Research article Special Issues

Hardy potential versus lower order terms in Dirichlet problems: regularizing effects

  • In this paper, dedicated to Ireneo Peral, we study the regularizing effect of some lower order terms in Dirichlet problems despite the presence of Hardy potentials in the right hand side.

    Citation: David Arcoya, Lucio Boccardo, Luigi Orsina. Hardy potential versus lower order terms in Dirichlet problems: regularizing effects[J]. Mathematics in Engineering, 2023, 5(1): 1-14. doi: 10.3934/mine.2023004

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  • In this paper, dedicated to Ireneo Peral, we study the regularizing effect of some lower order terms in Dirichlet problems despite the presence of Hardy potentials in the right hand side.



    En recuerdo de Ireneo: ‘Sed breves con las malas noticias’

    In the paper [8], Ireneo Peral and coauthors proved an existence and summability result on the solutions of the Dirichlet problem

    {div(M(x)u)=Bu|x|2+f(x)in Ω,u=0onΩ. (1.1)

    Here Ω is a bounded open subset of RN, N>2, such that 0 belongs to Ω, M:ΩRN2 is a measurable matrix such that

    M(x)ξξα|ξ|2,|M(x)|β, (1.2)

    for almost every x in Ω and for every ξ in RN, with 0<αβ, B>0 and f belongs to some Lebesgue space Lm(Ω).

    If B=0, the summability results by G. Stampacchia (see [11]) state that the weak solutions u in W1,20(Ω) of (1.1) are bounded if m>N2; while they belong to Lm(Ω), with m=NmN2m, when 2=2NN+2m<N2.

    If B>0, the differential operator

    A(v)=div(M(x)v)Bv|x|2,

    may no longer be coercive, so that both existence and summability results for (1.1) may not be true. However, we recall the following result due to Hardy:

    Proposition 1.1 (Hardy inequality). If v belongs to W1,20(Ω), then

    (N22)2Ω|v|2|x|2Ω|v|2. (1.3)

    Moreover H2=(N22)2 is optimal and is not achieved (for the proof, see [10] or [9]).

    Thanks to Hardy inequality, if 0<B<αH2, then the differential operator A(v) is coercive, so that existence and summability results for (1.1) can be proved. However, with respect to the case B=0, there is an important difference: the summability of the solution depends not only on the summability Lm(Ω) of the datum f, but also on the "size" of B. Indeed, in [8] it is proved that if 1<m<N2, and if

    0<B<αN(m1)(N2m)m2, (1.4)

    then there exists a (weak, or distributional, depending on whether m2 or m<2) solution u of (1.1), with u belonging to Lm(Ω). Note that if m tends to N2, or if m tends to 1, then B tends to zero, and that if m=2, then the condition on B becomes 0<B<αH2. In particular, observe that if f only belongs to L1(Ω), and B>0, neither existence nor summability results can be proved for Eq (1.1). Note also that, as it is proved in [8], if 0<B<αH2, and f belongs to Lm(Ω), with m>N2 (the classic threshold in order to have bounded solutions), then there exists a solution u in W1,20(Ω) of Eq (1.1), but such solution never belongs to L(Ω).

    In some recent papers (see [2], as well as [3]), the problem

    {div(M(x)u)+a(x)u=f(x)in Ω,u=0on Ω, (1.5)

    was studied when a(x)0 is a function in L1(Ω) such that there exists Q>0 such that

    |f(x)|Qa(x). (1.6)

    Under this assumption, the authors proved the existence of a weak solution u in W1,20(Ω) of (1.5), with the property that

    |u(x)|Q,

    so that u belongs to L(Ω), even though the datum f may only be a function in L1(Ω). This is clearly in sharp contrast with the existence results for the case a(x)0, where the solution u does not in general belong to W1,20(Ω), nor it is bounded.

