Standing Waves For A Coupled Nonlinear Hartree Equations With Nonlocal .

Standing waves for a coupled nonlinear Hartree equations Page 5 of 36168The proof of Theorem 1.1 is based on the celebrated moving plane method for cooperativeintegral-differential equations (see [10,37]). For the local interaction case (1.10), this propertyis well-known (also for the cooperative case β 0), see [7]. On the other hand, it is alsoknown that when β 0, a positive solution of (1.10) may not be radially symmetric, see[5,36]. For the existence and nonexistence of positive (ground state) solutions of (1.1), wehave the following main results.Theorem 1.2 Assume that μ, ν 0 and λ2 λ1 0 are fixed.(i) Let χ0 be the smaller root of the equation55λ 4 (2 λ 4 )y 2 (μ1 ν1 )y μ1 ν1 0,where33μ1 λ 4 μ, ν1 λ 4 ν, and λ λ2.λ1(1.14)(1.15)If β χ0 , then (1.1) possesses a positive radial ground state solution z Er .Moreover, if 0 β χ0 , then z is also a positive ground state solution.15(ii) If β max{λ2 μ, λ 2 ν} λ 4 max{μ1 , ν1 }, then (1.1) possesses a positive radial groundstate solution which is also a positive ground state solution.(iii) If β [ν, μ] and ν μ, then (1.1) has no positive solution.1(iv) For β ( , χ0 ) (max{λ2 μ, λ 2 ν}, ), there exists M M(μ, ν, λ1 , λ2 , β) suchthat any positive ground state solution (u, v) of (1.1) satisfies u v M.In part (i) of Theorem 1.2, the existence of positive radial ground state solution is shownfor β χ0 . Indeed we can prove the existence of positive radial ground state solutionfor β ( , χ1 ) where χ1 χ0 defined above (see Lemma 2.7), but the expressionof χ1 is more complicated so it is deferred to Sect. 2. In Sect. 2, we will also show thatχ0 χ1 min{μ1 , ν1 }. Note that β ( , χ1 ) covers all negative β value, whichcorresponds to the repulsive interaction case. For the repulsive case, whether the groundstate solution is radially symmetric is still not known as the method in Theorem 1.1 requiresβ 0. On the other hand, for the attractive case β 0, any positive solution is necessarilyradially symmetric from Theorem 1.1, hence any ground state solution must be radial.In the special case of λ1 λ2 , the ground state solution of (1.1) can be constructedfrom the solution of the scalar equation (1.12), and a more explicit expression of the positiveground state solutions can be obtained as follows.Theorem 1.3 Let w H 1 (R3 ) be the unique positive solution of (1.12) with σ τ 1.Assume that λ1 λ2 . (i) If β (0, min{μ, ν}) (max{μ, ν}, ), then ( κw, w) is a positive ground statesolution of (1.1), where κ 0 and 0 satisfyμκ β 1, βκ ν 1.(1.16)(ii) If β [min{μ, ν}, max{μ, ν}] and μ ν, then (1.1) does not have a positive solution.Some remarks on these results are in order:1. It is easy to see that the conditions in Theorem 1.2 reduce to the ones in Theorem 1.3when λ λ2 /λ1 1. While there is a gap between the existence and nonexistenceranges of β for λ 1 in Theorem 1.2, the β-range for the existence and nonexistenceof solution when λ 1 is optimal.123

