In this section, we get the regularity of solutions to (1.2) by using results to (1.3). Also, we obtain the monotonicity of cylindrical solutions to (1.2). The gradient estimate and Harnack inequality for classical solutions to (1.2) are provided by calculating the Harnack quantity, which is similar to Li [27] and Ma and Wei [28].
4.1 Regularity and symmetry of solutions to (1.2)
Lemma 4.1
Problem (1.2) allows a unique nonnegative solution
\(u\in H_{0}^{1}(\Omega)\) (\(\gamma\leq1\)) or
\(u\in H_{\mathrm{loc}}^{1}(\Omega)\) (\(\gamma> 1\)), and (3.4) is true for
\(u_{0}\), that is, \(u_{0}(\xi)\geq c_{\omega}\)
for
\(\xi\in\omega\subset\subset\Omega\).
Proof
For every n, since the solution \({{u}_{n}}\in H_{0}^{1}(\Omega)\) to (1.3) is bounded by Lemma 3.4 and \({{u}_{n}}\) converges pointwise to \({{u}_{0}}\) in Ω by Remark 3.1, it follows that \({{u}_{n}}\) converges weakly to \({{u}_{0}}\) in \(H_{0}^{1}(\omega )\), \(\omega\Subset\Omega\), that is, for any \(\varphi\in C_{0}^{1}(\Omega )\) such that \(\operatorname{supp}\varphi=\omega\),
$$ \lim_{n\to\infty} \int_{\omega}{{{\nabla }_{\mathbb{H}}} {{u}_{n}}\cdot{{ \nabla}_{\mathbb{H}}}\varphi}= \int _{\omega}{{{\nabla}_{\mathbb{H}}} {{u}_{0}} \cdot{{\nabla}_{\mathbb {H}}}\varphi}. $$
(4.1)
Noting that \({{u}_{n}}\) satisfies (3.4), we have that, for any \(\varphi\in C_{0}^{1}(\Omega)\),
$$0\le \biggl\vert \frac{\varphi}{{{({{u}_{n}}+\frac{1}{n})}^{\gamma}}} \biggr\vert \le\frac{{{ \Vert \varphi \Vert }_{{{L}^{\infty}}(\Omega )}}}{{c}_{\omega}^{\gamma}}. $$
By Lebesgue’s dominated convergence theorem,
$$ \lim_{n\to\infty} \int_{\Omega}{\frac{\varphi }{{({{u}_{n}}+\frac{1}{n})^{\gamma}}}= \int_{\Omega}{\frac{\varphi }{{u}_{0}^{\gamma}}}}. $$
(4.2)
This implies by (4.1) and (4.2) that problem (1.2) has a solution \({{u}_{0}}\).
Using \({{u}_{n}}\ge{{c}_{\omega}}\) on \(\omega\Subset\Omega\), this yields \({{u}_{0}}\ge0\) in Ω. The uniqueness can be proved as in the proof of Lemma 3.2. □
For the equation \({{\Delta}_{\mathbb{H}}}z=0\) in Ω, by [29] we have \(z\in {{C}^{2}}(\Omega)\).
Lemma 4.2
If
∂Ω satisfies the Wiener criterion and
\({{z}_{1}}\in H_{0}^{1}(\Omega)\cap{{C}^{2}}(\Omega)\)
solves
$$-{{\Delta}_{\mathbb{H}}} {{z}_{1}}=1 \quad \textit{in } \Omega, $$
then
\({{z}_{1}}\in{C}(\overline{\Omega} )\)
is positive with
\({{ {{z}_{1}}|}_{\partial\Omega}}=0\).
Proof
Let \({{z}_{2}}(\xi)=\frac{{x}_{1}^{2}}{2}\). Then \({{\Delta}_{\mathbb {H}}}{{z}_{2}}=1\). Since ∂Ω satisfies the Wiener criterion, it follows by [13] (p. 1158) that there exists \(z\in C(\overline{\Omega} )\cap{{C}^{2}}(\Omega)\) such that
$${{\Delta}_{\mathbb{H}}}z=0 \quad \mbox{in } \Omega \quad \mbox{and} \quad {{{{z}}|}_{\partial\Omega}}=z_{2}. $$
We easily see that \(z-{{z}_{2}}\in H_{0}^{1}(\Omega)\cap {{C}^{2}}(\Omega)\) satisfies
$$-{{\Delta}_{\mathbb{H}}}(z-{z}_{2})=1 \quad \mbox{in } \Omega, $$
and therefore \({{z}_{1}}=z-{{z}_{2}}\in{C}(\overline{\Omega} )\) is the required function with \(z_{1}>0\) and \({{ {{z}} |}_{\partial \Omega}}=0\). □
Lemma 4.3
If
∂Ω satisfies the Wiener criterion, then there exists a solution
\({{u}_{0}}\)
to problem (1.2), and
\({{u}_{0}}\in {C}(\overline{\Omega} )\).
