A Concise Introduction to Ordinary Differential Equations. David Protas

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1 A Concise Introduction to Ordinary Differential Equations David Protas

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3 A Concise Introduction to Ordinary Differential Equations 1 David Protas California State University, Northridge Please send any comments, corrections to david.protas@csun.edu This work is licensed under the Creative Commons Attribution-NonCommercial- NoDerivatives 4.0 International License. To view a copy of this license, visit

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5 Contents Preface iii Chapter 0. Introduction 1 1. Solutions of Ordinary Differential Equations 1 Chapter 1. First Order Equations 5 1. Preliminary Remarks 5 2. First Order Linear Equations 5 3. Some Other First Order Equations 9 4. Existence and Uniqueness Numerical Methods Direction Fields 39 Chapter 2. Linear Equations of Order Two and Higher Preliminary Remarks Definition of Linear Equations A First Look at Constant Coefficients Solution Spaces Linear Homogeneous Equations of Order Two Higher Order Linear Homogeneous Equations Nonhomogeneous Linear Equations and the Annihilator Method The Method of Variation of Parameters 69 Chapter 3. Series Solutions of Linear Equations Preliminary Remarks Power Series Solutions Regular Singular Points and the Euler Equation More on Equations with Regular Singular Points 91 Chapter 4. Systems of Differential Equations Preliminary Remarks Linear Systems and Vector Notation Constant Coefficients Complex Eigenvalues and Nonhomogeneous Systems Nonlinear Systems 117 Chapter 5. The Geometry of Systems Preliminary Remarks The Phase Plane Critical Points Perturbations and the Almost Linear Property 139 i

6 ii CONTENTS Answers to Selected Exercises 149 Index 155

7 Preface This book covers standard topics of a first (theoretically oriented) course in ordinary differential equations. It is assumed that the reader has had a class in linear algebra. Applications are mainly seen in the preliminary remarks to chapters and in the exercise sets. I have tried to create a work that allows most, if not all, of the material in it to be covered thoroughly in a one semester course. iii

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9 CHAPTER 0 Introduction 1. Solutions of Ordinary Differential Equations Many problems in science and engineering are represented in mathematical terms by equations involving a function of one variable along with one or more of its derivatives. Such an equation is called an ordinary differential equation. The word ordinary is used because the function of one variable has ordinary derivatives as opposed to partial derivatives. One simple example of an ordinary differential equation is x 2 y 6y = 0. Here, y is an unknown function of the independent variable x. We say that φ 1 (x) = x 3 is a solution of this equation on the interval (, ) since x 2 φ 1(x) 6φ 1 (x) = 0 for all x (, ). Specifically, x 2 6x 6 x 3 = 0 for all x. (How we find the solution φ 1 (x) = x 3 will be shown later.) Similarly, φ 2 (x) = x 2 is a solution of the equation on the interval (, 0) and on the interval (0, ). In general, φ is a solution of an ordinary differential equation F (x, y, y,..., y (n) ) = 0 on an interval I if F (x, φ(x), φ (x),..., φ (n) (x)) = 0 for all x I. In the example above, F (x, y, y, y ) = x 2 y 6y. For a second example, consider the ordinary differential equation y = 2xy 2. A solution of this equation on (, ) is φ(x) = 1/(x 2 + 1). We have let F (x, y, y ) = y + 2xy 2. Typically, an ordinary differential equation will have many solutions on an interval. We often need to find a solution of the equation that also satisfies certain extra conditions. For example, the function φ(x) = x 3 + 5x 2 is a solution of the equation x 2 y 6y = 0 that also satisfies the conditions φ(1) = 6 and φ (1) = 7. For this reason, we say that φ(x) = x 3 + 5x 2 is a solution of the initial value problem x 2 y 6y = 0 y(1) = 6, y (1) = 7. Definition 0.1. Let x 0 be a point in an interval I, and let β 1,..., β n be numbers. Then (1) F (x, y, y,..., y (n) ) = 0 y(x 0 ) = β 1, y (x 0 ) = β 2,..., y (n 1) (x 0 ) = β n is an initial value problem. A function φ is a solution of (1) on I if F (x, φ(x), φ (x),..., φ (n) (x)) = 0 1

10 2 0. INTRODUCTION for all x in I and φ(x 0 ) = β 1, φ (x 0 ) = β 2,..., φ (n 1) (x 0 ) = β n. The word initial is used because the independent variable often represents time and we want to use information about a given time x 0 to predict what will happen later on. Eventually, we will also be considering systems of ordinary differential equations. For now, however, we will deal with just one ordinary differential equation at a time. We are not going to be able to find one method of solution that can be applied to all ordinary differential equations. Instead, we will define certain types of equations and find methods of solution appropriate to these specific types. We start with the idea of order. The order of an ordinary differential equation is the highest order derivative that appears in the equation. For example, the order of is 2, while the order of y + 5x 3 y 6y = cos x yy + (y ) 4 + 6x 5 = 0 is 3. In the next chapter, we will study equations of order 1. Let us agree to say that two differential equations of the same order are equivalent if they have the same solution set. Often, when trying to solve a given differential equation, we will replace it by an equivalent equation that is in some sort of standard form or is easier to deal with for whatever reason. Most of the time, we will treat equivalent equations as though they were the same equation written in two different ways. Occasionally, we will replace an equation by a non-equivalent one when solving the second equation also gives us solutions of the first. It should be clear that y = 2xy 2 and y + 2xy 2 = 0 are equivalent. However, the ordinary differential equations y = 2xy 2 and y /y 2 = 2x are not equivalent since the constant function φ(x) 0 is a solution of the first but not of the second. Still, solutions of the second equation are also solutions of the first, and so solving the second equation is a way of finding solutions of the first. Exercises 1. Verify that φ(x) = x 2 is a solution of x 2 y 6y = 0 on (, 0) and (0, ). 2. Verify that φ(x) = e 3x is a solution of y 4y + 3y = 0 on (, ). 3. Verify that φ(x) = (6 3x) 1/3 is a solution of y = 1/y 2 on (, 2) and (2, ). 4. Verify that for each constant a, φ(x) = e ax is a solution of y = ay on (, ). 5. Verify that for each constant a, φ 1 (x) = cos ax and φ 2 (x) = sin ax are solutions of y + a 2 y = 0 on (, ). 6. Verify that φ(x) = x 3 + 5x 2 is a solution of the initial value problem x 2 y 6y = 0, y(1) = 6, y (1) = 7 on (0, ).

