u Àa
À
ð
Þ¼u Àa
þ
ð
Þ, T Àa
À
ð
Þ¼T Àa
þ
ð
Þ,
φ Àa
À
ð
Þ¼φ Àa
þ
ð
Þ, D Àa
À
ð
Þ¼D Àa
þ
ð
Þ,
n Àa
À
ð
Þ¼n Àa
þ
ð
Þ, J
n
Àa
À
ð
Þ¼J
n
Àa
þ
ð
Þ,
ð7:37Þ
u a
À
ð Þ ¼ u a
þ
ð Þ, T a
À
ð Þ ¼ T a
þ
ð Þ,
φ a
À
ð Þ ¼ φ a
þ
ð Þ, D a
À
ð Þ ¼ D a
þ
ð Þ,
n a
À
ð Þ ¼ n a
þ
ð Þ, J
n a
À
ð Þ ¼ J
n a
þ
ð Þ,
ð7:38Þ
T þ1
ð
Þ ¼0, D þ1
ð
Þ ¼ 0, J
n
þ1
ð
Þ ¼ 0:
ð7:39Þ
To determine the displacement and potential fields uniquely, we choose À1 as a
reference and set
u À1
ð
Þ ¼ 0, φ À1
ð
Þ ¼ 0:
ð7:40Þ
The axial electric polarization P and the related distributed effective polarization
charge density can be calculated from
P ¼ D À ε 0 E,
ρ
P
¼ ÀP k,k ¼ À
dP
dx
:
ð7:41Þ
Mathematically, we need to find solutions for each of the three regions in x < Àa,
|x| < a, and x > a, respectively, and apply boundary and continuity conditions. For
the linear solution, in each region, we have a system of ordinary differential
equations with constant coefficients. The final solution can be found in a straightforward manner. For x < Àa, the results are
u ¼
e
c
eλ þ pc
ð
Þ θ
k e 2 þ cε
ð
Þ
sinh ka exp kx
ð Þ,
S ¼
e
c
eλ þ pc
ð
Þ θ
e 2 þ cε
sinh ka exp kx
ð Þ,
ð7:42Þ
φ ¼ À
eλ þ pc
ð
Þ θ
k e 2 þ cε
ð
Þ
sinh ka exp kx
ð Þ,
E ¼
eλ þ pc
ð
Þ θ
e 2 þ cε
ð
Þ
sinh ka exp kx
ð Þ,
D ¼
eλ
c
þ p
θ sinh ka exp kx
ð Þ,
P ¼
eλ
c
þ p
θ sinh ka 1 À
cε 0
e 2 þ cε
exp kx
ð Þ,
ρ
P
¼ Àk
eλ
c
þ p
θ sinh ka 1 À
cε 0
e 2 þ cε
exp kx
ð Þ,
ð7:43Þ
184
7 Thermal Effects