168Page 6 of 36J. Wang, J. Shi2. In the proof of Theorems 1.2–1.3, the main difficulty is to exclude the semitrivial solutions of (1.1). In the local case, many work overcome this difficulty by using differentvariational methods, for instance, see ] and references therein. In the nonlocal case, some ideas of the papers [30,31,51]can still be adapted to our case. However, many difficulties arise due to the presence ofthe non-local terms, some new techniques and a more careful analysis of the interactiondepending on parameter β are required for the proof given here.3. To the best of our knowledge, Theorems 1.2–1.3 are the first rigorous results for theexistence of nontrivial solution of (1.1). These existence results of (1.1) could play animportant role in studying singular perturbation problem of (1.1) as in [62]. Results inthis nature for the local case (1.10) have been proved in [30,51]. For example, in [51,Theorem 2], the existence of a positive radial ground state solution was also shownunder a similar assumption as in Theorem 1.2 (i).For the uniqueness of positive solution or positive ground state solution of (1.1), we havethe following results.Theorem 1.4 Suppose that λ1 , λ2 , μ, ν 0 are fixed.(i) There exists β0 0 such that if 0 β β0 , up to a translation, (1.1) has a uniquepositive solution (u β , vβ ), which is radially symmetric and decreasing in the radialdirection, and is non-degenerate in Er . Moreover as β 0 , (u β , vβ ) (u 0 , v0 )strongly in Er , where (u 0 , v0 ) is the positive ground state solution of (1.1) with β 0and same center. (ii) When λ1 λ2 , if 0 β min{μ, ν} or β max{μ, ν}, then ( κw, w) is theunique positive ground state solution of (1.1) up to a translation.The first uniqueness result in Theorem 1.4 is of perturbation nature. At β 0, the nondegeneracy of the positive ground state solution is reduced to that of the scalar equation(1.12), and we follow the approach in [21,43,55] to prove the non-degeneracy. Together withthe a priori estimate in Theorem 1.2 part (iv), the uniqueness of the positive solution canbe shown. The second uniqueness result in Theorem 1.4 takes advantage of λ1 λ2 . Theuniqueness of positive solution or positive ground state solution for other cases is still open.For the local interaction case, the uniqueness of positive solution of (1.10) when λ1 λ2and β max{μ, ν} was proved in [60]. More partial uniqueness results for the case thatβ (0, min{μ, ν}) and λ1 λ2 in the local interaction situation were also proved in[11,12,60]. On the other hand, the uniqueness of positive solution of (1.12) proved in was[27,37]. We conjecture that under the conditions of (ii) of Theorem 1.4, ( κw, w) is theunique positive solution to (1.1). The uniqueness of positive solution of (1.1) when β 0is not expected, as for the local interaction case (1.10), multiple positive solutions have beenfound via bifurcation methods [5]. Showing the existence of multiple positive solutions of(1.1) is another interesting open question.Our last result concerns with the limiting behavior of the positive ground state solutionsof (1.1) as β .Theorem 1.5 Assume that μ, ν, λ1 , λ2 0 are fixed. Let {βn } be a sequence satisfyingβn 0 and βn as n , and let (u βn , vβn ) be any nonnegative nontrivial radialground state solution of (1.1) with β βn . Then as n , at least one of u βn 2λ1 andββ vβn 2λ2 goes to infinity, and Cr n as n where Cr n is the least energy level of(1.1) with β βn .123

