Quasi-periodic solution of quasi-linear fifth-order KdV equation

In this paper, we prove the existence of quasi-periodic small-amplitude solutions for quasi-linear Hamiltonian perturbation of the fifth-order KdV equation on the torus in presence of a quasi-periodic forcing.

not to list them here. In the present paper, we focus on the HPDEs with unbounded perturbations.
If the perturbation is unbounded, the homological equation in the KAM iteration reads as follows: where µ(ϕ) has zero average, and µ(ϕ) ≈ γ is usually of large magnitude. The equation of this type is called a small-denominator equation with large variable coefficient. It is crucial to get appropriate estimation of such equation. Assuming λ ≥ |ω|γ 1+β for some β > 0, Kuksin gave a valid estimate of the solution in [15], which is applied to the KdV equation and a whole hierarchy of so called higher order KdV equations, see [5], [16] and [13]. Subsequently, Liu-Yuan [20] gave a new estimate, including both β > 0 and β = 0, which extends the application of KAM theory to 1-dimensional derivative NLS (DNLS) and Benjamin-Ono equations. See [21], [31] and [28]. The case β < 0 is corresponding to quasi-linear or fully nonlinear equations, for which there has not yet any clue to get a required estimation of the solution for (1.1).
Recently, in a series of papers [1,2,4,6,11,12,23], Baldi-Berti-Feola-Montalto invented a sophisticated tool to deal with the case β < 0 for some quasi-linear or fully nonlinear partial differential equations, such as KdV and water wave equation. Take the fully nonlinear KdV equation (1.2) ∂ t u + u xxx + f (ωt, x, u, u x , u xxx ) = 0, x ∈ T = R/2πZ, as an example. By Nash-Moser iteration, the linearized homological equation can be seen as Lv = F , where (1.3) L = ω · ∂ ϕ + (1 + a 3 (ϕ, x))∂ 3 x + a 1 (ϕ, x)∂ x + a 0 (ϕ, x), ϕ ∈ T v . It is crucial to estimate the inverse of the linear operator L. Instead of directly reducing the linear operator L to a diagonal operator, Baldi-Berti-Feola used some sophisticated way to reduce the linear operator L to a diagonal operator plus a bounded perturbation by a set of regularization procedures. For example, the regularization procedures for the fully nonlinear KdV equation can be summarized as: • To eliminate the space variable dependence of the coefficients of ∂ 3 x by a ϕdependent changes of variable. Then, to eliminate the time dependence of the coefficients of ∂ 3 x by a quasi-periodic time re-parametrization(See [2] for detail). The linear operator L is thus reduced to (1.4) • The regularization procedure to dispose the coefficients of ∂ x can be divided into two steps.
Applying the delicate variable change as the abvoe to L, then, the linear operator L is reduced to (1.10) where m 5 ∈ R is constant and b i 's are function of (ϕ, x).
When one tries to reduce the coefficient b 3 = b 3 (ϕ, x) of ∂ 3 x to constant, a new difficulty arises: the variable change y = x + p(ϕ) to amend the coefficients of ∂ x can not be applied to the coefficients of ∂ 3 x . Therefore, a suitable regularization procedure for the higher-order KdV equations seems to be more complicated and maybe need some new technical method.
In this paper we make attempt to deal with the problem by combing regularization method by Baidi-Berti-Feola and the unbounded reduction method by Kuksin [17]. Consider the Hamiltonian quasi-linear fifth order KdV equation of the following form (1.11) ∂ t u = X H (u), with (1.12) X H (u) = ∂ x ∇ L 2 H(t, x, u, u x , u xx ), and (1. 13) The perturbation g(ωt, x, u, u x , u xx ) is unbounded, and quasi-periodic in time, periodic in space, a polynomial with regard to u, u x , u xx . Here and in other places in this paper, T d is short for the average(2π) −d T d , by a slight abuse of notation. The primary fifth-order KdV equation without perturbation g is (1.14) u t + ∂ 5 x u + 10u∂ 3 x u + 20∂ x u∂ 2 x u + 30u 2 ∂ x u = 0, which is a special case of the general fifth-order KdV equation (fKdV) of the following (1.15) u t + α∂ 5 x + βu∂ 3 x u + γ∂ x u∂ 2 x u + σu 2 ∂ x u = 0. The special case (1.14) is called the Lax case, which is characterized by β = 2γ and α = 3 10 γ 2 . This general fifth-order KdV equation describes motions of long waves in shallow water under gravity. In a one-dimensional nonlinear lattice, it is an important mathematical model with wide application in quantum mechanics and nonlinear optics. Typical examples are widely used in various fields such as solid state physics, plasma physics, fluid physics, and quantum field theory. The relevant research can be found in [27], [10] and [19].
The Banach space H s,p can be extended to the analytic functions defined on T b with any integer b > 0. If u(ϕ) = k∈Z b u k e ikϕ , ϕ ∈ T b , we can denote For notation convenience, when p = 0, · s,0 is simplified as · s .
Although just only the quasi-linear fifth-order KdV equation is investigated, the method in Theorem 1.1 also applies to other quasi-linear Hamiltonian higher order KdV equations, for example, the seventh order, even to some other quasi-linear or fully-nonlinear equations.

