The role of the power 3 for elliptic equations with negative exponents

Abstract

Let $\Omega \subset {\mathbb{R }}^{N}$ be a bounded regular domain of dimension $N\ge 3,\;h$ a positive $L^{1}$ function on $\Omega .$ Elliptic equations of singular growth like

$$\begin{aligned} -\Delta u=\frac{h(x)}{u^{p}}\quad \mathrm{in}\;\Omega , \quad u>0\quad \mathrm{in}\;\Omega , \quad u=0\quad \mathrm{on}\;\partial \Omega , \end{aligned}$$

have been the target of investigation for decades. A very nice result for existence of solutions of such an equation is due to Lazer–McKenna (Proc AMS 111:720–730, 1991) when $h$ is a positive continuous function on $\overline{\Omega }.$ In that paper the Lazer–McKenna obstruction was first presented: the equation has a $H_{0}^{1}$-solution if and only if $p<3.$ In this paper we provide an extension of the classical Lazer–McKenna obstruction and reveal the role of 3. Moreover, we give a local description of the solution set.

Appendix: the proof of Theorem 1

Note that, for any $u\in H_{0}^{1}(\Omega )$ with $\int _{\Omega }h(x)|u|^{1-p}dx<\infty ,$ the unique positive minimizer for $I(tu)$ is $t(u)u$ with

$$\begin{aligned} t(u)=\left[\frac{\int _{\Omega }h(x)|u|^{1-p}dx}{\Vert u\Vert ^{2}}\right]^{1/(1+p)}, \end{aligned}$$

(4)

that is $I(t(u)u)=\inf _{t>0}I(tu),$ and $t(u)u\in {\mathcal{N }^{*}}.$ Since $t(u_{0})u_{0}\in {\mathcal{N }^{*}}$ by (*), ${\mathcal{N }}(\supset {\mathcal{N }^{*}})$ and ${\mathcal{N }^{*}}$ are not empty. Clearly, since $tu_{0}\in {\mathcal{N }}$ for all $t\ge t({u_{0}}),$ ${\mathcal{N }}$ is unbounded. The closeness of ${\mathcal{N }}$ follows easily from Fatou’s Lemma. However, since $p>1,$ $\int _{\Omega }h(x)|u|^{1-p}dx$ is not continuous on $H_{0}^{1}(\Omega ),$ so ${\mathcal{N }^{*}}$ is not anymore a closed set in $H_{0}^{1}(\Omega ).$

Furthermore, unbounded ${\mathcal{N }}$ lies in the exterior of $H_{0}^{1}(\Omega )$ (i.e., stays away from a ball centered about 0). In fact suppose that there is a sequence $(u_{n})\subset \mathcal{N }$ with $u_{n}\rightarrow 0$ in $H_{0}^{1}(\Omega ).$ The reversed Hölder inequality then yields

$$\begin{aligned} \left(\int \limits _{\Omega }h^{\frac{1}{p}}(x)dx\right)^{p} \left(\int \limits _{\Omega }\left|u_{n}\right|dx\right)^{1-p}\le \int \limits _{\Omega }h(x)\left|u_{n}\right|^{1-p}dx\le \big \Vert u_{n}\big \Vert ^{2}\rightarrow 0, \end{aligned}$$

consequently,

$$\begin{aligned} \int \limits _{\Omega }\left|u_{n}\right|dx\rightarrow \infty , \end{aligned}$$

since $p>1,$ which is clearly impossible. Then there is a constant $c_{1}>0$ so that $\Vert u_{n}\Vert \ge c_{1}.$

We turn to $\inf _{\mathcal{N }}I.$ Since $\mathcal{N }$ is closed, we examine the best minimizing sequence for $\inf _{\mathcal{N }}I,$ that is, $(u_{n})\in \mathcal{N }$ satisfying (i) $I(u_{n})\le \inf _{\mathcal{N }}I +\frac{1}{n};$ (ii) $I(u_{n})\le I(w)+\frac{1}{n}\Vert u_{n}-w\Vert ,\;\forall w\in \mathcal{N }.$ Since $I(|u|)=I(u),$ we may assume that $u_{n}\ge 0$ in $\Omega .$ Since $u_{n}\in \mathcal{N }\;(\mathrm{i.e.},\; \int _{\Omega }h(x)|u_{n}|^{1-p}dx\le \Vert u_{n}\Vert ^{2}),$ $h(x)>0$ a.e. in $\Omega$ and $p>1,$ we have $u_{n}(x)>0$ a.e. in $\Omega ;$ and observe that as $n\rightarrow \infty :$

