In this section, we prove Theorems 1.2-1.3.
First, we need the following comparison principle for weak solutions to quasilinear equations (see [45] for a proof).
Lemma 5.1
(Weak comparison principle)
Let
\(D \subset\mathbb {R}^{N}\)
be a bounded domain, \(G: D \times\mathbb {R} \rightarrow\mathbb {R}\)
be nonincreasing in the second variable and continuous. Let
\(u, w\in W^{1,p}(D)\)
satisfy the respective inequalities
$$\begin{aligned} \int_{D}\vert \nabla u\vert ^{p-2}\nabla u \cdot \nabla\phi \leq& \int_{D}G(x,u) \phi \quad \textit{and} \\ \int_{D}\vert \nabla w\vert ^{p-2}\nabla w \cdot \nabla\phi \geq& \int_{D}G(x,w) \phi \end{aligned}$$
for all nonnegative
\(\phi\in W_{0}^{1,p}(D)\). Then the inequality
\(u \leq w\)
on
∂D
implies
\(u \leq w\)
in D.
Fix \(\varepsilon>0\). For any \(\delta>0\), we define \(\Omega_{\delta}=\{x\in\Omega: 0< d(x)<\delta\}\). Since Ω is \(C^{2}\)-smooth, choose \(\delta_{1}\in(0, \delta_{0})\) such that \(d\in C^{2}(\Omega_{\delta_{1}}) \) and
$$ \bigl\vert \nabla d(x) \bigr\vert = 1,\quad\quad \Delta d(x) =-(N-1)H( \bar{x})+o(1),\quad \forall x\in\Omega_{\delta_{1}}, $$
(5.1)
where, for \(x\in\Omega_{\delta_{1}}\), x̄ denotes the unique point of the boundary such that \(d(x) = \vert x - \bar{x} \vert \), and \(H(\bar{x}) \) denotes the mean curvature of the boundary at that point.
5.1 Proof of Theorem 1.2
Define \(r=d(x)\) and
$$\begin{aligned}& I_{1\pm}(r)=(A_{1}\pm\varepsilon)^{p-1}(p-1)q^{p-1} \biggl(p \frac {(A_{1}\pm\varepsilon)K^{q}(r)\phi''((A_{1}\pm \varepsilon)K^{q}(r))}{\phi'((A_{1}\pm \varepsilon)K^{q}(r))}+1+\frac{p}{q}\frac {K(r)k'(r)}{k^{2}(r)} \biggr) ; \\& I_{2}(x)=(A_{1}\pm\varepsilon)^{p-1}q^{p-1} \frac {K(r)}{k(r)}\Delta d(x)+\frac{b(x)}{k^{p}(r)}\frac {g (\phi((A_{1}\pm\varepsilon)K^{q}(r)) )}{ (\phi'((A_{1}\pm\varepsilon)K^{q}(r)) )^{p-1}}. \end{aligned}$$
By Lemmas 3.1 and 3.4, combined with the choices of \(A_{1}\) in Theorem 1.2, we get the following lemma.
Lemma 5.2
Suppose that
g
satisfies
\((\mathrm{g}_{1})\)-\((\mathrm{g}_{3})\)
and
b
satisfies
\((\mathrm{b}_{1})\)-\((\mathrm{b}_{4})\). Then
-
(i)
\(\lim_{r\rightarrow0} I_{1\pm}(r)=(A_{1}\pm \varepsilon)^{p-1}(p-1)q^{p-1}(q-q C_{g}-C_{k})\);
-
(ii)
\(\lim_{d(x) \rightarrow0} I_{2 }(x)=b_{0}=-A_{1}^{p-1}(p-1)q^{p-1}(q-q C_{g}-C_{k})\);
-
(iii)
\(\lim_{d(x)\rightarrow0} ( I_{1\pm }(r)+I_{2}(x) )=(p-1)q^{p-1}(q-q C_{g}-C_{k}) ((A_{1}\pm \varepsilon)^{p-1}-A_{1}^{p-1} )\).
