Asymptotic profile in a multi-dimensional stationary nonisentropic hydrodynamic semiconductor model

Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251 www.elsevier.com/locate/na

Asymptotic profile in a multi-dimensional stationary nonisentropic hydrodynamic semiconductor model Yeping Li∗ The Institute of Mathematics, Fudan University, Shanghai 200433, China Received 16 October 2005; accepted 30 June 2006

Abstract We study the irrotational subsonic stationary solutions of a multi-dimensional nonisentropic hydrodynamic model for semiconductor devices. This model consists of the continuous equations for the electron density, the electron current density and electron temperature, coupled the Poisson equation of the electrostatic potential. In some domain supplemented by the proper boundary conditions, we investigate the zero-electron-mass limit, the zero-relaxation-time limit and the Debye-length (quasi-neutral) limit, respectively. We show the strong convergence of the sequence of solutions and give the associated error estimates. 䉷 2006 Elsevier Ltd. All rights reserved.

MSC: 35M10; 35Q72; 76G75; 76N10 Keywords: Asymptotic profile; Zero-electron-mass limit; Zero-relaxation-time limit; Quasi-neutral limit; Hydrodynamic model; Semiconductor

1. Introduction In recent years, the hydrodynamic model for semiconductors has attracted a lots of attention and become an area of increasing interest in applied mathematics. This model is capable of capturing some important features of semiconductor devices that the traditional drift–diffusion model is unable to. For example, the hydrodynamic model can model modern submicron devices and hot electron effects. The full hydrodynamic model consists of the continuity equations expressing the conservation of mass, momentum, and energy, coupled self-consistently to the Poisson equation for the electric field. These models are derived from the moment equations of the Boltzmann transport (BTE) by suitable approximations. For more discussion on these models, we see [12,19]. In this paper, we consider a general multi-dimensional nonisentropic hydrodynamic model for semiconductor devices, derived by Anile and Pennisi in [2]. After proper normalization, the time-dependent version of this nonisentropic hydrodynamic model with heat sources in the multi-dimensional case

∗ Tel.: +86 2 65649953.

E-mail address: [email protected]. 1468-1218/$ - see front matter 䉷 2006 Elsevier Ltd. All rights reserved. doi:10.1016/j.nonrwa.2006.06.011

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reads as ⎧ nt + div(nu) = 0, ⎪ ⎪ ⎪ εu 1 ⎪ ⎪ , ⎨ ut + ε(u · ∇)u + ∇(nT ) = ∇ − n 1 2 T − TL (x) 2 22 − 1 ⎪ ⎪ Tt + u · ∇T + T div u − ε|u|2 − , ∇ · (∇T ) = ⎪ ⎪ 3 3n 31 2 2 ⎪ ⎩ 2   = n − b(x)

(1.1)

for x ∈ , which is a bounded semiconductor domain in Rd , d = 2, 3. Here n, u, T and  denote the electron density, the electric velocity, the carrier temperature and the electrostatic potential, respectively. The small physical parameters are the scaled electron mass ε > 0, the relaxation times 1 , 2 > 0, the Debye-length  > 0, and the thermal conductivity coefficient  > 0. In general, these physical coefficients may depend on n, TL (x) and T. In this paper, we only discuss the case that , 1 and 2 are positive constants. Without loss of generality, we take  = 1, 1 = 2 =  = const. The function TL (x) is the ambient device temperature, and b(x) stands for the prescribed density of positive charge background ions (doping profile). As to the more general case, we will investigate them in the future. In this paper, we are interested in the steady-state case nt = ut = Tt = 0. With these assumptions, the system (1.1) is prescribed as ⎧ div(nu) = 0, ⎪ ⎪ 1 εu ⎪ ⎪ , ⎨ ε(u · ∇)u + ∇(nT ) = ∇ − n  2 T − TL (x) 2 2 ⎪ ⎪ u · ∇T + T div u − T = ε|u|2 − , ⎪ ⎪ 3 3n 3  ⎩ 2  = n − b(x).

(1.2)

For the potential flow curl u = 0, we introduce the velocity potential  by u = −∇. Since 1 (u · ∇)u = ∇(|u|2 ) − u × curl u, 2 Eq. (1.2)2 can be reduced to ε T ε ∇(|∇|2 ) + ∇n + ∇T = ∇ + ∇. 2 n  Then for the smooth solutions, system (1.2) can be rewritten under the form: ⎧ div(n∇) = 0, ⎪ ⎪ T ε ε ⎪ ⎪ ⎨ ∇(|∇|2 ) + ∇n + ∇T = ∇ + ∇, 2 n  3n εn ⎪ 2 + 3n (T − T (x)), ⎪ T + ∇ · ∇T + nT  = − |∇| L ⎪ ⎪ 2 2 2 ⎩ 2   = n − b(x). Eliminating  from (1.3)2 and (1.3)4 , we have 

  T n − b(x) ε |∇|2 + div ∇n + T = + . 2 n  2

ε

Using (1.3)1 ( = −(∇n/n)∇) and (1.3)3 , (1.3)2 can be further reduced to  div

d T ε

b(x) ∇n − xi xj nxi xj + h(n, ∇n) = −εQ() − 2 , n n  i,j =1

(1.3)

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where Q() = (x1 x1 )2 + · · · + (xd xd )2 + 2(x1 x2 )2 + · · · + 2(x1 xd )2 + · · · + 2(xd−1 xd )2 and h(n, ∇n) =

d  ε ε

ε  2 (∇ · ∇n) −   n + ∇n · ∇ T + xi xi xj xj n n n2 i,j =1  1 3 3 ε − n 2 + ∇ · ∇T + |∇|2 − (T − TL (x)) . 2 2 2 

Moreover, we can pose the following Dirichlet boundary value conditions: n|j = nD ,

|j = D ,

T |j = TD

(1.4)

and |j = D .

