In our arguments, the proof of Theorem 2 contains the existence result which is stated as in Theorem 1. Therefore, for the sake of brevity, we will deal only with problem (6) in the case where f is not necessarily identical to zero.
An axially symmetric function \(u(y,z)=v( \vert y \vert ,z)\) solves problem (6) if, and only if, \(v:=v(r,z)\) (with \(r= \vert y \vert \)) solves
$$ \textstyle\begin{cases} \operatorname{div} (r^{map} \vert \nabla v \vert ^{p2}\nabla v )= r^{mbq} \vert v \vert ^{q2}v + r^{m} f & \mbox{in } S, \\ v= 0 & \mbox{on } \partial S, \end{cases} $$
(7)
where \(S:=(A,B)\times \mathbb {R}^{Nm1}\) and \(\partial S:= \{ A,B \} \times \mathbb {R}^{Nm1}\).
We denote by \(W_{0}^{1,p}(S)\) the subspace of axially symmetric functions of \(W_{0}^{1,p}(\Omega)\) with the norm defined by \(\Vert v \Vert = ( \int_{S} \vert \nabla v \vert ^{p} \,dx )^{\frac{1}{p}}\). This norm \(\Vert \cdot \Vert \) is equivalent to the standard norm on \(W_{0}^{1,p}(S)\) (see [24], pp.158159).
If \(G:=O(Nm1)\) is the group of isometries of \(\mathbb {R}^{Nm1}\), then
$$\operatorname{Fix}(G)=W_{0,G}^{1,p}(S)= \bigl\{ v\in W_{0}^{1,p}(S) : v(r,gz)=v(r,z), \forall g\in G \bigr\} $$
and
$$L^{q}(S)^{G}= \bigl\{ v\in L^{q}(S) : v(r,gz)=v(r,z), \forall g \in G \bigr\} $$
are the subspaces of invariant functions.
Since \(A< r< B\), then the norms on \(W_{0,G}^{1,p}(S)\) and \(L^{q}(S)^{G}\) given by
$$ \Vert v \Vert _{m,a,p}:=\biggl( \int_{S} r^{map} \vert \nabla v \vert ^{p}\, dx \biggr)^{\frac{1}{p}}\quad \mbox{and}\quad \vert v \vert _{m,b,q}:=\biggl( \int_{S} r^{mbq} \vert v \vert ^{q} \,dx \biggr)^{\frac{1}{q}} $$
(8)
are equivalent to the standard norms on \(W_{0}^{1,p}(S)\) and \(L^{q}(S)\), respectively.
Denote \(X:=W_{0}^{1,p}(S)\) and \(E:=W_{0,G}^{1,p}(S)\). Let \(I\colon X\to \mathbb {R}\) be the energy functional associated to problem (7) and defined by
$$I(v)= \frac{1}{p} \int_{S} r^{map} \vert \nabla v \vert ^{p}  \frac{1}{q} \int_{S} r^{mbq} \vert v \vert ^{q}  \int_{S} r^{m} f v, $$
where
$$I'(v) (\varphi)= \int_{S} r^{map} \vert \nabla v \vert ^{p2}\nabla v\nabla\varphi \int_{S} r^{mbq} \vert v \vert ^{q2}v \varphi  \int_{S} r^{m}f\varphi \quad \forall \varphi\in X . $$
Applying the principle of symmetric criticality (Lemma 1), we can look for critical points of the functional I constrained to E, which are weak solutions to problem (7).
Using Maz’ja’s inequality for the parameters in problem (7), for \(Nm>p\geq1\), \(q:=\frac{Np}{Np (a+1b)}\), \(a \frac{m}{Nm} \leq b\leq a+1\) and \(a < \frac{(m+1)p}{p}\), we get the existence of a positive constant C such that
$$\biggl( \int_{S} r^{mbq} \vert v \vert ^{q}\, dx \biggr)^{\frac{1}{q}} \leq C \biggl( \int_{S} r^{map} \vert \nabla v \vert ^{p}\, dx \biggr)^{\frac{1}{p}} $$
for every \(v\in X\). Therefore, the functional I is well defined for these parameters and the functions in the intervals and spaces previously mentioned.
