Singular and Degenerate Boundary Value Problems Related to the Electricity Theory

Abstract

The present paper draws attention to the weak solvability of a class of singular and degenerate problems with nonlinear boundary conditions. These problems derive from the electricity theory serving as mathematical models for physical phenomena related to the anisotropic media with “perfect” insulators or “perfect” conductors points. By introducing an appropriate weighted Sobolev space to the mathematical literature, we establish an existence and uniqueness result.

1. Introduction

During the last decades there has been a high interest in the study of the following class of problems: where is an arbitrary domain, is a nonnegative weight, and is a nonlinear function satisfying certain conditions. If the operator involved in (1) is a uniformly elliptic operator, then the existence of solutions can be ensured by classical critical point theory means; see, for example, [1, 2]. But when this is not the case and is a function that has zeros at some points or it is unbounded, the problem becomes much more mathematically complicated and thus the mathematical interest increases. However, the interest in differential operators with degeneracies or singularities is not purely mathematical, since they are used in the study of several physical phenomena. Indeed, the restrictions of the type for all , or for all , are not natural; thus it is worth struggling with all the difficulties that appear in the calculus otherwise. As for the methods utilized to overcome these inconveniences, they depend on the problem. Many papers discuss the existence of solutions for (1) when the weight has the particular form , so that, under some appropriate hypotheses on , a Caffarelli-Kohn-Nirenberg inequality can be applied; see, for example, [3–5] and the references therein.

The present paper approaches a different situation by preserving a higher degree of generality on the weight. Let be an open, bounded, connected subset, with Lipschitz continuous boundary partitioned in three measurable parts such that the Lebesgue measure of is positive. We are interested in the following problem.

Problem 1. Find such that where denotes, as usual, the outward unit normal to , satisfies , and a.e. in with , , and the other functions satisfy , , , and .

Working under the previous conditions, we want to prove that Problem 1 admits a unique weak solution. Our way of surpassing the difficulties created by the presence of the possibly singular or degenerate points is by conducting our study in a weighted Sobolev-type space that will be introduced in the next section. This space appears for the first time in the mathematical literature, although it is similar to the weighted Sobolev space used in [6]. These two studies share the same kind of space to work on because of the following common aspect: they both treat degenerate problems and in both cases we only require on a nonempty subset of the boundary instead of the usual zero Dirichlet boundary condition imposed on the entire boundary. This fact can be correlated with concrete applications and represents a novel trait of our research. Despite some similarities, there are some clear differences between problem (2)–(5) and the problem treated in [6]: the problem from here admits singular points, and, more importantly, it is addressed to a different range of applications. To be specific, while the problem from [6] is a frictional contact problem that arrives from the theory of elasticity, boundary value problems like (2)–(5) arrive from the theory of electricity. Also, we mention that equations of type (2) were introduced as models for various physical phenomena connected to the equilibrium of anisotropic media that possibly are somewhere “perfect” insulators or “perfect” conductors; see [7, page 79].

We will discuss Problem 1 in Section 3. But before that, we present the abstract framework and some preliminary results that serve our goal.

2. Functional Setting and Preliminaries

Let and be a bounded domain with Lipschitz continuous boundary . Let be a function verifying the hypotheses As an example of such function, one can take such that see also [8, Example 2.7].

Starting from the weight described by assumptions (6) we can introduce a weighted Sobolev-type space endowed with the inner product and the corresponding norm Following [9], this weighted Sobolev space associated with norm (10) is a Hilbert space. Using similar arguments to those from [6, Lemma 3.1], one can deduce that if , then , and, in addition, there exists such that On the other hand, according to the Gagliardo trace theorem (see [10]), there exists a linear continuous operator , called the Gagliardo trace operator, such that and there exists such that Everywhere below, by we will denote the Gagliardo trace operator. Taking into consideration the above said, In order to search for weak solutions for problem (2)–(5), we will work on the following subspace of where with . As a closed subspace of the Hilbert space , the space is also a Hilbert space. Besides, defines an inner product on the space . Let us denote by the norm induced by the inner product (16), By adapting the proof of [6, Lemma 3.2] to our present situation, one can see that the norms and are equivalent on . Consequently, is a Hilbert space too.

Let us proceed by recalling the following tool.

Theorem 2. Suppose is a Hilbert space. Let be a Lipschitz continuous operator with the property that Let be a proper, convex, and continuous functional. Then, given , there exists a unique element such that Moreover, the solution depends Lipschitz continuously on .

