Proof. The proof is standard by now, and we refer the reader to [Reference Szulkin and Weth66, Reference Szulkin and Weth67] for the details.

Proof. Since the functional I is invariant under isometries in $\mathcal{I}(\mathbb{B}^N)$, the proof is standard and we refer to [Reference Bhakta and Sandeep18, lemma 3.2].

Proof. First of all, by using Ekeland’s variational principle and Lemma 3.3, we can assume that $\{u_n\}$ is a $(PS)_d$ sequence for I. Moreover, $\{u_n\}$ is bounded by Lemma 3.2-(iv), then we shall further assume that for some subsequence, still denoted by itself, $ u_n \rightharpoonup u$ in $H^{1}(\mathbb{B}^N)$ for some $u \in H^{1}(\mathbb{B}^N)$. Exploiting Lemma 3.2-(iii), there holds d > 0, it then follows that

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}|u_n|^{q}\,dV_{\mathbb{B}^N}=\delta' \gt 0, \end{equation*}

because

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}} \gt 0 \end{equation*}

and $\{u_n\}$ is bounded in $L^{2}(\mathbb{B}^N)$. In what follows, we will divide our analysis into two cases:

Case 1: ${u} \not=0$.

In light of $\{u_n\}$ is a $(PS)_d$ sequence, we immediately conclude that $I'(u)=0$, and so, $u \in \mathcal{N}$. Thereby, it can be derived from Fatou’s lemma and $(f_4)$ that

\begin{equation*} d \leq I(u)=I(u)-\frac{1}{2}I'(u)u \leq \lim_{n\to +\infty}(I(u_n)-\frac{1}{2}I'(u_n)u_n)=\lim_{n\to +\infty}I(u_n)=d, \end{equation*}

showing that $I(u)=d$. This fact, jointly with $\{u_n\}$ is a $(PS)_d$ sequence, does imply that

\begin{equation*}\begin{gathered} \left(\frac{1}{2}-\frac{1}{\theta}\right)\|u_n\|^2+\int_{\mathbb{B}^N}\left(\frac{1}{\theta}f(u_n)u_n -F(u_n)\right)\,dV_{{\mathbb{B}^N}}\hfill\\ \ \ \ \ =\left(\frac{1}{2}-\frac{1}{\theta}\right)\|u\|^2+\int_{\mathbb{B}^N} \left(\frac{1}{\theta}f(u)u-F(u)\right)\,dV_{{\mathbb{B}^N}}+o_n(1),\hfill\\ \end{gathered} \end{equation*}

from where it yields that $u_n \to u$ in $H^{1}(\mathbb{B}^N)$. So, the proof is done.

Case 2: u = 0.

Arguing as the proof of [Reference Bhakta and Sandeep18, theorem 3.3], there are $\tau_n \in \mathcal{I}(\mathbb{B}^N$), $x_0 \in S_{\sqrt{3}}$ and $0 \lt \delta \lt S_{N,q}$ of such way that the sequence $\tilde{u}_n=u_n \circ \tau_{b_n}$ satisfies

\begin{equation*} \int_{A(x_0,\sqrt{3})}|\tilde{u}_n|^{q}\,dV_{{\mathbb{B}^N}}=\sup_{x \in S_{\sqrt{3}}}\int_{A(x,\sqrt{3})}|u_n|^{q}\,dV_{{\mathbb{B}^N}}=\delta. \end{equation*}

Furthermore, $\{\tilde{u}_n\}$ is also a $(PS)_d$ for I by Lemma 3.4. Using Lemma 3.2-(iv) again, for some subsequence, still denoted by $\{\tilde{u}_n\}$, there exists a $\tilde{u}\in H^{1}(\mathbb{B}^N)$ such that $\tilde{u}_n \rightharpoonup \tilde{u}$ in $H^{1}(\mathbb{B}^N)$.

Otherwise, we could suppose that $\tilde{u}=0$. Since $\{\tilde{u}_n\}$ is a $(PS)_d$ sequence for I, owing to the proof of [Reference Bhakta and Sandeep18, theorem 3.3] again, we use Lemma 2.1 to prove that

(3.4)\begin{equation} \int_{\mathbb{B}^N \cap [|x| \geq r]}|\tilde{u}_n|^q\,dV_{{\mathbb{B}^N}} \to 0, \end{equation}

for all $r \in (2-\sqrt{3},1)$. Recalling that the embedding $H^{1}(\mathbb{B}^N) \hookrightarrow L^{q}(\mathbb{B}^N \cap [|x| \lt r],dV_{{\mathbb{B}^N}})$ is compact, see Lemma A.1 in the Appendix, we have that

(3.5)\begin{equation} \int_{\mathbb{B}^N \cap [|x| \lt r]}|\tilde{u}_n|^q\,dV_{{\mathbb{B}^N}} \to 0. \end{equation}

From (3.4) and (3.5),

\begin{equation*} \int_{\mathbb{B}^N}|\tilde{u}_n|^q\,dV_{{\mathbb{B}^N}} \to 0, \end{equation*}

which contradicts with the limit below

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}|\tilde{u}_n|^{q}\,dV_{{\mathbb{B}^N}}=\liminf_{n \to +\infty}\int_{\mathbb{B}^N}|u_n|^{q}\,dV_{{\mathbb{B}^N}}=\delta' \gt 0, \end{equation*}

showing the claim 3.6. In the same spirit of Case 1, we see that $\tilde{u}_n \to \tilde{u}$ along a subsequence in $H^{1}(\mathbb{B}^N)$. This proves the theorem.

Proof of Theorem 3.1

As an immediate consequence of the compactness theorem exhibited in Theorem 3.5, we can prove that Problem (3.1) has a ground state solution $u\in H^1(\mathbb{B}^N)$. To finish the proof, we will prove that $u(x) \gt 0$ for all $x\in \mathbb{B}^N$.

