ESI The Erwin Schrodinger International Institute for Mathematical Physics Boltzmanngasse 9 A-1090 Wien, Austria Existence and Semi{Classical Limit of the Least Energy Solution to a Nonlinear Schrodinger Equation with Electromagnetic Fields Kazuhiro Kurata Vienna, Preprint ESI 562 (1998) Supported by Federal Ministry of Science and Transport, Austria Available via http://www.esi.ac.at June 16, 1998 Existence and semi-classical limit of the least energy solution to a nonlinear Schrodinger equation with electromagnetic elds Kazuhiro Kurata July 2, 1998 Keywords: nonlinear Schrodinger equation, magnetic eld, least energy solution, existence, semi-classical limit 1 Introduction and Main Result We study a standing wave (x; t) = e?iEth? u(x) to a time-dependent nonlinear Schrodinger equation with electromagnetic potentials: (1) ih@t = ( hi r ? A(x))2 + V (x) ? f (jj2); where h is positive constant, A(x) is a real-valued magnetic vector potential, V (x) is a real-valued electric potential, and f (jj2) is a nonlinear term. For simplicity, we treat the case n 3 and a special nonlinearity f (t) = t(p?1)=2 for some 1 < p < (n + 2)=(n ? 2), although we can also treat 2-dimensional case with some modications( see, Remark 2 and Remark 5). Then u satises (2) ( hi r ? A(x))2u + (V (x) ? E )u ? f (juj2)u = 0 in Rn: 1 Department of Mathematics, Tokyo Metropolitan University, Minami-Ohsawa 1-1, Hachioji-shi, Tokyo, Japan, e-mail: kurata@math.metro-u.ac.jp, 1 Put W (x) = V (x) ? E and assume that W is strictly positive; W (x) inf W W 0 > 0; x 2 Rn: Let F (t) = 0t f (s) ds and J (u) = 12 fjDh uj2 + W (x)juj2g dx ? F (juj2) dx; R R where Dh u = (D1h u; ; Dnh u) and Djh = i?1h@j ? Aj (x).The purpose of this paper is to show existence of the least energy solution (see Lemma 2) to (2) and to study a semi-classical limit of uh as h ! 0. We denote by H the Hilbert space dened by the closure of C01(Rn ) with respect to the norm R Z Z n n kuk2H = Z R n jDh uj2 + W (x)juj2 dx: Because of the super-nonlinearity of f (juj2)u = jujp?1u, there exists u0 2 H such that J (u0) 0. We look for the least energy solution by the mountain pass theorem. Let ? = f 2 C ([0; 1]; H ); (0) = 0; J ( (1)) 0g and let bh = inf sup J ( (t)): 2? t2[0;1] Let B = (Bjk ) be a magnetic eld dened by Bjk = @j Ak ? @k Aj . We say U 2 (RH )q ; 1 < q < +1, if U 2 Lqloc (Rn) and there exists a constant C such that 1=q 1 1 q dy j U ( y ) j C jB (x; r)j B(x;r) jB (x; r)j B(x;r) jU (y)j dy; and say U 2 (RH )1 if U 2 L1loc(Rn) and 1 sup jU (y)j C jB (x; r)j B(x;r) jU (y)j dy B (x;r) Z Z Z for every ball B (x; r) in Rn , respectively. For jB j + W 2 (RH )n=2 , we have jB j + W 2 (RH )n=2+ for some > 0 and hence we can dene m(x; jB j + W ) by 2 1 = inf fr > 0; r m(x; jB j + W ) jB (x; r)j B(x;r)(jB (y)j + W (y)) dy 1g: Z 2 We refer to [Sh1] for these facts and the properties of m(x; jB j+W ). Throughout this paper we assume the following conditions on A, B and W : A 2 2 (Rn ); W 2 C (Rn ) for some 2 (0; 1) and there exists a constant C Cloc loc such that jB j + W 2 (RH )n=2 ; jrB (x)j Cm(x; jB j + W )3: (3) Theorem 1 Under the assumption (3) and m(x; jB j+W ) ! 1 as jxj ! 