1 Introduction

1.1 Presentation of the problem

In this paper, we study the Cauchy problem for the general nonlinear higher order Camassa–Holm-type equation:

$$\begin{aligned} \left\{ \begin{array}{l} (1-m\partial _x^2)u_t + a_1(t,x,u)u_x +a_2(t,x,u,u_x)u_{xx}\\ \qquad +a_3(t,x,u) u_{xxx} +a_4(t,x) u_{xxxx}+a_5(t,x) u_{xxxxx}=f \\ \hbox { for} \quad (t,x) \in (0,T]\times \mathbb {R}\\ u_{\vert _{t=0}}=u^0, \end{array}\right. \end{aligned}$$
(1.1)

where \(u=u(t,x)\), from \([0,T]\times \mathbb {R}\) into \(\mathbb {R}\), is the unknown function of the problem, \(m>0\) and \(a_i\), \(1 \le i \le 5\), are real-valued smooth given functions where their exact regularities will be precised later. This equation covers several important unidirectional models for the water waves problems at different regimes which take into account the variations of the bottom. We have in view in particular the example of the Camassa–Holm equation (see [1, 2]), which is more nonlinear then the KdV equation (see for instance [3,4,5,6,7,8]). However, the most prominent example that we have in mind is the Kawahara-type approximation (see [9, 10]), in which case the coefficient \(a_5\) does not vanish. The presence of the fifth order derivative term is very important, so that the equation describes both nonlinear and dispersive effects as does the Camassa–Holm equation in the case of special tension surface values for the development of models for water waves problem was initiated in order to gain insight into wave breaking (see [11, 12]).

Looking for solutions of (1.1) plays an important and significant role in the study of unidirectional limits for water wave problems with variable depth and topographies. To our knowledge the problem (1.1) has not been analyzed previously. In the present paper, we prove the local well-posedness of the initial value problem (1.1) by a standard Picard iterative scheme and the use of adequate energy estimates under a condition of nondegeneracy of the higher dispersive coefficient \(a_5\).

1.2 Notations and main result

In the following, \(C_0\) denotes any nonnegative constant different than zero whose exact expression is of no importance. The notation \(a\lesssim b\) means that \(a\le C_0\ b\).

We denote by \(C(\lambda _1, \lambda _2,\dots )\) a nonnegative constant depending on the parameters \(\lambda _1\), \(\lambda _2\),...and whose dependence on the \(\lambda _j\) is always assumed to be nondecreasing.

For any \(s \in \mathbb {R}\), we denote [s] the integer part of s.

Let p be any constant with \(1\le p< \infty \) and denote \(L^p=L^p(\mathbb {R})\) the space of all Lebesgue-measurable functions f with the standard norm

$$\begin{aligned} \vert f \vert _{L^p}=\left( \int _{\mathbb {R}}\vert f(x)\vert ^p dx\right) ^{1/p}<\infty . \end{aligned}$$

The real inner product of any two functions \(f_1\) and \(f_2\) in the Hilbert space \(L^2(\mathbb {R})\) is denoted by

$$\begin{aligned} (f_1,f_2)=\int _{\mathbb {R}}f_1(x)f_2(x) dx. \end{aligned}$$

The space \(L^\infty =L^\infty (\mathbb {R})\) consists of all essentially bounded and Lebesgue-measurable functions f with the norm

$$\begin{aligned} \vert f\vert _{L^\infty }= \sup \vert f(x)\vert <\infty . \end{aligned}$$

We denote by \(W^{1,\infty }(\mathbb {R})=\{f, \text{ s.t. } f,\partial _x f\in L^{\infty }(\mathbb {R})\}\) endowed with its canonical norm.

For any real constant \(s\ge 0\), \(H^s=H^s(\mathbb {R})\) denotes the Sobolev space of all tempered distributions f with the norm \(\vert f\vert _{H^s}=\vert \Lambda ^s f\vert _{L^2} < \infty \), where \(\Lambda \) is the pseudo-differential operator \(\Lambda =(1-\partial _x^2)^{1/2}\). For any two functions \(u=u(t,x)\) and v(t, x) defined on \( [0,T)\times \mathbb {R}\) with \(T>0\), we denote the inner product, the \(L^p\)-norm and especially the \(L^2\)-norm, as well as the Sobolev norm, with respect to the spatial variable x, by \((u,v)=(u(t,\cdot ),v(t,\cdot ))\), \(\vert u \vert _{L^p}=\vert u(t,\cdot )\vert _{L^p}\), \(\vert u \vert _{L^2}=\vert u(t,\cdot )\vert _{L^2}\) and \( \vert u \vert _{H^s}=\vert u(t,\cdot )\vert _{H^s}\), respectively.

