谭玉鑫, 孙义静
中国科学院大学数学科学学院, 北京 100049
摘要: 证明-div（M（x）▽u）= $\frac{{f\left( x \right)}}{{{u^p}}}$正H₀¹-解的存在性，其中M（x）是有界椭圆矩阵（即存在0 < α ≤ β满足M（x）ξ·ξ ≥ α|ξ|²，|M（x）|≤ β，?x ∈ Ω，ξ ∈ Rⁿ）和-p < -1.本工作的关键点在于建立2个密切联系的集合，便于找到相应的能量泛函最小值。
关键词: 有界椭圆矩阵弱解强奇性
In this work, we consider the existence of solutions of the semilinear elliptic problem with a singular nonlinearity,

$\left\{ \begin{array}{l} - {\rm{div}}\left( {\begin{array}{*{20}{c}}{\mathit{\boldsymbol{M}}\left( x \right)}&{\nabla u}\end{array}} \right) = h\left( x \right){u^{ - p}}\;\;\;{\rm{in}}\;\Omega ,\\u > 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{in}}\;\Omega ,\\u = 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{on}}\;\Omega ,\end{array} \right.$

(1)

where Ω $\subset $Rⁿ is a bounded open set with smooth boundary $\partial $Ω, M(x) is a real symmetric matrix satisfying

$\begin{array}{*{20}{c}}{\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\xi }} \ge \alpha {{\left| \mathit{\boldsymbol{\xi }} \right|}^2},}\\{\left| {\mathit{\boldsymbol{M}}\left( x \right)} \right| \le \beta ,\forall x \in \Omega ,\mathit{\boldsymbol{\xi }} \in {{\bf{R}}^n},}\end{array}$

(2)

and h(x)>0 a.e.in Ω and -p < -1.By solutions we mean here weak solutions in H₀¹(Ω), i.e., u∈H₀¹(Ω) satisfying u(x)>0 in Ω and

$\begin{array}{*{20}{c}}{\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla \nu {\rm{d}}x} - \int_\Omega {\frac{{h\left( x \right)}}{{{u^p}}} \cdot \nu {\rm{d}}x} = 0,}\\{\forall \nu \in H_0^1\left( \Omega \right).}\end{array}$

Since the work by Stuart^[1], people have paid much attention to the existence and multiplicity of solutions for such singular equations

$ - {\rm{div}}\left( {\begin{array}{*{20}{c}}{\mathit{\boldsymbol{M}}\left( x \right)}&{\nabla u}\end{array}} \right) = f\left( {x,u} \right),$

where f(x, s) is singular at s=0. See Refs.[2-5] and the rich list of references provided by these papers for a survey. Recently, Boccardo and Orsina^[6] solved the problem with f(x, u)=h(x)u^-p, h(x)≥0, -p < -1 and provided the existence of an H_loc¹(Ω)-solution u by using approximation arguments when M(x) is a real symmetric matrix satisfying M(x)ξ·ξ≥α|ξ|², |M(x)|≤β, ?x∈Ω, ξ∈Rⁿ and ${{u}^{\frac{1+p}{2}}}$∈H₀¹(Ω). Then, under a superlinear perturbation of u^q with q>1, Boccardo^[7] also proved the existence of H_loc¹(Ω)-solution for each -p < -1 and ${{u}^{\frac{1+p}{2}}}$∈H₀¹(Ω). Recently, Boccardo and Casado-Dìaz^[8] studied some properties of the solution of problem (1). They showed that if M(x) is a bounded elliptic matrix, h(x)∈L^m(Ω), m≥(2^*)′, supp(h(x)) is compact, then the solution u of (1) obtained as the limit of the solution u_n of -div(M(x)Δu_n)= $\frac{h\left( x \right)}{1/n+u_{n}^{-p}}$ is in H₀¹(Ω). In this work, we will show a compatible condition on the couple (h(x), p), which is optimal for the existence of H₀¹-solutions.
We define the singular energy functional

$\begin{array}{l}I\left( u \right) = \frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u{\rm{d}}x} + \\\;\;\;\;\;\;\;\;\;\;\frac{1}{{p - 1}}\int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}{\rm{d}}x} ,\end{array}$

(3)

