The third order Benjamin-Ono equation on the torus : well-posedness, traveling waves and stability
aa r X i v : . [ m a t h . A P ] D ec The third order Benjamin-Ono equation on the torus :well-posedness, traveling waves and stability.
Louise Gassot
Abstract
We consider the third order Benjamin-Ono equation on the torus ∂ t u = ∂ x (cid:18) − ∂ xx u − uH∂ x u − H ( u∂ x u ) + u (cid:19) . We prove that for any t ∈ R , the flow map continuously extends to H sr, ( T ) if s ≥ , but doesnot admit a continuous extension to H − sr, ( T ) if < s < . Moreover, we show that the extensionis not weakly sequentially continuous in L r, ( T ) . We then classify the traveling wave solutionsfor the third order Benjamin-Ono equation in L r, ( T ) and study their orbital stability. Contents H sr, ( T ) , s ≥ . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Ill-posedness in H − sr, ( T ) , s > . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 A Appendices 15
A.1 About the hierarchy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15A.2 Equation for the fourth Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . 19A.3 Structure of the higher order Hamiltonians . . . . . . . . . . . . . . . . . . . . . . 21
We are interested in the third equation of the integrable Benjamin-Ono hierarchy on the torus ∂ t u = ∂ x (cid:18) − ∂ xx u − uH∂ x u − H ( u∂ x u ) + u (cid:19) . (1)The operator H is the Hilbert transform, defined as Hf ( x ) = X n ∈ Z \ − i sgn( n ) b f ( n ) e inx , f = X n ∈ Z b f ( n ) e inx , b f ( n ) = 12 π Z π f ( x ) e − inx d x. .1 Benjamin-Ono equations and integrability The Benjamin-Ono equation on the torus ∂ t u = H∂ xx u − ∂ x ( u ) , was introduced by Benjamin [2] and Ono [12] in order to describe long internal waves in atwo-layer fluid of great depth. This equation admits an infinite number of conserved quantities H k , k ≥ (see Nakamura [11] for a proof on the real line). The evolution equations associatedto the conservation laws ∂ t u = ∂ x ( ∇H k ( u )) (2)are the equations for the Benjamin-Ono hierarchy [8].From Nakamura [10] and Bock, Kruskal [3], we know that the Benjamin-Ono equation admitsa Lax pair dd t L u = [ B u , L u ] ,L u = Dh − T u , B u = iD + 2 iT D (Π u ) − iDT u . Here, D = − i∂ x and T u is the Toeplitz operator on the Hardy space L ( T ) = { h ∈ L ( T ) | ∀ n < , b h ( n ) = 0 } defined as T u : h ∈ L ( T ) Π( uh ) ∈ L ( T ) , and Π : L ( T ) → L ( T ) is the Szegő projector. The Hamiltonians H k ( u ) are defined from theLax operator L u as H k ( u ) = h L ku | i . (3)In particular, the Hamiltonian for equation (1) is H ( u ) + 12 H ( u ) = 12 π Z π (cid:18)
12 ( ∂ x u ) − u H∂ x u + 14 u (cid:19) d x. (4)In [5], Gérard and Kappeler constructed global Birkhoff coordinates for the Benjamin-Onoequation on the torus. In these coordinates, the evolution equations for the Benjamin-Onohierarchy are easier to understand. Indeed, denote by Φ the Birkhoff map Φ : u ∈ L r, ( T ) ( ζ n ( u )) n ≥ ∈ h + , where L r, ( T ) is the subspace of real valued functions in L ( T ) with zero mean, and h + = n ( ζ n ) n ≥ | X n ≥ n | ζ n | < + ∞ o . Then in the Birkhoff coordinates, equation (2) of the hierarchy associated to H k becomes ∂ t ζ n = iω ( k ) n ζ n , n ≥ when the frequencies ω ( k ) n = ∂ ( H k ◦ Φ − ) ∂ | ζ n | are well-defined. For instance, this formula is valid if the sequence ζ (0) = ( ζ n (0)) n ≥ only has afinite number of nonzero terms, or in other words, if Φ − ( ζ (0)) is a finite gap potential. In thiscase, the frequencies ω ( k ) n only depend on the actions | ζ p | , and the evolution simply reads ζ n ( t ) = ζ n (0) e iω ( k ) n ( ζ (0)) t , t ∈ R , n ≥ . or the third equation of the hierarchy (1), the frequencies write ω (4) n ( ζ ) = n + n X p ≥ p | ζ p | − X p ≥ min( p, n ) | ζ p | +3 X p,q ≥ min( p, q, n ) | ζ p | | ζ q | . (5)More details about the frequencies ω ( k ) n and formula (5) can be found in Appendix A.1.We refer to Saut [15] for a detailed survey of the Benjamin-Ono equation and of its hierarchy. Our first main result is the determination the well-posedness threshold for the third orderBenjamin-Ono equation. For s ∈ R , we use the notation H − sr, ( T ) = { u ∈ H s ( T , R ) | h u | i = 0 } . We prove that the flow map is globally C -well-posed (in the sense of Definitions 1 and 2 from [6])in H sr, ( T ) when s ≥ , but is not globally C -well-posed in H − sr, ( T ) when < s < . Theorem 1.1.
For all t ∈ R , the flow map for equation (1) S t : u u ( t ) , defined for finitegap potentials, admits a continuous extension to H sr, ( T ) for all s ≥ , but does not admit acontinuous extension to H − sr, ( T ) for < s < . Remark 1.2.
Note that if s ≥ , the maps t ∈ R u ( t ) constructed in this way are solutionsto equation (1) in the distribution sense. We also investigate the question of the sequential weak continuity for the flow map.
Theorem 1.3.
For all t ∈ R , the extension of flow map for equation (1) S t : u u ( t ) isweakly sequentially continuous in H sr, ( T ) for s > , but is not weakly sequentially continuous in L r, ( T ) . In [6], Gérard, Kappeler and Topalov proved that the flow map for the Benjamin-Ono equa-tion is globally C -well-posed in H sr, ( T ) for s > − , whereas from [13] there is no continuousextension of the flow map to H sr, ( T ) when s < − . We expect that the well-posedness thresholdon the torus increases by for each new equation in the hierarchy : for the equation correspond-ing to the k -th Hamiltonian H k , k ≥ , the threshold should be H k − r, ( T ) (see Remarks 2.6and A.1). Note that all the equations for the Benjamin-Ono hierarchy have critical Sobolevexponent − .Let us mention former approaches to the Cauchy problem for higher order Benjamin-Onoequations. Tanaka [17] considered more general third order type Benjamin-Ono equations onthe torus ∂ t u = ∂ x ( − ∂ xx u − c uH∂ x u − c H ( u∂ x u ) + u ) , and proved local well-posedness in H s ( T ) for s > . He deduced global well-posedness in H s ( T ) , s ≥ for the integrable case c = c = .On the real line, Feng and Han [4] proved local well-posedness in H s ( R ) , s ≥ for thethird equation of the Benjamin-Ono hierarchy (1). Considering more general third order typeBenjamin-Ono equations under the form ∂ t u − bH∂ xx u − a∂ xxx u = cv∂ x v − d∂ x ( vH∂ x v + H ( v∂ x v )) , Linares, Pilod and Ponce [7] established local well-posedness in H s ( R ) , s ≥ , then Molinet andPilod [9] proved global well-posedness in H s ( R ) , s ≥ .Concerning Benjamin-Ono equations of fourth order on the torus and on the real line,Tanaka [16] proved local well-posedness in H s , s > for a more general family of fourth order ype Benjamin-Ono equations, and deduced global well-posedness in H s , s ≥ in the integrablecase.Our second main result is the classification of the traveling waves for the third order Benjamin-Ono equation in L r, ( T ) , i.e. the solutions to (1) under the form u ( t, x ) = u ( x + ct ) , t ∈ R , x ∈ T , u ∈ L r, ( T ) . Definition 1.4.
