Volume 2008, Article ID 961045,14pages doi:10.1155/2008/961045
Research Article
Global Existence and Extinction of
Weak Solutions to a Class of Semiconductor
Equations with Fast Diffusion Terms
Bin Wu
Department of Information and Computing Sciences, College of Mathematics and Physics, Nanjing University of Information Science and Technology, Nanjing 210044, China
Correspondence should be addressed to Bin Wu,[email protected]
Received 2 March 2008; Accepted 13 August 2008
Recommended by Y. Giga
We consider the transient drift-diffusion model with fast diffusion terms. This problem is not only degenerate but also singular. We first present existence result for general nonlinear diffusivities for the Dirichlet-Neumann mixed boundary value problem. Then, the extinction phenomenon of weak solutions for the homogeneous Dirichlet boundary problem is studied. Sufficient conditions on the extinction and decay estimates of solutions are obtained by usingLp-integral model estimate method.
Copyrightq2008 Bin Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
The scaled semiconductor drift-diffusion model reads
−∇·∇ψ p−nCx, 1.1
nt− ∇ ·Jn rn, p1−np g, Jn ∇nm−n∇ψ, 1.2
pt∇ ·Jp rn, p1−np g, −Jp∇pmp∇ψ, 1.3
ifm 1exponents. The regimem > 1 describes a slow diffusion process in the electron
holedensity, whereas 0< m <1 is related to fast diffusion.
We supplement these equations with physically motivated boundary conditions 1:
ψ, n, p ψD, nD, pD, x, t∈ΣD≡ΓD×0, T, 1.4
∂ψ
∂η, ∂n ∂η,
∂p ∂η
0,0,0, x, t∈ΣN≡ΓN×0, T, 1.5
n, p n0, p0
, x∈Ω, t0, 1.6
where ∂Ω splits into two disjoint subsets ΓN and ΓD.ΓN models the union of insulating
boundary segments andΓDthe union of ohmic contacts.
The standard drift-diffusion model corresponding tom 1 has been mathematically and numerically investigated in many paperssee 1–4. Existence and uniqueness of weak solutions have been shown. Recently, the existence analysis of the bipolar drift-diffusion problem in the adiabatic case has been studied by many authors 5–9. But, as far as we know, few works are concerned with the mixed boundary value problem of the drift-diffusion model with the fast diffusion terms.
The phenomenon of extinction is an important property of solutions for many evolutionary equations which have been studied extensively by many authors; see, for example, 10,11. Especially, there are some papers concerning the extinction for fast diffusive equations. For instance, in 12, the extinction and positivity for the evolutionaryp-Laplacian equation without sources were studied by some authors, and Li and Wu 13investigated the extinction for a fast diffusive filtration equation with nonlinear source terms. But, most of the work about extinction results is limited to single equation.
We prove first the global existence of weak solutions to problem1.1–1.6in the sense of the definition below. Wheng ≡ 0 and the boundary condition is homogeneous and of Dirichlet type, we show that the solutionsnx, t, px, tvanish in finite time in the following two cases.
iThe initial data are appropriately small.
iiThe doping profile is sufficiently small and generation rate grows very slowly even if the initial data are very large.
Simultaneously, the decay estimates of solutions are obtained. For the proof of our result, we employLp-integral model estimate method and a crucial lemma on differential inequality. This technique has been successfully applied to the porous medium equation 14.
We make the following assumptions.
H1 Ω⊂RN N1,2,3is bounded and∂Ω∈C0,1, whose outward normal vector isη.
H2Cx, gx∈L∞Ωandgx≥0 for a.e.x∈Ω.
H3rn, pis a locally Lipschitz continuous function defined forn, pand 0≤rn, p≤
r <∞.
H4nmD, pmD, ψD∈H1Ω∩L∞Ω, andnD, pD≥0 inΩ.
Remark 1.1. It is obvious thatH4implies thatnD, pD∈H1Ω∩L∞Ω.
Let
Y ω∈H1Ω|ω|
ΓD 0
1.7
and letY∗denote the dual space ofY.
