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Partial Differential Equations
With Fourier Series And Boundary Value Problems
By Nakhle H. Asmar Dover Publications, Inc.
Copyright © 2017 Nakhle H. Asmar
All rights reserved.
ISBN: 978-0-486-80737-9
CHAPTER 1
A PREVIEW OF APPLICATIONS AND TECHNIQUES
Everything should be made as simple as possible, but no simpler.
-ALBERT EINSTEIN
Partial differential equations arise in modeling numerous phenomena in science and engineering. In this chapter we preview the ideas and techniques that will be studied in this book. Our goal is to convey something of the flavor of partial differential equations and their use in applications rather than to give a systematic development. In particular, by focusing on the vibration of a stretched string, we will see how partial differential equations arise in modeling a physical phenomenon and how the interpretation of their solutions helps us to understand this phenomenon. The chapter will culminate with a brief discussion of Fourier series that highlights their importance to the development of the applications of partial differential equations.
1.1 What Is a Partial Differential Equation?
Loosely speaking, a partial differential equation is an equation that involves an unknown function of several variables and its partial derivatives. You may recall from your study of ordinary differential equations that the description of specific classes of equations involves notions such as order, degree, linearity, and various other properties. In classifying and studying partial differential equations, other notions also arise, including the number of independent variables and the region in which we want to solve the equation. This entire book is devoted to a systematic development of certain classes of partial differential equations with an emphasis on applications. In this section, we confine ourselves to several elementary examples to give you a taste of the subject.
EXAMPLE 1 A first order partial differential equation Consider the equation
(1) [MATHEMATICAL EXPRESSION OMITTED]
where u = u(x, t) is the unknown function. If f is any differentiable function of a single variable and we set
u(x, t) = f (x - t),
then u is a solution of (1). Indeed, using the chain rule, we find that
[MATHEMATICAL EXPRESSION OMITTED]
and so (1) holds. Explicit examples of solutions of (1) are
[MATHEMATICAL EXPRESSION OMITTED]
The first three of these are shown in Figure 1.
The equation that we considered is called an advection equation from fluid dynamics (see Exercise 27, Section 1.2, for an interesting application). It is a first-order linear homogeneous partial differential equation. Here first order refers to the highest order derivative that appears in the equation, and linear homogeneous refers to the fact that any linear combination of solutions is again a solution. (These notions will be developed in greater detail in Chapter 3.) From Figure 1, it is clear that there is a great diversity of possible solutions to equation (1), which is a relatively simple partial differential equation. In contrast, you may recall that the family of solutions of a first order linear ordinary differential equation is much easier to describe. For example, the family of solutions of the equation
(2) y' - y = 0
is given by
y = Aex.
As illustrated in Figure 2, the solutions are all constant multiples of the solution ex. To single out a given solution of equation (2) we need to impose an initial condition, such as y(0) = 2, which gives the solution y = 2 ex.
To single out a given solution of the partial differential equation (1) it is evident from Figure 1 that much more is needed. Typically we will impose a condition specifying the values of u along some curve in the xt-plane, for example, along the x-axis. Such conditions are called boundary or initial conditions.
EXAMPLE 2 Boundary or initial conditions
Of all the solutions of (1), which one satisfies the condition [MATHEMATICAL EXPRESSION OMITTED] along the x-axis? Here we are looking for the solution that satisfies [MATHEMATICAL EXPRESSION OMITTED]. Using the solution u(x, t) = f (x - t) from Example 1, we find that
[MATHEMATICAL EXPRESSION OMITTED]
Hence the solution is given by [MATHEMATICAL EXPRESSION OMITTED]. (See Figure 1(b).) There is another interesting way to visualize the solution by thinking of t as a time variable. For any fixed value of t, the graph of u(x, t), as a function of x, represents a snapshot of a waveform. (See Figure 3.) Note that the waveform moves to the right without changing shape.
While in Example 1 we mentioned many solutions, we gave no indication of how to derive them. Also, it is not clear that we obtained all the solutions in that example. The following derivation will show that we have indeed found all of them and will illustrate how a judicious change of variables can be used to simplify a problem.