    The purpose of this paper is to prove existence and summability results for the boundary value problem

    {div(M(x)u)+a(x)u=Bu|x|2+f(x)in Ω,u=0on Ω, (1.7)

    with B>0, and a(x) and f(x) such that (1.6) holds. In other words, we will study whether assumption (1.6) (which yields existence of bounded solutions if B=0) allows to improve the results of [8] as far as existence and summability of solutions is concerned. As we will see, if no further assumptions on a(x) with respect to the function B/|x|2 are made, then existence of solutions in W1,20(Ω) for (1.7) follows for every 0<B<αH2, with solutions that become more and more summable as B approaches zero.

    In order to state our first result, let us define 2=2NN2 and the function F:[2,+)R by

    F(t)=αNt(N2Nt). (1.8)

    We remark that F(2)=αH2, that F is strictly decreasing (so that t=2 is a maximum for F on [2,+)), and that F(t) tends to zero as t tends to infinity (see Figure 1).

    Figure 1.  Summability of the solution u.

    The following result will be proved in Section 2.

    Theorem 1.2. Let a(x)0 and f(x) in L1(Ω) be such that (1.6) holds. IfB>0 is such that

    0<B<αH2, (1.9)

    then there exists a unique weak solution u of Eq (1.7), that is a function u in W1,20(Ω) such that a(x)u belongs to L1(Ω) and

    ΩM(x)uφ+Ωa(x)uφ=BΩuφ|x|2+Ωf(x)φ, (1.10)

    for every φ in W1,20(Ω)L(Ω). Furthermore, if pB>2 is the unique solution of the equation F(pB)=B on (2,+), then u belongs to Lp(Ω) for every 2p<pB.

    Remark 1.3 Observe that condition (1.6) allows the coefficient a(x) to vanish in a subset of Ω. The results of Theorem 1.2 can be compared with those of [1] and [4]. In [1] (dedicated to Ireneo Peral for his 70th birthday), existence and Lp-regularity of solutions for the equation

    div(M(x)u)+a|u|r2u=Bu|x|2+f(x),

    is proved for any B>0, under the assumption a>0 (so that a cannot vanish), and r>2: note that in this case the lower order term has a much stronger growth with respect to the one in Eq (1.7). These results were then generalized in [4,Theorem 2.1], where existence of solutions for the equation

    div(M(x)u)+a(x)|u|r2u=Bu|x|2+f(x),

    is proved for B even larger than αH2 and, roughly speaking, the nonnegative coefficient a(x) can vanish in a set of positive measure in the interior of Ω, under the assumptions r>2 and a(x)|f(x)|rr1 belongs to L1(Ω). Note that also in this case we have that the lower order term grows more than linearly, but that the datum f need not be "bounded" with respect to a(x).

    Remark 1.4. Note that if B tends to αH2, then pB tends to 2, while if B tends to zero then pB tends to infinity. Note also that, in contrast with what happens in the case B=0, the value of the constant Q in (1.6) has no influence on the summability of the solution.

    Remark 1.5. We remark the similarity between the summability result of Theorem 1.2, and the summability result of the paper [8] quoted before. In this latter paper, if m>2NN+2 and B>0 are such that (1.4) holds, then the weak solution u in W1,20(Ω) of Eq (1.1) belongs to Lm(Ω). In our Theorem 1.2, the weak solution u in W1,20(Ω) of Eq (1.7) belongs to Lp(Ω) for every p such that B<F(p) (this inequality is equivalent to inequality p<pB). If we choose p=m=NmN2m, the condition B<F(p) means

    B<F(m)=αNm(N2Nm)=αN(m1)(N2m)m2,

    which is exactly (1.4). Thus, the same assumption on B which yields solutions in Lm(Ω) for equation (1.1), yields solutions in Lm(Ω) for Eq (1.7): note however that in the case of Eq (1.7) the datum f only belongs to L1(Ω).

    We now remark that the function a(x) belongs to L1(Ω), while the function B/|x|2 belongs to Lm(Ω), for every 1m<N2, so that it is more summable than a(x). This means that it may happen that the function a(x) dominates the function B/|x|2. In this case, for every B>0 we are going to prove that there exist weak solutions u of Eq (1.7), which belong to L(Ω). Our result is the following, and will be proved in Section 3.

