\documentclass[reqno]{amsart} \usepackage{amssymb} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{ Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 164, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/164\hfil A Neumann problem] {A Neumann problem with the $q$-Laplacian on a solid torus in the critical of supercritical case} \author[A. Cotsiolis, N. Labropoulos \hfil EJDE-2007/164\hfilneg] {Athanase Cotsiolis, Nikos Labropoulos} % in alphabetical order \address{Athanase Cotsiolis \newline Department of Mathematics, University of Patras\\ Patras 26110, Greece} \email{cotsioli@math.upatras.gr} \address{Nikos Labropoulos \newline Department of Mathematics, University of Patras\\ Patras 26110, Greece} \email{nal@upatras.gr} \thanks{Submitted May 18, 2006. Published November 30, 2007.} \subjclass[2000]{35J65, 46E35, 58D19} \keywords{Neumann problem; $q$-Laplacian; solid torus; no radial symmetry; \hfill\break\indent critical of supercritical exponent} \begin{abstract} Following the work of Ding \cite{Din} we study the existence of a nontrivial positive solution to the nonlinear Neumann problem \begin{gather*} \Delta_qu+a(x)u^{q-1}=\lambda f(x)u^{p-1}, \quad u>0\quad \text{on } T,\\ \nabla u|^{q-2}\frac{\partial u}{\partial \nu}+b(x) u^{q-1} =\lambda g(x)u^{\tilde{p}-1} \quad\text{on }{\partial T},\\ p =\frac{2q}{2-q}>6,\quad \tilde{p}=\frac{q}{2-q}>4,\quad \frac{3}{2}0\quad \text{on } T,\\ |\nabla u|^{q-2}\frac{\partial u}{\partial \nu}+b(x) u^{q-1} =\lambda g(x)u^{\tilde{p}-1} \quad\text{on }{\partial T},\\ p =\frac{2q}{2-q}>6,\quad \tilde{p}=\frac{q}{2-q}>4,\quad \frac{3}{2}6$ and $\tilde{p}=\frac{q} {2-q}>4$, $\frac{3}{2}r>0\} $$ and $\mathcal{A}=\{(\Omega_i,\xi_i):i=1,2\}$ be an atlas on $T$ defined by \begin{gather*} \Omega_1=\{(x,y,z)\in T :(x,y,z)\notin H^+_{XZ}\},\\ \Omega_2=\{(x,y,z)\in T :(x,y,z)\notin H^-_{XZ}\} \end{gather*} where \begin{gather*} H_{XZ}^+=\{(x,y,z)\in \mathbb{R}^3 : x>0\, ,\, y=0 \}\\ H^-_{XZ}=\{(x,y,z)\in \mathbb{R}^3 : x<0\, ,\, y=0 \} \end{gather*} and $\xi_i:\Omega_i\to I_i\times D$, $i=1,2$, with $$ I_1=(0,2\pi), \quad I_2=(-\pi,\pi), \quad D=\{(t,s)\in \mathbb{R}^2 :t^2+s^2<1\}$$ and $\xi_i(x,y,z)=(\omega_i,t,s)$, $i=1,2$ with $$ \cos\omega_i=\frac{x}{\sqrt{x^2+y^2}},\quad \sin\omega_i=\frac{y}{\sqrt{x^2+y^2}},\quad i=1,2 $$ where $$ \omega_1= \begin{cases} \mathop{\rm arctan}\frac{y}{x},&x\neq 0 \\ \pi/2,&x=0,y>0 \\ 3\pi/2,&x=0\,,\,y<0 \end{cases}\quad \omega_2=\begin{cases} \mathop{\rm arctan}\frac{y}{x}, &x\neq 0 \\ \pi/2,&x=0,\;y>0 \\ -\pi/2,&x=0\,,\,y<0 \end{cases} $$ and $$ t=\frac{\sqrt{x^2+y^2}-l}{r}\,,\quad s=\frac{z}{r}. $$ The Euclidean metric $g$ on $(\Omega,\xi)\in \mathcal{A}$ can be expressed as $$ (\sqrt{g}\circ\xi^{-1})(\omega,t,s)=r^2(l+rt). $$ Consider the spaces of all $G$-invariant functions under the action of the group $G=O(2)\times I \subset O(3)$ $$ H_{1,G}^q=\{u\in H_1^q(T) : u\circ \tau=u\,,\; \forall \tau \in G\}, $$ where $H^q_1(T)$ is the completion of $C^\infty(T)$ with respect the to norm $$ \|u\|_{H^q_1}=\|\nabla u\|_q+ \|u\|_q. $$ For all $G$-invariants $u$ we define the functions $\phi(t,s)=(u\circ \xi^{-1})(\omega,t,s)$. Then we have \begin{gather}\label{eq2.1} \|u\|_{L^p(T)}^p=2\pi r^2 \int_D |\phi(t,s)|^p(l+rt)\,dt\,ds,\\ \label{eq2.2} \|\nabla u\|_{L^q(T)}^q=2\pi r^{2-q} \int_D |\nabla \phi(t,s)|^q (l+rt)\,dt\,ds, \\ \label{eq2.3} \|u\|_{L^p(\partial T)}^p=2\pi r \int_{\partial D }|\phi(t,0)|^p(l+rt)\,dt, \end{gather} where by $\phi$ we denote the extension of $\phi$ on $\partial D$. Let $K(2,q)$ be the best constant \cite{Aub} of the Sobolev inequality \[ \|\varphi\|_{L^p(\mathbb{R}^2)}\leq K(2,q)\|\nabla\varphi\|_{L^q(\mathbb{R}^2)} \] for the Euclidean space $\mathbb{R}^2$, where $1\leq q<2$, $p=2q/(2-q)$ and $\tilde{K}(2,q)$ be the best constant \cite{Lio} in the Sobolev trace embedding \[ \|\varphi\|_{L^{\tilde{p}}(\partial\mathbb{R}^2_+)} \leq \tilde{K}(2,q)\|\nabla\varphi\|_{L^q(\mathbb{R}^2_+)} \] for the Euclidean half-space $\mathbb{R}^2_+$, where $1\leq q<2$, $\tilde{p}=q/(2-q)$. Consider a point $P_j(x_j,y_j,z_j) \in\overline{T} $, and by $O_{P_j}$ denote the orbit of $P_j$ under the action of the group $G$. Let $l_j=\sqrt{x^2_j+y^2_j}\,$ be the horizontal distance of the orbit $O_{P_j}$ from the axis $z'z$. For $\varepsilon>0$ given and $\delta_j=l_j\varepsilon$, consider a finite covering $(T_j)_{j=1,\dots N}$ with $$ T_j=\{(x,y,z)\in \overline {T} : ( \sqrt{x^2+y^2}-l_j)^2+(z-z_j)^2< \delta^2_j\} $$ an open small solid torus (a tubular neighborhood of the orbit $O_{P_j}$). \subsection{Best constants on the solid Torus} \begin{theorem}\label{T3.1} Let $\overline{T}$ be the solid torus and $\tilde{p}$, $q$ be two positive real numbers such that $\tilde{p}=q/(2-q)$ with $10$ there exists a real number $B_\varepsilon$ such that for all $u\in H^q_{1,G}$ the following inequality holds: \begin{equation}\label{eq3.1} \|u\|^q_{L^{\tilde{p}}(\partial T)}\leq \Big[\frac{\tilde{K}^q(2,q)}{[2\pi(l-r)\Big]^{q-1}} +\varepsilon\Big]{\|\nabla u\|^q_{L^q(T)}} +B_\varepsilon\|u\|^q_{L^q(\partial T)} \end{equation} In addition the constant $\tilde{K}^q (2,q)/{[2\pi(l-r)]^{q-1}}$ is the best constant for the above inequality. \end{theorem} \subsection{Resolution of the problem} Consider the set \[ \Lambda=\{ c = ( {\alpha ,\beta } ) \in \mathbb{R}^2 :\alpha - \beta \geq \delta ,\;q \leq \alpha \leq p,\;q\leq \beta \leq \tilde{p} \}, \] with $\delta \in ({0,p - \tilde p} ) = \bigl( 0,q/(2 - q) \bigr)$, and define the functionals \begin{gather*} I(u) = \int_T {( {| {\nabla u} |^q + a| u |^q } )dV + \int_{\partial T} {b| u |^q dS}} \\ I_c (u) = \int_T {f| u |^\alpha dV + \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} } \end{gather*} for all $u \in H_{1,G}^q $ and for any $c \in\Lambda $. $I(u)$ and $I_c (u)$ are well defined because the imbeddings of $H^q_{1,G}$ onto $L^p$ and $L^{\widetilde{p}}$ are continue according to the Sobolev theorem. Define, also \begin{gather*} \Sigma _c = \{ {u \in H_{1,G}^q : I_c (u) = 1}\},\\ \mu _c = \inf \{ {I(u):u \in \Sigma _c } \},\\ c_0 = ( {p,\tilde