1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.11 Infinite Rod 195

Figure 9 Solution of example problem. Att=0, the temperature isT 0 >0for
−a<x<aand is 0 in the rest of the rod;u(x,t)is shown as a function ofx
on the interval− 3 a<x< 3 afor three times. The times are chosen so that the
dimensionless timekt/a^2 takes the values 0.01, 1, and 10 (to get a clear picture of
the changes inu). Note thatu(x,t)is positive everywhere for anyt>0. The values
T 0 =100 anda=1 have been used for convenience. Also see the CD.

In words, the rod has a center section of length 2awhose temperature is differ-
ent from that of the long sections to the left and right. We must compute the
coefficient functionsA(λ)andB(λ). The latter is identically 0 becausef(x)is
an even function; and

A(λ)=π^1

∫∞

−∞

f(x)cos(λx)dx

=^1

π

∫a

−a

T 0 cos(λx)dx

=

2 T 0

λπsin(λa).

Thus, the solution of the problem is

u(x,t)=

2 T 0

π

∫∞

0

sin(λa)

λ cos(λx)exp

(

−λ^2 kt

)

dλ. (6)

This function is graphed as a function ofxfor several values oftin Fig. 9
and animated on the CD. The figure suggests thatu(x,t)ispositivefor allx
whent>0. This is indeed true and illustrates an interesting property of the
solutions of the heat equation: the instantaneous transmission of information.
The“hot”sectionintheinterval−a<x<ainstantly raises the temperature
everywhere else from the initial value of 0 to a positive value.

Starting from the general form of a solution in Eq. (4), we can derive some
very interesting results. Change the variable of integration in Eq. (5) tox′and