GROUND STATES SOLUTIONS FOR A NON-LINEAR EQUATION INVOLVING A PSEUDO-RELATIVISTIC SCHRO¨DINGER OPERATOR 6 1 0 VINCENZO AMBROSIO 2 n a Abstract. Inthis paperwe provethe existence, regularityandsymmetry J of a ground state for a nonlinear equation in the whole space, involving a 5 pseudo-relativistic Schrödinger operator. 2 ] P A 1. Introduction . h Recently, the study of fractional and nonlocal operators of elliptic type has t a attracted the attentions of many mathematicians. These operators arises in m many different areas of research such as optimization, finance, minimal sur- [ faces, phasetransitions, quasi-geostrophicflows, crystaldislocation, anomalous 1 diffusion, conservation laws and ultra-relativistic limits of quantum mechanics. v For more details and applications we refer to [3, 6, 8, 10, 12, 17, 19, 25, 26, 27, 7 2 23, 33, 34, 37] and references therein. 8 In this paper we consider the following nonlinear fractional equation 6 0 [( ∆+m2)s m2s]u+µu = u p 2u in RN (1.1) − . − − | | 1 0 where s (0,1), N > 2s, p (2, 2N ), m 0 and µ > 0. ∈ ∈ N 2s ≥ 6 The fractional operator − 1 ( ∆+m2)s (1.2) : v − i which appears in (1.1) is defined in the Fourier space by setting X ( ∆+m2)su(k) = ( k 2 +m2)s u(k) r a F − | | F or via Bessel potential, that is for every u C (RN) ∈ c∞ u(x) u(y) (−∆+m2)su(x) = cN,smN+22sP.V.ZRN |x−−y|N+22s KN+22s(m|x−y|)dy+m2su(x) for every x RN: see [22]. Here P.V. stands for the Cauchy principal value, ∈ s(1 s) cN,s = 2−N+22s+1π−N2 22s − Γ(2 s) − and K denotes the modified Bessel function of the second kind with order ν. ν 1991 Mathematics Subject Classification. 35J60, 35Q55,35S05. Key words and phrases. Ground states, Schrödinger equation, Extension method. 1 When s = 1/2 the operator (1.2) has a clear meaning in quantum mechanic: it corresponds to the free Hamiltonian of a free relativistic particle of mass m. We remind that the study of √ ∆+m2 has been strongly influenced by − papers [28, 29] of Lieb and Yau on the stability of relativistic matter. For more recent works about this topic one can see [1, 16, 18, 24, 25, 27]. Let us point out that the operator (1.2) has a deep connection with the Sto- chastic Process theory: in fact ( ∆+m2)s m2s is an infinitesimal generator − − of a Levy process called the 2s-stable relativistic process: see [14, 31]. In the present paper we are interested in the study of the ground states of (1.1). Such problems are motivated in particular by the search for certain kinds of solitary waves (stationary states) in nonlinear equations of the Klein- Gordon or Schro¨dinger type. In the celebrated papers [4, 5] Berestycki and Lions studied the existence of ground state of the following problem ∆u = g(u) in RN u− H1(RN) and u 0 (1.3) (cid:26) ∈ 6≡ where N 3, g is a real continuous function such that ≥ (1) lim g(t) = µ < 0; t→0 t − (2) lim g(t) 0; t→+∞ tNN+22 ≤ (3) there exists−t > 0 such that G(t ) = t0 g(z)dz > 0. 