Singular Quasilinear Schrödinger Equations with Exponential Growth in Dimension Two

Abstract

In this work, we study the existence of positive solution for the following class of singular quasilinear Schrödinger equations:

$$\begin{aligned} -\text{ div } (g^2(u)\nabla u) + g(u)g'(u)|\nabla u|^2 + V(x)u = \frac{f(u)}{|x|^a} \quad \text{ in }\quad \mathbb {R}^2, \end{aligned}$$

where $a \in (0,2)$, $g:\mathbb {R}\rightarrow \mathbb {R}_+$ is a continuously differentiable function, V(x) is a positive potential and the nonlinearity f(u) can exhibit critical exponential growth. In order to prove our existence result, we combine minimax methods with a singular version of the Trudinger-Moser inequality.

1 Introduction and Main Result

In this paper, we consider quasilinear Schrödinger equations of the form

$$\begin{aligned} -\text{ div } (g^2(u)\nabla u) + g(u)g'(u)|\nabla u|^2 + V(x)u = \frac{f(u)}{|x|^a}\quad \text {in}\quad \mathbb {R}^2, \end{aligned}$$

(1.1)

where $a \in (0,2)$, $g: \mathbb {R} \rightarrow \mathbb {R}_+$ is a continuously differentiable function, $V: \mathbb {R}^2 \rightarrow \mathbb {R}$ is a positive potential and $f: \mathbb {R} \rightarrow \mathbb {R}$ is a continuous function that can exhibit critical exponential growth in sense of the Trudinger–Moser inequality (see (1.8)).

The study of equation (1.1) is related with the existence of solitary wave solutions for the nonlinear Schrödinger equation

$$\begin{aligned} i \partial _t w = -\Delta w + W(x)w - \tilde{p}(x,|w|^2)w - \Delta [\rho (|w|^2)]\rho '(|w|^2)w\quad \text {in}\quad \mathbb {R}^N,\nonumber \\ \end{aligned}$$

(1.2)

where $N\ge 1$, $w: \mathbb {R}\times \mathbb {R}^N \rightarrow \mathbb {C}$ is the unknown function, $W:\mathbb {R}^N \rightarrow \mathbb {R}$ is a given potential, $\rho :\mathbb {R}_+ \rightarrow \mathbb {R}$ and $\tilde{p}:\mathbb {R}^N \times \mathbb {R}_+ \rightarrow \mathbb {R}$ are real functions satisfying appropriate conditions. Equation (1.2) is called in the current literature as Generalized Quasilinear Schrödinger Equation and it has been accepted as model in many physical phenomena depending on the function $\rho$. For instance, if $\rho (s) = 1$ then we have the classical semilinear Schrödinger equation, see [25]. When $\rho (s) = s$, the equation arises from fluid mechanics, plasma physics and dissipative quantum mechanics, see [23, 27, 31]. For $\rho (s) = (1 + s)^{1/2}$, (1.2) models the propagation of a high-irradiance laser in a plasma as well as the self-channeling of a high-power ultrashort laser in matter, see [24]. For further physical applications, we quote [3, 32].

When we consider standing wave solutions for (1.2), that is, solutions of type $w(t,x) = \exp (-iEt)u(x)$, where $E \in \mathbb {R}$ and u is a real function, we know that w satisfies (1.2) if and only if the function u(x) solves the elliptic equation (see [8])

$$\begin{aligned} -\Delta u + V(x)u - \Delta [\rho (u^2)]\rho '(u^2)u = p(x,u)\quad \text{ in }\quad \mathbb {R}^N, \end{aligned}$$

(1.3)

where $V(x):=W(x)-E$ and $p(x,u):=\tilde{p}(x,u^2)$. Now, if we take

$$\begin{aligned} g^2(u) = 1 + \frac{[(\rho (u^2))']^2}{2}, \end{aligned}$$

then (1.3) turns into quasilinear elliptic equation (see [33])

$$\begin{aligned} -\text{ div } (g^2(u)\nabla u) + g(u)g'(u)|\nabla u|^2 + V(x)u = p(x,u)\quad \text{ in }\quad \mathbb {R}^N, \end{aligned}$$

(1.4)

which becomes (1.1) when $N=2$ and $p(x,u)=f(u)/|x|^a$. For example, when we have $g^2(s) = 1 + 2s^2$, that is, $\rho (s) = s$, we obtain the superfluid film equation in plasma physics

$$\begin{aligned} -\Delta u + V(x)u - \Delta (u^2)u = p(x,u)\quad \text {in}\quad \mathbb {R}^N, \end{aligned}$$

(1.5)

which has been extensively studied, see for example [7, 18, 29, 32]. More generally, if we put $g^2(s) = 1 + 2\gamma ^2 (s^2)^{2\gamma - 1}$, $\gamma >1/2$, that corresponds to $\rho (s)=s^{\gamma }$, we get the equation

$$\begin{aligned} -\Delta u + V(x)u - \gamma \Delta (|u|^{2\gamma })|u|^{2\gamma - 2}u = p(x,u)\quad \text{ in }\quad \mathbb {R}^N, \end{aligned}$$

(1.6)

which was addressed for instance in [1, 14, 28, 37]. Now, if we consider $\rho (s) = (1 + s)^{1/2}$, that is, $g^2(s)=1+s^2/[2(1+s^2)]$ we obtain

$$\begin{aligned} -\Delta u + V(x)u - \Delta [(1+u^2)^{1/2}]\frac{u}{2(1+u^2)^{1/2}} = p(x,u) \quad \text{ in }\quad \mathbb {R}^N, \end{aligned}$$

(1.7)

which was studied for instance in [6, 10].

Motivated by these physical aspects, Eq. (1.4) has attracted a lot of attention of many researchers and some existence and multiplicity results have been obtained (see [5, 10,11,12,13, 19, 26, 33,34,35,36]). In this work, more specifically, we intend to prove that equation (1.1) admits at least one positive solution. To achieve this goal, we shall apply variational methods in combination with a version singular of the Trudinger-Moser inequality.

As in the papers [10, 12, 33], we assume the following assumptions on the function g(s):

$(g_0)$:

$g \in C^1(\mathbb {R}, \mathbb {R}_+)$ is even, $g'(s) \ge 0$ for all $s \ge 0$ and $g(0) = 1$;

$(g_1)$:

there exists $\alpha \ge 1$ such that $(\alpha - 1)g(s) \ge g'(s)s$ for all $s \ge 0$;

$(g_2)$:

$\displaystyle \lim _{s \rightarrow +\infty } \frac{g(s)}{s^{\alpha - 1}} =: \beta > 0$.

Typical examples satisfying $(g_0){-}(g_2)$ are given by the functions:

• (a) $g(s)\equiv 1$ ($\alpha =1$ and $\beta =1$);

• (b) $g(s) =( 1 + 2s^2)^{1/2}$ ($\alpha =2$ and $\beta =\sqrt{2}$);

• (c) $g(s) = (1 + 2\gamma ^2 (s^2)^{2\gamma - 1})^{1/2}$ ($\alpha =2\gamma$ and $\beta =\sqrt{2}\gamma$),

which appear in the context of mathematical physics as indicated previously.

As it is known, the main difficulties in dealing with problem (1.4) is the lack of compactness, which is inherent to elliptic problems defined in unbounded domains and the fact that the energy functional associated to (1.4) is not generally well defined in the usual Sobolev space, because the presence of the integral $\int _{\mathbb {R}^N}g^2(u)|\nabla u|^2$ (see more details in Sect. 2). Hence, a direct variational approach is not possible.

To the best of our knowledge, the first existence result for generalized quasilinear elliptic problem of the type (1.4) in unbounded domains involving variational methods was due to [33]. The authors have used a change of variables and the Mountain-Pass Theorem to obtain positive solutions for (1.4) when p(x, u) is superlinear and has subcritical growth . Later on, by using change of variable, many authors proposed the critical problem when p(x, u) is the form $|u|^{\alpha 2^* - 2}u + f(u)$, see for instance [12, 13]. In [12], by using the semilinear dual equation, the authors postulated that the number $\alpha 2^*=2\alpha N/(N-2)$ must be the critical exponent for an equation of type (1.4) in $\mathbb {R}^N (N\ge 3)$. In the paper [26], Li and Wu studied the existence, multiplicity and concentration of solutions for the critical case ($N\ge 3$).

In the subcritical case, through change of variable, the authors in [19] studied problem (1.4) by using Orlicz space framework and proved the existence of positive solutions via minimax methods. Moreover, they considered the nonlinearity p(x, t) behaving like t at the origin and $t^3$ at infinity. Recently, by using the non-Nehari manifold method, Chen et al. in [5] proved that (1.4) admits a ground state solution under a monotonicity condition and some standard growth conditions on p(x, u). In [10], Deng and Huang proved the existence of ground state solutions for (1.4) by using Jeanjean’s monotonicity trick (see [21]).

Next, we assume that $V: \mathbb {R}^2 \rightarrow \mathbb {R}$ is a continuous function satisfying the condition

(V):

there exists a constant $V_0 > 0$ such that $V(x) \ge V_0$ for all $x \in \mathbb {R}^2$.

Unlike the articles cited above, this is the only condition imposed on the potential V. Here, we do not need another condition on V in order to guarantee some compactness result. Instead we exploit the fact that the embedding

$$\begin{aligned}&X := \left\{ v \in H^1(\mathbb {R}^2); \int _{\mathbb {R}^2} V(x)v^2\mathrm {d}x < \infty \right\} \hookrightarrow L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x)\\&\quad \text {is compact (see Section 2)}. \end{aligned}$$

About the nonlinearity f(u), we introduce the notion of criticality in dimension two for this class of problems. More precisely, we say that $f:\mathbb {R}\rightarrow \mathbb {R}$ has critical exponential growth at $+\infty$ if there exists $\varsigma _0 > 0$ such that

$$\begin{aligned} \lim _{s \rightarrow +\infty } f(s)e^{-\varsigma s^{2\alpha }} = \left\{ \begin{array}{ll} 0, &{} \text{ for } \text{ all } \ \varsigma > \varsigma _0, \\ + \infty , &{} \text{ for } \text{ all } \ \varsigma < \varsigma _0. \end{array}\right. \end{aligned}$$

(1.8)

As far as we know, this is the first work dealing with this class of quasilinear Schrödinger equations and involving exponential critical growth with singularity. We point out that (1.8) extends the definition founded in the papers [15, 17, 30]. Since the exponent $2\alpha$ can be bigger than 2, the growth (1.8) is better than the usual growth $e^{\varsigma s^{2}}$. This is possible due to the properties of the function g(s). Moreover, we assume that f satisfies the following conditions:

$(f_1)$:

$f(s) = o(s)$ as $s \rightarrow 0^+$ and $f(s) = 0$, for all $s \in (-\infty ,0]$;

$(f_2)$:

there exist $\theta > \alpha$ such that

$$\begin{aligned} 0 < 2\theta F(s) := 2\theta \int _0^s f(t)\mathrm {d}t \le sf(s), \ \ \text{ for } \text{ all } \ s \in (0,+\infty ); \end{aligned}$$

$(f_3)$:

there exist constants $s_0, M_0 > 0$ such that

$$\begin{aligned} F(s) \le M_0 f(s), \ \ \text{ for } \text{ all } \ \ s \ge s_0; \end{aligned}$$

$(f_4)$:

there exists $\xi _0 > 0$ such that

$$\begin{aligned} \liminf _{s \rightarrow +\infty } sf(s)e^{-\varsigma _0 s^{2\alpha }} \ge \xi _0. \end{aligned}$$

An elementary example of function satisfying $(f_1)-(f_4)$ is given by $f(s)= F'(s)$, where $F(s)= s^{3\alpha }e^{s^{2\alpha }}$ for $s\ge 0$ and $F(s) = 0$ for $s< 0$, with constant $\varsigma _0=1$.