1, there exists a least energy solution uh 2 H such that bh = J (uh ). Remark 1 When A 0, previous results require some condition for W at innity (e.g., liminf jxj!1 W (x) > inf W ) or the smallness of h(cf., e.g., [Oh1],[Ra], [DF1,2], [GU]). Note that Theorem 1 need to assume neither any condition at innity for W itself nor the smallness of h. If jB j+W 2 (RH )1 , then jB (x)j + W (x) Cm(x; jB j + W )2 holds for some constant C . It is known that U (x) = jP (x)j belongs to (RH )1 for any polynomial P (x) and > 0 ([Sh1,2]). Hence, if A(x) is polynomial and jB (x)j ! 1 for instance, we can apply Theorem 1. Moreover, we note that the condition m(x; jB j) ! 1 as jxj ! 1 holds in many cases, even if jB j itself do not have an homogeneous growth at innity. Next, we study the prole of the solution uh in the semi-classical limit of h ! 0. For the case A(x) 0, many authors studied the prole of the least energy solution uh, when h tends to zero( see, e.g., [FW], [Oh1], [Wa], [DN], [BW], [DF1], [Ra].) and showed that uh concentrates at the global minimum of W . Several authors ( [Oh2], [Gu], [DF2]) also studied the existence of multi-bump solutions which concentrate at local minimums of W . In such a case, the study of uh is decoupled into the scalar case and is reduced to the one of positive solutions of ?h2u + Wu = up in Rn : Now, we consider the general case. First, we observe that the transformation y = h?1 (x ? x0) for some x0 2 Rn and vh(y) = uh (hy + x0), vh satises ? y vh ? 2i A(hy + x0) ry vh(y) ? hi (divxA)(hy + x0)vh(y) + jA(hy + x0)j2vh(y) + W (hy + x0)vh(y) = jvh(y)jp?1vh(y): Since A(hy + x0) ! A(x0); W (hy + x0) ! W (x0) as h ! 0 on fjyj M g for each M > 0, vh might tend to some w in some sense and w satises ( 1i r ? A(x0))2w + W (x0)w = jwjp?1w: 3 Let (y) = es n A (x )y . j =1 i 0 j P Then A(x0) = r and w~(y) = e?i(y)w(y) satis- ?w~ + W 0w~ = jw~jp?1w: ~ Hence, one might conjecture that vh(y) ! ei(y)w~(y) as h ! 0. Actually, we prove an analogous result due to [Wa] in the case of A(x) 0. Namely, we show that fvhg has a subsequence fvk g, in which fvk g converges to a certain 2 (Rn ) and fjv jg converges to the least energy solution U function v~ in Cloc k for the scalar case in H 1(Rn ) and concentrates at a global minimum point of W . The magnetic potential A(x) contributes the phase factor of v~ (see also the part (c) of Theorem 2). Let U be the unique positive solution to ?U + W 0U = U p in Rn with max U = U (0). We use the notation A0 = A(x0) for the point x0 which will appear in the next theorem. Theorem 2 In addition to the assumptions in Theorem 1, suppose W = fx 2 Rn; W (x) = min W ( W 0)g is not empty and liminf jxj!1 W (x) > min W . (a) Let be a arbitrary small number. Then, there exist a sequence fxk g and a subsequence fhk g; hk ! 0(k ! 1), and positive constants C and R0 which do not depend on k such that fxk g converges to a point x0 2 W and ? pW 0 ?jx?x j k juh (x)j Ce holds on fjx ? xk j hk R0 g: In the estimate above, as xk we can take a maximum point of juh j and choose a large R0 to assure max ju (x)j > (W 0)1=(p?1): jx?x jh R0 h k hk k k k k (b) There exist a constant ! 