Definition 1.2. ψ, n, p is called the weak solution to problem 1.1–1.6 if n ∈ nD L20, T;Y, p∈pDL20, T;Y, ψ ∈ψDL20, T;Y, nt, pt∈L20, T;Y∗, n|t
0n0, p|t0 p0, and there hold
Ω∇ψ· ∇φdx
Ωp−nCφdx, ∀t∈0, T, ∀φ∈Y,
nt, v
T
0
Ω
∇nm−n∇ψ· ∇vdxdt
T
0
Ω rn, p1−np gvdxdt, ∀v∈L
20, T;Y,
pt, v
T
0
Ω
∇pmp∇ψ· ∇vdxdt
T
0
Ω
rn, p1−np gvdxdt, ∀v∈L20, T;Y.
1.8
2. Existence
This section is devoted to the proof of global existence of weak solutions to problem1.1–
1.6. We will prove the following existence theorem.
Theorem 2.1. Under hypotheses (H1)–(H4), there exists at least one weak solution of problem1.1–
1.6.
The following lemma will be used in the proof of existence result.
Lemma 2.2 see 9. Let (H1) hold,α, ν, F ∈ L∞QT, and F, ν ≥ 0 for a.e.x, t ∈ QT, θ0 ∈ L∞
Ω, θ∈H1Ω∩L∞Ω, γ∈L2QTN, γ·η0, divγi∈L∞QTweakly, andα≥ε >
0. Then, there exists a unique solutionθof the following problem:
θt− ∇ ·α∇θ−γθ νθF, x, t∈QT,
θ|ΣD θ,
∂θ ∂η
ΣN
0, θ0 θ0, x∈Ω,
2.1
such thatθ∈L20, T;Y, θt∈L20, T;Y∗
0, and
0≤θ≤leλt a.e.x, t∈QT, 2.2
Our main difficulty in the proof is that problem 1.1–1.6 is degenerate at points wheren, p ∞, and is singular at points wheren, p0. This difficulty leads us to consider the following auxiliary regularized problem with the initial and boundary conditions1.4–
1.6:
−∇ ·∇ψ pk−nkCx, 2.3
nt− ∇ ·
m
1
nkτ
1−m
∇n−n∇ψ
rnk, pk1−npkg, 2.4
pt− ∇ ·
m
1
pkτ
1−m
∇pp∇ψ
rnk, pk1−nkpg, 2.5
where 0< τ <1 andskmin{k,max{0, s}}.
Now, we use the Schauder fixed point theorem 15to prove the existence of weak solution to the above regularized problem.
Lemma 2.3. Under hypotheses (H1)–(H4), there exists at least one weak solution of problem2.3–
2.5, under conditions1.4–1.6.
Proof. Define the setκas
κx∈L2QT|x≥0 a.e. x∈QT,x
L2QT ≤R
. 2.6
It is obvious thatκis a closed convex set and weakly compact inL2QT. Givenn, p∈κ, we
consider the following problems:
−∇ ·∇ψ pk −nk Cx, 2.7
ψ|ΣD ψD,
∂ψ ∂η
ΣN
0, 2.8
nt− ∇ ·
m 1
nkτ
1−m
∇n−n∇ψ
rnk,pkpknrnk,pkg, 2.9
n|ΣD nD,
∂n ∂η
ΣN
0, n0 n0, x∈Ω, 2.10
pt− ∇ ·
m 1
pkτ
1−m
∇pp∇ψ
rnk,pknkprnk,pkg, 2.11
p|ΣD pD,
∂p ∂η
ΣN
0, p0 p0, x∈Ω. 2.12
We deduce the existence of a unique weak solution of 2.7 and 2.8 with the regularity
ν rnk, pk pk , andF rnk, pk g.Lemma 2.2ensures the existence and uniqueness of weak solution of problem2.9-2.10with the regularityn∈L20, T;H1Ω∩L∞QT, where
L∞ estimate ofnis dependent onk, τ and the known data, but independent ofR by2.2. Similar results hold for problem2.11-2.12.
