<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">OALibJ</journal-id><journal-title-group><journal-title>Open Access Library Journal</journal-title></journal-title-group><issn pub-type="epub">2333-9705</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/oalib.1111433</article-id><article-id pub-id-type="publisher-id">OALibJ-132396</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Biomedical&amp;Life Sciences</subject><subject> Business&amp;Economics</subject><subject> Chemistry&amp;Materials Science</subject><subject> Computer Science&amp;Communications</subject><subject> Earth&amp;Environmental Sciences</subject><subject> Engineering</subject><subject> Medicine&amp;Healthcare</subject><subject> Physics&amp;Mathematics</subject><subject> Social Sciences&amp;Humanities</subject></subj-group></article-categories><title-group><article-title>
 
 
  The Existence Result for a Fractional Kirchhoff Equation Involving Doubly Critical Exponents and Combined Nonlinearities
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Tianqing</surname><given-names>Zhang</given-names></name><xref ref-type="aff" rid="aff1"><sub>1</sub></xref></contrib></contrib-group><aff id="aff1"><label>1</label><addr-line>School of Mathematics, Liaoning Normal University, Dalian, China</addr-line></aff><pub-date pub-type="epub"><day>01</day><month>04</month><year>2024</year></pub-date><volume>11</volume><issue>04</issue><fpage>1</fpage><lpage>16</lpage><history><date date-type="received"><day>13,</day>	<month>March</month>	<year>2024</year></date><date date-type="rev-recd"><day>8,</day>	<month>April</month>	<year>2024</year>	</date><date date-type="accepted"><day>11,</day>	<month>April</month>	<year>2024</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  This paper zeroes in on the existence result of solutions to a fractional Kirchhoff equation with doubly critical exponents, mixed nonlinear terms and a continuous potential &lt;em&gt;V&lt;/em&gt;. After utilizing some energy estimates, one obtains the effect of exponents &lt;em&gt;p&lt;/em&gt; and &lt;em&gt;q&lt;/em&gt; on the existence of constrained minimizers, namely, the connection between the existence of normalized solutions and exponents &lt;em&gt;p&lt;/em&gt;, &lt;em&gt;q&lt;/em&gt;.
 
</p></abstract><kwd-group><kwd>Fractional Kirchhoff Equations</kwd><kwd> Constrained Minimizers</kwd><kwd> Doubly Critical Exponents</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>This paper is focused on the fractional Kirchhoff equation with combined nonlinearities as follows:</p><p>( a + b ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) ( − Δ ) s u + V ( x ) u = λ u + c | u | p − 2 u + d | u | q − 2 u   in   ℝ N (1.1)</p><p>where a is a positive constant, which will be defined specifically in the sequel, b &gt; 0 , 1 ≤ N ≤ 3 , 0 &lt; s &lt; 1 , 2 &lt; q &lt; p = 4 = 2 s ∗ and c &gt; 0 , d ≠ 0 , λ is a Lagrange constant. The fractional Laplacion ( − Δ ) s ( s ∈ ( 0,1 ) ) can be defined as</p><p>( − Δ ) s v ( x ) = C s P .V . ∫ ℝ N v ( x ) − v ( y ) | x − y | N + 2 s d y = C s l i m ε → 0 ∫ ℝ N \ B ε ( x ) v ( x ) − v ( y ) | x − y | N + 2 s d y</p><p>for v ∈ S ( ℝ N ) , where S ( ℝ N ) is the Schwartz space of rapidly decaying C ∞ function, B ε ( x ) denotes an open ball of radius ε centered at x ∈ ℝ N and the constant C s = ( ∫ ℝ N 1 − cos ( ξ 1 ) | ξ | N + 2 s ) − 1 .