Let Ω be a smooth bounded domain of $\mathbb{R}^N$, N≥1 and let H₀¹(Ω) be the standard Sobolev space consisting of functions which vanish on the boundary of Ω and whose gradient is in L²(Ω). We consider the following quasi-linear singular equation

$\left\{\begin{array}{l} -\Delta u-u \Delta\left(u^2\right)=a(x) \frac{1}{u} \text { in } \varOmega, \\ u>0 \text { in } \varOmega, u=0 \text { on } \partial \varOmega, \end{array}\right.$

(1)

where a(x)>0 a.e. in Ω and a∈L^p(Ω) with p>2. This type of equations is closely related to the standing wave solutions of the following quasilinear Schrodinger equation

$\begin{aligned} \mathrm{i} \partial_t \psi= & -\Delta \psi+V(x) \psi-h\left(x, |\psi|^2\right) \psi \\ & -\kappa \Delta\left[\rho\left(|\psi|^2\right)\right] \rho^{\prime}\left(|\psi|^2\right) \psi, \end{aligned}$

(2)

which has wide applications to physical models, suchas the superfluid film equation in plasma physics when ρ(s)=s^[1], the self-channelling of a high-power ultra short laser in matter when $\rho(s)=\sqrt{1+s}$^[2], fluid mechanics^[3], dissipative quantum mechanics, and condensed-matter theory^[4]. A lot of results have been obtained in, for example, Refs. [5-7] and references therein. Singular problems have been intensively studied since 1970s because of their wide applications: boundary layer phenomenon in fluid mechanics, chemical heterogenous catalysts, glacial advance, etc.; we refer the reader to the books by Agarwal and O'Regan^[8], and Hernández and Mancebo^[9] for an excellent introduction to the singular boundary theory.

For the quasilinear singular equation involving a singular function f

$\left\{\begin{array}{l} -\Delta u-u \Delta\left(u^2\right)=f(x, u) \text { in } \varOmega, \\ u>0 \text { in } \varOmega, u=0 \text { on } \partial \varOmega, \end{array}\right.$

(3)

there are only a few results available about the existence of solutions by now. In Ref. [10], Marcos do Ó and Moameni considered the existence of radially symmetric solutions when Ω is a ball centered at the origin and the nonlinearity f(u)=λu³-u-u^-γ with -γ∈(-1, 0). In Ref. [11], Santos, Yang and Zhou studied the case f(x, u)=λa(x)u^-γ+b(x)u^β with -γ∈(-1, 0), β∈(1, 2·2^*-1) and obtained the existence and multiplicity of solutions. For strongly singular quasilinear equations, in Ref. [12] Alves and Reis used the techniques developed in Ref. [13] by the second author for pure strongly singular equations and established the sufficient and necessary condition for the existence of solutions in the case when f(x, u)=h(x)u^-γ+g(x, u) with -γ < -1. When -γ=-1, Alves^[14] and Liu et al.^[15] established the existence results for all -γ < 0 and f(x, u)=a(x)u^-γ+λu^p with coefficient function a(x) satisfying: for each -γ < 0 there exists 0≤φ₀∈C(Ω) and p>N such that a(x)φ₀^-γ∈L^p(Ω), which is used to construct the upper-lower solutions to overcome the singularity. In this paper we study the case -γ=-1 and face it in the quasilinear singular problem (1) for general a(x)∈L^p(Ω) with p>2.

1 The main result

Theorem 1.1 Let $\varOmega \subset \mathbb{R}^N$, N≥1, be a bounded domain with smooth boundary ∂Ω. If a(x)>0 a.e. in Ω and a∈L^p(Ω) with p>2, then problem (1) admits a positive H₀¹(Ω)-solution.

By solutions of (1) we mean here solutions in H₀¹(Ω), i.e. u∈H₀¹(Ω) satisfying u(x)>0 in Ω and for all ψ∈H₀¹(Ω),

$\begin{gathered} \int_{\varOmega} \nabla u \nabla \psi+2 u^2 \nabla u \nabla \psi+2 u \psi|\nabla u|^2- \\ a(x) \frac{1}{u} \psi \mathrm{~d} x=0 . \end{gathered}$

We consider the following functional on H₀¹(Ω):

$J(u)=\frac{1}{2} \int_{\varOmega}\left(1+2 u^2\right)|\nabla u|^2 \mathrm{~d} x-\int_{\varOmega} a(x) \ln (u) \mathrm{d} x .$

