1.1. Introduction CHAPTER 1 BACKGROUND

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1 CHAPTER Introduction. BACKGROUND In the sciences and engineering, mathematical models are developed to aid in the understanding of phsical phenomena. These models often ield an equation that contains some derivatives of an unknown function. Such an equation is called a differential equation. Two eamples of models developed in calculus are the free fall of a bo and the deca of a radioactive substance. In the case of free fall, an object is released from a certain height above the ground and falls under the force of gravit. Newton s second law, which states that an object s mass times its acceleration equals the total force acting on it, can be applied to the falling object. This leads to the equation (see Figure.) where m is the mass of the object, h is the height above the ground, d 2 h/dt 2 is its acceleration, g is the (constant) gravitational acceleration, and mg is the force due to gravit. This is a differential equation containing the second derivative of the unknown height h as a function of time. Fortunatel, the above equation is eas to solve for h. All we have to do is divide b m and integrate twice with respect to t. That is, so and m d 2 h dt 2 mg, d 2 h dt 2 g, dh dt gt c h hatb gt2 2 c t c 2. mg h Figure. Apple in free fall We are assuming here that gravit is the onl force acting on the object and that this force is constant. More general models would take into account other forces, such as air resistance.

2 2 Chapter Introduction We will see that the constants of integration, c and c 2, are determined if we know the initial height and the initial velocit of the object. We then have a formula for the height of the object at time t. In the case of radioactive deca (Figure.2), we begin from the premise that the rate of deca is proportional to the amount of radioactive substance present. This leads to the equation da dt ka, k 0, where AA 0B is the unknown amount of radioactive substance present at time t and k is the proportionalit constant. To solve this differential equation, we rewrite it in the form da kdt A and integrate to obtain A da kdt ln A C kt C 2. Solving for A ields A AAtB e ln A e kt e C 2 C Ce kt, where C is the combination of integration constants e C 2 C. The value of C, as we will see later, is determined if the initial amount of radioactive substance is given. We then have a formula for the amount of radioactive substance at an future time t. Even though the above eamples were easil solved b methods learned in calculus, the do give us some insight into the stu of differential equations in general. First, notice that the solution of a differential equation is a function, like hatb or A AtB, not merel a number. Second, integration is an important tool in solving differential equations (not surprisingl!). Third, we cannot epect to get a unique solution to a differential equation, since there will be arbitrar constants of integration. The second derivative d 2 h/dt 2 in the free-fall equation gave rise to two constants, c and c 2, and the first derivative in the deca equation gave rise, ultimatel, to one constant, C. Whenever a mathematical model involves the rate of change of one variable with respect to another, a differential equation is apt to appear. Unfortunatel, in contrast to the eamples for free fall and radioactive deca, the differential equation ma be ver complicated and difficult to analze. A Figure.2 Radioactive deca For a review of integration techniques, see Appendi A.

3 Section. Background 3 R L emf + C Figure.3 Schematic for a series RLC circuit Differential equations arise in a variet of subject areas, including not onl the phsical sciences but also such diverse fields as economics, medicine, pscholog, and operations research. We now list a few specific eamples.. In banking practice, if PAtB is the number of dollars in a savings bank account that pas a earl interest rate of r% compounded continuousl, then P satisfies the differential equation () dp dt r P, t in ears A classic application of differential equations is found in the stu of an electric circuit consisting of a resistor, an inductor, and a capacitor driven b an electromotive force (see Figure.3). Here an application of Kirchhoff s laws leads to the equation (2) d 2 q L dt 2 R dq dt q EAtB, C where L is the inductance, R is the resistance, C is the capacitance, EAtB is the electromotive force, qatb is the charge on the capacitor, and t is the time. 3. In pscholog, one model of the learning of a task involves the equation (3) /dt 2p 3 / 2 A B 3 / 2 n. Here the variable represents the learner s skill level as a function of time t. The constants p and n depend on the individual learner and the nature of the task. 4. In the stu of vibrating strings and the propagation of waves, we find the partial differential equation 0 (4) 2 u 0t u c , where t represents time, the location along the string, c the wave speed, and u the displacement of the string, which is a function of time and location. We will discuss Kirchhoff s laws in Section 3.5. Historical Footnote: This partial differential equation was first discovered b Jean le Rond d Alembert (77 783) in 747.

4 4 Chapter Introduction To begin our stu of differential equations, we need some common terminolog. If an equation involves the derivative of one variable with respect to another, then the former is called a dependent variable and the latter an independent variable. Thus, in the equation d (5) 2 dt 2 a k 0, dt t is the independent variable and is the dependent variable. We refer to a and k as coefficients in equation (5). In the equation 0u 0u (6) 2, 0 0 and are independent variables and u is the dependent variable. A differential equation involving onl ordinar derivatives with respect to a single independent variable is called an ordinar differential equation. A differential equation involving partial derivatives with respect to more than one independent variable is a partial differential equation. Equation (5) is an ordinar differential equation, and equation (6) is a partial differential equation. The order of a differential equation is the order of the highest-order derivatives present in the equation. Equation (5) is a second-order equation because d 2 /dt 2 is the highest-order derivative present. Equation (6) is a first-order equation because onl first-order partial derivatives occur. It will be useful to classif ordinar differential equations as being either linear or nonlinear. Remember that lines (in two dimensions) and planes (in three dimensions) are especiall eas to visualize, when compared to nonlinear objects such as cubic curves or quadric surfaces. For eample, all the points on a line can be found if we know just two of them. Correspondingl, linear differential equations are more amenable to solution than nonlinear ones. Now the equations for lines a b c and planes a b cz d have the feature that the variables appear in additive combinations of their first powers onl. B analog a linear differential equation is one in which the dependent variable and its derivatives appear in additive combinations of their first powers. More precisel, a differential equation is linear if it has the format (7) a n AB dn n a n AB dn n p a AB a 0AB FAB, where a n AB, a n AB,..., a 0 AB and FAB depend onl on the independent variable. The additive combinations are permitted to have multipliers (coefficients) that depend on ; no restrictions are made on the nature of this -dependence. If an ordinar differential equation is not linear, then we call it nonlinear. For eample, d is a nonlinear second-order ordinar differential equation because of the 3 term, whereas t 3 dt t 3 is linear (despite the t 3 terms). The equation d 2 cos 2 is nonlinear because of the / term.

