Journal of Differential Equations

Transcription

1 J. Differential Equations 51 (011) Contents lists available at ScienceDirect Journal of Differential Equations Symmetry and uniqueness of minimizers of Hartree type equations with external Coulomb potential Vladimir Georgiev a,,1, George Venkov b a Dipartimento di Matematica, Università di Pisa, Largo Bruno Pintecorvo 5, 5617 Pisa, Italy b Faculty of pplied Mathematics and Informatics, Technical University of Sofia, Kliment Ohridski 8, 1756 Sofia, Bulgaria article info abstract rticle history: Received 4 ugust 010 Revised 0 February 011 vailable online pril 011 MSC: 35J50 35J60 35Q55 In the present article we study the radial symmetry and uniqueness of minimizers of the energy functional, corresponding to the repulsive Hartree equation in external Coulomb potential. To overcome the difficulties, resulting from the bad sign of the nonlocal term, we modify the reflection method and obtain symmetry and uniqueness results. 011 Elsevier Inc. ll rights reserved. Keywords: Hartree equations Minimizers Symmetry Variational methods Nonlinear solitary waves 1. Introduction Solitary waves associated with the Hartree type equation in external Coulomb potential are solutionsoftype χ(x)e iωt, x, t R, * Corresponding author. addresses: georgiev@dm.unipi.it (V. Georgiev), gvenkov@tu-sofia.bg (G. Venkov). 1 The author was supported by the Italian National Council of Scientific Research (project PRIN No. 008BLM8BB) entitled: nalisi nello spazio delle fasi per E.D.P /$ see front matter 011 Elsevier Inc. ll rights reserved. doi: /j.jde

2 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) where ω > 0 and χ satisfies the nonlinear elliptic equation χ(x) + χ(y) dy χ(x) χ(x) + ωχ(x) = 0. (1) x y The natural energy functional associated with this problem is (see [5,6]) E(χ) = 1 χ L ( χ ) 1 χ(x) dx, () where we shall denote ( f ) = f (x) f (y) dydx. (3) x y The corresponding minimization problem is associated with the quantity I N = min { E(χ); χ H 1, χ L = N }. (4) The existence of positive minimizers χ 0 (x), suchthat E(χ 0 ) = I N, χ 0 L = N, is established by Cazenave and Lions in [5] by the aid of the concentration compactness method. For a given ω > 0, the constrained minimization problem (4) can be compared with the unconstrained minimization problem S min ω = min{ S ω (χ); χ H 1}, where S ω (χ) is the corresponding action functional, defined by S ω (χ) = E(χ) + ω χ L. (5) There are different results on the symmetry (and uniqueness) of the minimizers. The basic result due to Gidas, Ni and Nirenberg [10] implies the radial symmetry of the minimizers associated with the semilinear elliptic equation u + f (u) = 0, provided suitable assumptions on the function f (u) are satisfied and the scalar function u is positive. s in the previous result due to Serrin [0], the proof is based on the maximum principle and the Hopf s lemma. Note, that the energy levels of the hydrogen atom are described by the eigenvalues ω k > 0ofthe eigenvalue problem e k (x) + e k(x) = ω k e k (x), e k (x) H. We have the following result.

