MAT 285: Introduction to Differential Equations James V. Lambers April 2, 27
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Contents Introduction 5. Some Basic Mathematical Models............................. 5.2 Solutions of Some Differential Equations......................... 6.3 Direction Fields....................................... 7.4 Classification of Differential Equations.......................... 9 2 First-Order Equations 2. Linear Equations...................................... 2.2 Separable Equations.................................... 8 2.2. Homogeneous Equations.............................. 2 2.3 Modeling with First-Order Equations........................... 22 2.4 Autonomous Equations.................................. 25 2.5 Exact Equations...................................... 28 2.5. Integrating Factors................................. 3 2.6 The Existence-Uniqueness Theorem........................... 33 3 Second-Order Equations 35 3. Homogeneous Equations with Constant Coefficients................... 35 3.2 The Wronskian....................................... 39 3.3 Complex Roots of the Characteristic Equation..................... 43 3.3. Derivation of Euler s Formula........................... 45 3.4 Reduction of Order and Repeated Roots......................... 45 3.5 Inhomogeneous Equations................................. 48 3.5. Structure of the Solution............................. 52 3.5.2 Appendix: Exponentiation of an Upper Triangular Matrix........... 53 3.5.3 Appendix: An Alternative Derivation...................... 54 3.6 The Method of Undetermined Coefficients........................ 56 3.7 Variation of Parameters.................................. 62 3.8 Mechanical Vibrations................................... 66 3.9 Forced Vibrations...................................... 7 4 Higher Order Linear Equations 73 4. General Theory of nth Order Linear Equations..................... 73 4.2 Homogeneous Equations with Constant Coefficients................... 75 4.3 The Method of Undetermined Coefficients........................ 76 3
4 CONTENTS 4.4 Variation of Parameters.................................. 77 4.4. Appendix: Determinants.............................. 8 4.4.2 Appendix: Derivation of Cramer s Rule..................... 82 5 Laplace Transforms 85 5. Definition of the Laplace Transform........................... 85 5.2 Solution of Initial Value Problems............................ 86 5.3 Step Functions....................................... 9 5.4 Differential Equations with Discontinuous Forcing Functions.............. 93
Chapter Introduction. Some Basic Mathematical Models This course is an introduction to differential equations, which are equations that depend on derivatives of unknown quantities. Differential equations arise in mathematical models of a wide variety of phenomena, such as propagation of waves, dissipation of heat energy, population growth, or motion of fluids. Solutions of differential equations yield valuable insight about such phenomena, and therefore techniques for solving differential equations are among the most essential methods of applied mathematics. We now illustrate mathematical models based on differential equations. Newton s Second Law states F ma m dv dt, where F, m, a and v represent force, mass, acceleration and velocity, respectively. We use this law to develop a mathematical model for the velocity of a falling object that includes a differential equation. The forces on the falling object include gravity and air resistance, or drag; to simplify the discussion, we neglect any other forces. The force due to gravity is equal to mg, where g is the acceleration due to gravity, and the drag force is equal to γv, where γ is the drag coefficient. We use downward orientation, so that gravity is acting in the positive (downward) direction and drag is acting in the negative (upward) direction. In summary, we have F mg γv. Combining with Newton s Second Law yields the differential equation m dv mg γv dt for the velocity v of the falling object. Another example of a mathematical model is a differential equation for the population p of a species, which can have the form dp rp d, dt where the constant r is the rate of reproduction of the species. In general, r is called a rate constant or growth rate. The constant d indicates the number of specimens that die per unit of time, perhaps due to predation or other causes. 5
6 CHAPTER. INTRODUCTION In general, a mathematical model is constructed by identifying all of the relevant quantities and how they relate to one another. Associating variables with these quantities facilities the development of an equation that relates them. Such an equation can be obtained from, for example, physical laws that govern the behavior of these quantities..2 Solutions of Some Differential Equations The differential equation from the preceding example is of the form dy dt ay b, where a and b are constants. In this section, we seek a solution of this more general differential equation. That is, we will find a function y(t) that satisfies the equation. We will see that a differential equation such as this one does not have a unique solution, as it does not include enough information. Typically, the differential equation is paired with an initial condition of the form y(t ) y, where t represents an initial time and y is an initial value. The differential equation, together with the intial condition, is called an initial value problem. Under certain assumptions, it can be proven that an initial value problem has a unique solution. We now solve the above differential equation. Rearranging yields dy y (b/a) dt a. Then, we can integrate both sides with respect to t: dy y (b/a) dt dt a dt. On the left side, we use a substitution to simplify the integral, which yields y y(t), dy dy dt dt y (b/a) dy a dt. Integrating both sides, we obtain ln y b a at + C, where C is an arbitrary constant. Exponentiating both sides yields y b a eat+c,
.3. DIRECTION FIELDS 7 which can be rearranged to obtain y b a + ceat, where c ±e C. This solution is known as the general solution. It is a function of the independent variable t as well as the parameter c, the value of which depends on the initial condition. Thus, the general solution consists of infinitely many curves, each one being the graph of a function of t corresponding to a different value of c. Each such curve is called an integral curve. We now use the general solution to obtain the solution of the initial value problem. Substituting the initial condition into the general solution, we have We then solve this equation for c to obtain c y b a + ceat. ( y b ) e at. a Substituting this expression for c into the general solution yields the solution of the inital value problem, y b ( a + y b ) e a(t t). a Example We revisit the equation dv dt 9.8 v 5. This is an equation of the form dy/dt ay b, with y v, a /5 and b 9.8. Applying the above formula for the general solution yields v(t) 49 + ce t/5, where c is an arbitrary constant. Now, if we add the initial condition v() 5, we obtain the unique solution v(t) 49 + (5 49)e (t )/5 49 + e t/5..3 Direction Fields It is possible to understand, in a qualitative sense, the behavior of solutions of a differential equation without actually computing such solutions explicitly. This understanding can be achieved through the construction of a direction field for the differential equation. Consider a differential equation of the form dy dt f(t, y).