11 1. SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS 3 7. Verify that φ(x) = xe 2x is a solution of the initial value problem y 4y + 4y = 0, y(0) = 0, y (0) = 1 on (, ). 8. Verify that φ(x) = 4 x 2 is a solution of the initial value problem y = x/y, y(0) = 2 on ( 2, 2). 9. State the order of y (4) = x 5 y State the order of y = e y y + x 4 (y ) 6.

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13 CHAPTER 1 First Order Equations 1. Preliminary Remarks Let A(t) be the number of ounces of a radioactive material left t months after a certain starting time. It is known that the rate of decay of the material is proportional to its amount. That is, A = ka, where k is a negative constant determined by the particular radioactive material under consideration. Let A 0 stand for the initial amount. We want to find the amount of material left at any time t. To do this we will solve an initial value problem involving a first order differential equation. If we replace A by y (and t by x) to conform to our usual choice of variables, we will be faced with the initial value problem y = ky, y(0) = y 0, which we will learn how to solve in the next two sections. There is no general method that enables us to solve all first order ordinary differential equations. We will look at a number special types of first order equations and develop methods for solving them; after that, we will see what we can do about the others. 2. First Order Linear Equations Definition 1.1. A first order ordinary differential equation is said to be linear if it is of the form y + a(x)y = g(x), where a, g are functions defined on some interval I. If g(x) = 0 for all x I, then the linear equation is called homogeneous. Otherwise, it is called nonhomogeneous. Example. (a) The equation y 4x 2 y = 0 is a first order linear homogeneous equation. (We will assume that the interval I is (, ) since we see no reason to avoid any values of x.) (b) The equation 3y = (ln x)y is equivalent to the first order linear homogeneous equation y ( 1 3 ln x)y = 0 (on the interval (0, )). (c) The linear equation y + e 3x y = sin x is nonhomogeneous. (d) The equation y + e 3x y 2 = sin x is not linear. Our goal now is to develop a method for solving first order linear equations. We will start by working out a couple of specific examples with the idea of generalizing what we learn from them. Example. Consider the equation y + 8x 3 y = 0 (on the interval (, )). Since the equation involves a derivative, it is tempting to try to integrate to solve for y. Integrating the right side of the equation presents no problem; we simply get a constant. However, we do not know how to integrate the left side since it 5

14 6 1. FIRST ORDER EQUATIONS involves an unknown function y and is not in any obvious way the derivative of something we can find. We will change the left side into a manageable expression by multiplying through the equation by an appropriately chosen function. Multiply both sides of the equation by e 2x4. (You should be wondering where this expression came from. The answer will be supplied shortly.) We get e 2x4 y + 8x 3 e 2x4 y = 0. Notice that (e 2x4 y) = e 2x4 y + 8x 3 e 2x4 y. Thus, our equation can be written (e 2x4 y) = 0. Now integration is easy. We get e 2x4 y = C, where C is an arbitrary constant. We conclude that if y is any solution of our equation, then it must be of the form y = Ce 2x4. Furthermore, since all of the steps we went through are reversible, every function given by y = Ce 2x4 is a solution. Now let s start to examine how we came up with e 2x4 as the function to multiply through the equation in the last example. Notice that 8x 3 dx = 2x 4 + K and so, e 2x4 was e raised to an antiderivative of the coefficient of y. We will see shortly that, in general, when solving y + a(x)y = b(x), multiplying through by e raised to an antiderivative of a(x) gives us an expression we can integrate. First however, let s examine what would have happened in the last example if we had chosen a different antiderivative. Let s multiply through y + 8x 3 y = 0 by e 2x4 +K, for any (non-zero) value of K. Since e 2x4 +K = e 2x4 e K, we get e K e 2x4 y + e K 8x 3 e 2x4 = 0. Now dividing by e K, gets us back exactly to where we were in the example when we did not have the constant K. This shows that the choice of the particular antiderivative, in other words the choice of K, doesn t matter. In practice, most people most of the time use K = 0 in this type of problem. To emphasize the idea that only one particular antiderivative is needed in the process we are developing, we will use the following somewhat unconventional notation. Definition 1.2. For any continuous function a, let the symbol a(x) dx stand for any one particular antiderivative of a. Example. Consider the equation y = 12x + 6xy (on the interval (, )). This equation is equivalent to the linear equation y 6xy = 12x. Compute 6x dx = 3x 2, and let µ = e 3x2. Multiplying our differential equation by µ gives e 3x2 y 6xe 3x2 y = 12xe 3x2. However, (e 3x2 y) = e 3x2 y 6xe 3x2 y, and so the equation becomes (e 3x2 y) = 12xe 3x2. Integrating, we get e 3x2 y = 2e 3x2 + C. Thus, if y is any solution of y 6xy = 12x, then it is of the form y = 2 + Ce 3x2. As in the last example, we can retrace our steps to see that every function of this form is a solution. Before leaving this example, let s see whether we can find a solution φ that satisfies the additional condition that φ(0) = 2. If φ is a solution, there must be

15 2. FIRST ORDER LINEAR EQUATIONS 7 a constant C such that φ(x) = 2 + Ce 3x2 for all x (, ). Then at x = 0, the additional condition says that 2 + Ce 0 = 2, and so φ satisfies this additional condition if and only if C = 4. In other words, the initial value problem y 6xy = 12x, y(0) = 2 has the unique solution φ(x) = 2 + 4e 3x2 on (, ). Now we are ready to solve first order linear equations in general. Theorem 1.1. Consider the first order linear differential equation y + a(x)y = g(x), where a and g are continuous functions on some interval I. Put A(x) = a(x) dx. Then the equation has infinitely many solutions on I, all of which are given by (1) φ(x) = e A(x) e A(x) g(x) dx. Furthermore, for any x 0 I and any number α, there is exactly one solution on I of the initial value problem y + a(x)y = g(x), y(x 0 ) = α. Proof. φ is a solution of y + a(x)y = g(x) φ (x) + a(x)φ(x) = g(x) e A(x) φ (x) + e A(x) a(x)φ(x) = e A(x) g(x) (e A(x) φ) = e A(x) g(x) e A(x) φ = e A(x) g(x) dx φ(x) = e A(x) e A(x) g(x) dx. Let s note that the indefinite integral e A(x) g(x) dx represents a family of functions, one of which is x x 0 e A(t) g(t) dt. Then we can write [ x ] (2) φ(x) = e A(x) e A(t) g(t) dt + C. x 0 From this it easily follows that there is exactly one solution of the differential equation that also satisfies the condition y(x 0 ) = α, and it is the one corresponding to C = αe A(x0). We should note that the function µ = µ(x) = e a(x) dx is often referred to as an integrating factor. Also, (1) is referred to as the general solution of the first order equation. It is the collection of all solutions of the equation and involves an arbitrary constant C, which is shown explicitly in (2). Example. Solve the initial value problem xy + 2y = 3x 4, y(1) = 8.