3 4 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Lemma 1. The eigenvalues ω k are given by 1 ω k =, k = 0, 1,... 4(k + 1) For the complete proof of the above statement one can see, for example, Chapter.14 and Chapter 15.1 in []. Moreover, the energy eigenfunction, corresponding to the first eigenvalue ω 0 has the form e 0 (x) = ce /, c > 0, while all eigenfunctions e k (x), k 1, are expressed in terms of Laguerre polynomials of, havingexactlyk roots. This fact guarantees that the maximum principle is not valid for ω = ω k. Deeper analysis on the question can show that the weak maximum principle is valid if and only if ω 1. This result can be compared with the existence of action minimizers for the corresponding 4 functional S ω, obtained by Lions for 0 < ω < 1/4 (see for details [14]). Theorem. We have the properties: (a) for any ω > 0, the inequality min S ω (χ) = Sω min > χ H 1 holds; (b) if 0 < ω < 1/4, thensω min < 0; (c) if 0 < ω < 1/4, then there exists a positive function χ(x) H 1,suchthat S ω (χ) = S min ω. Our main goal of this paper is to clarify if the positive minimizers of S ω are radially symmetric and unique. The above results show that we have to consider the domain 0 < ω < 1/4, where the key tool of Gidas, Ni and Nirenberg (i.e. the maximum principle for the corresponding linear operator) meets essential difficulties. The symmetry of the energy functional (even with constraint conditions) cannot imply, in general, the radial symmetry of the minimizers. This phenomena was discovered and studied in the works [7 9] in the scalar case. Some sufficient conditions that guarantee the symmetry of minimizers have been studied by Lopes in [15], by means of the reflection method that (for the case of plane x 1 = 0) uses the functions and { u(ˆx), ˆx = ( x1, x u 1 (x) =,...,x n ), if x 1 > 0; u(x), if x 1 < 0 { u(ˆx), ˆx = ( x1, x u (x) =,...,x n ), if x 1 < 0; u(x), if x 1 > 0. If the functional to be minimized has the form E(u) = 1 u + L ( ) F u(x) dx, R n then we have the relation E(u 1 ) + E(u ) = E(u)

4 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) and this enables one to obtain the symmetry of minimizer, when F (u) is a combination of functions of type u p, p. The reflection method works effectively when u(x) is a vector-valued function and constraint conditions (as in the problem (4)) are involved too. Recently, the reflection method was generalized in [3,16,17] for very general situations and one example of possible application is the functional of type E(u) = 1 u + L R n ( ) ( ) F u(x) dx u, involving nonlocal term as in (). This Choquard type functional has the specific property E(u 1 ) + E(u ) E(u), exploiting the negative sign of the nonlocal term ( u ). n analogous result for the scalar case can be obtained by means of the Schwarz symmetrization (or spherical decreasing rearrangement [13]) u () of the non-negative u H 1. Indeed, we have the equality ( ) F u(x) dx = ( F u (x) ) dx, R n R n as well as the inequalities u L u L, ( u ) ( u ), so, we get E ( u ) E(u) and one can use the property that u is minimizer. The functional in () is a typical example, when reflection method and Schwarz symmetrization meet essential difficulty to be applied directly. The main goal of this work is to find an approach to establish the symmetry of the minimizer for functionals of Hartree type (), involving nonlocal terms with bad sign. To state this main result, we shall try first to connect the minimizers of the constraint problem (4) (associated with the energy functional E(χ)) with the minimization of the action functional S ω (χ). Similar relation for local type interactions is discussed in Chapter IX of [4]. Then, we shall establish that the minimizer of Theorem is a radially symmetric function. Theorem 3. The solution χ(x) from Theorem is a radially symmetric function for 1 16 < ω < 1 4. Remark 1. The result of Theorem III.1 in [5] treats more general case of potentials of type V (x) = K j=1 Z x x j,

5 44 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) while in our case we have V (x) = Z. Therefore, the energy functional E(χ) is rotationally invariant in our case. From Theorems and 3 onecanseethatthesolutionχ 0 (x) of (4) is radially symmetric and unique (up to a multiplication with complex number z, with z =1). s it was mentioned above the energy (and therefore the action) is a functional involving the nonlocal term with bad sign. To explain the main idea to treat this case, we recall the rotational symmetry of the energy (and action) functional. Therefore, if χ is the action minimizer from Theorem, it is sufficient to show that the solution is symmetric with respect to x 1 -plane, for any choice of the x 1 -direction. In other words, we consider ˆχ(x) = χ(ˆx), withˆx = ( x 1, x, x 3 ) and we aim to prove that χ = ˆχ. To show this, we shall consider the two terms So, our goal is to verify the inequality χ ± = χ ± ˆχ. S ω (χ + ) + S ω (χ ) S ω (χ) (6) and see that the condition χ ˆχ implies S ω (χ )>0. The form of the functional S ω suggests one, in order to verify (6), to use an appropriate version of the Clarkson inequality for the quadratic form ( f ). Namely, we can prove that the following inequality (( ) ) f + g + (( f g ) ) ( f ) + (g ) holds true (see [18,19]). Unfortunately, the usual Clarkson inequality in the form given above, is too rough to serve as a tool for proving (6). Therefore, we shall use a refined version of Clarkson inequality (see Lemma 5 below) in the form (( ) ) f + g + (( f g ) ) ( f ) + (g ) ( f )(g ). 4 The final step is to treat the uniqueness of positive minimizers of the problem S min ω = min{ S ω (χ); χ H 1}. (7) Our proof cannot follow the Lieb s uniqueness proof for the ground state solution of the Choquard equation [1]. In general, the Lieb s proof strongly depends on the specific features of the nonlocal nonlinear equation (1) and differs from the corresponding results for semilinear elliptic equation given by Kwong in [11]. Indeed, once the radial symmetry is established, one can use Pohozaev identities and reduce the nonlocal nonlinear elliptic problem (1) to an ordinary differential equation of the type u (r) + W (r)u(r) + 4π r 0 ( 1 s 1 r ) u (s) ds u(r) = ωu(r),