For a proof of the above theorem, the reader can consult, for example, [11]. We end this section with three preliminary results on convex analysis.

Theorem 3 (see [12, Theorem .]). Let be a reflexive Banach space, and let be Gâteaux differentiable. Then the following conditions are equivalent:(i) is convex;(ii)one has where denotes the dual of the space .

Theorem 4 (see [11, Proposition 1.59]). Let be a real normed space and let be a proper convex lower semicontinuous functional. Then for all ; that is, is subdifferentiable on .

Theorem 5 (see [11, Proposition 1.60]). Let be a real normed space and let be convex. Assume and . If exists as a Gâteaux derivative, then .

Now we are ready to present our existence and uniqueness result.

3. Main Result

In what follows, we consider , where is the function from Problem 1. In order to deliver a weak formulation of Problem 1, we assume that is a strong solution. We multiply (2) by a function , and, after integration by parts, we obtain Due to boundary conditions (3)–(5), we infer that The density result implies that

This allows us to give the definition of the weak solution to our problem.

Definition 6. By a weak solution for Problem 1 one understands a function which verifies for all .

Let us introduce now the main result of our paper.

Theorem 7. There exists a unique weak solution to Problem 1.

To provide a proof for this theorem we rely on Theorem 2, so we prove several auxiliary lemmas.

Lemma 8. There exists such that

Proof. We introduce the form and we claim that is well defined, bilinear, and continuous. Indeed, since , and , function is well defined. Also, from the linearity of the gradient and of the integral operator, one can immediately see that is bilinear. Then, to show the continuity of , it is enough to notice that, by the Hölder inequality, we arrive at that is, With that, our claim is proved and we can apply Riesz’s representation theorem to deduce that there exists We have where, as usual, denotes the Euclidean norm on . Hence our proof is finished.

Let us define now the following application:

Lemma 9. The functional is well defined, continuous, Gâteaux differentiable, and convex.

Proof. Note that Since , we have that ; thus and consequently is well defined. Also, it is not difficult to see that is continuous.
In what concerns the Gâteaux differentiability of , for all we have By Lebesgue’s Theorem, the differentiation under the integral is allowed if, for close to zero, the term is dominated by a fixed function which does not depend on . Indeed, by (32), for . Therefore, It is clear that the application which maps into is linear, so we prove its continuity. We have Using (13), (11), and the equivalence of norms and on , there exists such that for all . This way, we have obtained that is Gâteaux differentiable and As for convexity, due to Theorem 3, it is enough to show that that is, for all . To do so, we start from the fact that for all , . This immediately leads to hence and (40) holds true. With this, we conclude the proof of the lemma.

Furthermore, the following result takes place.

Lemma 10. There exists a unique element such that

Proof. We introduce and we claim that is well defined, linear, and continuous. Indeed, the form is well defined because , , and . The linearity of follows easily from the linearity of the integral operator. Let us focus on the continuity of . First, by the Hölder inequality, for all . Then, by relation (10), we find that for all . At the same time, By (13) and (11) it follows that for all . We take into account the equivalence of the norms and and the linearity of , and then we put together (46), (48), and (50) to deduce that is continuous.
At this point, we have all the ingredients to apply Riesz’s representation theorem. So, we deduce that there exists a unique element such that and the proof is complete.

Keeping in mind all the above assumptions, notations, and results, we are able to introduce the following weak formulation.

Problem 11. Find such that where , , and are the ones from (29), (31), and (45), respectively.

Lemma 12. A function is a weak solution to Problem 1 if and only if it is a solution to Problem 11.

Proof. Let be a weak solution to Problem 1, meaning that verifies Definition 6. Consequently, Keeping in mind relations (26), (29), (39), and (45), by relation (40) it is obvious that is a solution to Problem 11.
Conversely, let us consider now to be a solution to Problem 11. Then . Moreover, due to Lemma 9, we can apply Theorems 4 and 5 to infer that . This yields Taking into consideration (26), (29), (39), and (45), we obtain that is a weak solution to Problem 1.

Finally, we can conclude our main result.

Proof of Theorem 7. Based on the previous results, the proof of Theorem 7 is immediate. Indeed, by Lemma 12, it is enough to prove that Problem 11 has a unique solution. To this end, we use Lemmas 8 and 9 and we apply Theorem 2.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.