Analysis of a low-profile, dual band patch antenna for wireless applications

1.   Introduction

In this paper, we are looking for solutions to the following Choquard equation

where $N \geq 3$, $\alpha \in (0, N)$, $F(s): = \int^{s}_{0}f(t)dt$, $\mu > 0$, $2 < q \leq \overline{q}: = 2+\frac{4}{N}$, $a > 0$ is a given mass, and $\lambda \in \mathbb{R}$ appears as an unknown Lagrange multiplier. $I_{\alpha}: \mathbb{R}^{N}\setminus \{0\} \mapsto \mathbb{R}$ is the Riesz potential defined by

The Choquard equation

has been studied extensively for its profound physical backgrounds. In particular, when $N = 3$, $\alpha = 2$, and $F(s) = \frac{s^{2}}{2}$, Eq (1.2) turns into

which was introduced by Pekar ^[1] to describe the quantum theory of a polaron at rest and then used by Choquard ^[2] to study steady states of the one one-component plasma approximation to the Hartree-Fock theory. Also, Eq (1.3) reemerged as a model of self-gravitating matter ^[3,4], and in that context it is known as the Schrödinger-Newton equation.

The pioneering mathematical research dates back to Lieb ^[2], in which the author proved the existence and uniqueness for Eq (1.3) by variational methods. Later, Lions ^[5] obtained the existence of normalized solutions. In the homogeneous nonlinearity case of Eq (1.2) with $F(s) = \frac{1}{p}|s|^{p}$, Moroz and Van Schaftingen in ^[6] established the existence of ground states to Eq (1.2) with an optimal range $\frac{N+\alpha}{N} < p < \frac{N+\alpha}{N-2}$. Moreover, Ghimenti and Van Schaftingen in ^[7] obtained solutions which are odd with respect to a hyperplane of $\mathbb{R}^{N}$. The existence of saddle type nodal solution for the Choquard equation was proven in the study ^[8]. For a more general nonlinearity, Moroz and Van Schaftingen ^[9] proved Eq (1.2) has a ground solution when the nonlinearity satisfies Berestycki-Lions type condition and $N \geq 3$, see also ^[10] for the case $N = 2$. For the Choquard equation with a local nonlinear perturbation

When $N = 3$, $0 < \alpha < 1$, $p = 2$, and $4 \leq q < 6$, Chen and Gao in ^[11] obtained the existence of solutions of Eq (1.4). For $q = 2^{*}$ in Eq (1.4), Seok in ^[12] constructed a family of nontrivial solutions. In ^[13], the authors studied Eq (1.4) with a general local nonlinearity $f(x, u)$ subcritical type instead of $|u|^{q-2}u$.

From a physical point of view, it is interesting to find solutions with prescribed $L^{2}-$ norm, since there is a conservation of mass. Solutions of this type are often referred to as normalized solutions. In recent years, normalized solutions to nonlinear elliptic problems have attracted much attention from researchers. In ^[14], Jeanjean using a mountain pass structure for a stretched functional to consider the equation

He proved the existence of at least one normalized solution of Eq (1.5) in a purely $L^{2}-$ supercritical case. Later, Bartsch and Valeriola ^[15] obtained the existence of infinitely many normalized solutions by a new linking geometry for the stretched functional. For more general nonlinearity $f(s)$ has $L^{2}-$ subcritical {growth, Shibata ^[16] obtained the existence and nonexistence of normalized ground state solution of} Eq (1.5) via minimizing method, Jeanjean and Lu in ^[17] showed the existence of nonradial normalized solutions for any $N\geq 4$. For more results on normalized solutions for Schrödinger equations by variational methods, we would like to refer ^{[18,19,20,21,22,23]}. Also, for the evolution equations with the singular potentials whose the steady state equations are the nonlinear elliptic equations, see ^[24].

Concerning the normalized solutions of Choquard equations, Li and Ye in ^[25] considered the existence of normalized solutions to the following equation

under a set of assumptions on $f$, where $f$ takes the special case $f(s) = C_{1}|s|^{r-2}s + C_{2}|s|^{p-2}s$ requires that $\frac{N+2+\alpha}{N} < r \leq p < \frac{N+\alpha}{N-2}$. Later, Yuan et al. ^[26] generalized the results in ^[25] to more general $f\in \mathcal{C}(\mathbb{R}, \mathbb{R})$. In ^[27], Bartsch et al. obtained the existence of infinitely many normalized solutions of (1.6). For $F(u) = \frac{|u|^{p}}{p}$, Ye in ^[28] studied the qualitative properties including existence and nonexistence of minimizers of the functional associated to the Eq (1.6). Yao et al. in ^[29] considered normalized solutions for the following problems

Under different assumptions on $\gamma, \mu, p$, and $q$, they proved several existence, multiplicity, and nonexistence results. For more related topics, we refer the reader to ^{[30,31,32,33]}.