For N ≥ , we say that u ∈ L r, ( T ) is a N gap potential if the set { n ≥ | ζ n ( u ) = 0 } , where Φ( u ) = ( ζ n ( u )) n ≥ , is finite and of cardinality N . Theorem 1.5.
A potential u ∈ L r, ( T ) defines a traveling wave for equation (1) if and only if• either u is a one gap potential ;• either u is a two gap potential, and the two nonzero indexes p < q satisfy, with γ p = | ζ p | and γ q = | ζ q | , < γ p < p + r p + 4 q p + q ! and γ q = q p + q − γ p + pγ p γ p + q . Note that from [5], the one gap potentials are the only traveling wave solutions to theBenjamin-Ono equation ; they have been characterized by Amick and Toland [1].Our last main result answers the question of orbital stability for these two types of travelingwaves.
Definition 1.6.
Let u ∈ L r, ( T ) be a one gap traveling wave. We say the u is orbitally stableif for all ε > , there exists δ > such that if v is a solution to (1) with initial condition v ∈ L r, ( T ) such that k v − u k L ( T ) ≤ δ , then sup t ∈ R inf θ ∈ T k v ( t ) − u ( · + θ ) k L ( T ) ≤ ε. Theorem 1.7.
The one gap traveling waves are orbitally stable, whereas the two gap travelingwaves are orbitally unstable.
For the Benjamin-Ono equation, Pava and Natali [14] proved the orbital stability of thetraveling wave solutions in H r, ( T ) . In [6], Gérard, Kappeler and Topalov improved the orbitalstability of these solutions to H − sr, ( T ) , ≤ s < . Plan of the paper
The paper is organized as follows. We first prove the well-posednessthreshold for the third order Benjamin-Ono equation (1) in Section 2. Finally, in Section 3, weclassify the traveling wave solutions and study their orbital stability properties.In Appendix A.1, we describe how to compute the Hamiltonians H k and frequencies ω ( k ) n = ∂ H k ◦ Φ − ∂ | ζ n | in terms of the action variables | ζ p | . In Appendix A.2, we retrieve the Hamiltonianand frequencies of the third order Benjamin-Ono equation (see formulas (4) and (5)) by startingfrom the definition (3) of the higher order Hamiltonians. In Appendix A.3, we provide analternative proof of a result from [18] about the structure of the higher order Hamiltonians byusing formula (3), which may be of independent interest. Acknowledgements
The author would like to thank her PhD advisor Professor P. Gérardfor introducing her to this problem, and for his continuous support and advices. Well-posedness threshold for the fourth Hamiltonian
Let N ∈ N and let U N be the set U N = { u ∈ L r, ( T ) | ζ N ( u ) = 0 , ζ j ( u ) = 0 ∀ j > N } . We know from [5], Theorem 3, that the restriction of the Birkhoff map Φ to U N is a real analyticdiffeomorphism onto some Euclidean space. In Birkhoff coordinates, the evolution along the flowof equation (1) for an initial data u ∈ U N writes ( ∂ t ζ n = iω (4) n ( u ) ζ n ζ n (0) = ζ n ( u ) , n ≥ , where for all n ≥ , the frequencies ω (4) n ( u ) are given by (5) ω (4) n ( u ) = n + n X p ≥ p | ζ p ( u ) | − X p ≥ min( p, n ) | ζ p ( u ) | +3 X p,q ≥ min( p, q, n ) | ζ p ( u ) | | ζ q ( u ) | . This implies that ζ n ( u ( t )) = ζ n ( u ) e iω (4) n ( u ) t , t ∈ R , n ≥ . Therefore, for any finite gap inifial data u , belonging to some of the sets U N , the flow map S t : u ∈ U N u ( t ) ∈ U N is well-defined.In part 2.1, we prove that for all t ∈ R , this flow map extends by continuity to H sr, ( T ) for s ≥ . We also show that the extension is sequentially weakly continuous in H sr, ( T ) for s > ,but not in L r, ( T ) . In part 2.2, we prove that the flow map does not extend by continuity to H − sr, ( T ) for s > . This gives a threshold for the global C -well-posedness of the third orderBenjamin-Ono equation in the sense of Definitions 1 and 2 from [6]. H sr, ( T ) , s ≥ Proposition 2.1.
Let s ≥ . For any u ∈ H sr, ( T ) , there exists a continuous map t ∈ R t ( u ) = u ( t ) ∈ H sr, ( T ) with u (0) = u such that the following holds.For any finite gap sequence ( u k ) k converging to u in H sr, ( T ) , for any t ∈ R , u k ( t ) = S t ( u k ) converges to u ( t ) in H sr, ( T ) as k goes to infinity.Moreover, the extension of the flow map S : u ∈ H sr, ( T ) ( t u ( t )) ∈ C ( R , H sr, ( T )) iscontinuous. Recall that from [6], as mentioned in the proofs of Proposition 2 and Theorem 8, we havethe following result. For s ≥ , the Birkhoff map Φ defines a homeomorphism between H sr, ( T ) and the space h + s + = n ( ζ n ) n ≥ | X n ≥ n s | ζ n | < + ∞ o . The proof of Proposition 2.1 therefore relies on the following sequential convergence result ob-tained after applying the Birkhoff map.
Lemma 2.2.
