Research Article | Open Access

Bin Wu, "Existence of Weak Solutions to a Class of Degenerate Semiconductor Equations Modeling Avalanche Generation", *Mathematical Problems in Engineering*, vol. 2009, Article ID 873683, 22 pages, 2009. https://doi.org/10.1155/2009/873683

# Existence of Weak Solutions to a Class of Degenerate Semiconductor Equations Modeling Avalanche Generation

**Academic Editor:**Shijun Liao

#### Abstract

We consider the drift-diffusion model with avalanche generation for evolution in time of electron and hole densities , coupled with the electrostatic potential in a semiconductor device. We also assume that the diffusion term is degenerate. The existence of local weak solutions to this Dirichlet-Neumann mixed boundary value problem is obtained.

#### 1. Introduction

In this paper, we consider the following degenerate semiconductor equations modeling avalanche generation: with initial and boundary conditions Here the unknowns and denote the electrostatic potential, the electron density, and the hole density, respectively. The boundary consists of two disjoint subsets and . The carrier densities and the potential are fixed at (Ohmic contacts), whereas models the union of insulating boundary segments. represents the electron current, and is the analogously defined physical quantity of the positively charged holes. Function denotes the doping profile (fixed charged background ions) characterizing the semiconductor under consideration, while the term models the effect of impact ionization (avalanche generation of charged particles) (cf. [1, 2] for details). is the net recombination-generation rate, where characterizes the mechanism of particle transition. The constant is the adiabatic or isothermal (if ) exponent. The regime describes a fast diffusion process in the electron (hole) density, whereas is related to slow diffusion.

The standard drift-diffusion model corresponding to has been mathematically and numerically investigated in many papers (see [3–6]). Existence and uniqueness of weak solutions have been shown. The standard model can be derived from Boltzmann's equation once assumed that the semiconductor device is in the low injection regime, that is, for small absolute values of the applied voltage. In [7] Jüngel showed that in the high-injection regime diffusion terms are no longer linear. A useful choice for is . In this case, the parabolic equations (1.2) and (1.3) become of degenerate type, and existence of solutions does not follow from standard theory. Recently, many authors [8–10] have studied the existence and uniqueness of weak solutions of this type of degenerate semiconductor equations without avalanche generation term. In [9], the degenerate semiconductor equations based on Fermi-Dirac statistics were introduced by Jüngel for the first time. The existence and uniqueness results are shown under the assumption that the solution of Poisson equation with Dirichlet-Neumann mixed boundary conditions had the regularity , this amounts to a geometric condition on , for example and . ([11, Theorem 3.29]). Then Guan and Wu [8] obtain similar results without the assumption above.

There are some papers concerning the semiconductor equations modeling avalanche generation. For instance, the existence of weak solutions of nondegenerate stationary problem has been investigated in [12, 13]. When , that is, the diffusion term is not degenerate, the authors [14] obtained the existence of local weak solutions of problem (1.1)–(1.6).

Our main goal in this paper is to study the existence of weak solutions of problem (1.1)–(1.6). In contrast to the above works, degeneration of diffusion term we are going to study introduces significant new technical difficulties to estimate the avalanche term.

We make the following assumptions:

(H1) is bounded and , whose outward normal vector is and , ;(H2);(H3) is a locally Lipschitz continuous function defined for and ;(H4), = const., ;(H5), , and , in ;(H6) and a.e. in .Let

*Definition 1.1. * is called the weak solution to the problem (1.1)–(1.6) if , , , , and there hold
Our main result in this paper is as follows.

Theorem 1.2. *Under hypotheses (H1)–(H6), there exists at least one local weak solution to the problem (1.1)–(1.6).*

#### 2. Approximate Problem

For simplicity, we assume that . As [14], we first construct the following bound approximate sequence of avalanche generation term : here . Obviously, .

Now we introduce the following approximate problem with the initial and boundary conditions (1.4)–(1.6):

This section is devoted to the proof of global existence of weak solutions to the above approximate problem (2.2)–(2.4), (1.4)–(1.6). We will prove the following existence theorem.

Theorem 2.1. *Under hypotheses (H1)–(H6), there exists at least one global weak solution to the problem (2.2)–(2.4), (1.4)–(1.6).*

The proof is based on Schauder's fixed pointed theorem. The main difficulty in the proof is that problem (2.2)–(2.4), (1.4)–(1.6) is degenerated at points where . This difficulty leads us to consider the following auxiliary regularized problem with the initial and boundary conditions (1.4)–(1.6): where and for any and .

Let and . It is obvious that is a closed convex set and weakly compact in .

The theory of linear elliptic boundary value problems [15] gives a unique such that where is dependent on and the norms for and , but not on .

Next, for the unique weak solution to problem (2.10)–(2.11), we consider the following problem:

Lemma 2.2. *Under hypotheses (H1)–(H6), there exists one global and unique weak solution to the problem (2.13)–(2.16). **Further there are bounds on and which depend on and the known data, but not on . Similar estimates also hold for . *

*Proof. *We begin by choosing a constant such that
and observe that satisfies (2.13)–(2.16) if and only if satisfies
or if and only if satisfies
where
Clearly, .

First, we prove the uniqueness of weak solution to the problem (2.13)–(2.16) which is equivalent to (2.18)–(2.21). Let be two weak solutions to the problem (2.18)–(2.21), then satisfies

here
such that
Take as test functions in (2.27), (2.28), respectively. By (2.32) and Hölder inequality,
Thus the uniqueness is established by Gronwall's inequality.

