1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

190 Chapter 2 The Heat Equation

we must use an integral — the continuous analogue of a sum or series — to
include all possibilities. Thusushould have the form

u(x,t)=

∫∞

0

B(λ)sin(λx)exp

(

−λ^2 kt

)

dλ. (9)

(We need not include negative values ofλ. They give no new solutions.) The
initial condition will be satisfied ifB(λ)is chosen to make

u(x, 0 )=

∫∞

0

B(λ)sin(λx)dλ=f(x), 0 <x.

We recognize this as a Fourier integral;B(λ)is to be chosen as

B(λ)=^2

π

∫∞

0

f(x)sin(λx)dx. (10)

IfB(λ)exists, then Eq. (9) is the solution of the problem. Notice that when
t>0, the exponential function makes the improper integral in Eq. (9) con-
verge very rapidly.
Some care must be taken in the interpretation of our solution. If the rod re-
ally is finite (say, lengthL) the expression in Eq. (9) is, of course, meaningless
forxgreater thanL. The presence of a boundary condition atx=Lwould
influence temperatures nearby, so Eq. (9) can be considered a valid approxi-
mation only forx L.

Example.
Solve the problem in Eqs. (1)–(4) using the initial temperature distribution

f(x)=

{

T 0 , 0 <x<b,

0 , b<x.

This means that a section of lengthbat the left end of the rod starts out at
temperatureT 0 , different from the temperature of the long right end, which
is at the same temperature as the left boundary. (We assumeT 0 >0.) The
solution is given by Eq. (9), withB(λ)calculated from Eq. (10):

B(λ)=π^2

∫∞

0

f(x)sin(λx)dx

=π^2

∫b

0

T 0 sin(λx)dx

=^2 λπT^0

(

1 −cos(λb)

)

.