A semilinear Schrödinger equation with magnetic field

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1 A semilinear Schrödinger equation with magnetic field Andrzej Szulkin Department of Mathematics, Stockholm University Stockholm, Sweden 1 Introduction In this note we describe some recent results by Gianni Arioli and the author [2] on the existence of nontrivial solutions of the semilinear stationary Schrödinger equation in a magnetic field (1) ( i + A) 2 u + V (x)u = g(x, u )u, x R. Here u : R C ( 2), V : R R is the scalar (or electric) potential and A = (A 1,..., A ) : R R is the vector (or magnetic) potential. Let B := curl A. For = 3 this is the usual curl operator and for general, B = (B jk ), 1 j, k, where B jk := j A k k A j. B represents the external magnetic field whose source is A. In what follows we always assume A L 2 loc (R, R ), V L 1 loc (R ) and V is bounded below. Denote A u = ( + ia) u and let := 2/( 2). Appropriate spaces in which we shall consider the problem (1) are H 1 A(R ) := {u L 2 (R ) : A u L 2 (R )} and, for 3, D 1,2 A (R ) := {u L 2 (R ) : A u L 2 (R )}. Both HA 1 (R ) and D 1,2 A (R ) are Hilbert spaces with inner product respectively A u A v + uv and A u A v (the bar denotes complex conjugation). By Section 2 of [6] and Theorem 7.22 of [10], C0 (R ) is dense in HA 1 (R ) and D 1,2 A (R ) (in [6] D 1,2 A (R ) has in fact been defined as the closure of C0 (R ) with respect to the norm corresponding to the inner product above). Supported in part by the Swedish Research Council. 1

2 Let g(x, u ) c(1 + u 2 2 ), F (x, u ) := u 0 g(x, s)s ds and consider the functional (2) J(u) := 1 A u 2 + V u 2 F (x, u ). 2 Suppose u HA 1 (R ). By the diamagnetic inequality u (x) A u(x) a.e. in R (see e.g. Theorem 7.21 in [10]), u H 1 (R ) and therefore u L p (R ) for any p [2, ] (for any p [2, + ) if = 2). It follows that whenever V L β (R ), where β /2 (β > 1 if = 2), then J C 1 (HA 1 (R ), R) and critical points of J are weak solutions of (1). We note also that J(e iθ u) = J(u) for any θ R, hence J is S 1 -invariant. Suppose A, à L α loc (R, R ) for some α [1, + ) and curl A = B = curl à (in the 1,α sense of distributions). Then à A = ϕ for some ϕ Wloc (R ), see Lemma 1.1 of [9]. It is easy to see that if ũ = e iϕ u, then Ãũ = e iϕ A u and hence Ãũ 2 = A u 2. It follows that if u H 1 A (R ), then ũ H 1 à (R ) and if u satisfies (1), then so does ũ with A replaced by Ã. The above properties are called the gauge invariance and they reflect the fact that the magnetic field B and not the particular choice of the vector potential A should be essential. The transformation u ũ is called the change of gauge. ote that there is a trivial change of gauge u ũ = e iθ u, where θ is a constant. Then à = A and it is a consequence of this property that J is S 1 -invariant. While there is a vast literature concerning the Schrödinger equation (1) with A = 0, to the best of our knowledge there are very few papers dealing with the magnetic case [6, 11, 12]. Also in [5, 8] the magnetic case has been considered, but from a very different point of view (semiclassical limits and related concentration phenomena). Denote A := ( i +A) 2. Since V is bounded below, so is the spectrum σ( A +V ) in L 2 (R ). Suppose F 0. If 0 / σ( A + V ), then either σ( A + V ) (0, + ) and the functional has a mountain pass geometry, or σ( A + V ) (, 0) and then J has a geometry of linking type. In what follows p will denote the usual L p -norm in R, σ( A + V ) the spectrum of A + V in L 2 (R ) and B(a, r) the open ball centered at a and having radius r. 2 The results First we consider a minimization problem in R, 3. Let A u (3) S 2 + V u 2 := inf u 2. u D 1,2 A ( )\{0} Our first result is a slight generalization of Theorem 3.7 in [6]. Theorem 1 If V 0, V L /2 loc (R ) and A L loc (R, R ), then the infimum in (3) is attained if and only if V 0 and B = curl A 0. 2