Now, let $C_0^{\infty } (\mathbb {R}^2)$ be the space of infinitely differentiable functions with compact support and $H^1(\mathbb {R}^2)$ the usual Sobolev space with the norm

$$\begin{aligned} \Vert u\Vert _{1,2} = \left[ \int _{\mathbb {R}^2} (|\nabla u|^2 + u^2)\mathrm {d}x\right] ^{1/2}. \end{aligned}$$

In this context, we say that a function $u:\mathbb {R}^2 \rightarrow \mathbb {R}$ is a weak solution of problem (1.1) if $u \in H^1(\mathbb {R}^2) \cap L_{loc}^{\infty } (\mathbb {R}^2)$ and for all $\varphi \in C_0^{\infty } (\mathbb {R}^2)$ it holds

$$\begin{aligned} \int _{\mathbb {R}^2} g^2 (u) \nabla u \nabla \varphi \mathrm {d}x + \int _{\mathbb {R}^2} g(u)g'(u)|\nabla u|^2 \varphi + \int _{\mathbb {R}^2} V(x)u\varphi \mathrm {d}x - \int _{\mathbb {R}^2} f(u)\varphi \mathrm {d}x = 0.\nonumber \\ \end{aligned}$$

(1.9)

Now, we may state our main result.

Theorem 1.1

Suppose that $(g_0)-(g_2)$, (V), (1.8) and $(f_1){-}(f_4)$ are satisfied. Then, problem (1.1) has a positive solution.

As already mentioned, the main difficulty in treating this class of Schrödinger equations in $\mathbb {R}^2$ is the possible lack of compactness as well as the critical exponential growth. Our result extends and improves the papers [14, 15, 17, 30] in the sense that we are considering a broader class of operators. In order, to prove Theorem 1.1, we use a change of variables and we transform equation (1.1) into a semilinear one. The functional energy, denoted by I, associated to this semilinear problem is well defined and it is differentiable in the subspace X of $H^1 (\mathbb {R}^2)$ (for details see Sect. 2). Therefore, we justify that critical points of I provide weak solutions to problem (1.1).

The hypotheses $(f_1)$ and $(f_2)$ are sufficient conditions to guarantee the geometry of a suitable version of the Mountain-Pass Theorem (see Theorem 2.7). Moreover, $(f_2)$ is important to prove that Cerami sequences are bounded (see Lemma 4.1). With respect to the hypothesis $(f_4)$, it is fundamental to prove an estimate for the minimax level of I, see Proposition 5.1. Furthermore, $(f_4)$ is more general than a similar condition found in [15], because here we do not require a lower bounded for the constant $\xi _0$. The hypothesis $(f_3)$ is central for the proof of the convergence in Lemma 4.4. These last two results allows us to obtain the estimate

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) ^2 \varsigma _0\Vert \nabla v_n\Vert _2^2 < 4\pi \end{aligned}$$

for n sufficiently large, where $(v_n)$ is a Cerami sequence at the minimax level. This estimate is fundamental for applying Corollary 4.7 and consequently is used in the proof of Theorem 1.1. The conditions of the type $(f_3)$ and $(f_4)$ were considered in the pioneering work due to de Figueiredo et al. [9].

The outline of the paper is as follows: in the forthcoming section is the reformulation of the problem and some preliminary results, including the appropriate variational setting to study the quasilinear problem, the regularity of the dual energy functional and properties of its critical points. Moreover, we present the singular Trudinger-Moser inequality due to [4]. In Sect. 3, we prove that the energy functional satisfies the geometric conditions of Theorem 2.7. Section 4 is dedicated to the proof of some technical results involving the Cerami sequences associated to the energy functional. In Sect. 5, we derive an important estimate for the mountain pass level and Sect. 6 is devoted to the proof of the main result of the work.

2 Variational Setting and Preliminaries

We begin this section by defining the following subspace X of $H^1(\mathbb {R}^2)$:

$$\begin{aligned} X = \left\{ v \in H^1(\mathbb {R}^2); \int _{\mathbb {R}^2} V(x)v^2\mathrm {d}x < \infty \right\} , \end{aligned}$$

which is a Hilbert space equipped with the inner product

$$\begin{aligned} \langle u, v\rangle _X = \int _{\mathbb {R}^2} (\nabla u \nabla v + V(x) uv)\mathrm {d}x, \ \ u,v \in X \end{aligned}$$

(2.1)

and its corresponding norm $\Vert v \Vert _X = \langle v, v \rangle ^{1/2}$. It is clear that the hypothesis (V) implies the continuity of the embedding $X \hookrightarrow H^1 (\mathbb {R}^2)$. Furthermore, by considering the weighted Lebesgue space

$$\begin{aligned} L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x) = \left\{ u: \mathbb {R}^2 \rightarrow \mathbb {R}: u \ \text{ is } \text{ mensurable } \text{ and } \ \int _{\mathbb {R}^2} \frac{|u|^p}{|x|^a}\mathrm {d}x < \infty \right\} , \end{aligned}$$

we have the following compactness lemma:

Lemma 2.1

Suppose $p \ge 2$ and $a \in (0,2)$. Then, the embedding $X \hookrightarrow L^p (\mathbb {R}^2, | x |^{- a} \mathrm {d} x)$ is compact.

Proof

See Theorem 1.2 in [38]. $\square$

Now, we are going to introduce our variational structure. As observed in the Introduction, formally (1.1) is the Euler-Lagrange equation associated to the energy functional

$$\begin{aligned} J(u) = \frac{1}{2} \int _{\mathbb {R}^2} g^2(u)|\nabla u|^2\mathrm {d}x + \frac{1}{2} \int _{\mathbb {R}^2} V(x)u^2\mathrm {d}x - \int _{\mathbb {R}^2} \frac{F(u)}{|x|^a}\mathrm {d}x. \end{aligned}$$

(2.2)

The first difficulty that we have to deal with is to find an appropriate variational setting in order to apply variational methods to study the existence of critical points for J, because $g^2(u)|\nabla u|^2$ is not necessary in $L^1(\mathbb {R}^2)$ if $u\in H^1(\mathbb {R}^2)$. To overcome this difficulty, we follow ideas introduced in [33] (see also [13]), that is, we make use of the change of variables

$$\begin{aligned} v = G(u) = \int _0^u g(s)\mathrm {d}s. \end{aligned}$$

Hence, after this change of variables, we obtain the new functional

$$\begin{aligned} I(v) = J(G^{-1}(u)) = \frac{1}{2}\int _{\mathbb {R}^2}(|\nabla v|^2 + V(x)[G^{-1}(v)]^2)\mathrm {d}x - \int _{\mathbb {R}^2} \frac{F(G^{-1}(v))}{|x|^a}\mathrm {d}x,\nonumber \\ \end{aligned}$$

(2.3)

which is well defined in the space X, under the conditions on g, V and f. For an easy reference, we list below the main properties of the function $G^{-1}$.

Lemma 2.2

Under conditions $(g_0)-(g_2)$, we have the following properties:

1.:

$G^{-1}$ is increasing; also G e $G^{-1}$ are odd functions;

2.:

$0 < [G^{-1}(t)]'= \frac{1}{g(G^{-1}(t))} \le 1 = \frac{1}{g(0)}$ for all $t \in \mathbb {R}$;

3.:

$|G^{-1}(t)| \le |t|$ for all $t \in \mathbb {R}$;

4.:

$\frac{G^{-1}(t)}{\alpha } \le \frac{t}{g(G^{-1}(t))} \le G^{-1}(t)$ for all $t \ge 0$ and $\frac{[G^{-1}(t)]^2}{\alpha } \le \frac{G^{-1}(t)t}{g(G^{-1}(t))} \le [G^{-1}(t)]^2$ for all $t \in \mathbb {R}$;

5.:

$\frac{|G^{-1}(t)|^{\alpha - 1}}{g(G^{-1}(t))} \le \frac{1}{\beta }$ for all $t \in \mathbb {R}$;

6.:

$|G^{-1}(t)|^{\alpha } \le \frac{\alpha }{\beta }|t|$ for all $t \in \mathbb {R}$;

7.:

$\frac{G^{-1}(t)}{t^{1/\alpha }} \rightarrow \left( \frac{\alpha }{\beta }\right) ^{1/\alpha }$ as $t \rightarrow +\infty$;

8.:

there exists a positive constant C such that

$$\begin{aligned} |G^{-1}(t)| \ge \left\{ \begin{array}{ll} C|t|, &{} |t| \le 1 ,\\ C|t|^{1/\alpha }, &{} |t| \ge 1. \end{array}\right. \end{aligned}$$

Proof

The item (1) follows from the monotonicity of G and since g is even. To prove (2), just to derive the equality $G(G^{-1}(t))=t$. For item (3), we use the Mean Value Theorem and (2) to conclude that $|G^{-1}(t)|=|G^{-1}(t)-G^{-1}(0)|= [G^{-1}(\xi )]'|t|\le |t|$ for some $\xi$ between 0 and t. Therefore this item is proved.

In order to show (4), consider $\sigma _1(t):=\alpha t-g(G^{-1}(t))G^{-1}(t)$ and $\sigma _2(t):=g(G^{-1}(t))G^{-1}(t)-t$. We have $\sigma _1(0)=\sigma _2(0)=0$ and by $(g_0)-(g_1)$

$$\begin{aligned} \sigma _1'(t)=\alpha -1-\frac{g'(G^{-1}(t))G^{-1}(t)}{g(G^{-1}(t))}\ge 0\quad \text {and}\quad \sigma _2'(t)=\frac{g'(G^{-1}(t))G^{-1}(t)}{g(G^{-1}(t))}\!\ge 0. \end{aligned}$$

Thus, $\sigma _1(t)\ge 0$, $\sigma _2(t)\ge 0$ for all $t\ge 0$ and the first part is done. For the second part, just to observe that $G^{-1}(t)t\ge 0$ for all $t\in \mathbb {R}$.