From the definition of the variational functional I, one can check that $I(|u|)=I(u)=d$ and $|u|\in \mathcal{N}$ which permit us to replace u with $|u|$. In order to prove that $u(x) \gt 0$ for all $x \in \mathbb{B}^N$, it is enough to apply the celebrated Harnack inequality (see e.g. [Reference Gilbarg and Trudinger38, theorem 8.20]). Indeed, by considering the functions

\begin{equation*} p(x)=\left(\frac{2}{1-|x|^2}\right)^{2} \quad \mbox{and} \quad v(x)=p^{\frac{N-2}{4}}(x)u(x), \quad x \in \mathbb{B}^N, \end{equation*}

a direct computation implies that v is a solution of the Euclidean type problem

\begin{equation*} \left\{ \begin{array}{l} -\Delta v= p^{\frac{N+2}{4}}\left(f(p(x)^{\frac{2-N}{4}}v)+ \hat{\lambda}p(x)^{\frac{2-N}{4}}v\right), \quad \mbox{in} \quad \mathbb{B}^N , \\ v=0, \quad \mbox{on} \quad \partial \mathbb{B}^N, \end{array} \right. \end{equation*}

where $\hat{\lambda}=\lambda-\frac{N(N-2)}{4}$ and $\mathbb{B}^N$ is endowed with Euclidean metric.

Assume by contradiction that there exists $x_0 \in \mathbb{B}^N$ such that $v(x_0)=0$. Since $v \not=0$, there exists $x_1 \in \mathbb{B}^N$ such that $v(x_1) \gt 0$. Have this in mind, fix $0 \lt R \lt 1$ such that $x_0,x_1 \in B_R(0)$. By means of [Reference Gilbarg and Trudinger38, theorem 8.20], there exists a constant C > 0 such that

\begin{equation*} \sup_{y \in B_R(0)}v(y) \leq C\inf_{y \in B_R(0)}v(y), \end{equation*}

which is absurd, because in this case

\begin{equation*} \sup_{y \in B_R(0)}v(y) \gt 0 \quad \mbox{and} \quad \inf_{y \in B_R(0)}v(y)=0. \end{equation*}

This completes the proof.

Proof. Assume by contradiction that u = 0 and let $t_n \gt 0$ such that $t_nu_n \in \mathcal{M}_\infty$. The standard calculations show that $\{t_n\}$ is bounded, and so,

\begin{align*} \tilde{d } +o_n(1)&=J(u_n)= \max_{t \geq 0}J(tu_n) \geq J(t_nu_n)=J_\infty(t_nu_n) +\frac{t_n^2}{2}\int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)\\ &\quad\times |u_n|^2\,dV_{{\mathbb{B}^N}}, \end{align*}

where we have used a similar property in Lemma 3.2-(i) in the second equality. From this,

\begin{equation*} \tilde{d} +o_n(1) \geq \tilde{d}_\infty + \frac{t_n^2}{2}\int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}}. \end{equation*}

In light of the fact that $\{t_n\}$ is bounded, then the proof is done if we can verify that

(3.12)\begin{equation} \int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)|u_n|^2dV_{{\mathbb{B}^N}} \to 0 \end{equation}

loading to

\begin{equation*} \tilde{d} \geq \tilde{d}_\infty, \end{equation*}

which contradicts with (3.11). Now, we focus on showing the validity of (3.12).

Owing to (2.3), one sees that $\tilde{A}(x)\to \tilde{A}_\infty$ as $d(x,0)\to\infty$ for all $x\in \mathbb{B}^N$ by $(\tilde{A}_2)$. Therefore, given a ɛ > 0, there is an $\tilde{r}\in(0,1)$ such that $|\tilde{A}(x)-\tilde{A}_\infty| \lt \varepsilon$ for all $\tilde{r} \lt |x| \lt 1$ and so

\begin{equation*} \int_{\mathbb{B}^N\cap[\tilde{r} \lt |x| \lt 1]}|\tilde{A}(x)-\tilde{A}_\infty||u_n|^2dV_{{\mathbb{B}^N}}\leq C \varepsilon, \end{equation*}

where C > 0 is independent of $n\in \mathbb{N}$. Consider the embedding $H^{1}(\mathbb{B}^N) \hookrightarrow L^{2}(\mathbb{B}^N \cap [|x|\leq\tilde{r}])$ is compact, see Lemma A.1 in the Appendix one has that

\begin{equation*} \int_{\mathbb{B}^N\cap[|x|\leq\tilde{r}]]}|\tilde{A}(x)-\tilde{A}_\infty||u_n|^2dV_{{\mathbb{B}^N}}\leq C_{\tilde{r}} \int_{\mathbb{B}^N\cap[|x|\leq\tilde{r}]]}|u_n|^2dV_{{\mathbb{B}^N}}=o_n(1), \end{equation*}

where $C_{\tilde{r}} \gt 0$ is independent of $n\in \mathbb{N}$. Letting $n\to\infty$ and then $\varepsilon\to0^+$, we finish the proof of this lemma.

Proof Theorem 3.7

With Lemma 3.8 in hand, we know that u is a nontrivial critical point for J, and so, $J(u) \geq \tilde{d}$. On the other hand, one has

\begin{equation*} J(u)=J(u)-\frac{1}{2}J'(u)u=\frac{1}{2}\int_{\mathbb{B}^N}(f(u)u-2F(u))dV_{{\mathbb{B}^N}}. \end{equation*}

By Fatou’s lemma,

\begin{align*} \frac{1}{2}\int_{\mathbb{B}^N}(f(u)u-2F(u))dV_{{\mathbb{B}^N}} &\leq \liminf_{n \to +\infty}\frac{1}{2}\int_{\mathbb{B}^N}(f(u_n)u_n-2F(u_u))dV_{{\mathbb{B}^N}}\\ &=\liminf_{n \to +\infty}J(u_n)=\tilde{d}, \end{align*}

from where it concludes $J(u)=\tilde{d}$, showing that u is indeed a nontrivial ground state solution of Problem (3.6). Concerning the positivity of u, the reader can refer to the proof of Theorem 3.1, we omit it here. The proof is completed.