Proof of Theorem 1.2
Let \(v \in C^{1+\alpha} (\Omega) \cap C^{1}(\bar{\Omega})\) be the unique solution of the problem
$$ -\Delta_{p} v=1,\quad v>0, x \in\Omega, v\vert _{\partial \Omega}=0. $$
(5.2)
Then, we see that
$$ \nabla v (x) \neq0, \quad \forall x \in\partial\Omega\quad \text{and}\quad c_{3} d(x)\leq v(x) \leq c_{4} d(x), \quad \forall x \in\Omega, $$
(5.3)
where \(c_{3}\), \(c_{4}\) are positive constants.
By Lemma 5.2, since \(K\in C[0, \delta_{0})\) with \(K(0)=0\), we see that there exist \(\delta_{1\varepsilon}, \delta_{2\varepsilon}\in (0, \min\{1, \delta_{0}\} )\) (which corresponds to ε) sufficiently small such that
-
(i)
\(0\leq K^{q}(r)\leq\delta_{1\varepsilon}\), \(r\in(0, \delta_{2\varepsilon})\);
-
(ii)
\(I_{1+}(r)+I_{2}(x)\leq0\), \(\forall (x,r)\in \Omega_{\delta_{1\varepsilon}}\times(0, \delta_{2\varepsilon}) \);
-
(iii)
\(I_{1-}(r)+I_{2}(x)\geq0\), \(\forall (x,r)\in \Omega_{\delta_{1\varepsilon}}\times(0, \delta_{2\varepsilon})\).
Now we define
$$ \bar{u}_{\varepsilon}=\phi \bigl((A_{1}+\varepsilon )K^{q} \bigl(d(x) \bigr) \bigr), \quad x\in\Omega_{\delta_{1\varepsilon}}. $$
Before we prove the theorem, let us note the following. Suppose that z is a \(C^{2}\) function on a domain Ω in \(\mathbb {R}^{N}\) and \(v = \phi(z)\), where ϕ is uniquely determined by (1.6). A direct computation shows that
$$\begin{aligned} \Delta_{p}v = &(p-1) \bigl\vert \phi'(z) \bigr\vert ^{p-2}\phi''(z)\vert \nabla z \vert ^{p}+ \bigl\vert \phi'(z) \bigr\vert ^{p-2}\phi'(z)\Delta_{p}z. \end{aligned}$$
(5.4)
Hence, by (5.4), Lemma 5.2, and a direct calculation we see that, for \(x\in\Omega_{\delta_{1\varepsilon}}\)
$$\begin{aligned}& \Delta_{p}\bar{u}_{\varepsilon}(x)+b(x) g \bigl( \bar{u}_{\varepsilon}(x) \bigr) \\& \quad = \bigl( \phi' \bigl(K^{q} \bigl(d(x) \bigr) \bigr) \bigr)^{p-1}k^{p} \bigl(d(x) \bigr) \bigl( I_{1+}(r)+I_{2}(x) \bigr) \leq0, \end{aligned}$$
where \(r=d(x)\), that is, \(\bar{u}_{\varepsilon}\) is a supersolution of problem (1.1) in \(\Omega_{\delta_{1\varepsilon}}\).
In a similar way, we show that
$$ \underline{u}_{\varepsilon}=\phi \bigl((A_{1}-\varepsilon )K^{q} \bigl(d(x) \bigr) \bigr),\quad x\in\Omega_{\delta_{1\varepsilon}}, $$
is a subsolution of problem (1.1) in \(\Omega_{\delta_{1\varepsilon}}\).
Let \(u\in C(\bar{\Omega})\cap C^{1,\alpha}(\Omega)\) be the unique solution to problem (1.1). We assert that there exists M large enough such that
$$ u(x)\leq M v (x)+\bar{u}_{\varepsilon}(x),\quad\quad \underline{u}_{\varepsilon}(x)\leq u(x)+M v(x), \quad x\in\Omega_{\delta _{1\varepsilon}}, $$
(5.5)
where v is the solution of problem (5.2).
