A variational characterization of 2-soliton profiles for the KdV equation
aa r X i v : . [ m a t h . A P ] J a n A variational characterization of2-soliton profiles for the KdV equation
John P. AlbertDepartment of Mathematics, University of OklahomaNorman, OK 73019 [email protected]
Nghiem V. NguyenDepartment of Mathematics and Statistics, Utah State UniversityLogan, UT 84322 [email protected]
January 27, 2021
Abstract
We use profile decomposition to characterize 2-soliton solutions of theKdV equation as global minimizers to a constrained variational probleminvolving three of the polynomial conservation laws for the KdV equation.
The variational properties of multi-soliton solutions of the Kortweg-de Vries(KdV) equation have played a central role in the study of this equation sinceshortly after the discovery of its remarkable properties in the 1960s. An earlymilestone was the paper [12] of Lax, in which it is pointed out that multi-soliton profiles are critical points of constrained variational problems in whichthe constraint functionals and the objective functional are conserved under theflow defined by the KdV equation. This suggested that it might be possible toestablish stability properties of multi-soliton solutions by using the conservedquantities as Lyapunov-type functionals.Benjamin [4] (see also Bona [5]) took a first step in this direction by showingthat solitary-wave profiles are local minimizers in H ( R ) for the conserved func-tional E , subject to the constraint that E be held constant (see Section 2 forthe definition of E j ), and deducing the orbital stability of solitary-wave solutionsas a consequence. Later, the work of Cazenave and Lions (see for example, [7],1nd the expository treatment in [1]) established that solitary-wave profiles areactually global minimizers of this variational problem in a strong sense: everyminimizing sequence for the variational problem has a subsequence which, afterappropriate translations, converges in H ( R ) to a solitary-wave profile. Orbitalstability of solitary waves is an immediate consequence.Maddocks and Sachs [18] generalized the theory of Benjamin and Bona toobtain a stability result for multi-soliton solutions of KdV. A key step in theirproof was to show that the profiles of N -soliton solutions are local minimizersin H N ( R ) of the conserved functional E N +2 , when the functionals E , E , . . . , E N +1 are held constant. Their proof, like that of Benjamin and Bona for singlesolitons, did not yield information about global minimizers of the variationalproblem.In this paper, we consider the special case N = 2 of the variational problemconsidered in [18]: that is, the problem of minimizing E when E and E areheld constant. In our main result, Theorem 2.6 below, we show that indeed2-soliton solutions represent the global minimizers for this variational problem.An easy consequence of Theorem 2.6 is a stability result for 2-soliton solu-tions in H ( R ), stated below as Corollary 2.7. Of course, this is only a specialcase of the stability result of [18], which was asserted for N -solitons for general N , not just for N = 2. Moreover, in recent years a number of papers haveappeared on the topic of stability of multi-solitons which have improved on theresult of [18]. In particular, Killip and Visan [11] have proved a stability re-sult for N -soliton solutions of KdV which is in some sense optimal: it assertsstability in H − , or more generally in any space H s with s ≥ −
1. Instead ofthe variational characterization of multi-solitons used here or in [18], they usea different variational characterization, which is motivated by the inverse scat-tering theory for KdV, yet which is well-adapted to potentials in low-regularitySobolev spaces where the classical inverse scattering theory does not apply. Wealso note the recent work of Le Coz and Wang [15], who by building on andelucidating the work of [18] obtain a stability result for N -soliton solutions ofthe modified KdV equation. Their methods should be transferrable to otherintegrable equations as well; and in particular it would be worth using them torevisit the stability theory for KdV multisolitons.We feel that the result and proof of Theorem 2.6 are interesting in theirown right, apart from the consequences for stability theory. The result settles,at least in the case N = 2, the question of whether multisolitons are actuallyglobal minimizers of a natural variational problem for KdV, expressed in termsof polynomial conservation laws which can be viewed as action variables in aformulation of KdV as an infinite-dimensional Hamiltonian system. It can thusbe viewed as a step towards obtaining an analogue for KdV on the real line of theelegant theory produced for the periodic KdV equation by Lax [13] and Novikov[21]. The proof has the advantage of simplicity: it shows that the result is astraightforward consequence of the profile decomposition, a general phenomenonunconnected with the KdV equation or its structure, once one shows that 2-soliton profiles are minimizers for the constrained variational problem whenconsideration is restricted to the set of multi-soliton profiles. In other words,2he fact that 2-soliton profiles are global minimizers in H can be shown to followfrom the profile decomposition, together with the fact that they are minimizerswithin the set of all multi-soliton profiles.An important caveat, however, is that the argument can only proceed be-cause of the uniqueness result for 2-solitons stated as Theorem 2.2 below; andsuch uniqueness results can be very difficult to prove in other settings. In fact,one of the main reasons we have restricted ourselves to 2-solitons in the presentpaper is that an analogue of Theorem 2.2 is not yet available for general N -solitons (see [2] for a discussion of what remains to be shown).The plan of the remainder of the paper is as follows. In Section 2 we reviewsome basic properties of N -soliton solutions and polynomial conservation lawsfor the KdV equation, state our main result Theorem 2.6, and sketch its proof.In Section 3, we review the profile decomposition, following [19]. In Section 4,we analyze a finite-dimensional minimization problem which arises from restrict-ing the admissible functions in (2.6) and (2.7) to N -soliton profiles. Section 5contains the proof of our main result, Theorem 2.6, and concludes with a proofof Corollary 2.7. Notation. If E is a measurable subset of R and 1 ≤ p < ∞ , we define L p ( E ) tobe the space of Lebesgue measurable real-valued functions u on E such that k u k L p ( E ) = (cid:0)R E | u | p dx (cid:1) /p is finite. In the case when E = R , we sometimesdenote L p ( R ) by simply L p , and denote the norm of u in L ( R ) by k u k L .When E is an open set in R , for l ∈ N , we define the L -based Sobolevspace H l = H l ( E ) to be the closure of the space C ∞ ( E ) of all infinitely smoothreal-valued functions on E with respect to the norm k u k H l ( E ) = l X i =0 Z E (cid:18) d i udx i (cid:19) dx ! / . Note that H ( E ) = L ( E ). In the case when E = R , we sometimes denote H l ( R ) by simply H l , and denote the norm of u in H l ( R ) by k u k H l .For x ∈ R and r > B ( x, r ) the open ball in R centered at x with radius r , or in other words the interval ( x − r, x + r ). Also, for any subset E of R , we denote by χ E the characteristic function of E , so that χ E ( x ) = 1for x ∈ E and χ E ( x ) = 0 for x / ∈ E . We begin by reviewing the definition and some basic properties of N -solitonsolutions of the Korteweg-de Vries (KdV) equation, and the associated sequenceof polynomial conservation laws. For more details and further references, thereader is referred to, for example, [8] and [9]; the early papers [12, 13, 14] ofLax are also very readable. 3uppose N ∈ N , 0 < C < · · · < C N and ( γ , . . . , γ N ) ∈ R N . An N -solitonprofile function is a function of the form ψ C ,...,C N ; γ ,...,γ N ( x ) = 3( D ′ /D ) ′ where D is defined as the N × N determinant of Wronskian form, D = D ( y , . . . , y N ) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) y . . . y N y ′ . . . y ′ N . . . . . . . . .y ( N − . . . y ( N − N (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) , with y j ( x ) = e √ C j ( x − γ j ) + ( − j − e − √ C j ( x − γ j ) , j = 1 , . . . , N. From N -soliton profile functions, we can construct N -soliton solutions of theKdV equation, u t + uu x + u xxx = 0 , (2.1)simply by defining u ( x, t ) = ψ C ,...,C N ; γ ( t ) ,...,γ N ( t ) ( x )where for j = 1 , . . . , N , γ j ( t ) = a j + C j t, and ( a , . . . , a N ) ∈ R N is arbitrary.In particular, a single-soliton profile is obtained by taking N = 1, in whichcase we have, for C > γ ∈ R , ψ C,γ ( x ) = 3( D ′ /D ) ′ where D = e √ C ( x − γ ) + e −√ C ( x − γ ) . In other words, ψ C,γ ( x ) = 3 C cosh ( √ C ( x − γ )) . Then a single-soliton solution of the KdV equation is obtained by taking u ( x, t ) = ψ C,γ ( t ) ( x )where γ ( t ) = a + Ct and a ∈ R is arbitrary.If the constants γ , . . . , γ N are widely separated, then the profile function ψ C ,...,C N ; γ ,...,γ N closely resembles a sum of single-soliton profiles P Ni =1 ψ C i ,γ i .A particular instance of this well-known fact that we will use below is thefollowing: Lemma 2.1.
Suppose < C < C ; γ , γ ∈ R ; and { x n } and { x n } aresequences such that lim n →∞ | x n − x n | = ∞ . Then lim n →∞ (cid:13)(cid:13) ψ C ,γ + x n + ψ C ,γ + x n − ψ C ,C ,γ + x n ,γ + x n (cid:13)(cid:13) H ( R ) = 0 . roof. This follows immediately from Lemma 3.6 of [3].The variational problem we are concerned with here has, as objective andconstraint functionals, polynomial conservation laws for the KdV equation.These are functionals of the form E j ( u ) = Z R P (cid:18) u, u x , u xx , . . . , ∂ j − u∂x j − (cid:19) dx where j ≥ P is a polynomial in its arguments. They are conservationlaws in the sense that, if u ( x, t ) is a solution of (2.1), then (at least formally), ddt [ E j ( u ( x, t ))] = 0for all t ∈ R . As shown for example in [14], the KdV equation has an infinitefamily of such conserved functionals, the first three of which are given by: E ( u ) = Z R u dx,E ( u ) = Z R (cid:18) u x − u (cid:19) dx,E ( u ) = Z R (cid:18) u xx − uu x + 532 u (cid:19) dx. Sobolev embedding theorems imply that for each N ≥ E N +2 definesa continuous functional on the Sobolev space H N = H N ( R ) of real-valuedfunctions whose derivatives up to order N are in L ( R ). For u ∈ H N , wedenote by ∇ E k ( u ) the Fr´echet derivative of E k at u , which coincides with theGateaux derivative of E k and is therefore defined as a linear functional on H N by ∇ E k ( u )[ v ] = lim ǫ → E k ( u + ǫv ) − E ( u ) ǫ , and identified with a function ∇ E k ( u )( x ) in the usual way, so that ∇ E k ( u )[ v ] = R R ∇ E k ( u )( x ) · v ( x ) dx for v ∈ H N . In particular, the Fr´echet derivatives of thefirst few functionals are given by ∇ E ( u ) = u, ∇ E ( u ) = − u xx − u, ∇ E ( u ) = u xxxx + 53 u xx + 56 ( u x ) + 58 u . For fixed 0 < C < · · · < C N , define S = S ( C , . . . , C N ) = (cid:8) ψ C ,...,C N ; γ ,...,γ N ( x ) : ( γ , . . . , γ N ) ∈ R N (cid:9) . (2.2)5t is well-known (for a proof, see for example Theorem 3.8 of [2]) that thereexist constants λ , . . . λ N +1 such that each ψ ∈ S satisfies the ordinary differ-ential equation ∇ E N +2 ( ψ ) = N +1 X k =2 λ k ∇ E k ( ψ ) . (2.3)Thus N -soliton profiles are (non-isolated) critical points for constrained varia-tional problems involving the functionals E j . The family of ordinary differentialequations (2.3) is collectively known as the stationary KdV hierarchy. Due towork of Novikov, Its/Matveev, and Gelfand/Dickey in the 1970’s (see [2] forreferences), it is known that each equation in the hierarchy has the structure ofa completely integrable Hamiltonian system, and indeed can be explicitly solvedby integration.In the case N = 2, equation (2.3) takes the form ∇ E ( ψ ) = λ ∇ E ( ψ ) + λ ∇ E ( ψ ) , (2.4)or ψ ′′′′ + 53 ψψ ′′ + 56 ( ψ ′ ) + 58 ψ = λ ψ − λ ( ψ ′′ + 12 ψ ) . A fact which is crucial for the proof of our main result below is that there is nochoice of the numbers λ , λ for which (2.4) has any solutions in H ( R ) besides1-soliton and 2-soliton profiles: Theorem 2.2 ([2]) . Suppose ψ ∈ H is a solution of (2.4) , in the sense ofdistributions, for some λ , λ ∈ R . Then ψ is either a 1-soliton or a 2-solitonprofile for the KdV equation. The proof given in [2] for Theorem 2.2 relies on the fact that, as mentionedabove, (2.4) can be explicitly integrated.We will also make crucial use of the fact that, for all k ≥ E k is constanton S , with its value on S given by E k ( C , . . . , C N ) := ( − k k − N X j =1 C (2 k − / j . (2.5)To prove (2.5), one first shows that it is valid in the case of a single-soliton profile,when N = 1 (cf. equation (3.18) of [18]). For a general N -soliton profile ψ ( x ) = ψ C ,...,C N ; γ ,...,γ N ( x ), one then proves (2.5) by considering a solution u ( x, t ) of(2.1) with u ( x,
0) = ψ ( x ). Since u ( x, t ) resolves into widely separated single-soliton profiles as t → ∞ , it follows that lim t →∞ E k ( u ( x, t )) = E k ( C , . . . , C N ).But since E k is a conserved functional for KdV, it then follows that E k ( ψ ) = E k ( C , . . . , C N ) as well.We now consider the constrained variational problem of minimizing the func-tional E over H ( R ), subject to the constraints that E and E be held con-stant. This is the same variational problem considered, in the case N = 2, in6he stability theory for N -solitons presented in [18]. Whereas it was shown in[18] that 2-soliton profiles are local minimizers of the variational problem, wewill show in Theorem 2.6 that they are global minimizers (and in fact, by theuniqueness result Theorem 2.2, every global minimizer is a 2-soliton profile).Moreover, every minimizing sequence for the problem must converge stronglyin H ( R ) to the set of minimizers.We begin by introducing some notation concerning the variational problem. Proposition 2.3.
