The effect of a positive bound state on the KdV solution. A case study
aa r X i v : . [ m a t h - ph ] M a y THE EFFECT OF A POSITIVE BOUND STATE ONTHE KDV SOLUTION. A CASE STUDY.
ALEXEI RYBKIN
Abstract.
We consider a slowly decaying oscillatory potentialsuch that the corresponding 1D Schr¨odinger operator has a posi-tive eigenvalue embedded into the absolutely continuous spectrum.This potential does not fall into a known class of initial data forwhich the Cauchy problem for the Korteweg-de Vries (KdV) equa-tion can be solved by the inverse scattering transform. We never-theless show that the KdV equation with our potential does admita closed form classical solution in terms of Hankel operators. Com-paring with rapidly decaying initial data our solution gains a newterm responsible for the positive eigenvalue. To some extend thisterm resembles a positon (singular) solution but remains bounded.Our approach is based upon certain limiting arguments and tech-niques of Hankel operators.
Contents
1. Introduction 22. Our analytic tools 52.1. Riesz projections 52.2. Reproducing kernels 62.3. Hankel operators 73. Our explicit potential and its short-range approximation 103.1. An explicit WvN type potential 103.2. Short-range approximation of Q Date : November, 2018.1991
Mathematics Subject Classification.
Key words and phrases.
KdV equation, embedded eigenvalues, Wigner-von Neu-mann potentials.The author is supported in part by the NSF grant DMS 1716975. Introduction
We are concerned with the initial value problem for the Korteweg-deVries (KdV) equation ∂ t u − u∂ x u + ∂ x u = 0 , −∞ < x < ∞ , t ≥ ,u ( x,
0) = q ( x ) . (1.1)As is a well-known, for smooth rapidly decaying q ’s (1.1) was solved inclosed form in the short 1967 paper [11] by Gardner-Greene-Kruskal-Miura (GGKM). This seminal paper introduces what we now call the inverse scattering transform (IST). Conceptually, it is similar to theFourier transform (see e.g. the classical books [1], [29]) but based onthe inverse scattering theory for the Schr¨odinger operator L q = − ∂ x + q ( x ) on L ( R ) . (1.2)Moreover, the solution q ( x, t ) to (1.1) for each t > u ( x, t ) = − ∂ x log τ ( x, t ) , (1.3)where τ is the so-called Hirota tau-function introduced in [17] whichadmits an explicit representation in terms of the scattering data of thepair ( L q , L ). The solution has a relatively simple and by now wellunderstood wave structure of running (finitely many) solitons accom-panied by radiation of decaying waves (see e.g. Grunert-Teschl [14]for a streamlined modern exposition). In about 1973, the IST was ex-tended to q ’s rapidly approaching different constants q ± as x → ±∞ (step initial profile). It appeared first in the physical literature [15] andwas rigorously treated in 1976 by Hruslov [18]. The formula (1.3) isalso available in this case with an explicit representation of the tau-function in terms of certain scattering data. We refer to our recent [13]and [32] where (1.3) is extended to essentially arbitrary q ’s with a rapiddecay only at + ∞ . The main feature of such initial profiles is infinitesequence of solitons emitted by the initial step. Note that a completerigorous investigation of all other asymptotic regimes and their gener-alizations was done only recently by Teschl with his collaborators (seee.g. [4], [9], [10]).Another equally important and explicitly solvable case is when q is periodic. The periodic IST is quite different from the GGKM oneand is actually the inverse spectral transform (also abbreviated as IST)since it relies on the Floquet theory for L q and analysis of Riemannsurfaces and hence is much more complex than the rapidly decayingcase. The solution u ( x, t ) is given essentially by the same formula Also transcripted as Khruslov.
DV EQUATION 3 (1.3), frequently referred to as the
Its-Matveev formula [19] (see also[8] by Dubrovin-Matveev-Novikov and the 2003 Gesztesy-Holden book[12] where a complete history is given), but τ is a multidimensional theta-function of real hyperelliptic algebraic curves explicitly computedin terms of spectral data of the associated Dirichlet problem for L q . Itis therefore very different from the rapidly decaying case. The mainfeature of a periodic solution is its quasi-periodicity in time t .We have outlined two main classes of initial data q in (1.1) for whicha suitable form of the IST was found during the initial boom followedby [11]. Such progress was possible due to well-developed inverse scat-tering/spectral theories for the underlying potentials q . However, whilewe have proven [13] that no decay at −∞ is required to do the ISTbut slower than x − decay at + ∞ results in serious complications.The main issue here is that the classical inverse scattering theory, thefoundation for the IST, has not been extended beyond short-range po-tentials, i.e. q ( x ) = O (cid:0) | x | − − ε (cid:1) , x → ±∞ . We emphasize that duringthe boom in scattering theory there was a number of results on (direct)scattering/spectral theory for a variety of long-range potentials but theinverse scattering theory is a different matter. It was shown in 1982 [2]that the short-range scattering data no longer determine the potentialuniquely even in the case when q ( x ) = O ( x − ) and it is not merely atechnical issue of adding some extra data. The problem appears to beopen even for L potentials (see Aktuson-Klaus [3]) for which all scat-tering quantities are well-defined but may exhibit an erratic behaviorat zero energy which is notoriously difficult to analyze and classify. Be-sides, a possible infinite negative spectrum begets an infinite sequenceof norming constants which can be arbitrary. Consequently, it is evenunclear how to state a (well-posed) Riemann-Hilbert problem whichwould solve the inverse scattering problem. Once we leave L then infi-nite embedded singular spectrum may appear leaving no hope to figureout what true scattering data might be. We note that any attempt totry the inverse spectral transform instead runs into equally difficultproblems (see, e.g. our [31] and the literature cited therein) as spectraldata evolve in time under the KdV flows by a simple law essentiallyonly for the so-called finite gap potentials. In addition, it makes senseto find a suitable IST for (1.1) if (1.1) is actually well-posed. The sem-inal 1993 Bourgain’s paper [6] says that (1.1) is well-posed if q is in L and not much better result should be expected regarding the decay at+ ∞ . Infinite dimensional in general.
