# Singular and Degenerate Boundary Value Problems Related to the Electricity Theory

Authors: Maria-Magdalena Boureanu; Andaluzia Matei
Journal: Mathematical Problems in Engineering
 (2015)

---

## 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.


---

## Body

## 1. Introduction

During the last decades there has been a high interest in the study of the following class of problems:(1)div(μ(x)∇u(x))+f(x)=0inΩ,u(x)=0on⁡∂Ω,
where Ω is an arbitrary domain, μ is a nonnegative weight, and f 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 μ 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 μ(x)≥const.>0 for all x, or μ(x)≤const.<∞ for all x, 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 μ(x)=|x|b, so that, under some appropriate hypotheses on b, 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. LetΩ⊂R2 be an open, bounded, connected subset, with Lipschitz continuous boundary Γ partitioned in three measurable parts Γ1,Γ2,Γ3 such that the Lebesgue measure of Γ1 is positive. We are interested in the following problem.Problem 1.
Findu:Ω¯→R such that
(2)div(β(x)∇u(x))+q0(x)=0inΩ,(3)u(x)=0on⁡Γ1,(4)β(x)∂u∂ν(x)=q2(x)on⁡Γ2,(5)β(x)∂u∂ν(x)=-kuxu2x+αon⁡Γ3,
where ν denotes, as usual, the outward unit normal to Γ, β satisfies β, β-1∈L2(Ω) and β>0 a.e. in Ω with inf⁡Ω¯β=0, sup⁡Ω¯β=∞, and the other functions satisfy k, α>0, q0∈L2(Ω), and q2∈L∞(Γ2).Working under the previous conditions, we want to prove that Problem1 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 u=0 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 Problem1 in Section 3. But before that, we present the abstract framework and some preliminary results that serve our goal.


## 2. Functional Setting and Preliminaries

LetN≥2 and Ω⊂RN be a bounded domain with Lipschitz continuous boundary Γ. Let ζ:Ω→R be a function verifying the hypotheses
(6)ζ∈L4(Ω),ζ-1∈L4(Ω),ζ(x)≠0a.e.on⁡Ω,inf⁡x∈Ω¯ζ2(x)=0,sup⁡x∈Ω¯ζ2(x)=∞.
As an example of such function, one can take ζ:(-1,1)×(-1,1)→R such that
(7)ζ(x1,x2)=1forx1≤0,1-x1x21/4forx1>0,x2>0,1-x11/4forx1>0,x2≤0;
see also [8, Example 2.7].Starting from the weightζ described by assumptions (6) we can introduce a weighted Sobolev-type space
(8)Hζ1(Ω)=v:Ω⟶R∣v∈L2Ω,ζ∇v∈L2Ω2,
endowed with the inner product
(9)u,vHζ1(Ω)=u,vL2(Ω)+ζ∇u,ζ∇vL2Ω2
and the corresponding norm
(10)vHζ1(Ω)=vL2Ω2+ζ∇vL2(Ω)221/2.
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 v∈Hζ1(Ω), then v∈W1,1(Ω), and, in addition, there exists cE=cE(Ω,ζ)>0 such that
(11)vW1,1(Ω)≤cEvHζ1(Ω),∀v∈Hζ1(Ω).
On the other hand, according to the Gagliardo trace theorem (see [10]), there exists a linear continuous operator γ:W1,1(Ω)→L1(Γ), called theGagliardo trace operator, such that
(12)γv=v|Γifv∈W1,1(Ω)∩C(Ω¯)
and there exists cG=cG(Ω)>0 such that
(13)γvL1(Γ)≤cGvW1,1(Ω)foreachv∈W1,1(Ω).
Everywhere below, by γ we will denote the Gagliardo trace operator. Taking into consideration the above said,
(14)foreveryv∈Hζ1(Ω)wededucethatγv∈L1(Γ).
In order to search for weak solutions for problem (2)–(5), we will work on the following subspace of Hζ1Ω:(15)Vζ=v∈Hζ1Ω∣γv=0a.e.on⁡Γ1,
where Γ1⊂Γ with meas⁡(Γ1)>0. As a closed subspace of the Hilbert space Hζ1(Ω),(·,·)Hζ1(Ω),·Hζ1(Ω), the space Vζ,(·,·)Hζ1(Ω),·Hζ1(Ω) is also a Hilbert space. Besides,
(16)u,vVζ=∫Ω‍ζ2(x)∇u(x)·∇v(x)dx
defines an inner product on the space Vζ. Let us denote by ·Vζ the norm induced by the inner product (16),
(17)vVζ=ζ∇vL2(Ω)2∀v∈Vζ.
By adapting the proof of [6, Lemma 3.2] to our present situation, one can see that the norms ·Hζ1(Ω) and ·Vζ are equivalent on Vζ. Consequently, (Vζ,(·,·)Vζ,·Vζ) is a Hilbert space too.Let us proceed by recalling the following tool.Theorem 2.
Suppose(X,(·,·)X,·X) is a Hilbert space. Let A:X→X be a Lipschitz continuous operator with the property that
(18)thereexistsm>0suchthatAu-Av,u-vX≥mu-vX2∀u,v∈X.
Let j:X→R¯ be a proper, convex, and continuous functional. Then, given f∈X, there exists a unique element u∈X such that
(19)Au,v-uX+j(v)-j(u)≥f,v-uX∀v∈X.
Moreover, the solution depends Lipschitz continuously on f.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 6.2.1.]).
LetX be a reflexive Banach space, and let f:X→R be Gâteaux differentiable. Then the following conditions are equivalent:(i)
f is convex;(ii)
one has(20)f(u)-f(v)≥f′v,u-vX⋆×X∀u,v∈X,where X⋆ denotes the dual of the space X.Theorem 4 (see [11, Proposition 1.59]).
LetX be a real normed space and let f:X→R¯ be a proper convex lower semicontinuous functional. Then ∂f(u)≠∅ for all u∈int(dom(f)); that is, f is subdifferentiable on int(dom(f)).Theorem 5 (see [11, Proposition 1.60]).
LetX be a real normed space and let f:X→R¯ be convex. Assume u∈X and f(u)∈R. If f′(u) exists as a Gâteaux derivative, then ∂f(u)={f′(u)}.Now we are ready to present our existence and uniqueness result.