Suppose ( a, b ) ∈ R . Then there exists a nonzero function g in H ( R ) such that E ( g ) = a and E ( g ) = b if and only if ( a, b ) ∈ Σ , where Σ = n ( a, b ) ∈ R : a > and b ≥ − ma / o , and m = 365 (cid:18) (cid:19) / . Proof.
From Cazenave and Lions’ variational characterization of solitary waves(see for example Theorem 2.9 and Proposition 2.11 of [1]), we know that forall a > g ∈ H ( R ) such that E ( g ) = a , we have E ( g ) ≥ − ma / ;and the minimum value E ( g ) = − ma / is attained when (and only when) g = ψ C,γ , where C = ( a/ / and γ ∈ R is arbitrary. It follows that if g is anonzero function in H ( R ) with E ( g ) = a and E ( g ) = b , then ( a, b ) ∈ Σ.Conversely, suppose ( a, b ) ∈ Σ. For α >
0, define g α ( x ) = √ αψ C,γ ( αx ),where C and γ are as above. Then E ( g α ) = a for every α > E ( g α ) = − ma / when α = 1, and E ( g α ) → + ∞ as α → ∞ . Therefore, since b ≥− ma / , the intermediate value theorem implies the existence of some α ∈ [1 , ∞ )for which E ( g α ) = b . Thus by taking g = g α , we can satisfy g ∈ H ( R ), E ( g ) = a , and E ( g ) = b .For ( a, b ) ∈ Σ, define Λ( a, b ) ⊆ R byΛ( a, b ) = { r ∈ R : for some g ∈ H ( R ), E ( g ) = a , E ( g ) = b , and E ( b ) = r } . (2.6)By the definition of Σ, the set on the right-hand side is nonempty, and we cantherefore define J ( a, b ) = inf Λ( a, b ) . (2.7)(Notice that we do not exclude here the possibility that J ( a, b ) = −∞ . However,as shown below at the beginning of Section 5, in fact J ( a, b ) > −∞ for all( a, b ) ∈ Σ.)
Definition 2.4.
Suppose ( a, b ) ∈ Σ . We say that a function φ ∈ H ( R ) is a minimizer for J ( a, b ) if E ( φ ) = a , E ( φ ) = b , and E ( φ ) = J ( a, b ) . We saythat a sequence { φ n } of functions in H ( R ) is a minimizing sequence for J ( a, b ) if lim n →∞ E ( φ n ) = a , lim n →∞ E ( φ n ) = b , and lim n →∞ E ( φ n ) = J ( a, b ) . J ( a, b ) exist, then they must of necessity be 1-soliton or 2-solitonprofiles: Proposition 2.5.
Suppose a and b are real numbers and u ∈ H is a minimizerfor J ( a, b ) . Then u must be either a 1-soliton profile or a 2-soliton profile. Inother words we have u ( x ) = ψ D ,D ; γ ,γ ( x ) for some real numbers D , D , γ , γ , with ≤ D < D .Proof. According to Theorem 2 on page 188 of [16], if u is a regular point of theconstraint functionals E and E , meaning that the Fr´echet derivatives ∇ E ( u )and ∇ E ( u ) are linearly independent, then there must exist real numbers λ and λ such that the equation ∇ E ( u ) = λ ∇ E ( u ) + λ ∇ E ( u ) holds, at leastin the sense of distributions. In this case, by Theorem 2.2, u is either a 1-solitonor a 2-soliton profile. On the other hand, if u is not a regular point of theconstrained functionals, then u satisfies the equation ∇ E ( u ) = λ ∇ E ( u ) forsome λ ∈ R , and it is an elementary exercise (see for example Theorem 4.2 of[2]) to show that the only possible solutions of this equation in H are 1-solitonprofiles.However, the preceding result of course leaves open the question of whetherany 1-soliton or 2-soliton profiles are in fact minimizers for J ( a, b ). Our mainresult determines the set of values of ( a, b ) within Σ for which minimizers for J ( a, b ) exist; and for such values of ( a, b ) determines the value of J ( a, b ), de-scribes all the minimizers for J ( a, b ), and describes the behavior of minimizingsequences for J ( a, b ). If S ⊆ H ( R ), we say that a sequence { φ n } converges to S in H ( R ) norm if d ( φ n , S ) = inf ψ ∈ S k φ n − ψ k H ( R ) → n → ∞ ,or, equivalently, if there exists a sequence { ψ n } of elements of S such thatlim n →∞ k φ n − ψ n k H ( R ) = 0 . Theorem 2.6.
Suppose ( a, b ) ∈ Σ ; that is, a > and b ≥ − ma / , where m = (cid:0) (cid:1) / .1. If b = − ma / , (2.8) then every minimizing sequence for J ( a, b ) converges to S ( C ) in H ( R ) norm, where S ( C ) is as defined in (2.2) , with C = ( a/ / = ( − b/ / .Every element of S ( C ) is a minimizer for J ( a, b ) , and J ( a, b ) = E ( C ) .2. If − ma / / > b > − ma / , (2.9)8 hen every minimizing sequence for J ( a, b ) converges to S ( C , C ) in H ( R ) norm, where ( C , C ) is the unique pair of numbers such that < C < C , E ( C , C ) = a , and E ( C , C ) = b . Every element of S ( C , C ) is aminimizer for J ( a, b ) , and J ( a, b ) = E ( C , C ) .3. If b ≥ − ma / / , (2.10) then there do not exist any minimizers for J ( a, b ) in H . We remark that the method used below to analyze the behavior of mini-mizing sequences under the assumptions (2.8) or (2.9) should also be applicablein case (2.10) holds, and suggests that in the latter case, minimizing sequences { φ n } should, as n → ∞ , come to resemble superpositions of widely separatedsingle-soliton profiles. Thus, for example, if b = − ma / / we expect that if { φ n } is a minimizing sequence for J ( a, b ), then there will exist a number C > { γ n } and { γ n } with lim n →∞ | γ n − γ n | = ∞ such thatlim n →∞ k φ n − ( ψ C,γ n + ψ C,γ n ) k H = 0 . Similarly, if b > − ma / / , we expect that the functions in a typical minimizingsequence for J ( a, b ) would resemble a superposition of three or more single-soliton profiles, two of which have equal amplitudes and whose distance fromeach other increases to infinity as n → ∞ . We do not pursue this topic furtherhere, however.We also remark that the method of proof of Theorem 2.6 should apply as wellto the variational problems satisfied by N -soliton profiles for general N ∈ N .One important obstacle we have encountered, however, is that of proving ananalogue of Theorem 2.2: i.e., of showing that for all possible choices of thenumbers λ , . . . , λ N +1 , the Euler-Lagrange equation ∇ E N +2 ( ψ ) = λ ∇ E ( ψ ) + · · · + λ N +1 ∇ E N +1 ( ψ ) (2.11)has no solutions in H N besides N -soliton profiles. As noted in [2], the explicitintegration of equation (2.11) can be carried out for general N just as it can for N = 2, whenever the value of ( λ , . . . , λ N +1 ) corresponds to that of an N -solitonprofile; but technical difficulties arise in proving that solutions corresponding toother values of ( λ , . . . , λ N +1 ) are singular.An immediate consequence of Theorem 2.6 is a stability result for 2-solitonsolutions of the KdV equation. This recovers (by a different proof) the specialcase N = 2 of the general result for N -soliton solutions given by Maddocks andSachs in [18]. 9 orollary 2.7. Suppose < C < C . Then every minimizing sequence { φ n } for J ( E ( C , C ) , E ( C , C )) converges strongly to S ( C , C ) in H ( R ) . More-over, S = S ( C , C ) is stable, in the sense that for every ǫ > there exists δ > such that if u ∈ H ( R ) and d ( u , S ) < δ , then d ( u ( · , t ) , S ) < ǫ for all t > .Remark. It follows from this stability result that there are C functions γ ( t )and γ ( t ) defined for t ≥ k u ( · , t ) − ψ C ,C ; γ ( t ) ,γ ( t ) k H ( R ) < ǫ and | γ ′ ( t ) − C | < ǫ , | γ ′ ( t ) − C | < ǫ for all t >
0. See [3].We conclude this section by sketching the idea of the proof of Theorem 2.6which is given in the succeeding sections.Suppose ( a, b ) ∈ Σ with − ma / ≤ b < − ma / / , and suppose { φ n } is aminimizing sequence for J ( a, b ). We can apply the profile decomposition, in theform of Corollary 3.4, to the sequence { ρ n } defined by ρ n = | φ n | + | φ ′ n | + | φ ′′ n | . It follows that we can decompose φ n as φ n = n X i =1 v in + w n , where the sequence { w n } is vanishing in the sense of Definition 3.1, and for each i ∈ N the sequence { v in } n ∈ N concentrates around some sequence { x in } n ∈ N , in thesense of Definition 3.2. From the concentration property of { v in } n ∈ N and thefact that { φ n } is a minimizing sequence, it follows that the sequence { v in } n ∈ N can be suitably translated so that it converges, weakly in H ( R ) and stronglyin H ( R ), to a minimizer g i for the variational problem E ( g i ) = inf (cid:8) E ( φ ) : φ ∈ H ( R ) , E ( φ ) = a i , E ( φ ) = b i (cid:9) for some real numbers a i , b i . As a critical point of this constrained variationalproblem, ψ = g i must satisfy the Euler-Lagrange equation (2.4). From Theorem2.2 it then follows that for each i we have g i = ψ D i ,D i for some numbers D i and D i with 0 ≤ D i < D i .In parts 1 and 2 of Theorem 2.6, our assumption on ( a, b ) implies that thereexist numbers C and C with 0 ≤ C < C such that E ( ψ C ,C ) = a and E ( ψ C ,C ) = b . Therefore, by definition of J ( a, b ), we have that J ( a, b ) ≤ E ( ψ C ,C ) = (36 / (cid:16) C / + C / (cid:17) . From the profile decomposition and the fact that { φ n } is a minimizing sequence,10e can obtain that ∞ X i =1 E ( g i ) = 36 ∞ X i =1 (cid:0) D i + D i (cid:1) ≤ lim n →∞ E ( φ n ) = 36 (cid:0) C + C (cid:1) ∞ X i =1 E ( g i ) = − ∞ X i =1 (cid:0) D i + D i (cid:1) ≤ lim n →∞ E ( φ n ) = − (cid:0) C + C (cid:1) ∞ X i =1 E ( g i ) = 367 ∞ X i =1 (cid:0) D i + D i (cid:1) ≤ lim n →∞ E ( φ n ) ≤ (cid:0) C + C (cid:1) . Permuting the terms of the sequence ( D / , D / , D / , D / , D / , D / , . . . )so that they form a decreasing sequence ( x , x , x , . . . ); and defining y = C / and y = C / , we thus have that ∞ X i =1 x i ≤ y + y ∞ X i =1 x i ≥ y + y ∞ X i =1 x i ≤ y + y . We analyze this system of inequalities in Section 4, where we show (cf. Lemma4.10) that it can only be satisfied if x = C , x = C , and x i = 0 for all i ≥ N -soliton profiles of the KdVequation, the only ones which could possibly solve the variational problem are1-soliton profiles (in the case when C = 0) and 2-soliton profiles (in the casewhen C > { v in } and w n afforded by the profile decomposition, is enough to allow us to deduce that thefunctions in the minimizing sequence { φ n } are either of the form φ n ( x ) = ψ C ,C ( x + x n ) + r n ( x )for some sequence { x n } of real numbers, where r n → H ( R ); or of the form φ n ( x ) = φ C ( x + x n ) + φ C ( x + x n ) + r n ( x )for some pair of sequences { x n } and { x n } of real numbers with | x n − x n | →∞ , where again r n → H ( R ). In either case, this shows that the set ofminimizing functions for the variational problem consists of the set S ( C , C ),and that { φ n } converges to S ( C , C ) in H ( R ) norm.Part 3 of Theorem 2.6 will follow from a simpler argument: under the givenassumptions on ( a, b ), no 1-soliton profile or 2-soliton profile ψ can exist sat-isfying the constraints E ( ψ ) = a and E ( ψ ) = b . But any minimizer for thevariational problem must satisfy the associated Euler-Lagrange equation (2.4),and therefore by Theorem 2.2 must be either a 1-soliton profile or a 2-solitonprofile. Hence no minimizers can exist.11 Profile decomposition
The idea of the proof of Theorem 2.6 is to use an elaboration of the method ofconcentration compactness, known as profile decomposition , which details theways in which a sequence of measures of bounded total mass can lose compact-ness.The technique of profile decomposition dates back to [10] and in some formeven earlier (see for example [6]). We will use a version which is due to Mari¸s [19].Actually, although the result of [19] is valid for arbitrary bounded sequences ofBorel measures on any metric space, for simplicity of notation we here restrictconsideration to bounded sequences of nonnegative functions in L ( R ). Definition 3.1.