Proof. According to the discussions in Theorem 3.1, there is a $U_0\in H^1(\mathbb{B}^N)$ such that

\begin{equation*} I_0(U_0)=d_0 \gt 0 \quad \mbox{and} \quad I'_0(U_0)=0. \end{equation*}

Let $t_\epsilon \gt 0$ be such that $t_\epsilon U_0 \in \mathcal{N}_\epsilon$. A simple argument shows that $\{t_\epsilon\}$ is bounded with respect to ϵ > 0 and so we can assume that $t_\epsilon \to t_0$ as $\epsilon\to0^+$. Having this in mind, we arrive at

\begin{equation*} d_\epsilon \leq I_\epsilon(t_\epsilon U_0)=\frac{t_\epsilon^2}{2}\int_{\mathbb{B}^N}(|\nabla_{\mathbb{B}^N} U_0|^2 +A(\epsilon g(x))|U_0|^2) \,dV_{{\mathbb{B}^N}}-\int_{\mathbb{B}^N}F(t_\epsilon U_0)dV_{{\mathbb{B}^N}}. \end{equation*}

Claim 3.11.

\begin{equation*} \lim_{\epsilon\to0^+}\int_{\mathbb{B}^N}|A(\epsilon g(x))-A(0)||U_0|^2dV_{{\mathbb{B}^N}}=0. \end{equation*}

Indeed, since $A \in L^{\infty}(\mathbb{R}^N)$, it follows that

\begin{equation*} |A(\epsilon g(x))-A(0)||U_0|^2 \leq 2|A|_\infty |U_0|^2 \in L^{1}(\mathbb{B}^N), \quad \forall \epsilon \gt 0. \end{equation*}

Now, using the fact that A is a continuous function, Lebesgue’s Dominated Convergence theorem yields that the claim above is true.

Applying again Lebesgue’s Dominated Convergence theorem, we also get

\begin{equation*} \lim_{\epsilon \to 0^+}\left(\frac{t_\epsilon^2}{2}\int_{\mathbb{B}^N}(|\nabla_{\mathbb{B}^N} U_0|^2+A(0)|U_0|^2) \,dV_{{\mathbb{B}^N}} -\int_{\mathbb{B}^N}F(t_\epsilon U_0)\,dV_{{\mathbb{B}^N}} \right)=I_0(t_0U_0), \end{equation*}

which together with claim 3.11 reveals the inequality below

(3.20)\begin{equation} \limsup_{\epsilon \to 0^+}d_\epsilon \leq I_0(t_0U_0)\leq \max_{t \geq 0}I_0(tU_0)=I_0(U_0)=d_0. \end{equation}

On the other hand, recalling $A(0)=\displaystyle \inf_{z \in \mathbb{R}^N}A(z)=A_0$, one finds

\begin{equation*} I_0(u) \leq I_\epsilon(u), \quad \forall u \in H^{1}(\mathbb{B}^N), \end{equation*}

implying that

\begin{equation*} d_0 \leq d_\epsilon, \quad \forall \epsilon \gt 0. \end{equation*}

Thereby,

(3.21)\begin{equation} d_0 \leq \liminf_{\epsilon \to 0} d_\epsilon. \end{equation}

From (3.20) and (3.21),

\begin{equation*} \lim_{\epsilon \to 0^+} d_\epsilon=d_0. \end{equation*}

The limit above combined with (4.30) yields that there exists a $\epsilon_0 \gt 0$ such that $d_\epsilon \lt d_\infty$ for all $\epsilon \in (0, \epsilon_0)$. The proof is completed.

Proof. Assume by contradiction that u = 0 and let $t_n \gt 0$ such that $t_nu_n \in \mathcal{M}_\infty$. It is easy to prove that $\{t_n\}$ is bounded. Since $\{u_n\}\subset \mathcal{M}_\epsilon$ is a $(PS)_{d_\epsilon}$ sequence of I_ϵ, one sees that

\begin{equation*} d_\epsilon +o_n(1)=I_\epsilon(u_n) \geq I_\epsilon(t_nu_n)=I_\infty(t_nu_n) +\frac{t_n^2}{2}\int_{\mathbb{B}^N}(A(\epsilon g(x))-A_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}}. \end{equation*}

From $(A_2)$, given a η > 0, there is a sufficiently large R > 0 such that

\begin{equation*} A(z)\geq A_\infty - \eta \quad \mbox{for} \quad |z| \geq R. \end{equation*}

Thereby,

\begin{equation*} d_\epsilon +o_n(1) \geq d_\infty -\eta \frac{t_n^2}{2}\int_{\mathbb{B}^N}|u_n|^2\,dV_{{\mathbb{B}^N}} +\frac{t_n^2}{2}\int_{\mathcal{O}_R}(A(\epsilon g(x))-A_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}} \end{equation*}

where $\mathcal{O}_R=\{x \in \mathbb{B}^N \,:\, |g(x)| \lt \epsilon^{-1}R\}$. According to the fixed $\epsilon\in(0,\epsilon_0)$, we would like to point out that condition (g) combined with (2.3) implies that there exists $\bar{r} \in (0,1)$ such that $\mathcal{O}_R \subset \{x \in \mathbb{B}^N \,:\, |x| \lt \bar{r}\}$. Since $\{u_n\}$ is bounded in $L^{2}(\mathbb{B}^N)$, $\{t_n\}$ is bounded in $\mathbb{R}$ and the embedding $H^{1}(\mathbb{B}^N) \hookrightarrow L^{2}(\mathcal{O}_R)$ is compact, see Lemma A.1 in the Appendix, one gets

\begin{equation*} d_\epsilon +o_n(1) \geq d_\infty -\eta C+ o_n(1), \end{equation*}

for some positive constant C that does not depend on η and $n \in \mathbb{N}$. Recalling η > 0 is arbitrary, we derive that

\begin{equation*} d_\epsilon \geq d_\infty, \end{equation*}

which contradicts with Lemma 3.10. The proof is completed.

Proof Theorem 3.9

Using some very similar arguments in the proof of Theorem 3.7 together with Lemma 3.12, we see that the proof is obvious and the detail is omitted.

Proof. The proof of the above inequality follows the following idea: By using interpolation on the Lebesgue space $L^{p}(\mathbb{B}^N)$, one has

(4.5)\begin{equation} |u|^{l}_{l} \leq C|u|^{(1-\beta_l )l}_{2} \vert u\vert^{\beta_l l}_{2^*} ,~\forall u\in H^1(\mathbb{B}^N) ~ (N\geq 2),\ \beta_l=N\left(\frac{1}{2}-\frac{1}{l}\right). \end{equation}

Applying Poincaré’s inequality (2.2), one finds

(4.6)\begin{align} & |u|^{l}_{l} \leq C|u|^{(1-\beta_l )l}_{2} \left(\int_{\mathbb{B}^N}\left[|\nabla_{\mathbb{B}^N}u|_{\mathbb{B}^N}^{2}-\frac{(N-1)^2}{4}|u|^2\right]\,dV_{{\mathbb{B}^N}}\right)^{\frac{\beta_l l}{2}} ,\nonumber\\ &\qquad\qquad\qquad\qquad ~\forall u\in H^1(\mathbb{B}^N) ~ (N\geq 2),\ \beta_l=N\left(\frac{1}{2}-\frac{1}{l}\right), \end{align}

from where it follows that

(4.7)\begin{align} |u|^{l}_{l} \leq C|u|^{(1-\beta_l )l}_{2} \left(\int_{\mathbb{B}^N}\left[|\nabla_{\mathbb{B}^N}u|_{\mathbb{B}^N}^{2}\right]\,dV_{{\mathbb{B}^N}}\right)^{\frac{\beta_l l}{2}} , &~\forall u\in H^1(\mathbb{B}^N) ~ (N\geq 2),\nonumber\\ & \quad\qquad\beta_l=N\left(\frac{1}{2}-\frac{1}{l}\right), \end{align}

showing the desired result.