</p><p>For the case of a &gt; 0 , b &gt; 0 , s = 1 , problem (1.1) is a classical Kirchhoff equation. And this type of equation has been associated with the following equation</p><p>u t t − ( a + b ∫ ℝ N | ∇ u | 2 d x ) Δ u = f ( x , u )   in   Ω &#215; ( 0, ∞ ) . (1.2)</p><p>Problem 1.2 was proposed by Kirchhoff [<xref ref-type="bibr" rid="scirp.132396-ref1">1</xref>] in 1883 at the outset, where he obtained the classical D’Alembert wave equation, where the nonlinearity f ( x , u ) is of general type. Besides, the physical and biological background of (1.2) can be found in [<xref ref-type="bibr" rid="scirp.132396-ref2">2</xref>] [<xref ref-type="bibr" rid="scirp.132396-ref3">3</xref>] and the references therein. And it has brought itself into notice after the seminal contribution of [<xref ref-type="bibr" rid="scirp.132396-ref4">4</xref>] . Next let’s study this type of equation:</p><p>( a + b ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) ( − Δ ) s u − λ u = f ( x , u )     in   ℝ N . (1.3)</p><p>When s = 1 , equation (1.3) is a type of typical Kirchhoff equation. And in the recent years, it has been studied by many authors. For example, in [<xref ref-type="bibr" rid="scirp.132396-ref5">5</xref>] , He and Zou obtained the existence and concentration behavior of positive solutions for a Kirchhoff equation. In [<xref ref-type="bibr" rid="scirp.132396-ref6">6</xref>] , Figueiredo et al. studied the existence and concentration results for a Kirchhoff type equation with general nonlinearities. For more results about the existence of solutions to the Kirchhoff type equation like (1.3) with s = 1 , we refer readers to [<xref ref-type="bibr" rid="scirp.132396-ref7">7</xref>] [<xref ref-type="bibr" rid="scirp.132396-ref8">8</xref>] and the references therein.</p><p>Moreover, for the case of 0 &lt; s &lt; 1 , namely, for the nonlocal operator ( − Δ ) s , its background can be found in several areas such as fractional quantum mechanics [<xref ref-type="bibr" rid="scirp.132396-ref9">9</xref>] , physics [<xref ref-type="bibr" rid="scirp.132396-ref10">10</xref>] and so on. About the fractional Kirchhoff problems, to the best of our knowledge, a lot of authors have obtained fruitful results. For example, in [<xref ref-type="bibr" rid="scirp.132396-ref11">11</xref>] , Caffarelli and Silvestre introduced the harmonic extension method changing this nonlocal problem into a local one in higher dimensions. In [<xref ref-type="bibr" rid="scirp.132396-ref12">12</xref>] , Gu and Yang studied a singular perturbation fractional Kirchhoff equation in the critical case. Furthermore, readers can refer to [<xref ref-type="bibr" rid="scirp.132396-ref13">13</xref>] [<xref ref-type="bibr" rid="scirp.132396-ref14">14</xref>] and the references therein for more results on the existence of solutions for the fractional Kirchhoff equation (1.3).</p><p>Motivated by Li and Chen ( [<xref ref-type="bibr" rid="scirp.132396-ref15">15</xref>] and [<xref ref-type="bibr" rid="scirp.132396-ref16">16</xref>] ), the aim of this paper is to generalize their results to the case of mixed nonlinearities.</p><p>By direct computation, it is easy to find that if N = 4 s , the critical Sobolev exponent 2 s ∗ = 2 N N − 2 s and the fractional Gagliardo-Nirenberg-Sobolev critical exponent 2 G N S ∗ = 2 + 8 s N are equal, moreover, 2 N N − 2 s = 2 N + 8 s N = 4 . And in this paper, we study the case of N = 4 s , namely, the doubly critical exponents case. Besides, we have</p><p>( if   N = 1,   then   s = 1 4 , if   N = 2,   then   s = 1 2 , if   N = 3,   then   s = 3 4 . (1.4)</p><p>It is customary that a (weak) solution of problem (1.1) is a critical point of the energy functional</p><p>E V c ( u ) = a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x + b 4 ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) 2   − c 4 ∫ ℝ N | u | p d x − d q ∫ ℝ N | u | q d x ,</p><p>constrained on</p><p>S V : = { u ∈ H V s : ∫ ℝ N | u | 2 d x = 1 } ,</p><p>where</p><p>H V s : = { u ∈ H s ( ℝ N ) : ∫ ℝ N     V ( x ) u 2 d x &lt; ∞ } .</p><p>The fractional Sobolev space H s ( ℝ N ) is defined as</p><p>H s ( ℝ N ) : = { u ∈ L 2 ( ℝ N ) : u ( x ) − u ( y ) | x − y | N + 2 s 2 ∈ L 2 ( ℝ N &#215; ℝ N ) }</p><p>with the norm</p><p>‖ u ‖ H s ( ℝ N ) = ( ∫ ℝ N ( | ( − Δ ) s 2 u | 2 + | u | 2 ) d x ) 1 2 ,</p><p>where</p><p>∫ ℝ N | ( − Δ ) s 2 u | 2 d x = ∫ ℝ 2 N | u ( x ) − u ( y ) | 2 | x − y | N + 2 s d x d y .