1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

2.5 Example: Different Boundary Conditions 163

12.This table gives values ofu( 0 ,t)for the functionufound in the example

and shown in Fig. 4. Make a graph ofu( 0 ,t)and describe the graph in

words.

kt/a^2 : 0. 001 0. 003 0. 01 0. 03 0. 1 0. 3 1

u( 0 ,t): 22. 9 24. 9 29. 0 35. 6 47. 9 58. 3 60. 0

13.Check that the partial differential equation and boundary conditions are

satisfied by the series in Eq. (9).

2.5 Example: Different Boundary Conditions

In many important cases, boundary conditions at the two endpoints will be
different kinds. In this section we shall solve the problem of finding the tem-
perature in a rod having one end insulated and the other held at a constant
temperature. The boundary value–initial value problem satisfied by the tem-
perature in the rod is

∂^2 u

∂x^2 =

1

k

∂u

∂t,^0 <x<a,^0 <t, (1)

u( 0 ,t)=T 0 , 0 <t, (2)

∂u

∂x(a,t)=^0 ,^0 <t, (3)

u(x, 0 )=f(x), 0 <x<a. (4)

It is easy to verify that the steady-state solution of this problem isv(x)=T 0.
Using this information, we can find the boundary value–initial value problem
satisfied by the transient temperaturew(x,t)=u(x,t)−T 0 :

∂^2 w

∂x^2

=^1

k

∂w

∂t

, 0 <x<a, 0 <t, (5)

w( 0 ,t)= 0 , ∂w

∂x

(a,t)= 0 , 0 <t, (6)

w(x, 0 )=f(x)−T 0 =g(x), 0 <x<a. (7)

Since this problem is homogeneous, we can attack it by the method of sep-
aration of variables. The assumption thatw(x,t)has the form of a product,
w(x,t)=φ(x)T(t), and insertion ofwin that form into the partial differential
equation (5) lead, as before, to

φ′′(x)

φ(x)

=T

′(t)

kT(t)

=constant. (8)