Proof. According to $(F_1)$, there are $C_1,C_2 \gt 0$ satisfying

\begin{equation*} |F(t)| \leq C_1|t|^{q}+C_2|t|^{p}, \quad \forall t \in \mathbb{R}. \end{equation*}

From which, we can apply (4.4) to see that

\begin{align*} \mathcal{J} (u) &=\frac{1}{2}\int_{\mathbb{B}^N } |\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}}-C_1\frac{C a^{(1-\beta_q)q}}{q} \left(\int_{\mathbb{B}^N}|\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2\,dV_{{\mathbb{B}^N}} \right)^{\frac{\beta_q q}{2} } \end{align*}

(4.8)\begin{align} &\ \ \ \ -\frac{C a^{(1-\beta_p)p}}{p} \left(\int_{\mathbb{B}^N}|\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2\,dV_{{\mathbb{B}^N}} \right)^{\frac{\beta_p p}{2} }. \end{align}

If $p,q \in (2,2+\frac{4}{N})$, clearly $\beta_p p,\beta_q q \lt 2$, which ensures the boundedness of $\mathcal{J}$ from below.

Proof. First of all, by $(F_1)-(F_2)$, there are three constants $C_1,C_2 \gt 0$ and $q_1 \in (2,2^*)$ such that

\begin{equation*} F(t) \geq C_1t^{q_1}-C_2t^{2} , \quad \forall t \geq 0. \end{equation*}

Now, given $u_0 \in C_0^{\infty}(\mathbb{B}^N)$ to be a nonnegative function with $|u_0|_{2}=1$, one has

\begin{equation*} \mathcal{J}(au_0) \leq\frac{a^2}{2}\int_{\mathbb{B}^N}|\nabla_{\mathbb{B}^N} u_0|^2\, dV_{{\mathbb{B}^N}}+C_2a^2 -C_1a^{q_1}\int_{\mathbb{B}^N}|u_0(x)|^{q_1}\,dV_{{\mathbb{B}^N}}. \end{equation*}

Since $q_1 \gt 2$, there is a constant $a^* \gt 0$ such that

\begin{equation*} \mathcal{J}(au_0) \lt 0, \quad a \gt a^*. \end{equation*}

Once $au_0 \in S(a)$, it follows that $\gamma(a) \lt 0$ for all $a \gt a^*$ finishing the proof.

Proof. Let ξ > 1 such that $a_2=\xi a_1$ and $\{u_n\} \subset S(a_1)$ be a minimizing sequence with respect to the number $\gamma(a_1)$, that is,

\begin{equation*} \mathcal{J}(u_n) \to \gamma(a_1) \,\,\text{as}\,\, n \to +\infty. \end{equation*}

Setting $v_n={\xi}{u_n}$, obviously $v_n \in S(a_2)$. Since by $(F_2)$, the function $t \mapsto \frac{F(t)}{t^{\vartheta}}$ is increasing on $(0,+\infty)$, we obtain the inequality

\begin{equation*} F(ts)\geq t^{\vartheta}F(s), \quad \forall t,s \gt 0 \quad \mbox{and} \quad t \geq 1, \end{equation*}

and so,

\begin{align*} \gamma(a_2) &\leq \mathcal{J}(v_n) = \xi^{2} \mathcal{J}(u_n)+\xi^{2}\displaystyle \int_{\mathbb{R}^N}F(|u_n|)\,dV_{{\mathbb{B}^N}}-\int_{\mathbb{R}^N}F(\xi |u_n|)\,dV_{{\mathbb{B}^N}} \\ &\leq \xi^{2} \mathcal{J}(u_n)+ (\xi^{2}-\xi^{\vartheta})\displaystyle \int_{\mathbb{R}^N}F(|u_n|)\,dV_{{\mathbb{B}^N}}. \end{align*}

Here we adopt the fact that F is even since f is odd, from where it follows that $F(s)=F(|s|)$ for all $s \in \mathbb{R}.$

Otherwise, there exists a subsequence of $\{u_n\}$, still denoted by itself, such that

\begin{equation*} \int_{\mathbb{B}^N}F(|u_n|)\,dV_{{\mathbb{B}^N}} \to 0, \,\,\text{as}\,\,n \to +\infty. \end{equation*}

Now, recalling that

\begin{equation*} 0 \gt \gamma(a_1)+o_n(1)=\mathcal{J}(u_n)\geq -\int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}}, \quad n \in \mathbb{N}, \end{equation*}

we arrive at a contradiction, and claim 4.6 is proved. Thanks to claim 4.6 and the fact that $\xi^2-\xi^{\vartheta} \lt 0$, we therefore obtain

\begin{equation*} \gamma(a_2)\leq \xi^2 \mathcal{J}(u_n)+{(\xi^{2}-\xi^{\vartheta})C}, \end{equation*}

for $n\in \mathbb{N}$ large. Letting $n \to +\infty$, it follows that

\begin{equation*} \gamma(a_2)\leq \xi^2 \gamma(a_1)+{(\xi^{2}-\xi^{\vartheta})C} \lt \xi^2\gamma(a_1), \end{equation*}

that is,

\begin{equation*} \frac{\gamma(a_2)}{a_2^2} \lt \frac{\gamma(a_1)}{a_1^2}, \end{equation*}

proving the lemma.