p} ), \quad t^ + = \sup ( {t,0} ), \end{gather*} for all $t \in \mathbb{R}$. Consequently for the problem \eqref{eP} we have the following theorem. \begin{theorem}\label{T3.2} Let $a$, $f$, $b$ and $g$ be four smooth functions, $G$-invariant and $q$, $p$, $\tilde{p}$ be three real numbers defined as in \eqref{eP}. Suppose that the function $f$ has constant sign (e.g. $f\geq0)$. The problem \eqref{eP} has a positive solution $u\in H_{1,G}^q$ if the following holds: \begin{equation}\label{eq3.2} (\sup_{T}f)\Big[\frac{K^q(2,q)\mu_{{c_0}}^+}{[\pi(l-r)]^{q/2}} \Big]^{p/2}+ \frac{p}{\widetilde{p}}(\sup_{{\partial T}}g)^+\Big[\frac{\widetilde{K}^q(2,q)\mu_{{c_0}}^+} {[2\pi(l-r)]^{q-1}}\Big]^{\widetilde{p}/2}<1 \end{equation} and if \begin{enumerate} \item $f>0$ everywhere and $g$ is arbitrary, or \item $f\geq0$, $g>0$ everywhere and $(-inf_{{T}}a)^+\kappa<1$, where \begin{equation}\label{eq3.3} \kappa=\inf\{A>0 : \exists B>0\; s.t.\|\psi\|^q_{L^q(T)}\leq A \|\nabla\psi\|^q_{L^q(T)}+B \|\psi\|^q_{L^q(\partial T)}\}. \end{equation} \end{enumerate} \end{theorem} In the rest of this paper we denote $K=K(2,q)$, $\tilde{K}=\tilde{K}(2,q)$ and $L=2\pi(l-r)$. \section{Proofs} \begin{proof}[Proof of Theorem \ref{T3.1}] The proof is carried out in two steps. \noindent\textbf{Step 1.} Suppose that there exist two real numbers $A$, $B$ such that for all $u\in H^q_{1,G}$ the following inequality holds: \[ \|u\|^q_{L^{\tilde{p}}(\partial T)}\leq A {\|\nabla u\|^q_{L^q(T)}}+B\|u\|^q_{L^q(\partial T)}. \] Then \[ A\geq \frac{\tilde{K}^q(2,q)}{|2\pi(l-r)|^{q-1}} \] Consider a transformation $F:D \to \mathbb{R}_ + ^2 $. Such a transformation, for example, is \[ F( {t,s} ) = ( {\frac{{4t}} {{t^2 + ( {1 + s} )^2 }},\frac{{2( {1 - t^2 - s^2 } )}} {{t^2 + ( {1 + s} )^2 }}} ), \] see \cite {Esc}. Denote by $({\tilde g_{ij}})$ the Euclidian metric on $D$, $\,dx\,dy$ the Euclidian metric on $\mathbb{R}_ + ^2 $ and $d\sigma $ the induced on $\partial \mathbb{R}_ + ^2 $. Choose a finite covering of $\bar D$ consisting of disks $D_k$, centered on $p_k$, such that: If $p_k \in D $, then entire $D_k$ lies in $D$ and if $p_k\in \partial D$, $D_k$ is a Fermi neighborhood. In these neighborhoods we have \begin{equation}\label{eq4.3} 1 - \varepsilon _0 \leqslant \sqrt {\det ( {\tilde g_{ij} } )} \leqslant 1 + \varepsilon _0 \end{equation} Suppose by contradiction that, there exists \[ A < \frac{\tilde K^q } {L^{q - 1}}\quad \text{and} \quad B\in\mathbb{R} \] such that the inequality \begin{equation}\label{eq4.12} \| u \|_{L_G^{\tilde p} (\partial T)}^q \leqslant A\| {\nabla u} \|_{L_G^q (T)}^q + B\| u \|_{L_G^q (\partial T)}^q \end{equation} holds for all $ u \in H_{1,G}^q (T)$. Fix a point $P_0 \in \partial T$, that belongs to the orbit of minimum range $l-r$. For any $\varepsilon_0>0$, we can choose $ \delta = \varepsilon_0 (l - r) < 1$ and $$ T_\delta = \{ {Q \in \mathbb{R}^3: d(Q,O_{P_0 } ) < \delta } \} $$ such that, if $I \times U \subset I \times D$ is the image of a neighborhood of $P_0 \in \partial T$ through the chart $\xi$ of $T$ and $V \subset \mathbb{R}_ + ^2 $ the image of $U$ through $F$, (\ref{eq4.3}) holds. It follows that, for any $u \in C_0^\infty (T_\delta )$, we have successively: \begin{gather*} \Big( {\int_{\partial T_\delta } {| u |} ^{\tilde p} dS} \Big)^{q/\tilde p} \leqslant A\int_{T_\delta } {| {\nabla u} |} ^q dV + B\int_{\partial T_\delta } {| u |} ^q dS, \\ \begin{aligned} &\Big( {2\pi \delta \int_{\partial D} {| \phi |} ^{\tilde p} (l - r + \delta t)dt } \Big)^{q/\tilde p}\\ & \leqslant 2\pi \delta ^{2 - q} A\int_{D} {| {\nabla \phi } |} ^q (l - r + \delta t)\,dt\,ds+ + 2\pi \delta B\int_{\partial D} {| \phi |} ^q (l - r + \delta t)dt, \end{aligned}\\ \begin{aligned} &\Big({( {1 - \varepsilon _0 } )\delta L\int_{F(\partial D)} {( {| \phi |^{\tilde p} \sqrt {\tilde g} } )} \circ F^{ - 1} d\sigma } \Big)^{q/\tilde p} \\ &\le ( {1 + \varepsilon _0 } )\delta ^{2 - q} AL\int_{F(D)} {( {| {\nabla \phi } |^q \sqrt {\tilde g} } ) \circ F^{ - 1} } \,dx\,dy\\ &\quad + ( {1 + \varepsilon _0 } )\delta LB\int_{F(\partial D)} {( {| \phi |^q \sqrt {\tilde g} } )} \circ F^{ - 1} d\sigma , \end{aligned}\\ \begin{aligned} &\Big( {( {1 - \varepsilon_0 } )^2 \delta L\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \\ &\leqslant ( {1 + \varepsilon_0 } )^2 \delta L \Big( {\delta ^{1 - q} A\int_{\mathbb{R}_ + ^2 } {| {\nabla \Phi } |} ^q \,dx\,dy + B\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^q d\sigma } \Big), \end{aligned}\\ \end{gather*} \begin{equation}\label{eq4.13} \Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant f( \varepsilon_0 ) L^{q - 1} A\int_{\mathbb{R}_ + ^2 } {| {\nabla \Phi } |} ^q \,dx\,dy + \tilde B\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^q d\sigma, \end{equation} where $f( \varepsilon_0 )= {{( {1 + \varepsilon_0 } )^2 }}/ {{( {1 - \varepsilon_0 })^{2q/\tilde p} }}$, $\tilde{B}=f(\varepsilon_0)(\delta L)^{q-1}B$ and $\tilde{p}=q/(2-q)$. Because of (\ref{eq4.12}) and since the above function $f:(0,1) \to (1, + \infty )$ with $$ f( t )=\frac {{( {1 + t } )^2 }} {{( {1 - t} )^{2q/\tilde p} }} $$ is monotonically increasing, we can choose $\varepsilon_0$ small enough, such that the following inequality holds \[ A < f(\varepsilon_0)A <\frac{\tilde K^q}{L^{q - 1}} \] hence $A'< \tilde K^q$ where $ A' = f(\varepsilon_0)L^{q - 1} A$. So for $\varepsilon_0$ small enough and for all $\Phi \in C_0^\infty ( D )$ we have \begin{equation}\label{eq4.14} \Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant A'\int_{ \mathbb{R}_ + ^2} {| {\nabla \Phi } |} ^q \,dx\,dy + \tilde{B}\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^q d\sigma \end{equation} On the other hand by H\"{o}lder's inequality, for all $\Phi \in C_0^\infty ( D_\delta )$, where $D_\delta\subset D $, we have \[ \int_{\partial D_\delta } {| \Phi |} ^q d\sigma _0 \leqslant [ {Vol( {\partial D_\delta } )} ]^{1 - ( {q/\tilde p} )} \Big( {\int_{\partial D_\delta } {( {| \Phi |^q } )} ^{\tilde p/q} d\sigma _0 } \Big)^{q/\tilde p} \] and since $\tilde{p}=q/(2-q)$, that is $1 - (q/\tilde