8 CHAPTER. INTRODUCTION The function f(t, y) is called the rate function, as its value is the instantaneous rate of change of y with respect to t. To obtain a direction field for this equation, we compute values of f(t, y) for several points in the ty-plane. Then, at each such point (t, y ), we plot an arrow emanating from (t, y ), whose slope is equal to f(t, y ). It follows that this arrow is tangent to the solution of the differential equation at (t, y ). Because the direction field consists of arrows that indicate the slope of the solution, it is also known as a slope field. If there is a value of y, denoted by y, such that f(t, y ) for all t, then a direction field that consists of points along the line y y includes arrows with a slope of zero, which means that the solution is not changing along this line, as dy/dt, and therefore y is constant. Such a value y is called an equilibrium solution of the differential equation. Example Consider the differential equation dv dt 9.8 v 5. This equation has an equilibrium solution of v 5(9.8) 49. This equilibrium solution manifests itself in the direction field in Figure. as a set of horizontal arrows on the line v 49. The direction field itself is obtained by plotting arrows with slope f(t, y ) 9.8 v /5 for several points (t, y ) in the ty-plane. Figure.: Direction field and equilibrium solution for the differential equation dv/dt 9.8 v/5.
.4. CLASSIFICATION OF DIFFERENTIAL EQUATIONS 9.4 Classification of Differential Equations There are many types of differential equations, and a wide variety of solution techniques, even for equations of the same type, let alone different types. We now introduce some terminology that aids in classification of equations and, by extension, selection of solution techniques. An ordinary differential equation, or ODE, is an equation that depends on one or more derivatives of functions of a single variable. Differential equations given in the preceding examples are all ordinary dfiferential equations, and we will consider these equations exclusively in this course. A partial differential equation, or PDE, is an equation that depends on one or more partial derivatives of functions of several variables. In many cases, PDE are solved by reducing to multiple ODE. Example The heat equation u t k2 2 u x 2, where k is a constant, is an example of a partial differential equation, as its solution u(x, t) is a function of two independent variables, and the equation includes partial derivatives with respect to both variables. The order of a differential equation is the order of the highest derivative of any unknown function in the equation. Example The differential equation dy dt ay b, where a and b are constants, is a first-order differential equation, as only the first derivative of the solution y(t) appears in the equation. On the other hand, the ODE y + 3y + 2y is a second-order differential equation, whereas the PDE known as the beam equation is a fourth-order differential equation. u t u xxxx A differential equation is linear if any linear combination of solutions of the equation is also a solution of the equation. A differential equation that is not linear is said to be nonlinear. Nonlinear equations are, in general, very difficult to solve, so in many cases one approximates a nonlinear equation by a linear equation, called a linearization, that is more readily solved. Example The ODE y + 3t 2 y e t, y + (sin t)y + ty
CHAPTER. INTRODUCTION are examples of linear differential equations. Note that coefficients of these equations may be functions of the independent variable t, but not of the dependent variable y. On the other hand, the PDE known as Burger s equation, u t + uu x, is a nonlinear differential equation. A linearization may be obtained by replacing the coefficient u of u x with a constant or a function of only x or t. A differential equation of any type, in conjunction with any other information such as an initial condition, is said to describe a well-posed problem if it satisfies three conditions, known as Hadamard s conditions for well-posedness: A solution of the problem exists. A solution of the problem is unique. The unique solution depends continuously on the problem data, which may include initial values or coefficients of the differential equation. That is, a small change in the data corresponds to a small change in the solution. Unfortunately, problems can easily fail to be well-posed by not satisfying any of these conditions. However, in this course we will learn to solve a variety of initial value problems that are well-posed.
Chapter 2 First-Order Equations In this chapter, we begin our study of the solution of first-order ODE. 2. Linear Equations Consider a first-order ODE of the form dy dt f(t, y), with an initial condition y(t ) y. The solution y(t) of such an initial value problem can be described explicitly, by an equation of the form y g(t) for some function g, or implicitly, through an equation of the form F (t, y) for some function F. An explicit equation is far more convenient, and in most cases we will obtain explicit solutions of ODE, but the implicit form is more generally applicable, as it is not always easy to solve for y in terms of t. With this in mind, we differentiate both sides of the equation F (t, y) with respect to t, and obtain, via the Chain Rule, F t + F dy y dt, from which it follows that the unknown function F must satisfy f(t, y) F t(t, y) F y (t, y), F y(t, y). Note that here we have used the standard notation F t F/ t, F y F/ y for partial derivatives. Unfortunately, it is generally not easy to find a function F that satisfies this equation. However, it can be simplified by keeping in mind that if y(t) satisfies the original ODE, then it also satisfies an ODE obtained by scaling (that is, multiplying both sides of) the original ODE by a nonzero function µ(t, y). The scaled ODE can be written in the form and then compared to the equation µ(t, y)f(t, y) + µ(t, y) dy dt, F t + F y dy dt