In [19, 20], we have an important result of compactness which ensures that the embedding \(W_{0,G}^{1,p}(S)\hookrightarrow L^{q}(S)\) is compact for \(1\leq m < Np\) and \(q\in(p,p_{N,m}^{*})\), where \(p_{N,m}^{*}:=\frac{p(Nm)}{Nmp}\). So, \(W_{0,G}^{1,p}(S)\) can be compactly embedded into \(L^{q}(S)^{G}\) for the norms defined in (8).
Note that when \(b=a+1\) and \(b= a \frac{m}{Nm}\), we have \(q=p\) and \(q=p_{N,m}^{*}\), respectively. Hence, we will consider \(a \frac{m}{Nm} < b < a+1\), so that the compactness result and Maz’ja’s inequality are both satisfied.
The following lemma shows that the functional I verifies the geometry conditions of the mountain pass theorem.
Lemma 3
Let
\(I\vert_{E}\)
be the energy functional associated to problem (7); then

(i)
there are
\(\overline{\varepsilon},\rho, \alpha>0\)
such that
\(I \vert_{\partial B_{\rho}}\geq\alpha\), since
\(0< \Vert f \Vert _{E^{1}}<\overline{\varepsilon}\);

(ii)
there is
\(e\in E\setminus\overline{B_{\rho}}\)
such that
\(I(e)<0\).
Proof
(i) For any \(\varepsilon>0\), we deduce that
$$ \begin{aligned} \biggl\vert \int_{S} r^{m}f v \biggr\vert &\leq C_{1} \biggl\vert \int_{S} f v \biggr\vert \leq \bigl(\varepsilon^{\frac{1}{p}} \Vert v \Vert _{m,a,p} \bigr)\cdot \biggl(\frac{C_{1}}{\varepsilon ^{\frac{1}{p}}} \Vert f \Vert _{E^{1}} \biggr) \\ &\leq \frac{\varepsilon}{p} \Vert v \Vert _{m,a,p}^{p} + \frac{C_{2}}{p' \varepsilon^{\frac{p'}{p}}} \Vert f \Vert _{E^{1}}^{p'} \end{aligned} $$
for all \(v\in E\), where \(\frac{1}{p} + \frac{1}{p'}=1\). Therefore,
$$ \begin{aligned}[b] I(v)&= \frac{1}{p} \Vert v \Vert _{m,a,p}^{p} \frac {1}{q} \vert v \vert _{m,b,q}^{q}  \int_{S} r^{m}f v \\ &\geq \frac{1}{p} \Vert v \Vert _{m,a,p}^{p}  \frac {1}{qS_{q}^{q}} \Vert v \Vert _{m,a,p}^{q} \frac{\varepsilon}{p} \Vert v \Vert _{m,a,p}^{p}  \frac{C_{2}}{p' \varepsilon^{\frac{p'}{p}}} \Vert f \Vert _{E^{1}}^{p'} \\ &= \biggl(\frac{1\varepsilon}{p}\frac{1}{qS_{q}^{q}} \Vert v \Vert _{m,a,p}^{qp} \biggr) \Vert v \Vert _{m,a,p}^{p}  \frac{C_{2}}{p' \varepsilon^{\frac{p'}{p}}} \Vert f \Vert _{E^{1}}^{p'}, \end{aligned} $$
(9)
where \(S_{q}\) is the best constant in the embedding \(W_{0,G}^{1,p}(S)\hookrightarrow L^{q}(S)^{G}\).
By fixing \(\varepsilon\in(0,1)\), we can find \(\rho>0\), with \(\Vert v \Vert _{m,a,p}=\rho\), \(\overline{\varepsilon}>0\) and \(\alpha >0\), such that the conclusion of the lemma holds true. For example, we can take
$$\rho= \bigl(MqS_{q}^{q} \bigr)^{\frac{1}{qp}}, \qquad \overline{ \varepsilon}= \frac{C_{2}^{\frac{1}{p'}} p^{\prime\frac{1}{p'}} \varepsilon^{\frac{1}{p}}}{2} M^{\frac{q}{p'(qp)}} \bigl( qS_{q}^{q} \bigr)^{\frac{p}{p'(qp)}},\qquad \alpha=\frac{1}{2} M^{\frac{q}{qp}} \bigl(qS_{q}^{q} \bigr)^{\frac{p}{qp}}, $$
where \(M=\frac{1}{2} (\frac{1\varepsilon}{p} )>0\).
