An example of stable excited state on nonlinear Schrödinger equation with nonlocal nonlinearity
aa r X i v : . [ m a t h . A P ] S e p An example of stable excited state on nonlinearSchr¨odinger equation with nonlocal nonlinearity
Masaya Maeda † and Satoshi Masaki ‡ † Mathematical institute, Tohoku University,Sendai, 980-8578, Japan ‡ Department of Mathematics, Gakushuin University,Toshima-ku Tokyo, 171-8588, Japan
Abstract
In this article, we consider nonlinear Schr¨odinger equation with nonlo-cal nonlinearity which is a generalized model of the Schr¨odinger-Poissonsystem (Schr¨odinger-Newton equations) in low dimensions. We first provethe global well-posedness in wider space than in previous result and showthe stability of standing waves including excites states. It turns out thatan example of stable excite states with high Morse index is contained.Several examples of traveling-wave-type solutions are also given.
In this article, we consider the following nonlinear Schr¨odinger equation withnonlocal nonlinearity: ( iu t = − ∆ u + λ | x | u + η (cid:0) | x | ∗ | u | (cid:1) u, ( t, x ) ∈ R d ,u (0 , x ) = u ( x ) ∈ Σ , (H)where λ, η ∈ R and Σ := { φ ∈ H | h x i φ ∈ L } . The operator ∗ stands for theusual convolution: f ∗ g ( x ) = R R d f ( x − y ) g ( y ) dy . It is shown in [11] that (H)is globally well-posed in Σ and further all the solution is written explicitly. Inthis article, we give several examples of standing waves and investigate stabilityof them. This contains a new type example of a stable excited state.The equation (H) is a generalized model of the Schr¨odinger-Poisson system(or the Schr¨odinger-Newton equations) iu t = − ∆ u + V u, ( t, x ) ∈ R d , − ∆ V = | u | , x ∈ R d ,u (0 , x ) = u ( x ) (SP)in low dimensions. Indeed, one sees that V = − (1 / | x | ∗ | u | ) when d = 1.Generalizing with respect to growth order of interaction potential and dimen-sion, we reach to ( iu t = − ∆ u + η ( | x | γ ∗ | u | ) u, ( t, x ) ∈ R d ,u (0 , x ) = u ( x ) . (1)1he equation (H) (with λ = 0) corresponds to the case γ = 2. For the well-posedness result on (SP), see [3] for d = 1 and see [9] for d = 2. Well-posednessresult of (1) for 0 < γ W ∗ | u | )( x ) behaveslike a linear potential W ( x ) k u k L near the spatial infinity. Hence, we put alinear potential λ | x | in (H) in order to investigate the competition between thispotential and the one created by the nonlinearity.An idea to treat this kind of nonlinearity is the following: We split thenonlinear potential of (H) as( | x | ∗ | u | ) = | x | Z | u ( y ) | dy − x · Z y | u ( y ) | dy + Z | y | | u ( y ) | dy. The point is that the first two terms are unbounded in x but time behavior ofthem can be specified thanks to the conservation laws. The last term can beabsorbed by gauge transformation. What requires the full regularity assumption u ∈ Σ is just the third term. It therefore can be expected that if this part isremoved then the regularity assumption can be loosen. From this respect, letus introduce iv t = − ∆ v + λ | x | v + ηv Z R d ( | x − y | − | y | ) | v ( y ) | dy, ( t, x ) ∈ R d ,v (0 , x ) = v ( x ) ∈ Σ / , (H ′ )where Σ / := { φ ∈ H / | h x i / φ ∈ L } . Notice that the modified nonlin-earity makes sense for v ∈ Σ / . When v ∈ Σ / \ Σ it turns out that thecorresponding solution has infinite energy. Existence of infinite energy solutionfor 2D Schr¨odinger-Poisson system is shown in [10] with the same modificationof nonlinear potential.In this article, we turn to the study of standing waves of to (H) or (H ′ ).Here, standing waves are solutions of the form e − iωt/ φ ( x ). The number ω ∈ R is referred to as the frequency. Standing waves of 1D Schr¨odinger-Poisson systemare studied in [1, 4]. The meaning “a standing wave is stable”, will be clear later,after we introduce the well-posedness results of (H) and (H ′ ). For the stabilityof standing waves of general Hamiltonian PDEs, there is an elegant theory byGrillakis, Shatah, and Strauss [5, 6]. Let E be a C functional which satisfies E ( e is u ) = E ( u ) for all s ∈ R , and consider the equation2 iu t = E ′ ( u ) . (2)Set Q ( u ) = || u || and S ω = E − ωQ , where S ω is called the “action”. Then, S ′ ω ( φ ω ) = 0 holds if and only if e − iωt/ φ ω is a standing wave solution. Weassume that there exists a frequency-to-standing-wave map ω φ ω such that S ′ ω ( φ ω ) = 0 and Ker S ′′ ω ( φ ω ) is spanned by iφ ω . Let n ( S ′′ ω ( φ ω )) be the numberof negative eigenvalues of S ′′ ω ( φ ω ). What Grillakis, Shatah, and Strauss [5, 6]have shown are summarized as follows(i) If n ( S ′′ ω ( φ ω )) = 0, then e − iωt/ φ ω is stable.(ii) If n ( S ′′ ω ( φ ω )) = 1 and d ′′ ( ω ) >
