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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Vespri, Vincenzo / Marano, Salvatore Angelo


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Volume 15, Issue 1

Issues

Volume 13 (2015)

Existence of entropy solutions for nonlinear elliptic degenerate anisotropic equations

Yuliya Gorban
Published Online: 2017-06-09 | DOI: https://doi.org/10.1515/math-2017-0064

Abstract

In the present article we deal with the Dirichlet problem for a class of degenerate anisotropic elliptic second-order equations with L1-right-hand sides in a bounded domain of ℝn(n ⩾ 2) . This class is described by the presence of a set of exponents q1,…, qn and a set of weighted functions ν1,…, νn in growth and coercitivity conditions on coefficients of the equations. The exponents qi characterize the rates of growth of the coefficients with respect to the corresponding derivatives of unknown function, and the functions νi characterize degeneration or singularity of the coefficients with respect to independent variables. Our aim is to investigate the existence of entropy solutions of the problem under consideration.

Keywords: Nonlinear elliptic degenerate anisotropic second-order equations; L1-data; Dirichlet problem; Existence of entropy solutions

MSC 2010: 35J25; 35J60; 35J70

1 Introduction

During the last twenty years the research on the existence and properties of solutions for nonlinear equations and variational inequalities with L1-data or measures as data were intensively developed. As is generally known, an effective approach to the solvability of second-order equations in divergence form with L1-right-hand sides has been proposed in [1]. In this connection we also mention a series of other close investigations for nondegenerate isotropic nonlinear second-order equations with L1-data and measures, entropy and renormalized solutions [210]. As for the solvability of nonlinear elliptic second-order equations with anisotropy and degeneracy (with respect to the independent variables), we note the following works. The existence of a weak (distributional) solution to the Dirichlet problem for a model nondegenerate anisotropic equation with right-hand side measure was established in [11]. The existence of weak solutions for a class of nondegenerate anisotropic equations with locally integrable data in ℝn(n ⩾ 2) was proved in [12], and an analogous existence result concerning the Dirichlet problem for a system of nondegenerate anisotropic equations with measure data was obtained in [13]. Moreover, in [14], the existence of weak solutions to the Dirichlet problem for nondegenerate anisotropic equations with right-hand sides from Lebesgues spaces close to L1 was established. Solvability of the Dirichlet problem for degenerate isotropic equations with L1-data and measures as data was studied in [1519]. Remark that in [15, 17], the existence of entropy solutions to the given problem was proved in the case of L1-data, and in [16], the existence of a renormalized solution of the problem for the same case was established. In [16, 18, 19], the existence of distributional solutions of the problem was obtained in the case of right-hand side measures.

Solvability of the Dirichlet problem for a class of degenerate anisotropic elliptic second-order equations with L1-right-hand sides was studied in [20]. This class is described by the presence of a set of exponents q1,…, qn and of a set of weighted functions ν1,…,νn in growth and coercitivity conditions on coefficients of the equations under consideration. The exponents qi characterize the rates of growth of the coefficients with respect to the corresponding derivatives of unknown function, and the functions νi characterize degeneration or singularity of the coefficients with respect to the independent variables. This is the most general situation in comparison with the above-mentioned works: the nondegenerate isotropic case means that νi ≡ 1 and qi = q1, i = 1,…,n; the nondegenerate anisotropic case means that νi ≡ 1, i = 1,…,n, and qi, i = 1,…,n, are generally different, and the degenerate isotropic case means that νi = ν1, i = 1,…,n, as in [1619] or νi, i = 1,…,n, are generally different as in [15] but qi = q1, i = 1,…,n.

In [20], the theorem on the existence and uniqueness of entropy solution to the Dirichlet problem for this class of the equations was proved. Moreover, the existence results of some other types of solutions to the given problem were also obtained. Observe that the proofs of these theorems are based on use of some results of [2123] on the existence and properties of solutions of second-order variational inequalities with L1-right-hand sides and sufficiently general constraints. Note that in [2023] right-hand sides to the investigated variational inequalities and equations depend on independent variables only, and belong to the class L1.

The present article is devoted to the Dirichlet problem for a same class of the nonlinear elliptic second-order equations in divergence form with degenerate anisotropic coefficients as in [20]. Here right-hand sides to the given equations depend on independent variables and unknown function. A model example of this class is an equation i=1nDi(νi(x)|Diu|qi2Diu)=nF(x,u),xΩ, where Ω is abounded domain in Rn(n2),1<qi<n,νi>0 a.e. in Ω,νiLloc1(Ω),(1/νi)1/qi1)L1(Ω), i = 1,…,n, F : Ω × ℝ → ℝ is a Carathéodory function.

The main result of this paper is a theorem on the existence of entropy solutions to the Dirichlet problem for the equations under consideration. We require an additional conditions to the function F in this theorem. Namely F(x,u) has an arbitrary growth with respect to the second variable, and F(x,u) belongs to L1(Ω) under the fixed value of the second variable. In our case we have no opportunity to use the results [2123] directly. We follow a general approach for proving the above-mentioned theorem. This approach has been proposed in [1] to the investigation on the existence and properties of solutions for nonlinear elliptic second-order equations with isotropic nondegenerate (with respect to the independent variables) coefficients and L1 -right-hand sides. In [21, 23] this approach has been taken to the anisotropic degenerate case. Also we use some ideas of [24].

2 Preliminaries

In this section we give some results of [23] which will be used in the sequel.

Let n ∈ ℕ, n ⩾ 2, Ω be a bounded domain in ℝn with a boundary ∂Ω, and for every i ∈{1,…,n} we have qi ∈(1,n).