Motivated by the above papers, it is natural to ask if the nonlinearity $f(u)$ in Eq (1.1) satisfies general growth assumptions, and if the normalized ground states still exist. In the present paper, we attempt to study this kind of problem. In order to prove the existence of normalized ground state solutions to Eq (1.1), assuming that nonlinearity $f \in \mathcal{C}(\mathbb{R}, \mathbb{R})$ satisfies the growth assumptions:

$(f_{1})$ $\lim_{s\rightarrow0 }\frac{f(s)}{|s|^{\frac{\alpha}{N}}} = 0$,

$(f_{2})$ $\lim_{|s|\rightarrow +\infty }\frac{f(s)}{|s|^{\frac{2+\alpha}{N}}} = 0$,

$(f_{3})$ $f$ is odd and $f$ does not change sign on $(0, +\infty)$,

$(f_{4})$ $\lim_{s\rightarrow 0}\frac{F(s)}{|s|^{\frac{N+2+\alpha}{N}}} = +\infty$,

$(f_{5})$ $\limsup_{s\rightarrow 0}\frac{F(s)}{|s|^{\frac{N+2+\alpha}{N}}} < +\infty$.

To find solutions of Eq (1.1), we define functional $I_{\mu}(u) : H^{1}(\mathbb{R}^{N}) \rightarrow \mathbb{R}$ by

For $a > 0$, set

Since $f \in \mathcal{C}(\mathbb{R}, \mathbb{R})$ satisfies $(f_{1})$ and $(f_{2})$, using the Hardy-Littlewood-Sobolev inequality, we see that $I_{\mu} \in \mathcal{C}^{1}(H^{1}(\mathbb{R}^{N}), \mathbb{R})$. Normalized solutions of Eq (1.1) can be obtained by looking for critical points of $I_{\mu}$ on the constraint $S(a)$. It is standard that each critical point $u\in S(a)$ of $I_{\mu}|_{S(a)}$, corresponds a Lagrangian multiplier $\lambda \in \mathbb{R}$ such that $(u, \lambda)$ solves Eq (1.1). We will be interested in ground state solutions, and following ^[34], we say that $u\in S(a)$ is a normalized ground state solution to Eq (1.1) if $(I_{\mu}|_{S(a)})'(u) = 0$ and

In particular, if $u\in S(a)$ is a minimizer of the minimization problem

then $u$ is a critical point of $I_{\mu}|_{S(a)}$ as well as a normalized ground state to Eq (1.1).

Our main results dealing with problem (1.1) are the following:

Theorem 1.1. Assume that ($f_{1}$)–($f_{3}$) hold. Let $\mu > 0$ and $2 < q < \overline{q}$. Then, for any $a > 0$, there exists a global minimizer $\widetilde{u}$ with respect to $m(a)$, which solves Eq (1.1) for some $\widetilde{\lambda} < 0$. Moreover, $\widetilde{u}$ is a ground state solution of Eq (1.1) which has constant sign, is radially symmetric with respect to some point in $\mathbb{R}^{N}$, and is decreasing.

Theorem 1.2. Assume that ($f_{1}$)–($f_{3}$) hold and $q = \overline{q}$. For any $\mu > 0$ small enough, there exist a $a_{0} = a_{0}(\mu) > 0$ and $a^{*} \in [0, a_{0})$ such that:

(i) for any $a\in (0, a^{*})$, there is no global minimizer with respect to $m(a)$.

(ii) for any $a\in (a^{*}, a_{0})$, there exists a global minimizer $u$ with respect to $m(a)$, which solves Eq (1.1) for some $\lambda < 0$. Moreover, $u$ is a ground state solution of Eq (1.1) which has constant sign, is radially symmetric with respect to some point in $\mathbb{R}^{N}$, and is decreasing.

(iii) if $(f_{4})$ holds, then $a^{*} = 0$, and if $(f_{5})$ holds, then $a^{*} > 0$.

Remark 1. The value $a_{0} = a_{0}(\mu) > 0$ is explicit and is given in Lemma 4.1. In particular, $a_{0} > 0$ can be taken arbitrary large by taking $\mu > 0$ small enough.

The following result positively answers the existence of global minimizer with respect to $m(a^{*})$ for the sharp threshold $a^{*} > 0$.

Theorem 1.3. Assume that $(f_{1})$–$(f_{3})$ and

($f_{6}$) $\limsup_{s\rightarrow 0}\frac{F(s)}{|s|^{\frac{N+2+\alpha}{N}}} = 0$ holds.

Let $q = \overline{q}$. For $a = a^{*}$, there exists a global minimizer $v$ with respect to $m(a^{*}) = 0$, which solves Eq (1.1) for some $\lambda < 0$. In particular, $v$ is a ground state solution of Eq (1.1) which has constant sign and is radially symmetric with respect to some point in $\mathbb{R}^{N}$.

To the best of our knowledge, the main results in this paper are new. This is a complement of the results for Choquard equations about the existence of normalized solutions. Our main theorems can be viewed as an extension of some results in ^[25,29] to more general cases. In our setting, we only consider the situation when the constraint functional $I_{\mu}|_{S(a)}$ is bounded from below and is coercive. As we will see, the existence of normalized states of Eq (1.1) are strongly effected by further assumptions on the exponent $q$. We are first devoted to prove the existence of ground state solutions of Eq (1.1) with $2 < q < 2+ \frac{4}{N}$ by application of the concentration-compactness principle ^[35]. In this case, $m(a) < 0$ for any $a > 0$ and the strict subadditivity inequality

holds, which permits us to exclude the dichotomy of the minimizing sequence. However, in the case of $q = \overline{q}: = 2 + \frac{4}{N}$, compare to ^[29], for a general $f$, the strict subadditivity inequality (1.8) does not hold, and $m(a) < 0$ for all $a > 0$ may not be satisfied. This prevents us from using the concentration-compactness principle in a standard way. In the proof, we adopt some ideas in ^[16,19] to recover the compactness of minimizing sequence with respect to $m(a)$.