2 CHAPTER 2. FIRST-ORDER EQUATIONS to obtain the conditions F t µf, F y µ. If we differentiate the first equation with respect to y and the second equation with respect to t, we obtain F ty (µ y f + µf y ), F yt µ t. Using the fact that F ty F yt, we then obtain an equation for µ, µ y f µf y µ t. The function µ is called an integrating factor. If it can be found, then the function F can be found via integration. Specifically, we can exploit the fact that F y µ to obtain F (t, y) µ(t, y) dy M(t, y) + c(t), where M(t, y) is an antiderivative of µ with respect to y, and c is an unknown function of t that plays the role of the constant of integration. Then, by differentiating both sides of this equation with respect to t, we obtain F t µf M t + c (t), which can then be solved for c(t) to complete the definition of F (t, y). The equation for µ is not always easy to solve, but it can be readily solved in certain special cases. For example, suppose that f(t, y) can be written in the form f(t, y) g(t)h(y), for some functions g and h that depend exclusively on t and y, respectively. When this is the case, we say that the ODE is separable. Then, by choosing µ(t, y) /h(y), we obtain the scaled ODE which yields the implicitly defined solution where M(y) h(y) dy, g(t) + dy h(y) dt, F (t, y) M(y) + c(t), c (t) µf M t g(t)h(y) h(y) We conclude that the solution is given by the equation F (t, y) h(y) dy g(t) dt. [M(y) g(t). t Example Consider the equation dy dt a(t)y,
2.. LINEAR EQUATIONS 3 which is separable, with g(t) a(t) and h(y) y. Then, the solution y(t) can be described by an equation of the form F (t, y), where F (t, y) y dy a(t) dt, or ln y a(t) dt + C, where C is an arbitrary constant. Exponentiating both sides, we obtain [ y(t) c exp a(t) dt, where c ±e C is a constant. Example From the previous example, the solution of the separable equation is where c is an arbitrary constant. dy dt 3t2 y y(t) ce 3t 2 dt ce t3, Another special case, which is of particular interest, is when f(t, y) is a linear function of t. That is, f(t, y) p(t)y + q(t), for given functions p and q that do not depend on y. An ODE of this form, dy dt + p(t)y q(t), is called a linear first-order ordinary differential equation. In this case, the integrating factor µ must satisfy the equation which simplifies to µ y ( py + q) µ( p) µ t, µ y py µ y q + µp µ t. By stipulating that µ is a function of only t, it follows that µ y and the equation for µ(t) simplifies to µ (t) p(t)µ(t). This is a separable equation of the kind we solved in the previous example, and therefore we have [ µ(t) c exp p(t) dt, where c is a constant. Since we are only using µ(t) to scale the original ODE so that it can be solved, we can simply set c.
4 CHAPTER 2. FIRST-ORDER EQUATIONS To obtain the solution y(t), we proceed as before and try to obtain the function F (t, y) by integration. We have F (t, y) µ(t) dy µ(t)y + c(t), where c(t) is unknown. From the condition F t µf, we obtain c (t) F t (t, y) [µ(t)y t µ(t)f(t, y) [µ(t)y t µ(t)( p(t)y + q(t)) µ (t)y µ(t)p(t)y µ(t)q(t) p(t)µ(t)y µ(t)q(t). It follows from integration that the solution of our linear ODE is F (t, y) µ(t)y µ(t)q(t) dt, which, from rearranging and using the formula for µ(t), yields the explicit solution y(t) µ(t)q(t) dt e p(t) dt p(t) dt e q(t) dt. µ(t) Example We will now solve the linear equation In this equation, p(t) t dy dt + y sin t, t >. t and q(t) sin t. The integrating factor µ(t) is given by µ(t) e t dt e ln t e ln t t, since t >. solution is It follows from the preceding solution formula, and integration by parts, that the y(t) t sin t dt [ t cos t + sin t + C, t t where C is an arbitrary constant. It is interesting to note that with this choice of µ(t), if we rewrite the scaled ODE as µ(t) dy dt then, because µ (t) p(t)µ(t), the left side becomes + µ(t)p(t)y µ(t)q(t), µ(t)y (t) + µ (t)y(t) d dt [µ(t)y(t), by the Product Rule for differentiation. It follows that d [µ(t)y(t) µ(t)q(t), dt
2.. LINEAR EQUATIONS 5 and therefore µ(t)y(t) µ(t)q(t) dt, which yields the same formula for the solution as before, y(t) µ(t)q(t) dt. µ(t) We now consider how best to take the initial condition y(t) y into account in our formula for the solution of a linear first-order ODE of the form dy/dt p(t)y(t) + q(t). To that end, we use definite integrals rather than indefinite integrals in obtaining our solution. First, we use the integrating factor t µ(t) e t p(s) ds. Note that the integral is with respect to s, rather than t. This is because t is used as a limit of integration, so it cannot be used as the variable of integration. Note that as a consequence of this formula, we have t µ(t ) e t p(s) ds e. Then, proceeding as before, our solution is defined by the equation F (t, y), where F (t, y) µ(t)y + c(t), c (t) µ(t)q(t). From these equations, we obtain an equation for the solution, F (t, y) µ(t)y t t µ(s)q(s) ds + K, where K is a constant to be determined. To obtain the value of K, we require that the initial condition be satisfied. We therefore set t t and y y, which yields F (t, y ) µ(t )y t t µ(s)q(s) ds K, which is satisfied by setting K µ(t )y y, since µ(t ). The solution y(t) therefore satisfies the equation t F (t, y) µ(t)y(t) y µ(s)q(s) ds, t which leads to the explicit formula for the solution, y(t) [ t y + µ(s)q(s) ds e [ t t t p(s) ds s y + e t p(z) dz q(s) ds. µ(t) t t We now verify that this function y(t) is actually a solution of the initial value problem. First, we set t t and obtain y(t ) e [ t t s t p(s) ds y + e t p(z) dz q(s) ds e [y + y, t
6 CHAPTER 2. FIRST-ORDER EQUATIONS so the initial condition is indeed satisfied. Then, we differentiate y(t) and obtain y (t) d dt [e t t p(s) ds [ y + e t t p(s) ds d dt d [ t p(s) ds dt t e t t p(s) ds d dt [ t y + [ t p(t)e t t p(s) ds t s e t s e t e t t p(s) ds s e t [ y + t p(z) dz q(s) ds + t p(z) dz q(s) ds t p(z) dz q(s) ds t s e t [ t s y + e t p(z) dz q(s) ds + t t p(z) dz q(s) ds p(t)y(t) + e t t p(s) ds e t t p(s) ds q(t) p(t)y(t) + q(t). + e [ t t p(s) ds t e t p(z) dz q(t) We conclude that our formula for y(t) produces the solution to the initial value problem. Example Consider the first-order linear equation y + y te t +. In this equation, p(t) and q(t) te t +. A direction field for this equation is shown in Figure 2.. From our solution formula, we obtain µ(t) e p(t) dt e dt e t, and y(t) µ(t)q(t) dt µ(t) e t e t (te t + ) dt e t t + e t dt [ t e t 2 2 + et + C t2 2 e t + + Ce t. Now, suppose that we impose the initial condition y() 2. Then, using our explicit solution formula that includes an initial condition, we have the integrating factor µ(t) e t ds e s t e t e t, which yields the solution y(t) [ t y + µ(s)q(s) ds µ(t) t