We set q = {qi : i = 1,…,n}, q=min{qi:i=1,,n},q¯=(1ni=1n1qi)1,q^=n(q¯1)(n1)q¯.

Let for every i ∈ {1,…,n} νi be a nonnegative function on Ω such that νi > 0 a.e. in Ω, νiLloc1(Ω),(1/νi)1/(qi1)L1(Ω).(1)

We set ν = {νi : i = 1,…,n}. We denote by W1,q(ν,Ω) the set of all functions uW1,1(Ω) such that for every i ∈{1,…,n} we have νi|Diu|qiL1(Ω).

Let ∥⋅∥1,q,ν be the mapping from W1,q(ν, Ω) into ℝ such that for every function uW1,q(ν, Ω) u1,q,ν=Ω|u|dx+i=1n(Ωνi|Diu|qidx)1/qi.

The mapping ∥⋅∥1,q,ν is a norm in W1,q(ν, Ω), and, in view of the second inclusion of (1), the set W1,q(ν, Ω) is a Banach space with respect to the norm ∥⋅∥1,q,ν. Moreover, by virtue of the first inclusion of (1), we have C0(Ω)W1,q(ν,Ω).

We denote by W1,q(ν,Ω) the closure of the set C0(Ω) in space W1,q(ν, Ω). Evidently, the set W1,q(ν,Ω) is a Banach space with respect to the norm induced by the norm ∥⋅∥1,q,ν. It is obvious that W1,q(ν,Ω)W1,1(Ω). Finally, we observe that W1,q(ν,Ω) is a reflexive space. The proof of the latter statement can be found in [21].

Note that the following assertion hold.

Proposition 2.1

If a sequence converges weakly in W1,q(ν,Ω), then it converges strongly in L1(Ω).

Further, let for every k > 0 Tk : ℝ → ℝ be the function such that Tk(s)=s,if |s|k,k sign s,if |s|>k.

By analogy with known results for nonweighted Sobolev spaces (see for instance [25]) we have: if uW1,q(ν,Ω) and k > 0, then Tk(u) ∈ W1,q(ν,Ω) and for every i ∈ {1,…,n} DiTk(u)=Diu1{|u|<k} a.e. in Ω.(2)

We denote by T1,q(ν,Ω) the set of all functions u : Ω → ℝ such that for every k > 0, Tk(u) ∈ W1,q(ν,Ω).

Clearly, W1,q(ν,Ω)T1,q(ν,Ω).(3)

For every u : Ω → ℝ and for every x ∈Ω we set k(u,x)=min{lN:|u(x)|l}.

Definition 2.2

Let uT1,q(ν,Ω), and i ∈{1,…,n}. Then δiu : Ω → ℝ is the function such that for every x ∈ Ω δiu(x) = DiTk(u,x)(u)(x).

Definition 2.3

If uT1,q(ν,Ω), then δ u : Ω → ℝn is the mapping such that for every x ∈Ω and for every i ∈{1,…,n} (δ u(x))i = δiu(x).

Now we give several propositions which will be used in the next sections.

Proposition 2.4

Let uT1,q(ν,Ω). Then for every k > 0 we have Di Tk(u) = δiu⋅ 1{|u| < k} a.e. in Ω, i = 1,…,n.

If uW1,q(ν,Ω), then for every i ∈{1,…,n} δiu = Diu a.e. in Ω.

Proposition 2.5

Let uT1,q(ν,Ω)andwW1,q(ν,Ω)L(Ω).ThenuwT1,q(ν,Ω), and for every i ∈{1,…,n} and for every k > 0 we have DiTk(uw)=δiuDiwa.e.in{|uw|<k}.

Proposition 2.6

There exists a positive constant c0 depending on n, q, and ∥ 1/νiL1/(qi − 1)(Ω), i = 1,…,n, such that for every function uW1,q(ν,Ω) (Ω|u|n/(n1)dx)(n1)/nc0i=1n(Ωνi|Diu|qidx)1/nqi.

3 Statement of the Dirichlet problem. The concept of its entropy solution

Let c1, c2 > 0, g1, g2L1(Ω), g1, g2 ⩾ 0 in Ω, and let for every i ∈ {1,…,n}ai : Ω × ℝn → ℝ be a Carathéodory function. We suppose that for almost every x ∈ Ω and for every ξ ∈ ℝn, i=1n(1/νi)1/(qi1)(x)|ai(x,ξ)|qi/(qi1)c1i=1nνi(x)|ξi|qi+g1(x),(4) i=1nai(x,ξ)ξic2i=1nνi(x)|ξi|qig2(x).(5)

Moreover, we assume that for almost every x ∈ Ω and for every ξ,ξ′ ∈ ℝn, ξξ′, i=1n[ai(x,ξ)ai(x,ξ)](ξiξi)>0.(6)

Now we give one result of [20] which will be used in the sequel.

Proposition 3.1

The following assertions hold:

  1. if u,wW1,q(ν,Ω) and i ∈ {1,…,n}, then ai(x,∇ u)DiwL1(Ω);

  2. if uT1,q(ν,Ω),wW1,q(ν,Ω)L(Ω), k > 0 and i ∈ {1,…,n}, then ai(x,δ u)Di Tk(uw) ∈ L1(Ω).

