STABILITY OF TRAVELING WAVES FOR THE BURGERS–HILBERT EQUATION

We consider smooth solutions of the Burgers–Hilbert equation that are a small perturbation δ from a global periodic traveling wave with small amplitude ϵ . We use a modified energy method to prove the existence time of smooth solutions on a time scale of 1 /(ϵδ) , with 0 < δ ≪ ϵ ≪ 1, and on a time scale of ϵ/δ 2 , with 0 < δ ≪ ϵ 2 ≪ 1. Moreover, we show that the traveling wave exists for an amplitude ϵ in the range ( 0 , ϵ ∗ ) , with ϵ ∗ ∼ 0 . 23, and fails to exist for ϵ > 2 / e .


The Burger-Hilbert equation (BH).
In this paper we study the size and stability of traveling waves of the Burgers-Hilbert equation (BH), for (x, t) ∈ Ω × R (1) f (x, 0) = f 0 (x). (2) where Ω is the real line R or the torus T = R/2πZ and Hf is the Hilbert transform which is defined for f : R (resp. T) −→ R by x − y dy resp. Hf (x) = 1 2π P.V.
2π 0 f (y) cot x − y 2 dy. 1 3 . Recently, with a different approach, Saut and Wang [25] proved shock formation in finite time for (BH) and Yang [27] constructed solutions that develop an asymptotic self-similar shock at one single point with an explicitly computable blowup profile for (BH).
In this paper we are concerned with the dynamics in the small amplitude regime where (BH) can be viewed as a perturbation of the linearized (BH) equation f t = H[f ]. Since the nonlinear term in (1) is quadratic and the Hilbert transform is orthogonal in L 2 , standard energy estimates yield a time of existence of smooth solutions T ∼ 1 ||f 0 || . Thanks to the effect of the Hilbert transform and using the normal form method, Hunter, Ifrim, Tataru and Wong (see [15] and [16]) were able to improve this time of existence. More precisely, if ǫ is the size of the initial data, they prove a lifespan T ∼ 1 ǫ 2 for small enough ǫ (see also [10] for a similar approach with a modified version of the (BH) equation). The proofs are based on the normal form method and in the modified energy method. Furthermore, Hunter [14] showed for 0 < ǫ ≪ 1 the existence of C ∞ -traveling wave solutions of the form Notice that, 1 n u ǫ (nx), 1 n v ǫ is also a C ∞ −traveling wave solution. Throughout the paper we will assume that the initial data f 0 has zero mean. Since the equation (1) preserves the mean, 2π 0 f (x, t)dx = 0 for all t. Since in the construction above u ǫ also has zero mean,