Proof. Setting the functional $\Psi:H^{1}(\mathbb{B}^N) \to \mathbb{R}$ given by

\begin{equation*} \Psi(u)=\frac{1}{2}\int_{\mathbb{B}^N}|u|^2\,dV_{{\mathbb{B}^N}}, \end{equation*}

we see that $S(a)=\Psi^{-1}(\{a^2/2\})$. Then, by Willem [Reference Willem69, chapter 8 and theorem 8.5], there exists $\{\mu_n\}\subset \mathbb{R}$ such that

(4.9)\begin{equation} \|\mathcal{J}'(u_n)-\mu_n \Psi'(u_n)\|_{(H^{1}(\mathbb{B}^N))^*} \to 0 \quad \mbox{as} \quad n \to +\infty. \end{equation}

Since $\{u_n\}$ is bounded in $H^{1}(\mathbb{B}^N)$, it concludes that $\{\mu_n\}$ is also a bounded sequence, then we can assume that $\mu_n\to \mu_a$ as $n \to +\infty$ along a subsequence. This together with (4.9) leads to

\begin{equation*} \mathcal{J}'(u_n)-\mu_a\Psi'(u_n)=o_n(1) \quad \mbox{in} \quad (H^{1}(\mathbb{B}^N))^*. \end{equation*}

Now, we are going to prove that $\mu_a \lt 0$. First of all, recall that

\begin{equation*} \int_{\mathbb{B}^N}|\nabla_{\mathbb{B}^N}u_n|^{2}_{{\mathbb{B}^N}}dV_{{\mathbb{B}^N}}-\int_{\mathbb{B}^N}f(u_n)u_n\,dV_{{\mathbb{B}^N}}=\mu_aa^2+o_n(1). \end{equation*}

Since $\mathcal{J}(u_n)=\gamma(a)+o_n(1)$, one gets

\begin{equation*} 2\gamma(a)+2\int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}}-\int_{\mathbb{B}^N}f(u_n)u_n\,dV_{{\mathbb{B}^N}}=\mu_aa^2+o_n(1). \end{equation*}

By $(F_2)$, it holds that

\begin{equation*} 2\gamma(a)+\left(\frac{2}{\theta}-1\right)\int_{\mathbb{B}^N}f(u_n)u_n\,dV_{{\mathbb{B}^N}}\geq\mu_aa^2+o_n(1). \end{equation*}

As $f(s)s \geq 0$ for all $s \in \mathbb{R}$, θ > 2, we get

\begin{equation*} 2\gamma(a) \geq \mu_aa^2. \end{equation*}

Now, recalling that $\gamma(a) \lt 0$ for $a \gt a^*$, it follows that $\mu_a \lt 0$ for all $a \gt a^*$. The proof is completed.

Proof. In this proof, we adopt some arguments found in the proof of [Reference Bhakta and Sandeep18, theorem 3.3]. Since $\{u_n\}$ is a minimizing sequence, we know that

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}|u_n|^{p_0}\,dV_{\mathbb{B}^N}=\delta' \gt 0, \end{equation*}

where $p_0=\max\{q,p\}$, because

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}} \gt 0, \end{equation*}

and $\{u_n\}$ is bounded in $L^{2}(\mathbb{B}^N)$.

Moreover, by corollary 4.8, $\{u_n\}$ is a $(PS)_d$ sequence for the functional $\bar{\mathcal{J}}$ given in (4.10). Arguing as in [Reference Bhakta and Sandeep18, theorem 3.3], there are $\tau_n \in \mathcal{I}(\mathbb{B}^N)$, $x_0 \in S_{\sqrt{3}}$ and $0 \lt \delta \lt S_{N,p_0}$ of such way that the sequence $\tilde{u}_n=u_n \circ \tau_n$ satisfies

\begin{equation*} \int_{A(x_0,\sqrt{3})}|\tilde{u}_n|^{p_0}\,dV_{{\mathbb{B}^N}}=\sup_{x \in S_{\sqrt{3}}}\int_{A(x,\sqrt{3})}|u_n|^{p_0}\,dV_{{\mathbb{B}^N}}=\delta. \end{equation*}

Since $\|\tilde{u}_n\|=\|u_n\|$ for all $n \in \mathbb{N}$, we have that $\{\tilde{u}_n\}$ is also a bounded sequence in $H^{1}(\mathbb{B}^N)$, and so, for some subsequence, still denoted by $\{\tilde{u}_n\}$, there exists $\tilde{u}\in H^{1}(\mathbb{B}^N)$ such that $\tilde{u}_n \rightharpoonup \tilde{u}$ in $H^{1}(\mathbb{B}^N)$. Moreover, $\{\tilde{u}_n\}$ is also a $(PS)_d$ sequence for $\bar{\mathcal{J}}$.

Otherwise, we must have $\tilde{u}=0$ that combines with the fact that $\{\tilde{u}_n\}$ is a $(PS)_d$ sequence for $\bar{\mathcal{J}}$ to give

(4.11)\begin{equation} \int_{\mathbb{B}^N \cap [|x| \geq r]}|\tilde{u}_n|^{p_0}\,dV_{{\mathbb{B}^N}} \to 0, \end{equation}

for $r \in (2-\sqrt{3},1)$. Recalling the embedding $H^{1}(\mathbb{B}^N) \hookrightarrow L^{p}(\mathbb{B}^N \cap [|x| \lt r],dV_{{\mathbb{B}^N}})$ is compact, see Lemma A.1 in the Appendix, we can derive that

(4.12)\begin{equation} \int_{\mathbb{B}^N \cap [|x| \lt r]}|\tilde{u}_n|^{p_0}\,dV_{{\mathbb{B}^N}} \to 0. \end{equation}

From (4.11) and (4.12),

\begin{equation*} \int_{\mathbb{B}^N}|\tilde{u}_n|^{p_0}\,dV_{{\mathbb{B}^N}} \to 0, \end{equation*}

which contradicts with the limit below

\begin{equation*} \liminf_{n \to +\infty}\int_{\mathbb{B}^N}|\tilde{u}_n|^{p_0}\,dV_{{\mathbb{B}^N}}=\liminf_{n \to +\infty}\int_{\mathbb{B}^N}|u_n|^{p_0}\,dV_{{\mathbb{B}^N}}=\delta' \gt 0, \end{equation*}

showing the claim 4.10, it finishes the proof.