</p><p>By remark 1.5.1 of [<xref ref-type="bibr" rid="scirp.132396-ref17">17</xref>] , we know the fact that a (local) point of minimum of a differentiable functional is a critical point. Then we study the minimization problem with respect to the fractional Kirchhoff functional on the L 2 -constrained manifold:</p><p>m V ( c ) : = inf u ∈ S V E V c ( u ) . (1.5)</p><p>This paper |   ⋅   | p denotes the norm of L p ( ℝ N ) defined by | u | p p = ∫ ℝ N | u | p d x . If V = 0 , we denote the space H V s by H 0 s , the set S V by S 0 , the functional E V c by E 0 c , and m V ( c ) by m 0 ( c ) respectively.</p><p>By the above notations, we are ready to give the main result of this paper, namely, a result about the minimization problem (1.5).</p><p>Theorem 1.1. Let 2 &lt; q &lt; p = 4 , a ≥ 2 d S s 2 − q | ( − Δ ) s 2 u | 2 2 ( q − 2 ) − ∫ ℝ N     V ( x ) u 2 d x | ( − Δ ) s 2 u | 2 2</p><p>and a &gt; 0 (where S s is defined in Lemma 2.1), c &gt; 0 , d &gt; 0 , and V ( ⋅ ) satisfies the following condition:</p><p>(C) V ∈ C ( ℝ N , [ 0 , ∞ ) ) , lim | x | → ∞ V ( x ) = ∞ , inf x ∈ ℝ N V ( x ) = 0 , and there exists a sufficiently small ε 0 &gt; 0 such that meas ( { V ( x ) ≤ ε 0 } ) ≤ ε 0 .</p><p>Then, there exists a positive constant c ∗ such that if 2 &lt; q &lt; p = 4 , then c ∗ = b S s 2 and</p><p>( m V ( c ) ∈ ( − ∞ ,0 ) , if   c ∈ ( 0, c ∗ ) , m V ( c ) = 0, if   c = c ∗ , m V ( c ) = − ∞ , if   c ∈ ( c ∗ , ∞ ) .</p><p>Furthermore, E V c ( u ) has an energy minimizer for c &lt; c ∗ , and has no minimizers for c ≥ c ∗ .</p><p>Remark 1.1. By Theorem 1.1, we get a threshold value of c &gt; 0 which separates the existence and nonexistence of minimizers, which improves ( [<xref ref-type="bibr" rid="scirp.132396-ref15">15</xref>] , Theorem 1.2), where the existence of minimizers for (1.1) with s = 1 and d = 0 is obtained. The main obstruction is to impose energy estimates to characterize the threshold value c ∗ and the infimum energy level m V ( c ) for 2 &lt; q &lt; p = 4 .</p><p>Remark 1.2. Theorem 1.1 is also true in the case of d &lt; 0 , with only a small change in the case of d &gt; 0 , which we omit here.</p><p>Remark 1.3. For V ( x ) = | x | 2 , readers can verify that it satisfies the condition (C) in Theorem 1.1.</p><p>This paper is organized as follows: We first list some preliminaries in Section 2, the main proof of Theorem 1.1 will be given in Section 3 and finally, we summarize the main contents of this paper in Section 4.</p></sec><sec id="s2"><title>2. Preliminaries</title><p>In this section, some results which will be used frequently throughout the rest of the paper are firstly listed below.</p><p>Lemma 2.1. ( [<xref ref-type="bibr" rid="scirp.132396-ref18">18</xref>] ) Let s ∈ ( 0,1 ) and p ∈ [ 1, + ∞ ) be such that s p &lt; N . Then, there exists a positive constant S s = S s ( N , p , s ) such that, for any measurable and compactly supported function u : ℝ N → ℝ , one has</p><p>S s | u | 2 s ∗ 2 ≤ ∫ ℝ 2 N | u ( x ) − u ( y ) | 2 | x − y | N + 2 s d x d y , (2.1)</p><p>where 2 s ∗ = 2 N N − 2 s is the so-called fractional critical Sobolev exponent. Moreover, equality (2.2) holds if and only if u ˜ = K ( μ 2 + | x − x 0 | 2 ) − N − 2 s 2 with K ∈ ℝ \ { 0 } , μ &gt; 0 , x 0 ∈ ℝ N fixed constants, S s is the best Sobolev embedding constant.</p><p>Lemma 2.2 ( [<xref ref-type="bibr" rid="scirp.132396-ref19">19</xref>] ) Let p ∈ [ 0,2 s ∗ − 2 ) and u ∈ H s ( ℝ N ) , then inequality</p><p>∫ ℝ N | u | p + 2 d x ≤ p + 2 | Q | 2 p α p β p ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) N p 4 s ( ∫ ℝ N | u | 2 d x ) 2 p s − N p + 4 s 4 s (2.2)</p><p>holds, where α p = 2 s 2 p s − N p + 4 s , β p = ( 2 p s − N p + 4 s N p ) N p 4 s and the function</p><p>Q ( x ) optimizes (2.2) and is the unique nonnegative radically solution of the fractional nonlinear equation</p><p>( − Δ ) s Q + Q − | Q | p Q = 0   in     ℝ N .