Let E be a given normed space we denote \(L^\infty ([0,T);E)\) the space of functions such that \(u(t,\cdot )\) is controlled in E, uniformly for \(t\in [0,T)\):

$$\begin{aligned} \left\| u\right\| _{L^\infty ([0,T);E)} \ = \ ess\sup _{t\in [0,T)}\vert u(t,\cdot )\vert _{E} \ < \ \infty . \end{aligned}$$

Let E be a given normed space we denote C([0, T); E) the space of functions such that \(u(t,\cdot )\) is controlled in E, uniformly for \(t\in [0,T)\):

$$\begin{aligned} \left\| u\right\| _{L^\infty ([0,T);E)} \ = \ \sup _{t\in [0,T)}\vert u(t,\cdot )\vert _{E} \ < \ \infty . \end{aligned}$$

Let X be a given space, we denote C([0, T); X) the space of functions such that \(u(t,\cdot )\) is in X.

Finally, \(C^k(\mathbb {R}^i)\), \(i\ge 1\) denote the space of k-times continuously differentiable functions over \(\mathbb {R}^i\).

For any closed operator T defined on a Banach space X of functions, the commutator [T, f] is defined by \([T,f]g=T(fg)-fT(g)\) with f, g and fg belonging to the domain of T. The same notation is used for f as an operator mapping the domain of T into itself.

Actually, we admit without proof this lemma that presents some properties for the commutator operator.

If \(f\in F\) and \(g\in G\), F and G being two Banach spaces, the notation \(\vert f\vert _{F}\lesssim \vert g\vert _{G}\) means that \(\vert f\vert _{F}\le C\vert g\vert _{G}\) for some constant C which does not depend on f nor g.

Here, \({\mathcal S}(\mathbb {R})\) denotes the Schwartz space of rapidly decaying functions, and for any distribution \(f\in {\mathcal S}'(\mathbb {R})\), we write \(\widehat{f}\) Fourier transform on \({\mathcal S}'(\mathbb {R})\).

We use the classical notation f(D) to denote the Fourier multiplier, namely, \(\widehat{f(D)u}(\cdot )=f(\cdot )\widehat{u}(\cdot )\).

We use the condensed notation

$$\begin{aligned} A_s=B_s +\left\langle C_s\right\rangle _{s> s_0} \end{aligned}$$
(1.2)

to say that \(A_s=B_s\) if \(s\le s_0\) and \(A_s=B_s+C_s\) if \(s> s_0\).

1.3 Product and commutator estimates in Sobolev spaces

Let us recall here some product as well as commutator estimates in Sobolev spaces, used throughout the present paper (see [11]).

Lemma 1.1

(product estimates) Let \(s\ge 0\), one has \(\forall f,g\in H^s(\mathbb {R})\bigcap L^\infty (\mathbb {R}),\) one has

$$\begin{aligned} \left| \ f \ g\ \right| _{H^s} \ \lesssim \ \left| \ f \ \right| _{L^\infty } \left| \ g\ \right| _{H^s}+\left| \ f \ \right| _{H^s} \left| \ g\ \right| _{L^\infty }. \end{aligned}$$

If \(s\ge s_0>1/2\), one deduces thanks to continuous embedding of Sobolev spaces,

$$\begin{aligned} \big \vert \ f \ g\ \big \vert _{H^s} \ \lesssim \ \big \vert \ f \ \big \vert _{H^s} \big \vert \ g\ \big \vert _{H^s} . \end{aligned}$$

More generally, for \(s\ge 0\) and \(s_0>1/2\), one has \(\forall f\in H^s(\mathbb {R})\bigcap H^{s_0}(\mathbb {R}),g\in H^s(\mathbb {R})\),

$$\begin{aligned} \big \vert \ f \ g\ \big \vert _{H^s}\lesssim \big \vert \ f \ \big \vert _{H^{s_{0}}}\big \vert \ g \ \big \vert _{H^s}+\left\langle \big \vert \ f \ \big \vert _{H^s} \big \vert \ g \ \big \vert _{H^{s_0}}\right\rangle _{s>{s_{0}}}. \end{aligned}$$

Let \(F\in C^\infty (\mathbb {R})\) be a smooth function such that \(F(0)=0\). If \(g\in H^s(\mathbb {R})\bigcap L^\infty (\mathbb {R})\) with \(s\ge 0\), one has \(F(g)\in H^s(\mathbb {R})\) and

$$\begin{aligned} \big \vert \ F(g) \ \big \vert _{H^s} \ \le \ C(\big \vert g\big \vert _{L^\infty }, F)\big \vert \ g \ \big \vert _{H^s}. \end{aligned}$$

We know recall commutator estimate, mainly due to the Kato–Ponce [13], and recently improved by Lannes [11] (see Theorems 3 and 6):

Lemma 1.2

(commutator estimates)

For any \(s\ge 0\), and \(\partial _x f, g\in L^\infty (\mathbb {R})\bigcap H^{s-1}(\mathbb {R}),\) one has

$$\begin{aligned} \big \vert \ [\Lambda ^s,f]g\ \big \vert _{L^2}\ \lesssim \ \big \vert \ \partial _x f\ \big \vert _{H^{s-1}}\big \vert \ g\ \big \vert _{L^\infty }+\big \vert \ \partial _x f\ \big \vert _{L^\infty }\big \vert \ g\ \big \vert _{H^{s-1}} . \end{aligned}$$

