Вычислительные технологии
Том 4, № 3, 1999
ABOUT STABILITY OF THE EQUILIBRIUM STATE
FOR A HYDRODYNAMICAL MODEL OF CHARGE
TRANSPORT IN SEMICONDUCTORS ∗
A. M. Blokhin
Institute of Mathematics SB RAS, Novosibirsk, Russia
e-mail: [email protected]
A. S. Bushmanova
Novosibirsk State University, Novosibirsk, Russia
V. Romano
Politecnico di Bari, sede di Taranto, Taranto, Italia
Доказана устойчивость и, при некоторых дополнительных предположениях на
плотность легирования, асимптотическая устойчивость (по Ляпунову) состояния равновесия гидродинамической модели переноса заряда в полупроводниках в линейном
приближении.
1. Preliminaries
Hydrodynamical models are widely used in mathematical modelling of physical phenomena
in modern semiconductor devices. Derivation of such models is based on the study of the
transport equation for the charge carriers density in an electric field. A conservation system with
infinite number of equations (i. e., a system of conservation laws) is obtained from the transport
equation with the help of a special technique of moments. Then, in a view of some physically
justified assumptions, the system is exposed to simplification. As the result, the system of
infinite number of moments equations reduces to one or another system of the hydrodynamical
type.
In this paper we take, as a basis, a hydrodynamical model of charge transport in semiconductors
suggested in the recently published paper [1]. In [2], this model is written as a quasilinear system
of dimensionless conservation laws. Using notation from [2], below we give a nondivergent
variant of this model
Rτ + uRs + Rus = 0,
ϑ
1
u
uτ + uus + Rs + ϑs + Σs = Q − ,
R
R
τp
¶
¾
µ
¶
½ µ
3(1 − ϑ)
2
2
1
Σ
2
1
2
+
,
θs =
−
ϑτ + uϑs +
ϑ+
us +
u
3
R
3R
3
τp 2τw
2τw
This work was supported by Russian Foundation for Basic Research, Grant No. 96–01–01560.
c A. M. Blokhin, A. S. Bushmanova, V. Romano, 1999.
°
∗
16
17
ABOUT STABILITY OF THE EQUILIBRIUM STATE
4Rϑ + 7Σ
8
4Ru2
Σ
us + θ s =
− ,
3
15
3τp
τσ
µ
¶
5
Σϑ
Σ
θ
16
θτ + uθs + (Rϑ + Σ)ϑs −
Rs + ϑ −
Σs + θus = − +
2
R
R
5
τq
Στ + uΣs +
+Σu
µ
1
1
1
+ −
τσ τp τq
¶
5
+ Ruϑ
2
µ
1
1
−
τp τq
¶
Ru3
+
2
µ
1
3
5
− −
3τw τq τp
¶
5
1−ϑ
− Ru
,
2
τw
ε2 ϕss = R − ρ.
(1.1)
(1.2)
Here R, u, ϑ, Σ, θ, ϕ are the density, velocity, temperature, stress, heat flux and electric
potential correspondingly; ε2 = 1/β, β is a positive constant (see [2]); Q = ϕs ; τp = τp (E), τw =
3
u2
+ ϑ; the doping density
τw (E), τσ = τσ (E), τq = τq (E) are the relaxation times; E =
2
2
ρ = ρ(s) is a function given on [0, 1]. In the sequel we will assume that the function (ρ(s) − 1)
is sufficiently smooth and finite, and
1 ≥ ρ(s) ≥ δ > 0,
s ∈ [0, 1].
A typical profile of the function taken in [2] is of the form:
Following [2], we will consider the well-known in physics of semiconductors problem on
ballistic diode and formulate boundary conditions at s = 0, 1 (see. [3, 4]):
R(τ, 0) = R(τ, 1) = ϑ(τ, 0) = ϑ(τ, 1) = 1, Σ(τ, 1) = 0,
ϕ(τ, 0) = A, ϕ(τ, 1) = A + B,
(1.3)
(1.4)
where A, B are constants and the bias across the diode B > 0. Without the loss of generality,
we assume that A = 0. Finally, as usual, at τ = 0 we formulate initial data.
Following [2], we give an equivalent formulation of mixed problem (1.1) – (1.4). We will
consider system (1.1) coupled with the relation
2
ε Qτ =
Z1
R(τ, ξ)u(τ, ξ) dξ − Ru
(1.20 )
0
instead of the Poisson equation (1.2). Equation (1.2) rewritten in the form
ε2 Qs = R − ρ
(1.200 )
18
A. M. Blokhin, A. S. Bushmanova, V. Romano
will be treated as an additional stationary law which the initial data, in particular, must meet.
From boundary conditions (1.4) it follows that the relation
Z1
(1.5)
Q(τ, ξ) dξ = B
0
is fulfilled and the initial data also satisfy this relation. The electric potential ϕ = ϕ(τ, s) is
found from the evident equality
Zs
ϕ(τ, s) = Q(τ, ξ) dξ.
(1.6)
0
Thus, instead of mixed problem (1.1) – (1.4) one can consider problem (1.1), (1.20 ), (1.3)
with additional requirements (1.200 ), (1.5), which actually are requirements on the initial data.
It is easily shown that these two formulations are equivalent, at least on smooth solutions.
Problem (1.1) – (1.4) has a stationary solution (the equilibrium state) at B = 0:
u(τ, s) = û = 0,
ϑ(τ, s) = ϑ̂ = 1,
Σ(τ, s) = Σ̂ = 0,
θ(τ, s) = θ̂ = 0,
R(τ, s) = R̂(s) = eϕ̂(s) ,
ϕ(τ, s) = ϕ̂(s),
(1.7)
where ϕ̂(s) satisfies the Poisson equation
ε2 ϕ̂00 = R̂ − ρ
(1.8)
ϕ̂(0) = ϕ̂(1) = 0.
(1.9)
with the boundary conditions
It is obvious that, at small ε, the solutions to boundary value problem (1.8), (1.9) can be
presented as
ϕ̂(s) = ln ρ(s) + O(ε).
(1.10)
By this remark, we will assume in the sequel that the function ϕ̂(s) is sufficiently smooth and
finite.
Remark 1.1. Let B = 0. From certain physical considerations (see [1]), the solution of
(1.1) – (1.4) tends to the equilibrium state at τ → ∞ for any initial data, i. e.,
u(τ, s) → 0, ϑ(τ, s) → 1, Σ(τ, s) → 0,
θ(τ, s) → 0,
R(τ, s) → R̂(s),
ϕ(τ, s) → ϕ̂(s).
Below we will prove this fact for a linearized mixed problem (though at a certain, very essential,
restriction on the doping density ρ(s)).
Remark 1.2. Note that we can consider as the equilibrium state the following functions:
û = 0, ϑ̂ = 1, Σ̂ = 0,
θ̂ = 0, R̂(s) = ρ(s), ϕ(τ, s) = ln ρ(s),
that are obviously solution of the system (1.1) and the Poisson equation
ε2 ϕ̂00 = R̂ − ρ̃,
where
ρ̃ = ρ − ε2 (ln ρ)00 .
It is evident that functions ρ and ρ̃ have the same profile with only the small difference on
1
1
5
5
some intervals − ∆ < s < + ∆, − ∆ < s < + ∆.
6
6
6
6
ABOUT STABILITY OF THE EQUILIBRIUM STATE
19
2. Linearization of mixed problem (1.1) – (1.4)
Let linearize original quasilinear problem (1.1) – (1.4) with respect to the equilibrium state
(1.7). As the result, omitting intermediate, quite cumbersome calculations, we come to a linear
system (small perturbations of the sought values are denoted with the same symbols):
rτ + us + ϕ̂0 u = 0,
uτ + rs + ϑs + σs + ϕ̂0 ϑ + ϕ̂0 σ + µu = Q,
3
3
ϑτ + us + qs + ϕ̂0 q + νϑ = 0,
2
2
2
2
3
3
στ + us + qs + ϕ̂0 q + χσ = 0,
4
5
5
4
2
2
2
qτ + ϑs + σs + γq = µ̄u,
5
5
5
(2.1)
Z1
(2.2)
ε2 Qτ = −R̂u +
R̂(ξ)u(τ, ξ) dξ.
0
1
1
1
R
Σ
θ
1
> 0, ν =
> 0, χ =
> 0, γ =
> 0, µ̄ = µ − γ,
Here r = , σ = , q = , µ =
τ̂
τ̂
τ̂
τ̂
p
w
σ
q
R̂
R̂
R̂
µ ¶
µ ¶
µ ¶
µ ¶
3
3
3
3
τ̂p = τp
, τ̂w = τw
, τ̂σ = τσ
, τ̂q = τq
. It follows from [2] that the constants
2
2
2
2
µ, ν, χ, γ, β are large parameters. Boundary conditions to system (2.1) become of the form
r(τ, 0) = r(τ, 1) = ϑ(τ, 0) = ϑ(τ, 1) = σ(τ, 1) = 0.
(2.3)
Linearization of relations (1.200 ), (1.5), (1.6) results in the following conditions
ε2 Qs = R̂r,
Z1
Q(τ, ξ) dξ = 0,
0
ϕ(τ, s) =
Zs
Q(τ, ξ) dξ, i. e., ϕs = Q.
(2.4)
0
The aim of this paper is the study of stability (by Lyapunov) of the trivial solution to linear
problem (2.1) – (2.3).
In the next section we will derive an a priori estimate which implies stability of the trivial
solution. Besides, in the last section we will prove asymptotic stability of this solution under
an additional condition on the function ρ(s).
We rewrite system (2.1) in the vector form
AUτ + BUs + DU = F ,
(2.10 )
20
A. M. Blokhin, A. S. Bushmanova, V. Romano
where