Proof. Since $\mathcal{J}$ is coercive on S(a), the sequence $\{u_n\}$ is bounded, and so, $u_n \rightharpoonup u$ in $H^1(\mathbb{B}^N)$ for some subsequence. If $u \not=0$ and $|u|_2=b \not=a$, we must have $b \in (0,a)$. By the Brézis–Lieb Lemma (see e.g. [Reference Willem69]),

\begin{equation*} |u_n|_2^{2}=|u_n-u|_{2}^{2}+|u|_{2}^{2}+o_n(1). \end{equation*}

Furthermore, as F is a C ¹ function and has a subcritical growth in the Sobolev sense, it is easy to verify that

\begin{equation*} \int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}}=\int_{\mathbb{B}^N}F(u_n-u)\,dV_{{\mathbb{B}^N}}+\int_{\mathbb{B}^N}F(u)\,dV_{{\mathbb{B}^N}}+o_n(1). \end{equation*}

Setting $v_n=u_n-u$, $d_n=|v_n|_2$ and supposing that $|v_n|_2 \to d$, we get $a^2=b^2+d^2$. From $d_n \in (0,a)$ for n large enough,

\begin{equation*} \gamma(a)+o_n(1)=\mathcal{J}(u_n)=\mathcal{J}(v_n)+\mathcal{J}(u)+o_n(1)\geq \gamma(d_n)+\gamma(b)+o_n(1). \end{equation*}

thereby, by Lemma 4.5,

\begin{equation*} \gamma(a)+o_n(1) \geq \frac{d_n^2}{a^2}\gamma(a)+\gamma(b)+o_n(1). \end{equation*}

Letting $n \to +\infty$, one finds

(4.13)\begin{equation} \gamma(a) \geq \frac{d^2}{a^2}\gamma(a)+\gamma(b). \end{equation}

Since $b \in (0,a)$, employing again Lemma 4.5 in (4.13), we arrive at the following inequality

\begin{equation*} \gamma(a) \gt \frac{d^2}{a^2}\gamma(a)+\frac{b^2}{a^2}\gamma(a)=\left(\frac{d^2}{a^2}+\frac{b^2}{a^2}\right)\gamma(a)=\gamma(a), \end{equation*}

which is absurd. This asserts that $|u|_2=a$, or equivalently, $u \in S(a)$.

As $|u_n|_2=|u|_2=a$, $u_n \rightharpoonup u$ in $L^{2}(\mathbb{B}^N)$ and $L^{2}(\mathbb{B}^N)$ is reflexive, it is well known that

(4.14)\begin{equation} u_n \to u \quad \mbox{in} \quad L^{2}(\mathbb{B}^N). \end{equation}

This combined with the interpolation theorem in the Lebesgue space and $(F_1)-(F_2)$ gives

(4.15)\begin{equation} \int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}} \to \int_{\mathbb{B}^N}F(u)\,dV_{{\mathbb{B}^N}}. \end{equation}

These limits together with $\gamma(a)=\displaystyle \lim_{n \to +\infty}\mathcal{J}(u_n)$ provide

\begin{equation*} \gamma(a) \geq \mathcal{J}(u). \end{equation*}

As $u \in S(a)$, we infer that $\mathcal{J}(u)=\gamma(a)$, then

\begin{equation*} \lim_{n \to +\infty}\mathcal{J}(u_n)= \mathcal{J}(u), \end{equation*}

that combines with (4.14) and (4.15) to give

\begin{equation*} \|u_n\|^{2} \to \|u\|^2, \end{equation*}

where $\|\cdot\|$ denotes the usual norm in $H^{1}(\mathbb{B}^N)$. The last limit permits to conclude that $u_n \to u$ in $H^{1}(\mathbb{B}^N)$.

Now, suppose that u = 0, that is, $u_n \rightharpoonup 0$ in $H^1(\mathbb{B}^N)$. Arguing as in claim 4.6, there exists a constant C > 0 such that

(4.16)\begin{equation} \int_{\mathbb{B}^N}F(u_n)\,dV_{{\mathbb{B}^N}}\geq C, \quad \text{for}\, n \in \mathbb{N}\,\,\text{large}, \end{equation}

then

\begin{equation*} \int_{\mathbb{B}^N}|u_n|^{p_0}\,dV_{{\mathbb{B}^N}}\geq C_1 \gt 0, \quad \text{for}\, n \in \mathbb{N}\,\,\text{large}, \end{equation*}

for some C ₁ and $p_0=\max\{q,p\}$. Via employing Lemma 4.9, there exists $\tau_n \in \mathcal{I}(\mathbb{B}^N)$ such that the sequence $\tilde{u}_n=u_n \circ \tau_n$ is also minimizing sequence associated with $\gamma(a)$, and for some subsequence, still denoted by itself, one has $ \tilde{u}_n \rightharpoonup \tilde{u}$ in $H^{1}(\mathbb{B}^N)$ for some $\tilde{u} \in H^{1}(\mathbb{B}^N) \setminus \{0\}$. Following as in the first part of the proof, we derive that $\tilde{u}_n \to \tilde{u}$ in $H^{1}(\mathbb{B}^N)$. This proves the theorem.

Proof of Theorem 4.1

By Lemma 4.3 there exists a bounded minimizing sequence $\{u_n\}\subset S(a)$ with respect to $\gamma(a)$, that is, $\mathcal{J}(u_n) \to \gamma(a)$. Thanks to Theorem 4.11, there is $u _a\in S(a)$ with $\mathcal{J}(u_a)=\gamma(a)$. Therefore, by the Lagrange multiplier theorem together with Lemma 4.7, there exists $\mu_a \lt 0$ such that

(4.17)\begin{equation} \mathcal{J}'(u_a)=\mu_a \Psi'(u_a) \quad \mbox{in} \quad (H^{1}(\mathbb{B}^N))^* \end{equation}

where $\Psi:H^{1}(\mathbb{R}^N) \to \mathbb{R}$ is given by

\begin{equation*} \Psi(u)=\int_{\mathbb{B}^N}|u|^2\,dV_{{\mathbb{B}^N}}, \quad u \in H^{1}(\mathbb{B}^N). \end{equation*}

Thereby, according to (4.17), there holds

\begin{equation*} -\Delta_{ \mathbb{B}^N} u_a=\mu_a u_a +f(u_a), \quad \mbox{in} \quad \mathbb{B}^N. \end{equation*}

Furthermore, arguing as in the proof of Theorem 3.1, we have that $u_a(x) \gt 0$ for all $x\in \mathbb{B}^N$.