In this situation, one must find the integrability of both $\int_{\varOmega}|\nabla u|^2 u^2 \mathrm{~d} x$ and $\int_{\varOmega} a(x) \ln u \mathrm{~d} x$ is obscure on the function space H₀¹(Ω), and so the functional I is not well defined on the entire space H₀¹(Ω). In addition, when one faces quasilinear terms, a change of functions relying upon a nonlinear ODE equation: h is the solution of the following problem

$\begin{cases}h^{\prime}(t)=\frac{1}{\sqrt{1+2 h^2(t)}}, & t \geqslant 0, \\ h(t)=-h(-t), & t \leqslant 0 .\end{cases}$

(c.f. Ref. [16]) and the change, defined by ω(x)=h^-1(u(x)), ∀x∈Ω, make singular terms more delicate. We give a way to deal with the singular functional after the function transform and establish the existence of solutions for more general coefficient functions of singular nonlinearities.

Notation. In the paper we make use of the following notation:

c denotes (possibly different) constants;

We denote by $\|u\|^2=\int_{\varOmega}|\nabla u|^2 \mathrm{~d} x$ the norm in H₀¹(Ω) and for u∈H₀¹(Ω), u⁺=max{u, 0}.

2 Reformulation of the problem (P)

First, we collect some properties of the function h(t) defined by the solution of the nonlinear equation:

$\begin{cases}h^{\prime}(t)=\frac{1}{\sqrt{1+2 h^2(t)}}, & t \geqslant 0, \\ h(t)=-h(-t), & t \leqslant 0 .\end{cases}$

Lemma 2.1 (c.f. Refs. [5, 16-18]) Let h be defined as above. Then h has the following properties:

1) h″(t)=-2h(t)(h′(t))⁴, t>0;

2) h is unique, invertible and $C^{\infty}(\mathbb{R})$-function;

3) 0≤h′(t)≤1, $\forall t \in \mathbb{R}$;

4) |h(t)|≤|t| and $|h(t)|^2 \leqslant \sqrt{2}|t|$, $\forall t \in \mathbb{R}$

5) $\frac{h(t)}{2} \leqslant t h^{\prime}(t) \leqslant h(t), t$, t≥0;

6) $|h(t)| \geqslant h(1)|t|$, |t|≤1 and $|h(t)| \geqslant h(1)|t|^{\frac{1}{2}}$, |t|>1;

7) $\lim \limits_{t \rightarrow 0} \frac{h(t)}{t}=1$, $\lim \limits_{t \rightarrow \infty} \frac{h(t)}{t}=0 \text { and } \lim \limits_{t \rightarrow \infty} \frac{h(t)}{\sqrt{t}}=2^{\frac{1}{4}}$;

8) $|h(t)| h^{\prime}(t) \leqslant \frac{1}{\sqrt{2}}, \forall t \in \mathbb{R}$, h^δ(t)h′(t) is increasing, t>0 and δ>1;

9) h(t)h′(t)/t, t>0 is decreasing, $\lim \limits_{t \rightarrow 0} \frac{h(t) h^{\prime}(t)}{t}=1$and$\lim \limits_{t \rightarrow \infty} \frac{h(t) h^{\prime}(t)}{t}=0$;

10) h(t)^-αh^′(t)t is strictly decreasing on (0, +∞) with α≥2.

Proof  We prove the point 10) of this lemma. The properties 1)-9) have been proved in Refs. [5, 16-18]. It is easily checked, thanks to point 5) of this lemma, that

$t \geqslant h\left(1+2 h^2\right)^{\frac{1}{2}} \frac{1}{\alpha} \geqslant h \frac{\left(1+2 h^2\right)^{\frac{3}{2}}}{\left(1+\left(2+\frac{2}{\alpha}\right) h^2\right)} \frac{1}{\alpha},$

since α≥2. Then it follows that

$\begin{gathered} \frac{\mathrm{d}}{\mathrm{~d} t}\left[h(t)^{-\alpha} h^{\prime}(t) t\right] \\ =\frac{h\left(1+2 h^2\right)^{\frac{3}{2}}-\left(\alpha+(2 \alpha+2) h^2\right) t}{h^{\alpha+1}\left(1+2 h^2\right)^2}<0 .□ \end{gathered}$

By the change of function ω(x)=h^-1(u(x)), we see that if ω∈H₀¹(Ω) is a solution of

$\left\{\begin{array}{l} -\Delta \omega=a(x) \frac{1}{h(\omega)} h^{\prime}(\omega) \text { in } \varOmega, \\ \omega>0 \text { in } \varOmega, \omega=0 \text { on } \partial \varOmega, \end{array}\right.$