3 ote that since V is only in L /2 loc (R ), V u 2 need not be finite for all u D 1,2 A (R ). However, it is finite for u C0 (R ) and therefore the minimization problem (3) makes sense. ote also that if S is attained at some u D 1,2 A (R ), then u is a solution of (1) with g(x, u ) = S u 2 2 ; hence v = S 1/(2 2) u solves (1) with g(x, u ) = u 2 2. Theorem 2 Suppose 4, V L 1 loc (R ), V := max{ V, 0} L /2 (R ), A L 2 loc (R, R ) and σ( A + V ) (0, + ). If there exists x R such that V (x) c < 0 in a neighborhood of x and A is continuous at x, then the infimum of (3) is attained for some u H 1 A (R ) \ {0}. Since σ( A +V ) (0, + ), it follows that if S is attained, then it is positive. Indeed, S 0 and if S = 0 is attained at some u 0, then u is an eigenvalue of A + V which is impossible. In the next theorems we shall need the following assumptions: A1 V L (R ), g C(R R +, R) and A L 2 loc (R, R ). A2 V, g and B = curl A are 1-periodic in x j, 1 j. A3 g(x, 0) = 0. A4 There are constants C > 0 and p (2, ) (p > 2 if = 2) such that g(x, u ) C(1 + u p 2 ) for all x, u. A5 There is a constant µ > 2 such that 0 < µf (x, u ) g(x, u ) u 2 whenever u 0. A6 There are constants C, ε 0 > 0 such that g(x, u + v )(u + v) g(x, u )u C v (1 + u p 1 ) whenever v ε 0, where p is as in (A4). ote that in view of the definition of F, (A5) is the usual superlinearity condition. Since B jk = j A k k A j in the sense of distributions, the periodicity of B should be interpreted as B( ) B( + e j ) being the zero distribution for any element e j of the standard basis in R. It is also clear that according to (A3), (1) has the trivial solution u = 0. Theorem 3 If 0 / σ( A + V ) and conditions (A1)-(A5) are satisfied, then equation (1) has a nontrivial solution u H 1 A (R ). A corresponding result is well-known for the Schrödinger equation with A = 0 (see e.g. [7, 14] and the references there). Finally we shall exploit the S 1 -invariance of J in order to show the existence of infinitely many solutions of (1). Theorem 4 If 0 / σ( A + V ) and conditions (A1)-(A6) are satisfied, then equation (1) has infinitely many geometrically distinct solutions. 3

4 By geometrically distinct we mean such u, v that v e iθ u for any θ R and v T z u for any z Z, where T z is a certain operator corresponding to the translation by elements of Z in the nonmagnetic case. A more precise definition will be given in the next section. The above result should be compared to the one contained in [3, 7], where A was equal to 0. We would also like to mention that each of the following conditions is sufficient for σ( A + V ) to be contained in (0, + ) (see [2]): P1 A L loc (R ), V L 1 loc (R ), where 3, and there exists a bounded set Ω R and a constant c > 0 such that V (x) c for all x / Ω and inf x Ω V (x) > Sµ(Ω) 2/. P2 A W 1, loc (R, R ), V L 1 loc (R ) and there exists a constant c > 0 such that inf x V (x) > c and either j A k k A j c a.e. in R or j A k k A j c a.e. in R for some j, k {1,..., }. Here (4) S := inf u D 1,2 ( )\{0} u 2 + V u 2 u 2 is the Sobolev constant for the embedding D 1,2 (R ) L 2 (R ) and µ(ω) is the measure of Ω. 3 Outline of proofs An important role in the proofs is played by the following two results: Proposition 1 (Diamagnetic inequality [10, Theorem 7.21]) If u H 1 A (R ) (resp. u D 1,2 A (R )), then u H 1 (R ) (resp. u D 1,2 (R )) and u (x) u(x) + ia(x)u(x) for a.e. x R. Proof (5) Since A is real-valued, ( u (x) = Re u ū ) = u ( Re ( u + iau) ū ) u + iau. u See [10] for more details. Proposition 2 Let A L 2 loc (R, R ) and suppose u n u in D 1,2 A (R ). Then, up to a subsequence, u n u a.e. in R and u n u in L q loc (R ) for any q [2, ). The same conclusion holds if u n u in H 1 A (R ). Proof By the diamagnetic inequality the injection D 1,2 A (R ) L 2 (R ) is continuous, hence u n u in L 2 (R ). Moreover, u n u is bounded in D 1,2 (R ). So passing to a subsequence, u n u a.e. and u n u 0 in D 1,2 (R ). It follows from the Rellich- Kondrachov theorem that u n u in L q loc (R ). The second part of the lemma is proved similarly. 4