</p><p>According to Lemma 2.1 and Lemma 2.2, when N = 4 s and | u | 2 = 1 , the fractional Sobolev inequality (2.1) and the Gagliardo-Nirenberg-Sobolev inequality (2.2) can be rewritten, in other words, the fractional Sobolev inequality (2.1) turns into</p><p>S s 2 ∫ ℝ N | u | 4 d x ≤ ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) 2 ,   u ∈ H s ( ℝ N ) . (2.3)</p><p>If we change p into p − 2 in (2.2), then the fractional Gagliardo-Nirenberg-Sobolev inequality (2.2) becomes</p><p>∫ ℝ N | u | p d x ≤ p | Q | 2 p − 2 α p β p ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) p − 2 ,   u ∈ H s ( ℝ N ) , (2.4)</p><p>with the equality holds when u = λ N 2 Q ( λ x ) / | Q | 2 , where α p = 1 4 − p and β p = ( 4 − p 2 ( p − 2 ) ) p − 2 .</p><p>Particularly, when U = u &#175; ( x S s 1 2 s ) , where u &#175; = u ˜ | u ˜ | 2 s ∗ , one has</p><p>S s 2 = | ( − Δ ) s 2 U | 2 4 | U | 4 4 ,   S s 2 = | ( − Δ ) s 2 U | 2 2 = | U | 4 4 .</p><p>Similar to ( [<xref ref-type="bibr" rid="scirp.132396-ref20">20</xref>] , Lemma 5.1), one gets the result about the embedding as follows:</p><p>Lemma 2.3. Assume that V ( x ) satisfies condition (C), then the embedding H V s ↪ L p ( ℝ N ) is compact for p ∈ [ 2,4 ) .</p><p>The proof of this lemma has already been given in ( [<xref ref-type="bibr" rid="scirp.132396-ref16">16</xref>] , Lemma 2.3), but for the readers’ convenience, we sketch it here again.</p><p>Proof. Step 1. We first show that H V s ↪ L p ( ℝ N ) holds for p = 2 .</p><p>By the Sobolev embedding theorem, one gets that L s ( ℝ N ) ↪ L 2 ( ℝ N ) continuously. Further, from H V s ↪ L s ( ℝ N ) continuously, we deduce that H V s ↪ L 2 ( ℝ N ) continuously.</p><p>Suppose that { u n } ⊂ H V s is a sequence such that u n ⇀ 0 in H V s . Then one gets u n ⇀ 0 in H s ( ℝ N ) and u n → 0 in L 2 ( B R ) , where B R is a ball in ℝ N with radius R centered at 0 ∈ ℝ N .</p><p>From condition (C), one gets lim | x | → ∞ V ( x ) = ∞ , it follows that for any ε &gt; 0 , there exists R &gt; 0 such that</p><p>| 1 V ( x ) | ≤ ε   for     | x | &gt; R   .</p><p>Thus,</p><p>∫ ℝ N | u n | 2 d x = ∫ B R | u n | 2 d x + ∫ B R c | u n | 2 d x ≤ ε + ε ∫ B R c | V ( x ) | | u n | 2 d x ≤ ε + C ε ( sup ∫ ℝ N | V ( x ) | | u n | 2 d x ) .</p><p>From this, we conclude that</p><p>u n → 0     in   L 2 ( ℝ N ) .</p><p>Thus, H V s ↪ L 2 ( ℝ N ) is compact.</p><p>Step 2. We prove the case of p &gt; 2 .</p><p>Since H V s ↪ L 2 ( ℝ N ) is compact by Step 1, one obtains that u n → 0 in L 2 ( ℝ N ) . By the following fractional Gagliardo-Nirenberg-Sobolev inequality ( p ∈ [ 2,4 ) ),</p><p>∫ ℝ N | u | p d x ≤ p | Q | 2 p − 2 α p β p ( ∫ ℝ N | ( − Δ ) s 2 u n | 2 d x ) N ( p − 2 ) 4 s ( ∫ ℝ N | u n | 2 d x ) 2 p s − N ( p − 2 ) + 4 s 4 s ,</p><p>we can deduce that u n → 0 in L p ( ℝ N ) for p ∈ [ 2,4 ) . <inline-formula><inline-graphic xlink:href="/html.scirp.org/file/132396x155.png" xlink:type="simple"/></inline-formula></p><p>The following lemma is adapted from ( [<xref ref-type="bibr" rid="scirp.132396-ref16">16</xref>] , Lemma 2.5), for readers’ convenience, we provide a brief proof.</p><p>Lemma 2.4. Assume that 2 &lt; q &lt; p = 4 ,   c &gt; 0 ,   d &gt; 0 and the energy functional E 0 c ( u ) is defined as</p><p>E 0 c ( u ) = a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + b 4 ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) 2 − c 4 ∫ ℝ N | u | 4 d x − d q ∫ ℝ N | u | q d x .</p><p>Set c ∗ = b S s 2 , m 0 ( c ) = inf u ∈ S 0 E 0 c ( u ) . Then m 0 ( c ) = − ∞ for c &gt; c ∗ , where</p><p>S 0 = { u ∈ H s ( ℝ N ) : ∫ ℝ N | u | 2 d x = 1 } .