Functional setting
In this section, we introduce some notations, definitions and technical tools, which will be used in section 3, 4, 5.
The phase space of (1.16) is endowed with non-degenerate symplectic form where ∂ −1 x u is the periodic primitive of u with zero average. The Hamiltonian vector field X H (u) = ∂ x ∇H(u) is the unique vector field satisfying the equality Recall Poisson bracket between two Hamiltonians F, G : The function in the present paper is quasi-periodic in the time variables and periodic in the space variable. This function is analytic for these variables in the domain of T v+1 s , where T v+1 s be the complexified torus with |Imφ i | ≤ s. So, the function will be of the form Now, we define some important norms: For analytic function u defined on T v+1 s , the max norm plays important role in our paper As a notation, we denote a ⋖ b as a ≤ Cb, where C is a constant depending on the form of equation, the number v of frequencies, the diophantine exponent τ in the non-resonance condition. a ≈ b means a ≤ C 1 b and C 2 a ≥ b.
When we consider a function f : Π → E, ω → F (ω), where (E, E ) is the Banach space and Π is the subset of R, we can define sup-norm and Lipschitz semi-norm below.
Then, the Lipschitz norm is and (2.11) Then, the lemma is proved.
The algebra properties of Banach space H s,p and H s s,p are also our concern.
If h 1 = h 1 (λ) and h 2 = h 2 (λ) depend in a Lipschitz way on the parameter λ ∈ Π ⊂ R, then [j] , By the Schwarz inequality, we have For the case s = 0, we have The case s > 0 is a simple variation.

Matrices with variable.
Let b ∈ N, and consider the exponential basis So, for parameter dependent matrices A := A(λ), λ ∈ Π ⊆ R, we can also define Lipschitz norms as |A| Lip s,p = |A| sup s,p + |A| lip s,p . We now show some properties of (s, p)-decay norm.
Moreover, if p = p(λ) is a Lipschitz family of functions, Proof. According to Definition 2.3, we see Then, the lemma is proved.
Proof. (2.30) are from the Taylor's series of e Ψ i and Lemma 2.7. To prove (2.31), we see and Then, use (2.30).

Linear time-dependent operator and Hamiltonian operators.
In this section, we give some definitions and properties of the linear time-dependent Hamiltonian systems which will be used in following section.
for some real linear operator G(t) which is selfadjoint with respect to the L 2 scalar product. The vector product is generated by the quadratic Hamiltonian If G(t) = G(ωt) is quasi-periodic in time, we say that the associate operator ω · ∂ ϕ − ∂ x G(ϕ) is Hamiltonian.
is a family of symplectic maps we say that the operator A defined by Ah(ϕ, x) = A(ϕ)h(ϕ, x), acting on the functions h : T v+1 → R, is symplectic.
Under a time dependent family of symplectic transformations u = Ψ(t)v the linear Hamiltonian equation transforms into the equation Note that E(T ) is self-adjoint with respect to the L 2 scalar product because Ψ T ∂ −1 , which is still Hamiltonian, according to the Definition 2.35.

