In the following, we suppose that \(E = W_{0}^{1, q}(\mathcal{M})\setminus \lbrace 0\rbrace \) is endowed with \(\|u\|_{E} = ( \int _{\mathcal{M}} | \nabla u |^{q} \,dv _{\mathfrak{g}} (z) )^{\frac{1}{q}}\).
Definition 4
We say that a function \(u\in E\) is a weak solution to problem (1), if
$$\begin{aligned}& \int _{\mathcal{M}} \bigl( \vert \nabla u \vert ^{p-2} \nabla u + \mu (z) \vert \nabla u \vert ^{q-2}\nabla u \bigr)\nabla \varphi \,dv _{ \mathfrak{g}} (z) + \int _{\mathcal{M}}V(z) \vert u \vert ^{p-2}u \varphi \,dv _{\mathfrak{g}} (z) \\& \quad = \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r-2}u\log \bigl( \vert u \vert \bigr)\varphi \,dv _{\mathfrak{g}} (z), \end{aligned}$$
for all \(\varphi \in D( \mathcal{M})\).
Consider the functional \(J_{\lambda}: E\rightarrow \mathbb{R}\) defined by
$$\begin{aligned} J_{\lambda}(u)={}&\frac{1}{p} \int _{\mathcal{M}} \vert \nabla u \vert ^{p}\,dv _{ \mathfrak{g}}(z)+ \frac{1}{q} \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}}(z)+ \frac{1}{p} \int _{ \mathcal{M}}V(z) \vert u \vert ^{p} \,dv _{\mathfrak{g}}(z) \\ &{} -\frac{\lambda}{r} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}}(z)+\frac{\lambda}{r^{2}} \int _{ \mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \end{aligned}$$
for all \(u\in E\).
Then \(J_{\lambda}\) is well defined and belongs to \(C^{1}(E)\). Furthermore, we have
$$\begin{aligned} \bigl\langle J'_{\lambda}(u), \varphi \bigr\rangle ={}& \int _{\mathcal{M}} \bigl( \vert \nabla u \vert ^{p-2} \nabla u+\mu (z) \vert \nabla u \vert ^{q-2} \nabla u \bigr) \nabla \varphi \,dv _{\mathfrak{g}} (z) \\ &{} + \int _{\mathcal{M}}V(z) \vert u \vert ^{p-2} u \varphi \,dv _{ \mathfrak{g}} (z) -\lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r-2} u \log \bigl( \vert u \vert \bigr) \varphi \,dv _{\mathfrak{g}} (z) \end{aligned}$$
for all \(u, \varphi \in E\).
Consider the Nehari set defined by
$$ \mathcal{N}= \bigl\{ u\in E: \bigl\langle J'_{\lambda}(u), u \bigr\rangle =0 \bigr\} .$$
We can deduce that the critical points of \(J_{\lambda}\) lie on \(\mathcal{N}\) and further that \(u\in \mathcal{N}\) if and only if u is a weak solution to problem (1). Let us define the maps \(\psi _{u}: \mathbb{R}^{+}\rightarrow \mathbb{R}\) by \(\psi _{u}(t)=J_{\lambda}(tu)\) and analyze \(\mathcal{N}\) in terms of the stationary points of fibering maps \(\psi _{u}\).
We have
$$\begin{aligned} \psi '_{u}(t) = {}& t^{p-1} \int _{\mathcal{M}} \vert \nabla u \vert ^{p} \,dv _{\mathfrak{g}}(z) + t^{q-1} \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z)+ t^{p-1} \int _{\mathcal{M}}V(z) \vert u \vert ^{p} \,dv _{\mathfrak{g}} (z) \\ &{} -\lambda t^{r-1} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) -\lambda t^{r-1} \log (t) \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{ \mathfrak{g}} (z) \end{aligned}$$
and
$$\begin{aligned} \psi ''_{u}(t) ={} & (p-1) t^{p-2} \int _{\mathcal{M}} \vert \nabla u \vert ^{p} \,dv _{\mathfrak{g}} (z) + (q-1) t^{q-2} \int _{ \mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) \\ &{} + (p-1) t^{p-2} \int _{\mathcal{M}}V(z) \vert u \vert ^{p} \,dv _{ \mathfrak{g}} (z) -\lambda (r-1) t^{r-2} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) \\ &{} -\lambda (r-1) t^{r-2} \log (t) \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) -\lambda t^{r-2} \int _{ \mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z). \end{aligned}$$
It is easy to verify that \(tu\in \mathcal{N} \Longleftrightarrow \psi ^{\prime }_{u}(t)=0\) for any \(u\in E\) and \(t>0\).