r
 u 
 

U =
 ϑ ,
 σ 
q

0 0
 0 µ


 0 0
D=

 0 0


0 0

1 0 0 0 0
 0 1 0 0 0


3
 0 0
0 0

2
A=
 0 0 0 3 0

4

2
0 0 0 0
5

0
0
0

0
0
0
0 



3ν
 0
0
0 
 + ϕ̂0 
2
 0

3χ

0 
0
 0

4

2γ
0
0
0
5





,





0 1 0 0
1 0 1 1
0 1 0 0




B=
 0 1 0 0


2
0 0 1
5

1 0 0 0
0 1 1 0 

0 0 0 1 
,
2 
0 0 0

5
0 0 0 0



F =


0
0
1
2
5
0





,



0
Q
0
0
µ̄u



.


System (2.10 ) is symmetric t-hyperbolic (by Friedrichs) (see [5]). The matrix B has two
positive, two negative and one zero eigenvalues. This means (see [5]) that one boundary
condition in (2.3) is redundant. In fact, condition σ(τ, 1) = 0 is automatically fulfilled if the
initial function σ0 (s) = σ(0, s) possesses the property
σ0 (1) = 0.
Indeed, from boundary conditions
r(τ, 0) = r(τ, 1) = ϑ(τ, 0) = ϑ(τ, 1) = 0
and from the first and the third equations of system (2.1) it follows that
us (τ, 0) = us (τ, 1) = qs (τ, 0) = qs (τ, 1) = 0.
(2.5)
While deriving (2.5), we took into account that ϕ̂(s) is a sufficiently smooth and finite function.
By (2.5), from the fourth equation of system (2.1) we obtain
στ (τ, 0) + χσ(τ, 0) = 0,
στ (τ, 1) + χσ(τ, 1) = 0
i. e.,
σ(τ, 0) = e−χτ σ0 (0),
σ(τ, 1) = e−χτ σ0 (1).
Thus, boundary condition σ(τ, 1) = 0 is fulfilled if σ0 (1) = 0. Later we will suppose also that
σ0 (0) = 0, i. e.,
σ(τ, 0) = 0.
(2.6)
The energy integral identity for system (2.10 ) in the differential form is (see [5]):
4
3
{R̂(AU, U )}τ + {R̂(BU, U )}s + R̂(2µu2 − 2µ̄uq + γq 2 + 3νϑ2 + χσ 2 ) = 2R̂uQ,
5
2
(2.7)
21
ABOUT STABILITY OF THE EQUILIBRIUM STATE
2
where (BU, U ) = 2(ur + uϑ + uσ + ϑq + σq), and, in a view of boundary conditions (2.3),
5
(2.8)
(BU, U )|s=0,1 = 0.
Integrating (2.7) by s from 0 to 1 (with account of (2.8)), we come to the energy integral
identity in the integral form for system (2.10 ):
d
||U (τ )||2A + 2
dτ
Z1
2γ
3ν
3χ 2
R̂(µu − µ̄uq + q 2 + ϑ2 +
σ ) dξ = 2
5
2
4
2
R1
R̂uQ dξ.
(2.9)
0
0
Here ||U (τ )||2A =
Z1
R̂(AU, U ) dξ.
0
Multiplying (2.2) by 2Q, integrating the obtained expression by s from 0 to 1 (accounting
R1
the relation Q(τ, ξ) dξ = 0, see (2.4)) and summing it up with (2.9), we finally have:
0
d (0)
I (τ ) + 2
dτ
Z1
R̂(µu2 − µ̄uq +
2γ 2 3ν 2 3χ 2
q + ϑ +
σ ) dξ = 0,
5
2
4
(2.10)
0
where I (0) (τ ) = ||U (τ )||2A + ε2
R1
Q2 (τ, ξ) dξ.
0
Fulfillment of (2.10) means well-posedness of linear mixed problem (2.1) – (2.3) since (2.10)
implies the a priori estimate:
d (0)
I (τ ) ≤ 0,
dτ
i. e., I (0) (τ ) ≤ I (0) (0),
τ >0
(2.11)
provided that
|µ̄| ≤
r
8
µγ.
5
(2.12)
In following we will assume that inequality (2.12) is fulfilled.
A priori estimate (2.11) means that
U (τ, s) ∈ L2 (0, 1),
Q(τ, s) ∈ L2 (0, 1)
at every τ > 0. From (2.4) it follows that
||ϕ(τ )||L2 (0,1) ≤ ||Q(τ )||L2 (0,1) ,
◦
i. e., ϕ(τ, s) ∈W21 (0, 1) at every τ > 0, and
||ϕ(τ )||
◦
W21 (0,1)
≤ ||Q(τ )||L2 (0,1) .
A priori estimate (2.11) also implies stability of the trivial solution to linear mixed problem
(2.1) – (2.3). In the subsequent sections we will obtain more delicate a priori estimates which
22
A. M. Blokhin, A. S. Bushmanova, V. Romano
allow to prove existence of a smooth solution to linear mixed problem (2.1) – (2.3) for all τ > 0
and stability and asymptotic stability (by Lyapunov) of its trivial solution.
Remark 2.1. Indeed, from (2.4), it follows that
Q(τ, s) ∈ W21 (0, 1),
◦
ϕ(τ, s) ∈W22 (0, 1)
for all τ > 0.