p) = 1 - ({2 - q} ) = q - 1$, we have \begin{equation}\label{eq4.15} \int_{\partial D_\delta} {| \Phi |} ^q d\sigma _0 \leqslant Vol( {\partial D_\delta} )^{q - 1} \Big( {\int_{\partial D_\delta} {| \Phi |} ^{\tilde p} d\sigma _0 } \Big)^{q/\tilde p} \end{equation} Hence, choosing $\varepsilon_0$ small enough, by (\ref{eq4.14}) and (\ref{eq4.15}), we get that there exists $A''< \tilde K^q$, such that for all $\Phi \in C_0^\infty ( D_\delta )$, \begin{equation}\label{eq4.16} \Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant A''\int_{ \mathbb{R}_ + ^2 } {| \nabla\Phi |} ^q \,dx\,dy\,. \end{equation} Let $\Psi \in C_0^\infty ( {\mathbb{R}_ + ^2 } )$ and set $\Psi _\lambda (x) = \lambda ^{1/\tilde p} \Psi (\lambda x),\,\lambda > 0$. For $\lambda>0$, sufficiently large, $\Psi_\lambda \in C_0^\infty ( D )$ and since $\| {\Psi _\lambda } \|_{L^{\tilde p} ( {\partial \mathbb{R}_ + ^2 } )} = \|\Psi \|_{L^{\tilde p} ( {\partial \mathbb{R}_ + ^2 } )}$ and $\| {\nabla \Psi _\lambda } \|_{L^q ( { \mathbb{R}_ + ^2 } )} = \| {\nabla \Psi } \|_{L^q ( { \mathbb{R}_ + ^2 } )} $, by (\ref{eq4.16}), the following inequality \[ \Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Psi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant A''\int_{ \mathbb{R}_ + ^2 } {|\nabla\Psi |} ^q \,dx\,dy \] holds for all $\Psi \in C_0^\infty ( {\mathbb{R}_ + ^2 } )$. This is a contradiction since $\tilde K$ is the best constant for the Sobolev inequality in $\mathbb{R}_ + ^2 $. \noindent\textbf{Step 2.} For all $\varepsilon>0$ there exists a real number $B_\varepsilon$ such that for all $u\in H^q_{1,G}$ the following inequality holds: $$ \|u\|^q_{L^{\tilde{p}}(\partial T)}\leq \big[\frac{\tilde{K}^q(2,q)}{[2\pi(l-r)]^{q-1}}+\varepsilon\big] {\|\nabla u\|^q_{L^q(T)}} +B_\varepsilon\|u\|^q_{L^q(\partial T)} $$ Assume by contradiction that there exists $\varepsilon _0 > 0$ such that for all $\alpha > 0$ we can find $u \in H_{1,G}^q (T)$ with \[ \| u \|_{L_G^{\tilde p} (\partial T)}^q > ( {\frac{{\tilde K^q }} {{L^{q - 1} }} + \varepsilon_0 } )\| {\nabla u} \|_{L_G^q (T)}^q + \alpha \| u \|_{L_G^q (\partial T)}^q \] or \[ \frac{{\| {\nabla u} \|_{L_G^q (T)}^q + \alpha \| u \|_{L_G^q (\partial T)}^q }} {{\| u \|_{L_G^{\tilde p} (\partial T)} }} < \big( {\frac{{\tilde K^q }} {{L^{q - 1} }} + \varepsilon _0 } \big)^{ - 1} \] It follows that, the above inequality remains true for all $\varepsilon \in ( {0,\varepsilon _0 } )$ and setting \[ I_\alpha = \inf_{u \in H_G^{1,q} (T)\backslash \{ 0\} } \frac{{\| {\nabla u} \|_{L_G^q (T)}^q + \alpha \| u \|_{L_G^q (\partial T)}^q }} {{\| u \|_{L_G^{\tilde p} (\partial T)}^q }} \] we conclude that for all $\alpha>1$, there exists $\theta _0 > 0$ independent of $\alpha$ such that \begin {equation}\label{eq4.17} I_\alpha < \big( {\frac{{\tilde K^q }} {{L^{q - 1} }} + \varepsilon _0 } \big)^{ - 1} = \frac{{L^{q - 1} }} {{\tilde K^q }} - \theta _0 \end {equation} As the quotient \[ \frac{{\| {\nabla u} \|_{L_G^q (T)}^q + \alpha \| u \|_{L_G^q (\partial T)}^q }} {{\| u \|_{L_G^{\tilde p} (\partial T)}^q }} \] is homogeneous, for any fixed $\alpha$ we can take a minimizing sequence $(u_k)\subset H^q_{1,G}(T)$ for it satisfying $\| u_k \|_{L_G^{\tilde p} (\partial T)}^q =1$. As \begin{equation}\label{eq4.18} \| {\nabla u_k } \|_{L_G^q (T)}^q + \alpha \| {u_k } \|_{L_G^q (\partial T)}^q \to I_\alpha\,, \end{equation} we conclude that $(u_k)$ is bounded in $L_G^q ( {\partial T} )$ and $(\nabla u_k)$ is bounded in $L_G^q ( {\partial T} )$. By the standard Sobolev trace inequality, we easily take, by contradiction, that there exists a constant $C$ such that \begin{equation}\label{eq4.19} \|u\|^q_{L^q(T)}\leq C(\|\nabla u\|^q_{L^q(T)}+\|u\|^q_{L^q(\partial T)}). \end{equation} These two facts together imply that $u_k \rightharpoonup u$ in $H_1^q ( T )$, $u_k \to u$ in $L^q ( T )$ and $u_k \to u$ in $L^q ( {\partial T} )$. Since convergence in $L^p$ spaces implies a.e. convergence, the function $u$ will be $G$-invariant. By theorem $4$ of \cite{Bie} we have $u_k \to u$ in $L_G^{\tilde p} ( {\partial T} )$, $\| u \|_{L_G^{\tilde p} (\partial T)}^q =1$ and \[ \| {\nabla u} \|_{L_G^q (T)}^q + \alpha \| u \|_{L_G^q (\partial T)}^q = I_\alpha, \] that is $u$ is minimizing of $I_\alpha$. Now, for each $\alpha>0$, let $u_\alpha \in H_{1,G}^q ( T )$ satisfy $\| u_\alpha \|_{L_G^{ \tilde p}(\partial T)} = 1$ and \begin {equation}\label{eq4.20} \| {\nabla u_\alpha} \|_{L_G^q (T)}^q + \alpha \| u_\alpha \|_{L_G^q (\partial T)}^q = I_\alpha \leqslant \frac{{L^{q - 1} }} {{\tilde K^q }} - \theta _0 \end{equation} Following arguments similar to the ones that proved $u$ is $G$-invariant minimizing of $I_\alpha$, we conclude that $(u_\alpha)$ is bounded in $H_{1,G}^q ( T )$, thus we can take a subsequence of $(u_\alpha)$, denoted $(u_\alpha)$ too, such that $u_\alpha \rightharpoonup u$ in $H_{1,G}^q ( T )$, $u_\alpha \to u$ in $L_G^q ( T )$ and $u_\alpha \to u$ in $L_G^q ( {\partial T} )$. Moreover, by (\ref{eq4.20}) we obtain \[ \| u_\alpha \|_{L_G^q (\partial T)}^q < \frac{1} {\alpha }\big( {\frac{{L^{q - 1} }} {{\tilde K^q }} - \theta _0 }\big), \] and sending $\alpha$ to $+\infty$ we have $u = 0$ on $\partial T$. Finally following the proof of theorem $4$ of \cite{Bie} we obtain that $\nabla u_\alpha \to \nabla u$ a.e. and so $( {\nabla u_\alpha } )$ is bounded in $L_G^q ( T )$. Because of (\ref{eq2.1}), (\ref{eq2.2}) and (\ref{eq2.3}) and since $1 < q < 2,\,\tilde p = q/\left( {2 - q} \right)$ we have \begin{align*} &\frac{{\| {\nabla u_\alpha } \|_{L^q (T)}^q + \alpha \| {u_\alpha } \|_{L^q (\partial T)}^q }} {{\| {u_\alpha } \|_{L^{\tilde p} (\partial T)}^q }}\\ &= \frac{{\int_T {| {\nabla u_\alpha } |^q dV + \alpha \int_{\partial T} {| {u_\alpha } |^q dS} } }} {{( {\int_{\partial T} {| {u_\alpha } |^{\tilde p} dS} } )^{q/\tilde p} }} \\ &= \frac{{2\pi \left( {\frac{\delta } {\lambda }} \right)^{2 - q} \int_D {\left| {\nabla \phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dtds + 2\pi \frac{\delta } {\lambda }\alpha \int_{\partial