(ii) Let \(v\in E\) such that \(\Vert v \Vert _{a,m,p}=1\). Then, for any \(t>1\), we have
$$I(tv)=\frac{1}{p} t^{p} \frac{1}{q} \vert v \vert _{m,b,q}^{q} t^{q}  t \int_{S} r^{m}fv. $$
Since \(q>p>1\), then we have \(\lim_{t\to\infty} I(tv)= \infty\). So, there is \(e\in E\setminus\overline{B_{\rho}}\) such that \(I(e)<0\). □
Lemma 4
The functional
I
satisfies the PalaisSmale condition in
E.
Proof
Let \(\{v_{n} \}\) be a PalaisSmale sequence for the functional I in E, i.e.,

1.
\(\vert I(v_{n}) \vert \leq M\) for some \(M>0\) and

2.
\(I'(v_{n})\to0\) in \(E^{1}\), where \(E^{1}\) is the dual space of E.
First we will show that \(\{v_{n} \}\) is bounded in E. Assume by contradiction that
$$ \Vert v_{n} \Vert _{m,a,p}\to\infty \quad \mbox{as } n\to\infty. $$
(10)
Given \(\varepsilon>0\), by items 1 and 2 we deduce that
$$ \biggl\vert I(v_{n})\frac{1}{q} I'(v_{n})v_{n} \biggr\vert \leq M + \frac{\varepsilon}{q} \Vert v_{n} \Vert _{m,a,p} $$
(11)
for \(n\in\mathbb{N}\) large enough.
Moreover, we also have
$$ \biggl\vert I(v_{n})\frac{1}{q} I'(v_{n}) (v_{n}) \biggr\vert \geq \biggl( \frac{1}{p} \frac{1}{q} \biggr) \Vert v_{n} \Vert _{m,a,p}^{p} C_{1} \biggl(1\frac{1}{q} \biggr) \Vert f \Vert _{E^{1}} \Vert v_{n} \Vert _{m,a,p} $$
(12)
for all \(n\in\mathbb{N}\).
Hence, for n large enough, we have
$$\biggl(\frac{1}{p}\frac{1}{q} \biggr) \Vert v_{n} \Vert _{m,a,p}^{p} \leq M + \biggl(\frac{\varepsilon}{q} + C_{1} \biggl(1\frac {1}{q} \biggr) \Vert f \Vert _{E^{1}} \biggr) \Vert v_{n} \Vert _{m,a,p}. $$
Letting \(n\to\infty\) in the previous inequality, we obtain a contradiction since \(1< p< q\). This implies that \(\{v_{n} \}\) is bounded in E.
Now we will prove that \(\{v_{n} \}\) is a Cauchy sequence in E. In [25] it is proved that the inequality
$$ \vert \xi\eta \vert ^{p} \leq \textstyle\begin{cases} ( \vert \xi \vert ^{p2} \xi \vert \eta \vert ^{p2}\eta ) (\xi\eta ), & \mbox{if } p\geq2; \\ ( \vert \xi \vert ^{p2} \xi \vert \eta \vert ^{p2}\eta ) (\xi\eta )^{\frac{p}{2}} ( \vert \xi \vert ^{p} + \vert \eta \vert ^{p} )^{\frac{2p}{2}}, & \mbox{if } 1< p< 2, \end{cases} $$
(13)
holds for all \(\xi, \eta\in \mathbb {R}^{N}\). Hence,
$$ \begin{aligned} \int_{S} r^{map} \vert \nabla v_{i} \nabla v_{j} \vert ^{p} &\leq \int_{S} r^{map} \bigl( \vert \nabla v_{i} \vert ^{p2}\nabla v_{i}  \vert \nabla v_{j} \vert ^{p2}\nabla v_{j} \bigr) (\nabla v_{i}\nabla v_{j} ) \\ &\leq \bigl\vert I'(v_{i}) (v_{i}v_{j}) \bigr\vert + \bigl\vert I'(v_{j}) (v_{i}v_{j}) \bigr\vert \\ &\quad {} + \biggl\vert \int_{S} r^{mbq} \bigl( \vert v_{i} \vert ^{q2}v_{i}  \vert v_{j} \vert ^{q2}v_{j} \bigr) (v_{i}v_{j} ) \biggr\vert \\ &:=I_{1} + I_{2}+ I_{3}. \end{aligned} $$
Since \(\{v_{n} \}\) is a PalaisSmale sequence, it follows that \(I_{1}=o ( \Vert v_{n} \Vert _{m,a,p} )\) and \(I_{2}=o ( \Vert v_{n} \Vert _{m,a,p} )\).