2 R such that for each R > 0 uh (x) ? U ( x ?h xk )e?i(!+h k k ?1 Pn A0 (x?xk )j ) k j =1 j !0 as hk ! 0 on fjx ? xk j Rhk g for each R > 0. (c) Fix the R0 as in the part (a). Suppose maxx2 jA(x)jR0 < 2. Then the maximum point xk of juh j is unique. Actually, we do not use the assumptions (3) and m(x; jB j + W ) ! 1 at innity in the proof of Theorem 2. We just assumed these assumptions to W k 4 assure the existence of the least solution, although we would obtain the existence of the least energy solution for small h, once we assume the condition liminf jxj!1 W (x) > min W or its local version (see [Ra], [DF1,2], [Gu]). Remark 2 Although we considered in Theorem 1 and 2 the simple model case f (juj2 )u = jujp?1u for the sake of simplicity, these results would be true for certain general nonlinear term f (juj2)u which was treated in [NT], [Gu], [DF1,2]. Remark 3 In the part (b), we show actually that vk (y) = uh (xk + hk y) A y ) in C 2 (Rn ) as h ! 0. We do not know in converges to U (y)e?i(!+ k loc part (c) whether we can remove the smallness condition on maxx2 jA(x)j Pn j =1 k 0 j j W or not. We have no further information on ! and xk . The construction of multi-bump solutions which concentrate near at the local minimum of W is not treated in this paper. This is a natural question and will be studied in near future. 2 The proof of Theorem 1 In this section, we give the proof of Theorem 1. Throughout this section, we use the notation D = Dh for the sake of simplicity, since the dependence on h is irrelevant here. The key step in the proof of Theorem 1 is the following Proposition. Proposition 1 Under the same assumption as in Theorem 1, the functional J (u) satises the Palais-Smale(PS) condition. Once we obtain this Proposition 1, we can prove Theorem 1 by the standard mountain pass theorem(see, e.g., [AR]). To show Proposition 1, we use the following inequality due to Shen[Sh2]. Lemma 1 Suppose n 3 and jB j+W 2 (RH )n=2 and jrB (x)j Cm(x; jB j+ W )3. There exists a constant C (which depends on h) such that Z Rn m(x; jB j + W juj dx C )2 2 for every u 2 C01(Rn). 5 Z Rn jDuj 2+ W juj dx 2 Remark 4 Even if W 0, we have the inequlity 2juj2 dx C 2 dx m ( x; j B j ) j Du j R R for every u 2 C01(Rn)([Sh3]). Hence, we can show still the exsitence of the least energy solution to (i?1 hr ? A(x))2u ? f (juj2)u = 0 under the condition that m(x; jB j) > 0 for some positive constant and m(x; jB j) ! +1 as jxj ! 1 in the same way as in the proof of Theorem 1. Here, we take the Hilbert space H as the completion of C01 (Rn ) with respect to the norm kukH = kDukL . However, we do not know the semiclassical limit in this Z Z n case. n 2 Remark 5 When n = 2 and B (x) 0, we know the following inequality due to Avron, Herbst and Simon [AHS, Theorem 2.9]: Z Rn B (x)ju(x)j2 dx Z Rn j(ir ? A(x))uj2 dx for every u 2 C01 (Rn ). Therefore, in the 2-dimensional case we can obtain existence of the least energy solution under the conditions B (x) B0 > 0 and jB (x)j ! +1 as jxj ! 