5 Below we shall prove Theorem 1 and briefly sketch the proofs of the other theorems. Proof of Theorem 1 ecessary condition. We first show that S = S. By the Sobolev and the diamagnetic inequalities, u 2 + V u 2 A u 2 + V u 2 S u 2 u 2, hence S S. Let ( (6) U ε (x) = (( 2)) ( 2)/4 ε ε 2 + x 2 ) ( 2)/2 and u ε (x) = ψ(x)u ε (x), where ψ C 0 (R, [0, 1]), ψ = 1 on B(0, 1/2) and ψ = 0 on R \ B(0, 1). Then (7) (ψu ε ) 2 2 u ε 2 2 = S /2 + O(ε 2 ) and u ε 2 = S/2 + O(ε ) (see e.g. [14], p. 35). Since u ε is bounded in L 2 (R ) and u ε 0 a.e., u ε 0 in L 2 (R ) as ε 0. Therefore V u ε 2 0 and Au ε 2 0 as ε 0 (recall V L /2 loc (R ) and A L loc (R, R )). Since u ε is real-valued, A u ε 2 + V u ε 2 u ε 2 + Au ε 2 + V u ε 2 u ε 2 = u ε 2 S as ε 0 and S S. ow assume that u is a minimizer normalized by u = 1. Then S = A u 2 + V u 2 A u 2 u 2 S and it follows that u(x) = U ε (x a)/ U ε for some a R (that the minimizer in (4) is unique up to translation and dilation may be seen e.g. from the proof of Theorem 1.42 in [14]). In particular, u > 0 for all x and therefore V 0. Moreover, the inequality of Proposition 1 must be an equality a.e. So by (5), the imaginary part of ( u + iau)ū must be zero which is equivalent to A = Im ( u/u). An easy computation shows that curl ( u/u) = 0. Sufficient condition. Assume V 0 and curl A = 0. Then A = ϕ for some ϕ W 1, loc (R ) according to [9] and it is easy to verify that u = U ε e iϕ is a minimizer for (3) for any ε > 0. Proof of Theorem 2 Assume without loss of generality that x = 0. Let θ(x) := j=1 A j(0)x j. Then (A + θ)(0) = 0 and by continuity (A + θ)(x) 2 c < c for all x < δ provided δ is small enough. Choosing a smaller δ if necessary we may also assume V (x) c whenever x < δ. Let U ε be as in (6) and let v ε (x) := ψ(x)u ε (x)e iθ(x), where ψ C0 (R, [0, 1]), ψ(x) = 1 in B(0, δ/2) and ψ(x) = 0 in R \ B(0, δ). By (7), A v ε 2 + V v ε 2 (ψu ε ) 2 + ψ 2 Uε 2 θ + A 2 cψ 2 Uε 2 S 2 + (c c) Uε 2 + O(ε 2 ) B(0,δ/2) 5

6 and v ε 2 = S( 2)/2 + O(ε ). Since B(0,δ/2) U 2 ε { Cε 2 log ε if = 4 Cε 2 if 5 for some C > 0 and all small ε > 0 (cf. e.g. [14], p. 35), an easy computation shows that A u S 2 + V u 2 = u 2 < S. inf u DA 1 ( )\{0} Having this, a usual argument based on the concentration-compactness lemma [14, Lemma 1.40] shows that if {u n } is a minimizing sequence such that u n = 1, then u n u in L 2 (R ) and in D 1,2 A (R ), possibly after passing to a subsequence. Hence u is a minimizer for (3). Moreover, since σ( A + V ) (0, + ), A u 2 + V u 2 ε u 2 2 for some ε > 0; therefore u D 1,2 A (R ) L 2 (R ), i.e. u HA 1 (R ). Finally we would like to point out that Lemma 1.40 in [14] must be adapted to the D 1,2 A (R )-setting. However, this is easily done by following the proof in [14] and employing Propositions 1 and 2 above. Proof of Theorem 3 Let E := H 1 A (R ) and let J be as in (2). Then J C 1 (E, R ) and J (u) = 0 if and only if u is a solution of (1). If σ( A + V ) (0, + ), then the quadratic form Q(u) := A u 2 + V u 2 is positive definite on E, otherwise E can be decomposed into the direct sum of two subspaces, E + and E, invariant with respect to A + V and such that Q is positive definite on E + and negative definite on E (cf. [13], Section 8). In the first case the functional J has the mountain pass geometry, and in the second one it has a geometry of linking type as described e.g. in [7]. Hence there exists a Palais-Smale sequence {u n } at some level c > 0 (cf. [7], Theorem 3.4 and [14], Theorems 2.9, 2.10, 6.10). Moreover, {u n } is bounded, so u n u after passing to a subsequence. By Lemma 1.7 in [7], either u n 0 in E (up to a subsequence) which is impossible because J(u n ) c > 0, or there exists a sequence {z n } in Z and r, η > 0 such that B(z n,r) u n(x) 2 η. We shall now use (A2) in order to construct a Palais-Smale sequence {v n } such that v n v 0. Since J (v n ) J (v), v is a critical point of J. By (A2), curl A(x + z) curl A(x) = B(x + z) B(x) = 0 for all z Z. It follows from Lemma 1.1 in [9] that (8) A(x + z) A(x) = ϕ z (x) for some ϕ z Hloc 1 (R ). Let (T z u)(x) := u(x+z)e iϕz(x). An explicit computation using (8) shows that T z is an isometry on E, J(T z u) = J(u) and J (T z u) = T z J (u) for each z Z. Denote v n := T zn u n. Then J(v n ) c and J (v n ) 0. Passing to a subsequence, v n v in E and, according to Proposition 2, v n v in L 2 loc (R ). Hence v is a critival point, and v 0 because v n (x) 2 = u n (x) 2 η. B(0,r) B(z n,r) 6