Fix s ≥ . Let ζ k = ( ζ kn ) n ≥ , k ∈ N , and ζ be elements of h + s + such that k ζ k − ζ k h
12 + s + −→ k → + ∞ . Then for all t ∈ R , k ( ζ kn e iω (4) n ( ζ k ) t ) n − ( ζ n e iω (4) n ( ζ ) t ) n k h
12 + s + −→ k → + ∞ , where the convergence is uniform on bounded time intervals. roof. Note that since ( ζ k ) k converges to ζ in h + s + , then for all n ∈ N , formula (5) for ω (4) n ( ζ k ) implies that ω (4) n ( ζ k ) converges to ω (4) n ( ζ ) as k goes to infinity.Let ε > . Fix K ∈ N such that for all k ≥ K , k ζ k − ζ k h
12 + s + ≤ ε. Using that ζ ∈ h + s + , fix N ∈ N such that (cid:16) X n ≥ N n s | ζ n | (cid:17) ≤ ε. Now, if k ≥ K , k ( ζ kn e iω (4) n ( ζ k ) t ) n − ( ζ n e iω (4) n ( ζ ) t ) n k h
12 + s + ≤ k ( ζ kn ) n − ( ζ n ) n k h
12 + s + + k ( ζ n ( e iω (4) n ( ζ k ) t − e iω (4) n ( ζ ) t )) n k h
12 + s + ≤ ε + N − X n =0 n s | ζ n ( e iω (4) n ( ζ k ) t − e iω (4) n ( ζ ) t ) | ! , which is less than ε for k large enough by convergence term by term of the elements in the sum.Moreover, this convergence is uniform on bounded time intervals. Proof of Proposition 2.1.
Let s ≥ and u ∈ H sr, ( T ) . Fix t ∈ R , and a sequence of finite gapinitial data ( u k ) k converging to u in H sr, ( T ) .We first establish that for all t ∈ R , ( u k ( t )) k has a limit in H sr, ( T ) as k goes to + ∞ . Byassumption, Φ( u k ) converges to Φ( u ) in h + s + . Define the sequence ζ ( t ) by ζ n ( t ) := ζ n ( u ) e iω (4) n ( u ) t , n ∈ N . Lemma 2.2 immediately implies that the sequence (Φ( u k ( t ))) k converges to ζ ( t ) in h + s + . Since Φ − defines a continuous application from h + s + to H sr, ( T ) , we deduce that u k ( t ) converges in H sr, ( T ) to u ( t ) := Φ − ( ζ ( t )) . Moreover, the convergence is uniform on bounded time intervals.We now prove the continuity of the flow map S t . Let u k ∈ H sr, ( T ) , k ∈ N , be a sequence ofinitial data converging to some u in H sr, ( T ) . Then Φ( u k ) converges to Φ( u ) in h + s + , and theabove Lemma 2.2 again implies that Φ( u k ( t )) converges to Φ( u ( t )) in h + s + . In other terms, u k ( t ) converges to u ( t ) in H sr, ( T ) , where again this convergence is uniform on bounded intervals. Corollary 2.3.
For all s > and all t ∈ R , the extension of the flow map restricted to H sr, ( T ) : u ∈ H sr, ( T ) u ( t ) ∈ H sr, ( T ) is sequentially weakly continuous.Proof. Let u k ∈ H sr, ( T ) , k ∈ N , be a sequence weakly converging in H sr, ( T ) to u ∈ H sr, ( T ) .Since the embedding H sr, ( T ) ֒ → L r, ( T ) is compact, ( u k ) k is strongly convergent to u in L r, ( T ) .By continuity of the flow map S t , one deduces that ( u k ( t )) k converges strongly to u ( t ) in L r, ( T ) .This implies that ( u k ( t )) k converges weakly to u ( t ) in H sr, ( T ) . Proposition 2.4.
For all t ∈ R ∗ , the extension of the flow map restricted to L r, ( T ) : u ∈ L r, ( T ) u ( t ) ∈ L r, ( T ) is not sequentially weakly continuous. roof. Fix t ∈ R ∗ and u ∈ L r, ( T ) \ { } . We construct a sequence ( u k ) k in L r, ( T ) weaklyconvergent to u in L r, ( T ) but such that u k ( t ) = S t ( u k ) is not weakly convergent to u ( t ) = S t ( u ) in L r, ( T ) .Let α > to be chosen later. For k ∈ N , we choose ( ζ p ( u k )) p converging weakly to ( ζ p ( u )) p in h + (so that u k converges weakly to u in L r, ( T ) ) and such that | ζ p ( u k ) | = | ζ p ( u ) | + αp δ k,p , p ≥ . For instance, for p = k we choose ζ p ( u k ) = ζ p ( u ) , and for p = k , we choose ζ k ( u k ) = p | ζ k ( u ) | + αk ζ k ( u ) | ζ k ( u ) | if ζ k ( u ) = 0 and ζ k ( u k ) = αk if ζ k ( u ) = 0 .Fix t = 0 . If u k ( t ) was weakly convergent to u in L r, ( T ) , then ( ζ p ( u k ( t )) p would convergeweakly to ( ζ p ( u ( t )) p in h + , and therefore component by component : ζ p ( u k ) e iω (4) p ( u k ) t −→ k → + ∞ ζ p ( u ) e iω (4) p ( u ) t , p ≥ . In particular, let p ≥ such that ζ p ( u ) = 0 . Then there exists a sequence ( n k ) k of integers suchthat ω (4) p ( u k ) t + 2 πn k −→ k → + ∞ ω (4) p ( u ) t. From the expression (5) of ω (4) p ( u k ) and the strong convergence of ( ζ p ( u k )) p to ( ζ p ( u )) p in ℓ = { ( ζ p ) p ≥ | P p ≥ | ζ p | < + ∞} by compactness, we get + ∞ X p =1 p | ζ p ( u ) | + α + 2 πn k t = + ∞ X p =1 p | ζ p ( u k ) | + 2 πn k t −→ k → + ∞ + ∞ X p =1 p | ζ p ( u ) | . We get a contradiction by choosing α πt Z . H − sr, ( T ) , s > Proposition 2.5.