0, then e − iωt/ φ ω is stable.2iii) If d ′′ ( ω ) = 0 and n ( S ′′ ω ( φ ω )) − max( d ′′ ( ω ) / | d ′′ ( ω ) | ,
0) is odd, then e − iωt/ φ ω is unstable.Obviously, there exist two more cases(iv) n ( S ′′ ω ( φ ω )) ≥ d ′′ ( ω ) = 0 and n ( S ′′ ω ( φ ω )) − max( d ′′ ( ω ) / | d ′′ ( ω ) | ,
0) is even.(v) The case d ′′ ( ω ) = 0.These cases are not discussed in [5, 6]. For the case (v), there are several results([2], [8], [12]). In the case (iv), the standing wave has high Morse indices. So, φ ω is not a minimizer of E under the constraint of Q . In this case, we say φ ω is unstable in energy. Standing waves which are unstable in energy seemto be unstable, and in many cases they are actually unstable. We shall showthat, however, there exist stable standing waves belonging to the case (iv) witharbitrary n ( S ′′ ω ( φ ω )) ≥ ′ ) are given in Theorems 1 and 2, respectively. We thendiscuss stability of standing waves of (H) and (H ′ ) in Theorem 3. In Theorem4, an example of excited state with high Morse index is introduced. At last, inTheorems 5 and 6, we consider some more example of standing waves includingtraveling-wave type solutions e − iωt/ e ig ( t ) e ix · g ( t ) φ ( x − g ( t )) , and superpositions of finite number of such traveling waves. Before stability results, we summarize well-posedness results for (H) and (H ′ ).Set M [ u ] = k u k L , X [ u ] := Z R d x | u | dx, P [ u ] := Im Z R d ¯ u ∇ u dx = Z R d ξ |F u | dξ. We also introduce energy E [ u ] = 12 k∇ u k L + λ k xu k L + η Z Z R d + d | x − y | | u ( x ) | | u ( y ) | dxdy. Set U κ ( t ) = e i t (∆ − κ | x | ) Define an R d -valued function g κ ( t ) = g κ ( t, a , b ) as asolution of an ODE g ′′ κ = − κg κ , g ′ κ (0) = a , and g κ (0) = b , where κ ∈ R is aparameter. Introduce the Galilean transform G κ ( t ) = G κ ( t, a , b ) by( G κ ( t, a , b ) φ )( x ) = e − i g κ ( t, a , b ) · g ′ κ ( t, a , b ) e ix · g ′ κ ( t, a , b ) φ ( x − g κ ( t, a , b )) . (3)Basic properties of Galilean transform are summarized in Section 2.3 heorem 1 ([11], Theorem 2.1) . (H) is globally well-posed in Σ . Uniquenessholds unconditionally. Further, let M = || u || , a = P [ u ] /M , b = X [ u ] /M ,and κ = λ + ηM . Mass M [ u ( t )] and energy E [ u ( t )] are conserved. Then thesolution is written as u ( t ) = e − i Ψ( t ) G λ ( t, a , b ) U κ ( t ) G κ (0 , a , b ) − u , = e − i Ψ( t ) G λ ( t, a , b ) G κ ( t, a , b ) − U κ ( t ) u , (4) where Ψ( t ) is given by Ψ( t ) = η R t || x U κ ( s ) G κ (0 , a , b ) − u || ds. Remark . One easily verifies that k u ( t ) k L p = kU κ ( t ) u k L p for all p and that X [ u ( t )] = X [ U λ ( t ) u ]. This means that dispersive properties of the solution isindicated by κ , while the motion of the whole system is by λ . Remark . Unconditional uniqueness means that the solution is unique in theclass C ( R : Σ) (without any additional assumption). Theorem 2. (H ′ ) is globally well-posed in Σ / . Uniqueness holds uncondition-ally. Further, let M = || v || , a = P [ v ] /M , b = X [ v ] /M , and κ = λ + ηM .Mass M [ u ( t )] is conserved. Energy E [ u ( t )] is finite and conserved if v ∈ Σ .Then the solution is written as v ( t ) = e − i Φ( t ) G λ ( t, a , b ) U κ ( t ) G κ (0 , a , b ) − v , = e − i Φ( t ) G λ ( t, a , b ) G κ ( t, a , b ) − U κ ( t ) v , (5) where Φ( t ) = − ηM R t | g λ ( s, a , b ) | ds .Remark . Theorem 2 is a generalization of Theorem 1 in such a sense that u = v ∈ Σ implies u ( t ) = e i (Ψ( t ) − Φ( t )) v ( t ). This can be seen from k xu k L =(2 /η )(Ψ ′ − Φ ′ ). Remark that if v ∈ Σ / \ Σ then the energy of the correspondingsolution is infinite.
Here we state the results for stability of standing waves including excited states.Let us first give a definition of the stability of standing waves.
Definition 1.
