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Existence of solutions for a semirelativistic Hartree equation with unbounded potentials Simone Secchi ∗ January 12, 2017 7 1 0 2 To Francesca, always n a J WeprovetheexistenceofasolutiontothesemirelativisticHartreeequation 1 1 ∆+m2u+V(x)u = A(x)(W up) up−2u − ∗| | | | ] p P under suitable growth assumption on the potential functions V and A. In A particular, both can be unbounded from above. . h t a m 1 Introduction [ 1 The mean field limit of a quantum system describing many self-gravitating, relativistic v 5 bosons with rest mass m > 0 leads to the time-dependent pseudo-relativistic Hartree 8 equation 8 ∂ψ 1 2 i = ∆+m2 m ψ ψ 2 ψ, x R3 (1.1) 0 ∂t − − − x ∗| | ∈ . (cid:16)p (cid:17) (cid:18)| | (cid:19) 1 where ψ : R R3 C is the wave field. Such a physical system is often referred to as 0 × → 7 a boson star in astrophysics (see [16, 20, 21]). Solitary wave solutions ψ(t,x) = e−itλφ, 1 λ R to equation (1.1) satisfy the equation : ∈ v Xi ∆+m2 m φ 1 φ2 φ = λφ in R3. − − − x ∗| | r (cid:16)p (cid:17) (cid:18)| | (cid:19) a For thenon-relativistic Hartreeequation drivenby thelocal differential operator ∆+ − m2, existence and uniqueness (modulo translations) of a minimizer were proved by Lieb [22] by usingsymmetric decreasing rearrangement inequalities. Within the same setting, always for the negative Laplacian, P.-L. Lions [25] proved existence of infinitely many ∗TheauthorissupportedbytheMIUR2015 PRINproject “Variational methods,with applicationsto problems in mathematical physics and geometry”. 1 sphericallysymmetricsolutionsbyapplication ofabstractcriticalpointtheorybothwith andwithoutconstraintsforamoregeneralradiallysymmetricconvolutionpotential. The non-relativistic Hartree equations is also known as theChoquard-Pekardor Schrödinger- Newton equation and recently a large amount of papers are devoted to the study of solitary states and its semiclassical limit: see [1, 4, 5, 7, 8, 9, 11, 12, 13, 14, 20, 23, 27, 26, 29, 30, 31, 38, 39] and references therein. Inthispaper,whichissomehowacontinuationoftheinvestigation webeganin[33,34], we consider the equation ∆+m2u+V(x)u = A(x)(W up) up−2u (1.2) − ∗| | | | p in the weighted space X = u L2(RN) u < , X ∈ | k k ∞ n o where 2 1/4 u 2 = ∆+m2 u dx+ V(x)u2dx. k kX ZRN (cid:12)(cid:12)(cid:16)− (cid:17) (cid:12)(cid:12) ZRN | | In the case A 1, equation(cid:12)(1.2) has been r(cid:12)ecently investigated in [12], where least- ≡ (cid:12) (cid:12) energy solutions are constructed for a bounded potential V. We collect our assumptions. (H1) N > 1, β > 1, 2 p < 2Nβ and r > max 1, 1−Nβ . ≤ β(N−1) 2+βN(p−2)−βp n o (H2) A L∞(RN) satisfies A(x) 1 for almost every x RN. ∈ loc ≥ ∈ (H3) V : RN R is continuous and → infV > B RN − for some B > 0; furthermore, 2 ( ∆+m2)1/4u + V u2 λ = inf RN − RN | | > 0. 1 u∈X\{0} R (cid:12)(cid:12)(cid:12) kuk22 (cid:12)(cid:12)(cid:12) R (H4) There exist C > 0 and R > 0 such that 0 0 2r 1 A(x)2r−1 C0 1+(max 0,V(x) )β ≤ { } (cid:16) (cid:17) for all x > R . 0 | | (H5) W: RN 0 R and W = W +W , where W Lr and W L∞. 1 2 1 2 \{ } → ∈ ∈ Equivalently, condition (H4) can be stated as 2r A(x)2r−1 limsup < + . 