The paper is organized as follows. In Section 2, we introduce the variational framework and give some preliminary results. In Section 3, we discuss the case of $2 < q < \overline{q}$ and prove Theorem 1.1. In Section 4, we deal with the case of $q = \overline{q}$ and prove Theorems 1.2 and 1.3.

In this paper, we will use the following notations:

$\bullet$ $H^{1}(\mathbb{R}^{N})$ is the usual Sobolev space endowed with the norm $\|u\|^{2} = \int_{\mathbb{R}^{N}}(|\nabla u |^{2} + |u|^{2}) dx$.

$\bullet$ $L^{s}(\mathbb{R}^{N})$, for $1 \leq s < \infty$, denotes the Lebesgue space with the norm $|u|_{s}^{s} = \int_{\mathbb{R}^{N}}|u |^{s} dx$.

$\bullet$ For any $r > 0$ and $x\in \mathbb{R}^{N}$, $B_{r}(x)$ denotes the ball of radius of $r$ centered at the $x$.

$\bullet$ The letters $C$, $C_{0}, C', C'', C_{1}, C_{2}\cdot\cdot\cdot$ are positive (possibly different) constants.

$\bullet$ $o_{n}(1)$ denotes the vanishing quantities as $n\rightarrow \infty$.

2.   Preliminaries

In this section, we give some results which will be useful in forthcoming sections. First, let us recall the following Hardy-Littlewood-Sobolev inequality which will be frequently used throughout the paper.

Lemma 2.1. (See [^[36], Theorem4.3]). Suppose $\alpha \in (0, N)$, and $s, r > 1$ with $\frac{1}{r} + \frac{1}{s} = 1+\frac{\alpha}{N}$. Let $f\in L^{s}(\mathbb{R}^{N})$ and $g\in L^{r}(\mathbb{R}^{N})$. Then, there exists a constant $C(N, \alpha, s, r) > 0$ such that

In particular, if $r = s = \frac{2N}{N+\alpha}$, then

Let $N \geq 1$ and $\frac{N+\alpha}{N} < p < 2^{*}_{\alpha}: = \frac{N+\alpha}{N-2}$. Then, we introduce the following Gagliardo-Nirenberg inequality of Hartree type (^[28])

We also recall the following Gagliardo-Nirenberg inequality. For $p\in (2, 2^{*})$ and $u\in H^{1}(\mathbb{R}^{N})$,

Lemma 2.2. Assume that ($f_{1}$)-($f_{2}$) hold. Let $\{u_{n}\} \subset H^{1}(\mathbb{R}^{N})$ be a bounded sequence. If either $\lim_{n \rightarrow \infty} |u_{n}|_{2} = 0$ or $\lim_{n \rightarrow \infty} |u_{n}|_{\frac{2(N+\alpha + 2)}{N + \alpha}} = 0$ holds, then

Proof. By $(f_{1})$ and $(f_{2})$, for any $\varepsilon > 0$, there exists $C_{\varepsilon} > 0$ such that

Then, using Lemma 2.1, for $u\in H^{1}(\mathbb{R}^{N})$ we obtain

By Eq (2.2), we have

If $\lim_{n \rightarrow \infty} |u_{n}|_{2} = 0$, by Eqs (2.3), (2.4), and the boundedness of $\{u_{n}\}$, we get

If $\lim_{n \rightarrow \infty} |u_{n}|_{\frac{2(N+\alpha + 2)}{N + \alpha}} = 0$, by Eq (2.3), we have

Since $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$ and $\varepsilon > 0$ is arbitrary, the conclusion holds. The proof is completed.

In our subsequent arguments, we will use the following nonlocal version of the Brezis-Lieb lemma.

Lemma 2.3. (See [^[37], Lemma 2.2]). Assume $\alpha \in (0, N)$ and there exists a constant $C > 0$ such that

Let $\{u_{n}\} \subset H^{1}(\mathbb{R}^{N})$ be such that $u_{n} \rightharpoonup u$ weakly in $H^{1}(\mathbb{R}^{N})$ and almost everywhere in $\mathbb{R}^{N}$ as $n \rightarrow \infty$. Then,

Using a similar argument as the proof of (^[9] Theorem 3), we have the following Poho$\check{\mbox{z}}$aev of Eq (1.1).

Lemma 2.4. Assume that $N \geq 3$ and $\alpha \in (0, N)$. If $f\in \mathcal{C}(\mathbb{R}, \mathbb{R})$ satisfies ($f_{1}$) and ($f_{2}$), and if $(u, \lambda)\in H^{1}(\mathbb{R}^{N})\times \mathbb{R}$ solves problem (1.1), then

For each $u\in S(a)$ and $t > 0$, we define the scaling function

It is clear that $u^{t}\in S(a)$ and

3.   $L^{2}$-subcritical perturbation

In this section, we deal with the case $2 < q < \overline{q}$ and prove Theorem 1.1.