16 8 1. FIRST ORDER EQUATIONS Dividing through by x, we get y + 2 x y = 3x3, which we will think of as the equation in standard form. The largest interval containing x = 1 for which this equation is well defined is (0, ), and we will look for a solution in this interval; however, it could be that we might only want to consider x 1 so that x = 1 is an initial point. Rather than just plugging into the formula given in the last theorem, we will find it more instructive to recreate the process that is described in its proof. Since x > 0, 2 x dx = 2 ln x, and so the integrating factor is µ = e 2 ln x = e ln(x2) = x 2. Multiplication through the equation in standard form by the integrating factor gives us x 2 y + 2xy = 3x 5, and so, (x 2 y) = 3x 5. Integrating this, we get x 2 y = 1 2 x6 + C. Thus, the differential equation has as its general solution y = 1 2 x4 + Cx 2. Now we use the initial condition y(1) = 8. Plugging x = 1 and y = 8 into the general solution, we get 8 = C, and so C = 2. The solution to the initial value problem is therefore for all x > 0. y = 1 2 x x 2 We should note that in the next chapter, the notion of a first order linear differential equation will be extended slightly to include equations of the form a 0 (x)y + a(x)y = g(x). This will have no effect on what we have done so far; wherever a 0 (x) is not zero, we can, as we did in the last example, divide through by it to get an equation in the standard form we have been dealing with in this section. Exercises Find the general solution for each of the following equations. 1. y + 5y = e 2x 2. y 4y = 2e 3x 3. 2y 4xy = 3e x2 4. 3y + 12xy = e 2x2 5. y = 2 x y + 5x7 6. y = 3 x y + x2

17 3. SOME OTHER FIRST ORDER EQUATIONS 9 Find the solution of each of the following initial value problems. 7. y = 2xy + e x2, y(0) = 4 8. y = 8xe x3 3x 2 y, y(1) = 0 9. y + 1 2xy = x, y(0) = y 6xy = 5x, y(0) = Let A(t) be the number of ounces of a radioactive material left t months after a certain starting time, and let A 0 be the initial amount. Find A(t) by solving the initial value problem A = ka, A(0) = A Consider an electrical circuit containing only a resistor and an inductor. Let i(t) be the current. If R is the resistance and L is the inductance (both constant) and there is a constant impressed voltage E, then i satisfies the differential equation If i(0) = i 0, solve for i. L di + Ri = E. dt 13. Suppose that φ 1 is a solution of y + a(x)y = g 1 (x) and φ 2 is a solution of y + a(x)y = g 2 (x) on an interval I. (a) If φ = φ 1 +φ 2, prove that φ is a solution of y +a(x)y = g 1 (x)+g 2 (x). (b) Use the method indicated by part (a) to solve y + 3y = x + e x. 14. If φ is a solution of y +a(x)y = g(x), find a solution of y +a(x)y = 2g(x). More generally, if k is any number, find a solution of y + a(x)y = kg(x). 15. Prove that the solutions of y +a(x)y = 0 form a one dimensional subspace of the vector space of all differentiable functions on I. 3. Some Other First Order Equations We will now look at some equations that are not linear. Of the many possible types, we will study a few of the most standard ones. The first type will involve separating variables. Example. Consider the equation y = 6x/y 2 for x (, ). Let s assume that φ is a solution of this equation. Then φ (x) = 6x/[φ(x)] 2. We would like to integrate both sides of the equation, but we are not ready to do so, yet. First multiply through by φ(x). We get Now we can say [φ(x)] 2 φ (x) = 6x. [φ(x)] 2 φ (x) dx = 6x dx. Evaluating these integrals (and using the substitution u = φ(x), du = φ (x) dx for the left-hand one), we get [φ(x)] 3 = 3x 2 + C, 3

18 10 1. FIRST ORDER EQUATIONS and so if φ is a solution, then φ(x) = (9x 2 + 3C) 1/3. Furthermore, it is easy to check that every function of this form is, in fact, a solution. Note that since 3C can stand for any number just as C can, most people would write the answer as φ(x) = (9x 2 + C) 1/3. The process we just went though can be given more succinctly as follows. Starting with y = 6x/y 2, we first write the derivative in Leibnitz notation so that we have dy dx = 6x/y2. Then we separate the variables (including the differentials), getting y 2 dy = 6x dx, and so, y 2 dy = 6x dx. From this, we get y 3 /3 = 3x 2 + C. Therefore, y = (9x 2 + C) 1/3, as before. If integrating with respect to y on one side of the equation while integrating with respect to x on the other side seems suspicious to you, just note how it corresponds to integration by substitution when we went through the problem originally. Definition 1.3. A first order differential equation is said to be a separable equation if it is of the form y = M(x)/N(y). Note that a separable equation can be rewritten as N(y) dy = M(x) dx, which helps explain why the word separable is used. The only reason for the minus sign is to make the notation consistent with something else coming shortly. As in the case of the first example in this section, our method of solution will be to separate the variables, integrate, and then solve for the dependent variable. Instead of formalizing this as a theorem, we will just work through a few more examples. There are some pitfalls in this process, as the examples will show. Example. Consider the equation y + 2xy = 0. This equation is linear and can be solved as such. However, it is possible to separate the variables, and it will be instructive to attack the problem in this way. We have dy dx = 2xy, and so, y 1 dy = 2x dx. Then, y 1 dy = 2x dx. From this we get ln y = x 2 + C and then e ln y = e x2 +C, y = e C e x2, and finally y = ±e C e x2. Put K = ±e C. Since C can be any number, it follows that K can be any non-zero number. Thus, we have y = Ke x2 for any K 0. Notice that we have missed finding the constant solution y 0, (which corresponds to K = 0). This happened because when we were separating the variables, we divided by y, which is only allowed if y 0. If we had tried to write the equation in standard form for a separable equation, we would have had early warning of a possible problem involving y = 0 because we would have had N(y) = 1/y. In the next two examples, we will need to be concerned about the domain a solution can have.