Let F : Ω × ℝ → ℝ be a Carathéodory function. We consider the following Dirichlet problem: i=1nxiai(x,u)=F(x,u)in Ω,(7) u=0on Ω.(8)

Definition 3.2

An entropy solution of problem (7), (8) is a function uT1,q(ν,Ω) such that: F(x,u)L1(Ω);(9) for every function wW1,q(ν,Ω)L(Ω) and for every k ⩾ 1 Ω{i=1nai(x,δu)DiTk(uw)}dxΩF(x,u)Tk(uw)dx.(10)

Note that the left-hand integral in (10) is finite. It follows from assertion b) of Proposition 3.1. The right-hand integral in (10) is also finite. It follows from the boundedness of the function Tk and inclusion (9).

4 Main result

Next theorem is the main result of this paper.

Theorem 4.1

Suppose the following conditions are satisfied:

  1. for a.e. x ∈ Ω the function F(x,⋅) is nonincreasing on ℝ;

  2. for any s ∈ ℝ the function F (., s) belongs to L1(Ω).

Then there exists an entropy solution of the Dirichlet problem (7), (8).

Proof

According to the approach from [1], we will consider a sequence of the approximating problems for the equations with smooth right-hand sides. Then we will obtain special estimates of the solutions of these problems. Finally, we will pass to the limit. The proof is in 9 steps.

Step 1. We set f = F (⋅, 0). Let for every l ∈ ℕ, Fl : Ω × ℝ → ℝ be the function such that Fl(x,s)=Tl(f(x)F(x,s)),(x,s)Ω×R.

By virtue of condition 1), we have: if lN, then for a.e. xΩ the function Fl(x,) is nondecreasing on R.(11)

Further, in view of condition 2), we have fL1(Ω). Hence there exists {fl} ⊂ C0(Ω) such that: limlflfL1(Ω)=0,(12) lNflL1(Ω)fL1(Ω).(13)

Using the inequalities (4)(6), property (11), and well-known results on the solvability of the equations with monotone operators (see for instance [26]), we obtain: if l ∈ ℕ, then there exists the function ulW1,q(ν,Ω) such that for every function wW1,q(ν,Ω) Ω{i=1nai(x,ul)Diw+Fl(x,ul)w}dx=Ωflwdx.(14)

It means that the function ulW1,q(ν,Ω) is a generalized solution of the Dirichlet problem: i=1nxiai(x,u)+Fl(x,u)=flin Ω,u=0 on Ω.

We denote by ci, i = 3,4,…, the positive constants depending only on n, q, c1, c2, ∥g1L1(Ω), ∥g2L1(Ω), ∥fL1(Ω), ∥F(⋅,−1)∥L1(Ω), ∥F (⋅, 1)∥L1(Ω), ∥1/νiL1/(qi − 1)(Ω), i = 1,…,n, and meas Ω.

Let us show that for every k ⩾ 1 and l ∈ ℕ the following inequalities hold: {|ul|<k}{i=1nνi|Diul|qi}dxc3k,(15) {|ul|k}|Fl(x,ul)|dxc4.(16)

In fact, let k ⩾ 1 and l ∈ ℕ. As ulW1,q(ν,Ω) we have Tk(ul) ∈ W1,q(ν,Ω). In view of (14) and (13) we obtain Ω{i=1nai(x,ul)DiTk(ul)+Fl(x,ul)Tk(ul)}dxkfL1(Ω).

Using (2) and (5) in the left-hand side of this inequality, we get c2{|ul|<k}{i=1nνi|Diul|qi}dx+ΩFl(x,ul)Tk(ul)dxkfL1(Ω)+g2L1(Ω).(17)

Assertion (11) and properties of the function Tk imply that Fl(x,ul)Tk(ul)0 a.e. in Ω,(18) Fl(x,ul)Tk(ul)=k|Fl(x,ul)| a.e. in {|ul|k}.(19)

The estimate (15) follows from (18) and (17). Finally, the inequality (16) follows from (19) and (17).

Step 2. Now we show that for every k ⩾ 1 and l ∈ ℕ meas{|ul|k}c5kq^,(20) meas{νi1/qi|Diul|k}c6kqiq^/(1+q^),i=1,,n.(21)

In fact, let k ⩾ 1 and l ∈ ℕ. We have |Tk(ul)| = k on {|ul| ⩾ k}; then kn/(n1)meas{|ul|k}Ω|Tk(ul)|n/(n1)dx.(22)

Using Proposition 2.6, (2) and (15), we obtain (Ω|Tk(ul)|n/(n1)dx)(n1)/nc0i=1n({|ul|<k}νi|Diul|qidx)1/nqic0(c3k)1/q¯.

The inequality (20) follows from the latter estimate and (22).

Next, we fix i ∈ {1,…,n}, and set k=kqi/(1+q^),G={|ul|<k,νi1/qi|Diul|k}.

We have meas{νi1/qi|Diul|k}meas{|ul|k}+meas G.(23)

From (20) it follows that meas{|ul|k}c5kq^.(24)

Moreover, in view of the set’s G definition and (15) we get kqimeas G{|ul|<k}νi|Diul|qidxc3k.

The inequality (21) follows from the latter estimate and (23), (24).

Step 3. Assertions (2) and (15) imply that for every k ⩾ 1 the sequence {Tk(ul)} is bounded in W1,q(ν,Ω). As the space W1,q(ν,Ω) is reflexive, then there exist an increasing sequence {lh} ⊂ ℕ, and sequence {zk} ⊂ W1,q(ν,Ω) such that for every k ∈ ℕ we have a weak convergence Tk(ulh) → zk in W1,q(ν,Ω). Without loss of generality it can be assumed that kNTk(ul)zk weakly in W1,q(ν,Ω).(25)

Step 4. Let us show that the sequence {ul} is fundamental on measure.