We say that a sequence { f n } of nonnegative functions in L ( R ) is vanishing if for every r > , we have lim n →∞ sup y ∈ R Z B ( y,r ) f n = 0 . Definition 3.2. If { f n } is a sequence of nonnegative functions in L ( R ) and { x n } is a sequence of real numbers, we say that { f n } concentrates around { x n } if for every ǫ > , there exists r ǫ > such that Z R \ B ( x n ,r ǫ ) f n < ǫ for every n ∈ N . Theorem 3.3 ([19]) . Suppose { ρ n } is a sequence of nonnegative functions whichis bounded in L ( R ) . Then either { ρ n } is vanishing, or there exists a subse-quence of { ρ n } (which we continue to denote by { ρ n } ), which satisfies one ofthe following two properties: either(1) there exist k ∈ N and for each i ∈ { , . . . , k } a number m i > and sequenceof balls { B ( x in , r in ) } n ∈ N in R with lim n →∞ r in = ∞ , such that(a) B ( x in , r in ) ∩ B ( x jn , r jn ) = ∅ for all n ∈ N and all i, j ∈ { , . . . , k } with i = j ,(b) for each i ∈ { , . . . , k } , lim n →∞ Z B ( x in ,r in / ρ n = m i ,(c) for each i ∈ { , . . . , k } , lim n →∞ Z B ( x in ,r in ) \ B ( x in ,r in / ρ n = 0 ,(d) for each i ∈ { , . . . , k } , the sequence { ρ n χ B ( x in ,r in ) } n ∈ N concentratesaround { x in } n ∈ N ,(e) the sequence { ρ n χ R \∪ ki =1 B ( x in ,r in ) } n ∈ N is vanishing;or
2) for each i ∈ N there is a number m i > and a sequence of balls { B ( x in , r in ) } n = i,i +1 ,i +2 ,... in R , with lim n →∞ r in = ∞ , such that(a) B ( x in , r in ) ∩ B ( x jn , r jn ) = ∅ for all i, j ∈ N with i = j , and all n ∈ N with n ≥ i and n ≥ j ,(b) for each i ∈ N , lim n →∞ Z B ( x in ,r in / ρ n = m i ,(c) for each i ∈ N , ∞ X n = i Z B ( x in ,r in ) \ B ( x in ,r in / ρ n ≤ i ,(d) for each i ∈ N , the sequence { ρ n χ B ( x in ,r in ) } n ≥ i concentrates around { x in } n ≥ i ,(e) the sequence { ρ n χ R \∪ ni =1 B ( x in ,r in ) } n ∈ N is vanishing, and(f ) if for each N ∈ N and each n ≥ N , we define g Nn = ρ n χ R \∪ Ni =1 B ( x in ,r in ) ,and define the increasing function q Nn ( r ) for r > by q Nn ( r ) = sup y ∈ R Z B ( y,r ) g Nn , then lim N →∞ (cid:18) lim r →∞ (cid:18) lim sup n →∞ q Nn ( r ) (cid:19)(cid:19) = 0 . (3.1)One can view (3.1) as saying that although, for any given value of N , thesequence { g Nn } n ∈ N is not necessarily vanishing, it does come closer, in somesense, to being a vanishing sequence as N → ∞ .To shorten our proof of Theorem 2.6, we observe that the two cases inTheorem 3.3 can be combined into one, if we drop the requirement that m i > i : Corollary 3.4.
Suppose { ρ n } is a sequence of nonnegative functions whichis bounded in L ( R ) , and suppose { ρ n } is not vanishing. Then there exists asubsequence of { ρ n } (which we continue to denote by { ρ n } ), a sequence { m i } i ∈ N of nonnegative numbers, and for each i ∈ N a sequence of balls { B ( x in , r in ) } n ∈ N in R , with lim n →∞ r in = ∞ , such that(a) B ( x in , r in ) ∩ B ( x jn , r jn ) = ∅ for all i, j ∈ N with i = j , and all n ∈ N ,(b) for each i ∈ N , lim n →∞ Z B ( x in ,r in / ρ n = m i ,(c) for each i ∈ N , ∞ X n = i Z B ( x in ,r in ) \ B ( x in ,r in / ρ n ≤ i ,(d) for each i ∈ N , the sequence { ρ n χ B ( x in ,r in ) } n ∈ N concentrates around { x in } n ∈ N , e) the sequence { ρ n χ R \∪ ni =1 B ( x in ,r in ) } n ∈ N is vanishing, and(f ) if for each N ∈ N and each n ∈ N , we define g Nn = ρ n χ R \∪ Ni =1 B ( x in ,r in ) , anddefine the increasing function q Nn ( r ) for r > by q Nn ( r ) = sup y ∈ R Z B ( y,r ) g Nn , then lim N →∞ (cid:18) lim r →∞ (cid:18) lim sup n →∞ q Nn ( r ) (cid:19)(cid:19) = 0 . Proof.
To obtain Corollary 3.4 from Theorem 3.3, we observe that if (2) holdsin Theorem 3.3, then the statements in Corollary 3.4 will also hold if we simplydefine B ( x in , r in ) = ∅ when i < n . So we need only consider the case when (1)holds in Theorem 3.3.Define E n = R \ ∪ ki =1 B ( x in , r in ) for n ∈ N . Since { ρ n χ E n } n ∈ N is vanishing,then for each fixed j ∈ N , lim n →∞ sup y ∈ R Z B ( y,j ) ρ n χ E n = 0 . Therefore we can define a sequence n < n < n < . . . such thatsup y ∈ R Z B ( y,j ) ρ n j χ E nj ≤ j +1 for all j ∈ N . If we now pass to the subsequence { ρ n j } j ∈ N , continuing to denotethis subsequence by { ρ n } n ∈ N , we have thatsup y ∈ R Z B ( y,n ) ρ n χ E n ≤ n +1 (3.2)for all n ∈ N . Also, because of part (1)(c) of Theorem 3.3, by passing to afurther subsequence we can guarantee that Z B ( x in ,r in ) \ B ( x in ,r in / ρ n ≤ n +1 (3.3)holds for all i ∈ { , , . . . , k } and all n ∈ N as well.For each i ≥ k + 1, we set m i = 0, and for all n ∈ N we define r in = n if n ≥ i and r in = 0 if n < i . For each fixed n ∈ N , we define a sequence { x jn } inductively for all i ≥ k + 1 by choosing x in to be any real number such that B ( x in , r in ) is disjoint from ∪ i − j =1 B ( x jn , r jn ). Then we have that lim n →∞ r in = ∞ foreach i ∈ N , and part (a) of the Corollary holds.For all i ≥ k + 1 and for all n ∈ N , since B ( x in , r in ) ⊂ E n , it follows from(3.2) that Z B ( x in ,r in ) ρ n ≤ n +1 . (3.4)14his implies that parts (b) and (c) of the Corollary hold for all i ≥ k + 1. For1 ≤ i ≤ k we already know that part (b) holds, and part (c) follows from (3.3).To prove part (d), we fix i such that i ≥ k + 1, and observe that by (3.4),for every ǫ > N ∈ N such that Z R ρ n χ B ( x in ,r in ) < ǫ for all n > N .Also, for each n ∈ { , . . . , N } , since ρ n χ B ( x in ,r in ) ∈ L ( R ), we can find r ǫ,n > Z R \ B ( x in ,r ǫ,n ) ρ n χ B ( x in ,r in ) < ǫ. Then if we set r ǫ = max { r ǫ, , . . . , r ǫ,N } , we have that Z R \ B ( x in ,r ǫ ) ρ n χ B ( x in ,r in ) < ǫ for all n ∈ N . This proves part (d) for all i such that i ≥ k + 1, and we alreadyknow that part (d) holds for 1 ≤ i ≤ k .Finally, part (e) of the Corollary follows immediately from part (1)(e) ofTheorem 3.3, as does part (f) of the Corollary; since (1)(e) of Theorem 3.3implies that lim n →∞ q Nn ( r ) = 0 for every r > N ≥ k .We record the following important feature of vanishing sequences. Lemma 3.5.
Suppose < p ≤ ∞ . Then there exists a constant C p > suchthat for all u ∈ H ( R ) , k u k L p ( R ) ≤ C p sup y ∈ R Z B ( y, | u ′ | + | u | dx ! − p k u k p H ( R ) (3.5) Proof.
This lemma is standard; a proof can be found, for example, in [19].
Corollary 3.6.
Suppose { g n } is a bounded sequence in H ( R ) . If {| g n | + | g ′ n | } is vanishing, then lim n →∞ k g n k L p = 0 for all p > . N -solitons Lemma 4.1.
Suppose
A, B > and k ∈ N . If the system of equations k X i =1 x i = A k X i =1 x i = B (4.1) has a solution ( x , . . . , x k ) with x k ≥ for i = 1 , . . . , k ; then (cid:18) k (cid:19) / ≤ BA ≤ . (4.2)15 roof. Suppose the system (4.1) has a solution ( x , . . . , x k ) with x i ≥ i = 1 , . . . , k . Defining p i = x i /A for each i , we have that 0 ≤ p i ≤
1, so p / i ≤ p i . Therefore ( B/A ) = k X i =1 p / i ≤ k X i =1 p i = 1 , which implies that B/A ≤
1. Also, by H¨older’s inequality we have A = k X i =1 x i ≤ k X i =1 x i ! / k X i =1 ! / = B k / , which implies that (1 /k ) / ≤ B/A . Lemma 4.2.
Suppose
A, B > and consider the systems y + y = A y + y = B (4.3) and y + y = A y + y = B (4.4) for ( y , y ) in the first quadrant U = (cid:8) ( y , y ) ∈ R : y ≥ , y ≥ (cid:9) .1. If B/A = 1 , then (4.3) has exactly two solutions in U , given by (0 , A ) and ( A, , and and (4.4) has exactly one solution in U , given by (0 , A ) .2. If (1 / / < B/A < , then (4.3) has exactly two solutions in U , whichare of the form ( α, β ) and ( β, α ) ,where < α < β ; and (4.4) has exactlyone solution in U , which is of the form ( γ, δ ) where < γ < δ .3. If B/A = (1 / / , then (4.3) has exactly one solution in U , which isgiven by ( A/ / , A/ / ) ; and (4.4) has exactly two solutions in U : onegiven by ( A/ / , , and one of the form ( γ, δ ) where < γ < δ .4. If (1 / / < B/A < (1 / / , then (4.3) has no solutions in U , and (4.4) has exactly two solutions in U , which are of the form ( γ , δ ) and ( γ , δ ) , where < γ < δ and < δ < γ .5. If B/A = (1 / / , then (4.3) has no solutions in U , and (4.4) has exactlyone solution in U , which is given by ( A/ / , A/ / ) .Proof. Suppose ( y , y ) solves (4.3) and y , y >
0. Letting θ = y /y , we obtainthat y = A/ (1 + θ ) / and g ( θ ) = ( B/A ) , where g ( t ) := (1 + t ) (1 + t ) . (4.5)16onversely, for each choice of θ > g ( θ ) = ( B/A ) , we have asolution ( y , y ) of (4.3) given by the positive numbers y = A/ (1 + θ ) / and y = θy . Therefore, for a given choice of B/A , the number of solutions ( y , y )of (4.3) with y > y > θ > g ( θ ) = ( B/A ) .We have that g (0) = 1, g (1) = 1 /
4, lim t →∞ g ( t ) = 1, and g ′ ( t ) = 15(1 + t ) ( t − t )(1 + t ) , so g ( t ) is monotone decreasing for 0 ≤ t ≤ ≤ t < ∞ . Therefore the equation g ( θ ) = ( B/A ) has exactly two solutionswhen B/A ∈ ((1 / / , B/A = (1 / / .The assertions of the lemma concerning (4.3) then follow.On the other hand, when y >
0, we have that ( y , y ) solves (4.4) if andonly if y = A/ (2 + θ ) / , y = θy , and h ( θ ) = ( B/A ) , where h ( t ) := (2 + t ) (2 + t ) . We have h (0) = 1 / h (1) = 1 / t →∞ h ( t ) = 1, and h ′ ( t ) = 30(2 + t ) ( t − t )(2 + t ) , so that h ( t ) is monotone decreasing for 0 ≤ t ≤ ≤ t < ∞ . When B/A = (1 / / , the equation h ( θ ) = ( B/A ) hasexactly one solution, namely θ = 1. When B/A ∈ ((1 / / , (1 / / ), theequation h ( θ ) = ( B/A ) has exactly two solutions, one of which is greater thanone and one of which is less than one. When B/A = (1 / / there are againexactly two solutions, one of which is θ = 0 and the other of which is a value θ >
1. Finally, when
B/A ∈ ((1 / / , h ( θ ) = ( B/A ) , and it satisfies θ >
1. These statements imply the assertionsof the lemma concerning (4.4).
Definition 4.3.
Let D = { ( A, B ) ∈ R : A > , B > , (1 / / ≤ B/A ≤ } . For each ( A, B ) ∈ D , we define m ( A, B ) = y + y , where ( y , y ) is the unique solution to (4.3) satisfying ≤ y ≤ y and y > ,guaranteed by Lemma 4.2. Lemma 4.4. . For all ( A, B ) ∈ D , and for every λ > , we have m ( λA, λB ) = λ m ( A, B ) . (4.6)
2. The function m ( A, B ) is continuous on the set D .Proof. The homogeneity property (4.6) of m ( A, B ) is an easy consequence of thedefinition of m ( A, B ). To see that m ( A, B ) is continuous on D , observe first thatfor given ( A, B ) ∈ D , the numbers y and y given in Definition 4.3 are givenby y = A/ (˜ θ + 1) / and y = ˜ θy , where ˜ θ = ˜ θ ( A, B ) is the unique solution in[0 ,
1] of the equation g (˜ θ ) = ( B/A ) , and g is the function defined in (4.5). Since g is continuous and monotone decreasing on [0 , g ([0 , / , h : [1 / , → [0 ,
1] defined by h ( g ( t )) = t is also continuous.Therefore ˜ θ ( A, B ) = h (( B/A ) ) is continuous on D , so y and hence also y depend continuously on ( A, B ). So m ( A, B ) = y + y is continuous on D aswell. Lemma 4.5.
