3.8 Mechanical and Electrical Vibrations
C H A P T E R
Introduction
In this chapter we try in several different ways to give perspective to your study of differential equations. First, we use two problems to illustrate some of the basic ideas that we will return to and elaborate upon frequently throughout the remainder of the book. Later, we indicate several ways of classifying equations, in order to provide organizational structure for the book. Finally, we outline some of the major figures and trends in the historical development of the subject. The study of differential equations has attracted the attention of many of the world’s greatest mathematicians during the past three centuries. Nevertheless, it remains a dynamic field of inquiry today, with many interesting open questions.
1.1 Some Basic Mathematical Models; Direction Fields
Before embarking on a serious study of differential equations (for example, by reading this book or major portions of it) you should have some idea of the possible benefits
ODE
to be gained by doing so. For some students the intrinsic interest of the subject itself is enough motivation, but for most it is the likelihood of important applications to other fields that makes the undertaking worthwhile.
Many of the principles, or laws, underlying the behavior of the natural world are statements or relations involving rates at which things happen. When expressed in mathematical terms the relations are equations and the rates are derivatives. Equations ODE
containing derivatives are differential equations. Therefore, to understand and to investigate problems involving the motion of fluids, the flow of current in electric
1
Chapter 1. Introduction
circuits, the dissipation of heat in solid objects, the propagation and detection of seismic waves, or the increase or decrease of populations, among many others, it is necessary to know something about differential equations.
A differential equation that describes some physical process is often called a math- ematical model of the process, and many such models are discussed throughout this book. In this section we begin with two models leading to equations that are easy to solve. It is noteworthy that even the simplest differential equations provide useful models of important physical processes.
Suppose that an object is falling in the atmosphere near sea level. Formulate a differ E X A M P L E ential equation that describes the motion.
1
We begin by introducing letters to represent various quantities of possible interest in this problem. The motion takes place during a certain time interval, so let us use t A Falling to denote time. Also, let us use v to represent the velocity of the falling object. The
Object velocity will presumably change with time, so we think of v as a function of t; in other words, t is the independent variable and v is the dependent variable. The choice of units of measurement is somewhat arbitrary, and there is nothing in the statement of the problem to suggest appropriate units, so we are free to make any choice that seems reasonable. To be specific, let us measure time t in seconds and velocity v in
meters/second. Further, we will assume that v is positive in the downward direction, that is, when the object is falling.
The physical law that governs the motion of objects is Newton’s second law, which states that the mass of the object times its acceleration is equal to the net force on the object. In mathematical terms this law is expressed by the equation F = ma.
(1)
In this equation m is the mass of the object, a is its acceleration, and F is the net force exerted on the object. To keep our units consistent, we will measure m in kilograms, a
in meters/second2, and F in newtons. Of course, a is related to v by a = dv/dt, so we can rewrite Eq. (1) in the form F = m(dv/dt).
(2) Next, consider the forces that act on the object as it falls. Gravity exerts a force equal to the weight of the object, or mg, where g is the acceleration due to gravity. In the units we have chosen, g has been determined experimentally to be approximately equal to 9.8 m/sec2 near the earth’s surface. There is also a force due to air resistance, or drag, which is more difficult to model. This is not the place for an extended discussion of the drag force; suffice it to say that it is often assumed that the drag is proportional to the velocity, and we will make that assumption here. Thus the drag force has the magnitude γ v, where γ is a constant called the drag coefficient.
In writing an expression for the net force F we need to remember that gravity always acts in the downward (positive) direction, while drag acts in the upward (negative) direction, as shown in Figure 1.1.1. Thus F = mg − γ v
(3) and Eq. (2) then becomes
dv
m
= mg − γ v.
(4)
dt
Equation (4) is a mathematical model of an object falling in the atmosphere near sea level. Note that the model contains the three constants m, g, and γ . The constants m
and γ depend very much on the particular object that is falling, and usually will be different for different objects. It is common to refer to them as parameters, since they may take on a range of values during the course of an experiment. On the other hand, the value of g is the same for all objects.
γ υ
m
mg
FIGURE 1.1.1
Free-body diagram of the forces on a falling object.
To solve Eq. (4) we need to find a function v = v(t) that satisfies the equation. It is not hard to do this and we will show you how in the next section. For the present, however, let us see what we can learn about solutions without actually finding any of them. Our task is simplified slightly if we assign numerical values to m and γ , but the procedure is the same regardless of which values we choose. So, let us suppose that m = 10 kg and γ = 2 kg/sec. If the units for γ seem peculiar, remember that γ v must have the units of force, namely, kg-m/sec2. Then Eq. (4) can be rewritten as
dv
v
= 9.8 − .
(5)
dt
5
Investigate the behavior of solutions of Eq. (5) without actually finding the solutions E X A M P L E in question.