Standing waves for a coupled nonlinear Hartree equations Page 7 of 36168Our result here implies that when β goes to negative infinity, at least one of componentof the ground state solution blow up, hence the separation of phases does not occur in thecase of nonlocal interaction. For the Eq. (1.10) with local interaction, the phase separationbehavior when β has been proved in, for example, [39,44,57,58]. In that case, theprofile of components of solution of the limiting equation tend to separate in different regionsof the underlying domain.The paper is structured this way: In Sect. 2, we provide preliminary results, including theproof of Theorem 1.3; we prove the existence of positive ground state when β χ0 (part (i)of Theorem 1.2) in Sect. 3, and we prove the existence of positive ground state for large β(part (ii) of Theorem 1.2) and nonexistence for intermediate β (part (iii) of Theorem 1.2) inSect. 4; the a priori estimate of the positive ground state solutions (part (iv) of Theorem 1.2)and the asymptotic behavior of the ground state solutions as β (Theorem 1.5) areproved in Sect. 5, and the uniqueness of positive solution (Theorem 1.4) is shown in Sect. 6.The proof of the radial symmetry property in Theorem 1.1 is not directly related to otherparts, so we prove it in Sect. 7.2 Preliminary resultsThroughout the paper, we use the following notation: ( u 2 u 2 ); · is the norm of H 1 (R3 ) defined by u 2 R3 ( u 2 M u 2 ), for a · M is an equivalent norm of H 1 (R3 ) defined by u 2M R3positive function or constant M; E H 1 (R3 ) H 1 (R3 ), and Er Hr1 (R3 ) Hr1 (R3 ) where Hr1 (R3 ) {u H 1 (R3 ) :u(x) u( x )}; For z (u, v) E H 1 (R3 ) H 1 (R3 ), z 2E u 2λ1 v 2λ2 , where λ1 , λ2 0are the parameters in (1.1); 1/ p u pfor 0 p . · p is the norm of L p (R3 ) defined by u p S infu H 1 (R3 )\{0}R3 u 22where 2 6 here as N 3. u 22 We first recall the following classical Hardy–Littlewood–Sobolev inequality (see [28,Theorem 4.3]).Lemma 2.1 Assume that f L p (R3 ) and g L q (R3 ). Then one has R3 R3f (x)g(y)d xd y c( p, q, t) f p g q , x y twhere 1 p, q , 0 t 3 and1t1 2.pq3The following basic inequality is of fundamental importance for considering (1.1). It iswell-known but we include a proof here for reader’s convenience.123

168Page 8 of 36J. Wang, J. Shi12Lemma 2.2 For u, v L 5 (R3 ), we have that R3R3u 2 (x)v 2 (y)d xd y x y R3 R3u 2 (x)u 2 (y)d xd y x y 21 R3 R3 21v 2 (x)v 2 (y)d xd y. x y (2.1)In particular, if u, v H 1 (R3 ), then (2.1) holds.12Proof From Lemma 2.1, when u, v L 5 (R3 ), the integrals in (2.1) are all convergent. Firstwe claim that for any x, y R3 , there exists a constant K 0 independent of x, y such that 111·dz.