T ÀL
ð Þ ¼ 0, D ÀL
ð Þ ¼ 0,
J
p
ÀL
ð Þ ¼ 0, J
n
ÀL
ð Þ ¼ 0,
ð7:53Þ
u 0
À
ð Þ ¼ u 0
þ
ð Þ,
T 0
À
ð Þ ¼ T 0
þ
ð Þ,
φ 0
À
ð Þ ¼ φ 0
þ
ð Þ, D 0
À
ð Þ ¼ D 0
þ
ð Þ,
ð7:54Þ
p 0
À
ð Þ ¼ p 0
þ
ð Þ, J
p 0
À
ð Þ ¼ J
p 0
þ
ð Þ,
n 0
À
ð Þ ¼ n 0
þ
ð Þ, J
n 0
À
ð Þ ¼ J
n 0
þ
ð Þ,
ð7:55Þ
T L
ð Þ ¼ 0, D L
ð Þ ¼ 0,
J
p L
ð Þ ¼ 0, J
n L
ð Þ ¼ 0:
ð7:56Þ
Δp and Δn satisfy the following global charge neutrality conditions:
Z L
ÀL
Δpdx ¼ 0,
Z L
ÀL
Δndx ¼ 0:
ð7:57Þ
Only one of Eq. (7.57) 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:
ð7:58Þ
We need to find solutions in the two regions with –L < x < 0 and 0 < x < L separately
and apply boundary and continuity conditions.
For the usual built-in fields in the junction, we perform a linear analysis analytically using the linearized constitutive relations in Eq. (7.32). In each region we have
a system of 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 in the left half and a double prime for
those in the right half. The general solution for –L < x < 0 is
Δp À Δn ¼ A 1 sinh k
0 x À L
ð
ÞþB 1 sinh k
0 x þ L
ð
Þ,
ð7:59Þ
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 ,
ð7:60Þ
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 ,
ð7:61Þ
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
þC 5 x þ C 6 ,
ð7:62Þ
where A 1 , B 1 , and C 1 -C 6 are undetermined constants and
190
7 Thermal Effects