    Theorem 1.6. Let a(x)0 and f(x) in L1(Ω) be such that (1.6) holds. If B>0 and ρ>1 are such that

    a(x)ρB|x|2, (1.11)

    then there exists a unique weak solution u in W1,20(Ω)L(Ω) of (1.7).

    Suppose now that a(x)=C/|x|2, with C>0. If C>B, the result of Theorem 1.6 states that there exist bounded weak solutions u of (1.7); if C=B then any weak solution u of (1.7) is also a solution of

    div(M(x)u)=f(x),

    with |f(x)|QB/|x|2: it is well known from the results of Stampacchia that in this case u may not be in L(Ω). This shows that condition (1.11) (with ρ>1) is somehow necessary in order to have bounded solutions.

    If C<B<αH2 one can only apply Theorem 1.2 to deduce the existence of weak solutions u of Eq (1.7), with u in Lp(Ω) for every p<pB. In Section 4 we are going to prove that the result of Theorem 1.2 is in some sense sharp: for every p>pB there exists Cp<B, such that for a(x)=Cp/|x|2 (and a suitable function f(x) satisfying (1.6)) there exists a weak solution u of Eq (1.7) such that u does not belong to Lp(Ω).

    In the final section of this paper, we will study the boundary value problem associated to a nonlinear quasilinear equation with a lower order term with quadratic growth with respect to the gradient, namely

    {div(M(x)u)+g(u)|u|2=Bu|x|2+f(x)in Ω,u=0on Ω, (1.12)

    where f(x) is a function in L1(Ω) and g:RR is a continuous function such that g(0)=0 and g(t)t is increasing on (0,+) (and decreasing on (,0)). Also in this case, as in the case of Theorem 1.6, we will prove that the lower order term g(u)|u|2 "dominates" the term Bu/|x|2, so that existence of solutions in W1,20(Ω) will follow for every B>0.

    In what follows, we will denote by C any constant depending on the data of the problem (such as N, Ω, α, β, ) but never on the approximation parameter n.

    Proof. Let n in N be fixed, and define

    an(x)=a(x)1+Qna(x),fn(x)=f(x)1+1n|f(x)|, (2.1)

    with Q>0 given by (1.6). Note that, since the function tt1+Qnt is increasing for t>0, from (1.6) it follows that

    |fn(x)|=|f(x)|1+1n|f(x)|Qa(x)1+1nQa(x)=an(x),

    so that (1.6) is satisfied by fn(x) and an(x) for every n in N. A straightforward application of the Schauder theorem yields that for every n in N there exists a weak solution un in W1,20(Ω) of the equation

    {div(M(x)un)+an(x)un=B|x|2+1nun1+1n|un|+fn(x)in Ω,un=0on Ω. (2.2)

    Furthermore, since the right hand side is bounded by Bn2+n, and since an(x)0, the function un belongs to L(Ω) thanks to the results by G. Stampacchia (see [11]).

    We are going to prove that if B>0 satisfies assumption (1.9) then the sequence {un} is bounded in W1,20(Ω). In order to do that, we choose un as test function in the weak formulation for Eq (2.2) to deduce that

    ΩM(x)unun+Ωan(x)un2=ΩB|x|2+1nun21+1n|un|+Ωfn(x)un.

    Using (1.2) and (1.6), we obtain from the previous identity that

    αΩ|un|2+Ωan(x)un2BΩun2|x|2+QΩan(x)|un|, (2.3)

    which implies, thanks to Hardy inequality (1.3), that

    αΩ|un|2+Ωan(x)|un|(|un|Q)BH2Ω|un|2. (2.4)

    We now observe that since t(tQ)Q2 for every 0tQ and that a(x)0, we have

    Ωan(x)|un|(|un|Q)={|un|Q}an(x)|un|(|un|Q)+{|un|>Q}an(x)|un|(|un|Q){|un|Q}an(x)|un|(|un|Q)Q2Ωan(x)Q2Ωa(x).

    Using this inequality in (2.4) we obtain that

    (αBH2)Ω|un|2Q2Ωa(x). (2.5)

    Thanks to assumption (1.9), and to the fact that a(x) belongs to L1(Ω), from (2.5) we obtain that the sequence {un} is bounded in W1,20(Ω), as desired. Therefore, there exists a function u in W1,20(Ω) such that, up to subsequences, the sequence {un} converges to u weakly in W1,20(Ω), weakly in L2(Ω), and almost everywhere in Ω.

