1. Introduction
We investigate the existence of a non-trivial solution of the elliptic problem
with $g(s)=\mu f(s) + |s|^{2^*-2}s$, $N\geq 3$, µ > 0, under no periodicity condition on V that changes sign, and $s\mapsto f(s)/s$ is bounded.
This problem has been extensively studied considering several potentials V and non-linearities g. For the case where V and $g(s)=g(x,s)$ are periodic functions in the variable x and g have a subcritical growth, we refer the reader to [Reference Jeanjean7–Reference Li and Szulkin9, Reference Pankov and Pflüger13, Reference Stuart and Zhou16, Reference Szulkin and Weth17 ] and references therein.
The case when g possesses critical growth and potential V changes sign, we refer the works of [Reference Alves and Germano2, Reference Chabrowski and Szulkin4, Reference Schechter and Zou14, Reference Zhang, Xu and Zhang18 ], which are more closely related to this article. In all of them, the condition of periodicity is crucial in order to overcome the lack of compactness in $\mathbb{R}^N$.
On the other side, Maia e Soares in [Reference Maia and Soares12] considered the case when V is not periodic (namely, $V(x)\to V_\infty \gt 0$ as $|x|\to \infty$) and g is asymptotically linear at infinity, without any monotonicity condition on $s\mapsto g(s)/s$. The framework that the authors deal with the problem makes possible to apply the celebrated result due to Berestycki and Lions in [Reference Berestycki and Lions3] and ensures that the limit problem associated to $(P_\mu)$
has a non-trivial ground state solution $u_0\in C^2(\mathbb{R}^N,\mathbb{R})$, which is positive, radially symmetric and decays exponentially, namely,
As quoted by the authors in page 21, after some interactions between the energy functionals associated to $(P_\mu)$ and $(P_{\mu,\infty})$, property (1.1) of the solution u 0 was strongly needed in order to prove that the weak solution of $(P_\mu)$ is non-trivial (see Section 5 in [Reference Maia and Soares12], p. 19).
Our work complements all results cited above. Differently from them, the potential V is non-periodic and changes sign and the non-linearity $g(s)=\mu f(s) + |s|^{2^*-2}s$ possesses a critical growth, with $s\mapsto f(s)/s$ bounded. This scenario brings several difficulties.
The central idea of our approach is to apply the version of the Linking Theorem due to Maia and Soares in [Reference Maia and Soares12, Theorem 1.2] and, for that purpose, takes a positive ground state solution of $(P_{\mu,\infty})$ that has an exponential decay and makes some suitable interactions with problem $(P_\mu)$. Since our problem is critical, we cannot apply [Reference Berestycki and Lions3] directly as the authors do in [Reference Maia and Soares12]. Therefore, how to guarantee that $(P_{\mu,\infty})$ has some non-trivial ground state solution? And, then, is it possible to show that this solution has exponential decay as (1.1)?
In this paper, to prove that problem $(P_\mu)$ has a non-trivial solution, we first answer these two question, considering the elliptic problem
with $N\geq 3$, µ > 0 and $u\in E:= H^1(\mathbb{R}^N)$ and $V:\mathbb{R}^N\to\mathbb{R}$ is a potential satisfying the conditions:
$(V_1)$ $V\in L^\infty(\mathbb{R}^N)$;
$(V_2)$ $\displaystyle\lim_{|x|\rightarrow +\infty}V(x)=V_\infty \gt 0$;
$(V_3)$$0\notin \sigma(L)$ and $\inf\sigma(L) \lt 0$, where $\sigma(L)$ is the spectrum of the operator $L=-\Delta+V$.
$(V_4)$$V(x) \leq V_\infty - Ce^{-\gamma_1|x|^{\gamma_2}}$, with $\gamma_1 \gt 0$ and $\gamma_2\in (0,1)$.
The conditions that we consider on the non-linearity $f\in C(\mathbb{R},\mathbb{R})$ are the following:
$(f_1)$ $\displaystyle\lim_{s\rightarrow 0}\displaystyle\frac{f(s)}{s}=0$ and $f(s)=0$, for all $s\in(-\infty, 0]$ ;
$(f_2)$ We have $f(s)=0$ for $s\leq 0$ and $\displaystyle\frac{|f(s)|}{|s|} \lt m$ for all s ≠ 0;
$(f_3)$ If $F(s):=\displaystyle\int_{0}^{s}f(t)\,{\rm d}t$ and $Q(s):=\displaystyle\frac{1}{2}f(s)s-F(s)$, then for all $s\in\mathbb{R}\setminus\{0\}$,
An important consequence of assumptions $(f_1)$ and $(f_2)$ is that, given ɛ > 0 and $2\leq p\leq 2^*$, there exists $C_\varepsilon \gt 0$ such that
for all $s\in\mathbb{R}$.
One of our main results is the next theorem.
Theorem 1.1. Suppose that assumptions $(V_1)-(V_4)$ and $(f_1)-(f_3)$ hold. Then, there exists $\mu^* \gt 0$ such that, if $\mu\geq \mu^*$, problem $(P_\mu)$ has a nontrivial and nonnegative solution uµ in $H^1(\mathbb{R}^N)$. Moreover, the limit problem
has a ground state solution u 0 such that, if $\nu\in\left(0,\sqrt{V_\infty}\right)$, then there exists $C=C(m,\nu) \gt 0$ satisfying
We stress here that, in order to prove Theorem 1.1, we do not need any monotonicity hypothesis on function $s\mapsto f(s)/s$, as one can find in the literature about similar problems. One example of non-linearity that satisfies our assumptions but $f(s)/s$, s ≠ 0, is not increasing is $f(s)=\dfrac{s^7-1,5s^5+2s^3}{1+s^6}$ for $s\in\mathbb{R}$. Then f satisfies our hypotheses; however, $f(s)/s$, s ≠ 0, is not increasing.
This paper is organized into seven sections as follows. In § 2, we focus on providing the appropriate variational setting for the problem. In § 3, we obtain the geometry of a version of the Linking Theorem, and as a result, we obtain an appropriate Cerami sequence. This sequence is proven to be bounded in § 4. In § 5, we present the first part of the proof of Theorem 1.1, and in § 6, Appendix A, we present the second part of the proof and we also discuss the limit problem $(P_{\mu,\infty})$ and its ground state solution in detail. Finally, in Appendix B, § 7, we present a technical result on significant convergences.