1 ∞ |x|→+∞ 1+(max 0,V(x) )β { } 2 Remark 1.1. We highlight that both the potential V and the potential A in front of the nonlinear term in the right-hand side can be unbounded. To the best of our knowledge, this case is treated hereforthefirsttime. Inthe very recent paper[6], asimilar equation has been studied under an assumption that, in our framework, reads as V 1 and ≡ lim A(x) = 1. |x|→+∞ To state our last assumption, we define, for any open subset Ω RN and any t 2, ⊂ ≥ the quantity (see Section 2 for the definition of the space L1/2,2(Ω)) 2 ( ∆+m2)1/4u + V u2 ν(t,Ω)= inf Ω − Ω | | if Ω = u∈L1/2,2(Ω) R (cid:12)(cid:12) ( ut)2(cid:12)(cid:12)/t R 6 ∅ u6=0 (cid:12) Ω| | (cid:12) R = + if Ω = . ∞ ∅ We assume that (ν) there exists t [2,2N(N 1)) such that 0 ∈ − lim ν(t ,RN B )= + , 0 R R→+∞ \ ∞ where B = x RN x < R . R ∈ | | | n o The number λ in assumption (H3) can be actually seen as an eigenvalue, as the 1 following result shows. Proposition 1.2. Assume that V : RN R is continuous and infRN V > B for some → − B > 0. Assume also that (ν) holds with t = 2. Then there exists ϕ X such that 0 1 ∈ ∆+m2ϕ +V(x)ϕ = λ ϕ in RN. 1 1 1 1 − p Furthermore ϕ can be assumed to be positive in RN. 1 Proof. It follows immediately from the assumptions that λ > B > 0. Let u be a 1 n n − { } minimizing sequence for λ , in the sense that 1 1 2 lim ∆+m2 4 u + V u 2 = λ (1.3) n n 1 n→+∞ZRN (cid:12)(cid:12)(cid:16)− (cid:17) (cid:12)(cid:12) ZRN | | (cid:12) (cid:12) u 2 = 1. (1.4) (cid:12) (cid:12) n RN | | Z As in [10, Lemma 2.1], we can assume that u is non-negative. It follows from (1.3) that n ( ∆+m2)41un 2 λ1+1+B. RN − ≤ Z (cid:12) (cid:12) (cid:12) (cid:12) Let u be the weak limit of u(cid:12) in L1/2,2(RN(cid:12) ). Since u u strongly in L2(B ) for n n n R { } → any R > 0, we may assume that u converges to u almost everywhere, and consequently n 3 u 0. We fix a smooth cut-off function ψ such that ψ 0 on B and ψ 1 on R ≥ ≡ ≡ RN B . Then R+1 \ u u 2 (1 ψ)(u u) 2+ ψ(u u) 2 k n− k2 ≤ k − n − k2 k n − k2 1 1 2 o (1)+ ∆+m2 4 (u u) + V u u2 ≤ n ν(2,RN \BR) ZRN (cid:12)(cid:12)(cid:16)− (cid:17) n − (cid:12)(cid:12) ZRN | n− | ! (cid:12) (cid:12) o (1)+o (1), (cid:12) (cid:12) ≤ n R where lim o (1) = 0 and lim o (1) = 0: this proves that u u strongly in n→+∞ n R→+∞ R n → L2(RN). In particular, u2 = 1 due to (1.4). Plainly, RN | | R ( ∆+m2)14u 2 liminf ( ∆+m2)14un 2. RN − ≤ n→+∞ − Z (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Set G= x RN V(x(cid:12)) > 1 . Since V(cid:12) L∞(RN(cid:12) G), we have (cid:12) ∈ | ∈ \ n o lim V u 2 = V u2. n n→+∞ZRN\G | | ZRN\G | | On the other hand, by [19, Theorem 6.54] we deduce that v L2(G) V v 2 ∈ 7→ | | ZG is weakly lower semicontinuous. To summarize, liminf V u 2 = lim V u 2+liminf V u 2 n n n n→+∞ZRN | | n→+∞ZRN\G | | n→+∞ZRN | | V u2. ≥ RN | | Z Hence u is a minimizer for λ . The strict positivity of u follows from [17, Proposition 1 2]. Remark 1.3. In the previous Proposition, we did not assume that λ > 0. Under this 1 additional assumption, the proof would be much easier. Itisimportanttoremarkthatcondition(ν)isrelatedtosomeotherpopularconditions. In Proposition 3.2 we prove that the coercivity assumption lim V(x) = + |x|→+∞ ∞ implies (ν). Similarly, the condition (ν′) For every b> 0, the set Vb has finite Lebesgue measure also implies (ν), see Proposition 3.4. Finally, Sirakov’s