Sketch of the proof of theorem 1
Now we briefly describe the proof of theorem 1. Assume that the solution is a small perturbation around the traveling wave u ǫ (x + v ǫ t). Then the linearization of the Burgers-Hilbert equation (1) is L ǫ g := −v ǫ g x + Hg + (u ǫ (x)g) x = 0 so to the first order, the perturbation g solves the equation g t = L ǫ g, with solution g(x, t) = e tLǫ g(x, 0).
Therefore the linear evolution of g is determined by the eigenvalues of L ǫ . The full nonlinear evolution of g is g t = L ǫ g + N(g, g) where N(g, g) is a nonlinearity that is (at least) quadratic in g. We plug in the linear solution to get g t = e tLǫ L ǫ g(x, 0) + N(e tLǫ g(x, 0), e tLǫ g(x, 0)) to second order, which integrates to g(x, t) = e tLǫ g(x, 0) + e tLǫ t 0 e −sLǫ N(e sLǫ g(x, 0), e sLǫ g(x, 0))ds.
These perturbations generate translations and variations along the bifurcation curve. We treat them separately using a more sophisticated ansatz We will show in Proposition 5 that if |ǫ 0 | and f − u ǫ 0 H 2 /|ǫ 0 | are sufficiently small, then f can always be put in the form above, with |ǫ − ǫ 0 |/|ǫ 0 | also small, and the expansion of g not involving any eigenvector with eigenvalue 0. This way we removed the resonance caused by the eigenvalue 0 from the evolution of g.
We also need to analyze the other eigenvalues of L ǫ , a first order differential operator with variable coefficients, and a quasilinear perturbation from L 0 = ∂ x + H, whose eigenvectors are the Fourier modes e inx . Just like the Schrödinger operator with potential −∆ + V , with a basis of eigenvectors known as the "Jost functions", giving rise to the "distorted Fourier transform" (see Agmon [1]), L ǫ can also be diagonalized using a combination of conjugation and perturbative analysis. More precisely, let g = h x . Then where c ǫ → 1 as ǫ → 0, and R ǫ is a small smoothing remainder (i.e., it gains derivatives of arbitrarily high orders). Thus L ǫ is conjugate to c ǫ ∂ x + H + R ǫ , whose eigenvalues can be approximated by those of c ǫ ∂ x + H, which are ±(nc ǫ i−i), n = 1, 2, . . . . The general theory of unbounded analytic operators developed in [21] allows us to justify this approximation up to O(ǫ 6 ) (see Corollary 2), and to relate the eigenvectors of L ǫ to the Fourier modes (see Lemma 5), in the sense that another linear maph → h conjugates L ǫ into a Fourier multiplier whose action on e i(n+sgn n)x is multiplication by λ n (n = 0). At the end of the day we have the following estimate for small ǫ: |λ l + λ m − λ n | > 1/2, l + m = n, ǫ 2 /5, l + m = n, see Proposition 4. Because this value appears in the denominator in (5), if g has size δ, a direct application of the normal form transformation yields a lifespan comparable to ǫ 2 /δ 2 . To improve on this, we will make use of the structure of the nonlinearity: The first term is the usual product-style nonlinearity, which imposes the restriction l + sgn l + m + sgn m = n + sgn n, and implies that l + m − n = ±1 = 0, so the normal form transformation can be carried out as before. The second term is of size |ǫ| and gains a factor of 1/|ǫ| in the lifespan. Thus the usual energy estimate can show a lifespan comparable to 1/|ǫδ|, and the normal form transformation can show a lifespan comparable to |ǫ|/δ 2 . This decomposition of the nonlinearity into one part satisfying classical additive frequency restrictions, and another part enjoying better estimates analytically was first used in Germain-Pusateri-Rousset [13] to show global wellposedness of the 1D Schrödinger equation with potential (see also Chen-Pusateri [7]). Our result shows that this approach can be adapted to quasilinear equations and to the case of discrete spectrum.

The outline of the paper.
In section 2 we study the traveling waves solutions for (1). For sake of completeness we sketch the proof of existence which follows from bifurcation theory. In addition we analyze the size of the traveling waves. In section 3 we study the linearization of equation (1) around the traveling waves. In section 4, we introduce a new frame of references which will help us to avoid the resonances found in section 3. Finally, in section 5 we prove theorem 1.