    Thanks again to the boundedness of {un} in W1,20(Ω), and to Hardy inequality (1.3), the sequence {un2|x|2} is bounded in L1(Ω); therefore, from (2.3) (dropping the positive first term) and from Young inequality we have that there exists C>0 such that

    Ωan(x)un2C+QΩan(x)|un|C+12Ωan(x)un2+CΩan(x)C+12Ωan(x)un2.

    Therefore, the sequence {an(x)un2} is bounded in L1(Ω). Let now E be a measurable subset of Ω. Then, for k>0 we have

    Ean(x)|un|=E{|un|k}an(x)|un|+E{|un|>k}an(x)|un|kEan(x)+1kΩan(x)un2kEa(x)+Ck

    where we have used the boundedness of {an(x)un2} in L1(Ω) in the last passage. Let now ε>0 be fixed, and choose k>0 large enough to have that Ck<ε. Once k>0 has been chosen, let meas(E) be small enough in order to have

    kEa(x)<ε.

    Such a choice of E is possible since a(x) belongs to L1(Ω). We have thus proved that if meas(E) is small enough, then

    Ean(x)|un|<2ε,nN,

    that is, that the sequence {an(x)un} is uniformly equi-integrable. Since it is almost everywhere convergent to a(x)u, Vitali theorem implies that

    the sequence{an(x)un}stronglyconvergestoa(x)uinL1(Ω).

    This convergence, the convergences already proved on the sequence {un}, and the strong convergence of B|x|2+1n to B|x|2 in Ls(Ω), for every s<N2, imply that one can pass to the limit in the identities

    ΩM(x)unφ+Ωan(x)unφ=ΩB|x|2+1nunφ1+1n|un|+Ωfn(x)φ,

    for every φ in W1,20(Ω)L(Ω), to have (1.10) holds true.

    Once existence of a weak solution has been proved, we turn now to uniqueness. Suppose that u and v are two weak solutions of (1.7), and define w=uv. Since w belongs to W1,20(Ω), from Hardy inequality (1.3) it follows that w2|x|2 belongs to L1(Ω). For k>0, and t in R, let us define

    Tk(t)=max(k,min(t,k)),Gk(t)=tTk(t)=(|t|k)+sgn(t), (2.6)

    and consider

    Sk(x)=BTk(w(x))Gk(w(x))|x|2.

    Since we have that Sk tends to zero almost everywhere in Ω, and since

    0Sk(x)=BTk(w(x))Gk(w(x))|x|2B[w(x)]2|x|2L1(Ω),

    by Lebesgue theorem we have that the sequence {Sk} tends to zero strongly in L1(Ω). Observe now that w is a weak solution of

    div(M(x)w)+a(x)w=Bw|x|2,

    that is, we have

    ΩM(x)wφ+Ωa(x)wφ=BΩwφ|x|2,

    for every φ in W1,20(Ω)L(Ω). Choosing φ=Tk(w) we obtain, using (1.2), dropping a positive term, and recalling that w=Tk(w)+Gk(w), that

    αΩ|Tk(w)|2BΩwTk(w)|x|2=BΩ[Tk(w)]2|x|2+BΩTk(w)Gk(w)|x|2.

    Recalling the definition of Sk, and using Hardy inequality (1.3), the previous inequality implies that

    αΩ|Tk(w)|2BH2Ω|Tk(w)|2+ΩSk(x),

    which yields that

    (αBH2)Ω|Tk(w)|2ΩSk(x).

    Recalling that 0<B<αH2, and letting k tend to infinity, we obtain from the above inequality, using that the sequence {Sk} tends to zero in L1(Ω), that

    0(αBH2)Ω|w|20,

    which then implies that w=0, and so u=v.