The Regularization of the linearized operator
In this section, we perform a regularization procedure, which conjugates the linearized operator L(u n ) defined in (3.4) to the operator L(u n ) defined in (3.64), the coefficients of the highest order spatial derivative operator are constant. The method has been used in [1,2,4,6,11,12]. Our existence proof is based on a modified Newton iteration. The main step concerns the invertibility of the linearized operator h, obtained by linearizing (1.16) at any approximate (or exact) solution u. The coefficients a i = a i (ϕ, x) = a i (u, ε)(ϕ, x) are periodic functions of (ϕ, x), depending on u and ε. Then, we have . In the Hamiltonian case (1.11), the linearized operator (3.1) also has the form The coefficients a i , together with their derivative ∂ u a i [h] with respect to u in the direction h, satisfy the following estimates: Moreover, if λ → u(λ) is a Lipschitz family, and satisfying u Lip s,p+2 ≤ 1, then, we have Then, these estimates are straightforward.

Change of space variable.
We consider a ϕ-dependent family of space variable change of the form where β is a (small) real analytic function, 2π-periodic in all its arguments. The change of variables (3.9) induces on the space of functions the linear operator The operator T is invertible, with inverse (3.11) (T −1 h)(ϕ, y) = h(ϕ, y +β(ϕ, y)).
where y → y +β(ϕ, y) is the inverse of (3.9), namely In the Hamiltonian case, in order to keep the Hamiltonian structure of linear operator, the operator T needs a slight change. The modified linear operator is Remark 3.1. By [2, remark 4.1.3], the modified change of variable and its inverse (3.12) are symplectic, for each ϕ ∈ T v . Also, both A and A −1 are maps from H 1 0 to H 1 0 , for each ϕ ∈ T v . Now, we calculate the conjugate A −1 LA of the linearized operator L in (3.1). The conjugate A −1 aA of any multiplication operator a : h(ϕ, x) → a(ϕ, x)h(ϕ, x) is the multiplication operator (T −1 a) that maps v(ϕ, y) → (T −1 a)v(ϕ, y). The conjugate of differential operators are x β] , where all the coefficients {T −1 [..]} are periodic functions of (ϕ, y).
Remark 3.2. we give out some calculation tricks which have been used above.
x (a∂ 2 x )} are slightly more finicky. Let's take ∂ x a as an example, we see Now, we get Since T only make changes on the space variable, Then, we have a solution (with zero average) of (3.19) x is defined by linearity as In other words, ∂ −1 x is the primitive of h with zero average in x . Thus we obtain the operator L 1 in (3.15), that b 2 (ϕ, x) = 1 + b(ϕ).

Estimates of b(ϕ)
We prove b(ϕ) satisfies the following estimates: If u is small enough, we have Since u is small enough, ψ(t) and g(t) can be well defined by its power series expansion, i.e. g(t) = 1 − 1 5 t + 3 25 t 2 + · · · . Hence we have The first and last inequality of (3.26) can be proved in such way. The second inequality is a direct result of M g s,p ≤ C g s,p . Proof of (3.25): The derivative of c with respect to u in the direction h is Using the same method as (3.26), by (3.5) and (3.6), we can get (3.25).

Estimates of
Using the same way as (3.23), we can get Use the same way as (3.26), the bounds (3.7), (3.8) and (3.26) imply The inverse function y → y +β(ϕ, y) is also under our consideration. By Lemma 6.7, one gets .

Estimates of the coefficients b i
Consider the coefficients b * 1 , b * 0 , which are given in (3.16). We have b i 99s 100 ,p ⋖ u s,p+2s 0 +6 , and Ultimately, (3.48), (3.50) and (3.51) imply that By the same way as b 1 , we can get

Time reparametrization.
In this section, we will make constant the coefficient of the highest order spatial derivative operator of L 1 , by a quasi-periodic reparametrization of time. The change of variables has the form where α is a (small) real analytic function, 2π-periodic in all its arguments. The induced linear operator on the space of functions is where ϕ = θ + ωα(θ) is the inverse of θ = ϕ + ωα(ϕ). Then, the time derivative operator becomes The spatial derivative operator dose not have any change. Thus, see (3.58) We look for α such that the coefficients of the highest order derivatives are proportional, namely for some constant m ∈ R. This is equivalent to require that Integrating on T v determines the value of the constant m, We can find the unique solution of (3.60) with zero average With this choice of α, we have Suppose u Lip s,p+4s 0 +τ 0 +9 ≪ 1 100 , we have these estimation below: 1. Estimates of m The coefficient m, defined in(3.61), satisfies the following estimates: Similarly we get the Lipschitz part of (3.66). The estimates (3.67) follows by (3.25), since

