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

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lim u (x, t) = 0 (7)

t

for all x regardless of the initial condition. This is in accord with the result expected from physical intuition.

We now consider two other problems of one-dimensional heat conduction that can be handled by the method developed in Section 10.5.

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

u(0, t) = Tv 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 ^<x>, 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 ofEq. (9) satisfying Eqs. (10) is

x

v(x) = (T2 — T1) L + Tv (11)

10.6 Other Heat Conduction Problems

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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)

a2wxx = wt, (13)

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

w(0, t) = u(0, t) — v(0) = T — T = 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

OO

x v > *,2_2 ,2+it2 nnx

u(x, t) = (T2 — Tx)- + T + ?cne~n n a t/L si^^, (16)

n=1

where

cn = L

fL Ã x

io f (x) — T — Ti) L — T1

• nn x

sin dx. (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.

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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

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