2.. LINEAR EQUATIONS 7 Figure 2.: Direction field for y + y te t + e t [2 + t t e s (se s + ) ds s + e s ds e [2 t + ( ) s e [2 t 2 t + 2 + es [ e t 2 + t2 2 + et 2 2 e [ e t + t2 2 + et t2 2 e t + + e t. It can be seen that as t, the solution y(t) approaches, which is consistent with what can be observed from the direction field. We will see that solutions of ODE often contain functions form e at, where a is a constant, or of the form sin bt or cos bt, where b is a constant. The presence of e at in a solution y(t), where a >, generally causes exponential growth, meaning that y(t) ± more rapidly than any power of t. On the other and, exponential decay occurs when a < ; that is, e at more rapidly than any power of /t for negative a. Factors of the form sin bt or cos bt cause oscillation. The higher the value of b, the more rapid the oscillation. It is important to note that the behavior of the solution of an ODE may depend on its initial conditions. In the preceding example, with y() 2, exponential decay occurred, which caused y(t)
8 CHAPTER 2. FIRST-ORDER EQUATIONS to approach very rapidly. However, if the initial condition had been y(), then the solution would have been y(t), and there would have been no exponential decay. This is an example of an equilibrium solution that was discussed previously. 2.2 Separable Equations We now revisit the solution of separable first-order ODE of the form dy dt g(t)h(y) for some functions g and h that depend only on t and y, respectively. First, we scale (multiply both sides of) the equation by /h(y), and then rewrite the equation in the form M(t) + N(y) dy dt, where M(t) g(t) and N(y) /h(y). Then, we compare this equation to the one we previously obtained by differentiation the equation F (t, y) with respect to t, F t + F y dy dt, and infer that the equation F (t, y) implicitly describes the solution of this ODE provided that F satisfies the conditions F t M(t), F y N(y). We can readily obtain F by integrating the first equation with respect to t, which yields F (t, y) M(t) dt + c(y), where c is an unknown function of y, and then differentiating this equation with respect to y to obtain F y c (y). Because we also know that F y N(y), we must have c (y) N(y), and we conclude that the solution y(t) satisfies the equation F (t, y) M(t) dt + N(y) dy. Once these integrals are evaluated, it may then be possible to solve the resulting equation for y. It is essential that a constant of integration be included in the result. Example Consider the separable equation t 2 + y 2 dy dt. In this problem, M(t) t 2 and N(t) /y 2. It follows that the solution satisfies t 2 dt + dy. y2
2.2. SEPARABLE EQUATIONS 9 Evaluating these integrals yields the equation t 3 3 y + C, where C is an arbitrary constant. Solving this equation for y yields y t 3 3 + C. Now, we add an initial condition y(t ) y, and take it into account in the solution process. Starting from the equation M(t) + N(y) dy dt, we integrate both sides of this equation with respect to t, from t to T (where T is a variable denoting final time ), and obtain T t M(t) dt + T t N(y) dy dt. dt The right-hand side must be equal to, rather than a constant, because the left side is equal to zero when T t. Then, in the second integral, we use the u-substitution to obtain T u y, t M(t) dt + du dy dt dt y(t ) y N(u) du. The modified limits in the second integral are obtained by substituting the original limits into the u-substitution, and enforcing the initial condition. Changing variable names, we obtain the following equation for the solution y(t), t t M(s) ds + y(t) Example We solve the separable initial value problem te t + y dy dt y N(u) du., y(). In this problem, M(t) te t and N(y) y. The solution y(t) is given by the equation Evaluating the integrals yields t se s ds + y(t) (s )e s t + u2 2 u du. y(t).
2 CHAPTER 2. FIRST-ORDER EQUATIONS Substituting the limits, we obtain (t )e t ( )e + y(t)2 2 2. Simplifying and rearranging yields y(t) 2 2( t)e t. In view of the initial condition y(), we choose the positive square root and conclude y(t) 2( t)e t. Example Consider the separable initial value problem dy dt Rearranging, we can write the ODE in the form 2t y + t 2, y() 2. y 2t + t 2 y dy dt, which yields M(t) 2t/( + t 2 ), N(y) y. It follows from this and the initial condition that the solution y(t) is given by the equation t 2s + s 2 ds y(t) 2 u du. Evaluating the integrals yields Substituting the limits, we obtain ln + s 2 t u2 2 y(t) 2. ln( + t 2 ) ln y(t)2 2 + 2. Simplifying and rearranging yields y(t) 2 2 ln( + t 2 ) + 4. In view of the initial condition y() 2, we choose the negative square root and conclude y(t) 2 ln( + t 2 ) + 4.
2.2. SEPARABLE EQUATIONS 2 2.2. Homogeneous Equations Now we will see that our approach to solving separable equations can be applied to certain problems that, in their original form, are not necessarily separable. Suppose a first-order ODE can be written in the form dy dx f(y/x). Generally, such an equation is not separable with respect to the variables x and y. An equation of this form is called homogeneous. To solve this kind of equation, we use a change of dependent variable, v y/x. Then, we have dy dx f(v). However, before we can proceed, we must eliminate y from the equation entirely. To that end, we differentiate the equation y vx with respect to x, using the Product Rule, and obtain Then, our transformed equation is separable, as we can rearrange it to obtain dy dx x dv dx + v. x dv dx + v f(v) dv dx f(v) v, x from which we can clearly see that dv/dx is the product of a function of x and a function of v. Rewriting this equation in the usual form for separable equations, we can apply our solution formula to obtain x dx + x + v f(v) dv dx, dv. v f(v) We can then solve this equation for v, and use the relation v y/x to obtain the solution y of the original equation. Example Consider the equation (x 2 + 3xy + y 2 ) dx x 2 dy, which can be rewritten in the more conventional form dy dx x2 + 3xy + y 2 x 2.