We are now in a position to prove the existence result. Define , . Clearly, is a Hilbert space with respect to the scalar product

Set, for ,
The operator is well defined and bounded (because ). To prove the existence result by using [16, Theorem 30.A], it suffices to verify that the operator is hemicontinuous, monotone, and coercive.

Note that

for any The hemicontinuity of is easily obtained by the standard method.

For the monotone, we first notice that

Hence
By the choice of , we can easily obtain the monotone of . Moreover, from (2.38) we also know that the operator is coercive.

Therefore, there exists a unique with such that in and

Especially,
That is, is a weak solution to the problem (2.22)–(2.25).

Finally, noting that we can easily establish the bounds on and by the standard energy estimate.

Lemma 2.3. *Under hypotheses (H1)–(H6), there exists at least one global weak solution to the problem (2.5)–(2.7), (1.4)–(1.6). *

*Proof. *We define the mapping as
with solution of (2.13)–(2.16). From Lemma 2.2, we know that is well defined and compact. Indeed, lies in a bounded sunsets of and lies in a bounded subset of . Since the injection is compact, we conclude from Aubin's lemma that is relatively compact in . And for given and , holds if we choose large enough.

To apply Schauder's fixed point theorem, we still need to prove that the mapping is continuous. Consider any sequence strongly in and let . Since is relatively compact in and bounded in , we can extract subsequences such that

We only have to show . To do this, we only need to prove
where is a test function, and are solutions of (2.10)–(2.11) corresponding to , respectively. The reminder of convergence proof is standard (details see [8] or [9]). Use as test function in a modification of (2.13) in which the functions have been replaced by respectively. Then we have
A similar estimate holds for . Then adding the two inequalities and using strongly in (details see [9]) and (2.42), we conclude that This implies that a.e. in . Then we can easily prove (2.43) by using Vitali's theorem.

Now existence of a fixed point of follows which is a solution of (2.5)–(2.7), (1.4)–(1.6).

To obtain the existence result of problem (2.2)–(2.4), (1.4)–(1.6), the following estimates on uniformly in are necessary.

Lemma 2.4. *The solutions of problem (2.5)–(2.7), (1.4)–(1.6) satisfy the estimates
**
where is dependent on and the known data, but not on . *

*Proof. *By taking as test function in (2.6), we have
By taking into account in and the nonnegativity of and we obtain
and thus a.e. in . Similarly, we have a.e. in .

To obtain the upper bound set

and use as test function in (2.6), then
Adding the equality (2.49) for and a similar inequality for , we get
Noticing that
and applying Hölder inequality, we further have
where is independent of and . Gronwall's inequality then implies that
for all and . Since the right-hand side of this equality does not depend on , we can let and then to obtain the desired upper bound.

Thus, taking large enough, we see that solves subject to the initial and boundary conditions (1.4)–(1.6).

*Proof of Theorem 2.1. *Noticing that the function is bound, we can obtain the following convergence properties by using the same method as the proof in [8, Theorem 1.1] and Lemma 3.2:
In addition, a standard elliptic estimate gives
from which we conclude
and furthermore
Next, using as test function in (2.55), we get
as , where we have used
and (2.57)–(2.60). The same argument shows that as .

Thus, there exists a subsequence (not relabeled) such that almost everywhere in as . Then it follows from Vitali's theorem that

Now we can conclude that is the solution of the problem (2.2)–(2.4), (1.4)–(1.6) from the above convergence by standard method and then complete the proof of Theorem 2.1.

#### 3. Proof of the Main Result

In the last section, we prove that there is at least one global weak solution to the problem (2.2)–(2.4), (1.4)–(1.6) for every given . In the following what we need to do is to prove that the limit of is a solution of (1.1)–(1.6). To this end, we first give some uniform estimates for the problem (2.2)–(2.4), (1.4)–(1.6). For simplicity, we drop the subscript of and set

Lemma 3.1. *, there holds
**
where , satisfy
*

This is the well-known Gagliardo-Nirenberg Inequality [15].

Lemma 3.2. *If is the solution of the problem (2.2)–(2.4), (1.4)–(1.6), the following estimate holds:
**
where is a sufficiently small constant.*

*Proof. *First of all, the following estimate of
can follow from the standard techniques in [17]. Then by taking as test function in (2.2), we have
The assumptions we have made and Poincaré inequality yield
Using as test function for (2.2), and noting that we get
A similar estimate for follows from a same procedure. Then the proof is completed.

Lemma 3.3. *If is the solution of the problem (2.2)–(2.4), (1.4)–(1.6), there holds that
**
for some sufficiently small ; here positive constant is independent of .*

*Proof. *Without loss of generality, we assume and the or case is easier.*Case 1. *. In this case, from Hölder inequality and (3.3), we have
Noting that , we can choose a constant such that . Then by taking and as test function in (2.3) and (2.4), respectively, and adding them together, we get
We estimate the right-hand side term by term. Due to the equation of and (3.6), we obtain
where we use that and when . and can be bounded as
where we have used the nonnegativity of and . Inserting (3.13)–(3.15) into (3.12) we conclude that
Similarly, taking and as test function in (2.3) and (2.4), we conclude that
Choosing sufficiently small, and then summing (3.16) and (3.17), we have
Now we estimate the term (or ). Due to Lemma 2.2 we get
where satisfy
Consequently we obtain, taking into account (3.11) and the choice of ,
for sufficiently small and such that , where only depends on .

This proves that

for some sufficiently small by Lemma 2.2 of the appendix in [14] , and thus
Using the above estimates and (3.4)–(3.6), we can obtain (3.10).*Case 2. *