ALEXEI RYBKIN
In the current paper we look into a specific representative of theimportant class of continuous potentials asymptotically behaving like q ( x ) = ( c/x ) sin 2 x + O (cid:0) x − (cid:1) , x → ±∞ . (1.4)In the half line context such potentials first appeared in 1929 in thefamous paper [25] by Wigner-von Neumann where they explicitly con-structed a potential of type (1.4) with c = − q of type (1.4) with | c | > Wigner-vonNeumann resonance under a small perturbation. If | c | > / √ L q is infinite in general[20]. While there is a very extensive literature on potentials of type(1.4) (commonly referred to as Wigner-von Neumann type potentials)but, as Matveev puts it in [7], ”The related inverse scattering problemis not yet solved and the study of the related large times evolution isa very challenging problem”. Observe that since any Wigner-von Neu-mann potential is clearly in L , the Bourgain Theorem [6] guaranteeswell-posedness of (1.1) and the good open problem is if we can solve itby a suitable IST. Our goal here is to investigate a specific case of (1.4)which can be done by the IST. Namely, we consider an even potential Q ( x ) defined for x ≥ Q ( x ) = − d dx log (cid:16) ρx − ρ x (cid:17) , where ρ is an arbitrary positive constant. One can easily check that Q is continuos and behaves like (1.4) with c = −
4. The main feature of Q is that L Q admits an explicit spectral analysis and consequently thescattering problem for the pair ( L Q , L ) can also be solved explicitly. Inparticular, +1 is a positive bound state of L Q but its negative spectrumconsists of just one bound state. We show that for (1.1) with initial data Q the tau-function in (1.3) can be explicitly calculated. The formulahowever is expressed in the language of Hankel operators (which isnot commonly used in integrable systems) and we have to postponeit till Section 4. We only mention here that, comparing to the shortrange case, the tau-function τ gains an extra factor responsible for thepositive bound state. Unfortunately, we were unable to find the ISTeven in this case but we able to detour it by means of suitable limitingarguments. Our limiting arguments are based on certain short rangeapproximations of Q combined with techniques of Hankel operatorsdeveloped in our [13]. In fact, for 3D radially symmetric potentials.
DV EQUATION 5
The reader will see that our approach is not restricted to just oneinitial condition and should work for a whole class of initial data (atleast [28] gives some hopes). We however do not make an attempt tobe more general for two reasons. First of call, our consideration wouldcomplicate a great deal due to numerous extra technicalities. But themain reason is that the scattering theory, the backbone of our approach,is not developed well enough outside of short-range potentials. (At leastnot to our satisfaction). For instance, there are only some results onregularity properties of scattering data for Wigner-von Neumann typepotentials (see [21]) but almost nothing is known about their smallenergy behavior. The latter was posed as an open question in [21]but, to the best of our knowledge, there has been no progress in thisdirection since then. This is a major impediment to our approach as itrequires a careful control of the scattering matrix at all energy regimes.2.
Our analytic tools
To translate our problem into the language of Hankel operators somecommon definitions and facts are in order [26], [30].2.1.
Riesz projections.
Recall, that a function f analytic in the up-per half plane C ± := { z | ± Im z > } is in the Hardy space H ± of C ± if sup h> Z R ± ih | f ( z ) | | dz | < ∞ . It is a fundamental fact of the theory of Hardy spaces that any f ∈ H ± has non-tangential boundary values f ( x ± i
0) for almost every(a.e.) x ∈ R and H ± are subspaces of L := L ( R ). Thus, H ± areHilbert spaces with the inner product induced from L : h f, g i H ± = h f, g i L = h f, g i = 12 π Z ∞−∞ f ( x ) ¯ g ( x ) dx. It is well-known that L = H ⊕ H − , the orthogonal (Riesz) projection P ± onto H ± being given by( P ± f )( x ) = ± πi lim ε → Z ∞−∞ f ( s ) dss − ( x ± iε ) (2.1)= ± πi Z ∞−∞ f ( s ) dss − ( x ± i . In what follows, we set H = H . Notice that for any f ∈ H P − (cid:18) · − λ f (cid:19) = 1 · − λ f ( λ ) , λ ∈ C + . (2.2) ALEXEI RYBKIN
Besides H ± , we will also use H ∞± , the algebra of uniformly bounded in C ± functions.2.2. Reproducing kernels.
Recall that, a given fixed λ ∈ C ± thefunction k λ ( z ) := iz − λ , λ ∈ C ± (2.3)is called the reproducing (or Cauchy-Szego) kernel for H ± . Clearly, k k λ k = p h k λ , k λ i = 1 √ λ (2.4)and hence k λ ∈ H ± if λ ∈ C ± . The main reason why reproducingkernels are convenient is the following f ∈ H , λ ∈ C + = ⇒ f ( λ ) = h f, k λ i (Cauchy’s formula) (2.5a) f ∈ L , λ ∈ R = ⇒ ( P ± f ) ( λ ) = ±h f, k λ ± i i . (2.5b)Let B be a Blaschke product with finitely many simple zeros z n ∈ C + ,i.e., B ( z ) = Y n b n ( z ) , b n ( z ) = z − z n z − z n . Introduce K B = span { k z n } . It is an easy but nevertheless fundamentally important fact in interpo-lation of analytic functions, the study of the shift operator, so-calledmodel operators, etc. that K B = H ⊖ BH , where BH := (cid:8) Bf : f ∈ H (cid:9) . (2.6) Lemma 2.1.