6 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) where W χ (r) = 1 r 4π χ (s)sds. The positive sign in front of the nonlinear term is the main obstacle to apply Sturm type argument and derive the uniqueness of positive solutions to this ordinary differential equation. However, for 1 16 < ω < 1 we can apply the approach based on the refined Clarkson inequality and using the orthogonal projection on the eigenspace of the first eigenvalue of the operator + 1/, we can establish 4 the following result. Theorem 4. Let 1 16 < ω < 1. Then, the solution χ of minimization problem (7) is unique. 4 Let us mention that the results in Theorems 3 and 4 can be compared with the results in [1], where the uniqueness of minimizers for the constrained variational problem (4) is studied. To show the relations between action minimization and (4) one has to apply the uniqueness of action minimizers or alternatively, the uniqueness of minimizers of constrained variational problem. The plan of the work is the following. In Section, by the aid of a refined version of Clarkson inequality, we prove Theorem 3, stating that the minimizers are radially symmetric. In Section 3 we establish the Pohozaev integral relation, corresponding to Eq. (1), and in Section 4 we prove uniqueness Theorem 4. Finally, in ppendix we prove for completeness the existence of positive action minimizers, stated in Theorem, while in ppendix B the connection between energy and action minimizers is discussed.. Radial symmetry of action minimizers Even in the nonlocal case, the problem that action and energy minimizers are non-negative functions, is easy to be proved. Indeed, if χ(x) H 1 is a real-valued minimizer of the functional S ω (χ) = 1 χ L ( χ ) 1 0 χ(x) dx + ω χ, L (8) then χ(x) satisfies the inequality χ L χ L, as well as the identities ( χ ) = ( χ ), χ(x) dx = χ(x) dx, so χ(x) 0 is also a minimizer of S ω. Let us define the bilinear form ( L ω (χ,ψ)= 1 ) + ω χ,ψ, L ω > 0 (9) and the corresponding quadratic form L ω (χ) = ( 1 + ω ) χ,χ L. (10)

7 46 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) The quadratic form (χ) defined in (3) generates the corresponding bilinear form (χ,ψ)= Then, the action functional S ω can be written as χ(x)ψ( y) dydx. (11) x y S ω (χ) = 1 L ω(χ) ( χ ). (1) lso,foranyfunctionχ we shall denote our x 1 -axis. It is easy to check that ˆχ(x) = χ(ˆx), where ˆx = ( x 1, x, x 3 ) for any choice of S ω (χ) = S ω ( ˆχ), L ω (χ) = L ω ( ˆχ). (13) With our next result, we shall establish Clarkson type inequalities for the forms and L ω.infact, we shall prove the lemma. Lemma 5. The following inequalities hold ( f + g L ω (( f + g ) ) + Proof. It is easy to verify the relation (( f + g Note that from equality (16) becomes (( f + g ) ) + ) ) + ) ( f g + L ω (( f g (( f g ) = L ω( f ) + L ω (g), (14) ) ) ( f ) + (g ) ( f )(g ). (15) 4 ) ) = 1 16 ( f + g + fg ) ( f + g fg ). (16) (a + b) + (a b) = (a) + (b), (( f g ) ) = 1 [ ( f + g ) + 4( fg) ] 8 = 1 8 [ ( f ) + ( g ) + ( f, g ) + 4( fg) ]. (17) On the other hand, it is easy to check that the following inequalities ( f, g ) ( f ) ( g ) (18) and ( fg) ( f ) ( g ) (19)