Lemma 3.1. Assume that ($f_{1}$) and ($f_{2}$) hold. For any $2 < q < \overline{q}$ and $a, \mu > 0$, we have

Proof. By $(f_{1})$ and $(f_{2})$, for any $\varepsilon > 0$, there exists $C_{\varepsilon} > 0$ such that

Using Lemma 2.1 and Eq (2.2), we obtain, for $u\in H^{1}(\mathbb{R}^{N})$

Then, for any $u\in S(a)$, by Eqs (2.2) and (3.1), we get

Choosing $\varepsilon = (4C_{4}a^{\frac{2+\alpha}{N}})^{-\frac{1}{2}}$, it follows from Eq (3.2) that

for every $u\in S(a)$. Since $2 < q < \overline{q}$, we see that $0 < q\gamma_{q} < 2$, and hence $I_{\mu}$ is coercive on $S(a)$, which provides that $m(a) > -\infty$.

On the other hand, for $u\in S(a)$

Noticing that $2 < q < \overline{q}$, we have $2-\frac{N(q-2)}{2} > 0$, and hence $I_{\mu}(u^{t}) < 0$ for every $u\in S(a)$ with $t > 0$ small enough. Therefore, we have that $m(a) < 0$ for any $a > 0$.

Since $m(a) < 0$ for any $a > 0$, we can give the following strict sub-additivity.

Lemma 3.2. Let $a_{1}, a_{2} > 0$ be such that $a_{1} + a_{2} = a$. Then,

Proof. For $u\in S(a)$ and $\theta > 1$, we set $\overline{u}(x) = u(\theta^{-\frac{1}{N}}x)$. Then, $\overline{u}(x) \in S(\theta a)$. Let $\{u_{n}\} \subset S(a)$ be a minimizing sequence for $m(a)$. Since $\theta > 1$, we have

As a consequence,

with equality if and only if

But this is can not occur. Otherwise, by Eq (3.3), we find

By Lemma 3.1, $I_{\mu}$ is coercive on $S(a)$ and we have $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. It follows from $\theta > 1$ that

Combining Eq (2.2) we get $\lim_{n \rightarrow \infty}|u_{n}|^{q}_{q} = 0$. Then, by Lemma 3.1, we obtain

a contradiction. Thus, we have the strict inequality

Next, we show that $m(a) < m(a_{1}) + m(a_{2})$. We may assume that $a_{1}\geq a_{2}$, by Eq (3.4) we have

The proof is completed.

Lemma 3.3. Assume that $(f_{1})$ and $(f_{2})$ hold and $a, \mu > 0$. Let $\{u_{n}\} \subset H^{1}(\mathbb{R}^{N})$ be a sequence such that

Then, taking a subsequence if necessary, there exist $\widetilde{u} \in S(a)$ and a family $\{y_{n}\} \subset \mathbb{ R}^{N}$ such that $u_{n}(\cdot + y_{n}) \rightarrow \widetilde{u}$ in $H^{1}(\mathbb{R}^{N})$ as $n \rightarrow \infty$. Specifically, $\widetilde{u}$ is a global minimizer.

Proof. Since $\{u_{n}\} \subset H^{1}(\mathbb{R}^{N})$ satisfies Eq (3.5), it is easy to see that $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. From the concentration-compactness lemma ^[35], there exists a subsequence of $\{u_{n}\}$ (denoted in the same way) satisfying one of the three following possibilities:

vanishing: for all $R > 0$

dichotomy: there exists a constant $b\in (0, a)$, sequences $\{u^{(1)}_{n}\}$, $\{u^{(2)}_{n}\}$ bounded in $H^{1}(\mathbb{R}^{N})$ such that as $n \rightarrow \infty$

compactness: there exists a sequence $\{y_{n}\} \subset \mathbb{R}^{N}$ with the following property: for any $\varepsilon > 0$, there exists $R > 0$ such that

Claim 1. Vanishing does not occur.

Otherwise, by Lemma I.1 of ^[38], we get $u_{n} \rightarrow 0$ strongly in $L^{p}(\mathbb{R}^{N})$ for $2 < p < 2^{*}$. Since $2 < \frac{2(N+2+\alpha)}{N+\alpha} < 2^{*}$, it follows from Lemma 2.2 that

Then, by Lemma 3.1 and Eq (3.5) we have

This contradiction proves Claim 1.

Claim 2. Dichotomy does not occur.

Otherwise, if dichotomy occurs, there exist $b\in (0, a)$ and sequences $\{u^{(1)}_{n}\}$, $\{u^{(2)}_{n}\}$ satisfying Eq (3.6). Furthermore, we may assume

Then, we have

and

By $(f_{1})$, $(f_{2})$, and Lemma 2.1, we obtain

By the definition of $v_{n}$ and Eq (3.6), we obtain that $|v_{n}|_{p}\rightarrow 0$ for $2\leq p < 2^{*}$. It follows from the above inequality and the boundedness of $\{u_{n}\}$ that

Using Eq (3.6) and Lemma 2.2, we get

Moreover, by the Young's inequality (^[36] Theorem 4.2) and $dist(supp\; u^{(1)}_{n}, supp \; u^{(2)}_{n}) \rightarrow +\infty$, we infer that

Thus, Eqs (3.6) and (3.8)–(3.12) imply that

which contradicts to Lemma 3.2, and proves Claim 2.