Remark 2.2. Treating the Poisson equation (see (2.4))
ϕss = β R̂r,
as an ordinary differential equation for the unknown function ϕ with the boundary conditions
ϕ(τ, 0) = ϕ(τ, 1) = 0,
we arrive at
ϕ(τ, s) = β
Z1
g(s, ξ)R̂(ξ)r(τ, ξ) dξ
(2.13)
0
and
Q(τ, s) = ϕs (τ, s) = β
Zs
R̂(ξ)r(τ, ξ) dξ − β
0
Z1
(1 − ξ)R̂(ξ)r(τ, ξ) dξ.
(2.14)
0
Here g(s, ξ) is the Green function
g(s, ξ) =
½
ξ(s − 1) for 0 < ξ ≤ s,
s(ξ − 1) for s < ξ < 1.
It is seen that the functions ϕ, Q, given by (2.13), (2.14), satisfy conditions (2.4).
Remark 2.3. Differentiate (2.10 ) by τ . Then, for Uτ , we have the system
A(Uτ )τ + B(Uτ )s + DUτ = F τ ,
(2.15)
where
F τ = (0, Qτ , 0, 0, µ̄uτ )∗
∗ stands for transposition.
For system (2.15) it is easy to derive the energy integral identity in the integral form:
d (1)
I (τ ) + 2
dτ
Z1
R̂(µu2τ − µ̄uτ qτ +
0
where I (1) (τ ) = ||Uτ (τ )||2A + ε2
R1
0
Q2τ (τ, ξ) dξ.
2γ 2 3ν 2 3χ 2
q + ϑτ +
σ ) dξ = 0,
5 τ
2
4 τ
(2.16)
23
ABOUT STABILITY OF THE EQUILIBRIUM STATE
The a priori estimate follows from (2.16)
I (1) (τ ) ≤ I (1) (0),
(2.17)
τ > 0.
Differentiating system (2.15) once again by τ , we finally arrive at the estimate of the form:
I (2) (τ ) ≤ I (2) (0),
R1
Here I (2) (τ ) = ||Uτ τ (τ )||2A + ε2
0
etc.
τ >0
(2.18)
Q2τ τ (τ, ξ) dξ.
Remark 2.4. Let introduce a value
Z1
t2 = ||U0 ||2W 2 (0,1) =
2
R̂[(U0 , U0 ) + (U00 , U00 ) + (U000 , U000 )] dξ,
0
i. e., the squared norm of the vector of initial data U0 (s) = U (0, s) = (r0 (s), u0 (s), ϑ0 (s), σ0 (s),
q0 (s))* in the space W22 (0, 1).
We can prove the following estimates
Z1
Z1
Q2 (0, ξ) dξ ≤ K1 t2 ,
Q2τ (0, ξ) dξ ≤ K2 t2 ,
Q2τ τ (0, ξ) dξ ≤ K3 t2 ,
(2.19)
0
0
0
Z1
where K1 , K2 , K3 are the positive constants.
Indeed, by the formula (2.14) we have
|Q(0, s)|2 = |β
≤ [2β
R1
0
≤ 4β
Rs
0
R1
0
R1
R̂(ξ)|r(0, ξ)| dξ]2 = 4β 2 ( R̂1/2 |r0 |R̂1/2 dξ)2 ≤
2
R1
R̂ dξ
0
where K0 =
R1
R̂(ξ)r(0, ξ) dξ − β (1 − ξ)R̂(ξ)r(0, ξ) dξ|2 ≤
R1
0
0
R̂r02
2
dξ = 4β K0
R1
0
R̂r02 dξ ≤ K1 t2 ,
R̂ dξ > 0, K1 > 0. Thus, the first of estimates (2.19) is fulfilled. Later, since
0
Rs
R1
|Qτ (0, s)| ≤ β(| R̂(ξ)rτ (0, ξ) dξ| + | (1 − ξ)R̂(ξ)rτ (0, ξ) dξ| ≤
≤ 2β
R1
0
0
0
R1
R1
R̂(ξ)|rτ (0, ξ)| dξ ≤ 2β( R̂ dξ)1/2 ( R̂rτ2 dξ)1/2 ,
0
0
than, if the inequality
Z1
R̂(ξ)rτ2 (0, ξ) dξ ≤ K4 t2 ,
0
is held, than the estimate
Z1
0
Q2τ (0, ξ) dξ ≤ K2 t2
K4 > 0
(2.20)
24
A. M. Blokhin, A. S. Bushmanova, V. Romano
took place.
By the analogy, under condition
Z1
R̂(ξ)rτ2τ (0, ξ) dξ ≤ K5 t2 ,
(2.21)
K5 > 0
0
we can show correctness of the third of estimates (2.19). The proof of inequalities (2.20), (2.21)
is given in section 3.
Finally note that
Z1
R̂Q2s (0, ξ) dξ
≤
Z1
2
2 2
2
2
R̂β R̂ r (0, ξ) dξ ≤ β max |R̂ (s)|
s∈[0,1]
R̂r02 dξ ≤ K6 t2 .
(2.22)
0
0
0
Z1
Remark 2.5. Unfortunately, we did not succeed in obtaining the additional entropy conservation
law for system (1.1). Such a law is easily derived for linearized system (2.1) by exclusion of the
derivatives of us , qs from the first, third, fourth equations of system (2.1):
5
5
(r + ϑ − σ)τ + ϕ̂0 u + νϑ − χσ = 0.
4
4
(2.23)
Remark 2.6. We will say that the trivial solution of linear problem (2.1)–(2.3) is stable, if
for any ε̂ > 0 there exists δ̂(ε̂) > 0 such that from the inequality
||U0 || ≤ δ̂
will follow the inequality
||U (τ )|| ≤ ε̂
for all τ > 0.
Remark 2.7. We will say that the trivial solution to linear problem (2.1) – (2.3) is asymptotically