where -p < -1. The main difficulty is the absence of integrability of u^-p for u∈H₀¹(Ω) when -p < -1 and any inequality that relates u∈H₀¹(Ω) will not be of much help. It should be noted also that there is a sharp contrast between the case -1 < -p < 0, for which the energy functional is continuous, and the case -p < -1. Generally, the sub-supersolution method is very effective in dealing with singularity. However, the method cannot be used for such general measurable h(x)>0. To reverse this situation, we use constrained sets to restore integrability and recast problem (1) into a variational framework in the spirit of our earlier works^[9-12]. We defined constrained sets N₁ and N₂ as follows: N₁:={u∈H₀¹(Ω):u≥0 in Ω and ∫_ΩM(x)▽u·▽u≥∫_Ωh(x)|u|^1-p}, N₂:={u∈H₀¹(Ω):u≥0 in Ω and ∫_ΩM(x)▽u·▽u=∫_Ωh(x)|u|^1-p}. Here, special care must be taken to establish the validity and connection of the two constraints which simplify the existence of a minimizer for the singular functional I. It should also be noted that for -p < -1, N₂ is not closed as usual (certainly not weakly closed) in H₀¹(Ω).
In this paper we will use the notation,
C, C_i, c_i, i=1, 2, …, denoting (possibly different) constants.
We denote the Dirichlet norm in H₀¹(Ω) by ‖u‖²=∫_Ω|▽u|²dx, |M(x)|=det M(x), and M(x)ξ·η:=ξ^TM(x)η.
1 Main resultsTheorem 1.1?? Let Ω $\subset $Rⁿ be bounded open set with smooth boundary $\partial $Ω, M(x) be the real symmetric matrix satisfying M(x)ξ·ξ≥α|ξ|², |M(x)|≤β, ?x∈Ω, ξ∈Rⁿ with 0 < α≤β, h(x)∈L¹(Ω), h(x)>0 a.e.in Ω and -p < -1, then

$ - {\rm{div}}\left( {\begin{array}{*{20}{c}}{\mathit{\boldsymbol{M}}\left( x \right)}&{\nabla u}\end{array}} \right) = h\left( x \right){u^{ - p}}$

admits an H₀¹-solution if there exists u₀∈H₀¹(Ω) such that

$\int_\Omega {h\left( x \right){{\left| {{u_0}} \right|}^{1 - p}}{\rm{d}}x} < + \infty .$

(4)

2 Proof of Theorem 1.1Proof?? It should be noted that the topology on H₀¹(Ω) which was generated by the norm

${\left( {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} } \right)^{\frac{1}{2}}}$

is equivalent to the one that was generated by the norm

${\left( {\int_\Omega {{{\left| {\nabla u} \right|}^2}} } \right)^{\frac{1}{2}}},$

since

$\begin{array}{*{20}{c}}{\alpha \int_\Omega {{{\left| {\nabla u} \right|}^2}} \le \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} \le }\\{\frac{\beta }{{{\alpha ^{n - 1}}}}\int_\Omega {{{\left| {\nabla u} \right|}^2}} .}\end{array}$

Hence

$\begin{array}{*{20}{c}}{{{\left( {H_0^1\left( \Omega \right),{{\left( {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} } \right)}^{\frac{1}{2}}}} \right)}^ * } = }\\{{{\left( {H_0^1\left( \Omega \right),{{\left( {\int_\Omega {{{\left| {\nabla u} \right|}^2}} } \right)}^{\frac{1}{2}}}} \right)}^ * }.}\end{array}$

Then, it follows that

${u_n} \to u\;{\rm{weakly}}\;{\rm{in}}\;\left( {H_0^1\left( \Omega \right),{{\left( {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} } \right)}^{\frac{1}{2}}}} \right)$

is equal to

${u_n} \to u\;{\rm{weakly}}\;{\rm{in}}\;\left( {H_0^1\left( \Omega \right),{{\left( {\int_\Omega {{{\left| {\nabla u} \right|}^2}} } \right)}^{\frac{1}{2}}}} \right).$

The key to prove (1) depends on a natural interpolation between the constrained sets N_i, i=1, 2. Taking u∈H₀¹(Ω) with

$\int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}} < \infty ,$

the function

$U\left( t \right): = {t^{1 + p}}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u{\rm{d}}x} - \int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}} {\rm{d}}x$

is increasing on t>0 with $\mathop {\lim }\limits_{t \to + \infty } \, U\left( t \right)=+\infty $ and $\mathop {\lim }\limits_{t \to {0^ + }} \, U\left( t \right) < 0$. Since

$\begin{array}{*{20}{c}}{\frac{{{\rm{d}}I\left( {tu} \right)}}{{{\rm{d}}t}} = t\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u{\rm{d}}x} - {t^{ - p}}\int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}} {\rm{d}}x}\\{ = {t^{ - p}}\left( {{t^{1 + p}}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u{\rm{d}}x} - \int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}} {\rm{d}}x} \right),}\end{array}$

then it follows that there exists the unique positive minimizer t(u)u such that

$I\left( {tu} \right) \ge I\left( {t\left( u \right)u} \right),\forall t > 0.$

(5)

In particular, assumption (4) of Theorem 1.1 implies the existence of t(u₀)>0 such that t(u₀)u₀∈N₂ and hence N₁($\supset $N₂) and N₂ are not empty. Clearly, since tu₀∈N₁ for all ≥t(u₀), N₁ is unbounded in H₀¹(Ω). The closeness of N₁ follows easily from Fatou's lemma. However, it should be noted that N₂ is not anymore a closed set in H₀¹(Ω) since ∫_Ωh(x)|u|^(1-p)dx is not continuous in H₀¹(Ω) as -p < -1. Furthermore, unbounded N₁ lies in the exterior of H₀¹(Ω) (i.e., it stays away from a ball centered at zero). Indeed, since -p < -1, the reversed H?lder inequality

