S 3 ¼
∂u 3
∂x 3
,
E 3 ¼ À
∂φ
∂x 3
:
ð3:6Þ
Consider the special case of a rod in static equilibrium without body force, i.e.,
f 3 ¼ 0. In this case, from Eq. (3.5) 1 ,
T 3 ¼ C 1 ,
ð3:7Þ
where C 1 is an integration constant. From Eq. (3.2) 1 ,
S 3 ¼
1
c 33
C 1 þ e 33 E 3
ð
Þ :
ð3:8Þ
Substituting Eq. (3.8) and Eq. (3.6) 2 into Eq. (3.2) 2 , we obtain
D 3 ¼
e 33
c 33
C 1 þ ε 33 þ
e
2
33
c 33
E 3 ¼
e 33
c 33
C 1 À ε 33 φ ,3 :
ð3:9Þ
For equilibrium, from Eq. (3.5) 3,4 , J
p
3 and J
n
3 are constants. We consider the case
when these constants are zero. Then
J
p
3 ¼ Àqpμ
p
33
∂φ
∂x 3
À qD
p
33
∂p
∂x 3
¼ 0,
J
n
3 ¼ Àqnμ
n
33
∂φ
∂x 3
þ qD
n
33
∂n
∂x 3
¼ 0:
ð3:10Þ
With the use of Eq. (1.4), we rewrite Eq. (3.10) as
1
p
∂p
∂x 3
¼ À
q
k B T
∂φ
∂x 3
,
1
n
∂n
∂x 3
¼
q
k B T
∂φ
∂x 3
:
ð3:11Þ
Equation (3.11) can be integrated to produce
p ¼ p
0 exp À
q
k B T
φ
,
n ¼ n
0 exp
q
k B T
φ
,
ð3:12Þ
where p
0 and n
0 are integration constants. Physically they are the values of p and n at
φ ¼ 0. We note that the definitions of p
0 and n
0 here in Eq. (3.12) are different from
3.1 One-Dimensional Equations
33