Next, from $(g_0)-(g_2)$ we deduce that $g(s) \ge \beta |s|^{\alpha - 1}$ for all $s \in \mathbb {R}$ and taking $s=G^{-1}(t)$ we obtain (5). From item (5) and using integration, the proof of item (6) follows.

Now, let us check (7). By the limit in $(g_2)$, given $\varepsilon > 0$ there exists $R> 0$ such that $g(s) \le 1 + \beta _{\varepsilon }s^{\alpha - 1}$ for $s \ge R$, where $\beta _{\varepsilon } = \beta + \varepsilon$. By using (6), $(g_0)$ and the Mean Value Theorem, for $t_0\ge R$ we get

$$\begin{aligned} \begin{aligned} G^{-1}(t) - G^{-1}(t_0)&= \int _{t_0}^t \frac{1}{g(G^{-1}(s))}\mathrm {d}s \ge \int _{t_0}^t \frac{1}{g\left( \left( \frac{\alpha }{\beta }\right) ^{1/\alpha }s^{1/\alpha }\right) }\mathrm {d}s \\&\ge \int _{t_0}^t \frac{1}{ 1 + \beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}s^{\frac{\alpha - 1}{\alpha }}}\mathrm {d}s \\&\ge \int _{t_0}^t \frac{1}{\beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}s^{\frac{\alpha - 1}{\alpha }}}\mathrm {d}s - \int _{t_0}^t \frac{1}{\beta _{\varepsilon }^2\left( \frac{\alpha }{\beta }\right) ^{\frac{2(\alpha - 1)}{\alpha }}s^{\frac{2(\alpha - 1)}{\alpha }}}\mathrm {d}s. \end{aligned} \end{aligned}$$

If $\alpha > 2$ and by calculating the last two integrals, there exists a positive constant $C_1$ such that

$$\begin{aligned} \begin{aligned} G^{-1}(t)&\ge G^{-1}(t_0) - \frac{\alpha }{\beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}}t_0^{1/\alpha } + \frac{\alpha }{\beta _{\varepsilon }^2(\alpha - 2)\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}}(t_0^{\frac{2 - \alpha }{\alpha }} - t^{\frac{2 - \alpha }{\alpha }}) \\&\quad +\frac{\alpha }{\beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}}t^{1/\alpha } \\&\ge -C_1 + \frac{\alpha }{\beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}}t^{1/\alpha }. \end{aligned} \end{aligned}$$

As $\beta _{\varepsilon } \rightarrow \beta$ when $\varepsilon \rightarrow 0^+$, we conclude that

$$\begin{aligned} \liminf _{t \rightarrow +\infty } \frac{G^{-1}(t)}{t^{1/\alpha }} \ge \left( \frac{\alpha }{\beta }\right) ^{1/\alpha }. \end{aligned}$$

Using again (6) we establish the desired limit for $\alpha > 2$. If $\alpha = 2$, for all $t > t_0 + 1 \ge R + 1$ there exists a positive constant $C_2$ satisfying

$$\begin{aligned} G^{-1}(t)&\ge G^{-1}(t_0) + \frac{2}{\beta _{\varepsilon }\left( \frac{2}{\beta }\right) ^{1/2}}(t^{1/2} - t_0^{1/2}) - \frac{1}{\beta _{\varepsilon }^2\left( \frac{2}{\beta }\right) } \int _{t_0}^t \frac{1}{s}\mathrm {d}s \\&\ge -C_2 \log t + \frac{2}{\beta _{\varepsilon }\left( \frac{2}{\beta }\right) ^{1/2}}t^{1/2}, \end{aligned}$$

from where we reach

$$\begin{aligned} \liminf _{t \rightarrow +\infty } \frac{G^{-1}(t)}{t^{1/2}} \ge \left( \frac{2}{\beta }\right) ^{1/2}. \end{aligned}$$

which is the desired limit. Finally, for $1< \alpha < 2$ we have the estimate

$$\begin{aligned} G^{-1}(t) \ge - \left( \frac{\alpha }{\beta }\right) ^{1/\alpha }t_0^{1/\alpha } - \frac{\alpha ^{\frac{2 - \alpha }{\alpha }}}{\beta ^{2/\alpha }(2 - \alpha )}t^{\frac{2 - \alpha }{\alpha }} + \frac{\alpha }{\beta _{\varepsilon }\left( \frac{\alpha }{\beta }\right) ^{\frac{\alpha - 1}{\alpha }}}t^{1/\alpha } \end{aligned}$$

and similarly we get the result. To conclude, item (8) follows directly from (7). $\square$

The next proposition presents an important compactness result.

Proposition 2.3

Suppose that (V) is satisfied. Then, the map $v \rightarrow G^{-1}(v)$ from X into $L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x)$ is compact for $2 \le p < \infty$.

Proof

Let $(v_n) \subset X$ be a bounded sequence in X. By Lemma 2.2-(2),(3) we have $\Vert G^{-1}(v_n)\Vert \le \Vert v_n\Vert$. Thus, $(G^{-1}(v_n))$ is bounded in X and since the embedding $X \hookrightarrow L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x)$ is compact for $2 \le p < \infty$, up to a subsequence, there exists $w \in L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x)$ such that $G^{-1}(v_n) \rightarrow w$ in $L^p(\mathbb {R}^2, |x|^{-a}\mathrm {d}x)$ and the proof is done. $\square$

It is standard to see that under the assumptions on V, g and f, the functional I is of class $C^1$ on X with

$$\begin{aligned} I'(v)\varphi = \int _{\mathbb {R}^2} \left( \nabla v \nabla \varphi + V(x) \frac{G^{-1}(v)}{g(G^{-1}(v))}\varphi \right) \mathrm {d}x - \int _{\mathbb {R}^2} \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a}\varphi \mathrm {d}x,\nonumber \\ \end{aligned}$$

(2.4)

for $v, \varphi \in X$ and therefore critical points of I turn out to be weak solutions of the semilinear equation

$$\begin{aligned} -\Delta v + V(x)\frac{G^{-1}(v)}{g(G^{-1}(v))} = \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a} \ \ \text{ in }\,\ \ \mathbb {R}^2. \end{aligned}$$

(2.5)

We also observe that given $\varepsilon > 0$, $q \ge 1$ and $\varsigma > \varsigma _0$, by $(f_1)$ and (1.8) there exists a constant $C_{\varepsilon } > 0$ satisfying

$$\begin{aligned} |f(s)| \le \varepsilon |s| + C_{\varepsilon }|s|^{q - 1}(e^{\varsigma s^{2\alpha }}- 1)\quad \text {for all}\,\,\ s \in \mathbb {R}. \end{aligned}$$

(2.6)

We will see in Proposition 2.6 that if $v \in H^1(\mathbb {R}^2)$ is a critical point of the functional I, then $u = G^{-1}(v)$ is a weak solution of (1.1). Therefore, to obtain weak solutions of (1.1), it will be sufficient to look for critical points of I.

At first, let us recall the following Trudinger-Moser inequality due to [16]:

Lemma 2.4

If $\varsigma > 0$, $a \in (0,2)$ and $u \in H^1(\mathbb {R}^2)$, then

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{(e^{\varsigma u^2} - 1)}{|x|^a}\mathrm {d}x < \infty . \end{aligned}$$

(2.7)

Moreover, if $0< \varsigma < 2\pi (2 - a)$ and $\Vert u\Vert _2 \le M$, then there exists a positive constant $C=C(\varsigma ,a, M)$, which depends only on M, a and $\varsigma$, such that

$$\begin{aligned} \sup _{\Vert \nabla u\Vert _2 \le 1} \int _{\mathbb {R}^2} \frac{(e^{\varsigma u^2} - 1)}{|x|^a}\mathrm {d}x \le C. \end{aligned}$$

(2.8)

In many arguments, we will need of the following lemma:

Lemma 2.5

Let $\varsigma > 0$ and $r \ge 1$. Then

$$\begin{aligned} (e^{\varsigma s^2} - 1)^r \le e^{r\varsigma s^2} - 1, \ \ \text{ for } \text{ all } \ s \in \mathbb {R}. \end{aligned}$$

Proof

Just analyze the limits of the function $\xi (s) = (e^{\varsigma s^2} - 1 )^r/(e^{r\varsigma s^2} - 1)$ at the origin and at infinity applying the L’Hôpital rule. $\square$

Proposition 2.6

(Critical points of I and solutions of (1.1)). Every critical point v of I belongs to $C_{loc}^{0,\vartheta } (\mathbb {R}^2)$ for some $\vartheta \in (0,1)$ and $u = G^{-1}(v)$ is a weak solution of (1.1).

Proof

Every critical point v of I satisfies the equation $-\Delta v = w$ in $\mathbb {R}^2$ in weak sense, where

$$\begin{aligned} w(x) = \frac{1}{g(G^{-1}(v))} \left[ \frac{f(G^{-1}(v))}{|x|^a} - V(x)G^{-1}(v)\right] . \end{aligned}$$

From this, for $t > 1$, according to (2.6), (5) and (10) of Lemma 2.2, Lemma 2.5, for almost everywhere $x \in B_R \equiv B_R(0)$, we obtain

$$\begin{aligned} \begin{aligned} |w(x)|^t&\le \left[ \frac{|G^{-1}(v)}{g(G^{-1}(v))}\right] ^t\left[ \frac{C_1}{|x|^a} + \frac{C_{2}}{|x|^a}(e^{\varsigma [G^{-1}(v)]^{2\alpha }} - 1)+ V(x)\right] ^t \\&\le C_3\left[ \frac{1}{|x|^{at}} + \frac{1}{|x|^{at}}\left( e^{t\left( \frac{\alpha }{\beta }\right) ^2\varsigma v^{2}} - 1\right) + M_R^t\right] \end{aligned} \end{aligned}$$

where $M_R := \sup \{V(x): x \in \overline{B_R}\}$. Now, considering $t> 1$ such that $0<at <2$ and using Lemma 2.4 we conclude that $w \in L^t (B_R)$. So, applying Schauder regularity theory, it follows that $v \in C_{loc}^{0,\vartheta }(\mathbb {R}^2)$ to some $\vartheta \in (0,1)$. In particular, $v \in L^{\infty }_{loc}(\mathbb {R}^2)$. The rest of the argument follows in a similar way to the proof of Proposition 2.9 in [14]. $\square$

To conclude this section, we present a version of the Mountain-Pass Theorem, which is a consequence of the Ekeland Variational Principle as developed in [2]. We will also need to establish a local version of the same theorem.