$$\begin{aligned} \frac{1}{2}\big \Vert u_{n}\big \Vert ^{2} + \frac{1}{p-1}\int \limits _{\Omega }h(x) \left|u_{n}\right|^{1-p}dx\rightarrow \inf _{\mathcal{N }}I\in (-\infty ,\,+\infty ), \end{aligned}$$

we have $\Vert u_{n}\Vert \le c_{2}$ for suitable constant $c_{2}>0,$ so (a subsequence of) $u_{n}\rightarrow u^{*}$ weakly in $H_{0}^{1}(\Omega ),$ strongly in $L^{2}(\Omega ),$ and pointwise a.e. in $\Omega .$

To take advantage of those constraints involved, we distinguish two cases according to $u_{n}$ belonging to $\mathcal{N }$ or $\mathcal{N ^{*}}.$

(1) Case 1. Suppose that $(u_{n})\subset \mathcal{N }{\backslash } \mathcal{N ^{*}}$ for all $n$ large. Fix $\varphi \in H_{0}^{1}(\Omega ),$ $\varphi \ge 0$ and $n$ by now, there holds

$$\begin{aligned} \int \limits _{\Omega }h(x)\left(u_{n}+t\varphi \right)^{1-p}dx\le \int \limits _{\Omega }h(x)u_{n}^{1-p}dx<\big \Vert u_{n}\big \Vert ^{2},\quad \forall t\ge 0, \end{aligned}$$

thanks to $u_{n}\in \mathcal{N }{\backslash } {\mathcal{N }^{*}}$ and $p>1.$ Subsequently, choose $t>0$ sufficiently small such that

$$\begin{aligned} \int \limits _{\Omega }h(x)\left(u_{n}+t\varphi \right)^{1-p}dx<\big \Vert u_{n}+t\varphi \big \Vert ^{2}\quad \left(\mathrm{i.e.,}\;u_{n}+t\varphi \in \mathcal{N }\right), \end{aligned}$$

and by (ii) conclude:

$$\begin{aligned}&\frac{1}{n}\Vert t\varphi \Vert \ge I\left(u_{n}\right)-I\left(u_{n}+t\varphi \right) =\frac{1}{2}\big \Vert u_{n}\big \Vert ^{2} + \frac{1}{p-1}\int \limits _{\Omega }h(x) u_{n}^{1-p}dx-\frac{1}{2}\big \Vert u_{n}+t\varphi \big \Vert ^{2}\\&\quad - \frac{1}{p-1}\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}dx. \end{aligned}$$

Thus, dividing by $t>0$ and passing to the liminf as $t\rightarrow 0,$ by Fatou’s Lemma we have

$$\begin{aligned}&\frac{\Vert \varphi \Vert }{n}+\int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx \ge \int \limits _{\Omega }\liminf _{t\rightarrow 0}\frac{h(x) (u_{n}+t\varphi )^{1-p}-h(x) u_{n}^{1-p}}{(1-p)t}dx\\&\quad =\int \limits _{\Omega }h(x)u_{n}^{-p}\varphi dx, \end{aligned}$$

whence, using Fatou’s Lemma again and letting $n$ tend to infinity, we obtain

$$\begin{aligned} \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \varphi dx\ge \int \limits _{\Omega }h(x)u^{*-p}\varphi . \end{aligned}$$

(5)

By (5), we know immediately that $u^{*}\in \mathcal{N }.$ Furthermore, we may estimate:

$$\begin{aligned} \inf _{\mathcal{N }}I=\lim _{n\rightarrow \infty }I\left(u_{n}\right)&\ge \liminf _{n\rightarrow \infty }\left[\frac{1}{2}\int \limits _{\Omega }\big \Vert \nabla u_{n}\big \Vert ^{2}dx\right]+\liminf _{n\rightarrow \infty }\left[\frac{1}{p-1}\int \limits _{\Omega }h(x)u_{n}^{1-p}dx\right]\\&\ge \frac{1}{2}\int \limits _{\Omega }\mid \nabla u^{*}\mid ^{2}dx+\liminf \left[ \frac{1}{p-1}\int \limits _{\Omega }h(x) u_{n}^{1-p}dx\right]\\&\ge \frac{1}{2}\int \limits _{\Omega }\mid \nabla u^{*}\mid ^{2}dx+\frac{1}{p-1}\int \limits _{\Omega }h(x) u^{*1-p}dx\quad (\mathrm{Fatou^{\prime }s Lemma})\\&= I(u^{*})\ge I(t(u^{*})u^{*})\\&\ge \inf _{\mathcal{N ^{*}}}I\ge \inf _{\mathcal{N }}I, \end{aligned}$$