For all t > , there is no continuous local extension of the flow map S t to H − sr, ( T ) for < s < in the distribution sense.Proof. Let us fix < s < and an initial data u ∈ H − sr, ( T ) \ L r, ( T ) . From [6], Theorem 5,the Birkhoff map extends by continuity as an homeomorphism Φ : u ∈ H − sr, Φ( u ) = ( ζ n ( u )) n ≥ ∈ h − s + where h − s + = n ( ζ n ) n ≥ | X n ≥ n − s | ζ n | < + ∞ o . Therefore, ( ζ n ( u )) n ≥ := Φ( u ) ∈ h − s + is well defined. Let u k , k ∈ N , be a sequence finite gapinitial data, to be chosen later, such that u k converges in H − sr, ( T ) to u . Write Φ( u k ) = ( ζ n ( u k ) n ≤ N k ) n , k ∈ N . Since u k is a finite gap potential, it belongs to L r, ( T ) . Recall that ω (4) n ( u k ) − n N k X p =1 p | ζ p ( u k ) | = f ω n ( u k ) here f ω n ( u k ) = n − N k X p =1 min( p, n ) | ζ p ( u k ) | +3 N k X p =1 N k X q =1 min( p, q, n ) | ζ p ( u k ) | | ζ q ( u k ) | . Since u k converges to u in H − sr, ( T ) , the series P p ≥ | ζ p ( u ) | is convergent, and X p ≥ | ζ p ( u k ) | −→ k → + ∞ X p ≥ | ζ p ( u ) | . In particular, the term f ω n ( u k ) converges as k goes to infinity to f ω n ( u ) = n − + ∞ X p =1 min( p, n ) | ζ p ( u ) | +3 + ∞ X p =1 + ∞ X q =1 min( p, q, n ) | ζ p ( u ) | | ζ q ( u ) | . For k ∈ N , let τ k := N k X p =1 p | ζ p ( u k ) | = 12 k u k k L ( T ) and v k ( t, · ) := u k ( t, · − τ k t ) . We use the following identity from the proof of Proposition B.1. in [5] : ζ n ( u ( · + τ )) = ζ n ( u ) e iτn , τ ∈ R , u ∈ L r, ( T ) , to deduce that for n ∈ N , ζ n ( v k ( t )) = ζ n ( u k ( t )) e − inτ k t = ζ n ( u k ) e i ( ω (4) n ( u k ) − nτ k ) t . Since ω (4) n ( u k ) − nτ k −→ k → + ∞ f ω n ( u ) , the sequence ( ζ n ( v k ( t ))) k is convergent : ζ n ( v k ( t )) −→ k → + ∞ ζ n ( u ) e i f ω n ( u ) t . (6)Let t > . If there was a local extension of the flow map S t in the distribution sense, then u k ( t ) would be weakly convergent to u ( t ) in H − sr, ( T ) . Applying the Birkhoff map, which isweakly sequentially continuous (see [6], Theorem 6), Φ( u k ( t )) would converge weakly to Φ( u ( t )) in h − s + . In particular, for all n , ζ n ( v k ( t )) e iτ k nt = ζ n ( u k ( t )) −→ k → + ∞ ζ n ( u ( t )) . (7)We deduce from (6) and (7) that if ζ n ( u ) = 0 , then e iτ k nt −→ k → + ∞ ζ n ( u ( t )) ζ n ( u ) e − i f ω n ( u ) t . (8)We construct the sequence ( u k ) k in order to contradict this latter point. Let n ∈ N such that ζ n ( u ) = 0 . Fix k ∈ N . From the fact that u does not belong to L r, ( T ) , X p>k p | ζ p ( u ) | = + ∞ , herefore one can choose N k ≥ k + 1 such that N k X p = k +1 p | ζ p ( u ) | ≥ πnt . Let < α k < such that there exists an integer m k such that X p ≤ k p | ζ p ( u ) | + α k N k X p = k +1 p | ζ p ( u ) | = 1 nt ( kπ + 2 πm k ) . We define u k by ζ p ( u k ) = ζ p ( u ) if p ≤ k √ α k ζ p ( u ) if k < p ≤ N k if N k < p , p ∈ N . By construction, u k is finite gap and converges to u in H − sr, ( T ) . However, τ k = X p ≤ k p | ζ p ( u ) | + α k N k X p = k +1 p | ζ p ( u ) | = 1 nt ( kπ + 2 πm k ) , which implies that e iτ k nt = ( − k . In particular, the sequence ( e iτ k nt ) k is not convergent, and we get a contradiction with (8). Remark 2.6.
We expect that with a similar argument, one can prove the following fact. For thehigher equations of the hierarchy, the well-posedness threshold increases by for each equation(see Remark A.1). In this part, we classify all traveling wave solutions to equation (1) ∂ x ( − cu − ∂ xx u − u H∂ x u − H ( u∂ x u ) + u ) = 0 . A traveling wave of speed c ∈ R is a solution to (1) of the form u ( t, x ) = u ( x + ct ) .The argument of [5], Proposition B.1., applies for the higher equations of the Benjamin-Onohierarchy, implying that a potential u is a traveling wave solution to (1) of speed c ∈ R if andonly if for all t ∈ R and n ≥ , e icnt ζ n ( u ) = e iω (4) n ( u ) t ζ n ( u ) , or in other words : ∀ n ≥ , if ζ n ( u ) = 0 , then cn = ω (4) n ( u ) . (9)In particular, all one gap potentials are traveling wave solutions. Note that these potentials arethe only traveling wave solutions to the Benjamin-Ono equation and write (see [5], Appendix B) u ( x ) = pw e ipx − w e ipx + pw e − ipx − w e − ipx ith the nonzero gap being at index p ≥ and w = ζ p ( u ) √ p + γ p ( u ) . As we will see in this section,the one gap potentials are not the only traveling wave solutions for equation (1).In part 3.1, we first show that the traveling waves are necessarily one gap and two gappotentials, then provide a classification of the two gap traveling waves in term of their actions.In part 3.2, we prove that the one gap traveling waves are orbitally stable whereas the two gaptraveling waves are orbitally unstable. In the following, it will be more convenient to work with the actions γ p = | ζ p ( u ) | . Formula (5)for ω (4) n ( u ) ω (4) n ( u ) = n + n X p ≥ pγ p − X p ≥ min( p, n ) γ p + 3 X p,q ≥ min( p, q, n ) γ p γ q shows that ω (4) n ( u ) n is equivalent to n as n goes to infinity, therefore, from condition (9), thetraveling waves for the third equation of the hierarchy (1) are necessarily finite gap solutions. Proposition 3.1.
Let u be a two gap potential, and p < q be the indices of the two nonzerogaps with gaps γ p > and γ q > . Then u is a traveling wave for equation (1) if and only if < γ p < p + r p + 4 q p + q ! and γ q = q p + q + pγ p − γ p γ p + q . Proof.
Let u be such a two gap potential. Then u is a traveling wave if and only if ω (4) p ( u ) p = ω (4) q ( u ) q , where ω (4) p ( u ) p = p + ( pγ p + qγ q ) − p ( γ p + γ q ) + 3( γ p + 2 γ p γ q + γ q ) and ω (4) q ( u ) q = q + ( pγ p + qγ q ) − p q γ p + qγ q ) + 3( pq γ p + 2 pq γ p γ q + γ q ) . Taking the difference of the two terms, ω (4) p ( u ) p = ω (4) q ( u ) q , if and only if q − p + 3 (cid:18) − p ( pq − γ p − ( q − p ) γ q + ( pq − γ p + 2( pq − γ p γ q (cid:19) . Dividing by − pq ) , this necessary and sufficient condition becomes q p + q pγ p − qγ q − γ p − γ p γ q , i.e. (2 γ p + q ) γ q = q p + q pγ p − γ p . (10) ix γ p > and γ q satisfying this latter equality. We get that γ q > if and only if the left-handside of the equality is positive, i.e. < γ p < p + r p + 4 q p + q ! . (11)Conversely, any two gap solution u satisfying (10) and (11) verifies ω (4) p ( u ) p = ω (4) q ( u ) q , thereforeis a traveling wave solution.Let us give an idea of the form of a two gap potential u with gaps at indices p < q . ByTheorem 3 in [5], the extension of Π u as an holomorphic function on the unit disc { z ∈ C || z | < } satisfies Π u ( z ) = − z Q ′ ( z ) Q ( z ) where Q ( z ) = det(Id − zM ) and M = ( M nm ) ≤ n,m ≤ q − is a q × q matrix defined by M nm = δ m,n +1 if ζ n +1 = 0 √ µ n +1 q κ m κ n +1 ζ m ( u ) ζ n +1 ( u )( λ m − λ n − if ζ n +1 = 0 . A precise definition of µ n , κ n and λ n can be found in [5]. Therefore, Q and Q ′ respectively write Q ( z ) = 1 − z p M p − , − z q M q − , M p − ,p and Q ′ ( z ) = − pz p − M p − , − qz q − M q − , M p − ,p This leads to u ( x ) = p e ipx M p − , + q e iqx M q − , M p − ,p − e ipx M p − , − e iqx M q − , M p − ,p + p e − ipx M p − , + q e − iqx M q − , M p − ,p − e − ipx M p − , − e − iqx M q − , M p − ,p . Proposition 3.2.
