# Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

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where arbitrary constants of proportionality have been dropped. Each of these functions satisfies the differential equation (1) and the boundary conditions (24). Because both the differential equation and boundary conditions are linear and homogeneous, any finite linear combination of the fundamental solutions satisfies them. We will assume that this is true for convergent infinite linear combinations of fundamental solutions as well. Thus, to satisfy the initial condition (3), we assume that u(x, t) has the form

c “

u(x, t) = -0u0(x, t) + Y2cnun(x, t)

2 n=1

cë v > „,2_2 ,2+n 2 nnx

= 7 + E cne~n Ï à /L cos —. (35)

ï=1

The coefficients cn are determined by the requirement that

cn nn x

u(x, 0) = — + Ó cn cos l = f (x). (36)

n=1

Thus the unknown coefficients in Eq. (35) must be the coefficients in the Fourier cosine series of period 2L for f. Hence

2 fL . nnx

cn = — f (x) cos —-— dx, n = 0, 1,2,.... (37)

L Jq l

With this choice of the coefficients cQ, c1, c2,..., the series (35) provides the solution to the heat conduction problem for a rod with insulated ends, Eqs. (1), (3), and (24).

It is worth observing that the solution (35) can also be thought of as the sum of a steady-state temperature distribution (given by the constant c0/2), which is independent of time t, and a transient distribution (given by the rest of the infinite series) that vanishes in the limit as t approaches infinity. That the steady-state is a constant is consistent with the expectation that the process of heat conduction will gradually smooth out the temperature distribution in the bar as long as no heat is allowed to escape to the outside. The physical interpretation of the term

c 1 ÃL

~2 = L Jo f (x) dx (38)

is that it is the mean value of the original temperature distribution.

10.6 Other Heat Conduction Problems

587

EXAMPLE

2

Find the temperature u (x, t) in a metal rod of length 25 cm that is insulated on the ends as well as on the sides and whose initial temperature distribution is u (x, 0) = x for 0 < x < 25.

The temperature in the rod satisfies the heat conduction problem (1), (3), (24) with L = 25. Thus, from Eq. (35), the solution is

i(x, t) = C2 + J]

cne~n2 n 2(A/625 cos

n1

nn x

’

where the coefficients are determined from Eq. (37). We have

2 '25

x dx = 25

and, for n 1,

³

25

nn x

c = — I x cos-------------------------dx

n 25 Ë 25

= 50(cosnn — 1)/(nn)2 = — 10°°/(nn) , n °ddj

n even.

Thus

25 100 ^ 1 — ï2ÿ 2Ë/625 cos(nn x/25)

(39)

(40)

(41)

(42)

is the solution of the given problem.

For a2 = 1 Figure 10.6.2 shows plots of the temperature distribution in the bar at several times. Again the convergence of the series is rapid so that only a relatively few terms are needed to generate the graphs.

C0 =

u

FIGURE 10.6.2 Temperature distributions at several times for the heat conduction problem of Example 2.

588

Chapter 10. Partial Differential Equations and Fourier Series

PROBLEMS

More General Problems. The method of separation of variables can also be used to solve heat conduction problems with other boundary conditions than those given by Eqs. (8) and Eqs. (24). For example, the left end of the bar might be held at a fixed temperature T while the other end is insulated. In this case the boundary conditions are

u(0, t) = T, ux(L, t) = 0, t > 0. (43)

The first step in solving this problem is to reduce the given boundary conditions to homogeneous ones by subtracting the steady-state solution. The resulting problem is solved by essentially the same procedure as in the problems previously considered. However, the extension of the initial function f outside of the interval [0, L ] is somewhat different from that in any case considered so far (see Problem 15).

A more general type of boundary condition occurs when the rate of heat flow through the end of the bar is proportional to the temperature. It is shown in Appendix A that the boundary conditions in this case are of the form

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