22 CHAPTER 2. FIRST-ORDER EQUATIONS By decomposing the numerator into 3 separate terms, we obtain dy dx x2 x 2 + 3xy x 2 + y2 x 2 + 3 y ( y ) 2 x + f(y/x), x where f(v) + 3v + v 2. We can see that this equation is homogeneous. Applying our solution formula, we obtain an equation for v y/x, x dx + v ( + 3v + v 2 dv. ) Evaluating the first integral and simplifying the second yields ln x v 2 dv. + 2v + As v 2 + 2v + (v + ) 2, we can use the Power Rule for the second integral and obtain where C is an arbitrary constant. Solving for v yields ln x + v + + C, v ln x + C. Finally, we use the relation v y/x to obtain the solution of the original equation, y x x ln x + C. It is worth noting that this approach is useful for any equation of the form where a, b, c and d are constants. dy ay + bx dx cy + dx, 2.3 Modeling with First-Order Equations We have learned how to solve certain types of first-order differential equations, but it is just as important to be able to derive such equations from a verbal description of phenomena to be modeled. This process requires the identification of all relevant quantities and the relationships between them, which typically include physical laws or other similar principles. By formulating these principles mathematically and relating them to identified quantities (known or unknown), a differential equation for a particular unknown quantity of interest can be obtained. The solution of the differential equation then allows understanding of how this quantity behaves, usually over time. Different laws or principles generally lead to different types of differential equations; we will limit ourselves to those that yield first-order ODE.
2.3. MODELING WITH FIRST-ORDER EQUATIONS 23 Example Consider a tank used in hydrodynamic experiments. After one experiment the tank contains 2 L of a dye solution with a concentration of g/l. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 L/min, with the well-stirred solution flowing out at the same rate. Find the time that will elapse before the concentration of the dye in the tank reaches % of its initial value. First, we let Q(t) denote the amount of dye in the tank at time t, measured in g. From the problem statement, we will derive a first-order differential equation for Q(t) and solve it. Then, we will use our solution to determine how long it will take for the concentration of dye to be reduced to the required level. Our guiding principle in deriving the differential equation is dq dt flow rate in flow rate out. Because only fresh water is flowing into the tank, the inflow rate of the dye is zero. The rate of dye leaving the tank is obtained using the fact that solution (water and dye) is leaving the tank at the same rate at which fresh water is entering, and the fact that the total amount of solution is 2 L. It follows that the concentration of the dye at time t is Q(t) g/2 L Q(t)/2 g/l. Because water is flowing in at the rate of 2 L/min, the outflow rate of the dye at time t is 2 L/min (Q(t)/2) g/l Q(t)/ g/min. We conclude that our differential equation is dq dt.q, with initial condidtion Q() 2 g, due to the given initial concentration of g/l and the initial amount of 2 L of solution. Solving this equation yields the solution Q(t) 2e.t. The concentration reaches % of its initial value, which is (.)2 2 g, when 2e.t 2, or e.t.. It follows that t. ln. ln ln 46.5 min. We continue our discussion of modeling with an example involving financial applications. The value of an investment as a function of time can be modeled using a linear first-order ODE with coefficients obtained from the assumed annual rate of return r, and an assumed annual rate of investment k which is positive for a deposit, and negative for a withdrawal. Example Suppose that k dollars per year is invested, with an annual rate of return r. Let S(t) denote the value of the investment at time t, where t is measured in years. Our goal is for S(t) to reach $ million in 4 years for a given amount of annual investment k, or a given annual return rate r. For simplicity, we assume that investments are made continuously and that the return on the investment is compounded continuously. It follows that S(t) satisfies the differential equation ds dt rs + k, with initial condition S() S. For this problem, we assume no initial capital, so S.