stable, if, at an arbitrary initial data U0 (s) from a Sobolev space, the solution U (τ, s) tends to
zero at τ → ∞, also in the Sobolev space.
3. Stability of trivial solution to problem (2.1) – (2.3)
Since we assume that the function ϕ̂(s) is sufficiently smooth (see section 1), henceforth we
can suggest that |ϕ̂0 (s)|, |ϕ̂00 (s)| < C0 , where C0 is the positive constant.
From the first equation of system (2.1) it follows that
rτ2 = u2s + (ϕ̂0 )2 u2 + 2us ϕ̂0 u ≤ 2(u2s + C02 u2 ),
so
Z1
0
R̂(ξ)rτ2 (0, ξ) dξ ≤ K4 t2 .
25
ABOUT STABILITY OF THE EQUILIBRIUM STATE
Remark 3.1. The constants K1 , K2 , K3 , K4 (as well as other positive constants Ki , i =
5, . . . , 7, Nj , j = 1, . . . , 3, Mk , k = 1, . . . , 18, which appear in the sequel in this section) are
determined finally via the constants µ, ν, χ, γ, β and the function ρ(s).
Let us differentiate the second equation of (2.1) by s:
uτ s = Qs − (rss + ϑss + σss + ϕ̂0 (ϑs + σs ) + µus + ϕ̂00 (ϑ + σ)).
Then, with account to relation (2.22) it yields the inequality
Z1
R̂(ξ)u2τ s (0, ξ) dξ ≤ K7 t2 .
0
Now, differentiating the first equation of (2.1) by τ :
rτ τ + uτ s + ϕ̂0 uτ = 0,
we arrive at the estimate
Z1
R̂(ξ)rτ2τ (0, ξ) dξ ≤ K5 t2 .
0
Thus the inequalities (2.20), (2.21) from Remark 2.4 are valid, so the estimates (2.19) are
proved.
Note that
(AUτ , Uτ ) ≤ N1 {(U, U ) + (Us , Us ) + Q2 }
(3.1)
since at the desired functions in (2.1) we have either the constants or the known bounded
functions.
From the differentiated by s system (2.1) we obtain
(AUτ s , Uτ s ) ≤ N2 {(U, U ) + (Us , Us ) + (Uss , Uss )}.
And differentiating (2.1) by τ , we finally have
(AUτ τ , Uτ τ ) ≤ N2 {(U, U ) + (Us , Us ) + (Uss , Uss ) + Q2 + Q2τ }.
(3.2)
Gathering estimates (2.11), (2.17), (2.18), (2.19), (3.1) and (3.2), we can write that
Z1
R̂[r2 (τ, ξ) + u2 (τ, ξ) + ϑ2 (τ, ξ) + σ 2 (τ, ξ) + q 2 (τ, ξ)+
0
+rτ2 (τ, ξ) + u2τ (τ, ξ) + ϑ2τ (τ, ξ) + στ2 (τ, ξ) + qτ2 (τ, ξ)+
+rτ2τ (τ, ξ) + u2τ τ (τ, ξ) + ϑ2τ τ (τ, ξ) + στ2τ (τ, ξ) + qτ2τ (τ, ξ)] dξ ≤ M1 t2 ,
τ > 0.
(3.3)
Completing the derivation of an a priory estimate, we deduce some auxiliary inequalities.
First, we will derive estimates for the first derivatives by s of the desired functions. From the
first equation of system (2.1) it follows that
u2s ≤ rτ2 + C02 u2 ,
26
A. M. Blokhin, A. S. Bushmanova, V. Romano
so
Z1
R̂(ξ)u2s (τ, ξ) dξ ≤ M2 t2 ,
τ > 0.
(3.4)
0
Evaluate the derivative qs with the help of the third equation of system (2.1):
µ
¶
3
3ν
0
qs = −
ϑτ + us + ϕ̂ q + ϑ ,
2
2
i. e.,
Z1
R̂qs2 dξ ≤ M3 t2 ,
τ > 0.
(3.5)
0
While deriving similar results for ϑs , σs , rs it is necessary to have estimates for derivatives
uτ s , rτ s , ϑτ s , στ s . Differentiating the first equation of (2.1) by τ , we will have
rτ τ + uτ s + ϕ̂0 uτ = 0.
Whence we easily obtain
Z1
R̂u2τ s dξ ≤ M4 t2 ,
τ > 0.
(3.6)
0
Now let us use the second and fifth equation of system (2.1) differentiated by τ :
rτ s + ϑτ s + στ s = F1 = Qτ − uτ τ − ϕ̂0 ϑτ − ϕ̂0 στ − µuτ ,
(3.7)
2
2γ
2
ϑτ s + στ s = F2 = µ̄uτ − qτ τ − qτ ,
5
5
5
(3.8)
with the help of (3.7) and (3.8) we transform the additional relation (2.23) into:
µ
¶
¶
µ
5
5
rτ s + ϑτ s − στ s + ν̄ rτ s + ϑτ s − στ s = F3 = ν̄F1 − νF2 − ϕ̂0 uτ s − ϕ̂00 uτ ,
(3.9)
4
4
τ
µ
¶
5
8
5
where ν̄ = χ + ν. Multiplying (3.9) by 2R̂ rτ s + ϑτ s − στ s and integrating by s from 0
9
45
4
to 1, we get
 1