19 3. SOME OTHER FIRST ORDER EQUATIONS 11 Example. Consider the equation xy = y 2. Separating the variables leads to y 2 dy = x 1 dx, which in turn gives 1/y = ln x + C after integration. Thus, y = 1/(ln x + C). Of course, we have to exclude x = 0 from the domain of any such solution because of the logarithm. Furthermore, we must avoid any x such that ln x + C = 0. In other words, both x = 0 and x = ±e C must be excluded from the domain of the solution y = 1/(ln x + C). Also, our method has missed finding y 0, which is obviously a solution. Example. Let s solve the equation y = a 2 + y 2, where a is any constant. Separating the variables gives us dy a 2 + y 2 = dx. If a 0, we are led to 1 a tan 1 y a = x+c, after integration. Then y a = tan(ax+ac), and so y = a tan(ax + ac). We can simplify the form of the answer, writing y = a tan(ax + C) since ac represents an arbitrary constant. Note that because of the tangent function, we should restrict x so that ax + C does not equal odd integer multiples of π/2. On the other hand, if a = 0, the reader can check that y = 1/(x + C) for x C and y 0 are solutions. Our next two examples show that initial value problems involving separable equations do not necessarily have unique solutions. In the first example, we have no solutions; in the second, we have more than one. Example. Consider the initial value problem y 2 y = x 2, y(1) = 0. This problem obviously has no solution as can be seen by trying to plug x = 1 and y = 0 into the equation simultaneously. Let s apply the method of separation of variables to this equation anyway to see what happens. We get y 2 dy = x 2 dx, and then y 3 /3 = x 3 /3 + C. The initial condition implies 0 = 1/3 + C. Thus, C = 1/3 and we wind up with the function y = (x 3 1) 1/3, which is not differentiable at x = 1. Example. Consider the initial value problem y = 3xy 1/3, y(0) = 0. Separating the variables of the differential equation leads to y 1/3 dy = 3x dx, which in turn gives (3/2)y 2/3 = (3/2)x 2 +C after integration. Even though N(y) = y 1/3 does not exist at y = 0, we will find solutions for this example by using the initial condition y(0) = 0 along with what we have so far. We get 0 = 0 + C and so, C = 0. Then y 2/3 = x 2. Solving for y carefully gives us y = x 3 and y = x 3. As can be checked directly, both y = x 3 and y = x 3 are solutions of the initial value problem on the interval (, ). Note that y 0 is yet another solution of the initial value problem on (, ). The difficulties we have encountered in solving separable equations should, perhaps, make us appreciate the situation for linear equations, where there are no surprises involving the domain of a solution, and where every initial value problem has a unique solution. We will now move on to another type of first order equation that has been studied extensively.

20 12 1. FIRST ORDER EQUATIONS or, Definition 1.4. Consider a first order differential equation in the form y = M(x, y)/n(x, y), M(x, y) + N(x, y)y = 0. The equation is said to be exact in a rectangle R = {(x, y): x x 0 a, y y 0 b} if M and N are continuous for all (x, y) R and there exists a function F such that F x = M and F y = N throughout R. Let s start by showing that a particular example is exact. Our method is identical to the process of finding a function F of two variables whose gradient F is a given vector field. Example. Consider the differential equation y = 2xy/(x 2 + 2y). If you like, this can be rewritten as 2xy + (x 2 + 2y)y = 0. In any event, we can set M(x, y) = 2xy and N(x, y) = x 2 + 2y. To show that the equation is exact, we must find a function F such that F #1 x = 2xy #2 F y = x2 + 2y. Starting with the first of these two conditions, we can integrate with respect to x while holding y constant. This gives F (x, y) = x 2 y + K(y), where K(y) is constant with respect to the variable of integration x but might depend on y. This, in turn, leads to F y = x2 + K (y). Comparing this to #2 gives us x 2 + K (y) = x 2 + 2y, and so, K (y) = 2y. Since it is enough to find just one function F, we can simply chose K(y) = 2y dy = y 2. We wind up with F (x, y) = x 2 y + y 2, which certainly has the desired first partial derivatives. Furthermore, it has the desired derivatives at all points (x, y). Therefore, the equation is considered to be exact in the entire plane R 2, not just on a bounded rectangle. Notice that we still haven t solved the differential equation. However, having found a function F that shows exactness, that will be easy. The next theorem when applied to our example says that if y is any function of x implicitly defined by the equation, x 2 y + y 2 = C, then y is a solution of the exact equation. Here, C is an arbitrary constant. Since it is possible in this example, let s solve for y explicitly.

21 3. SOME OTHER FIRST ORDER EQUATIONS 13 Using the quadratic formula on y 2 + x 2 y C = 0 to solve for the unknown y, we get y = x2 ± x 4 + 4C. 2 We should note that solutions corresponding to negative values of C will have restricted domains. Theorem 1.2. Suppose the equation M(x, y) + N(x, y)y = 0 is exact in a rectangle R, with F being a function such that F x = M and F y = N in R. Then, every differentiable function φ defined implicitly by F (x, y) = C, where C is a constant, is a solution of the exact equation. Also, every solution of the exact equation whose graph lies in R, can be found this way. Proof. Suppose φ is a differentiable function defined implicitly by F (x, y) = C. In other words, F (x, φ(x)) = C for all x in an interval I, (where R = I I ). Put W (x) = F (x, φ(x)). Then, 0 = W (x) (since W (x) = C) = F x (x, φ(x)) 1 + F y (x, φ(x)) φ (x) (by the chain rule) = M(x, φ(x)) + N(x, φ(x)) φ (x) Therefore, φ is a solution of M(x, y) + N(x, y)y = 0. Conversely, suppose φ is a solution of M(x, y) + N(x, y)y = 0. In other words, M(x, φ(x)) + N(x, φ(x)) φ (x) = 0. Then F x (x, φ(x)) + F y (x, φ(x)) φ (x) = 0. With W defined as above, this says that W (x) = 0, and so, W is constant. In other words, F (x, φ(x)) = C. Any equation can be written using differentials as M(x, y) + N(x, y)y = 0 M(x, y) dx + N(x, y) dy = 0. In particular, any exact equation can be written as F F dx + dy = 0. x y The above proof can then be summarized by df = 0 if and only if F = C. Example. Solve the equation 3x/y + y/x + 2y = 0. This can be rewritten as 3x 2 + y 2 + 2xyy = 0. To try to solve this, let s look for a function F as above, in the hope that the equation is exact. We want to find F (x, y) such that #1 F x = 3x2 + y 2 #2 F y = 2xy.