The orthogonal projections P B of H onto K B and P ⊥ B = I − P B are given by P B = B P − B, P ⊥ B = B P + B. (2.7) Furthermore, if A is a linear bounded operator in H then the matrixof P B A P B with respect to ( k z n ) is given by ( P B AP B ) mn = (cid:10) Ak z n , k ⊥ z m (cid:11) , (2.8) where k ⊥ z n ( z ) := 2 Im z n B n ( z n ) B n ( z ) k z n ( z ) , B n := B/b n (2.9) form a bi-orthogonal basis for ( k z n ) . I.e., (cid:10) k ⊥ z n , k z m (cid:11) = δ nm . It can also be infinite but it doesn’t concern us.
DV EQUATION 7
Proof. (2.7) are proven in [27]. To show (2.8) we first explicitly evaluate P B . By (2.1) for f ∈ H we have P − Bf = − πi Z ∞−∞ f ( s ) B ( s ) dss − ( x − i (cid:0) P − Bf (cid:1) ( x ) = − X n Res (cid:18) f ( z ) /B ( z ) z − x , z n (cid:19) = X n i Im z n B n ( z n ) f ( z n ) x − z n = X n i Im z n B n ( z n ) h f, k z n i x − z n (by (2.5a)).Hence, by (2.7), P B f = X n h f, k z n i i Im z n B n ( z n ) B n · − z n = X n h f, k z n i k ⊥ z n , where k ⊥ z n is given by (2.9). It remains to verifies that (cid:0) k ⊥ z n (cid:1) forms abi-orthogonal basis for K B . Indeed, (cid:10) k ⊥ z n , k z m (cid:11) = (cid:28) z n B n ( z n ) B n k z n , k z m (cid:29) = 2 Im z n B n ( z n ) h B n k z n , k z m i = 2 Im z n B n ( z n ) B n ( z m ) k z n ( z m ) . If n = m then B n ( z m ) = 0. If n = m then by (2.4) (cid:10) k ⊥ z n , k z m (cid:11) = 2 Im z n k z n ( z n ) = 1 . The formula (2.8) easily follows now. (cid:3)
Hankel operators.
A Hankel operator is an infinitely dimen-sional analog of a Hankel matrix, a matrix whose ( j, k ) entry dependsonly on j + k . In the context of integral operators the Hankel opera-tor is usually defined as an integral operator on L ( R + ) whose kerneldepends on the sum of the arguments( H f )( x ) = Z ∞ h ( x + y ) f ( y ) dy, f ∈ L ( R + ) , x ≥ ALEXEI RYBKIN and it is this form that Hankel operators typically appear in the inversescattering formalism. It is much more convenient for our purposes toconsider
Hankel operators on H (cf. [26], [30]).Let ( J f )( x ) = f ( − x )be the operator of reflection on L and let ϕ ∈ L ∞ . The operators H ( ϕ ) defined by H ( ϕ ) f = JP − ϕf, f ∈ H , (2.11)is called the Hankel operator with the symbol ϕ .It is clear that H ( ϕ ) is bounded from H to H and H ( ϕ + h ) = H ( ϕ ) for any h ∈ H ∞ . (2.12)It is also straightforward to verify that H ( ϕ ) is selfadjoint if J ϕ = ¯ ϕ. The following elementary lemma on Hankel operators with analyticsymbols will be particularly useful.
Lemma 2.2.
Let a function ϕ be meromorphic on C and subject to ϕ ( − z ) = ¯ ϕ ( z ) (symmetry). (2.13) If ϕ has finitely many simple poles { z n } Nn = − N in C + , is bounded on R ,and for any h ≥ ϕ ( x + ih ) = O (cid:0) x − (cid:1) , x → ±∞ , (2.14) then the Hankel operator H ( ϕ ) is selfadjoint, trace class, and admitsthe decomposition H ( ϕ ) = H ( φ ) + H (Φ) , (2.15) where φ is a rational function and Φ is an entire function given respec-tively by φ ( x ) = X − N ≤ n ≤ N Res ( ϕ, z n ) x − z n , Φ ( x ) = − πi Z R + ih ϕ ( s ) s − x ds, h > max n Im z n . (2.16) Moreover, H ( φ ) = X − N ≤ n ≤ N i Res ( ϕ, z − n ) (cid:10) · , k z − n (cid:11) k z n , (2.17) H (Φ) = Z R + ih dz π ϕ ( z ) h· , k z i k − z = Z R + ih dz π ϕ ( − z ) h· , k − z i k z , (2.18) where k λ ( z ) = iz − λ is the reproducing kernel of H . DV EQUATION 9
Proof.
The selfadjointness follows from (2.13). By (2.12) H ( ϕ ) = H ( P − ϕ )and hence we have to worry only about P − ϕ . By by the residue theorem( h > max k Im z k ), we have( P − ϕ ) ( x ) = − πi Z R ϕ ( s ) s − ( x − i ds = X − N ≤ n ≤ N Res ( ϕ, z n ) x − z n − πi Z R + ih ϕ ( s ) s − z ds = φ ( x ) + Φ ( x ) , and (2.15) follows. Apparently Φ is analytic (and bounded) belowthe line R + ih . Since h is arbitrary, Φ is then entire. Moreover, allderivatives of Φ are bounded on R and therefore H (Φ) is at least traceclass (in any Shatten-von Neumann ideal).It follows from (2.2) that for any z ∈ C + H ( 1 · − z ) f = if ( z ) k − z and (2.17)-(2.18) follow. (cid:3) Corollary 2.3. If ϕ has no poles in C + then H ( ϕ ) = H (Φ) . Corollary 2.4.
If (2.14) holds uniformly in h ≥ h > max n Im z n then Φ = 0 . A very important feature of analytic symbols is that H ( ϕ ) is well-defined outside of H . In particular, H ( ϕ ) k x + i is a smooth elementof H for any x ∈ R while k x + i H . We will need the followingstatement. Corollary 2.5.