2. Variational setting
Let $E:=H^1(\mathbb{R}^N)$. The energy functional $I:E\rightarrow\mathbb{R}$ associated with equation (P) is given by
with $u\in E$. It is well known from conditions $(V_2)$ and $(V_3)$ that the eigenvalue problem
has a sequence of eigenvalues $\lambda_1 \lt \lambda_2\leq \cdots\leq \lambda_k \lt 0$ (see [Reference Egorov and Kondratiev6], Theorem 30, p. 150). We denote by ϕi the eigenfunction corresponding to λi, $i\in\{1,2,\ldots,k\}$ in $H^1(\mathbb{R}^N)$. Setting
we see that $E=E^+ \oplus E^-$. According to Stuart in [Reference Stuart15], Theorem 3.15, the essential spectrum of $-\Delta +V$ is the interval $[V_\infty,+\infty)$, and this implies that $\dim E^- \lt \infty$. Having made these considerations, every function $u\in E$ may be written as $u=u^+ + u^-$ uniquely, where $u^+\in E^+$ and $u^-\in E^-$. Hence, by using the arguments in Lemma 1.2 of [Reference Costa and Tehrani5], we may introduce the new inner product $\langle \cdot, \cdot\rangle$ in E, namely,
such that the corresponding norm $\|\cdot\|$ is equivalent to $\|\cdot\|_E$, the usual norm in $E=H^1(\mathbb{R}^N)$. In addition, the functional I may be written as
for every function $u=u^++u^-\in E$. We call the attention to the fact that, since $\lambda_i\neq 0$ for all $i\in\{1,2,\ldots,k\}$, it follows from (2.1) and the definition of ϕi that
for all functions $u^+\in E^+$ and $v^-\in E^-$.
To deal with compactness issues, we will prove several auxiliary results concerning the limit problem associated to $(P_\mu)$, namely,
Hereafter, let us denote by $J_{\mu}: H^1(\mathbb{R}^N) \to \mathbb{R}$ the associated functional given by
Also, let us consider the level $d_\mu:=\inf\limits_{u\in\mathcal{N}_\mu}J_\mu(u)$, where
3. Geometry of the Linking Theorem
In this section, we are going to show that functional Iµ satisfies the geometry of the following version of the Linking Theorem.
Theorem 3.1. [Reference Maia and Soares12, Theorem 1.2] Let E be a real Banach space with $E=V \oplus X$, where V is finite dimensional. Suppose there exist real constants $R \gt \rho \gt 0,\ \alpha \gt \beta$ and there exists an $e \in \partial B_{1} \cap X$ such that $I \in C^{1}(E, \mathbb{R})$ satisfies
(I$\left._{1}\right)$ $\left.\quad I\right|_{\partial B_{\rho} \cap X} \geq \alpha ;$
(I2) Setting $M:=\left(\bar{B}_{R} \cap V\right) \oplus\{$re $: 0 \leq r \leq R\}$, there exists an h$_{0} \in C(M, E)$ such that
(i) $\sup\limits_{w \in M} I\left(h_{0}(w)\right) \lt +\infty$,
(ii) $\sup\limits_{w \in \partial M} I\left(h_{0}(w)\right)=\beta$,
(iii) $h_{0}(\partial M) \cap\left(\partial B_{\rho} \cap X\right)=\emptyset$,
(iv) There exists a unique $u\in h_{0}(M) \cap\left(\partial B_{\rho} \cap X\right)$ such that
\begin{equation*}\operatorname{deg}\left(h_{0}, \operatorname{int}(M), u\right) \neq 0.\end{equation*}
Then I possess a Cerami sequence on a level $c \geq \alpha$, which can be characterized as
where $\Gamma:=\left\{h \in C(M, E):\left.h\right|_{\partial M}=h_{0}\right\}$.
To prove that functional Iµ has the geometry of Theorem 3.1, let $u_0\in H^1(\mathbb{R}^N)$ be a non-trivial ground state solution of problem $(P_{\mu,\infty})$ given by Proposition 6.7 in Appendix A. By hypothesis $(f_2)$, u 0 is nonnegative.
Given $w\in E$ and $y\in\mathbb{R}^N$, to simplify the notation, we write $w^+(\cdot-y)$ (or $w^-(\cdot - y)$) referring to the projection in $E^+$ (respectively, in $E^-$) of the translated function $w(\cdot-y)$.
Proceeding as Claim 4.5 in [Reference Maia, Junior and Ruviaro11], we may prove that $u_0^+(\cdot-y)$ is a non-trivial function just choosing $y\in\mathbb{R}^N$ with norm sufficiently large. Now, let us consider R > 0, any non-trivial function $e\in E^+$ with $\|e\|=1$ and the sets
and
Defining
and $h_0(v^-) = |v^-|$, where $u_0\in E$ is the non-trivial solution to the limit problem $(P_\infty)$ found before and L > 0 to be chosen, we have the following lemmas. The first one proves item $(I_1)$ from Theorem 3.1.
Lemma 3.2. There exists ρ > 0 such that
Proof. For ρ > 0, let $w^+\in E^+$ with $\|w^+\|=\rho$. Then, from (1.2), for all ɛ > 0, there exists $C_\varepsilon \gt 0$ such that
where we used Sobolev embeddings and the equivalence of the norms. It follows that, for $\varepsilon \lt \dfrac{1}{4\mu C}$,
Now, choosing $0 \lt \rho \lt \left( \dfrac{2^*}{4(1+2^*\mu C_\varepsilon C)} \right)^{\frac{1}{2^*-2}}$, the result follows.
The next result shows that item (ii) from Theorem 3.1 holds, choosing β = 0. Before stating this result, we remember an important result from spectral theory that characterizes the functions that belong to $E^-$ as follows: $u\in E^-$ if and only if
Thus, if $u\in E^-$, then $|u|\in E^-$.