Thus, the map
S:κ2−→L2QT2, n, p−→n, p 2.13
is well defined and compact. For the above given R, Sκ2 → κ2 holds if we choose R
sufficiently large. By using the standard methodfor details, see 5or 9, we can obtain the continuity of mapS. Thus, existence of a fixed point ofS follows the weak solution of problem2.3–2.5, under conditions1.4–1.6.
Lemma 2.4. The weak solutions of problem2.3–2.5, under conditions1.4–1.6, satisfy the estimates
0≤nx, t, px, t≤k, a.e.x, t∈QT, 2.14
if one choosesksuch that
k≥M:max
sup
Ω
n0, p0
,sup
ΣD
nD, pDergL∞ΩCxL∞ΩT. 2.15
Proof. SetNne−βtandP pe−βt; hereβ >0 is to be determined. Then,N, Psatisfies
NtβN− ∇ ·
m
1
eβtN kτ
1−m
∇N−N∇ψ
e−βtrn
k, pk1−npkg,
2.16
PtβP− ∇ ·
m
1
eβtP kτ
1−m
∇PP∇ψ
e−βtrnk, pk1−nkpg, x, t∈QT,
2.17
eβtN, eβtPnD, pD, x, t∈ΣD, 2.18
∂N
∂η, ∂P ∂η
0,0, x, t∈ΣN, 2.19
N, P n0, p0
, x∈Ω, t0. 2.20
To obtain the upper bound, we compareNandPwithzt:k0eμ−βt, where
k0≥max
sup
Ω
n0, p0
,sup
ΣD
and μ such that μ > β is a large constant which can be determined by the following calculations. Then, we could takeN−zandP−zas test functions in2.16and2.17, respectively, and obtain
1 2
Ω
N−z2
t P−z2
tβ
t
0
Ω
N−z2
P−z2
t 0 Ωm 1
eβtN kτ
1−m
∇N−z2t 0
Ωm
1
eβtP kτ
1−m
∇P−z2
≤ −μ
t
0
Ωz
N−z P−z
t
0
Ω
N· ∇N−z−P· ∇P−z· ∇ψ
t
0
Ωe
−βtrn
k, pkgN−z P−z
≤ −μ
t
0
Ωz
N−z P−N1
2 t 0 Ω
pk−nkCxN−z2
−P−z2
t 0 Ω
rgzN−zP−Nt
0
Ωz
pk−nkCxN−z P−z
≤−μrgL∞ΩCxL∞Ω
t
0
Ωz
N−z P−z
1
2CxL∞Ω t
0
Ω
N−z2
P−z2
,
2.22
where in the last inequality we have used
pk−nkN−z−P−ze−βtpk−nkn−k0eμt
−p−k
0eμt
≤
0. 2.23
Choosingβ 1/2CxL∞Ωandμ≥rgL∞ΩCxL∞Ω, we get
1 2
Ω
N−z2
t P−z2
tm
1
kτ
1−mt
0
Ω
∇N−z2∇P−z2≤0.
2.24
Thus, we obtain a weak solution of the following problem:
−∇ ·∇ψ p−nCx, 2.25
nt− ∇ ·
m
1
nτ
1−m
∇n−n∇ψ
rn, p1−np g, 2.26
pt− ∇ ·
m
1
pτ
1−m
∇pp∇ψ
rn, p1−np g, 2.27
with the initial and boundary conditions1.4–1.6.
In what follows, we give some estimates of the weak solutions to the above problem uniformly inτ, which are necessary in the proof ofTheorem 2.1.
Lemma 2.5. The solutions of problem 2.25–2.27, under conditions 1.4–1.6, satisfy the estimates m2 T 0 Ω 1
nτ
2−2m
∇n2 1
pτ
2−2m
∇p2
≤C, 2.28
whereCis independent ofτ.
