0.3 Boundary Value Problems 29
Figure 5 Cylinder of heat-conducting material.This is approximately true for a suspension bridge. The boundary value prob-
lem to be solved is then
d^2 u
dx^2=w
T, 0 <x<a,
u( 0 )=h 0 , u(a)=h 1. (7)The general solution of the differential equation (7) can be found by the
procedures of Sections 1 and 2. It is
u(x)=(w
2 T)
x^2 +c 1 x+c 2 ,wherec 1 andc 2 are arbitrary. The two boundary conditions require
u( 0 )=h 0 : c 2 =h 0 ,u(a)=h 1 :(
w
2 T)
a^2 +c 1 a+c 2 =h 1.Thesetwoaresolvedforc 1 andc 2 in terms of given parameters. The result,
after some beautifying algebra, is
u(x)= w
2 T(
x^2 −ax)
+h^1 −h^0
ax+h 0. (8)Clearly, this function specifies the cable’s shape as part of a parabola opening
upward.
Example: Heat Conduction in a Rod.
A long rod of uniform material and cross section conducts heat along its axial
direction (see Fig. 5). We assume that the temperature in the rod,u(x),does
not change in time. A heat balance (“what goes in must come out”) applied to
a slice of the rod betweenxandx+ x(Fig. 6) shows that the heat flow rate
q, measured in units of heat per unit time per unit area, obeys the equation
q(x)A+g(x)A x=q(x+ x)A, (9)