Proof. The first part is trivial and we omit it here. Suppose by the contradiction that $u_n \rightharpoonup 0$ in $H^{1}(\mathbb{B}^N)$. Then,

\begin{equation*} \tilde{\gamma}(a)+o_n(1)=\mathbb{J}(u_n)=\mathbb{J}_\infty(u_n) + \displaystyle \int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)|u_n|^2 \,dV_{{\mathbb{B}^N}}+o_n(1) \end{equation*}

and so,

(4.24)\begin{equation} \tilde{\gamma}(a)+o_n(1)\geq \tilde{\gamma}_\infty(a) + \displaystyle \int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)|u_n|^2 \,dV_{{\mathbb{B}^N}}+o_n(1). \end{equation}

Arguing as in the study of ground state solution, namely (3.12), we know that

\begin{equation*} \int_{\mathbb{B}^N}(\tilde{A}(x)-\tilde{A}_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}} \to 0. \end{equation*}

Taking the limit $n \to +\infty$ in (4.24) and using the above limit, we derive that

\begin{equation*} \tilde{\gamma}(a) \geq \tilde{\gamma}_\infty(a) \end{equation*}

which contradicts with (4.23). Thus, we must have $u \not= 0$. The proof is completed.

Proof of Theorem 4.12

Due to the constant $\hat{a}^* \gt 0$ mentioned above, for every $a \gt \hat{a}^*$, the existence of solution $(\hat{u}_a,\hat{\mu}_a)$ then follows directly by lemmas 4.13 and 4.14, where $\hat{\mu}_a \lt 0$ is analogous to Lemma 4.7. In view of the proof of Theorem 3.1, $\hat{u}_a(x) \gt 0$ for all $x\in \mathbb{B}^N$.

Proof. The proof is very similar to that of Lemma 3.10 and so we omit it here.

Proof. The first part is trivial and we omit it here. Suppose by the contradiction that $u_n \rightharpoonup 0$ in $H^{1}(\mathbb{B}^N)$. Then,

\begin{equation*} \Upsilon_\epsilon(a)+o_n(1)=\mathfrak{J}_\epsilon(u_n)=\mathfrak{J}_\infty(u_n)+\frac{1}{2}\int_{\mathbb{B}^N}(A(\epsilon g(x))-A_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}}. \end{equation*}

From $(A_2)$, given η > 0, there is a sufficiently large R > 0 such that

\begin{equation*} A(z)\geq A_\infty - \eta \quad \mbox{for} \quad |z| \geq R. \end{equation*}

Thereby,

\begin{equation*} \Upsilon_\epsilon(a) +o_n(1) \geq \Upsilon_\infty(a) -\eta\frac{t_n^2}{2} \int_{\mathbb{B}^N}|u_n|^2\,dV_{{\mathbb{B}^N}}+\frac{t_n^2}{2}\int_{\mathcal{O}_R}(A(\epsilon g(x))-A_\infty)|u_n|^2\,dV_{{\mathbb{B}^N}}, \end{equation*}

where $\mathcal{O}_R=\{x \in \mathbb{B}^N \,:\, |g(x)| \lt \epsilon^{-1}R\}$. Proceeding as the same way exhibited in Lemma 3.12, we can derive that

\begin{equation*} \Upsilon_\epsilon(a)\geq \Upsilon_\infty(a), \end{equation*}

which contradicts with Lemma 4.18. The proof of the lemma is complete.

Proof of Theorem 4.15

Due to the constant $a_* \gt 0$ and $\epsilon_* \gt 0$ mentioned above, for every $a \gt a_*$ and $\epsilon\in(0,\epsilon_*)$, the existence of solution $({u}_a,{\mu}_a)$ then follows directly by lemmas 4.16 and 4.18, where ${\mu}_a \lt 0$ is similar to Lemma 4.7. By the proof of Theorem 3.1, ${u}_a(x) \gt 0$ for all $x\in \mathbb{B}^N$.

Proof. Arguing as in Lemma 4.3, we see that (4.4) combined with (5.9) gives

\begin{align*} E_\mathcal{R}(u) \geq \frac{1}{2}\int_{{\mathbb{B}^N}} |\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}}-(1+\beta \mathcal{R}^{p-q})a^{(1-\beta_q)q}&\left(\int_{{\mathbb{B}^N}}|\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}} \right)^{\frac{\beta_q q}{2} },\\ & \qquad\qquad\quad \forall u \in S(a). \end{align*}

Fixing $\underline{\beta}^*=\underline{\beta}^*(\mathcal{R})=\frac{1}{\mathcal{R}^{p-q}}$ and $\beta \in (0,\underline{\beta}^*)$, one gets

\begin{equation*} E_\mathcal{R}(u) \geq \frac{1}{2}\int_{{\mathbb{B}^N}} |\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}}-2a^{(1-\beta_q)q}\left(\int_{{\mathbb{B}^N}}|\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}} \right)^{\frac{\beta_q q}{2} }, \quad \forall u \in S(a). \end{equation*}

Due to $\beta_q q \lt 2$, exploiting Young’s inequality, there is a constant $C_1 \gt 0$ independent of $\mathcal{R}$ such that

\begin{equation*} \left(\int_{{\mathbb{B}^N}}|\nabla_{\mathbb{B}^N} u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}} \right)^{\frac{\beta_q q}{2} } \leq C_1+ \frac{1}{8a^{(1-\beta_q)q}}\int_{{\mathbb{B}^N}} |\nabla_{ \mathbb{B}^N } u|_{{\mathbb{B}^N}}^2 \,dV_{{\mathbb{B}^N}}, \quad \forall u \in H^{1}({\mathbb{B}^N}). \end{equation*}

Hence, there is a constant $C_2 \gt 0$ independent of $\mathcal{R} \gt 0$ such that

\begin{equation*} |\nabla_{{\mathbb{B}^N}} u |_2^{2} \leq 4E_\mathcal{R}(u) +C_2, \quad \forall a \gt \underline{a}^*,~ \beta \in (0,\underline{\beta}^*), ~\mathcal{R} \gt 0 \quad \mbox{and} \quad u \in S(a). \end{equation*}

Fixing $\phi \in S(a)$, we derive that $E_\mathcal{R}(u_\mathcal{R})\leq E_\mathcal{R}(\phi) \leq \frac{1}{2}|\nabla_{{\mathbb{B}^N}} \phi|_2^{2}$, and so,

\begin{equation*} |\nabla_{\mathbb{B}^N} u_\mathcal{R}|_2 \leq \sqrt{2|\nabla_{\mathbb{B}^N} \phi|_2^{2} +C_2}=C, \quad \forall a \gt \underline{a}^*,\quad \beta \in (0, \underline{\beta}^*) \quad\mbox{and} \quad \mathcal{R} \gt 0, \end{equation*}

showing the desired result.