D} {\left| {\phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } }} {{\left( {2\pi \frac{\delta } {\lambda }\int_{\partial D} {\left| {\phi _\alpha } \right|^{\tilde p} \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } \right)^{q/\tilde p} }} \\ &= \frac{{2\pi \left( {\frac{\delta } {\lambda }} \right)^{2 - q} \int_D {\left| {\nabla \phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dtds} }} {{\left( {2\pi \frac{\delta } {\lambda }\int_{\partial D} {\left| {\phi _\alpha } \right|^{\tilde p} \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } \right)^{q/\tilde p} }} + \frac{{2\pi \frac{\delta } {\lambda }\alpha \int_{\partial D} {\left| {\phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} }} {{\left( {2\pi \frac{\delta } {\lambda }\int_{\partial D} {\left| {\phi _\alpha } \right|^{\tilde p} \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } \right)^{q/\tilde p} }} \\ &= \left( {\frac{1} {\lambda }} \right)^{2 - q - \frac{q} {{\tilde p}}} \frac{{2\pi \delta ^{2-q}\int_D {\left| {\nabla \phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dtds} }} {{\left( {2\pi\delta \int_{\partial D} {\left| {\phi _\alpha } \right|^{\tilde p} \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } \right)^{q/\tilde p} }} \\ &+ \left( {\frac{1} {\lambda }} \right)^{1 - \frac{q} {{\tilde p}}} \frac{{2\pi \delta \alpha\int_{\partial D} {\left| {\phi _\alpha } \right|^q \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} }} {{\left( {2\pi \delta \int_{\partial D} {\left| {\phi _\alpha } \right|^{\tilde p} \left( {l - r + \delta \frac{t} {\lambda }} \right)dt} } \right)^{q/\tilde p} }} \\ &\leq \frac{{2\pi \delta ^{2 - q} \int_D {| {\nabla \phi _\alpha } |^q ( {l - r + \delta \frac{t} {\lambda }} )\,dt\,ds + 2\pi \delta \alpha \int_{\partial D} {| {\phi _\alpha } |^q ( {l - r + \delta \frac{t} {\lambda }} )dt } } }} {{( {2\pi \delta \int_{\partial D} {| {\phi _\alpha } |^{\tilde p} ( {l - r + \delta \frac{t} {\lambda }} )dt } } )^{q/\tilde p} }} \end{align*} and as $\lambda \to + \infty $ the above inequality yields \begin{align*} & \frac{{\| {\nabla u_\alpha } \|_{L^q (T)}^q + \alpha \| {u_\alpha } \|_{L^q (\partial T)}^q }} {{\| {u_\alpha } \|_{L^{\tilde p} (\partial T)}^q }} \\ &\leq \frac{{L\delta ^{2 - q} \int_D {| {\nabla \phi _\alpha } |^q \,dt\,ds + L\delta \alpha \int_{\partial D} {| {\phi _\alpha } |^q dt } } }} {{( {L\delta \int_{\partial D} {| {\phi _\alpha } |^{\tilde p} dt } } )^{q/\tilde p} }} \\ &= L^{q - 1} \frac{{\int_D {| {\nabla \phi _\alpha } |^q \,dt\,ds + \delta ^{1 - ( {q/\tilde p} )} \alpha \int_{\partial D} {| {\phi _\alpha } |^q dt } } }} {{( {\int_{\partial D} {| {\phi _\alpha } |^{\tilde p} dt } } )^{q/\tilde p} }} \\ &< L^{q - 1} \frac{{\int_D {| {\nabla \phi _\alpha } |^q \,dt\,ds + \alpha \int_{\partial D} {| {\phi _\alpha } |^q dt } } }} {{( {\int_{\partial D} {| {\phi _\alpha } |^{\tilde p} dt } } )^{q/\tilde p} }} \end{align*} From the above inequality and (\ref{eq4.20}) we obtain \[ L^{q - 1} \frac{{\int_D {| {\nabla \phi _\alpha } |^q \,dt\,ds + \alpha \int_{\partial D} {| {\phi _\alpha } |^q dt } } }} {{( {\int_{\partial D} {| {\phi _\alpha } |^{\tilde p} dt} } )^{q/\tilde p} }} < \frac{{L^{q - 1} }} {{\tilde K}} - \theta _0 \] or \begin {equation}\label{eq4.21} \frac{{\int_D {| {\nabla \phi _\alpha } |^q \,dt\,ds + \alpha \int_{\partial D} {| {\phi _\alpha } |^q dt} } }} {{( {\int_{\partial D} {| {\phi _\alpha } |^{\tilde p} dt} } )^{q/\tilde p} }} < \frac{1} {{\tilde K^q }} - \theta \end {equation} According to \cite[Theorem 4]{Bie} such a function satisfying inequality (\ref{eq4.21}) does not exist, and the theorem is proved. \end{proof} \subsection{Proof of the main theorem } \begin{proof}[Proof of Theorem \ref{T3.2}] The proof is based on ideas from \cite{Che}. We recall, in this point, some notation: \[ \Lambda=\{ c = ( {\alpha ,\beta } ) \in \mathbb{R}^2 : \alpha - \beta \geq \delta ,\;q \leq \alpha \leq p,\;q\leq \beta \leq \tilde{p} \}, \] where $\delta \in ({0,p - \tilde p} ) = \bigl( 0,q/(2 - q) \bigr)$, \begin{gather*} I(u) = \int_T {( {| {\nabla u} |^q + a| u |^q } )dV + \int_{\partial T} {b| u |^q dS}},\\ I_c (u) = \int_T {f| u |^\alpha dV + \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} } \end{gather*} where $u \in H_{1,G}^q $ and $c \in\Lambda $, \begin{gather*} \Sigma _c = \{ u \in H_{1,G}^q :\;I_c (u) = 1 \},\\ \mu _c = \inf \{ {I(u):u \in \Sigma _c } \},\\ c_0 = ( {p,\tilde p} ),\quad t^ + = \sup ( {t,0} ),\; t \in \mathbb{R}. \end{gather*} Because the imbeddings of $ H_{1,G}^q (T)$ in $L_G^p (T)$ and $L_G^{\tilde p} (\partial T)$ are continuous but not compact, we adopt the procedure of solving an approximating equation and then we pass to the limit as in \cite{Aub}. \noindent\textbf{1.} The proof of this part is carried out in six steps. \noindent\textbf{Step 1.} A real $t_c\in \Sigma_c$. Define on $[ {0, + \infty } )$ the continuous function $h_c $ with \[ h_c (t) = t^\alpha \int_T {fdV} + \frac{\alpha } {\beta }t^\beta \int_{\partial T} {gdS} \] Since $\alpha > \beta $ we have $h_c (0) = 0$, $\lim_{t \to \infty } h_c (t) = + \infty $ and there exists $t_c > 0$ such that $h(t_c ) = 1$. Hence the constant function, which in every point is equal to $t_c $, belongs to $\Sigma _c $, and then $\Sigma _c \ne \emptyset$. \noindent\textbf{Step 2.