1. Before giving a proof of Proposition 1, we mention the meaning of the least energy solution. We say u is the least energy solution, if u is a solution and satises J (u) J (v) for every solutions v. Let M = fu 2 H ; u 6= 0; jDuj2 + W juj2 dx = jujp+1 dxg. Note that any non-trivial solution to (2) belongs to M. The next lemma is rather well-known (see, e.g., [NT], [Wa], [DF1]) and shows that the montain pass solution obtained by Theorem 1 is the least energy solution. Lemma 2 We have bh = inf v2M J (v) > 0. Proof: Let bh = inf u2H (supt>0 J (tu)): It is easy to see that for each u 6= 0 2 H there exists a unique positive number t(u) such that supt>0 J (tu) = J (t(u)u). Actually, t(u) is determined by R R Z Z t(u)2 jDuj2 + W juj2 dx = t(u)p+1 jujp+1 dx: Now, we can see M = fv = t(u)u; u 6= 0 2 H g and bh = inf v2M J (v). We claim bh = bh which conclude Lemma 2. For each v 2 M, since J (tv) ! ?1 6 as t ! 1, there exists t0 > 0 such that J (t0v) 0. Hence, (t) = tt0v belongs to ? and it follows that bh bh. On the other hand, we claim that for each 2 ?, there exists 2 [0; 1] such that ( ) 2 M. This claim yields bh bh and complete the proof. We show the claim by contradiction. If (t) 62 M for every t 2 [0; 1]. Then Z Z jD (t)j2 + W j (t)j2 dx > j (t)jp+1 dx (4) holds for every t 2 [0; 1]. By the Sobolev's inequality, we see Z Z Z jDuj2 + W juj2 dx = jujp+1 dx C ( jDuj2 + W juj2 dx)(p+1)=2 for some positice constant C and for v 2 M. Hence, there exists a constant such that if kvkH and v = 6 0, then v 62 M and jDvj2 + W jvj2 dx > p +1 jvj dx hold. We may assume (t) =6 0 for t =6 0 and hence (t) 62 M for small t and (4) holds. By the continuity of (t), (4) holds for all t 2 [0; 1]. It follows that J ( (t)) > 0 for all t 2 [0; 1]. This contradicts to J ( (1)) 0. 2 Proof of Proposition 1: Take a sequence fun g H such that J (un) ! c and J 0(un) ! 0 as n ! 1 for some constant c. Then, we have R R hJ 0(un); i = Re Z Rn Dun D + Wun ? junjp?1un dx = o(kkH ) (5) for any 2 H and 1 jDu j2 + W ju j2 dx ? 1 p+1 n n 2 R p + 1 R junj dx ! c: Putting = un in (5), we obtain Z Z n n kunkH ? 2 Z ju jp+1 dx = o(kun kH ): Rn n (6) and (7) imply p?1 p+1 p + 1 R junj dx = 2c + o(1) + o(kunkH ): Substituting (8) to (7), we obtain the boundedness of fung in H and Z n kun kH = 2 Z Rn junjp+1 dx + o(1): 7 (6) (7) (8) (9) Hence, we can take a subsequence fun g fung such that un converges weakly to a u 2 H . We write uk = un for the sake of simplicity. Taking = uk ? u in (5), we get k k k Z Re Duk D(uk ? u) + Wuk (uk ? u) ? juk jp?1uk (uk ? u) dx = o(1): This yields Z kuk ? uk2H + Re uD(uk ? u) + Wu(uk ? u) dx ? Re jukjp?1uk (uk ? u) dx = o(1): Z The second term converges to zero, because of the weak convergence of uk to u in H . We claim the following which concludes the strong convergence kuk ? ukH ! 0. Claim: Jk juk jp juk ? uj dx ! 0 as k ! 1. Holder's inequality and (8) yield R Jk Z C p=(p+1)Z Rn Z ju ? ujp+1 dx Rn k Rn juk ? ujp+1 dx =(p+1) 1 =(p+1) 1 juk jp+1 dx : Now, by Lemma 1 we have Z Rn m(x; jB j + W )juk (x)j2 dx M for some constant M . Since m(x; jB j + W ) ! 1, for any > 0 there exists R > 0 such that m(x; jB j + W ) M= on jxj R. This imples Z fjxjRg for every k and hence Z also holds. Since 2 < p + 1 that kuk ? ukL juk (x)j2 dx ju(x)j2 dx 2n=(n ? 