µ
¶2 
¶2
Z
Z1 µ

5
5
∂
R̂ rτ s + ϑτ s − στ s
dξ + (2ν̄ − 1) R̂ rτ s + ϑτ s − στ s
dξ ≤ M5 t2 ,

∂τ 
4
4
0
then for ν̄ >
0
1
we have
2
Z1
0
µ
5
R̂ rτ s + ϑτ s − στ s
4
¶2
dξ ≤ M6 t2 .
27
ABOUT STABILITY OF THE EQUILIBRIUM STATE
Since from (3.7) it follows that
Z1
R̂ (rτ s + ϑτ s + στ s )2 dξ ≤ M7 t2 ,
0
than we receive
Z1
R̂στ2s dξ ≤ M8 t2 ,
0
and after that from (3.8) and (3.7) we sequentially deduce analogous estimates for ϑ2τ s and rτ2s .
Finally
Z1
R̂[rτ2s + ϑ2τ s + στ2s ] dξ ≤ M9 t2 ,
τ > 0.
(3.10)
0
Later, from the first equation of system (2.1), we obtain
uss = −rτ s − (ϕ̂0 u)s ,
i. e.,
Z1
R̂u2ss dξ ≤ M10 t2 ,
τ > 0.
(3.11)
0
In order to evaluate the derivative qss , we differentiate by s the third and the fourth
equations of system (3.1). From these equations and from the last equation of system (3.1), we
find
µ
¶
¶
µ
¶
µ
2
1
2
2
16
2
16
16
2
2
qss = − ϑτ s − σss −
uss −
(ϕ̂0 u)s + qτ ++ γq − µ̄u.
+
+
+
3ν 75χ
ν
5χ
3ν 75χ
3ν 75χ
5
5
So, if we assume that
0 < ν0 < ν,
0 < χ0 < χ,
than we have
Z1
2
R̂qss
dξ ≤ M11 t2 ,
τ > 0.
(3.12)
0
To evaluate the derivatives ϑs , σs , we differentiate the third and forth equations of system
(2.1) by s and multiply them by 2R̂ϑs and 2R̂σs , correspondingly. Finally, we have the
inequalities:

 1
Z1
Z


3 ∂
2
R̂ϑs dξ + (3ν − 1) R̂ϑ2s dξ ≤ M12 t2 ,

2 ∂τ 
0
0
28
A. M. Blokhin, A. S. Bushmanova, V. Romano
 1
Z
i. e.,
3 ∂
4 ∂τ 
R̂σs2 dξ