Let e − iωt/ φ ( x ) be a solution of ( H ) . We say e − iωt/ φ ( x ) isstable in a Banach space X if for all ε > , there exists δ > which satisfiesthe following. If || u − φ || X < δ , then sup t> inf s ∈ R ,y ∈ R d || φ − e is u ( t, · − y ) || X < ε, where u ( t ) ∈ X is the solution of (H) with u (0) = u ∈ X . Stability of standing waves of (H ′ ) is also defined by replacing (H) anda solution u in the above definition with (H ′ ) and a solution v , respectively.Notice that the definition of stability implicitly requires well-posedness of theCauchy problem. In our case, they are already established in Theorems 1 and2. For a multi-index n = ( n , · · · , n d ), we introduceΩ n ( x ) := 1 p π d/ | n | n ! e | x | (cid:18) − ∂∂x (cid:19) n e −| x | , (6)4here n ! = Q dj =1 n j !. Also set Ω n,κ ( x ) = κ d/ Ω n ( κ / x ) for κ >
0. It is wellknown that, for any fixed κ > { Ω n,κ } n is a CONS of L ( R d ), and each Ω n,κ is an eigenfunction of − ∆ + κ | x | associated with an eigenvalue κ ( | n | + d ).For s >
0, set Σ s := D (( − ∆ + | x | ) s/ ) and define Σ − s as a dual of Σ s . SetΣ = L . Norm of Σ s ( s ∈ R ) is given by k f k Σ s := X n (cid:18) | n | + d (cid:19) s | ( f, Ω n ) | . The space Σ is identical to Σ . Our stability result is the following. Theorem 3.
Let d ≥ . Suppose λ ≥ , η ∈ R and M > satisfy κ = λ + ηM > . For any multi-index n , the following hold.1. Set ω = (1 + ηM κ − ) κ ( | n | + d ) . Then, e − iω t/ M / Ω n,κ ( x ) is a stand-ing wave solution of (H) and is stable in Σ s for s ≥ .2. Set ω = κ ( | n | + d ) . Then, e − iω t/ M / Ω n,κ ( x ) is a standing wavesolution of (H ′ ) and is stable in Σ s for s ≥ . One sees that the above standing wave solution is Ground state when n = 0,and is an excited state otherwise (see Lemma 5.3).Next theorem is concerned with stable excited states with high Morse indices.Since our purpose is to show that such a state does exist, let us restrict ourselvesto the simplest situation. So, we only treat the one dimensional case of (H),and the case where a perturbation is supposed to be an even function. Theorem 4.
Let d = 1 and set L r = { u ∈ L | u ( x ) = u ( − x ) , x ∈ R } . (i) Let λ = 0 and η = 1 . Then for any m ∈ N , there exists a stable standingwave e − iωt/ φ ω with φ ω ∈ Σ r , d ′′ ( ω ) < and n ( S ′′ ω ( φ ω ) | L r ) = 2 m . (ii) Let λ = 2 and η = − . Then for any m ∈ N , there exists a stable standingwave e − iωt/ φ ω with d ′′ ( ω ) > and n ( S ′′ ω ( φ ω ) | L r ) = 2 m + 1 . Let us show some more example of standing wave solutions. The first one is atraveling-wave-type solution with one peak.
Theorem 5 (Single-peak standing wave) . Let d > . Suppose λ, η ∈ R and M > satisfy κ = λ + ηM > . Suppose u ( x ) = v ( x ) = M G κ (0 , a , b )Ω n,κ ( x ) , where n is a multi-index and a , b ∈ R d
1. The corresponding solutions of (H) becomes u ( t, x ) = e − itω / M G λ ( t, a , b )Ω n,κ ( x ) (7) with ω = ( κ + ηM κ − )( | n | + d ) .2. If | a | = λ | b | then the corresponding solutions of (H ′ ) becomes v ( t, x ) = e − itω / M G λ ( t, a , b )Ω n,κ ( x ) , (8) with ω = ( κ − ηM | b | )( | n | + d ) . Theorem 6 (Multi-peak standing wave) . Let d > . Suppose λ, η ∈ R and M > satisfy κ = λ + ηM > . Let L be a positive integer. For α j ∈ C , a j , b j ∈ R d , and n j ∈ ( N ∪ { } ) d ( j = 1 , , . . . , L ), set u ( x ) = v ( x ) = µ L X j =1 α j G κ (0 , a j , b j )Ω n j ,κ ( x ) , where µ > is chosen so that M [ u ] = M . Then, the solutions of (H) and (H ′ ) are u ( t, x ) = µ L X j =1 e α j e − i Ψ j ( t ) G λ ( t, a , b ) G κ ( t, a j − a , b j − b )Ω n j ,κ ( x ) (9) v ( t, x ) = µ L X j =1 e α j e − i Φ j ( t ) G λ ( t, a , b ) G κ ( t, a j − a , b j − b )Ω n j ,κ ( x ) , (10) respectively, where a = P [ u ] /M , b = X [ u ] /M , e α j = α j e i ( a j · b − b j · a ) , Ψ j ( t ) =Ψ( t ) + κ ( | n j | + d ) t , and Φ j ( t ) = Φ( t ) + κ ( | n j | + d ) t .Remark . One sees from (9) that each peak keeps its shape for all time.The motion of a peak in physical and Fourier spaces is described by a sum oftwo Galilean transform. The first one is G κ ( t, a j − a , b j − b ). Recall that a and b denote mean momentum and mean position of the whole system, andso a j − a and b j − b are relative momentum and relative position of eachpeak, respectively. This transform therefore suggests that all peaks are rotatingaround the center of mass with the same period 2 πκ − / . The second transform G λ ( t, a , b ) corresponds to the motion of the center of mass. More precisely, X [ u ( t )] = M g λ ( t, a , b ). If λ = 0 (no external potential) then the motion is astraight line, and if λ > πλ − / . Notice that, as long as η = 0 (in presence of nonlinearity),the period in which each peak rotates around the center of mass and the periodin which the center of mass does around the origin do not coincide.The rest of paper is organized as follows. We summarize in Section 2 severalproperties on Galilean transform which we use, and then prove Theorem 2 inSection 3. Section 4 is devoted to the proof of Theorems 5 and 6. Then, stabilityof standing waves is considered in Section 5. Let us summarize briefly the properties on the Galilean transform.