Thanks to continuous embedding of Sobolev spaces, one has for \(s\ge s_0+1, \ s_0>\frac{1}{2},\)

$$\begin{aligned} \big \vert \ [\Lambda ^s,f]g\ \big \vert _{L^2}\ \lesssim \ \big \vert \ \partial _x f\ \big \vert _{H^{s-1}}\big \vert \ g\ \big \vert _{H^{s-1}} . \end{aligned}$$

More generally, for any \(s\ge 0\) and \(s_0>1/2\), \(\partial _x f, g\in H^{s_0}(\mathbb {R})\bigcap H^{s-1}(\mathbb {R}),\) one has

$$\begin{aligned} \big \vert \ [\Lambda ^s,f]g\ \big \vert _{L^2}\ \lesssim \big \vert \ \partial _x f\ \big \vert _{H^{s_0}}\big \vert \ g\ \big \vert _{H^{s-1}}+\left\langle \big \vert \ \partial _x f\ \big \vert _{H^{s-1}}\big \vert \ g\ \big \vert _{H^{s_0}}\right\rangle _{s>{s_{0}+1}}. \end{aligned}$$

We conclude this section with the following remark Also, let us remark these continuous embedding.

Remark 1.1

Let \(s>\frac{3}{2} \), then:

  • \(H^s(\mathbb {R})\hookrightarrow W^{1,\infty }(\mathbb {R})\)

  • \( H^{s-1}(\mathbb {R})\hookrightarrow L^{\infty }(\mathbb {R})\)

  • \( H^s(\mathbb {R})\hookrightarrow H^{s-1}(\mathbb {R})\).

Moreover, we define the following operators for \(s>0\): \(\displaystyle {\Lambda _m^s=(1-m\partial _x^2)^{\frac{s}{2}}}\) and its inverse \(\displaystyle {\Lambda _m^{-s}}\) such that the Fourier Transform is given as following:

$$\begin{aligned} \displaystyle {\displaystyle {\widehat{\Lambda _m^{-s}(u)}}= (1+m\xi ^2)^{-\frac{s}{2}}\hat{u}}. \end{aligned}$$

Finally, we will study the local well-posedness of the initial value problem (1.1) in \(H^{s}(\mathbb {R})\) endowed with canonical norm.

1.4 Main results

Let us now state our main result:

Theorem 1.1

Let \(s>\frac{5}{2}\) and \(f\in C([0,T];H^{s}(\mathbb {R}))\). We suppose that:

  • \(a_1, a_2, a_3\) are smooth mappings such that \(a_1, a_3\) in \(C([0,T],C^{[s]+1}(\mathbb {R}^2))\) and \(a_2\) in \(C([0,T],C^{[s]+1}(\mathbb {R}^3))\).

  • \(a_4\in C([0,T];H^{s+1}(\mathbb {R}))\), \(\partial _t a_4 \in L^{\infty }(0,T,L^{\infty }(\mathbb {R}))\),

  • \(a_5\in C([0,T],L^{\infty }(\mathbb {R}))\), \(\partial _x a_5\in C([0,T];H^{s+2}(\mathbb {R}))\), with \( \partial _ta_5\in L^{\infty }(0,T;L^{\infty }(\mathbb {R}))\),

  • \(F(t,x):=\int _0^x\frac{a_4}{a_5}dy \in C([0,T];L^{\infty }(\mathbb {R})) ~\text {and} ~\partial _tF \in L^{\infty }(0,T;L^{\infty }(\mathbb {R}))\),

Assume moreover that there is a positive constant \( c_1>0 \) such that \(c_1\le \vert a_5(t,x)\vert \;\forall \,(t,x) \in [0,T]\times \mathbb {R}.\) Then for all \(u^0 \in H^{s}(\mathbb {R})\), there exist a time \(T^{\star }>0\) and a unique solution u to (1.1)

in \(C([0,T^{\star }];H^{s}(\mathbb {R}))\).

Remark 1.2

There is no restriction on the signs of the coefficients \(a_2\) and \(a_4\); this means that our result handles also the case of the anti-diffusive terms, in which case these terms are controlled by dispersion.

2 Proof of the main results

Before we start the proof, we give the following useful lemma:

Lemma 2.1

Let \(m>0\), \(s>0\) then the linear operator

\(\Lambda ^2_m\): \(H^{s+2}(\mathbb {R})\rightarrow H^s(\mathbb {R})\) is well defined, continuous, one-to-one and onto. If we suppose that \(u=\Lambda ^{-2}_mf \) for \(f\in H^s(\mathbb {R})\) then:

$$\begin{aligned} \vert u\vert _{H^{s+2}}\\le & {} \frac{1}{m}\vert f\vert _{H^s}\quad \hbox {if}\quad 0<m\le 1 \end{aligned}$$
(2.1)
$$\begin{aligned} \vert u\vert _{H^{s+2}}\\le & {} \vert f\vert _{H^s} \quad \hbox {if}\quad m\ge 1. \end{aligned}$$
(2.2)