Proof of Theorem 5.1

Thanks to (5.9), following the proof of Theorem 4.1 in Section 4, we can claim that $u_\mathcal{R}$ is a solution of the problem

\begin{align*} \left\{ \begin{aligned} &-\Delta_{ \mathbb{B}^N} u=\mu_\mathcal{R} u+f_\mathcal{R}(u) \quad \hbox{in } \quad \mathbb{B}^N,\\ &u(x) \gt 0 \quad \hbox{in} \quad \mathbb{B}^N, \\ \end{aligned} \right. \end{align*}

for some $\mu_\mathcal{R} \lt 0$. Proceeding as in the proof of Lemma 4.7, we can prove that $\mu_\mathcal{R} \lt -\frac{2\Lambda}{a^2} \lt 0$. On the other hand, since $\beta\in (0,\underline{\beta}^*)$, the definition of $f_\mathcal{R}$ together with (5.9) leads to

\begin{equation*} 0 \leq f_R(t) \leq 2t^{q-1} , \quad \forall t \geq 0 \quad \mbox{and} \quad \mathcal{R} \gt 0. \end{equation*}

Thereby, $u_\mathcal{R}\geq 0$ and $\{u_\mathcal{R}\}$ is bounded in $L^{s}(\mathbb{B}^N)$ for all $\mathcal{R} \gt 0$ and $s \in [2,2^*]$, we could apply the Nash-Moser iteration technique as done in [Reference Alves and Figueiredo3, lemma 4.1] and [Reference Bhakta and Sandeep18, theorem 4.1] to get a constant M > 0 independent of $\mathcal{R} \gt 0$ such that

\begin{equation*} |u_\mathcal{R}|_\infty \leq M, \quad \forall \mathcal{R} \gt 0. \end{equation*}

Let us fix $\mathcal{R} \gt M$, then the function $u_\mathcal{R}$ is a positive solution for Problem (5.1) if $a \gt \underline{a}^*$ and $\beta\in (0, \underline{\beta}^*)$. This finishes the proof of Theorem 5.1.

Proof of Theorem 5.3

Thanks to (5.9), following the proof of Theorem 4.1 in Section 4, we can claim that $u_\mathcal{R}$ is a solution of the problem

\begin{align*} \left\{ \begin{aligned} &-\Delta_{ \mathbb{B}^N} u+\tilde{A}(x)u=\mu_\mathcal{R} u+f_\mathcal{R}(u) \quad \hbox{in } \quad \mathbb{B}^N,\\ &u(x) \gt 0 \quad \hbox{in} \quad \mathbb{B}^N, \\ \end{aligned} \right. \end{align*}

for some $\mu_\mathcal{R} \lt 0$. Proceeding as in the proof of Lemma 4.7, we can prove that $\mu_\mathcal{R} \lt -\frac{2\Lambda}{a^2} \lt 0$. On the other hand, since $\beta\in (0,\underline{\beta}^*)$, the definition of $f_\mathcal{R}$ together with (5.9) leads to

\begin{equation*} 0 \leq f_R(t) \leq 2t^{q-1} , \quad \forall t \geq 0 \quad \mbox{and} \quad \mathcal{R} \gt 0. \end{equation*}

Thereby, $u_\mathcal{R}\geq 0$ and $\{u_\mathcal{R}\}$ is bounded in $L^{s}(\mathbb{B}^N)$ for all $\mathcal{R} \gt 0$ and $s \in [2,2^*]$. As $\tilde{A}(x) \geq A_0$ for all $x \in \mathbb{B}^N$, we could apply the Nash-Moser iteration technique as done in [Reference Alves and Figueiredo3, lemma 4.1] and [Reference Bhakta and Sandeep18, theorem 4.1] to get a constant M > 0 independent of $\mathcal{R} \gt 0$ such that

\begin{equation*} |u_\mathcal{R}|_\infty \leq M, \quad \forall \mathcal{R} \gt 0. \end{equation*}

Let us fix $\mathcal{R} \gt M$, then the function $u_\mathcal{R}$ is a positive solution for Problem (5.12) if $a \gt \underline{a}^*$ and $\beta\in (0, \underline{\beta}^*)$. This finishes the proof of Theorem 5.1.

Proof of Theorem 5.4

Thanks to (5.9), following the proof of Theorem 4.1 in Section 4, we can claim that $u_\mathcal{R}$ is a solution of the problem

\begin{align*} \left\{ \begin{aligned} &-\Delta_{ \mathbb{B}^N} u+A(\epsilon g(x))u=\mu_\mathcal{R} u+f_\mathcal{R}(u) \quad \hbox{in } \quad \mathbb{B}^N,\\ &u(x) \gt 0 \quad \hbox{in} \quad \mathbb{B}^N, \\ \end{aligned} \right. \end{align*}

for some $\mu_\mathcal{R} \lt 0$. Proceeding as in the proof of Lemma 4.7, we can prove that $\mu_\mathcal{R} \lt -\frac{2\Lambda}{a^2} \lt 0$. On the other hand, since $\beta\in (0,\underline{\beta}^*)$, the definition of $f_\mathcal{R}$ together with (5.9) leads to

\begin{equation*} 0 \leq f_R(t) \leq 2t^{q-1} , \quad \forall t \geq 0 \quad \mbox{and} \quad \mathcal{R} \gt 0. \end{equation*}

Thereby, $u_\mathcal{R}\geq 0$ and $\{u_\mathcal{R}\}$ is bounded in $L^{s}(\mathbb{B}^N)$ for all $\mathcal{R} \gt 0$ and $s \in (2,2^*)$. As ${A}(x) \geq A_0$ for all $x \in \mathbb{B}^N$, we could apply the Nash-Moser iteration technique as done in [Reference Alves and Figueiredo3, lemma 4.1] and [Reference Bhakta and Sandeep18, theorem 4.1] to get a constant M > 0 independent of $\mathcal{R} \gt 0$ such that

\begin{equation*} |u_\mathcal{R}|_\infty \leq M, \quad \forall \mathcal{R} \gt 0. \end{equation*}

Let us fix $\mathcal{R} \gt M$, then the function $u_\mathcal{R}$ is a positive solution for Problem (5.15) if $a \gt \underline{a}^*$ and $\beta\in (0, \underline{\beta}^*)$. This finishes the proof of Theorem 5.4.