0
Z1


+
µ
¶ Z1
3χ
−1
R̂σs2 dξ ≤ M13 t2 ,
2
0
R̂[ϑ2s + σs2 ] dξ ≤ M14 t2 ,
τ > 0.
(3.13)
0
1
2
and χ > that is true (see [2]).
3
3
From the second equation of system (2.1) it follows that
While deriving (3.13) we assumed that ν >
rs = F4 = Q − (uτ + ϑs + σs + ϕ̂0 ϑ + ϕ̂0 σ + µu),
so
Z1
R̂rs2 dξ ≤ M15 t2 ,
(3.14)
τ > 0.
0
From the third equation of (2.1) differentiated by τ it follows that
9ν 2 2
9
ϑ ,
qτ2s ≤ ϑ2τ τ + u2τ s + C02 qτ2 +
3
4 τ
i. e.,
Z1
R̂qτ2s dξ ≤ M16 t2 ,
(3.15)
τ > 0.
0
Finally to estimate the derivatives ϑss , rss , σss we will use again the method we have used
while deriving the derivatives ϑτ s , rτ s , στ s . To this end we differentiate the second and fifth
equations of system (2.1) by s:
rss + ϑss + σss = G1 = β R̂r − uτ s − (ϕ̂0 ϑ)s − (ϕ̂0 σ)s − µus ,
2
2
2γ
ϑss + σss = G2 = µ̄us − qτ s − qs ,
5
5
5
and (2.23) by s twice:
µ
µ
¶
¶
5
5
11
11
rss + ϑss − σss + ν rss + ϑss − σss = G3 = −νG2 + νG1 − (ϕ̂0 u)ss .
4
15
4
15
τ
(3.16)
(3.17)
(3.18)
Then it follows from (3.16), (3.17), (3.18) that
Z1
0
2
2
R̂[rss
+ ϑ2ss + σss
] dξ ≤ M11 t2 ,
τ > 0.
(3.19)
29
ABOUT STABILITY OF THE EQUILIBRIUM STATE
Gathering estimates (3.3) – (3.6), (3.10) – (3.15), (3.19), we can write down the desired a
priori estimate:
Z1
R̂ [(U, U )+(Uτ , Uτ )+(Us , Us )+(Uτ τ , Uτ τ )+(Uτ s , Uτ s ) + (Uss , Uss )] dξ ≤ M12 t2 , τ > 0. (3.20)
0
From (3.20) it follows that
◦
U (τ, s) ∈ W22 (0, 1),
Q(τ, s) ∈ W23 (0, 1),
ϕ(τ, s) ∈W24 (0, 1) for all τ > 0.
It also implies stability (by Lyapunov) of the equilibrium state in the linear approximation.
Remark 3.2. More exactly,
◦
r(τ, s), ϑ(τ, s), σ(τ, s) ∈W22 (0, 1).
Besides,
||U (τ )||2W 2 (0,1) ≤ M12 t2 ,
τ > 0,
2
which means (see Remark 2.5) stability (by Lyapunov) of the trivial solution to mixed problem
(2.1) – (2.3). Here
||U (τ )||2W 2 (0,1)
2
=
Z1
R̂[(U, U ) + (Us , Us ) + (Uss , Uss )] dξ.
0
4. Asymptotic stability of trivial solution
Evidently, the first equation of system (2.1) can be rewritten as
(R̂r)τ + (R̂u)s = 0.
It is expedient to introduce into consideration a potential Ψ = Ψ(τ, s) such that
R̂u = Ψτ ,
(4.1)
R̂r = −Ψs .
First two boundary conditions in (2.3) imply
Ψs (τ, 0) = Ψs (τ, 1) = 0.
(4.2)
The remained conditions in (2.3) and assumption (2.6) give
(R̂ϑ)(τ, 0) = (R̂ϑ)(τ, 1) = 0,
(R̂σ)(τ, 0) = (R̂σ)(τ, 1) = 0.
Note that equation (2.2) with regard to (4.1) transforms into


Z1
ε2 Q + Ψ − Ψ dξ  = 0,
0
τ
(4.3)
(4.4)
30
A. M. Blokhin, A. S. Bushmanova, V. Romano
i. e.,
2
ε Q+Ψ−
Z1
Ψ dξ = A0 (s),
0
where A0 (s) is an arbitrary function. On the other hand, by (2.4), we have
ε2 Qs = R̂r,
i. e., A00 (s) = 0 and A0 = const;
R1
since Q dξ = 0, A0 = 0.
0
So,

Q=β
Z1
0

Ψ dξ − Ψ = βh(τ, s).
(4.5)
Rewrite the second condition of system (2.1) as
(R̂u)τ + (R̂r)s − ϕ̂0 R̂r + (R̂ϑ)s + (R̂σ)s + µR̂u = R̂Q,
and then, with the use of relations (4.1), in the form
Ψτ τ − Ψss + (R̂ϑ)s + (R̂σ)s + µΨτ + ϕ̂0 Ψs = R̂βh.
(4.6)
By the analogy, the third, fourth, fifth equations from (2.1) can be rewritten as follows:
3
3ν
(R̂ϑ)τ + Ψτ s − ϕ̂0 Ψτ + (R̂q)s + (R̂ϑ) = 0,
2
2
(4.7)
2
3χ
3
(R̂σ)τ + Ψτ s − ϕ̂0 Ψτ + (R̂q)s +
(R̂σ) = 0,
4
5
4
(4.8)
2
2
2γ
2
(R̂q)τ + (R̂ϑ)s − ϕ̂0 (R̂ϑ) + (R̂σ)s − ϕ̂0 (R̂σ) + (R̂q) − µ̄Ψτ = 0.
5
5
5
5
(4.9)
Later we will use equations obtained from (4.6) – (4.9) by differentiating by s:
Hτ τ − Hss + (R̂ϑ)ss + (R̂σ)ss + µHτ + ϕ̂0 Hs + β(2R̂ − ρ)H = R̂0 βh,
(4.10)
3
3ν
(R̂ϑ)τ s + Hτ s − (ϕ̂0 Ψτ )s + (R̂q)ss + (R̂ϑ)s = 0,
2
2
(4.11)
2
3χ
3
(R̂σ)τ s + Hτ s − (ϕ̂0 Ψτ )s + (R̂q)ss +
(R̂σ)s = 0,
4
5
4
(4.12)
2
2
2γ
2
(R̂q)τ s + (R̂ϑ)ss − (ϕ̂0 R̂ϑ)s + (R̂σ)ss − (ϕ̂0 R̂σ)s + (R̂q)s − µ̄Hτ = 0,
5
5
5
5
(4.13)
31
ABOUT STABILITY OF THE EQUILIBRIUM STATE
where H = Ψs .
Remark 4.1. The aggregate h (see (4.5)) can be rewritten
h(τ, s) =
Z1
Ψ(τ, ξ) dξ − Ψ(τ, s) =
0
Z1
[Ψ(τ, ξ) − Ψ(τ, s)] dξ =
0


Z1 Zξ
Z1 Zξ
=  Ψz (τ, z) dz dξ =
H(τ, z) dz dξ.
s
0
0
(4.14)
s
Now we proceed to derivation of the desired estimate. With this purpose, we multiply
equation (4.6) by 2Ψτ , equation (4.7) by 2(R̂ϑ), equation (4.8) by 2(R̂σ), equation (4.9) by
2(R̂q), sum them up and integrate it by s from 0 to 1 (accounting boundary conditions (4.2) –
(4.4)):
 1