In the sequel, we will denote $\|w\|^2_{V_\infty} = \displaystyle\int_{\mathbb{R}^N} (|\nabla w|^2+V_\infty w^2)\,{\rm d}x$ for any $w\in H^1(\mathbb{R}^N)$.
Lemma 3.3. There exist R > 0 sufficiently large, which does not depend on µ, such that, for all µ > 0,
Proof. Denoting by $\Upsilon^+(t)(x) = u_0^+\left(\frac{x - y}{tL}\right)$, let us separate this proof in three possible cases. If $w=tRe+v^-$ with $t\in [0,1]$ and $\|v^-\|=R$, we have
for R > 0 large enough. Second, if $w=v^- \in \bar{B}_{R}(0) \cap E^-$, one has
To finish the proof, if $w=Re + v^-$, with $\|v^-\|\leq R$, then
where we used the facts that, for L > 0 large enough, it holds that $J_\mu(\Upsilon(1)) \lt 0$; we also have for $|y|$ sufficiently large that $\left\|\Upsilon^{-}(1)\right\|$ is small enough such that $J_\mu(\Upsilon(1)) \lt 0$ implies $J_\mu\left(\Upsilon^{+}(1)\right) \leq 0$ and
Moreover, we applied the non-decreasing condition for function F since $f(s)\geq 0$ for all s and for function $s\mapsto s^{2^*}$, for s > 0.
Now, let us demonstrate that item (iii) from Theorem 3.1 also is valid.
Lemma 3.4. It holds that $h_{0}(\partial M) \cap\left(\partial B_{\rho} \cap E^+\right)=\emptyset$.
Proof. Observe that
and that
In addition, to guarantee that $\left(\bar{B}_{R} \cap E^-\right) \oplus\left\{\Upsilon^{+}(1)\right\} \cap E^+ \cap \partial B_{\rho}=\emptyset$, it is enough to choose a sufficiently large $L,|y| \gt 0$ such that $I_\mu(\Upsilon^{+}(1))\approx I_\mu(\Upsilon(1))\leq J_\mu(\Upsilon(1)) \lt -1$. Therefore, by Lemma 3.2, necessarily
where $\Upsilon^{+}(1)=u_0^{+}\left(\frac{x-y}{L}\right)$. Then, we conclude that
The lemma follows.
Finally, let us prove that item (iv) from Theorem 3.1 is true.
Lemma 3.5. There exists a unique $u\in h_{0}(M) \cap\left(\partial B_{\rho} \cap E^+\right)$ such that
Proof. Consider the function $\psi:[0,1] \rightarrow \mathbb{R}$, given by $\psi(t)=\left\|\Upsilon^{+}(t)\right\|$, is strictly increasing and hence injective. Moreover, ψ is continuous, $\psi(0)=0$, and from (3.1), we have $\psi(1) \gt \rho$. Thus, from the Intermediate Value Theorem, there exists some (unique, since ψ is injective) $t_{0} \in(0,1)$ such that $\psi\left(t_{0}\right)=\rho$. Hence,
and there exists an unique $w=\Upsilon^{+}\left(t_{0}\right) \in h_{0}(M) \cap\left(\partial B_{\rho} \cap E^+\right)$. Since $Rte \mapsto h_{0}( Rte )=\Upsilon^{+}(t)$ is injective, there exists a unique $u_{0}=R t_{0} e \in \operatorname{int}(M)$ such that $h_{0}\left(u_{0}\right)=\Upsilon^{+}\left(t_{0}\right)$. Therefore, $\operatorname{deg}\left(h_{0}, \operatorname{int}(M), w\right) \neq 0$, proving (iv).
4. Boundedness of Cerami sequences
We say that a sequence $(u_n)\subset E$ is a Cerami sequence at level c for functional Iµ if
as $n\rightarrow +\infty$. Before we state the next result, we note that, if $(v_n)$ is a bounded sequence in E, then $(v_n)$ satisfies either
(i) vanishing: for all r > 0, $\displaystyle\limsup_{n\rightarrow \infty} \sup_{y\in\mathbb{R}^N}\int_{B(y,r)}|v_n|^2\,{\rm d}x = 0$
or
(ii) non-vanishing: there exist $r,\eta \gt 0$ and a sequence $(y_n)\subset\mathbb{R}^N$ such that
\begin{equation*} \displaystyle\limsup_{n\rightarrow \infty} \int_{B(y_n,r)}|v_n|^2\,{\rm d}x \gt \eta. \end{equation*}
Lemma 4.1. Let $(u_n)\subset E$ be a Cerami sequence at level c > 0. Then, $(u_n)$ has a bounded subsequence.
Proof. Suppose by contradiction that $1\leq \|u_n\|\rightarrow \infty$ as $n\to+\infty$. Consider
and note that $\|v_n\|=1$. The sequence $(v_n)$ is bounded; however, we will show that neither (i) or (ii) is true. First, notice that from hypothesis $(f_3)$,
which shows the boundedness of the sequence $\left(|u_n|_{2^*}\right)$. It will be used in the calculations that follows.