Suppose y ≥ y ≥ and z ≥ z ≥ , and z + z ≤ y + y z + z ≥ y + y . (4.7) Then z + z ≥ y + y . (4.8) Equality holds in (4.8) only if y = z and y = z .Proof. We may assume y > z >
0, or otherwise there is nothing to prove.Let A = y + y , B = y + y , and C = y + y . By Lemmas 4.1 and 4.2,we have (1 / / ≤ B/A ≤
1. Define θ = y /y ∈ [0 ,
1] and ˜ θ = z /z ∈ [0 , g as in (4.5). Then from the definitions of A and B wededuce that g ( θ ) = ( B/A ) ∈ [1 / , B (1 + ˜ θ ) / ≤ z ≤ A (1 + ˜ θ ) / , (4.9)which implies that g (˜ θ ) ≥ ( B/A ) = g ( θ ). Since, as shown in the proof ofLemma 4.2, g is monotone decreasing on [0 , θ ≤ θ . Now define k ( t ) := (1 + θ ) (1 + θ ) . Then as in the proof of Lemma 4.2, an elementary computation (whose detailswe omit) shows that k ( t ) is, like g ( t ), strictly decreasing on [0 ,
1] and strictlyincreasing on [1 , ∞ ). Therefore k (˜ θ ) ≥ k ( θ ) = ( C/B ) , and so C (1 + ˜ θ ) / ≤ B (1 + ˜ θ ) / . (4.10)18aken with (4.9), this implies that C (1 + ˜ θ ) / ≤ z , (4.11)which yields (4.8).If equality holds in (4.8), then equality also holds in (4.11), so from (4.9) and(4.10) we have that equality holds in (4.10). Therefore k (˜ θ ) = k ( θ ). Since k isstrictly decreasing on [0 , θ = θ , and hence g (˜ θ ) = g ( θ ) andso ˜ B/ ˜ A = B/A . But from (4.7) we have that ˜ A ≤ A and ˜ B ≥ B , so it followsthat ˜ A = A and ˜ B = B . Hence z and z satisfy the same equation (4.3) as y and y , so by Lemma 4.2, we must have z = y and z = y . Lemma 4.6.
Suppose
A, B > and (1 / / < B/A < . For x = ( x , x , x ) ∈ R , define g ( x ) = x + x + x g ( x ) = x + x + x f ( x ) = x + x + x , and define Γ = (cid:8) x ∈ R : g ( x ) = A and g ( x ) = B (cid:9) Ω = (cid:8) x ∈ R : x > , x > , and x > (cid:9) . (4.12) Then Γ ∩ Ω is nonempty, and is a smooth one-dimensional submanifold of R .If we assume further that B/A ≥ (1 / / , then Γ ∩ Ω must consist of threenonempty connected components Γ , Γ , and Γ . For each i = 1 , , , let Γ i denote the closure of Γ i , and ∂ Γ i the boundary of Γ i , in the topology of R .Then the restriction of f to Γ i takes its maximum value at a single point in Γ i ,and takes its minimum value f min on ∂ Γ i . For each ( x , x , x ) ∈ Γ ∩ Ω , wehave f ( x , x , x ) > f min .Proof. Since (1 / / < B/A <
1, then by Lemma 4.2, there exists a solution( y , y ) = ( γ, δ ) to (4.4) with 0 < γ < δ . Setting x = x = γ and x = δ then defines a point Q = ( γ, γ, δ ) ∈ Γ ∩ Ω, and shows that Γ ∩ Ω is nonempty.The fact that Γ ∩ Ω is a smooth one-dimensional submanifold of R followsfrom the implicit function theorem (see, for example, Theorem 1.38 of [22]) andthe fact that the gradients ∇ g ( x ) and ∇ g ( x ) are linearly independent at all x ∈ Γ ∩ Ω. Indeed, if for some c , c ∈ R , with c , c not both zero, we have c ∇ g ( x ) + c ∇ g ( x ) = 0, then it follows easily that x = x = x . But then g ( x ) = A and g ( x ) = B imply that B/A = (1 / / , contradicting ourassumption about B/A .Now suppose we are at a point x = ( x , x , x ) ∈ Γ ∩ Ω where x = x . (Note that the point Q defined above is such a point.) Then from theimplicit function theorem it follows that there exists a neighborhood I of x in R such that for all t ∈ I , there are unique numbers x ( t ) and x ( t ) so that19 t, x ( t ) , x ( t )) ∈ Γ ∩ Ω. Moreover, x ( t ) and x ( t ) are smooth functions of t ∈ I ,with dx dt = t ( x − t ) x ( x − x ) dx dt = t ( t − x ) x ( x − x ) (4.13)on I .In particular, this analysis when applied to the point Q shows that thatthere are functions x ( t ) and x ( t ) defined for t in a neighborhood I of γ suchthat ( t, x ( t ) , x ( t )) ∈ Γ ∩ Ω for all t ∈ I , and equations (4.13) hold on I . From(4.13) we have that dx dt < t = γ , so there exists an ǫ > < x ( t ) < t < x ( t ) for all t such that γ < t < γ + ǫ .Assume now further that B/A ≥ (1 / / . Then by Lemma 4.2, the point( γ, δ ) defined above is the only solution ( y , y ) to (4.4) with y > y > S be the set of all t > γ such that there exist smooth functions x ( t ), x ( t )defined for all t ∈ ( γ, t ) such that ( t, x ( t ) , x ( t )) ∈ Γ ∩ Ω and0 < x ( t ) < t < x ( t )for all t ∈ ( γ, t ). Then S is nonempty and bounded, since ǫ ∈ S and t ≤ A for all t ∈ S . Therefore S has a finite supremum, which we denote by t m .Equations (4.13) imply that dx dt ≤ dx dt ≤ t ∈ [ γ, t m ), so x ( t )and x ( t ) have limits as t approaches t m from the left; we denote these limitsby x ( t m ) and x ( t m ) respectively.We have that 0 ≤ x ( t m ) ≤ t m ≤ x ( t m ). It cannot be the case that0 < x ( t m ) < t m < x ( t m ), for then an application of the implicit functiontheorem would allow us to extend x ( t ) and x ( t ) to an open interval containing t = t m , contradicting the maximality of t m . Since x ( t ) is nonincreasing on[ γ, t m ) and x ( γ ) = γ , we have x ( t m ) < t m . Also, we cannot have 0 < x ( t m )
0, so it follows that x ( t m ) = 0.We have shown that Γ ∩ Ω contains the smooth arc { ( t, x ( t ) , x ( t )) : γ ≤ t ≤ t m } whose endpoints are Q and P = ( t m , , x ( t m )) ∈ ∂ Ω. By symmetry, Γ ∩ Ω alsocontains a smooth arc whose endpoints are Q and P = (0 , t m , x ( t m )) ∈ ∂ Ω.The interior of the union of these two arcs is a connected component Γ of Γ ∩ Ω.We now consider the problem of maximizing or minimizing f ( x ) subjectto the constraint x ∈ Γ . If the maximum or minimum occurs at an interiorpoint x = ( x , x , x ) ∈ Γ , then x must be a critical point of the constrainedvariational problem, in the sense that ∇ f ( x ) = λ ∇ g ( x ) + λ ∇ g ( x )20or some λ , λ ∈ R . This implies that7 x i = 3 λ x i + 5 λ x i for i = 1 , ,
3. Letting z i = x i , we have for all i, j ∈ { , , } that5 λ z i − z j ) = z i − z j . Therefore either z i = z j or z i + z + j = 5 λ /
7. It follows that the set { z , z , z } cannot consist of three distinct numbers: if, for example, z = z and z = z ,then we must have that z + z = z + z = 5 λ /
7, so z = z . It followsthen from Lemma 4.2 that the only possible critical points of f on Γ ∩ Ω are Q = ( γ, γ, δ ), ( γ, δ, γ ), and ( γ, δ, γ ). But since δ > γ , and x > x at all pointson Γ except Q , we conclude that Q is the only critical point of f on Γ .We have now shown that either f takes its maximum value over Γ at Q andits minimum value at P and P , or f takes its minimum value over Γ at Q andits maximum value at P and P . To decide between these two alternatives, itsuffices to determine whether the restriction of f to Γ has a local maximum ora local minimum at Q . For this purpose, we use the second derivative test forconstrained extrema, as expounded for example in [20].Consider the Lagrangian L ( x ) defined by L ( x ) = f ( x ) − λ ( g ( x ) − A ) − λ ( g ( x ) − B ) , and form the “augmented Hessian”, a 5 × H defined by H = (cid:20) (cid:21) , where B is the 2 × B = (cid:20) − ( g ) x − ( g ) x − ( g ) x − ( g ) x − ( g ) x − ( g ) x (cid:21) , C = B T is the transpose of B , and D is the 3 × L , given by D ij = L x i x j for i, j ∈ { , , } . Here and in what follows we use to denote matrices of various sizes (in thiscase, a 2 × H of H at x = Q . Calculationsshow that at x = Q , we have B = (cid:20) − γ − γ − δ − γ − γ − δ (cid:21) and D = − γ ( γ + δ ) 0 00 − γ ( γ + δ ) 00 0 − δ ( γ + δ ) ;21rom which one finds thatdet( − BD − C ) = − γδ D = − ( γ − δ ) δ γ . Let I be the 2 × I the 3 × (cid:20) I − D − C I (cid:21) has determinant equal to one, and so we can writedet H = det (cid:20) (cid:21) (cid:20) I − D − C I (cid:21) = det (cid:20) A − BD − C B0 D (cid:21) == det( A − BD − C ) det D = 14 · γ δ ( γ − δ ) . Since γ < δ , we have shown that det H < x = Q . It is easy to checkthat B has full rank at x = Q . Therefore, according to Theorem 36 on p. 58 of[20], we have that v T Dv < v ∈ R satisfying Bv = . In other words, the Hessian D of L is negative definite in all directions v which are tangent to both the surfaces { x : g ( x ) = A } and { x : g ( x ) = B } at Q . From a classical result in the calculus of variations (see for example page334 of [17]), it follows that f ( x ) has a local maximum at Q subject to therestriction x ∈ Γ .We have now proved that the restriction of f takes its maximum over Γ at Q = ( γ, γ, δ ) ∈ Γ and its minimum value at the endpoints P = ( t m , , x ( t m ))and P = (0 , t m , x ( t m )) of Γ . Let us define f max = f ( Q ) and f min = f ( P ) = f ( P ). Since the restriction of f to Γ has no critical points in Γ \ Q , we musthave f ( x ) > f min for all x ∈ Γ .By symmetry, it follows that Γ ∩ Ω also contains a component Γ whichincludes the point ( γ, δ, γ ) and whose closure has endpoints (0 , x ( t m ) , t m ) and( t m , x ( t m ) , which includes the point ( δ, γ, γ ) and whoseclosure has endpoints ( x ( t m ) , , t m ) and ( x ( t m ) , t m , f on Γ is attained at ( γ, δ, γ ), and is equal to f max ;the minimum value of f on Γ is attained at the boundary points of Γ , and isequal to f min ; and f ( x ) > f min for all x ∈ Γ . Similar statements hold for Γ .To complete the proof of the Lemma, it remains only to show that Γ ∩ Ωcontains no other components besides Γ , Γ , and Γ . To prove this, assume Q = ( x , x , x ) ∈ Γ ∩ Ω; we wish to show that Q ∈ Γ i for some i ∈{ , , } . We know x , x , and x cannot all be equal (for this would imply B/A = (1 / / ); and if any two of x , x , x are equal, then by Lemma4.2, Q must be one of the points ( γ, γ, δ ), ( γ, δ, γ ), or ( δ, γ, γ ), and thereforelies in one of the Γ i . We may therefore assume without loss of generality that x < x < x . Then the analysis above shows that there exists some ǫ > x ( t ) = ( t, x ( t ) , x ( t )) mapping I = ( x − ǫ, x + ǫ ) into22 ∩ Ω, such that x (0) = Q , and satisfying t < x ( t ) < x ( t ) on I . Moreover, x ( t ) satisfies equations (4.13), which imply that dx dt < dx dt > I .Now let S be the set of all t > x such that there exist smooth functions x ( t ), x ( t ) defined for all t ∈ ( x , t ) such that ( t, x ( t ) , x ( t )) ∈ Γ ∩ Ω and0 < t < x ( t ) < x ( t )for all t ∈ ( x , t ). Again we define t m = sup S and let x ( t m ) denote thelimit of x ( t ) as t approaches t m from the left. The implicit function theoremand the maximality of t m imply that we must have x ( t m ) = t m . But thisthen implies that the point ( t m , x ( t m ) , x ( t m )) = ( γ, γ, δ ) ∈ Γ . Thereforethe set S = { t ∈ [ x , t m ] : ( t, x ( t ) , x ( t )) ∈ Γ } is non-empty. The uniquenessassertion in the implicit function theorem tells us that for every t ∈ [ x , t m ], theequations g ( x ) = A and g ( x ) = B determine x and x uniquely as functionsof x in some open neighborhood of ( t, x ( t ) , x ( t )). Therefore S is open. Onthe other hand, S is clearly closed, by the continuity of x ( t ) and x ( t ) andthe fact that Γ is a closed subset of R . So we must have S = [ x , t m ], andtherefore Q ∈ Γ . Remark:
In the case (1 / / < B/A < (1 / / , a similar analysisshows that Γ ∩ Ω is homeomorphic to a circle, and contains all six of thepoints P ( γ , γ , δ ), P ( γ , δ , γ ), P ( δ , γ , δ ), P ( γ , γ , δ ), P ( γ , δ , γ ),and P ( δ , γ , δ ), where ( γ , δ ) and ( γ , δ ) are as described in part 4 of Lemma4.2. Moreover, points P , P , and P are local maxima for the restriction of f to Γ ∩ Ω; while points P , P , and P are local minima. However, we will notneed these facts in what follows. Lemma 4.7.