Proof
Taking \({u}_{0}^{*}={((\gamma+1){{z}_{1}})^{\frac{1}{\gamma+1}}}\), we have by Lemma 4.2 that
$$\begin{aligned} & {{\Delta}_{\mathbb{H}}} {u}_{0}^{*} + \bigl({u}_{0}^{*}\bigr)^{-\gamma} \\ &\quad =-\gamma{\bigl((\gamma+1){{z}_{1}}\bigr)^{\frac{-2\gamma-1}{\gamma +1}}} {{ \vert {{\nabla}_{\mathbb{H}}} {{z}_{1}} \vert }^{2}}+{{\bigl((\gamma+1){{z}_{1}}\bigr)}^{\frac{-\gamma}{\gamma+1}}} {{\Delta }_{\mathbb{H}}} {{z}_{1}}+{{\bigl((\gamma+1){{z}_{1}} \bigr)}^{\frac{-\gamma }{\gamma+1}}} \\ &\quad ={-}\gamma{{\bigl((\gamma+1){{z}_{1}}\bigr)}^{\frac{-2\gamma-1}{\gamma +1}}} {{ \vert {{\nabla}_{\mathbb{H}}} {{z}_{1}} \vert }^{2}}\le0. \end{aligned}$$
(4.3)
Noting that \({{u}_{0}}\) is nonnegative, we know by (4.3) that
$$0\le{{u}_{0}}\le{u}_{0}^{*}={{\bigl(( \gamma+1){{z}_{1}}\bigr)}^{\frac {1}{\gamma+1}}}, $$
that is, \({{u}_{0}^{*}}\) and 0 are the super- and subsolutions of (1.2), respectively. This implies the existence of a solution \({{u}_{0}}\) to (1.2). Using \({{z}_{1}}\in{C}(\overline{\Omega} )\) in Lemma 4.2, we have \({{u}_{0}}\in{C}(\overline{\Omega} )\). □
Lemma 4.4
If
∂Ω satisfies the Wiener criterion, \({{u}_{0}}\in {C}(\overline{\Omega} )\cap H_{0}^{1}(\Omega)\) (\(\gamma\leq1\)) or
\(u_{0}\in{C}(\overline{\Omega} )\cap H_{\mathrm{loc}}^{1}(\Omega)\) (\(\gamma> 1\)) is a cylindrical solution to problem (1.2), then for any
\({{\lambda}_{0}}<\lambda<{{\lambda}_{1}}\), we have
$$ {{u}_{0}}(\xi)< {({u}_{{{0}}})}_{\lambda}( \xi)\quad \textit{for all } \xi \in{{\Sigma}_{\lambda}} $$
(4.4)
with respect to
t.
Proof
Step 1. Suppose that λ is sufficiently close to \(\lambda_{0}\). Then by Lemma 3.3 we have that \(u_{n}>0\) in \({\Sigma}_{\lambda}\), and then \(u_{n}\leq({u_{n}})_{\lambda}\) on \(\partial{\Sigma }_{\lambda}\). Since \(u_{n}=({u_{n}})_{\lambda}\) on the hyperplane \(T_{\lambda}\), it follows that \(u_{n}\leq ({u_{n}})_{\lambda}\) in \({{\Sigma}_{\lambda}}\), and then \(u_{n}<({u_{n}})_{\lambda}\) on \({{\Sigma}_{\lambda}}\) by Lemma 3.5.
Step 2. Since Step 1 provides a starting point, we can move the plane \(T_{\lambda}\) to the right keeping \(u_{n}\leq ({u_{n}})_{\lambda}\) on \({{\Sigma}_{\lambda}}\) until its limiting position \(\lambda_{1}\). Denoting
$$\overline{\lambda}=\sup \bigl\{ \lambda| u_{n}\leq ({u_{n}})_{\lambda}\mbox{ in }{{\Sigma}_{\mu}}, \mu\le\lambda \bigr\} , $$
we claim that \(\overline{\lambda}=\lambda_{1}\). Indeed, suppose \(\overline{\lambda}<\lambda_{1}\); we will show that the plane \(T_{\lambda}\) can be moved further to the right or, to be more rigorous, there exists \(\varepsilon>0\) such that, for any \(\lambda\in (\overline{\lambda},\overline{\lambda}+\varepsilon)\),
$$ u_{n}\leq({u_{n}})_{\lambda}. $$
(4.5)
This is a contradiction to the definition of λ̅. Hence \(\overline{\lambda}=\lambda_{1}\).