Lemma 2.1.
Let G κ be the Galilean transform given in (3) . Then, for any κ, t ∈ R and a , b , a , a , b , b ∈ R d , we have the following properties.1. G κ ( t, ,
0) = Id and G κ (0 , a , b ) = G (0 , a , b ) .2. X [ G κ ( t, a , b ) u ] = X [ u ] + g κ ( t ) M [ u ] , P [ G κ ( t, a , b ) u ] = P [ u ] + g ′ κ ( t ) M [ u ] . . G κ ( t, a , b ) − = G κ ( t, − a , − b ) .4. (2 i∂ t + ∆ − κ | x | ) G κ ( t, a , b ) = G κ ( t, a , b )(2 i∂ t + ∆ − κ | x | ) .
5. If N ( u ) is either F ( | u | ) u or u R R d K ( x − y ) | u ( y ) | dy , where F : R → R and K : R d → R , then N ( G κ ( t, a , b ) u ) = G κ ( t, a , b ) N ( u ) . U κ ( t ) G κ (0 , a , b ) = G κ ( t, a , b ) U κ ( t ) . G κ ( t, a , b ) G κ ( t, a , b ) u ( x ) is equal to e i ( g ′ · g − g · g ′ )( t ) e − i ( g + g )( t ) · ( g + g ) ′ ( t ) e ix · ( g + g ) ′ ( t ) u ( x − ( g + g )( t )) , where g j ( t ) = g κ j ( t, a j , b j ) . In particular, G κ ( t, a , b ) G κ ( t, a , b ) = e i ( a · b − a · b ) G κ ( t, a + a , b + b ) . Proof.
The first two properties are obvious. The forth and fifth are well known.The sixth is just a rephrase of Proposition 2.5 in [11]. One can show the last oneby a simple computation. Note that if κ = κ then the Wronskian g ′ · g − g · g ′ is independent of time and so that G κ ( t, a , b ) G κ ( t, a , b ) = e i ( a · b − a · b ) G κ ( t, a + a , b + b )follows. Letting a = − a and b = − b , we obtain the third. Lemma 2.2.
Let d > and s ∈ R . There exists a constant C = C ( d, s ) suchthat (cid:13)(cid:13) e ix · a f (cid:13)(cid:13) Σ s + k f ( · − a ) k Σ s C h| a |i | s | k f k Σ s Proof.
This inequality is easily verified when s = 2 m ( m = 0 , , , . . . ) since k f k Σ s is written as (cid:13)(cid:13) ( − ∆ + | x | ) m f (cid:13)(cid:13) L . Then, the case s > s <
0, we use a duality argument.
Lemma 2.3.
Let s ∈ R and f ∈ Σ s . Then, (cid:13)(cid:13) e ix · a f − f (cid:13)(cid:13) Σ s + k f ( · − a ) − f k Σ s → as | a | → .Proof. Set f N = P | n | N ( f, Ω n )Ω n . Then, f N → f in Σ s as N → ∞ . ByLemma 2.2, for any ε > N such that k f − f N k Σ s ε/ (cid:13)(cid:13) e ix · a ( f − f N ) (cid:13)(cid:13) Σ s + k ( f − f N )( · − a ) k Σ s ε | a |
1. Take m ∈ N so that s m . For any fixed multi-index n , wededuce from Lebesgue’s convergence theorem that (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) −
12 ∆ + 12 | x | (cid:19) m ( e ix · a Ω n − Ω n ) (cid:13)(cid:13)(cid:13)(cid:13) L + (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) −
12 ∆ + 12 | x | (cid:19) m (Ω n ( · − a ) − Ω n ) (cid:13)(cid:13)(cid:13)(cid:13) L → | a | →
0. Since k u k Σ s k u k Σ m = (cid:13)(cid:13) ( − ∆ + | x | ) m u (cid:13)(cid:13) L for u ∈ Σ s andsince f N is a finite combination of Ω n , there exists δ ∈ (0 ,
1] such that if | a | δ then (cid:13)(cid:13) e ix · a f N − f N (cid:13)(cid:13) Σ s + k f N ( · − a ) − f N k Σ s ε . Lemma 2.4.
Let s ∈ R . For any t, κ ∈ R and a , b , the Galilean transform G κ ( t, a , b ) is a bounded linear map from Σ s to itself. The operator norm has thefollowing estimate: kG κ ( t, a , b ) k B (Σ s ) (1 + C ( | a | + | b | )) | s | κ > , (1 + C ( | a | t + | a | + | b | )) | s | κ = 0 , (1 + C ( | a | + | b | ) e | t √ κ | ) | s | κ < . In particular, for each u ∈ Σ s , a mapping R d + d ∋ ( t, κ, a , b )
7→ G κ ( t, a , b ) u ∈ Σ s is continuous. (H ′ ) In this section, we prove Theorem 2.
Lemma 3.1.