2
We will proceed by looking at Eq. (5) from a geometrical viewpoint. Suppose that v has a certain value. Then, by evaluating the right side of Eq. (5), we can find the A Falling corresponding value of dv/dt. For instance, if v = 40, then dv/dt = 1.8. This means
Object that the slope of a solution v = v(t) has the value 1.8 at any point where v = 40. We
(continued) can display this information graphically in the tv-plane by drawing short line segments, or arrows, with slope 1.8 at several points on the line v = 40. Similarly, if v = 50, then dv/dt = −0.2, so we draw line segments with slope −0.2 at several points on the line v = 50. We obtain Figure 1.1.2 by proceeding in the same way with other values of v. Figure 1.1.2 is an example of what is called a sometimes a slope
field.
The importance of Figure 1.1.2 is that each line segment is a tangent line to the graph of a solution of Eq. (5). Thus, even though we have not found any solutions, and Chapter 1. Introduction
υ
60
56
52
48
44
40
2
4
6
8
10
t
FIGURE 1.1.2 A direction field for Eq. (5).
no graphs of solutions appear in the figure, we can nonetheless draw some qualitative conclusions about the behavior of solutions. For instance, if v is less than a certain critical value, then all the line segments have positive slopes, and the speed of the falling object increases as it falls. On the other hand, if v is greater than the critical value, then the line segments have negative slopes, and the falling object slows down as it falls. What is this critical value of v that separates objects whose speed is increasing from those whose speed is decreasing? Referring again to Eq. (5), we ask what value of v will cause dv/dt to be zero? The answer is v = (5)(9.8) = 49 m/sec.
In fact, the constant function v(t) = 49 is a solution of Eq. (5). To verify this statement, substitute v(t) = 49 into Eq. (5) and observe that each side of the equation is zero. Because it does not change with time, the solution v(t) = 49 is called an
It is the solution that corresponds to a balance between gravity and drag. In Figure 1.1.3 we show the equilibrium solution v(t) = 49 superimposed on the direction field. From this figure we can draw another conclusion, namely, that all other solutions seem to be converging to the equilibrium solution as t increases.
υ
60
56
52
48
44
40
2
4
6
8
10
t
FIGURE 1.1.3
Direction field and equilibrium solution for Eq. (5).
1.1Some Basic Mathematical Models; Direction Fields The approach illustrated in Example 2 can be applied equally well to the more general Eq. (4), where the parameters m and γ are unspecified positive numbers. The results are essentially identical to those of Example 2. The equilibrium solution of Eq. (4) is v(t) = mg/γ . Solutions below the equilibrium solution increase with time, those above it decrease with time, and all other solutions approach the equilibrium solution as t becomes large.
Direction Fields.
Direction fields are valuable tools in studying the solutions of differential equations of the form d y = f (t, y),
(6)
dt where f is a given function of the two variables t and y, sometimes referred to as the The equation in Example 2 is somewhat simpler in that in it f is a function of the dependent variable alone and not of the independent variable. A useful direction field for equations of the general form (6) can be constructed by evaluating f
at each point of a rectangular grid consisting of at least a few hundred points. Then, at each point of the grid, a short line segment is drawn whose slope is the value of f
at that point. Thus each line segment is tangent to the graph of the solution passing through that point. A direction field drawn on a fairly fine grid gives a good picture of the overall behavior of solutions of a differential equation. The construction of a direction field is often a useful first step in the investigation of a differential equation.
Two observations are worth particular mention. First, in constructing a direction field, we do not have to solve Eq. (6), but merely evaluate the given function f (t, y)
many times. Thus direction fields can be readily constructed even for equations that may be quite difficult to solve. Second, repeated evaluation of a given function is a task for which a computer is well suited and you should usually use a computer to draw a direction field. All the direction fields shown in this book, such as the one in Figure 1.1.2, were computer-generated.
Field Mice and Owls.
Now let us look at another quite different example. Consider a population of field mice who inhabit a certain rural area. In the absence of predators we assume that the mouse population increases at a rate proportional to the current population. This assumption is not a well-established physical law (as Newton’s law of motion is in Example 1), but it is a common initial hypothesis 1 in a study of population growth. If we denote time by t and the mouse population by p(t), then the assumption about population growth can be expressed by the equation d p = rp,
(7)
dt
where the proportionality factor r is called the rate constant or growth rate. To be specific, suppose that time is measured in months and that the rate constant r has the value 0.5/month. Then each term in Eq. (7) has the units of mice/month.
Now let us add to the problem by supposing that several owls live in the same neighborhood and that they kill 15 field mice per day. To incorporate this information 1A somewhat better model of population growth is discussed later in Section 2.5.
Chapter 1. Introduction into the model, we must add another term to the differential equation (7), so that it becomes d p = 0.5p − 450.