Theorem 2.7

(Mountain-Pass Theorem) Let X be a Banach space and $\Phi \in C^1(X; \mathbb {R})$ with $\Phi (0) = 0$. Let $\mathcal {S}$ be a closed subset of X which disconnects (archwise) X. Let $v_0 = 0$ and $v_1 \in X$ be points belonging to distinct connected components of $X \backslash \mathcal {S}$. Suppose that

$$\begin{aligned} \inf _{\mathcal {S}} \Phi \ge \sigma > 0 \ \ \text{ and } \ \ \Phi (v_1) \le 0 \end{aligned}$$

(2.9)

and let

$$\begin{aligned} \Gamma = \{\gamma \in C([0,1];X): \gamma (0) = 0 \ \ \text{ and } \ \ \gamma (1) = v_1\}. \end{aligned}$$

(2.10)

Then

$$\begin{aligned} c = \inf _{\gamma \in \Gamma } \max _{t \in [0,1]} \Phi (\gamma (t)) \ge \sigma \end{aligned}$$

and there exists a Cerami sequence¹ for $\Phi$ at the level c. The number c is called the mountain-pass level of $\Phi$.

3 Geometric Properties

In this section, we are going to show that the functional I satisfies the geometric conditions (2.9). For this, we need to obtain some technical lemmas.

Lemma 3.1

Assume that (V) and $(g_0)-(g_2)$ hold. If $v \in X$, $\varsigma > 0$, $t > 0$ and $\Vert v\Vert _2 \le M$ with $\left( \frac{\alpha }{\beta }\right) ^2\varsigma \Vert \nabla v\Vert _2^2 < 2\pi (2 - a)$, then there exists $C = C(a,\alpha , \varsigma , M, t) > 0$ such that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{e^{\varsigma |G^{-1}(v)|^{2\alpha }}-1}{|x|^a}|G^{-1}(v)|^t \mathrm {d}x \le C\Vert G^{-1}(v)\Vert ^t. \end{aligned}$$

Proof

Consider $r > 1$ close to 1 such that $\left( \frac{\alpha }{\beta }\right) ^2r\varsigma \Vert \nabla v\Vert _2^2 < 2\pi (2 - ar)$, $ar < 2$ and $ts \ge 2$, where $s = r/(r - 1)$. Using (5) of Lemma 2.2 and Holder’s inequality, we have

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{e^{\varsigma |G^{-1}(v)|^{2\alpha }} - 1}{|x|^a}|G^{-1}(v)|^t\mathrm {d}x \le \left[ \int _{\mathbb {R}^2} \frac{(e^{\left( \frac{\alpha }{\beta }\right) ^2\varsigma v^2} - 1)^r}{|x|^{ar}}\mathrm {d}x\right] ^{1/r}\Vert G^{-1}(v)\Vert _{ts}^t \end{aligned}$$

and by Lemmas 2.4, 2.5 and the continuous embedding $H^1(\mathbb {R}^2) \hookrightarrow L^{ts}(\mathbb {R}^2)$, we conclude

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^2} \frac{e^{\varsigma |G^{-1}(v)|^{2\alpha }} - 1}{|x|^a}|G^{-1}(v)|^t\mathrm {d}x&\le \left[ \int _{\mathbb {R}^2} \frac{e^{\left( \frac{\alpha }{\beta }\right) ^2r\varsigma \Vert \nabla v\Vert _2^2 \left( \frac{v}{\Vert \nabla v\Vert _2}\right) ^2} - 1}{|x|^{ar}}\mathrm {d}x\right] ^{\frac{1}{r}}\Vert G^{-1}(v)\Vert _{ts}^t \\&\le C_1\Vert G^{-1}(v)\Vert _{ts}^t \le C \Vert G^{-1}(v)\Vert ^t, \end{aligned} \end{aligned}$$

which proves the lemma. $\square$

Lemma 3.2

Assume that (V) holds. If $v \in H^1(\mathbb {R}^2)$ and $t \ge 2$, then there exists $C = C(t) > 0$ such that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{|G^{-1}(v)|^{t}}{|x|^a}\mathrm {d}x \le C \Vert G^{-1}(v)\Vert ^t. \end{aligned}$$

Proof

Let $r> 1$ be close to 1 such that $ar <2$ and $s = r / (r - 1)$. Using Hölder’s inequality and the continuous embedding $X \hookrightarrow L^{q} (\mathbb {R}^2)$ for all $2 \le q <\infty$, we obtain

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^2} \frac{|G^{-1}(v)|^t}{|x|^a}\mathrm {d}x&\le \int _{|x| > 1} |G^{-1}(v)|^t\mathrm {d}x + \left( \int _{|x| \le 1}\frac{1}{|x|^{ar}}\mathrm {d}x\right) ^{1/r}\left( \int _{|x| \le 1}|G^{-1}(v)|^{ts}\mathrm {d}x\right) ^{1/s} \\&\le \Vert G^{-1}(v)\Vert _t^t + C_1\Vert G^{-1}(v)\Vert _{ts}^t \\&\le C\Vert G^{-1}(v)\Vert ^t \end{aligned} \end{aligned}$$

and the proof follows. $\square$

In view of the last estimates, we can prove that the functional I has the mountain-pass geometry. For this purpose, for $\rho > 0$, we define

$$\begin{aligned} S_{\rho }= \left\{ v \in X: \int _{\mathbb {R}^2} |\nabla v|^2\mathrm {d}x + \int _{\mathbb {R}^2} V(x)[G^{-1}(v)]^2\mathrm {d}x = \rho ^2 \right\} . \end{aligned}$$

Since $Q: X \rightarrow \mathbb {R}$, defined by

$$\begin{aligned} Q(v) = \int _{\mathbb {R}^2} \{|\nabla v|^2 + V(x)[G^{-1}(v)]^2\}\mathrm {d}x, \end{aligned}$$

is a continuous function, it follows that $S_{\rho }$ is a closed subset that disconnects the space X.

Lemma 3.3

Suppose that (V), $(g_0)$ and $(f_1)$ are satisfied. Then, there exist $\rho > 0$ and $\sigma > 0$ satisfying

$$\begin{aligned} I(v) \ge \sigma , \ \ \text{ for } \text{ all }\ \ v \in S_{\rho }. \end{aligned}$$

Proof

From the estimate (2.6), given $\varepsilon > 0$ there is $C_{\varepsilon } > 0$ such that

$$\begin{aligned} |F(s)| \le \frac{\varepsilon }{2}s^2 + C_{\varepsilon }|s|^t (e^{\varsigma s^{2\alpha }} -1), \quad \text{ for } \text{ all } \quad s \in \mathbb {R}, \ t > 2. \end{aligned}$$

(3.1)

Now, if $\left( \frac{\alpha }{\beta }\right) ^2\varsigma \rho ^2 < 2\pi (2 - a)$, by using (3.1), Lemma 3.1, Lemma 3.2, Lemma 2.2-(2) and the continuous embedding $H^1(\mathbb {R}^2) \hookrightarrow L^t(\mathbb {R}^2)$, we obtain

$$\begin{aligned} \begin{aligned} I(v)&\ge \frac{1}{2}Q(v) - \frac{\varepsilon }{2}C\Vert G^{-1}(v)\Vert ^2 - C_1\Vert G^{-1}(v)\Vert ^t \\&\ge \left( \frac{1}{2} - \frac{\varepsilon }{2}C\right) Q(v) - C_1Q(v)^{t/2}. \end{aligned} \end{aligned}$$

Taking $0< \varepsilon < 1/C$ and since $t > 2$, we may choose $0< \rho < \frac{\beta }{\alpha }\left( \frac{2\pi (2 - a)}{\varsigma }\right) ^{1/2}$ such that $\left( \frac{1}{2} - \frac{\varepsilon }{2}C\right) \rho ^2 - C_1\rho ^{t} > 0$. Thus, considering $\sigma = \left( \frac{1}{2} - \frac{\varepsilon }{2}C\right) \rho ^2 - C_1\rho ^{t} > 0$ we conclude $I(v) \ge \sigma$ for all $v \in S_{\rho }$. $\square$

Lemma 3.4

Suppose that (V), $(g_0) - (g_2)$ and $(f_2)$ are satisfied. Then, there exists $e \in X$ such that $Q(e) > \rho ^2$ and

$$\begin{aligned} I(e)< 0 < \sigma \le \inf _{v \in S_{\rho }} I(v). \end{aligned}$$

Proof

First, consider $\varphi \in C_0^{\infty }(\mathbb {R}^2,[0,1]) \backslash \{0\}$ such that $\text{ supp }(\varphi ) = \overline{B_1}$. From $(f_2)$, there are positive constants $C_1$ and $C_2$ such that $F(s) \ge C_1|s|^{2\theta } - C_2$ for all $s \in \mathbb {R}$. Thus, for $t > 0$ we have

$$\begin{aligned} \begin{aligned} I(t\varphi )&= \frac{1}{2}\int _{\overline{B_1}} (|\nabla (t\varphi )|^2 + V(x)[G^{-1}(t\varphi )]^2)\mathrm {d}x - \int _{\overline{B_1}} \frac{F(G^{-1}(t\varphi ))}{|x|^a}\mathrm {d}x \\&\le \frac{t^2}{2}\int _{\overline{B_1}} (|\nabla \varphi |^2 + V(x)\varphi ^2)\mathrm {d}x - C_1 \int _{\overline{B_1}} \frac{|G^{-1}(t\varphi )|^{2\theta }}{|x|^a}\mathrm {d}x + C_2\int _{\overline{B_1}} \frac{1}{|x|^a}\mathrm {d}x \\&\le t^2\left[ \frac{\Vert \varphi \Vert ^2}{2} - C_1 \int _{\overline{B_1}} \frac{|G^{-1}(t\varphi )|^{2\theta }}{t^2|x|^a} \mathrm {d}x + \frac{C_2}{t^2}\int _{\overline{B_1}} \frac{1}{|x|^a}\mathrm {d}x\right] . \end{aligned} \end{aligned}$$

Since $2\theta - 2\alpha > 0$, for $x \in \overline{B_1}$, by using Lemma 2.2-(7), it follows that

$$\begin{aligned} \frac{|G^{-1}(t\varphi (x))|^{2\theta }}{t^2}= & {} \left( \frac{G^{-1}(t\varphi (x))}{\root \alpha \of {t\varphi (x)}}\right) ^{2\alpha }|G^{-1}(t\varphi (x))|^{2\theta - 2\alpha }\varphi (x)^2 \rightarrow \\&+\infty \ \ \text{ as } \ \ t \rightarrow +\infty . \end{aligned}$$

Thus, according to Fatou’s Lemma, we obtain

$$\begin{aligned} \int _{\overline{B_1}} \frac{|G^{-1}(t\varphi )|^{2\theta }}{t^2|x|^a} \mathrm {d}x \rightarrow +\infty \ \ \text{ as } \ \ t \rightarrow +\infty . \end{aligned}$$

and therefore $I(t\varphi ) \rightarrow -\infty$. Setting $e := t\varphi$ with t large enough, the proof is finished. $\square$

4 On Cerami Sequences for I

The purpose of this section is to prove some results about the Cerami sequences for the functional I. The first one is the following:

Lemma 4.1

Suppose that (V), $(g_0)-(g_1)$ and $(f_2)$ are satisfied. Let $(v_n)$ be in X such that $I(v_n) \rightarrow c \in \mathbb {R}$ and $I'(v_n)v_n \rightarrow 0$ as $n \rightarrow +\infty$. Then, $Q(v_n)$ is bounded and $(v_n)$ is bounded in $H^1(\mathbb {R}^2)$.