</p><p>Proof. Choosing x 0 = 0 in u ˜ . Then</p><p>U = u ˜ | u ˜ | 2 s ∗ = K ( μ 2 + | x S s 1 2 s | 2 ) − N − 2 s 2 | u ˜ | 2 s ∗ = θ ( μ 2 + | x S s 1 2 s | 2 ) − N − 2 s 2 ,</p><p>where θ = K | u ˜ | 2 s ∗ . By the assumption of N = 4 s , it yields that for R &gt; 1 ,</p><p>∫ R &lt; | x | &lt; 2 R     U 2 d x = ∫ R &lt; | x | &lt; 2 R [ θ ( μ 2 + | x S s 1 2 s | 2 ) − N − 2 s 2 ] 2 d x ≤ S s 2 θ 2 ω N ∫ R &lt; r &lt; 2 R     r N − 1 r − 4 s d r &lt; 2 S s 2 θ 2 ω N : = A 1 , (2.5)</p><p>where ω N is the surface area of unit sphere in ℝ N , and</p><p>∫ | x | &gt; R | ( − Δ ) s 2 U | 2 d x ≤ ∫ ℝ N | ( − Δ ) s 2 U | 2 d x = S s 2 . (2.6)</p><p>∫ | x | &gt; R     U 4 d x = ∫ | x | &gt; R [ θ ( μ 2 + | x S s 1 2 s | 2 ) − N − 2 s 2 ] 4 d x ≤ θ 4 ω N ∫ r &gt; R     r N − 1 ( | r S s 1 2 s | 2 ) − 4 s d r = S s 4 θ 4 ω N N R N : = A 2 R N . (2.7)</p><p>For 2 &lt; q &lt; 4 , there is</p><p>∫ | x | &gt; R     U q d x = ∫ | x | &gt; R [ θ ( μ 2 + | x S s 1 2 s | 2 ) − N − 2 s 2 ] q d x ≤ θ q ω N ∫ r &gt; R     r N − 1 ( | r S s 1 2 s | 2 ) − q s d r = S s q θ q ω N N R N : = A 3 R N . (2.8)</p><p>Since 0 &lt; s &lt; 1 and N = 4 s , we consider the computation of ∫ | x | &lt; R     U 2 d x in three cases as below:</p><p>1) If N = 1 , then s = 1 4 and</p><p>∫ | x | &lt; R     U 2 d x = θ 2 ω 1 S s 2 ( ln | R + R 2 + ( μ S s 2 ) 2 | − ln | μ S s 2 | ) .</p><p>2) If N = 2 , then s = 1 2 and</p><p>∫ | x | &lt; R     U 2 d x = θ 2 ω 2 S s 2 2 ( ln | R 2 + ( μ S s ) 2 | − ln | ( μ S s ) 2 | ) .</p><p>3) If N = 3 , then s = 3 4 and</p><p>∫ | x | &lt; R     U 2 d x = θ 2 ω 3 ∫ r &lt; R     r 2 ( μ 2 + | r S s 2 3 | 2 ) − 3 2 d r = S s 2 θ 2 ω 3 ( ln | R + R 2 + ( S s 2 3 μ ) 2 | − ln | S s 2 3 μ | − R R 2 + ( S s 2 3 μ ) 2 ) .</p><p>So, there exists a ρ &gt; 0 such that</p><p>∫ | x | &lt; R     U 2 d x ≥ ρ ln ( R 2 + μ S s 2 ) . (2.9)</p><p>Then we consider a radially symmetric cut-off function ϕ ∈ C 0 ∞ ( ℝ N ) , which satisfies ϕ = 1 in B R = { x ∈ ℝ N : | x | ≤ R } , ϕ = 0 in B 2 R c = { x ∈ ℝ N : | x | &gt; 2 R } and 0 ≤ ϕ ≤ 1 , | ∇ ϕ | ≤ 2 R .</p><p>Set U &#175; = ϕ U | ϕ U | 2 , U λ = λ N 2 U &#175; ( λ x ) . Then it’s easy to get that U &#175; , U λ ∈ S 0 and</p><p>E 0 c ( U λ ) = a λ 2 s 2 | ϕ U | 2 2 ∫ ℝ N | ( − Δ ) s 2 ( ϕ U ) | 2 d x + b λ 4 s 4 | ϕ U | 2 4 ( ∫ ℝ N | ( − Δ ) s 2 ( ϕ U ) | 2 d x ) 2     − c λ N 4 | ϕ U | 2 4 ∫ ℝ N | ϕ U | 4 d x − d λ 2 s ( q − 2 ) q | ϕ U | 2 q ∫ ℝ N | ϕ U | q d x .</p><p>According to the definition of function ϕ , it follows that</p><p>∫ ℝ N | ( − Δ ) s 2 ( ϕ U ) | 2 d x = ( ∫ | x | ≤ R + ∫ R &lt; | x | ≤ 2 R ) | ( − Δ ) s 2 ( ϕ U ) | 2 d x = ∫ | x | ≤ R | ( − Δ ) s 2 U | 2 d x + ∫ R &lt; | x | ≤ 2 R | ( − Δ ) s 2 ( ϕ U ) | 2 d x ,</p><p>where</p><p>∫ R &lt; | x | ≤ 2 R | ( − Δ ) s 2 ( ϕ U ) | 2 d x = ∫ ℝ N ∫ R &lt; | x | ≤ 2 R ϕ ( x ) 2 | U ( x ) − U ( y ) | 2 | x − y | N + 2 s d x d y       + ∫ ℝ N ∫ R &lt; | x | ≤ 2 R U ( y ) 2 | ϕ ( x ) − ϕ ( y ) | 2 | x − y | N + 2 s d x d y       + 2 ∫ ℝ N ∫ R &lt; | x | ≤ 2 R ϕ ( x ) U ( y ) ( U ( x ) − U ( y ) ) ( ϕ ( x ) − ϕ ( y ) ) | x − y | N + 2 s d x d y .</p><p>In addition, we get the following estimate</p><p>∫ ℝ N ∫ R &lt; | x | ≤ 2 R ϕ ( x ) 2 | U ( x ) − U ( y ) | 2 | x − y | N + 2 s d x d y ≤ ∫ R &lt; | x | ≤ 2 R | ( − Δ ) s 2 U | 2 d x ,</p><p>∫ ℝ N ∫ R &lt; | x | ≤ 2 R U ( y ) 2 | ϕ ( x ) − ϕ ( y ) | 2 | x − y | N + 2 s d x d y ≤ C R 2 ∫ ℝ N ∫ { x ∈ ℝ N : R &lt; | x | ≤ 2 R , | x − y | ≤ R } U ( y ) 2 | x − y | N + 2 s − 2 d x d y         + 4 ∫ ℝ N ∫ { x ∈ ℝ N : R &lt; | x | ≤ 2 R , | x − y | ≥ R } U ( y ) 2 | x − y | N + 2 s d x d y ≤ A 4 R 2 s .</p><p>And it follows from H&#246;lder inequality that</p><p>2 ∫ ℝ N ∫ R &lt; | x | ≤ 2 R ϕ ( x ) U ( y ) ( U ( x ) − U ( y ) ) ( ϕ ( x ) − ϕ ( y ) ) | x − y | N + 2 s d x d y ≤ C ( ∫ ℝ N ∫ R &lt; | x | ≤ 2 R U ( y ) 2 | ϕ ( x ) − ϕ ( y ) | 2 | x − y | N + 2 s d x d y ) 1 2         &#215; ( ∫ ℝ N ∫ R &lt; | x | ≤ 2 R ϕ ( x ) 2 | U ( x ) − U ( y ) | 2 | x − y | N + 2 s d x d y ) 1 2 ≤ A 5 R s .</p><p>Choosing L = max { A 4 , A 5 } , we obtain that</p><p>∫ ℝ N | ( − Δ ) s 2 ( ϕ U ) | 2 d x ≤ ∫ | x | &lt; 2 R | ( − Δ ) s 2 U | 2 d x + 2 L R s .</p><p>In addition,</p><p>∫ ℝ N | U ϕ | q d x ≥ ∫ | x | ≤ R | U | q d x ≥ ∫ ℝ N | U | q d x − A 3 R N ,</p><p>where ∫ ℝ N | U | q d x is bounded from above by (2.8) and H&#246;lder inequality. Then one gets</p><p>E 0 c ( U λ ) ≤ a λ 2 s 2 | ϕ U | 2 2 ( S s 2 + 2 L R s ) + λ 4 s 4 | ϕ U | 2 4 ( ( b S s 2 − c ) S s 2 + 4 b L S s 2 R s + 4 b L 2 R 2 s + c A 2 R N )   − d λ 2 s ( q − 2 ) q | ϕ U | 2 q ∫ ℝ N | U | q d x + d λ 2 s ( q − 2 ) q | ϕ U | 2 q A 3 R N . (2.10)</p><p>When c &gt; c ∗ = b S s 2 and R is sufficiently large, we get</p><p>4 b L S s 2 R s + 4 b L 2 R 2 s + c A 2 R N &lt; 1 2 ( c − b S s 2 ) S s 2 .