(4)

namely,

$\int_{\varOmega} \nabla \omega \nabla \psi-a(x) \frac{h^{\prime}(\omega)}{h(\omega)} \psi \mathrm{d} x=0, \quad \forall \psi \in H_0^1(\varOmega)$

then h(ω)∈H₀¹(Ω) is the solution of (1). Now we write that

$I(\omega)=\frac{1}{2} \int_{\varOmega}|\nabla \omega|^2 \mathrm{~d} x-\int_{\varOmega} a(x) \ln h(\omega(x)) \mathrm{d} x .$

Note that, after the transformation of u(x)∈H₀¹(Ω) into h^-1(u(x))=ω(x)∈H₀¹(Ω), the difficulty for the integrability of $\int_{\varOmega} u^2|\nabla u|^2 \mathrm{~d} x$ has transferred to how to control the singular term lnh(ω(x)), ∀x∈Ω, since h(0)=0, in which h(t) the solution of an nonlinear O.D.E. can not be expressed. We first restrict our attention to the well defined set of functions in H₀¹(Ω) for the singular functional I(ω): X and $\mathcal{N}$ as follows

$\begin{aligned} & X=\left\{\omega \in H_0^1(\varOmega): \omega>0 \text { a. e. in } \varOmega, \right. \\ & \left.a(x) \ln \omega(x) \in L^1(\varOmega)\right\}, \\ & \mathcal{N}=\left\{\omega \in X:\|\omega\|^2-\int_{\varOmega} a(x) \frac{h^{\prime}(\omega)}{h(\omega)} \omega \mathrm{d} x=0\right\} . \end{aligned}$

Indeed, since h(0)=0 and h(t) is strictly increasing on [0, +∞), thanks to properties 4) and 6) of Lemma 2.1, we get

$\begin{aligned} \mid & \int_{A_{+}} a(x) \ln h(\omega(x)) \mathrm{d} x\left|\leqslant \int_{A_{+}} a(x)\right| \ln h(\omega(x)) \mid \mathrm{d} x \\ = & \int_{A_{+}} a(x) \ln h(\omega(x)) \mathrm{d} x \leqslant \int_{A_{+}} a(x) \ln \omega(x) \mathrm{d} x<+\infty \\ & \left|\int_{A_{-}} a(x) \ln h(\omega(x)) \mathrm{d} x\right| \leqslant \int_{A_{-}} a(x)|\ln h(\omega(x))| \mathrm{d} x \\ = & \int_{A_{-}} a(x) \ln \left(\frac{1}{h(\omega(x))}\right) \mathrm{d} x \leqslant \\ & \int_{A_{-} \cap[\omega<1]} a(x)\left[\ln \left(\frac{1}{h(1)}\right)+\ln \left(\frac{1}{\omega(x)}\right)\right] \mathrm{d} x+ \\ & \int_{A_{-} \cap[\omega \geqslant 1]} a(x)\left[\ln \left(\frac{1}{h(1)}\right)+\frac{1}{2} \ln \left(\frac{1}{\omega(x)}\right)\right] \mathrm{d} x \\ \leqslant & 2|\ln h(1)| \int_{\varOmega} a(x) \mathrm{d} x+\frac{3}{2} \int_{\varOmega} a(x)|\ln \omega(x)| \mathrm{d} x<+\infty \end{aligned}$

where

$\begin{aligned} & A_{+}=\{x \in \varOmega: \ln h(\omega(x))>0\}, \\ & A_{-}=\{x \in \varOmega: \ln h(\omega(x)) \leqslant 0\}, \end{aligned}$

which clearly implies that the singular functional I is well defined on X. In addition, for any t>0 and ω∈X,

$I(t \omega)=t^2 \frac{1}{2}\|\omega\|^2-\int_{\varOmega} a(x) \ln h(t \omega) \mathrm{d} x,$

so that

$\frac{\mathrm{d}}{\mathrm{~d} t} I(t \omega)=t\|\omega\|^2-\int_{\varOmega} a(x) \frac{h^{\prime}(t \omega)}{h(t \omega)} \omega \mathrm{d} x .$

Note that, thanks to property 5) of Lemma 2.1, we get

$\frac{h^{\prime}(t \omega(x))}{h(t \omega(x))} \omega(x) \leqslant(t \omega(x))^{-1} \omega(x)=t^{-1}, \forall x \in \varOmega .$

Thus $\int_{\varOmega} a(x) \frac{h^{\prime}(t \omega)}{h(t \omega)} \omega \mathrm{d} x$ makes sense.

Claim 2.1  The set X is not empty.