$\begin{array}{*{20}{c}}{\beta {{\left\| u \right\|}^2} \ge \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} \ge \int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}} }\\{ \ge {{\left( {\int_\Omega {h{{\left( x \right)}^{1 - p}}} } \right)}^p}{{\left( {\int_\Omega {\left| u \right|} } \right)}^{1 - p}}}\end{array}$

and Poincaré inequality

${\left( {\int_\Omega {\left| u \right|} } \right)^{1 - p}} \ge {C_1}{\left\| u \right\|^{1 - p}},$

imply that ‖u‖≥C for all u∈N₁.
It should also be noted that

$\begin{array}{*{20}{c}}{\left| {\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }}} \right| \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left| \mathit{\boldsymbol{\xi }} \right| \cdot \left| \mathit{\boldsymbol{\eta }} \right|,}\\{\forall \mathit{\boldsymbol{\xi }},\mathit{\boldsymbol{\eta }} \in {{\bf{R}}^n},x \in \Omega .}\end{array}$

(6)

Indeed, since M(x) is a real symmetric matrix, there exists an orthogonal matrix Q(x) such that

${\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{MQ}} = \left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right),$

where λ_i, i=1, …, n, the eigenvalues of M(x), satisfy λ_i≥α since M(x)ξ·ξ≥α|ξ|², ?x∈Ω, ξ∈Rⁿ. Then it yields that

$\mathit{\boldsymbol{M}} = \mathit{\boldsymbol{Q}}\left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right){\mathit{\boldsymbol{Q}}^{\rm{T}}}$

and

$\begin{array}{l}\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }} = {\mathit{\boldsymbol{\xi }}^{\rm{T}}}\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\eta = }}{\mathit{\boldsymbol{\xi }}^{\rm{T}}}\mathit{\boldsymbol{Q}}\left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right){\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }}\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathit{\boldsymbol{ = }}{\left( {{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }}} \right)^{\rm{T}}}\left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right){\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }}.\end{array}$

Hence, if one defines x:=Q^Tξ, y:=Q^Tη, one can obtain

$\begin{array}{l}\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }} = {\mathit{\boldsymbol{x}}^{\rm{T}}}\left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right)y\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \left( {\begin{array}{*{20}{c}}{{x_1}}& \cdots &{{x_n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\lambda _1}}&{}&{}\\{}& \ddots &{}\\{}&{}&{{\lambda _n}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{y_1}}\\ \vdots \\{{y_n}}\end{array}} \right)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = {\lambda _1}{x_1}{y_1} + {\lambda _2}{x_2}{y_2} + \cdots + {\lambda _n}{x_n}{y_n},\end{array}$

which implies that

$\begin{array}{l}\left| {\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }}} \right| = \left| {\sum\limits_{i = 1}^n {{\lambda _i}{x_i}{y_i}} } \right|\\\;\;\;\;\; = \left| {\prod\limits_{i = 1}^n {{\lambda _i}\left( {\frac{{{\lambda _1}}}{{\prod {{\lambda _i}} }}{x_1}{y_1} + \frac{{{\lambda _2}}}{{\prod {{\lambda _i}} }}{x_2}{y_2} + \cdots + \frac{{{\lambda _n}}}{{\prod {{\lambda _i}} }}{x_n}{y_n}} \right)} } \right.\\\;\;\;\;\; \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left| x \right| \cdot \left| y \right|.\end{array}$

Since

$\begin{array}{l}{\left| x \right|^2} = {\mathit{\boldsymbol{x}}^{\rm{T}}} \cdot \mathit{\boldsymbol{x}} = {\left( {{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }}} \right)^{\rm{T}}} \cdot {\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }} = {\mathit{\boldsymbol{\xi }}^{\rm{T}}}\mathit{\boldsymbol{Q}} \cdot {\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\xi }}\\\;\;\;\;\; = {\mathit{\boldsymbol{\xi }}^{\rm{T}}} \cdot \mathit{\boldsymbol{\xi }} = {\left| \mathit{\boldsymbol{\xi }} \right|^2},\end{array}$

$\begin{array}{l}{\left| y \right|^2} = {\mathit{\boldsymbol{y}}^{\rm{T}}} \cdot \mathit{\boldsymbol{y}} = {\left( {{\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }}} \right)^{\rm{T}}} \cdot {\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }} = {\mathit{\boldsymbol{\eta }}^{\rm{T}}}\mathit{\boldsymbol{Q}} \cdot {\mathit{\boldsymbol{Q}}^{\rm{T}}}\mathit{\boldsymbol{\eta }}\\\;\;\;\;\; = {\mathit{\boldsymbol{\eta }}^{\rm{T}}} \cdot \mathit{\boldsymbol{\eta }} = {\left| \mathit{\boldsymbol{\eta }} \right|^2},\end{array}$

it follows that

$\left| {\mathit{\boldsymbol{M}}\left( x \right)\mathit{\boldsymbol{\xi }} \cdot \mathit{\boldsymbol{\eta }}} \right| \le \frac{\beta }{{{\alpha ^{n - 1}}}}\left| \mathit{\boldsymbol{x}} \right| \cdot \left| \mathit{\boldsymbol{y}} \right| = \frac{\beta }{{{\alpha ^{n - 1}}}}\left| \mathit{\boldsymbol{\xi }} \right| \cdot \left| \mathit{\boldsymbol{\eta }} \right|.$