To prove (4.5), we separate \({{\Sigma}_{\overline{\lambda }+\varepsilon}}\) into two parts, \({{\Sigma}_{\overline{\lambda }-\varepsilon}}\) and \({{\Sigma}_{\overline{\lambda}+\varepsilon }}\setminus {{\Sigma}_{\overline{\lambda}-\varepsilon}}\) for \(\varepsilon>0\).
Clearly, \({{({{u}_{n}})}_{{\bar{\lambda}}}}-{{u}_{n}}\ge {{c}_{\varepsilon}}>0\) in \({{\Sigma}_{\overline{\lambda}-\varepsilon }}\) for any \(\varepsilon>0\). Since \({{({{u}_{n}})}_{\lambda }}-{{u}_{n}}\) depends continuously on λ, there exists \(\varepsilon>0\) sufficiently small such that, for all \(\lambda\in (\overline{\lambda},\overline{\lambda}+\varepsilon)\), we have
$$ {{({{u}_{n}})}_{\lambda}}-{{u}_{n}}\ge0 \quad \mbox{in } {{\Sigma }_{\overline{\lambda}-\varepsilon}}. $$
(4.6)
On the other hand, since \({{({{u}_{n}})}_{\lambda}}-{{u}_{n}}\ge0\) on \(\partial({{\Sigma}_{\overline{\lambda}+\varepsilon}} \setminus {{\Sigma}_{\overline{\lambda}-\varepsilon}})\) for fixed \(\varepsilon >0\), it follows by Lemma 3.5 that
$$ {{({{u}_{n}})}_{\lambda}}-{{u}_{n}}\ge0 \quad \mbox{in } {{\Sigma }_{\overline{\lambda}+\varepsilon}} \setminus {{\Sigma}_{\overline {\lambda}-\varepsilon}}. $$
(4.7)
Combining (4.6) and (4.7), we prove (4.5).
Using the equality \(\overline{\lambda}=\lambda_{1}\) and the convergence of \(u_{n}\) to \(u_{0}\) a.e., we get that \({{u}_{0}}\le {{({{u}_{0}})}_{\lambda}}\) on \({{\Sigma}_{\lambda}}\). So (4.4) is proved by Lemma 2.3. □
4.2 Harnack inequality for a solution to (1.2)
Results in the subsection are not used in the sequel, but they are of independent interest.
Lemma 4.5
Let
\({{u}_{0}}\in{C}^{3}(\Omega)\)
be a positive solution to (1.2). Then, for any
\(R=R({{\xi}_{0}})>0\), \({{\xi}_{0}}\in\Omega\), and
\(B({{\xi}_{0}},2R)\subset\Omega\), there exists a positive constant
\(C=C(R)\)
such that
$$ {{ \vert {{\nabla}_{\mathbb{H}}} {{u}_{0}} \vert }^{2}}\le C{u}_{0}^{2} \quad \textit{in } B({{ \xi}_{0}},R). $$
(4.8)
Proof
Let \(w_{0}=\log{{u}_{0}}\). Then
$$ {{\nabla}_{\mathbb{H}}}w_{0}={u}_{0}^{-1}{{ \nabla}_{\mathbb{H}}} {{u}_{0}} $$
(4.9)
and
$$ {{\Delta}_{\mathbb{H}}}w_{0}=-{{ \vert {{ \nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}- \frac{1}{{u}_{0}^{\gamma+1}}. $$
(4.10)
Taking a smooth cut-off function ϕ with \(\phi=1\) in \(B(\xi_{0},R)\) and \(\phi=0\) outside \(B(\xi_{0},2R)\), define
$$P=\phi{{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}, $$
which as usual is called the Harnack quantity.