(2.2) K2 y z 2 x y R3 x z Indeed the right hand side of (2.2) can be considered as a function h(x, y). Then h is translation and rotation invariant in (x, y), it follows that h depends only on x y . Furthermoreh(s x y ) s 1 x y and the left hand side of (2.2) satisfies the same scaling. Thereforethey must agree up to a constant factor.From (2.2), Fubini’s theorem and the Cauchy–Schwarz inequality, we obtain that u 2 (x)v 2 (y)d xd y x y R3 R3 u 2 (x)v 2 (y)dxdy dz K22R3R3 x z R3 y z 2 21 2 21u 2 (x)v 2 (y)d x dzdy dz KK22R3R3 x z R3R3 y z 21u 2 (y) Kdy dz2R3R3R3 y z 21v 2 (x)v 2 (y) Kdxdy dz22R3R3 x z R3 y z 21 21u 2 (x)u 2 (y)v 2 (x)v 2 (y) .d xd yd xd y x y x y R3 R3R3 R3 u 2 (x)dx x z 2 In the following, to simplify our notation, we define f 2 (y)dy, for f H 1 (R3 ).φ f (x) R3 x y Apparently we have the following symmetry property of φ f for u, v H 1 (R3 ): u 2 (x)v 2 (y)φu (x)v 2 (x)d x φv (x)u 2 (x)d x.d xd y x y R3R3 R3R3(2.3)(2.4)To study the solutions of (1.1) in a variational framework, we consider the following twocases: (1) β is negative or β is positive and small; (2) β is positive and large.For the case (1) we consider the energy functional Lλ1 λ2 on the setNλ1 λ2 {z (u, v) E : u 0, v 0, F1 (z) 0 and F2 (z) 0},123(2.5)

Standing waves for a coupled nonlinear Hartree equations Page 9 of 36 ( u 2 λ1 u 2 )d x μφu u 2 d x βφu v 2 d x,R3R3R3 222F2 (z) ( v λ2 v )d x νφv v d x βφv u 2 d x.168 whereF1 (z) R3R3(2.6)R3We defineC infz N λ1 λ2Lλ1 λ2 (z) and Cr infz N λr λLλ1 λ2 (z),(2.7)1 2where Nλr1 λ2 Nλ1 λ2 Er .For the case (2), we also consider the problem on the entire Nehari manifoldMλ1 λ2 {z (u, v) E \ {(0, 0)} : Lλ 1 λ2 (z)[z] F1 (z) F2 (z) 0},(2.8)and corresponding critical values areC0 infz M λ1 λ2Lλ1 λ2 (z), and Cr0 infr Lλ1 λ2 (z),z M λ λ1 2(2.9)where Mλr1 λ2 Mλ1 λ2 Er . Some standard arguments show that any nontrivial solutionof (1.1) is on both the sets Nλ1 λ2 and Mλ1 λ2 . Following the idea of [30,51], we shall provethe infimums C and C 0 are attained by a nontrivial solution of (1.1). It is also clear thatNλ1 λ2 Mλ1 λ2 and Nλr1 λ2 Mλr1 λ2 , which imply that C 0 C and Cr0 Cr . From theabove definitions, we know that if z (u, v) Nλ1 λ2 (or Mλ1 λ2 , and u 0 and v 0)satisfies Lλ1 λ2 (z) C (or C 0 ), and z is a solution of (1.1), then z is a ground statesolution of (1.1). Similarly if z (u, v) Nλr1 λ2 (or Mλr1 λ2 , and u 0, v 0) satisfiesLλ1 λ2 (z) Cr (or Cr0 ), and z is a solution of (1.1), then z is a radial ground state solutionof (1.1).If C 0 or Cr0 is attained by z Mλ1 λ2 , then z is a solution of (1.1) (see, for example, [51, Proposition 3.5] or [61, Chapter 4]). The following lemma shows that when β μν, if Cor Cr is attained by z Nλ1 λ2 , then z is also a solution of (1.1). Lemma 2.3 Suppose that β μν. If C (or Cr ) is attained by z (u, v) Nλ1 λ2(or Nλr1 λ2 ), then z is a solution of (1.1).Proof We only show that if C is attained by z (u, v) Nλ1 λ2 , then z is a solution of (1.1).One needs to prove that any minimizer of Lλ1 λ2 restricted to Nλ1 λ2 satisfies Lλ 1 λ2 (z)[φ] 0for any φ E.Let Fi (i 1, 2) be defined as in (2.6). We claim that if z Nλ1 λ2 satisfying Lλ1 λ2 (z) C, then F1 (z) and F2 (z) are linear independent. Assume that for K 1 , K 2 R such thatK 1 F1 (z) K 2 F2 (z) 0. Since F1 (z) 0, it follows from (K 1 F1 (z) K 2 F2 (z))[(u, 0)] 0that K1μφu u 2 d x K 2 βφu v 2 d x 0.(2.10)R3R3Similarly it follows from F2 (z) 0 and (K 1 F1 (z) K 2 F2 (z))[(0, v)] 0 that K1βφv u 2 d x K 2 νφv v 2 d x 0.R3Set R3 φu u 2 d x βμφu v 2 d x R3A R3 .φv u 2 d x νφv v 2 d xβR3(2.11)(2.12)R3123