0 0 0 In [4], solutions are constructed by solving the minimization problem R inf u 2dx : G(u)dx = 1 . |∇ | RN RN nZ Z o Then a rescaling from u to a suitable u(t 1x) produces a solution of (1.3) by − absorbing the Lagrange multiplier. More recently, in [19] authors studied, in a similar way, a non-local version of the above problem with a particular nonlinearity: ( ∆)su = µu+ u p 2u in RN u− Hs(RN−) | | − and u 0 . (1.4) (cid:26) ∈ 6≡ Our aim is to prove that their result holds again when we replace ( ∆)s by − ( ∆+m2)s m2s. Unfortunately the method used in [19] does not work in − − the case of the Bessel operator (1.2). The most important difference between ( ∆)s and (1.2) is that the first has some scaling properties that the latter − does not have. In fact, it is not hard to check that one can solve the minimum problem inf ( ∆+m2)s/2u 2dx : G(u)dx = 1 | − | RN RN nZ 2 Z o in the space of radially symmetric functions. But, in this case, we are not able to absorb the Lagrange multiplier because of the lack of scaling invariance for the Bessel fractional operator. It is worth remembering that, by using different variational techniques and a Pohozaev type identity, Chang and Wang [15] proved the existence of a ground statetotheproblem(1.4)withamoregeneralnonlinearity. Inafutureworkwe will try to extend their result to the more general operator ( ∆+m2)s m2s. − − Our first result can be stated as follows Theorem 1. Let m > 0 and µ > 0. Then there exists a positive ground state solution u Hs (RN) to (1.1) which is radially symmetric. ∈ m The main difficulty of studying of problem (1.1) is the non-local character of the involved operator. To overcome this difficulty, we use a technique very commoninthe recent literature: the Caffarelli-Silvestre extension method[11]. Such method consists to write a given nonlocal problem in a local way via the Dirichlet-Neumann map: this allow us to apply known variational techniques to these kind of problems; see for instance [2, 13, 9, 36]. More precisely, for any u Hs (RN) there exists a unique v H1(RN+1,y1 2s) weakly solving ∈ m ∈ m + − div(ya v)+m2yav = 0 in RN+1 − ∇ + v(x,0) = u(x) on ∂RN+1 (cid:26) + and such that ∂v ∂v := limya (x,y) = ( ∆+m2)su(x) in H s(RN) ∂νa −y 0 ∂y − m− → where a = 1 2s ( 1,1). − ∈ − We will exploit this fact, and we will prove the existence, regularity results and qualitative properties of solutions to div(ya v)+m2yav = 0 in RN+1 − ∇ + . (1.5) ∂v = κ [m2su µu+ u p 2u] on ∂RN+1 (cid:26) ∂νa s − | | − + When we assume m sufficiently small we are able to rediscover the result obtained by [19] (and [15]). In fact, we are able to pass to the limit in (1.1) as m 0 and we find a nontrivial solution to → ( ∆)su+µu = u p 2u in RN. − − | | More precisely we obtain Theorem 2. There exists a nontrivial ground state solution u Hs(RN) to ∈ (1.4) such that u is radially symmetric. 3 2. preliminaries In this section we collect some notations e basic facts we will use later. Let s (0,1) and m > 0. ∈ Let RN+1 = (x,y) RN+1 : y > 0 . + { ∈ } We use the notation (x,y) = x 2 +y2 to denote the euclidean norm in | | | | RN+1. Let R > 0. We denote by + p B+ = (x,y) RN+1 : (x,y) < R , R { ∈ + | | } Γ0 = (x,0) ∂RN+1 : x < R R { ∈ + | | } and Γ+ = (x,y) RN+1 : (x,y) = R . R { ∈ + | | } Let q [1, ]. We denote by the Lq-norm in A RN and Lq(A) Lq(B) ∈ ∞ |·| ⊂ ||·|| the Lq-norm in B RN+1, respectively. ⊂ + We define Hs (RN) as the completion of C (RN) with respect to the norm m c∞ u 2 := ( k 2+m2)s u(k) 2dk < , | |Hms(RN) RN | | |F | ∞ Z where u(k) is the Fourier transform of u. When m = 1 we write Hs(RN) = Hs(RNF) and = . 