First, suppose that hypothesis (i) is satisfied for sequence $(v_n)$. Since the sequence $(u_n)$ is a Cerami sequence, we have
and
Subtracting equation (4.2) from (4.1), we have
Thus,
provided $f(s)=0$ if $s\leq 0$, where we define $\Omega_n^+=\{x\in\mathbb{R}^N; \ u_n(x) \gt 0\}$. By equivalence of the norms, there exists a constant $\nu_0 \gt 0$ such that
for any $w\in E$. Given $0 \lt \varepsilon \lt \frac{1}{2}\nu_0$, by hypothesis $(f_1)$, there exists δ > 0 such that
For each $n\in\mathbb{N}$, consider the set
Thus, from (4.4) and by Hölder’s inequality,
From (4.3), we conclude that
Since the function $\displaystyle\frac{|f(\cdot)|}{|\cdot|}$ is bounded by $(f_2)$, by Hölder’s inequality with exponent $\displaystyle\frac{p}{2} \gt 1$, we obtain
Assumption (i) and Lions’s Lemma ensure that $\|v_n\|_{L^p(\mathbb{R}^N)}\rightarrow 0$. Therefore, up to a subsequence, from (4.5), we obtain
Now we consider two disjoint subsets of $\Omega_n^+\setminus\tilde{\Omega}_n$. Hypothesis $(f_3)$ implies that there exists R > 0 such that, if s > R,
Without loss of generality, we assume $0 \lt \delta \lt R$. For each $n\in\mathbb{N}$, let
and thus, by (4.1),
which implies that the sequence $(|A_n|)$ is bounded. Consider also
Since $B_n = (\Omega_n^+\setminus\tilde{\Omega}_n)\setminus A_n$, we have
It follows from (4.7) and the boundedness of the sequence $(|A_n|)$ that
Since the interval $[\delta, R]$ is compact and the functions f and F are continuous, we have by hypothesis $(f_3)$ that $\overline{\delta}:=\displaystyle\inf_{s\in[\delta,R]}\left(\displaystyle\frac{1}{2}f(s)s - F(s)\right) \gt 0$. Thus, from (4.8),
We have a contradiction with the fact that
Therefore, (i) does not hold for sequence $(v_n)$. Now, suppose that (ii) holds for sequence $(v_n)$. By equivalence of the norms, there exist constants $C_1,C_2 \gt 0$ such that
Let $(y_n)\subset\mathbb{R}^N$ be the sequence given by hypothesis (ii). Consider $\tilde{v}_n(x)=v_n(x+y_n)$ and $\tilde{u}_n(x)=u_n(x+y_n)$. Note that $(\tilde{v}_n)$ is bounded in E. In fact, from (4.9), it follows that
Thus, up to a subsequence,
We note that $\tilde{v}\neq 0$, since by (ii) and (4.10),
By (4.9), $\|\tilde{u}_n\|\geq \displaystyle\frac{C_1}{C_2}\|u_n\|$, which goes to infinity as $n\rightarrow \infty$. It follows from (4.10) that
with $|\Omega| \gt 0$ and $\Omega\subset B(0,r)$. Since $\|\tilde{u}_n\|\rightarrow\infty$, we have $|\tilde{u}_n(x)|\rightarrow\infty$ a.e. in Ω. Thus, Fatou’s Lemma yields
However, this contradicts the fact that $(|u_n|_{2^*})$ is bounded. This implies that hypothesis (ii) does not hold for sequence $(v_n)$. We conclude that, up to a subsequence, $(u_n)$ is bounded.
In the sequel, since M is a closed and bounded subset of a finite dimensional space, we have that M is compact. Therefore, by the continuity of h 0 and Iµ, for each µ > 0, there exist $t_\mu \gt 0$ and $v_\mu^-\in E^-$ such that
with $t_\mu Re+v_\mu^-\in M$. Let us prove that in fact $t_\mu \gt 0$. Otherwise, if $t_\mu=0$, then by the definition of Iµ, we have $I_\mu(|v_\mu^-|)\leq 0$. However, by the proof of Lemma 3.2, we may choose several small values of t > 0 such that $tRe \in M$ and $\left\|u_0^+\left(\dfrac{\cdot-y}{tL} \right)\right\| \lt \rho$, satisfying $I_\mu(h_0(tRe)) = I_\mu\left( u_0^+\left(\dfrac{\cdot-y}{tL} \right)\right) \gt 0$, contradicting the maximality of $I_\mu(|v_\mu^-|)$. Therefore, $t_\mu \gt 0$.
Lemma 4.2. It holds that $c_\mu\to 0$ as $\mu\to+\infty$, where $ c_\mu=\displaystyle\inf _{h \in \Gamma} \max _{w \in M} I_\mu(h(w)), $ with $\Gamma:=\left\{h \in C(M, E):\left.h\right|_{\partial M}=h_{0}\right\}$.
Proof. First, remember that $t_\mu \gt 0$ is not equal to zero. We claim that $t_\mu Re+v_\mu^- \to 0$ as $\mu\to +\infty$. In fact, since $t_\mu Re+v_\mu^-\in M$ and M is a compact set, passing to a subsequence if necessary, we may suppose that $t_\mu Re+v_\mu^-\to w_0:=t_0Re+v^-_0\in M$ strongly as $\mu\to +\infty$. Let us show that $w_0=0$. Otherwise, suppose $w_0\neq 0$ and note that, by the equivalence of the norms,
After a change of variables, since $0\leq t_\mu\leq 1$, it is possible to find a constant C > 0, which does not depend on µ, such that
for all µ > 0. At this moment, supposing $t_0 \gt 0$, since u 0 is positive and $u_0^-(\cdot - y)\to 0$ strongly in E as $|y|\to \infty$, we choose $y\in\mathbb{R}^N$ with norm large enough to get $F(h_0(w_0))=F(h_0(t_0Re+v^-_0))=F(u^+_0((\cdot - y)/t_0L)+|v_0^-|)\approx F(u_0((\cdot-y)/t_0L)+|v_0^-|) \gt 0$. Therefore, Fatou’s lemma provides
Then, we obtain for µ > 0 sufficiently large that
which is an absurd. This contradiction shows that $w_0=0$ and proves our claim.
It follows from the continuity of the norm and of the function h 0 (using also that $h_0(0)=0$) that
as $\mu\to \infty$, and the result follows.
5. A nontrivial solution for $(P_\mu)$
We begin with a technical result.
Lemma 5.1. If $\mu_{2} \gt \mu_{1}\geq0,$ there exists C > 0 such that, for all $x_{1},x_{2}\in\mathbb{R}^{N}$,
Proof. Since
we have
The proof follows.
Lemma 5.2. For every µ > 0, it holds that $c_\mu \lt d_\mu$.
Proof. For simplicity, C will denote a positive constant, not necessarily the same one. By the definitions of the functionals Iµ and Jµ and fixing $u_{\mu}=v_\mu^- + t_\mu R e$, we have
provided $|y|$ is large enough and F and $s\mapsto s^{2^*}$, for s > 0, are non-decreasing.