Furthermore, for u∈N₁,

$\begin{array}{l}\int {h\left( x \right){{\left| u \right|}^{1 - p}}} \le \int {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \frac{\beta }{{{\alpha ^{n - 1}}}}\int {{{\left| {\nabla u} \right|}^2} < \infty } .\end{array}$

(7)

Then there exists t_u>0 such that t_uu∈N₂, I(u)≥I(t_uu)≥ $\mathop {\inf }\limits_{{N_2}} \, I$, and therefore

$\mathop {\inf I}\limits_{{N_1}} \ge \mathop {\inf I}\limits_{{N_2}} .$

(8)

However, since N₁ $\supset $ N₂, it follows that

$\mathop {\inf I}\limits_{{N_1}} \le \mathop {\inf I}\limits_{{N_2}} .$

(9)

In view of (8) and (9), it yields that

$\mathop {\inf I}\limits_{{N_1}} = \mathop {\inf I}\limits_{{N_2}} .$

Now, we turn our attention to $\mathop {\inf }\limits_{{N_1}} \, I$. For N₁, we can assert that it is closed in H₀¹(Ω). Indeed, as u_n→u in H₀¹(Ω), u_n→u in L²(Ω), u_n→u a.e.in Ω, and ∫M(x)▽_n·▽u_n≥∫h(x)|u_n|^1-p. Since h(x)>0 a.e.in Ω, it follows that u_n>0 a.e. in Ω. By (6), we have

$\begin{array}{l}\mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \int {h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \int {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \frac{\beta }{{{\alpha ^{n - 1}}}}\int {{{\left| {\nabla {u_n}} \right|}^2} < \infty } ,\end{array}$

and based on Fatou' lemma, we obtain

$\begin{array}{l}\int {h\left( x \right){{\left| u \right|}^{1 - p}}} = \int {\mathop {\lim \;\rm{inf}}\limits_{n \to \infty } h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} \\ \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \int {h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \int {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} \\ = \int {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} .\end{array}$

On the other hand, we can claim that I(u) is weakly lower semi-continuous, that is,

$\begin{array}{*{20}{c}}{I\left( u \right) \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } I\left( {{u_n}} \right)}\\{{\rm{as}}\;{u_n} \to u\;{\rm{weakly}}\;{\rm{in}}\;\left( {H_0^1\left( \Omega \right),{{\left( {\int {{{\left| {\nabla {u_n}} \right|}^2}} } \right)}^{\frac{1}{2}}}} \right).}\end{array}$

as u_n→u weakly in $\left( {H_0^1\left( \Omega \right),{{\left( {\int {|\nabla u{|^2}} } \right)}^{\frac{1}{2}}}} \right)$. Indeed, there holds that

$\int {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} \to \int {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} ,$

and by Fatou's lemma one can also obtain

$\int {h\left( x \right){{\left| u \right|}^{1 - p}}} \le \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \int {h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} $

if u_n→u weakly in $\left( {H_0^1\left( \Omega \right),{{\left( {\int {|\nabla u{|^2}} } \right)}^{\frac{1}{2}}}} \right)$. Now we can use Ekeland's principle^[13] to exploit the property of the best minimizing sequence for $\mathop {\inf }\limits_{{N_1}} \, I$, that is, (u_n)∈N₁ satisfying

$\begin{align} & \left( \text{ⅰ} \right)I\left( {{u}_{n}} \right) < \underset{{{N}_{1}}}{\mathop{\inf }}\,I+\frac{1}{n} \\ & \left( \text{ⅱ} \right)I\left( {{u}_{n}} \right)\le I\left( v \right)+\frac{1}{n}||{{u}_{n}}-v||,\forall v\in {{N}_{1}} \\ \end{align}$

since N₁ is a closed set in H₀¹(Ω). We may assume u_n≥0 as I(u)=I(|u|). Since -p < -1, I(u) is coercive on N₁ and therefore (u_n) is bounded in H₀¹(Ω). Indeed, by -p < -1,

$\begin{array}{*{20}{c}}{I\left( u \right) = \frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u{\rm{d}}x} + }\\{\frac{1}{{p - 1}}\int_\Omega {h\left( x \right){{\left| u \right|}^{1 - p}}{\rm{d}}x} \ge \frac{\alpha }{2}{{\left\| u \right\|}^2}.}\end{array}$