2), there exists 2 (0; 1) such fjxjRg < 2 = p+1 kuk ? ukL kuk ? uk1L? : 2 8 2 By Sobolev's inequality, we have kuk ? ukL C krjuk ? ujkL kD(uk ? u)kL M: Here we also used the inequality jrjuj(x)j jDu(x)j a.e. x (see, e.g., [RS]). Note that, since jruj jDuj + jAuj, fuk g is bounded also on H 1(K ) for any bounded domain K , and hence uk converges strongly to u in L2(K ). Thus, we obtain lim sup juk ? uj2 dx lim sup juk ? uj2 dx 2: 2 2 Z k!1 2 Z Rn k!1 fjxjRg Therefore, we have kuk ? ukL ! 0 and Jk ! 0. 2 2 3 Proof of Theorem 2 First, we give an estimate of the energy of uh (see, e.g., [NT], [Wa], [DF1]). Let W 0 = min W and let U be the least energy solution associated to the energy on M0;r; (10) J0(v) = 21 jrvj2 + W 0jvj2 dx ? p +1 1 jvjp+1 dx; where M0;r = fv 6= 0 2 H 1(Rn; R); jrvj2 + W 0jvj2 dx = jvjp+1 dxg. It is well-known that U is a radially symmetric and unique (up to translation) positive solution to the equation: ?U + W 0U = jU jp?1U in Rn: (11) We put b0;r = J0(U ). In a similar way, we dene the class M0;c = fv 6= 0 2 H 1(Rn; C); jrvj2 + W 0jvj2 dx = jvjp+1 dxg and denote by b0;c the least energy of J0(v) on M0;c. Lemma 3 We have hnb0;r bh = J (uh) hnb0;r(1 + o(1)) as h ! 0. Proof: Let x0 2 W be a point and 2 C01 (B (x0; 2R)) satisfy (x) 1 on B (x0; R), where R > 0 is a some xed number. Put u(x) = (x)U ((x ? x0)=h)ei((x?x )=h). By Lemma 2, we have bh = J (uh) sup J (tu) = J (thu): (12) Z Z R R R 0 t>0 9 R Here, the positive constant th is uniquely determined by Z Z t2h jDuj2 + W juj2 dx = tph+1 jujp+1 dx: It follows that (13) J (thu) = 21 ? p +1 1 tph+1 jujp+1 dx: By using the transformation y = (x ? x0)=h and the exponential decay of U , it is easy to see Z and jujp+1 dx = hn Z Z U p+1 dy(1 + o(1)) Z Z W juj2 dx = hn W 0 U 2 dy(1 + o(1)) as h ! 0. Let M = max(kkL1 ; krkL1 ). We see jDuj2 dx = h(r)(x)U ( x ?h x0 )ei((x?x )=h) + (x)(rU )( x ?h x0 )ei((x?x )=h) 2 + (x)(A(x0) ? A(x0 + hx))U ( x ?h x0 )ei((x?x )=h) dx Z Z 0 0 0 Z = hn (x0 + hy)jrU (y)j2 dy + J; where J 2hn+1 M 2 + 2hn M 2 + + Z Z Z f2R=hjyjR=hg f2R=hjyjg jU (y)jjrU (y)j dy + hn+2M 2 jU (y)j2 dy U (y)jA(x0) ? A(x0 + hy)jjrU (y)j dy j U (y)j2jA(x0) ? A(x0 + hy)j dy f2R=hjyjg 2hn+2 M 2 Z Z j U (y)j2jA(x0) ? A(x0 + hy)j dy: f2R=hjyjg Note that jA(x0) ? A(x0 + hy)j maxz2B(x ;2R) jrA(z)jhjyj Chjyj on jyj 2R=h. Hence, by using the exponential decay of U we have 0 Z jU (y)j2jA(x0) ? A(x0 + hy)j dy Ch f2R=hjyjg 10 Z Rn jU (y)j2jyj dy O(h): By controlling other terms in a similar way by using also the exponential decay of rU , we obtain Z Z jDuj2 dx = hn jrU j2 dy(1 + o(1)): Hence, by substituting these into (13) we have Z Z t2h jrU j2 + W 0U 2 dy(1 + o(1)) = tph+1 U p+1 dy(1 + o(1)): Since jrU j2 + W 0U 2 dy = U p+1 dy, we obtain th ! 1 as h ! 0. Hence, it follows J (thu) = ( 21 ? p +1 1 )hn U p+1 dy(1 + o(1)) = b0;rhn (1 + o(1)) (14) and (12) and (14) yield the desired upper bound. Since it is easy to see hn J0(jvj) J (u) for each u 2 H and v(y) = u(hy), we have hnb0;r bh by the characterization bh = bh = inf u2H (supt>0 J (tu)) (see Lemma 2). 2 Lemma 4 There exist a sequence fxk g converges to some point x0 2 W a subsequence fvh g; hk ! 0(k ! 