Suppose x , . . . , x n are numbers such that x ≥ x ≥ · · · ≥ x n ≥ , with x > , and for each m ∈ { , . . . , n } define A m = m X i =1 x i ! / B m = m X i =1 x i ! / . Then for each m ∈ { , . . . , n } , B m − A m − ≥ B m A m , (4.14) and the inequality is strict if x m > .Proof. The statement is obvious if x m = 0, so we may assume x m >
0. Let f ( x ) = ( B m − x ) ( A m − x ) . Then f ( x m ) = ( B m − /A m − ) and f (0) = ( B m /A m ) ,23o it suffices to show that f ( x m ) > f (0). Now f ′ ( x ) = 15 x ( B m − A m x )( B m − x ) ( A m − x ) . So f ′ ( x ) > ≤ x < x , where x = p B m /A m . But since x m ≤ x i for all i ∈ { , . . . , m } , we have A m x m = m X i =1 x i ! x m ≤ m X i =1 x i = B m , so x m ≤ x . Therefore f ( x n ) > f (0), as desired. Lemma 4.8.
Let
A, B > be such that (1 / / ≤ B/A ≤ , and let n ∈ N ,with n ≥ . Suppose x , . . . , x n are numbers such that x ≥ · · · ≥ x n ≥ , with x > , and n X i =1 x i = A n X i =1 x i = B . (4.15) Then n X i =1 x i ≥ m ( A, B ) + E, (4.16) where E = E ( x , x , x ) is defined by E ( x , x , x ) := x + x + x − m (( x + x + x ) / , ( x + x + x ) / ) . (4.17) In particular, E ( x , x , x ) > . (4.18) Proof.
Let ˜ A = ( x + x + x ) / and ˜ B = ( x + x + x ) / . If we defineΓ and Ω as in (4.12) with A replaced by ˜ A and B replaced by ˜ B , then since x ≥ x ≥ x >
0, the point x = ( x , x , x ) lies in Γ ∩ Ω. The inequality (4.18)thus follows from Lemma 4.6.To prove (4.16), we use induction on n . When n = 3, the result is trivial.Suppose n ≥ n −
1; wewish to prove it for n .Suppose that x ≥ · · · ≥ x n ≥
0, with x >
0, and that (4.15) holds.If x n = 0, then we are done by the inductive hypothesis, so we may assume x n >
0. Let A n − = ( A − x n ) / x n B n − = ( B − x n ) / x n , y i = x i /x n for 1 ≤ i ≤ n −
1. Then y ≥ · · · ≥ y n − , and n − X i =1 y i = A n − n − X i =1 y i = B n − . From Lemma 4.7 it follows that B n − /A n − > B/A , and from Lemma 4.1 wehave that B n − /A n − ≤
1. Hence (1 / / ≤ B n − /A n − ≤
1, and we maytherefore apply the inductive hypothesis to the numbers y ≥ y ≥ · · · ≥ y n − ≥
0. There results the inequality n − X i =1 y i ≥ m ( A n − , B n − ) + E , (4.19)where E = y + y + y − m (( y + y + y ) / , ( y + y + y ) / ) . From (4.6), however, it follows that E = E/x n , so multiplying (4.19) by x n ,we conclude that n − X i =1 x i ≥ x n m ( A n − , B n − ) + E. (4.20)From Lemma 4.2 we have that there exist w , w with 0 ≤ w < w suchthat w + w = A n − w + w = B n − . By definition of the function m , we have w + w = m ( A n − , B n − ) . (4.21)Letting z = x n w , z = x n w , and z = x n , we see that z + z + z = A z + z + z = B . Therefore ( z , z , z ) is in the closure of the set Γ ∩ Ω defined in Lemma 4.6.From Lemma 4.2 we see that the boundary of Γ ∩ Ω consists exactly of the sixpoints (0 , α, β ), (0 , β, α ), ( α, , β ), ( β, , α ), ( α, β, β, α, f defined in Lemma 4.6 takes the samevalue α + β , which by definition is equal to m ( A, B ). Hence, by Lemma 4.6,we have that f ( z , z , z ) ≥ m ( A, B ) . f ( z , z , z ) = x n (1 + w + w ), using (4.21) we deduce that x n + x n m ( A n − , B n − ) ≥ m ( A, B ) . Therefore, by (4.20), n X i =1 x i = x n + n − X i =1 x i ≥ x n + x n m ( A n − , B n − ) + E ≥ m ( A, B ) + E, as was desired.We are now ready for the main results of this section. Lemma 4.9.
Suppose y ≥ y ≥ and y > . Let n ∈ N , and suppose x , . . . , x n are numbers such that x ≥ · · · ≥ x n ≥ , and n X i =1 x i ≤ y + y n X i =1 x i ≥ y + y . (4.22)
1. If n ≥ and y = 0 , then x = 0 and x = y .2. If n ≥ and x > , then n X i =1 x i ≥ y + y + E ( x , x , x ) , (4.23) where E ( x , x , x ) > is as in (4.17) and (4.18) .Proof. Define A = (cid:0) y + y (cid:1) / , B = (cid:0) y + y (cid:1) / , A n = (cid:0)P ni =1 x i (cid:1) / , and B n = (cid:0)P ni =1 x i (cid:1) / . From Lemma 4.1, Lemma 4.7, and (4.22), we have that1 = B A ≥ B A ≥ B A ≥ · · · ≥ B n A n ≥ BA ≥ (cid:18) (cid:19) / . (4.24)To prove part 1 of the Lemma, we simply observe that if y = 0 then B/A = 1, and if x > B /A > B /A by Lemma 4.7. Thus (4.24)immediately gives a contradiction. So if y = 0, we must have x = 0 and hence x = y .To prove part 2 of the Lemma, we first observe that from (4.24), Lemma4.2, and the definition of the function m , we obtain that there exist z and z with 0 ≤ z ≤ z and z > z + z = A n z + z = B n (4.25)26nd z + z = m ( A n , B n ).Now if n ≥ x >
0, then from Lemma 4.8 it follows that n X i =1 x i ≥ m ( A n , B n ) + E ( x , x , x ) = z + z + E ( x , x , x ) . (4.26)But since z + z ≤ A z + z ≥ B , it follows from Lemma 4.5 that z + z ≥ y + y . This, combined with (4.26),gives (4.23).Note that an interesting, and immediate, consequence of Lemma 4.9 is thefollowing: among the set of all N -soliton profiles for the KdV equation, the oneswhich minimize E subject to the constraints that E and E be held constantare precisely the 1-soliton and 2-soliton profiles. Lemma 4.10.
Suppose y ≥ y ≥ and y > . Let { x n } n ∈ N be a sequencesuch that x ≥ x ≥ x ≥ · · · ≥ , and ∞ X i =1 x i ≤ y + y ∞ X i =1 x i ≥ y + y . (4.27)
1. If y = 0 , then x = 0 and x = y .2. If x > , then ∞ X i =1 x i ≥ y + y + E ( x , x , x ) , (4.28) where E ( x , x , x ) > is as in (4.17) and (4.18) .Proof. Define A = (cid:0) y + y (cid:1) / , B = (cid:0) y + y (cid:1) / , A ∞ = (cid:0)P ∞ i =1 x i (cid:1) / , and B ∞ = (cid:0)P ∞ i =1 x i (cid:1) / ; and for n ∈ N define A n = (cid:0)P ni =1 x i (cid:1) / and B n = (cid:0)P ni =1 x i (cid:1) / . Thus lim n →∞ A n = A and lim n →∞ B n = B . From Lemmas 4.1 and 4.7and our assumptions, we have that1 ≥ B A ≥ B A ≥ · · · ≥ B n A n ≥ · · · ≥ B ∞ A ∞ ≥ BA ≥ (cid:18) (cid:19) / . Part 1 of the Lemma is now proved by the same argument as part 1 ofLemma (4.9).To prove part 2 of the Lemma, we suppose n ≥ x >
0, and considerfirst the case when the first inequality in (4.27) is strict: that is, when P ∞ i =1 x i < + y . For each i ∈ N and n ∈ N , define α n = B n /B ∞ and x in = x i /α n . Thenlim n →∞ α n = 1, and for each n ∈ N we have n X i =1 x in = ∞ X i =1 x i ≥ y + y . Also, lim n →∞ n X i =1 x in = ∞ X i =1 x i < y + y , so by choosing n sufficiently large we have P ni =1 x in ≤ y + y . We can thusapply Lemma 4.9 to x n ≥ · · · ≥ x nn ≥ n ∈ N , andobtain that n X i =1 x in ≥ y + y + E ( x n , x n , x n ) , or 1 α n n X i =1 x i ≥ y + y + 1 α n E ( x , x , x ) . Taking the limit as n → ∞ then gives the desired result (4.28).Next, consider the case when P ∞ i =1 x i > y + y . Then for sufficientlylarge n , (4.22) holds, so by Lemma 4.9 we conclude that (4.23) holds, whichimmediately implies (4.28).It remains then only to consider the case when A ∞ = A and B ∞ = B .In this case we argue as follows. For each n ∈ N , since B n /A n ≥ (1 / / ,by Lemma 4.2 we can choose z n ≥ z n ≥ A n = z n + z n and B n = z n + z n ; we then have that m ( A n , B n ) = z n + z n . From Lemma 4.9,we have that n X i =1 x i ≥ m ( A n , B n ) + E ( x , x , x ) . (4.29)But, by Lemma 4.6, we have lim n →∞ m ( A n , B n ) = m ( A ∞ , B ∞ ) = m ( A, B ). Tak-ing the limit on both sides of (4.29) as n → ∞ then gives the desired result. We first prove part 3 of Theorem 2.6, which is an easy consequence of theresults of the preceding sections. Suppose ( a, b ) ∈ Σ, and suppose that (2.10)holds. Assume for contradiction that there exists a minimizer u ∈ H ( R ) for J ( a, b ). Then by Proposition 2.5, there must exist real numbers D , D , γ , γ with 0 ≤ D < D such that u = ψ D ,D ; γ ,γ . Since E ( ψ ) = a and E ( ψ ) = b ,it follows from (2.5) that 12 (cid:16) D / + D / (cid:17) = a − (cid:16) D / + D / (cid:17) = b. A = ( a/ / , B = ( − b/ / , k = 2, x = D / , and x = D / . Therefore, by Lemma 4.1, we must have that B/A ≥ (1 / / . Further, we cannot have that B/A = (1 / / , for by part3 of Lemma 4.2, this would imply that x = x , contradicting the fact that D < D . Hence B/A > (1 / / . But this means that b < − ma / / , which contradicts our assumption (2.10). This then completes the proof of part3 of the Theorem.Turning to the proof of parts 1 and 2 of Theorem 2.6, we now suppose that( a, b ) ∈ Σ and either (2.8) or (2.9) holds. In particular we must have that b < { φ n } be any minimizing sequence for J ( a, b ), so that lim n →∞ E ( φ n ) = a ,lim n →∞ E ( φ n ) = b , and lim n →∞ E ( φ n ) = J ( a, b ). (Note that minimizing sequencesalways exist. For example, from the definition of J ( a, b ) it follows that we canchoose { r n } to be any sequence in Λ( a, b ) such that r n → J ( a, b ), and then take { φ n } such that E ( φ n ) = a , E ( φ n ) = b , and E ( φ n ) = r n for each n ∈ N .)Since { E ( φ n ) } converges, then { φ n } is bounded in L . Also, since bySobolev embedding and interpolation we have Z R ( φ ′ n ) dx = 2 E ( φ n ) + 13 Z R u ≤ E ( φ n ) + C k φ n k H / ≤ E ( φ n ) + C k φ n k / L k φ n k / H , it follows that k φ n k H ≤ C (1 + k φ n k / H ) , which implies that { φ n } is bounded in H . Finally, we have Z R ( φ ′′ n ) dx = 2 E ( φ n ) + Z R (cid:18) uu x − u (cid:19) dx, (5.1)and since { φ n } is bounded in H , it follows from Sobolev inequalities thatthe integral on the right is bounded. Since { E ( φ n ) } is bounded above and { R R ( φ ′′ n ) dx } is bounded below, it follows from (5.1) that both these sequencesare in fact bounded. Therefore { φ n } is bounded in H , and J ( a, b ) > −∞ .Define ρ n := φ n + ( φ ′ n ) + ( φ ′′ n ) . Since { φ n } is bounded in H , then { ρ n } is bounded in L , and we can applyCorollary 3.4 to { ρ n } .We observe that { ρ n } is not a vanishing sequence in the sense of Definition3.1. Indeed, if { ρ n } did vanish, then it would follow from Lemma 3.5 thatlim n →∞ k φ n k L = 0, which in turn implies thatlim inf n →∞ E ( φ n ) = 12 lim inf n →∞ Z R ( φ ′ n ) ≥ . n →∞ E ( φ n ) = b < , giving a contradiction.Since { ρ n } does not vanish, then from Corollary 3.4 we obtain a sequence ofballs { B ( x in , r in ) } n ∈ N for each i ∈ N , satisfying properties (a) to (f).Define η to be a smooth function on R such that η ( x ) = 1 for | x | ≤ / η ( x ) = 0 for | x | ≥
1; and for
R >
0, define η R ( x ) = η ( x/R ). For each i ∈ N and n ∈ N , define v in ( x ) = φ n ( x ) η r in ( x − x in ) . (5.2)We then have the decomposition φ n = n X i =1 v in + w n , (5.3)where for each n ∈ N we define w n ( x ) = φ n ( x ) e η n ( x ) , with e η n ( x ) = 1 − n X i =1 η r in ( x − x in ) . For all n ∈ N and i ∈ N , define A in = B ( x in , r in ) \ B ( x in , r in / Z in = R \ B ( x in , r in / W n = R \ ∪ ni =1 B ( x in , r in ) . Then supp v in ⊆ B ( x in , r in )supp w n ⊆ W n ∪ (cid:0) ∪ ni =1 A in (cid:1) supp w ′ n ⊆ ∪ ni =1 A in . (5.4)We will need some preliminary results on the behavior of the decomposition(5.3), which we state in the next few lemmas. Lemma 5.1.