(1.5)

In addition, in order to make a precisely presentation, we give the following assumptions: (H1) (H2) (H3) (H4)

 is a bounded convex C 2, -domain in Rd for some 0 < < 1, d = 2, 3. b(x), TL (x) ∈ L∞ (), there exist b1 , b2 , TL1 , TL2 such that 0 < b1 b(x) b2 , 0 < TL1 TL (x) TL2 for x ∈ . nD , TD ∈ H 2,q (), q > d/(1 − ), 0 < nD1 nD (x) nD2 , 0 < TD1 TD (x) TD2 for x ∈ j. D ∈ C 2, (), D ∈ H 2,q ().

In practice application, such as physical experiments and numerical simulations, etc., the zero-electron-mass limit ε → 0, the zero-relaxation-time limit  → 0 and the Debye-length limit  → 0 have been extensively used (see [3,23]). Mathematically, there exists a wide literature on these limits in various hydrodynamical models. For example, we can refer [5,13,11,14,17] to the time-dependent hydrodynamical models and [4,20] to the classical drift-diffusion models. Moreover, we can find the corresponding results for the stationary hydrodynamic semiconductor models [21,24]. It is worth to noticing that there exist many results on the stationary solutions for these hydrodynamic semiconductor models [6–8,15,18,26], and the existence of the global smooth solutions and the weak solutions for the time-dependent hydrodynamic models [1,10,16,22,27]. The purpose of this paper is to give a justification of the above three limits in the irrotational steady-state nonisentropic hydrodynamic semiconductor model for subsonic case. We assume that the boundary data are smooth and in the subsonic region. In the zero-relaxation-time limit we use a scaling similar to the one for the time-dependent hydrodynamic model and we additionally assume that TL (x) has more regularity and boundary data are in relaxation state. In the quasi-neutral limit we assume also that the boundary data are in equilibrium state. We first show the existence and uniqueness of solutions to problem (1.3)–(1.5), which are also in the subsonic region. Next, we prove the strong convergence of the sequence of solutions for each limit with error estimates. It is meaningful to point out that once the above three limits are proved, the problem of commutativity of the limits arises. At a formal level, it is clear that the zero-electron-mass limit and the quasineutral limit commute. When the zero-relaxation-time limit is involved, either pair of these limits still commute provided that the scaling covers the boundary conditions. See the limit equations (2.21)–(2.22), (3.13)–(3.14) and (4.9)–(4.10). The details of the statement and the rigorous justification of these results will be given in a forthcoming work. The paper is organized as follows. In the next section, we prove the zero-electron-mass limit by establishing some uniform estimates for the sequence of solutions. The uniform convergence with rate O(ε) is given, too. In Section 3, we give a simple description of convergence result for the zero-relaxation-time limit with the additional assumption of TL (x), by a similar method used in Section 2. The convergence rate is obtained if TL (x)2,q ∼ . Section 4 is devoted to the quasi-neutral limit with boundary data in equilibrium and b(x)2,q + TD 2,q ∼ 2 . The convergence result with rate is obtained with a slightly different proof from those used in Section 2. Notation: We denote C k, () to the Hölder space with norm

U C k, = U C k + [D j U ]C , |j |=k

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where U C k =



sup |U |, [U ] =

|j |  k 

|U (x) − U (y)| . |x − y|

x =y,x,y∈ sup

By H k,p , k 0, 1p ∞, we denote the Sobolev space with norm

D j U p , U k,p = |j |  k

where  · p denotes the norm of Lp -space. The integration domain  can be omitted without any ambiguity. C represents a generic constant even if it may vary from line to line, C(·) means that C depends on ·, and the summation convection is used throughout this paper. 2. The zero-electron-mass limit In this section we study the zero-electron-mass limit ε → 0 in (1.3)–(1.5). The Debye-length  and the relaxationtime  are supposed to be constants independent of ε. Since the solution of (1.3)–(1.5) is dependent on ε, we may rewrite (1.3)–(1.5) as: ⎧ div(nε ∇ε ) = 0, ⎪ ⎪ ⎪ ε Tε ε ⎪ ⎪ ⎨ ∇(|∇ε |2 ) + ∇nε + ∇Tε = ∇ε + ∇ε , nε  2 (2.1) 3nε εnε ⎪ 2 + 3nε (T − T (x)), ⎪ + · ∇T − ∇n · ∇ T = − | T ∇ |∇ ⎪ ε ε ε ε ε L ε ε ε ⎪ ⎪ 2 2 2 ⎩ 2  ε = nε − b(x) and the following Dirichlet boundary conditions nε |j = nD ,

ε |j = D ,

Tε |j = TD ,

ε |j = D .

