16 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS shows that they are continuous in the upper half-plane defined by y>0.Hence Theorem 1.2.1 enables us to conclude that through any point (xo,yo),yo>0 in the upper half-plane there is some interval centered at xo on which the given differential equation has a unique solution.Thus,for example,even without solving it,we know that there exists some interval centered at 2 on which the initial-value problem dy/dx=xyi2,y(2)=I has a unique solution. ■ In Example 1,Theorem 1.2.1 guarantees that there are no other solutions of the initial-value problems y'=y,y(0)=3 and y'=y,y(1)=-2 other than y=3e and y=-2e-1,respectively.This follows from the fact that f(x,y)=y and af/ay=1 are continuous throughout the entire xy-plane.It can be further shown that the interval on which each solution is defined is(,) INTERVAL OF EXISTENCE/UNIQUENESS Suppose y(x)represents a solution of the initial-value problem(2).The following three sets on the real x-axis may not be the same:the domain of the function y(x),the interval over which the solution y(x)is defined or exists,and the interval lo of existence and uniqueness.Example 2 of Section 1.1 illustrated the difference between the domain of a function and the interval of definition.Now suppose (xo.yo)is a point in the interior of the rectan- gular region R in Theorem 1.2.1.It turns out that the continuity of the function f(x,y)on R by itself is sufficient to guarantee the existence of at least one solution of dy/dx =f(x,y),y(xo)=yo,defined on some interval I.The interval I of defini- tion for this initial-value problem is usually taken to be the largest interval contain- ing xo over which the solution y(x)is defined and differentiable.The interval I depends on both f(x,y)and the initial condition y(xo)=yo.See Problems 31-34 in Exercises 1.2.The extra condition of continuity of the first partial derivative af/ay on R enables us to say that not only does a solution exist on some interval lo con- taining xo,but it is the only solution satisfying y(xo)=yo.However,Theorem 1.2.1 does not give any indication of the sizes of intervals I and Io;the interval I of definition need not be as wide as the region R,and the interval lo of existence and uniqueness may not be as large as I.The number h>0 that defines the interval Io:(xo-h,xo+h)could be very small,so it is best to think that the solution y(x) is unique in a local sense-that is,a solution defined near the point (xo,yo).See Problem 44 in Exercises 1.2. REMARKS ()The conditions in Theorem 1.2.1 are sufficient but not necessary.This means that when f(x,y)and af/ay are continuous on a rectangular region R,it must always follow that a solution of(2)exists and is unique whenever (xo yo)is a point interior to R.However,if the conditions stated in the hypothesis of Theorem 1.2.1 do not hold,then anything could happen:Problem (2)may still have a solution and this solution may be unique,or(2)may have several solu- tions,or it may have no solution at all.A rereading of Example 5 reveals that the hypotheses of Theorem 1.2.1 do not hold on the line y=0 for the differential equation dy/dx =xy2,so it is not surprising,as we saw in Example 4 of this section,that there are two solutions defined on a common interval-h<x<h satisfying y(0)=0.On the other hand,the hypotheses of Theorem 1.2.1 do not hold on the line y=1 for the differential equation dy/dx=y-1. Nevertheless it can be proved that the solution of the initial-value problem dy/dx=ly-1,y(0)=1,is unique.Can you guess this solution? (ii)You are encouraged to read,think about,work,and then keep in mind Problem 43 in Exercises 1.2