8 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) hold true. For the proof of (18) one can refer to Theorem 9.8 in [13], while (19) follows from Hölder inequality (see, for instance, p. 354 in [19]). Hence, (17) implies (( ) ) f + g + (( f g ) ) ( f ) + (g ) ( f )(g ), 4 which proves (15). The first relation (14) in the lemma, follows directly. The next result will play the crucial role in the present study. We shall prove the following lemma. Lemma 6. If L ω ( f ) = L ω (g) and μ, ν 0 satisfy (μ + ν ) = 1,then L ω (μ f + ν g) + L ω (μ f ν g) = L ω ( f ). (0) If ( f ) = (g ) and μ, ν 0 satisfy (μ + ν ) = 1,thenwehave ( (μ f + ν g) ) + ( (μ f ν g) ) ( f ). (1) Proof. Setting μ 1 = μ, ν 1 = ν, we apply (14) with f, g replaced by μ 1 f and ν 1 g respectively. Thus, we get ( ) ( ) μ1 f + ν 1 g μ1 f ν 1 g L ω + L ω = μ 1 L ω( f ) + ν 1 L ω(g). () From L ω ( f ) = L ω (g) and μ 1 + ν 1 =, we complete the proof of (0). Similarly, applying (15) and the assumption ( f ) = (g ),wefind or, equivalently (( μ1 f + ν 1 g ) ) + (( μ1 f ν 1 g ) ) μ4 1 + ν μ 1 ν ( 1 f ) 8 ( (μ f + ν g) ) + ( (μ f ν g) ) ( μ 4 + ν 4 + 6μ ν ) ( f ). (3) Consider now the homogeneous quartic polynomial ( μ 4 + ν 4 + 6μ ν ) (4) on the circle μ + ν = 1. Substituting ν = 1 μ, we obtain the following estimate ( μ 4 + ν 4 + 6μ ν ) = (( μ + ν ) + 4μ ν ) = 1 + 4μ 8μ 4 = 1 (1 4μ ) 1. (5) Then, from (3) and (5) follows the proof of the lemma.

9 48 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Turning back to the minimization problem of the action functional S ω, we observe the following fact. If χ(x) is a minimizer of the problem min S ω (χ), (6) χ H 1 then ˆχ(x) and ˆχ(x) are also minimizers of S ω (χ). Moreover, we have the property. Lemma 7. ssume that χ(x) is a minimizer of the problem (6) and one of the following alternatives: 1. L ω (χ ˆχ) 0;. L ω (χ + ˆχ) 0 holds. Then χ = ˆχ. Proof. For simplicity, we shall consider the first case only. Suppose χ ˆχ, then from (14) we have ( ) χ + ˆχ L ω ( ) χ ˆχ + L ω = L ω (χ), (7) implying ( ) χ + ˆχ L ω L ω (χ). (8) pplying now (15), we obtain (( χ + ˆχ ) ) + (( χ ˆχ which, together with the assumption χ ˆχ gives that (( χ + ˆχ ) ) (χ ) + ( ˆχ ) + 3 (χ )( ˆχ ) 8 4 (χ ) + ( ˆχ ) = ( χ ), (9) ) ) < ( χ ). (30) Thus, from (8), (30) and the definition (1) it follows ( ) χ + ˆχ S ω < S ω (χ), (31) which contradicts to the assumption that χ is a minimizer. This proves the lemma. Lemma 8. Let us assume that g e 0 in L.Then ( L ω (g) ω 1 ) g. 16 L