Hence, the compactness holds, namely, there exists a subsequence $\{y_{n}\} \subset \mathbb{R}^{N}$ such that $\widetilde{u}_{n} = u_{n}(x + y_{n}) \rightarrow \widetilde{u}$ in $L^{2}(\mathbb{R}^{N})$, $\widetilde{u}_{n} \rightharpoonup \widetilde{u}$ in $H^{1}(\mathbb{R}^{N})$ and $\widetilde{u} \in S(a)$. It follows from interpolation inequality and Sobolev inequality that

Then, Lemma 2.2 implies

Hence,

Thus, we have $I_{\mu}(\widetilde{u}) = m(a)$ and $\|\widetilde{u}_{n}\| \rightarrow \|\widetilde{u}\|$ as $n \rightarrow \infty$. Moreover, $u_{n}(x + y_{n}) \rightarrow \widetilde{u}$ in $H^{1}(\mathbb{R}^{N})$.

Proof. [Proof of Theorem 1.1] By Lemmas 3.1 and 3.3, there exists a global minimizer $\widetilde{u}$ for $I_{\mu}$ on $S(a)$ with $m(a) = I_{\mu}(\widetilde{u}) < 0$. Furthermore, $\widetilde{u}$ is a ground state solution of Eq (1.1) for some $\widetilde{\lambda} \in \mathbb{R}$. Then, by Lemma 2.4, we have

Then,

Since $\int\limits_{\mathbb{R}^{N}}(I_{\alpha}*F(\widetilde{u}))F(\widetilde{u})dx > 0$, by Eq (3.13) we have $\widetilde{\lambda} < 0$.

By $(f_{3})$, without loss of generality, we may assume that $f\geq 0$ on $(0, +\infty)$. Since $f$ is odd, then $F$ is even and thus for every $u\in H^{1}(\mathbb{R}^{N})$, $I_{\mu}(|u|) = I_{\mu}(u)$. From this one easily obtain that the function $|\widetilde{u}|$ is also a ground state solution of Eq (1.1). By regularity properties of ^[9], $|\widetilde{u}|$ is continuous, we can apply the strong maximum principle get $|\widetilde{u}| > 0$ on $\mathbb{R}^{N}$ and thus $\widetilde{u}$ has constant sign.

Finally, we prove the symmetry of $\widetilde{u}$. Assume that $H\subset \mathbb{R}^{N}$ is a closed half-space and that $\sigma_{H}$ denotes the reflection with respect to $\partial H$. The polarization $\widetilde{u}^{H}(x): \mathbb{R}^{N}\rightarrow \mathbb{R}$ of $\widetilde{u}$ is defined for $x\in \mathbb{R}^{N}$ by

By the properties of polarization (^[9] Lemma 5.4), we observe that

and then

Moreover, since $F$ is nondecreasing on $(0, +\infty)$, we have $F(\widetilde{u}^{H}) = F(\widetilde{u})^{H}$. Therefore, by Lemma 5.5 of ^[9], we get

with equality if and only if

On the other hand, since $\widetilde{u} \in S(a)$ and $|\widetilde{u}|^{2}_{2} = |\widetilde{u}^{H}|^{2}_{2}$, we have $I_{\mu}(\widetilde{u}^{H}) \geq m(a)$. It follows from Eq (3.14) that $I_{\mu}(\widetilde{u}^{H}) = I_{\mu}(\widetilde{u})$, and thus Eq (3.15) holds. If $F(\widetilde{u})^{H} = F(\widetilde{u})$, for every $x\in H$,

Since $F$ is nondecreasing on $(0, +\infty)$, we get $\widetilde{u}(\sigma_{H}(x)) \leq \widetilde{u}(x)$. In particular, $f(\widetilde{u}^{H}) = f(\widetilde{u})$ on $\mathbb{R}^{N}$, hence $\widetilde{u}^{H} = \widetilde{u}$. If $F(\widetilde{u}(\sigma_{H})) = F(\widetilde{u})^{H}$, we similarly get $\widetilde{u}^{H} = \widetilde{u}(\sigma_{H})$. Since the hyperplane $H$ is arbitrary, in either case we conclude that the function $\widetilde{u}$ is radially symmetric with respect to some point $x_{0}\in \mathbb{R}^{N}$, and is radially decreasing.

4.   $L^{2}$-critical perturbation

In this section, we consider the case $q = \overline{q}$ and prove Theorems 1.2 and 1.3. Set $\overline{a}_{N}: = \frac{\overline{q}}{2C^{\overline{q}}_{N, \overline{q}}} > 0$.

Lemma 4.1. Assume that $(f_{1})$ and $(f_{2})$ hold. For any $\mu > 0$, there exists $a_{0}(\mu) = (\overline{a}_{N}\mu^{-1})^{\frac{N}{2}}$, such that for any $a\in (0, a_{0})$, we have

Proof. For every $u\in S(a)$, by Eqs (2.2) and (3.1), we get

Since $a < a_{0}$, we have $\frac{1}{2} - \frac{\mu C^{\overline{q}}_{N, \overline{q}}}{\overline{q}}a^{\frac{2}{N}} > 0$. We choose $\varepsilon > 0$ small such that $C_{4}\varepsilon^{2}a^{\frac{\alpha+2}{N}} = \frac{1}{2}(\frac{1}{2} - \frac{\mu C^{\overline{q}}_{N, \overline{q}}}{\overline{q}}a^{\frac{2}{N}})$, it follows that

This implies $m(a) > -\infty$.