Moreover,

$$\begin{aligned} \Lambda ^s\Lambda ^{-2}_m=\Lambda ^{s-2}\Lambda ^{0}_m=\Lambda ^{0}_m\Lambda ^{s-2}, \end{aligned}$$

where \(\Lambda ^0_m\): \(H^{s}(\mathbb {R})\rightarrow H^s(\mathbb {R})\) is linear continuous one-to-one and onto operator defined by

$$\begin{aligned} \displaystyle {\widehat{\Lambda ^0_m u(\xi )}=(1+\xi ^2)(1+m\xi ^2)^{-1}\hat{u}(\xi ),} \end{aligned}$$

with

$$\begin{aligned} \vert \Lambda ^0_m \vert _{H^{s}\rightarrow H^s}\le&\max {(\frac{1}{m},1)}, \end{aligned}$$
(2.3)
$$\begin{aligned} \vert (\Lambda ^0_m)^{-1} \vert _{H^{s}\rightarrow H^s} \le&\max {(m,1)}. \end{aligned}$$
(2.4)

Proof

We have \(\Vert \Lambda ^{-2}_mf\Vert _{H^{s+2}} = \Vert (1+\xi ^2)^{\frac{s}{2}+1}(1+m\xi ^2)^{-1}\hat{f}\Vert _{L^2}\). If \( m \ge 1\), then \(1+m\xi ^2 \ge 1+ \xi ^2\) and \(\frac{ 1+ \xi ^2}{ 1+ m\xi ^2} \le 1\), therefore

$$\begin{aligned}&\Vert (1+\xi ^2)^{\frac{s}{2}+1}(1+m\xi ^2)^{-1}\hat{f}\Vert _{L^2}= \Vert (1+\xi ^2)^{\frac{s}{2}}(1+\xi ^2)(1+m\xi ^2)^{-1}\hat{f}\Vert _{L^2}\nonumber \\&\quad \le \Vert (1+\xi ^2)^{\frac{s}{2}}\hat{f}\Vert _{L^2}. \end{aligned}$$

If \(0< m < 1 \), we have \(\frac{ 1+ \xi ^2}{ 1+ m\xi ^2}= 1 + (1-m)\frac{\xi ^2}{ 1+ m\xi ^2} \le 1 + \frac{(1-m)}{m} = \frac{1}{m}\), then

$$\begin{aligned} \Vert \Lambda ^{-2}_mf\Vert _{H^{s+2}} \le \frac{1}{m} \Vert f\Vert _{H^{s}}. \end{aligned}$$

Now we have

$$\begin{aligned} \Vert \Lambda ^{0}_mf\Vert _{H^s} =\Vert \Lambda ^2\Lambda ^{-2}_mf\Vert _{H^s}=\Vert \Lambda ^{-2}_mf\Vert _{H^{s+2}} \le \max (1,\frac{1}{m})\Vert f\Vert _{H^s}. \end{aligned}$$

and

$$\begin{aligned} \Vert (\Lambda ^{0}_m)^{-1}f\Vert _{H^s} = \Vert (1+m\xi ^2)(1+\xi ^2)^{-1}(1+\xi ^2)^{{\frac{s}{2}}}\hat{f}\Vert _{L^2}. \end{aligned}$$

If \( m \ge 1\), then \((1+m\xi ^2)(1+\xi ^2)^{-1} = 1 + (m-1)\frac{\xi ^2}{1+\xi ^2} \le m\), therefore

$$\begin{aligned} \Vert (\Lambda ^{0}_m)^{-1}f\Vert _{H^s} \le m \Vert f\Vert _{H^s}. \end{aligned}$$

If \(0< m < 1\), \((1+m\xi ^2)(1+\xi ^2)^{-1} \le 1\), then

$$\begin{aligned} \Vert (\Lambda ^{0}_m)^{-1}f\Vert _{H^s} \le \Vert f\Vert _{H^s}. \end{aligned}$$

Finally \(\Vert (\Lambda ^{0}_m)^{-1}f\Vert _{H^s} \le \max (1,m) \Vert f\Vert _{H^s}\).

We will start the proof of Theorem 1.1 by studying a linearized problem associated to (1.1). \(\square \)

2.1 Linear analysis

For any smooth enough v, we define the “linearized” operator:

$$\begin{aligned} {\mathcal L}(v,\partial )= & {} \Lambda ^2_m\partial _t+a_1(t,x,v)\partial _x+a_2(t,x,v,v_x)\partial _x^2\\\\+ & {} a_3(t,x,v) \partial _x^3+a_4(t,x) \partial _x^4+a_5 (t,x)\partial _x^5. \end{aligned}$$

and the following initial value problem:

$$\begin{aligned} \left\{ \begin{array}{l} {\mathcal L}(v,\partial )u=f,\\ u_{\vert _{t=0}}=u^0. \end{array}\right. \end{aligned}$$
(2.5)