10 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Proof. Note that g e 0 in L implies g = k 1 c k e k + h, where h is in the absolutely continuous space of the self-adjoint operator + 1 in L, while e k are eigenvectors of the same operator in {g L ; g e 0 } with eigenvalues ω k 1/16. On the absolutely continuous space the operator has spectrum on (, 0) and it is non-positive, so ( + 1 ) h, h 0. and Hence, we have ( + 1 ) g, g c k ω k 1 ( ) ck = g L ( L ω (g) = + 1 ) ( g, g + ω g ω 1 ) g. L 16 L This completes the proof of the lemma. Now, we are ready to prove the radial symmetry of the action minimizer, stated in Theorem 3. Proof of Theorem 3. Taking into account Lemma 7, we shall take a minimizer χ(x) 0ofS ω and shall show that the condition implies that χ = ˆχ or Let 1 16 < ω < 1 4, L ω (χ ˆχ)>0. (3) χ(x) = e 0 (x) + f (x), where e 0 (x) = ce /, c > 0 is the eigenvector corresponding to the first eigenvalue of the operator + 1/, while f, e 0 L = 0. Since e 0 is a radial function, we have ê 0 = e 0,so pplying now Lemma 8, we find χ ˆχ = f ˆf = g, g, e 0 L = 0, g 0. ( L ω (χ ˆχ) = L ω (g) ω 1 ) g > 0, 16 L since ω > 1/16 and g 0. Hence, (3) is fulfilled and the proof of the theorem is complete.

11 430 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Identities for minimizers In this part we shall establish simple identities for (1). More precisely, we shall prove the following Lemma 9. If χ H 1 ( ) and satisfies (1) in H 1 ( ), then the following identity holds χ + ω χ = L L χ(x) dx ( χ ). (33) Proof. To prove (33) we multiply Eq. (1) by χ, take the real part and integrate over. This completes the proof of the lemma. The above lemma is useful to treat the uniqueness of the minimizers (modulo multiplication by complex constant z with z =1). Indeed, let χ 1 and χ be minimizers of the problem S min ω = min{ S ω (χ); χ H 1( )}. (34) Since S ω (χ) = 1 χ L + ω χ L 1 χ(x) dx ( χ ), we can apply the identity (33) of Lemma 9. In this way we find S ω (χ) = 1 4 ( χ ) (35) and ( χ 1 ) = ( χ ), L ω (χ 1 ) = L ω (χ ), (36) where L ω (χ) is defined according to (10). 4. Uniqueness of minimizers In this section we shall prove the uniqueness result of Theorem 4. The classical approach for proving the uniqueness of minimizers is to reduce the initial nonlinear equation to an ordinary differential equation, using the radial symmetry. Uniqueness of positive ground state solutions for nonlinear Schrödinger equation on R n with local nonlinearities of the form u p u for 0 < p < 4, is a wellknown fact, due to Kwong [11]. The proof in this case relies on Sturm comparison theorems, but it n cannot be applied directly to nonlocal equations, such as (1). For the attractive Choquard equation, Lieb in [1] prove uniqueness of energy minimizer by using Newton s theorem for radial function f (x) = f (), that is f ( y ) dy = x y n f ( y ) dy. (37) max{, y } The repulsive sign of the Hartree term in (1) is again the main obstacle for applying directly the standard technique.

12 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Proof of Theorem 4. Let χ 1 and χ be non-negative minimizers of the problem S min ω = min{ S ω (χ); χ H 1}. Since they are radial functions, one can rewrite the elliptic equation (1), using Newton s theorem (37), as an ordinary differential equation of the form where χ (r) r χ (r) χ(r) r + 4π The above equation can be rewritten in the form χ (r) r χ (r) W (r)χ(r) + 4π r 0 0 W (r) = 1 r 4π χ (s)sds. χ (s)s ds χ(r) + ωχ(r) = 0. (38) max{r, s} ( 1 χ (s) r 1 ) s dsχ(r) + ωχ(r) = 0, s 0 If we set u(r) = rχ(r), then from the identity the last equation becomes u (r) + W (r)u(r) 4π χ (r) + χ (r) = u (r), r r r 0 ( 1 r 1 s ) u (s) ds u(r) = ωu(r). (39) This observation shows that the assumption χ(x) is a non-negative minimizer implies u(r) >0for r > 0. Hence χ 1 (x) and χ (x) are positive functions. Our goal is to use the projection of χ 1 and χ on the one-dimensional eigenspace E 0 = { αe /, α (, ) } corresponding to the first eigenvalue ω 0 = 1/4 of the operator + 1/. First,wehavetoobserve that χ 1 is not orthogonal to E 0.Indeed,ifχ 1 E 0,thenLemma8implies ( L ω (χ 1 ) ω 1 ) χ 1 > L The relation (1) guarantees now S ω (χ 1 )>0 and this contradicts the relation (35). The contradiction shows that χ 1 (and also χ ) is not orthogonal to E 0. Let χ 1 = μ 1 αe / + f 1, χ = μ αe / + f,