Traveling waves
The existence of traveling waves for (1) was shown in [14]. Here we will study their size after we give some details about the existence proof. We look for solutions of (1) in the form thus we have to find (u ǫ , v ǫ ) solving If (u ǫ , v ǫ ) is a solution, so is (u n ǫ (x), v n ǫ ) = (u ǫ (nx)/n, v ǫ /n). Thus from one solution we can get n-fold symmetric solutions for all n ≥ 1.
To solve (6) we can apply the Crandall-Rabinowitz theorem (see [8]) to where H k,+ r (T) = {2π-periodic, mean zero, and even functions, analytic in the strip {|Im(z)| < r}} endowed with the norm and H k,− r (T) = {2π-periodic, and odd functions, analytic in the strip {|Im(z)| < r}} endowed with the norm Here || · || H k (T) is the usual Sobolev norm, and it is enough to take k ≥ 1 and r = 1. We just notice that F (0, µ) = 0 and the derivative of F at u = 0, µ = 0, just has a non trivial element in its kernel belonging to H k,+ r (T), namely, h = cos(x).
Thus, the application of the C-R theorem allows to show the existence of a branch of solutions (u ǫ , v ǫ ) ∈ (H 1,+ 1 , R), bifurcating from (0, −1), for (6) with the assymptotic We remark that we obtain a bifurcation curve which is differentiable and hence analytic on B δ for δ small enough. The rest of this section is devoted to proving further properties of these solutions.
Introducing the asymptotic expansion taking u 1 = cos(x), λ 0 = −1 and comparing the coefficient in ǫ n we obtain that for n = 2, 3, ... We notice that in order to solve the equation Hu + u ′ = f we need (f, sin(x)) = 0. Therefore we have to choose v n−1 = 1 π (sin(x), f n ). This gives us a recurrence for (u n , v n−1 ), n ≥ 2, in terms of {(u m , v m−1 )} n−1 m=1 . In order to study this recurrence we will introduce the ansatz u n = n k=2 u n,k cos(kx).
By induction, one can check that the rest of coefficients in the expansion on cosines of u n must be zero. In addition, if u ǫ (x) solves (6), u −ǫ (x + π) is also a bifurcation curves in the direction of cos(x), and then by uniqueness, u ǫ (x) = u −ǫ (x + π), which yields u n,k = 0 if n − k = 1, mod (2).
Comparing the coefficient of sin(kx), with k = n mod(2), and 2 ≤ k ≤ n, we have that Up to order O(ǫ 4 ) we find that The recurrence (10)-(11) allows us to prove the following result.
Theorem 2. The radius of convergence of the series (8), with the coefficients given by the expression (9), (11) and (10) is not bigger than 2 e . Proof. From (10) and (11) we have that Then y − xy ′ = 1 2 yxy ′ , (2x + xy)y ′ = 2y, (2 + y)y ′ 2y = 1 x , ln y + y/2 = ln x + C, Since y ∼ x for small x, C = 1, so where W is the Lambert W-function. Since the radius of convergence of W at 0 is 1/e, the radius of convergence of y at 0 is 2/e, so the radius of convergence of (10) and (11) is at most 2/e.
In addition we can get a bound for how large the traveling wave can be.
Proof. This proof is based on the implicit funtion theorem. Firsly we will introduce the spaces L 2,− = {odd functions f ∈ L 2 (T)}, H 1,+ = {even functions f ∈ H 1 (T)}. The space X will the orthogonal complement of the span of cos(x) in H 1,+ . We will equip L 2,− with the norm in such a way that || sin(nx)|| L 2,− = 1, for n ≥ 1. We also define Thus || cos(nx)|| X = n − 1, for n ≥ 2. The reason why we take these norms is technical and it will arise below. Finally we define Because the existence of traveling waves we already know that there exists ǫ * such that for every ǫ ∈ [0, ǫ * ), there existũ ǫ and µ ǫ satisfying G(ǫ,ũ ǫ , µ ǫ ) = 0.
Proof. Let And we have that

In addition
Since, for n ≥ 6, we have that (n+1) 2 (n−1) 2 ≤ 49 25 we finally obtain that We The minimum in σ of max 10 + 9σ, is attached when 10+9σ = 59 9 + 1 σ and it is 11.373. Then minimum in γ of max 5 + 4γ, 5.5 + 1 γ less than 7.5. Finally 2(1 + 49/25) = 5.92. Therefore En . D n = 0 only when n ≥ 4. We have that D 4 = 16 and when n ≥ 5, For E n , by partial fraction decomposition and when 1 ≤ n ≤ 3, For n = 4 we have and finally We have that We also can bound We take σ n = 1 for n ≥ 6 to get One can compute that (n + 2) 2 + (n − 2) 2 = 2(n 2 + 4) and that Then the maximum of these all numbers is the first one which is ≤ 17. Thus Now, with the lemmas 1, 2 and 3 we are ready to bound the right hand side of (15). Indeed, Turning to the other side, we have that Since By the comparison principle, r ǫ is bounded from above by the solution to Since When x > 0, the quadratic coefficient and the constant is positive, so this equation has a non-negative root iff Hence the solution can be extended to ǫ = x * ≈ 0.23. In order to achieve this last conclusion we just notice that the solution to (2), with y(0) = 0 can be extended only if A ǫ < 1, since 1 − A ǫ arises in the denominator. The above argument shows that for ǫ ∈ (−x * , x * ), the bifurcation curve produces a traveling wave u ǫ = ǫ cos x − ǫ 2 2 cos 2x + ǫ 2ũ ǫ which travels at speed v ǫ = −1 − ǫµ ǫ . Since all the operators involved are analytic in all its arguments, the bifurcation curve is analytic in ǫ on (−x * , x * ). It may be the case, however, that the power series for u ǫ and v ǫ around ǫ = 0 has a smaller radius of convergence than x * (for example, the function f (x) = (x 2 + 1) −1 is analytic on the whole real line, but the radius of convergence of its power series around 0 is only 1.) We now show that the radius of convergence of the power series for u ǫ and v ǫ are indeed at least x * .
We note that the above argument also works if ǫ is replaced with ǫe ia (a ∈ R), so the bifurcation curve (u ǫ , v ǫ ) is also analytic in a neighborhood of {ǫe ia : ǫ ∈ (−x * , x * )}. Hence the curve is analytic in the disk of radius x * centered at 0, so the radius of convergence of its power series around 0 is at least x * .