(2.2)

Since ε → 0, we may suppose that ε ∈ (0, 1]. Now, we give the existence and uniqueness of the irrotational subsonic stationary solutions for (2.1)–(2.2): Theorem 2.1. Let the assumptions (H1)–(H4) hold. Then problem (2.1)–(2.2) has a unique strong solution (nε , ε , Tε , ε ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q (), which satisfies n11 nε n12 , nε H 2,q A1 ,

T11 Tε T12 , x ∈ , ε C 2, A2 , Tε H 2,q A3 ,

ε H 2,q A4

(2.3) (2.4)

if |TL1 − TL2 | + |TL1 − TD2 | + |TL2 − TD1 | + D C 2, () <  holds, where n1i , T1i (i = 1, 2) and Aj (j = 1, 2, 3, 4) are positive constants independent of ε. Proof. The proof of the existence of solutions is based on Schauder fixed point theorem. For the sake of completeness, we give the details, especially the uniform estimates, through we can find them in [15]. For the convenience, we omit the subscript ε. We define a close convex set B = {(m, ) ∈ C 1, () × C 1, () : n11 m n12 , T11  T12 , mC 1, k1 , C 1, k2 }, where the positive numbers n1i , T1i , ki (i = 1, 2) will be defined below. Further, choosing (m, ) ∈ B, we can set up a map S : (m, ) → (n, T ) which is defined as follows: Solve  ∇m ∇ = 0,  + (2.5) m |j = D

Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

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for , then T +

3m εm 3m |∇|2 + (T − TL (x)) ∇ · ∇T − ∇m · ∇T = − 2 2 2

(2.6)

with T |j = TD

(2.7)

and  div

d   ε  ε

ε ∇n − +  ∇ · ∇n xi xj nxi xj + ∇ · ∇m + m m m2 m i,j =1

 d ε

ε 3 1 3 b(x) 2 − xi xi xj nxj − n 2 + ∇ · ∇ + |∇| − ( − TL (x)) = −εQ() − 2 2 2 2 m   i,j =1

(2.8)

with n|j = nD .

(2.9)

Obvious, by the standard Hölder-estimates for the second-order linear elliptic equations, we know that there exists a unique  satisfying (2.5) C 2, C(n11 , n12 , k1 ).

(2.10)

Moreover, using the Lp -theory of the second-order linear elliptic equations, we can know that there are n ∈ H 2,q () for (2.8)–(2.9) and T ∈ H 2,q () for (2.6)–(2.7), that is, S(B) = (n, T ) ∈ H 2,q () × H 2,q (). Obviously, S(B) is precompact in C 1, () × C 1, () by Sobolev’s imbedding theorems and q > d/(1 − ). Moreover, the continuity of S, regarded as a map of a subset of C 1, () × C 1, () into itself can be proved by standard arguments based on H 2,q -estimates for solutions of linear elliptic equations. We can omitted the details here. In order to apply the Schauder fixed-point theorem, it remains to prove that S(B) ⊂ B. Indeed, taking T = max(TL2 , TD2 ) and using (T − T )+ = max(T − T , 0) as a test function in the weak formulation of (2.6), we have 3m ∇ · ∇(T − T )+ (T − T )+ dx + ∇m · ∇T (T − T )+ dx |∇(T − T )+ |2 dx − 2 3m εm 2 + (2.11) = |∇| (T − T ) dx − (T − TL (x))(T − T )+ dx. 2 2 Applying Cauchy–Schwartz’s and Young’s inequalities, one get 3m + + ∇ · ∇(T − T ) (T − T ) dx C (|∇(T − T )+ |2 + |(T − T )+ |2 ) dx, − 2 εm |∇|2 (T − T )+ dx − ∇m · ∇T (T − T )+ dx + 2 εm = ∇m · ∇(T − T + T )(T − T )+ + |∇|2 (T − T )+ dx 2 + C (T − T )(T − T ) dx + C(n12 ) (T − T )+ dx, and

3m (T − TL (x))(T − T )+ dx = − 2 3n12 + − (T − T )(T − T ) dx − 2

−

3m (T − T + T − TL (x))(T − T )+ dx 2 3m (T − TL (x))(T − T )+ dx. 2

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Furthermore, we have

|∇(T − T )+ |2 dx C(meas(T > T ))1/2



|(T − T )+ |2 dx

1/2 .

Using Poincaré’s inequality and choosing  such that n12 1, we have ∇(T − T )+ 2 C(meas(T > T ))1/2 .

(2.12)

It follows from Sobolev’s imbedding theorem that the imbedding H01 () → Lr () is continuous for r ∈ [2, ∞] if d = 2, for r ∈ (2, 6] if d = 3. And it is well known for T > T and r > 2, the inequality (meas(T > T ))1/r (T − T ) C(T − T )+ 1,2

(2.13)

holds. Therefore, we know from (2.12), (2.13) and Poincaré inequality that, there exists another positive constant C such that meas(T > T ) 

Cr (meas(T > T ))r/2 . (T − T )r

Choosing r/2 > 1 and applying the Stampacchia’s Lemma in [25], we know that there is a constant C such that T T + C,

x ∈ .

So, we can take T12 = 2T so that T T12 ,

x ∈ .

Similarly, using (−T + T )+ as a test function in the weak formulation of (2.6), and repeating the above procedures, we can obtain the estimate ∇(−T + T )+ 2 C(meas(−T > − T ))1/2 .

(2.14)

Hence, there exists a constant C such that −T  − T + C, for x ∈ . Hence, taking T11 = 21 T , we also can get T T11 ,

x ∈ .