24 CHAPTER 2. FIRST-ORDER EQUATIONS This ODE is a linear equation of the form dy p(t)y + q(t), dt where y S, p(t) r, and q(t) k. It follows that the integrating factor is µ(t) e t p(s) ds e t r ds e rs t e rt. Using the solution formula for general linear equations, y(t) [ t y + µ(s)q(s) ds, µ(t) t we obtain [ t S(t) e rt S + e rs k ds t e [S rt + k e rs ds S e rt + ke rt e rs r S e rt k r ert (e rt ) S e rt + k r (ert ). With our assumption of no initial capital, which means S, our solution reduces to S(t) k r (ert ). Now, we assume that r 7.5%.75. We wish to determine how much should be invested each year so that after 4 years, the value of the investment will be $ million. This entails solving the equation S(4) k.75 (e.75(4) ) $,, for k. Rearranging this equation yields.75.75 k $,, e.75(4) $,, $3, 93. 9.855 Now, we assume that k $2, /year, and we wish to determine what rate of return r will yield $,, after 4 years. This entails solving the equation S(4) 2 r (e4r ) $,, for r. This equation cannot be solved analytically. However, using numerical methods such as Newton s Method or the Bisection Method, an approximate solution of r 9.77% can be obtained. This same model can be used to model the balance of a loan, where r is the interest rate and k is the annual rate of payment, which is negative in this context. The amount of the loan serves as the initial value. Once the loan is paid off, then the difference between the initial value and the total amount of the payments is equal to the total interest paid. t
2.4. AUTONOMOUS EQUATIONS 25 2.4 Autonomous Equations A first-order ODE is said to be autonomous if it is of the form dy dt f(y). That is, the derivative of the dependent variable y can be expressed as a function of only y, and not the independent variable t. An autonomous equation can easily be seen to be separable, and its solution satsifies the equation y(t) y f(u) du t t, where y(t ) y is the initial condition. However, even if the integral of /f is not easily computed, it is still possible to obtain a detailed understanding of the qualitative behavior of the solution for different initial values. We recall that an equilibrium solution is a value of y such that dy/dt for all t in the domain of the solution. Therefore, any value of y such that f(y) is an equilibrium solution. Once all equilibrium solutions are found, we can then determine how the solution behaves for nonequilibrium values of y. For example, suppose that y is the only equilibrium solution. If f(y) < for y > y, and f(y) > for y < y, then it follows that for any initial value y, the solution y(t) will converge to y as t, because the sign of dy/dt f(y) will cause y(t) to approach the line y y. We say that y is a stable equilibrium solution, as the solution y(t) does not deviate significantly from the equilibrium state when the initial value y deviates in either direction from y. Now, suppose that the opposite situation occurs: f(y) > for y > y, and f(y) < for y < y. Then, we say that y is an unstable equilibrium solution, because for any initial value y other than y, the solution y(t) will not converge to y as t ; rather, it will exhibit exponential deviation from y. If there is another equilbrium solution, then y(t) may approach that value instead. The reason why y is called an unstable solution is that if y deviates from y at all, then no matter how small the deviation, the solution y(t) will exhibit further deviation from y as t increases. Finally, it is possible that an equilibrium solution y may be neither stable nor unstable, but rather semi-stable. This is the case if solutions approach y when the initial value y deviates from y in one direction, but not in the other. That is, y appears to be stable for y > y and unstable for y < y, or vice versa. This occurs when f(y ) and f (y ) but f (y ), which causes f to touch, but not cross, the y-axis at y. Therefore, dy/dt has the same sign for y on either side of y, leading to stability on one side of the line y y but instability on the other. To help determine whether equilbrium solutions are stable, unstable, or semi-stable, it is helpful to draw a phase line. A phase line is a representation of the real number line, drawn vertically and corresponding to the y-axis with the positive y-axis pointing up. First, all equilibrium solutions are marked on the number line. Then, in the intervals that have equilibrium solutions as endpoints (including semi-infinite intervals), arrows are used to indicate the sign of f(y): arrows point downward where f(y) < and upward where f(y) >. Then, the arrows can be used to determine stability of the equilibrium solutions. If the arrows above and below an equilibrium solution y both point toward y, then y is stable. If both arrows point away from y, then it is unstable. If one arrow points toward y and the other points away from y, then y is semi-stable.
26 CHAPTER 2. FIRST-ORDER EQUATIONS Example We consider an initial value problem involving exponential growth, dy dt ry, y() y, where r is a positive constant. The ODE in this problem can be used as a simple model of population growth. The initial value problem has the solution y(t) y e rt. With f(y) ry, the only equilibrium solution is y. When y >, f(y) >, and when y <, f(y) <, so in either case, the solution y(t) is diverging from the equilibrium solution. Therefore, y is an unstable equilibrium point. The phase line for this problem features an upward arrow on the positive y-axis and a downward arrow on the negative y-axis, both diverging from the point y that is marked on the phase line as an equilibrium solution. Example We consider the logistic equation dy ( dt r y ) y, y() y, K where r and K are positive constants. This is a modification of the previous equation that attempts to model population growth more realistically. The constant K is referred to as a saturation level, or environmental carrying capacity, of a species. The growth rate, rather than being a constant r, is now r( y/k), so that when y reaches K, the growth rate becomes, and if y > K, the growth rate becomes negative. We will solve this equation, but first we will determine the behavior of its equilibrium solutions, which are y and y K. When y <, dy/dt <, so solutions with y < diverge from y. When < y < K, dy/dt >, so solutions with < y < K diverge from y but converge toward y K. Finally, when y > K, dy/dt <, so solutions with y > K converge toward y K. We conclude that y is an unstable equilibrium point, and that y K is a stable equilibrium point. Now, we solve the equation. First, we rewrite it as r + K dy y(y K) dt, and then integrate with respect to t from to T to obtain T y(t ) K r dt + du, y u(u K) where we have used the substitution u y(t) in the second integral. Using partial fraction decomposition, we obtain y(t ) rt + y u K du, u which yields rt + ln y(t ) K ln y(t ) ln y K + ln y.