5 Section. Background 5 Although the majorit of equations one is likel to encounter in practice fall into the nonlinear categor, knowing how to deal with the simpler linear equations is an important first step (just as tangent lines help our understanding of complicated curves b providing local approimations).. EXERCISES In Problems 2, a differential equation is given along with the field or problem area in which it arises. Classif each as an ordinar differential equation (ODE) or a partial differential equation (PDE), give the order, and indicate the independent and dependent variables. If the equation is an ordinar differential equation, indicate whether the equation is linear or nonlinear. d (Hermite s equation, quantum-mechanical harmonic oscillator) d cos 3t dt 2 4 dt (mechanical vibrations, electrical circuits, seismolog) 0 2 u u (Laplace s equation, potential theor, electricit, heat, aeronamics) A2 3B 4. A 3B (competition between two species, ecolog) 5. ka4 BA B, where k is a constant dt (chemical reaction rates) 6. c a 2, where C is a constant b d C (brachistochrone problem, calculus of variations) 7. d (Kidder s equation, flow of gases through a porous medium) dp 8. kpap pb, where k and P are constants dt (logistic curve, epidemiolog, economics) 9. 8 d 4 A B 4 (deflection of beams) d (aeronamics, stress analsis) 0N. where k is a constant 0t 02 N 0r 2 0N kn, r 0r (nuclear fission) d A 2 B 9 0 (van der Pol s equation, triode vacuum tube) In Problems 3 6, write a differential equation that fits the phsical description. 3. The rate of change of the population p of bacteria at time t is proportional to the population at time t. 4. The velocit at time t of a particle moving along a straight line is proportional to the fourth power of its position. 5. The rate of change in the temperature T of coffee at time t is proportional to the difference between the temperature M of the air at time t and the temperature of the coffee at time t. 6. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t. 7. Drag Race. Two drivers, Alison and Kevin, are participating in a drag race. Beginning from a standing start, the each proceed with a constant acceleration. Alison covers the last /4 of the distance in 3 seconds, whereas Kevin covers the last /3 of the distance in 4 seconds. Who wins and b how much time? Historical Footnote: In 630 Galileo formulated the brachistochrone problem A bráísto shortest, róno time B, that is, to determine a path down which a particle will fall from one given point to another in the shortest time. It was reproposed b John Bernoulli in 696 and solved b him the following ear. 5

6 6 Chapter Introduction.2 SOLUTIONS AND INITIAL VALUE PROBLEMS An nth-order ordinar differential equation is an equalit relating the independent variable to the nth derivative (and usuall lower-order derivatives as well) of the dependent variable. Eamples are 2 d2 2 3 B ad2 dt2b 0 d 4 t 4 dt (second-order, independent, dependent) (second-order, t independent, dependent) (fourth-order, t independent, dependent). Thus, a general form for an nth-order equation with independent, dependent, can be epressed as () F a,,,..., dn n b 0, where F is a function that depends on,, and the derivatives of up to order n; that is, on,,..., d n / n. We assume that the equation holds for all in an open interval I ( a 6 6 b, where a or b could be infinite). In man cases we can isolate the highest-order term d n / n and write equation () as (2) d n f a,, n,..., d n n b, which is often preferable to () for theoretical and computational purposes. Eplicit Solution Definition. A function fab that when substituted for in equation () [or (2)] satisfies the equation for all in the interval I is called an eplicit solution to the equation on I. Eample Show that fab 2 is an eplicit solution to the linear equation (3) d , but cab 3 is not. Solution The functions fab 2, f AB 2 2, and f AB are defined for all 0. Substitution of fab for in equation (3) gives A2 2 3 B 2 2 A2 B A2 2 3 B A2 2 3 B 0. 6

7 Section.2 Solutions and Initial Value Problems 7 Since this is valid for an 0, the function fab 2 is an eplicit solution to (3) on A q, 0B and also on A0,. For cab we have qbc 3 AB 3 2, c AB 6, and substitution into (3) gives , which is valid onl at the point 0 and not on an interval. Hence c() is not a solution. Eample 2 Show that for an choice of the constants c and c 2, the function fab c e c 2 e 2 is an eplicit solution to the linear equation (4) 2 0. Solution We compute f AB c and f AB c e 4c 2 e 2 e 2c 2 e 2. Substitution of f, f, and f for,, and in equation (4) ields Ac e 4c 2 e 2 B A c e 2c 2 e 2 B 2Ac e c 2 e 2 B Ac c 2c Be A4c 2 2c 2 2c 2 Be 2 0. Since equalit holds for all in A q, qb, then fab c e c 2 e 2 is an eplicit solution to (4) on the interval A q, qb for an choice of the constants c and c 2. As we will see in Chapter 2, the methods for solving differential equations do not alwas ield an eplicit solution for the equation. We ma have to settle for a solution that is defined implicitl. Consider the following eample. Eample 3 Show that the relation (5) implicitl defines a solution to the nonlinear equation (6) 32 2 on the interval A2, qb. Solution When we solve (5) for, we obtain Let s tr f() to see if it is an eplicit solution. Since df/ 3 2 / A22 3 8B, both f and df/ are defined on A2, qb. Substituting them into (6) ields A2 3 8B, which is indeed valid for all in A2, qb. [You can check that cab is also an eplicit solution to (6).]