Indeed, let k ⩾ 1, l,j ∈ ℕ. We fix t > 0, and set G′ = {|ul| < k, |uj| < k,|uluj| ⩾ t}. It is clear that meas{|uluj|t}meas{|ul|k}+meas{|uj|k}+meas G.(26)

As t ⩽ |Tk(ul) − Tk(uj)| on G′, we obtain t meas GΩ|Tk(ul)Tk(uj)|dx.

This inequality, (20) and (26) imply that for every k ⩾ 1, and l,j ∈ ℕ meas{|uluj|t}2c5kq^+t1Ω|Tk(ul)Tk(uj)|dx.(27)

Let ε > 0. We fix k ∈ ℕ such that 2c5kq^ε/2.(28)

Taking into account (25) and Proposition 2.1, we infer a strong convergence Tk(ul) → zk in L1(Ω). Then there exists N ∈ ℕ such that for every l,j ∈ ℕ, l,jN Ω|Tk(ul)Tk(uj)|dxεt/2.

From this inequality, (27), and (28) we deduce that for every l,j ∈ ℕ, l,jN meas{|uluj|t}ε.

This means that the sequence {ul} is fundamental on measure.

Step 5. Now we show that for every i ∈ {1,…,n} the sequence {νi1/qiDiul} is fundamental on measure.

For every t > 0 and l,j ∈ ℕ we put Nt(l,j)=meas{i=1nνi1/qi|DiulDiuj|t}.

Besides, for every t > 0, h,k ⩾ 1, and l,j ∈ ℕ we set Et,h,k(l,j)={i=1nνi1/qi|DiulDiuj|t,i=1nνi1/qi|Diul|h,i=1nνi1/qi|Diuj|h,|uluj|<1k}.

Using (21), we establish that for every t > 0, hn, k ⩾ 1, and l,j ∈ ℕ Nt(l,j)2c6nn+1hqq^/(1+q^)+meas{|uluj|1/k}+meas Et,h,k(l,j).(29)

Further, we get one estimate for some integrals over Et,h,k(l, j). So we introduce now auxiliary functions and sets.

Let for every x ∈ Ω Φx : ℝn × ℝn → ℝ be a function such that for every pair (ξ,ξ′) ∈ ℝn × ℝn Φx(ξ,ξ)=i=1n[ai(x,ξ)ai(x,ξ)](ξiξi).

Recall that ai,i = 1,…,n, are Carathéodory functions, and inequality (6) holds for almost every x ∈ Ω and every ξ,ξ′ ∈ ℝn, ξξ′. Then there exists a set E ⊂ Ω, meas E = 0, such that:

  1. for every x ∈ Ω ∖ E the function Φx is continuous on ℝn × ℝn;

  2. for every x ∈ Ω ∖ E and ξ,ξ′ ∈ ℝ,nξξ′, we have Φx(ξ,ξ′) > 0.

Put for every t > 0, h > t, and x ∈ Ω Gt,h(x)={(ξ,ξ)Rn×Rn:i=1nνi1/qi(x)|ξi|h,i=1nνi1/qi(x)|ξi|h,i=1nνi1/qi(x)|ξiξi|t}.

As νi > 0 a.e. in Ω,i = 1,n, then there exists a set E~ ⊂ Ω, meas E~ = 0, such that the set Gt,h(x) is nonempty for every t > 0,h > t, and x ∈ Ω ∖ E~.

Let for every t > 0 and h > t μt,h : Ω → ℝ be a function such that μt,h(x)=minGt,h(x)Φx,if xΩ(EE~),0,if xEE~.

For every t > 0 and h > t we have μt,h > 0 a.e. in Ω, and μt,hL1(Ω).

Let t > 0, ht + 1, k ⩾ 1, and l,j ∈ ℕ. We fix xEt,h,k(l, j) ∖ (EE~), and set ξ = ∇ ul(x), ξ′ = ∇ uj(x).

As (ξ; ξ′) ∈ Gt,h(x), then μt,h(x) ⩽ Φx(ξ,ξ′). This inequality and function’s Φx definition imply that μt,h(x)i=1n[ai(x,ul(x))ai(x,uj(x))](Diul(x)Diuj(x)).

Then, taking into account (6) and (2), we obtain Et,h.k(l,j)μt,hdxΩ{i=1n[ai(x,ul)ai(x,uj)]DiT1/k(uluj)}dx.(30)

In view of (14) we have Ω{i=1nai(x,ul)DiT1/k(uluj)}dx=ΩflT1/k(uluj)dxΩFl(x,ul)T1/k(uluj)dx,Ω{i=1nai(x,uj)DiT1/k(ujul)}dx=ΩfjT1/k(ujul)dxΩFj(x,uj)T1/k(ujul)dx.

From these equalities and (30) it follows that Et.h,k(l,j)μt,hdx1kΩ|flfj|dx+1kΩ|Fl(x,ul)Fj(x,uj)|dx.(31)

Using (16) and conditions 1), 2), we find that for every l,j ∈ ℕ Ω|Fl(x,ul)Fj(x,uj)|dxc7.

From the latter estimate and (31) we deduce that for every t > 0,ht + 1,k ⩾ 1, and l,j ∈ ℕ the following inequality holds: Et,h.k(l,j)μt,hdx1kΩ|flfj|dx+c7k.(32)

The sequence {ul} is fundamental on measure. Then there exists an increasing sequence {nk} ⊂ ℕ such that for every k ∈ ℕ and l,j ∈ ℕ, l,jnk, we have meas{|uluj|1/k}1/k.(33)

Let t > 0 and ε > 0. We fix ht + n such that 2c6nn+1hqq^/(1+q^)ε/4.(34)

Put for every k ∈ ℕ αk=supl,jnkmeas Et,h,k(l,j).