Proof. First of all, a standard elliptic estimate gives
∇ψL2QT≤C
1nL2QTpL2QT
≤C. 2.29
Let
ϕn, nDm
n
nD
1
sτ
1−m
ds. 2.30
Sinceϕn, nD|ΓD 0 and T
0
Ω
∇ϕ
n, nD2≤
2m2 T 0 Ω 1
nτ
2−2m
∇n2 1
nDτ
2−2m
∇nD2
≤2m2
1
τ
2−2m
n2
L20,T;H1ΩnD
2
L20,T;H1Ω
≤Cτ,
2.31
we haveϕn, nD∈L20, T;Y. Then,
m2 T 0 Ω 1
nτ
2−2m
|∇n|2m2
T 0 Ω 1
nτ
1−m
∇n·
1
nDτ
1−m
∇nD − T 0 Ωntϕ n, nD T 0
Ωn∇ψ· ∇ϕ
n, nD T 0 Ω
rn, p1−np gϕn, nD
I1I2I3I4.
By Young’s inequality and2.29, we have
I1≤ m2 4 0 Ω 1
nτ
2−2m
|∇n|2
T 0 Ω ∇ nm
D2, 2.33
I3≤
1 8 T 0 Ω
∇ϕn, nD24
T
0
Ω|
n∇ψ|2
≤ m2 4 T 0 Ω 1
nτ
2−2m
|∇n|2 m2
4 T 0 Ω ∇ nm
D24M2∇ψ2L2QT
≤ m2 4 T 0 Ω 1
nτ
2−2m
|∇n|2C.
2.34
Furthermore,
I4≤mr1M2gL∞Ω
T 0 Ω n 0 1
sτ
1−m
ds nD
0
1
sτ
1−m
ds
≤2mr1M2gL∞ΩQT M
0
1
sτ
1−m
ds≤C.
2.35
For the termI2, we see that
I2≤
T 0 Ωnϕt n, nD
ΩnTϕ
n, nDT
ΩnTϕ
n, nD0
≤m T 0 Ω nnt 1
nτ
1−m
4M|Ω| M
0
1
sτ
1−m
ds ≤m T 0 Ω ∂ ∂t n 0
s 1
sτ
1−m
ds C ≤m Ω
nx,T
n0x
s
1
sτ
1−m
ds
C≤C.
2.36
Inserting2.33–2.36into2.32, we finally get
m2 T 0 Ω 1
nτ
2−2m
|∇n|2≤C. 2.37
A similar estimate holds forp, and then we complete the proof.
Lemma 2.6. The weak solutions nand p of problem 2.25–2.27, under conditions1.4–1.6, satisfy the following estimate:
Proof. We will only prove that2.38holds forn. The rest can be derived similarly. Without loss of generality, we assume 0< τ <1. Then, we conclude that
T
0
Ω|∇
n|2
T
0
Ω
nτ2−2m 1
nτ
1−m
|∇n| 2
≤M12−2m T
0
Ω
1
nτ
1−m
|∇n| 2
≤C.
2.39
Following Lemmas2.5and2.6, we can easily get the following lemma.
Lemma 2.7. The weak solutions nand p of problem 2.25–2.27, under conditions1.4–1.6, satisfy the following estimate:
ntL2
0,T;Y∗, ptL20,T;Y∗≤C. 2.40
Proof ofTheorem 2.1. Let {ψτ, nτ, pτ} be the sequence of solutions of problem 2.25–2.27, under conditions 1.4–1.6. By passing to a subsequence, if necessary, from Lemmas 2.4– 2.7, we infer that
nεt,pεt−→nt, ptweakly inL20, T;Y∗,
ψε, nε, pε−→ψ, n, pstrongly inL2QT, a.e.inQT,
ψε, nε, pε−→ψ, n, pweakly inL20, T;H1Ω,
∇nτm,∇pτm−→∇nm,∇pmweakly inL2QT.
2.41
Now, we can conclude thatψ, n, p is a weak solution of problem1.1–1.6by standard method, and then complete the proof ofTheorem 2.1.