In fact, we can choose M large enough such that
$$u(x)\leq \bar{u}_{\varepsilon}(x)+M v(x) \quad \text{and}\quad \underline{u}_{\varepsilon}(x)\leq u(x)+M v(x) \quad \text{on } \bigl\{ x\in \Omega: d(x)=\delta_{1\varepsilon}\bigr\} . $$
We see by \((\mathrm{g}_{1})\) that \(\bar{u}_{\varepsilon}(x)+M v(x)\) and \(u(x)+M v(x)\) are also supersolutions of problem (1.1) in \(\Omega_{\delta_{1\varepsilon}}\). Since \(u= \bar{u}_{\varepsilon}+M v=u+Mv=\underline{u}_{\varepsilon}=0 \) on ∂Ω, (5.5) follows by \((\mathrm{g}_{1})\) and the weak comparison principle (Lemma 5.1). Hence, for \(x\in \Omega_{\delta_{1\varepsilon}}\)
$$\frac{u(x)}{\phi((A_{1}+\varepsilon) K^{q}(d(x)))}\leq\frac{M v(x)}{\phi((A_{1}+\varepsilon) K^{q}(d(x)))}+1 $$
and
$$1-\frac{M v(x)}{\phi((A_{1}-\varepsilon) K^{q}(d(x)))}\leq\frac{u(x)}{\phi((A_{1}-\varepsilon) K^{q}(d(x)))}. $$
Consequently, by (5.3) and Lemma 3.4(iii),
$$1\leq\liminf_{d(x) \rightarrow0 } \frac{u(x)}{\phi((A_{1}-\varepsilon) K^{q}(d(x)))} \leq \limsup _{d(x) \rightarrow0 } \frac{u(x)}{\phi((A_{1}+\varepsilon) K^{q}(d(x)))} \leq1 $$
and
$$\lim_{d(x) \rightarrow0 } \frac{\phi((A_{1}-\varepsilon) K^{q}(d(x)))}{\phi( K^{q}(d(x)))}=(A_{1}- \varepsilon)^{1-C_{g}}. $$
Thus, letting \(\varepsilon\rightarrow0\), we obtain (1.8). □
5.2 Proof of Theorem 1.3
As before, fix \(\varepsilon>0\). For any \(\delta>0\), we define \(\Omega_{\delta}=\{x\in\Omega: 0< d(x)<\delta\}\). Since Ω is \(C^{2}\)-smooth, choose \(\delta_{1}\in(0, \delta_{0})\) such that \(d\in C^{2}(\Omega_{\delta_{1}})\) and (5.1) holds.
Define \(r=d(x)\) and
$$\begin{aligned} &I_{1\pm}(r)=(A_{2}\pm\varepsilon)^{p-1}(p-1) \biggl( \frac{(A_{2}\pm \varepsilon)h(r)\phi''((A_{2}\pm\varepsilon)h(r))}{\phi'((A_{1}\pm \varepsilon)h(r))}\frac{(h'(r))^{p}}{h(r)r^{-p}L(r)} \\ &\hphantom{I_{1\pm}(r)=}{}+ \frac {(h'(r))^{p-2}h''(r)}{r^{-p}L(r)} \biggr) , \\ &I_{2}(x)=(A_{2}\pm\varepsilon)^{p-1} \frac{(h'(r))^{p-1}}{r^{-p}L(r)}\Delta d(x)+\frac {b(x)}{k^{p}(r)}\frac{g (\phi((A_{2}\pm \varepsilon)h(r)) )}{ (\phi'((A_{2}\pm \varepsilon)h(r)) )^{p-1}}. \end{aligned}$$
By Lemmas 3.2 and 3.4, combined with the choices of \(A_{2}\) in Theorem 1.3, we get the following lemma.
Lemma 5.3
Suppose that
g
satisfies
\((\mathrm{g}_{1})\)-\((\mathrm{g}_{3})\), b
satisfies
\((\mathrm{b}_{1})\)-\((\mathrm{b}_{3})\), and
\((\mathrm{b}_{5})\)
holds. Then
-
(i)
\(\lim_{r\rightarrow0} I_{1\pm }(r)=-(p-1)(A_{2}\pm \varepsilon)^{p-1}\);
-
(ii)
\(\lim_{d(x) \rightarrow0} I_{2 }(x)=b_{1}=(p-1)A_{2}^{p-1}\);
-
(iii)
\(\lim_{d(x)\rightarrow0} ( I_{1\pm }(r)+I_{2}(x) )=-(p-1) ((A_{2}\pm \varepsilon)^{p-1}-A_{2}^{p-1} )\).