EXAMPLE 3 General solution of (1)
The plan is to reduce the partial differential equation (1) to an ordinary differential equation by means of a linear change of variables
α = ax + bt, β = cx + dt,
where a, b, c, and d will be chosen appropriately. The chain rule in two dimensions gives
[MATHEMATICAL EXPRESSION OMITTED]
Similarly,
[MATHEMATICAL EXPRESSION OMITTED]
Plugging into (1) and simplifying, we obtain
[MATHEMATICAL EXPRESSION OMITTED]
Now we select the values for a, b, c, and d so as to get rid of one partial derivative. Let us get rid of [partial derivative]u/[partial derivative]β by taking a = 1, b = 0, c = 1, d = — 1.
a = 1, b = 0, c = 1, d = -1.
With this choice, the equation becomes
[MATHEMATICAL EXPRESSION OMITTED]
This simple ordinary differential equation in a has the solution u = C, where C is a function of β (which is constant with respect to α). Thus the general solution is u = C(β) = C(x - t), which matches our answer to Example 1. In closing we note that other choices of the constants a, b, c, and d can be used to derive the solution. (See Exercise 4.)
Example 3 shows us how the great diversity of solutions arises from partial differential equations. When we integrate an equation with two independent variables, the constant of integration that appears must be allowed to be a function of the remaining variable. Thus after integrating the equation we get an arbitrary function of one variable, in contrast with the case of an ordinary differential equation where we get only an arbitrary constant.
The change of variables in Example 3 reduced the partial differential equation to an ordinary differential equation, which was then easily solved. This reduction to an ordinary differential equation from a partial differential equation is a recurring theme throughout the book. (See Exercise 10 for another illustration.)
Exercises 1.1
1. Suppose that u1 and u2 are solutions of (1). Show that c1u1 + c2u2 is also a solution, where c1 and c2 are constants. (This shows that (1) is a linear equation.)
2. (a) Which solution of (1) is equal to [MATHEMATICAL EXPRESSION OMITTED] on the x-axis?
(b) Plot the solution as a function of x and t, and describe the image of the x-axis.
3. (a) Find the solution of (1) that is equal to t on the t-axis.
(b) Plot the solution as a function of x and t, and describe the image of the t-axis.
4. Refer to Example 3 and solve (1) using a = 1, b = 1, c = 1, d = -1.
In Exercises 5-8, derive the general solution of the given equation by using an appropriate change of variables, as we did in Example 3.
9. (a) Find the solution in Exercise 5 that is equal to 1/1 + x2 along the x-axis.
(b) Plot the graph of the solution as a function of x and t.
(c) In what direction is the waveform moving as t increases?
The method of characteristic curves This method allows you to solve first order, linear partial differential equations with nonconstant coefficients of the form
(3) [MATHEMATICAL EXPRESSION OMITTED]
where p(x, y) is a function of x and y. We begin by reviewing two important concepts from calculus of several variables and ordinary differential equations.
Suppose that v = (v1, v2) is a unit vector and u(x, y) is a function of two variables. The directional derivative of u (in the direction of v) at the point (x0, y0) is given by
[MATHEMATICAL EXPRESSION OMITTED]
The directional derivative measures the rate of change of u at the point (x0, y0) as we move in the direction of v. Consequently, if (a, b) is any nonzero vector, then the relation
[MATHEMATICAL EXPRESSION OMITTED]
states that u is not changing in the direction of (a, b).
Moving to ordinary differential equations, consider the equation
(4) dy/dx = p(x, y),
and think of the solutions of (4) as curves in the xy-plane. Equation (4) is telling us that the slope of the tangent line to a solution curve at the point (x0, y0) is equal to p(x0, y0), or that the solution curve at the point (x0, y0) is changing in the direction of the vector (1, p(x0, y0)). If we plot the vectors (1, p(x, y)) for many points (x, y), we obtain the direction field for the differential equation (4). The direction field is very important, since it gives us a quick way for visualizing the qualitative behavior of the solutions of (4). In particular, we can trace a solution curve through a point (x, y) by constantly moving in the direction of the vector field (1, p(x, y)).
We are now ready to present the method of characteristic curves. Going back to (3), this equation is telling us that, at any point (x, y), the directional derivative of u in the direction of the vector (1, p(x, y)) is 0. Hence u(x, y) remains constant if from the point (x, y) we move in the direction of vector (1, p(x, y)). But the vector (1, p(x, y)) determines the direction field for (4). Consequently, u is constant on the solution curves of (4). These curves are called the characteristic curves of (3) (see Figure 4 for the case p(x, y) = x, treated below). For convenience, let us write the solutions of (4) (the characteristic curves) in the form
φ (x, y) = C,
where C is arbitrary. Since u is constant on each such curve, the values of u depend only on C. Thus u(x, y) = f(C), where f is an arbitrary function, or
u(x, y) = f(φ(x, y),
which yields the solution of (3).