8 8 Chapter Introduction Implicit Solution Definition 2. A relation GA, B 0 is said to be an implicit solution to equation () on the interval I if it defines one or more eplicit solutions on I. Eample 4 Show that (7) e 0 is an implicit solution to the nonlinear equation (8) A e B e 0. Solution First, we observe that we are unable to solve (7) directl for in terms of alone. However, for (7) to hold, we realize that an change in requires a change in, so we epect the relation (7) to define implicitl at least one function AB. This is difficult to show directl but can be rigorousl verified using the implicit function theorem of advanced calculus, which guarantees that such a function AB eists that is also differentiable (see Problem 30). Once we know that is a differentiable function of, we can use the technique of implicit differentiation. Indeed, from (7) we obtain on differentiating with respect to and appling the product and chain rules, d A e B e a b 0 or A e B e 0, which is identical to the differential equation (8). Thus, relation (7) is an implicit solution on some interval guaranteed b the implicit function theorem. Eample 5 Verif that for ever constant C the relation C is an implicit solution to (9) 4 0. Graph the solution curves for C 0,, 4. (We call the collection of all such solutions a one-parameter famil of solutions.) Solution When we implicitl differentiate the equation C with respect to, we find 8 2 0, See Vector Calculus, 5th ed, b J. E. Marsden and A. J. Tromba (Freeman, San Francisco, 2004).

9 Section.2 Solutions and Initial Value Problems 9 C = 0 C = C = 4 C = 0 2 C = C = C = 4 C = 4 2 Figure.4 Implicit solutions C which is equivalent to (9). In Figure.4 we have sketched the implicit solutions for C 0,, 4. The curves are hperbolas with common asmptotes 2. Notice that the implicit solution curves (with C arbitrar) fill the entire plane and are nonintersecting for C 0. For C 0, the implicit solution gives rise to the two eplicit solutions 2 and 2, both of which pass through the origin. For brevit we hereafter use the term solution to mean either an eplicit or an implicit solution. In the beginning of Section., we saw that the solution of the second-order free-fall equation invoked two arbitrar constants of integration c, c 2 : hatb gt2 2 whereas the solution of the first-order radioactive deca equation contained a single constant C: AAtB Ce kt. It is clear that integration of the simple fourth-order equation d c t c 2, brings in four undetermined constants: AB c 3 c 2 2 c 3 c 4. It will be shown later in the tet that in general the methods for solving nth-order differential equations evoke n arbitrar constants. In most cases, we will be able to evaluate these constants if we know n initial values A 0 B, A 0 B,..., A 0 B. An B

10 0 Chapter Introduction Initial Value Problem Definition 3. B an initial value problem for an nth-order differential equation F a,,,..., dn n b 0, we mean: Find a solution to the differential equation on an interval I that satisfies at 0 the n initial conditions A 0 B 0, A 0 B, d n n A 0 B n, where 0 I and 0,,..., n are given constants. In the case of a first-order equation, the initial conditions reduce to the single requirement A 0 B 0, and in the case of a second-order equation, the initial conditions have the form A 0 B 0, A 0 B. The terminolog initial conditions comes from mechanics, where the independent variable represents time and is customaril smbolized as t. Then if t 0 is the starting time, At 0 B 0 represents the initial location of an object and At 0 B gives its initial velocit. Eample 6 Solution Show that fab sin cos is a solution to the initial value problem (0) d ; A0B, A0B. Observe that fab sin cos, df/ cos sin, and d 2 f/ 2 sin cos are all defined on A q, qb. Substituting into the differential equation gives A sin cos B Asin cos B 0, which holds for all A q, qb. Hence, fab is a solution to the differential equation in (0) on A q, qb. When we check the initial conditions, we find fa0b sin 0 cos 0, df cos 0 sin 0, A0B which meets the requirements of (0). Therefore, fab is a solution to the given initial value problem.

11 Section.2 Solutions and Initial Value Problems Eample 7 As shown in Eample 2, the function fab c e c 2 e 2 d is a solution to for an choice of the constants c and c 2. Determine c and c 2 so that the initial conditions are satisfied. A0B 2 and A0B 3 Solution To determine the constants c and c 2, we first compute df/ to get df/ c e 2c 2 e 2. Substituting in our initial conditions gives the following sstem of equations: fa0b c e0 c 2e0 2, df A0B c e 0 2c 2 e 0 3, or c c 2 2, c 2c 2 3. Adding the last two equations ields 3c 2, so c 2 /3. Since c c 2 2, we find c 7/3. Hence, the solution to the initial value problem is fab A7/3Be A/3Be 2. We now state an eistence and uniqueness theorem for first-order initial value problems. We presume the differential equation has been cast into the format f A, B. Of course, the right-hand side, f A, B, must be well defined at the starting value 0 for and at the stipulated initial value 0 A 0 B for. The hpotheses of the theorem, moreover, require continuit of both f and 0f/0 for in some interval a b containing 0, and for in some interval c d containing 0. Notice that the set of points in the -plane that satisf a b and c d constitutes a rectangle. Figure.5 on page 2 depicts this rectangle of continuit with the initial point A 0, 0 B in its interior and a sketch of a portion of the solution curve contained therein. Eistence and Uniqueness of Solution Theorem. Consider the initial value problem f A, B, A 0 B 0. If f and 0f/0 are continuous functions in some rectangle R E A, B: a 6 6 b, c 6 6 df that contains the point A 0, 0 B, then the initial value problem has a unique solution fab in some interval 0 d 0 d, where d is a positive number.