Let u ∈ Σ / . Then, X [ u ] and P [ u ] are well-defined. Further, if { u n } n ⊂ Σ / converges to u in Σ / as n → ∞ then X [ u n ] and P [ u n ] convergesto X [ u ] and P [ u ] , respectively, as n → ∞ .Proof. For each i ∈ [1 , d ], one verifies that Z R d x i | u ( x ) | dx = Re X n p n i + 1) a n a n + e i , Im Z R d u ( x ) ∂ i u ( x ) dx = Im X n p n i + 1) a n a n + e i , (11)where a n = ( u, Ω n ) and e i is a multi-index such that the i -th component isone and the others are zero. Therefore, | X [ u ] | + | P [ u ] | C k u k Σ / holds.Convergence part follows from | X [ u ] − X [ v ] | + | P [ u ] − P [ v ] | C ( k u k Σ / + k v k Σ / ) k u − v k Σ / . Lemma 3.2.
Let V ( t ) : R → R be a continuous function of time. If u ∈ C ( R ; L ) ∩ C ( R ; Σ − ) solve iu t + ∆ u − V ( t ) | x | u = 0 in Σ − sense. then k u ( t ) k L is constant.Proof. By an abstract theory by Kato [7], there exists a unique propagator { U ( t, s ) } t,s ∈ R with the following properties;1. U ( t, s ) is unitary on Σ − and strongly continuous in s, t ;2. U ( t, t ) = 1 and U ( t, s ) = U ( t, r ) U ( r, s ) for any t, s, r ∈ R ;3. U ( t, s ) L ⊂ L . Further, U ( t, s ) | L is unitary on L and strongly contin-uous in s, t ; 8. dds U ( t, s ) y = U ( t, s ) i ( − ∆ + V ( s )2 | x | ) y ∈ Σ − for any y ∈ L ;5. ddt U ( t, s ) y = i ( ∆ − V ( t )2 | x | ) U ( t, s ) y ∈ Σ − for any y ∈ L (See Theorems 4.1 and 5.1 and Remarks 5.3 and 5.4 of [7]). Set w ( s ) = U ( t, s ) u ( s ). By the forth property of { U ( t, s ) } and by assumption on u , weobtain dds w ( s ) = U ( t, s ) i (cid:18) −
12 ∆ + V ( s )2 | x | (cid:19) u ( s ) + U ( t, s ) u t ( s ) = 0 in Σ − for any 0 < s < t . Hence, u ( t ) = w ( t ) = w (0) = U ( t, u (0) for t >
0. Thisholds also for t < { U ( t, s ) } .We are now in a position to prove Theorem 2. Proof of Theorem 2.
Let v ∈ Σ / and let M , a , b , and κ be as in the statementof Theorem 2. We shall first prove that v ( t ) given by the first line of (5) isa solution of (H ′ ). Set w ( t ) = U κ ( t ) G κ (0 , a , b ) − v . It holds that w is in C ( R ; Σ / ) ∩ C ( R ; Σ − / ) and solves 2 iw t = − ∆ w + κ | x | w in Σ − / sense with w (0) = G κ (0 , a , b ) − v . Moreover, M [ w ( t )] = M and X [ w ( t )] ≡ P [ w ( t )] ≡ z ( t ) = G λ ( t, a , b ) w ( t ). Then, z ( t ) ∈ C ( R ; Σ / ) by Lemma 2.4.Similarly, z ( t ) ∈ C ( R ; Σ − / ) follows from2 iz t = ( | g ′ λ ( t ) | + g λ ( t ) · g ′′ λ ( t )) z − g ′′ λ ( t ) · xz − ig ′ λ ( t ) · G λ ( t ) ∇ w + G λ ( t )2 iw t . We have M [ z ( t )] = M . It also holds from the second property of Lemma 2.1 that X [ z ( t )] = M g λ ( t ). The forth property of Lemma 2.1 implies 2 iz t +∆ z − λ | x | z = G λ ( t, a , b )(2 iw t + ∆ w − λ | x | w ) = G λ ( t, a , b ) ηM | x | w . Since G λ ( t ) ηM | x | w = ηM | x | z − ηM g λ ( t ) · xz + ηM | g λ ( t ) | z = η | x | M [ z ( t )] z − ηx · X [ z ( t )] z + 2Φ ′ ( t ) z = η Z ( | x − y | − | y | ) | z ( t, y ) | dyz + 2Φ ′ ( t ) z, we conclude that v ( t ) := z ( t ) e − i Φ( t ) is a solution of (H ′ ). Statements on energyconservation follows from Lemma 5.3, below.The data-to-solution map v v ( t ) is continuous as a map from Σ / to L ∞ loc ( R ; Σ / ) due to Lemmas 2.4 and 3.1.Let us proceed to the uniqueness. Assume e v ( t ) ∈ C ( R ; Σ / ) ∩ C ( R ; Σ − / )solves (H ′ ) in Σ − / sense with the same initial condition e v (0) = v . Since M [ e v ( t )] and X [ e v ( t )] are continuous, there exists an R d -valued function H ( t ) ∈ C ( R ) such that H ′′ ( t ) = − ( λ + ηM [ e v ( t )]) H ( t ) + ηX [ e v ( t )] (12)and H (0) = H ′ (0) = 0. Define e w ( t ) by e v ( t, x ) = e − i e Φ( t ) e − i H ( t ) · H ′ ( t ) e ix · H ′ ( t ) e w ( t, x − H ( t )) , (13)9here e Φ( t ) = − η R t H ( s ) · X [ e v ( s )] ds. It is easy to see e w ∈ C ( R ; Σ / ) ∩ C ( R ; Σ − / ) since H ( t ) ∈ C ( R ). We have M [ e w ] = M [ e v ] and X [ e w ] = X [ e v ] − M [ e v ] H . By