On the other hand, taking n = max(nD2 , b2 ), and using (n − n)+ = max(n − n, 0) as the test function in the weak formulation of (2.8), we have

d ε

 + |∇(n − n)+ |2 dx − xi xj (n − n)+ xi (n − n)xj dx m m i,j =1 ⎛ ⎞ d d 

ε

ε + ⎠ − ⎝   (n − n)+ xi xi xj (n − n)+ xi − xj (n − n) dx m xi xj xj m i,j =1 i,j =1   ε ε − +  ∇ · ∇(n − n)+ (n − n)+ dx ∇ · ∇m + 2 m m   ε 3 1 3 2 + (n − b(x)) − εQ() (n − n)+ dx. + n ∇ · ∇ + |∇| − ( − TL (x)) (n − n) dx = − 2 2 2 2 (2.15)

Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

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For all the integral terms in (2.15), we handle them in order as follows:

d  ε

+ 2 + |∇(n − n) | dx − xi xj (n − n)+ xi (n − n)xj dx m m i,j =1 ⎛ d d  

ε ε

+ xi xi xj (n − n)+ − ⎝ − xi xj (n − n)+ xj (n − n) dx xi m m xj i=1 i,j =1  T1  |∇(n − n)+ |2 dx − C |(n − n)+ |2 dx, − C2 n2 ⎞  ε 1 +⎠ + (n − n) dx C(k1 ) |(n − n)+ |2 dx, +  ∇ · ∇(n − n) − ∇ · ∇m + m m2  3 3 ε 2 + n ∇ · ∇ + |∇| − ( − TL (x)) (n − n) dx C (n − n)+ dx 2 2 2

and



−

n

b(x)



− εQ() − 2 (n − n)+ dx 2  1 = − 2 (n − n + n − b(x))(n − n)+ dx + εQ()(n − n)+ dx  1 +  − 2 (n − n + n − b(x))(n − n) dx + C (n − n)+ dx. 

Hence, inserting the above four inequalities into (2.15) and taking  such that k1 1, one get  1/2 |(n − n)+ |2 dx , |∇(n − n)+ |2 dx C(meas(n > n))1/2 which yields ∇(n − n)+ 2 C(meas(n > n))1/2 . Similar to (2.13) and (2.14), we know that there is a positive constant C such that n n + C. Hence, we can take n12 = 2n, so that n n12 ,

x ∈ .

For the lower bound, taking n = min(nD1 , b1 ) and using (−n + n)+ as a test function in the weak formulation of (2.8), and repeating the above procedures, we can obtain a similar estimate ∇(−n + n)+ 2 C(meas(−n > − n))1/2 , and in an analogous way we conclude the existence of a constant C such that −n  − n + C,

x ∈ .

Hence, n11 = 21 n, we also can get n n11 ,

x ∈ .

Finally, in terms of the standard Lp -theory as previous stated, we have n2,q C(nD 2,q , n11 , n12 , T11 , T12 , k1 )

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and T 2,q C(TD 2,q , n11 , n12 , T11 , T12 , k1 ). Keeping k1 1 in mind, we have (n, T )C 1, C(n, T )2,p C(n11 , n12 , TL1 , TL2 , b1 , b2 , T11 , T12 , 1) := (k1 , k2 ). Hence, Schauder’s Theorem can guarantee the existence of (n, T ) for (2.1)–(2.2). Now we can compute  by solving (2.5) replacing m with n. So, we have established the existence of the strong solution for (2.1)–(2.2). Next, we are going to prove the uniqueness of the strong irrotational and subsonic stationary solutions obtained as above. Let (n(1) , (1) , T (1) , (1) ) and (n(2) , (2) , T (2) , (2) ) be two strong solutions for (2.1)–(2.2), satisfying (2.3)–(2.4). Then taking the difference of Eq. (2.1)3 satisfying by (n(1) , (1) , T (1) ) and (n(2) , (2) , T (2) ), respectively, and using T (1) − T (2) as the test function in the weak formulation of the difference equation, we obtain    2 3   (1)  n(1)∇(1) ∇T (1) − n(2) ∇(1) ∇T (2) T (1) − T (2) dx  T − T (2)  dx − 2    (1) (1) (1) − n  T − n(2) (1) T (2) T (1) − T (2) dx        3 =− T (1) − T (2) dx n(1) T (1) − TL (x) − n(2) T (2) − TL (x) 2    ε + (2.16) n(1) |∇(1) |2 − n(2) |∇(2) |2 T (1) − T (2) dx. 2 For all the integrals in (2.16), we deal with them as in the following order. Obviously, the direct calculation leads to        3 T (1) − T (2) dx n(1) T (1) − TL (x) − n(2) T (2) − TL (x) − 2        3 =− n(1) T (1) − T (2) − T (2) − TL (x) (n(1) − n(2) ) T (1) − T (2) dx  2   2 2 3n11 (1) (2)  − dx + C n(1) − n(2) dx + C T −T 2 and ε 2



  n(1) |∇(1) |2 − n(2) |∇(2) |2 T (1) − T (2) dx        1 = T (1) − T (2) dx n(1) |∇(1) |2 − |∇(2) |2 + |∇(2) |2 n(1) − n(2) 2   2  2  2 C n(1) − n(2) + T (1) − T (2) + ∇(1) − ∇(2) dx,

with the aid of Young’s inequality. Similarly, we can get   3  (1) (1) n ∇ ∇T (1) − n(2) ∇(1) ∇T (2) T (1) − T (2) dx − 2    3 =− n(1) ∇(1) ∇T (1) − ∇T (2) T (1) − T (2) dx 2       3  (1) − T (1) − T (2) dx n ∇T (2) ∇(1) − ∇(2) + ∇T (2) ∇(2) n(1) − n(2) 2  2  2  2 (1) (2) (1) (2) (1) (2) C n −n dx + T −T + ∇T − ∇T   2 2 + C T (1) − T (2) dx + C() ∇(1) − ∇(2) dx

Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

and

1243

   − n(1) (1) T (1) − n(2) (1) T (2) T (1) − T (2) dx          = (1) T (1) n(1) − n(2) + n(2) T (1) (1) − (2) + ∇n(2) ∇(2) T (1) − T (2) T (1) − T (2) dx    2  2 2 2 (1) (2) (1) (2) (1) (2) C n −n dx + C T −T + T −T dx + C() ∇(1) − ∇(2) dx.