## 3. Main Result

In what follows, we considerζ=β, where β is the function from Problem 1. In order to deliver a weak formulation of Problem 1, we assume that u is a strong solution. We multiply (2) by a function v∈C∞(Ω¯), and, after integration by parts, we obtain
(21)∫Ω‍βx∇ux·∇vxdx=∫Ω‍q0(x)v(x)dx+∫Γ‍β(x)∂u∂ν(x)v(x)dΓ.
Due to boundary conditions (3)–(5), we infer that
(22)∫Ω‍βx∇ux·∇vxdx=∫Ω‍q0(x)v(x)dx+∫Γ1‍β(x)∂u∂ν(x)v(x)dΓ+∫Γ2‍q2(x)v(x)dΓ-k∫Γ3‍uxvxu2x+αdΓ∀v∈C∞(Ω¯).
The density result C∞(Ω¯)¯=W1,1(Ω) implies that
(23)∫Ω‍β(x)∇u(x)·∇v(x)dx=∫Ω‍q0(x)v(x)dx+∫Γ1‍β(x)∂u∂ν(x)γv(x)dΓ+∫Γ2‍q2(x)γv(x)dΓ-k∫Γ3‍γuxγvxγu2x+αdΓ∀v∈W1,1(Ω).This allows us to give the definition of the weak solution to our problem.Definition 6.
By a weak solution for Problem1 one understands a function u∈Vζ which verifies
(24)∫Ω‍βx∇ux·∇vxdx=∫Ω‍q0(x)v(x)dx+∫Γ2‍q2(x)γv(x)dΓ-k∫Γ3‍γu(x)γv(x)γu2(x)+αdΓ,
for all v∈Vζ.Let us introduce now the main result of our paper.Theorem 7.
There exists a unique weak solution to Problem1.To provide a proof for this theorem we rely on Theorem2, so we prove several auxiliary lemmas.Lemma 8.
There existsA:Vζ→Vζ such that
(25)Au-Av,u-vVζ=u-vVζ2∀u,v∈Vζ.Proof.
We introduce the form(26)a:Vζ×Vζ⟶R,a(u,v)=∫Ω‍β(x)∇u(x)·∇v(x)dx
and we claim that a is well defined, bilinear, and continuous. Indeed, since u, v∈Hζ1(Ω) and β=ζ2, function a is well defined. Also, from the linearity of the gradient and of the integral operator, one can immediately see that a is bilinear. Then, to show the continuity of a, it is enough to notice that, by the Hölder inequality, we arrive at
(27)au,v≤ζ∇uL2Ωζ∇vL2Ω;
that is,
(28)au,v≤uVζvVζ.
With that, our claim is proved and we can apply Riesz’s representation theorem to deduce that there exists
(29)A:Vζ⟶Vζ,Av1,v2Vζ=a(v1,v2).
We have
(30)Au-Av,u-vVζ=∫Ω‍β(x)∇u(x)-∇v(x)2dx=u-vVζ2∀u,v∈Vζ,
where, as usual, · denotes the Euclidean norm on R2. Hence our proof is finished.Let us define now the following application:(31)j:Vζ⟶R,j(u)=∫Γ3‍kγu2(x)+α2-αdΓ.Lemma 9.
The functionalj is well defined, continuous, Gâteaux differentiable, and convex.Proof.
Note that(32)γu2(x)+α2-α=γuxγu2x+α2+αγux≤γux.
Since u∈Vζ, we have that γu∈L1(Γ); thus
(33)∫Γ3‍γu2x+α2-αdΓ<∞,
and consequently j is well defined. Also, it is not difficult to see that j is continuous.
In what concerns the Gâteaux differentiability ofj, for all u,v∈Vζ we have
(34)limt→0+j(u+tv)-j(u)t=ddtj(u+tv)t=0=kddt∫Γ3‍γu+tv2(x)+α2-αdΓt=0.
By Lebesgue’s Theorem, the differentiation under the integral is allowed if, for t close to zero, the term (γ(u+tv))2(x)+α2-α is dominated by a fixed L1 function which does not depend on t. Indeed, by (32),
(35)γu+tv2(x)+α2-α≤γu+tvx≤γux+γvx,
for |t|<1. Therefore,
(36)limt→0+j(u+tv)-j(u)t=k∫Γ3‍∂∂tγux+tγvx2+α2-αt=0dΓ=k∫Γ3‍γu(x)γv(x)γu2(x)+α2dΓ.
It is clear that the application which maps v∈Vζ into k∫Γ3‍γu(x)γv(x)/(γu)2(x)+α2dΓ is linear, so we prove its continuity. We have