At the maximum point of P, we have \({\nabla}_{\mathbb{H}}P=0\) and \({\Delta}_{\mathbb{H}}P\leq0\). Then
$$\begin{aligned}& {{\nabla}_{\mathbb{H}}}\bigl({{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\bigr)=-{{\phi}^{-1}} {{\nabla}_{\mathbb{H}}} \phi{{ \vert {{\nabla }_{\mathbb{H}}}w_{0} \vert }^{2}}=-{{\phi}^{-2}}P{{\nabla}_{\mathbb {H}}}\phi, \end{aligned}$$
(4.11)
$$\begin{aligned}& {{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}} {{\Delta}_{\mathbb {H}}}\phi+2{{\nabla}_{\mathbb{H}}}\phi \cdot{{\nabla}_{\mathbb {H}}}\bigl({{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\bigr)+\phi{{\Delta }_{\mathbb{H}}}\bigl({{ \vert {{ \nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\bigr)\le0. \end{aligned}$$
(4.12)
Using the basic formula
$$ {{\Delta}_{\mathbb{H}}} {{ \vert {{\nabla}_{\mathbb{H}}}v \vert }^{2}}=2\bigl(({{\nabla}_{\mathbb{H}}} {{ \Delta}_{\mathbb{H}}}v,{{\nabla }_{\mathbb{H}}}v)+{{ \bigl\vert {D}_{\mathbb{H}}^{2}v \bigr\vert }^{2}}\bigr), $$
(4.13)
for any smooth function v in \(\mathbb{H}^{n}\), it follows
$$ \phi{{\Delta}_{\mathbb{H}}}\bigl({{ \vert {{ \nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\bigr)=2\phi({{ \nabla}_{\mathbb{H}}} {{\Delta}_{\mathbb {H}}}w_{0},{{ \nabla}_{\mathbb{H}}}w_{0})+2\phi{{ \bigl\vert {D}_{\mathbb {H}}^{2}w_{0} \bigr\vert }^{2}}. $$
(4.14)
Let us estimate the right-hand side in (4.14). For the first term, we have
$$\begin{aligned} 2\phi({{\nabla}_{\mathbb{H}}} {{\Delta}_{\mathbb{H}}}w_{0},{{ \nabla }_{\mathbb{H}}}w_{0})&=2\phi\biggl({{\nabla}_{\mathbb{H}}} \biggl(-{{ \vert {{\nabla }_{\mathbb{H}}}w_{0} \vert }^{2}}-\frac{1}{{u}_{0}^{\gamma +1}}\biggr),{{\nabla}_{\mathbb{H}}}w_{0} \biggr) \\ & =-2\phi\bigl({{\nabla}_{\mathbb{H}}}\bigl({{ \vert {{\nabla}_{\mathbb {H}}}w_{0} \vert }^{2}}\bigr),{{\nabla}_{\mathbb{H}}}w_{0} \bigr)+2(\gamma +1)\phi\biggl(\frac{{{\nabla}_{\mathbb{H}}}{{u}_{0}}}{{u}_{0}^{\gamma +2}},{{\nabla}_{\mathbb{H}}}w_{0} \biggr) \\ & =-2\phi\bigl(-{{\phi}^{-2}}P{{\nabla}_{\mathbb{H}}}\phi,{{\nabla }_{\mathbb{H}}}w_{0}\bigr)+2(\gamma+1)\phi\frac{{{ \vert {{\nabla}_{\mathbb {H}}}w_{0} \vert }^{2}}}{{u}_{0}^{\gamma+1}} \\ & =2{{\phi}^{-1}}({{\nabla}_{\mathbb{H}}}\phi,{{ \nabla}_{\mathbb {H}}}w_{0})P+\frac{2(\gamma+1)}{{u}_{0}^{\gamma+1}}P \\ &\ge-{{\phi}^{-1}} {{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}P-{{\phi}^{-1}} {{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}P+\frac{2(\gamma+1)}{{u}_{0}^{\gamma+1}}P \\ &=-{{\phi}^{-1}} {{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}P-{{\phi}^{-2}} {{P}^{2}}+ \frac{2(\gamma+1)}{{u}_{0}^{\gamma+1}}P. \end{aligned}$$
(4.15)
Due to (4.10), the second term of (4.14) satisfies
$$\begin{aligned} 2\phi{{ \bigl\vert {D}_{\mathbb{H}}^{2}w_{0} \bigr\vert }^{2}}&\ge2\phi{{ \vert {{\Delta}_{\mathbb{H}}}w_{0} \vert }^{2}} \\ & =2\phi{{ \biggl\vert -{{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}-\frac{1}{{u}_{0}^{\gamma+1}} \biggr\vert }^{2}} \\ & =2\phi{{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{4}}+\frac {4\phi{{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}}{{u}_{0}^{\gamma+1}}+\frac{2\phi}{{u}_{0}^{2(\gamma+1)}} \\ &\ge\frac{2}{\phi}{{P}^{2}}+\frac{4}{{u}_{0}^{\gamma+1}}P. \end{aligned}$$
