\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\} $$an open small solid torus (a tubular neighborhood of the orbit O_{P_j}). \subsection{Best constants on the solid Torus} \begin{theorem}\label{T3.1} Let \overline{T} be the solid torus and \tilde{p}, q be two positive real numbers such that \tilde{p}=q/(2-q) with 10 there exists a real number B_\varepsilon such that for all u\in H^q_{1,G} the following inequality holds: $$\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)}$$ In addition the constant \tilde{K}^q (2,q)/{[2\pi(l-r)]^{q-1}} is the best constant for the above inequality. \end{theorem} \subsection{Resolution of the problem} Consider the set $\Lambda=\{ c = ( {\alpha ,\beta } ) \in \mathbb{R}^2 :\alpha - \beta \geq \delta ,\;q \leq \alpha \leq p,\;q\leq \beta \leq \tilde{p} \},$ with \delta \in ({0,p - \tilde p} ) = \bigl( 0,q/(2 - q) \bigr), and define the functionals \begin{gather*} I(u) = \int_T {( {| {\nabla u} |^q + a| u |^q } )dV + \int_{\partial T} {b| u |^q dS}} \\ I_c (u) = \int_T {f| u |^\alpha dV + \frac{\alpha } {\beta }\int_{\partial T} {g| u |^\beta dS} } \end{gather*} for all u \in H_{1,G}^q  and for any c \in\Lambda . I(u) and I_c (u) are well defined because the imbeddings of H^q_{1,G} onto L^p and L^{\widetilde{p}} are continue according to the Sobolev theorem. Define, also \begin{gather*} \Sigma _c = \{ {u \in H_{1,G}^q : I_c (u) = 1}\},\\ \mu _c = \inf \{ {I(u):u \in \Sigma _c } \},\\ c_0 = ( {p,\tilde p} ), \quad t^ + = \sup ( {t,0} ), \end{gather*} for all t \in \mathbb{R}. Consequently for the problem \eqref{eP} we have the following theorem. \begin{theorem}\label{T3.2} Let a, f, b and g be four smooth functions, G-invariant and q, p, \tilde{p} be three real numbers defined as in \eqref{eP}. Suppose that the function f has constant sign (e.g. f\geq0). The problem \eqref{eP} has a positive solution u\in H_{1,G}^q if the following holds: $$\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$$ and if \begin{enumerate} \item f>0 everywhere and g is arbitrary, or \item f\geq0, g>0 everywhere and (-inf_{{T}}a)^+\kappa<1, where $$\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)}\}.$$ \end{enumerate} \end{theorem} In the rest of this paper we denote K=K(2,q), \tilde{K}=\tilde{K}(2,q) and L=2\pi(l-r). \section{Proofs} \begin{proof}[Proof of Theorem \ref{T3.1}] The proof is carried out in two steps. \noindent\textbf{Step 1.} Suppose that there exist two real numbers A, B such that for all u\in H^q_{1,G} the following inequality holds: $\|u\|^q_{L^{\tilde{p}}(\partial T)}\leq A {\|\nabla u\|^q_{L^q(T)}}+B\|u\|^q_{L^q(\partial T)}.$ Then $A\geq \frac{\tilde{K}^q(2,q)}{|2\pi(l-r)|^{q-1}}$ Consider a transformation F:D \to \mathbb{R}_ + ^2 . Such a transformation, for example, is $F( {t,s} ) = ( {\frac{{4t}} {{t^2 + ( {1 + s} )^2 }},\frac{{2( {1 - t^2 - s^2 } )}} {{t^2 + ( {1 + s} )^2 }}} ),$ see \cite {Esc}. Denote by ({\tilde g_{ij}}) the Euclidian metric on D, \,dx\,dy the Euclidian metric on \mathbb{R}_ + ^2  and d\sigma  the induced on \partial \mathbb{R}_ + ^2 . Choose a finite covering of \bar D consisting of disks D_k, centered on p_k, such that: If p_k \in D , then entire D_k lies in D and if p_k\in \partial D, D_k is a Fermi neighborhood. In these neighborhoods we have $$\label{eq4.3} 1 - \varepsilon _0 \leqslant \sqrt {\det ( {\tilde g_{ij} } )} \leqslant 1 + \varepsilon _0$$ Suppose by contradiction that, there exists $A < \frac{\tilde K^q } {L^{q - 1}}\quad \text{and} \quad B\in\mathbb{R}$ such that the inequality $$\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$$ holds for all  u \in H_{1,G}^q (T). Fix a point P_0 \in \partial T, that belongs to the orbit of minimum range l-r. For any \varepsilon_0>0, we can choose  \delta = \varepsilon_0 (l - r) < 1 and$$ 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*} $$\label{eq4.13} \Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant f( \varepsilon_0 ) L^{q - 1} A\int_{\mathbb{R}_ + ^2 } {| {\nabla \Phi } |} ^q \,dx\,dy + \tilde B\int_{\partial \mathbb{R}_ + ^2 } {| \Phi |} ^q d\sigma,$$ 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} }} $$is monotonically increasing, we can choose \varepsilon_0 small enough, such that the following inequality holds $A < f(\varepsilon_0)A <\frac{\tilde K^q}{L^{q - 1}}$ hence A'< \tilde K^q where  A' = f(\varepsilon_0)L^{q - 1} A. So for \varepsilon_0 small enough and for all \Phi \in C_0^\infty ( D ) we have $$\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$$ On the other hand by H\"{o}lder's inequality, for all \Phi \in C_0^\infty ( D_\delta ), where D_\delta\subset D , we have $\int_{\partial D_\delta } {| \Phi |} ^q d\sigma _0 \leqslant [ {Vol( {\partial D_\delta } )} ]^{1 - ( {q/\tilde p} )} \Big( {\int_{\partial D_\delta } {( {| \Phi |^q } )} ^{\tilde p/q} d\sigma _0 } \Big)^{q/\tilde p}$ and since \tilde{p}=q/(2-q), that is 1 - (q/\tilde p) = 1 - ({2 - q} ) = q - 1, we have $$\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}$$ Hence, choosing \varepsilon_0 small enough, by (\ref{eq4.14}) and (\ref{eq4.15}), we get that there exists A''< \tilde K^q, such that for all \Phi \in C_0^\infty ( D_\delta ), $$\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\,.$$ Let \Psi \in C_0^\infty ( {\mathbb{R}_ + ^2 } ) and set \Psi _\lambda (x) = \lambda ^{1/\tilde p} \Psi (\lambda x),\,\lambda > 0. For \lambda>0, sufficiently large, \Psi_\lambda \in C_0^\infty ( D ) and since \| {\Psi _\lambda } \|_{L^{\tilde p} ( {\partial \mathbb{R}_ + ^2 } )} = \|\Psi \|_{L^{\tilde p} ( {\partial \mathbb{R}_ + ^2 } )} and \| {\nabla \Psi _\lambda } \|_{L^q ( { \mathbb{R}_ + ^2 } )} = \| {\nabla \Psi } \|_{L^q ( { \mathbb{R}_ + ^2 } )} , by (\ref{eq4.16}), the following inequality $\Big( {\int_{\partial \mathbb{R}_ + ^2 } {| \Psi |} ^{\tilde p} d\sigma } \Big)^{q/\tilde p} \leqslant A''\int_{ \mathbb{R}_ + ^2 } {|\nabla\Psi |} ^q \,dx\,dy$ holds for all \Psi \in C_0^\infty ( {\mathbb{R}_ + ^2 } ). This is a contradiction since \tilde K is the best constant for the Sobolev inequality in \mathbb{R}_ + ^2 . \noindent\textbf{Step 2.} For all \varepsilon>0 there exists a real number B_\varepsilon such that for all u\in H^q_{1,G} the following inequality holds:$$ \|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 $$\label{eq4.18} \| {\nabla u_k } \|_{L_G^q (T)}^q + \alpha \| {u_k } \|_{L_G^q (\partial T)}^q \to I_\alpha\,,$$ 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 $$\label{eq4.19} \|u\|^q_{L^q(T)}\leq C(\|\nabla u\|^q_{L^q(T)}+\|u\|^q_{L^q(\partial T)}).$$ 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 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 $$\label{eq4.24} \| u\|_{q,T} \leqslant \varepsilon _1 \| {\nabla u} \|_{q,T} + C$$ 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 \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} 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 $$\label{eq4.26} \| u \|_{q,T}^q \le 2^{q - 1} \varepsilon _1^q \| {\nabla u} \|_{q,T}^q + 2^{q - 1} C^q$$ 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 \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} 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 $$\label{eq4.30} \| {\nabla u_c } \|_{q,T}^q \leqslant ( {1 + \varepsilon } )I(u_c ) + A(\varepsilon )\| {u_c } \|_{q,T}^q$$ 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 $$\label{eq4.31} ( {x + y} )^{\rho /q} \leqslant ( {1 + \varepsilon } )x^{\rho /q} + Dy^{\rho /q}$$ 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 \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} Similarly by (\ref{eq4.29}) because of (\ref{eq4.30}), (\ref{eq4.31}) we can write, $$\label{eq4.33} \| {u_c } \|_{\tilde p,\partial T}^\beta \leqslant ( {1 + \varepsilon } )^q \tilde {\mathcal{K}}^\beta ( {\mu _c^ + } )^{\beta /q} + D( {A\tilde {\mathcal{K}}^q + C} )^{\beta /q} \| {u_c } \|_{q,T}^\beta$$ So by (\ref{eq4.32}), (\ref{eq4.33}) and H\"{o}lder inequality we have \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} 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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