condition [35] 4 (ν′′) Thereexistst [2,2N/(N 1))suchthat,foranyr > 0andanysequence x of 1 n n ∈ − { } pointsinRN suchthatlim x = + ,thereresultslim ν(t ,B(x ,r)) = n→+∞ n n→+∞ 1 n | | ∞ + ∞ is equivalent to (ν): see Proposition 3.5. Remark 1.4. While dealing with the semirelativistic Hartree equation, it is customary to rewritetheoperator √ ∆+m2+V as √ ∆+m2 m+(V +m)in orderto exploit the − − − factthat√ ∆+m2 m > 0inthesenseoffunctionalcalculus. Thenaturalassumption − − from a variational point of view is therefore inf V > m. Our assumption (H3) is, in RN − general, less restrictive as it only requires a suitable lower bound for V. We can state the main result of this paper. Theorem 1.5. Suppose that (H1), (H2), (H3), (H4), (H5), and (ν) are satisfied. Then equation (1.2) has infinitely many distinct solutions. The proof is based on the ideas developed in [35] for a local equation and a pointwise nonlinearity. Remark 1.6. It should be noticed that our results continue to hold if we replace (1.2) with ( ∆+m2)su+V(x)u = (W up) up−2u − ∗| | | | with 0 < s < 1 and N > 2s, for instance N 3. Of course some numbers must be ≥ replaced: 2N/(N 1)shouldbecome 2N/(N 2s), andsoon. We preferto work outthe − − details for s= 1/2, which corresponds to the physical model of the Hartree equation. In Section 2 we introduce the necessary preliminaries on function space. In Section 3 we compare several assumptions the ensure a compact embedding result. In Section 4 we prove a compactness theorem that is used in Section 5 to prove our main existence result. Notation 1. The letters c and C will stand for a generic positive constant that may vary from line to line. 2. Theoperator dwillbereserved forthe(Fréchet) deriva- tive, so thatdI denotes theFréchet derivative of a func- tion I. 3. The symbol N will be reserved for the Lebesgue N- L dimensional measure. 4. The Fourier transform of a function u will be denoted by u. F 5 2 Variational setting Let us recall the definition of the Bessel function space defined for α >0 by Lα,2(RN) = f: f = G ⋆g for some g L2(RN) , α ∈ n o where the Bessel convolution kernel is defined by Gα(x) = 1 ∞exp π x2 exp t tα−2N−1dt (4π)α/2Γ(α/2) −t| | −4π Z0 (cid:18) (cid:19) (cid:18) (cid:19) The norm of this Bessel space is f = g if f = G ⋆g. The operator 2 α k k k k (I ∆)−αu= G ⋆u 2α − is usually called Bessel operator of order α. In Fourier variables the same operator reads −α/2 G = −1 1+ ξ 2 , α F ◦ | | ◦F (cid:18)(cid:16) (cid:17) (cid:19) so that f = (I ∆)α/2f . k k − 2 (cid:13) (cid:13) For more detailed information, see [2(cid:13), 36] and th(cid:13)e references therein. The use of (cid:13) (cid:13) ( ∆+m2)α/2 instead of ( ∆+I)α/2 is clearly harmless. We summarize the embedding − − properties of Bessel spaces. For the proofs we refer to [18, Theorem 3.2], [36, Chapter V, Section 3] and [37, Section 4]. Theorem 2.1. 1. Lα,2(RN) = Wα,2(RN) = Hα(RN). 