3. Extinction
Besides the assumptionsH1–H5, we need for our extinction result the following structural condition onrn, p.
H6rn, p≤ρ0npfor some positive constantρ0.
Becausern, pdenotes the generation term in the recombination-generation rate, this condition means that the generation rate grows very slowly inn, pwhenρ0is very small.
Additionally, we replace boundary conditions1.4-1.5with homogeneous Dirichlet conditions
npψ0, x∈∂Ω×0, T. 3.1
It is obvious that the existence of weak solutions of problem1.1–1.3, under conditions
Before stating our main result in this section, we list the following important condi-tions inTheorem 3.1:
CxL∞
Ω2ρ0≤
1 2C
−2
0 |Ω|−2m/m1N−2/N
×n0mLm11Ωp0mLm11Ω
m−1/m1
,
3.2
n0m1
Lm1Ωp0
m1 Lm1Ω≤
C−2
0 |Ω|−2m/m1N−2/N
CxL∞Ω2ρ0
1m/1−m
, 3.3
CxL∞Ω2ρ0≤
2ms
ms2C
−2 0
n0sLs11Ωp0sLs11Ω
m−1/s1
, 3.4
n0s1
Ls1Ωp0
s1 Ls1Ω≤
4msC−02
ms2Cx
L∞Ω2ρ0
1s/1−m
, 3.5
wheres N1−m/2−1 andC0is the embedding constant of the following inequality: uL2N/N−2Ω≤C0∇uL2Ω, ∀u∈H01Ω. 3.6 Theorem 3.1. Let (H1)–(H6) be fulfilled, letgx≡0, and letψ, n, pbe a weak solution of problem
1.1–1.3, under conditions3.1and1.6. Then, one has the following.
iIf 3.2or 3.3below is satisfied, then ux, tand px, tvanish in the finite time for
N−2/N2≤m <1, and n·, tm1
Lm1Ωp·, t
m1
Lm1Ω≤B1e−α1t, t∈
0, T01
,
n·, tm1
Lm1Ωp·, t
m1 Lm1Ω
≤n·, T01Lmm11Ωp
·, T01mLm11Ω
1−m/1m
−C11−m
1m t
1m/1−m
,
t∈T01, T1
,
n·, tm1
Lm1Ωp·, t
m1
Lm1Ω≡0, t∈
T1,∞
.
3.7
iiIf 3.4or 3.5below is satisfied, then ux, tand px, tvanish in the finite time for 0< m <N−2/N2, and
n·, ts1
Ls1Ωp·, t
s1
Ls1Ω≤B2e−α2t, t∈
0, T02, n·, ts1
Ls1Ωp·, t
s1
Ls1Ωn·, t
s1
Ls1Ωp·, t
s1 Ls1Ω
≤n·, T02s1 Ls1Ωp
·, T02s1 Ls1Ω
1−m/1s
−C21−m 1s t
1s/1−m
,
t∈T02, T2
,
n·, ts1
Ls1Ωp·, t
s1
Ls1Ω≡0, t∈
T2,∞
.
3.8
To obtain the above result, we will use the following lemma which is of crucial impor-tance to the proof.
Lemma 3.2. Let0< k < q≤1, and letyt≥0be a solution of the differential inequality
dy dt βy
k≤σyq t≥0, y0 y
0≥0, 3.9
whereβandσare positive constants. If
iy0< β
σ
1/q−k
or (iiσ≤ βy
k−q 0
2 3.10
holds, then there existα >0andB >0such that
0≤yt≤Be−αt, t≥0. 3.11
Proof. We will use upper and lower solutions’ methods to complete the proof. The proof of caseiis basically the same as method of caseii. So, we are only devoted to the proof of casei. In fact, we only need to chooseα, Bproperly such thatBe−αtis an upper solution of
3.9. That is,αandBmust satisfy
−αBe−αtβBke−αkt≥σBqe−αqt t≥0, B≥y
0. 3.12
ChooseBsuch that lnB <lnβ/σ/q−kandαsufficiently small. Then,
σ βBq−k
α
βB1−k≤1, 3.13
which implies that
αq−kt≥ln
σ βBq−k
α βB1−k
, ∀t≥0. 3.14
Noting that 0< q≤1, therefore we have
e−αk−qt≥ σ βBq−k
α
βB1−ke−1−qαt, ∀t≥0, 3.15
which is equivalent to3.12ify0≤B <β/σ1/q−k. The lemma is proved.