Let us estimate the integral $\displaystyle\frac{1}{2}\displaystyle\int_{\mathbb{R}^N} \left(V(x) - V_\infty \right)(\Upsilon^+(t_\mu))^2\,{\rm d}x$. We begin remembering that
for all $x,y\in\mathbb{R}^N$. Thus, replacing x by x + y, we get
for all $x,y\in\mathbb{R}^N$. Since $\|\Upsilon^-(t)(x+y)\|_2=\|\Upsilon^-(t)\|_2\to 0$ as $|y|\to\infty$ uniformly on $t\in [0,1]$ (see (3.8) and (3.9) in [Reference Maia and Soares12] and note that we are considering for t = 0, $\Upsilon^-(0)=0$), it yields the pointwise convergence
We have from assumption $(V_4)$ that
where we used that, for all $x,y\in\mathbb{R}^N$, $|t_\mu L x+y|^{\gamma_2}\leq Ct_\mu^{\gamma_2} L^{\gamma_2} |x|^{\gamma_2} + C|y|^{\gamma_2}\leq C|x|^{\gamma_2}+C|y|^{\gamma_2}$ and the function $s\mapsto -e^{-\gamma_1 s}, \ s \gt 0$, is increasing.
Now using (5.2) and the Lebesgue Theorem, we obtain as $|y|\to \infty$ that
with $\alpha_0 \gt 0$ does not depending on y. It follows from this and (5.3) that
for $|y|$ large enough.
Therefore, from (5.1), we obtain
By Mean Value Theorem (choosing the points $a:=\Upsilon(t_\mu)$ and $h:=-\Upsilon^-(t_\mu)$ that implies $a+h=\Upsilon^+(t_\mu)$) and the growth of f, we have for $2 \lt p \lt 2^*$ that
where $r_t(x) \in (0,1)$. Following the same arguments, we arrive at
Now using the exponential decay of $\Upsilon(t_\mu)$ given by Proposition 6.10 in Appendix A, with ν > 0 to be chosen, we get from Lemma 5.1 that
just choosing ν > 0 small enough. By the same arguments, we also have
and
Now applying (5.8)–(5.10) in (5.6) and (5.7), and defining $\|\Upsilon^-(t_\mu)\|_{V_\infty}:=\beta_y$, one has
and
Choosing $0 \lt \varepsilon \lt \frac{1}{2}$, inequalities (5.11) and (5.12) provide
Since $\beta_y\to 0$, we have
taking $|y|$ sufficiently large. It follows that
Thus, returning to (5.1) with (5.4) and (5.13), we get
with $|y|$ large enough. Since the function ${\rm e}^{-C|y|^{\gamma_2}}$ decays more slowly than the other terms of exponential functions (because $0 \lt \gamma_2 \lt 1$), we conclude from (5.14) that
To finish the proof, we proceed as follows.
We used the fact that the maximum $\displaystyle\max_{t\in [0,1]} J_{\mu}(\Upsilon(t))$ is achieved at the unique point t 0, where $J_{\mu}^{\prime}(\Upsilon(t_0))\Upsilon(t_0) =0.$ This point t 0 is unique once $J_{\mu}^{\prime}(\Upsilon(t_0))\Upsilon(t_0) =0$ is equivalent to
which has only one solution in t 0. Since u 0 is a non-trivial solution of $(P_{\mu,\infty})$, we can infer that the value of t 0 is $t_0=\dfrac{1}{L}$. The proof is complete.
Lemma 5.3. Consider $(u_n)\subset E$, a $(Ce)_c$ sequence for functional Iµ such that
where C 1 is given by (4.9). Then, there exist a sequence $(y_n)\subset E$ and $\rho,\eta \gt 0$, satisfying
Proof. Suppose, by contradiction, that (5.15) does not hold. Then, by [Reference Lions10, Lemma 8.4], we have, for $p\in(2,2^*)$, that
Consequently, by (1.2) and $(f_3)$,
Since $u_n\rightharpoonup 0$ in $H^1(\mathbb{R}^N)$ and $\dim E^- \lt \infty$, then $\|u^-_n\|\rightarrow 0$. Thus, using that $I_\mu^{\prime}(u_n)u_n=o_n(1)$ and $(u_n)$ is bounded in $H^1(\mathbb{R}^N)$, we obtain by (5.16)
Hence, by $(5.16)$ and inequality above,
This implies that $l=Nc \gt 0$. Now considering S the best constant of the Sobolev embedding $D^{1,2}(\mathbb{R}^N)\hookrightarrow L^{2^*}(\mathbb{R}^N$), we can conclude, by (4.9), that
Using that l = Nc and inequality above, we obtain $c\geq\dfrac{1}{N{C_1^{\frac{N}{2}}}}S^\frac{N}{2}$, which contradicts our hypothesis.
Proposition 5.4. Let $(u_n)\subset E$ be a $(Ce)_c$ sequence for functional Iµ. If
then problem $(P_\mu)$ has a non-trivial solution uµ.
Proof. Using Lemma 4, there is $u_\mu\in E$ such that $u_n\rightharpoonup u_\mu$ in E and $I^{\prime}_\mu(u_\mu)=0$. Suppose, by contradiction, that $u_\mu\equiv 0$. Then, by Lemma 5.3, there are $(y_n)\subset\mathbb{R}^N$ and $\rho,\;\eta \gt 0$ such that (5.15) holds. Note that since $\dim E^-$ is finite, $\|u_n^-\|\rightarrow 0$ and, consequently, for all $v\in E$,
which imply $I_\mu(u_n)=I_\mu(u^+_n)+o_n(1)$ and $I^{\prime}_\mu(u_n^+)=o_n(1)$.
Let us prove that $\lim\limits_{n\to+\infty}|y_n|\rightarrow+\infty$ occurs. In fact, if there exists $\overline{R} \gt 0$ such that $B_{\rho}(y_n)\subset B_{\overline{R}}(0)\subset\mathbb{R}^N$, for all $n\in\mathbb{N}$, then, since $(u_n)$ converges strongly to 0 in $L^2_{\rm loc}(\mathbb{R}^N)$, we see that
contradicting (5.15).
Now define $w_n(x):=u_n^+(x+y_n)$, for all $x\in \mathbb{R}^N$. Therefore, $(w_n)$ is bounded in E, and there exists $w_n \rightharpoonup w_\mu\in E$. Using the arguments above, we have
We claim that $w_\mu\neq0$. In fact, by (5.15),
Hence,
Therefore, uµ may not be trivial, and the proof of the lemma is complete.
