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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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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) = k1 sinh jx + k2cosh jx. (31)
In this case the boundary conditions can be satisfied only by choosing k1 = 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) = k1 x + k2. (32)
The boundary conditions (29) and (30) require that k1 = 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) = k1 sin jx + k2cos jx. (33)
The boundary condition (29) requires that k1 = 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) of Eq. (28) are proportional to exp(-n2n2a2t/L2).
586
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):
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
Thus the unknown coefficients in Eq. (35) must be the coefficients in the Fourier cosine series of period 2L for f. Hence
With this choice of the coefficients c0, 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
u0(x, t) = 1, un (x, t) = e-
nn x
L ’
n = 1, 2,...,
(34)
n=1
(35)
The coefficients cn are determined by the requirement that
n=1
(36)
n = 0, 1, 2,....
(37)
(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) = C-2 +
cne-"2 n Vt/625 cos
n=1
nn x
~i5 ’
where the coefficients are determined from Eq. (37). We have
2 ^
2 r2 25 Jo
x dx = 25
and, for n > 1,
o
25
nn x
c = — I x cos dx
n 25 Jo 25
= 50(cosnn - 1)/(nn)2 = \~^/(nn) , n odd;
n even.
Thus
25 100 ^ 1 -n2n Vt/625 cos(nn x/25)
(x ’t ) = Y - e
n=1,3,5,... 1
(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.
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