13 43 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) where αe / E 0, withα > 0 and f 1, f E 0. Note that μ 1, μ > 0, since χ 1, χ and e 0 are positive functions. We can choose α > 0, such that ( μ 1 + μ ) = 1, (40) used as assumption in Lemma 6. The other assumption ( χ 1 ) = ( χ ), L ω (χ 1 ) = L ω (χ ), is already established in (36). pplying Lemma 6, we find the identity L ω (μ χ 1 + μ 1 χ ) + L ω (μ χ 1 μ 1 χ ) = L ω (χ 1 ), as well as the inequality ( (μ χ 1 + μ 1 χ ) ) + ( (μ χ 1 μ 1 χ ) ) ( χ 1 ). Then, we have the relation μ χ 1 μ 1 χ = μ f 1 μ 1 f = g E 0. If g = 0, then χ 1 = μ 1 χ /μ and one can use the ODE (39) and the corresponding integral identity (33), to show that χ 1 = χ.ifg 0, then one can apply Lemma 8 and find ( ) L ω (μ χ 1 μ 1 χ ) ω 1 16 g > 0. L Hence, S(μ χ 1 + μ 1 χ ) = 1 L ω(μ χ 1 + μ 1 χ ) ( (μ χ 1 + μ 1 χ ) ) < S ω (χ 1 ) and this is a contradiction. The contradiction shows that χ 1 = χ and this completes the proof of Theorem 4. cknowledgments The authors are grateful to Louis Jeanjean for important discussions and remarks on symmetry of minimizers as well as to the referee for pointing out a gap in the proof of Theorem 3. ppendix. Existence of action minimizers The existence of action minimizers for Hartree type equation is already established in [14]. For completeness, we shall sketch the proof. To show the boundedness from below of S ω, we shall prove the following inequalities involving homogeneous Sobolev norms f Ḣ s ( ) = ( ) s/ f L ( ), s > 3/.

14 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Lemma 10. For any p 1 [3, 6] and p [, 3] we have the estimates where ( 1 ( 1 χ(x) p1 dx) 1/p1 C χ θ 1 Ḣ 1 χ (1 θ 1)/ Ḣ 1, (.1) χ(x) p dx) 1/p C χ θ L χ θ 3 θ 1 = p 1, θ = 4(3 p ) p, θ 3 = p p. Ḣ1 χ (1 θ θ 3 )/, (.) Ḣ 1 Remark. The assumptions p 1 [3, 6] and p [, 3] guarantee that all parameters θ 1, θ, θ 3, θ + θ 3 are in the interval [0, 1]. Remark 3. The relation f = ( ) 1 f, f = 1 Ḣ 1 L 4π f (x) f (y) dydx x y implies χ Ḣ 1 = 1 4π ( χ ). Proof of Lemma 10. For p 1 = 6 the inequality (.1) becomes ( χ(x) 6 dx) 1/6 C χ Ḣ1 1 and this is the standard Sobolev embedding. For p 1 = 3 we have to verify the following estimate ( χ(x) 3 dx) 1/3 C χ 1/3 Ḣ 1 χ 1/3 Ḣ 1. (.3) This inequality follows from f (x)g(x) dx f Ḣ 1 g Ḣ 1 with f (x) = χ(x), g(x) = χ(x) = χ (x) and the observation that χ Ḣ1 = χ Ḣ1. Interpolation between p 1 = 6 and p 1 = 3 proves (.1). The inequality (.) for p = 3 follows from (.3).