Let us show that αk → 0. Assume the converse. Then there exist τ > 0, an increasing sequence {ks} ⊂ ℕ, and sequences {ls},{js} ⊂ ℕ such that for every s ∈ ℕ we have ls,jsnks and meas Et,h,ks(ls,js)τ.(35)

Let Gs = Et,h,ks(ls,js), s ∈ ℕ.

In view of (32) and (13) for every s ∈ ℕ we get Gsμt,hdxc7+2ks.

It follows that limsGsμt,hdx=0.

From this assertion, taking into account μt,hL1(Ω) and μt,h > 0 a.e. in Ω, we infer that meas Gs → 0. This fact is in contradiction to (35). Hence, we conclude that αk → 0.

Finally, we fix k ∈ ℕ such that the inequalities hold: 1/kε/4,αkε/2.(36)

Let l,j ∈ ℕ,l,jnk. From (29), (33), (34) and (36) it follows that Nt(l,j) ⩽ ε.

This means that for every i ∈ {1,…,n} the sequence {νi1/qiDiul} is fundamental on measure.

Step 6. From results of the Steps 4 and 5, and F. Riesz’s theorem we get the following facts: there exist measurable functions u : Ω → ℝ and v(i) : Ω → ℝ,i = 1,…,n, such that the sequence {ul} converges to u on measure, and for every i ∈ {1,…,n} the sequence {νi1/qiDiul} converges to v(i) on measure. As is generally known, we can extract subsequences converging almost everywhere in Ω to the corresponding functions. We may assume without loss of generality that ulu a.e. in Ω,(37) i{1,,n}νi1/qiDiulv(i) a.e. in Ω.(38)

From (37), (25) and Proposition 2.1 we deduce that for every k ∈ ℕ Tk(u)W1,q(ν,Ω),(39) Tk(ul)Tk(u) weakly in W1,q(ν,Ω).(40)

Let us show that uT1,q (ν, Ω). Indeed, let k > 0. Take h ∈ ℕ, h > k. In view of (39) we have Th(u)∈ W1,q (ν, Ω). Hence, by inclusion (3) we obtain Tk(Th(u)) ∈ W1,q (ν, Ω). This fact and the equality Tk(u) = Tk(Th(u)) imply that Tk(u)∈ W1,q (ν, Ω). Therefore, uT1,q (ν, Ω).

Step 7. Now we show that i{1,,n}Diulδiu a.e. in Ω.(41) In fact, let i ∈ {1, …, n}. In view of (37) there exists a set E′ ⊂ Ω, meas E′ = 0, such that xΩEul(x)u(x),(42) and in view of (38) there exists a set E″ ⊂ Ω, meas E″ = 0, such that xΩEνi1/qi(x)Diul(x)v(i)(x).(43)

Fix k ∈ ℕ. By (2) we have: if l ∈ ℕ, then there exists a set E(l) ⊂ Ω, meas E(l) = 0, such that x{|ul|<k}E(l)DiTk(ul)(x)=Diul(x).(44)

We denote by Ê a union of sets E′, E″ and E(l), lN. Clearly, meas Ê = 0. Let x ∈ {|u| < k} \ Ê. In view of (42) there exists l0 ∈ ℕ such that for every l ∈ ℕ, ll0, we have |ul(x)| < k. Let l ∈ ℕ, ll0. Then x ∈ {|ul| < k}\ E(l) and according to (44) we get νi1/qi(x)DiTk(ul)(x)=νi1/qi(x)Diul(x). From this equality and (43) we deduce that νi1/qiDiTk(ul)(x)v(i)(x). Thus, νi1/qiDiTk(ul)v(i)a.e. in {|u|<k}.(45)

Besides, in view of (2) and (15) for every l ∈ ℕ Ωνi|DiTk(ul)|qidxc3k.(46) Using Fatou’s lemma, from (45) and (46) we infer that the function |v(i)|qi is summable in {|u| < k}.

Further, let φ : Ω → ℝ be a measurable function such that |φ| ⩽ 1 in Ω, and let ε > 0. As the function |v(i)| is summable on {|u| < k}, then there exists ε1 ∈ (0, ε) such that for every measurable set G ⊂ {|u| < k}, meas Gε1, we have Ω|v(i)|dxε.(47) Moreover, in view of (45) and Egorov’s theorem, there exists a measurable set Ω′ ⊂ {|u| < k} such that meas ({|u|<k}Ω)ε1,(48) νi1/qiDiTk(ul)ν(i) uniformly in Ω.(49)

From (47) and (48) we infer that {|u|<k}Ω|v(i)|dxε,(50) and from (49) we deduce that there exists l1 ∈ ℕ such that for every l ∈ ℕ, ll1, Ω|νi1/qiDiTk(ul)v(i)|dxε.(51) Let l ∈ ℕ, ll1. Using (50), (51), Hölder’s inequality, (48), and (46), we get |{|u|<k}[νi1/qiDiTk(ul)v(i)]φdx|2ε+{|u|<k}Ωνi1/qi|DiTk(ul)|dx2ε+ε(qi1)/qi(Ωνi|DiTk(ul)|qidx)1/qi2ε+ε(qi1)/qi(c3k)1/qi. Since ε is an arbitrary constant, from the latter estimate it follows that liml{|u|<k}[νi1/qiDiTk(ul)v(i)]φdx=0.(52)

On the other hand, let F : W1,q (ν, Ω) → ℝ be a functional such that for every function vW1,q (ν, Ω) F,v={|u|<k}νi1/qiDivφdx. It is easy to see that F ∈ ( W1,q (ν, Ω))*. Hence, by virtue of (40), we have F,Tk(ul)F,Tk(u).