</p><p>Thus, it follows from (2.10) that m 0 ( c ) ≤ E 0 c ( u ) → − ∞ as λ → ∞ , i.e., m 0 ( c ) = − ∞ for c &gt; c ∗ . <inline-formula><inline-graphic xlink:href="/html.scirp.org/file/132396x219.png" xlink:type="simple"/></inline-formula></p><p>In the following lemma, we give the estimates of c ∗ for p = 4 .</p><p>Lemma 2.5. Suppose 2 &lt; q &lt; p = 4 , d is a positive constant and V ( x ) satisfies condition (C). Let c ∗ = b S s 2 , then</p><p>( m V ( c ) &gt; 0,   if   c ∈ ( 0, c ∗ ) , m V ( c ) = 0,   if   c = c ∗ , m V ( c ) = − ∞ ,   if   c ∈ ( c ∗ , ∞ ) .</p><p>Proof. 1) If c &lt; c ∗ = b S s 2 , we set γ = min { ε 0 , b S s 2 − c − ε 0 } and choose ε 0 &lt; b S s 2 − c satisfying condition (C) in Theorem 1.1. Firstly, by H&#246;lder inequality and u ∈ S V , we have</p><p>∫ ℝ N | u | q d x = ∫ ℝ N | u | 2 ( q − 2 ) | u | q − 2 ( q − 2 ) d x ≤ ( ∫ ℝ N ( | u | 2 ( q − 2 ) ) 2 q − 2 d x ) q − 2 2 ( ∫ ℝ N ( | u | q − 2 ( q − 2 ) ) 2 4 − q d x ) 4 − q 2 ≤ ( ∫ ℝ N | u | 4 d x ) q − 2 2 ( ∫ ℝ N | u | 2 d x ) 4 − q 2 = ( ∫ ℝ N | u | 4 d x ) q − 2 2 .</p><p>Then by Sobolev inequality, Young’s inequality, H&#246;lder inequality and condition (C), we get for any u ∈ S V ,</p><p>E V c ( u ) = a 2 | ( − Δ ) s 2 u | 2 2 + b 4 | ( − Δ ) s 2 u | 2 4 + 1 2 ∫ ℝ N     V ( x ) u 2 d x − c 4 | u | 4 4 − d q | u | q q ≥ b 4 S s 2 | u | 4 4 − c 4 | u | 4 4 + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d ( | u | 4 4 ) q − 2 2 ≥ γ 2 − 1 2 ∫ V ( x ) ≤ γ ( γ − V ( x ) ) u 2 d x + b 4 S s 2 | u | 4 4 − c 4 | u | 4 4 − d ( | u | 4 4 ) q − 2 2 = γ 2 − 1 2 ∫ V ( x ) ≤ γ [ 1 γ ( γ − V ( x ) ) 2 ] 1 2 [ γ u 4 ] 1 2 d x + 1 4 ( b S s 2 − c ) | u | 4 4 − d ( | u | 4 4 ) q − 2 2 ≥ γ 2 − 1 2 ∫ V ( x ) ≤ γ [ 1 γ ( γ − V ( x ) ) 2 2 + γ u 4 2 ] d x + 1 4 ( b S s 2 − c ) | u | 4 4 − d ( | u | 4 4 ) q − 2 2 = γ 2 − 1 4 γ ∫ V ( x ) ≤ γ ( γ − V ( x ) ) 2 d x + b S s 2 − c − γ 4 | u | 4 4 − d ( | u | 4 4 ) q − 2 2 .</p><p>Let f ( t ) = b S s 2 − c − γ 4 t − d t q − 2 2 , by direct computation, one gets</p><p>f ′ ( t ) = b S s 2 − c − γ 4 − ( q − 2 ) d 2 t q − 4 2 .</p><p>When f ′ ( t ) = 0 , there is</p><p>t = ( 2 ( q − 2 ) d b S s 2 − c − γ ) 2 4 − q</p><p>and</p><p>min t ≥ 0 f ( t ) = d ( q − 2 2 − 1 ) ( 2 ( q − 2 ) d b S s 2 − c − γ ) q − 2 4 − q .</p><p>Thus, we get</p><p>E V c ( u ) ≥ γ 2 − γ 4 ε 0 + d ( q − 2 2 − 1 ) ( 2 ( q − 2 ) d b S s 2 − c − γ ) q − 2 4 − q = γ 4 ( 2 − ε 0 ) + d ( q − 2 2 − 1 ) ( 2 ( q − 2 ) d b S s 2 − c − γ ) q − 2 4 − q ≥ γ 4 + d ( q − 2 2 − 1 ) ( 2 ( q − 2 ) d b S s 2 − c − γ ) q − 2 4 − q = γ [ 1 4 + d ( q − 2 2 − 1 ) [ 2 d ( q − 2 ) ] q − 2 4 − q 1 γ ( 1 b S s 2 − c − γ ) q − 2 4 − q ] .</p><p>If γ = ε 0 , namely ε 0 &lt; b S s 2 − c − ε 0 , we set d &gt; 0 satisties the condition that</p><p>d ( 1 − q − 2 2 ) [ 2 d ( q − 2 ) ] q − 2 4 − q 1 ε 0 ( 1 b S s 2 − c − ε 0 ) q − 2 4 − q &lt; 1 8 ;</p><p>and if γ = b S s 2 − c − ε 0 , i.e., b S s 2 − c − ε 0 &lt; ε 0 , we set d &gt; 0 satisties the condition that</p><p>d ( 1 − q − 2 2 ) [ 2 d ( q − 2 ) ] q − 2 4 − q 1 b S s 2 − c − ε 0 ( 1 ε 0 ) q − 2 4 − q &lt; 1 8 .</p><p>Then we get E V c ( u ) ≤ γ 8 . This indicates that m V ( c ) &gt; 0 for all c &lt; c ∗ = b S s 2 .</p><p>2) If c &gt; c ∗ = b S s 2 , then by (2.10) of Lemma 2.4, we obtain</p><p>E V c ( U λ ) = E 0 c ( U λ ) + 1 2 ∫ ℝ N     V ( x ) U λ 2 d x ≤ a λ 2 s 2 | ϕ U | 2 2 ( S s 2 + 2 L R s ) − λ 4 s 4 | ϕ U | 2 4 ( ( c − b S s 2 ) S s 2 − 4 b L S s 2 R s   − 4 b L 2 R 2 s − c A 2 R N − 2 | ϕ U | 2 2 ∫ ℝ N     V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x )   − d λ 2 s ( q − 2 ) q | ϕ U | 2 q ∫ ℝ N | U | q d x + d λ 2 s ( q − 2 ) q | ϕ U | 2 q A 3 R N .