We have lim n →∞ n X i =1 Z A in ρ n = 0 . (5.5) Proof.
For given ǫ >
0, choose N ∈ N so that ∞ X i = N i < ǫ n ≥ N then wecan use part (c) of Corollary 3.4 to write n X i = N Z A in ρ n ≤ n X i = N ∞ X j = i Z A ij ρ j ≤ n X i = N i < ǫ . i ∈ N ,lim n →∞ Z A in ρ n = 0. Therefore, once N has been chosen, we can find N ∈ N sothat if n ≥ N then N − X i =1 Z A in ρ n < ǫ n ≥ max( N , N ) wehave n X i =1 Z A in ρ n < ǫ , as desired. Lemma 5.2.
Suppose
B > , and suppose { φ n } is any sequence of functions in H ( R ) satisfying k φ n k H ≤ B for all n ∈ N . Let v in and w n be defined for n ∈ N and i ∈ N by (5.2) and (5.3) . Then there exists a constant C > dependingonly on η and B (and in particular, not on k or n ) such that for all n ∈ N , andfor m = 2 , , , (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E m ( φ n ) − n X i =1 E m ( v in ) − E m ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C n X i =1 Z A in ρ n . (5.6) Proof.
We substitute (5.3) into E ( φ n ) = R R φ n , and expand, expressing E ( φ n ) as a sum of integrals. Since v in and v jn have disjoint supports for i = j ,all integrals in this expression whose integrands contain two factors of v in withdistinct values of i will vanish. We thus obtain E ( φ n ) = n X i =1 E ( v in ) + E ( w n ) + n X i =1 Z R v in w n . Since the intersection of the supports of v in and w n is contained in A in , and | v in | ≤ | φ n | and | w n | ≤ | φ n | everywhere on R , we have Z R | v in w n | = Z A in | v in w n | ≤ Z A in | φ n | ≤ Z A in ρ n . (5.7)This then establishes (5.6) for m = 2.Substituting (5.3) into the expression for E ( φ n ), we obtain the estimate (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E ( φ n ) − n X i =1 E ( v in ) − E ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n X i =1 (cid:18)Z R | v in ′ || w ′ n | + Z R | v in | | w n | + Z R | v in || w n | (cid:19) . Then we write Z R | v in ′ || w ′ n | = Z A in | v in ′ || w ′ n | ≤ C Z A in (cid:0) | φ n | + | φ ′ n | (cid:1) ≤ C Z A in ρ n , Z R | v in | | w n | ≤ k v n k L ∞ Z A in | v in w n | ≤ C k φ n k L ∞ Z A in ρ n ≤ C Z A in ρ n , (5.8)31nd similarly for R R | v in || w n | . (Here and in what follows we use C to stand forvarious constants which depend only on η and B .) This establishes (5.6) for m = 3.Finally, to prove (5.6) for m = 4, we substitute (5.3) into E ( φ n ) and write (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E ( φ n ) − n X i =1 E ( v in ) − E ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C Z R n X i =1 | v in ′′ w ′′ n | + | v in v in ′ w ′ n | + | v in || w ′ n | + | v in ′ | | w n | + X s,t ≥ s + t ≤ | v in | s | w n | t . (5.9)All the terms on the right side of (5.9) can be estimated like the terms in thepreceding paragraphs. For example, we have Z R | v in | | w n | ≤ k v n k L ∞ Z A in | v in w n | ≤ C k φ n k L ∞ Z A in ρ n ≤ C k φ n k H Z A in ρ n Z R | v in ′ | | w n | ≤ k w n k L ∞ Z A in | v in ′ | ≤ C k φ n k L ∞ Z A in ρ n ≤ C k φ n k H Z A in ρ n . Clearly, similar estimates hold for the remaining integrals; we omit the details.
Lemma 5.3.
In addition to the assumptions of Lemma 5.2, assume that f ∈ H with k f k H ≤ B . For each n ∈ N and each i ∈ { , . . . , n } , define ˜ φ n = f + n X j =1 j = i v jn + w n . Then there exists a constant
C > , depending only on η and B , such that (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E m ( ˜ φ n ) − E m ( f ) − n X j =1 j = i E m ( v jn ) − E m ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C n X j =1 j = i Z A jn ρ n + C k f k H m − ( Z in ) , (5.10) for m = 2 , , and .Proof. Proceeding as in the proof of Lemma 5.2, we write (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E ( ˜ φ n ) − E ( f ) − n X j =1 j = i E ( v jn ) − E ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ C Z R | f w n | + C n X j =1 j = i (cid:18)Z R | f v jn | + Z R | v jn w n | (cid:19) n X j =1 j = i Z R | v jn w n | is estimated as in (5.7). Also, since the support of w n iscontained in Z in , and the same is true of the support of v jn whenever j = i , then Z R | f w n | ≤ k f k L ( Z in ) k w n k L ≤ C k f k L ( Z in ) k φ n k L ≤ C k f k L ( Z in ) . Finally, since the supports of the functions { v jn } j ∈ N are mutually disjoint, andfor j = i are all contained in Z in , there exists C depending only on η and B (and not on k , n or i ) such that n X j =1 j = i Z R | f v jn | ≤ C Z Z in | f φ n | ≤ C k f k L ( Z in ) . This proves (5.10) for m = 2.Similarly, (5.10) is proved for m = 3 by expanding E ( φ n ) as in Lemma 5.2,then using estimates such as (5.8), together with the estimates Z R | f ′ | X j = i | v jn ′ | + | w ′ n | ≤ C k f ′ k L ( Z in ) k φ n k H ≤ C k f k H ( Z in ) k φ n k H , Z R | f | X j = i | v jn | + | w n | ≤ k f k L ∞ k f k L ( Z in ) k φ n k L ≤ C k f k L ( Z in ) , and a similar estimate for R R | f | ( P j = i | v jn | + | w n | ).Finally, (5.10) is proved for m = 4 by expanding E ( φ n ) to obtain an ex-pression similar to (5.9), but with additional terms on the right-hand side of theform X j = i Z R (cid:0) | f ′′ || v jn ′′ | + | f || v jn ′ | + | f | | v jn ′ | + | f || f ′ || v jn ′ | (cid:1) ++ Z R (cid:0) | f ′′ || w ′′ n | + | f || w ′ n | + | f | | w ′ n | + | f || f ′ || w ′ n | (cid:1) + X s,t ≥ s + t ≤ Z R | f | s X j = i | v jn | + | w n | t . These can each be estimated by the terms on the right-hand side of (5.10). Forexample, we have Z R | f | | w n | ≤ k f k L ∞ k w n k L ∞ Z Z in | f | ≤ C k f k H k w n k H k f k L ( Z in ) Z R | f || f ′ || w ′ n | ≤ k f k L ∞ k w ′ n k L Z Z in | f ′ | ! / ≤ C k f k H k w n k H k f k H ( Z in ) . The remaining terms are estimated similarly. We omit the details, which arestraightforward.
Lemma 5.4.
Suppose < p ≤ ∞ . Then lim n →∞ k w n k L p ( R ) = 0 .Proof. For each n ∈ N , we have from (5.4) that w n = w n (cid:0) χ W n + P ni =1 χ A in (cid:1) and w ′ n = w ′ n (cid:0) χ W n + P ni =1 χ A in (cid:1) ; and from the definition of w n we see thatthere exists a constant C > n ∈ N and x ∈ R , w n ( x ) ≤ Cφ n ( x ) ≤ Cρ n ( x ), and ( w ′ n ) ( x ) ≤ C (cid:0) φ n ( x ) + ( φ ′ n ( x )) (cid:1) ≤ Cρ n ( x ). Therefore,we can writesup y ∈ R Z B ( y, (cid:0) w n + ( w ′ n ) (cid:1) = sup y ∈ R "Z B ( y, (cid:0) w n + ( w ′ n ) (cid:1) χ W n + n X i =1 Z B ( y, (cid:0) w n + ( w ′ n ) (cid:1) χ A in ≤ C sup y ∈ R Z B ( y, ρ n χ W n + C n X i =1 Z A in ρ n . (5.11)But lim n →∞ sup y ∈ R Z B ( y, ρ n χ W n = 0 , (5.12)by part (e) of Corollary 3.4. Combining Lemma 5.1, (5.11), and (5.12) giveslim n →∞ sup y ∈ R Z B ( y, (cid:0) w n + ( w ′ n ) (cid:1) dx = 0 . Since { w n } , like { φ n } , is a bounded sequence in H ( R ), the proof is then com-pleted by applying Lemma 3.5. Lemma 5.5.
We have lim N →∞ lim sup n →∞ n X i = N Z R | v in | = 0 (5.13) and lim N →∞ lim sup n →∞ n X i = N Z R (cid:20)(cid:12)(cid:12)(cid:12)(cid:12) v in (cid:16) v in ′ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) + | v in | (cid:21) = 0 . (5.14)34 roof. For a fixed value of n ∈ N , the supports of the functions { v in } i =1 ,...,n aremutually disjoint. Therefore, if we define f N,n = n X i = N v in , we can write, for all N, n such that
N < n , n X i = N Z R (cid:12)(cid:12) v in (cid:12)(cid:12) = Z R | f N,n | . Now Z R | f N,n | ≤ C k f N,n k H ( R ) sup y ∈ R Z B ( y, (cid:0) | f ′ N,n | + | f N,n | (cid:1)! / ≤ C sup y ∈ R Z B ( y, ρ n χ R \∪ Ni =1 B ( x in ,r in ) ! / = C (cid:0) q Nn (1) (cid:1) / , (5.15)where q Nn ( r ) is the function defined in part (f) of Corollary 3.4. (In obtaining(5.15), we used Lemma 3.5 along with the facts that the support of f N,n liesoutside ∪ Ni =1 B ( x in , r in ), and that | f N,n | and | f ′ N,n | are majorized pointwiseby Cρ n , where C depends only on the cutoff function η .) Since q Nn ( r ) is anincreasing function of r , it follows from part (f) of Corollary 3.4 thatlim N →∞ lim sup n →∞ q Nn (1) = 0 . (5.16)Therefore (5.13) follows from (5.15).Similarly, we can use Lemma 3.5 to write n X i = N Z R | v in | = Z R ( f N,n ) ≤ C k f N,n k H ( R ) sup y ∈ R Z B ( y, (cid:0) | f ′ N,n | + | f N,n | (cid:1)! ≤ Cq Nn (1) , (5.17)and by H¨older’s inequality, Sobolev embedding, and (5.15), we have n X i = N Z R (cid:12)(cid:12)(cid:12)(cid:12) v in (cid:16) v in ′ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) = Z R (cid:12)(cid:12) f N,n ( f ′ N,n ) (cid:12)(cid:12) ≤ (cid:18)Z R (cid:12)(cid:12) f ′ N,n (cid:12)(cid:12) / (cid:19) / (cid:18)Z R | f N,n | (cid:19) / ≤ C k f N,n k / H ( R ) (cid:0) q Nn (1) (cid:1) / . (5.18)Estimates (5.17) and (5.18) together with (5.16) then imply (5.14).Fix i ∈ N , and for n ∈ N define θ in ( x ) = v in ( x + x in ). Since { θ in } n ∈ N is abounded sequence in H ( R ), then by passing to a subsequence, we may assume35hat it converges weakly in H ( R ) to some function g i ∈ H ( R ). By a diag-onalization argument, and replacing { φ n } n ∈ N by an appropriate subsequence,we may assume that for every i ∈ N , the sequence { θ in } n ∈ N converges weaklyin H ( R ) to g i . (In what follows we will often replace sequences by subse-quences without changing notation.) Further, by again passing to appropriatesubsequences, we may assume that the sequences { E ( θ in ) } n ∈ N and { E ( θ in ) } n ∈ N converge. Define a i = lim n →∞ E ( θ in ) = lim n →∞ E ( v in ) b i = lim n →∞ E ( θ in ) = lim n →∞ E ( v in ) . (5.19) Lemma 5.6.