22 14 1. FIRST ORDER EQUATIONS Starting with #1, we integrate with respect to x while holding y constant to get F (x, y) = x 3 + xy 2 + K(y). When we then take the partial derivative of this with respect to y and compare the result to #2, we get 2xy + K (y) = 2xy. Thus K is any constant, and we might as well choose K = 0 since we only need to find one function F. Therefore, we have F (x, y) = x 3 + xy 2, which satisfies both condition #1 and condition #2. We are not finished, however. We were supposed to solve a differential equation, and all we have so far is a function of two variables, which does not tell us what y is. The solutions of 2y = 3x/y y/x are given implicitly by x 3 + xy 2 = C. Therefore, y = ± (x 3 + C)/x. We should note that for each C, the corresponding solution has a restricted domain. Example. Solve the equation x 2 +y +y 2 y = 0. Proceeding as above, we want to find F (x, y) such that F #1 x = x2 + y F #2 y = y2. Starting with #1, we integrate with respect to x while holding y constant to get F (x, y) = x 3 /3 + xy + K(y). When we then take the partial derivative of this with respect to y and compare the result to #2, we get x + K (y) = y 2. This implies that K (y) = x + y 2, which is clearly impossible since K (y) cannot vary with x. We conclude that no function F satisfies both conditions #1 and #2. In other words, the differential equation is not exact. Since it also is neither linear nor separable, we will have to leave it unsolved, for now. The last example shows that it would be desirable to have a quick method for determining whether or not a first order differential equation is exact. The following result, which gives us such a method, parallels the standard condition for determining whether or not a 2-dimensional vector field is conservative. Theorem 1.3. Let M and N be real valued functions of two real variables that have continuous first partial derivatives on a rectangle R = {(x, y): x x 0 a, y y 0 b}. Then the equation M(x, y) + N(x, y)y = 0 is exact if and only if (3) in R. M y = N x Proof. First, suppose that M(x, y) + N(x, y)y = 0 is exact. There is a function F such that F x = M and F y = N in R. Differentiate to get 2 F y x = M y and 2 F x y = N x.

23 3. SOME OTHER FIRST ORDER EQUATIONS 15 But F has continuous second partial derivatives since M and N have continuous first partial derivatives, and so by a standard result of calculus, 2 F y x = 2 F x y. Therefore, (3) holds throughout R. Conversely, assume that (3) holds in R. We need to find a function F such that F x = M and F y = N in R. If such a function F exists, then using the Fundamental Theorem of Calculus, we will have F (x, y) F (x 0, y 0 ) = F (x, y) F (x 0, y) + F (x 0, y) F (x 0, y 0 ) = = x x 0 x Using this as a guide, define F on R by F (x, y) = x It then follows immediately that F (s, y) ds + x x 0 M(s, y) ds + x 0 M(s, y) ds + y y 0 y y F (x, y) = M(x, y) x for all (x, y) R. Also, for each (x, y) R, F (x, y) = y = x x 0 x x 0 F y (x 0, t) dt y 0 N(x 0, t) dt. y 0 N(x 0, t) dt. M y (s, y) ds + N(x 0, y) N x (s, y) ds + N(x 0, y) = N(x, y) N(x 0, y) + N(x 0, y) = N(x, y). (In the preceding, bringing the partial derivative with respect to y inside the integral with respect to s is allowed since M has continuous first partial derivatives.) Therefore, the equation is exact in R. The hypothesis that domain under consideration was a rectangle R was used to ensure that the integrations defining F were over lines in the domain. However, using some more advanced techniques, it could be shown that the theorem is true for any simply connected domain. Roughly speaking, this means that the domain has no holes. The reader should test the last theorem on all of the examples that preceded it. In particular, for x 2 + y + y 2 y = 0, we have M(x, y) = x 2 + y and N(x, y) = y 2. This implies that M y (x, y) = 1 while N x (x, y) = 0, which confirms that fact that the differential equation is not exact. On the other hand, the equation 3x/y + y/x + 2y = 0 involves some subtleties that will be discussed immediately after the next example.

24 16 1. FIRST ORDER EQUATIONS Example. Solve the initial value problem y = (2xy + 1)/(x 2 + cos y), y(0) = 0. The differential equation can be rewritten as or 2xy (x 2 + cos y)y = 0, (2xy + 1)dx + (x 2 + cos y)dy = 0. In any event, M(x, y) = 2xy + 1 and N(x, y) = x 2 + cos y. Before going any further, we can check whether the equation is exact by computing M y (x, y) = 2x and N x (x, y) = 2x. Now that we know that the equation is exact, let s look for a function F such that F #1 x = 2xy + 1 F #2 y = x2 + cos y. Just for the sake of variety, we will start with condition #2 this time. Integrating with respect to y while holding x constant gives us F (x, y) = x 2 y + sin y + K(x). When this, in turn, is differentiated with respect to the other variable, x, we get F x = 2xy + K (x). This, combined with #1, implies that K (x) = 1, and so we can choose K(x) = x. Thus, F (x, y) = x 2 y + sin y + x. The solutions of the differential equation are given by x 2 y + sin y + x = C. Plugging in the initial condition leads to C = 0. Therefore, the solution of the initial value problem is given implicitly by x 2 y + sin y + x = 0. In this example, as is often the case, there is no way of exhibiting the solution y explicitly. At times, an equation that is not exact can be replaced by an equivalent one that is exact. We have seen this happen already. Consider again the equation 3x/y + y/x + 2y = 0. This equation is not exact since M(x, y) = 3x/y + y/x and N(x, y) = 2 imply M y (x, y) N x (x, y). However, when we multiplied through by xy, we got 3x 2 + y 2 + 2xyy = 0, which we showed to be exact. From this we see that changing the form of a differential equation in an apparently inconsequential way can, in fact, have major consequences in terms of exactness. The method of solving first order linear equations that was presented in the last section parallels the process of replacing a differential equation with an equivalent one. Let s look back at linear equations with this in mind. We start with the equation y + a(x)y = g(x). Rewrite it as g(x) + a(x)y + y = 0, which is exact only in the very special case of a being identically zero. As before, multiply through the equation by the integrating factor µ = µ(x) = e a(x) dx.