    We now turn to the second part of the result. Since we already know that there exists a solution u in W1,20(Ω), in order to show that u belongs to Lp(Ω), for every 2p<pB, it is enough to prove that the sequence {un} is bounded in Lp(Ω) for every 2p<pB. To this aim, let γ1, and choose |un|2γ2un as test function in the weak formulation of Eq (2.2) (this can be done since un belongs to L(Ω) for every n in N). We have

    (2γ1)ΩM(x)unun|un|2γ2+Ωan(x)|un|2γ=ΩB|x|2+1n|un|2γ1+1n|un|+Ωfn(x)|un|2γ2un.

    Using (1.2) and (1.6) we obtain from the previous identity that

    α(2γ1)Ω|un|2|un|2γ2+Ωan(x)|un|2γBΩ|un|2γ|x|2+QΩan(x)|un|2γ1. (2.7)

    We now remark that

    |un|2|un|2γ2=1γ2||un|γ|2,

    so that (2.7) can be rewritten as

    α2γ1γ2Ω||un|γ|2+Ωan(x)|un|2γ1(|un|Q)BΩ(|un|γ)2|x|2. (2.8)

    Thus, using Hardy inequality (1.3), from (2.8) we deduce that

    (α2γ1γ2BH2)Ω||un|γ|2+Ωan(x)|un|2γ1(|un|Q)0. (2.9)

    Since t2γ1(tQ)Q2γ for every 0tQ, we have

    Ωan(x)|un|2γ1(|un|Q){|un|Q}an(x)|un|2γ1(|un|Q)Q2γΩan(x)Q2γΩa(x),

    so that from (2.9) we obtain that

    (α2γ1γ2BH2)Ω||un|γ|2Q2γΩa(x). (2.10)

    If we now assume that γ1 is such that

    α2γ1γ2BH2>0, (2.11)

    from (2.10) it follows that the sequence {|un|γ} is bounded in W1,20(Ω). Thanks to Sobolev embedding, this implies that the sequence {|un|γ} is bounded in L2(Ω), so that the sequence {un} is bounded in L2γ(Ω).

    Summing up, we have that if (2.11) holds, that is if

    B<α2γ1γ2H2=α2γ1(2γ)2(N2)2, (2.12)

    then the sequence {un} is bounded in L2γ(Ω). Setting p=2γ we have, after some straightforward simplifications, that

    α2γ1(2γ)2(N2)2=αNp(N2Np)=F(p).

    Recalling that by definition F(pB)=B, we have from (2.12) that the sequence {un} is bounded in Lp(Ω) for every p2 such that F(pB)<F(p); since F is decreasing on [2,+), we have that the sequence {un} is bounded in Lp(Ω) for every 2p<pB, as desired.

    Proof. In this case, by assumption (1.11), for any n in N, we slightly modify the approximate problems (2.2) and we consider the solution un of

    {div(M(x)un)+an(x)un=B|x|2+ρQBnun1+1n|un|+fn(x)in Ω,un=0onΩ. (3.1)

    Observe that since the function tt1+Qnt is increasing, and since (1.11) holds, we have

    an(x)=a(x)1+Qna(x)ρB|x|21+QnρB|x|2=ρB|x|2+ρQBn=ρB|x|2+ρQBn,

    so that

    B|x|2+ρQBn1ρan(x),nN. (3.2)

    Let k>0 and choose Gk(un) as test function in the weak formulation of (3.1) (recall that the function Gk(t) is defined by (2.6)). We have

    ΩM(x)unGk(un)+Ωan(x)unGk(un)=ΩB|x|2+ρQBnunGk(un)1+1n|un|+Ωfn(x)Gk(un).

    Using (1.2) and (1.6), as well as (3.2), from the above identity we obtain that

    αΩ|Gk(un)|2+Ωan(x)|un||Gk(un)|ΩB|x|2+ρQBn|un||Gk(un)|+QΩan(x)|Gk(un)|1ρΩan(x)|un||Gk(un)|+QΩan(x)|Gk(un)|.

    From the above inequality we obtain, dropping a positive term, that

    Ωan(x)[(11ρ)|un|Q]|Gk(un)|0.