Estimates of the coefficients c i
The coefficients c i defined in (3.65), for i = 0, 1, satisfy the following estimates: Differentiating c i with respect to u in the direction h gives

Estimates on L.
Recall the procedure performed in the previous subsection, we have conjugated the operator L to L, that is In the following lemma, we summarize the estimates for the linear operator L and U 1 , U 2 , also define constants (2) : The constant coefficient m ,defined in (3.61), satisfies The detail of these estimates can be found in the previous subsection, we just give a summary here. In this section, we make a reduction to eliminate the unbounded perturbation of linear operator L obtained in (3.64). The goal is to conjugate it to a diagonal operator J plus a sufficient small unbounded remainder R. Before we apply the reducibility scheme, we will make some definition, recall and revise some important lemmas.
Now, we recall the classical Kuksin's lemma.
Lemma 4.4 (Kuksin). Consider the following first order partial differential equation for the unknown function u defined on the torus T v , where ω = (ω 1 , · · · , ω n ) ∈ R v and d ∈ R. We make the following assumption.
The inverse of L(u n ) is our main concern. However, the estimate of L(0), a diagonal operator, is straightforward. In order to make the structure of this paper much more simplicity, the initial approximate solution is u 1 other than u 0 . So, we will estimate the inverse of L(0) and set the initial parameter .
has a unique solution v with zero average, satisfying v Lip s,p ′ ≤ 1 Thus, (4.20) can be transformed to Applying the operator ∆v = v(λ 1 ) − v(λ 2 ) to L(0)v = F (0), we have Again using (4.19), we see Combining (4.24) with (4.25), one gets x v 1 , one gets KAM step: In this section, we will give the outline of the reducibility and show in detail one key step of the KAM iteration. The purpose is to define a transformation operator Φ m conjugating L m , a diagonal operator J m plus a ε m remainder R m , to L m+1 , a diagonal operator J m+1 plus a ε m+1 remainder R m+1 . Now, we have already got the regularized linear operator L(u n ) at the approximate solution u n , which is (4.30) L = ω · ∂ θ + m∂ 5 y + ∂ y {∂ y [c 1 (θ, y)∂ y )] + c 0 (θ, y)}. Now, the linear operator L can be denote as where (4.32) D = m∂ 5 y , R = ∂ y {∂ y [(c 1 (θ, y)∂ y )] + c 0 (θ, y)}. According to Definition 4.2, we see |R| ς s,s 0 ≤ c 1 (θ, y) s,p + c 0 (θ, y) s,p . By Lemma 3.2 and Lemma 6.6, the coefficients of perturbation term R can be divided into n parts, which is ) is as small as v m , the function space of (c i (u m ) − c i (u m − 1 )) can also be controlled by v m . Now, the linear operator L(u n ) can be seen as   The purpose of reducibility is to make the reminder of the linear operator L m much more small. If the the reminder of L m can be divided into Q m and R m , R m lies in a much more general analytical space and Q m is much more smaller. We can consider the homological equation to eliminate R m . Thus, the transformation operator Ψ m can lies in a much more general Banach space.
In order to exhibit the outline of our reducibility, we will give the outline of the one step of reduction. The transformation Ψ m = e Φm acting on the operator L m : Then, we can get If we solve the homological equation L m+1 can be denote as Before we give the iteration lemmas, we need the following iteration constants and domains.
iteration parameters: Set n ≥ 1, m ≥ 1. Then, (m, n) indicates the m th step KAM reduction for the linear operator L(u n ). • s n = ( 10 11 ) n−1 s 1 , s 1 = s, which dominate the width of the u n .
• s ′ n = 99 101 s n , s 1 = s, which dominate the width of the coefficients c i (u n ) and R n .
• σ m = 1 200 s m , which serve as a bridge from s m to s m+1 .
[H2] m Assume we have get the following operator after (m − 1) th step KAM reduction for the linear operator L(u n ), which satisfies the following hypothesis: (S1) m : defined for all λ ∈ Λ m.n , where Λ 0.n = Λ(u n ) (is the domain of u n ), and, for m ≤ n, is Lipschitz − continuous in λ, and fulfills the estimate: (4.54) sup µ k (λ, θ) is real analytic in θ and Lipschitz − continous in λ of zero average. It also satisfies (4.46), and Q n+1 = 0. (S2) m : The reminder R m is Lipschitz − continous in λ, and satisfies the estimate: C is an constant only depend on v. Moreover, for m ≥ 1, we have  Corollary 4.7. ∀λ ∈ Λ n,m , the sequence  Proof. For convenience, let L, R refer to L m , R m , L + , R + refer to L m+1 , R m+1 .
(Part1: Homological equation) ). Multiply −i on both side of the following equation: Then, the homological equation is equivalent to Now, applying Kuksin's lemma to (4.63), we get  we have Then, consider the infinite matrices of elements , .