Therefore, adding the above four inequalities to (2.16), we deduce that    2  2 2  |∇ T (1) −T (2) |2 dx C() dx+C n(1) −n(2) dx. ∇(1) −∇(2) + (1) −(2)

(2.17)

Further, taking the difference of (2.1)1 satisfying by (n(1) , (1) ) and (n(2) , (2) ), multiplying it by ((1) − (2) ) and integrating the resultant equation over , we discover    n(1) ∇(1) − n(2) ∇(2) ∇(1) − ∇(2) dx = 0, which implies  2 |∇(1) − ∇(2) |2 dx C n(1) − n(2) dx.

(2.18)

Similarly, we also have  2  2 (2) 2 (1) (1) (2) (1) (2) | − | dx C dx+C |∇(1) −∇(2) |2 dx. ∇n −∇n + n −n

(2.19)

 Finally, to derive an estimate for n(1) −n(2) and ∇(n(1) −n(2) ) in L2 -norm, we are going to treat (div(2.1)2 (n(1) , (1) , T (1) , (1) ) − div(2.1)2 (n(2) , (2) , T (2) , (2) ))(ln n(1) − ln n(2) ) dx, then, we have    (1)     T (2) (2) T (1) (2) (1) (2) (1) (1) (2) n −n ∇ ln n ∇n − ∇n − ln n ln n − ln n dx + dx n(1) n(2)   ε     =− ∇T (1) − ∇n(1) − ln T (2) − ∇ |∇(1) |2 − |∇(2) |2 ∇ ln n(2) dx 2    ε  (1) (2) (1) + ∇ ln n − ln n(2) dx. ∇ − ∇  Employing the above same arguments and techniques, we can deduce   2  (1) (2) (1) (2) 2 ln n − ln n + |∇ ln n − ln n | dx    2  2 C T (1) − T (2) + |∇ T (1) − T (2) |2 + ∇(1) − ∇(2) dx.

(2.20)

Further, inserting (2.18)–(2.20) into (2.17), with the aid of Poincaré’s inequality, we obtain       ∇ T (1) − T (2) 2 C n(1) − n(2) 2 + ∇ n(1) − n(2) 2     C ∇ T (1) − T (2) 2 + T (1) − T (2) 2   C∇ T (1) − T (2) 2 . Thus, we have T (1) = T (2) . Further, we can deduce that n(1) = n(2) , (1) = (2) and (1) = (2) . This complete the proof. 

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Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

Now we consider the convergence of the sequence (nε , ε , Tε , ε )ε>0 . Before stating the result, let us perform formally the zero-electron-mass limit ε → 0 in (2.1)–(2.2) to derive the limit problem. Set ε = 0, then the formal zero-electron-mass limit (n, , T , ) satisfies the problem: ⎧ div(n∇) = 0, ⎪ ⎪ T ⎪ ⎪ ⎨ ∇n + ∇T = ∇, n (2.21) 3n 3n ⎪ ⎪ T + (x)) = 0, ∇ · ∇T − ∇n · ∇T − (T − T L ⎪ ⎪ 2 2 ⎩ 2 xx = n − b(x) for x ∈ , and the following boundary conditions: n|j = nD ,

|j = D ,

T |j = TD ,

|j = D .

(2.22)

Now we prove rigorously the zero-electron-mass limit and give the convergence rate of (nε , ε , Tε , ε ) to (n, , T , ) in the space H 2,q () × C 2, () × H 2,q () × H 2,q (). Theorem 2.2. Let all the conditions in Theorem 2.1 be satisfied. Then, problem (2.21)–(2.22) admits a unique solution (n, , T , ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q () and the whole sequence (nε , ε , Tε , ε ) of solutions to (2.1)–(2.2) convergence to (n, , T , ). Moreover, there is a constant A5 independent of ε such that, as ε → 0, nε − n2,q A5 ε,

ε − C 2, A5 ε,

Tε − T 2,q A5 ε,

ε − 2,q A5 ε.

(2.23)

Proof. From the uniform estimates (2.4) and the Ascoli theorem, it is easy to see that there is a subsequence (nε , ε , Tε , ε )ε>0 (not relabeled) and function (n, , T , ) such that nε → n, Tε → T , ε →  in H 2,q () weakly and ε →  in C 2 () uniformly, nε → n, Tε → T , ε →  in C 1 () uniformly. It is obvious that these convergences are sufficient to the limit in Eqs. (2.1) and the boundary condition (2.2). This shows the existence of solutions (n, , T , ) to the problem (2.21)–(2.22). In order to get the convergence of the whole sequence (nε , ε , Tε , ε ), we only need to establish the uniqueness of solutions (n, , T , ) of (2.21)–(2.22), which can be obtained by the same way of the proof in Theorem 2.1. Finally, the regularity of the solution in H 2,q ()×C 2, ()×H 2,q ()×H 2,q () is a direct consequence of problem (2.21)–(2.22). To prove the convergence rates (2.23), let us denote the equations of ε − , nε − n and Tε − T by (ε − ) + ∇ ln nε · (∇ε − ∇) + ∇ · (∇ ln nε − ∇ ln n) = 0, nε − n div(Tε ∇(ln nε − ln n)) − 2 ε = −(Tε − T ) − div(∇ ln n(Tε − T )) − ∇nε · ∇ε − εQ(ε ) nε d d ε

ε ε

+ εxi εxj nεxi xj − 2 (∇ε · ∇nε )2 + εxi εxi xj nεxj nε nε nε i,j =1

i,j =1

and 3 3 (Tε − T ) + nε ∇ε · ∇(Tε − T ) − ∇nε · ∇ε (Tε − T ) − nε (Tε − T ) 2   2 εnε 3 3 2 |∇ε | − nε ∇T − ∇nε T ∇(ε − ) − ∇T ∇ − T ∇ ∇(nε − n) =− 2 2 2 3 + (T − TL (x))(nε − n). 2

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Then, for the above three equations, we obtain from Lemma 9.17 in [9] that, there is a constant C > 0 independent of ε such that ε − C 2, C ln nε − ln n2,q ,

(2.24)

 ln nε − ln n2,q C + C(Tε − T )q + C∇(Tε − T )q ,

(2.25)

Tε − T 2,q Aε + C(∇(n − nε )q + nε − nq ) + C∇(ε − )q .