In addition, for $u\in S(a)$ we have

Noticing that $a < a_{0}$, by Eq (2.2), we have $\frac{1}{2} |\nabla u|^{2}_{2} - \frac{\mu}{\overline{q}} |u|^{\overline{q}}_{\overline{q}} > 0$, and

Moreover, $|u^{t}|^{\frac{2(N+2+\alpha)}{N+\alpha}}_{\frac{2(N+2+\alpha)}{N+\alpha}} = t^{\frac{2N}{N+\alpha}}|u|^{\frac{2(N+2+\alpha)}{N+\alpha}}_{\frac{2(N+2+\alpha)}{N+\alpha}} \rightarrow 0$ as $t\rightarrow 0$. By Lemma 2.2, we have

Then, by Eq (4.1) we infer that

Thus, $m(a) \leq 0$. The proof is completed.

Next, we give some properties of $m(a)$.

Lemma 4.2. It holds that

(i) Let $a\in (0, a_{0})$. Then, we have for all $b\in (0, a)$

and if $m(b)$ or $m(a-b)$ is reached, then the inequality is strict.

(ii) Taking $\mu > 0$ small enough, there exists $b_{0} > 0$ such that $0 < b_{0} < a_{0}(\mu)$. Then, for any $a \in (b_{0}, a_{0})$, we have $m(a) < 0$.

(iii) $a\in (0, a_{0}) \mapsto m(a)$ is continuous.

Proof. (i) Fix $b\in (0, a)$, we first show that

and that if $m(b)$ is reached, the inequality is strict. By the definition of $m(b)$, for any $\varepsilon > 0$ sufficiently small, there exists a $u\in S(b)$ such that

Now set $\overline{u}(x) = u(\theta^{-\frac{1}{N}}x)$. Note that $\overline{u}(x) \in S(\theta b)$. It follows from Eq (4.3) that

Since $\varepsilon > 0$ is arbitrary, we have that $m(\theta b) \leq \theta m(b)$. If $m(b)$ is reached, we can let $\varepsilon = 0$ in Eq (4.3), and thus Eq (4.4) implies $m(\theta b) < \theta m(b)$.

Then, by Eq (4.2) we have

with a strict inequality if $m(b)$ is reached.

(ii) By $(f_{3})$, there exists $v\in H^{1}(\mathbb{R}^{N})$ such that

For any $b > 0$, set $v_{b}(x) = v(b^{-\frac{1}{N}}|v|^{\frac{2}{N}}_{2}x)$. Obviously $v_{b}\in S(b)$. Then, we have

Since $g(b)\rightarrow -\infty$ as $b\rightarrow +\infty$ and by choosing $b_{0} > 0$ large such that $I_{\mu}(v_{b_{0}}) < 0$, it follows that

Now, taking $\mu > 0$ small enough such that $b_{0} < a_{0}(\mu)$. For any $a\in (b_{0}, a_{0})$, by Lemma 4.1 and $(i)$, we obtain

(iii) Let $a \in (0, a_{0})$ be arbitrary and $\{ a_{n}\} \subset (0, a_{0})$ be such that $a_{n} \rightarrow a$ as $n \rightarrow \infty$. By the definition of $m(a_{n})$, for every $n$ there exists $u_{n} \in S(a_{n})$ such that

Since $m(a_{n})\leq 0$, by the proof of Lemma 4.1, the sequence $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. Set $v_{n} = \sqrt{\frac{a}{a_{n}}}u_{n}$. It is clear that $v_{n} \in S(a)$. Then, we can write

where

By the boundedness of $\{u_{n}\}$, we have $\{ |u_{n}|^{\overline{q}}_{\overline{q}}\}$ is bounded. Thus, we have

Moreover,

Since $a_{n} \rightarrow a$ as $n \rightarrow \infty$, then $\{\sqrt{\frac{a}{a_{n}}}\}$ is bounded. It follows from $(f_{1})$ and $(f_{2})$ that

Thus, using Lemma 2.1 and the Sobolev imbedding inequality, we get

Thus,

Similarly,

Then, in view of Eq (4.9), we have

By Eqs (4.7), (4.8), and (4.10), we have

It follows from Eqs (4.5) and (4.6) that

On the other hand, for any $\varepsilon > 0$, there exists $u\in S(a)$ such that

Set $\widetilde{u}_{n}(x) = \sqrt{\frac{a_{n}}{a}}u(x)$. Then, $\widetilde{u}_{n} \in S(a_{n})$. Similar to Eq (4.11), we have

It follows from Eq (4.12) that

Since $\varepsilon > 0$ is arbitrary, we get

Thus, we infer that $m(a_{n}) \rightarrow m(a)$ as $n \rightarrow \infty$.

Now, for any fixed $\mu > 0$ small enough, we define

By Lemma 4.1, $a^{*} \in [0, a_{0}(\mu))$ is well defined and satisfying

Lemma 4.3. Assume that ($f_{1}$)–($f_{3}$) hold. For any $a\in (0, a_{0})$, if ($f_{4}$) holds, then $a^{*} = 0$; if ($f_{5}$) holds, then $a^{*} > 0$.

