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

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wxx = Wt, (21)

the homogeneous boundary conditions

w(0, t) = 0, w(30, t) = 0, (22)

and the modified initial condition

w(x, 0) = 60 - 2x - (20 + x) = 40 - 3x. (23)

Note that this problem is of the form (1), (2), (3) with f (x) = 40 — 3x, a2 = 1, and L = 30. Thus the solution is given by Eqs. (4) and (6).

Figure 10.6.1 shows a plot of the initial temperature distribution 60 — 2x, the final temperature distribution 20 + x, and the temperature at two intermediate times found by solving Eqs. (21) through (23). Note that the intermediate temperature satisfies the boundary conditions (19) for any t > 0. As t increases, the effect of the boundary conditions gradually moves from the ends of the bar toward its center.

FIGURE 10.6.1 Temperature distributions at several times for the heat conduction problem of Example 1.

Bar with Insulated Ends. A slightly different problem occurs if the ends of the bar are insulated so that there is no passage of heat through them. According to Eq. (2) in Appendix A, the rate of flow of heat across a cross section is proportional to the rate of change of temperature in the x direction. Thus in the case of no heat flow the boundary conditions are

ux (0, t) = 0, ux (L, t) = 0, t > 0. (24)

The problem posed by Eqs. (1), (3), and (24) can also be solved by the method of separation of variables. If we let

u(x, t) = X(x)T(t),

(25)

10.6 Other Heat Conduction Problems

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2 rr = ~X, (26)

and substitute for u in Eq. (1), then it follows as in Section 10.5 that

X"_ 1 T'

~X = a2 T

where X is a constant. Thus we obtain again the two ordinary differential equations

X" + X X = 0, (27)

T' + a2XT = 0. (28)

For any value of X a product of solutions of Eqs. (27) and (28) is a solution of the partial differential equation (1). However, we are interested only in those solutions that also satisfy the boundary conditions (24).

If we substitute for u(x, t) from Eq. (25) in the boundary condition at x = 0, we obtain X'(0)T(t) = 0. We cannot permit T(t) to be zero for all t, since then u(x, t) would also be zero for all t. Hence we must have

X (0) = 0. (29)

Proceeding in the same way with the boundary condition at x = L, we find that

X'(L) = 0. (30)

Thus we wish to solve Eq. (27) subject to the boundary conditions (29) and (30). It is possible to show that nontrivial solutions of this problem can exist only if X is real. One way to show this is indicated in Problem 18; alternatively, one can appeal to a more

general theory to be discussed later in Section 11.2. We will assume that X is real and

consider in turn the three cases X < 0, X = 0, and X > 0.

If X < 0, it is convenient to let X = —/j2, where j is real and positive. Then Eq. (27) becomes X" — j2X = 0 and its general solution is

X(x) = kx sinh jx + k2cosh jx. (31)

In this case the boundary conditions can be satisfied only by choosing kx = k2 = 0. Since this is unacceptable, it follows that X cannot be negative; in other words, the problem (27), (29), (30) has no negative eigenvalues.

If X = 0, then Eq. (27) is X" = 0, and therefore

X (x) = kx x + k2. (32)

The boundary conditions (29) and (30) require that kx = 0 but do not determine k2. Thus X = 0 is an eigenvalue, corresponding to the eigenfunction X (x) = 1. For X = 0 it follows from Eq. (28) that T(t) is also a constant, which can be combined with k2. Hence, for X = 0, we obtain the constant solution u(x, t) = k2.

Finally, if X > 0, let X = j2, where j is real and positive. Then Eq. (27) becomes X" + j2X = 0, and consequently

X(x) = kx sin jx + k2cos jx. (33)

The boundary condition (29) requires that kx = 0, and the boundary condition (30) requires that j = nn/L for n = 1, 2, 3,..., but leaves k2 arbitrary. Thus the problem (27), (29), (30) has an infinite sequence of positive eigenvalues X = n2n2/L2 with the corresponding eigenfunctions X(x) = cos(nnx/L). For these values of X the solutions T (t) ofEq. (28) are proportional to exp(-n2n 2a2t/L 2).

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Chapter 10. Partial Differential Equations and Fourier Series

Combining all these results, we have the following fundamental solutions for the problem (1), (3), and (24):

u0(x, t) = 1,

un(x, t) = e-2* 2“2t/L2 cos ïï^, ï = 1, 2,..., (34)

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