There are no three gap traveling waves.Proof.
Let p < q < r the indices for the nonzero gaps for a three gap potential u . The speedsfor each mode write ω (4) p ( u ) p = p + ( pγ p + qγ q + rγ r ) − p ( γ p + γ q + γ r )+ 3( γ p + 2 γ p ( γ q + γ r ) + γ q + 2 γ q γ r + γ r ) ,ω (4) q ( u ) q = q + ( pγ p + qγ q + rγ r ) − p q γ p + q ( γ q + γ r ))+ 3( pq γ p + 2 pq γ p ( γ q + γ r ) + γ q + 2 γ q γ r + γ r ) and ω (4) r ( u ) r = r + ( pγ p + qγ q + rγ r ) − p r γ p + q r γ q + rγ r )+ 3( pr γ p + 2 pr γ p ( γ q + γ r ) + qr γ q + 2 qr γ q γ r + γ r ) . et us now subtract the equalities. ω (4) q ( u ) q − ω (4) p ( u ) p = q − p − pq − pγ p + ( q − p )( γ q + γ r ))+ 3(( pq − γ p + 2( pq − γ p ( γ q + γ r )) . Dividing by − pq ) , we get that if ω (4) q ( u ) q = ω (4) p ( u ) p , then q p + q pγ p − q ( γ q + γ r ) − γ p − γ p ( γ q + γ r ) , or equivalently ( q + 2 γ p )( γ q + γ r ) = q p + q pγ p − γ p . (12)Doing the same for indices p and r , ω (4) r ( u ) r − ω (4) p ( u ) p = r − p − pr − pγ p + ( q r − p ) γ q + ( r − p ) γ r )+ 3(( pr − γ p + 2( pr − γ p ( γ q + γ r ) + ( qr − γ q + 2( qr − γ q γ r ) . Dividing by − pr ) , if ω (4) r ( u ) r = ω (4) p ( u ) p , then r r + p pγ p + pr − q r − p γ q − rγ r − γ p − γ p ( γ q + γ r ) − r − qr − p γ q − r − qr − p γ q γ r . (13)Subtracting (13) and (12), if ω (4) p ( u ) p = ω (4) q ( u ) q = ω (4) r ( u ) r , then ( r − q + 2 r − qr − p γ q ) γ r = r − q + p ( r − q )3 + ( pr − q r − p + q ) γ q − r − qr − p γ q . Since pr − q r − p + q = pr − q + qr − pqr − p = ( p + q ) r − qr − p , by multiplication by r − pr − q , we get ( r − p + 2 γ q ) γ r = ( r − p ) r + q + p p + q ) γ q − γ q . (14)Now, if u is a traveling wave, the first equality (12) implies γ q + γ r = q p + q + pγ p − γ p q + 2 γ p = p + q − γ p p + q + pγ p − γ p q + 2 γ p = p + q γ p q + 2 γ p ) ( p − q − γ p ) , in particular, since p < q and γ p > , necessarily γ q + γ r < p + q . (15) owever, the second equality (14) implies that γ r = ( r − p ) r + q + p + ( p + q ) γ q − γ q r − p + 2 γ q = ( r − p ) q + p + ( p + q ) γ q − γ q r − p + 2 γ q + ( r − p ) r r − p + 2 γ q = p + q − γ q p + q + ( p + q ) γ q − γ q r − p + 2 γ q + ( r − p ) r r − p + 2 γ q = p + q γ q p + q − γ q r − p + 2 γ q + ( r − p ) r r − p + 2 γ q ) . Since from (15), γ q < p + q , we get from this latter equality for γ r that γ r > p + q , but this is a contradiction with (15). Corollary 3.3.
There are no N gap traveling waves for N ≥ .Proof. The proof is the same as for the three gap traveling waves case, but with some additionalterms which might hinder understanding for a first reading. We explain here how to adapt theproof.Let u be a N gap potential, N ≥ , and p < q < r < r < · · · < r N be the indices for thenonzero gaps. Let Γ r := γ r + N X k =4 γ r k . The speeds for the three smallest modes at indices p, q and r write ω (4) p ( u ) p = p + ( pγ p + qγ q + rγ r + N X k =4 r k γ r k ) − p ( γ p + γ q + Γ r )+ 3( γ p + 2 γ p ( γ q + Γ r ) + γ q + 2 γ q Γ r + γ r ) ,ω (4) q ( u ) q = q + ( pγ p + qγ q + rγ r + N X k =4 r k γ r k ) − p q γ p + q ( γ q + Γ r ))+ 3( pq γ p + 2 pq γ p ( γ q + Γ r ) + γ q + 2 γ q Γ r + Γ r ) and ω (4) r ( u ) r = r + ( pγ p + qγ q + rγ r + N X k =4 r k γ r k ) − p r γ p + q r γ q + r Γ r )+ 3( pr γ p + 2 pr γ p ( γ q + Γ r ) + qr γ q + 2 qr γ q Γ r + Γ r ) . The rest of the proof is identical up to replacing γ r by Γ r everywhere from this point on. .2 Orbital stability Proposition 3.4.
The one gap traveling waves in L r, ( T ) are orbitally stable.Proof. Let u be a one gap traveling wave, u k a sequence of initial data converging to u in L r, ( T ) , and t k a sequence of times. We prove that up to some subsequence, inf θ ∈ T k u k ( t k ) − u ( · + θ ) k L r, ( T ) −→ k → + ∞ . It is enough to show that there exists θ ∈ T such that in L r, ( T ) , u k ( t k ) −→ k → + ∞ u ( · + θ ) , i.e. such that in h + , Φ( u k ( t k )) −→ k → + ∞ Φ( u ( · + θ )) . Recall that Φ( u k ( t k )) = ( ζ n ( u k ) e iω (4) n ( u k ) t k ) n . Let p ≥ be the index for which ζ p ( u ) = 0 . Up to some subsequence, there exists θ ∈ T suchthat e iω (4) p ( u k ) t k −→ k → + ∞ e iθ . Moreover, since u k converges to u in L r, ( T ) , p | ζ p ( u k ) − ζ p ( u ) | + X n = p n | ζ n ( u k ) | −→ k → + ∞ . We deduce that k Φ( u k ( t k )) − Φ( u ( · + θ )) k h = p | ζ p ( u k ) e iω (4) n ( u k ) t k − ζ p ( u ) e iθ | + X n = p n | ζ n ( u k ) | −→ k → + ∞ . Proposition 3.5.