This fact and functional’s F definition imply that liml{|u|<k}νi1/qiDiTk(ul)φdx={|u|<k}νi1/qiDiTk(u)φdx.(53)

From (52) and (53) we deduce that {|u|<k}[v(i)νi1/qiDiTk(u)]φdx=0. In turn, from this equality and Proposition 2.1 we infer that v(i)=νi1/qiδiu a.e. in {|u|<k}.

Since k ∈ ℕ is an arbitrary number, from the latter assertion it follows that v(i)=νi1/qiδiu a.e. in Ω.(54) Taking into account that νi > 0 a.e. in Ω, from (38) and (54) we obtain that Di ulδiu a.e. in Ω. Thus, (41) is proved.

Assertion (41) along with the fact that ai, i = 1, …, n, are Carathéodory functions implies that i{1,,n}ai(x,ul)ai(x,δu) a.e. in Ω.(55)

Step 8. Let us show that the following assertions are fulfilled: F(x,u)L1(Ω);(56) Fl(x,ul)fF(x,u) strongly in L1(Ω).(57) Indeed, in view of (37) we have Fl(x,ul)fF(x,u) a.e. in Ω.(58) Moreover, using (16) and conditions 1) and 2), we get for every l ∈ ℕ Ω|Fl(x,ul)|dxc8. From this fact, (58), and Fatou’s lemma we obtain inclusion (56).

Now let us prove (57). Firstly, we establish that for every k, l ∈ ℕ the following estimate holds {|ul|2k}|Fl(x,ul)|dx{|ul|k}|f|dx+flfL1(Ω)+2g2L1(Ω)k1.(59) Let zC1(ℝ) be a function such that 0 ⩽ z ⩽ 1 on ℝ, z = 0 on [−1;1], z = 1 on (−∞;−2] ∪[2;+∞), and for every s ∈ ℝ z′(s) sign s ⩾ 0, |z′(s)|⩽ 2.

We fix arbitrary k, l ∈ ℕ. We denote by zk : ℝ→ ℝ a function such that for every s ∈ ℝ zk(s)=T1(sk)z(sk).(60) From the properties of the functions T1 and z it follows that for every s ∈ ℝ |zk(s)|1.(61) Besides, sR,|s|k,zk(s)=0;(62) sR,|s|2k,|zk(s)|=1.(63) Definition (60) implies that zk(ul)∈ W1,q (ν, Ω) and Dizk(ul)=k1z(ulk)T1(ulk)Diula.e. in Ω,i=1,,n.(64) Substituting w = zk (ul) into (14), and using (61), (62), we get Ω{i=1nai(x,ul)Dizk(ul)}dx+ΩFl(x,ul)zk(ul)dx{|ul|k}|f|dx+flfL1(Ω).(65) We denote by Ik,l the first integral in the left-hand side of (65). In view of the function’s z definition sR,|s|k, or |s|2k,|z(s)|=0.(66) Using (64), (66), and (5), we establish that Ik,l=k1{k|ul|2k}[z(ulk)T1(ulk){i=1nai(x,ul)Diul}]dxk1{k|ul|2k}[z(ulk)T1(ulk){c2i=1nvj|Diul|qig2}]dx.(67) From the truncated function’s property and our condition z′(s) sign s ⩾ 0, ∀ s ∈ ℝ, it follows that almost everywhere in {k ⩽ |ul | ⩽ 2k} z(ulk)T1(ulk)=z(ulk) sign (ulk)0.

Taking into account this fact and our condition |z′(s)|⩽ 2, ∀ s ∈ ℝ, we deduce from (67) Ik,l2k1{k|ul|2k}g2dx. This and (65) imply ΩFl(x,ul)zk(ul)dx{|ul|k}|f|dx+flfL1(Ω)+2g2L1(Ω)k1.(68)

Note that in view of (11) and the function’s zk definition we have Fl(x,ul)zk(ul)0 a.e. in Ω, and in view of (63) we get Fl(x,ul)zk(ul)=|Fl(x,ul)| a.e. in {|ul|2k}. Then ΩFl(x,ul)zk(ul)dx{|ul|2k}|Fl(x,ul)|dx. Finally, assertion (59) is derived from the latter inequality and (68).

Next, we fix an arbitrary ε > 0. It is clear that there exists ε1 > 0 such that for every measurable set G ⊂Ω, meas Gε1, G(|f|+|F(x,u)|)dxε. We fix k ∈ ℕ such that the following inequalities hold: 2g2L1(Ω)k1ε,(69) c5kq^ε1.(70) By condition 2), we infer that the functions F(⋅, −2k) and F(⋅, 2k) belong to L1(Ω). Hence, there exists ε2 > 0 such that for every measurable set G ⊂ Ω, meas Gε2, G(|F(,2k)|+|F(,2k)|)dxε. In view of (58) there exists a measurable set Ω1⊂Ω such that meas (ΩΩ1)min(ε1,ε2),(71) and Fl(x, ul) → fF (x, u) uniformly in Ω1. Then there exists L1∈ ℕ such that for every l ∈ ℕ, lL1, Ω1|Fl(x,ul)(fF(x,u))|dxε.(72) Besides, in view of (12) there exists L2 ∈ ℕ such that for every l ∈ ℕ, lL2, flfL1(Ω)ε.(73) Now fix l ∈ ℕ, l ⩾ max(L1, L2). Using (71) and (72), we obtain Fl(x,ul)(fF(x,u))L1(Ω){|ul|2k}|Fl(x,ul)|dx+.(ΩΩ1){|ul|<2k}|Fl(x,ul)|dx+2ε.(74) By virtue of (20) and (70) we get meas {|ul| ⩾ k} ⩽ ε1. Then {|ul|k}|f|dxε.(75)