Proof of Claim 2.1  We let φ₁ be the first positive eigenfunction of-Δ in Ω with Dirichlet boundary condition, that is, -Δφ₁=λ₁φ₁, φ₁| _∂Ω=0, with λ₁ the first Dirichlet eigenvalue of-Δ. It is well known that

$\int_{\varOmega} \varphi_1^{-\gamma}(x) \mathrm{d} x<\infty$

(5)

if and only if -γ>-1 (c.f. Ref. [19]). By the result of Ref. [19](Theorem 1 and 2), we also know that: for any -α∈(-3, -1), there exist two positive constants d₀, d₁ and ω_α(x)∈C²(Ω)∩C(Ω)∩H₀¹(Ω), ω_α(x)>0 in Ω such that

$d_0 \varphi_1^{\frac{2}{1+\alpha}}(x) \leqslant \omega_\alpha(x) \leqslant d_1 \varphi_1^{\frac{2}{1+\alpha}}(x), \forall x \in \Omega.$

(6)

We choose $\alpha_0=\frac{2 p-1}{p-1}$ so that -α₀∈(-3, -1) and $-\frac{2 p^{\prime}}{1+\alpha_0}>-1$ since p>2, where $p^{\prime}=\frac{p}{p-1}$. Then it is possible to find the corresponding function ω_α0(x)∈H₀¹(Ω), ω_α0(x)>0, ∀x∈Ω verifying $d_0 \varphi_1^{\frac{2}{1+\alpha_0}}(x) \leqslant \omega_{\alpha_0}(x) \leqslant d_1 \varphi_1^{\frac{2}{1+\alpha_0}}(x), \forall x \in \varOmega$. For such a ω_α0∈H₀¹(Ω), ω_α0(x)>0 ∀x∈Ω, we divide the domain Ω into three parts:

$\begin{aligned} A_{\omega_{\alpha_0}} & =\left\{x \in \varOmega: \ln \omega_{\alpha_0}<-1\right\}, \\ B_{\omega_{\alpha_0}} & =\left\{x \in \varOmega:\left|\ln \omega_{\alpha_0}\right| \leqslant 1\right\}, \\ D_{\omega_{\alpha_0}} & =\left\{x \in \varOmega: \ln \omega_{\alpha_0}>1\right\}, \end{aligned}$

and then we have

$\begin{aligned} & \left|\int_{B_{\omega_{\alpha_0}}} a(x) \ln \omega_{\alpha_0} \mathrm{~d} x\right| \leqslant \int_{B_{\omega_{\alpha_0}}} a(x)\left|\ln \omega_{\alpha_0}\right| \mathrm{d} x \\ & \leqslant \int_{B_{\omega_{\alpha_0}}} a(x) \mathrm{d} x \leqslant \int_{\varOmega} a(x) \mathrm{d} x<+\infty, \\ & \left|\int_{D_{\omega_{\alpha_0}}} a(x) \ln \omega_{\alpha_0} \mathrm{~d} x\right|=\int_{D_{\omega_{\alpha_0}}} a(x) \ln \omega_{\alpha_0} \mathrm{~d} x \\ & \leqslant \int_{\varOmega} a(x) \omega_{\alpha_0} \mathrm{~d} x \leqslant c\left(\int_{\varOmega} a(x)^p \mathrm{~d} x\right)^{\frac{1}{p}}\left\|\omega_{\alpha_0}\right\|<+\infty, \\ & \left|\int_{A_{\omega_{\alpha_0}}} a(x) \ln \omega_{\alpha_0} \mathrm{~d} x\right| \leqslant \int_{A_{\omega_{\alpha_0}}} a(x)\left|\ln \omega_{\alpha_0}\right| \mathrm{d} x \\ & \leqslant \int_{A_{\omega_{\alpha_0}}} a(x) \frac{1}{\omega_{\alpha_0}} \mathrm{~d} x \leqslant \int_{\varOmega} a(x) \frac{1}{\omega_{\alpha_0}} \mathrm{~d} x<+\infty, \end{aligned}$

where we have used the fact that

$\begin{aligned} & \int_{\varOmega} a(x) \omega_{\alpha_0}^{-1} \mathrm{~d} x \leqslant d_0^{-1} \int_{\varOmega} a(x) \varphi_1^{\frac{2}{1+\alpha_0}}(x) \mathrm{d} x \\ & \leqslant d_0^{-1}\left(\int_{\varOmega} a^p(x) \mathrm{d} x\right)^{\frac{1}{p}}\left(\int_{\varOmega} \varphi_1^{-\frac{2 p^{\prime}}{1+\alpha_0}}(x) \mathrm{d} x\right)^{\frac{1}{p^{\prime}}}, \end{aligned}$

so that a(x)lnω_α0(x)∈L¹(Ω) and the set X is non-empty. This ends the proof of Claim 2.1.□

Claim 2.2  For every ω∈X there exists some t(ω)>0 (which may be not unique) such that $t(\omega) \omega \in \mathcal{N}$and I(t(ω)ω)≤I(tω), ∀t>0.