</p><p>Similar to the proof of Lemma 2.4, we choose R large enough such that</p><p>4 b L S s 2 R s + 4 b L 2 R 2 s + c A 2 R N &lt; 1 2 ( c − b S s 2 ) S s 2 .</p><p>For λ &gt; 1 , we obtain that</p><p>ϕ ( λ x ) U ( λ x ) ≤ ϕ ( x ) U ( x ) ,   ϕ ( λ x ) U ( λ x ) → 0     as   λ → ∞ .</p><p>It follows from the Lebesgue dominated convergence theorem that</p><p>lim λ → ∞ ∫ ℝ N     V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x = 0.</p><p>Therefore, according to the above inequalities and the definition of infimum, we deduce that</p><p>m V ( c ) ≤ lim λ → ∞ E V c ( U λ ) = − ∞ .</p><p>3) If c = c ∗ = b S s 2 , then using the Sobolev inequality and H&#246;lder inequality as in case (1), we have</p><p>E V c ( u ) = a 2 | ( − Δ ) s 2 u | 2 2 + b 4 | ( − Δ ) s 2 u | 2 4 − c 4 | u | 4 4 + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d q | u | q q ≥ a 2 | ( − Δ ) s 2 u | 2 2 + 1 2 ∫ ℝ N     V ( x ) u 2 d x + 1 4 ( b S s 2 − c ) | u | 4 4 − d ( | u | 4 4 ) q − 2 2 = a 2 | ( − Δ ) s 2 u | 2 2 − d S s 2 − q | ( − Δ ) s 2 u | 2 2 ( q − 2 ) + 1 2 ∫ ℝ N     V ( x ) u 2 d x .</p><p>Choosing appropriate a &gt; 0 (Actually, we set a ≥ 2 d S s 2 − q | ( − Δ ) s 2 u | 2 2 ( q − 2 ) − ∫ ℝ N     V ( x ) u 2 d x | ( − Δ ) s 2 u | 2 2 and a &gt; 0 in Theorem), we conclude m V ( c ) ≥ 0 for c = c ∗ .</p><p>Further, we prove that m V ( c ) ≤ 0 for c = c ∗ . Analogous to the proof of Lemma 2.4, we get that there exists a R ( ε ) &gt; 0 satisfying for R ≥ R ( ε ) ,</p><p>| ϕ U | 2 2 ≥ ∫ | x | &lt; R | U | 2 d x ≥ ρ ln ( R 2 + μ S s 2 ) ≥ 1 ε ,   | ( − Δ ) s 2 U | 2 2 = S s 2 ≥ 2 L R s .</p><p>Repeating the previous proof, we get</p><p>lim λ → ∞ ∫ ℝ N V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x = 0.</p><p>So, there exists λ ( ε ) &gt; 0 such that</p><p>∫ ℝ N     V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x ≤ ε 2   for   λ ≥ λ ( ε ) .</p><p>According to the previous analysis, we obtain</p><p>m V ( c ∗ ) ≤ E V c ∗ ( U λ ) ≤ a λ 2 s 2 | ϕ U | 2 2 ( S s 2 + 2 L R s ) + d λ 2 s ( q − 2 ) q | ϕ U | 2 q A 3 R N − d λ 2 s ( q − 2 ) q | ϕ U | 2 q ∫ ℝ N | U | q d x</p><p>  + λ 4 s 4 | ϕ U | 2 4 ( 4 b L S s 2 R s + 4 b L 2 R 2 s + c A 2 R N + 2 | ϕ U | 2 2 ∫ ℝ N     V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x ) ≤ C 1 ε λ 2 s + λ 4 s 4 | ϕ U | 2 4 ( 4 b L S s 2 R s + 4 b L 2 R 2 s + c A 2 R N )   + λ 4 s 2 | ϕ U | 2 2 ∫ ℝ N     V ( x ) ϕ ( λ x ) 2 U ( λ x ) 2 d x + C 2 ε 2 λ 2 s ( q − 2 ) − C 3 ε λ 2 s ( q − 2 ) ≤ C ˜ ε λ 2 s + C ′ ε 3 λ 4 s + C ′ 2 ε 2 λ 2 s ( q − 2 ) − C ′ 3 ε λ 2 s ( q − 2 ) ,</p><p>where C 1 , C 2 , C 3 , C ˜ , C ′ , C ′ 2 , C ′ 3 are all positive constants.</p><p>Setting λ = max { λ ( ε ) , ε − 1 4 s } , we get m V ( c ∗ ) ≤ C ε 1 2 , namely, m V ( c ∗ ) ≤ 0 . Thus, we deduce m V ( c ∗ ) = 0 . <inline-formula><inline-graphic xlink:href="/html.scirp.org/file/132396x279.png" xlink:type="simple"/></inline-formula></p></sec><sec id="s3"><title>3. Proof of the Main Result</title><p>In this section, we prove the main result of this paper.</p><p>Proof of Theorem 1.1. By Lemma 2.5, we only need to prove that E V c ( u ) has no minimizers for c = c ∗ and it has an energy minimizer for all c &lt; c ∗ when 2 &lt; q &lt; p = 4 .</p><p>If p = 4 and c = c ∗ , using the fractional Sobolev inequality (2.3) and the H&#246;lder inequality as in the proof of Lemma 2.5 and the assumption about a in Theorem 1.1, we deduce</p><p>E V c ∗ ( u ) ≥ a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d q | u | q q   + b 4 ( 1 − c ∗ c ∗ ) ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) 2 = a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d q | u | q q ≥ a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d ( | u | 4 4 ) q − 2 2 ≥ a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d S s 2 − q | ( − Δ ) s 2 u | 2 2 ( q − 2 ) ≥ 0.</p><p>Arguing by contradiction that m V c can be obtained for c = c ∗ . Then, by Lemma 2.5, we deduce that</p><p>0 = m V ( c ∗ ) = E V c ∗ ( u ) ≥ a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x − d S s 2 − q | ( − Δ ) s 2 u | 2 2 ( q − 2 ) .