From (59), (69), (73) and (75) we deduce {|ul|2k}|Fl(x,ul)|dx3ε.(76) Note that by condition 1), we have |Fl(x,ul)|2(|F(x,2k)|+|F(x,2k)|) a.e. in {|ul|<2k}.(77) In view of (71) we get ΩΩ1(|F(x,2k)|+|F(x,2k)|)dxε.(78)

From (77) and (78) it follows that (ΩΩ1){|ul|<2k}|Fl(x,ul)|dx2ε.(79) Using (74), (76), and (79), we infer Fl(x,ul)(fF(x,u))L1(Ω)5ε.

Now we can conclude that ∥ Fl(x, ul)−(fF(x, u))∥L1(Ω) → 0. Thus, assertion (57) is proved.

Step 9. Let wW1,q (ν, Ω) ∩ L(Ω), and k ⩾ 1. Now we show that Ω{i=1nai(x,δu)DiTk(uw)}dxΩF(x,u)Tk(uw)dx.(80) Put H={|uw|<k},H0={|uw|=k}, and let for every l ∈ ℕ Hl={|ulw|<k}H0,El={|ulw|<k}H0. First of all we prove that for every function, ∈ L1(Ω) HlφdxHφdx.(81)

Indeed, let, φL1 (Ω). For every j ∈ ℕ put H(j)={|uw|<k1/j},H~(j)={|uw|>k+1/j}. We have meas (HH(j))0,meas ({|uw|>k}H~(j))0.(82) We fix an arbitrary ε > 0. In view of the property of Lebesgue integral’s absolute continuity and (82) there exists j ∈ ℕ such that HH(j)|φ|dxε/4,{|uw|>k}H~(j)|φ|dxε/4.(83) Moreover, in view of the property of Lebesgue integral’s absolute continuity, (37), and Egorov’s theorem there exists a measurable set Ω′ ⊂ Ω such that ΩΩ|φ|dxε/4,(84) ulu uniformly in Ω.(85) Assertion (85) means that we can find l0 ∈ ℕ such that for every l ∈ ℕ, ll0, and x ∈ Ω′ |ul(x)u(x)|<1/j.(86)

Let l ∈ ℕ, ll0. From (86) it follows that (H(j)Hl)Ω=,{|ulw|<k}H~(j)Ω=. Then HHl(HH(j))(ΩΩ),HlH({|uw|>k}H~(j))(ΩΩ).

These facts, (83), and (84) imply that HHl|φ|dxε/2,HlH|φ|dxε/2. Hence, |HlφdxHφdx|ε. The latter estimate means that (81) is true.

Further, put k1=k+wL(Ω),φ1=i=1nai(x,Tk1+1(u))Diw, and let for every l ∈ ℕ ψl=i=1nai(x,ul)Diul+g2,Sl=Hl{i=1n[ai(x,ul)ai(x,Tk1+1(u))]Diw}dx,Sl=El{i=1nai(x,w)[DiulDiw]}dx. We fix an arbitrary l ∈ ℕ. In view of (14) we have Ω{i=1nai(x,ul)DiTk(ulw)}dx=Ω(flFl(x,ul))Tk(ulw)dx.(87) Using (2) and (6), we get Ω{i=1nai(x,ul)DiTk(ulw)}dxHl{i=1nai(x,ul)[DiulDiw]}dx+Sl. From this inequality and (87) we obtain Hl{i=1nai(x,ul)Diul}dxHl{i=1nai(x,ul)Diw}dx+Ω(flFl(x,ul))Tk(ulw)dxSl.

Hence, for every l ∈ ℕ HlψldxΩ(flFl(x,ul))Tk(ulw)dx+Hl(φ1+g2)dx+SlSl.(88)

Note that by virtue of (12) and (37) we get fl Tk(ulw) → fTk(uw) strongly in L1(Ω). Therefore, ΩflTk(ulw)dxΩfTk(uw)dx.(89) Besides, in view of (57) and (37) we obtain Fl(x, ul)Tk(ulw) → (fF(x, u))Tk(uw) strongly in L1(Ω). Hence, ΩFl(x,ul)Tk(ulw)dxΩ(fF(x,u))Tk(uw)dx.(90) As uT1,q (ν, Ω), then we have Tk(u) ∈ W1,q (ν, Ω). Therefore, assertion a) of Proposition 3.1 implies an inclusion φ1 ∈ L1(Ω). Besides, we have g2L1(Ω). Thus, using (81), we deduce that Hl(φ1+g2)dxH(φ1+g2)dx.(91) Now we prove that Sl0.(92)

Indeed, let ε ∈ (0, 1). In view of the property of Lebesgue integral’s absolute continuity, (37), (55), and Egorov’s theorem there exists a measurable set Ω1⊂Ω such that ΩΩ1{|φ1|+i=1nνi|Diw|qi}dxεn,(93) ulu uniformly in Ω1,(94) i=1nai(x,ul)Diwi=1nai(x,δu)Diw iniformly in Ω1.(95)