a computation with an identity η Z ( | x − y | − | y | ) | e v ( y ) | dy e v = η | x | M [ e v ] e v − ηx · X [ e v ] e v = e − i e Φ( t ) e − i H · H ′ e ix · H ′ ( ηM [ e v ] | x | e w )( t, x − H ( t )) + 2 ηM [ e v ] H · x e v − ηM [ e v ] | H | e v − ηX [ e v ] · x e v and (12), one verifies that e w solves 2 i e w t + ∆ e w − ( λ + ηM [ e w ]) | x | e w = 0 inΣ − / with e w (0) = v . Since V ( t ) := λ + ηM [ e w ( t )] is continuous in time byassumption, M [ e w ( t )] ≡ M [ v ] follows from Lemma 3.2. Set M := M [ v ] and κ = λ + ηM . e w is written as e w ( t ) = U κ ( t ) v and so X [ e w ( t )] = M g κ ( t, a , b ).The equation which H solves now becomes H ′′ ( t ) = − λH ( t ) + ηM g κ ( t ) and H (0) = H ′ (0) = 0. Solving this, we see H ( t ) = g λ ( t ) − g κ ( t ). Consequently, X [ e v ( t )] = M g λ ( t ) and e Φ( t ) = − ηM Z t ( g λ − g κ )( s ) · g λ ( s ) ds = Φ( t ) + 12 ( g λ ( t ) · g ′ κ ( t ) − g ′ λ ( t ) · g κ ( t )) . Substituting these formulas to (13) and using the third and the seventh prop-erties of Lemma 2.1, we see that e v ( t ) coincides with v ( t ). We now turn to the proof of Theorems 5 and 6.
Lemma 4.1.
Let u ∈ Σ and κ = λ + ηM > . The phase Ψ( t ) in Theorem 1is written as Ψ( t ) = η κ (cid:16) k A κ w k L + (cid:13)(cid:13) A † κ w (cid:13)(cid:13) L (cid:17) t + ηκ sin( √ κt ) Re (cid:16) e − i √ κt ( A κ w , A † κ w ) (cid:17) , where w = G κ (0 , a , b ) − u , A κ = √ ( κ / x + κ − / ∇ ) , and A † κ = √ ( κ / x − κ − / ∇ ) .Proof. It is well known that x U κ ( t ) = U κ ( t ) (cid:18) cos( √ κt ) x − i sin( √ κt ) √ κ ∇ (cid:19) . By definitions of A κ and A † κ , the right hand side is written as1 √ κ / ( e − i √ κt A κ + e i √ κt A † κ ) . It therefore holds that k x U κ ( t ) w k L = 12 κ / (cid:16) k A κ w k L + (cid:13)(cid:13) A † κ w (cid:13)(cid:13) L (cid:17) + 1 κ / Re (cid:16) e − i √ κt ( A κ w , A † κ w ) (cid:17) . Thus, integration in time gives us the desired result.10 roof of Theorem 5.
Let u = M G κ (0 , a , b )Ω n,κ ( x ). Then, M [ u ] = M holds. P [ u ] = M a and X [ u ] = M b follow from the second property ofLemma 2.1 since P [Ω n,κ ] = X [Ω n,κ ] = 0. One sees that G κ (0 , a , b ) − u =Ω n,κ . Therefore, the formula (4) tells us that u ( t ) = e − i Ψ( t ) G λ ( t, a , b ) U κ ( t )Ω n,κ = e − i Ψ( t ) G λ ( t, a , b ) e − i κ ( | n | + d ) t Ω n,κ . Moreover, we deduce from Lemma 4.1 that Ψ( t ) = ηM κ / ( | n | + d ) t . The secondpart is shown by a similar argument. We only note that Φ( t ) is proportional to t if and only if | a | = λ | b | , and that Φ( t ) = − ηM | b | t under this condition. Proof of Theorem 6.
Let a and b be as in assumption. We only consider (H).Since the Galilean transform is a linear map, G κ (0 , a , b ) − u = µ L X j =1 e α j G κ (0 , a j − a , b j − b )Ω n j ,κ , where we have applied the third and the seventh properties of Lemma 2.1.Thanks to the sixth properties of Lemma 2.1, we see U κ ( t ) G κ (0 , a , b ) − u = µ L X j =1 e α j G κ ( t, a j − a , b j − b )( e − i κ ( | n j | + d ) t Ω n j ,κ ) . Now the result is obvious by (4).
This section is devoted to the proof of Theorems 3 and 4. We prepare severallemmas for the proof of theorems.
Lemma 5.1.
Let { u n } ⊂ Σ / . If || u n − M / Ω n,κ || Σ / → , then P [ u n ] → and X [ u n ] → .Proof. This immediately follows from Lemma 3.1 and the fact that P [ M / Ω n,κ ] =0 and X [ M / Ω n,κ ] = 0. Lemma 5.2.
Let α ≥ , then sup t> | g ′ α ( t, a , b ) | → as | a | + | b | → .Proof. For the case α = 0, we have g ( t, a , b ) = a t + b and for the case α >
0, wehave g α ( t ) = a − / a sin α / t + b cos α / t . Thus, the conclusion of the lemmais obvious.We prove Theorem 3 by using Lemmas 2.2, 2.3, 5.1, and 5.2. Proof of Theorem 3.
We consider standing waves of (H). Fix n and set Q n,M = e − iω t/ M / Ω n,κ . Let s ≥ u ∈ Σ s with k u k L = M . Since { Ω n,κ } is a CONS of L , u is decomposed as u = X m a m Ω m,κ .