1) and U~ such that vh (y) = uh (xk + 2 (Rn ) as h ! 0. Here (y ) = n A (x )y and hk y) ! ei(y)U~ (y) in Cloc k j =1 j 0 j ~U 2 H 1(Rn ) is a solution to ?U~ + W (x0)U~ = jU~ jp?1U~ in Rn. Moreover, v~ = jU~ j is a positive solution to ?~v + W 0v~ = v~p in Rn . Proof: We follow the argument as in [Wa]. (Step 1) By Lemma 3, we have ( 21 ? p +1 1 )kuhk2H hn b0;r(1 + o(1)): Let wh (y) = uh(hy). Then wh satises (r ? iA(hy))2wh (y) ? W (hy)wh(y) + jwh(y)jp?1wh(y); y 2 Rn: Since kuhk2H = hn( j(r?iA(hy))wh(y)j2 dy + W (hy)jwh(y)j2 dy), we obtain ( 21 ? p +1 1 ) jrjwh(y)jj2 dy + W (hy)jwh(y)j2 dy ( 12 ? p +1 1 ) j(r ? iA(hy))wh(y)j2 dy + W (hy)jwh(y)j2 dy b0;r(1 + o(1)): R R Z k Pk R R Z Z Z Z 11 k In particular, we have the boundedness of H 1(Rn)-norm of jwhj. Note that since W 0 = min W we have b0;r 21 jrjwhj(y)j2 + W 0jwh(y)j2 dy ? p +1 1 jwhj(y)p+1 dy 21 j(r ? iA(hy))wh(y)j2 + W (hy)jwh(y)j2 dy ? p +1 1 jwh j(y)p+1 dy = ( 12 ? p +1 1 ) jwhj(y)p+1 dy = b0;r(1 + o(1)): (15) Then there exists a sequence fyhg and positive constants L and such that Z Z Z Z Z lim hinf !0 Z B (yh ;L) jwh(x)j2 dx > 0: (16) Otherwise, by using Lemma 2.18 in [CR] (see also [Li]) we have kwhkL (R ) ! 0 for every 2 < q < 2n=(n ? 2). This contradicite to (15). (Step 2) Let vh(y) = wh(yh + y). Then fjvhjg is bounded in H 1(Rn) and satises jvhj W (hyh + hy)jvhj ? jvhjp (17) in the distribution ( and weak ) sense by Kato's inequality. By taking a subsequence fvh g, we may assume that jvk j = jvh j converges to a nonnegative v~ 2 H 1(Rn) weakly in H 1(Rn) and converges strongly in Lqloc (Rn) for 2 q < 2n=(n ? 2). By (16) v~ 6= 0. We now claim that fhk yh g is a bounded sequence. Otherwise, there exists a sequence hk ! 0 and such that xk hk yh ! 1 as k ! 1. Multiplying v~ to (17), we have q k n k k k Z ? rjvkj rv~ dy Z Z fjyjRg W (xk + hk y)jvkjv~ dy ? jvkjpv~ dy for any xed R > 0. By the assumption W~ 0 liminf jxj!1 W (x) > min W , there exsits a constant > 0 (e.g., = (W~ 0 ? W 0)=2) such that Z jrjv~jj dy + (W 2 0+ ) Z fjyjRg Z jv~j dy jv~jp+1 dy: 2 By letting R ! 1, we obtain Z jrjv~jj dy + (W 2 0+ Z Z ) jvj dy jv~jp+1 dy: 12 ~2 This means there exists 2 (0; 1) such that v~ 2 M0;r. Now we have b0;r J0(v~) = ( 21 ? p +1 1 ) (v~)p+1 dy < ( 21 ? p +1 1 ) (~v)p+1 dy Z Z Z lim kinf jvk jp+1 dy = b0;r(1 + o(1)); !1 and this is a contradiction. (Step 3) Taking a subsequence again, we may assume that fvkg converges to v~(6= 0) 2 H 1(Rn ) and xk = hk yk ! x0 for some x0. We claim x0 2 W . By the argument above, we have Z Z Z jrv~j2 + W (x0)~v2 dy v~p+1 dy: Since W 0 = min W , it follows that Z Z Z jrv~j2 + W 0v~2 dy v~p+1 dy: Hence, there exists 2 (0; 1] such that v~ 2 M0;r. Now we have ?n b = lim ( 1 ? 1 ) jv jp+1 dy b0;r = klim h k !1 k h k!1 2 p + 1 ( 21 ? p +1 1 ) jv~jp+1 dy ( 21 ? p +1 1 ) jv~jp+1 dy = J0(v~) b0;r: This implies = 1 (i.e., v~ 2 M0;r) and W (x0) = W 0, i.e., x0 2 W , and J0(~v) = b0;r. Thus v~ is a non-negative and non-trivial solution to ?