12 2 Chapter Introduction d 0 = () c a b Figure.5 Laout for the eistence uniqueness theorem The preceding theorem tells us two things. First, when an equation satisfies the hpotheses of Theorem, we are assured that a solution to the initial value problem eists. Naturall, it is desirable to know whether the equation we are tring to solve actuall has a solution before we spend too much time tring to solve it. Second, when the hpotheses are satisfied, there is a unique solution to the initial value problem. This uniqueness tells us that if we can find a solution, then it is the onl solution for the initial value problem. Graphicall, the theorem sas that there is onl one solution curve that passes through the point A 0, 0 B. In other words, for this first-order equation, two solutions cannot cross anwhere in the rectangle. Notice that the eistence and uniqueness of the solution holds onl in some neighborhood A 0 d, 0 db. Unfortunatel, the theorem does not tell us the span A2dB of this neighborhood (merel that it is not zero). Problem 8 elaborates on this feature. Problem 9 gives an eample of an equation with no solution. Problem 29 displas an initial value problem for which the solution is not unique. Of course, the hpotheses of Theorem are not met for these cases. When initial value problems are used to model phsical phenomena, man practitioners tacitl presume the conclusions of Theorem to be valid. Indeed, for the initial value problem to be a reasonable model, we certainl epect it to have a solution, since phsicall something does happen. Moreover, the solution should be unique in those cases when repetition of the eperiment under identical conditions ields the same results. The proof of Theorem involves converting the initial value problem into an integral equation and then using Picard s method to generate a sequence of successive approimations that converge to the solution. The conversion to an integral equation and Picard s method are discussed in Group Project B at the end of this chapter. A detailed discussion and proof of the theorem are given in Chapter 3. At least this is the case when we are considering a deterministic model, as opposed to a probabilistic model. All references to Chapters 3 refer to the epanded tet Fundamentals of Differential Equations and Boundar Value Problems, 6th ed.

13 Section.2 Solutions and Initial Value Problems 3 Eample 8 Solution Eample 9 Solution For the initial value problem () 3 2 3, () 6, does Theorem impl the eistence of a unique solution? Dividing b 3 to conform to the statement of the theorem, we identif f A, B as A 2 3 B/3 and 0f/ 0 as 2. Both of these functions are continuous in an rectangle containing the point (, 6), so the hpotheses of Theorem are satisfied. It then follows from the theorem that the initial value problem () has a unique solution in an interval about of the form A d, db, where d is some positive number. For the initial value problem (2) / 32 3, A2B 0, does Theorem impl the eistence of a unique solution? Here f A, B 3 2 / 3 and 0f/0 2 / 3. Unfortunatel 0f/0 is not continuous or even defined when 0. Consequentl, there is no rectangle containing A2, 0B in which both f and 0f/0 are continuous. Because the hpotheses of Theorem do not hold, we cannot use Theorem to determine whether the initial value problem does or does not have a unique solution. It turns out that this initial value problem has more than one solution. We refer ou to Problem 29 and Group Project G of Chapter 2 for the details. In Eample 9 suppose the initial condition is changed to A2B. Then, since f and 0f/0 are continuous in an rectangle that contains the point A2, B but does not intersect the -ais sa, R E A, B: , F it follows from Theorem that this new initial value problem has a unique solution in some interval about 2..2 EXERCISES. (a) Show that is an implicit solution (c) Show that fab 2 is an eplicit solution to 2 d 2 / 2 2 on the interval A0, qb. to / / A2B on the interval A q, 3B. (b) Show that 3 3 sin = is an implicit solution to In Problems 3 8, determine whether the given function is A cos sin B a solution to the given differential equation. 3A sin B 3. 2 cos t 3 sin t, 0 on the interval A0, p/2b. d 2 2. (a) Show that fab 2 4. sin, is an eplicit solution to cos 2t, t sin 2t dt on the interval A q, qb. (b) Show that fab e is an eplicit solution to 2 e 2 A 2Be 2 on the interval A q, qb sin 2 e, 8. u 2e 3t e 2t, e2 3e, d 2 u dt 2 u du 3u 2e2t dt 4 5e d

14 4 Chapter Introduction In Problems 9 3, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship does define implicitl as a function of and use implicit differentiation. 9. ln 2, ,., e e e 2. 2 sina B, 2 seca B 3. sin 3 2, 6 A B3 sin 2 A B Show that fab c sin c 2 cos is a solution to d 2 / 2 0 for an choice of the constants c and c 2. Thus, c sin c 2 cos is a two-parameter famil of solutions to the differential equation. 5. Verif that fab 2/ A ce B, where c is an arbitrar constant, is a one-parameter famil of solutions to A 2B. 2 Graph the solution curves corresponding to c 0,, 2 using the same coordinate aes. 6. Verif that 2 c 2, where c is an arbitrar nonzero constant, is a one-parameter famil of implicit solutions to 2 and graph several of the solution curves using the same coordinate aes. 7. Show that fab Ce 3 is a solution to / 3 3 for an choice of the constant C. Thus, Ce 3 is a one-parameter famil of solutions to the differential equation. Graph several of the solution curves using the same coordinate aes. 8. Let c 7 0. Show that the function fab (c 2 2 ) is a solution to the initial value problem / 2 2, (0) /c 2, on the interval c 6 6 c. Note that this solution becomes unbounded as approaches c. Thus, the solution eists on the interval ( d, d) with d c, but not for larger d. This illustrates that in Theorem the eistence interval can be quite small (if c is small) or quite large (if c is large). Notice also that there is no clue from the equation / 2 2 itself, or from the initial value, that the solution will blow up at c. 9. Show that the equation A/B has no (real-valued) solution. 20. Determine for which values of m the function fab e m is a solution to the given equation. d 2 (a) d 3 (b) 3 3 d Determine for which values of m the function fab m is a solution to the given equation. (a) 3 2 d (b) 2 d Verif that the function fab c e c 2 e 2 is a solution to the linear equation for an choice of the constants c and c 2. Determine c and c 2 so that each of the following initial conditions is satisfied. (a) (b) In Problems 23 28, determine whether Theorem implies that the given initial value problem has a unique solution. 23. A0B 7 4 4, d A0B 2, A0B AB, AB 0 dt t sin2 t, 3 4t 0, dt ApB 5 A2B p 26. cos sin t, ApB 0 dt 27. AB 0, 28. A2B 3 23,