Proof of Claim 2.2  Fix ω∈X. We set

$f(t)=t\|\omega\|^2-\int_{\varOmega} a(x) \frac{1}{h(t \omega)} h^{\prime}(t \omega) \omega \mathrm{d} x, \forall t>0 .$

Thanks to property 5) of Lemma 2.1, we have that

$\frac{1}{2} \leqslant \frac{h^{\prime}(t \omega(x))}{h(t \omega(x))} t \omega(x) \leqslant 1, \quad \forall x \in \varOmega .$

We set

$g_{i, \omega}(t)=t\|\omega\|^2-\frac{1}{i t} \int_{\varOmega} a(x) \mathrm{d} x, \forall t>0, i=1, 2,$

(7)

so that f(t) verifies

$g_{1, \omega}(t) \leqslant f(t) \leqslant g_{2, \omega}(t), \forall t>0 .$

Clearly, g_{i, ω}(t), i=1, 2, is strictly increasing for all t>0 and satisfies

$g_{i, \omega}(t) \rightarrow+\infty \text { as } t \rightarrow+\infty, g_{i, \omega}(t) \rightarrow-\infty \text { as } t \rightarrow 0^{+},$

then it follows that

$f(t) \rightarrow+\infty \text { as } t \rightarrow+\infty, f(t) \rightarrow-\infty \text { as } t \rightarrow 0^{+} .$

Moreover, since $f(t)=\frac{\mathrm{d}}{\mathrm{~d} t} I(t \omega)$, ∀t>0, I(tω) is decreasing on t>0 small and increasing on t>0 large, so that there must exists some t(ω)>0 (may be not unique) such that I(tω)≥I(t(ω)ω), ∀t>0, which gives

$f(t(\omega))=\left.\frac{\mathrm{d}}{\mathrm{~d} t}\right|_{t=t(\omega)} I(t \omega)=0 .$

(8)

We thus obtain, thanks to ω∈X and (8), that

$a(x) \ln (t(\omega) \omega(x)) \in L^1(\varOmega), t(\omega) \omega(x) \in \mathcal{N}.□$

Proof of Theorem 1.1  We know from the property 4) of Lemma 2.1 that there exists $c_X \in \mathbb{R}$ such that for any function ω∈X,

$\begin{aligned} & I(\omega)=\frac{1}{2} \int_{\varOmega}|\nabla \omega|^2 \mathrm{~d} x-\int_{\varOmega} a(x) \ln h(\omega(x)) \mathrm{d} x \\ \geqslant & \frac{1}{2}\|\omega\|^2-\int_{\varOmega} a(x) h(\omega(x)) \mathrm{d} x\\ & \geqslant \frac{1}{2}\|\omega\|^2-\int_{\Omega} a(x) \omega(x) \mathrm{d} x \geqslant \frac{1}{2}\|\omega\|^2- \\ & \quad\left(\int_{\varOmega} a(x)^p \mathrm{~d} x\right)^{\frac{1}{p}}\left(\int_{\varOmega} \omega^{p^{\prime}}(x) \mathrm{d} x\right)^{\frac{1}{p^{\prime}}} \\ & \geqslant \frac{1}{2}\|\omega\|^2-c\|\omega\| \geqslant c_X, \quad \forall \omega \in X, \end{aligned}$

(9)

where we have used the fact that lns≤s, ∀s>0. This, with Claim 2.1 implies $\inf \limits_{\forall \omega \in X} I(\omega)$ is a finite number. Claim 2.2 independently gives that

$\inf \limits_{\forall \omega \in X} I(\omega) \leqslant \inf \limits_{\forall \omega \in \mathcal{N}} I(\omega) \leqslant I(t(\omega) \omega) \leqslant I(\omega), \forall \omega \in X .$

Thus, we have obtained

$\inf _X I=\inf _N I \in \mathbb{R} .$

We let {ω_n} ⊂X be the sequence such that $I\left(\omega_n\right) \rightarrow \inf \limits_X I$, which gives with $\inf \limits_X I$ a finite number and (8), that {ω_n} ⊂X is bounded in H₀¹(Ω). Then there exist ω₀∈H₀¹(Ω), g∈L²(Ω) and a subsequence, still denoted by ω_n, such that