1 |·|Hs(RN) |·|H1s(RN) We denote by H1(RN+1,y1 2s) the completion of C (RN+1) with respect m + − c∞ + to the norm v 2 := y1 2s( v 2 +m2v2)dxdy < . || ||Hm1(RN++1,y1−2s) RN+1 − |∇ | ∞ ZZ + Asbefore,whenm = 1,weusethenotationH1(RN+1,y1 2s) = H1(RN+1,y1 2s) + − 1 + − and = . ||·||H1(RN++1,y1−2s) ||·||H11(RN++1,y1−2s) It is possible to prove (see [22]) that it holds the following result: Theorem 3. There exists a trace operator Tr : H1(RN+1,y1 2s) Hs (RN) m + − → m such that (1) Tr(v) = v(x,0) for any v C (RN+1) ∈ c∞ + (2) κ Tr(v) 2 v 2 for any v H1(RN+1,y1 2s), s| |Hms(RN) ≤ || ||Hm1(RN++1,y1−2s) ∈ m + − where κs = 21−2sΓ(Γ1(−s)s). Equality holds in (2) for some function v H1(RN+1,y1 2s) if and ∈ m + − only if v is a weak solution to div(y1 2s v)+m2y1 2sv = 0 in RN+1. − − ∇ − + As a consequence we can deduce 4 Theorem4. Letu Hs (RN). Thenthere exists aunique v H1(RN+1,y1 2s) ∈ m ∈ m + − which solves div(y1 2s v)+m2y1 2sv = 0 in RN+1 − − ∇ − + . (2.1) v = u on ∂RN+1 (cid:26) + In particular ∂v limy1 2s (x,y) = κ ( ∆+m2)su(x) in H s(RN). −y 0 − ∂y s − m− → Since Hs (RN) Lq(RN) for any q 2N , we can prove that m ⊂ ≤ N 2s − Theorem 5. For any v H1(RN+1,y1 2s) and for any q 2, 2N ∈ m + − ∈ N 2s − h i C u 2 κ ( k 2 +m2)s u(k) 2dk q,s,N| |Lq(RN) ≤ s | | |F | RN Z y1 2s( v 2 +m2v2)dxdy (2.2) − ≤ |∇ | RN+1 ZZ + where u(x) = v(x,0) is the trace of v on ∂RN+1. + In the sequel we will exploit Theorem 4 and we will look for the solutions to the following problem div(ya v)+m2yav = 0 in RN+1 − ∇ + (2.3) ∂v = κ [m2su µu+ u p 2u] on ∂RN+1 (cid:26) ∂νa s − | | − + where ∂v ∂v := limya (x,y) ∂νa −y 0 ∂y → and a = 1 2s ( 1,1). − ∈ − For simplicity we will assume that κ = 1. Finally we recall the following s compact embedding (see [30]): Theorem 6. Let Xm = v H1(RN+1,y1 2s) : v is radially symmetric with respect to x . rad { ∈ m + − } Then Xm is compactly embedded in Lq(RN) for any q 2, 2N . rad ∈ N 2s − (cid:16) (cid:17) 3. Some results on elliptic problems involving ( ∆+m2)s: − Schauder estimates and maximum principles In this section we give some results about local Schauder estimates and maximum principle for problems involving the operator div(y1 2s v)+m2y1 2sv. − − − ∇ Firstly we give the following definition: 5 Definition 1. Let R > 0 and h L1(Γ0). We say that v H1(B+) is a weak ∈ R ∈ m R solution to div(y1 2s v)+m2y1 2sv = y1 2sd in B+ − − ∇ − − R ∂v = h on Γ0 (cid:26) ∂ν1−2s R if y1 2s[ v φ+m2vϕ]dxdy = hϕdx − ∇ ∇ ZZBR+ ZΓ0R for any ϕ C1(B+) such that ϕ = 0 on Γ+. ∈ R R Now, we state the following several regularity results whose proof can be found in [22]. Theorem 7. Let a,b Lq(Γ0) for some q > N and c,d Lr(B+,y1 2s) for ∈ R 2s ∈ R − some r > N+2 2s. Let v H1(B+,y1 2s) be a weak solution to 2− ∈ m R − div(y1 2s v)+m2y1 2sv = y1 2sd in B+ − − ∇ − − R . ∂v = a(x)v +b(x) on Γ0 (cid:26) ∂ν1−2s R Then v C0,α(B+ ). ∈ R/2 Theorem 8. Let a,b Ck(Γ0) for some q > N and c, d L (B+) for ∈ R 2s ∇x ∇x ∈ ∞ R some k 1. Let v H1(B+,y1 2s) be a weak solution to ≥ ∈ m R − div(y1 2s v)+m2y1 2sv = y1 2sd in B+ − − ∇ − − R . ∂v = a(x)v +b(x) on Γ0 (cid:26) ∂ν1−2s R Then v Ci,α(B+ ) for i = 1,...,k. ∈ R/2 Theorem 9. Let g C0,γ(Γ0) for some γ [0,2 2s). Let v H1(B+,y1 2s) ∈ R ∈ − ∈ m R − be a weak solution to div(y1 2s v)+m2y1 2sv = 0 in B+ − − ∇ − R . ∂v = g on Γ0 (cid:26) ∂ν1−2s R Then, for any t > 0 sufficiently small, y1 2s∂ v C0,α([0,t ) Γ ). 