shows that they are continuous in the upper half-plane defined by y 0. Hence Theorem 1.2.1 enables us to conclude that through any point (x0, y0), y0 0 in the upper half-plane there is some interval centered at x0 on which the given differential equation has a unique solution. Thus, for example, even without solving it, we know that there exists some interval centered at 2 on which the initial-value problem dydx xy1/2, y(2) 1 has a unique solution. In Example 1, Theorem 1.2.1 guarantees that there are no other solutions of the initial-value problems y y, y(0) 3 and y y, y(1) 2 other than y 3ex and y 2ex1 , respectively. This follows from the fact that f(x, y) y and f y 1 are continuous throughout the entire xy-plane. It can be further shown that the interval I on which each solution is defined is ( , ). INTERVAL OF EXISTENCE/UNIQUENESS Suppose y(x) represents a solution of the initial-value problem (2). The following three sets on the real x-axis may not be the same: the domain of the function y(x), the interval I over which the solution y(x) is defined or exists, and the interval I0 of existence and uniqueness. Example 2 of Section 1.1 illustrated the difference between the domain of a function and the interval I of definition. Now suppose (x0, y0) is a point in the interior of the rectangular region R in Theorem 1.2.1. It turns out that the continuity of the function f(x, y) on R by itself is sufficient to guarantee the existence of at least one solution of dydx f(x, y), y(x0) y0, defined on some interval I. The interval I of definition for this initial-value problem is usually taken to be the largest interval containing x0 over which the solution y(x) is defined and differentiable. The interval I depends on both f(x, y) and the initial condition y(x0) y0. See Problems 31–34 in Exercises 1.2. The extra condition of continuity of the first partial derivative f y on R enables us to say that not only does a solution exist on some interval I0 containing x0, but it is the only solution satisfying y(x0) y0. However, Theorem 1.2.1 does not give any indication of the sizes of intervals I and I0; the interval I of definition need not be as wide as the region R, and the interval I0 of existence and uniqueness may not be as large as I. The number h 0 that defines the interval I0: (x0 h, x0 h) could be very small, so it is best to think that the solution y(x) is unique in a local sense—that is, a solution defined near the point (x0, y0). See Problem 44 in Exercises 1.2. REMARKS (i) The conditions in Theorem 1.2.1 are sufficient but not necessary. This means that when f(x, y) and f y are continuous on a rectangular region R, it must always follow that a solution of (2) exists and is unique whenever (x0, y0) is a point interior to R. However, if the conditions stated in the hypothesis of Theorem 1.2.1 do not hold, then anything could happen: Problem (2) may still have a solution and this solution may be unique, or (2) may have several solutions, or it may have no solution at all. A rereading of Example 5 reveals that the hypotheses of Theorem 1.2.1 do not hold on the line y 0 for the differential equation dydx xy1/2, so it is not surprising, as we saw in Example 4 of this section, that there are two solutions defined on a common interval h x h satisfying y(0) 0. On the other hand, the hypotheses of Theorem 1.2.1 do not hold on the line y 1 for the differential equation dydx y 1. Nevertheless it can be proved that the solution of the initial-value problem dydx y 1, y(0) 1, is unique. Can you guess this solution? (ii) You are encouraged to read, think about, work, and then keep in mind Problem 43 in Exercises 1.2. 16 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS
1.2 INITIAL-VALUE PROBLEMS.17 EXERCISES 1.2 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1 and 2,y=1/(1 cie-x)is a one-parameter dy二y 19. 20. dx -y=x family of solutions of the first-order DEy'=y-y2.Find a solution of the first-order IVP consisting of this differential 2L.(4-y2y'=x2 22.(1+y3)y'=x2 equation and the given initial condition. 23.(x2+y2y'=y2 24.(y-x)y'=y+x 1.y(0)=- 2.y(-1)=2 In Problems 25-28 determine whether Theorem 1.2.1 guar- antees that the differential equation y'=vy2-9 pos- In Problems 3-6,y=1/(x2+c)is a one-parameter family sesses a unique solution through the given point. of solutions of the first-order DE y'+2xy2=0.Find a solution of the first-order IVP consisting of this differential 25.