Proof

Using Lemma 2.2-(4) and $(f_2)$, we obtain

$$\begin{aligned} I(v_n) - \frac{\alpha }{2\theta }I'(v_n)v_n= & {} \left( \frac{1}{2} - \frac{\alpha }{2\theta }\right) \int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x + \frac{1}{2}\int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^2\mathrm {d}x \\&- \frac{\alpha }{2\theta }\int _{\mathbb {R}^2} V(x)\frac{G^{-1}(v_n)}{g(G^{-1}(v_n))}v_n\mathrm {d}x - \int _{\mathbb {R}^2} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \\&+ \frac{\alpha }{2\theta }\int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n))|x|^a}v_n\mathrm {d}x \\\ge & {} \left( \frac{1}{2} - \frac{\alpha }{2\theta }\right) Q(v_n) \\&+ \frac{1}{2\theta }\int _{\{G^{-1}(v_n) > 0\}}\frac{f(G^{-1}(v_n))G^{-1}(v_n) - 2\theta F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \\\ge & {} \left( \frac{1}{2} - \frac{\alpha }{2\theta }\right) Q(v_n). \end{aligned}$$

Since $I(v_n) = c + o_n(1)$ and $I'(v_n)v_n = o_n(1)$, as $n \rightarrow +\infty$, it follows that

$$\begin{aligned} \left( \frac{1}{2} - \frac{\alpha }{2\theta }\right) Q(v_n) \le c + o_n(1). \end{aligned}$$

(4.1)

Now, since $\theta > \alpha$, for some constant $C>0$ we have

$$\begin{aligned} Q(v_n) = \int _{\mathbb {R}^2} \{|\nabla v_n|^2 + V(x)[G^{-1}(v_n)]^2\}\mathrm {d}x \le C. \end{aligned}$$

(4.2)

In view of (4.1), it remains to show that $\int _{\mathbb {R}^2} v_n^2\mathrm {d}x$ is bounded. By condition (V) and Lemma 2.2-(8) there exists a constant $C_1 > 0$ such that

$$\begin{aligned} \begin{aligned} \int _{\mathbb {R}^2} v_n^2 \mathrm {d}x&= \displaystyle \int _{\{|v_n| \le 1\}} v_n^2\mathrm {d}x + \int _{\{|v_n|> 1\}} v_n^2\mathrm {d}x \\&\le \displaystyle \frac{1}{C_1^2V_0} \int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^2\mathrm {d}x + \frac{1}{C_1^{2\alpha }}\int _{\mathbb {R}^2} [G^{-1}(v_n)]^{2\alpha }\mathrm {d}x. \end{aligned} \end{aligned}$$

(4.3)

Next, we will use the Gagliardo-Nirenberg inequality (see [22], p. 31), which asserts

$$\begin{aligned} \Vert u\Vert _q \le C(\vartheta )\Vert u\Vert _r^{1 - \vartheta }\Vert \nabla u\Vert _2^{\vartheta } \end{aligned}$$

(4.4)

for all $u \in H^1(\mathbb {R}^2) \cap L^r(\mathbb {R}^2)$, where $1 \le r < \infty$, $0 < \vartheta \le 1$ and $\frac{1}{q} = \frac{1 - \vartheta }{r}$. Setting $u = G^{-1}(v_n)$, $\vartheta = 1 - \frac{1}{\alpha }$ and $r = 2$, we have $q = 2\alpha$. Hence, by using (V) and (4.4), we get

$$\begin{aligned} \int _{\mathbb {R}^2} |G^{-1}(v_n)|^{2\alpha }\mathrm {d}x \le \frac{C(\vartheta )^{2\alpha }}{V_0}\left( \int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^{2}\mathrm {d}x\right) \left( \int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x\right) ^{\alpha - 1}.\nonumber \\ \end{aligned}$$

(4.5)

From (4.2), (4.3) and (4.5), it follows that $\int _{\mathbb {R}^2} v_n^2\mathrm {d}x$ is bounded and the lemma is proved. $\square$

Corollary 4.2

Suppose that (V), $(g_0)-(g_1)$ and $(f_2)$ are satisfied. Let $(v_n)$ be a Cerami sequence for I in X. Then, there exists $C > 0$ such that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))v_n|}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x \le C. \end{aligned}$$

Proof

By Lemma 2.2-(4) and since $I'(v_n)v_n \rightarrow 0$ as $n \rightarrow +\infty$, we have

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))v_n}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x\le & {} \int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x + \int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^2\mathrm {d}x + o_n(1)\\\le & {} Q(v_n) + o_n(1). \end{aligned}$$

By the previous lemma, $Q(v_n)$ is bounded and the above estimate shows the result. $\square$

Lemma 4.3

Suppose that (V), $(g_0)-(g_1)$ and $(f_1)-(f_2)$ are satisfied. Let $(v_n)$ be a Cerami sequence for I. Then, $(v_n)$ has a subsequence, still denoted by $(v_n)$, such that $v_n \rightharpoonup v$ in $H^1(\mathbb {R}^2)$ such that $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$ and

$$\begin{aligned} \frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n))|x|^a} \rightarrow \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a} \quad \text{ in } \quad L^1_{loc}(\mathbb {R}^2), \ \ \text{ as } \ n \rightarrow +\infty . \end{aligned}$$

Proof

According to Lemma 4.1, $(v_n)$ is bounded in $H^1(\mathbb {R}^2)$. Thus, up to a subsequence, $v_n \rightharpoonup v$ in $H^1(\mathbb {R}^2)$. Furthermore, the function v satisfies $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$, because $Q(v_n)$ is bounded and by Fatou’s Lemma

$$\begin{aligned} \int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x\le \liminf _{n \rightarrow +\infty }\int _{\mathbb {R}^2}V(x)|G^{-1}(v_n)|^2 \mathrm {d}x\le C. \end{aligned}$$

Now, it is sufficient to prove that

$$\begin{aligned} \int _{B_R} \frac{|f(G^{-1}(v_n))|}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x \rightarrow \int _{B_R} \frac{|f(G^{-1}(v))|}{g(G^{-1}(v))|x|^a}\mathrm {d}x,\ \ \text{ as } \ n \rightarrow +\infty . \end{aligned}$$

By using Lemma 4.1, Lemma 2.2-(3) and since the embedding $H^1(\mathbb {R}^2) \hookrightarrow L_{loc}^t(\mathbb {R}^2)$, for all $t \ge 1$, is compact, we can assume that $G^{-1}(v_n) \rightarrow G^{-1}(v)$ strongly in $L^t (B_R)$ for any $t \in [1,+\infty )$. Moreover, by using items (2) and (3) of Lemma 2.2, Lemma 2.4, Corollary 4.2, estimate (2.6) and Holder’s inequality, we obtain

$$\begin{aligned} |G^{-1}(v)| \in L^1(B_R), \ \ \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a} \in L^1(B_R) \ \ \text{ and } \ \ \int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))v_n|}{g(G^{-1}(v_n))|x|^a} \le C. \end{aligned}$$

The rest of the argument follows the same steps as in the proof of Lemma 4.3 in [14]. $\square$

Lemma 4.4

Suppose that (V), $(g_0)-(g_1)$ and $(f_1) - (f_3)$ are satisfied. Let $(v_n)$ be a Cerami sequence for I in X. Then, $(v_n)$ has a subsequence, still denoted by $(v_n)$, such that

$$\begin{aligned} \frac{F(G^{-1}(v_n))}{|x|^a} \rightarrow \frac{F(G^{-1}(v))}{|x|^a} \ \ \text{ in } \ \ L^1(\mathbb {R}^2), \ \ \text{ as } \ \ n \rightarrow +\infty , \end{aligned}$$

where v is the weak limit of $(v_n)$ in $H^1(\mathbb {R}^2)$ with $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$.

Proof

From Lemma 2.2-(4) and Corollary 4.2 we have

$$\begin{aligned} \frac{1}{\alpha }\int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))G^{-1}(v_n)|}{|x|^a}\mathrm {d}x \le \int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))v_n|}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x \le C. \end{aligned}$$

Thus, similarly to Lemma 4.3, we get

$$\begin{aligned} \frac{f(G^{-1}(v_n))}{|x|^a} \rightarrow \frac{f(G^{-1}(v))}{|x|^a} \ \ \text{ in } \ \ L^1_{loc}(\mathbb {R}^2), \ \ \text{ as } \ \ n \rightarrow + \infty . \end{aligned}$$

(4.6)

Next, by using $(f_2)$ and $(f_3)$, for each $R > 0$, there exists $C > 0$ such that $F(G^{-1}(v_n)) \le C[f(G^{-1}(v_n))]$ in $\overline{B_R}$. This together with (4.6) and the generalized Lebesgue dominated convergence theorem, up to a subsequence, implies that

$$\begin{aligned} \frac{F(G^{-1}(v_n))}{|x|^a} \rightarrow \frac{F(G^{-1}(v))}{|x|^a} \ \ \text{ in } \ \ L^1(B_R), \ \ \text{ for } \text{ all } \ R > 0. \end{aligned}$$

To conclude the convergence of the lemma, it is sufficient to prove that given $\delta > 0$, there exists $R > 0$ such that

$$\begin{aligned} \int _{B_R^c} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \le \delta \ \ \text{ and } \ \ \int _{B_R^c} \frac{F(G^{-1}(v))}{|x|^a}\mathrm {d}x \le \delta . \end{aligned}$$