25 3. SOME OTHER FIRST ORDER EQUATIONS 17 We get µ(x)g(x) + µ(x)a(x)y + µ(x) y = 0. }{{}}{{} M(x,y) N(x,y) Since µ (x) = µ(x)a(x), M y (x, y) = µ(x)a(x) = N x (x, y), and so we now have an exact equation. Following our usual procedure, we can find F (x, y) = µ(x)y µ(x)g(x) dx, and setting this equal to a constant gives us the general solution of the first order linear equation, in agreement with what we got before. There are many other first order differential equations that when multiplied by an appropriate integrating factor µ give an equivalent equation that is exact (an then can be solved). Unfortunately, it is usually quite difficult to find µ. This is especially truly when µ is a function of both x and y. For more about this, see the exercises. We have seen how to solve three types of first order differential equations: linear, separable, and exact. Of course, there are many first order equations of importance that do not fit any of these patterns. In the exercises, we will briefly investigate a few other standard types of first order equations. There are still other equations for which there are no known methods for finding solutions expressible either explicitly or implicitly, as we have been able to do so far. In the next two sections we will show that many of these equations do have solutions, and we will use numerical methods for approximating the solutions. Exercises 1. For each of the following equations, state which of the methods of this section and the last section apply. (For some equations, more than one method is possible; for others, none may be possible.) Do not solve the equations. (a) y + xy 2 = 0 (b) y + xy 2 = 1 (c) y + x 2 y = 1 (d) y + x 2 y = 3x 2 (e) x 2 y + 2xy = 0 (f) x 2 y + 3xy = 1 (g) x 2 y + 2xy = x (h) x 2 y + 2xy = x 2 Using the methods of this section and the last section, solve the following equations or initial value problems. 2. y = 3x 2 e y e y, y(0) = 0 3. (2x + y 2 ) + (2xy + 1)y = 0 4. y = 4x 3 y + 2xe x4 5. y = 2xe y /(x 2 e y + 1)

26 18 1. FIRST ORDER EQUATIONS 6. xy 2 + (x 2 + 1)y = 0 7. (2xy 3 + 2) + 3x 2 y 2 y = 0, y(1) = 3 8. y + 2x 1 y = 2x 4 9. y = (2x sin y)/(1 + x cos y), y(0) = y = a/y (a + y 3 ) + 3xy 2 y = Try to solve the initial value problem (2xy 3 + 2) + 3x 2 y 2 y = 0, y(0) = 3. What is wrong with your answer? 13. Prove that every separable equation can be made to be exact. 14. An equation of the form y = (a by)y is called a logistic equation. If the number N (in thousands) at time t (in years) of a species that has just been introduced to a certain island is assumed to satisfy the logistic equation N = (5 2N)N, and if now at time t = 0 the population is 1000, what will the population of this species be six months from now? Also, what will the population tend toward in the long run? 15. Consider an object of mass m at the end of a horizontal spring with stiffness coefficient k. Let x be the (variable) distance the object is away from its resting position. (Stretching the spring corresponds to x > 0, while compressing it corresponds to x < 0.) If friction is ignored and the spring is assumed massless, it can be shown that the velocity v of the object obeys the differential equation dv dx = k x m v. Show that 1 2 mv kx2 is constant, which says that the sum of the kinetic energy and the potential energy of the system remains constant over time. 16. Given an equation M(x, y) + N(x, y)y = 0 (which is assumed not exact), show that if the expression 1 N ( M y N x ) is a continuous function of x alone (or is constant), then multiplying the equation by µ = µ(x) = e leads to an equivalent exact equation. 1 N ( M y N x ) dx 17. Given an equation M(x, y) + N(x, y)y = 0 (which is assumed not exact), show that if the expression 1 M ( N x M y ) is a continuous function of y alone (or is constant), then multiplying the equation by µ = µ(y) = e leads to an equivalent exact equation. 1 M ( N x M y ) dy 18. Use a one variable integrating factor to solve y 2 + (3xy + 1)y = Use a one variable integrating factor to solve 2y 2 + 2x yy = 0.

27 4. EXISTENCE AND UNIQUENESS Verify that µ(x, y) = 1/(xy 2 ) is an integrating factor for 3y 2 xy+x 2 y = 0 and then use it to solve the equation. A function of two variables f is said to be homogeneous of degree zero if f(tx, ty) = f(x, y) for all real numbers x, y, t. In this case, the differential equation y = f(x, y) is said to be homogeneous. (This use of the word homogeneous has no relation to its use in connection with linear equations.) To solve a homogeneous equation, make the substitution v = y/x. In other words, let y = xv, and so dy dx = x dv dx + v. This will lead to a separable equation in the variables x, v, which you then solve. Use this technique on the following equations and initial value problems. 21. y = (x 3 + y 3 )/xy y = (x 2 y y 3 )/x xy + y xy = 0, y(1) = 1 (x > 0) 24. y = (xy + y 2 )/x 2, y(1) = 1/2 (x > 0) An equation of the form y + a(x)y = b(x)y n is called a Bernoulli equation. To solve it, make the substitution v = y 1 n. In other words, let y = v 1/(1 n) and then dy dx = 1 dv 1 nvn/(1 n) dx. This will lead to a linear equation in the variables x, v, which you then solve. Use this technique on the following equations. 25. y + 2y/x = y 3 /x y y/x = 1/y 2 4. Existence and Uniqueness We are going to show that first order differential equations under rather general conditions have solutions. This will require a number of steps, including replacing an initial value problem by an equivalent integral equation. We start by carefully stating what it means for a function to be a solution of an initial value problem and to be a solution of an integral equation. Let f be a continuous real valued function on a rectangle R = {(x, y): x x 0 a, y y 0 b}. A function φ is said to be a solution of the initial value problem (4) y = f(x, y) y(x 0 ) = y 0 on an interval I containing x 0 if φ is differentiable on I, (x, φ(x)) R for all x I, φ (x) = f(x, φ(x)) for all x I, and φ(x 0 ) = y 0. Correspondingly, φ is said to be a solution of the integral equation (5) y = y 0 + x x 0 f(t, y) dt

