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(b) aluminum, or (c) cast iron.
> 19. For the rod of Problem 18 find the time that will elapse before the center of the bar cools
to a temperature of 5°C if the bar is made of (a) silver, (b) aluminum, or (c) cast iron.
20. In solving differential equations the computations can almost always be simplified by the use of dimensionless variables. Show that if the dimensionless variable ? = x/L is introduced, the heat conduction equation becomes
d 2 u L 2 d u
—r = —r —, 0 < ? < 1, t > 0.
ä?2 a2 dt’
Since L2/a2 has the units of time, it is convenient to use this quantity to define a dimensionless time variable ò = (a2/L2)t. Then show that the heat conduction equation reduces to
d u d u
d?2 äò ’
0 < ? < 1, ò > 0.
21. Consider the equation
auxx — but + cu = 0, (i)
where a, b, and c are constants.
10.6 Other Heat Conduction Problems
(a) Let u(x, t) = estw(x, t), where & is constant, and find the corresponding partial differential equation for w.
(b) If b = 0, show that & can be chosen so that the partial differential equation found in part (a) has no term in w. Thus, by a change of dependent variable, it is possible to reduce Eq. (i) to the heat conduction equation.
22. The heat conduction equation in two space dimensions is
a2(uxx + uyy) = ut ¦
Assuming that u(x, y, t) = X(x)Y(y)T(t), find ordinary differential equations satisfied by X(x), Y(y), and T(t).
23. The heat conduction equation in two space dimensions may be expressed in terms of polar coordinates as
a2[urr + (1/r )ur + (1/r 2)ugg] = ut.
Assuming that u(r, 0,t) = R(r)®(0)T(t), find ordinary differential equations satisfied by R(r), ®(9), and T(t).
10.6 Other Heat Conduction Problems
In Section 10.5 we considered the problem consisting of the heat conduction equation
a2uxx = ut, 0 < x < L, t > 0, (1)
the boundary conditions
u(0, t) = 0, u(L, t) = 0, t > 0, (2)
and the initial condition
u(x, 0) = f (x), 0 < x < L. (3)
We found the solution to be
u(x, t) = ? cne—nlnlalt/Ll sin^-, (4)
where the coefficients c are the same as in the series
cn sin-^. (5)
The series in Eq. (5) is just the Fourier sine series for f; according to Section 10.4 its coefficients are given by
2 fL nnx
cn = L Jo f (x) sin— dx. (6)
Hence the solution of the heat conduction problem, Eqs. (1) to (3), is given by the series in Eq. (4) with the coefficients computed from Eq. (6).
We emphasize that at this stage the solution (4) must be regarded as a formal solution; that is, we obtained it without rigorous justification of the limiting processes involved.
Chapter 10. Partial Differential Equations and Fourier Series
Such a justification is beyond the scope of this book. However, once the series (4) has been obtained, it is possible to show that in0 < x < L, t > 0 it converges to a continuous function, that the derivatives u xx and ut can be computed by differentiating the series (4) term by term, and that the heat conduction equation (1) is indeed satisfied. The argument rests heavily on the fact that each term of the series (4) contains a negative exponential factor, and this results in relatively rapid convergence of the series. A further argument establishes that the function u given by Eq. (4) also satisfies the boundary and initial conditions; this completes the justification of the formal solution.
It is interesting to note that although f satisfies the conditions of the Fourier convergence theorem (Theorem 10.3.1), it may have points of discontinuity. In this case the initial temperature distribution u(x, 0) = f (x) is discontinuous at one or more points. Nevertheless, the solution u(x, t) is continuous for arbitrarily small values of t > 0. This illustrates the fact that heat conduction is a diffusive process that instantly smooths out any discontinuities that may be present in the initial temperature distribution. Finally, since f is bounded, it follows from Eq. (6) that the coefficients cn are also bounded. Consequently, the presence of the negative exponential factor in each term of the series (4) guarantees that