# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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Nonhomogeneous Boundary Conditions. Suppose now that one end of the bar is held at a constant temperature T1 and the other is maintained at a constant temperature T2. Then the boundary conditions are

u(0, t) = Tj, u(L, t) = T2, t > 0. (8)

The differential equation (1) and the initial condition (3) remain unchanged.

This problem is only slightly more difficult, because of the nonhomogeneous boundary conditions, than the one in Section 10.5. We can solve it by reducing it to a problem having homogeneous boundary conditions, which can then be solved as in Section

10.5. The technique for doing this is suggested by the following physical argument.

After a long time, that is, as t ^ro, we anticipate that a steady temperature distribution v(x) will be reached, which is independent of the time t and the initial conditions. Since v(x) must satisfy the equation of heat conduction (1), we have

v"(x) = 0, 0 < x < L. (9)

Hence the steady-state temperature distribution is a linear function of x. Further, v(x) must satisfy the boundary conditions

v(0) = T, v( L) = T2, (10)

which are valid even as t ^ro. The solution of Eq. (9) satisfying Eqs. (10) is

x

v(x) = (T2 - Tx)L + Tv (11)

10.6 Other Heat Conduction Problems

583

Returning to the original problem, Eqs. (1), (3), and (8), we will try to express u(x, t) as the sum of the steady-state temperature distribution v(x) and another (transient) temperature distribution w(x, t); thus we write

u(x, t) = v(x) + w(x, t). (12)

Since v(x) is given by Eq. (11), the problem will be solved provided we can determine w(x, t). The boundary value problem for w(x, t) is found by substituting the expression in Eq. (12) for u (x, t) in Eqs. (1), (3), and (8).

From Eq. (1) we have

a2(v + w) = (v + w) ;

it follows that

(14)

2wxx = Wt, (13)

since vxx = 0 and vt = 0. Similarly, from Eqs. (12), (8), and (10),

w(0, t) = u(0, t) v(0) = T1 T1 = 0,

w(L, t) = u(L, t) v(L) = T2 T2 = 0.

Finally, from Eqs. (12) and (3),

w(x, 0) = u(x, 0) v(x) = f (x) v(x), (15)

where v(x) is given by Eq. (11). Thus the transient part of the solution to the original

problem is found by solving the problem consisting of Eqs. (13), (14), and (15). This latter problem is precisely the one solved in Section 10.5 provided that f (x) v(x) is now regarded as the initial temperature distribution. Hence

CO

x v > ,2_2 ,2*/j-2 nn x

u(x, t) = (T2 T,)L + T + ?cne~n n a t/L sin, (16)

n1

where

= L

fL r x

J0 f (x) - (T2 - Ti) L - Ti

nn x

sin ax. (17)

This is another case in which a more difficult problem is solved by reducing it to a simpler problem that has already been solved. The technique of reducing a problem with nonhomogeneous boundary conditions to one with homogeneous boundary conditions by subtracting the steady-state solution is capable of wide application.

2

EXAMPLE

1

Consider the heat conduction problem

uxx = ut, 0 < x < 30, t > 0, (18)

u(0, t) = 20, u(30, t) = 50, t > 0, (19)

u(x, 0) = 60 2x, 0 < x < 30. (20)

Find the steady-state temperature distribution and the boundary value problem that determines the transient distribution.

584

Chapter 10. Partial Differential Equations and Fourier Series

The steady-state temperature satisfies v"(x) = 0 and the boundary conditions v(0) = 20 and v(30) = 50. Thus v(x) = 20 + x. The transient distribution w(x, t) satisfies the heat conduction equation

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 = (26)

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

X"_ 1 T'

X = a T

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

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