For every x, s ∈ RH (Φ) k x ( s ) = lim ε → H (Φ) k x + iε ( s )= − Z R + ih Φ ( z )( z − x ) ( z + s ) dz π = − Z R + ih ϕ ( z )( z − x ) ( z + s ) dz π (2.19)=: K x ( s ) ∈ C ∞ ( R ) ∩ H . Moreover, if ϕ ε → ϕ uniformly on R + ih then for every x, s ∈ R lim ε → H (Φ ε ) k x + iε ( s + iε ) = K x ( s ) . (2.20) Convergence in (2.19) and (2.20) also holds in L . Proof.
It follows from (2.18) that H (Φ) k x + iε ( s ) = Z R + ih dz π ϕ ( z ) h k x + iε , k z i k − z ( s )= Z R + ih dz π ϕ ( z ) h k x + iε , k z i k − z ( s ) (by (2.5a))= − Z R + ih dz π ϕ ( z ) k x + iε ( z ) k − z ( s ) → − Z R + ih ϕ ( z )( z − x ) ( z + s ) dz π = − Z R + ih Φ ( z )( z − x ) ( z + s ) dz π ,ε → , where we have used two obvious facts: (a) k x + iε ( z ) → k x ( z ) uniformlyon R + ih , and (b) by the Lebesgue dominated convergence Z R + ih φ ( z )( z − x ) ( z + s ) dz π = lim h →∞ Z R + ih φ ( z )( z − x ) ( z + s ) dz π = 0 . Thus (2.19) is proven. (2.20) is proven similarly. (cid:3) Our explicit potential and its short-rangeapproximation
In this section we explicitly construct a symmetric Wigner-von Neu-mann type potentials supporting one negative and one positive boundstate. Our construction is base upon a classical Gelfand-Levitan ex-ample [22] of an explicit potential of a half-line Schr¨odinger operatorwhich spectral measure has one positive pure point. The symmetricextension of this potential to the whole line will be our initial condi-tion. We then find its explicit short range approximation, which willbe crucial to our consideration.3.1.
An explicit WvN type potential.
Consider the function m ( λ ) = i √ λ + 2 ρ − λ , Im λ ≥ , (3.1)where ρ is some positive number. This is a Herglotz function (i.e. ana-lytic function mapping C + to C + ) which coincides with the Titchmarsh-Weyl m − function of the (Dirichlet) Schr¨odinger operators − d /dx + We recall that the problem − ∂ x u + q ( x ) u = λu, x ∈ (0 , ±∞ ) , u ( ± , λ ) = 1has a unique square integrable ( Weyl ) solution Ψ ± ( x, λ ) for any Im λ > q ’s (called limit point case ). Define then the (Titchmarsh-Weyl)m-function m ± for (0 , ±∞ ) as follows: m ± ( λ ) = ± ∂ x Ψ ± ( ± , λ ). DV EQUATION 11 q ( x ) on L (0 , ∞ ) with a Dirichlet boundary condition at 0. The po-tential q has the following explicit form q ( x ) = − d dx log τ ( x ) , x ≥ , (3.2)where τ ( x ) = 1 + 2 ρ Z x sin s ds = 1 + ρx − ( ρ/
2) sin 2 x. (3.3)Introduce Q ( x ) = (cid:26) q ( x ) , x ≥ q ( − x ) , x < , (3.4)i.e., Q is an even extension of q . One can easily see that the function Q is continuous and Q (0) = 0 but not continuously differentiable. Infact, Q is as smooth at x = 0 as | sin x | . Moreover, one has Q ( x ) = − xx + O (cid:18) x (cid:19) , x → ±∞ , (3.5)and hence Q ∈ L ( R ) but (1 + | x | ) Q ( x ) is not in L ( R ). Thus, Q isnot short-range. Also note that Z ∞−∞ Q ( x ) dx = 0 . The main feature of Q is that L Q admits an explicit spectral and scat-tering theory. Theorem 3.1.
The Schr¨odinger operator L Q on L ( R ) with Q givenby (3.4) has the following properties: (1) (Spectrum) The spectrum of L Q consists of the two fold abso-lutely continuos part filling (0 , ∞ ) , one negative bound state − κ found from the real solution of κ + κ = 2 ρ (3.6) and one positive (embedded) bound state +1 . (2) (Scattering quantities) For the norming constant c of − κ wehave c = − i Res ( T ( k ) , iκ ) = − i Res ( R ( k ) , iκ ) = 2 ρ κ + 1 (3.7) and for the scattering matrix we have S ( k ) = (cid:18) T ( k ) R ( k ) R ( k ) T ( k ) (cid:19) , k ∈ R , (3.8) where T and R are, respectively, the transmission and reflectioncoefficients given by T ( k ) = P ( k ) P ( k ) + 2 iρ , R ( k ) = − iρP ( k ) + 2 iρ , (3.9) P ( k ) := k − k. Proof.
Due to symmetry m − = m + = m it follows from the generaltheory [33] that the eigenvalues of the Schr¨odinger operator L Q are the(necessarily simple) poles of m and 1 /m . Thus, L Q has one positivebound state +1 (the pole of m ( λ )) and one negative bound state − κ (the zero of m ( λ )). Clearly (3.6) holds. The fact about the absolutelycontinuos spectrum also follows from the general theory (as well asfrom (2) below) and therefore (1) is proven.Turn to (2). By a direct computation one verifies that f ± ( x, k ) = (cid:26) ± (cid:18) e ± ix k + 1 − e ∓ ix k − (cid:19) ρ sin x ρ | x | − ( ρ/
2) sin 2 | x | (cid:27) e ± ikx , ± x ≥ , solve the Schr¨odinger equation L Q f = k f for ± x ≥ k = ±
1. Sinceclearly f ± ( x, k ) = (1 + o (1)) e ± ikx , x → ±∞ , we can claim that f ± are Jost solution corresponding to ±∞ . By thegeneral formulas (see e.g. [16]) T ( k ) = 1 f − ( k ) f + ( k ) 2 ikm + ( k ) + m − ( k ) (transmission coefficient),(3.10a) R ( k ) = − f + ( k ) f + ( k ) m + ( k ) + m − ( k ) m + ( k ) + m − ( k ) (right reflection coefficient),(3.11a) L ( k ) = − f − ( k ) f − ( k ) m + ( k ) + m − ( k ) m + ( k ) + m − ( k ) (left reflection coefficient)(3.12a)and f ± ( k ) := f ± (0 , k ) are Jost functions. Since in our case m ± = m and f ± ( k ) = 1, we immediately see that L = R and arrive at (3.8).It remains to demonstrate (3.7). Recall the general fact (see e.g. [3])that for any short-range q Res (
T, iκ n ) = i ( − n − p c + n c − n , (3.13) DV EQUATION 13 where c ± n are right/left norming constant associated with the boundstates − κ n ( n = 1 , , ... ) enumerated in the increasing order. If q iseven then c + n = c − n = c n and hence in our case of a single bound state − κ we have Res ( T, iκ ) = ic and the first equation in (3.7) follows. The second and third equationsin (3.7) can be verified by a direct computation. (cid:3) Remark 3.2.