(1.4) 26.(5,3) equation and the given initial condition.Give the largest interval over which the solution is defined. 27.(2,-3) 28.(-1,1) 29.(a)By inspection find a one-parameter family of solu- 3.y(2)= 4.y(-2)= tions of the differential equation xy'=y.Verify that 5.y(0)=1 6.y周=-4 each member of the family is a solution of the initial-value problem xy'=y,y(0)=0. In Problems 7-10,x cI cos t c2 sin t is a two-parameter (b)Explain part(a)by determining a region R in the family of solutions of the second-order DEx"+x=0.Find xy-plane for which the differential equation xy'=y a solution of the second-order IVP consisting of this differ- would have a unique solution through a point (xo yo) ential equation and the given initial conditions. in R. (c)Verify that the piecewise-defined function 7.x(0)=-1,x'(0)=8 8.x(T/2)=0,x'(π/2)=1 0,x<0 y= x,x≥0 9.x(π/6)=x'(π/6=0 10.x(π/4)=V2,x'(π/4)=2V2 satisfies the condition y(0)=0.Determine whether this function is also a solution of the initial-value In Problems 11-14,y=cie*+c2e-x is a two-parameter problem in part (a). family of solutions of the second-order DE y"-y=0.Find a solution of the second-order IVP consisting of this differ- 30.(a)Verify that y tan (x+c)is a one-parameter family of solutions of the differential equation y'=1 +y2. ential equation and the given initial conditions. (b)Since f(x,y)=1 +y2 and af/ay 2y are continu- 11.y(0)=1,y'(0)=2 ous everywhere,the region R in Theorem 1.2.1 can be taken to be the entire xy-plane.Use the family of 12.y(1)=0,y'(1)=e solutions in part (a)to find an explicit solution of the first-order initial-value problem y'=1 +y2, 13.y(-1)=5,y'(-1)=-5 y(0)=0.Even though xo=0 is in the interval 14.y(0)=0,y'(0)=0 (-2,2),explain why the solution is not defined on this interval. In Problems 15 and 16 determine by inspection at least two (c)Determine the largest interval of definition for the solutions of the given first-order IVP. solution of the initial-value problem in part(b). 15.y'=3y2B,y(0)=0 31.(a)Verify that y=-1/(x+c)is a one-parameter family of solutions of the differential equation 16.xy'=2y,y(0)=0 y'=y2. (b)Since f(x,y)=y2 and af/ay=2y are continuous In Problems 17-24 determine a region of the xy-plane for everywhere,the region R in Theorem 1.2.1 can be which the given differential equation would have a unique taken to be the entire xy-plane.Find a solution from solution whose graph passes through a point (xo,yo)in the the family in part (a)that satisfies y(0)=1.Then region. find a solution from the family in part (a)that satisfies y(0)=-1.Determine the largest interval I 17. 18.=V of definition for the solution of each initial-value dx dx problem
1.2 INITIAL-VALUE PROBLEMS ● 17 EXERCISES 1.2 Answers to selected odd-numbered problems begin on page ANS-1. In Problems 1 and 2, y 1(1 c1ex ) is a one-parameter family of solutions of the first-order DE y y y2 . Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. 1. 2. y(1) 2 In Problems 3–6, y 1(x2 c) is a one-parameter family of solutions of the first-order DE y 2xy2 0. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution is defined. 3. 4. 5. y(0) 1 6. In Problems 7–10, x c1 cos t c2 sin t is a two-parameter family of solutions of the second-order DE x x 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 7. x(0) 1, x(0) 8 8. x(2) 0, x(2) 1 9. 10. In Problems 11–14, y c1ex c2ex is a two-parameter family of solutions of the second-order DE y y 0. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions. 11. 12. y(1) 0, y(1) e 13. y(1) 5, y(1) 5 14. y(0) 0, y(0) 0 In Problems 15 and 16 determine by inspection at least two solutions of the given first-order IVP. 15. y 3y2/3, y(0) 0 16. xy 2y, y(0) 0 In Problems 17–24 determine a region of the xy-plane for which the given differential equation would have a unique solution whose graph passes through a point (x0, y0) in the region. 17. 