(4.16)
Substituting (4.15) and (4.16) into (4.14), we obtain
$$ \phi{{\Delta}_{\mathbb{H}}}\bigl({{ \vert {{ \nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\bigr)\ge- \frac{{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{ \phi}P-\frac{ {{P}^{2}}}{{{\phi}^{2}}}+\frac{2\gamma +6}{{u}_{0}^{\gamma+1}}P+\frac{2}{\phi}{{P}^{2}}. $$
(4.17)
From (4.12), \({{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}=\frac{P}{\phi}\), (4.11), and (4.17) we have that
$$\begin{aligned} 0&\ge\phi{{\Delta}_{\mathbb{H}}}\bigl({{ \vert {{\nabla}_{\mathbb {H}}}w_{0} \vert }^{2}}\bigr)+\frac{{{\Delta}_{\mathbb{H}}}\phi}{\phi }P-\frac{2{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{\phi }P \\ & \ge\frac{{{\Delta}_{\mathbb{H}}}\phi}{\phi}P-\frac{2{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{\phi^{2} }P-\frac{{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{ \phi}P- \frac {{{P}^{2}}}{{{\phi}^{2}}}+\frac{2\gamma+6}{{u}_{0}^{\gamma +1}}P+\frac{2}{\phi}{{P}^{2}}, \end{aligned}$$
(4.18)
and so
$$-\frac{{{\Delta}_{\mathbb{H}}}\phi}{\phi}+\frac{2{{ \vert {{\nabla }_{\mathbb{H}}}\phi \vert }^{2}}}{\phi^{2} }+\frac{{{ \vert {{\nabla }_{\mathbb{H}}}\phi \vert }^{2}}}{ \phi}-\frac{2\gamma +6}{{u}_{0}^{\gamma+1}}\ge \frac{2\phi-1}{\phi^{2} }P. $$
This gives
$$\frac{2\phi-1}{\phi^{2} }P\le-\frac{{{\Delta}_{\mathbb{H}}}\phi }{\phi}+\frac{2{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{\phi^{2}}+\frac{{{ \vert {{\nabla}_{\mathbb{H}}}\phi \vert }^{2}}}{\phi} \quad \mbox{on } B(\xi_{0},2R). $$
Using the bounds \(\vert {{\nabla}_{\mathbb{H}}}\phi \vert \le C{{R}^{-1}}\) and \(\vert {{\Delta}_{\mathbb{H}}}\phi \vert \le C{{R}^{-2}}\), we get
$$ (2\phi-1)P\le-\phi{{\Delta}_{\mathbb{H}}}\phi+2{{ \vert {{ \nabla }_{\mathbb{H}}}\phi \vert }^{2}}+\phi{{ \vert {{ \nabla}_{\mathbb {H}}}\phi \vert }^{2}}\le\frac{C}{{{R}^{2}}}. $$
(4.19)
Since \(\phi=1\) on \(B({{\xi}_{0}},R)\), by (4.19) we have that
$$ {{ \vert {{\nabla}_{\mathbb{H}}}w_{0} \vert }^{2}}\le C(R) \quad \mbox{on } B(\xi_{0},R), $$
(4.20)
which implies (4.8). □
Theorem 4.6
Let
\({{u}_{0}}\in{C}^{3}(\Omega)\)
be a positive solution to (1.2). Then, for any
\(R=R({{\xi}_{0}})>0\), \({{\xi}_{0}}\in\Omega\), and
\(B({{\xi}_{0}},2R)\subset\Omega\), there exists
\(C=C(R)>0\)
such that, for all
\(\xi,\eta\in B({{\xi}_{0}},\frac{R}{2})\),
$$ {{u}_{0}}(\xi)\le C{{u}_{0}}(\eta). $$
(4.21)
Proof
Let I be the shortest curve in \(B({{\xi}_{0}},R)\) joining ξ to η. Then the length of I is clearly at most 2R. Integrating the quantity \(\vert {{\nabla}_{\mathbb{H}}}\log{{u}_{0}} \vert \) along I yields
$$ \log{{u}_{0}}(\xi)-\log{{u}_{0}}(\eta)\le \int_{I }{ \vert {{\nabla }_{\mathbb{H}}} \log{{u}_{0}} \vert }. $$
(4.22)
Applying (4.20), we get
$$ \int_{I }{ \vert {{\nabla}_{\mathbb{H}}} \log{{u}_{0}} \vert }\le \int _{I}{C}\le C. $$
(4.23)
Now (4.21) is obtained from (4.22) and (4.23). □