Z L
ÀL
Δpdx ¼ 0,
Z L
ÀL
Δndx ¼ 0:
ð3:200Þ
Only one of Eq. (3.200) is independent. To determine the displacement and potential
fields uniquely, we choose a reference point, e.g., x ¼ a, and set
u a
ð Þ ¼ 0, φ a
ð Þ ¼ 0:
ð3:201Þ
Mathematically, we need to find solutions in the two regions in –L separately and apply boundary and continuity conditions.
For the case of a small F and small carrier concentration perturbations, we
perform a theoretical analysis using the linearized constitutive relations for currents
in Eq. (3.193). In each region we have a system of linear ordinary differential
equations with constant coefficients. The procedure for finding a general solution
is straightforward. The junction may be heterogeneous. We use a prime for the
material parameters of the left half and a double prime for those of the right half. The
general solution for –L Δp À Δn ¼ A 1 sinh k
0 x À L
ð
ÞþB 1 sinh k
0 x þ L
ð
Þ,
ð3:202Þ
φ ¼ À
q
k
0
ð Þ
2 ε 0 T
33
A 1 sinh k
0 x À L
ð
ÞþB 1 sinh k
0 x þ L
ð
Þ
½
Š þ C 1 x þ C 2 ,
ð3:203Þ
u ¼
e
0
c 0
q
k
0
ð Þ
2 ε 0 T
33
A 1 sinh k
0 x À L
ð
ÞþB 1 sinh k
0 x þ L
ð
Þ
½
Š þ C 3 x þ C 4 ,
ð3:204Þ
Δp ¼
p
0
0 μ
0 p
D
0 p
q
k
0
ð Þ
2 ε 0 T
33
A 1 sinh k
0 x À L
ð
ÞþB 1 sinh k
0 x þ L
ð
Þ
½
Š þ C 5 x þ C 6 , ð3:205Þ
where A 1 , B 1 , and C 1 –C 6 are undetermined constants, and
k
0
ð Þ
2 ¼
p
0
0 μ
0 p
D
0 p þ
n
0
0 μ
0 n
D
0 n
q
ε 0 T
33
:
ð3:206Þ
Similarly, for 0 Δp À Δn ¼ A 2 sinh k
00 x À L
ð
ÞþB 2 sinh k
00 x þ L
ð
Þ,
ð3:207Þ
φ ¼ À
q
k
00
ð Þ
2 ε 00 T
33
A 1 sinh k
00 x À L
ð
ÞþB 1 sinh k
00 x þ L
ð
Þ
½
Š þ C 7 x þ C 8 , ð3:208Þ
u ¼
e
00
c 00
q
k
00
ð Þ
2 ε 00 T
33
A 2 sinh k
00 x À L
ð
ÞþB 2 sinh k
00 x þ L
ð
Þ
½
Š þ C 9 x þ C 10 , ð3:209Þ
78
3 Extension of Rods