(8)
dt
Observe that the predation term is −450 rather than −15 because time is measured in months and the monthly predation rate is needed.
Investigate the solutions of Eq. (8) graphically.
E X A M P L E A direction field for Eq. (8) is shown in Figure 1.1.4. For sufficiently large values
3
of p it can be seen from the figure, or directly from Eq. (8) itself, that d p/dt is positive, so that solutions increase. On the other hand, for small values of p the opposite is the case. Again, the critical value of p that separates solutions that increase from those that decrease is the value of p for which d p/dt is zero. By setting dp/dt equal to zero in Eq. (8) and then solving for p, we find the equilibrium solution p(t) = 900 for which the growth term and the predation term in Eq. (8) are exactly balanced. The equilibrium solution is also shown in Figure 1.1.4.
p
1000
950
900
850
800
1
2
3
4
5
t
FIGURE 1.1.4 A direction field for Eq. (8).
Comparing this example with Example 2, we note that in both cases the equilibrium solution separates increasing from decreasing solutions. However, in Example 2 other solutions converge to, or are attracted by, the equilibrium solution, while in Example 3 other solutions diverge from, or are repelled by, the equilibrium solution. In both cases the equilibrium solution is very important in understanding how solutions of the given differential equation behave.
A more general version of Eq. (8) is d p = rp − k,
(9)
dt
where the growth rate r and the predation rate k are unspecified. Solutions of this more general equation behave very much like those of Eq. (8). The equilibrium solution of Eq. (9) is p(t) = k/r. Solutions above the equilibrium solution increase, while those below it decrease.
You should keep in mind that both of the models discussed in this section have their limitations. The model (5) of the falling object ceases to be valid as soon as the object hits the ground, or anything else that stops or slows its fall. The population model (8) eventually predicts negative numbers of mice (if p < 900) or enormously large numbers (if p > 900). Both these predictions are unrealistic, so this model becomes unacceptable after a fairly short time interval.
Constructing Mathematical Models.
In applying differential equations to any of the numerous fields in which they are useful, it is necessary first to formulate the appropriate differential equation that describes, or models, the problem being investigated. In this section we have looked at two examples of this modeling process, one drawn from physics and the other from ecology. In constructing future mathematical models yourself, you should recognize that each problem is different, and that successful modeling is not a skill that can be reduced to the observance of a set of prescribed rules. Indeed, constructing a satisfactory model is sometimes the most difficult part of the problem.
Nevertheless, it may be helpful to list some steps that are often part of the process:
1.
Identify the independent and dependent variables and assign letters to represent them. The independent variable is often time.
2.
Choose the units of measurement for each variable. In a sense the choice of units is arbitrary, but some choices may be much more convenient than others. For example, we chose to measure time in seconds in the falling object problem and in months in the population problem.
3.
Articulate the basic principle that underlies or governs the problem you are investigating. This may be a widely recognized physical law, such as Newton’s law of motion, or it may be a more speculative assumption that may be based on your own experience or observations. In any case, this step is likely not to be a purely mathematical one, but will require you to be familiar with the field in which the problem lies.
4.
Express the principle or law in step 3 in terms of the variables you chose in step 1. This may be easier said than done. It may require the introduction of physical constants or parameters (such as the and the determination of appropriate values for them. Or it may involve the use of auxiliary or intermediate variables that must then be related to the primary variables.
5.
Make sure that each term in your equation has the same physical units. If this is not the case, then your equation is wrong and you should seek to repair it. If the units agree, then your equation at least is dimensionally consistent, although it may have other shortcomings that this test does not reveal.
6.
In the problems considered here the result of step 4 is a single differential equation, which constitutes the desired mathematical model. Keep in mind, though, that in more complex problems the resulting mathematical model may be much more complicated, perhaps involving a system of several differential equations, for example.
Chapter 1. Introduction
PROBLEMS
In each of Problems 1 through 6 draw a direction field for the given differential equation. Based on the direction field, determine the behavior of y as t → ∞. If this behavior depends on the initial value of y at t = 0, describe this dependency.
15. A pond initially contains 1,000,000 gal of water and an unknown amount of an undesirable (a) Write a differential equation whose solution is the amount of chemical in the pond at 16. A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time.
17. A certain drug is being administered intravenously to a hospital patient. Fluid containing (a) Assuming that the drug is always uniformly distributed throughout the bloodstream, 䉴 18. For small, slowly falling objects the assumption made in the text that the drag force is
(d) Using the data in part (c), draw a direction field and compare it with Figure 1.1.3.
2See Lyle N. Long and Howard Weiss, “The Velocity Dependence of Aerodynamic Drag: A Primer for Mathematicians,” Amer. Math. Monthly 106, 2 (1999), pp. 127–135.