15 Section.3 Direction Fields (a) For the initial value problem (2) of Eample 9, show that f AB 0 and f 2 AB A 2B 3 are solutions. Hence, this initial value problem has multiple solutions. (See also Group Project G in Chapter 2.) (b) Does the initial value problem 3 2 / 3, (0) 0 7, have a unique solution in a neighborhood of 0? 30. Implicit Function Theorem. Let GA, B have continuous first partial derivatives in the rectangle R E A, B: a 6 6 b, c 6 6 df containing the point A 0, 0 B. If GA 0, 0 B 0 and the partial derivative G A 0, 0 B 0, then there eists a differentiable function fab, defined in some interval I A 0 d, 0 db, that satisfies G A, fabb 0 for all I. The implicit function theorem gives conditions under which the relationship GA, B 0 defines implicitl as a function of. Use the implicit function theorem to show that the relationship e 0, given in Eample 4, defines implicitl as a function of near the point A0, B. 3. Consider the equation of Eample 5, (3) 4 0. (a) Does Theorem impl the eistence of a unique solution to (3) that satisfies A 0 B 0? (b) Show that when 0 0, equation (3) can t possibl have a solution in a neighborhood of 0 that satisfies A 0 B 0. (c) Show that there are two distinct solutions to (3) satisfing A0B 0 (see Figure.4 on page 9)..3 DIRECTION FIELDS The eistence and uniqueness theorem discussed in Section.2 certainl has great value, but it stops short of telling us anthing about the nature of the solution to a differential equation. For practical reasons we ma need to know the value of the solution at a certain point, or the intervals where the solution is increasing, or the points where the solution attains a maimum value. Certainl, knowing an eplicit representation (a formula) for the solution would be a considerable help in answering these questions. However, for man of the differential equations that we are likel to encounter in real-world applications, it will be impossible to find such a formula. Moreover, even if we are luck enough to obtain an implicit solution, using this relationship to determine an eplicit form ma be difficult. Thus, we must rel on other methods to analze or approimate the solution. One technique that is useful in visualizing (graphing) the solutions to a first-order differential equation is to sketch the direction field for the equation. To describe this method, we need to make a general observation. Namel, a first-order equation f A, B specifies a slope at each point in the -plane where f is defined. In other words, it gives the direction that a graph of a solution to the equation must have at each point. Consider, for eample, the equation () 2. The graph of a solution to () that passes through the point A 2, B must have slope A 2B 2 3 at that point, and a solution through A, B has zero slope at that point. A plot of short line segments drawn at various points in the -plane showing the slope of the solution curve there is called a direction field for the differential equation. Because the

16 6 Chapter Introduction 0 0 (a) (b) Figure.6 (a) Direction field for / 2 (b) Solutions to / (a) = 2 (b) = Figure.7 (a) Direction field for / 2 (b) Direction field for / / direction field gives the flow of solutions, it facilitates the drawing of an particular solution (such as the solution to an initial value problem). In Figure.6(a) we have sketched the direction field for equation () and in Figure.6(b) have drawn several solution curves in color. Some other interesting direction field patterns are displaed in Figure.7. Depicted in Figure.7(a) is the pattern for the radioactive deca equation / 2 (recall that in Section. we analzed this equation in the form da/dt ka). From the flow patterns, we can see that all solutions tend asmptoticall to the positive -ais as gets larger. In other words, an material decaing according to this law eventuall dwindles to practicall nothing. This is consistent with the solution formula we derived earlier, A Ce kt, or Ce 2.

17 Section.3 Direction Fields (a) Figure.8 (a) A solution for / 2 (b) (b) A solution for / / From the direction field in Figure.7(b), we can anticipate that all solutions to / / also approach the -ais as approaches infinit (plus or minus infinit, in fact). But more interesting is the observation that no solution can make it across the -ais; 0 AB 0 blows up as goes to zero from either direction. Eception: On close eamination, it appears the function AB 0 might just make it through this barrier. As a matter of fact, in Problem 9 ou are invited to show that the solutions to this differential equation are given b C/, with C an arbitrar constant. So the do diverge at 0, unless C 0. Let s interpret the eistence uniqueness theorem of Section.2 for these direction fields. For Figure.7(a), where / f A, B 2, we select a starting point 0 and an initial value A 0 B 0, as in Figure.8(a). Because the right-hand side f A, B 2 is continuousl differentiable for all and, we can enclose an initial point A 0, 0 B in a rectangle of continuit. We conclude that the equation has one and onl one solution curve passing through A 0, 0 B, as depicted in the figure. For the equation f A, B, the right-hand side f A, B / does not meet the continuit conditions when 0 (i.e., for points on the -ais). However, for an nonzero starting value 0 and an initial value A 0 B 0, we can enclose A 0, 0 B in a rectangle of continuit that ecludes the -ais, as in Figure.8(b). Thus, we can be assured of one and onl one solution curve passing through such a point. The direction field for the equation 32 / 3 is intriguing because Eample 9 of Section.2 showed that the hpotheses of Theorem do not hold in an rectangle enclosing the initial point 0 2, 0 0. Indeed, Problem 29 of that section demonstrated the violation of uniqueness b ehibiting two solutions, AB 0