</p><p>From this, we get u = 0 in D s ,2 ( ℝ N ) . So u = 0 in L 2 ( ℝ N ) , which is in contradiction with | u | 2 = 1 . Therefore, E V c ( u ) has no minimizers for c = c ∗ .</p><p>If p = 4 and c &lt; c ∗ , let { u n } ⊂ S V be a minimizing sequence for m V ( c ) , then by the Fractional Sobolev inequality (2.3), we deduce that</p><p>E V c ( u n ) = a 2 ∫ ℝ N | ( − Δ ) s 2 u n | 2 d x + 1 2 ∫ ℝ N     V ( x ) u n 2 d x   + b 4 ( ∫ ℝ N | ( − Δ ) s 2 u n | 2 d x ) 2 − c 4 | u n | 4 4 − d q | u n | q q ≥ a 2 ∫ ℝ N | ( − Δ ) s 2 u n | 2 d x + 1 2 ∫ ℝ N     V ( x ) u n 2 d x   + b S s 2 − c 4 ∫ ℝ N | u n | 4 d x − d q ∫ ℝ N | u n | q d x .</p><p>From this, we can deduce { u n } is bounded in H V s . According to Lemma 2.3, we may assume that there exists a u ∈ H V s such that u n ⇀ u in H V s and u n → u in L p ( ℝ N ) for p ∈ ( 2,4 ) , then u ∈ S V and there exists ξ n ∈ ℝ such that</p><p>lim n → ∞ ( E V c ' ( u n ) − ξ n u n ) = 0.</p><p>Set l i m n → ∞ | ( − Δ ) s 2 u n | 2 2 = B . Then</p><p>lim n → ∞ ξ n = lim n → ∞ ( E V c ' ( u n ) , u n ) = lim n → ∞ [ ( E V c ' ( u n ) , u n ) − 4 E V c ( u n ) + 4 m V ( c ) ] = lim n → ∞ ( 4 m V ( c ) − a | ( − Δ ) s 2 u n | 2 2 − ∫ ℝ N     V ( x ) u n 2 d x − d ( q − 4 ) q ∫ ℝ N | u n | q d x ) : = ξ .</p><p>Choosing ϕ ∈ H V , we get</p><p>0 = lim n → ∞ ( E V c ' ( u n ) − ξ n u n , ϕ ) = lim n → ∞ ( ( E V c ' ( u n ) , ϕ ) − ξ n ∫ ℝ N     u n ϕ d x ) = lim n → ∞ ( ( a + b | ( − Δ ) s 2 u n | 2 2 ) ∫ ℝ N ( − Δ ) s 2 u n ( − Δ ) s 2 ϕ d x + ∫ ℝ N     V ( x ) u n ϕ d x   − c ∫ ℝ N     u n 3 ϕ d x − d ∫ ℝ N     u n q − 1 ϕ d x − ξ n ∫ ℝ N     u n ϕ d x ) = ( a + b B ) ∫ ℝ N ( − Δ ) s 2 u ( − Δ ) s 2 ϕ d x + ∫ ℝ N     V ( x ) u ϕ d x − c ∫ ℝ N     u 3 ϕ d x − d ∫ ℝ N     u q − 1 ϕ d x − ξ ∫ ℝ N     u ϕ d x . (3.1)</p><p>If we choose ϕ = u in (3.1), then</p><p>ξ = ( a + b B ) ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + ∫ ℝ N     V ( x ) u 2 d x − c ∫ ℝ N     u 4 d x − d ∫ ℝ N     u q d x .</p><p>Since u ∈ S V and | ( − Δ ) s 2 u | 2 2 ≤ lim inf n → ∞ | ( − Δ ) s 2 u n | 2 2 = B , we obtain</p><p>m V ( c ) ≤ E V c ( u ) = a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x     + b 4 ( ∫ ℝ N | ( − Δ ) s 2 u | 2 d x ) 2 − c 4 | u | 4 4 − d 4 ∫ ℝ N     u q d x − d ( 4 − q ) 4 q ∫ ℝ N     u q d x ≤ a 2 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 2 ∫ ℝ N     V ( x ) u 2 d x + b B 4 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x     − c 4 | u | 4 4 − d 4 ∫ ℝ N     u q d x − d ( 4 − q ) 4 q ∫ ℝ N     u q d x</p><p>≤ a 4 ∫ ℝ N | ( − Δ ) s 2 u | 2 d x + 1 4 ∫ ℝ N     V ( x ) u 2 d x + 1 4 ξ − d ( 4 − q ) 4 q | u | q q ≤ lim inf n → ∞ ( a 4 ∫ ℝ N | ( − Δ ) s 2 u n | 2 d x + 1 4 ∫ ℝ N     V ( x ) u n 2 d x + 1 4 ξ n + d ( q − 4 ) 4 q | u n | q q ) = m V ( c ) .</p><p>Therefore, E V c ( u ) = m V ( c ) and E V c ( u ) has an energy minimizer u.</p></sec><sec id="s4"><title>4. Conclusion</title><p>In summary, in the previous sections, combining the fractional Gagliardo-Nirenberg-Sobolev inequality, and fractional Sobolev inequality with some energy estimates, we obtained the existing result of the fractional Kirchhoff equation with doubly critical exponents and mixed nonlinearities. <inline-formula><inline-graphic xlink:href="/html.scirp.org/file/132396x325.png" xlink:type="simple"/></inline-formula></p></sec><sec id="s5"><title>Conflicts of Interest</title><p>The author declares no conflicts of interest.</p></sec><sec id="s6"><title>Cite this paper</title><p>Zhang, T.Q. (2024) The Existence Result for a Fractional Kirchhoff Equation Involving Doubly Critical Exponents and Combined Nonlinearities. Open Access Library Journal, 11: e11433. http://doi.org/10.4236/oalib.1111433</p></sec></body><back><ref-list><title>References</title><ref id="scirp.132396-ref1"><label>1</label><mixed-citation publication-type="other" xlink:type="simple">Kirchhoff, G. 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