At this moment, the first part of the proof of Theorem 1.1 may be presented. In § 2, we proved that Iµ satisfies all conditions in Theorem 3.1, what implies the existence of a Cerami sequence $(u_n)$ at level $c_\mu \gt 0$, where $ \displaystyle c_\mu=\inf _{h \in \Gamma} \max _{w \in M} I_\mu(h_0(w)) $ with $\Gamma:=\left\{h \in C(M, E):\left.h\right|_{\partial M}=h_{0}\right\}$. This sequence is bounded by Lemma 4 and then $(u_n)$ converges weakly to a solution uµ of $(P_\mu)$. To show that uµ is non-trivial, we have from Lemma 5.2 that $c_\mu \lt d_\mu$ and, from Lemma 4.2, we choose µ > 0 large enough to obtain $c_\mu \lt \dfrac{1}{{C_1^{\frac{N}{2}}}N}S^\frac{N}{2}$ and apply Proposition 5.4 to get a non-trivial solution uµ of problem $(P_\mu)$, which is non-negative because of hypothesis $(f_2)$, as we wished to prove.
6. Appendix A: One ground state solution to the Limit Problem $(P_{\mu,\infty})$ and some of its properties
To prove the existence of ground state solution to the limit problem $(P_{\mu,\infty})$, we consider
Hereafter, let us denote by $J_{\mu}: H^1(\mathbb{R}^N) \to \mathbb{R}$ the associated functional given by
In this section, we consider $H^1(\mathbb{R}^N)$ endowed with the following norm
Notice that weak solutions of problem $(P_{\mu,\infty})$ in $H^1(\mathbb{R}^N)$ are critical points of functional $J_\mu\in C^1(H^1(\mathbb{R}^N),\mathbb{R})$.
Let us show that functional Jµ has a mountain pass geometry.
Proposition 6.1. The following statements hold.
(i) There exist $\alpha,\;\rho \gt 0$ such that
\begin{equation*} J_\mu(u)\geq\alpha,\ \text{for all} \ u\in H^1(\mathbb{R}^N), \ \text{with} \ \|u\|_{H^1}=\rho. \end{equation*}(ii) For all $u\in H^1(\mathbb{R}^N)\setminus\{0\}$, we have
\begin{equation*} \limsup\limits_{t\to+\infty}J_\mu(tu)\leq -\infty. \end{equation*}
Proof. Using $(1.2)$ and Sobolev embeddings, we obtain
Choosing $\epsilon\in\left(0,\dfrac{1}{\mu C}\right)$ and taking $\|u\|_{H^1}$ small enough, we can determine positive numbers α and ρ such that
To prove the second item, let us consider u ≠ 0 and t > 0. Then, by $(f_3)$,
Letting $t\to+\infty$, we obtain
We say that a sequence $(u_{n})\subset H^1(\mathbb{R}^N)$ is a Palais-Smale sequence at level bµ for the functional Jµ if
as $n\to\infty$, where
and
Notice that Proposition 6.1 implies the existence of a Palais-Smale sequence at level bµ for the functional Jµ. Using this Palais-Smale sequence, we show the existence of non-trivial critical point for Jµ, but we need to show some technical results. First, let S > 0 be the best constant to the Sobolev embedding $D^{1,2}(\mathbb{R}^N)\hookrightarrow L^{2^*}(\mathbb{R}^N)$.
Lemma 6.2. Consider $(u_n)\subset H^1(\mathbb{R}^N)$ a Palais-Smale sequence at level bµ for the functional Jµ such that
Then, there exist a sequence $(y_n)\subset\mathbb{R}^N$ and $\rho,\eta \gt 0$ such that
Proof. One proceeds exactly as in Proposition 5.4.
Lemma 6.3. If $\mu \to +\infty$, then $b_\mu\to 0$.
Proof. Consider a function $\varphi\in C^\infty_0(\mathbb{R}^N)$ such that $\|\varphi\|^2_{H^1}=2$. Then, by Lemma 6.1, there exists $t_\mu \gt 0$ such that
We are going to show that, up to subsequence, $t_\mu\to 0$ when $\mu\to+\infty$. First, using the characterization of bµ and $(f_3)$, we have
and this implies that $(t_\mu)\subset\mathbb{R}$ is bounded. Hence, up to a subsequence, $(t_\mu)$ converges to some $t_0\geq0$. To prove that $t_0=0$, let us suppose, by contradiction, that $t_0 \gt 0$. Then, by $(f_3)$,
which implies that
Therefore, up to a subsequence, $(t_\mu)$ converges to 0. Hence, by $(f_3)$ one more time,
for all µ > 0. This proves the lemma.
Proposition 6.4. There exists $\mu^* \gt 0$ such that the limit problem $(P_{\mu, \infty})$ has a nontrivial solution.
Proof. By Proposition 6.1, we get a Palais-Smale sequence $(u_n)\subset H^1(\mathbb{R}^N)$ at level $b_\mu \gt 0$. Using the proof of Lemma 4 with a slight modification, we can show that the sequence $(u_n)$ is bounded in $H^1(\mathbb{R}^N)$. Then, there exists $u_\infty\in H^1(\mathbb{R}^N)$ such that, up to a subsequence, $u_n\rightharpoonup u_0 \in H^1(\mathbb{R}^N)$.
If $u_0\neq 0$ then, using a density argument, we have a nontrivial solution of $(P_{\mu,\infty})$. On the other hand, if $u_0\equiv 0$, we can find $\mu^* \gt 0$ such that, by Lemma 6.3,
Then, by Lemma 6.2, there exist a sequence $(y_n)\subset\mathbb{R}^N$ and $\rho,\eta \gt 0$ such that
Setting $w_n:=u_n(x+y_n)$ and, since the problem is invariant under translations, we have that $(w_n)\subset\mathbb{R}^N$ is a bounded Palais-Smale sequence. Hence, there exists $w_0\in H^1(\mathbb{R}^N)$ such that, up to a subsequence, $w_n\rightharpoonup w_0 \in H^1(\mathbb{R}^N)$ and, by $(6.2)$ and Sobolev embeddings,
which implies that $w_0\neq 0$ and, using a density argument one more time, we have a non-trivial solution of $(P_{\mu,\infty})$.