Equation (2.5) is a linear equation which can be solved by a standard method (see [14]) in any time interval in which its coefficients are defined and regular enough. We first establish some precise energy-type estimates of the solution. We define the “energy” norm,

$$\begin{aligned} E^s(u)^2=\vert w \Lambda ^s u\vert _{L^2}^2, \end{aligned}$$

where w is a weight function that will be chosen later. For the moment, we just require that there exists two positive numbers \(w_1,w_2\) such that for all (t, x) in \((0,T]\times \mathbb {R}\),

$$\begin{aligned} w_1\le w(t,x)\le w_2, \end{aligned}$$

so that \(E^s(u)\) is uniformly equivalent to the standard \(H^s\)-norm. Differentiating \(\frac{1}{2}e^{-\lambda t}E^s(u)^2\) with respect to time, one gets using (2.5)

$$\begin{aligned} \frac{1}{2}e^{\lambda t}\partial _t(e^{-\lambda t} E^s(u)^2)= & {} -\frac{\lambda }{2} E^s(u)^2 - \big (\Lambda _m^0\Lambda ^{s-2} (a_1u_x) ,w^2\Lambda ^s u\big )\\&-\big (\Lambda _m^0\Lambda ^{s-2}( a_2u_{xx}) ,w^2\Lambda ^su\big )-\big (\Lambda _m^0\Lambda ^{s-2}(a_3 u_{xxx}),w^2\Lambda ^s u\big )\\&-\big (\Lambda _m^0\Lambda ^{s-2}( a_4u_{xxxx}),w^2\Lambda ^s u\big )\\&- \big (\Lambda _m^0\Lambda ^{s-2}(a_5u_{xxxxx}),w^2\Lambda ^s u\big )\\&+\big (\Lambda _m^0\Lambda ^{s-2}f ,w^2\Lambda ^su\big )+\big (ww_t\Lambda ^su ,\Lambda ^su\big ). \end{aligned}$$

We now turn to estimating the different terms of the r.h.s of the previous identity by using the needed estimates provided from Sect. 1.3

\(\bullet \) Estimate of \(\big (\Lambda ^{s-2} (a_1u_x ),\Lambda ^0_m w^2\Lambda ^su\big )\). By the Cauchy-Schwarz inequality and the Sect. 1.3 on the composite functions we have

$$\begin{aligned} \vert \big (\Lambda ^{s-2} (a_1u_x ),\Lambda ^0_m w^2\Lambda ^{s}u\big )\vert&\le \frac{1}{m}\vert a_1(t,x,v)-a_1(t,x,0)\vert _{H^{s-2}}\times \nonumber \\&\qquad \vert a_1(t,x,0) u_x \vert _{H^{s-1}} \vert w^2\Lambda ^s u \vert _{L^2}\nonumber \\&\le C(m^{-1}, a_1 , \Vert v\Vert _{H^s},\vert w\vert _{L^{\infty }}) E^s(u)^2. \end{aligned}$$
(2.6)

\(\bullet \) Estimate of \(\big (\Lambda ^{s-2} (a_2u_{xx} ),\Lambda ^0_m w^2\Lambda ^su\big )\). Similarly as the above estimation, we have

$$\begin{aligned} \vert \big (\Lambda ^{s-2} (a_2u_{xx} ),\Lambda ^0_m w^2\Lambda ^su\big )\vert&\le C(m^{-1},a_2 , \Vert v\Vert _{H^s},\vert w\vert _{L^{\infty }}) E^s(u)^2. \end{aligned}$$

\(\bullet \)Estimate of \(\big (\Lambda ^{s-2} (a_3u_{xxx} ),\Lambda ^0_m w^2\Lambda ^su\big )\). Since we have more than s derivative on u, we remark that one can write:

$$\begin{aligned} a_3u_{xxx} = \partial _x^2(a_3\partial _xu) - \partial _x^2a_3\partial _xu-2a_3\partial _x^2u, \end{aligned}$$

then

$$\begin{aligned} \Lambda ^{s-2} (a_3u_{xxx}) = \Lambda ^{s-2}(\partial _x^2(a_3\partial _xu)) - \Lambda ^{s-2}(\partial _x^2a_3\partial _xu)-2\Lambda ^{s-2}(\partial _xa_3\partial _x^2u). \end{aligned}$$