Hence, up to subsequence (still denoted by u_n), u_n $\rightharpoonup$ u^* weakly in H₀¹(Ω), strongly in L²(Ω), and pointwise a.e. in Ω. Therefore u^*≥0. More precisely,

${u^ * } > 0\;{\rm{a}}{\rm{.}}\;{\rm{e}}{\rm{.}}\;{\rm{in}}\;\Omega $

(10)

as ∫_Ωh(x)|u^*|^1-p < ∞ by Fatou's lemma. Moreover, we shall show that u^*∈N₂ by evaluating the best minimizing sequence (u_n)∈N₁.
Case 1. Suppose that (u_n)N₁\N₂ for all n large. Fix φ∈H₀¹(Ω), φ≥0 and n by now. Note that, as (u_n)N₁\N₂ and p>1, there holds that $\int {\mathit{\boldsymbol{M}}\left( x \right)} \nabla {{u}_{n}}\cdot \nabla {{u}_{n}}>\int_{\Omega }{h\left( x \right)}|{{u}_{n}}{{|}^{1-p}}\ge \int_{\Omega }{h\left( x \right)}|{{u}_{n}}+t\varphi {{|}^{1-p}}$ for t≥0. Subsequently, choose t>0 sufficiently small such that

$\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right)} > \int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}{\rm{d}}x} ,$

that is,

${u_n} + t\varphi \in {N_1}.$

In virtue of (ⅰ) and (ⅱ), we obtain that

$\begin{array}{l}\frac{t}{n}\left\| \varphi \right\| + \frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\left( {{{\left| {\nabla \left( {{u_n} + t\varphi } \right)} \right|}^2} - {{\left| {\nabla {u_n}} \right|}^2}} \right){\rm{d}}x} \\ \ge \frac{1}{{1 - p}}\int_\Omega {h\left( x \right)\left( {{{\left| {\left( {{u_n} + t\varphi } \right)} \right|}^{1 - p}} - {{\left| {{u_n}} \right|}^{1 - p}}} \right){\rm{d}}x} .\end{array}$

Dividing by t>0, passing to the liminf as t→0⁺, we obtain

$\begin{array}{l}\frac{{\left\| \varphi \right\|}}{n} + \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla \varphi } \\\;\;\;\;\;\;\; \ge \int_\Omega {\mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \frac{{h\left( x \right)}}{{1 - p}}\frac{{{{\left( {{u_n} + t\varphi } \right)}^{1 - p}} - u_n^{1 - p}}}{t}} \\\;\;\;\;\;\;\; = \int_\Omega {h\left( x \right)u_n^{ - p}\varphi } .\end{array}$

Using Fatou's lemma again and letting n tend to infinity, we have

$\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \varphi } \ge \int_\Omega {h\left( x \right){u^{ * - p}}\varphi } ,\forall \varphi \ge 0.$

In view of (10), we obtain that u^*∈N₁, and by the above argument (5) there exists a unique t(u^*) such that I(t(u^*)u^*)= $\mathop {\min }\limits_{t > 0} \, I\left( t{{u}^{*}} \right)$. So

$\begin{array}{l}\mathop {\inf }\limits_{{N_1}} I = \mathop {\lim }\limits_{n \to \infty } I\left( {{u_n}} \right)\\\; = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} + \frac{1}{{p - 1}}\int_\Omega {h\left( x \right)u_n^{1 - p}} } \right]\\\; \ge \mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \left[ {\frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} } \right] + \\\;\;\;\;\mathop {\lim \;\rm{inf}}\limits_{n \to \infty } \left[ {\frac{1}{{p - 1}}\int_\Omega {h\left( x \right)u_n^{1 - p}} } \right]\\\; \ge \frac{1}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla {u^ * }} + \frac{1}{{p - 1}}\int_\Omega {h\left( x \right){u^{ * 1 - p}}} \\\; = I\left( {{u^ * }} \right) \ge I\left( {t\left( {{u^ * }} \right){u^ * }} \right) \ge \mathop {\inf }\limits_{{N_2}} I \ge \mathop {\inf }\limits_{{N_1}} I,\end{array}$

and thus t(u^*)=1, which means that

$\mathop {\min }\limits_{t > 0} I\left( {t{u^ * }} \right) = I\left( {{u^ * }} \right),{u^ * } \in {N_2}.$

Case 2. There exists a subsequence of (u_n) (still denoted by u_n), which belongs to N₂.
Let φ∈H¹₀(Ω),φ≥0, be fixed. Since -p < -1,

$\begin{array}{*{20}{c}}{\int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}{\rm{d}}x} \le \int_\Omega {h\left( x \right)u_n^{1 - p}{\rm{d}}x} < \infty ,}\\{\forall t \ge 0.}\end{array}$

By the previous argument (5), the function f_{n, φ}(t):=t(u_n+tφ), ?t≥0 exists, and, moreover, using the notation therein, f_{n, φ}(0)=1 and f_{n, φ}(t)(u_n+tφ)∈N₂. The continuity of f_{n, φ}(t), t>0 depends on ∫_Ωh(x)|u_n|^1-p < ∞ and dominates convergence. Indeed,