4.3.
The estimation of F(u n+1 ) and v n+1 . Review the definition of F (u), which is (4.118) Proof.
Now, the whole necessary estimates has been prepared. We will prove the last piece (P2) n of iteration Lemma [H1] n .

Measure estimation
For notational convenience, we extend the eigenvalues d m i (u n ), which are defined Although Θ mn and Γ n seem different, the following two lemmas can applying them both.
Proof. We see Obviously, we have | Γ n | ≤ Cα 0 . Consider the set Θ mn in a different view. Set where Λ m is set removed from the m-step reduction for all L(u n ),n ≥ m. By Lemma 5.1, R m ijℓ (u n ) = ∅ are confined in the ball i 4 + j 4 ≤ 16|ω||ℓ|. Then, we have n .

Technical lemmas
Suppose the function in this paper real analytic on T n s , and s 0 an integer greater than n 2 . Definition 6.1. Let p be an integer, the max norm of D p u on T n s is |D p u| s = α∈Z n ,|α|=p |D α u| s . Lemma 6.1 ([7]). For σ > 0 and v > 0, the following inequalities holds : Proof. The proof can be found in [7, p22].
Lemma 6.2 (Appendix A. [18]). If s ≥ 0 and p > n 2 , then uv(x) s,p ≤ c u(x) s,p v(x) s,p with a finite constant c depending on p and n.
Proof. For n = 1, the detail of the proof can be found in [18]. n > 1 is a simple variation. Lemma 6.3. For u be analytic on T n s , we have the following inequalities, Since e −|k|σ [k] ν ≤ e −ν ( ν σ ) ν , we get Considering (6.4), since u is analytic on T n s , the Fourier coefficients u k satisfy D α u is also an analytic function on T n s , the Fourier coefficients (D α u) k = u k (ik) α , k α = k α 1 1 · · · k αn n , satisfy If |α| = p + s 0 , by Definition (6.1), we have

Now, we have
To prove the left part of (6.4), we see 22]). If f is analytic from the segment joining z 1 and z 0 defined on C n to C n . Then, there are point w 1 , w 2 , · · · w 2n on the segment such that where λ i ≥ 0 and 2n i=1 λ i = 1. Proof. The detail of the proof can be found in [22]. Lemma 6.5. (Change of variable) Let f be a real analytic function on T n s , with |f | s.p ≤ 1 100 . Then, there are an constant C depending on n and p, such that (i). If u is a real analytic function on T n s , u(x + f (x)) is also a real analytic function on T n 100s 101 and satisfies (6.7) |u(x + f (x))| 100s 101 ,p ≤ C|u(x)| s,p . (ii). Considering another analytic function g on T n s with |g| s.p+s 0 ≤ 1 100 , we have (6.8) |u in a Lipschtiz way by a parameter λ ∈ π ⊂ R, and |f λ | s,p+s 0 ≤ 1 100 , for all λ. Then, we have (6.9) |u Proof. For symbolic simplicity, the following calculations only consider n = 1 in form.
With regard to general situation, there is no difference.
. Collecting all the term in the sum, we have Finally, (6.11), (6.13) and (6.15) imply Differentiating the left of inequality (6.17), we have Then, we can get )| 100s 101 ≤|u| s,2 |f − g| 100s 101 ,1 . (i) f is invertible, its inverse is f −1 (y) = g(y) = y + q(y), where q be real analytic on T n 99 where the constant C depends on n and m.
The constant C depends on n and m.
Proof. For symbolic simplicity, the following calculations only consider n = 1 in form.
With regard to general situation, there is no difference.