For this, we also note that by $(f_2)$ and $(f_3)$, there exists $C_1 > 0$ satisfying

$$\begin{aligned} |F(x,s)| \le C_1|f(x,s)|, \ \ \text{ for } \text{ all }\ \ (x,s) \in \mathbb {R}^2 \times \mathbb {R}. \end{aligned}$$

Thus, for each $A > 0$, we obtain

$$\begin{aligned} \begin{aligned} \int _{|x|> R\atop {|G^{-1}(v_n)|> A}} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x&\le C_1 \int _{|x|> R\atop {|G^{-1}(v_n)| > A}} \frac{|f(G^{-1}(v_n))|}{|x|^a}\mathrm {d}x \\&\le \frac{C_1}{A}\int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))G^{-1}(v_n)|}{|x|^a}\mathrm {d}x. \end{aligned} \end{aligned}$$

Since

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))G^{-1}(v_n)|}{|x|^a}\mathrm {d}x \le C, \end{aligned}$$

given $\delta > 0$, we may choose $A > 0$ such that

$$\begin{aligned} \frac{C_1}{A}\int _{\mathbb {R}^2} \frac{|f(G^{-1}(v_n))G^{-1}(v_n)|}{|x|^a}\mathrm {d}x < \frac{\delta }{2}. \end{aligned}$$

Thus,

$$\begin{aligned} \int _{|x|> R\atop {|G^{-1}(v_n)| > A}} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \le \frac{\delta }{2}. \end{aligned}$$

(4.7)

Moreover, since f has critical exponential growth and satisfies $(f_1)$ and $(f_2)$, there exists $C(A) > 0$ such that

$$\begin{aligned} F(x,G^{-1}(s)) \le C(A) |G^{-1}(s)|^2, \ \ \text{ for } \text{ all }\ \ (x,G^{-1}(s)) \in \mathbb {R}^2 \times [-A,A]. \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \int _{|x|> R \atop {|G^{-1}(v_n)| \le A}} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x&\le C(A)\int _{|x|> R\atop {|G^{-1}(v_n)| \le A}} \frac{|G^{-1}(v_n)|^2}{|x|^a}\mathrm {d}x \\&\le 2C(A)\int _{|x|> R\atop {|G^{-1}(v_n)| \le A}} \frac{|G^{-1}(v_n) - G^{-1}(v)|^2}{|x|^a}\mathrm {d}x \\&\quad + \ 2C(A)\int _{|x| > R\atop {|G^{-1}(v_n)| \le A}} \frac{|G^{-1}(v)|^2}{|x|^a}\mathrm {d}x. \end{aligned} \end{aligned}$$

Hence, by using Proposition (2.3), given $\delta > 0$, we may choose $R > 0$ satisfying

$$\begin{aligned} \int _{|x| > R \atop {|G^{-1}(v_n)| \le A}} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \le \frac{\delta }{2}. \end{aligned}$$

(4.8)

From (4.7) and (4.8), given $\delta > 0$, there exists $R > 0$ such that

$$\begin{aligned} \int _{|x| > R} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \le \delta . \end{aligned}$$

Similarly, we obtain

$$\begin{aligned} \int _{|x| > R} \frac{F(G^{-1}(v))}{|x|^a}\mathrm {d}x \le \delta . \end{aligned}$$

Combining all the above estimates and since $\delta > 0$ is arbitrary, it follows that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x \rightarrow \int _{\mathbb {R}^2} \frac{F(G^{-1}(v))}{|x|^a}\mathrm {d}x, \ \ \text{ as } \ n \rightarrow + \infty , \end{aligned}$$

and this completes the proof. $\square$

Lemma 4.5

Suppose that (V), $(g_0)-(g_1)$ and $(f_1)-(f_2)$ are satisfied. If $(v_n) \subset X$ is a Cerami sequence for I such that $v_n \rightharpoonup v$ weakly in $H^1(\mathbb {R}^2)$ with $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$, then

$$\begin{aligned}&\int _{\mathbb {R}^2} \nabla v \nabla \varphi \mathrm {d}x +\int _{\mathbb {R}^2} \frac{V(x)G^{-1}(v)}{g(G^{-1}(v))}\varphi \mathrm {d}x\\&\quad = \int _{\mathbb {R}^2} \frac{f(G^{-1}(v))}{g(G^{-1}(v)|x|^a}\varphi \mathrm {d}x, \ \ \text{ for } \text{ all }\ \ \varphi \in C_0^{\infty }(\mathbb {R}^2). \end{aligned}$$

Proof

First, we have that $I'(v)\varphi$ is well defined for $\varphi \in C_0^{\infty }(\mathbb {R}^2)$ and therefore just prove that $I'(v)\varphi = 0$ for all $\varphi \in C_0^{\infty }(\mathbb {R}^2)$. Note that

$$\begin{aligned}&I'(v_n)\varphi - I'(v)\varphi - \int _{\mathbb {R}^2} (\nabla v_n - \nabla v)\nabla \varphi \mathrm {d}x \nonumber \\&\quad = \int _{\mathbb {R}^2} \left[ \frac{G^{-1}(v_n)}{g(G^{-1}(v_n))} - \frac{G^{-1}(v)}{g(G^{-1}(v))}\right] V(x)\varphi \mathrm {d}x \nonumber \\&\qquad + \int _{\mathbb {R}^2} \left[ \frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n))|x|^a} - \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a}\right] \varphi \mathrm {d}x . \end{aligned}$$

(4.9)

In view of $v_n \rightharpoonup v$ weakly in $H^1(\mathbb {R}^2)$, we have $v_n \rightarrow v$ in $L^p_{loc} (\mathbb {R}^2)$, with $p \ge 1$. Then, up to a subsequence,

$$\begin{aligned} \begin{array}{lll} v_n(x) \rightarrow v(x) \ \ \text{ a.e. } \ \text{ in } \ \mathcal {K}:= \text{ supp } \ \varphi , \ \text{ as } \ n \rightarrow +\infty , \\ |v_n(x)| \le |w_p(x)| \ \ \text{ for } \text{ every } \ n \in \mathbb {N} \ \text{ and } \text{ a.e. } \text{ in } \ \mathcal {K}, \ \text{ with } \ w_p \in L^{p}(\mathcal {K}). \end{array} \end{aligned}$$

Consequently,

$$\begin{aligned} \frac{G^{-1}(v_n)}{g(G^{-1}(v_n))} \rightarrow \frac{G^{-1}(v)}{g(G^{-1}(v))} \ \ \text{ a.e. } \text{ in } \ \ \mathcal {K}, \ \text{ as } \ n \rightarrow +\infty . \end{aligned}$$

Furthermore, by the continuity of V and Lemma 2.2-(2) and (3), there exists a constant $C > 0$ such that

$$\begin{aligned} \frac{|V(x)G^{-1}(v_n)\varphi |}{g(G^{-1}(v_n))} \le |V(x)v_n \varphi | \le C|w_2\Vert \varphi | \in L^1(\mathcal {K}). \end{aligned}$$

Using these estimates, Lebesgue Dominated Convergence Theorem and the weak convergence $v_n \rightharpoonup v$ in $H^1(\mathbb {R}^2)$, we obtain

$$\begin{aligned} \int _{\mathbb {R}^2} (\nabla v_n - \nabla v)\nabla \varphi \mathrm {d}x \rightarrow 0 \ \ \text{ and } \ \ \int _{\mathbb {R}^2} \left[ \frac{G^{-1}(v_n)}{g(G^{-1}(v_n))} - \frac{G^{-1}(v)}{g(G^{-1}(v))}\right] V(x)\varphi \mathrm {d}x \rightarrow 0, \end{aligned}$$

as $n \rightarrow +\infty$. In addition, by Lemma 4.3, we have

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n))|x|^a}\varphi \mathrm {d}x \rightarrow \int _{\mathbb {R}^2} \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a}\varphi \mathrm {d}x. \end{aligned}$$

Hence, taking the limit in (4.9), we get $I'(v_n)\varphi - I'(v)\varphi \rightarrow 0$ for all $\varphi \in C_0^{\infty }(\mathbb {R}^2)$ and once $I'(v_n) \rightarrow 0$, we conclude $I'(v)\varphi = 0$ for all $\varphi \in C_0^{\infty }(\mathbb {R}^2)$. This finalizes the proof. $\square$

Lemma 4.6

Suppose that (V), $(g_0){-}(g_1)$ and $(f_1){-}(f_2)$ are satisfied. Let $(v_n)$ be a Cerami sequence for I in X such that $\left( \frac{\alpha }{\beta }\right) ^2\varsigma _0\Vert \nabla v_n\Vert _2^2 < 2\pi (2 - a)$. Then, $(v_n)$ has a subsequence, still denoted by $(v_n)$, such that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))(v - v_n)}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x \rightarrow 0, \end{aligned}$$

as $n \rightarrow +\infty$, where v is the weak limit of $(v_n)$ in $H^1(\mathbb {R}^2)$ with $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$.

Proof

By (2.6), given $\varepsilon > 0$, there exists $C_{\varepsilon } > 0$ such that

$$\begin{aligned} \left| \frac{f(G^{-1}(v_n))(v - v_n)}{g(G^{-1}(v_n))}\right| \le \varepsilon |G^{-1}(v_n)\Vert v - v_n| + C_{\varepsilon }[e^{(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1]|v - v_n|. \end{aligned}$$

Hence, by Lemma 2.2-(5), one has

$$\begin{aligned} \begin{aligned}&\left| \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))(v - v_n)}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x\right| \\&\quad \le \varepsilon C_1\int _{\mathbb {R}^2} \frac{|G^{-1}(v_n)\Vert G^{-1}(v - v_n)|}{|x|^a}\mathrm {d}x \\&\qquad + \varepsilon C_1 \int _{\mathbb {R}^2} \frac{|G^{-1}(v_n)\Vert G^{-1}(v - v_n)|^{\alpha }}{|x|^a}\mathrm {d}x \\&\qquad + C_{\varepsilon } \int _{\mathbb {R}^2} \frac{[e^{(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1]|G^{-1}(v - v_n)|}{|x|^a}\mathrm {d}x \\&\qquad + C_{\varepsilon } \int _{\mathbb {R}^2} \frac{[e^{(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1]|G^{-1}(v - v_n)|^{\alpha }}{|x|^a}\mathrm {d}x. \end{aligned} \end{aligned}$$