18. dy dx 1xy dy dx y2/3 y(0) 1, y(0) 2 x(> 4) 12, x(> 4) 212 x(> 6) 1 2, x(> 6) 0 y(1 2) 4 y(2) 1 2 y(2) 1 3 y(0) 1 3 19. 20. 21. (4 y2)y x2 22. (1 y3 )y x2 23. (x2 y2)y y2 24. (y x)y y x In Problems 25–28 determine whether Theorem 1.2.1 guarantees that the differential equation possesses a unique solution through the given point. 25. (1, 4) 26. (5, 3) 27. (2, 3) 28. (1, 1) 29. (a) By inspection find a one-parameter family of solutions of the differential equation xy y. Verify that each member of the family is a solution of the initial-value problem xy y, y(0) 0. (b) Explain part (a) by determining a region R in the xy-plane for which the differential equation xy y would have a unique solution through a point (x0, y0) in R. (c) Verify that the piecewise-defined function satisfies the condition y(0) 0. Determine whether this function is also a solution of the initial-value problem in part (a). 30. (a) Verify that y tan (x c) is a one-parameter family of solutions of the differential equation y 1 y2 . (b) Since f(x, y) 1 y2 and f y 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Use the family of solutions in part (a) to find an explicit solution of the first-order initial-value problem y 1 y2 , y(0) 0. Even though x0 0 is in the interval (2, 2), explain why the solution is not defined on this interval. (c) Determine the largest interval I of definition for the solution of the initial-value problem in part (b). 31. (a) Verify that y 1(x c) is a one-parameter family of solutions of the differential equation y y2 . (b) Since f(x, y) y2 and f y 2y are continuous everywhere, the region R in Theorem 1.2.1 can be taken to be the entire xy-plane. Find a solution from the family in part (a) that satisfies y(0) 1. Then find a solution from the family in part (a) that satisfies y(0) 1. Determine the largest interval I of definition for the solution of each initial-value problem. y 0, x 0 x, x 0 y 1y2 9 dy dx x y x dy dx y
18.CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS (c)Determine the largest interval of definition for the 36. solution of the first-order initial-value problem y'=y2,y(0)=0.[Hint:The solution is not a mem- ber of the family of solutions in part (a). 32.(a)Show that a solution from the family in part (a) of Problem 31 that satisfies y'=y2,y(1)=1,is y=1/2-x). (b)Then show that a solution from the family in part(a) of Problem 31 that satisfies y'=y2,y(3)=-1,is -5 y=1/(2-x). (c)Are the solutions in parts(a)and(b)the same? FIGURE 1.2.8 Graph for Problem 36 33.(a)Verify that 3x2-y2 =c is a one-parameter fam- 37. ily of solutions of the differential equation y dy/dx 3x. (b)By hand,sketch the graph of the implicit solution 3x2-y2 =3.Find all explicit solutions y=(x)of the DE in part(a)defined by this relation.Give the interval of definition of each explicit solution. (c)The point(-2,3)is on the graph of 3x2-y2=3, but which of the explicit solutions in part (b)satis- fesy(-2)=3? 34.(a)Use the family of solutions in part(a)of Problem 33 to find an implicit solution of the initial-value FIGURE 1.2.9 Graph for Problem 37 problem y dy/dx 3x,y(2)=-4.Then,by hand, sketch the graph of the explicit solution of this 38. problem and give its interval of definition. (b)Are there any explicit solutions of y dy/dx =3x that pass through the origin? In Problems 35-38 the graph of a member of a family of solutions of a second-order differential equation d2y/dx2=f(x,y.y')is given.Match the solution curve with at least one pair of the following initial conditions. -5 (a)y(1)=1,y'(1)=-2 b)y(-1)=0,y'(-1)=-4 FIGURE 1.2.10 Graph for Problem 38 (c)y(1)=1,y'(1)=2 (d④y(0)=-1,y'(0)=2 Discussion Problems (e)y(0)=-1,y'(0)=0 In Problems 39 and 40 use Problem 51 in Exercises 1.1 and ()y(0)=-4,y'(0)=-2 (2)and (3)of this section. 35. 39.Find a functiony =f(x)whose graph at each point (x,y) has the slope given by 8e2x+6x and has the y-intercept (0,9). 40.Find a function y=f(x)whose second derivative is y"=12x-2 at each point (x,y)on its graph and y=-x+5 is tangent to the graph at the point corre- sponding tox =1. 41.Consider the initial-value problem y'=x-2y, -5十 y(0)=.Determine which of the two curves shown in Figure 1.2.11 is the only plausible solution curve. FIGURE 1.2.7 Graph for Problem 35 Explain your reasoning