We shall split \(\mathcal{N} \) into three subsets which correspond to local minima, local maxima, and points of inflection of fibering maps, that is,
$$\begin{aligned}& \begin{aligned} \mathcal{N}^{+}&=\bigl\lbrace u\in \mathcal{N}:\psi ''_{u}(1)>0 \bigr\rbrace \\ &= \biggl\lbrace u\in E: (q-p) \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) +\lambda (p-r) \int _{ \mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{ \mathfrak{g}} (z) \\ &\quad {}> \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \biggr\rbrace , \end{aligned} \\& \begin{aligned}\mathcal{N}^{0}&=\bigl\lbrace u\in \mathcal{N}:\psi ''_{u}(1) =0 \bigr\rbrace \\ &= \biggl\lbrace u\in E: (q-p) \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) +\lambda (p-r) \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) \\ &\quad {}= \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{ \mathfrak{g}} (z) \biggr\rbrace , \end{aligned} \\& \begin{aligned}\mathcal{N}^{-}&=\bigl\lbrace u\in \mathcal{N}:\psi ^{\prime \prime }_{u}(1) < 0 \bigr\rbrace \\ &= \biggl\lbrace u\in E: (q-p) \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) + \lambda (p-r) \int _{ \mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{ \mathfrak{g}} (z) \\ &\quad {}< \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \biggr\rbrace . \end{aligned} \end{aligned}$$
Lemma 4
Let \(u_{0}\notin \mathcal{N}^{0}\). Then \(u_{0}\) is a critical point of \(J_{\lambda}\) if \(u_{0}\) is a local minimizer of \(J_{\lambda}\) on \(\mathcal{N}\).
Proof
We remark that \(u_{0}\) is a solution to the optimization problem to minimize \(J_{\lambda}\) subject to \(I(u)=0\), where
$$\begin{aligned} I(u) =& \int _{\mathcal{M}} \bigl( \vert \nabla u \vert ^{p} + \mu (z) \vert \nabla u \vert ^{q} \bigr) \,dv _{\mathfrak{g}} (z) + \int _{ \mathcal{M}}V(z) \vert u \vert ^{p} \,dv _{\mathfrak{g}} (z) \\ &{}- \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z), \end{aligned}$$
and, since \(u_{0}\) is a local minimizer of \(J_{\lambda}\) on \(\mathcal{N}\), we have
$$ I(u_{0})=\bigl\langle J'_{\lambda}(u_{0}),u_{0} \bigr\rangle . $$
(2)
Then, there exists a Lagrange multiplier \(\alpha \in \mathbb{R}\) such that \(J'_{\lambda}(u_{0})=\alpha I'(u_{0})\), namely \(0=\langle J'_{\lambda}(u_{0}), u_{0}\rangle =\alpha \langle I'(u_{0}), u_{0}\rangle \).
Furthermore, \(\langle I'(u_{0}), u_{0}\rangle \neq 0\) since \(u_{0}\notin \mathcal{N}^{0}\) which implies \(\alpha =0\) and, actually, that \(u_{0}\) is a critical point of \(J_{\lambda}\). □
Lemma 5
There exists a positive constant \(\lambda _{0}\) such that, for any \(0<\lambda <\lambda _{0}\), the functional \(J_{\lambda}\) is bounded and coercive on \(\mathcal{N}\).
Proof
Letting \(u\in E\) with \(\Vert u\Vert _{E} > 1 \), we obtain
$$\begin{aligned} J_{\lambda}(u)\geq {}& \frac{1}{q} \biggl( \int _{\mathcal{M}} \vert \nabla u \vert ^{p} \,dv _{\mathfrak{g}} (z) + \int _{\mathcal{M}} \mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) + V_{0} \int _{\mathcal{M}} \vert u \vert ^{p} \,dv _{\mathfrak{g}} (z) \biggr) \\ &{} -\frac{\lambda}{r} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) + \frac{\lambda}{r^{2}} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z), \end{aligned}$$
and we know that
$$ \log (s)\leq \frac{s^{\alpha}}{\alpha e} \quad \text{for all } \alpha >0 \text{ and } s>0, $$
(3)
thus
$$ J_{\lambda}(u)\geq \frac{\mu _{0}}{q} \Vert u \Vert _{E}^{q}- \frac{\lambda \Vert a \Vert _{\infty}}{r(q-r) e} \int _{\mathcal{M}} \vert u \vert ^{q} \,dv _{\mathfrak{g}} (z)$$
with \(\alpha =q-r\).