18 8 Chapter Introduction () = ( 2) 3 0 () = (a) (b) Figure.9 (a) Direction field for / 3 2 / 3 (b) Solutions for / 3 2 / 3, A2B 0 and AB A 2B 3, passing through A2, 0B. Figure.9(a) displas this direction field, and Figure.9(b) demonstrates how both solution curves can successfull negotiate this flow pattern. Clearl, a sketch of the direction field of a first-order differential equation can be helpful in visualizing the solutions. However, such a sketch is not sufficient to enable us to trace, unambiguousl, the solution curve passing through a given initial point A 0, 0 B. If we tried to trace one of the solution curves in Figure.6(b) on page 6, for eample, we could easil slip over to an adjacent curve. For nonunique situations like that in Figure.9(b), as one negotiates the flow along the curve A 2B 3 and reaches the inflection point, one cannot decide whether to turn or to (literall) go off on the tangent A 0B. Eample The logistic equation for the population p (in thousands) at time t of a certain species is given b (2) dp dt pa2 pb. (Of course, p is nonnegative. The interpretation of the terms in the logistic equation is discussed in Section 3.2.) From the direction field sketched in Figure.0 on page 9, answer the following: (a) If the initial population is that is, pa0b 3 4, what can ou sa about the limiting population lim ts q patb? (b) Can a population of 000 ever decline to 500? (c) Can a population of 000 ever increase to 3000? Solution (a) The direction field indicates that all solution curves 3 other than patb 0 4 will approach the horizontal line p 2 as t S q; that is, this line is an asmptote for all positive solutions. Thus, lim ts q patb 2.

19 Section.3 Direction Fields 9 p (in thousands) t Figure.0 Direction field for logistic equation (b) The direction field further shows that populations greater than 2000 will steadil decrease, whereas those less than 2000 will increase. In particular, a population of 000 can never decline to 500. (c) As mentioned in part (b), a population of 000 will increase with time. But the direction field indicates it can never reach 2000 or an larger value; i.e., the solution curve cannot cross the line p 2. Indeed, the constant function patb 2 is a solution to equation (2), and the uniqueness part of Theorem, page, precludes intersections of solution curves. Notice that the direction field in Figure.0 has the nice feature that the slopes do not depend on t; that is, the slopes are the same along each horizontal line. The same is true for Figures.8(a) and.9. This is the ke propert of so-called autonomous equations f AB, where the right-hand side is a function of the dependent variable onl. Group Project C, page 32, investigates such equations in more detail. Hand sketching the direction field for a differential equation is often tedious. Fortunatel, several software programs have been developed to obviate this task. When hand sketching is necessar, however, the method of isoclines can be helpful in reducing the work. The Method of Isoclines An isocline for the differential equation f A, B is a set of points in the -plane where all the solutions have the same slope /; thus, it is a level curve for the function f A, B. For eample, if (3) f A, B, the isoclines are simpl the curves (straight lines) c or c. Here c is an arbitrar constant. But c can be interpreted as the numerical value of the slope / of ever solution curve as it crosses the isocline. (Note that c is not the slope of the isocline itself; the latter is, obviousl,.) Figure.(a) on page 20 depicts the isoclines for equation (3). An applet, maintained on the web at sketches direction fields and automates most of the differential equation algorithms discussed in this book.

20 20 Chapter Introduction c = 5 c = 4 0 c = 3 c = 2 0 c = c = 5 c = 4 c = 3 c = 2 c = (a) c = 0 (b) 0 (c) Figure. (a) Isoclines for (b) Direction field for (c) Solutions to To implement the method of isoclines for sketching direction fields, we draw hash marks with slope c along the isocline f A, B c for a few selected values of c. If we then erase the underling isocline curves, the hash marks constitute a part of the direction field for the differential equation. Figure.(b) depicts this process for the isoclines shown in Figure.(a), and Figure.(c) displas some solution curves. Remark. The isoclines themselves are not alwas straight lines. For equation () at the beginning of this section (page 5), the are parabolas 2 c. When the isocline curves are complicated, this method is not practical.

21 Section.3 Direction Fields 2.3 EXERCISES. The direction field for / = 2 + is shown in Figure.2. (a) Sketch the solution curve that passes through A0, 2B. From this sketch, write the equation for the solution. (b) Sketch the solution curve that passes through A, 3B. (c) What can ou sa about the solution in part (b) as S q? How about S q? = = Figure.3 Direction field for / 4/ and 5. Wh is the value 8 called the terminal velocit? 4. If the viscous force in Problem 3 is nonlinear, a possible model would be provided b the differential equation dt 3 8. Figure.2 Direction field for / 2 2. The direction field for / 4/ is shown in Figure.3. (a) Verif that the straight lines 2 are solution curves, provided 0. (b) Sketch the solution curve with initial condition A0B 2. (c) Sketch the solution curve with initial condition A2B. (d) What can ou sa about the behavior of the above solutions as S q? How about S q? 3. A model for the velocit at time t of a certain object falling under the influence of gravit in a viscous medium is given b the equation dt 8. From the direction field shown in Figure.4, sketch the solutions with the initial conditions A0B 5, 8, Redraw the direction field in Figure.4 to incorporate this 3 dependence. Sketch the solutions with initial conditions A0B 0,, 2, 3. What is the terminal velocit in this case? 8 0 υ Figure.4 Direction field for dt 8 t