Using Hölder’s inequality, we have
$$ \begin{aligned} \biggl\vert \int_{S} r^{mbq} \bigl( \vert v_{i} \vert ^{q2}v_{i}  \vert v_{j} \vert ^{q2}v_{j} \bigr) (v_{i}v_{j} ) \biggr\vert &\leq \int_{S} r^{mbq} \bigl( \vert v_{i} \vert ^{q1} + \vert v_{j} \vert ^{q1} \bigr) \vert v_{i}v_{j} \vert \\ &= \int_{S} \bigl(r^{\frac{mbq}{q}} \vert v_{i} \vert \bigr)^{q1} r^{\frac{mbq}{q}} \vert v_{i}v_{j} \vert \\ &\quad {} + \int_{S} \bigl(r^{\frac{mbq}{q}} \vert v_{j} \vert \bigr)^{q1} r^{\frac{mbq}{q}} \vert v_{i}v_{j} \vert \\ &\leq \bigl( \vert v_{i} \vert _{m,b,q}^{q1} + \vert v_{j} \vert _{m,b,q}^{q1} \bigr) \vert v_{i} v_{j} \vert _{m,b,q}. \end{aligned} $$
It follows from the previous inequality and from the compact embedding \(W_{0,G}^{1,p}(S)\hookrightarrow L^{q}(S)^{G}\) that \(I_{3}= o ( \Vert v_{n} \Vert _{a,m,p} )\). Therefore, \(\{v_{n} \}\) is a Cauchy sequence and the functional I satisfies the PalaisSmale condition. □
Lemma 5
The functional
I
is weakly lower semicontinuous in
E, i.e., if
\(\{v_{n} \}\)
converges weakly to
v
in
E, then
\(I(v)\leq\liminf I(v_{n})\).
Proof
Let a sequence \(\{v_{n} \}\subset E\) be weakly convergent to v in E. Since the norm \(\Vert \cdot \Vert _{m,a,p}\) is weakly lower semicontinuous in E, it follows that
$$ \frac{1}{p} \int_{S} r^{map} \vert \nabla v \vert ^{p} \leq \liminf_{n\to\infty} \biggl(\frac{1}{p} \int_{S} r^{map} \vert \nabla v_{n} \vert ^{p} \biggr). $$
(14)
We can conclude from the compact embedding \(W_{0,G}^{1,p}(S)\hookrightarrow L^{q}(S)^{G}\) that \(\{v_{n} \}\) converges strongly to v in \(L^{q}(S)^{G}\). Therefore,
$$ \lim_{n\to\infty}\frac{1}{q} \int_{S} r^{mbq} \vert v_{n} \vert ^{q} = \frac{1}{q} \int_{S} r^{mbq} \vert v \vert ^{p}. $$
(15)
By hypothesis, the sequence \(\{v_{n} \}\) converges weakly to v in E and \(f\in W_{0}^{1,p}(\Omega)\subset E^{1}\); hence,
$$ \lim_{n\to\infty} \int_{S} r^{m} fv_{n} = \int_{S} r^{m} f v. $$
(16)
Finally, combining (14), (15) and (16), we deduce that
$$ \begin{gathered} \liminf_{n\to\infty} I(v_{n}) \\ \quad = \liminf_{n\to\infty} \biggl[ \frac{1}{p} \int_{S} r^{map} \vert \nabla v_{n} \vert ^{p}  \biggl( \lim_{n\to\infty}\frac{1}{q} \int_{S} r^{mbq} \vert \nabla v_{n} \vert ^{q} + \lim_{n\to\infty} \int_{S} r^{m} fv_{n} \biggr) \biggr] \\ \quad \geq I(v); \end{gathered} $$
therefore, the functional I is weakly lower semicontinuous in E. □
Proof of Theorem 2
By Lemmas 3 and 4, all the assumptions of the mountain pass theorem in [22] are satisfied. Hence, we deduce the existence of \(v_{1}^{*} \in W_{0,G}^{1,p}(S)\) which is a weak solution to problem (7) and \(I(v_{1}^{*})=\overline{c}>0\).
