30 Chapter 0 Ordinary Differential Equations
Figure 6 Section cut from heat-conducting cylinder showing heat flow.in whichAis the cross-sectional area andgis the rate at which heat enters the
slice by means other than conduction through the two faces. For instance, if
heat is generated in the slice by an electric currentI, we might have
g(x)A x=I^2 R x, (10)whereRis the resistance of the rod per unit length. If heat is lost through
the cylindrical surface of the rod by convection to a surrounding medium at
temperatureT,theng(x)would be given by “Newton’s law of cooling,”
g(x)A x=−h(u(x)−T)C x, (11)whereCis the circumference of the rod andhis the heat transfer coefficient.
(This minus sign appears because, ifu(x)>T, heat actually leaves the rod.)
Equation (9) may be altered algebraically to read
q(x+ x)−q(x)
x
=g(x),and application of the limiting process leaves
dq
dx=g(x). (12)The unknown functionu(x)does not appear in Eq. (12). However, a well-
known experimental law (Fourier’s law) says that the heat flow rate through a
unit area of material is directly proportional to the temperature difference and
inversely proportional to thickness. In the limit, this law takes the form
q=−κdu
dx. (13)The minus sign expresses the fact that heat moves from hotter toward cooler
regions.
Combining Eqs. (12) and (13) gives the differential equation
−κd(^2) u
dx^2
=g(x), 0 <x<a, (14)
whereais the length of the rod and the conductivityκis assumed to be con-
stant.