11y using the formula (4), one sees that u ( t ) = e i ˜Ψ( t ) e ix · ( g ′ κ ( t, − a , − b )+ g ′ λ ( t, a , b )) × X m e − itκ / ( | m | + d ) a m Ω m,κ ( x − g κ ( t, − a , − b ) − g λ ( t, a , b )) , where ˜Ψ( t ) is a function independent of x . Therefore, we have e i ˜Φ( t ) u ( t, x + g κ ( t, − a , − b ) + g λ ( t, a , b ))= e ix · ( g ′ κ ( t, − a , − b )+ g ′ λ ( t, a , b )) a n Ω n,κ + e ix · ( g ′ κ ( t, − a , − b )+ g ′ λ ( t, a , b )) X m = n e − itκ / ( | m |−| n | + d ) a m Ω m,κ , (14)where ˜Φ is another function independent of x . The difference between Q n,M and (14) is hence calculated as follows; (cid:13)(cid:13)(cid:13) Q n,M − e i ˜Φ( t )+ iω t/ u ( t, x + g κ ( t, − a , − b ) + g λ ( t, a , b )) (cid:13)(cid:13)(cid:13) Σ s ≤ | M / − a n | (cid:13)(cid:13) Ω n,κ (cid:13)(cid:13) Σ s + | a n | (cid:13)(cid:13)(cid:13) (1 − e ix · ( g ′ κ ( t, − a , − b )+ g ′ λ ( t, a , b )) )Ω n,κ (cid:13)(cid:13)(cid:13) Σ s + (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) e ix · ( g ′ κ ( t, − a , − b )+ g ′ λ ( t, a , b )) X m = n a m Ω m (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)(cid:13) Σ s . (15)It is obvious from Lemma 2.2 that the first term and the third term in (15) aresmall when (cid:13)(cid:13) u − M / Ω m,κ (cid:13)(cid:13) Σ s is sufficiently small. We deduce from Lemmas5.1 and 5.2 that if (cid:13)(cid:13) u − M / Ω m,κ (cid:13)(cid:13) Σ s is small then sup t> | g ′ κ ( t, − a , − b ) + g ′ λ ( t, a , b )) | ≪
1. It then follows from Lemma 2.3 that the second term in (15)is also small.For the case k u k L = M , we can show that u ( t ) is near Q n,M ′ where M ′ = k u k L . Since for u sufficiently near Q n,M , we have | M − M ′ | ≪
1. So, Q n,M and Q n,M ′ is also near to each other (up to phase). Therefore, we havethe conclusion. The proof for the standing waves of (H ′ ) is similar. We use (5)instead of (4).Next lemma is concerned with energy. By this lemma, one sees that, inTheorem 3, the standing wave solution is a ground state if n = 0 and an excitedstate otherwise. This is an immediate consequence of the representation of thesolutions, so we omit details of the proof. Lemma 5.3.
Let d > , λ, η ∈ R , and u , v ∈ Σ with M [ u ] = M [ v ] = M . Let κ = λ + ηM . Let u ( t ) and v ( t ) be solutions to (H) and (H ′ ) given in Theorems1 and 2, respectively. Then, the energy E [ u ( t )] (resp. E [ v ( t )] ) is conserved andgiven by (cid:18)(cid:18) −
12 ∆ + κ | x | (cid:19) w , w (cid:19) + M | a | + λ | b | ) , where w = G κ (0 , a , b ) − u (resp. w = G κ (0 , a , b ) − v ). Further, suppose κ > . Set e ( M ) := inf u ∈ Σ ,M [ u ]= M E [ u ] . The following holds. . If λ > then E [ u ] = e ( M ) holds if and only if w = M / Ω ,κ and a = b = 0 .2. If λ = 0 then E [ u ] = e ( M ) holds if and only if w = M / Ω ,κ and a = 0 .3. If λ < then e ( M ) = −∞ . We now move to the proof of Theorem 4. Before the proof, let us introducesome notations. In what follows, we shall consider (H) in the following cases.(I) d = 1, λ = 0 and η = 1,(II) d = 1, λ = 2 and η = − I and (H) II , respectively.By Theorem 5, the following functions are solutions of (H) I and (H) II . Q I ,n ( t, x ; M ) := exp (cid:18) − M / (cid:18) n + 12 (cid:19) t (cid:19) M + Ω n ( M / x ) ,Q II ,n ( t, x ; M ) := exp (cid:18) − (cid:18) n + 12 (cid:19) (cid:18) κ / − M κ − / (cid:19) t (cid:19) M κ Ω n ( κ / x ) , where κ = 2 − M . We can also write Q I ,n = e − ω I ,n ( M )2 t M + Ω n ( M / x ) ,Q II ,n = e − ω II ,n ( M )2 t M κ Ω n ( κ / x ) , where ω I ,n ( M ) := 3 M / (cid:18) n + 12 (cid:19) ,ω II ,n ( M ) := 2 (cid:18) n + 12 (cid:19) (cid:18) κ / − M κ − / (cid:19) . Let S ω, I , S ω, II be the action of (H) I and (H) II , respectively. Further, set d I ( ω ) := S ω, I (cid:16) M + Ω n ( M / x ) (cid:17) ,d II ( ω ) := S ω, II (cid:16) M κ Ω n ( κ / x ) (cid:17) . Since Q I ,n and Q II ,n are stable (Theorem 3), it suffices to show the followinglemma. Lemma 5.4.