22 22 Chapter Introduction 5. The logistic equation for the population (in thousands) of a certain species is given b dp dt 3p 2p2. (a) Sketch the direction field b using either a computer software package or the method of isoclines. (b) If the initial population is that is, pa0b 3 4, what can ou sa about the limiting population lim ts q p AtB? (c) If pa0b 0.8, what is lim ts q p AtB? (d) Can a population of 2000 ever decline to 800? 6. Consider the differential equation sin. (a) A solution curve passes through the point A, p/2b. What is its slope at this point? (b) Argue that ever solution curve is increasing for. (c) Show that the second derivative of ever solution satisfies d 2 2 cos sin 2. 2 (d) A solution curve passes through A0, 0B. Prove that this curve has a relative minimum at A0, 0B. 7. Consider the differential equation dp pa p BA2 pb dt for the population p (in thousands) of a certain species at time t. (a) Sketch the direction field b using either a computer software package or the method of isoclines. (b) If the initial population is that is, pa0b 4 4, what can ou sa about the limiting population lim ts q p AtB? (c) If pa0b.7, what is lim ts q p AtB? (d) If pa0b 0.8, what is lim ts q p AtB? (e) Can a population of 900 ever increase to 00? 8. The motion of a set of particles moving along the -ais is governed b the differential equation dt t3 3, where AtB denotes the position at time t of the particle. (a) If a particle is located at when t 2, what is its velocit at this time? (b) Show that the acceleration of a particle is given b d 2 dt 2 3t2 3t (c) If a particle is located at 2 when t 2.5, can it reach the location at an later time? 3 Hint: t 3 3 At BAt 2 t 2 B Let fab denote the solution to the initial value problem, A0B. (a) Show that f AB f AB fab. (b) Argue that the graph of f is decreasing for near zero and that as increases from zero, fab decreases until it crosses the line, where its derivative is zero. (c) Let * be the abscissa of the point where the solution curve fab crosses the line. Consider the sign of f A*B and argue that f has a relative minimum at *. (d) What can ou sa about the graph of fab for *? (e) Verif that is a solution to / and eplain wh the graph of fab alwas stas above the line. (f) Sketch the direction field for / b using the method of isoclines or a computer software package. (g) Sketch the solution fab using the direction field in part AfB. 0. Use a computer software package to sketch the direction field for the following differential equations. Sketch some of the solution curves. (a) / sin (b) / sin (c) / sin sin (d) / (e) / In Problems 6, draw the isoclines with their direction markers and sketch several solution curves, including the curve satisfing the given initial conditions.. / /, A0B 4 2. /, A0B 3. / 2, A0B 4. / /, A0B 5. / 2 2, A0B 0 6. / 2, A0B 22

23 Section.4 The Approimation Method of Euler From a sketch of the direction field, what can one sa about the behavior as approaches q of a solution to the following? 3 8. From a sketch of the direction field, what can one sa about the behavior as approaches q of a solution to the following? 9. B rewriting the differential equation / / in the form integrate both sides to obtain the solution C/ for an arbitrar constant C. 20. A bar magnet is often modeled as a magnetic dipole with one end labeled the north pole N and the opposite end labeled the south pole S. The magnetic field for the magnetic dipole is smmetric with respect to rotation about the ais passing lengthwise through the center of the bar. Hence we can stu the magnetic field b restricting ourselves to a plane with the bar magnet centered on the -ais. For a point P that is located a distance r from the origin, where r is much greater than the length of the magnet, the magnetic field lines satisf the differential equation (4) and the equipotential lines satisf the equation (5) (a) Show that the two families of curves are perpendicular where the intersect. [Hint: Consider the slopes of the tangent lines of the two curves at a point of intersection.] (b) Sketch the direction field for equation (4) for 5 5, 5 5. You can use a software package to generate the direction field or use the method of isoclines. The direction field should remind ou of the eperiment where iron filings are sprinkled on a sheet of paper that is held above a bar magnet. The iron filings correspond to the hash marks. (c) Use the direction field found in part (b) to help sketch the magnetic field lines that are solutions to (4). (d) Appl the statement of part (a) to the curves in part (c) to sketch the equipotential lines that are solutions to (5). The magnetic field lines and the equipotential lines are eamples of orthogonal trajectories. (See Problem 32 in Eercises 2.4, pages ).4 THE APPROXIMATION METHOD OF EULER Euler s method (or the tangent-line method) is a procedure for constructing approimate solutions to an initial value problem for a first-order differential equation () f A, B, A 0 B 0. It could be described as a mechanical or computerized implementation of the informal procedure for hand sketching the solution curve from a picture of the direction field. As such, we will see that it remains subject to the failing that it ma skip across solution curves. However, under fairl general conditions, iterations of the procedure do converge to true solutions. Equations (4) and (5) can be solved using the method for homogeneous equations in Section 2.6 (see Eercises 2.6, Problem 47).