Now use the identity \(\Lambda ^2=1-\partial _x^2\) to get that

$$\begin{aligned} \Lambda ^{s-2}(\partial _x^2(a_3\partial _xu))= & {} \Lambda ^{s-2}\big ((1-\Lambda ^2)(a_3\partial _xu)\big )\\= & {} \Lambda ^{s-2}(a_3\partial _xu)-\Lambda ^{s}(a_3\partial _xu)\\= & {} \Lambda ^{s-2}(a_3\partial _xu)-[\Lambda ^s,a_3]\partial _xu-a_3\Lambda ^s\partial _xu, \end{aligned}$$

then we obtain:

$$\begin{aligned}&\big (\Lambda ^{s-2} (a_3u_{xxx} ),\Lambda ^0_m w^2\Lambda ^su\big )\\&\quad = \big (\Lambda ^{s-2}(a_3\partial _xu),\Lambda ^0_m w^2\Lambda ^su\big )-\big ([\Lambda ^s,a_3]\partial _xu,\Lambda ^0_m w^2\Lambda ^su\big )\\&\qquad -\big (a_3\Lambda ^s\partial _xu,\Lambda ^0_m w^2\Lambda ^su\big )-\big (\Lambda ^{s-2}(\partial _x^2a_3\partial _xu),\Lambda ^0_m w^2\Lambda ^su\big )\\&\qquad -2\big (\Lambda ^{s-2}(\partial _xa_3\partial _x^2u),\Lambda ^0_m w^2\Lambda ^su\big ). \end{aligned}$$

By integration by parts, the third term of the last equality becomes:

$$\begin{aligned} \big (a_3\Lambda ^s\partial _xu,\Lambda ^0_m w^2\Lambda ^su \big )= -\frac{1}{2}\big (\partial _x(\Lambda ^0_m w^2a_3),(\Lambda ^su)^2\big ), \end{aligned}$$

Now by Cauchy Schwarz we have:

$$\begin{aligned} \vert \big (\Lambda ^{s-2} (a_3u_{xxx} ),\Lambda ^0_m w^2\Lambda ^su\big )\vert\le & {} \frac{1}{m}\big (\Vert a_3\partial _xu\Vert _{H^{s-2}}E^s(u) \\&+\Vert \partial _xa_3\Vert _{H^{s-1}}\Vert \partial _x u\Vert _{H^{s-1}}E^{s}(u) \\&+\Vert w^2a_3\Vert _{W^1{,\infty }}E^s(u)^2 \\&+\Vert \partial _x^2a_3\partial _xu\Vert _{H^{s-2}}E^s(u)+ \Vert a_3\partial _x^2u\Vert _{H^{s-2}}E^s(u)\big )\\\le & {} C(m^{-1},a_3,\Vert v\Vert _{H^s},\Vert w\Vert _{W^{1,\infty }})E^s(u)^2. \end{aligned}$$

\(\bullet \) Estimate of \(\big ([\Lambda ^{s-2},a_4]\partial _x^4u,\Lambda _m^0 w^2 \Lambda ^s u\big ) + \big (a_4\Lambda ^{s-2}\partial _x^4u,\Lambda _m^0 w^2 \Lambda ^s u\big ):\)

$$\begin{aligned} a_4\Lambda ^{s-2}\partial _x^4u= & {} a_4\Lambda ^{s-2}(1-\Lambda ^2)\partial ^2_xu= a_4(\Lambda ^{s-2}-\Lambda ^s)\partial ^2_xu\\&=a_4\Lambda ^{s-2}\partial ^2_xu - a_4\Lambda ^s\partial ^2_xu, \end{aligned}$$

then:

$$\begin{aligned} \big (a_4\Lambda ^{s-2}\partial _x^4u,\Lambda _m^0 w^2\Lambda ^s u\big )=\big (a_4\Lambda ^{s-2}\partial ^2_xu,\Lambda _m^0w^2\Lambda ^su\big )-\big (a_4\Lambda ^s\partial ^2_xu,\Lambda _m^0w^2 \Lambda ^s u\big ) \end{aligned}$$

By Cauchy Schwarz, the first term of the last equality is controlled by:

$$\begin{aligned} \vert \big (a_4\Lambda ^{s-2}\partial ^2_xu,\Lambda _m^0w^2\Lambda ^su\big )\vert \le \frac{1}{m} \vert a_4\Lambda ^{s-2}\partial ^2_xu\vert _{L^2}E^s(u)\le C(m^{-1},\vert a_4 \vert _{L^\infty })E^s(u)^2. \end{aligned}$$

\(\big (a_4\Lambda ^s\partial ^2_xu,\Lambda _m^0w^2 \Lambda ^s u\big )=-\big (a_4\Lambda _m^0w^2,(\partial _x\Lambda ^s u)^2\big )+ Q_1,\) where

$$\begin{aligned} \vert Q_1 \vert \le C(m,s,\vert w\vert _{W^{1,\infty }},\vert \partial _x a_4\vert _{L^\infty })E^{s}(u)^2. \end{aligned}$$

Now, using the first order Poisson brackets : (see [15] for more details)

$$\begin{aligned} \{\Lambda ^{s-2},a_4\}_{1}= -(s-2)\partial _x(a_4)\Lambda ^{s-2}\partial _x, \end{aligned}$$

we get:

$$\begin{aligned} ([\Lambda ^{s-2},a_4]\partial _x^4u,\Lambda _m^0w^2\Lambda ^su) =(s-2)(\partial _x(a_4)\Lambda ^s\partial _xu,\Lambda _m^0w^2\Lambda ^su)+ Q_2, \end{aligned}$$