    Choosing k>0 such that (11ρ)k>Q, we therefore have that

    0Ωan(x)[(11ρ)|un|Q]|Gk(un)|0,

    from which it follows that Gk(un)=0; that is, |un|k almost everywhere in Ω, which implies that the sequence {un} is bounded in L(Ω). Once this boundedness has been proved, choosing un as test function in the weak formulation of Eq (3.1), and using (1.2), one has (dropping a positive term) that

    αΩ|un|2BΩun2|x|2+Ω|fn(x)||un|Cun2L(Ω)C,

    so that the sequence {un} is bounded in W1,20(Ω). From these estimates, and reasoning as in the proof of Theorem 1.2, it follows that the weak limit u of the sequence {un} in W1,20(Ω) is a weak solution u of (4.1) that belongs to L(Ω). Uniqueness is then proved as in the proof of Theorem 1.2.

    Remark 3.1. Note that any weak solution of (1.7) is also a weak solution of

    div(M(x)u)+b(x)u=f(x),

    where

    b(x)=(a(x)B|x|2).

    Since under assumption (1.11) we have that

    b(x)(11ρ)a(x)=A(x),

    and since if |f(x)|Qa(x) one also has that |f(x)|QA(x), with

    Q=Q11ρ,

    the boundedness result of Theorem 1.6 can also be obtained using the boundedness result of [2]. It is by the convenience of the reader that we have given a self contained proof of the above theorem.

    As stated in the Introduction, we are going to prove that if α=1, then for every 0<B<H2, and for every p>pB there exist ap(x)0, with ap(x)B/|x|2, and fp(x), such that (1.6) holds, for which there exists a weak solution u in W1,20(Ω) of

    Δu+ap(x)u=Bu|x|2+fp(x),

    with u that does not belong to Lp(Ω). Therefore, Theorem 1.2 is sharp since the summability of the solution u can be at most LpB(Ω), and not better.

    In order to prove the result, let Ω=B1(0), let 0<B<H2, and let p>pB>2; since F(pB)=B, and F is decreasing, we have that B>F(p). Define

    u(x)=1|x|Np1,

    and observe that u is the weak solution in W1,20(Ω) of the equation

    {Δu=F(p)u|x|2+F(p)|x|2in Ω,u=0onΩ.

    Define

    Cp=BF(p),ap(x)=Cp|x|2andQp=F(p)BF(p),fp(x)=Qp|x|2.

    Thanks to these definitions, we have that 0ap(x)B/|x|2, that |fp(x)|Qpap(x), and that u, which does not belong to Lp(Ω), is a weak solution of

    {Δu+ap(x)u=Bu|x|2+fp(x)in Ω,u=0on Ω. (4.1)

    Since such weak solution is unique by Theorem 1.2, we have proved that the result of that theorem is sharp.

    The result of Theorem 1.6 states that if the lower order term a(x) dominates the Hardy potential B/|x|2, then existence of bounded solutions follows for any B>0. The same result is true if one considers gradient dependent lower order terms having quadratic growth. Our result is the following.

    Theorem 5.1. Let B>0, and let f(x) be a function in L1(Ω). Let g:RR be a continuous functionsuch that g(0)=0 and that g(t)t is increasing on (0,+) and decreasing on (,0). Then there exists a weak solution u in W1,20(Ω) of the boundary value problem (1.12), that is: g(u)|u|2 belongs to L1(Ω) and

    ΩM(x)uφ+Ωg(u)|u|2φ=BΩuφ|x|2+Ωf(x)φ, (5.1)

    for every φ in W1,20(Ω)L(Ω).

    Proof. Let n in N, let fn(x)=Tn(f(x)), and let un in W1,20(Ω) be a solution of

    {div(M(x)un)+unn+g(un)|un|2=B|x|2+1nun1+1n|un|+fn(x)inΩ,un=0onΩ. (5.2)

    The existence of un follows from the results of [7], where it is also proved that un belongs to L(Ω) for every n in N (note that the right hand side of the equation is bounded by Bn2+n).

    We now follow [5] (see also [6]) and choose T1(un) as test function in the weak formulation of (5.2). We obtain, dropping a positive term and using (1.2),

    αΩ|T1(un)|2+ΩT1(un)g(un)|un|2ΩB|x|2+1nunT1(un)1+1n|un|+Ω|fn(x)||T1(un)|.