Let n be an even integer. Then, we have the following. (i) d ′′ I ( ω ) | ω = ω (1) < and n (cid:16) S ′′ ω, I (Ω n ) (cid:12)(cid:12) L r (cid:17) = n . (ii) d ′′ II ( ω ) | ω = ω (1) > and n (cid:16) S ′′ ω, II (Ω n ) (cid:12)(cid:12) L r (cid:17) = n + 1 .Proof of Lemma 5.4. We first calculate d ′′ for the cases (I) and (II). Since d ( ω ) = S ω ( φ ω ), one obtains d ′ ( ω ) = − Q ( φ ω ). Thus, d ′′ ( ω ) = − dMdω . For13he case (I), we have ω I ,n ( M ) = 3 M / (cid:0) n + (cid:1) . Differentiate ω I ,n with respectto M to yield dω I ,n dM (cid:12)(cid:12)(cid:12)(cid:12) M =1 = 32 (cid:18) n + 12 (cid:19) > , which implies d ′′ I ( ω ) | M =1 <
0. On the other hand, we have ω II ,n ( M ) = (2 n + 1) (2 − M ) − / (cid:0) − M (cid:1) since κ = 2 − M . One hence verifies that dω II ,n dM (cid:12)(cid:12)(cid:12)(cid:12) M =1 = − (cid:18) n + 12 (cid:19) < , which yields d ′′ II ( ω ) | M =1 > S ′′ ω (Ω n ) for the cases (I) and (II). Weexpress S ′′ ω (Ω n ) by a 2 × L i,j ) i,j =1 , as follows (cid:18) Re (cid:0) S ′′ I ,ω (Ω n ) u (cid:1) Im (cid:0) S ′′ I ,ω (Ω n ) u (cid:1)(cid:19) = (cid:18) L L L L (cid:19) (cid:18) Re u Im u (cid:19) . For the case (I), we have E = R R |∇ u | + R R R R | x − y | | u ( x ) | | u ( y ) | . So, wehave S ′′ I ,ω (Ω n )= (cid:18) − ∆ + | x | + 2( | x | ∗ (Ω n · ))Ω n − (cid:0) n + (cid:1) − ∆ + | x | − (cid:0) n + (cid:1)(cid:19) . It is easy to see that L = − ∆ + | x | − (cid:0) n + (cid:1) has n/ L = − ∆+ | x | +2( | x | ∗ (Ω n · ))Ω n − (cid:0) n + (cid:1) = L +2( | x | ∗ (Ω n · ))Ω n . Let h = P n ∈ N ∪{ } a l Ω l .Then, it holds that (cid:10)(cid:0) | x | ∗ (Ω n h ) (cid:1) Ω n , h (cid:11) = Z R Z R | x − y | Ω n ( x )Ω n ( y ) h ( x ) h ( y ) dydx = 2 Z R | x | Ω n ( x ) h ( x ) dx Z R Ω n ( y ) h ( y ) dy = p ( n + 1)( n + 2)2 a n +2 a n + (cid:18) n + 12 (cid:19) a n + p n ( n − a n a n − Therefore, the term 2( | x | ∗ (Ω n · ))Ω n affects only the frame { Ω n − , Ω n , Ω n +2 } .In this frame, the action of L can be represented as the following matrix. A I = ( A ij ) i,j =1 , , = − √ n ( n − √ n ( n − n + √ ( n +1)( n +2)4 √ ( n +1)( n +2)4 , where L ( a Ω n − + a Ω n + a Ω n +2 ) = (cid:16)P j =1 , , A ij a j (cid:17) Ω n − i . The char-acteristic polynomial of A I becomes F I ( λ ) = λ − (cid:18) n + 12 (cid:19) λ − (cid:0) n + n + 33 (cid:1) λ + 72 (cid:18) n + 12 (cid:19) . F I (0) > > F I ( n ) holds for any even integer n , A I has one negativeeigenvalue and two positive eigenvalues. Therefore, we have the conclusion.Now we consider the case (II). As in the case (I), since E = R R |∇ u | + R R | x | | u | − R R R R | x − y | | u ( x ) | | u ( y ) | , one obtains S ′′ II ,ω (Ω n )= (cid:18) − ∆ + | x | + 2( | x | ∗ (Ω n · ))Ω n − (cid:0) n + (cid:1) − ∆ + | x | − (cid:0) n + (cid:1)(cid:19) . In this case, the representation matrix of S ′′ II ,ω (Ω n ) (cid:12)(cid:12) { Ω n − , Ω n , Ω n +2 } is A II = − − √ n ( n − − √ n ( n − − (cid:0) n + (cid:1) − √ ( n +1)( n +2)4 − √ ( n +1)( n +2)4 , and the characteristic polynomial becomes as F II ( λ ) = λ + (cid:18) n + 12 (cid:19) λ − (cid:0) n + n + 33 (cid:1) λ − (cid:18) n + 12 (cid:19) . Since F II ( λ ) = − F I ( − λ ), we see that A II has two negative eigenvalues and onepositive eigenvalue.Therefore, we have the conclusion of the Lemma. Acknowledgment
The authors express their deep gratitude to Professor Kenji Yajima for fruitfuldiscussions and for his leading the authors to the work of Kato. The secondauthor is supported by Japan Society for the Promotion of Science (JSPS)Grant-in-Aid for Research Activity Start-up (22840039).
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