2.4. AUTONOMOUS EQUATIONS 27 Using the properties of logarithmic functions, and then exponentiating, we obtain y (y(t ) K) y(t )(y K) e rt. Rearranging to solve for y(t ) yields y (y(t ) K) e rt y(t )(y K), Assume that < y < K. Then < y(t ) < K as well, which allows us to drop the absolute value signs. We then have y y(t ) y K e rt y(t )(y K), which can be rearranged to obtain y(t ) y K y + e rt (K y ). It can be seen from this explicit solution that as T, y(t ) K, which is consistent with our previous analysis. If we assume that y <, or y > K, then we actually obtain the same formula for y(t ). In the case of y > K, the solution will approach K as T increases, as expected since y K is a stable equilibrium point. However, if y <, then there is a finite time T at which the denominator will equal zero, resulting in a vertical asymptote at which y. In Figure 2.2, the solution is plotted for several values of y, in the case of r and K, and the phase line is shown as well. Figure 2.2: Solutions (left plot) and phase line (right plot) for y (y )y
28 CHAPTER 2. FIRST-ORDER EQUATIONS 2.5 Exact Equations We recall that a first-order ODE of the form M(t, y) + N(t, y) dy dt has a solution described by an equation of the form F (t, y) if F t M, F y N, as can be seen by comparing the ODE to the equation F t + F y dy dt obtained by differentiating the equation F (t, y) with respect to t. As discussed previously, if we differentiate F t M with respect to y and F y N with respect to t, we obtain F ty M y, F yt N t. From the equality of mixed partial derivatives when they are continuous, we obtain the equation M y N t. When this is the case, we say that the ODE is exact. To solve an exact equation, we integrate both sides of the equation F t M with respect to t to obtain F (t, y) M(t, y) dt + g(y), where g is an unknown function of y. Differentiating this equation with respect to y yields F y N(t, y) [ M(t, y) dt + g (y). y This yields an equation for g (y), which can then be integrated to obtain a formula for F (t, y). We conclude that the solution y(t) of an exact equation can be described by the equation { F (t, y) M (t, y) + N(t, y) } y M (t, y) dy, where M (t, y) M(t, y) dt. where M (t, y) does not include a constant of integration. When an initial condition y(t ) y is specified, this solution formula can be modified accordingly to obtain y { M (t, y) + N(t, u) } y u M (t, u) du,
2.5. EXACT EQUATIONS 29 where M (t, y) t t M(s, y) ds. Note that the variables of integration are changed, because the variables t and y are limits of integration. It should be noted that an equation of the form M(t, y) + N(t, y) dy dt may be written in the form M(t, y) dt + N(t, y) dy, but the same solution technique applies. Also, the dependent variable t may be x instead. Example Consider the equation (y/t + 6t) + (ln t 2) dy dt. In this equation, M(t, y) y/t + 6t and N(t, y) ln t 2. We compute M y y (y/t + 6t) t, N t t (ln t 2) t. Since M y N t, we know that the equation is exact. Applying the solution formula for an exact equation, we first compute M (t, y) M(t, y) dt y/t + 6t dt y ln t + 3t 2. Note that we do not include a constant of integration. Then, we compute N(t, y) y M (t, y) (ln t 2) (y ln t + 3t 2 ) y ln 2 2 ln 2 2. It is worth noting that this expression is not a function of t; this is true in general when solving exact equations. Finally, our solution satisfies the equation [ M (t, y) + N(t, y) y M (t, y) dy, or y ln t + 3t 2 + 2 dy. Integrating yields the equation y ln t + 3t 2 2y + C, which implicitly defines the solution y as a function of t.
3 CHAPTER 2. FIRST-ORDER EQUATIONS Example Consider the initial value problem (2x y) dx + (2y x) dy, y() 3. In this problem, M(x, y) 2x y and N(x, y) 2y x. From we find that M y N x, so the equation is exact. We then compute M (x, y) Next, we compute x M y (2x y) y, N x (2y x) x, M(s, y) ds x 2s y ds (s 2 ys) x x2 xy + y. N(x, u) u M (x, u) (2u x) (x 2 xu + u) u (2u x) ( x + ) 2u. It is worth noting that this expression is not a function of x. Finally, our solution satisfies the equation y [ M (x, y) + N(x, u) u M (x, u) du, or y x 2 xy + y + 2u du. 3 Integrating yields the equation which simplifies to 3 x 2 xy + y + (u 2 u) y 3 x2 xy + y + y 2 y 9 + 3, y 2 xy + x 2 7. This is a quadratic equation in y. Applying the quadratic formula yields y x ± x 2 4(x 2 7) 2 x ± 28 3x 2. 2 Only the solution with the plus sign satisfies the initial condition y() 3. We conclude that the solution is y x + 28 3x 2. 2 It is worth noting that a separable equation of the form M(t) + N(y) dy dt is also exact, as M y N t. It follows that separable equations are merely special cases of exact equations, and therefore the solution formula for exact equations can be applied to separable equations as well.
2.5. EXACT EQUATIONS 3 2.5. Integrating Factors If a first-order equation of the form M dx + N dy is not exact, then it can often be made exact by scaling the equation by a function µ(x, y), known as an integrating factor. We have previously used this approach for linear equations that are not separable. Now, we generalize this technique to nonlinear equations. We need to find µ so that the scaled equation µm dx + µn dy is exact. This requires that (µm) y (µn) x, or, by the Product Rule, µm y + µ y M µn x + µ x N, or, upon rearranging, µ(m y N x ) µ x N µ y M. In general, this equation is difficult to solve for µ. However, we can solve it in two special cases. First, suppose that we require that µ is a function of only x, so that µ y. Then, µ satisfies the equation µ (x) M y N x µ(x), N where (M y N x )/N must be a function of only x. It follows that µ(x) e g(x) dx, g(x) M y(x, y) N x (x, y), N(x, y) where g(x), in its simplest form, does not depend on y. Similarly, suppose that µ must be a function of only y, so that µ x. Then, µ satisfies the equation µ (y) N x M y M µ(y), where (N x M y )/M must be a function of only y. It follows that µ(y) e h(y) dy, h(y) N x(x, y) M y (x, y), M(x, y) where h(y), in its simplest form, does not depend on x. Example We will solve the equation y dx + (2xy e 2y ) dy. In this case, M(x, y) y and N(x, y) 2xy e 2y. Because M y, N x 2y,
32 CHAPTER 2. FIRST-ORDER EQUATIONS the equation is not exact. To find an integrating factor, we examine M y N x N 2y 2xy e 2y, N x M y M 2y. y Because the second function depends only on y, we can use the integrating factor [ 2y µ(y) exp dy y [ exp 2 y dy Now, the scaled equation e 2y ln y e 2y ln y e e2y y. e 2y e2y y dx + y y (2xy e 2y ) dy should be exact. First, we simplify the equation to obtain e 2y dx + (2xe 2y /y) dy. In this modified equation, M(x, y) e 2y and N(x, y) 2xe 2y /y. From M y 2e 2y, N x 2e 2y, we confirm that it is indeed exact. To solve the equation, we first compute M (x, y) M(x, y) dx e 2y dx xe 2y. Note that we do not add a constant of integration. Then, we compute N(x, y) y M (x, y) 2xe 2y y 2xexy y. Note that this function does not depend on x. The solution y therefore satisfies the equation [ M (x, y) + N(x, y) y M (x, y) dy xe 2y y dy xe2y ln y + C, where C is an arbitrary constant. It should be noted that by rewriting the original equation in the form dy/dx f(x, y), it can be seen that y is also a solution.