Proof. (i) Let $a\in (0, a_{0})$, we choose $u\in S(a)\cap C^{\infty}_{0}(\mathbb{R}^{N})$ and set

By the assumption $(f_{4})$, there exists $\delta > 0$ such that

Since $|u^{t}|_{\infty} \leq \delta$ for sufficiently small $t > 0$, we have

Thus, $m(a) \leq I_{\mu}(u^{t}) < 0$ for any $a\in (0, a_{0})$. By the definition of $a^{*}$, we get $a^{*} = 0$.

(ii) By $(f_{2})$ and $(f_{5})$, there exists $C' > 0$ such that

For $u \in S(a)$, by Eqs (2.1) and (4.14), we get

It follows from Eq (2.2) that

Taking $a > 0$ small enough such that $\frac{C''}{2}a^{\frac{2+\alpha}{N}} + \frac{\mu}{\overline{q}}C^{\overline{q}}_{N, \overline{q}}a^{\frac{2}{N}} \leq \frac{1}{2}$, we get $I_{\mu}(u) \geq 0$. This implies $m(a) \geq 0$ for a small $a > 0$. From Lemma 4.1 that $m(a) = 0$ for a small $a > 0$. Hence, we have $a^{*} > 0$.

Lemma 4.4. Assume that $(f_{1})$–$(f_{3})$ hold. Let $\mu > 0$ and $a\in (0, a_{0})$. Let $\{ u_{n}\} \subset S(a)$ be a minimizing sequence for $m(a)$. Then, one of the following holds:

(i).

(ii). Taking a subsequence if necessary, there exist $u \in S(a)$ and a family $\{y_{n}\} \subset \mathbb{R}^{N}$ such that $u_{n}(\cdot + y_{n}) \rightarrow u$ in $H^{1}(\mathbb{R}^{N})$ as $n \rightarrow \infty$. Specifically, $u$ is a global minimizer.

Proof. Suppose that $\{ u_{n}\} \subset S(a)$ is a minimizing sequence which does not satisfy (i). Then, we have

Taking a subsequence if necessary, we may assume there exists a family $\{y_{n}\} \subset \mathbb{R}^{N}$ such that

Let us consider $u_{n}(\cdot+y_{n})$. Since $\{ u_{n}\} \subset S(a)$ is a minimizing sequence of $m(a)$, by the proof of Lemma 4.1, $\{ u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$, Hence, $\{u_{n}(\cdot+y_{n})\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. Then, there exists $u\in H^{1}(\mathbb{R}^{N})$ such that $u_{n}(\cdot + y_{n})\rightharpoonup u$ in $H^{1}(\mathbb{R}^{N})$, $u_{n}(\cdot+y_{n})\rightarrow u$ in $L^{p}_{loc}(\mathbb{R}^{N})$ for $2\leq p < 2^{*}$, and $u_{n}(\cdot+y_{n})\rightarrow u$ almost everywhere in $\mathbb{R}^{N}$. It follows that

Then, by $u_{n}(\cdot+y_{n})\rightarrow u$ in $L^{2}_{loc}(\mathbb{R}^{N})$, we obtain that $u \neq 0$ and $|u|_{2}^{2} > 0$. Set $v_{n}(x) = u_{n}(x + y_{n}) -u$. Then, $v_{n} \rightharpoonup 0$ in $H^{1}(\mathbb{R}^{N})$, and we have

Moreover, by Lemma 2.3, we have

Since $I_{\mu}(u_{n}) = I_{\mu}(u_{n}(\cdot+y_{n})) = I_{\mu}(v_{n}+u)$, by Eqs (4.16)–(4.18), we obtain

Claim. $\lim_{n\rightarrow \infty} |v_{n}|^{2}_{2} = 0$.

Denote $b = |u|^{2}_{2} > 0$. By Eq (4.15), we see that $b \leq a$, and if we prove that $b = a$, then the claim holds. We suppose by contradiction that $b < a$. Since $v_{n} \in S(|v_{n}|^{2}_{2})$, then

It follows from $I_{\mu}(u_{n}) \rightarrow m(a)$ and Eq (4.19) that

By Lemma 4.2 (iii) and Eq (4.15), we obtain

Note that $b = |u|^{2}_{2}$, we have $u \in S(b)$ and $I_{\mu}(u) \geq m(b)$. If $I_{\mu}(u) > m(b)$, then Lemma 4.2 (i) and Eq (4.20) imply

which is impossible. Hence, $I_{\mu}(u) = m(b)$, that is $u$ is a minimizer of $m(b)$ on $S(b)$. It follows from Lemma 4.2 (i) and Eq (4.20) that

which is a contradiction. Thus, the claim holds and Eq (4.15) implies $|u|^{2}_{2} = a$.

Finally, we show that $\lim_{n \rightarrow \infty}|\nabla v_{n}|^{2}_{2} = 0$. Since $u\in S(a)$, we have $I_{\mu}(u) \geq m(a)$. It follows from Eq (4.19) that

this implies

On the other hand, by Lemma 2.2, Eq (2.2), and $|v_{n}|^{2}_{2} \rightarrow 0$, we infer that

In view of Eq (4.21), we get

Then, we have $\lim_{n \rightarrow \infty}|\nabla v_{n}|^{2}_{2} = 0$. It follows from $|v_{n}|^{2}_{2} \rightarrow 0$ that $v_{n} \rightarrow 0$ in $H^{1}(\mathbb{R}^{N})$. Hence, $u_{n}(\cdot + y_{n}) \rightarrow u$ in $H^{1}(\mathbb{R}^{N})$ as $n \rightarrow \infty$.