$\begin{array}{*{20}{c}}{f_{n,\varphi }^2\left( t \right)\int {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right)} }\\{f_{n,\varphi }^{1 - p}\left( t \right)\int {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} ,}\end{array}$

that is,

${f_{n,\varphi }}\left( t \right) = {\left[ {\frac{{\int {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} }}{{\int {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right)} }}} \right]^{\frac{1}{{p + 1}}}}.$

The key to showing that u^*∈N₂ hinges on the estimation of f′_{n, φ}(0) defined as

${{f'}_{n,\varphi }}\left( 0 \right) = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {{f_{n,\varphi }}\left( t \right) - 1} \right)}}{t} \in \left[ { - \infty , + \infty } \right].$

If the limit does not exist, we let t_k→0 (instead of t→0) with t_k>0 chosen in such a way that $\mathop {\lim }\limits_{k \to \infty } \frac{{\left( {{f_{n, \varphi }}\left( {{t_k}} \right) - 1} \right)}}{{{t_k}}} \in \left[{-\infty, + \infty } \right]$. We deduce that f_{n, φ}(t) has uniform behavior at zero with respect to n, i.e., |f′_{n, φ}(0)|≤C for suitable C>0 independent of n. In fact, with f_{n, φ}(t)(u_n+tφ)∈N₂, u_n∈N₂, we have

$\begin{array}{l}0 = f_{n,\varphi }^2\left( t \right)\int {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right)} - \\\;\;\;\;\;f_{n,\varphi }^{1 - p}\left( t \right)\int {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} ,\end{array}$

$0 = \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} - \int_\Omega {h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} .$

By the continuity of f_{n, φ}(t), t>0, it holds

$\begin{array}{l}0 = \left\{ {\left( {{f_{n,\varphi }}\left( t \right) + 1} \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right)} - } \right.\\\left. {\left( {1 - p} \right){{\left[ {{f_{n,\varphi }}\left( 0 \right) + o\left( 1 \right)} \right]}^{ - p}}\int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} } \right\} \cdot \\\frac{{\left( {{f_{n,\varphi }}\left( t \right) - 1} \right)}}{t} - \frac{1}{t}\left\{ {\int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} + h\left( x \right)u_n^{1 - p}{\rm{d}}x - } \right.\\\left. {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right) - \mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}{\rm{d}}x} } \right\},\end{array}$

and by letting t→0⁺, then

$\begin{array}{l}0 \ge {{f'}_{n,\varphi }}\left( 0 \right)\left\{ {2\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} + \left( {p - 1} \right)\int_\Omega {h\left( x \right)u_n^{1 - p}} } \right\} + \\\;\;\;\;\;2\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi } ,\end{array}$

which implies that f′_{n, φ}(0)≠+∞. Indeed, due to u_n∈N₂ $ \subseteq $ N₁ and B(0, r₀)∩N₁=?, it follows that

$\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} \ge \alpha {\left\| {{u_n}} \right\|^2} > \alpha r_0^2 > 0.$

(11)

Since -p < -1, by the reversed H?lder inequality it yields that

$\begin{array}{*{20}{c}}{\int_\Omega {h\left( x \right)u_n^{1 - p}} \ge {{\left( {\int_\Omega {h{{\left( x \right)}^{1/p}}} } \right)}^p}{{\left( {\int_\Omega {{u_n}} } \right)}^{1 - p}}}\\{ \ge {C_2}{{\left( {\int_\Omega {h{{\left( x \right)}^{1/p}}} } \right)}^p}{{\left\| {{u_n}} \right\|}^{1 - p}} > 0.}\end{array}$

In addition,

$\begin{array}{*{20}{c}}{\left| {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi } } \right| \le \frac{\beta }{{{\alpha ^{n - 1}}}} \cdot }\\{\int_\Omega {\left| {\nabla {u_n}} \right| \cdot \left| \varphi \right| \le {C_3}\left\| {{u_n}} \right\| \cdot \left\| \varphi \right\|} .}\end{array}$

(12)

Furthermore, since r₀ is independent of n, it follows that

${{f'}_{n,\varphi }}\left( 0 \right) \le {c_1}\;{\rm{uniformly}}\;{\rm{in}}\;n.$

(13)

On the other hand, we will show that f′_{n, φ}(0) cannot go to -∞ as n→∞, that is, f′_{n, φ}(0) is bounded from below uniformly for all n large. Indeed, by the fact that u∈N₂, we have