Linearization around traveling waves
In this section we will analyse the spectrum of the operator corresponding to the linearization of equation (1) around the traveling wave (u ǫ , v ǫ ) bifurcating from zero in the direction of the cosine studied in the previous section. Actually and ( Putting these in (1), we get the equation for g(x, t): The linearization around g = 0 is

The eigenvalue 0
The action of L on the Fourier modes is with eigenvalues 0 (double), ±i, ±2i, . . . (on L 2 (T) with zero mean). We first study the perturbation of the eigenspace corresponding to the double eigenvalue of 0. By translational symmetry, for any δ ∈ R, u ǫ (x + δ) is also a solution to Differentiation with respect to δ then shows that Also, since u ǫ lies on a bifucation curve, we can differentiate

Simplifying the linearized operator
We want to solve the eigenvalue problem Let g = h ′ . Then the antiderivative of the above is Then By the change of variable z = φ ǫ (y), The convolution kernel of the operator and the ǫ-derivative of the kernel is Near x = 0, csc x − 1/x 2 and cot x − 1/x are smooth, and (φ −1 ǫ ) ′ is smooth everywhere, so when x − z is small enough, up to a smooth function in (x, z), where the dot over H means that the norm does not measure frequency zero.

Definition 1.
We say an operator is of class S if it satisfies (22). We say a family of operators is of class S uniformly if for each k and m there is an implicit constant that makes (22) true for all operators in the family.
Thus ∂ ǫ R ǫ is of class S uniformly in ǫ. Since R 0 = 0, R ǫ /ǫ is also of class S uniformly in ǫ.
Now the eigenvalue problem forh is of the form or, equivalently, where and R ǫ /ǫ is of class S uniformly in ǫ. Note that since u ǫ and v ǫ are analytic functions of ǫ on a neighborhood of 0, with u 0 = 0 and v 0 = −1, so are φ ǫ , R ǫ and c ǫ with φ 0 = I, R 0 = 0 and c 0 = 1.
so by (25), Moreover the projection is the projection on the span of e i(n+sgn n)x and the projection is the projection on the span of e ix and e −ix . Now when ǫ is small enough and z ∈ Γ n , we have that is analytic in ǫ near 0, with R ′ 0 = 0, thanks to the analyticity of c ǫ . Taking the inverse gives that and the Neumann series converges because when ǫ is small enough (depending on k). Moveover, exist and satisfy Q n (ǫ) − P n k ǫ, n ∈ Z (31) uniformly in n. Then by Chapter I, Section 4.6 of [21], when ǫ is small enough, Q n (ǫ) is conjugate to P n . Thus dim ran Q n (ǫ) = 1 for n = 0 and dim ran Q 0 (ǫ) = 2. So ∂ x + c −1 ǫ H + c −1 ǫ R ǫ has a single eigenvalue enclosed by Γ n for n = 0. In section 3.1 we showed that the action on the range of Q 0 (ǫ) is given by a nonzero nilpotent 2 by 2 matrix. If z is outside all these circles, then (27) still holds and the Neumann series (29) still converges to show that ∂ x + c −1 ǫ H + c −1 ǫ R ǫ − z is invertible, so it has no other eigenvalues.