7 Proof of Theorem 4 Let J and E, E +, E be as in the preceding proof (E = {0} if σ( A + V ) (0, + )). Denote S θ u = e iθ u, θ S 1 = R/2πZ. Then J(S θ u) = J(u) and (as we already have seen) J(T z u) = J(u) for all θ S 1, z Z, u E. ote that S θ T z = T z S θ. Let (u) := {S θ T z u : θ S 1, z Z }. O S 1 Two solutions u, v of (1) are called geometrically distinct if they belong to different orbits, i.e. if O S 1 (u) O S 1 (v). The proof of Theorem 4 is a straightforward adaptation of that in [1]. Suppose (1) has only finitely many geometrically distinct solutions. Denote the set of critical points of J by K J, let C be a set consisting of arbitrarily chosen representatives of the orbits O S 1 (u), u K J \ {0} and K := O S 1(C) = {S θ u : θ S 1, u C}. Then K is a compact set, (9) K J \ {0} = O (K) and if F := P E +(K), where P E + is the orthogonal projection on E +, then (10) T z1 F T z2 F = whenever z 1, z 2 Z, z 1 z 2. These conditions correspond to (9) and (10) in [1]. Clearly, J is even and K, F are symmetric (i.e. K = K, F = F). From now on we consider J as an even functional and disregard the S 1 -invariance. Let U δ := E {u + E + : d(u +, T z F) < δ}, z where d(u, A) denotes the distance from u to the set A and let H be the class of mappings f : E E such that f is a homeomorphism, f( u) = f(u) for all u and f(j c ) J c for all c 1 (J c := {u E : J(u) c}). One can show that if J satisfies (9), (10) and c inf KJ \{0} J, then there exists a mapping f H such that f(j c+ε \ U δ ) J c ε provided δ and ε are small enough. This is a variant of the deformation lemma which will be needed in the minimax argument below. Fix a small ρ > 0. For a closed and symmetric A, define γ (A) = min f H γ(f(a) B(0, ρ) E+ ), where γ denotes the Krasnoselskii genus (γ is a variant of Benci s pseudoindex [4]). Let d k := inf sup J(u). γ (A) k If ρ is small enough, then J B(0,ρ) E + d > 0 and it follows that d k d. Moreover, it can be shown that there are sets of arbitrarily large pseudoindex, so d k is defined for all k 1. Since d 0 := sup Uδ J <, it follows from the deformation lemma that d k d 0 for all k. Hence d k d d 0. Since γ(s 1 ) = 2, it is easy to see that γ(f) = 2 and γ(ūδ) = 2 provided δ is small enough. Using the deformation lemma once more we obtain u A k γ (J d k+ε ) γ (J d k+ε \ U δ ) + γ(ūδ) γ (J d k ε ) + 2. Therefore γ (J dk ε ) k 2, so d k ε d k 2 and d ε d, a contradiction. Hence there is no compact set K satisfying (9) and (10) and the number of geometrically distinct solutions of (1) must be infinite. 7

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