15 434 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) For p = (.) reduces to the simple inequality ( χ(x) dx) 1/ C χ L. 1 n interpolation argument implies (.) and completes the proof of the lemma. fter this lemma we can show that the action functional is bounded from below. Lemma 11. For any ω > 0 the inequality min S ω (χ) = Sω min > χ H 1 holds. For 0 < ω < 1/4 we have S min ω < 0. Proof. The only negative term in S ω is 1 χ(x) dx. Decomposing the integration domain into 1 and > 1 we apply Hölder inequality and obtain χ(x) ( dx C 1 χ(x) /p1 ( dx) p1 + C >1 χ(x) p dx) /p, where p 1 > 3 > p. pplying Lemma 10 as well as the Young inequality X θ 1 Y θ Z θ 3 ε X + εy + C ε Z, with θ j (0, 1), θ 1 + θ + θ 3 = 1, we get χ(x) dx ε χ + ε χ + L L C ε ( χ ). This estimate implies S ω (χ) 1 ε χ + ω ε χ + 1 L L 4 ( χ ) C ε ( χ ). Choosing ε > 0 so small that ε < min(1, ω), wefind S ω (χ) 1 4 ( χ ) C ε ( χ ) C ε.

16 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) Then Since To finish the proof we take χ δ (x) = δe /, such that ( + 1 ) χ δ = 1 χ δ. 4 S ω (χ δ ) = (ω 1/4) χ δ + ( ) L χδ /. χ δ L = C 0 δ, ( ) ( χ ) δ / = O δ 4, the condition ω (0, 1/4) implies S ω (χ δ )<0 and this completes the proof. Proof of Theorem. Take a minimizing sequence χ k H 1, so that lim S ω(χ k ) = Sω min < 0. (.4) k The argument of the proof of Lemma 11 guarantees that there exists a constant C > 0, so that χ k H 1 C. (.5) One can find χ (x) H 1 so that (after taking a subsequence) χ k tends weakly in H 1 to χ.usingthe inequality χ(x) dx C R, >R and the compactness of the embedding L p ( < R) H 1 ( < R), when p < 6, we see that (choosing a suitable subsequence) lim k Then we introduce ϕ k, ϕ so that χ k (x) dx = χ (x) dx. (.6) ϕ k = 4πχ k (x), ϕ = 4πχ (x). One can show that ϕ k tends weakly to ϕ in Ḣ 1. We have also the identities ( ) χ k = ϕ k (x)χ 1 k (x) dx = 4π ϕ k L and ( ) χ = ϕ (x)χ 1 (x) dx = 4π ϕ L so we obtain

17 436 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) S ω (χ k ) = 1 χ k L + ω χ k L ( χ k Using (.4) and (.6), we get lim S ω(χ k ) + 1 k ) 1 = 1 χ k + ω χ L k + 1 L 16π ϕ k 1 L χ k (x) dx χ k (x) dx. χ k (x) 1 dx = lim χ k + ω χ k L k + 1 L 16π ϕ k L = Sω min + 1 χ (x) dx. It is well known that for any sequence f k in a Hilbert space H tending weakly (in H) to f H, one has lim inf k f k H f H (.7) and lim f k f H = 0 lim f k H = f H. k k (.8) From (.7) we have Sω min + 1 χ (x) dx χ L + ω χ L π ϕ L and a strict inequality is impossible since this will contradicts the definition of Sω min.hence 1 lim χ k + ω χ k L k + 1 L 16π ϕ k = 1 χ L + ω χ L + 1 L 16π ϕ L and applying (.8) we conclude that lim χ k χ H 1 = 0. k This completes the proof of the theorem. ppendix B. Connection between the action and energy minimization problems Consider the minimization problem S min ω = min{ S ω (χ); χ H 1}, (B.1) associated with the action functional S ω (χ) and the Lions Cazenave minimization problem I N = min { E(χ); χ H 1, χ L = N }. (B.)