Analyticity of eigenvalues and eigenvectors
By (25) and (26), (∂ x + H − z) −1 is analytic in (z, ǫ) for z in a neighborhood U of ∪ n∈Z Γ n , and ǫ near 0. By (28), R ′ ǫ is analytic in ǫ near 0, so the series (29) shows that (∂ x + c −1 ǫ H + c −1 ǫ R ǫ − z) −1 is analytic in (z, ǫ) for z ∈ U and ǫ near 0, and the integral (30) shows that all the projections Q n (ǫ) (n ∈ Z) are analytic in a neighborhood of 0 independent of n.
Let ψ n (ǫ) be the corresponding eigenvectors to Q n (ǫ) for n = 0. Thanks to (31), a good choice is ψ n (ǫ) = Q n (ǫ)e i(n+sgn n)x , which is nonzero and analytic in a neighborhood of 0 independent of n. Then by (23), On the other hand, the left-hand side equals which is another vector analytic in ǫ near 0. Then by the next lemma, all the eigenvalues c −1 ǫ λ n (ǫ), and hence λ n (ǫ), are analytic in a neighborhood of 0 independent of n. Then λ(ǫ) is analytic in ǫ ∈ U.
Proof. Without loss of generality assume that 0 ∈ U. Since the result is local in ǫ, it suffices to show that λ(ǫ) is analytic in a smaller neighborhood of 0.
Since u(0) = 0, we can find a linear functional f such that f (u(0)) = 0. Then f (u(ǫ)) = 0 in a neighborhood of 0, and so is analytic in a neighborhood of 0.
Regarding the double eigenvalue 0, in section 3.1 we showed that u ′ ǫ and ∂ ǫ u ǫ are two generalized eigenvectors of the operator L ǫ . Using the relation given in section 3.2, they correspond to two generalized eigenvectors ψ − 0 (ǫ) and ψ + 0 (ǫ) of the operator ∂ x +c −1 Then clearly ψ ± 0 (ǫ) are both analytic in ǫ. From the analyticity of the eigenvalues c −1 ǫ λ n (ǫ), it is easy to derive bounds on their Taylor coefficients. Proposition 1. For k ≥ 1 and n = 0, the coefficient of ǫ k in c −1 ǫ λ n (ǫ) is bounded in absolute value by C k for a constant C > 0 independent of n, Proof. At the end of section 3.3 we showed that when ǫ is in a neighborhood of 0 independent of n, the eigenvalues c −1 ǫ λ n (ǫ) are enclosed in the circle Γ n . Then |c −1 ǫ λ n (ǫ) − ni| < 1/2, n = ±1, ±2, . . . . The result follows from Cauchy's integral formula for Taylor coefficients.
Corollary 1. For k ≥ 0 and n = 0, the coefficient of ǫ k in λ n (ǫ) is bounded in absolute value by |n|C k for a constant C > 0 independent of n, Proof. Since c ǫ is analytic in ǫ near 0 with c 0 = 1, and λ n (0) = ni, the result follows from Leibniz's rule.

Conjugation to a Fourier multiplier
We have conjugated the eigenspaces of T = ∂ x + c −1 ǫ H + c −1 ǫ R ǫ (and also of c ǫ ∂ x + H + R ǫ ) to Fourier modes via the operator where P 0 is the projection onto the span of e ±ix , Q 0 (ǫ) is the projection onto the span of ψ ± 0 (ǫ), P n is the projection onto the span of e i(n+sgn n)x , Q n (ǫ) is the projection onto the span of ψ n (ǫ), n = ±1, ±2, . . . . We will view T as a perturbation of ∂ x + c −1 ǫ H and follow the proof of Chapter V, Theorem 4.15a in [21]. In the process we will extract more information from the fact that R ǫ is of class S. Since we have that and W 0 = 0.
Proposition 2. W ǫ /ǫ is of class S uniformly in ǫ.
Proof. We bound each term on the right-hand side separately. By Chapter V, (4.38) of [21], We now bound the operator norms of the right-hand side, with uniformity in ǫ and decay in n, in order to show that the sum in n converges. First note that it is clear from the frequency side that when ǫ is in a neighborhood of 0 independent of n and z ∈ ∪ n∈Z Γ n , for all m ≥ 0, the operator (∂ x + c −1 ǫ H − z) −1 is bounded from H m to H m , uniformly in ǫ and z. Since R ǫ /ǫ is of class S uniformly in ǫ (see (22) and notice that R 0 = 0), it follows from the Neumann series that (T − z) −1 Ḣm →Ḣ m is finite and only depends on m. Since |z − (n + (1 − c −1 ǫ ) sgn n)i| and |z − c −1 ǫ λ n (ǫ)| are uniformly bounded from below, both Z n (ǫ) and Z ′ n (ǫ) are bounded froṁ H m toḢ m , uniformly in ǫ and n. Since Q n (ǫ) is given by a similar integral (30), it also has this property, which is also trivially true for P n . Now for all n, m, k ∈ Z, m, k ≥ 0 andh ∈ L 2 , because P n is the projection onto very specific Fourier modes. For the first term we have To introduce the action of Q n (ǫ), note that the image of Q n (ǫ) lies in the eigenspace of the operator c ǫ ∂ x + H + R ǫ , with eigenvalue λ n (ǫ), so for n = 0 and u ∈ Im Q n (ǫ) we have This also holds for n = 0 because R ǫ /ǫ is of class S uniformly, so W ǫ /ǫ is of class S uniformly in ǫ thanks to the convergence of n∈Z (1 + |n|) −2 .
Now for k = 0, 1, . . . , there is a neighborhood of 0 such that when ǫ is in this neighborhood, W ǫ Ḣk →Ḣ k < 1, so 1 + W ǫ :Ḣ k →Ḣ k is invertible. By (32) and (33) it follows easily that so the eigenspaces of T is conjugated to the (span of) Fourier modes, and hence T is conjugated to a Fourier multiplier. We have proven the following lemma: Lemma 5. For ǫ small enough, there exists an operator W ǫ , such that W ǫ /ǫ is of class S, uniformly in ǫ. Moreover, 3. If ψ is in the closed linear span of the eigenvectors ψ n (ǫ) (n = 0) of c ǫ ∂ x + H + R ǫ , then where Λ ǫ is a multiplier such that Λ ǫ e i(n+sgn n)x = λ n (ǫ)e i(n+sgn n)x , n = ±1, ±2, . . .