~v + W 0v~ = v~p in Rn, and hence v~ is positive by the strong maximum principle. (Step 4) For each compact set K Rn, there exists a constant CK such that supy2K jA(xk + hk y)j2 CK for every 0 < h < 1 and large k. Therefore, we obtain the boundedness of H 1(K )-norm of fvk g for each K . By the subsolution estimate, we can also obtain the boundedness L1 (K )-norm, and 2; n hence Schauder's estimate leads to the uniform boundedness in Cloc (R ). Z k Z Z 13 Hence, taking a subsequence again if necessary, fvh g converges to certain v 2 and also converges weakly in Lp+1 (Rn ), and v satises in Cloc ?v ? 2i A0 rv + jA0j2v + W 0v = jvjp?1v in Rn ; where A0 = A(x0) and W 0 = W (x0). Let U~ (y) = e?i(y)v(y). Then U~ satises jU~ j = v~ 6= 0 and ?U~ + W 0U~ = jU~ jp?1U~ in Rn : k 2 Lemma 5 Let be arbitrary small and fvh g be the subsequence as above. k Then there exist positive constants C; R0 which do not depend on k such that jvh (y)j Ce? k p 0 W ?jyj on jyj R0 . Proof: (Step 1) We write vk = vh . By Kato's inequality( see, e.g., [RS]), we have jvkj W (xk + hk y)jvkj ? jvkjp ?jvk jp?1jvkj in the distribution sense. Boundedness of k(r ? iA (x0 + h))vhkL (R ) and Sobolev's inequality imply the boundedness of L2 (Rn)-norm of jvhj. Let ch(x) = jvh(x)jp?1, then there exists a constant C such that kch kL ? (R ) C . Since 2=(p ? 1) > n=2, by the subsolution estimate (see, e.g., [GT]) we obtain Z 1=2 2 jvk (y)j C0 jvkj dz (18) k 2 2 =(p 1) B (y;1) n n for jyj 1 and a constant C0 which does not depend on k. Now we claim jvk j converges to v~ strongly in H 1(Rn), since Z Z jrv~j2 + W 0v~2 dy lim kinf jrjvkjj2 + W 0jvkj2 dy !1 lim kinf j(r ? iA(xk + hk y))vkj2 + W (xk + hk y)jvk j2 dy !1 1 ? 1 )?1b = ( 1 ? 1 )?1b = klim ( h !1 2 p + 1 2 p + 1 0;r Z k = Z jrv~j2 + W 0v~2 dy: 14 Thus there exists a constant C1 which does not depend on k and R such that Z lim ( R!1 j vkj2 dy)1=2 C1kjvk j ? v~k1L=2(R ): fjyjRg 2 n Now, take an arbitrary small positive constant a so that c0 = W 0 ? ap?1 > 0. Choose k0 so that kjvkj ? v~k1L=2(R ) a(2C0C1)?1 holds for every k k0. Then we can take a suciently large R1 so that ( j vkj2 dy)1=2 2Ca fjyjR g 0 for k k0. Now we can take a sucientlly large R0 R1 such that ( fjyjR0 g jvkj2 dy)1=2 2Ca 0 for k 1. Combining with the subsolution estimate (18), we get jvk (y)j a on R0 + 1 jyj for every k. (Step 2) Let ?0(y) = ?0(y; 0) be a fundamental solution to ? + c0. We choose ?0 so that jvk(y)j c0?0(y) holds on jyj = R0 . Let w = jvk j ? c0?0. Then we have by Kato's inequality w = jvk j ? c0?0 (sign vk )D2 (vk ) ? (c0)2?0 (W 0 ? jvkjp?1)jvk j ? (c0)2?0 c0(jvk j ? c0?0) = c0w: By the maximum principle, we can conclude w(y) 0 on jyj R0 . pSince it is well-known that there exists a constant C such that ?0(y) Ce? c jyj on jyj 1 and we can take c0 to be arbitrary close to W 0, we obtain the desired estimate. 2 Lemma 6 There exist a constant ! 2 R and a point y0 2 Rn such that U~ (y) = ei! v~(y) = ei! U (y ? y0). Moreover, U~ is a non-trival least energy solution for b0;c . Proof: Since we already know jU~ j = v~ > 0. Let R00 be a xed number. Then, we have bh = J (uh ) n 2 Z 1 Z 0 k k 15 1 j(r ? iA(x + h y))v (y)j2 + W (x + h y)jv (y)j2 dy ? 