(37)k∫Γ3‍γu(x)γv(x)γu2(x)+α2dΓ≤k∫Γ3‍γvxdΓ.
Using (13), (11), and the equivalence of norms ·Vζ and ·Hζ1(Ω) on Vζ, there exists k0 such that
(38)k∫Γ3‍γu(x)γv(x)γu2(x)+α2dΓ≤k0vVζ
for all v∈Vζ. This way, we have obtained that j is Gâteaux differentiable and
(39)j′u,vVζ=k∫Γ3‍γu(x)γv(x)γu2(x)+α2dΓ.
As for convexity, due to Theorem 3, it is enough to show that
(40)j(u)-j(v)≥j′v,u-vVζ∀u,v∈Vζ
that is,
(41)∫Γ3‍γv2x+α2-αdΓ-∫Γ3‍γu2x+α2-αdΓ≥∫Γ3‍γuxγvx-γuxγu2(x)+αdΓ
for all u,v∈Vζ. To do so, we start from the fact that
(42)[γv2(x)+α2][γu2(x)+α2]≥γu(x)γv(x)+α2,
for all u, v∈Vζ. This immediately leads to
(43)γv2x+α2][γu2x+α2-α2-γu2x≥γuxγvx-γu2x;
hence
(44)γv2(x)+α2-γu2(x)+α2≥γuxγvx-γuxγu2(x)+α2
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 elementf∈Vζ such that
(45)∫Ω‍q0(x)v(x)dx+∫Γ2‍q2(x)γv(x)dΓ=f,vVζ∀v∈Vζ.Proof.
We introduce(46)Φ:Vζ⟶R,Φ(v)=∫Ω‍q0(x)v(x)dx+∫Γ2‍q2(x)γv(x)dΓ
and we claim that Φ is well defined, linear, and continuous. Indeed, the form Φ is well defined because q0,v∈L2(Ω), q2∈L∞(Γ2), and γv∈L1(Γ). The linearity of Φ follows easily from the linearity of the integral operator. Let us focus on the continuity of Φ. First, by the Hölder inequality,
(47)∫Ω‍q0xvxdx≤q0L2(Ω)vL2(Ω)
for all v∈Vζ. Then, by relation (10), we find that
(48)∫Ω‍q0xvxdx≤q0L2(Ω)vHζ1(Ω)
for all v∈Vζ. At the same time,
(49)∫Γ2‍q2xγvxdΓ≤q2L∞(Γ2)γvL1(Γ).
By (13) and (11) it follows that
(50)∫Γ2‍q2xγvxdΓ≤cGcEq2L∞(Γ2)vHζ1(Ω)
for all v∈Vζ. We take into account the equivalence of the norms ·Vζ and ·Hζ1(Ω) and the linearity of Φ, and then we put together (46), (48), and (50) to deduce that Φ is continuous.
At this point, we have all the ingredients to apply Riesz’s representation theorem. So, we deduce that there exists a unique elementf∈Vζ such that
(51)Φ(v)=f,vVζ∀v∈Vζ
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.
Findu∈Vζ such that
(52)Au,v-uVζ+j(v)-j(u)≥f,v-uVζ∀v∈Vζ,
where A, j, and f are the ones from (29), (31), and (45), respectively.Lemma 12.
A functionu∈Vζ is a weak solution to Problem 1 if and only if it is a solution to Problem 11.Proof.
Letu∈Vζ be a weak solution to Problem 1, meaning that u verifies Definition 6. Consequently,
(53)∫Ω‍βx∇ux·∇ux-∇vxdx=∫Ω‍q0xux-vxdx+∫Γ2‍q2xγux-γvxdΓ-k∫Γ3‍γuxγux-γvxγu2x+αdΓ∀v∈Vζ.
Keeping in mind relations (26), (29), (39), and (45), by relation (40) it is obvious that u is a solution to Problem 11.
Conversely, let us consider nowu∈Vζ to be a solution to Problem 11. Then f-Au∈∂j(u). Moreover, due to Lemma 9, we can apply Theorems 4 and 5 to infer that f-Au=j′(u). This yields
(54)f,vVζ-Au,vVζ=j′u,vVζ∀v∈Vζ.
Taking into consideration (26), (29), (39), and (45), we obtain that u is a weak solution to Problem 1.Finally, we can conclude our main result.Proof of Theorem7.
Based on the previous results, the proof of Theorem7 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.



---

*Source: 865261-2015-02-23.xml*