168Page 10 of 36J. Wang, J. Shi It follows from Lemma 2.2 and 0 β μν that φu u 2 d xφv v 2 d x β 2det (A) μνR3R3R3 2φu v 2 d x 0.(2.13)That is, A is positively definite, which implies that K 1 K 2 0. Thus F1 (z) and F2 (z) arelinear independent. Since z is a minimizer of Lλ1 λ2 restricted on Nλ1 λ2 , then according to[8, Corollary 4.1.2], there exist two Lagrange multipliers H1 , H2 R such thatLλ 1 λ2 (z) H1 F1 (z) H2 F2 (z) 0.(2.14)So, by using the same arguments as in (2.10)–(2.13), one can prove that H1 H2 0.For the case of β 0 we can use the idea of the proof of [31, Lemma 2.1] to prove theconclusion. Here we omit the details. It follows from Lemma 2.3 that in order to prove the conclusion (i) of Theorem 1.2, weneed to show that Cr is attained by a positive z Nλr1 λ2 for β χ0 , and C Cris attained by a positive z Nλr1 λ2 for 0 β χ0 , where χ0 is given in Theorem 1.2. Forthis purpose we shall make good use of the unique positive solution of (1.12). Let w be theunique positive solution of (1.12) with σ τ 1. Define σwσ,τ (x) w( σ x), wσ (x) wσ,1 (x).(2.15)τThen wσ,τ is the unique positive solution of (1.12).Since wσ,τ is the unique positive solution of (1.12), one can verify the following facts(see [37, Theorem 2], and [51, Section 3.4] or [30, Lemma 2]).Lemma 2.4 Consider the the minimization problems u 2σ11 u 2σ 24 and Tσ,τ infτ φu u , (2.16)1u M0R3( R3 τ φu u 2 ) 2!where M0 u H 1 (R3 ) : u 0, u 2σ R3 τ φu u 2 . Then the function wσ,τ (x) is aminimizer of Tσ,τ and the unique positive solution of (1.12). Moreover, we haveSσ,τ infu H 1 (R3 )\{0}3Tσ,τ R331 2σ4σ4Sand Sσ,τ S1,1 S1 ,4 σ,τττ where S1,1 S1 2(2.17) 1φw w 22.We introduce a function θ : [1, ) R defined by φw (x)wλ2 (x)d x3θ (λ) R,φw (x)w 2 (x)d x(2.18)R3The following lemma gives some estimates of θ (λ).Lemma 2.5 Let θ (λ) be defined as in (2.18). Then for any λ 1 we have13λ 2 θ (λ) λ 4 .123(2.19)

Standing waves for a coupled nonlinear Hartree equations Page 11 of 36168Proof Since w Hr1 (R3 ) is radial and is strictly decreasing in r x , it follows that w(x) w( λx) for λ 1, x R3 .So we infer from Lemma 2.2 and change of variables that w 2 (x)w 2 ( λy)φw (x)wλ2 (x)d x λ2d xd y x y R3R3 R3 1 21 22 (x)w 2 (y)2 ( λx)w 2 ( λy)wwd xd yd xd y λ2 x y x y R3 R3R3 R3 λ34 λ34 R3 R3R3R3w 2 (x)w 2 (y)d xd y x y 21 R3R3w 2 (x)w 2 (y)d xd y x y 21w 2 (x)w 2 (y)d xd y. x y 3That is, θ (λ) λ 4 . Furthermore we see that R3 φw (x)wλ2 (x)d x λ2 R31 λ 2R3 w 2 (y)w 2 ( λx)d xd y x y w 2 ( h )w 2 (z)λdhdz h z 11w 2 (h)w 2 (z)φw (x)w 2 (x)d x, λ 2dhdz λ 2 h z R3 R3R3(2.20)which implies the lower bound of θ (λ). R3R3Next we use the function w to provide some estimates for C and Cr .Lemma 2.6 Let θ (λ) be defined as in (2.18). If κ, 0 satisfy"μκ βθ (λ) 1,3(2.21)3βθ (λ)κ λ 2 ν λ 2 , then we have ( κwλ1 , wλ2 ) Nλr1 λ2 . That is, Nλr1 λ2 and Nλ1 λ2 . Moreover,there exists ρ0 0 such that 3 3 1κλ12 λ22 w 20 ρ0 C Cr Lλ1 λ2 ( κwλ1 , wλ2 ) 4(2.22) 3 31222 φw w d x.κλ1 λ24R3 Proof To prove ( κwλ1 , wλ2 ) Nλr1 λ2 , it suffices to show that (u, v) ( κwλ1 , wλ2 )satisfy φu u 2 βφv u 2 , v 2λ2 νφv v 2 βφu v 2 .(2.23) u 2λ1 μR3R3R3R3123