According to Theorem 2 and Poincaré inequality, there exists a positive constant \(C_{q}\) such that
$$ J_{\lambda}(u) \geq \frac{\mu _{0}}{q} \Vert u \Vert _{E}^{q}- \frac{\lambda \Vert a \Vert _{\infty}}{r(q-r) e}C_{q} \Vert u \Vert _{E}^{q} \geq \biggl( \frac{\mu _{0}}{q}- \frac{\lambda \Vert a \Vert _{\infty}}{r(q-r) e}C_{q} \biggr) \Vert u \Vert _{E}^{q}. $$
Choosing \(0<\lambda <\lambda _{0}= \frac{r(q-r) e}{qC_{q}\Vert a\Vert _{\infty}}\) implies that \(J_{\lambda}\) is coercive.
Moreover, we have
$$\begin{aligned} J_{\lambda}(u)& \leq \frac{1}{p} \biggl(\varrho (u)+ \int _{\mathcal{M}}V(z) \vert u \vert ^{p} \,dv _{\mathfrak{g}} (z) -\lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) \biggr) \\ &\quad {}+ \frac{\lambda}{r^{2}} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \\ & =\frac{\lambda}{r^{2}} \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \leq \frac{\lambda}{r^{2}} \Vert a \Vert _{ \infty} \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z). \end{aligned}$$
Thanks to Theorem 2, there exists \(C_{r} > 0\) such that
$$ J_{\lambda}(u)\leq C_{r} \frac{\lambda}{r^{2}} \Vert a \Vert _{ \infty} \Vert u \Vert _{E}^{r}. $$
□
Lemma 6
Let \(\lambda _{1} = \frac{(q-p) \mu _{0} }{\Vert a\Vert _{L^{\infty}}C_{1}}\) where \(C_{1} (r, q, \mathcal{M} )\) is a constant to be specified later. Then, for any λ such that \(0<\lambda <\lambda _{1}\), we have \(\mathcal{N}^{0}\cup \mathcal{N}^{-}= \emptyset \) and \(\mathcal{N}^{+}\neq \emptyset \).
Proof
We proceed by contradiction to prove that \(\mathcal{N}^{0}\cup \mathcal{N}^{-}\neq \emptyset \).
Indeed, let \(u\in \mathcal{N}^{0}\cup \mathcal{N}^{-}\), then we get
$$\begin{aligned}& (q-p) \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{ \mathfrak{g}} (z) +\lambda (p-r) \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \log \bigl( \vert u \vert \bigr) \,dv _{\mathfrak{g}} (z) \\& \quad \leq \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{ \mathfrak{g}} (z), \end{aligned}$$
thus
$$ (q-p) \int _{\mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{ \mathfrak{g}} (z) \leq \lambda \int _{\mathcal{M}} a(z) \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \leq \lambda \Vert a \Vert _{L^{ \infty}} \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z).$$
Using the fact that \(L^{q}(\mathcal{\mathcal{M}})\subset L^{r}(\mathcal{\mathcal{M}})\) and Poincaré inequality, there exists a positive constant \(C_{1} (r, q, \mathcal{M} )\) such that
$$ \int _{\mathcal{M}} \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) \geq C_{1} \biggl( \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{ \mathfrak{g}} (z) \biggr)^{\frac{q}{r}},$$
hence
$$\begin{aligned} (q-p) \mu _{0} C_{1} \biggl( \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \biggr)^{\frac{q}{r}}&\leq (q-p) \int _{ \mathcal{M}}\mu (z) \vert \nabla u \vert ^{q} \,dv _{\mathfrak{g}} (z) \\ &\leq \lambda \Vert a \Vert _{L^{\infty}} \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z), \end{aligned}$$
thus
$$ \biggl( \int _{\mathcal{M}} \vert u \vert ^{r} \,dv _{\mathfrak{g}} (z) \biggr)^{\frac{q}{r}-1}\leq \lambda \frac{ \Vert a \Vert _{L^{\infty}} }{(q-p) \mu _{0} C_{1}}, $$
and, when \(\lambda \rightarrow 0\), we have \(u=0\), which is a contradiction.
Now, according to Lemma 5, the set \(\mathcal{N}^{+}\neq \emptyset \). □