that is,

$$\begin{aligned} u^{*}\in {\mathcal{N }^{*}}\quad (\mathrm{i.e.,}\;t(u^{*})=1). \end{aligned}$$

(6)

(2) Case 2. There exists a subsequence of $(u_{n})$ (which we still call $u_{n}$), which belongs to $\mathcal{N ^{*}}.$

Fix $\varphi \in H_{0}^{1}(\Omega ),\;\varphi \ge 0$ and $n$ by now. Then for all $t\ge 0,$ $\int _{\Omega }h(x)(u_{n}+t\varphi )^{1-p}\le \int _{\Omega }h(x)u_{n}^{1-p}<\infty$ which ensures the existence of the corresponding unique positive number for the function $u_{n}(x)+t\varphi (x),$ denoted by $f_{n,\varphi }(t),$ so that $f_{n,\varphi }(t) (u_{n}+t\varphi )\in \mathcal{N ^{*}}$ as in (4), that is,

$$\begin{aligned} f_{n,\varphi }(t)=\left[\frac{\int _{\Omega }h(x)(u_{n}+t\varphi )^{1-p}dx}{\Vert u_{n}+t\varphi \Vert ^{2}}\right]^{1/(1+p)},\quad \forall t\ge 0. \end{aligned}$$

Obviously, $f_{n,\varphi }(0)=1$ as $u_{n}\in \mathcal{N ^{*}}.$ The continuity of $f_{n,\varphi }(t)$ follows from the fact $p>1$ and dominated convergence. For the sake of simplicity, we assume henceforth that

$$\begin{aligned} f_{n,\varphi }^{\prime }(0):= \lim _{t\rightarrow 0}\frac{f_{n,\varphi }(t)-1}{t}\in [-\infty ,\,+\infty ]. \end{aligned}$$

If the limit does not exist, we let $t_{k}\rightarrow 0$ (instead of $t\rightarrow 0$) with $t_{k}>0$ chosen in such a way that $q_{n}:= \lim _{k\rightarrow \infty }\frac{f_{n,\varphi }(t_{k})-1}{t_{k}}\in [-\infty ,\,+\infty ],$ and replace $f_{n,\varphi }^{\prime }(0)$ by $q_{n}.$ We deduce that $f_{n,\varphi }(t)$ has uniform behavior at zero with respect to $n,$ i.e., $|f^{^{\prime }}_{n,\varphi }(0)|\le C$ for suitable $C>0$ independent of $n.$ In fact, from $f_{n,\varphi }(t) (u_{n}+t\varphi ),\,u_{n}\in {\mathcal{N }^{*}}$ we have

$$\begin{aligned} 0&= f_{n,\varphi }^{2}(t)\big \Vert u_{n}+t\varphi \big \Vert ^{2} - f_{n,\varphi }^{1-p}(t)\int \limits _{\Omega } h(x) \left(u_{n}+t\varphi \right)^{1-p} dx,\\ 0&= \big \Vert u_{n}\big \Vert ^{2}-\int \limits _{\Omega } h(x) u_{n}^{1-p} dx. \end{aligned}$$

Consequently, $f^{^{\prime }}_{n,\varphi }(0)$ can be estimated by

$$\begin{aligned} 0&\ge \{\left[f_{n,\varphi }(t)+1\right]\big \Vert u_{n}+t\varphi \big \Vert ^{2}\\ \quad&-(1-p)\left[f_{n,\varphi }(0)+\theta \left(f_{n,\varphi }(t)-f_{n,\varphi }(0)\right)\right]^{-p} \int \limits _{\Omega } h(x) \left(u_{n}+t\varphi \right)^{1-p}\}\\ \quad&\times \left[f_{n,\varphi }(t)-f_{n,\varphi }(0)\right]+\big \Vert u_{n}+t\varphi \big \Vert ^{2}-\big \Vert u_{n}\big \Vert ^{2}\quad ( \mathrm{since}\,p>1). \end{aligned}$$