¸ 
Z ·

d
3
3
2
Ψ2τ + Ψ2s + (R̂ϑ)2 + (R̂σ)2 + (R̂q)2 dξ +

dτ 
2
4
5
0
Z1 ·
3χ
2γ
3ν
(R̂σ)2 + (R̂q)2 + ϕ̂0 Ψτ Ψs − ϕ̂0 Ψτ (R̂ϑ) − ϕ̂0 Ψτ (R̂σ)−
µΨ2τ + (R̂ϑ)2 +
+2
2
4
5
0
¸
Z1
2 0
−µ̄Ψτ (R̂q) − ϕ̂ (R̂ϑ)(R̂q) − ϕ̂ (R̂σ)(R̂q) dξ = 2β R̂Ψτ h dξ.
5
0
(4.15)
0
By analogy, multiplying (4.10) by 2Hτ , (4.11) by 2(R̂ϑ)s , (4.12) by 2(R̂σ)s , (4.13) by 2(R̂q)s ,
summing up, integrating by s (accounting (4.2) – (4.4) and also (2.5)), we have:

 1
¸ 
Z ·

d
3
3
2
Hτ2 + Hs2 + (R̂ϑ)2s + (R̂σ)2s + (R̂q)2s + β(2R̂ − ρ)H 2 dξ +

dτ 
2
4
5
0
Z1 ·
3ν
3χ
2γ
µHτ2 + (R̂ϑ)2s +
+2
(R̂σ)2s + (R̂q)2s + ϕ̂0 Hτ Hs − (ϕ̂0 Ψτ )s (R̂ϑ)s − (ϕ̂0 Ψτ )s (R̂σ)s −
2
4
5
0
¸
Z1
2
−µ̄Hτ (R̂q)s − (ϕ̂0 R̂ϑ)s (R̂q)s − (ϕ̂0 R̂σ)s (R̂q)s dξ = 2β R̂0 Hτ h dξ.
5
(4.16)
0
We will use the result of integration by s from 0 to 1 (with account to boundary conditions)
of equation (4.10) multiplied by 2H:
 1

Z
Z1 ·

£
¤ 
d
2
−Hτ2 + Hs2 − Hs (R̂ϑ)s −
2HHτ + µH dξ + 2

dτ 
0
0
32
A. M. Blokhin, A. S. Bushmanova, V. Romano
¸
Z1
β
−Hs (R̂σ)s + (3R̂ − ρ)H 2 dξ = 2β R̂0 Hh dξ.
2
(4.17)
0
Summing up (4.15)–(4.17), we obtain the expression:
d (0)
J + J (1) = 2β
dτ
Z1
R̂Ψτ h dξ + 2β
Z1
0
R̂ Hτ h dξ + 2β
R̂0 Hh dξ ≤
0
0
0
Z1
 12  1
 1
 12  1
 21
Z
Z
Z
≤ 2β  R̂2 dξ   (R̂u)2 dξ   (R̂r)2 dξ  +
0
0
0
 21 1
 21  1
 12
 1
 21  1
 1
Z
Z
Z
Z
Z
0 2
2
2
0 2








(R̂ ) dξ
(R̂r)τ dξ
(R̂r) dξ
(R̂r)2 dξ. (4.18)
+ 2β
(R̂ ) dξ
+2β
0
0
0
0
0
Note that the estimate of the right hand part in (4.18) is derived with the help of the inequalities
of Hölder and Cauchy, and formula (4.14). Besides, in (4.18)
J (0)
Z1 ·
3
3
2
(R̂u)2 + (R̂ϑ)2 + (R̂σ)2 + (R̂q)2 + (R̂rτ )2 +
=
2
4
5
0
3
3
2
+ (R̂r)2s + (R̂ϑ)2s + (R̂σ)2s + (R̂q)2s +
2
4
5
i
2
+ (β(2R̂ − ρ) + µ + 1)(R̂r) + 2(R̂r)(R̂rτ ) dξ,
J (1)
Z1 ·
3χ
2γ
3ν
(R̂σ)2 + (R̂q)2 +
µ(R̂u)2 + (R̂ϑ)2 +
=2
2
4
5
0
+ (µ − 1)(R̂r)2τ + (R̂r)2s +
3ν
3χ
2γ
(R̂ϑ)2s +
(R̂σ)2s + (R̂q)2s +
2
4
5
β
(3R̂ − ρ)(R̂r)2 − ϕ̂0 (R̂u)(R̂r) − ϕ̂0 (R̂u)(R̂ϑ) − ϕ̂0 (R̂u)(R̂σ) −
2
2
− µ̄(R̂u)(R̂q) − ϕ̂0 (R̂ϑ)(R̂q) − ϕ̂0 (R̂σ)(R̂q) + ϕ̂0 (R̂r)τ (R̂r)s −
5
+
− (ϕ̂0 R̂u)s (R̂ϑ)s − (ϕ̂0 R̂u)s (R̂σ)s + µ̄(R̂r)τ (R̂q)s − (ϕ̂0 R̂ϑ)s (R̂q)s −
¸
2 0
− (ϕ̂ R̂σ)s (R̂q)s + (R̂r)s (R̂ϑ)s + (R̂r)s (R̂σ)s dξ.
5
Rewrite (4.18) with account to the estimate of the right hand part as the following inequality
d (0)
J + J (2) ≤ 0,
dτ
(4.19)
33
ABOUT STABILITY OF THE EQUILIBRIUM STATE
where