2.6. THE EXISTENCE-UNIQUENESS THEOREM 33 2.6 The Existence-Uniqueness Theorem So far, we have learned how to solve certain specific types of first-order ODE. We now prove that under certain assumptions, a general first-order ODE has a unique solution. Theorem (Existence-Uniqueness) Let the function f and f/ y be continuous in some rectangle α t β, γ y δ containing the point (t, y ). Then, in some interval t h t t + h contained in (α, β), the initial value problem has a unique solution y(t). y f(t, y), y(t ) y Proof We first note that if we integrate both sides of the equation y (s) f(s, y(s)) with respect to s, from t to t, then we obtain t y(t) y + f(s, y(s)) ds. t Then, we define a sequence of functions, known as the Picard iterates, as follows: y (t) y, y n+ (t) y + t t f(s, y n (s)) ds, n,, 2,.... Our goal is to prove that for some interval t t h contained within [α, β, the Picard iterates converge to a function y(t) that is a solution of the initial value problem. To that end, we first note that from the continuity of f on R, we have t y (t) y f(s, y ) ds M t t, t where f(t, y ) M on [t h, t + h. Then, we let n and examine the difference t t y n+ (t) y n (t) f(s, y n (s)) f(s, y n (s)) ds f(s, y n (s)) f(s, y n (s)) ds. t t Because of the continuity of f/ y in the rectangle R [α, β [γ, δ, we have, for a, b [γ, δ and s t h, by the Mean Value Theorem, f(s, b) f(s, a) f (s, ψ)(b a) y K b a, where ψ lies between a and b and f/ y K on R. It follows that t y n+ (t) y n (t) K y n (s) y n (s) ds. t Setting n yields t t y 2 (t) y (t) K y (s) y (s) ds KM s t ds KM t t 2. t t 2
34 CHAPTER 2. FIRST-ORDER EQUATIONS Continuing by induction, we obtain y n+ (t) y n (t) MKn t t n+ (n + )! MKn h n+. (n + )! It follows that as n, y n+ (t) y n (t), thus establishing convergence of the Picard iterates to a solution y(t) lim n y n(t). Now that we have proved that a solution exists, we show that it is also unique. We assume that there are two distinct solutions, y(t) and ỹ(t). Then we have t t y(t) ỹ(t) f(s, y(s)) f(s, ỹ(s)) ds K y(s) ỹ(s) ds. t t Let z(t) y(t) ỹ(t). Then, z(t), z(t ), and for t t, we have, by the Fundamental Theorem of Calculus, z (t) K y(t) ỹ(t) Kz(t). Rearranging and using the integrating factor µ(t) e Kt, we obtain e Kt z (t) Ke Kt z(t) [e Kt z(t). Integration from t to t yields e Kt z(t), and therefore z(t). However, we also have z(t). We conclude that z(t), which implies that y(t) ỹ(t), contradicting our assumption that these solutions are distinct. We conclude that the solution is unique. We now make two noteworthy observations about the preceding proof. The proof relies on the fundamental result that a continuous function on a compact set, such as a closed interval [a, b or a closed rectangle [a, b [c, d, has a maximum and minimum. The use of Picard iterates is an example of fixed-point iteration, in which an equation of the form g(x) x is solved by choosing an initial guess x and then computing x n+ g(x n ) for n,, 2,.... This iteration provides both a practical and theoretical foundation for various methods of solving nonlinear equations or systems of equations.
Chapter 3 Second-Order Equations We have learned various techniques for solving first-order ODE, that only involve the first derivative of the solution. We now consider second-order ODE. The most general second-order ODE is an equation of the form F (t, y, y, y ) for some function F of 4 variables. Such an equation can be either linear or nonlinear. We will focus on linear second-order equations of the form y + p(t)y + q(t)y g(t), with coefficients p(t) and q(t) and a right-hand side g(t). If g(t), we say that the equation is homogeneous (not to be confused with homogeneous first-order equations considered previously); otherwise, it is said to be inhomogeneous. If p(t) and q(t) are constants, we say that the equation is constant-coefficient; otherwise, it is variable-coefficient. 3. Homogeneous Equations with Constant Coefficients We now consider how to solve a linear homogeneous second-order ODE with constant coefficients, y + py + qy. Such an equation is normally part of an initial value problem that has two initial conditions, y() y, y () z, due to the second derivative in the ODE. In order to solve this problem, we can leverage our knowledge of first-order linear equations. First, we define the variables u y, u 2 y. Then our second-order ODE can be rewritten as a system of first-order ODE, u u 2, u 2 + pu 2 + qu, 35
36 CHAPTER 3. SECOND-ORDER EQUATIONS with accompanying initial conditions u () y, u 2 () z. This initial value problem can be expressed in matrix-vector form as follows: where u [ u u 2 u Au, u (t ) u, [, A q p [ y, u(t ) z Recall that the solution of the first-order initial value problem y ay, y() y is y(t) e at y. Similarly, the solution of the preceding first-order system is where the matrix exponential e At is defined by u(t) e At u e At (At) j, j! j based on the Maclaurin series for the exponential function e x j However, this form of the solution is not convenient for describing the solution y(t) of our original second-order problem, so we must use an alternative approach. To that end, we consider the question of when the system u Au, for a general 2 2 matrix A, is easy to solve. This is certainly the case when A is a diagonal matrix of the form [ λ A. λ 2 In this case, the equations in the system, x j j!. u λ u, u 2 λ 2 u 2, are decoupled, and therefore can be solved independently. This yields the solutions u (t) e λ t y, u 2 (t) e λ 2t z. Therefore, we consider whether our system that is obtained from a second-order ODE can somehow be transformed into a system in which the equations are decoupled. We define new variables w and w 2 through a linear transformation of u and u 2 : u p w + p 2 w 2, u 2 p 2 w + p 22 w 2,.