Proof ofTheorem 3.1. We consider first the caseN−2/N2≤m <1. Multiplying1.2and
1.3bynmandpm, respectively, and employing the Poisson equation1.1, we get
1
m1
d dt
Ω
nm1pm1dx
Ω
∇
nm2∇pm2
dx
m
m1
Ω
p−nCxnm1pm1dx
Ωrn, p1−np
nmpmdx
I1I2.
The embedding theorem gives
nm
Lm1Ω≤ |Ω|m/m1−N−2/2NnmL2N/N−2Ω
≤C0|Ω|m/m1−N−2/2N∇
nm
L2Ω,
3.17
and similarly forp.
By Young’s inequality, we have
I2≤ρ0
Ωnp
nmpmdx≤2ρ 0
Ω
nm1pm1dx. 3.18
By3.16–3.18, taking into accountp−nnm1−pm1≤0, we obtain
d
dtf1t m1C −2
0 |Ω|−2m/m1N−2/Nf1t
2m/m1≤m1Cx
L∞Ω2ρ0
f1t,
3.19
where
f1t nmLm11ΩpmLm11Ω. 3.20
ByLemma 3.2, there existα1andB1such that
0≤f1t≤B1e−α1t, t≥0, 3.21
if3.2or3.3holds.
Furthermore, there existT01such that
1mC−2
0 |Ω|−2m/m1N−2/N−CxL∞Ω2ρ0
f1t1−m/1m
≥1mC0−2|Ω|−2m/m1N−2/N−CxL∞Ω2ρ0
B1e−α1T01
1−m/1m
:C1 >0
3.22
holds fort∈ T01,∞. Thus, whent∈ T01,∞,3.19turns to
d
dtf1t C1f1t
2m/m1≤0, t≥T
01. 3.23
By a direct calculation, we deduce that
f1t≤
f1
T01
1−m/1m− C11−m
1m t
1m/1−m
, t∈T01, T1
,
f1t≡0, t∈
T1,∞
,
3.24
where
T1
1m
C11−m
n·, T01mLm11Ωp
·, T01mLm11Ω
1−m/1m
For the case 0< m < N−2/N2, usingns andpsas test functions in1.2and
1.3, respectively, we obtain
1
s1
d dt
Ω
ns1ps1dx 4ms
ms2
Ω
∇
nms/22∇pms/22
dx
ss
1
Ω
p−nCxns1ps1dx
Ωrn, p1−np
nspsdx.
3.26
To simplify, we denote
f2t ns1
Ls1ΩpLss11Ω. 3.27
Hence, we have
d dtf2t
4s1ms
ms2 C− 2 0 f2t
ms/1s≤s1Cx
L∞Ω2ρ0
f2t, 3.28
where we have used
uLms1Ωs/2
Ωu
ms/2·2N/N−2dx
N−2/2N
≤C0∇
ums/2
L2Ω, ∀u∈H01Ω.
3.29
ByLemma 3.2, there existα2andB2such that
0≤f2t≤B2e−α2t, t≥0, 3.30
if3.4or3.5holds.
By a simple analysis similar to that of the caseN−2/N2≤m <1, we obtain
f2t≤
f2
T02
1−m/1s− C21−m
1s t
1s/1−m
, t∈T02, T2
,
f2t≡0, t∈
T2,∞
,
3.31
where
C2 s1
4ms
ms2C− 2
0 −CxL∞Ω2ρ0
B2e−α2T021−m/1s
>0, 3.32
T2
1s
C21−m
n·, T02sLs11Ωp
·, T02sLs11Ω
1−m/1s
. 3.33
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