The two gap traveling waves in L r, ( T ) are orbitally unstable.Proof. Let u be a two gap traveling wave such that the nonzero terms of the sequence Φ( u ) =( ζ n ( u )) n ≥ are ζ p ( u ) and ζ q ( u ) . We define the sequence u k of two gap initial data by theirnonzero gaps at indices p and q , denoted ζ p ( u k ) and ζ q ( u k ) , as follows. We fix ζ p ( u k ) := ζ p ( u ) and choose any sequence of nonzero complex numbers ( ζ q ( u k )) k such that ζ q ( u k ) −→ k → + ∞ ζ q ( u ) but for all k ∈ N , ε k := | ζ q ( u k ) | −| ζ q | = 0 . Then we construct t k ∈ R in order to negate theorbital stability of u .Assume by contradiction that inf θ ∈ T k u k ( t k ) − u ( · + θ ) k L r, ( T ) −→ k → + ∞ . Then there exists a sequence θ k ∈ T , k ∈ N , such that in L r, ( T ) , k u k ( t k , · − θ k ) − u k L r, ( T ) −→ k → + ∞ . Applying the Birkhoff map, which is continuous on L r, ( T ) , ζ p ( u k ) e iω (4) p ( u k ) t k − ipθ k = ζ p ( u k ( t k )) e − ipθ k −→ k → + ∞ ζ p ( u ) nd ζ q ( u k ) e iω (4) q ( u k ) t k − iqθ k = ζ q ( u k ( t k )) e − iqθ k −→ k → + ∞ ζ q ( u ) . This implies by taking the arguments that for some integers n p,k and n q,k , ω (4) p ( u k ) t k − pθ k + 2 πn p,k −→ k → + ∞ and ω (4) q ( u k ) t k − qθ k + 2 πn q,k −→ k → + ∞ , therefore pqt k ω (4) q ( u k ) q − ω (4) p ( u k ) p ! + 2 π ( pn q,k − qn p,k ) −→ k → + ∞ . (16)However, writing ε k = | ζ q ( u k ) | −| ζ q ( u ) | = γ q ( u k ) − γ q ( u ) , we get that the speeds of thetwo modes p and q for the initial data u k are given by ω (4) p ( u k ) p = ω (4) p ( u ) p + qε k − pε k + 6 ε k ( γ p ( u ) + γ q ( u )) + 3 ε k and ω (4) q ( u k ) q = ω (4) q ( u ) q + qε k − qε k + 6 ε k ( pq γ p ( u ) + γ q ( u )) + 3 ε k . Since u is a traveling wave, ω (4) p ( u ) p = ω (4) q ( u ) q and therefore ω (4) q ( u k ) q − ω (4) p ( u k ) p = − q − p + 2(1 − pq ) γ p ( u )) ε k . Since ε k = 0 , then ω (4) q ( u k ) q = ω (4) p ( u k ) p . It is therefore possible to choose a sequence t k such thatthe limit (16) does not hold and get a contradiction. For instance, we can choose t k such that − pqt k ( q − p + 2(1 − pq ) γ p ) ε k = π. A Appendices
A.1 About the hierarchy
The aim of this Appendix is to provide a way to compute the Hamiltonians H k and frequencies ω ( k ) n for the higher order Benjamin-Ono equations in terms of the actions γ p = | ζ p ( u ) | . Inparticular, we establish formula (5) ω (4) n ( u ) = n + n X p ≥ pγ p − X p ≥ min( p, n ) γ p + 3 X p,q ≥ min( p, q, n ) γ p γ q for finite gap potentials u .We need to recall first some notation. Given u ∈ L r, ( T ) , we consider its Lax operator L u = − i∂ x − T u acting on L ( T ) , with domain H ( T ) = H ( T ) ∩ L ( T ) . The spectrum of L u is discrete, with eigenvalues λ ( u ) < λ ( u ) < · · · < λ n ( u ) < · · · . oreover (see [5]), γ n ( u ) = λ n ( u ) − λ n − ( u ) − , n ≥ is non-negative and satisfies γ n ( u ) = | ζ n ( u ) | . We also define f n ( u ) ∈ H ( T ) as the L -normalized eigenfunction for L u associated to the eigenvalue λ n ( u ) .Let u be a finite gap potential and use the above notation. From [5] (3.8), a variant of thegenerating function, denoted by f H ε , is defined as f H ε = + ∞ X n =0 |h | f n i| ελ n . From the decomposition (2.12) in [5] : Π u = − P + ∞ n =1 λ n h | f n i f n , we know using formula (3)that H k ( u ) = + ∞ X n =0 |h | f n i| λ kn = ( − k k ! d k d ε k | ε =0 f H ε . We now make use of the generating function to derive a recurrence formula for the H k . Set g ε := − dd ε log f H ε = − f H ε dd ε f H ε , then from [5] (3.11), g ε writes g ε = λ ελ + + ∞ X n =1 γ n (1 + ε ( λ n − + 1))(1 + ελ n ) . Using the identity − d k +1 d ε k +1 f H ε = d k d ε k (cid:16) g ε f H ε (cid:17) = k X l =0 (cid:18) kl (cid:19) d l d ε l ( g ε ) d k − l d ε k − l ( f H ε ) . and defining P l := ( − l l ! d l d ε l | ε =0 ( g ε ) , we get the recurrence relation H k +1 = 1 k + 1 k X l =0 P l H k − l . (17)Moreover, the frequencies ω ( k ) n = ∂ H k ∂γ n , satisfy the recurrence formula ω ( k +1) n = 1 k + 1 k X l =0 ∂P l ∂γ n H k − l + P l ω ( k − l ) n . (18)We now simplify P l and ∂P l ∂γ n : P l = λ l +10 + X n ≥ γ n l X m =0 ( λ n − + 1) m λ l − mn , nd since λ n − + 1 = λ n − γ n , P l = λ l +10 + X n ≥ λ l +1 n − ( λ n − γ n ) l +1 . From (3.13), λ n = n − s n +1 where s n = P ∞ k = n γ n for n ≥ , therefore P l = ( − l +1 s l +11 + X n ≥ ( n − s n +1 ) l +1 − ( n − s n ) l +1 . (19)We deduce l + 1 ∂P l ∂γ n = ( − l +1 s l + ( n − s n ) l − X p Note that formulas (5) and (22) for ω (4) n ( u ) and ω (5) n ( u ) , which have beenestablished for finite gap potentials u , still make sense for ω (4) n ( u ) if u ∈ L r, ( T ) and for ω (5) n ( u ) if u ∈ H r, ( T ) , but diverge if u ∈ H sr, ( T ) for s right below these respective exponents( s < and s < ).One can actually show by induction the following facts. In the formula (19) for P k , there isone term c P + ∞ p =1 p k γ p , the other terms being convergent if P + ∞ p =1 p k − γ p < + ∞ . This impliesthat in the formula (20) for H k appears one term c