24 24 Chapter Introduction (, ) ( 2, 2 ) ( 0, 0 ) Slope f( 0, 0 ) Slope f(, ) Slope f( 2, 2 ) ( 3, 3 ) Figure.5 Polgonal-line approimation given b Euler s method The method is illustrated in Figure.5. Starting at the initial point A 0, 0 B, we follow the straight line with slope f A 0, 0 B, the tangent line, for some distance to the point A, B. Then we reset the slope to the value f A, B and follow this line to A 2, 2 B. In this wa we construct polgonal (broken line) approimations to the solution. As we take smaller spacings between points (and thus emplo more points), we ma epect to converge to the true solution. To be more precise, assume that the initial value problem () has a unique solution fab in some interval centered at 0. Let h be a fied positive number (called the step size) and consider the equall spaced points n J 0 nh, n 0,, 2,.... The construction of values n that approimate the solution values fa n B proceeds as follows. At the point A 0, 0 B, the slope of the solution to () is given b / f A 0, 0 B. Hence, the tangent line to the solution curve at the initial point A 0, 0 B is 0 A 0 B f A 0, 0 B. Using this tangent line to approimate fab, we find that for the point 0 h fa B J 0 hfa 0, 0 B. Net, starting at the point A, B, we construct the line with slope given b the direction field at the point A that is, with slope equal to. If we follow this line, B f A, B 3 namel, A B f A, B 4 in stepping from to 2 h, we arrive at the approimation fa 2 B 2 J hfa, B. Repeating the process (as illustrated in Figure.5), we get fa 3 B 3 J 2 hfa 2, 2 B, fa 4 B 4 J 3 hfa 3, 3 B, etc. The smbol J means is defined to be. Because is an approimation to fa B, we cannot assert that this line is tangent to the solution curve fab.

25 Section.4 The Approimation Method of Euler 25 This simple procedure is Euler s method and can be summarized b the recursive formulas (2) (3) n n h, n n h f A n, n B, n 0,, 2,.... Eample Solution Use Euler s method with step size h 0. to approimate the solution to the initial value problem (4) 2, AB 4 at the points.,.2,.3,.4, and.5. Here 0, 0 4, h 0., and f A, B 2. Thus, the recursive formula (3) for n is n n h f A n, n B n A0.B n 2 n. Substituting n 0, we get , 0 A0.B A0.BAB Putting n ields , 2 A0.B A0.BA.B Continuing in this manner, we obtain the results listed in Table.. For comparison we have included the eact value (to five decimal places) of the solution fab A 2 7B 2 /6 to (4), which can be obtained using separation of variables (see Section 2.2). As one might epect, the approimation deteriorates as moves farther awa from. TABLE. Computations for 2, () 4 Euler s n n Method Eact Value Given the initial value problem () and a specific point, how can Euler s method be used to approimate fab? Starting at 0, we can take one giant step that lands on, or we can take several smaller steps to arrive at. If we wish to take N steps, then we set h A 0 B/N so that the step size h and the number of steps N are related in a specific wa. For eample, if 0.5 and we wish to approimate fa2b using 0 steps, then we would take h A2.5B/ It is epected that the more steps we take, the better will be the approimation. (But keep in mind that more steps mean more computations and hence greater accumulated roundoff error.) 25

26 26 Chapter Introduction Eample 2 Use Euler s method to find approimations to the solution of the initial value problem (5), A0B at, taking, 2, 4, 8, and 6 steps. Remark. Observe that the solution to (5) is just fab e, so Euler s method will generate algebraic approimations to the transcendental number e Solution Here f A, B, 0 0, and 0. The recursive formula for Euler s method is n n h n A hb n. To obtain approimations at with N steps, we take the step size h /N. For N, we have fab A BAB 2. For N 2, fa 2 B fab 2. In this case we get A 0.5BAB.5, fab 2 A 0.5BA.5B For N 4, fa 4 B fab 4, where A 0.25BAB.25, 2 A 0.25BA.25B.5625, 3 A 0.25BA.5625B.9533, fab 4 A 0.25BA.9533B (In the above computations, we have rounded to five decimal places.) Similarl, taking N 8 and 6, we obtain even better estimates for fab. These approimations are shown in Table.2. For comparison, Figure.6 on page 27 displas the polgonal-line approimations to e using Euler s method with h /4 AN 4B and h /8 AN 8B. Notice that the smaller step size ields the better approimation. TABLE.2 Euler s Method for =, (0) = Approimation N h for FAB e

27 Section.4 The Approimation Method of Euler = e 2.25 h = / h = / / 8 / 4 3/ 8 / 2 5/ 8 3/ 4 7/ 8 Figure.6 Approimations of e using Euler s method with h = /4 and /8 How good (or bad) is Euler s method? In judging a numerical scheme, we must begin with two fundamental questions. Does the method converge? And, if so, what is the rate of convergence? These important issues are discussed in Section 3.6, where improvements in Euler s method are introduced (see also Problems 2 and 3 of this section). Eample 3 Suppose vatb satisfies the initial value problem dv dt 3 2v2, v(0) 2. B eperimenting with Euler s method, determine to within one decimal place value of va0.2b and the time it will take vatb to reach zero. A 0.B the Solution Determining rigorous estimates of the accurac of the answers obtained b Euler s method can be quite a challenging problem. The common practice is to repeatedl approimate va0.2b and the zero crossing, using smaller and smaller values of h, until the digits of the computed values stabilize at the required accurac level. For this eample, Euler s algorithm ields the following values: h 0. h 0.05 h h h va0.2b va0.2b va0.2b va0.2b va0.2b va0.3b va0.35b va0.375b va0.4b va0.4b va0.4b Acknowledging the remote possibilit that finer values of h might reveal aberrations, we state with reasonable confidence that va0.2b The Intermediate Value Theorem would impl that vat 0 B 0 at some time t 0 satisfing t , if the computations were perfect; the clearl provide evidence that t