Same way as we did in the proof, one can find an analogof Theorem 3.1 for the truncated potentials Q | R ± . There will be nopositive bound state but the formulas (3.10a)-(3.12a) immediately yieldsame (3.9) where ρ is replaced with ρ . Indeed, for Q | R + we have m + (cid:0) k (cid:1) = m (cid:0) k (cid:1) = ik + 2 ρ − k , m − (cid:0) k (cid:1) = ik, f ± ( k ) = 1 , and the claim follows. Moreover, (3.7) also holds for the truncated Q with the same substitution. This demonstrates clearly that the standardtriple ( R, κ, c ) no longer constitutes scattering data. Short-range approximation of Q . The simples short range ap-proximation is based upon a truncation but the limiting procedure willnot be simple. We instead approximate the scattering data. Whilemuch more complicated than truncation, the limiting procedure be-comes easier to track.If you recall the famous characterization of the scattering matrix [23]of a short-range potential, one of the conditions is that T ( k ) can vanishon C + only at k = 0. But in our case this occurs if P ( k ) = 0 whichhappens also for k = ±
1. This prompts to replace P ( k ) in T ( k ) givenby (3.9) with P ( k ) + iε with some small ε >
0. Clearly P ( k ) + iε = k − k + iε = ( k − µ ε ) ( k + µ ε ) ( k − iν ε ) , where µ ε = 1 − iε/ O (cid:0) ε (cid:1) , ν ε = ε + O (cid:0) ε (cid:1) , ε → . (3.14)Thus two real zeros ± C − . Form the Blaschke product B ε with zeros z − = − µ, z = µ, z = iν ∈ C + . I.e., B ε = b − b b , b n ( k ) = k − z n k − z n . It follows from (3.14) that as ε → z n = n + iε/ | n | + O (cid:0) ε (cid:1) , n = 0 , ± . The Blaschke product B ε will be a building block in our approximation.Apparently, B ε → ε → C + and a.e.on R . We are now ready to present our approximation. Theorem 3.3.
Let ( ε > ) T ε ( k ) = ( P ( k ) + iε ) /b ( k ) P ( k ) + iρ (1 + a ) 1 P ( k ) + iρ (1 − a ) ,R ε ( k ) = − iaρP ( k ) + iρ (1 + a ) P ( k ) P ( k ) + iρ (1 − a ) 1 B ε ( k ) , (3.15) a := q − ( ε/ρ ) . Then (1)
The matrix S ε = (cid:18) T ε R ε R ε T ε (cid:19) is the scattering matrix of a short-range potential having twobound states − (cid:0) κ ε ± (cid:1) , κ ε + > κ ε − , subject to κ ε + = κ + O (cid:0) ε (cid:1) , κ ε − = O (cid:0) ε (cid:1) , ε → . (3.16)(2) If we choose the left and right norming constants associated with − (cid:0) κ ε ± (cid:1) equal to each other and to satisfy c ε ± = ∓ i Res (cid:0) T ε , iκ ε ± (cid:1) , (3.17) then the unique potential Q ε ( x ) corresponding to the scatteringdata (cid:8) R ε , κ ε ± , c ε ± (cid:9) is even and everywhere Q ε ( x ) → Q ( x ) , ε → . (3.18) Proof.
To prove part 1 of the statement one needs to check all theconditions of the Marchenko characterization [23]. Is is straightforwardbut quite involved and we omit it. By the general theory, the boundstates are the squares of the (simple) poles of T ε in C + , i.e. the solutionsof two P ( k ) + iρ (1 ± a ) = 0 , B ε ( k ) = 0 . Since each equation has only one imaginary solution iκ ε ± , we have ex-actly two bound states − (cid:0) κ ε ± (cid:1) which are clearly subject to (3.16).Note that z , the zero of b , is a removable singularity by our veryconstruction of B ε . DV EQUATION 15
Turn now to part 2. Consider the reflection coefficient R ε . Appar-ently, R ε is a rational function with five simple poles. Two imaginarypoles iκ ε ± are shared with T ε plus z n , n = 0 , ±
1, the zeros of B ε ( k ). Bya direct computation, one checksRes (cid:0) R ε , iκ ε ± (cid:1) = ± Res (cid:0) T ε , iκ ε ± (cid:1) = ic ε ± (since (3.17)). (3.19)We now solve the inverse scattering problem for the data (cid:8) R ε , κ ε ± , − i Res (cid:0) R ε , iκ ε ± (cid:1)(cid:9) , basing upon our Hankel operator approach [13]. To this end, form thesymbol ϕ εx ( k ) = − Res (cid:0) R ε , iκ ε + (cid:1) k − iκ ε + e − κ ε + x + − Res (cid:0) R ε , iκ ε − (cid:1) k − iκ ε − e − κ ε − x + R ε ( k ) e ikx . (3.20)One immediately sees that ϕ εx is subject to the conditions of Lemma2.2 with three (symmetric) poles z n , n = 0 , ±
1. By condition, the leftand right scattering data are identical and hence Q ε must be even andit enough to recover it only on (0 , ∞ ). Therefore we can assume that x > H ( ϕ εx ) = X − ≤ n ≤ i Res ( ϕ εx , z − n ) (cid:10) · , k z − n (cid:11) k z n . Thus, our Hankel operator is rank 3 and by the Dyson formula [13] wehave Q ε ( x ) = − ∂ x log det ( I + H ( ϕ εx )) , x > . Note that our Q ε has an exponential decay and can be explicitly evalu-ated. We however don’t really need it. We will take the limit as ε → (cid:3) We emphasize that Part 2 of Theorem 3.3 is essential because, dueto nonuniqueness, it is a priori unclear if our approximations indeedconverges to the original potential.Note also, that Q ε ( x ) all have the property that T ε (0) = 0. Suchpotentials are called exceptional because generically T (0) = 0 . Main results
Through this section ξ x,t ( k ) = exp { i (8 k t + 2 kx ) } . Theorem 4.1.