Assertion (94) means that we can find l0 ∈ ℕ such that for every l ∈ ℕ, ll0, and x ∈ Ω1 |ul(x)u(x)|ε.(96) Moreover, in view of (95) there exists l1∈ ℕ such that for every l ∈ ℕ, ll1, we get Ω1i=1nai(x,ul)Diwi=1nai(x,δu)Diwdxε.(97) Let l ∈ ℕ, l ⩾ max(l0, l1). As wL (Ω), there exists a set Ê⊂Ω, meas Ê = 0, such that for every x ∈ Ω \ Ê we have |w(x)|⩽∥ wL(Ω). From this fact and (96) it follows that (Hl ∩Ω1)\ Ê⊂ {|u| < k1 + 1}. Using this inclusion, Proposition 2.4, and (97), we obtain HlΩ1|i=1nai(x,ul)Diwφ1|dxε. The latter inequality and (93) imply that |Sl|2ε+i=1nHlΩ1|ai(x,ul)||Diw|dx.(98)

Taking into account Hölder inequality, (4), an inclusion Hl\ Ê⊂ {|ul| < k1}, (15), and (93), we established that for every i ∈ {1, …, n} HlΩ1|ai(x,ul)||Diw|dx(c1c3k1+1+g1L1(Ω))ε. From this and (98) we deduce |Sl|2ε+n(c1c3k1+1+g1L1(Ω))ε. Thus, (92) is true.

Further, we show that Sl0.(99) It suffices to take meas H0 > 0. Let i ∈ {1, …, n}. Since uT1,q (ν, Ω) and wW1,q (ν, Ω) ∩ L1(Ω), by virtue of Proposition 2.5, we have uwT1,q (ν, Ω). Hence, from Proposition 2.4 it follows that DiTk(uw)=0a.e. inH0.(100)

On the other hand, for almost every xH0 the inequality |u(x)| < k1 + 1 holds. So, Tk(uw) = Tk1+1(u) − w a.e. in H0. Therefore, DiTk(uw)=DiTk1+1(u)Diwa.e. inH0. Then, taking into account (100), we get Di Tk1+1(u) = Di w a.e. in H0. This and Proposition 2.4 imply that δiu = Di w a.e. in H0. From this result and (41) we infer that for every i ∈ {1, …, n} Di ulDi w a.e. in H0. Hence, i=1nai(x,w)[DiulDiw]0a.e. inH0.(101) Next, we put φ2=i=1n|ai(x,w)||Diw|,φ3=i=1n(1/νi)1/(qi1)|ai(x,w)|qi/(qi1). In view of (4) the functions φ2 and φ3 are summable in Ω.

We fix an arbitrary ε > 0. In view of the property of Lebesgue integral’s absolute continuity, (101), and Egorov’s theorem there exists a measurable set Ω2H0 such that H0Ω2(φ2+φ3)dxε,i=1nai(x,w)[DiulDiw]0iniformly in Ω2.(102)

The latter property means that me can find l0 ∈ ℕ such that for every l ∈ ℕ, ll0, Ω2|i=1nai(x,w)[DiulDiw]|dxε.(103) Let l ∈ ℕ, ll0. Using (102) and (103), we infer that |Sl|2ε+i=1nElΩ2|ai(x,w)||Diul|dx.(104) By the virtue of Hölder inequality, (102), and (15) we deduce that for every i ∈ {1, …, n} ElΩ2|ai(x,w)||Diul|dx(ElΩ2φ3dx)(qi1)/qi({|ul|<k1}νi|Diul|qidx)1/qiε(qi1)/qi(c3k1)1/qi. This fact along with (104) and an arbitrariness of ε implies that (99) is true.

Further, let χ : Ω → ℝ be a characteristic function of the set H, and let for every l ∈ ℕ χl : Ω → ℝ be a characteristic function of the set Hl. We have lim_lχlχa.e. inΩ.(105) Indeed, in view of (37) there exists a set E0 ⊂ Ω, meas E0 = 0, such that for every x ∈ Ω \ E0 ul(x)→ u(x). Let x ∈ Ω \ E0. If xH, then χ(x) = 0. Hence, χ(x)⩽χl(x), ∀ lN. Let xH. As ul(x)→ u(x), there exists l1∈ ℕ such that for every l ∈ ℕ, ll1, we have |ul(x)− u(x) | < k − |u(x)− w(x)|. Then for arbitrary l ∈ ℕ, ll1, we get |ul(x) − w(x)| < k. Therefore, xHl and χl(x) = 1 = χ(x). Thus, in any case we have χ(x)⩽ lim_lχl(x) and assertion (105) holds.

From (105), (41), (55), and (5) it follows that lim_l(ψlχl)(i=1nai(x,δu)δiu+g2)χa.e. inΩ.(106) Using (5), (88)(92), (99), Fatou’s lemma, and (106), we established that the function (i=1nai(x,δu)δiu+g2)χ is summable in Ω and Ω{i=1nai(x,δu)δiu+g2}χdxΩF(x,u)Tk(uw)dx+H(φ1+g2)dx. From the latter inequality and Propositions 2.4 and 2.5 we obtain (80).

So, we proved that uT1,q (ν, Ω), and properties (9) and (10) of Definition 2.2 are satisfied. Thus, u is an entropy solution to the Dirichlet problem (7), (8). The theorem is proved. □

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About the article

Received: 2016-11-06

Accepted: 2017-04-05

Published Online: 2017-06-09


Citation Information: Open Mathematics, Volume 15, Issue 1, Pages 768–786, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2017-0064.

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© 2017 Gorban. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License. BY-NC-ND 3.0

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