144 Chapter 2 The Heat Equation
Example.
For the preceding problem,v(x)should be the solution to the problem
d^2 v
dx^2 =^0 ,^0 <x<a, (5)
v( 0 )=T 0 ,v(a)=T 1. (6)On integrating the differential equation twice, we finddv
dx=A,v(x)=Ax+B.The constantsAandBaretobechosensothatv(x)satisfies the boundary
conditions:
v( 0 )=B=T 0 ,v(a)=Aa+B=T 1.When the two equations are solved forAandB, the steady-state distribution
becomes
v(x)=T 0 +(T 1 −T 0 )xa. (7)Of course, Eqs. (5) and (6), which together form the steady-state problem
corresponding to Eqs. (1)–(4), could have been derived from scratch, as was
done in Chapter 0, Section 3. Here, however, we see it as part of a more com-
prehensive problem.
We can establish this rule for setting up the steady-state problem corre-
sponding to a given heat conduction problem: Take limits in all equations that
are valid for larget(the partial differential equation and the boundary condi-
tions), replacinguand its derivatives with respect toxbyvand its derivatives,
and replacing∂u/∂tby 0.
Example.
Find the steady-state problem and solution of Eqs. (13)–(16) of Section 2.1,
which were
∂^2 u
∂x^2=^1
k∂u
∂t, 0 <x<a, 0 <t, (8)
u( 0 ,t)=T 0 , 0 <t, (9)−κ∂u
∂x(a,t)=h(
u(a,t)−T 1)
, 0 <t, (10)u(x, 0 )=f(x), 0 <x<a. (11)