For each i ∈ N , { θ in } n ∈ N converges strongly to g i in H ( R ) .Proof. First note that by part (d) of Corollary 3.4, for every ǫ > R ǫ > Z B ( x in ,r in ) \ B ( x in ,R ǫ ) ρ n < ǫ for all n ∈ N . By taking R ǫ larger if necessary, we may assume as well that Z R \ B (0 ,R ǫ ) (cid:0) ( g ′′ i ) + ( g ′ i ) + g i (cid:1) < ǫ. From the definition of θ in , it is easy to see that there exists a constant C suchthat for all n , k θ in k H ( R \ B (0 ,R ǫ )) ≤ C Z B ( x in ,r in ) \ B ( x in ,R ǫ ) ρ n , and therefore k θ in − g i k H ( R \ B (0 ,R ǫ )) ≤ k θ in k H ( R \ B (0 ,R ǫ )) + k g i k H ( R \ B (0 ,R ǫ )) < ǫ. On the other hand, since the inclusion of H ( B (0 , R ǫ )) into H ( B (0 , R ǫ )) iscompact, then { θ in } n ∈ N has a subsequence { θ in k } k ∈ N that converges strongly to g i in H ( B (0 , R ǫ )). Then for all sufficiently large k , k θ in k − g i k H ( B (0 ,R ǫ )) < ǫ ,and therefore k θ in k − g i k H ( R ) < ǫ .It follows from the preceding that for every ǫ >
0, there exists a subsequence { θ in k } k ∈ N of { θ in } n ∈ N such that k θ in k − g i k H ( R ) < ǫ for all k ∈ N . By nowtaking a sequence of values of ǫ tending to zero and using a diagonalizationargument, we obtain a subsequence of { θ in } n ∈ N which converges to g i stronglyin H ( R ). Since the same argument shows that every subsequence of { θ in } n ∈ N has a subsubsequence which converges to g i in H ( R ), it follows that { θ in } n ∈ N itself converges to g i in H ( R ). Lemma 5.7.
For each i ∈ N , we have lim n →∞ Z R | v in | p = Z R | g i | p (5.20)36 or all p such that ≤ p ≤ ∞ , and lim n →∞ Z R v in (cid:16) v in ′ (cid:17) = Z R g i ( g ′ i ) . (5.21) In particular, lim n →∞ E ( v in ) = E ( g i ) = a i lim n →∞ E ( v in ) = E ( g i ) = b i . (5.22) Proof.
Equation (5.20) follows from Lemma 5.6 and the fact that, by standardSobolev embedding theorems, L p embeds continuously in H ( R ) when 2 ≤ p ≤∞ . For (5.21), we can write Z R v in ( v in ′ ) − Z R g i ( g ′ i ) = Z R ( v in − g i ) (cid:16) v in ′ (cid:17) + Z R g i ( v in ′ − g ′ i )( v in ′ + g ′ i ) . Using H¨older’s inequality and Sobolev embedding, we can majorize the integralson the right-hand side by k v in − g i k L ∞ ( R ) k v in k H ( R ) + k g i k L ∞ ( R ) k v in − g i k H ( R ) (cid:0) k v in k H ( R ) + k g i k H ( R ) (cid:1) ≤ C k v in − g i k H ( R ) (cid:16) k v in k H ( R ) + k g i k H ( R ) (cid:17) . Since g i ∈ H ( R ), and { v in } n ∈ N converges to g i in H ( R ), the preceding expres-sion has limit zero as n → ∞ , proving (5.21). Finally, (5.22) follows immediatelyfrom Lemma 5.6 and (5.20). Lemma 5.8.
For each i ∈ N , E ( g i ) ≤ lim inf n →∞ E ( θ in ) = lim inf n →∞ E ( v in ) . (5.23) Proof.
Choose a subsequence { θ in k } k ∈ N of { θ in } n ∈ N such that lim k →∞ E ( θ in k ) =lim inf n →∞ E ( θ in ). By the weak compactness of bounded sets in Hilbert space, wecan find a further subsequence, also denoted by { θ in k } , which converges weaklyin H ( R ) to g i . By the lower semicontinuity of the norm in Hilbert space, wehave that k g i k H ≤ lim inf k →∞ k θ in k k H Since { θ in k } k ∈ N converges strongly in H ( R ) to g i , this implies that k g ′′ i k L ≤ lim inf k →∞ k ( θ in k ) ′′ k L . On the other hand, by Lemma 5.7, we have that Z R (cid:18) − g i g ix + 532 g i (cid:19) = lim k →∞ Z R (cid:18) − θ in k ( θ in k ) x + 532 ( θ in k ) (cid:19) . Combining the last two statements, we obtain (5.23).37or each i ∈ N , if g i ≡
0, then obviously a i = b i = 0. If on the other hand g i is not identically zero, then by Proposition 2.3, ( a i , b i ) ∈ Σ, and so J ( a i , b i )is well-defined by (2.7). In that case, we have: Lemma 5.9.
For each i ∈ N , if g i is not identically zero, then g i is a minimizerfor J ( a i , b i ) .Proof. We prove the lemma by contradiction. If g i is not a minimizer for J ( a i , b i ), then there must exist a function h ∈ H such that E ( h ) = a i , E ( h ) = b i , and E ( h ) < E ( g i ). Define, for n ∈ N , h n ( x ) = h ( x − x in )and ˜ φ n = h n + n X j =1 j = i v jn + w n . To obtain the desired contradiction, we will show thatlim n →∞ E ( ˜ φ n ) = a lim n →∞ E ( ˜ φ n ) = b lim inf n →∞ E ( ˜ φ n ) < J ( a, b ) . To begin with, use the triangle inequality to write (cid:12)(cid:12)(cid:12) E ( φ n ) − E ( ˜ φ n ) (cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E ( φ n ) − n X j =1 E ( v jn ) − E ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ++ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) E ( ˜ φ n ) − E ( h n ) − n X j =1 j = i E ( v jn ) − E ( w n ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12) E ( v in ) − E ( h n ) (cid:12)(cid:12) . We can use Lemma 5.2 and Lemma 5.3 with f = h n to estimate the first twoterms on the right-hand side of the preceding inequality, and thus obtain that (cid:12)(cid:12)(cid:12) E ( φ n ) − E ( ˜ φ n ) (cid:12)(cid:12)(cid:12) ≤ | E ( v in ) − E ( h n ) | + C n X j =1 Z A jn ρ n + C k h n k L ( Z in ) . Since k h n k L ( Z in ) = Z R \ B (0 ,r in / h ( x ) dx ! / , and h ∈ L ( R ) and lim n →∞ r in = ∞ , it follows that lim n →∞ k h n k L ( Z in ) = 0. Finally,we have thatlim n →∞ ( E ( v in ) − E ( h n )) = lim n →∞ ( E ( v in ) − E ( h )) = lim n →∞ ( E ( v in ) − a i ) = 0 . n →∞ ( E ( φ n ) − E ( ˜ φ n )) = 0, so lim n →∞ E ( ˜ φ n ) = a . Similar arguments apply to E ( φ n ) − E ( ˜ φ n ) and E ( φ n ) − E ( ˜ φ n ). FromLemmas 5.2 and 5.3 we obtain that | E ( φ n ) − E ( ˜ φ n ) | ≤ | E ( v in ) − E ( h n ) | + C n X j =1 Z A jn ρ n + C k h n k H ( Z in ) (5.24)and E ( φ n ) − E ( ˜ φ n ) ≥ E ( v in ) − E ( h n ) − C n X j =1 Z A jn ρ n − C k h n k H ( Z in ) . (5.25)The same considerations as in the preceding paragraph show that it follows from(5.24) that lim n →∞ E ( ˜ φ n ) = b . Also, from (5.25) and (5.23) we obtain thatlim inf n →∞ [ E ( φ n ) − E ( ˜ φ n )] ≥ lim n →∞ [ E ( v in ) − E ( h n )] ≥ E ( g i ) − E ( h ) > , and hence lim sup n →∞ E ( ˜ φ n ) < lim n →∞ E ( φ n ) = J ( a, b ) . In particular, it follows that there exists some sufficiently large n for which E ( ˜ φ n ) < J ( a, b ). But since E ( ˜ φ n ) = a and E ( ˜ φ n ) = b , this contradicts thedefinition of J ( a, b ).From Proposition 2.5 and Lemma 5.9, we conclude that for each i ∈ N , thereexist D i , D i , γ i , γ i ∈ R with 0 ≤ D i ≤ D i such that g i ( x ) = ψ D i ,D i ; γ i ,γ i ( x ) . (5.26)Here we follow the conventions that if D i = 0, then ψ D i ,D i ; γ i ,γ i = ψ D i ; γ i ;and if D i = D i = 0, then ψ D i ,D i ≡
0. Also, in what follows we will occa-sionally omit the subscripts γ i and γ i , referring to g i simply as ψ D i ,D i . Lemma 5.10.
For the numbers D i and D i defined for i ∈ N by (5.26) , wehave ∞ X i =1 (cid:16) D / i + D / i (cid:17) ≤ a ∞ X i =1 (cid:16) D / i + D / i (cid:17) ≥ − b ∞ X i =1 (cid:16) D / i + D / i (cid:17) ≤ J ( a, b ) . (5.27)39 roof. For m = 2 , ,
4, if we define ǫ mn for n ∈ N by ǫ mn = E m ( φ n ) − n X i =1 E m ( v in ) − E m ( w n ) , then we have from Lemmas 5.2 and 5.1 that lim n →∞ ǫ mn = 0.In case m = 2, we have E ( f ) ≥ f ∈ L . Therefore we have, for all N, n ∈ N such that n > N , N X i =1 E ( v in ) = E ( φ n ) − n X i = N +1 E ( v in ) − E ( w n ) − ǫ n ≤ E ( φ n ) − ǫ n . Holding N fixed and taking the limit on both sides as n → ∞ , and recalling(5.19), we obtain that N X i =1 E ( g i ) = 12 N X i =1 (cid:16) D / i + D / i (cid:17) ≤ a. Then taking the limit as N → ∞ yields the first inequality in (5.27).Next, we consider the case m = 3. We have, for all N, n ∈ N such that n > N , N X i =1 E ( v in ) = E ( φ n ) − n X i = N +1 E ( v in ) − E ( w n ) − ǫ n = E ( φ n ) − n X i = N +1 Z R (cid:18)
12 ( v in ′ ) −
16 ( v in ) (cid:19) − Z R (cid:18)
12 ( w n ′ ) − w n (cid:19) − ǫ n ≤ E ( φ n ) + 16 n X i = N +1 Z R ( v in ) + 16 Z R w n − ǫ n . (5.28)Let ǫ > N ∈ N such thatlim sup n →∞ n X i = N +1 Z R (cid:12)(cid:12) v in (cid:12)(cid:12) < ǫ. For this fixed value of N , by taking the limit as n goes to infinity of both sidesof (5.28) and using Lemma 5.4, we obtain that − N X i =1 (cid:16) D / i + D / i (cid:17) = N X i =1 b i ≤ b + ǫ , and hence 365 ∞ X i =1 (cid:16) D / i + D / i (cid:17) ≥ N X i =1 (cid:16) D / i + D / i (cid:17) ≥ − b − ǫ . ǫ >
0, we have proved the second inequalityin (5.27).In case m = 4, we have N X i =1 E ( v in ) ≤ E ( φ n ) + n X i = N +1 Z R (cid:20) (cid:12)(cid:12)(cid:12) v in ( v in ′ ) (cid:12)(cid:12)(cid:12) + 532 (cid:12)(cid:12) v in (cid:12)(cid:12) (cid:21) + 56 Z R w n ( w ′ n ) − Z R w n − ǫ n , (5.29)for all N, n ∈ N such that n > N . Using Lemma 5.4 we see that lim n →∞ Z R w n = 0and lim n →∞ (cid:12)(cid:12)(cid:12)(cid:12)Z R w n ( w ′ n ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ lim n →∞ k w n k L ∞ k w n k H = 0 . For given ǫ >
0, by (5.14), we can choose N ∈ N such that for all N ≥ N ,lim sup n →∞ n X i = N +1 Z R (cid:20) (cid:12)(cid:12)(cid:12) v in ( v in ′ ) (cid:12)(cid:12)(cid:12) + 532 (cid:12)(cid:12) v in (cid:12)(cid:12) (cid:21) < ǫ. For each fixed value of N ≥ N , taking the limit on both sides of (5.29) as n goes to infinity and using (5.23), we then obtain367 N X i =1 (cid:16) D / i + D / i (cid:17) = N X i =1 E ( g i ) ≤ J ( a, b ) + ǫ. Since this is true for all N ≥ N , it follows that367 ∞ X i =1 (cid:16) D / i + D / i (cid:17) ≤ J ( a, b ) + ǫ, and since ǫ > D i and D i can begreater than any fixed positive number. Therefore it is possible to re-order thenumbers in the sequence (cid:16) D / , D / , D / , D / , D / , D / . . . (cid:17) (5.30)so that they form a non-increasing sequence, whose terms we denote by { x n } ,with x ≥ x ≥ x ≥ . . . . Proof of part 1 of Theorem 2.6 . Suppose that (2.8) holds. We let C = ( a/ / = ( − b/ / >
0. For every γ ∈ R , we have E ( ψ C,γ ) = 12 C / = a,E ( ψ C,γ ) = − (36 / C / = b. (5.31)41rom the definition of J ( a, b ) it therefore follows that J ( a, b ) ≤ E ( ψ C,γ ) = E ( C ) = (36 / C / . (5.32)Let y = C and y = 0. From (5.27) and (5.31), it follows that the inequali-ties (4.22) are satisfied by the numbers { x n } defined after (5.30). Therefore, bypart 1 of Lemma 4.10, we must have x = 0 and x = y . Thus g = ψ C,γ forsome γ ∈ R , and g i ≡ a i = b i = 0 for all i ≥ n →∞ E ( φ n ) = a = 12 C / = E ( g )lim n →∞ E ( φ n ) = b = − C / = E ( g ) . Also, from (5.27) we have that (36 / C / ≤ J ( a, b ), and combined with (5.32),this gives lim n →∞ E ( φ n ) = J ( a, b ) = 367 C / = E ( g ) . (5.33)In particular, we have now shown that g , and hence also every element of S ( C ),is a minimizer for J ( a, b ).From Lemmas 5.1 and 5.2, we have that, for m = 2 , , E m ( φ n ) = E m ( v n ) + n X i =2 E m ( v in ) + E m ( w n ) + ǫ mn , where lim n →∞ ǫ mn = 0. From Lemma 5.6 we conclude thatlim n →∞ E ( v n ) = lim n →∞ E ( θ n ) = E ( g ) = a = lim n →∞ E ( φ n )lim n →∞ E ( v n ) = lim n →∞ E ( θ n ) = E ( g ) = b = lim n →∞ E ( φ n ) . Therefore for m = 2 and m = 3 we havelim n →∞ " E m ( w n ) + n X i =2 E m ( v in ) = 0 . (5.34)When m = 2, (5.34) immediately implies thatlim n →∞ " k w n k L ( R ) + n X i =2 k v in k L ( R ) = 0 . (5.35)We claim that when m = 3, (5.34) implies thatlim n →∞ " k ( w n ) ′ k L ( R ) + n X i =2 k ( v in ) ′ k L ( R ) = 0 . (5.36)42o prove this, since lim n →∞ Z R | w n | = 0 by Lemma 5.4, it is enough to show thatlim n →∞ n X i =2 Z R | v in | = 0 . (5.37)Let ǫ > N such thatlim sup n →∞ n X i = N Z R | v in | < ǫ. Therefore, there exists N such that for all n ≥ N , n X i = N Z R | v in | < ǫ. For each fixed i ≥
2, since g i ≡
0, it follows from Lemma 5.6 that lim n →∞ k v in k H ( R ) =0, and hence by Sobolev embedding that lim n →∞ k v in k L p ( R ) = 0 for all p ≥
2. Sothere exists N such that for all n ≥ N , N − X i =2 Z R | v in | < ǫ. Then for all n ≥ max( N , N ), n X i =2 Z R | v in | < ǫ, proving (5.37) and (5.36).Define ˜ φ n ( x ) := φ n ( x + x n ) for n ∈ N . From (5.3) we have˜ φ n = θ n + n X i =2 ˜ v in + ˜ w n for all n ∈ N , where ˜ v in ( x ) := v in ( x + x n ) and ˜ w n ( x ) := w n ( x + x n ). Therefore k ˜ φ n − g k H ( R ) ≤ k θ n − g k H ( R ) + n X i =2 k ˜ v in k H ( R ) + k ˜ w n k H ( R ) , and so from Lemma 5.6, (5.35), and (5.36), we conclude that ˜ φ n ( x ) convergesstrongly to g in H ( R ).In particular, since { ˜ φ n } is bounded in H ( R ), it follows by the same argu-ments used to prove Lemma 5.7 thatlim n →∞ Z R ˜ φ n ( ˜ φ ′ n ) = Z R g ( g ′ ) n →∞ Z R ˜ φ n = Z R g . Therefore lim n →∞ Z R ( ˜ φ ′′ n ) = lim n →∞ (cid:18) E ( ˜ φ n ) + 53 Z R ˜ φ n ( ˜ φ ′ n ) − Z R ˜ φ n (cid:19) = E ( g ) + 53 Z R g ( g ′ ) − Z R g = Z R ( g ′′ ) . Hence we have that lim n →∞ k ˜ φ n k H ( R ) = k g k H ( R ) . (5.38)But, from the weak compactness of the unit sphere in Hilbert space, we mayassume by passing to a further subsequence that { ˜ φ n } converges weakly in H ( R ), and the limit must be g . From (5.38) it then follows that { ˜ φ n } mustconverge strongly to g in H ( R ). This implies thatlim n →∞ k φ n − ψ C,γ + x n k H ( R ) = 0 , which, since ψ C,γ + x n ∈ S ( C ) for all n ∈ N , shows that { φ n } converges stronglyto S ( C ) in H ( R ). This then completes the proof of part 1 of Theorem 2.6. Proof of part 2 of Theorem 2.6.