Where

$$\begin{aligned} \vert Q_2\vert \le C(m,s,\vert w\vert _{W^{2,\infty }},\vert a_4\vert _{H^{s+1}})E^s(u)^2. \end{aligned}$$

Now, by integration by parts we have:

$$\begin{aligned} (s-2)(\partial _x(a_4)\Lambda ^s\partial _xu,\Lambda _m^0w^2\Lambda ^su)= -\frac{(s-2)}{2}(\partial _x(\partial _x(a_4)\Lambda _m^0w^2)\Lambda ^su,\Lambda ^su), \end{aligned}$$

then

$$\begin{aligned} \vert ([\Lambda ^{s-2},a_4]\partial _x^4u,\Lambda _m^0w^2\Lambda ^su)\vert \le C(m,s,\vert w\vert _{W^{2,\infty }},\vert a_4\vert _{H^{s+1}})E^s(u)^2. \end{aligned}$$

\(\bullet \) Estimate of \(\big ([\Lambda ^{s-2},a_5]\partial _x^5u,\Lambda _m^0w^2\Lambda ^su\big ) + \big (a_5\Lambda ^{s-2}\partial _x^5u,\Lambda _m^0w^2\Lambda ^su\big ):\)

$$\begin{aligned} a_5\Lambda ^{s-2}\partial _x^5u= & {} a_5\Lambda ^{s-2}(1-\Lambda ^2)\partial _x^3u=a_5\Lambda ^{s-2}\partial _x^3u-a_5\Lambda ^s\partial _x^3u=\\&a_5\Lambda ^{s-2}\partial _xu-a_5\Lambda ^{s}\partial _xu-a_5\Lambda ^s\partial _x^3u. \end{aligned}$$

Therefore,

$$\begin{aligned} \big (a_5\Lambda ^{s-2}\partial _x^5u,\Lambda _m^0w^2\Lambda ^su\big ) = \big (a_5\Lambda ^{s-2}\partial _xu,\Lambda _m^0w^2\Lambda ^su\big )-\big (a_5\Lambda ^{s}\partial _xu,\Lambda _m^0w^2\Lambda ^su\big )-\\ \qquad \qquad \big (a_5\Lambda ^s\partial _x^3u,\Lambda _m^0w^2\Lambda ^su\big ). \end{aligned}$$

The first two terms can be easily controlled by \(E^s(u)^2\) as above. Now,

$$\begin{aligned} \big (a_5\partial _x^3\Lambda ^s u,\Lambda _m^0w^2\Lambda ^s u\big )= & {} - \frac{1}{2}\big (\partial _x^3(a_5 \Lambda _m^0 w^2)\Lambda ^s u,\Lambda ^s u\big ) \\&- \frac{3}{2}\big (\partial _x^2(w^2\Lambda _m^0 a_5)\Lambda ^{s}\partial _x u,\Lambda ^s u\big )\\&-\frac{3}{2}\big (\partial _x(\Lambda _m^0 w^2a_5)\Lambda ^s u,\Lambda ^{s} \partial _x^2 u\big ). \end{aligned}$$

By integration by parts, we obtain

$$\begin{aligned} -\frac{3}{2}\big (\partial _x(\Lambda _m^0w^2a_5)\Lambda ^s u,\Lambda ^{s}\partial _x^2 u\big )= & {} \frac{3}{2} \big (\partial _x^2(\Lambda _m^0w^2a_5)\Lambda ^s u,\Lambda ^{s}\partial _x u\big ) \\&+\frac{3}{2} \big (\partial _x(a_5 \Lambda _m^0w^2),(\Lambda ^{s}\partial _x u)^2\big ). \end{aligned}$$

Now:

$$\begin{aligned}{}[\Lambda ^{s-2},a_5]\partial _x^5u=\{\Lambda ^{s-2},a_5\}_2\partial _x^5u+Q_3\partial _x^5u, \end{aligned}$$

where \(\{\cdot ,\cdot \}_2\) stands for the second order Poisson brackets,

$$\begin{aligned} \{\Lambda ^{s-2},a_5\}_2= & {} -(s-2)\partial _x(a_5)\Lambda ^{s-4}\partial _x+\frac{1}{2}[(s-2)\partial _x^2(a_5)\Lambda ^{s-4}\\&-(s-4)(s-2)\partial _x^2(a_5)\Lambda ^{s-6}\partial _x^2] \end{aligned}$$

and \(Q_3\) is an operator of order \(s-5\) that can be controlled by the general commutator estimates (see [15]). We thus get

$$\begin{aligned} \vert \big (Q_3\partial _x^5u,\Lambda _m^0w^2\Lambda ^s u\big )\vert \le C(m,\vert \partial _xa_5\vert _{H^{s+1}})E^s(u)^2. \end{aligned}$$