Dividing by $t>0$ and passing to the limit as $t\rightarrow 0,$ we obtain:

$$\begin{aligned} 0&\ge \int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx+ \left[2\big \Vert u_{n}\big \Vert ^{2}+(p-1)\int \limits _{\Omega } h(x) u_{n}^{1-p} dx\right]f^{^{\prime }}_{n,\varphi }(0)\\&= \int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx +\left[(p+1)\big \Vert u_{n}\big \Vert ^{2}\right]f^{^{\prime }}_{n,\varphi }(0)\quad (\mathrm{since}\,u_{n}\in {\mathcal{N }^{*}}). \end{aligned}$$

Since $(u_{n})\subset {\mathcal{N }^{*}} (\subset {\mathcal{N }})$ is bounded and ${\mathcal{N }}$ stays away from some ball centered about 0, we conclude that

$$\begin{aligned} f^{^{\prime }}_{n,\varphi }(0)\in [-\infty ,\,+\infty ),\quad \mathrm{and}\quad f^{^{\prime }}_{n,\varphi }(0)\le c_{3}\quad \mathrm{uniformly\;in}\;{ n} \end{aligned}$$

(7)

for suitable constant $c_{3}>0.$

On the other hand, $f^{^{\prime }}_{n,\varphi }(0)\ne -\infty$ and bounded uniformly from below with respect to $n$. Indeed, by (ii)

$$\begin{aligned}&\frac{1}{n}\left|f_{n,\varphi }(t)-1\right|\big \Vert u_{n}\big \Vert +\frac{1}{n}t f_{n,\varphi }(t)\Vert \varphi \Vert \ge \frac{1}{n} \big \Vert u_{n}-f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)\big \Vert \ge I\left(u_{n}\right)\\&\quad -I\left(f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)\right) = \left(\frac{1}{2}+\frac{1}{p-1}\right)\big \Vert u_{n}\big \Vert ^{2}\\&\quad -\left(\frac{1}{2}+\frac{1}{p-1}\right)\big \Vert f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)\big \Vert ^{2}, \end{aligned}$$

that is,

$$\begin{aligned}&\frac{1}{n}t f_{n,\varphi }(t)\Vert \varphi \Vert +\left(\frac{1}{2}+\frac{1}{p-1}\right)\left[\big \Vert u_{n}+t\varphi \big \Vert ^{2}-\big \Vert u_{n}\big \Vert ^{2}\right] \\&\ \ge \left\{ -\left(\frac{1}{2}+\frac{1}{p-1}\right) \left[f_{n,\varphi }(t)+1\right]\big \Vert u_{n}+t\varphi \big \Vert ^{2}-\frac{\Vert u_{n}\Vert }{n}\,\mathrm{sgn}\left[f_{n,\varphi }(t)-1\right]\right\} \left[f_{n,\varphi }(t)-1\right]. \end{aligned}$$

Noting that $-[f_{n,\varphi }(t)+1]\Vert u_{n}+t\varphi \Vert ^{2}\rightarrow -2\Vert u_{n}\Vert ^{2}\le -2c_{1}^{2}<0$ as $t\rightarrow 0,$ from the construction of coefficient we see that $f^{^{\prime }}_{n,\varphi }(0)\ne -\infty$ and cannot diverge to $-\infty$ as $n\rightarrow \infty ,$ that is,

$$\begin{aligned} f^{^{\prime }}_{n,\varphi }(0)\in (-\infty ,\,+\infty ),\quad \mathrm{and}\quad f^{^{\prime }}_{n,\varphi }(0)\ge c_{4}\quad \mathrm{uniformly\;in\;all}\;{ n}\; \mathrm{large}, \end{aligned}$$

(8)

for suitable real number $c_{4}\in \mathbb{R },$ as claimed.