(2.26)

and

Moreover, applying Lemma 9.17 in [9] to the equation (nε − n) = eln nε (ln nε − ln n) +  ln n(eln nε − eln n ) + eln nε ((∇ ln nε )2 − (∇ ln n)2x ) + (∇ ln n)2 (eln nε − eln n ), we have nε − n2,q C ln n − ln nε 2,q .

(2.27)

Combining (2.24)–(2.25) and (2.27), we can get the first three estimates of (2.23). Further, through treating the difference equation of (2.1)4 and (2.21)4 , we also obtain the last inequality of (2.23). This ends the proof.  3. The zero-relaxation-time limit In the study of the zero-relaxation-time limit  → 0 in (1.3)–(1.5), we assume that the Debye-length  and ε are constants independent of . Because of the consistence of the steady-state temperature and the ambient lattice temperature function TL (x) in the relaxation limit, we have to assume that TL (x) is more smooth and the boundary conditions are in relaxation state. More precisely, we need the following hypothesis: TL (x) ∈ H 2,q (),

TL (x)|j = TD .

(3.1)

Motivated by the physical consideration, a scaling is need to study the zero-relaxation-time limit in the transient Euler–Poisson system as in [13,16,17,14]. The scaling can be found by a formal asymptotic expansion in power of . This enables us to consider the same scaling as in the time-independent problem, i.e.,  , T = T ,  = .  In this situation, problem (1.3)–(1.5) reads ⎧ div(n ∇ ) = 0, ⎪ ⎪ ⎪ T ⎪ ε2 ⎪ ⎨ ∇(|∇ |2 ) + ∇n + ∇T = ∇ + ε∇ , 2 n 3n εn 3n ⎪ ⎪ ⎪ T + ∇ · ∇T − ∇n · ∇ T = − |∇|2 + (T − TL (x)), ⎪ ⎪ 2 2 2 ⎩ 2   = n − b(x) n = n,

 =

(3.2)

(3.3)

for x ∈  with Dirichlet boundary conditions n |j = nD ,

 |j = D ,

T |j = TD ,

 |j = D .

(3.4)

Then the results of the existence and uniqueness of solutions, and the uniform estimates with respect to  can be stated as follows: Theorem 3.1. Let the assumptions (H1)–(H4) hold. In addition, assume that TL (x) satisfies (3.1). Then problem (3.3)–(3.4) has a unique strong solution (n ,  , T ,  ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q (), which satisfies n21 n n22 , T21 T T22 , x ∈ , n 2,q B1 ,  C 2, B2 , T 2,q B3 ,

 2,q B4

(3.5) (3.6)

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Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

if |TL1 − TL2 | + |TL1 − TD2 | + |TL2 − TD1 | + D C 2, () <  holds, where n2i , T2i (i = 1, 2) and Bj (j = 1, 2, 3, 4) are positive constants independent of . Proof. The proof is similar to that of Theorem 2.1 except for the derivation of uniform bound for T C 1, . As in Section 2, we only need to control T 2,q . Omitting the subscript , Eq. (3.3)3 can be rewritten as  3n 3n ∇(T − TL (x)) − ∇n + ∇(T − TL (x)) (T − TL (x)) + 2 2 εn 3n (3.7) ∇∇TL (x) + ∇n∇TL (x). =− |∇|2 − TL (x) − 2 2 Multiplying (3.7) by (T − TL (x))|T − TL (x)|q−2 and integrating over , we have  3n − ∇n∇ |T − TL (x)|q dx 2 3n = ((T − TL (x)) + ∇∇(T − TL (x)))(T − TL (x))|T − TL (x)|q−2 dx 2  εn 3 + |∇|2 + TL (x) + n∇∇TL (x) − ∇n∇TL (x) (T − TL (x))|T − TL (x)|q−2 dx. (3.8) 2 2 For the integration terms above, applying the integration by parts, we have (T − TL (x))(T − TL (x))|T − TL (x)|q−2 dx 4(q − 1) q−1 = − ∇|T − TL (x)| ∇|T − TL (x)| dx = − |∇|T − TL (x)|q/2 |2 dx q2 and



(3.9)

3n ∇∇(T − TL (x))(T − TL (x))|T − TL (x)|q−2 dx 2 C C



(|T − TL (x)|q + |∇|T − TL (x)|q/2 |2 ) dx |∇|T − TL (x)|q/2 |2 dx.

