146 Chapter 2 The Heat Equation

It is easy to see thatv(x)=T(any constant) is a solution to this problem.
However, there is no information to tell what valueTshould take. Thus, this
boundary value problem has infinitely many solutions.

It should not be supposed that every steady-state temperature distribution
has a straight-line graph. This is certainly not the case in the problem of Exer-
cise 1.
While the steady-state solution gives us some valuable information about
the solution of an initial value–boundary value problem, it also is important
as the first step in finding the complete solution. We now isolate the “rest” of
the unknown temperatureu(x,t)by defining thetransient temperature distri-
bution,

w(x,t)=u(x,t)−v(x).

The nametransientis appropriate because, according to our assumptions
about the behavior ofufor large values oft,weexpectw(x,t)to tend to zero
asttends to infinity.
In general, the transient also satisfies an initial value–boundary value prob-
lem that is similar to the original one but is distinguished by having a homo-
geneous partial differential equation and boundary conditions. To illustrate
this point, we shall treat the problem stated in Eqs. (1)–(4) whose steady-state
solution is given by Eq. (7).
By using the equalityu(x,t)=w(x,t)+v(x)and what we know aboutv—
that is, Eqs. (5) and (6) — we make the original problem foru(x,t)into a new
problem forw(x,t),asshowninwhatfollows.

∂^2 u

∂x^2 =

∂^2 w

∂x^2 +

d^2 v

dx^2 by rules of calculus,

=∂

(^2) w
∂x^2 because of Eq. (5),
∂u
∂t=
∂w
∂t +
dv
dt by rules of calculus,

∂w
∂t v(x)does not depend ont,
∂^2 w
∂x^2

=^1

k

∂w

∂t

by substituting into Eq. (1),

u( 0 ,t)=w( 0 ,t)+v( 0 ) by definition of the transient,

T 0 =w( 0 ,t)+T 0 from Eqs. (2) and (6),

u(a,t)=w(a,t)+v(a) by definition of the transient,

T 1 =w(a,t)+T 1 from Eqs. (3) and (6),

u(x, 0 )=w(x, 0 )+v(x) by definition of the transient,

f(x)=w(x, 0 )+T 0 +(T 1 −T 0 )x/a from Eqs. (4) and (7).