Assume that (2.9) holds. Applyingpart 2 of Lemma 4.2 with A = ( a/ / and B = ( − b/ / , we obtain thatthere exists a unique pair of numbers y and y such that 0 < y < y and (4.3)holds. Define C = y and C = y ; then we have 0 < C < C and E ( C , C ) = 12 (cid:16) C / + C / (cid:17) = aE ( C , C ) = − (cid:16) C / + C / (cid:17) = b. (5.39)Therefore, for every pair ( γ , γ ) ∈ R , we have E ( ψ C ,C ; γ ,γ ) = a and E ( ψ C ,C ; γ ,γ ) = b ; and hence from the definition of J ( a, b ) we have that E ( ψ C ,C ; γ ,γ ) = E ( C , C ) = 367 (cid:16) C / + C / (cid:17) ≥ J ( a, b ) . (5.40)Let { x n } be the numbers defined after (5.30). From (5.27), (5.39), and (5.40)it follows that, for these numbers, the inequalities (4.27) hold, along with theinequality ∞ X i =1 x i ≤ y + y . Therefore, by part 2 of Lemma 4.10, we must have that x i = 0 for all i ≥ x + x ≤ y + y x + x ≥ y + y x + x ≤ y + y .
44t therefore follows from Lemma 4.5 that x = y and x = y .We thus see that (after relabelling the numbers D i and D i if necessary),we can reduce consideration to two possible cases: Case I in which0 < D = C < D = C , and D i = D i = 0 for all i ≥
2, and Case II in which0 = D < D = C , D < D = C , and D i = D i = 0 for all i ≥ g = ψ C ,C ,γ ,γ for some ( γ , γ ) ∈ R . Thenlim n →∞ E ( φ n ) = a = E ( g )lim n →∞ E ( φ n ) = b = E ( g ) , and from (5.27) and (5.40) we have thatlim n →∞ E ( φ n ) = J ( a, b ) = E ( g ) . In particular, this implies that g , along with every other element of S ( C , C ),is a minimizer for J ( a, b ).The same argument as in the paragraphs following equation (5.33) now showsthat the translated sequence ˜ φ n ( x ) = φ n ( x + x n ) converges strongly in H ( R )to g . Hence lim n →∞ k φ n − ψ C ,C ,γ + x n ,γ + x n k H ( R ) = 0 , which shows that { φ n } converges strongly to S ( C , C ) in H ( R ). This com-pletes the proof of part 2 of Theorem 2.6 in Case I.We turn now to Case II. In this case, we have that g = ψ C ,γ and g = ψ C ,γ for some ( γ , γ ) ∈ R ; and g i ≡ i ≥
3. Then from (5.39) wehave that lim n →∞ E ( φ n ) = a = E ( g ) + E ( g )lim n →∞ E ( φ n ) = b = E ( g ) + E ( g ) , and from (5.27) and (5.40) we have thatlim n →∞ E ( φ n ) = J ( a, b ) = E ( g ) + E ( g ) . (5.41)In particular, this implies again that every element of S ( C , C ) is a minimizerfor J ( a, b ). However, now it is no longer the case that one can translate thefunctions in the sequence { φ n } to obtain a strongly convergent sequence in H ( R ). Instead, we must modify the argument in the proof of part 1 of theTheorem, as follows.Repeating the argument used above to obtain (5.34), we obtain in this casethat lim n →∞ " E m ( w n ) + n X i =3 E m ( v in ) = 0 (5.42)45or m = 2 and m = 3. When m = 2, (5.42) immediately implies thatlim n →∞ " k w n k L ( R ) + n X i =3 k v in k L ( R ) = 0 . Also, by the same proof used above to prove (5.37), we have in this case thatlim n →∞ n X i =3 Z R | v in | = 0 , (5.43)and together with Lemma 5.4 and (5.42) for m = 3, this implies thatlim n →∞ " k w n k H ( R ) + n X i =3 k v in k H ( R ) = 0 . (5.44)Since, by the Sobolev embedding theorem, Z R (cid:12)(cid:12) w n ( w ′ n ) (cid:12)(cid:12) ≤ k w n k L ∞ ( R ) Z R ( w ′ n ) ≤ C k w n k H ( R ) , it follows that lim n →∞ (cid:12)(cid:12)(cid:12)(cid:12) E ( w n ) − Z R ( w ′′ n ) (cid:12)(cid:12)(cid:12)(cid:12) = 0 , (5.45)and hence lim inf n →∞ E ( w n ) ≥ . (5.46)From (5.6) and (5.41), we havelim n →∞ E ( φ n ) = lim n →∞ n X i =1 E ( v in ) + E ( w n ) ! = E ( g ) + E ( g ) . Now observe that, by the same argument used to deduce (5.37) and (5.43) from(5.13), it follows from (5.14) thatlim n →∞ n X i =3 Z R (cid:20)(cid:12)(cid:12)(cid:12)(cid:12) v in (cid:16) v in ′ (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) + | v in | (cid:21) = 0 . (5.47)Therefore we can writelim n →∞ E ( φ n ) = lim n →∞ " E ( v n ) + E ( v n ) + 12 n X i =3 Z R (cid:0) v i ′′ n (cid:1) + E ( w n ) . (5.48)For every sequence { n k } k ∈ N of integers approaching infinity, it follows from(5.47), (5.48), and Fatou’s Lemma thatlim k →∞ E ( φ n k ) ≥ lim inf k →∞ E ( v n k ) + lim inf k →∞ E ( v n k ) + 12 ∞ X i =3 lim inf k →∞ Z R (cid:0) v i ′′ n k (cid:1) + lim inf k →∞ E ( w n k )= lim inf k →∞ E ( v n k ) + lim inf k →∞ E ( v n k ) + ∞ X i =3 lim inf k →∞ E ( v in k ) + lim inf k →∞ E ( w n k ) . (5.49)46e claim now that lim n →∞ E ( w n ) = 0 (5.50)and, for every i ∈ N , lim n →∞ E ( v in ) = E ( g i ) . (5.51)For otherwise, we would have either lim sup n →∞ E ( w n ) ≥ ǫ for some ǫ >
0, orlim sup n →∞ E ( v i n ) ≥ E ( g i ) + ǫ for some i ∈ N and some ǫ >
0. In either case itwould follow from Lemma 5.8 and (5.46) that there exists a sequence of integers { n k } k ∈ N approaching infinity for which (5.49) implieslim k →∞ E ( φ n k ) ≥ ∞ X i =1 E ( g i ) + ǫ. But then from (5.41), since g i ≡ i ≥
3, we obtain thatlim k →∞ E ( φ n k ) ≥ lim n →∞ E ( φ n ) + ǫ. This contradiction proves our claim.From (5.45) and (5.50) we conclude that lim n →∞ Z R ( w ′′ n ) = 0, and togetherwith (5.44) this gives that lim n →∞ k w n k H ( R ) = 0 . (5.52)For all i ∈ N , since { θ in } n ∈ N converges to g i strongly in H ( R ) and weaklyin H ( R ), and since (5.51) implies that lim n →∞ E ( θ in ) = E ( g i ) as well, it followsfrom the same argument used to prove (5.38) thatlim n →∞ k θ in − g i k H ( R ) = 0 . Therefore, we havelim n →∞ k v in − ψ C i ,γ i + x in k H ( R ) = 0 for i = 1 ,
2; (5.53)and lim n →∞ k v in k H ( R ) = 0 for i ≥ . (5.54)From Corollary 3.4, we see that lim n →∞ | x n − x n | = ∞ , since | x n − x n | ≥ r n + r n and lim n →∞ r n = lim n →∞ r n = ∞ . From (5.3) and the triangle inequality, we have k φ n − ψ C ,C ,γ + x n ,γ + x n k H ( R ) ≤ k v n − ψ C ,γ + x n k H + k v n − ψ C ,γ + x n k H + (cid:13)(cid:13) ψ C ,γ + x n + ψ C ,γ + x n − ψ C ,C ,γ + x n ,γ + x n (cid:13)(cid:13) H + k X i =3 k v in k H + k w n k H . n goes to infinity. This thencompletes the proof of part 2 of Theorem 2.6. Proof of Corollary 2.7.
Corollary 2.7 follows from Theorem 2.6 by astandard argument, which we include here for the reader’s convenience. Suppose C and C are given such that 0 < C < C , and define a = E ( C , C ) = 12( C / + C / ) b = E ( C , C ) = −
365 ( C / + C / ) . Then from Lemma 4.2 it follows that a and b satisfy (2.9), and so the assertionabout convergence of minimizing sequences to S ( C , C ) follows from Theorem2.6.To prove stability, we argue by contradiction: if S were not stable, thenthere would exist a sequence of initial data { u n } n ∈ N in H ( R ) such thatlim n →∞ d ( u n , S ) = 0 and a number ǫ > { t n } n ∈ N suchthat the solutions u n ( x, t ) of KdV with initial data u n ( · ,
0) = u n would satisfy d ( u ( · , t n ) , S ) ≥ ǫ (5.55)for all n ∈ N . Let φ n = u ( · , t n ) for n ∈ N . Since E , E , and E are continuousfunctionals on H ( R ), and are conserved under the time evolution of the KdVequation, we havelim n →∞ E ( φ n ) = lim n →∞ E ( u n ) = E ( C , C ) = a lim n →∞ E ( φ n ) = lim n →∞ E ( u n ) = E ( C , C ) = b lim n →∞ E ( φ n ) = lim n →∞ E ( u n ) = E ( C , C ) = J ( a, b ) . Hence { φ n } is a minimizing sequence for J ( a, b ), and so must converge stronglyto S in H ( R ). But this then contradicts (5.55). The authors would like to thank Jerry Bona for introducing us to the problemsconsidered in this paper, and more generally for his help and guidance overmany years.
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