Therefore, we obtain  1/q 1 |T − TL (x)|q dx C(nC 1, + ∇TL (x)q + TL (x)q ), 

(3.10)

(3.11)

which yields by applying Theorem 9.15 and Lemma 9.17 in [9] to (3.7) that T − TL (x)2,q C(nC 1, + ∇TL (x)q + TL (x)q ),

(3.12)

where C is a constant independent of . Now we investigate the zero-relaxation-time limit in (3.3)–(3.4). Set =0, the limit (n, , T , ) of (n ,  , T ,  )>0 is formally governed by the following drift–diffusion equations ⎧ div(n∇) = 0, ⎪ ⎪ ⎨T ∇n + ∇T = ∇ + ε∇, (3.13) n ⎪ ⎪ ⎩ T = TL (x), 2  = n − b(x)

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for x ∈  with boundary conditions n|j = nD ,

|j = D ,

|j = D .

(3.14)

Then the convergence result of (n ,  , T ,  )>0 to (n, , T , ) with convergence rate O() in H 2,q () × C 2, () × H 2,q () × H 2,q () can be stated as follows. Theorem 3.2. Let the conditions in Theorem 3.2 hold and (n ,  , T ,  )>0 be a sequence of solutions to (3.3)–(3.4). Then problem (3.13)–(3.14) has a unique solution (n, , T , ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q () which is the limit of (n ,  , T ,  ). Moreover, there is a constant B5 independent of  such that, as  → 0, n − n2,q B5 ,

 − C 2, B5 ,

T − T 2,q B5 ,

 − 2,q B5 ,

(3.15)

if TL (x)q = O(1). Proof. It is clear that the uniform estimates (3.5)–(3.6) are sufficient to pass to the limit in Eqs. (3.3) and the boundary conditions (3.4) to obtain a solution (n, , T , ) for the problem (3.13)–(3.14), just like proof of Theorem 2.2. The regularity of the solution (n, , T , ) in H 2,q () × C 2, () × H 2,q () × H 2,q () is a direct consequence of problem (3.13)–(3.14). Now we derive the convergence rates (3.15). First, T − TL (x)2,q C( + TL (x)q )

(3.16)

holds from a similar inequality to (3.10). On the other hand, let us denote the equations of  −  and n − n by ( − ) + ∇ ln n (∇ − ∇) + ∇(∇ ln n − ∇ ln n) = 0

(3.17)

and div(T ∇ ln n − T ∇ ln n) +

n − n

= −(T − T ) + ε( − ) −

ε2 (|∇ |2 ). 2

(3.18)  Then, for (3.17) and (3.18), we know from Lemma 9.17 in [9] that, there is a constant C > 0 independent of  such that 2

 − C 2, C ln n − ln n2,q

(3.19)

 ln n − ln n2,q C + C(T − T )q + CT − T q + C( − )q .

(3.20)

and

Keeping (2.27) in mind, and combining (3.16)–(3.20), we can get the first three estimates of (3.15). Further, the straightforward treat the difference equation of (3.3)4 and (3.13)4 , we also obtain the last inequality of (3.15). This completes the proof.  4. The quasi-neutral limit In this section, we treat the quasi-neutral limit  → 0 in (1.3)–(1.5). In order to attain the aim, we assume that the relaxation time  and the mass ε are constants independent of , and b(x) is more smooth and the boundary conditions are in equilibrium. More precisely, we have to assume that b(x) ∈ H 2,q (),

b(x)|j = nD .

(4.1)

Since  → 0 in the quasi-neutral limit we may assume  ∈ (0, 1], and we rewrite problem (1.3)–(1.5) under the form ⎧ div(n ∇ ) = 0, ⎪ ⎪ ⎪ ε T ε ⎪ ⎪ ⎨ ∇(|∇ |2 ) + ∇n + ∇T = ∇ + ∇ , 2 n  (4.2) 3n εn ⎪ 2 + 3n (T − T (x)), ⎪ T ∇ |∇ + · ∇T − ∇n · ∇ T = − | ⎪ L         ⎪ ⎪ 2 2 2 ⎩ 2   = n − b(x)

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Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

with the following Dirichlet boundary conditions: n |j = nD ,

 |j = D ,

T |j = TD ,

 |j = D .

(4.3)

Theorem 4.1. Let the assumptions (H1)–(H4) hold. In addition, assume that b(x) satisfies (4.1). Then problem (1.3)–(1.5) has a unique strong solution (n ,  , T ,  ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q (), which satisfies n31 n n32 , T31 T T32 , x ∈ , n 2,q C1 ,  C 2, C2 , T 2,q C3 ,

 2,q C4

(4.4) (4.5)

if |TL1 − TL2 | + |TL1 − TD2 | + |TL2 − TD1 | + D C 2, () <  holds, where n3i , T3i (i = 1, 2) and Cj (j = 1, 2, 3, 4) are positive constants independent of . Proof. Since the proof is similar to that of Theorem 2.1 except for the uniform bound of n C 1, with regard to , from the Sobolev’s imbedding inequality, we only need to control n 2,q here. We can rewrite (4.2) as n − b(x) 2

d ε

T xi xj (n − b(x))xi xj = div ∇(n − b(x)) − n n i,j =1 ⎞ ⎛  d

ε ε ε ∇ · ∇(n − b(x)) − + ⎝ 2 ∇ · ∇n + xi xi xj (n − b(x))xj ⎠ n n n i,j =1 ⎛    d T ε

ε ε ⎝ + εQ( )+div ∇ ·∇b(x) ∇b(x) +T − xi xj b(x)xi xj + ∇ ·∇n + n n n n2 i,j =1 ⎞ d ε