1 p+1 = k k k k k k 2 p + 1 jvk j dy hnk 12 fjyjR00g j(r ? iA(xk + hk y))vk (y)j2 + W 0jvk (y)j2 dy ? p +1 1 R jvk jp+1 dy hnk 12 fjyjR00g jrU~ (y)j2 + W 0jU~ (y)j2 dy ? p +1 1 fjyjR00 g jU~ (y)jp+1 dy ? p +1 1 fjyjR00 g jvk (y)jp+1 dy + o(1) hnk 12 fjyjR00g jrU~ (y)j2 + W 0jU~ (y)j2 dy ? p +1 1 fjyjR00 g jU~ (y)jp+1 dy ? p +1 1 fjyjR00 g jvk (y)jp+1 dy + o(1) ; where W 0 = W (x0) for some x0 2 W . For any > 0, by Lemma 5 we can take large R00 enough to assure 1 p+1 p + 1 fjyjR00g jvk (y)j dy < for every k. By letting k ! 1, we have 1 ~ (y)j2 + W 0jU~ (y)j2 dy ? 1 ~ p+1 jr U 2 fjyjR00g p + 1 fjyjR00g jU (y)j dy ? b0;r: Taking the limit of R00 ! 1, we obtain J0(U~ ) b0;r. On the other hand, we know J0(jU~ j) = J0(~v) = b0;r. This implies J0(U~ ) = b0;r and jrjU~ j(x)j jrU~ (x)j a.e. x. Hence, we obtain the desired estimate. The last statement in Lemma 6 follows from Lemma 7. 2 Although we do not need the following lemma in our proof, it is interesting. We do not know whether Lemma 7 is also true for general nonlinearity f (t) or not. Lemma 7 The equality b0;c = b0;r holds, and the least energy solution Uc for b0;c has the form ei! U (y ? y0) for some point y0 and a constant ! 2 R. Proof of Theorem 2: It is easy to show max jvk j > (W 0 )1=(p?1) by the maximum principle. We already know that the part (a) of Theorem 2 holds for some xk converges to xo 2 W . Because of this uniform decay estimate, we may take R0 large enough to assure max ju (x)j = fjmax jv (y)j > (W 0)1=(p?1): fjx?x jh R0 g h yjR0 g k hnk Z Z Z Z Z n Z Z Z Z Z Z Z Z k k k 16 This and Lemma 5 give the part (a) except the last statement. The part (b) is a consequence of Lemma 4 and Lemma 6. We prove the part (c) and the last statement of (a) in Theorem 2. We may take x0k 2 B (xk; hk R0 ) as the maximum point for juh j. Since xk ! x0, x0k ! x0 as k ! 1. Hence we obtain p ?j ? 0j ? juh (x)j Ce on fjx ? x0k j hk R0g by replaceing the constant R0 and C if necessary. Since A y in C 2 (Rn ) and jv (y )j = v~(y ) > 0, vk (y) converges v(y) = v~(y)ei!ei loc if maxx2 jA(x)j is small enough to assure supjyjR0 jA(x0) yj < 2 we may assume that vk (y) can be written vk (y) = jvkj(y)e?i! (y) on jyj R0 and that satises j!k (y) ? !0j < 2 for some !0. Hence jvkj ! v~ and !k (y) ! ! + nj=1 A0j yj in C 2(B (O; R0)). Then the argument by using wellknown properties of U as in [Wa] yields the uniqueness of the maximum point x0k of juh j and y0 = O. However, y0 must be the origin without the smallness assumption on maxx2 jA(x)j, because we can emply the same argument as above in a region jyj R for suciently small R. 2 Proof of Lemma 7: We dene jruj2 + W 0juj2 dx ; c = u2H inf (R ;C) ( jujp+1 dx)2=(p+1) jruj2 + W 0juj2 dx : r = u2H inf (R ;R) ( jujp+1 dx)2=(p+1) Since jrjuj(x)j jru(x)j a.e. x, we have c r , and the opposite inequality is easy. Hence, c = r holds. Similaly, b0;c b0;r is easy since M0;c M0;r . Let Uc and ur be the least energy solution associated to b0;c and b0;r, respectively. Suppose b0;c < b0;r. Then we have k W0 x x k hk k Pn 0 j j j =1 W k P k W R 1 R n R 1 Z c = jUc R Z jUc jp+1 dx < jur jp+1 dx: Thus Z n p?1)=(p+1) ( jp+1 dx < Z jur jp+1 dx p?1)=(p+1) ( = r by the uniqueness to the real-valued case. This is a contradiction and hence b0;c = b0;r holds. 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