By Hölder’s inequality and choosing $t > 1$ such that $t' = t/(t - 1) \ge 2$, we get

$$\begin{aligned} \begin{aligned}&\left| \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))(v - v_n)}{g(G^{-1}(v_n))|x|^a}\mathrm {d}x\right| \\&\quad \le \displaystyle \varepsilon C_1\left( \int _{\mathbb {R}^2} \frac{|G^{-1}(v_n)|^2}{|x|^a}\mathrm {d}x\right) ^{\frac{1}{2}}\left( \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^2}{|x|^a}\mathrm {d}x\right) ^{\frac{1}{2}} \\&\qquad + \varepsilon C_1\left( \int _{\mathbb {R}^2} \frac{|G^{-1}(v_n)|^2}{|x|^a}\mathrm {d}x\right) ^{\frac{1}{2}}\left( \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^{2\alpha }}{|x|^a}\mathrm {d}x\right) ^{\frac{1}{2}} \\&\qquad + C_{\varepsilon } \left\{ \int _{\mathbb {R}^2} \frac{[e^{t(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1]}{|x|^a}\mathrm {d}x\right\} ^{\frac{1}{t}}\left\{ \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^{t'}}{|x|^a}\mathrm {d}x\right\} ^{\frac{1}{t'}} \\&\qquad + C_{\varepsilon } \left\{ \int _{\mathbb {R}^2} \frac{[e^{t(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1]}{|x|^a}\mathrm {d}x\right\} ^{\frac{1}{t}}\left\{ \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^{\alpha t'}}{|x|^a}\mathrm {d}x\right\} ^{\frac{1}{t'}} . \end{aligned}\nonumber \\ \end{aligned}$$

(4.10)

Next, note that there exists $t > 1$ sufficiently close to 1, $\varepsilon > 0$ sufficiently small and $C > 0$ such that

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{e^{t(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1}{|x|^a}\mathrm {d}x \le C. \end{aligned}$$

(4.11)

Indeed, we can infer that for n sufficiently large, there exists $t > 1$, sufficiently close to 1, and $\varepsilon > 0$ sufficiently small so that $\left( \frac{\alpha }{\beta }\right) ^2t(\varsigma _0 + \varepsilon )\Vert \nabla v_n\Vert _2^2 < 2\pi (2 - a)$. Hence, by Lemma 2.2-(7) and Lemma 2.4, we get

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{e^{t(\varsigma _0 + \varepsilon )|G^{-1}(v_n)|^{2\alpha }} - 1}{|x|^a}\mathrm {d}x \le \int _{\mathbb {R}^2} \frac{e^{t(\varsigma _0 + \varepsilon )\left( \frac{\alpha }{\beta }\right) ^2\Vert \nabla v_n\Vert _2^2 \left( \frac{|v_n|}{\Vert \nabla v_n\Vert _2}\right) ^2} - 1}{|x|^a}\mathrm {d}x \le C, \end{aligned}$$

which proves (4.11). Since $G^{-1}(v_n - v)$ is a bounded sequence in X and for $p \in [2, +\infty )$ the embedding $X \hookrightarrow L^p (\mathbb {R}^2; |x|^{-a}\mathrm {d}x)$ is compact, up to a subsequence, we have

$$\begin{aligned} \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^{t'}}{|x|^a}\mathrm {d}x \rightarrow 0 \quad \text{ and } \quad \int _{\mathbb {R}^2} \frac{|G^{-1}(v - v_n)|^{\alpha t'}}{|x|^a}\mathrm {d}x \rightarrow 0. \end{aligned}$$

Therefore, from (4.10) and (4.11) we conclude the proof of the theorem. $\square$

We recall that the minimax level of I is given by

$$\begin{aligned} 0 < c_m = \inf _{\gamma \in \Gamma } \max _{t \in [0,1]} I(\gamma (t)), \end{aligned}$$

(4.12)

where $\Gamma = \{\gamma \in C([0,1];X): \gamma (0) = 0 \ \text{ and } \ \gamma (1) = e\}$ and e was given in Lemma 3.4.

As a consequence of Lemma 4.6, we have the following result, which is essential for the proof of Theorem 1.1.

Corollary 4.7

Suppose that (V), $(g_0)-(g_2)$ and $(f_1)-(f_2)$ are satisfied. Let $(v_n)$ be a Cerami sequence for I in X at the level $c_m$ satisfying $\left( \alpha /\beta \right) ^2 \varsigma _0\Vert \nabla v_n\Vert ^2_2 < 2\pi (2 - a)$ and $v_n \rightharpoonup 0$ weakly in X. Then $c_m = 0$, where $c_m$ is given in (4.12).

Proof

Indeed, since $I'(v_n)v_n \rightarrow 0$,

$$\begin{aligned} \int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x + \int _{\mathbb {R}^2} \frac{V(x)G^{-1}(v_n)}{g(G^{-1}(v_n)}v_n = \int _{\mathbb {R}^2}\frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n)|x|^a}v_n + o_n(1). \end{aligned}$$

Hence, by Lemma 2.2-(4) we have

$$\begin{aligned} \begin{aligned}&\int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x + \frac{1}{\alpha } \int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^2\mathrm {d}x\\&\quad \le \int _{\mathbb {R}^2}\frac{f(G^{-1}(v_n))}{g(G^{-1}(v_n)|x|^a}v_n + o_n(1) \le \int _{\mathbb {R}^2} \frac{f(G^{-1}(v_n))G^{-1}(v_n)}{|x|^a}\mathrm {d}x + o_n(1). \end{aligned}\nonumber \\ \end{aligned}$$

(4.13)

Moreover, as $I(v_n) \rightarrow c_m$ we get

$$\begin{aligned} c_m= & {} \frac{1}{2} \int _{\mathbb {R}^2} |\nabla v_n|^2\mathrm {d}x + \frac{1}{2}\int _{\mathbb {R}^2} V(x)[G^{-1}(v_n)]^2\mathrm {d}x\nonumber \\&- \int _{\mathbb {R}^2} \frac{F(G^{-1}(v_n))}{|x|^a}\mathrm {d}x + o_n(1). \end{aligned}$$

(4.14)

Then, by (4.13), (4.14) and Lemma 4.6-(2),(3), we conclude that $c_m = 0$ as we desired. $\square$

5 Minimax Level Estimate

In this section, we obtain an estimate for the mountain pass level of I, which will be crucial to study the behavior of Cerami sequences for I. For this, let $r > 0$ and consider the Moser’s sequence defined by

$$\begin{aligned} M_n(x,r) = \frac{1}{\sqrt{2\pi }} \left\{ \begin{array}{lll} \sqrt{\log n}, &{} \text{ if } \ |x| \le \frac{r}{n}, \\ \dfrac{\log (r/|x|)}{\sqrt{\log n}}, &{} \text{ if } \ \frac{r}{n} \le |x| \le r, \\ 0, &{} \text{ if } \ |x| > r, \end{array}\right. \end{aligned}$$

which satisfies $M_n \in H_0^1(B_r)$, $\Vert \nabla M_n\Vert _2 = 1$ for all $n \in \mathbb {N}$ and

$$\begin{aligned} \Vert M_n\Vert _2^2 = \frac{r^2}{4 \log n} - \frac{r^2}{2n^2} - \frac{r^2}{4n^2 \log n}. \end{aligned}$$

Proposition 5.1

Assume that (V), $(g_0)-(g_2)$, $(f_1)$, $(f_2)$ and $(f_4)$ are satisfied. Then, the minimax level $c_m$ satisfies

$$\begin{aligned} c_m < \frac{(2 - a)\pi }{(\frac{\alpha }{\beta })^2 \varsigma _0}. \end{aligned}$$

(5.1)

Proof

To prove (5.1), it is sufficient to obtain $n \in \mathbb {N}$ such that

$$\begin{aligned} \max _{t \ge 0} I(t\widetilde{M}_n) < \frac{(2 - a)\pi }{(\frac{\alpha }{\beta })^2 \varsigma _0}, \end{aligned}$$

where $\widetilde{M}_n = M_n/\Vert M_n\Vert$. Suppose, for the sake of contradiction, that for all $n \in \mathbb {N}$, we have

$$\begin{aligned} \max _{t \ge 0} I(t\widetilde{M}_n) \ge \frac{(2 - a)\pi }{(\frac{\alpha }{\beta })^2 \varsigma _0}. \end{aligned}$$

(5.2)

In view of Lemma 3.3 and Lemma 3.4, for all $n \in \mathbb {N}$, there exists $t_n > 0$ such that

$$\begin{aligned} I(t_n\widetilde{M}_n) = \max _{t \ge 0} I(t\widetilde{M}_n). \end{aligned}$$

(5.3)

By Lemma 2.2-(3), (5.2), (5.3), $(f_2)$ and $\Vert \widetilde{M}_n\Vert = 1$, it follows that

$$\begin{aligned} t_n^2 \ge \frac{2(2 - a)\pi }{(\frac{\alpha }{\beta })^2 \varsigma _0}, \end{aligned}$$

(5.4)

because

$$\begin{aligned} \begin{aligned} \frac{t_n^2}{2}&= \frac{t_n^2}{2}\int _{\mathbb {R}^2} \left( |\nabla \widetilde{M}_n|^2 + V(x)\widetilde{M}_n^2\right) \mathrm {d}x \\&\ge \frac{1}{2}\int _{\mathbb {R}^2} \left\{ |\nabla (t_n\widetilde{M}_n)|^2 + V(x)[G^{-1}(t_n\widetilde{M}_n)]^2\right\} \mathrm {d}x\\&\quad - \int _{\mathbb {R}^2} \frac{F(G^{-1}(t_n\widetilde{M}_n))}{|x|^a}\mathrm {d}x\ge \frac{(2 - a)\pi }{\left( \frac{\alpha }{\beta }\right) ^2\varsigma _0}. \end{aligned} \end{aligned}$$

Next, we will show that the sequence $(t_n)$ is bounded. To achieve this goal, let us remember that $\frac{d}{\mathrm {d}t} I(t\widetilde{M}_n) = 0$ at $t = t_n$, that is, $I'(t_n\widetilde{M}_n)\cdot \widetilde{M}_n = 0$. Thus,

$$\begin{aligned}&t^2_n \int _{\mathbb {R}^2} \left[ |\nabla \widetilde{M}_n|^2 + t_n^{-2} V(x) \frac{G^{-1}(t_n \widetilde{M}_n)}{g(G^{-1}(t_n \widetilde{M}_n))}t_n \widetilde{M}_n\right] \mathrm {d}x\\&\quad - \int _{\mathbb {R}^2} \frac{f(G^{-1}(t_n\widetilde{M}_n))}{g(G^{-1}(t_n\widetilde{M}_n))|x|^a}t_n\widetilde{M}_n\mathrm {d}x = 0. \end{aligned}$$

