c
d
2 u
dx
2
þ e
d
2
φ
dx
2
¼ 0,
e
d
2 u
dx
2
À ε
d
2
φ
dx
2
¼ q Àn þ N
þ
D
À
Á :
ð3:87Þ
Equation (3.87) can be rewritten as
d
2 u
dx
2
¼ À
e
c
d
2
φ
dx
2
,
n ¼
ε
T
33
q
d
2
φ
dx
2
þ N
þ
D :
ð3:88Þ
In the following we restrict ourselves to the case of uniform doping with a constant
N
þ
D . From the static form of Eq. (3.82), J
n is a constant. We consider an electrically
isolated rod with J
n
¼ 0 at its ends. Hence J
n is identically zero throughout the rod.
Then, from Eqs. (3.84) 2 and (3.85), we obtain
Ànμ
n dφ
dx
þ D
n dn
dx
¼ 0:
ð3:89Þ
In summary, we have three equations in Eqs. (3.88) and (3.89) for φ, n, and u. The
boundary conditions are
T ÆL
ð Þ ¼ f , D ÆL
ð Þ ¼ 0, J
n
ÆL
ð Þ ¼ 0:
ð3:90Þ
The charge neutrality condition is
Z L
ÀL
Àn þ N
þ
D
À
Á
dx ¼ 0,
ð3:91Þ
which in this case of electrons only without holes is not an independent condition
and is implied by Eq. (3.81) 2 and the boundary conditions on D in Eq. (3.90). It is
kept here formally only. To make the mechanical displacement and the electric
potential fields unique, we need to impose the following conditions:
u 0
ð Þ ¼ 0, φ 0
ð Þ ¼ 0:
ð3:92Þ
We look for a perturbation solution in the following form in terms of zero-, first-,
and second-order fields:
50
3 Extension of Rods