− xi xi xj b(x)xj ⎠ =: I1 + I2 + I3 + I4 . (4.6) n 

i,j =1

Multiplying (4.6) by (n − b(x))|n − b(x)|q−2 and integrating the resultant equation over , we find 1 q (n − b(x)) dx = (I1 + I2 + I3 + I4 )(n − b(x))|n − b(x)|q−2 dx. 2 Applying integration by parts as in Section 3, we have T 4(q − 1) q−2 I1 (n − b(x))|n − b(x)| dx = − |∇|n − b(x)|q/2 |2 dx q2 n and I2 (n − b(x))|n − b(x)|q−2 dx

d    ε

xi xj jxi |n − b(x)|q/2 jxj |n − b(x)|q/2 dx n i,j =1 ⎛ ⎞ d  

2 ε + jxi ⎝ xi xj ⎠ |n − b(x)|q/2 jxj |n − b(x)|q/2 dx. q n

4(q − 1) = q2

i,j =1

Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

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Therefore, by Young’s and Poincaré’s inequalities, there exists constant C > 0 depending on q, , C 1 norm of n and C 2 norm of  such that I2 (n − b(x))|n − b(x)|q−2 dx   |n − b(x)|q + |∇|n − b(x)|q/2 |2 dx C  C 2, C  C 2, |∇|n − b(x)|q/2 |2 dx. Similarly, we have I3 (n − b(x))|n − b(x)|q−2 dx C  C 2, (|n − b(x)|q + |∇|n − b(x)|q/2 |2 ) dx C  C 2, |∇|n − b(x)|q/2 |2 dx and

I4 (n − b(x))|n − b(x)|q−2 dx q−1

C( C 2, + T q + b(x)2,q )n − b(x)q

.

On the other hand, as in Section 3, we have T 2,q C(T ∞ , TL (x)∞ ,  C 1, )(TD 2,q + n q ) and C 2, Cn C 1, Cn 2,q . Subsequently, we obtain n − b(x)q C2 (T ∞ , TL (x)∞ , n ∞ , n q , TD 2,q , b(x)2,q )

(4.7)

n − b(x)2,q C(T ∞ , n ∞ , TL (x)∞ )(TD 2,q + b(x)2,q +  2,q ).

(4.8)

and

Taking into account the proper smallness of , we can establish the uniform bound of n 2,q . This completes the proof.  It is clear that the uniform estimates (4.4)–(4.5) are sufficient to pass to the limit in Eqs. (4.2) and the boundary conditions (4.3). Let (n, , T , ) be the formal limit of (n ,  , T ,  ). Then (n, , T , ) is a solution of the problem ⎧ div(n∇) = 0, ⎪ ⎪ ⎪ ⎪ ε ∇(|∇|2 ) + T ∇n + ∇T = ∇ + ε ∇, ⎨ 2 n  (4.9) ⎪ ⎪ T + 3n ∇ · ∇T − ∇n · ∇T = − εn |∇|2 + 3n (T − TL (x)), ⎪ ⎪ ⎩ 2 2 2 n = b(x) and the following Dirichlet boundary conditions: |j = D ,

T |j = TD ,

|j = D .

(4.10)

It is easy to see that the solution of (4.9)–(4.10) is also unique. A result of error estimates between (n ,  , T ,  ) and (n, , T , ) is stated as follows.

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Y. Li / Nonlinear Analysis: Real World Applications 8 (2007) 1235 – 1251

Theorem 4.2. Let the conditions in Theorem 3.2 hold and (n ,  , T ,  )>0 be a sequence of solutions to (4.2)–(4.3). Then problem (4.9)–(4.10) has a unique solution (n, , T , ) ∈ H 2,q () × C 2, () × H 2,q () × H 2,q () which is the limit of (n ,  , T ,  ). Meanwhile, we have n − nq C5 2 ,

(4.11)

where C5 is a constant independent of . Moreover, if b(x)2,q + TD 2,q = O(1)2 , we can obtain n − n2,q C5 2 ,

 − C 2, C5 2 ,

T − T 2,q C5 2 ,

 − 1,q C5 2 .

(4.12)

Proof. Similar to Theorems 2.2 and 3.2, we only need to derive the convergence rates (4.11)–(4.12). First, from the above analysis, we have n − nq C2

(4.13)

n − b(x)2,q C(T ∞ , n ∞ , TL (x)∞ )(TD 2,q + b(x)2,q +  2,q ).

(4.14)

and

Therefore, we obtain that (4.11) and the first estimate in (4.12). On the other hand, let us denote the equations of  − , T − T and  −  by ( − ) + ∇ ln n (∇ − ∇) + ∇(∇ ln n − ∇ ln n) = 0, 3n 3 ∇ ∇(T − T ) − ∇n ∇ (T − T ) − n (T − T ) 2 2  εn ε 3 3n 2 2 2 = (|∇ | − |∇| ) + |∇| − ∇∇T (n − n) −  ∇T ∇( − ) 2 2 2 2 3 + ∇n T ∇( − ) + T ∇∇(n − n) + (n − n)(T − TL (x)) 2

(4.15)

(T − T ) +

and

(4.16)

 ε T T ε 2 2 ∇n − ∇n + (T − T ) − ∇( − ). ( − ) = (|∇ | − |∇| ) + div 2 n n  (4.17)

Then, we obtain from Lemma 9.17 in [9] that, there is a constant C > 0 independent of  such that  − C 2, Cn − n2,q ,

(4.18)

T − T 2,q C(∇n − ∇nq + n − nq + ∇ − ∇q )

(4.19)

 − 2,q C(n − n2,q + T − T 2,q ),

(4.20)

and

combining (4.13) and (4.18)–(4.20), we can deduce the estimates (4.11) and (4.12). This complete the proof.



Acknowledgments I express my deep thanks to my tutor, Professor Zhouping Xin, whose support and encouragement have been necessary to the completion of this work. The paper is partially granted financial support from China Postdoctoral Science Foundation (2005037481).

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