By Lemma 2.2-(4), $(f_2)$ and $\Vert \nabla \widetilde{M}_n\Vert _2 \le 1$, one has

$$\begin{aligned} t_n^2= & {} t_n^2\int _{\mathbb {R}^2} \left[ |\nabla \widetilde{M}_n|^2 + V\frac{t_n^2\widetilde{M}_n^2}{t_n^2} \right] \mathrm {d}x \ge t_n^2 \int _{\mathbb {R}^2} \left[ |\nabla \widetilde{M}_n|^2 + \frac{G^{-1}(t_n \widetilde{M}_n)t_n \widetilde{M}_n}{t_n^2 g(G^{-1}(t_n\widetilde{M}_n))}\right] \mathrm {d}x \nonumber \\= & {} \int _{\mathbb {R}^2} \frac{f(G^{-1}(t_n\widetilde{M}_n))}{g(G^{-1}(t_n\widetilde{M}_n))|x|^a}t_n\widetilde{M}_n\mathrm {d}x \ge \int _{B_\frac{r}{n}(0)} \frac{f(G^{-1}(t_n\widetilde{M}_n))}{g(G^{-1}(t_n\widetilde{M}_n))|x|^a}t_n\widetilde{M}_n\mathrm {d}x \nonumber \\\ge & {} \frac{1}{\alpha } \int _{B_\frac{r}{n}(0)} \frac{f(G^{-1}(t_n\widetilde{M}_n))G^{-1}(t_n\widetilde{M}_n)}{|x|^a}\mathrm {d}x. \end{aligned}$$

(5.5)

According to $(f_4)$, given $\varepsilon > 0$ there exists $R_{\varepsilon } > 0$ such that

$$\begin{aligned} sf(s) \ge (\xi _0 - \varepsilon )e^{\varsigma _0 |s|^{2\alpha }}, \ \ \text{ for } \text{ all } \ s \ge R_{\varepsilon }. \end{aligned}$$

(5.6)

Since $G^{-1}(t_n \widetilde{M}_n) > R_{\varepsilon }$ in $B_{\frac{r}{n}}(0)$ for n sufficiently large, using (5.5) and (5.6), we obtain

$$\begin{aligned} t_n^2 \ge \frac{\xi _0 - \varepsilon }{\alpha }\int _{B_{\frac{r}{n}(0)}} \frac{e^{\varsigma _0|G^{-1}(t_n \widetilde{M}_n)|^{2\alpha }}}{|x|^a}\mathrm {d}x. \end{aligned}$$

(5.7)

In view of Lemma 2.2-(7), given $\eta > 0$ there exists $R_{\eta } > 0$ such that

$$\begin{aligned} |G^{-1}(s)|^{2\alpha } \ge \left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] s^2, \ \ \text{ for } \text{ all } \ s \ge R_{\eta }. \end{aligned}$$

(5.8)

Thus, for n sufficiently large (without loss of generality we can assume $R_{\varepsilon } > R_{\eta }$), using (5.7) and (5.8) we get

$$\begin{aligned} t_n^2\ge & {} \frac{\xi _0 - \varepsilon }{\alpha } \int _{B_{\frac{r}{n}(0)}} \frac{e^{\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] t_n^2 \widetilde{M}_n^2}}{|x|^a}\mathrm {d}x \nonumber \\= & {} \frac{\xi _0 - \varepsilon }{\alpha } e^{\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] \frac{1}{2\pi }\frac{\log n}{\Vert M_n\Vert ^2}t_n^2} \frac{2\pi }{2 - a}\left( \frac{r}{n}\right) ^{2 - a} \nonumber \\= & {} \frac{\xi _0 - \varepsilon }{\alpha } e^{\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] \frac{1}{2\pi }\frac{\log n}{\Vert M_n\Vert ^2}t_n^2 - (2 - a)\log n} \frac{2\pi }{2 - a}r^{2 - a} . \end{aligned}$$

(5.9)

Hence,

$$\begin{aligned} 1 \ge \frac{\xi _0 - \varepsilon }{\alpha } e^{\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] \frac{1}{2\pi }\frac{\log n}{\Vert M_n\Vert ^2}t_n^2 - (2 - a)\log n - 2\log t_n} \frac{2\pi }{2 - a}r^{2 - a}, \end{aligned}$$

(5.10)

which implies

$$\begin{aligned} \varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] \frac{1}{2\pi }\frac{\log n}{\Vert M_n\Vert ^2}t_n^2 - (2 - a)\log n - 2\log t_n \le C. \end{aligned}$$

This estimate shows that $(t_n)$ is bounded, otherwise, once $\Vert M_n\Vert ^2 \le 1 + \Vert V\Vert _{L^{\infty }(B_r)}\Vert M_n\Vert _2^2$, we have

$$\begin{aligned} \begin{aligned}&\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] \frac{1}{2\pi }\frac{\log n}{\Vert M_n\Vert ^2}t_n^2 - (2 - a)\log n - 2\log t_n \\&\quad \ge \displaystyle t_n^2\log n \left\{ \frac{\varsigma _0\left[ \left( \frac{\alpha }{\beta }\right) ^2 - \eta \right] }{2\pi \left( 1 + \Vert V\Vert _{L^{\infty }(B_r)\Vert M_n\Vert _2}\right) } - \frac{2 - a}{t_n^2} - \frac{2\log t_n}{t_n^2 \log n}\right\} \\&\quad \rightarrow +\infty , \ \ \text{ as } \ n \rightarrow +\infty , \end{aligned} \end{aligned}$$

which is a contradiction with (5.10). Thus, by (5.4), (5.9) and since $(t_n)$ is bounded, there are constants $C_1 = C_1(a,\varsigma _0, \alpha , \beta , \eta )>0$ and $C_2 > 0$ such that

$$\begin{aligned} C_1 \frac{\log n}{\Vert M_n\Vert ^2} - \log n \le C_2. \end{aligned}$$

(5.11)

However,

$$\begin{aligned} \begin{aligned} C_1 \frac{\log n}{\Vert M_n\Vert ^2} - \log n&= \frac{C_1\log n - \Vert M_n\Vert ^2 \log n}{\Vert M_n\Vert ^2} \\&\ge \frac{C_1 \log n - \left[ 1 + \Vert V\Vert _{L^{\infty }(B_r)}\left( \frac{r^2}{4 \log n} - \frac{r^2}{2n^2} - \frac{r^2}{4n^2 \log n}\right) \right] \log n}{1 + \Vert V\Vert _{L^{\infty }(B_r)}\left( \frac{r^2}{4 \log n} - \frac{r^2}{2n^2} - \frac{r^2}{4n^2 \log n}\right) } \\&= \frac{(C_1 - 1)\log n + \Vert V\Vert _{L^{\infty }(B_r)}\left( \frac{r^2}{4n^2} + \frac{r^2 \log n}{2n^2} - \frac{r^2}{4}\right) }{1 + \Vert V\Vert _{L^{\infty }(B_r)}\left( \frac{r^2}{4 \log n} - \frac{r^2}{2n^2} - \frac{r^2}{4n^2 \log n}\right) } \longrightarrow +\infty , \end{aligned} \end{aligned}$$

as $n \rightarrow +\infty$, which contradicts (5.11). The proposition is proved. $\square$

6 Proof of Theorem 1.1

According to Lemma 3.3 and Lemma 3.4, the hypotheses of Theorem 2.7 are satisfied. Thus, the minimax level $c_m$ of I is positive and there is a Cerami sequence $(v_n)$ for I at the level $c_m$. Applying Lemma 4.1 and 4.3, we may assume, without loss generality, that $v_n \rightharpoonup v$ weakly in $H^1(\mathbb {R}^2)$ for some $v \in H^1(\mathbb {R}^2)$ with $\int _{\mathbb {R}^2}V(x)|G^{-1}(v)|^2 \mathrm {d}x<\infty$. From Lemma 4.5, v is a weak solution of equation (2.5). Now, suppose by contradiction, that v is zero. In view of Lemma 4.4 and since $I(v_n) \rightarrow c_m$ as $n \rightarrow +\infty$, we reach

$$\begin{aligned} \frac{1}{2}\int _{\mathbb {R}^2} \left\{ |\nabla v_n|^2 + V(x)\left[ G^{-1}(v_n)\right] ^2\right\} \mathrm {d}x = c_m + o_n(1). \end{aligned}$$

(6.1)

From Proposition 5.1, we have

$$\begin{aligned} c_m < (2 - a)\pi /(\frac{\alpha }{\beta })^2 \varsigma _0. \end{aligned}$$

(6.2)

Using condition (V), (6.1) and (6.2), there exists $n_0 \in \mathbb {N}$ such that

$$\begin{aligned} \left( \frac{\alpha }{\beta }\right) ^2 \varsigma _0 \Vert \nabla v_n\Vert _2^2 < 2\pi (2 - a), \ \ \text{ for } \text{ all } \ n \ge n_0. \end{aligned}$$

Thus, in view of Corollary 4.7, we get $c_m = 0$, which is a contradiction. Therefore, $v \ne 0$.

Next, we prove that v is nonnegative. Indeed, if $v^- = \max \{-v,0\}$ then $v^{-} \in H^1(\mathbb {R}^2)$ and by density we get

$$\begin{aligned} \int _{\mathbb {R}^2} |\nabla v^-|^2\mathrm {d}x + \int _{\mathbb {R}^2} V(x)\frac{G^{-1}(v)}{g(G^{-1}(v))}(-v^-)\mathrm {d}x = \int _{\mathbb {R}^2} \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a}(-v^-)\mathrm {d}x \le 0. \end{aligned}$$

On the other hand, we know that $\frac{G^{-1}(v)}{g(G^{-1}(v))}(-v^-) \ge 0$ and this implies that $\int _{\mathbb {R}^2} |\nabla v^-|^2\mathrm {d}x = 0$. Thus, $v^- = 0$ almost everywhere in $\mathbb {R}^2$ and therefore $v \ge 0$. In order to prove that $v > 0$ in $\mathbb {R}^2$, we suppose, otherwise, that there exists $x_0 \in \mathbb {R}^2$ such that $v(x_0) = 0$. Notice that 2.5 can be written in the form

$$\begin{aligned} -\Delta v + c(x)v = V(x)\frac{v - G^{-1}(v)}{g(G^{-1}(v))} + \frac{f(G^{-1}(v))}{g(G^{-1}(v))|x|^a} \ge 0 \end{aligned}$$

where $c(x) = V(x)\frac{v}{g(G^{-1}(v))} > 0$ for all $x \in \mathbb {R}^2$. Recalling that $v \in C_{loc}^{0,\vartheta }(\mathbb {R}^2)$, using Strong Maximum Principle (see [20], Theorem 8.19) in an arbitrary ball centered in $x_0$, we can conclude that $v \equiv 0$, which is impossible. Therefore, v has to be strictly positive. In view of Proposition 2.6 we reach $u = G^{-1}(v)$ is a positive solution of (1.1) and the proof of Theorem 1.1 is complete.