Taylor expansion of eigenvalues
Now we Taylor expand the eigenvalues λ n (ǫ) for n = 0. To do so it is more convenient to study the eigenvalue problem (19) for h: Recall the operator L = L 0 = ∂ x + H whose action on the Fourier modes is with eigenvalues 0 (double), ±i, ±2i, . . . (g mean zero).
Since (u ǫ , v ǫ ) is analytic in ǫ on a neighborhood of 0, and by Chapter VII, Theorem 2.6 in [21], L ǫ is a holomorphic family of operators of type (A), so by Chapter VII, Section 2.3, all the results in Chapter II, Sections 1 and 2 apply, and we can compute the Taylor coefficients of λ(ǫ) as if L ǫ acted on a finite dimensional vector space. We start with computing the resolvent of L: whose action on the Fourier modes is Around the eigenvalue ni (n = ±1, ±2, . . . ) we have the expansion where P n is the projection on the span of e i(n+sgn n)x and , m = n + sgn n.
By [21] (II.2.33), where S (0) n = −P n and for h ≥ 1, S (h) n = S h n , with S n defined in (37), and L (v) is the coefficient of ǫ v in the Taylor expansion of L ǫ . Note that the constraints in the summation imply that there is some j ∈ {1, . . . , p} such that h j = 0 and so S (h j ) n = −P n , so every summand is a finite-rank operator whose trace is thus well defined. Lemma 6. If A is a finite-rank operator, then Tr AB = Tr BA.
Proof. By linearity we can assume A has the form A(·) = f (·)v for some (not necessarily continuous) linear functional f . Then Tr A = f (v). Since AB(·) = f (B·)v and BA(·) = f (·)Bv, it follows that Tr AB = f (Bv) = Tr BA.
Using the lemma above, we can simplify the sum in λ · · · L (v 1 ) on e i(n+sgn n)x and take the (n + sgn n)-th mode to compute the trace. Thus (38) Let us compute some terms λ Put s = sgn n. We extract the (n + s)-th mode of each term: We can further compute that for n = ±1, ±2, ±3, . . . .
When k ≥ 2|n| + 4, λ (k) n is still purely imaginary but the formula λ (k) n = ic (k) (n + sgn n) does not hold in general.
Proof. Firstly we notice that, for n = ±1 the coefficient of ǫ 6 in λ ±1 (ǫ) is, which does not hold λ (6) ±1 = ±2ic (6) . Next, we prove the fist part of the lemma. In each summand of (38), all the coefficients are real, except that each operator L brings a factor of i to the Fourier coefficients (via the operator ∂ x ), and each operator S n removes a factor of i (see (37)). Hence each summand is purely imaginary, and so is λ In each summand of (38), the operator S (h j ) n is a Fourier multiplier that does not shift the modes, while the operator L (m) g = (u (m) − v (m) )g ′ shifts the modes by at most m because u (m) only contains modes up to e ±imx . Also the amount of shift = m (mod 2). Thus when acting the sequence L (vp) S (hp) n · · · L (v 1 ) on e i(n+s)x , the mode is consecutively shifted by at most v 1 , v 2 , . . . , v p , and the total amount of shifts = j v j = k (mod 2). Since in the end we are taking the (n + s)-th mode, the total amount of shifts must be 0 in order to count, so when k is odd λ (k) n = 0. When k is even, the mode e i(n+s)x can only be shifted as far as e i(n+s±k/2)x ; otherwise it can never be shifted back. Hence when k ≤ 2|n| + 2 = 2|n + s|, the frequency always has the same sign as n or becomes 0. In the former case we can take sgn m = sgn n in (37), while in the latter case the derivative in L kills it, so it does not hurt if we still take sgn m = sgn n in (37). Either way we can take sgn m = sgn n in (37). Thus the action of S n is the same as that of S ′ n , where , m = n + sgn n.
For n > 0, the operator S ′ n is the analog of S n for L + with i.e., n remains the same if we replace L with L + . Now we have that Using the same change of variable as in Section 3.2, the problem above can be transformed toh ′ − ic −1 ǫh = c −1 ǫ λ + (ǫ)h whose eigenvalues are λ + n ′ (ǫ) = n ′ c ǫ i − i. Since when ǫ → 0, λ n (ǫ) → ni and c ǫ → 1, we must have that n ′ = n + 1, and so λ n (ǫ) = (n + 1)c ǫ i − i + O n (ǫ 2n+4 ).
When m + n + l = 0 and mnl = 0, since λ n (ǫ) is odd in n, it follows that λ m (ǫ) + λ n (ǫ) + λ l (ǫ) = 0. We do have time resonance in this case. We will eliminate this case by choosing a new frame of reference.