28 20 1. FIRST ORDER EQUATIONS on an interval I containing x 0 if φ is continuous on I, (x, φ(x)) R for all x I, and x φ(x) = y 0 + f(t, φ(t)) dt x 0 for all x I. We have already seen numerous examples of initial value problems and their solutions. Here is an example of an integral equation and its solution. Example. Put f(x, y) = 3y/x, x 0 = 1, y 0 = 2. A solution of is y = φ(x) = 2x 3 since 2 + x 1 3φ(t) t y = 2 + dt = 2 + x 1 x 1 3y t dt 6t 2 dt = 2x 3 = φ(t). Theorem 1.4. A function φ is a solution of the initial value problem (4) on an interval I if and only if it is a solution of the integral equation (5) on I. Proof. Suppose φ satisfies (5) on I. Since f(t, φ(t)) is a continuous function of t, we immediately get that φ is a differentiable function of x and φ (x) = f(x, φ(x)) for all x I. Clearly, φ(x 0 ) = y 0. Thus, φ is a solution of (4) on I. Conversely, suppose φ is a solution of (4) on I. Then, φ (x) = f(x, φ(x)) and φ(x 0 ) = y 0. Putting G(x) = f(x, φ(x)), we have φ = G, and so, In other words, for all x I, and φ satisfies (5). φ(x) φ(x 0 ) = x x 0 G(t) dt. x φ(x) = y 0 + f(t, φ(t)) dt x 0 We now set out to try to solve (5). We will use the method of successive approximations, also known as the method of Picard iterations. An outline of the method is as follows. Our first attempt at a solution is the function φ 0 given by for all x I. Next try φ 1 defined by Then try φ 2 defined by In general, put φ 1 (x) = y 0 + (6) φ k (x) = y 0 + φ 0 (x) = y 0 x x 0 f(t, φ 0 (t)) dt. x φ 2 (x) = y 0 + f(t, φ 1 (t)) dt. x 0 x x 0 f(t, φ k 1 (t)) dt.

29 4. EXISTENCE AND UNIQUENESS 21 Our hope is that by letting φ(x) = lim k φ k (x) and then taking the limit of both sides of (6), we will wind up with φ(x) = y 0 + x x 0 f(t, φ(t)) dt, and so φ will be a solution of (5). There are, however, a couple of non-trivial concerns with this. It is not certain that lim k φ k (x) will exist, and it is not certain that when taking the limit of the right side of (6), we can bring the limit inside the integral. Putting aside these concerns for the moment, however, let s try out the method of successive approximations on a simple example. Example. Consider the initial value problem y = y, y(0) = 1. We have f(x, y) = y, x 0 = 0, and y 0 = 1. Thus, the equivalent integral equation is Furthermore, φ 0 (x) 1 and y = 1 + φ k (x) = 1 + x 0 x 0 y dt. φ k 1 (t) dt for k = 1, 2,.... Applying this formula repeatedly, we get φ 1 (x) = 1 + φ 2 (x) = 1 + φ 3 (x) = 1 + x 0 x 0 x 0 1 dt = 1 + x (1 + t) dt = 1 + x + x2 2 (1 + t + t2 2 ) dt = 1 + x + x2 2 + x3 3 2 and so on. An easy proof by induction shows that, in general, φ k (x) = 1 + x + x2 2! + + xk k! for all x. We recognize that we are getting partial sums for the Maclaurin series of e x, and so the sequence of successive approximations φ k (x) converges to φ(x) = e x for all x. We can immediately check that e x is, in fact, the solution of the initial value problem. In the last example, everything worked out as we would hope. It is now time to face the task of determining conditions that will ensure this sort of nice behavior. First, let us note the following standard result, which is commonly stated in calculus courses and proved in a course in (multivariable) advanced calculus. Since R is a closed and bounded set and f is continuous on R, f is bounded on R, i.e., there is a constant M such that f(x, y) M for all (x, y) R. Using the notation that has been introduced in this section, we are ready for the following lemma, which among other things tells us that all the successive approximations φ k exist, at least near the point x 0. Lemma 1.1. Let α = min{a, b/m}, and put I = {x: x x 0 α}. Then for each k = 0, 1,..., φ k is continuous on I and (x, φ k (x)) R for all x I. Moreover, (7) φ k (x) y 0 M x x 0

30 22 1. FIRST ORDER EQUATIONS for all x I. Proof. Our proof will be by mathematical induction. First note that all the results are obviously true for φ 0. Next, assume inductively that the lemma is true for k = m 1, where m is an arbitrary, fixed positive integer. Then f(t, φ m 1 (t)) is a continuous function of t I. Thus, it follows from (6) that φ m is continuous on I (and, in fact, is differentiable). Also, x φ m (x) y 0 = f(t, φ m 1 (t)) dt M x x 0 x 0 since (t, φ m 1 (t)) R for all t I. It remains only to show that (x, φ m (x)) R for all x I. For this, first note that for x I, x x 0 α a. Also, if x I, φ m (x) y 0 M x x 0 Mα M b/m = b. Thus, (x, φ m (x)) R for all x I. The lemma is, therefore, true for k = m. To be sure that the successive approximations converge to a solution, we will need a further condition on f. Definition 1.5. Let f be a function of two variables defined on a set S. We say that f satisfies a Lipschitz condition on S if there exists a constant K such that for all (x, y 1 ) and (x, y 2 ) in S. f(x, y 1 ) f(x, y 2 ) K y 1 y 2 Example. The function f(x, y) = x 3 y 2 satisfies a Lipschitz condition on the rectangle R = [ 1, 3] [ 5, 5] = {(x, y): x 1 2, y 5} since for any (x, y 1 ) and (x, y 2 ) in R, x 3 y 2 1 x 3 y 2 2 = x 3 y 1 + y 2 ) y 1 y y 1 y 2 = 270 y 1 y 2. You should notice that the Lipschitz condition, as defined here, says much more about f(x, y) as a function of y than as a function of x. In particular, for each fixed x, f(x, y) will be continuous with respect to y, but for a fixed value of y, f(x, y) is not necessarily continuous with respect to x. You are asked to show this in the exercises. Also, it should be mentioned that there exists an analogous notion of a Lipschitz condition for functions of one variable, which is of interest in other settings. Next comes a result that, for many examples, gives us an easy way of determining that a function satisfies a Lipschitz condition. Theorem 1.5. Let S be either a rectangle or a strip {(x, y): x x 0 a, y y 0 b} {(x, y): x x 0 a, < y < }. Suppose there is a constant K such that f y (x, y) K for all (x, y) S. Then, f satisfies a Lipschitz condition on S with Lipschitz constant K.