$\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} = \int_\Omega {h\left( x \right){{\left| {{u_n}} \right|}^{1 - p}}} ,$

which imples

$I\left( u \right) = \left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} ,$

and by condition (ⅱ) we have the additional condition

$\begin{array}{l}\frac{1}{n}\left| {\frac{{1 - {f_{n,\varphi }}\left( t \right)}}{t}} \right| \cdot \left\| {{u_n}} \right\| + \frac{1}{n}{f_{n,\varphi }}\left( t \right)\left\| \varphi \right\|\\ \ge \frac{1}{n}\left\| {{u_n} \cdot {f_{n,\varphi }}\left( t \right)\left( {{u_n} + t\varphi } \right)} \right\|\frac{1}{t}\\ \ge \left[ {I\left( {{u_n}} \right) - I\left( {{f_{n,\varphi }}\left( t \right)\left( {{u_n} + t\varphi } \right)} \right.} \right]\frac{1}{t},\end{array}$

that is,

$\begin{array}{l}\frac{{\left\| \varphi \right\|}}{n}{f_{n,\varphi }}\left( t \right) \ge \frac{{{f_{n,\varphi }}\left( t \right) - 1}}{t}\left\{ { - \left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\left[ {{f_{n,\varphi }}\left( t \right) + 1} \right] \cdot } \right.\\\left. {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right){{\left| {\nabla \left( {{u_n} + t\varphi } \right)} \right|}^2}} - \frac{{\left\| {{u_n}} \right\|}}{n} \cdot {\mathop{\rm sgn}} \left( {{f_{n,\varphi }}\left( t \right) - 1} \right)} \right\} - \\\frac{1}{t}\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\left[ {{{\left| {\nabla \left( {{u_n} + t\varphi } \right)} \right|}^2} - {{\left| {\nabla {u_n}} \right|}^2}} \right]{\rm{d}}x} .\end{array}$

Letting t→0⁺, we obtain that

$\begin{array}{l}\frac{{\left\| \varphi \right\|}}{n} \ge - {{f'}_{n,\varphi }}\left( 0 \right)\left\{ {\left( {\frac{1}{2} + \frac{1}{{p - 1}}} \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} + } \right.\\\left. {\frac{{\left\| {{u_n}} \right\|}}{n} \cdot {\mathop{\rm sgn}} {{f'}_{n,\varphi }}\left( 0 \right)} \right\} - \left( {1 + \frac{1}{{p - 1}}} \right)\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi } .\end{array}$

By (2) and (11) it yields that

$\beta {\left\| u \right\|^2} \ge \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla u \cdot \nabla u} \ge \alpha {\left\| {{u_n}} \right\|^2} > \alpha r_0^2 > 0,$

and in view of (12) it holds that f′_{n, φ} is bounded below. More precisely,

${{f'}_{n,\varphi }}\left( 0 \right) \ge {c_2}\;{\rm{uniformly}}\;{\rm{in}}\;{\rm{all}}\;n\;{\rm{large}}$

(14)

as r₀ is independent of n.
Now, applying condition (ⅱ) again, we have that

$\begin{array}{l}\frac{1}{n}\left[ {\frac{{\left| {{f_{n,\varphi }}\left( t \right) - 1} \right|}}{t}\left\| {{u_n}} \right\| + {f_{n,\varphi }}\left( t \right)\left\| \varphi \right\|} \right]\\\;\;\; \ge \frac{1}{n}\left\| {{f_{n,\varphi }}\left( t \right)\left( {{u_n} + t\varphi } \right) - {u_n}} \right\|\frac{1}{t}\\\;\;\; \ge \left[ {I\left( {{u_n}} \right) - I\left( {{f_{n,\varphi }}\left( t \right)\left( {{u_n} + t\varphi } \right)} \right.} \right]\frac{1}{t},\end{array}$

that is,

$\begin{array}{l}\frac{{\left\| {{u_n}} \right\|}}{n}\frac{{\left| {{f_{n,\varphi }}\left( t \right) - 1} \right|}}{t} + \frac{{\left\| \varphi \right\|}}{n}{f_{n,\varphi }}\left( t \right)\\ \ge \left\{ { - \frac{{\left[ {{f_{n,\varphi }}\left( t \right) + 1} \right]}}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right) + } } \right.\\\left. {{{\left[ {{f_{n,\varphi }}\left( 0 \right) + o\left( 1 \right)} \right]}^{ - p}}\int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}{\rm{d}}x} } \right\} \cdot \\\frac{{{f_{n,\varphi }}\left( t \right) - 1}}{t} + \frac{1}{{1 - p}}\int_\Omega {\frac{{h\left( x \right)\left[ {{{\left( {{u_n} + t\varphi } \right)}^{1 - p}} - u_n^{1 - p}} \right]}}{t} - } \\\frac{{\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right) - \mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} }}{{2t}}.\end{array}$