Proof of Theorem 1.3
By Lemma 5.3, since \(h\in C[0, \delta_{0})\) with \(h(0)=0\), we see that there exist \(\delta_{1\varepsilon}, \delta_{2\varepsilon}\in (0, \min\{1, \delta_{0}\} )\) (which corresponds to ε) sufficiently small such that
-
(i)
\(0\leq h(r)\leq\delta_{1\varepsilon}\), \(r\in(0, \delta_{2\varepsilon})\);
-
(ii)
\(I_{1+}(r)+I_{2}(x)\leq0\), \(\forall (x,r)\in \Omega_{\delta_{1\varepsilon}}\times(0, \delta_{2\varepsilon}) \);
-
(iii)
\(I_{1-}(r)+I_{2}(x)\geq0\), \(\forall (x,r)\in \Omega_{\delta_{1\varepsilon}}\times(0, \delta_{2\varepsilon})\).
As in the proof of Theorem 1.2, we define
$$ \bar{u}_{\varepsilon}=\phi \bigl((A_{1}+\varepsilon)h \bigl(d(x) \bigr) \bigr), \quad x\in\Omega_{\delta_{1\varepsilon}}, $$
where
$$ h(t)= \int_{0}^{t}s^{-1} \bigl(L(s) \bigr)^{\frac{1}{p-1}}\,ds. $$
By (5.4), Lemma 5.3, and a direct calculation we see that, for \(x\in\Omega_{\delta_{1\varepsilon}}\)
$$\begin{aligned}& \Delta_{p}\bar{u}_{\varepsilon}(x)+b(x) g \bigl( \bar{u}_{\varepsilon}(x) \bigr) \\& \quad = \bigl( \phi' \bigl(h(r) \bigr) \bigr)^{p-1}r^{-p}L(r) \bigl( I_{1+}(r)+I_{2}(x) \bigr) \leq0, \end{aligned}$$
where \(r=d(x)\), that is, \(\bar{u}_{\varepsilon}\) is a supersolution of problem (1.1) in \(\Omega_{\delta_{1\varepsilon}}\).
In a similar way, we show that
$$ \underline{u}_{\varepsilon}=\phi \bigl((A_{2}-\varepsilon)h \bigl(d(x) \bigr) \bigr), \quad x\in\Omega_{\delta_{1\varepsilon}}, $$
is a subsolution of problem (1.1) in \(\Omega_{\delta_{1\varepsilon}}\).
As in the proof of Theorem 1.2, we obtain, for \(x\in \Omega_{\delta_{1\varepsilon}}\)
$$\frac{u(x)}{\phi((A_{2}+\varepsilon) h(d(x)))}\leq\frac{M v(x)}{\phi((A_{2}+\varepsilon) h(d(x)))}+1 $$
and
$$1-\frac{M v(x)}{\phi((A_{2}-\varepsilon) h(d(x)))}\leq\frac{u(x)}{\phi((A_{2}-\varepsilon)h(d(x)))}. $$
Consequently, by (5.3) and Lemma 3.4(iv),
$$1\leq\liminf_{d(x) \rightarrow0 } \frac{u(x)}{\phi((A_{2}-\varepsilon) h(d(x)))} \leq\limsup _{d(x) \rightarrow0 } \frac{u(x)}{\phi((A_{2}+\varepsilon) h(d(x)))} \leq 1 $$
and
$$\lim_{d(x) \rightarrow0 } \frac{\phi((A_{2}-\varepsilon) h(d(x)))}{\phi( h(d(x)))}=(A_{2}- \varepsilon)^{1-C_{g}}. $$
Thus, letting \(\varepsilon\rightarrow0\), we obtain (1.9). □