$\left\{\begin{array}{l} \omega \rightarrow \omega_0 \text { in } H_0^1(\varOmega), \\ \omega_n \rightarrow \omega_0 \text { a.e. in } \varOmega, \\ \omega_n \rightarrow \omega_0 \text { in } L^2(\varOmega), \\ \left|\omega_n(x)\right| \leqslant g(x), \left|\omega_0(x)\right| \leqslant g(x) \text { a. e. in } \varOmega . \end{array}\right.$

(10)

Since {ω_n} ⊂X and ω_n(x)>0 a.e. x∈Ω, we know ω₀(x)≥0, a.e. x∈Ω. We prove now ω₀(x)>0 a.e. x∈Ω and a(x)lnω₀(x)∈L¹(Ω) i.e. ω₀∈X. Since {ω_n} ⊂X, using (9), we write that

$\begin{aligned} & \frac{1}{2}\left\|\omega_n\right\|^2-c_X \geqslant \int_{\varOmega} a(x) \omega_n(x) \mathrm{d} x \\ \geqslant & \int_{\varOmega} a(x) \ln \omega_n(x) \mathrm{d} x \geqslant \int_{\varOmega} a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x \\ = & \frac{1}{2}\left\|\omega_n\right\|^2-I\left(\omega_n\right), \end{aligned}$

where we have used property 4) of Lemma 2.1 for h. By the boundedness of {ω_n} in H₀¹(Ω) and $I\left(\omega_n\right) \rightarrow \inf \limits_X I$ which is a finite number, we write

$\begin{aligned} & \int_{\varOmega} a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x=O(1), \\ & \int_{\varOmega} a(x) \ln \omega_n(x) \mathrm{d} x=O(1), \\ & \int_{\varOmega} a(x) \omega_n(x) \mathrm{d} x=O(1) . \end{aligned}$

(11)

Taking advantage of a(x)(lnω_n(x)-ω_n(x))≤0 a.e. in Ω and using Fatou's Lemma, one also gets $a(x) \limsup _{n \rightarrow+\infty}\left(\ln \omega_n(x)-\omega_n(x)\right)$ is integrable on Ω, and

$\begin{aligned} & \limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x)\left(\ln \omega_n(x)-\omega_n(x)\right) \mathrm{d} x \\ \leqslant & \int_{\varOmega} a(x) \limsup _{n \rightarrow+\infty}\left(\ln \omega_n(x)-\omega_n(x)\right) \mathrm{d} x, \end{aligned}$

(12)

so that, since

$\begin{aligned} & a(x) \limsup _{n \rightarrow+\infty}\left(\ln \omega_n(x)-\omega_n(x)\right) \\ = & \begin{cases}-\infty, & \omega_0(x)=0, \\ a(x)\left[\ln \omega_0(x)-\omega_0(x)\right], & \omega_0(x)>0, \end{cases} \end{aligned}$

we thus obtain, thanks to (11) and (12) and a(x)[lnω₀(x)-ω₀(x)]≤0, ∀x∈Ω, that

$\omega_0(x)>0 \text { a. e. } x \in \varOmega .$

Since p>2 (so p′ < 2) and ω_n→ω₀ in L²(Ω), using Hölder's inequality, we get

$\lim \limits_{n \rightarrow+\infty} \int_{\varOmega} a(x) \omega_n(x) \mathrm{d} x=\int_{\varOmega} a(x) \omega_0(x) \mathrm{d} x .$

(13)

We write now, with ω₀(x)>0 a.e. in Ω, lnω_n(x)-ω_n(x)≤0 a.e. in Ω, and using Fatou's lemma, that

$\begin{aligned} & \limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \ln \omega_n(x) \mathrm{d} x-\int_{\varOmega} a(x) \omega_0(x) \mathrm{d} x \\ & =\limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \ln \omega_n(x) \mathrm{d} x-\limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \omega_n(x) \mathrm{d} x \\ & \leqslant \limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x)\left(\ln \omega_n(x)-\omega_n(x)\right) \mathrm{d} x \\ & \leqslant \int_{\varOmega} a(x) \limsup _{n \rightarrow+\infty}\left(\ln \omega_n(x)-\omega_n(x)\right) \mathrm{d} x \\ & =\int_{\varOmega} a(x)\left(\ln \omega_0(x)-\omega_0(x)\right) \mathrm{d} x \\ & =\int_{\varOmega} a(x) \ln \omega_0(x) \mathrm{d} x-\int_{\varOmega} a(x) \omega_0(x) \mathrm{d} x . \end{aligned}$