At this moment, we will concentrate in showing that the limit problem has a ground state solution. For this, let us consider
Before stating the results, observe that Lemmas 5.4 and 6.2 are still valid when we consider a sequence $(u_n)\subset\mathcal{N}_\mu$ instead of a $(Ce)_c$ sequence for functional Jµ.
Lemma 6.5. Consider $\mu^* \gt 0$ given by Proposition 6.4. If $\mu\geq\mu^*$, then the following proprieties hold.
(i) $\mathcal{N}_\mu\neq \emptyset$.
(ii) There exists $\rho_{\mu} \gt 0$ such that
\begin{equation*} \int_{\mathbb{R}^N}|u|^{2^*}\,{\rm d}x\geq\rho_{\mu} \gt 0,\quad \forall\ u\in\mathcal{N}_\mu. \end{equation*}(iii) If $u\in\mathcal{N}_\mu$, then $J_\mu (u)\geq\dfrac{1}{N}\displaystyle\int_{\mathbb{R}^N}|u|^{2^*}\,{\rm d}x\geq \dfrac{\rho_\mu}{N} \gt 0$.
Proof. We deduce immediately item (i) from the existence of a non-trivial solution obtained by Proposition 6.4. To prove the second item, we are going to use $(1.2)$ to ensure that
Then, considering $u\in\mathcal{N}_\mu$ and using Sobolev embeddings,
Choosing $\epsilon\in\left(0,\dfrac{1}{\mu} \right) $, then there exists $K_\mu \gt 0$ such that
Using again (6.3) and (6.4), we can find $\rho_\mu \gt 0$ such that
Let us show item (iii). Using $(f_3)$ and item (ii), we obtain, for $u\in\mathcal{N}_\mu$,
The above Lemma ensures that
The next result establishes the existence of a non-trivial ground state solution $w_\infty$ to problem $(P_{\mu,\infty})$.
Lemma 6.6. If $\mathcal{N}_\mu\neq \emptyset$, let $(u_n)\subset \mathcal{N}_\mu$ be a minimizing sequence for Jµ. Then $(u_n)$ is bounded in $H^1(\mathbb{R}^N)$. Moreover, there exists $\mu^* \gt 0$ such that the infimum of Jµ on $\mathcal{N}_\mu$ is attained for all $\mu \gt \mu^*$.
Proof. The existence of a minimizing sequence $(u_n)\subset\mathcal{N}_\mu$ is assured by Lemma 6.5. Using the same arguments as in Lemma 4.1, we show that $(u_n)$ is bounded in $H^1(\mathbb{R}^N)$. Hence, by usual arguments and Sobolev embeddings, there exists $w_{\infty}\in H^1(\mathbb{R}^N)$ such that, up to a subsequence,
The pointwise convergence of the gradient in (6.6) is guaranteed by Lemma 7.1 in the Appendix B.
By density arguments, $J_{\mu}^{\prime}(w_\infty)(w_\infty)=0$. Observe that, if $w_\infty\neq 0$, then $w_\infty\in\mathcal{N}_\mu$, and hence, by Fatou’s Lemma and $(f_3)$,
Then, we conclude that $(u_n)$ converges strongly to $w_\infty$ in $H^1(\mathbb{R}^N)$ and hence $J_{\mu}(w_\infty)=d_\mu$.
On the other hand, if $w_\infty\equiv 0$, we define $w_n(x):=u_n(x+y_n)$. Then, arguing as we did in Lemmas 5.3, 5.4 and 6.3, there exists $\mu^* \gt 0$ such that if $\mu \gt \mu^*$, then there is $\widetilde{w_{\mu,\infty}}\in\mathcal{N}_{\mu,\infty}$ satisfying, up to a subsequence,
Then, by Fatou’s Lemma and $(f_3)$,
which completes the proof.
Proposition 6.7. There exists $\mu^* \gt 0$ such that problem $(P_{\mu,\infty})$ has a ground state solution $u_0\in H^1(\mathbb{R}^N)$ for all $\mu \gt \mu^*$.
Without loss of generality, we may suppose that $u_0\geq 0$ in $\mathbb{R}^N$. To see this, it is enough to truncate the functional Jµ, considering $(\max\{w,0\})^{2^*}$ in place of $|w|^{2^*}$ and using hypothesis $(f_2)$.
The next lemma is an important consequence of the standard regularity arguments. It will be used to apply the Divergence Theorem in the sequel.
Lemma 6.8. It holds that $u_0\in H^2(\mathbb{R}^N)$.
We also may suppose that $u_0\in H^1_{\text{rad}}(\mathbb{R}^N)$. Indeed, it is enough to solve problem $(P_{\mu,\infty})$ in $H^1_{\text{rad}}(\mathbb{R}^N)$ and use the Symmetric Criticality Principle. This implies the following lemma whose proof is an immediate consequence of Strauss inequality.
Lemma 6.9. There exists $C=C(N) \gt 0$ such that, for all x ≠ 0,
The next result is the most important concerning the qualitative properties of u 0. It will guarantee that the solution u 0 has an appropriate exponential decay, which was used before to relate two important levels in order to obtain a strong convergence of a Cerami sequence. Recall that $\dfrac{|f(s)|}{|s|} \lt m$ for all $s\in\mathbb{R}$ from hypothesis $(f_2)$.