18 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) s we have seen before, for every ω (1/16, 1/4), there exists (at most one) solution χ ω H 1 ( ), which is positive and radially symmetric, and such that S ω (χ ω ) = S min ω. (B.3) Let us denote N(ω) = χ ω L. (B.4) The above definition of the function N(ω) poses the question if Sω min = I N(ω) + ω N(ω). For completeness, in this section we shall prove the following lemma. Lemma 1. If χ 1 is a solution of (B.) with N = N(ω), thenχ 1 satisfies the equation χ 1 (x) + χ 1 (y) dy χ 1 (x) χ 1(x) x y + ωχ 1 (x) = 0 (B.5) and S ω (χ 1 ) = min { S ω (χ); χ H 1}. Proof. To prove the lemma we shall follow the idea of the proof of Corollary in [4]. It is obvious, that the relation S ω (χ 1 ) = E(χ 1 ) + ω N(ω), guarantees that χ 1 is a minimizer of the problem min S ω (χ). χ L =N(ω) Since, S ω (χ 1 ) = min S ω (χ) min S ω (χ) = S ω (χ ω ), χ L =N(ω) we can use (B.4) and see that this inequality becomes equality, so S ω (χ 1 ) = min S ω (χ) = min S ω (χ) = S ω (χ ω ). χ L =N(ω) Now, the uniqueness result of Theorem 4 implies χ 1 = χ ω and completes the proof.

19 438 V. Georgiev, G. Venkov / J. Differential Equations 51 (011) References [1] N. Bazley, R. Seydel, Existence and bounds for critical energies of the Hartree equation, Chem. Phys. Lett. 4 (1974) [] D. Bohm, Quantum Theory, Prentice Hall, New York, [3] J. Byeon, L. Jeanjean, M. Mariş, Symmetry and monotonicity of least energy solutions, Calc. Var. Partial Differential Equations 36 (009) [4] T. Cazenave, Semilinear Schrödinger Equations, Courant Lect. Notes, vol. 10, Courant Institute of Mathematical Sciences, merican Mathematical Society, Providence, 003. [5] T. Cazenave, P.L. Lions, Orbital stability of standing waves for some nonlinear Schrödinger equations, Comm. Math. Phys. 85 (198) [6] G. Coclite, V. Georgiev, Solitary waves for Maxwell Schrödinger equations, Electron. J. Differential Equations 94 (004) [7] V. Coti-Zelati, M. Esteban, Symmetry breaking and multiple solutions for a Neumann problem in an exterior domain, Proc. Roy. Soc. Edinburgh Sect. 116 (1990) [8] M. Esteban, Nonsymmetric ground states of symmetric variational problems, Comm. Pure ppl. Math. 44 (1991) [9] M. Esteban, Rupture de symétrie pour des problémes de Neumann surlinéaires dans les ouverts extérieur, Note C.R..S. Sér. I 308 (1989) [10] B. Gidas, W. Ni, L. Nirenberg, Symmetry of positive solutions of nonlinear elliptic equations in R N, Math. nal. ppl. 7 (1981) [11] M. Kwong, Uniqueness of positive solutions of u u + u p = 0inR n, rch. Ration. Mech. nal. 105 (1989) [1] E. Lieb, Existence and uniqueness of the minimizing solution of Choquard s nonlinear equation, Stud. ppl. Math. 57 (1977) [13] E. Lieb, M. Loss, nalysis, Grad. Stud. Math., vol. 14, merican Mathematical Society, Providence, [14] P.L. Lions, Some remarks on Hartree equation, Nonlinear nal. 5 (1981) [15] O. Lopes, Radial and non radial minimizers for some radially symmetric functionals, Electron. J. Differential Equations 3 (1996) [16] O. Lopes, M. Mariş, Symmetry of minimizers for some nonlocal variational problems, J. Funct. nal. 54 (008) [17] M. Mariş, On symmetry of minimizers, rch. Ration. Mech. nal. 19 (009) [18] D. Ruiz, The Schrödinger Poisson equation under the effect of a nonlinear local term, J. Funct. nal. 37 (006) [19] D. Ruiz, On the Schrödinger Poisson Slater system: behavior of minimizers, radial and nonradial cases, rch. Ration. Mech. nal. 198 (010) [0] J. Serrin, symmetry problem in potential theory, rch. Ration. Mech. nal. 43 (1971)