Putting together (7) and (8) we find that

$$\begin{aligned} f^{^{\prime }}_{n,\varphi }(0)\in (-\infty ,\,+\infty ),\quad \mathrm{and}\quad |f^{^{\prime }}_{n,\varphi }(0)|\le C\quad \mathrm{uniformly\;in\;all}\;{ n}\;\mathrm{large}. \end{aligned}$$

Thus, we can locate $u^{*}$ in Case 2 (as (6) in Case 1). Fix $\varphi \in H_{0}^{1}(\Omega ),\;\varphi \ge 0$ and $n$ by now and write (ii)

$$\begin{aligned}&\frac{1}{n}\left[\left|f_{n,\varphi }(t)-1\right|\cdot \big \Vert u_{n}\big \Vert +t f_{n,\varphi }(t)\Vert \varphi \Vert \right] \ge \frac{1}{n} \big \Vert f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)-u_{n}\big \Vert \\&\quad \ge I\left(u_{n}\right)-I\left(f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)\right) =\frac{1}{2}\big \Vert u_{n}\big \Vert ^{2} + \frac{1}{p-1}\int \limits _{\Omega }h(x) u_{n}^{1-p}dx\\&\qquad -\frac{1}{2}\big \Vert f_{n,\varphi }(t)\left(u_{n}+t\varphi \right)\big \Vert ^{2}-\frac{f_{n,\varphi }^{1-p}(t)}{p-1}\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}dx\\&\quad =-\frac{1}{2}\left[f_{n,\varphi }^{2}(t)-1\right]\big \Vert u_{n}+t\varphi \big \Vert ^{2}-\frac{1}{2} \left[\big \Vert u_{n}+t\varphi \big \Vert ^{2}-\big \Vert u_{n}\big \Vert ^{2}\right]\\&\qquad -\frac{1}{p-1}\left[f_{n,\varphi }^{1-p}(t)-1\right]\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}dx\\&\qquad -\frac{1}{p-1}\left[\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}-h(x) u_{n}^{1-p}dx\right]. \end{aligned}$$

Thus, dividing by $t>0$ and passing to the liminf as $t\rightarrow 0,$ by the above assertion (i.e., $f^{^{\prime }}_{n,\varphi }(0)\in (-\infty ,\,+\infty )$) we get

$$\begin{aligned} h(x)u_{n}^{-p}\varphi \quad \mathrm{is\;integrable\;in}\;\Omega , \end{aligned}$$

and

$$\begin{aligned}&\frac{1}{n}\left[\left|f^{^{\prime }}_{n,\varphi }(0)\right|\cdot \big \Vert u_{n}\big \Vert +\Vert \varphi \Vert \right] \ge f^{^{\prime }}_{n,\varphi }(0)\left[-\big \Vert u_{n}\big \Vert ^{2}+\int \limits _{\Omega }h(x) u_{n}^{1-p}dx\right]-\int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx\\&\qquad +\liminf _{t\rightarrow 0}\left[\frac{1}{1-p}\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}-h(x) u_{n}^{1-p}dx\right]\\&\quad =-\int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx+\liminf _{t\rightarrow 0}\left[\frac{1}{1-p}\int \limits _{\Omega }h(x) \left(u_{n}+t\varphi \right)^{1-p}-h(x) u_{n}^{1-p}dx\right]\\&\quad \ge -\int \limits _{\Omega }\nabla u_{n}\cdot \nabla \varphi dx+\int \limits _{\Omega }h(x)u_{n}^{-p}\varphi dx. \end{aligned}$$

Furthermore, by the above assertion (i.e., $|f^{^{\prime }}_{n,\varphi }(0)|\le C$ uniformly in all $n$ large) again, we obtain

$$\begin{aligned} h(x)u^{*-p}\varphi \quad \mathrm{is\;integrable\;in}\;\Omega , \end{aligned}$$

and

$$\begin{aligned} \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \varphi dx\ge \liminf _{n\rightarrow \infty } \int \limits _{\Omega }h(x)u_{n}^{-p}\varphi dx\ge \int \limits _{\Omega }h(x)u^{*-p}\varphi dx, \end{aligned}$$

as $n\rightarrow \infty .$ That is,

$$\begin{aligned} \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \varphi dx-\int \limits _{\Omega }h(x)u^{*-p}\varphi dx\ge 0,\quad \forall \varphi \in H_{0}^{1}(\Omega ),\quad \varphi \ge 0, \end{aligned}$$

(9)

which gives $u^{*}\in \mathcal{N }.$ By taking the same argument as in Case 1 we see also

$$\begin{aligned} u^{*}\in \mathcal{N ^{*}}. \end{aligned}$$

(10)