168Page 12 of 36J. Wang, J. ShiA direct computation shows that ## κwλ1 2λ1 κ λ31 w( λ1 x) 2 d x λ31w 2 ( λ1 x)d xR3R3 332222 κλ1 w(y) dy w (y)dy κλ1φw w 2 d x.R3R3On the other hand, by the changes of variables x (2.24)R3 xλ1and y y ,λ1one sees that φ κwλ ( κwλ1 )2 βφ wλ ( κwλ1 )21233RR w 2 ( λ1 x)w 2 ( λ1 y)w 2 ( λ1 x)w 2 ( λ2 y)2 4 μκ λ1d yd x βκ λ21 λ22d yd x x y x y R3 R3R3 R3 w 2 (x)w 2 ( λ2 y)3λ2w 2 (x)w 2 (y)λ1 μκ 2 λ12d yd x βκ 2d yd x x y x y λ1 R 3 R 3R3 R3 3φw w 2 d x. κλ12 [μκ β θ (λ)]μR3(2.25)So if μκ βθ (λ) 1, then one has that the quantity in (2.24) equals to the one in (2.25).3That is, the first equality in (2.23) is satisfied. Similarly, by using βθ (λ)λ 2 κ ν 1, thesecond equality in (2.23) is also satisfied.Next we prove the second part of the lemma. Since for each z (u, v) Nλ1 λ2 , we have R3( u 2 λ1 u 2 v 2 λ2 v 2 ) R3(μφu u 2 βφu v 2 βφv u 2 νφv v 2 ). (2.26)By Lemma 2.1, for some c1 0 independent of u, v, one has that R3φu (x)u 2 d x R3R3u 2 (y)u 2 (x)d yd x c1 x y 125 53 c1 u 412 .(2.27)u 2 (x)v 2 (y)d yd x c1 u 212 v 212 . x y 55(2.28)R3u5Similarly, one can also prove that R3 φv (x)v 2 d x c1 u 412 and5 R3R3Substituting (2.27)–(2.28) into (2.26) and using Sobolev embedding, we obtain u 2λ1 v 2λ2 c1 μ u 412 c1 ν v 412 2c1 β u 212 v 212555 c2 ( u 4λ1 v 4λ2 2 u 2λ1 v 2λ2 ),5(2.29)for some c2 0. Furthermore, for each z (u, v) Nλ1 λ2 , one has1Lλ1 λ2 (u, v) 4 R3 u 2 λ1 u 2 v 2 λ2 v 2 which implies that C ρ0 0 for some ρ0 0.12311,( u 2λ1 v 2λ2 ) 44c2(2.30)

Standing waves for a coupled nonlinear Hartree equations Page 13 of 36168 Finally, since κ, 0 satisfy (2.21), we have that ( κwλ1 , wλ2 ) Nλ1 λ2 , and Cr Lλ1 λ2 ( κwλ1 , wλ2 ) κ ( wλ1 2 λ1 wλ21 ) ( wλ2 2 λ2 wλ22 )(2.31)4 R34 R3 333311 (κλ12 λ22 ) w 2 (κλ12 λ22 )φw w 2 .44R3 This finishes the proof of lemma.In the following we shall discuss the solvability of (2.21). From elementary calculation,we know that κ 0 and 0 if either3333det (Aλ ) λ 2 μν β 2 θ 2 (λ) 0 and βθ (λ) min{ν, λ 2 μ},ordet (Aλ ) λ 2 μν β 2 θ 2 (λ) 0 and βθ (λ) max{ν, λ 2 μ},where Aλ (2.32)(2.33) μ βθ (λ).3βθ (λ) λ 2 νBy Lemma 2.5 and further direct computation, we have that (2.32) is satisfied if33 μν β λ 4 min{ν, λ 2 μ} min{ν1 , μ1 },(2.34)where μ1 and ν1 are defined in (1.15). Similarly (2.33) is satisfied if11β max{λμ, λ 2 ν} λ 4 max{ν1 , μ1 }.(2.35)When (2.34) or (2.35) is satisfied, we can solve that3κ Defineλ 2 (ν βθ (λ))3λ 2 μν β 2 θ 2 (λ)3and λ 2 μ βθ (λ)3λ 2 μν β 2 θ 2 (λ)3a(λ) h(λ)(2 h(λ)) where h(λ) λ 4 θ (λ).(2.36)(2.37)From Lemma 2.5 we obtain that5λ 4 h(λ) 1,and55λ 4 (2 λ 4 ) a(λ) 1for λ 1.(2.38)Now we are ready to study the behaviour of minimizing sequences of Lλ1 λ2 on Nλ1 λ2 byusing some ideas from [51].Lemma 2.7 Suppose that λ λ2 /λ1 1.

unique positive solution wσ,τ H1(R3) that is radially symmetric for any σ,τ 0. We look for solutions of (1.1) which are different from the preceding ones. A solution (u,v)of (1.1) is nontrivial if u 0andv 0. A solution (u,v)with u 0andv 0 is called a positive solution. A solution is called a ground state solution (or positive ground