A new frame of reference
Recall that the traveling wave solution i.e., v ǫ u ′ ǫ = Hu ǫ + u ǫ u ′ ǫ . Now we aim to find a new reference frame. Let P ± 0 (ǫ) be the projection on the 1 dimensional space spanned by the eigenvector ϕ + 0 (ǫ) = ∂ ǫ u ǫ and ϕ − 0 (ǫ) = −ǫ −1 u ′ ǫ , respectively. Then we aim to rewrite where ǫ, a ∈ R and P ± 0 (ǫ(t))g = 0. We first show that it is always possible, provided that f is close to a traveling wave.
Proposition 5. Let k ≥ 2. Then there is r = r(k) > 0 such that if |ǫ 0 | < r and f − u ǫ 0 H k < r|ǫ 0 |, then there is ǫ ∈ R, a ∈ R/2πZ and g ∈ H k such that Moreover, ǫ, a and g depend smoothly on f .
Let g = f (x − a) − u ǫ . Then (39) and (40) clearly hold. Moreover, showing (41). The smooth dependence of ǫ, a and g on f is also clear.
By translation symmetry, if f is r|ǫ 0 |-close to u ǫ 0 (x + a) for some a ∈ R/2πZ, we can reach a similar conclusion. Then we can write f (x, t) = u ǫ(t) (x + a(t)) + g(x + a(t), t).
We will obtain an energy estimate for g. Combined with local wellposedness of the equation (1) and Proposition 5, we can show that the solution extends as long as the energy estimate closes, see the end of section 5.2.
To get the energy estimate, we first need to derive an evolution equation for g. Since f is differentiable in t, so are ǫ(t), a(t) and g, and we get

Diagonalization
To find the evolution of other modes, we diagonalize the equation for g. (20). Recall from (18) that On the other hand, By the chain rule, Using the operator W ǫ from lemma 5 we have that where Λ ǫ is a Fourier multiplier whose action on the Fourier mode e i(n+sgn n)x is multiplication by λ n (ǫ). Since W ǫ /ǫ is of class S, uniformly in ǫ, for any smooth function F , the operator is of class S, with the implicit constants depending on the C k norms of F . Let h = (1 + W ǫ )h. Then /ǫ are of class S, uniformly in ǫ when ǫ is small. Moreover, since W ǫ is analytic in ǫ with W 0 = 0, so is R ǫ (1) with R 0 (1) = 0. Hence R ǫ (1)/ǫ is of class S uniformly in ǫ, and so is Since ∂ ǫ u ǫ and u ′ ǫ are in the generalized eigenspace of L ǫ associated with the eigenvalue 0, we have that are in the space spanned by sin x and cos x, according to Lemma 5. Now we have where N ǫ [h, h] is given by (48) and is also of class S uniformly in ǫ when ǫ is small. Recall that ǫ ′ (t) and a ′ (t) are chosen such that P 0 (ǫ)g(t) = 0 for all t, where P 0 (ǫ) is the projection onto the span of ∂ ǫ u ǫ and u ′ ǫ . This implies that Q 0 (ǫ)h(t) = 0 for all t, where Q 0 (ǫ) is the projection onto the span of to the span of sin x and cos x, we have thatĥ(1) =ĥ(−1) = 0 for all t.