2. If α 0 and 2 q 2∗ = 2N/(N 2α), then Lα,2(RN) is continuously embedded ≥ ≤ ≤ α − into Lq(RN); if 2 q <2∗ then the embedding is locally compact. ≤ α 3. Assume that 0 α 2 and α > N/2. If α N/2 > 1 and 0 < µ α N/2 1, ≤ ≤ − ≤ − − then Lα,2(RN) is continuously embedded into C1,µ(RN). If α N/2 < 1 and − 0 < µ α N/2, then Lα,2(RN) is continuously embedded into C0,µ(RN). ≤ − Remark 2.2. AlthoughtheBesselspaceLα,2(RN)istopologically undistinguishablefrom the Sobolev fractional space Hα(RN), we will not systematically confuse them, since our equation involves the Bessel norm. For a general open subset Ω of RN, the Bessel space Lα,2(Ω) is defined ad the space of the restrictions to Ω of functions in Lα,2(RN). In this paper we will always take α = 1/2. As we said in the Introduction, we work in the weighted space X = u L1/2,2(RN) u < X ∈ | k k ∞ n o where 1/4 2 u 2 = ∆+m2 u dx+ V(x)u2dx k kX ZRN (cid:12)(cid:12)(cid:16)− (cid:17) (cid:12)(cid:12) ZRN | | (cid:12) (cid:12) (cid:12) (cid:12) 6 Lemma 2.3. If assumption (H3) holds, then there exists a constant > 0 such that ℵ u u k kX ≥ ℵk kL1/2,2(RN) for every u X. ∈ Proof. We proceed by contradiction. Assume that for some sequence u in X there n n { } results 1 u 2 = 1, u 2 < . k nkL1/2,2 k nkX n By definition of λ > 0, the last inequality entails u = o(1). But then the contra- 1 n L2 k k diction 1 o(1) = B u 2 V u 2 < 1 − k nkL2 ≤ RN | n| n − Z arises as n + . → ∞ Solutions to (1.2) correspond to critical points of the functional I: X R defined by → 1 1 I(u) = u 2 W(x y)A(x)u(x)p u(y)pdxdy. (2.1) 2k kX − 2p ZRN×RN − | | | | We need to prove that I is well-defined on X. Proposition 2.4. The space X is continuously embedded into the weighted Lebesgue space Lt(RN,A2r2−r1d N) L 2(Nβ−1) for every 2 t . ≤ ≤ β(N−1) Proof. Let us decompose A(x)2r2−r1 u(x)tdx = A(x)2r2−r1 u(x)tdx ZRN | | ZB(0,R0) | | + A(x)2r2−r1 u(x)tdx, (2.2) ZRN\B(0,R0) | | where R > 0 is the number defined in assumption (H4). Now, 0 A(x)2r2−r1 u(x)tdx C0 1+(max 0,V(x) )β1 u(x)tdx ZRN\B(0,R0) | | ≤ ZRN\B(0,R0)(cid:16) { } (cid:17)| | C0 u(x)tdx+C0 1+V(x)β1 u(x)tdx, ≤ | | | | ZΩ1 ZΩ2(cid:16) (cid:17) where Ω = x RN V(x) < 0 , Ω = x RN V(x) 0 . 1 2 ∈ | ∈ | ≥ n o n o 7 As a consequence, inserting this into (2.2), ZRN A(x)2r2−r1|u(x)|tdx ≤ 2C0kuktLt +kAkL∞(B(0,R0))kuktLt +ZΩ2V(x)β1|u(x)|tdx. If we apply Hölder’s inequality to the last integral, we obtain V(x)β1 u(x)tdx = V(x)β1 u(x) β2 u(x)t−β2 dx | | | | | | ZΩ2 ZΩ2 1 β−1 V(x)u(x)2dx β u(x) t−β2 β−β1 dx β . ≤ | | | | (cid:18)ZΩ2 (cid:19) (cid:18)ZΩ2 (cid:0) (cid:1) (cid:19) But V(x)u(x)2dx = V(x)u(x)2dx V(x)u(x)2dx ZΩ2 | | ZRN | | −Z{x∈RN|V(x)<0} | | B u 2 +B u(x)2dx 1+ u 2 . ≤ k kX ZRN | | ≤ (cid:18) λ1(cid:19)k kX In conclusion, RN A(x)2r2−r1|u(x)|tdx ≤ C1kuktLt +C2kukXβ2kuk(Lt(βt−β−22))//β(β−1)). (2.3) Z It is elementary to check that tβ 2 2N 2 − ≤ β 1 ≤ N 1 − − whenever 2(Nβ 1) 2 t − , ≤ ≤ β(N 1) − sothatwecaninvokethecontinuousembeddingofX intoLs(RN)for2 s 2N/(N 1) ≤ ≤ − and conclude from (2.3) that there exists a positive constant C such that A(x)2r2−r1 u(x)tdx C u t for every u X. RN | | ≤ k kX ∈ Z This completes the proof. Remark 2.5. Observe that when infRN V > 0 and A= 1 identically, we can let β + → ∞ and recover the weaker assumption 2 t 2N/(N 1). ≤ ≤ − To proceed further, we need the following inequality due to Hardy, Littlewood and Sobolev. We firstly recall that a function h belongs to the weak Lq space Lq(RN) if w there exists a constant C > 0 such that, for all t > 0, Cq N x RN h(x) > t . L ∈ | | | ≤ tq (cid:16)n o(cid:17) Its norm is then 1/q h = supt N x RN h(x) >t . q,w k k L ∈ | | | t>0 (cid:16) (cid:16)n o(cid:17)(cid:17) 8 Proposition 2.6 ([24]). Assume that p, q and t lie in (1,+ ) and p−1+q−1+t−1 = 2. ∞ Then, for some constant N > 0 and for any f Lp(RN), g Lt(RN) and h p,q,t ∈ ∈ ∈ Lq (RN), we have the inequality w f(x)h(x y)g(y)dxdy N f g h . (2.4) p,q,t p t q,w (cid:12)ZRN×RN − (cid:12) ≤ k k k k k k (cid:12) (cid:12) Writing W =(cid:12) W + W Lr(RN) + L∞(RN(cid:12)) and using (2.4) we can estimate the (cid:12) 1 2 (cid:12) ∈ convolution term as follows by means of Proposition 2.4: (W up)Aup = u(x)pA(x)W(x y)u(y)pdxdy ZRN ∗| | | | ZRN×RN | | − | | = u(x)pA(x)W (x y)u(y)pdxdy 1 ZRN×RN | | − | | + u(x)pA(x)W (x y)u(y)pdxdy 2 ZRN×RN | | − | | 2r−1 2r−1 2rp 2r 2r 2rp 2r C u 2r−1A2r−1 W1 r u 2r−1 ≤ RN | | k k RN | | (cid:18)Z (cid:19) (cid:18)Z (cid:19) + W u(x)pA(x)dx u(y)pdy 2 ∞ k k RN | | RN | | (cid:18)Z (cid:19)(cid:18)Z (cid:19) 2r−1 2r−1 2rp 2r 2r 2r 2rp 2r C u 2r−1A2r−1 W1 r A2r−1 u 2r−1 ≤ RN | | k k RN | | (cid:18)Z (cid:19) (cid:18)Z (cid:19) + W2 ∞ upA2r2−r1 upA2r2−r1 k k RN | | RN | | (cid:18)Z (cid:19)(cid:18)Z (cid:19) 2p 2p C u W + W u , ≤ k kL22rr−p1(RN,A2r2−r1dLN)k 1kr k 2k∞k kLp(RN,A2r2−r1dLN) (2.5) where we have used several times the fact that A(x) 1 for almost every x RN. Since ≥ ∈ 1 Nβ 2Nβ r > max 1, − and p < , 2+βN(p 2) βp β(N 1) (cid:26) − − (cid:27) − we can use Proposition 2.4 and from (2.5) we see that the convolution term in I is finite. It is easy to check, by the same token and taking into account the assumption p 2, ≥ that I C1(X). ∈ Remark 2.7. As before, if inf V > 0 and A = 1 identically, we can recover the assump- RN tion N r > max 1, p+(2 p)N (cid:26) − (cid:27) used in [12]. 3 Comparison between conditions on V In this section we prove that condition (ν) is actually weaker than both the coercivity of V and of condition (ν′). We start with a preliminary technical result. 9 Lemma 3.1. Let Ω RN be an open set. If 2 t < r < 2N/(N 1), then there exist ⊂ ≤ − a constant C = C(r,t,N,λ ) >0 and a number 0 < θ < 1 such that 1 1 ν(r,Ω) C(ν(t,Ω))θ1. ≥ If 2 < r < t < 2N/(N 1), then there exist a constant C = C(r,t,N,λ ) > 0 and a 1 − number 0 < θ < 1 such that 2 ν(r,Ω) C(ν(t,Ω))θ2. ≥ In particular, lim ν(r,RN B )= + R R→+∞ \ ∞ if lim ν(t,RN B ) = + . R R→+∞ \ ∞ Proof. We write 1 1 θ θ = − + r t 2N N−1 for some0 < θ < 1. Then u u 1−θ u θ for everyu L1/2,2(Ω). TheGagliardo- k kr ≤ k kt k kN2−N1 ∈ Nirenberg inequality (see [28, Theorem 2.1] together with Theorem 2.1) yields u r k k ≤ C u θ u 1−θ for every u L1/2,2(Ω). We invoke Lemma 2.3 to get k kL1/2,2k kt ∈ u 2 C u 2θ u 2(1−θ) k kr ≤ k kXk kt As a consequence, u 2 1 u 2 ν(r,Ω) = inf k kX inf k kX u∈L1/2,2(Ω)\{0} kuk2r ≥ C u∈L1/2,2(Ω)\{0} kuk2Vθkuk2t(1−θ) 2(1−θ) 1 u 1 inf k kX (ν(t,Ω))1−θ. ≥ C u∈L1/2,2(Ω)\{0} u 2(1−θ) ≥ C k kt The second inequality follows by the same token. Proposition 3.2. If lim V(x) = + , then (ν) holds true. |x|→+∞ ∞ Proof. By Lemma 3.1 it is sufficient to prove the validity of (ν) with t = 2. For any 0 u X and any R > 0, ∈ ( ∆+m2)14u 2+ V u2 V u2 inf V u2. ZRN\BR(cid:12) − (cid:12) ZRN\BR | | ≥ ZRN\BR | | ≥ RN\BR ZRN\BR| | (cid:12) (cid:12) Therefore ν(cid:12)(2,RN B ) (cid:12)inf V, and we conclude by letting R + . \ R ≥ RN\BR → ∞ 10

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