Since $\int_{\varOmega} a(x) \omega_0(x) \mathrm{d} x$ is finite, using (11), we have that

$\begin{aligned} & -\infty<\limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \ln \omega_n(x) \mathrm{d} x \\ \leqslant & \int_{\varOmega} a(x) \ln \omega_0(x) \mathrm{d} x \leqslant \int_{\varOmega} a(x) \omega_0(x) \mathrm{d} x<+\infty, \end{aligned}$

so that we obtain a(x)lnω₀(x)∈L¹(Ω). It is worthy remarking here that the main difference between dealing with the singularities u^-1 and u^-ν(ν≠1) lies in the energy controlling: $\int_{\varOmega} \ln u(x) \mathrm{d} x$ should be controlled from both sides since lns < 0, ∀s∈(0, 1), lns→-∞ as s→0⁺ and lns>0, ∀s>1, and $\int_{\varOmega} u^{1-\nu} \mathrm{d} x$ just need to be controlled form one side since s^1-ν>0, ∀s>0.

Then we prove that $I\left(\omega_0\right)=\inf _X I$. Moreover, $\omega_0 \in \mathcal{N}$. Indeed, since

$\begin{aligned} & a(x) \ln h\left(\omega_n(x)\right) \leqslant a(x) \ln \left(\omega_n(x)\right) \\ \leqslant & a(x) \omega_n(x) \leqslant a(x) g(x) \end{aligned}$

and a(x)g(x)∈L¹(Ω), using Fatou's lemma and ω_n(x)→ω₀ a.e. in Ω, we get

$\begin{gathered} \limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x \leqslant \\ \int_{\varOmega} a(x) \ln h\left(\omega_0(x)\right) \mathrm{d} x, \end{gathered}$

(14)

so that

$\begin{aligned} & \inf _X I=\lim _{n \rightarrow+\infty} I\left(\omega_n\right)=\liminf _{n \rightarrow+\infty} \int_{\varOmega} \frac{1}{2}\left|\nabla \omega_n\right|^2- \\ & a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x \geqslant \liminf _{n \rightarrow+\infty} \frac{1}{2} \int_{\Omega}\left|\nabla \omega_n\right|^2 \mathrm{~d} x+ \\ & \liminf _{n \rightarrow+\infty} \int_{\varOmega}-a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x \\ & \geqslant \frac{1}{2}\left\|\omega_0\right\|^2+\int_{\varOmega}-a(x) \ln h\left(\omega_0(x)\right) \mathrm{d} x=I\left(\omega_0\right) \\ & \geqslant I\left(t\left(\omega_0\right) \omega_0\right) \geqslant \inf _N I=\inf _X I, \end{aligned}$

where we have used the weak lower semicontinuity of norm, (14), and

$\begin{aligned} & \liminf _{n \rightarrow+\infty} \int_{\varOmega}-a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x \\ = & -\limsup _{n \rightarrow+\infty} \int_{\varOmega} a(x) \ln h\left(\omega_n(x)\right) \mathrm{d} x . \end{aligned}$

This proves that

$I\left(\omega_0\right)=\inf _X I$

(15)

This leads to

$\frac{{\rm{d}}}{{{\rm{d}}t}}\left| {_{t = 1}I\left( {t{\omega _0}} \right)} \right. = 0,$

that is, $\omega_0 \in \mathcal{N}$.

We now show that h(ω₀) is a solution of problem (1). Suppose φ∈H₀¹(Ω), φ(x)≥0, ∀x∈Ω and ε>0. Set v=ω₀+εφ, we divide the domain Ω into three parts:

$\begin{gathered} A_v=\{x \in \varOmega: \ln v<-1\}, \\ B_v=\{x \in \varOmega:|\ln v| \leqslant 1\}, \\ D_v=\{x \in \varOmega: \ln v>1\}, \end{gathered}$