} $\sup \{ \mu _c : c \in \Lambda\} < + \infty $. We will prove that there exists $\tilde t \in \mathbb{R}$ such that $t_c \leqslant \tilde t$ for all $c \in \Lambda $ and the following holds \[ \mu _c \leqslant {\rm I}(t_c ) = \Big( {\int_T {adV + \int_{\partial T} {bdS} } } \Big)t_c^q \leqslant \Big( {\int_T {| a |dV + \int_{\partial T} {| b |dS} } } \Big)\tilde t^q \] $\bullet$ If $\int_{\partial T} {gdS \geqslant 0} $, since $f> 0$ and $\alpha > \beta $, by equality \[ 1 = I_c (t_c ) = t_c^\alpha \int_T {fdV + \frac{\alpha } {\beta }t_c^\beta \int_{\partial T} {gdS} } \] arises \[ t_c = \Big( {\int_T {fdV + \frac{\alpha } {\beta }t_c^{\beta - \alpha } \int_{\partial T} {gdS} } } \Big)^{ - 1/\alpha } \geqslant \sup \{ {\big( {\int_T {fdV} } \big)^{ - 1/\alpha } : q \leqslant \alpha \leqslant p} \} \] However, if $\int_T {fdV} < 1$, since $q \leqslant \alpha \leqslant p$, the following holds \[ \big( {\int_T {fdV} } \big)^{ - 1/q} \leqslant \big( {\int_T {fdV} } \big)^{ - 1/\alpha } < 1 \] while, if $\int_T {fdV} \geqslant 1$, we have the inequality \[ \big( {\int_T {fdV} } \big)^{ - 1/q} \geqslant \big( {\int_T {fdV} } \big)^{ - 1/\alpha } \geqslant 1 \] Therefore, in this case, we set \[ \tilde t = \max \{ {1,\,\big( {\int_T {fdV} } \big)^{ - 1/q} } \} \geqslant \sup \{ {\big( {\int_T {fdV} } \big)^{ - 1/\alpha } : q \leqslant \alpha \leqslant \frac{{2q}} {{2 - q}}} \} \] $\bullet$ If $\int_{\partial T} {gdS} < 0$, let $\tilde t_0 \in \mathbb{R}$ such that \[ \tilde t_0 \geqslant \max \big\{ {1,\,\big( {p\frac{{| {\int_{\partial T} {gdS} } |}} {{\int_T {fdV} }}} \big)^{1/\delta } } \big\} \] When $t \geqslant \tilde t_0 $, because of $q \leqslant \alpha \leqslant p$, $q \leqslant \beta \leqslant \tilde p$ and $\beta \leqslant \alpha - \delta $, we get $(\alpha /\beta ) \leqslant (p/q)$, and then \begin{align*} h_c (t) &= t^\alpha \int_T {fdV} + \frac{\alpha } {\beta }t^\beta \int_{\partial T} {gdS} \\ & \geqslant t^\alpha \int_T {fdV} + \frac{p} {q}t^{\alpha - \delta } \int_{\partial T} {gdS}\\ & = t^\alpha \Big( {1 + \frac{p} {q}\frac{{| {\int_{\partial T} {gdS} } |}} {{\int_T {fdV} }}t^{ - \delta } } \Big)\int_T {fdV} \\ & \geqslant t^q \Big( {1 + \frac{p} {q}\frac{{| {\int_{\partial T} {gdS} } |}} {{\int_T {fdV} }}\tilde t_0^{ - \delta } } \Big)\int_T {fdV} \\ & \geqslant \frac{{t^q }} {q}\int_T {fdV} \end{align*} Hence, for \[ \tilde t = \max \{ {q^{1/q} \big( {\int_T {fdV} } \big)^{- 1/q} ,\, \tilde t_0 } \} \] we have $h_c (\tilde t) \geqslant 1$ and then $t_c \leqslant \tilde t$. \noindent\textbf{Step 3.} $\inf \{ { {\mu _c } : c \in \Lambda }\} > - \infty$. Since $f>0$ everywhere, $m = \inf _T f > 0$ and because of $ H_{1,G}^q (T)\hookrightarrow L^{\tilde p}_G (\partial T)$ is continuous (see \cite[lemma 2.1]{Cot-Lab}) there exists $C \geqslant 1$ such that for all $ \psi \in H_{1,G}^q (T)$, \[ \| \psi \|_{\tilde p,\partial T} \leqslant C( {\| {\nabla \psi } \|_{q,T} + \| \psi \|_{q, T} } ) \] We set \begin{gather*} C_1 = \sup_{q \leqslant \alpha \leqslant p} \big[ {m^{ - 1/q} \big( {\int_T f dV} \big)^{( {1/q} ) - ( {1/\alpha } )} } \big], \\ C_2 = \sup_{q \leqslant \beta \leqslant \tilde p} \big[ {2^{\beta - 1}q^{-1} p\, C^\beta \| g \|_\infty [ {Vol(\partial T)} ]^{1 - ( {\beta /\tilde p} )} } \big], \\ C_3 = \sup_{q \leqslant \alpha \leqslant p} \big[ {2^{1/\alpha } C_1 ( {C_2 + 1} )^{1/\alpha } } \big]. \end{gather*} We recall, for reals $x,y\geq 0$, the elementary inequalities \begin{gather}\label{eq4.22} ( {x + y} )^\beta \leqslant 2^{\beta - 1} ({x^\beta + y^\beta } ), \\ \label{eq4.23} ( {x + y} )^{1/\alpha } \leqslant 2^{1/\alpha } ( {x^{1/\alpha } + y^{1/\alpha } } ). \end{gather} Let $u \in \Sigma _c $ with $\| u \|_{q,T} > 1$. Since $u \in \Sigma _c $ we have \[ \int_T {f| u |^\alpha dV + \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} } = 1 \] and then \[ \int_T {f| u |^\alpha dV = } 1 - \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} \] By (\ref{eq4.22}), (\ref{eq4.23}) and since $q \leqslant \alpha \leqslant p$, $q \leqslant \beta \leqslant \tilde p$ we obtain \begin{align*} \| u \|_{q,T} & \leqslant m^{ - 1/q} \Big( {\int_T {f| u |} ^q dV} \Big)^{1/q} \\ & \leq m^{ - 1/q} \Big( {\int_T f dV} \Big)^{({1/q} ) - ( {1/\alpha } )} \big( {\int_T {f| u |} ^\alpha dV} \Big)^{1/\alpha } \\ & \leqslant C_1 ( {1 - \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} } )^{1/\alpha } \\ & \leqslant C_1 ( {1 + \frac{p} {q}\| g \|_\infty [ {Vol(\partial T)} ]^{1 - ( {\beta /\tilde p} )} \| u \|_{\tilde p,\partial T}^\beta } )^{1/\alpha }\\ & \leqslant C_1 [ {1 + \frac{p} {q}\| g \|_\infty [ {Vol(\partial T)} ]^{1 - ( {\beta /\tilde p} )} C^\beta ( {\| {\nabla u} \|_{q,T} + \| u \|_{q,\partial T} } )^\beta } ]^{1/\alpha }\\ & \leqslant C_1 [ {1 + 2^{\beta - 1}q^{-1} p\| g \|_\infty [ {Vol(\partial T)} ]^ {1 - ( {\beta /\tilde p} )} C^\beta ( {\| {\nabla u} \|_{q,T}^\beta + \| u \|_{q, T} ^\beta } )} ]^{1/\alpha }\\ &\leqslant C_1 [ {C_2 \| {\nabla u} \|_{q,T}^\beta + ( {C_2 + 1} )\| u \|_ {q, T}^\beta } ]^{1/\alpha } \\ & \leqslant 2^{1/\alpha } C_1[ C_2^{1/\alpha } \| {\nabla u} \|_{q,T}^{\beta /\alpha } + ( {C_2 + 1} )^{1/\alpha } \| u \|_{q, T}^{\beta /\alpha } ] \\ & \leqslant C_3 ( {\| {\nabla u} \|_{q,T}^{\beta /\alpha } + \| u \|_{q, T}^{\beta /\alpha } } ). \end{align*} Since $$ \frac{\beta}{\alpha} \leqslant 1 - \frac{\delta}{\alpha} \leqslant 1 - \frac{\delta}{p} , $$ if $\varepsilon \in ( {0,1} )$ and $C_4 = C_4 ({\varepsilon ,p,\delta ,C_3 })$ is a constant such that $$ t^{1 - (\delta/p)} \leqslant \frac{\varepsilon}{C_3 }t + C_4, $$ for any $t \geqslant 0$, the latter inequality implies \begin{align*} \| u \|_{q,T} & \leqslant C_3 ( {1 + \| {\nabla u} \|_{q,T}^{1 - ( {\delta /p} )} + \| u \|_{q,T}^{1 - ( {\delta /p} )} } ) \\ & \leqslant \varepsilon ( {\| {\nabla u} \|_{q,T} + \| u \|_{q,T} } ) + C_3 ( {1 + 2C_4 } ) \end{align*} Hence if $\varepsilon _1 = \varepsilon /( {1 - \varepsilon } )$ and $C = C_3 ( {1 + 2C_4 } )/( {1 - \varepsilon } )$ is a constant depending on $\varepsilon_1$, but not on $c \in \Lambda $, by the last inequality we obtain \begin{equation}\label{eq4.24} \| u\|_{q,T} \leqslant \varepsilon _1 \| {\nabla u} \|_{q,T} + C \end{equation} Since we can take $C \geqslant 1$, (\ref{eq4.24}) holds for $c \in \Lambda $ and for all $u \in \Sigma _c $. If $b\not \equiv 0$, since $H_{1,G}^q (T)\hookrightarrow L^q_G (\partial T)$ and $H_{1,G}^q (T)\hookrightarrow L^q_G (T)$ are compact, for any $\varepsilon ' \in ( {0,1} )$ we can find a constant $C' = C'( {\varepsilon ' ,b} )$ such that \[ \| \psi \|_{q,\partial T}^q \leqslant \| b \|_\infty ^{ - 1} ( {\varepsilon '\| {\nabla \psi } \|_{q,T}^q + C'\| \psi \|_{q,T}^q } ) \] for all $ \psi \in H_{1,G}^q (T)$, and then we obtain \begin{equation}\label{eq4.25} \begin{aligned} I(\psi) &= \int_T {( {| {\nabla \psi} |^q + a| \psi |^q } )dV + \int_{\partial T} {b| \psi |^q dS} } \\ & \ge \| {\nabla \psi} \|_{q,T}^q - \| a \|_{\infty ,T} \| \psi \|_{q,T}^q - \| b \|_{\infty ,T} \| \psi \|_{q,\partial T}^q \\ & \ge ( {1 - \varepsilon '} )\| {\nabla \psi} \|_{q,T}^q - \| a \|_{\infty ,T} \| \psi \|_{q,T}^q - C'\| \psi \|_{q,T}^q \\ & = ( {1 - \varepsilon '} )\| {\nabla \psi} \|_{q,T}^q - A( {\varepsilon '} )\| \psi \|_{q,T}^q \end{aligned} \end{equation} where $A(\varepsilon ') = \| a \|_{\infty ,T} + C'$. An inequality of the type (\ref{eq4.25}) is always true if $b \equiv 0$, because of inequality \[ I(\psi) \geqslant \| {\nabla \psi } \|_{q, T}^q - \| a \|_{\infty ,T} \| \psi \|_{q,T}^q\,. \] By (\ref{eq4.24}) because of (\ref{eq4.22}) we obtain \begin{equation}\label{eq4.26} \| u \|_{q,T}^q \le 2^{q - 1} \varepsilon _1^q \| {\nabla u} \|_{q,T}^q + 2^{q - 1} C^q \end{equation} Thus if we choose $\varepsilon _1 $, small enough, such that $1 -\varepsilon ' - 2^{q-1}A(\varepsilon ') \varepsilon _1^q > 0$, for all $u \in \Sigma_c$, by (\ref{eq4.25}) and (\ref{eq4.26}) we obtain \begin{align*} I(u) &\geqslant \bigl( 1 - \varepsilon ' - 2^{q-1}A(\varepsilon ') \varepsilon _1^q \bigr)\| {\nabla u} \|_{q, T}^q - 2^{q-1}A(\varepsilon ') C^q \\ &\geqslant - 2^{q-1}A(\varepsilon ') C^q \end{align*} Hence for all $c \in \Lambda$, $\mu\geq - 2^{q-1}A(\varepsilon ')C^q$. We observe that if $\Gamma$ is a subset of $ \cup _{c \in \Lambda } \Sigma _c $, such that $\sup \{ {I(u):u \in \Gamma } \} = L < + \infty $, $\Gamma$ is bounded in $H^q_{1,G}(T)$. Indeed, according to the former, for any $u\in \Gamma$ we have \[ \| {\nabla u}\|_{q,T}^q \leqslant \frac{{I(u) + 2^{q-1}A (\varepsilon ')C^q }} {1 - \varepsilon ' - 2^{q-1}A(\varepsilon ') \varepsilon _1^q } \leqslant \frac{{L + 2^{q-1}A(\varepsilon ') C^q }} {{1 - \varepsilon ' - 2^{q-1}A(\varepsilon ')\varepsilon _1^q }} = c \] and \[ \| u \|_{q,T} \leqslant \varepsilon _1 \| {\nabla u} \|_{q,T} + C \leqslant c ', \] thus \[ \mathop {\sup _{u \in \Gamma } ( {\| {\nabla u} \|_{q,T} + \| u \|_{q,T} } ) < + \infty}. \] \noindent\textbf{Step 4.} $\mu _c = \inf \{ {\,{\rm I}(u):u \in \Sigma _c\,} \}$ is attained. Suppose that $c = ( {\alpha ,\beta } ) \in \Lambda ,\,\,\alpha < p$ , $\beta < \tilde p$. Let a sequence $( {u_j } )\in \Sigma_c$ such that $\lim_{j \to \infty } I(u_j ) = \mu _c $. Since $| {\gamma u_j } | = \gamma ( {| {u_j } |} )$, ($\gamma u_j$ is the trace of $u_j$ on $\partial T$), a.e. on $\partial T$ and $I( {| {u_j } |} ) = I( {u_j } )$, $I_c ( {| {u_j } |} ) = I_c ( {u_j } )$ a.e. in $T$ , we conclude that $| {u_j } | \in \Sigma _c $; in a way similar to the one that employs $( {| {u_j } |} )$ in place of $( {u_j } )$, ($\gamma ( {| {u_j } |} )$ in place of $\gamma u_j $ respectively) or we can consider $u_j $'s nonnegative a.e. (the same for $\gamma u_j $'s). The sequence $( {I( {u_j } )} )$ is bounded in $H_{1,G}^q (T)$ implies that $\sup _{j \in \mathbb{N}} ( {\| {u_j } \|_{H_1^q (T)} } ) < + \infty $. Since the imbeddings of $H_{1,G}^q (T)$ in $L^q_G (T)$, $L^q_G (\partial T)$, $L^\alpha_G (T)$ and $L^\beta_G (\partial T)$ are compacts, there is a function $u_c \in H_{1,G}^q (T)$ and a subsequence $( {u_j } )$ of $( {u_j } )$ such that $( {u_j } )\rightharpoonup u_c $ in $H_{1,G}^q (T)$, $( {u_j } )\to u_c $ in everyone of the previous $L^r$ spaces, $( {u_j } )\to u_c $ a.e. in $T$. (The same holds and for traces on $\partial T$). Hence $u_c $ and $\gamma u_c $ are nonnegative. We also have $I_c ( {u_c } ) = \lim_{j \to \infty } I_c ( {u_j } ) = 1$ and then $u_c \in \Sigma _c $ and $I_c ( {u_c } ) \geqslant \mu _c $. Moreover since \[ \|\nabla {u_c } \|_q \leqslant \mathop {\underline {\lim } }_{j \to + \infty } \| {\nabla u_j } \|_q \] and \[ \int_T {au_c^q dV + \int_{\partial T} {bu_c^q dS} } = \lim_{j \to \infty } \Big(\int_T {au_j^q dV + \int_{\partial T} {bu_j^q dS} }\Big) \] holds, we conclude that $I( {u_c } ) \leqslant \lim _{j \to \infty } I( {u_j } ) = \mu _c $ and $I( {u_c } ) = \mu _c $. \noindent\textbf{Step 5.} There exists a week solution $u_{c_0}\geq 0$. We observe that the deferential $DI_c (u)$ of $I_c $ is $\not =0$ for all $u \in \Sigma _c $. (Because if $DI_c (u) = 0$ for all $\psi \in C_0^\infty (T)$ then $\int_T {fu} | u |^{\alpha - 2} \psi dV = 0$. This implies that $fu| u |^{\alpha - 2} \psi = 0$ in $( {C_0^\infty ( T )} )^\prime $ and since $f>0$, $u=0$. Then $u=0$ in $H_{1,G}^q (T)$, which is impossible since $I_c (u) = 1$). After this a Lagrange multiplicand $\lambda _c $ exists such that, for all $ \psi \in H_{1,G}^q (T)$, it satisfies the next Euler equation \begin{equation}\label{eq4.27} \begin{aligned} &\int_T {( {| {\nabla u_c } |^{q - 2} \nabla u_c \nabla \psi + au_c^{q - 1} \psi } )dV + \int_{\partial T} b u_c^{q - 1} } \psi dS \\ &=\lambda _c \Big( {\int_T {fu_c^{\alpha - 1} \psi dV} + \int_{\partial T} {gu_c^{\beta - 1} \psi dS} } \Big) \end{aligned} \end{equation} In the following we suppose that $c = ( {\alpha ,\beta }) \to c_0 = ( {p,\tilde p} )$ and we will prove that there are a real $\lambda _{c_0 } $ and a function $u_{c_0 }$ such that \begin{align*} &\int_T {( {| {\nabla u_{c_0 } } |^{q - 2} \nabla u_{c_0 } \nabla \psi + au_{c_0 }^{q - 1} \psi } )dV + \int_{\partial T} b u_{c_0 }^{q - 1} } \psi dS \\ &=\lambda _{c_0 } ( {\int_T {fu_{c_0 }^{p - 1} \psi dV} + \int_{\partial T} {gu_{c_0 }^{\tilde p - 1} \psi dS} } ) \end{align*} that is $u_{c_0 } $ is a week solution of \eqref{eP}. Substituting $\psi = u_c $ in (\ref{eq4.27}) we obtain \[ \int_T {( {| {\nabla u_c } |^q + au_c^q } )dV + \int_{\partial T} b u_c^q } dS = \lambda _c \Big( {\int_T {fu_c^\alpha dV} + \int_{\partial T} {gu_c^\beta dS} } \Big) \] or \begin{align*} \lambda _c \Big( {\int_T {fu_c^\alpha dV} + \int_{\partial T} {gu_c^\beta dS} } \Big) = I( {u_c } ) = \mu _c \end{align*} Moreover, we have \begin{align*} 1 = I_c ( {u_c } ) &= \int_T {fu_c^\alpha dV + \frac{\alpha } {\beta }\int_{\partial T} {gu_c^\beta dS} } \\ & = \frac{\alpha } {\beta }( {\int_T {fu_c^\alpha dV + \int_{\partial T} {gu_c^\beta dS} } } ) + ( {1 - \frac{\alpha } {\beta }} )\int_T {fu_c^\alpha dV} \end{align*} and since $$ ( {1 - \frac{\alpha } {\beta }})\int_T {fu_c^\alpha dV} < 0, $$ we obtain \[ \int_T {fu_c^\alpha dV + \int_{\partial T} {gu_c^\beta dS} } > \frac{\beta } {\alpha } \geqslant \frac{q} {p} > 0 \] Hence $\lambda_c$ and $\mu _c $ have the same sign and since the set $\{ {\mu _c } \}_{c\, \in \Lambda } $ is bounded a constant $C$ exists, such that \[ | {\lambda _c } | \leqslant \frac{p} {q}| {\mu _c } | \leqslant C \] Since $\sup \{ {I( {u_c } ): c\, \in \Lambda}\} < \infty$, we have (step 2) $$ \sup \{ {\| {u_c } \| _{H_1^q }: c\, \in \Lambda } \} < \infty. $$ Moreover, because the embeddings of $H_{1,G}^q (T)$ in $L^p_G(T)$ and $L^{\tilde p}_G (\partial T)$ are continuous, we have \begin{gather*} \sup \{ {\| {u_c } \| _{p, T} : c\, \in \Lambda} \} < \infty, \\ \sup \{ {\| {u_c } \| _{{\tilde p}, \partial T} :\,c\, \in \Lambda } \} < \infty \end{gather*} We observe that $\frac{{\beta - 1}} {{\tilde p - 1}} < 1$ and then \begin{align*} \| {u_c^{\beta - 1} } \|_{\tilde p/( {\tilde p - 1} ),\partial T} & \leqslant [ {Vol( {\partial T} )} ]^{1 - ( {\beta /\tilde p} )} \| {u_c } \|_{\tilde p,\partial T}^{\beta - 1} \\ & \leqslant [ {\max ( {1,[ {Vol( {\partial T} )} ]^{1 - ( {\beta /\tilde p} )} } )}] [ {\max ( {1,\| {u_c } \|_{\tilde p, \partial T}^{\tilde p - 1} } )} ] \leqslant ct \end{align*} that is, the sets $\{ {u_c^{\alpha - 1} }\}$, (respectively $\{ {u_c^{\beta - 1} } \}$) are bounded in the Banach reflexive spaces $L^{p/( {p - 1} )} (T)$, (respectively $L^{\tilde p/( {\tilde p - 1})} (\partial T)$), of which the dual $L^p (T)$ (respectively $L^{\tilde p} (\partial T)$) contain $ H_{1,G}^q (T)$. The above implies the existence of a sequence $c_j = ( {\alpha _j ,\beta _j } ) \in \Lambda $, which converges to $c_0$, of a real $\lambda _{c_0 }$ which is the limit of $( {\lambda _{c_j } } )$ and of a function $u_{c_0 } $ with the following properties: \begin{itemize} \item[(a)] $ u_{c_j } \rightharpoonup u_{c_0 }$ on $ H_1^q (T)$, (by Banach's theorem), \item[(b)] $u_{c_j } \to u_{c_0 }$ on $ L^q (T)$ (resp. $L^q(\partial T)$, (by Kondrakov's theorem), \item[(c)] $u_{c_j } \to u_{c_0 }$ a.e. in $T$ , (resp. $\partial T$) (by proposition 3.43 of \cite{Aub}) and \item[(d)] $ u_{c_j }^{\alpha _j - 1} \rightharpoonup u_{c_0 }^{p - 1}$ in $L^{p/( {p - 1} )} (T)$ (resp. $ u_{c_j }^{\beta _j - 1} \rightharpoonup u_{c_0 }^{\tilde p - 1} $ in $L^{\tilde p/( {\tilde p - 1} )} (\partial T)$) (by Banach theorem). \end{itemize} From (a) arises that \[ \int_T {| {\nabla u_{c_j } } |^{q - 2} \nabla u_{c_j } \nabla } \psi dV \to \int_T {| {\nabla u_{c_0 } } |^{q - 2} \nabla u_{c_0 } \nabla } \psi dV \] for all $ \psi \in H_1^q (T)$.\\ From $(c)$ arises that $u_{c_0 } $ is $G$-invariant and $u_{c_0 } \geqslant 0$ on $T$ (resp. $\gamma u_{c_0 } \geqslant 0$ in $\partial T$). We may also assume that the sequence $( {\mu _{c_j } })$ converges, with limit $\mu _0 \leqslant \mu _{c_0 } $. Indeed, let $u \in \Sigma _{c_0 } $ be a non-negative function such that $\mu _{c_0 } \leqslant I(u) \leqslant \mu _{c_0 } + \varepsilon $, with $\varepsilon > 0$. Then for all $j$, there exists a unique real number $t_j > 0$ such that $I_{c_j } ({t_j u} ) = 1$ and $\lim_{j \to \infty }t_j = 1$. If this is not the case, there will exist a subsequence $t_{j_k } $ with $\lim_{j \to \infty } t_{j_k } = l \ne 1$. But in $T$ the following holds, \[ 0 \leqslant fu^{\alpha _{j_k } } \leqslant f( {1 + u^p } ) \in L^1 (T) \] and on $\partial T$ the following also holds, \[ 0 \leqslant | g |u^{\alpha _{j_k } } \leqslant | g |( {1 + u^{\tilde p} } ) \in L^1 (\partial T). \] According to the dominated convergence theorem we have \[ I_{c_{j_k } } ( {t_{j_k } u} ) \to l^p \int_T {fu^p } dV + \frac{p} {{\tilde p}}l^{\tilde p} \int_{\partial T} {gu^{\tilde p} } dS \] namely, $I_{c_0 } ( {lu} ) = 1$. This is contradiction since $l \ne 1$ and $I_{c_0 } ( u ) = 1$, whereas we know that there exists unique real number $r > 0$ such that $I_{c_0 } ( {ru} ) = 1$. Since $\mu _{c_j } \leqslant I(t_j u) = t_j^q I(u)$ we conclude that $\lim \sup _{j \to + \infty } \mu _{c_j } \leqslant \mu _{c_0 } $. Now we can write equation (\ref{eq4.24}) for $u_{c_j } $, and as $j \to + \infty $ by Lebesgue's theorem, we find that for all $ \psi \in H_{1,G}^q (T)$ the following holds \begin{align*} &\int_T {( {| {\nabla u_{c_0 } } |^{q - 2} \nabla u_{c_0 } \nabla \psi + au_{c_0 }^{q - 1} \psi } )dV + \int_{\partial T} b u_{c_0 }^{q - 1} } \psi dS\\ &= \lambda _{c_0 } \Big( {\int_T {fu_{c_0 }^{p - 1} \psi dV} + \int_{\partial T} {gu_{c_0 }^{\tilde{p} - 1} \psi dS} }\Big) \end{align*} which implies that $( {\lambda _{c_0 } ,u_{c_0 } } )$ is a week solution of the problem \eqref{eP}. \noindent\textbf{Step 6.} $u_{c_0 } >0$ everywhere. We proved in step 5 that $u_{c_0 }\geq 0 $. By the maximum principle \cite{Vaz}, the function $u_{c_0 } $ is identically equal to $0$ or $u_{c_0 }>0 $ everywhere in $T$, and finally in $\bar T$: since every point $P$, where $u_{c_0 } $ attains it's minimum in $\bar T$, belongs to $\partial T$, assume that $u_{c_0 }$ is regular and that there exists $P_0 \in \partial T$ such that $u_{c_0 }(P_0)=0$. By Hopf's lemma \cite{Vaz}, we have that the normal derivative has strict sign, $\bigl(\partial u_{c_0 } / {\partial \nu}\bigr)( P_0 ) < 0$, but the boundary condition imposes $$ {| {\nabla u_{c_0 }} |^{q - 2} } \frac{{\partial u_{c_0 } }} {{\partial \nu}}( P_0 ) = ( { - bu_{c_0 }^{q-1} + \lambda _{c_0 } gu_{c_0 }^{\tilde p - 1} } )( P_0 ) = 0, $$ a contradiction which proves that $u_{c_0 } ( P )> 0$ in $\bar{T}$. For the solution $u_{c_0 } $ to be strictly positive it suffices $u_{c_0 } \not \equiv 0$, which implies that $\| {\,u_{c_0 } } \|_{q,T} = \lim_{j\to \infty } \| {\,u_{c_j } } \|_{q,T} > 0$. By \cite[theorem 3.1]{Cot-Lab} and theorem \ref{T3.1} of this paper, we conclude that for any $\mathcal{K} > K /\sqrt {L/2} $ and for any $\tilde {\mathcal{K}} > \tilde { K } / L ^{( {q - 1} )/q} $ there exists a constant $C( {\mathcal{K},\tilde {\mathcal{K}}} )$ such that