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Several generalizations of the heat equation (10) also occur in practice. First, the bar material may be nonuniform and the cross section may not be constant along the length of the bar. In this case, the parameters k, p, s, and A may depend on the axial variable x. Going back to Eq. (2) we see that the rate of heat transfer from left to right across the cross section at x = x0 is now given by
H(x0, t) = -k(x0)A(x0)ux(x0, t)
with a similar expression for H (x0 + Ax, t). Ifwe introduce these quantities into Eq. (4) and eventually into Eq. (9), and proceed as before, we obtain the partial differential equation
(k Aux )x = sp Aut. (18)
We will usually write Eq. (18) in the form
r (x )ut = [p(x )ux ]x, (19)
where p(x) = k(x) A(x) and r (x) = s (x)p(x) A(x). Note that both these quantities are intrinsically positive.
A second generalization occurs if there are other ways in which heat enters or leaves the bar. Suppose that there is a source that adds heat to the bar at a rate G(x, t, u) per unit time per unit length, where G (x, t, u) > 0. In this case we must add the term G(x, t, u) Ax At to the left side of Eq. (9), and this leads to the differential equation
r (x)ut = [p(x)ux]x + G(x, t, u). (20)
If G (x , t, u) < 0, then we speak of a sink that removes heat from the bar at the rate G(x, t, u) per unit time per unit length. To make the problem tractable we must restrict the form of the function G. In particular, we assume that G is linear in u and that the coefficient of u does not depend on t. Thus we write
G(x, t, u) = F(x, t) — q(x)u. (21)
The minus sign in Eq. (21) has been introduced so that certain equations that appear
later will have their customary forms. Substituting from Eq. (21) into Eq. (20), we
r (x)ut = [p(x)ux]x — q(x)u + F(x, t). (22)
This equation is sometimes called the generalized heat conduction equation. Boundary value problems for Eq. (22) will be discussed to some extent in Chapter 11.
Finally, if instead of a one-dimensional bar, we consider a body with more than one significant space dimension, then the temperature is a function of two or three space
coordinates rather than of x alone. Considerations similar to those leading to Eq. (10)
can be employed to derive the heat conduction equation in two dimensions,
a2(uxx + uyy) = ut > (23)
or in three dimensions,
a2(uxx + uyy + uzz) = ut. (24)
The boundary conditions corresponding to Eqs. (12) and (13) for multidimensional problems correspond to a prescribed temperature distribution on the boundary, or to an insulated boundary. Similarly, the initial temperature distribution will in general be a function of x and y for Eq. (23), and a function of x, y, and z for Eq. (24).
Derivation of the Wave Equation. In this appendix we derive the wave equation in one space dimension as it applies to the transverse vibrations of an elastic string, or cable; the elastic string may be thought of as a violin string, a guy wire, or possibly an electric power line. The same equation, however, with the variables properly interpreted, occurs in many other wave problems having only one significant space variable.
Chapter 10. Partial Differential Equations and Fourier Series
Consider a perfectly flexible elastic string stretched tightly between supports fixed at the same horizontal level (see Figure 10.B.1a). Let the x-axis lie along the string with the endpoints located at x = 0 and x = L. If the string is set in motion at some initial time t = 0 (by plucking, for example) and is thereafter left undisturbed, it will vibrate freely in a vertical plane provided that damping effects, such as air resistance, are neglected. To determine the differential equation governing this motion we will consider the forces acting on a small element of the string of length Ax lying between the points x and x + Ax (see Figure 10.B.1b). We assume that the motion of the string is small, and as a consequence, each point on the string moves solely in a vertical line. We denote by u(x, t) the vertical displacement of the point x at the time t. Let the tension in the string, which always acts in the tangential direction, be denoted by T(x, t), and let p denote the mass per unit length of the string.
Newton’s law, as it applies to the element Ax of the string, states that the net external force, due to the tension at the ends of the element, must be equal to the product of the mass of the element and the acceleration of its mass center. Since there is no horizontal acceleration, the horizontal components must satisfy
T(x + Ax, t) cos(d + Ad) — T(x, t) cos d = 0. (1)
If we denote the horizontal component of the tension (see Figure 10.B.1c) by H, then Eq. (1) states that H is independent of x.
On the other hand, the vertical components satisfy
T(x + Ax, t) sin(d + Ad) — T(x, t) sin d = p Ax utt(x, t), (2)
where x is the coordinate of the center of mass of the element of the string under consideration. Clearly, x lies in the interval x < x < x + Ax. The weight of the string,