J (2) = J (1) − β 
Z1
0
 12
R̂2 dξ  ε1
Z1
(R̂u)2 dξ−
0
 12
 1

 1
Z
Z1
Z
Z1
1
0
2
2
2
(R̂r) dξ − β  (R̂ ) dξ  ε2 (R̂r)2τ dξ−
−β  R̂ dξ 
ε1
1
2
0
0
0
0
 21
 1
 12 1
 1
Z
Z
Z
Z1
1
0
2
0
2
2
−β  (R̂ ) dξ 
(R̂r)2 dξ.
(R̂r) dξ − 2β  (R̂ ) dξ 
ε2
0
0
0
0
(2)
While deriving J we used the Cauchy inequality with some positive constants ε1 , ε2 .
In J (0) and J (2) the expressions under the integral sign are positive definite squared forms
of the variables R̂r, R̂u, R̂ϑ, R̂σ, R̂q, (R̂r)s , (R̂ϑ)s , (R̂σ)s , (R̂q)s , R̂rτ , since the parameters
1
µ, ν, χ, γ, β are sufficiently large, with µ > γ and if we choose constants ε1 , ε2 such that
5
εi <
3µ − γ
,
β
i = 1, 2,
1
1
+
< 2l.
ε1 ε2
Here positive constant l is founded from the inequality
 12
 1
Z
δ −  (ρ0 (ξ))2 dξ  > l,
(4.20)
0
which exactly is the essential restriction on the function ρ(s) first mentioned in Remark 1.1.
Note, that µ̄ < 0 (see [1]) and with high accuracy, by (1.10),
3R̂ − ρ = 2ρ.
Under fulfillment of (4.20) there exists a constant M0 > 0 such that
J (2) ≥ M0 J (0) .
(4.21)
By (4.21) inequality (4.19) transforms into
d (0)
J + M0 J (0) ≤ 0,
dτ
i. e.,
J (0) (τ ) ≤ e−M0 τ J (0) (0).
Remind relations (4.1) and rewrite (4.22) as
Z1
0
£ 2
r (τ, s) + u2 (τ, s) + ϑ2 (τ, s) + σ 2 (τ, s) + q 2 (τ, s) + rτ2 (τ, s) +
(4.22)
34
A. M. Blokhin, A. S. Bushmanova, V. Romano
¤
+ rs2 (τ, s) + u2s (τ, s) + ϑ2s (τ, s) + σs2 (τ, s) + qs2 (τ, s) dξ ≤ M1 e−M0 τ t2 .
Here M1 > 0 is a constant, t2 = ||U0 ||2W 1 (0,1) =
2
R1
0
(4.23)
[(U0 , U0 ) + (U00 , U00 )] dξ is the squared norm of
the vector of initial data U0 (s) = U (0, s) = (r0 (s), u0 (s), ϑ0 (s), σ0 (s), q0 (s)) * in the Sobolev’s
space W21 (0, 1).
Remark 4.2. Constants M0 , M1 , as well as positive constants M2 , M3 , are finally determined
via the constants µ, ν, χ, γ, β and the function ρ(s).
Derivatives uτ , ϑτ , στ and qτ are estimated with the help of system (2.1):
uτ = Q − (rs + ϑs + σs + ϕ0 ϑ + ϕ̂0 σ + µu),
2
3
ϑτ = − (us + qs + ϕ̂0 q + νϑ),
3
2
4
2
2 0
3
στ = − (us + qs + ϕ̂ q + νϑ),
3
5
5
4
2
2
5
qτ = − (ϑs + σs + γq − µ̄u).
2
5
5
Consequently,
Z1
0
£ 2
¤
uτ (τ, s) + ϑ2τ (τ, s) + στ2 (τ, s) + qτ2 (τ, s) dξ ≤ M2 e−M0 τ t2 .
(4.24)
Combining estimates (4.23) and (4.24), we come to the desired a priori estimate:
R1
[(U, U ) + (Uτ , Uτ ) + (Us , Us )] dξ ≤ M3 e−M0 τ t2 ,
τ > 0.
(4.25)
0
From (4.25) it follows that
U (τ, s) ∈ W21 (0, 1),
Q(τ, s) ∈ W22 (0, 1),
◦
ϕ(τ, s) ∈W23 (0, 1), for all τ > 0,
and the equilibrium state in the linear approximation is asymptotically stable (by Lyapunov).
Remark 4.3. Precisely,
◦
r(τ, s), ϑ(τ, s), σ(τ, s) ∈W21 (0, 1).
Besides,
||U (τ )||2W 1 (0,1) ≤ M3 e−M0 τ t2 ,
2
τ > 0,
just this means (see Remark 2.6) asymptotic stability (by Lyapunov) of the trivial solution to
mixed problem (2.1)–(2.3). Remind that
||U (τ )||2W 1 (0,1) =
2
Z1
0
[(U, U ) + (Us , Us )] dξ.
ABOUT STABILITY OF THE EQUILIBRIUM STATE
35
5. Conclusions
The analysis, carried out in the paper, states a very important (from the applications point of
view) fact on asymptotic stability of the equilibrium state for the antidemocratic hydrodynamical
model (see [1]) of charge transport in semiconductors. Indeed, in absence of the bias across the
real semiconductor devices, transport of charge carriers (i. e., electric flow) must be absent.
Consequently, applying hydrodynamical models in description of physical phenomena of charge
transport in semiconductors, we must require of them the adequate description of these phenomena
(including correct description of the transition process in semiconductor devices in absence of
the bias across the diode).
Unfortunately, the fact of asymptotical stability of the equilibrium state is proved under
essential restriction (4.20) on the doping density ρ(s) and in the linear approximation as yet.
It should be noted at the same time that proof of stability of the equilibrium state does not
contain any restrictions on the doping density.
We gratefully thank Prof. A. M. Anile for many helpful discussions. We also appreciate
A. A. Iohrdanidy for efficient cooperation.
References
[1] Anile A. M., Muscato O. Improved hydrodynamical model for carrier transport in
semiconductors. Physical Rev. B., 51, No. 23, 1995, 16728–16740.
[2] Blokhin A. M., Iohrdanidy A. A., Merazhov I. Z. Numerical investigation of a
hydrodynamical model of charge transport in semiconductors. Preprint No. 33, Institute
of Mathematics SB RAS, Novosibirsk, 1996.
[3] Gardner C. L., Jerome J. W., Rose D. J. Numerical methods for hydrodynamic device
model: subsonic flow. IEEE Transactions on Computer-aided Design, 1989, 8, No. 5, 501–
507.
[4] Gardner C. L. Numerical simulation of a steady-state electron shock wave in a
submicrometer semiconductor device. IEEE Transactions on Electron Devices, 1991, 38,
No. 2, 392–398.
[5] Blokhin A. M. Energy Integrals and Their Applications to Gas Dynamic Equations.
Nauka, Novosibirsk, 1986.
[6] Blokhin A. M. Elements of Theory of Hyperbolic Systems and Equations. Izd-vo NGU,
Novosibirsk, 1995.
Received for publication November 24, 1998,
in revised form January 6, 1999