Energy estimates
Sinceĥ(1) =ĥ(−1) = 0 for all t, for k = 0, 1, . . . we define the energy and aim to control its growth. Using the evolution equation (49) for h and the anti-selfadjointness of Λ ǫ we get When ǫ is small enough, the last two are bounded operators with bounded inverse betweeṅ H k , k = 0, 1, . . . , so To bound E Φ we use (47) and (50) to get For the sake of bounding this term, since the inner product is taken in the space L 2 /(1), we can without loss of generality assume thatĥ(0) = 0 (which is not true in general) and integrate by parts to get so again by (47) and (50), Combining (51), (52) and (53) shows that

Normal form transformation
To bound E N we recall the expression of N ǫ from (48). Since N ǫ does not depend on the constant mode of h, we can also assume without loss of generality thatĥ(0) = 0. We further decompose On the other hand, the integral vanishes unless m + n + l + sgn m + sgn n + sgn l = 0 (61) in which case m + n + l is an odd number, so is non-zero. Then by Case 1 of Proposition 4, m + n + l = 0, the denominator is uniformly bounded from below thanks to (62). Unless j = 2 in (57) and (58), we can integrate by parts if necessary to ensure that at most k − 1 derivatives in x hit each factor of j. Then similarly to (64) it follows that for k ≥ 5, For j = 2, by symmetry of D ǫ it is clear that which according to (49) equals Similarly to (52), By definition of D ǫ , x e i(m+sgn m)x ×ĥ(n + sgn n, t)∂ k x e i(n+sgn n)xĥ (l + sgn l, t)∂ 2 x e i(l+sgn l)x dx where m ′ + sgn m ′ = p + m + sgn m = 0, ±1. We break the summation into several parts. Part 1: |p| ≥ |m + sgn m|/3. Then we can transfer the extra derivative from h to Φ ǫ , and compute as in (52) to get Part 2: |p| < |m+sgn m|/3 but |p| ≥ |n+sgn n|/3. If |n+sgn n| ≥ |m|/3 then |p| ≥ |m|/9, and we get the same bound as before. Otherwise, since the integral vanishes unless p + m + n + l + sgn m + sgn n + sgn l = 0 in which case we have |l + sgn l| > |n + sgn n|/3, and we can transfer the extra derivative to the factor ∂ 2 x h to get (note that Φ ǫ C k k g 2 provided that g(t) H 1 /|ǫ| is small enough. Part 3: |p| < |m + sgn m|/3 and |p| < |n + sgn n|/3. Then sgn(m ′ + sgn m ′ ) = sgn(m+sgn m), i.e., sgn m ′ = sgn m, so m ′ = m+p. By symmetry, x e i(l+sgn l)x ×Φ ǫ (p)e ipxĥ (m + sgn m, t)∂ k+1 x e i(m+sgn m)xĥ (n + sgn n, t)∂ k x e i(n+sgn n)x dx x e i(l+sgn l)x ×Φ ǫ (p)e ipxĥ (m + sgn m, t)∂ k x e i(m+sgn m)xĥ (n + sgn n, t)∂ k+1 x e i(n+sgn n)x dx.
We now turn to . Similarly to (52), is of the same form as the left-hand side of (70), so we trace the same argument to get provided that ǫ is small enough. Hence provided that ǫ is small enough. Combining (68), (75) and (79) shows that, for k ≥ 4, provided that ǫ and g(t) H 1 /|ǫ| are small enough. Finally, combining (65), (66), (67) and (80) shows that, for k ≥ 5, provided that ǫ and g(t) H 1 /|ǫ| are small enough.

Lifespan when δ ≪ ǫ
In this section we will obtain a preliminary bound for E N 2 = E N − E N 1 and show a lifespan of 1 ǫδ when ||g 0 || H 5 (T) = δ ≪ ǫ, i.e., δ ≤ cǫ for some c > 0 independent of ǫ.

Longer lifespan when δ ≪ ǫ 2
When the perturbation g is very small compare to ǫ 2 , i.e., ||g 0 || H 5 (T) = δ ≪ ǫ 2 , we can obtain a longer lifespan by applying the normal form transformation to where c kj , c ′ kj and c kji ∈ R are constants and we integrated by parts to get rid of the terms with k + 1 derivatives falling on a single factor of h, except for the term with j = 1 in E N 21 , in which the k + 1 derivatives do not matter in view of the fact that the operator (φ ′ ǫ • φ −1 ǫ )((1 + W ǫ ) −1 − 1) is of class S.
Assume δ/ǫ 2 is small. Then the second term on the right-hand side δ 5/2 , so we close the estimate for a time t k |ǫ|/δ 2 , which is also the lifespan in this case.