Proposition 6.10. If $\nu\in\left(0,\sqrt{V_\infty}\right)$ and $\mu\geq \mu^*$, then there exists $C=C(m,\nu) \gt 0$ such that
Proof. For x ≠ 0, we have that
Define $C:= {\rm e}^{\nu R} \gt 0$, where R > 0 to be chosen, and
By the definition of C, we get $w(x)\leq 0$ for $|x|\leq R$. Let us prove that this inequality still holds for $|x| \gt R$. For this end, consider the set
where $w_+(x)=\max\{w(x),0\}$. Suppose, by contradiction, that $\Omega\neq \emptyset$. Since $w_+\in H^1(\mathbb{R}^N)$, it follows that Ω is a Lebesgue measurable set, which satisfies
Therefore, by the Divergence Theorem (note that $w_+ =0$ on $\partial D(R)$), we get
Using the definition of w and that u 0 is a solution of $(P_{\beta,\infty})$, we obtain
where we used (1.2). Now, choosing ɛ > 0 such that $\mu\varepsilon+\nu^2-V_\infty \lt 0$, which implies in particular that $\mu\varepsilon-V_\infty \lt 0$, we obtain from Lemma 6.9 that
for some constants $c_1,c_2 \gt 0$ that do not depend on R > 0. We choose R > 0 sufficiently large such that, for $|x| \gt R$,
Noting that $u_0(x) \gt C|u_0|_\infty {\rm e}^{-\nu|x|}$ in Ω, we get
If necessary, we take R > 0 even large so that
what is possible in view of $\mu\varepsilon-V_\infty+\nu^2 \lt 0$. Then, we arrive at
an absurd. Thus, $\Omega=\emptyset$ and the proposition follows.
All this section proves the second and final part of Theorem 1.1.
7. Appendix B: A technical result
Lemma 7.1. Let $(u_n)\subset \mathcal{N}_\mu$ be a sequence satisfying $u_n \rightharpoonup w_\infty$ in $H^1(\mathbb{R}^N)$ for some $w_\infty\in H^1(\mathbb{R}^N)$. Then, passing to a subsequence, $\nabla u_n \to \nabla w_\infty$ strongly in $[L^2_{\rm loc}(\mathbb{R}^N)]^N$ and $\nabla u_n(x)\to \nabla w_\infty(x)$ almost everywhere $x\in\mathbb{R}^N$.
Proof. We will adapt some ideas found in [Reference Alves and Figueiredo1]. Since $u_n\rightharpoonup w_\infty$ in $H^1(\mathbb{R}^N)$, then, up to a subsequence,
Given any R > 0, let $\psi\in C_0^\infty (\mathbb{R}^N,[0,1])$ be such that $\psi\equiv 1$ in $B_R(0)$ and $\psi\equiv 0$ in $\mathbb{R}^N\backslash B_{2R}(0)$. Then, by Cauchy–Schwarz inequality and the fact that $\int_{\mathbb{R}^N} \psi \nabla w_\infty \nabla \left(u_n-w_\infty \right) \,{\rm d}x = o_n(1)$ (because of the weak convergence $u_n \rightharpoonup w_\infty$), we have the following facts:
and
where in the last convergence, we used the growth of f. Moreover, let $2^*-1 \lt s \lt 2^*$ and consider $r=\dfrac{s}{2^*-1} \gt 1$. Then, $r^{\prime}:=\dfrac{r}{r-1}$ satisfies $r^{\prime}=\dfrac{s}{s-2^*+1} \lt 2^*$. By Hölder inequality with exponents r and r ʹ, we get from (7.1) and from the boundedness of $(u_n)$ in $L^s(\mathbb{R}^N)$ that
Therefore, remembering that $u_n\in \mathcal{N}_\mu$, one obtains
Since R > 0 is arbitrary, this implies that, passing to a subsequence, $\nabla u_n \to \nabla w_\infty$ in $[L^2_{\rm loc}(\mathbb{R}^N)]^N$.
To complete the proof of the lemma, let us show that $\nabla u_n(x)\to \nabla w_\infty(x)$ a.e. $x\in\mathbb{R}^N$ by using a diagonal process. Since $\nabla u_n \rightarrow \nabla w_\infty$ in $[L^2(B_1(0))]^N$, there exists an infinite subset $\mathbb{N}_1 \subset \mathbb{N}$ such that the subsequence $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}_1}$ converges to $\nabla w_\infty(x)$ a.e. $x\in B_1(0)$. Since $\left(\nabla u_n\right)_{n \in \mathbb{N}_1}$ converges to $\nabla w_\infty$ in $[L^2(B_2(0))]^N$, we obtain a subsequence $\left(\nabla u_n\right)_{n \in \mathbb{N}_2} \operatorname{with} \mathbb{N}_2 \subset \mathbb{N}_1$ such that $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}_2}$ converges to $\nabla w_\infty(x)$ a.e. $x\in B_2(0)$. Proceeding in this way, we find infinite subsets of indexes $\mathbb{N}_{k+1} \subset \mathbb{N}_k \subset \mathbb{N}$ such that $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}_k}$ converges to $\nabla w_\infty(x)$ a.e. $x\in B_k(0)$. Consider $\mathbb{N}^*=\left\{n_1^*, n_2^*, \ldots, n_k^*, \ldots\right\} \subset \mathbb{N}$ with $n_k^*$ being the kth element of $\mathbb{N}_k$. Therefore, $\left(\nabla u_n(x)\right)_{n \in \mathbb{N} *}$ is, from its kth element, a subsequence of $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}_k}$ and hence converges to $\nabla w_\infty(x)$ a.e. $x\in B_k(0)$. For each $k \in \mathbb{N}$, there exists $Z_k \subset B_k(0)$ of zero measure such that $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}_k}$ converges to $\nabla w_\infty(x)$ for all $x \in B_k(0) \backslash Z_k$. Take $Z:=\cup_{m=1}^{\infty} Z_m$. Then, Z has zero measure and, for all $x \in \mathbb{R}^N \backslash Z$, we have $x \in B_k(0) \backslash Z_k$ for some $k \in \mathbb{N}$ and $\left(\nabla u_n(x)\right)_{n \in \mathbb{N}^*}$ converges to $\nabla w_\infty(x)$. This shows that, up to a subsequence, $\nabla u_n(x)\to\nabla w_\infty(x)$ a.e. $x\in \mathbb{R}^N$ and completes the proof.
Funding Statement
Gustavo S. A. Costa was supported by CNPq, Conselho Nacional de Desenvolvimento Científico e Tecnológico, Brazil (grant number 163054/2020-7). Giovany M. Figueiredo was supported by FAPDF – Demanda Espontânea 2021 and CNPq Produtividade 2019.