Let Q be the initial condition (3.4) in the KdV equation(1.1), ϕ x,t ( k ) = R ( k ) ξ x,t ( k ) − Res ( Rξ x,t , iκ ) k − iκ , and H x,t := H ( ϕ x,t ) , the associated Hankel operator. Then (1.1) hasthe (unique) classical solution given by u ( x, t ) = u ( x, t ) + u ( x, t ) (4.1) where u ( x, t ) = − ∂ x log det { I + H x,t } , (4.2) and u ( x, t ) = − ∂ x log τ ( x, t ) ,τ ( x, t ) = 1 + ρ ( x + 12 t ) − ρ x + 8 t )+ ρ I + H x,t ) − ( H x,t k i − ξ x,t (1) H x,t k − i ) (cid:12)(cid:12) i . Here, as before, k λ ( s ) = is − λ is the reproducing kernel.Proof. Since our approximation Q ε ( x ) decays exponentially, the (clas-sical) solution to the KdV equation can be found in closed form byDyson’s formula Q ε ( x, t ) = − ∂ x log det (cid:0) I + H (cid:0) ϕ εx,t (cid:1)(cid:1) , (4.3)where ϕ εx,t ( k ) = − Res (cid:0) R ε , iκ ε + (cid:1) k − iκ ε + ξ x,t (cid:0) iκ ε + (cid:1) + − Res (cid:0) R ε , iκ ε − (cid:1) k − iκ ε − ξ x,t (cid:0) iκ ε − (cid:1) (4.4)+ R ε ( k ) ξ x,t ( k ) . Note that due to the Bourgain theorem [6] the limit lim ε → Q ε ( x, t )does exist but we cannot pass to the limit in (4.3) under the determi-nant sign since, as we will see later, H (cid:0) ϕ εx,t (cid:1) doesn’t converge in thetrace norm to H ( ϕ x,t ), where ϕ x,t ( k ) = − Res (
R, iκ ) k − iκ ξ x,t ( iκ ) + R ( k ) ξ x,t ( k ) . To detour this circumstance we split our determinant as follows. Con-sider K ε = span { k z n } n = − , and decompose H into the orthogonal sum (see Subsection 2.2) H = K ε ⊕ K ⊥ ε , K ⊥ ε = B ε H . (4.5) DV EQUATION 17
The decomposition (4.5) induces the block representation H (cid:0) ϕ εx,t (cid:1) = (cid:18) H H H ∗ H (cid:19) , where H := P B ε H (cid:0) ϕ εx,t (cid:1) P B ε , H = P ⊥ B ε H (cid:0) ϕ εx,t (cid:1) P ⊥ B ε H := P ⊥ B ε H (cid:0) ϕ εx,t (cid:1) P B ε , H ∗ = P B ε H (cid:0) ϕ εx,t (cid:1) P ⊥ B ε , and ϕ εx,t ( k ) = − Res (cid:0) R ε , iκ ε + (cid:1) k − iκ ε + ξ x,t (cid:0) iκ ε + (cid:1) + − Res (cid:0) R ε , iκ ε − (cid:1) k − iκ ε − ξ x,t (cid:0) iκ ε − (cid:1) + R ε ( k ) ξ x,t ( k ) . Examine the block H first. It follows from (4.4) that the poles of ϕ εx,t coincide with zeros ( z n ) of B ε and therefore by Lemma 2.2 ( h > κ ε + ) H (cid:0) ϕ εx,t (cid:1) = X − ≤ n ≤ i Res ( ξ x,t R ε , z − n ) (cid:10) · , k z − n (cid:11) k z n + Z R + ih dz π ϕ εx,t ( z ) h· , k z i k − z = X − ≤ n ≤ i Res ( ξ x,t R ε , z − n ) (cid:10) · , k z − n (cid:11) k z n + H (cid:0) Φ εx,t (cid:1) . One immediately sees that H = P ⊥ B ε H (cid:0) Φ εx,t (cid:1) P ⊥ B ε . Since ϕ εx,t → ϕ x,t uniformly on R + ih , we obviously haveΦ εx,t ( x ) = − πi Z R + ih ϕ εx,t ( s ) s − x ds → − πi Z R + ih ϕ x,t ( s ) s − x ds = Φ x,t ( x ) , ε → , in C n ( R ) for any n which in turn implies [30] that lim ε → H (cid:0) Φ εx,t (cid:1) = H (Φ x,t ) in the trace norm (in fact in all S p , p > ϕ x,t ( k ) = − Res (
R, iκ ) k − iκ ξ x,t ( iκ ) + R ( k ) ξ x,t ( k ) , we see that iκ is a removable singularity for ϕ x,t and hence by Corollary2.3 H (Φ x,t ) = H ( ϕ x,t ) . Since B ε → P ⊥ B ε = B ε P + B ε → I, ε → . (4.6)But [5], if H n → H in trace norm, A n is self-adjoint, sup n k A n k < ∞ ,and A n → A strongly, then A n H n A n → AHA in trace norm. Therefore,we can conclude that in trace norm H → H ( ϕ x,t ) , ε → . (4.7)We now make use of a well-known formula from matrix theory:det (cid:18) A A A A (cid:19) = det A det (cid:0) A − A A − A (cid:1) , (4.8)which yieldsdet (cid:0) I + H (cid:0) ϕ εx,t (cid:1)(cid:1) (4.9)= det { I + H } · det (cid:8) I + H − H ∗ ( I + H ) − H (cid:9) . Our goal is to study what happens to (4.9) as ε →