31 4. EXISTENCE AND UNIQUENESS 23 Proof. For any fixed x [x 0 a, x 0 + a] choose (x, y 1 ) and (x, y 2 ) in S. The mean value theorem applied to f thought of as a function of the single variable y on the interval between y 1 and y 2 states that there exists a number c (which can depend on x) between y 1 and y 2 such that f(x, y 1 ) f(x, y 2 ) = f y (x, c)(y 1 y 2 ). Thus, f(x, y 1 ) f(x, y 2 ) K y 1 y 2 for all (x, y 1 ) and (x, y 2 ) in S. Example. Let s look at the previous example f(x, y) = x 3 y 2 on the rectangle R = [ 1, 3] [ 5, 5] again. On R, f y (x, y) = 2x 3 y 270. Thus, f satisfies a Lipschitz condition on R (with Lipschitz constant 270). The fact that we got the same Lipschitz constant as before is of no importance; in other examples, this might not happen. Theorem 1.6. (Existence Theorem) Let f be a continuous function on a rectangle R = {(x, y): x x 0 a, y y 0 b}. Suppose f(x, y) M for all (x, y) R and f satisfies a Lipschitz condition on R. Put α = min{a, b/m} and φ 0 (x) = y 0 φ k (x) = y 0 + x x 0 f(t, φ k 1 (t)) dt, k = 1, 2,... Then, on I = {x: x x 0 α}, the sequence {φ k } converges to a solution of the initial value problem y = f(x, y), y(x 0 ) = y 0. Before proving the theorem, we will see how it applies to a specific example. Example. Consider the initial value problem y = x 3 y 2 + 1, y(0) = 1. We cannot find a solution to this problem since the equation is not of a type we have studied, but we will be able to show that a solution exists. Let R = [ 2, 2] [0, 2]. (The choice of R is arbitrary, except that the center must be at (x 0, y 0 ) = (0, 1).) We have f(x, y) = x 3 y and a = 2, b = 1. Then, M = 33 and α = min{2, 1/33} = 1/33. Furthermore, f satisfies a Lipschitz condition on R since f y (x, y) = 2x 3 y 32 for all (x, y) R. Therefore, the initial value problem has a solution on I = [ 1/33, 1/33], and this solution is the limit of the sequence of successive approximations. Note that we could have gotten a solution on a larger interval I if we had chosen R with more care. Proof of the Existence Theorem. We will start by getting an idea of how much successive approximations differ from each other. Specifically, we want to prove (8) φ k (x) φ k 1 (x) MKk 1 x x 0 k, k! where K is a Lipschitz constant, for all x I, k = 1, 2,.... We will use mathematical induction. When k = 1, (8) is true since if x I, φ 1 (x) φ 0 (x) = φ 1 (x) y 0 M x x 0

32 24 1. FIRST ORDER EQUATIONS by the last lemma. Now assume inductively that (8) is true for k = m, where m is an arbitrary positive integer. If x I and x > x 0, x φ m+1 (x) φ m (x) = [f(t, φ m (t)) f(t, φ m 1 (t))] dt x 0 x x 0 f(t, φ m (t)) f(t, φ m 1 (t)) dt x x 0 K φ m (t) φ m 1 (t) dt x K MKm 1 t x 0 m dt x 0 m! = [MK m (t x 0) m+1 ] x (m + 1)m! = MKm x x 0 m+1, (m + 1)! x 0 (Lipschitz condition) (inductive assumption) and so (8) holds for k = m + 1. If x I and x < x 0, a similar computation shows that (8) holds for k = m + 1. Thus, (8) is true for all k = 1, 2,.... Due the tools we have available to us, it turns out that it will be convenient deal with a series rather than the given sequence of successive approximations. To this end, let s note that φ k = φ 0 + (φ 1 φ 0 ) + (φ 2 φ 1 ) + + (φ k φ k 1 ). In other words, φ k (x) is a partial sum of the series (9) φ 0 (x) + [φ k (x) φ k 1 (x)]. k=1 This series converges on I by the comparison test and (8) since MK k 1 x x 0 k k=0 converges for all x (and has sum (M/K)e K x x0 ). Call the sum of (9), φ(x). Then, the sequence {φ k } converges to φ on I. Next, we will show that φ is continuous on I. Since we get φ k (x 1 ) φ k (x 2 ) = = x1 x 0 x1 x 2 k! f(t, φ k 1 (t)) dt f(t, φ k 1 (t)) dt, x2 x 0 φ k (x 1 ) φ k (x 2 ) M x 1 x 2 for all k. It then follows from letting k that (10) φ(x 1 ) φ(x 2 ) M x 1 x 2 f(t, φ k 1 (t)) dt for all x 1 and x 2 in I. From this, we easily get that φ is continuous on I (just as we noted earlier that the Lipschitz condition for a function of two variables implies continuity in y).

33 4. EXISTENCE AND UNIQUENESS 25 It is clear that φ(x 0 ) = y 0. Also, letting x 1 = x and x 2 = x 0 in (10), we have φ(x) y 0 ) M x x 0 Mα b, and so (x, φ(x)) R for all x I. Now all that remains is to show that φ satisfies the integral equation (5). As mentioned before, it is not elementary that taking the limit as k of both sides of (6) proves that φ is a solution of (5). We will need an inequality that describes in detail the convergence of φ k (x) to φ(x). For any x I, φ(x) φ k (x) = i=k+1 i=k+1 i=k+1 i=k+1 MKk α k+1 (k + 1)! = M K [φ i (x) φ i 1 (x)] φ i (x) φ i 1 (x) MK i 1 x x 0 i i! MK i 1 α i i! K j α j j=0 (Kα) k+1 (k + 1)! ekα. Setting ɛ k = (Kα) k+1 /(k + 1)!, we have j! (by (8)) (11) φ(x) φ k (x) M K ɛ ke Kα for all x I. Note that ɛ k 0 as k, as can be seen from e Kα = k ɛ k. Now, we have x φ k (x) = y 0 + f(t, φ k 1 (t)) dt, x 0 and from this we want to show that x φ(x) = y 0 + f(t, φ(t)) dt x 0 for all x I. Let k. Clearly φ k (x) φ(x). Thus, we will be done if x x 0 f(t, φ k 1 (t)) dt x x 0 f(t, φ(t)) dt,