and then we have

$\begin{aligned} & \left|\int_{R_v} a(x) \ln \left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x\right| \\ & \leqslant \int_{B_v} a(x)\left|\ln \left(\omega_0+\varepsilon \varphi\right)\right| \mathrm{d} x \\ & \leqslant \int_{B_v} a(x) \mathrm{d} x \leqslant \int_{\Omega} a(x) \mathrm{d} x<+\infty, \\ & \left|\int_{D_v} a(x) \ln \left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x\right|=\int_{D_v} a(x) \ln \left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x \\ & \leqslant \int_{\varOmega} a(x)\left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x \\ & \leqslant c\left(\int_{\varOmega} a(x)^p \mathrm{~d} x\right)^{\frac{1}{p}}\left\|\omega_0+\varepsilon \varphi\right\|<+\infty, \\ & \left|\int_{A_v} a(x) \ln \left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x\right| \\ & \leqslant \int_{A_v} a(x)\left|\ln \left(\omega_0+\varepsilon \varphi\right)\right| \mathrm{d} x \\ & =\int_{A_v} a(x) \ln \frac{1}{\left(\omega_0+\varepsilon \varphi\right)} \mathrm{d} x \leqslant \int_{A_v} a(x) \ln \frac{1}{\omega_0} \mathrm{~d} x \\ & \leqslant \int_{A_v} a(x)\left|\ln \omega_0\right| \mathrm{d} x \\ & \leqslant \int_{\varOmega} a(x)\left|\ln \omega_0\right| \mathrm{d} x<+\infty \end{aligned}$

so ω₀+εφ∈X. By Claim 2.2 and (15) we can conclude that

$\begin{aligned} I\left(\omega_0+\varepsilon \varphi\right) & \geqslant I\left(t\left(\omega_0+\varepsilon \varphi\right)\left(\omega_0+\varepsilon \varphi\right)\right) \\ & \geqslant \inf _N I=I\left(\omega_0\right), \end{aligned}$

and then we have

$\begin{aligned} & \frac{\left\|\omega_0+\varepsilon \varphi\right\|^2-\left\|\omega_0\right\|^2}{2} \\ \geqslant & \int_{\varOmega} a(x) \ln h\left(\omega_0+\varepsilon \varphi\right) \mathrm{d} x-\int_{\varOmega} a(x) \ln h\left(\omega_0\right) \mathrm{d} x \end{aligned}$

(16)

Dividing (16) by ε>0 and letting ε→0⁺ we conclude with Fatou's Lemma that

$\begin{aligned} & \int_{\varOmega} \nabla \omega_0 \nabla \varphi \mathrm{~d} x \\ \geqslant & \liminf _{\varepsilon \rightarrow 0^{+}} \frac{\int_{\varOmega} a(x)\left[\ln h\left(\omega_0+\varepsilon \varphi\right)-\ln h\left(\omega_0\right)\right] \mathrm{d} x}{\varepsilon} \\ \geqslant & \int_{\varOmega} \liminf _{\varepsilon \rightarrow 0^{+}} \frac{a(x)\left[\ln h\left(\omega_0+\varepsilon \varphi\right)-\ln h\left(\omega_0\right)\right]}{\varepsilon} \mathrm{d} x \\ = & \int_{\varOmega} a(x) \frac{h^{\prime}\left(\omega_0\right)}{h\left(\omega_0\right)} \varphi \mathrm{d} x . \end{aligned}$

(17)

Suppose ψ∈H₀¹(Ω) and t>0. Inserting φ=(ω₀+tψ)⁺ into (17), we get

$\begin{aligned} & 0 \leqslant \frac{1}{t} \int_{\varOmega} \nabla \omega_0 \nabla\left(\omega_0+t \psi\right)^{+}- \\ & a(x) \frac{h^{\prime}\left(\omega_0\right)}{h\left(\omega_0\right)}\left(\omega_0+t \psi\right)^{+} \mathrm{d} x=\frac{1}{t}\left(\int_{\varOmega}-\int_{\varOmega \cap\left[\omega_{0}+t \psi<0\right]}\right) \\ & \left(\nabla \omega_0 \nabla\left(\omega_0+t \psi\right)-a(x) \frac{h^{\prime}\left(\omega_0\right)}{h\left(\omega_0\right)}\left(\omega_0+t \psi\right)\right) \mathrm{d} x \\ & \leqslant \int_{\varOmega} \nabla \omega_0 \nabla \psi-a(x) \frac{h^{\prime}\left(\omega_0\right)}{h\left(\omega_0\right)} \psi \mathrm{d} x- \\ & \int_{\varOmega \cap\left[\omega_0+t \psi<0\right]} \nabla \omega_0 \nabla \psi \mathrm{~d} x . \end{aligned}$

(18)

Since meas[ω₀+tψ]→0 as t→0⁺, by passing to the limit as t→0⁺ we can conclude that

$0 \leqslant \int_{\varOmega} \nabla \omega_0 \nabla \psi-a(x) \frac{h^{\prime}\left(\omega_0\right)}{h\left(\omega_0\right)} \psi \mathrm{d} x .$

The same conclusion can be drawn for -ψ in place of ψ, thus it follows that h(ω₀) is indeed a H₀¹(Ω)-solution of (1).□