(c) Determine the largest interval I of definition for the solution of the first-order initial-value problem y y2 , y(0) 0. [Hint: The solution is not a member of the family of solutions in part (a).] 32. (a) Show that a solution from the family in part (a) of Problem 31 that satisfies y y2 , y(1) 1, is y 1(2 x). (b) Then show that a solution from the family in part (a) of Problem 31 that satisfies y y2 , y(3) 1, is y 1(2 x). (c) Are the solutions in parts (a) and (b) the same? 33. (a) Verify that 3x2 y2 c is a one-parameter family of solutions of the differential equation y dydx 3x. (b) By hand, sketch the graph of the implicit solution 3x2 y2 3. Find all explicit solutions y (x) of the DE in part (a) defined by this relation. Give the interval I of definition of each explicit solution. (c) The point (2, 3) is on the graph of 3x2 y2 3, but which of the explicit solutions in part (b) satis- fies y(2) 3? 34. (a) Use the family of solutions in part (a) of Problem 33 to find an implicit solution of the initial-value problem y dydx 3x, y(2) 4. Then, by hand, sketch the graph of the explicit solution of this problem and give its interval I of definition. (b) Are there any explicit solutions of y dydx 3x that pass through the origin? In Problems 35–38 the graph of a member of a family of solutions of a second-order differential equation d2 ydx2 f(x, y, y) is given. Match the solution curve with at least one pair of the following initial conditions. (a) y(1) 1, y(1) 2 (b) y(1) 0, y(1) 4 (c) y(1) 1, y(1) 2 (d) y(0) 1, y(0) 2 (e) y(0) 1, y(0) 0 (f) y(0) 4, y(0) 2 35. 18 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Discussion Problems In Problems 39 and 40 use Problem 51 in Exercises 1.1 and (2) and (3) of this section. 39. Find a function y f(x) whose graph at each point (x, y) has the slope given by 8e2x 6x and has the y-intercept (0, 9). 40. Find a function y f(x) whose second derivative is y 12x 2 at each point (x, y) on its graph and y x 5 is tangent to the graph at the point corresponding to x 1. 41. Consider the initial-value problem y x 2y, . Determine which of the two curves shown in Figure 1.2.11 is the only plausible solution curve. Explain your reasoning. y(0) 1 2 FIGURE 1.2.7 Graph for Problem 35 y x 5 −5 5 FIGURE 1.2.10 Graph for Problem 38 y x 5 −5 5 36. 37. 38. FIGURE 1.2.8 Graph for Problem 36 FIGURE 1.2.9 Graph for Problem 37 y x 5 −5 5 y x 5 −5 5
1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS.19 (2.1) (a) (b) FIGURE 1.2.12 Two solutions of the IVP in Problem 44 FIGURE 1.2.11 Graphs for Problem 41 42.Determine a plausible value of xo for which the dy/dx xy12,y(2)=1 on the interval (-) graph of the solution of the initial-value problem Resolve the apparent contradiction between this fact y'+2y 3x -6,y(xo)=0 is tangent to the x-axis at and the last sentence in Example 5. (xo,0).Explain your reasoning. 43.Suppose that the first-order differential equation Mathematical Model dy/dx=f(x,y)possesses a one-parameter family of 45.Population Growth Beginning in the next section solutions and that f(x,y)satisfies the hypotheses of Theorem 1.2.1 in some rectangular region R of the we will see that differential equations can be used to describe or model many different physical systems.In xy-plane.Explain why two different solution curves this problem suppose that a model of the growing popu- cannot intersect or be tangent to each other at a point lation of a small community is given by the initial-value (xo,yo)in R. problem 44.The functions y(x)= -oo<<o and dp dt =0.15P0)+20,P(0)=100, 0, x<0 y(x)= 品4，x≥0 where P is the number of individuals in the community and time t is measured in years.How fast-that is,at have the same domain but are clearly different.See what rate-is the population increasing at t=0?How Figures 1.2.12(a)and 1.2.12(b),respectively.Show that fast is the population increasing when the population both functions are solutions of the initial-value problem is500? 