for all $\psi \in H_{1,G}^q (T)$ the following inequalities hold \begin{gather}\label{eq4.28} \| \psi \|_{p,T}^q \leqslant \mathcal{K}^q \| {\nabla \psi } \|_{q,T}^q + C\| \psi \|_{q,T}^q, \\ \label{eq4.29} \| \psi \|_{\tilde p,\partial T}^q \leqslant \tilde {\mathcal{K}}^q \| {\nabla \psi } \|_{q,T}^q + C\| \psi \|_{q,T}^q \end{gather} By (\ref{eq4.25}) we have \begin{equation}\label{eq4.30} \| {\nabla u_c } \|_{q,T}^q \leqslant ( {1 + \varepsilon } )I(u_c ) + A(\varepsilon )\| {u_c } \|_{q,T}^q \end{equation} where $1 + \varepsilon = 1/(1 - \varepsilon ')$, $A(\varepsilon ) = A(\varepsilon ')/(1 - \varepsilon ')$. Let $\varepsilon >0$ and $D( {\varepsilon ,p} )$ be a constant such that for any $x,y \geqslant 0$ and $\rho \in [{0,p} ]$ the following holds \begin{equation}\label{eq4.31} ( {x + y} )^{\rho /q} \leqslant ( {1 + \varepsilon } )x^{\rho /q} + Dy^{\rho /q} \end{equation} By (\ref{eq4.28}) because of (\ref{eq4.30}), (\ref{eq4.31}) and since $I(u_c ) = \mu _c $ for $\alpha < p$, $\beta < \tilde p$ we can write \begin{equation}\label{eq4.32} \begin{aligned} \| {u_c } \|_{p,T}^\alpha & \leqslant ( {\mathcal{K}^q \| {\nabla u_c } \|_{q,T}^q + C\| {u_c } \|_{q,T}^q } )^{\alpha /q} \\ & \leqslant [ {( {1 + \varepsilon } )^{q/p} \mathcal{K}^q \mu _c^ + + ( {A\mathcal{K}^q + C} )\| {u_c } \|_{q,T}^q } ]^{\alpha /q} \\ & \leqslant ( {1 + \varepsilon } )^q \mathcal{K}^\alpha ( {\mu _c^ + } )^{\alpha /q} + D( {A\mathcal{K}^q + C} )^{\alpha /q} \| {u_c } \|_{q,T}^\alpha \\ \end{aligned} \end{equation} Similarly by (\ref{eq4.29}) because of (\ref{eq4.30}), (\ref{eq4.31}) we can write, \begin{equation}\label{eq4.33} \| {u_c } \|_{\tilde p,\partial T}^\beta \leqslant ( {1 + \varepsilon } )^q \tilde {\mathcal{K}}^\beta ( {\mu _c^ + } )^{\beta /q} + D( {A\tilde {\mathcal{K}}^q + C} )^{\beta /q} \| {u_c } \|_{q,T}^\beta \end{equation} So by (\ref{eq4.32}), (\ref{eq4.33}) and H\"{o}lder inequality we have \begin{equation}\label{eq4.34} \begin{aligned} 1 = I_c (u_c ) & = \int_T {fu_c^\alpha } dV + \frac{\alpha } {\beta }\int_{\partial T} {gu_c^\beta } dS \\ & \leqslant ( {\mathop {\sup f}_T } )[ {Vol( T )} ]^{1 - ( {a/p} )} \| {u_c } \|_{p,T}^\alpha \\ &\quad + \frac{\alpha } {\beta }( {\mathop {\sup g}_{\partial T} } )^+ [ {Vol( {\partial T} )} ]^ {1 - ( {\beta /\tilde p} )} \| {u_c } \|_{\tilde p,\partial T}^\beta \\ & \leqslant ( {1 + \varepsilon } )^q [ {( {\mathop {\sup f}_T } )[ {Vol( T )} ]^{1 - ( {a/p} )} \mathcal{K}^\alpha ( {\mu _c^ + } )^{\alpha /q} } ] \\ &\quad + ( {1 + \varepsilon } )^q [ {\frac{\alpha } {\beta }( {\mathop {\sup g}_{\partial T} } )^+[ {Vol( {\partial T} )} ]^{1 - ( {\beta /\tilde p} )} \tilde {\mathcal{K}}^\beta ( {\mu _c^ + } )^{\beta /q} } ] \\ &\quad + C_1 \| {u_c } \|_{q,T}^\alpha + C_2 \| {u_c } \|_{q,T}^\beta \end{aligned} \end{equation} where the constants \begin{gather*} C_1 = D( {A\mathcal{K}^q + C} )^{\alpha /q} ( {\mathop {\sup f}_T } )[ {Vol( T )} ]^{1 - ( {a/p} )}, \\ C_2 = ( {\alpha /\beta } )D( {A\tilde {\mathcal{K}}^q + C} )^{\beta /q} ( {\mathop {\sup g}_{\partial T} } )^+[ {Vol( {\partial T} )} ]^{1 - ( {\beta /\tilde p} )} \end{gather*} are bounded by a constant $\tilde C(\varepsilon,\mathcal{K},\tilde {\mathcal{K}})> 0$ independent of $\alpha$ and $\beta$. By (\ref{eq4.34}) for $c = c_j $, since $\lim_{j \to \infty } \mu _{c_j } \leqslant \mu _{c_0 } $ for $j \to + \infty $, we obtain \begin{align*} 1 & \leqslant ( {1 + \varepsilon } )^q \big[ {( {\mathop {\sup f}_T } )\mathcal{K}^p ( {\mu _{c_0 }^ + } )^{p/q} + \frac{p} {{\tilde p}}( {\mathop {\sup g}_{\partial T} } )^+\tilde {\mathcal{K}}^{\tilde p} ( {\mu _{c_0 }^ + } )^{\tilde p/q} } \big] \\ &\quad + \tilde{C}( \| {u_{c_0} } \|_{q,T}^p + \| {u_{c_0} } \|_{q, T}^{\tilde p}) \end{align*} because the sequence $( {u_{c_j } } )\to u_{c_0 } $ strongly into $L^q (T)$. Thus if the condition (\ref{eq3.2}) of the theorem is satisfied and if we choose $\varepsilon>0$ small enough and $\mathcal{K},\tilde {\mathcal{K}}$ close enough to $K/\sqrt {L/2}$, $\tilde K / L^{( q - 1) / q} $, respectively, we obtain $\| {u_{c_0 } } \|_{q,T}> 0$ and hence we proved the first part of the theorem. \noindent\textbf{2.} If $f$ becomes $0$, $f \geqslant 0$, we impose the condition $g > 0$, namely $\nu = \inf_{\partial T} g > 0$, the proof follows along similar lines as in case 1. Thus we have to find two positive constants $A_1 ,A_2 > 0$ such that if $u \in \cup _{c\, \in \Lambda } \Sigma _c $ the following will hold, \[ I(u) \geqslant A_1 \| {\nabla u} \|_q^q - A_2 \] Since $g > 0$ we have $\sup_{u \in \cup _{c\in \Lambda } \Sigma _c } \| u \|_{q,\partial T} = D < + \infty $, because for such a $u$ the following holds \[ 1 = I_c (u) \geqslant \int_{\partial T} {gu^\beta dS} \geqslant \nu\| u \|_{\beta ,\partial T}^\beta \geqslant \nu[ {Vol(\partial T)} ]^{1 - ( {\beta /q} )} \| u \|_{q,\partial T}^\beta \] Let $\kappa$ be the best constant of the inequality (\ref{eq3.3}), namely of \[ \| \psi \|_{q,T}^q \leqslant A\| {\nabla \psi } \|_{q,T}^q + B\| \psi \|_{q,\partial T}^q ,\psi \in H_{1,G}^q \] If $\breve{a}\kappa < 1$ where $\breve{a}= ( { - \inf a} )^ + $ we choose $A$ close to $\kappa$ such that $A _1 = 1 - A\breve{a} > 0$ and then we have \begin{align*} I(u) &\ge \int_T {( {| {\nabla u} |^q - \breve{a} u^q } )dV + \int_{\partial T} {bu^q } } dS \\ & \ge \| {\nabla u} \|_q^q -\breve{a} \| u \|_{q,T}^q - \| b \|_\infty \| u \|_{q,\partial T}^q \\ & \ge \| {\nabla u} \|_q^q - \breve{a} A\| {\nabla u} \|_q^q - \breve{a} B\| u \|_{q,\partial T}^q - \| b \|_\infty \| u \|_{q,\partial T}^q \\ & = ( {1-\breve{a} A} )\| {\nabla u} \|_q^q - ( {\breve{a} B + \| b \|_\infty } )\| u \|_{q,\partial T}^q \\ & \ge A_1 \| {\nabla u} \|_q^q - A_2 \, \end{align*} where $A_1=1-\breve{a}A$, $A_2 = ({\breve{a}B + \| b \|_\infty } )D^q $ and the theorem is proved. \end{proof} \subsection*{Acknowledgments} The authors would like to thank the anonymous referee for the valuable suggestions that improved this article. \begin{thebibliography}{00} \bibitem{Aub} Th. 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