P + ∞ p =1 p k − γ p , the other terms being conver-gent if P + ∞ p =1 p k − γ p < + ∞ . Consequently, in formula (21) for ω ( k ) n , k ≥ , appears one term c P + ∞ p =1 p k − γ p , the other terms being convergent if P + ∞ p =1 p k − γ p < + ∞ .From these facts, one can see that the formula for ω ( k ) n can be extended by continuity topotentials in H s k r, ( T ) where s k = k − , however there is no continuous extension to H s k r, ( T ) when s k − < s < s k . This explains why the well-posedness threshold for the equation associatedto the Hamiltonian H k in the hierarchy should be H s k r, ( T ) . A.2 Equation for the fourth Hamiltonian From formula (3) and the decomposition (2.12) in [5] : Π u = − P + ∞ n =1 λ n h | f n i f n , we see thatfor k ≥ , H k ( u ) = h L k − u Π u | Π u i , (23)where L u ( h ) = Dh − Π( uh ) , D = − i∂ x , h ∈ H ( T ) . For instance, H ( u ) = 12 π Z π uH∂ x u − u d x leads to the Benjamin-Ono equation ∂ t u = H∂ x u − ∂ x ( u ) . Proposition A.2. The Hamiltonian for the third order equation of the Benjamin-Ono hierar-chy (1) is H ( u ) = 12 π Z π (cid:18) 12 ( ∂ x u ) − u H∂ x u + 14 u (cid:19) − k u k L ( T ) , therefore the third order equation of the Benjamin-Ono hierarchy writes ∂ t u = ∂ x ( − ∂ xx u − uH∂ x u − H ( u∂ x u ) + u ) Proof. Let u ∈ L r, ( T ) . We develop H ( u ) = k D Π u k L ( T ) − Re h D Π u | Π( u Π u )) i + k Π( u Π u ) k L ( T ) , and study each term separately. irst, since u is real, b u ( − n ) = b u ( n ) , therefore k D Π u k L ( T ) = X n ≥ | n | | b u ( n ) | = 12 X n ∈ Z | n | | b u ( n ) | = 12 k ∂ x u k L ( T ) . Then, u being with average zero, u = Π u + Π u , leading to h D Π u | Π( u Π u )) i = 12 π Z π D (Π u ) u Π u d x = 12 π Z π D ( u ) u Π u d x − π Z π D (Π u ) u Π u d x so that h D Π u | Π( u Π u )) i = 12 π Z π D ( u )Π u d x − π Z π D (Π u ) u d x = − π Z π u D (Π u ) d x + 12 π Z π Π u Du d x. Taking the real part, Re h D Π u | Π( u Π u )) i = 2 (cid:16) h D Π u | Π( u Π u )) i + h D Π u | Π( u Π u )) i (cid:17) = − π Z π u ( D (Π u ) + D Π u ) d x + 12 π Z π Π u Du + (Π u ) Du d x. Using that Df = − Df , Re h D Π u | Π( u Π u )) i = − π Z π u ( D (Π u ) − D (Π u )) d x + 12 π Z π (Π u − (Π u ) ) Du d x = − π Z π u ( D (Π u ) − D (Π u )) d x + 12 π Z π (Π u − Π u ) uDu d x = − π Z π u ( D (Π u ) − D (Π u )) d x − 12 12 π Z π D (Π u − Π u ) u d x = − 32 12 π Z π u ( D (Π u ) − D (Π u )) d x. It now remains to remark that D (Π u ) − D (Π u ) = − H∂ x u in order to conclude the identity Re h D Π u | Π( u Π u )) i = 34 h H∂ x u | u i . Finally, we treat the last term k Π( u Π u ) k L ( T ) . Note that by decomposing u = Π u + Π u , Π( u Π u ) = (Π u ) + Π(Π u Π u ) , therefore | Π( u Π u ) | = | Π u | +(Π u ) Π(Π u Π u ) + Π u Π(Π u Π u ) + | Π(Π u Π u ) | . By removing the useless projections, k Π( u Π u ) k L ( T ) = 12 π Z π | Π u | +(Π u ) Π u Π u + Π u Π u Π u + | Π(Π u Π u ) | d x = 12 π Z π (Π u ) Π u + (Π u ) Π u + Π u Π u + | Π(Π u Π u ) | d x. ut if we take the fourth power of the identity u = Π u + Π uu = (Π u ) + Π u + 4(Π u ) Π u + 4Π u Π u + 6Π u (Π u ) , and make use of the fact that the mean of u is zero, we get k u k L ( T ) = 12 π Z π u ) Π u + 4Π u Π u + 6(Π u ) Π u d x. By subtraction, the following cancellations happen : k Π( u Π u ) k L ( T ) − k u k L ( T ) = 12 π Z π | Π(Π u Π u ) | d x − 12 12 π Z π (Π u ) Π u d x. To conclude, since Π u Π u is real, we can use the identity k f k + |h f | i| = 2 k Π f k for real valuedfunctions f ∈ L ( T ) to get π Z π | Π(Π u Π u ) | d x = 12 12 π Z π | Π u Π u | d x − |h Π u Π u, i| , leading to k Π( u Π u ) k L ( T ) − k u k L ( T ) = − |h Π u Π u, i| = − k Π u k L ( T ) = − k u k L ( T ) . A.3 Structure of the higher order Hamiltonians The aim of this Appendix is to give an alternative proof of Proposition 2.2 in [18].We first recall the notation introduced in [18] for the sake of completeness. For a smoothfunction u ∈ C ∞ ( T ) , define by induction the sets P n ( u ) as P ( u ) = { H ε ∂ α x u | ε ∈ { , } , α ∈ N } , P ( u ) = { ( H ε ∂ α x u )( H ε ∂ α x u ) | ε , ε ∈ { , } , α , α ∈ N } and for n ≥ , P n ( u ) = ( k Y l =1 H ε l p j l ( u ) | k ∈ J , n K , ε , . . . , ε k ∈ { , } , k X l =1 j l = n, p j l ( u ) ∈ P j l ( u ) ) . Moreover, for p n ( u ) ∈ P n ( u ) , the term f p n ( u ) is uniquely defined from p n ( u ) by removing all thesymbols H in the expression of p n ( u ) and only keeping the symbols ∂ α i x u . In this case, if f p n ( u ) = n Y i =1 ∂ α i x u, the maximal order of derivative involved and the sum of these orders are respectively denoted | p n ( u ) | = sup i ∈ J ,n K α i and k p n ( u ) k = n X i =1 α i . We now retrieve a proof of the following result (Proposition 2.2 in [18]). roposition A.3. Let k = 2( m + 1) be an even integer. Then there exists c ∈ R such that the k -th Hamiltonian H k writes, for all u ∈ C ∞ ( T ) , H k +2 ( u ) = 12 k u k H m +1 ( T ) + c Z π u ( H∂ mx u )( ∂ m +1 x u ) d x + R, where for some real numbers c ( p ) , R = m +4 X j =3 X p ( u ) ∈P j ( u ) k p ( u ) k =2 m +4 − j | p ( u ) |≤ m c ( p ) Z π p ( u ) d x. Note that H k +2 ( u ) = E k/ ( u ) with the notation from [18]. Proof. Recall formula (23) H k +2 ( u ) = h L ku Π u | Π u i = h L m +1 u Π u | L m +1 u Π u i = h ( D − T u ) m +1 Π u | ( D − T u ) m +1 Π u i , where D = − i∂ x and T u : h ∈ L ( T ) Π( uh ) .We expand H k +2 ( u ) as a sum of terms depending on whether we applied the operator D orthe operator T u when applying L u .It is possible to decompose H k +2 ( u ) as follows : H k +2 ( u ) = 12 k u k H m +1 ( T ) + A + B, where k Π u k H m +1 ( T ) = k u k H m +1 ( T ) is obtained when one only applies operator D , A is obtainedwhen one applies only once the operator T u and (2 m + 1) times the operator DA = − Re m X j =0 h D m − j Π( uD j Π u ) | D m +1 Π u i = − Re m X j =0 h D m − j ( uD j Π u ) | D m +1 Π u i , and B is obtained when we apply at least twice in total the operator T u . • We first prove that one can decompose A as A = c Z π u ( H∂ mx u )( ∂ m +1 x u ) d x + e A where for some real numbers c ( p ) , e A = X p ( u ) ∈P ( u ) k p ( u ) k =2 m +1 | p ( u ) |≤ m c ( p ) Z π p ( u ) d x. (24)Let j ∈ J , m K . By integration by parts and Leibniz’ formula, h D m − j ( uD j Π u ) | D m +1 Π u i = h D m +1 − j ( uD j Π u ) | D m Π u i = h uD m +1 (Π u ) | D m Π u i + h D m +1 − j ( u ) D j (Π u ) | D m Π u i + m − j X k =1 (cid:18) m + 1 − jk (cid:19) h D k ( u ) D m +1 − k (Π u ) | D m Π u i . e take the real part and sum over the indices j . When distinguishing the cases j = 0 and j ≥ , we see that for some suitable e A as in (24), A decomposes as A = − m + 1) Re ( h uD m +1 (Π u ) | D m Π u i ) − Re ( h D m +1 ( u )Π u | D m Π u i ) + e A. Write Π u = u + iHu , then there exists some real constants c ( ε , ε ) such that A = X ε ,ε ∈{ , } c ( ε , ε ) Z π u∂ mx ( H ε u ) ∂ m +1 x ( H ε u ) d x − Re ( h D m +1 ( u ) iHu | D m (Π u ) i ) + e A. On the one hand, the terms in the sum are simplified as follows (see the remark from Tzvetkovand Visciglia [18]). When ε = ε , Z π u∂ mx ( H ε u ) ∂ m +1 x ( H ε u ) d x = 12 Z π u∂ x (( ∂ mx ( H ε u )) ) d x = − Z π ∂ x ( u )( ∂ mx ( H ε u )) d x so this term is a remainder term to be added to e A . Moreover, by integration by parts, Z π u∂ mx ( u ) ∂ m +1 x ( Hu ) d x = − Z π ∂ x ( u ) ∂ mx ( u ) ∂ mx ( Hu ) d x − Z π u∂ m +1 x ( u ) ∂ mx ( Hu ) d x. Therefore, the sum can be written as a linear combination of the term R π u ( H∂ mx u ) ∂ m +1 x u d x and other terms that can be added to the remainder e A .On the other end,Re ( h D m +1 ( u ) iHu | D m (Π u ) i ) = Re ( h ∂ m +1 x ( u ) Hu | ∂ mx (Π u ) i )= 12 (cid:0) Re ( h ∂ m +1 x ( u ) Hu | ∂ mx ( u ) i ) + Re ( h ∂ m +1 x ( u ) Hu | i∂ mx ( Hu ) i ) (cid:1) Since u is real valued, so is Hu , thereforeRe ( h ∂ m +1 x ( u ) Hu | i∂ mx ( Hu ) i ) = 0 . By integration by parts, we then writeRe ( h D m +1 ( u ) iHu | D m (Π u ) i ) = 14 π Z π ∂ m +1 x ( u ) Hu∂ mx ( u ) d x = − π Z π ∂ x ( Hu )( ∂ mx ( u )) d x as a remainder term to be added to e A . • We now tackle term B , for which we have applied T u at least twice. We show that it canbe written for some real numbers c ( p ) as a sum B = m +4 X j =4 X p ( u ) ∈P j ( u ) k p ( u ) k =2 m +4 − j | p ( u ) |≤ m c ( p ) Z π p ( u ) d x. Let e B be one of the terms in B obtained by applying T u ( j − times on the left side and ( k − times on the right side. ssume that we have applied T u at least once in each side of the brackets, i.e. j − ∈ J , m +1 K and k − ∈ J , m + 1 K . Then we can apply Leibniz’ rule and decompose the left side as a complexlinear combination of terms of the form p ( u ) where p ( u ) ∈ P j ( u ) , k p ( u ) k = m +2 − j and | p ( u ) |≤ m (for the right side we just replace j by k ). The term e B is therefore a complex linear combinationof terms R π p ( u ) d x , where p ( u ) ∈ P l ( u ) for some l = j + k ∈ J , m + 4 K , k p ( u ) k = 2 m + 4 − l and | p ( u ) |≤ m .Otherwise, we have applied T u at least twice in the same side of the brackets, let us say theleft, and we only have applied the operator D on the other side : j − ∈ J , m + 1 K and k − .Again by Leibniz’ rule, e B decomposes as a sum e B = m +2 X j =3 X p ( u ) ∈P j ( u ) k p ( u ) k = m +2 − j | p ( u ) |≤ m − c ( p ) h p ( u ) | D m +1 Π u i . But then by integration by parts and Leibniz’ rule again, e B = m +2 X j =3 X p ( u ) ∈P j ( u ) k p ( u ) k = m +2 − j | p ( u ) |≤ m − c ( p ) h Dp ( u ) | D m Π u i = m +2 X j =3 X p ( u ) ∈P j ( u ) k p ( u ) k = m +3 − j | p ( u ) |≤ m c ′ ( p ) h p ( u ) | D m Π u i = m +3 X j =4 X p ( u ) ∈P j ( u ) k p ( u ) k =2 m +4 − j | p ( u ) |≤ m c ′′ ( p ) Z π p ( u ) d x, which is of the desired form. References [1] C. J. Amick and J. F. Toland. 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International Mathematics Research Notices , 2014(17):4679–4714, 052013. Département de mathématiques et applications, École normale supérieure, CNRS, PSLUniversity, 75005 Paris, FranceUniversité Paris-Saclay, CNRS, Laboratoire de mathématiques d’Orsay, 91405, Orsay,France E-mail address : [email protected]@math.u-psud.fr