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS REVIEW MATERIAL Units of measurement for weight,mass,and density Newton's second law of motion ·Hooke'slaw 。Kirchhoff's laws .Archimedes'principle INTRODUCTION In this section we introduce the notion of a differential equation as a mathematical model and discuss some specific models in biology,chemistry,and physics.Once we have studied some methods for solving DEs in Chapters 2 and 4,we return to,and solve,some of these models in Chapters 3 and 5. MATHEMATICAL MODELS It is often desirable to describe the behavior of some real-life system or phenomenon,whether physical,sociological,or even eco- nomic,in mathematical terms.The mathematical description of a system of phenom- enon is called a mathematical model and is constructed with certain goals in mind. For example,we may wish to understand the mechanisms of a certain ecosystem by studying the growth of animal populations in that system,or we may wish to date fossils by analyzing the decay of a radioactive substance either in the fossil or in the stratum in which it was discovered
42. Determine a plausible value of x0 for which the graph of the solution of the initial-value problem y 2y 3x 6, y(x0) 0 is tangent to the x-axis at (x0, 0). Explain your reasoning. 43. Suppose that the first-order differential equation dydx f(x, y) possesses a one-parameter family of solutions and that f(x, y) satisfies the hypotheses of Theorem 1.2.1 in some rectangular region R of the xy-plane. Explain why two different solution curves cannot intersect or be tangent to each other at a point (x0, y0) in R. 44. The functions and have the same domain but are clearly different. See Figures 1.2.12(a) and 1.2.12(b), respectively. Show that both functions are solutions of the initial-value problem y(x) 0, 1 16 x4 , x 0 x 0 y(x) 1 16 x4 , x 1.3 DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS ● 19 FIGURE 1.2.11 Graphs for Problem 41 (0, ) 1 2 1 1 x y dydx xy1/2, y(2) 1 on the interval ( , ). Resolve the apparent contradiction between this fact and the last sentence in Example 5. Mathematical Model 45. Population Growth Beginning in the next section we will see that differential equations can be used to describe or model many different physical systems. In this problem suppose that a model of the growing population of a small community is given by the initial-value problem where P is the number of individuals in the community and time t is measured in years. How fast—that is, at what rate—is the population increasing at t 0? How fast is the population increasing when the population is 500? dP dt 0.15P(t) 20, P(0) 100, FIGURE 1.2.12 Two solutions of the IVP in Problem 44 (a) (2, 1) y x (b) (2, 1) y x DIFFERENTIAL EQUATIONS AS MATHEMATICAL MODELS REVIEW MATERIAL ● Units of measurement for weight, mass, and density ● Newton’s second law of motion ● Hooke’s law ● Kirchhoff’s laws ● Archimedes’ principle INTRODUCTION In this section we introduce the notion of a differential equation as a mathematical model and discuss some specific models in biology, chemistry, and physics. Once we have studied some methods for solving DEs in Chapters 2 and 4, we return to, and solve, some of these models in Chapters 3 and 5. 1.3 MATHEMATICAL MODELS It is often desirable to describe the behavior of some real-life system or phenomenon, whether physical, sociological, or even economic, in mathematical terms. The mathematical description of a system of phenomenon is called a mathematical model and is constructed with certain goals in mind. For example, we may wish to understand the mechanisms of a certain ecosystem by studying the growth of animal populations in that system, or we may wish to date fossils by analyzing the decay of a radioactive substance either in the fossil or in the stratum in which it was discovered.