Collecting (5), (6), (9) and (10), we conclude that in either case,

$$\begin{aligned} u^{*}\in {\mathcal{N }^{*}},\quad \mathrm{and}\quad \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \varphi -\int \limits _{\Omega }h(x)u^{*-p}\varphi \ge 0,\quad \forall \varphi \in H_{0}^{1}(\Omega ),\quad \varphi \ge 0. \end{aligned}$$

The strict positivity $u^{*}(x)\ge c_{5} dist(x,\,\partial \Omega ),\; \forall x\in \Omega$ with suitable constant $c_{5}>0$ follows from the maximum principle.

Now it is to prove that $u^{*}\in H_{0}^{1}(\Omega )$ is a solution to ($E_{h,p}$) for all $p>1.$

Set $\varphi (x)\equiv (u^{*}(x)+\varepsilon \phi (x))^{+}\in H_{0}^{1}(\Omega )$ for arbitrary $\phi \in H_{0}^{1}(\Omega ),$ $\varepsilon >0,$ and apply the above inequalities to find

$$\begin{aligned} 0&\le \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \varphi -h(x)u^{*-p}\varphi \\&= \int \limits _{[u^{*}+\varepsilon \phi \ge 0]}\nabla u^{*}\cdot \nabla (u^{*}+\varepsilon \phi )- h(x)u^{*-p}(u^{*}+\varepsilon \phi )\\&= \left(\int \limits _{\Omega } - \int \limits _{[u^{*}+\varepsilon \phi <0 ]}\right) \nabla u^{*}\cdot \nabla (u^{*}+\varepsilon \phi )- h(x)u^{*-p}(u^{*}+\varepsilon \phi )\\&= \Vert \nabla u^{*}\Vert _{2} ^{2}-\int \limits _{\Omega } h(x) u^{*1-p} dx +\varepsilon \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \phi - h(x)u^{*-p}\phi \\ \quad&-\int \limits _{[u^{*}+\varepsilon \phi < 0]} \nabla u^{*}\cdot \nabla (u^{*}+\varepsilon \phi )- h(x)u^{*-p}(u^{*}+\varepsilon \phi )\\&= \varepsilon \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \phi - h(x)u^{*-p}\phi \quad (\mathrm{since}\,u^{*}\in \mathcal{N ^{*}})\\ \quad&-\int \limits _{[u^{*}+\varepsilon \phi < 0]} \nabla u^{*}\cdot \nabla (u^{*}+\varepsilon \phi )- h(x)u^{*-p}(u^{*}+\varepsilon \phi )\\&\le \varepsilon \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \phi - h(x)u^{*-p}\phi - \varepsilon \int \limits _{[u^{*}+\varepsilon \phi < 0]} \nabla u^{*}\cdot \nabla \phi dx. \end{aligned}$$

Since the measure of the domain of integration $[u^{*}+\varepsilon \phi < 0]$ tends to zero as $\varepsilon \rightarrow 0,$ we then divide the above expression by $\varepsilon >0$ to obtain

$$\begin{aligned} \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \phi dx-\int \limits _{\Omega }h(x)u^{*-p}\phi dx\ge 0, \end{aligned}$$

as $\varepsilon \rightarrow 0.$ Replacing $\phi$ by $-\phi$ we conclude:

$$\begin{aligned} \int \limits _{\Omega }\nabla u^{*}\cdot \nabla \phi dx-\int \limits _{\Omega }h(x)u^{*-p}\phi dx= 0,\quad \forall \phi \in H_{0}^{1}(\Omega ), \end{aligned}$$

and therefore $u^{*}$ is a $H_{0}^{1}$-solution to ($E_{h,p}$).

In order to prove uniqueness, assume that $u^{*}$ and $v^{*}$ both solve ($E_{h,p}$). Subtracting the equations we deduce that $w\equiv u^{*}-v^{*}\in H_{0}^{1}(\Omega )$ satisfies

$$\begin{aligned} \int \limits _{\Omega }|\nabla (u^{*}-v^{*})|^{2}dx=\int \limits _{\Omega }h(x)\left(u^{*-p}-v^{*-p}\right)(u^{*}-v^{*})dx\le 0. \end{aligned}$$

This shows that $w\equiv 0$ and completes the proof.$\square$