    From this inequality it follows, using that g(t)t is increasing on (0,+) and decreasing on (,0), that

    α{|un|1}|un|2+max{g(1),|g(1)|}{|un|>1}|un|2BΩ|un||x|2+Ω|f(x)|.

    Defining μ=min(α,max{g(1),|g(1)|}), from the above inequality it follows that

    μΩ|un|2BΩ|un||x|2+Ω|f(x)|. (5.3)

    We now observe that by Hölder, Sobolev and Young inequalities we have

    Ω|un||x|2(Ω|un|2)12(Ω1|x|4NN+2)N+22NC(Ω|un|2)12CΩ|un|2+C,

    where in the second to last passage we have used that N>2 so that 1/|x|4NN+2 belongs to L1(Ω). Using this inequality in (5.3) we have that

    μΩ|un|2BCΩ|un|2+Ω|f(x)|+C,

    from which it follows that

    the sequence{un}isbounded in W1,20(Ω).

    Passing to a subsequence if necessary, we may assume the sequence {un} converges to a function u weakly in W1,20(Ω), strongly in Lρ(Ω) for every ρ<2, and almost everywhere in Ω. From these convergences it follows that

    B|x|2+1nun1+1n|un|stronglyconvergesinLσ(Ω),for everyσ<2.

    In particular, it converges in L1(Ω), so that one can repeat the proof of [5] to have that

    thesequence{un}stronglyconvergesinW1,20(Ω), (5.4)

    which in turn implies that

    thesequence {un}almosteverywhereconvergesto u. (5.5)

    In order to pass to the limit in the weak formulation of (5.2), we need to deal with the lower order term with quadratic growth with respect to the gradient: we are going to prove that

    thesequence {g(un)|un|2} strongly converges in L1(Ω) tog(u)|u|2. (5.6)

    Since we already know that g(un)|un|2 almost everywhere converges to g(u)|u|2 as a consequence of the almost everywhere convergence of un, of the continuity of g(t), and of (5.5), to prove (5.6), by Vitali theorem, it suffices to show the equi-integrability of the sequence {g(un)|un|2}. In order to do that, let h>0, k>0 and choose 1hTh[Gk(un)] as test function in the weak formulation of (5.2). Dropping the positive first term, and letting h tend to 0, we obtain (see [5])

    {|un|>k}|g(un)||un|2B{|un|>k}|un||x|2+{|un|>k}|f(x)|. (5.7)

    Since |un|/|x|2 is compact in L1(Ω), and since meas({|un|>k}) tends to zero as k tends to infinity uniformly in n in N, from (5.7) we have that

    limk+{|un|>k}|g(un)||un|2=0,uniformlyinnin N. (5.8)

    Let now E be a measurable subset of Ω; for every k>0 we have

    E|g(un)||un|2=E{|un|k}|g(un)||un|2+E{|un|>k}|g(un)||un|2max[k,k]|g(t)|E|un|2+{|un|>k}|g(un)||un|2.

    We now fix ε>0 and use (5.8) to choose k>0 large enough in order to have

    {|un|>k}|g(un)||un|2<ε.

    Once k>0 has been chosen, we use (5.4) in order to choose meas(E) small enough to have

    max[k,k]|g(t)|E|un|2<ε.

    Therefore, for every ε>0 we have that if meas(E) is small enough, then

    E|g(un)||un|2<2ε,nN,

    which proves the equi-integrability of the sequence {g(un)|un|2}, which implies that (5.6) holds true.

    Having proved all these convergences, we can pass to the limit in the identities

    ΩM(x)unφ+Ωg(un)|un|2φ=ΩB|x|2+1nunφ1+1n|un|+Ωfn(x)φ,

    which hold for every φ in W1,20(Ω)L(Ω), to have that u is such that (5.1) holds.

    The first author is supported by FEDER-MINECO (Spain) grant PGC2018-096422-B-I00 and Junta de Andalucía (Spain) grant FQM-116.

    The authors declare no conflict of interest.



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