0. The determinantson the right hand side of (4.9) behave very differently and we treat themseparately. It follows from (4.7) thatlim ε → det { I + H } = det { I + H ( ϕ x,t ) } . (4.10)Turn now to the second determinant in (4.9). It is clearly a 3 × O ( ε ). To this end, we explicitly evaluate it in the basis( k z n )det (cid:8) I + H − H ∗ ( I + H ) − H (cid:9) (4.11)= det h − − + d h − + d − h − + d − h − + d − h + d h + d h − + d − h + d h − − + d , where h mn and d mn are the matrix entries of X − ≤ n ≤ i Res ( ξ x,t R ε , z − n ) (cid:10) · , k z − n (cid:11) k z n and P B H (cid:0) Φ εx,t (cid:1) P B − H ∗ ( I + H ) − H DV EQUATION 19 respectively. By Lemma 2.1 h mn = * X − ≤ j ≤ i Res ( ξ x,t R ε , z − j ) (cid:10) k z n , k z − j (cid:11) k z j , k ⊥ z m + (4.12)= i Res ( ξ x,t R ε , z − m ) (cid:10) k z n , k z − m (cid:11) = i Res ( ξ x,t R ε , z − m ) k z n ( z − m )= ξ x,t ( z − m ) Res ( R ε , z − m ) z m + z n . Incidentally, (4.12) implies h − = h − , h − − = h . Recall that z n are chosen so that P ( z n ) − iε = 0 if n = ± P ( z n ) + iε = 0 if n = 0. Rewriting (3.15) as R ε ( k ) = aR ( k ) P ( k ) + 2 iρP ( k ) + iρ (1 + a ) P ( k ) P ( k ) + iρ (1 − a ) 1 B ε ( k ) , for the residues we then haveRes ( R ε , z n )= aR ( z n ) P ( z n ) + 2 iρP ( z n ) + iρ (1 + a ) P ( z n ) P ( z n ) + iρ (1 − a ) 2 i Im z n B εn ( z n ) . One now readily verifies that P ( z n ) + 2 iρP ( z n ) + iρ (1 + a ) = 1 + O (cid:0) ε (cid:1) ,P ( z n ) P ( z n ) + iρ (1 − a ) = 1 + ( − n ε ρ + O (cid:0) ε (cid:1) ,B εn ( z n ) − = 1 + 5 inε/ O (cid:0) ε (cid:1) , and thusRes ( R ε , z n ) (4.13)= 2 i Im z n R ( z n ) (cid:20) iε (cid:18) n + ( − n ρ (cid:19) + O (cid:0) ε (cid:1)(cid:21) . Inserting (4.13) into (4.12) yields h mn = 2 i Im z m z m + z n ( ξ x,t R ) ( z − m ) (cid:20) iε (cid:18) m + ( − m ρ (cid:19) + O (cid:0) ε (cid:1)(cid:21) . Observe, that h n,m = O ( ε ) if n = − m and h m, − m doesn’t vanish as ε → only h − and h matter. Recalling that R (1) = − h − (4.14)= ξ x,t (1) (cid:26) − ε (cid:20) ρ + iξ x,t (1) ( Rξ x,t ) ′ (1) + 5 i (cid:21) + O (cid:0) ε (cid:1)(cid:27) ,h − − = iε ξ x,t (1) [1 + O ( ε )] . (4.15)Similarly, for the matrix ( d mn ) we have d mn = (cid:10) H (cid:0) Φ εx,t (cid:1) k z n , k ⊥ z m (cid:11) − (cid:10) ( I + H ) − H k z n , H k ⊥ z m (cid:11) (4.16)= 2 Im z m B εm ( z m ) (cid:8)(cid:10) H (cid:0) Φ εx,t (cid:1) k z n , B εm k z m (cid:11) − (cid:10) ( I + H ) − H k z n , H B εm k z m (cid:11)(cid:9) . = εD mn + O ( ε ) , where D mn will be computed later. For the determinant in (4.11) weclearly havedet (cid:8) I + H − H ∗ ( I + H ) − H (cid:9) (4.17)= (1 + h + d ) det (cid:18) h − − + d h − + d − h − + d − h − − + d (cid:19) + O (cid:0) ε (cid:1) = 2 n(cid:12)(cid:12) h − − + d (cid:12)(cid:12) − | h − + d − | o + O (cid:0) ε (cid:1) (by (4.15)-(4.16))= 2 (cid:0) − | h − | + 2 Re h − − (cid:1) + 2 ε Re (cid:2) D − ξ x,t (1) D − (cid:3) + O (cid:0) ε (cid:1) . Evaluate each term in the right hand side of (4.17) separately. By(4.14)-(4.15) one has1 − | h − | + 2 Re h − − (4.18)= ε n /ρ + Re iξ x,t (1) (cid:2) ( Rξ x,t ) ′ (1) − (cid:3) + O ( ε ) o = 2 ερ n ρ ( x + 12 t ) − ρ x + 8 t ) + O ( ε ) o and D mn = lim ε → (cid:8)(cid:10) H (cid:0) Φ εx,t (cid:1) k z n , B εm k z m (cid:11) − (cid:10) ( I + H ) − H k z n , H B εm k z m (cid:11)(cid:9) =: D (1) mn + D (2) mn . DV EQUATION 21
Since H (cid:0) Φ εx,t (cid:1) is a self-adjoint operator, by Corollary 2.5 we have ( m = ± , n = 1) D (1) mn = lim ε → (cid:10) H (cid:0) Φ εx,t (cid:1) B m k z m , k z n (cid:11) (4.19)= lim ε → H (cid:0) Φ εx,t (cid:1) B m k z m (cid:12)(cid:12) z n (by (2.2))= K m ( n ) , where K m ( n ) = − Z R + ih ϕ x,t ( z )( z − m ) ( z + n ) dz π . Similarly, by (4.6), (4.7), and Corollary 2.5 we have D (2) mn = − H ( ϕ x,t ) ( I + H ( ϕ x,t )) − K m (cid:12)(cid:12) n + i . (4.20)Therefore, combining (4.19) and (4.20) we have D mn = K m ( n ) − H ( ϕ x,t ) ( I + H ( ϕ x,t )) − K m (cid:12)(cid:12) n + i = ( I + H ( ϕ x,t )) − K m (cid:12)(cid:12) n + i . Substituting this and (4.18) into (4.17) yields ρ ε det (cid:8) I + H − H ∗ ( I + H ) − H (cid:9) = 1 + ρ ( x + 12 t ) − sin (2 x + 8 t )+ ρ I + H ( ϕ x,t )) − ( K − ξ x,t (1) K − ) (cid:12)(cid:12) i + O ( ε ) (4.21)We have now prepared all the ingredients to find the solution to theKdV equation with the initial data Q ε by the Dyson formula. Indeed, Q ε ( x, t ) = − ∂ x log det (cid:8) I + H (cid:0) ϕ εx,t (cid:1)(cid:9) (by (4.17)) (4.22)= 2 ∂ x log det ( I + H ) − ∂ x log det (cid:8) I + H − H ∗ ( I + H ) − H (cid:9) = − ∂ x log det { I + H ( ϕ x,t ) } (by (4.10) and (4.21)) − ∂ x log n ρ ( x + 12 t ) − ρ x + 8 t )+ ρ I + H ( ϕ x,t )) − ( K − ξ x,t (1) K − ) (cid:12)(cid:12) i o + O ( ε ) . We are now able to fill the gap left in the proof of Theorem 3.3, i.e.(3.18). To this end, set t = 0 in (4.22) and take x >
0. In this case ξ x, ∈ H ∞ and hence ϕ x, ∈ H ∞ . Therefore, H ( ϕ x,t ) = 0 and by the Lebesgue dominated convergence theorem (or by Corollary 2.5) we alsohave K m ( s ) = − Z R + ih ϕ x, ( z )( z − m ) ( z + s ) dz π = − lim h →∞ Z R + ih ϕ x, ( z )( z − m ) ( z + s ) dz π = 0 . Eq. (4.22) simplifies now to read Q ε ( x,
0) = − ∂ x log (cid:16) ρx − ρ x (cid:17) + O ( ε ) , x > . Recalling (3.2), we conclude that Q ε ( x ) = Q ε ( x, → q ( x ) for x > Q ε ( x ) is even, (3.18) follows.Pass now in (4.22) to the limit as ε →
0. Apparently,lim ε → Q ε ( x, t ) = − ∂ x log det { I + H ( ϕ x,t ) }− ∂ x log n ρ ( x + 12 t ) − ρ x + 8 t )+ ρ I + H ( ϕ x,t )) − ( K − ξ x,t (1) K − ) (cid:12)(cid:12) i o . By the Bourgain theorem Q ( x, t ) = lim ε → Q ε ( x, t ) is the (unique)solution to the KdV equations with data Q ( x ). Recalling Corollary2.5, we see that K n = H ( ϕ x,t ) k n + i , n = ± . This completes the proof of the theorem. (cid:3)
Note that the first term u ( x, t ) in the solution (4.1) is given bythe same Dyson formula (4.2) as in the short-range case but of course u ( x,
0) is not a short range potential. The second term u ( x, t ) in(4.1) is responsible for the bound state +1 and if ρ = 1 it resemblesthe so-called positon solution u pos ( x, t ) = − ∂ x log (cid:26) x + 12 t −
12 sin 2 ( x + 4 t ) (cid:27) . (4.23)Such solutions seem to have appeared first in the late 70s earlier 80sbut a systematic approach was developed a decade later by V. Matveev(see his 2002 survey [24]).The formula (4.23) readily yields basic properties of one-positionsolutions. (1) As a function of the spatial variable u pos ( x, t ) has adouble pole real singularity which oscillates in the 1 / x = − t −
1. (2) For a fixed t ≥ u pos ( x, t ) = − x + 4 t ) x + O (cid:0) x − (cid:1) , x → ±∞ . (4.24) DV EQUATION 23
Observe that u pos ( x,
0) = − ∂ x log (cid:18) x −
12 sin 2 x (cid:19) , which coincides on (0 , ∞ ) with our Q ( x ) for ρ = 1. Moreover, com-paring (3.5) with (4.24) one can see that the asymptotic behaviors for x → −∞ of our Q ( x ) with ρ = 1 and u pos ( x,
0) differ only by O ( x − ).But, of course, Q ( x ) is bounded on ( −∞ ,
0) while u pos ( x,
0) is not.Note also that the positon is somewhat similar to the soliton given by u sol ( x, t ) = − ∂ x log cosh ( x − t ) . (4.25)As opposed to the soliton, the positon has a square singularity (not asmooth hump) moving in the opposite direction three times as fast.We note that multi-positon as well as soliton-positon solutions havebeen studied in great detail (see [24] the references cited therein). In[24] Matveev also raises the equation if there is a bounded positon, i.e.a solution having all properties of a positon but is regular. We areunable to tell if our solution is a bounded positon or not. References [1] Ablowitz, M. J.; Clarkson , P. A.
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