Let Then satisfies and its solution is Equivalently with and an arbitrary constant, is the general solution. EXERCISE In Example 7.4.6, show that 2. Find the genral solution of the following: 23

24 Solve the following IVP's: Let be a solution of and be a solution of Then show that is a solution of 5. Reduce the following to linear equations and hence solve: Find the solution of the IVP 24

25 Miscellaneous Remarks In Section 7.4, we have learned to solve the linear equations. There are many other equations, though not linear, which are also amicable for solving. Below, we consider a few classes of equations which can be solved. In this section or in the sequel, denotes or A word of caution is needed here. The method described below are more or less ad hoc methods. 1. EQUATIONS SOLVABLE FOR Y: Consider an equation of the form (7.5.1) Differentiating with respect to we get (7.5.2) The Equation (7.5.2) can be viewed as a differential equation in and We now assume that (7.5.2) can be solved for and its solution is (7.5.3) If we are able to eliminate between (7.5.1) and (7.5.3), then we have an implicit solution of the (7.5.1). 8. Solve Solution: Differentiating with respect to and replacing by we get 25

26 So, either That is, either or Eliminating from the given equation leads to an explicit solution The first solution is a one-parameter family of solutions, giving us a general solution. The latter one is a solution but not a general solution since it is not a one parameter family of solutions. 15. EQUATIONS IN WHICH THE INDEPENDENT VARIABLE MATHEND000# IS MISSING: These are equations of the type If possible we solve for and we proceed. Sometimes introducing an arbitrary parameter helps. We illustrate it below. Solve where is a constant. Solution: We equivalently rewrite the given equation, by (arbitrarily) introducing a new parameter by from which it follows and so Therefore, a general solution is 26

27 16. EQUATIONS IN WHICH MATHEND000# (DEPENDENT VARIABLE OR THE UNKNOWN) IS MISSING: We illustrate this case by an example. Find the general solution of Solution: Recall that Now, from the given equation, we have Therefore, (regarding given by as a parameter). The desired solution in this case is in the parametric form, where is an arbitrary constant. Remark The readers are again informed that the methods discussed in are more or less ad hoc methods. It may not work in all cases. EXERCISE Find the general solution of Hint: Differentiate with respect to to get ( a linear equation in ). Express the solution in the parametric form 2. Solve the following differential equations: 27

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29 Initial Value Problems As we had seen, there are no methods to solve a general equation of the form (7.6.1) and in this context two questions may be pertinent. 1. Does (7.6.1) admit solutions at all (i.e., the existence problem)? 2. Is there a method to find solutions of ( 7.6.1) in case the answer to the above question is in the affirmative? The answers to the above two questions are not simple. But there are partial answers if some additional restrictions on the function are imposed. The details are discussed in this section. For with we define DEFINITION (Initial Value Problems) Let be a continuous function on a The problem of finding a solution of (7.6.2) in a neighbourhood of (or an open interval containing ) is called an Initial Value Problem, henceforth denoted by IVP. The condition in (7.6.2) is called the INITIAL CONDITION stated at and is called the INITIAL VALUE. Further, we assume that and are finite. Let Such an exists since is a closed and bounded set and is a continuous function and let The ensuing proposition is simple and hence the proof is omitted. PROPOSITION A function is a solution of IVP (7.6.2) if and only if satisfies 29

30 (7.6.3) In the absence of any knowledge of a solution of IVP (7.6.2), we now try to find an approximate solution. Any solution of the IVP (7.6.2) must satisfy the initial condition a crude approximation to the solution of IVP (7.6.2), we define Hence, as Now the Equation (7.6.3) appearing in Proposition 7.6.2, helps us to refine or improve the approximate solution with a hope of getting a better approximate solution. We define and for we inductively define As yet we have not checked a few things, like whether the point or not. We formalise the theory in the latter part of this section. To get ourselves motivated, let us apply the above method to the following IVP. EXAMPLE Solve the IVP Solution: From Proposition 7.6.2, a function is a solution of the above IVP if and only if We have and So, 30

31 By induction, one can easily verify that Note: The solution of the given IVP is This example justifies the use of the word approximate solution for the 's. We now formalise the above procedure. DEFINITION (Picard's Successive Approximations) Consider the IVP (7.6.2). For with define inductively (7.6.4) Then are called Picard's successive approximations to the IVP (7.6.2). Whether ( 7.6.4) is well defined or not is settled in the following proposition. PROPOSITION The Picard's approximates 's, for the IVP ( 7.6.2) defined by (7.6.4) is well defined on the interval Proof. We have to verify that for each i.e., for belongs to the domain of definition of for This is needed due to the reason that appearing as integrand in ( 7.6.4) may not be defined. For it is obvious that as and For we notice that, if then So, whenever 31

32 The rest of the proof is by the method of induction. We have established the result for namely Assume that for whenever Now, by definition of we have But then by induction hypotheses and hence This shows that whenever Hence for holds and therefore the proof of the proposition is complete. height6pt width 6pt depth 0pt Let us again come back to Example in the light of Proposition EXAMPLE Compute the successive approximations to the IVP (7.6.5) Solution: Note that and The set on which we are studying the differential equation is By Proposition 7.6.2, on this set Therefore, the approximate solutions 's are defined only for the interval if we use Proposition Observe that the exact solution and the approximate solutions 's of Example exist on But the approximate solutions as seen above are defined in the interval 32

33 That is, for any IVP, the approximate solutions 's may exist on a larger interval as compared to the interval obtained by the application of the Proposition We now consider another example. EXAMPLE Find the Picard's successive approximations for the IVP (7.6.6) where Solution: By definition and A similar argument implies that for all and Also, it can be easily verified that is a solution of the IVP (7.6.6). Also is a solution of ( 7.6.6) and the 's do not converge to Note here that the IVP (7.6.6) has at least two solutions. The following result is about the existence of a unique solution to a class of IVPs. We state the theorem without proof. THEOREM (Picard's Theorem on Existence and Uniqueness) Let and Let be such that as well as are continuous on Also, let be constants such that Let Then the sequence of successive approximations (defined by (7.6.4)) for the IVP (7.6.2) uniformly converges on to a solution of IVP (7.6.2). Moreover the solution to IVP (7.6.2) is unique. 33

34 Remark The theorem asserts the existence of a unique solution on a subinterval of the given interval In a way it is in a neighbourhood of and so this result is also called the local existence of a unique solution. A natural question is whether the solution exists on the whole of the interval beyond the scope of this book. The answer to this question is Whenever we talk of the Picard's theorem, we mean it in this local sense. EXERCISE Compute the sequence of the successive approximations to the IVP 2. Show that the solution of the IVP is 3. The IVP has solutions as well as Why does the existence of the two solutions not contradict the Picard's theorem? 4. Consider the IVP for any 1. Compute the interval of existence of the solution of the IVP by using Theorem

35 2. Show that is the solution of the IVP which exists on whole of This again shows that the solution to an IVP may exist on a larger interval than what is being implied by Theorem