20 CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS Construction of a mathematical model of a system starts with (i)identification of the variables that are responsible for changing the system.We may choose not to incorporate all these variables into the model at first.In this step we are specifying the level of resolution of the model. Next (ii)we make a set of reasonable assumptions,or hypotheses,about the system we are trying to describe.These assumptions will also include any empirical laws that may be applicable to the system. For some purposes it may be perfectly within reason to be content with low- resolution models.For example,you may already be aware that in beginning physics courses,the retarding force of air friction is sometimes ignored in modeling the motion of a body falling near the surface of the Earth,but if you are a scientist whose job it is to accurately predict the flight path of a long-range projectile, you have to take into account air resistance and other factors such as the curvature of the Earth Since the assumptions made about a system frequently involve a rate of change of one or more of the variables,the mathematical depiction of all these assumptions may be one or more equations involving derivatives.In other words,the mathemat- ical model may be a differential equation or a system of differential equations. Once we have formulated a mathematical model that is either a differential equa- tion or a system of differential equations,we are faced with the not insignificant problem of trying to solve it.If we can solve it,then we deem the model to be reason- able if its solution is consistent with either experimental data or known facts about the behavior of the system.But if the predictions produced by the solution are poor, we can either increase the level of resolution of the model or make alternative as- sumptions about the mechanisms for change in the system.The steps of the model- ing process are then repeated,as shown in the following diagram: Express assumptions in terms Mathematical Assumptions of differential equations formulation If necessary, alter assumptions Solve the DEs or increase resolution of model Check model Display model predictions predictions with Obtain (e.g.,graphically) solutions known facts Of course,by increasing the resolution,we add to the complexity of the mathemati- cal model and increase the likelihood that we cannot obtain an explicit solution A mathematical model of a physical system will often involve the variable time t. A solution of the model then gives the state of the system;in other words,the values of the dependent variable(or variables)for appropriate values of t describe the system in the past,present,and future. POPULATION DYNAMICS One of the earliest attempts to model human popula- tion growth by means of mathematics was by the English economist Thomas Malthus in 1798.Basically,the idea behind the Malthusian model is the assumption that the rate at which the population of a country grows at a certain time is proportional"to the total population of the country at that time.In other words,the more people there are at time t, the more there are going to be in the future.In mathematical terms,if P(t)denotes the 'If two quantitiesu andare proportional,we writeThis means that one quantity is a constant multiple of the other:u=kv
Construction of a mathematical model of a system starts with (i) identification of the variables that are responsible for changing the system. We may choose not to incorporate all these variables into the model at first. In this step we are specifying the level of resolution of the model. Next (ii) we make a set of reasonable assumptions, or hypotheses, about the system we are trying to describe. These assumptions will also include any empirical laws that may be applicable to the system. For some purposes it may be perfectly within reason to be content with lowresolution models. For example, you may already be aware that in beginning physics courses, the retarding force of air friction is sometimes ignored in modeling the motion of a body falling near the surface of the Earth, but if you are a scientist whose job it is to accurately predict the flight path of a long-range projectile, you have to take into account air resistance and other factors such as the curvature of the Earth. Since the assumptions made about a system frequently involve a rate of change of one or more of the variables, the mathematical depiction of all these assumptions may be one or more equations involving derivatives. In other words, the mathematical model may be a differential equation or a system of differential equations. Once we have formulated a mathematical model that is either a differential equation or a system of differential equations, we are faced with the not insignificant problem of trying to solve it. If we can solve it, then we deem the model to be reasonable if its solution is consistent with either experimental data or known facts about the behavior of the system. But if the predictions produced by the solution are poor, we can either increase the level of resolution of the model or make alternative assumptions about the mechanisms for change in the system. The steps of the modeling process are then repeated, as shown in the following diagram: Of course, by increasing the resolution, we add to the complexity of the mathematical model and increase the likelihood that we cannot obtain an explicit solution. A mathematical model of a physical system will often involve the variable time t. A solution of the model then gives the state of the system; in other words, the values of the dependent variable (or variables) for appropriate values of t describe the system in the past, present, and future. POPULATION DYNAMICS One of the earliest attempts to model human population growth by means of mathematics was by the English economist Thomas Malthus in 1798. Basically, the idea behind the Malthusian model is the assumption that the rate at which the population of a country grows at a certain time is proportional* to the total population of the country at that time. In other words, the more people there are at time t, the more there are going to be in the future. In mathematical terms, if P(t) denotes the Assumptions Mathematical formulation Obtain solutions Check model predictions with known facts Express assumptions in terms of differential equations Display model predictions (e.g., graphically) Solve the DEs If necessary, alter assumptions or increase resolution of model 20 ● CHAPTER 1 INTRODUCTION TO DIFFERENTIAL EQUATIONS * If two quantities u and v are proportional, we write u v. This means that one quantity is a constant multiple of the other: u kv