0 − y 0 R/8 ∈ × In the spirit of the paper [7] we can prove the following maximum principles: Theorem 10. (weak maximum principle) Let v H1(B+,y1 2s) be a weak ∈ m R − solution to div(y1 2s v)+m2y1 2sv 0 in B+ − − ∇ − ≥ R ∂v 0 on Γ0 . (3.1)  v∂ν1−20s ≥ on ΓR+  ≥ R Then v 0 in B+. ≥ R Proof. It is enough to multiply the weak formulation of above problem by v . (cid:3) − 6 Remark 1. We can deduce also the strong maximum principle: either v 0 or ≡ v > 0 in B+ Γ0. In fact, v can’t vanish at an interior point by the classical R ∪ R strong maximum principle for strictly elliptic operators. Finally the fact that v can’t vanish at a point in Γ0 follows by the Hopf principle that we will proved R below. Theorem 11. Let C = B (0) (0,1) and v H1(C ,y1 2s) C(C ) be a R R × ∈ m R − ∩ R weak solution to div(y1 2s v)+m2y1 2sv 0 in C − − R − ∇ ≤ v > 0 in C . (3.2) R  v(0,0) = 0  Then  v(0,y) limsup y1 2s < 0. − − y y 0+ → If y1 2sv C(C ) then − y R ∈ ∂v (0,0) < 0. ∂y1 2s − Proof. We consider the function w (x,y) = y 1+2s(y +Ay2)ϕ(x) A − where A > 0 is a constant that will be chosen later and ϕ(x) is the first eigenfunction of ∆ +m2 in B (0) with zero boundary condition. Then we x R/2 can conclude pro−ceeding as in the proof of Proposition 4.11 in [7]. (cid:3) Theorem 12. Let d C0,α(Γ0) and v H1(B+,y1 2s) be a weak solution to ∈ R ∈ m R − div(y1 2s v)+m2y1 2sv = 0 in B+ − − ∇ − R ∂v +d(x)v = 0 on Γ0 . (3.3)  v∂ν1−20s on BR+  ≥ R Then v > 0 in B+ Γ0 unless v 0 in B+. R ∪ R ≡ R Proof. By using Theorem 7 and Theorem 9 we know that v and y1 2sv are − y C0,α up to the boundary. So the equation ∂v + d(x)v = 0 is satisfied ∂ν1 2s pointwise on Γ0. If u is not identically 0 in B+ th−en u > 0 n B+ by the strong R R R maximum principle for the operator L(v) = div(y1 2s v) + m2y1 2sv. If − − − ∇ u(x ,0) = 0 at some point (x ,0) Γ0, then ∂v (x ,0) < 0. This gives a 0 0 ∈ R ∂y1−2s 0 (cid:3) contradiction. 7 4. existence of ground state In this section we prove the existence of a ground state to (1.1). Let us consider the following functional 1 m2s µ 1 (v) = y1 2s( v 2+m2v2)dxdy v2+ v2dx v pdx m − I 2 |∇ | − 2 2 −p | | RN+1 RN RN RN ZZ + Z Z Z (4.1) defined for any v H1(RN+1,y1 2s). ∈ m + − Firstly we note that y1 2s( v 2 +m2v2)dxdy +(µ m2s) v2dx − |∇ | − RN+1 RN ZZ + Z is equivalent to the standard norm in H1(RN+1,y1 2s) m + − v 2 = y1 2s( v 2 +m2v2)dxdy. || ||Hm1(RN++1,y1−2s) RN+1 − |∇ | ZZ + In fact, if µ m2s then we have ≥ y1 2s( v 2+m2v2)dxdy +(µ m2s) v2dx v 2 RN+1 − |∇ | − RN ≥ || ||Hm1(RN++1,y1−2s) ZZ + Z and by using (2.2) we get µ m2s y1 2s( v 2+m2v2)dxdy+(µ m2s) v2dx 1+ − v 2 . RN+1 − |∇ | − RN ≤ m2s || ||Hm1(RN++1,y1−2s) ZZ + Z (cid:16) (cid:17) Now, we suppose µ < m2s. Then y1 2s( v 2+m2v2)dxdy +(µ m2s) v2dx v 2 RN+1 − |∇ | − RN ≤ || ||Hm1(RN++1,y1−2s) ZZ + Z and by (2.2) µ m2s y1 2s( v 2+m2v2)dxdy+(µ m2s) v2dx 1+ − v 2 . RN+1 − |∇ | − RN ≥ m2s || ||Hm1(RN++1,y1−2s) ZZ + Z (cid:16) (cid:17) Thus C (m,s,µ) v 2 v 2 +(µ m2s) v 2 1 || ||Hm1(RN++1,y1−2s) ≤ || ||Hm1(RN++1,y1−2s) − | |L2(RN) C (m,s,µ) v 2 (4.2) ≤ 2 || ||Hm1(RN++1,y1−2s) and we set v 2 := y1 2s( v 2 +m2v2)dxdy +(µ m2s) v2dx. || ||e,m − |∇ | − RN+1 RN ZZ + Z 8 Now, in order to prove Theorem 1, we minimize on the following Nehari m I manifold = v H1(RN+1,y1 2s) 0 : (v) = 0 . Nm { ∈ m + − \{ } Jm } where (v) = (v)v that is Jm Im′ (v) = y1 2s( v 2 +m2v2)dxdy m2s v2 +µ v2dx v pdx m − J |∇ | − − | | RN+1 RN RN RN ZZ + Z Z Z = v 2 v pdx. || ||e,m− | | RN Z Finally we define c = min (v). m m v mI ∈N Proof. (proof of Theorem 1) We divide the proof into several steps. Step 1 The set is not empty. m Fix v H1(RNN+1,y1 2s) 0 . Then ∈ m + − \{ } t2 tp h(t) := (tv) = v 2 v p Im 2|| ||e,m− p| |Lp(RN) achieves its maximum in some τ > 0. By differentiating h with respect to t we have (τv)v = 0 and τv . Im′ ∈ Nm Step 2 Selection of an adequate minimizing sequence. We prove that there exists a radially symmetric function v such that m ∈ N (v) = c . m m I Let (v ) be a minimizing sequence for and u its trace. Let u˜ be j m m j j ⊂ N I the symmetric-decreasing rearrangement of u . It is known (see [32]) that j ( k 2 +m2)s u˜ (k) 2dk ( k 2 +m2)s u (k) 2dk (4.3) j j | | |F | ≤ | | |F | RN RN Z Z and u˜ = u for any q 1. Now, let v˜ be the unique solution j Lq(RN) j Lq(RN) j | | | | ≥ to div(y1 2s v˜ )+m2v˜ = 0 in RN+1 − − ∇ j j + . (4.4) v˜ (x,0) = u˜ (x) on ∂RN+1 (cid:26) j j + We recall that v˜ = u˜ , (4.5) || j||Hm1(RN++1,y1−2s) | j|Hms(RN) by using (2) in Theorem 3. 9 Taking into account (2.2), (4.3) and (4.5) we get y1 2s( v˜ 2 +m2v˜2)dxdy = ( k 2 +m2)s u˜ (k) 2dk − |∇ j| j | | |F j | RN+1 RN ZZ + Z ( k 2 +m2)s u (k) 2dk j ≤ | | |F | RN Z y1 2s( v 2 +m2v2)dxdy, ≤ − |∇ j| j RN+1 ZZ + so we deduce that (v˜ ) (v ) = 0 and (v˜ ) (v ). m j m j m j m j J ≤ J I ≤ I Proceeding as inthe proof ofStep 1, we canfind t > 0 such that (t v˜ ) = 0. j m j j J By using the fact that (v˜ ) 0 we can see that t 1. m j j J ≤ ≤ Since (v) = (v)v we have Jm Im′ 0 = (t v˜ ) m j j J 1 1 = (t v˜ )+ tv˜ pdx t v˜ pdx m j j j j j I p | | − 2 | | RN RN h Z Z i and by using the fact that 0 < t 1 we obtain j ≤ 1 1 (t v˜ ) = t v˜ pdx m j j j j I 2 − p | | RN (cid:16) (cid:17)Z 1 1 v˜ pdx j ≤ 2 − p | | RN (cid:16) (cid:17)Z 1 1 = v pdx j 2 − p | | RN (cid:16) (cid:17)Z = (v ). j I Then w := t v˜ is a minimizing sequence, radially symmetric with respect to j j j x, of on . By using (4.2) we get m m I N 1 1 1 1 C (m,s,µ) w 2 w 2 = (w ) < C 2 − p 1 || j||Hm1(RN++1,y1−2s) ≤ 2 − p || j||e,m Im j (cid:16) (cid:17) (cid:16) (cid:17) then, by using Theorem 6 we can assume that w ⇀ w in H1(RN+1,y1 2s) (4.6) j m + − 2N w ( ,0) w( ,0) in Lq(RN) q 2, . (4.7) j · → · ∀ ∈ N 2s (cid:16) − (cid:17) Then we deduce that (w) c and (w) 0. Now we claim that w is not m m I ≤ J ≤ identically zero. We check this, we assume by contradiction that w = 0. By 10

Ground states solutions for a non-linear equation involving a pseudo-relativistic Schr\"odinger operator

Similar Ground states solutions for a non-linear equation involving a pseudo-relativistic Schr\"odinger operator