Proof. [Proof of Theorem 1.2] (i) For the case $0 < a < a^{*}$, we assume by contradiction that there exists a global minimizer with respect to $m(a)$. By Eq (4.13), we know $m(a) = 0$. By Lemma 4.2 (i) with the strict inequality, we infer that

which is a contradiction.

(ii) For the case $a^{*} < a < a_{0}$. By Eq (4.13), we have $m(a) < 0$. Let $\{u_{n}\} \subset S(a)$ satisfying $\lim_{n \rightarrow \infty}I_{\mu}(u_{n}) = m(a)$. It is easy to see that $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. Then, one of the two alternatives in Lemma 4.4 occurs. Let assume that (i) of Lemma 4.4 take place. The Lemma 1.1 of ^[38] implies $u_{n} \rightarrow 0$ in $L^{t}(\mathbb{R}^{N})$ for $2 < t < 2^{*}$. Moreover, since $2 < \frac{2(N + 2 + \alpha)}{N +2} < 2^{*}$, by Lemma 2.2 we have

From this, we infer that

which contradicts to $m(a) < 0$. Thus, Lemma 4.4 (ii) holds, namely, there exist $u\in S(a)$ and a family $\{y_{n}\} \subset \mathbb{R}^{N}$ such that $u_{n}(\cdot + y_{n}) \rightarrow u$ in $H^{1}(\mathbb{R}^{N})$ as $n \rightarrow \infty$, and $u$ is a global minimizer to $m(a)$. Moreover, $u$ is a ground state solution to Eq (1.1) for some $\lambda \in \mathbb{R}$. By Lemma 2.4, Eq (3.13), and $m(a) < 0$, we have $\lambda < 0$. Similar as the proof of Theorem 1.1, $u$ has constant sign and is radially symmetric about a point in $\mathbb{R}^{N}$.

Proof. [Proof of Theorem 1.4] By $(f_{6})$ and (iii) of Theorem 1.2, we have $a^{*} > 0$. Let $a_{n} = a^{*} + \frac{l}{n}$ for any $n\in \mathbb{N}$, where $0 < l < a_{0} - a^{*}$. By Eq (4.13) and Theorem 1.2 (ii), for every $n$, there exists a global minimizer $u_{n} \in S(a_{n})$ such that

Then, $\{u_{n}\}$ is bounded in $H^{1}(\mathbb{R}^{N})$. Since $a_{n} \rightarrow a^{*}$, we have $\lim_{n \rightarrow \infty}|u_{n}|^{2}_{2} = a^{*}$. Set $v_{n} = \frac{\sqrt{a^{*}}}{|u_{n}|_{2}}\cdot u_{n}$. Clearly, $v_{n} \in S(a^{*})$ and $\frac{\sqrt{a^{*}}}{|u_{n}|_{2}} \rightarrow 1$ as $n \rightarrow \infty$. By the similar proof of Lemma 4.2 (iii), we obtain

Thus, $\{v_{n}\}$ is a minimizing sequence with respect to $m(a^{*})$.

Next, we prove that $\{v_{n}\}$ is non-vanishing, that is

Otherwise, $\delta = 0$. By the definition of $v_{n}$ and $\frac{\sqrt{a^{*}}}{|u_{n}|_{2}} \rightarrow 1$, we get

By (ii) of Theorem 1.2, $u_{n}$ satisfies Eq (1.1) and we may assume that $u_{n}$ is radially symmetric with respect to the origin and decreasing for any $n \in \mathbb{N}$. Using the elliptic regularity theory, we see that $\{u_{n}\}$ is bounded in $\mathcal{C}^{1}(B(0, 1))$. Thus, by Eq (4.23) we get

It follows from $(f_{6})$ that for any $\varepsilon > 0$ and sufficiently large $n$,

Choosing $\varepsilon > 0$ small enough such that $C\varepsilon^{2}(a^{*} + l)^{\frac{2+\alpha}{N}} \leq \frac{1}{2} - \frac{\mu}{\overline{q}} C^{\overline{q}}_{N, \overline{q}}(a^{*} + l)^{\frac{2}{N}}$, then by Eq (2.2) we have

which contradicts Eq (4.22), and hence $\delta > 0$.

Then Lemma 4.4 (ii) holds, that is, up to a subsequence, there exist $v \in S(a^{*})$ and a family $\{y_{n}\} \subset \mathbb{R}^{N}$ such that $v_{n}(\cdot + y_{n}) \rightarrow v$ in $H^{1}(\mathbb{R}^{N})$ as $n \rightarrow \infty$. Then, $v$ is a global minimizer to $m(a^{*})$. Moreover, $v$ is a ground state solution to (1.1) for some $\lambda \in \mathbb{R}$. By Lemma 2.4 and Eq (3.13), we have

which implies $\lambda < 0$. By the same arguments of Theorem 1.1, we obtain the symmetric of $v$.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to express sincere thanks to the anonymous referees for their carefully reading this paper and valuable useful comments. This work was supported by Qinglan Project of Jiangsu Province (2022).

Conflict of interest

The authors declare there is no conflicts of interest.