Now, we will prove that there is a second weak solution \(v_{2}^{*} \in W_{0,G}^{1,p}(S)\) such that \(v_{1}^{*}\neq v_{2}^{*}\). For \(\rho>0\) given as in Lemma 3, we define the number \(\underline{c}\) by
$$\underline{c}:=\inf_{ \{v\in E: \Vert v \Vert _{m,a,p}\leq\rho \}} I(v). $$
It is clear that \(\underline{c}\leq I(0)=0\). If \(\underline{c}=I(0)\), then 0 is a minimum value for I; hence,
$$0= I'(0) (\varphi) =  \int_{S} r^{m} f \varphi, \quad \forall\varphi\in E, $$
which contradicts the fact that \(f\neq0\). Therefore, \(\underline{c}< I(0)=0\).
Denote by \(\overline{B}_{\rho}\) the closed ball of radius ρ centered at the origin in E, i.e.,
$$\overline{B}_{\rho}= \bigl\{ v\in E : \Vert v \Vert _{m,a,p} \leq\rho \bigr\} . $$
It follows that the set \(\overline{B}_{\rho}\) is a complete metric space with respect to distance defined by \(d(u,v):=\Vert uv \Vert _{m,a,p}\) for all \(u, v\in\overline{B}_{\rho}\).
By Lemma 5, the functional I is weakly lower semicontinuous and bounded from below by relation (9).
Let ε such that \(0<\varepsilon< \inf_{\partial B_{\rho}} I  \inf_{B_{\rho}} I\). Using Ekeland’s variational principle [23] for the functional \(I: \overline{B}_{\rho}\rightarrow \mathbb {R}\), there is a function \(v_{\varepsilon}\in\overline{B}_{\rho}\) such that
$$I(v_{\varepsilon})< \inf_{\overline{B}_{\rho}} I + \varepsilon, I(v_{\varepsilon}) < I(v) + \varepsilon \Vert vv_{\varepsilon} \Vert _{m,a,p},\quad v\neq v_{\varepsilon}. $$
Since
$$I(v_{\varepsilon})\leq \inf_{\overline{B}_{\rho}} I + \varepsilon\leq \inf _{B_{\rho}} I + \varepsilon< \inf_{\partial B_{\rho}} I, $$
it follows that \(v_{\epsilon}\in B_{\rho}\).
We now define the functional \(K: \overline{B_{\rho}}\rightarrow \mathbb {R}\) by \(K(v)=I(v) + \varepsilon \Vert vv_{\varepsilon} \Vert _{m,a,p}\). It is immediate that \(v_{\varepsilon}\) is a minimum point of K, and so
$$ \frac{K(v_{\varepsilon}+ t\varphi)  K(v_{\varepsilon})}{t} \geq0 $$
(17)
for \(t>0\) small enough and \(\varphi\in B_{\rho}\). From inequality (17) we deduce that
$$ \frac{I(v_{\varepsilon}+ t\varphi)  I(v_{\varepsilon})}{t} + \varepsilon \Vert \varphi \Vert _{m,a,p} \geq0. $$
(18)
It follows from (18) by letting \(t\to0^{+}\) that \(I'(v_{\varepsilon})\varphi+\varepsilon \Vert \varphi \Vert _{m,a,p} \geq 0\). Note that −φ also belongs to \(B_{\rho}\). So, replacing φ by −φ, we have
$$I'(v_{\varepsilon}) (\varphi) + \varepsilon \Vert {}\varphi \Vert _{m,a,p} \geq0 $$
or simply \(I'(v_{\varepsilon})(\varphi)\leq\varepsilon \Vert \varphi \Vert _{m,a,p}\). In this way, we can deduce that \(\Vert I'(v_{\epsilon}) \Vert _{E^{1}} \leq\varepsilon\).
Therefore, from the previous information we can conclude that there is a sequence \(\{v_{k} \}\subset B_{\rho}\) such that
$$I(v_{k})\to\underline{c} \quad \mbox{and}\quad I'(v_{k}) \to0 \quad \mbox{in } E^{1} \mbox{ as } k\to\infty. $$
Using Lemma 4 we can show that up to a subsequence, \(\{v_{k} \}\) converges strongly to some \(v_{2}^{*}\in E\). Thus, \(v_{2}^{*}\) is a weak solution of (7) and \(v_{2}^{*}\) is a nontrivial solution since \(I(v_{2}^{*})=\underline{c}<0\). Finally, since \(I(v_{1}^{*})=\overline{c}>0>\underline{c}=I(v_{2}^{*})\), then \(v_{2}^{*}\neq v_{1}^{*}\). □