As an illustration, let us solve the equation
[MATHEMATICAL EXPRESSION OMITTED]
We have p(x, y) = x. Solving dy/dx = x, we find the characteristic curves y = 1/2 x2 + C, or y - 1/2 x2 = C. Hence φ(x, y) = y - 1/2 x2, and the general solution of the partial differential equation is u(x, y) = f (φ(x, y)), where f is an arbitrary function. The validity of this solution can be checked directly by plugging it back into the equation.
It is worth noting that in solving the partial differential equation (3) by the method of characteristic curves, all we had to do is solve the ordinary differential equation (4). Thus, the method of characteristic curves is yet another method that reduces a partial differential equation to an ordinary differential equation.
10. Consider the equation [MATHEMATICAL EXPRESSION OMITTED], where a, b are nonzero constants.
(a) What is the equation saying about the directional derivative of u?
(b) Determine the characteristic curves.
(c) Solve the equation using the method of characteristic curves.
In Exercises 11-14, (a) solve the given equation by the method of characteristic curves, and (b) check your answer by plugging it back into the equation.
11. [MATHEMATICAL EXPRESSION OMITTED]
12. [MATHEMATICAL EXPRESSION OMITTED]
13. [MATHEMATICAL EXPRESSION OMITTED]
14. [MATHEMATICAL EXPRESSION OMITTED]
Solving and Interpreting a Partial Differential Equation
With the discovery of Newtonian mechanics in the late seventeenth century, it became clear that many laws of physics and engineering are best described in terms of relations involving rates of change. When translated into mathematical language, these relations lead to differential equations, since rates of change correspond to derivatives. Consider Newton's second law of motion, F = ma. This fundamental law tells us that if we can describe explicitly the force, then we obtain a differential equation for the position. (Recall that a stands for acceleration and, as such, it is the second derivative of position.)
For example, in a mass-spring system, the mass is subject to a linear restoring force, F = -kx (Hooke's law). In this case, the motion of the mass is determined by the differential equation mx" + kx = 0. Here x represents the position of the mass as a function of the time t. This equation allows us to solve for the specific motion of the mass if we are also supplied with its position and velocity at a given time t0. This data is called the initial data and is usually specified at time t0 = 0.
Moving to a situation where a partial differential equation is needed, we will consider a familiar phenomenon: the vibrating string. The partial differential equation that arises in this case is called the one dimensional wave equation. Its analytical solution, which will be treated in full detail in Chapter 3, is based on the same ideas that are required for the treatment of more sophisticated problems. In this section we only intend to preview these ideas in order to introduce you to them in a relatively simple setting.
The Vibrating String
Consider a string stretched along the x-axis between x = 0 and x = L and free to vibrate in a fixed plane. We want to describe the motion of each point of the string as time progresses. For that purpose we use the function u(x, t) to denote the displacement of the string at time t at the point x (see Figure 1). Note that already the unknown function here is a function of two variables, in contrast to the simple mass-spring system. Based on this representation, the velocity of the string at position x is [partial derivative]u/[partial derivative]t and its acceleration there is [partial derivative]2u/[partial derivative]t2. The equation that governs the motion of the stretched string will be derived in Section 3.2. Here we shall only write it down and develop an intuitive understanding of it. This equation is known as the one dimensional wave equation and is given by
(1) [MATHEMATICAL EXPRESSION OMITTED]
The constant c depends on the physical parameters of the stretched string — in particular, its linear mass density and its tension. The key to understanding this equation from a physical point of view is to interpret it in light of Newton's second law. The left side represents the acceleration of a small portion of the string centered at a point x, while the right side is telling us that this small portion feels a force whose sign depends on the concavity of the string at that point.
Thus, if we view Figure 2 as giving us the initial displacement of a string that is released from rest, then immediately following release those portions of the string where it is concave up will start to move up, and those where it is concave down will start to move down (see arrows in Figures 2 and 3). This point of view can be applied to any snapshot of the string's position, except that in general each portion of the string will then have a nonzero velocity associated with it and this will also have to be taken into account.
(Continues...)
Excerpted from Partial Differential Equations by Nakhle H. Asmar. Copyright © 2017 Nakhle H. Asmar. Excerpted by permission of Dover Publications, Inc..
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