In other words,

$\begin{array}{l}\frac{1}{{p - 1}}\int_\Omega {\frac{{h\left( x \right)\left[ {u_n^{1 - p} - {{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} \right]}}{t}} \\ \le \left\{ {\frac{{\left[ {{f_{n,\varphi }}\left( t \right) + 1} \right]}}{2}\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right) - } } \right.\\\left. {{{\left[ {{f_{n,\varphi }}\left( 0 \right) + o\left( 1 \right)} \right]}^{ - p}}\int_\Omega {h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}{\rm{d}}x} } \right\} \cdot \\\frac{{{f_v}\left( t \right) - 1}}{t} + \frac{{\left\| {{u_n}} \right\|}}{n}\frac{{\left| {{f_{n,\varphi }}\left( t \right) - 1} \right|}}{t} + \frac{{\left\| \varphi \right\|}}{n}{f_{n,\varphi }}\left( t \right) + \\\frac{{\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla \left( {{u_n} + t\varphi } \right) \cdot \nabla \left( {{u_n} + t\varphi } \right) - \mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} }}{{2t}}.\end{array}$

In view of (13)and (14), it holds that

$\begin{array}{l}\mathop {\lim \inf }\limits_{t \to {0^ + }} \frac{1}{{p - 1}}\int_\Omega {\frac{{h\left( x \right)\left[ {u_n^{1 - p} - {{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} \right]}}{t}} \\ \le {{f'}_{n,\varphi }}\left( 0 \right)\left\{ {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla {u_n}} - h\left( x \right){{\left( {{u_n} + t\varphi } \right)}^{1 - p}}{\rm{d}}x} \right\} + \\\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi {\rm{d}}x} + \frac{1}{n}\left[ {{{f'}_{n,\varphi }}\left( 0 \right)\left\| {{u_n}} \right\| + \left\| \varphi \right\|} \right]\\ = \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi {\rm{d}}x} + \frac{1}{n}\left[ {{{f'}_{n,\varphi }}\left( 0 \right)\left\| {{u_n}} \right\| + \left\| \varphi \right\|} \right]\\ < \infty .\end{array}$

On the other hand, since -p < -1, φ≥0, h(x)>0, and t>0, we have

$\frac{{h\left( x \right)\left[ {u_n^{1 - p} - {{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} \right]}}{t} \ge 0,$

and by Fatou' lemma, we have

$\begin{array}{l}\int_\Omega {h\left( x \right)u_u^{1 - p}\varphi {\rm{d}}x} \\ \le \mathop {\lim \inf }\limits_{t \to {0^ + }} \frac{1}{{p - 1}}\int_\Omega {\frac{{h\left( x \right)\left[ {u_n^{1 - p} - {{\left( {{u_n} + t\varphi } \right)}^{1 - p}}} \right]}}{t}} \\ \le \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u_n} \cdot \nabla \varphi } + \frac{1}{n}\left[ {{{f'}_{n,\varphi }}\left( 0 \right)\left\| {{u_n}} \right\| + \left\| \varphi \right\|} \right].\end{array}$

Hence, using Fatou's lemma again and n→∞, we obtain

$\int_\Omega {h\left( x \right){u^{ * 1 - p}}{\rm{d}}x} \le \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \varphi ,\forall \varphi } \ge 0.$

In other words,

$\int_\Omega {h\left( x \right){u^{ * 1 - p}}{\rm{d}}x} - \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot la\varphi \ge 0,\forall \varphi } \ge 0.$

By the same reasoning as in case 1 we derive that

${u^ * } \in {N_2}.$

Now it remains to show that u^*∈H₀¹(Ω) is a weak solution for problem (1) for all -p < -1. Letting ψ∈H₀¹(Ω) be fixed and applying the above inequalities one finds

$\begin{array}{l}0 \le \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla {{\left( {{u^ * } + t\psi } \right)}^ + }} - \int_\Omega {h\left( x \right){u^{ * - p}}{{\left( {{u^ * } + t\psi } \right)}^ + }} \\ = \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \left( {{u^ * } + t\psi } \right)} - \int_\Omega {h\left( x \right){u^{ * - p}}\left( {{u^ * } + t\psi } \right)} - \\\;\;\;\int_{{u^ * } + t\psi < 0} {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \left( {{u^ * } + t\psi } \right)} + \\\;\;\;\int_\Omega {h\left( x \right){u^{ * - p}}\left( {{u^ * } + t\psi } \right)} \\ \le t\left\{ {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \psi } - \int_\Omega {h\left( x \right){u^{ * - p}}\psi } } \right\} - \\\;\;\;\;\int_{{u^ * } + t\psi < 0} {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla {u^ * }} - t\int_{{u^ * } + t\psi < 0} {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \psi } \\ \le t\left\{ {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \psi } - \int_\Omega {h\left( x \right){u^{ * - p}}\psi } - } \right.\\\;\;\;\;\left. {\int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \psi } } \right\}.\end{array}$

Since meas[u^*+tψ < 0]→0 as t→0, we may divide the inequality by t>0 and pass to the limit as t→0, and we conclude that

$0 \le \int_\Omega {\mathit{\boldsymbol{M}}\left( x \right)\nabla {u^ * } \cdot \nabla \psi } + \int_\Omega {h\left( x \right){u^{ * - p}}\psi {\rm{d}}x} .$

By the arbitrariness of ψ∈H₀¹(Ω), u^* is indeed a H₀¹(Ω)-solution of problem (1).
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