1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

320 Chapter 5 Higher Dimensions and Other Coordinates

n αn J 1 (αn) α^2

nJ 1 (αn)

1 2.405 + 0. 5191 + 1. 6020

2 5.520 − 0. 3403 − 1. 0647

3 8.654 + 0. 2715 + 0. 8512

4 11.792 − 0. 2325 − 0. 7295

Table 2 Values for Eq. (18)

Figure 8 Graphs of the solution of the example problem. The functionv(r,t)as
giveninEq.(18)isshownvsrfor times chosen so thatkt/a^2 takes the values 0,
0 .01, 0.1, and 0.5.T 0 =100 anda=1.

Putting together the numerator and denominator from Eqs. (15) and (16),
we find that the coefficients we need are

an=α^2 T^0

nJ 1 (αn)

. (17)

Thus, the solution to the heat conduction problem is

v(r,t)=T 0

∑∞

n= 1

2

αnJ 1 (αn)J^0 (λnr)exp

(

−λ^2 nkt

)

. (18)

InTable2arelistedthefirstfewvaluesoftheratio2/[αnJ 1 (αn)]. Figure 8
shows graphs ofv(r,t)as a function ofrfor several times. Also, see Exercise 1.
An animation is shown on the CD.