We now use the fact that \(H^1(\mathbb {R})\) is continuously embedded in \(L^{\infty }(\mathbb {R})\) to get

$$\begin{aligned}&\vert \big ([s\partial _x^2(a_5)\Lambda ^{s-4}- (s-4)(s-2)\partial _x^2(a_5)\Lambda ^{s-6}\partial _x^2]\partial _x^5u,\Lambda _m^0w^2\Lambda ^s u\big )\vert \\&\le C(m,s,\vert \partial _xa_5\vert _{H^{s+1}},\vert w\vert _{W^{1,\infty }})E^s(u)^2. \end{aligned}$$

This leads to the expression

$$\begin{aligned} \big ([\Lambda ^{s-2},a_5]\partial _x^5 u,\Lambda _m^0w^2\Lambda ^su\big )= -(s-2)\big (\partial _x (a_5) \Lambda ^{s}\partial _x^2 u,\Lambda _m^0w^2\Lambda ^s u\big )+Q_4, \end{aligned}$$

where \(\vert Q_4 \vert \le C(m,s,\vert w\vert _{W^{1,\infty }}, \vert a_5\vert _{H^{s+1}} )E^s(u)^2\). Remarking now, by integration by parts

$$\begin{aligned} -(s-2)\big (\partial _x(a_5)\Lambda ^{s}\partial _x^2 u ,\Lambda _m^0w^2\Lambda ^s u\big )&=(s-2)\big (\partial _x (\partial _x(a_5)\Lambda _m^0w^2) \Lambda ^{s} \partial _xu,\Lambda ^su\big )\nonumber \\&\quad +(s-2)\big (\partial _x (a_3) \Lambda _m^0w^2,(\Lambda ^s \partial _x u)^2\big ).\nonumber \\ \end{aligned}$$
(2.7)

We now choose w such that

$$\begin{aligned}&-(s-2)\big (\partial _x (a_5)\Lambda _m^0 w^2,(\Lambda ^s \partial _x u)^2\big )+\frac{3}{2} \big (\partial _x(a_5\Lambda _m^0w^2),(\Lambda ^{s}\partial _x u)^2\big )\nonumber \\&+ \big (a_4\Lambda _m^0w^2,(\partial _x\Lambda ^s u)^2\big )=0; \end{aligned}$$
(2.8)

therefore, if we take \(w=(\Lambda _m^0)^{-1}\Big (\vert a_5\vert ^ {\big ({\frac{2s-7}{6}\big )}}\displaystyle {\exp (-\frac{1}{3}\int _0^x\frac{a_4}{a_5}dy)}\Big )\) we easily obtain (2.8). Finally, one has

$$\begin{aligned}&\big ([\Lambda ^{s-2},a_5]\partial _x^5 u,\Lambda _m^0w^2\Lambda ^s u\big )+\big (a_5\partial _x^5\Lambda ^{s-2} u,\Lambda _m^0w^2\Lambda ^s u\big )\\&\quad =Q_4+(s-2)\big (\partial _x (\partial _x(a_5)\Lambda _m^0w^2) \Lambda ^s\partial _x u,\Lambda ^s u\big ) -\frac{1}{2}\big (\partial _x^3(a_5\Lambda _m^0w^2)\Lambda ^s u,\Lambda ^s u\big )\\&\qquad - \frac{3}{2}\big (\partial _x^2(a_5\Lambda _m^0w^2)\Lambda ^{s}\partial _x u,\Lambda ^s u\big )+ \frac{3}{2} \big (\partial _x^2(a_5\Lambda ^0_mw^2)\Lambda ^{s}\partial _x u,\Lambda ^su\big ); \end{aligned}$$

therefore,

$$\begin{aligned}&\vert \big ([\Lambda ^{s-2},a_5]\partial _x^5 u,\Lambda _m^0w^2\Lambda ^s u\big )+\big (a_5\partial _x^5\Lambda ^{s-2} u,\Lambda _m^0w^2\Lambda ^s u\big )\vert \\&\quad \le C(s,m,\vert \partial _xa_5\vert _{H^{s+1}}) E^s(u)^2. \end{aligned}$$

\(\bullet \) Estimate of \(\big (w_t\Lambda ^{s-2}u ,\Lambda _m^0w\Lambda ^su\big )\): Using the Cauchy-Schwarz inequality we obtain

$$\begin{aligned} \vert \big (w_t\Lambda ^su ,w\Lambda ^su\big )\vert \le C(m,\vert w_t\vert _{L^{\infty }},\vert w\vert _{L^{\infty }}) E^s(u)^2. \end{aligned}$$

Gathering the information provided by the above estimates, since one has

$$\begin{aligned} \vert \big (\Lambda ^{s-2}f ,\Lambda _m^0w^2\Lambda ^su\big )\vert \le \frac{1}{m}E^s(f)E^s(u). \end{aligned}$$

If we assemble the previous estimates and using Gronwall’s lemma we obtain the following estimate:

$$\begin{aligned} e^{\lambda t}\partial _t (e^{-\lambda t}E^s(u)^2) \le \big (C(E^s(v))-\lambda \big )E^s(u)^2+2 E^s(f)E^s(u). \end{aligned}$$

Taking \(\lambda =\lambda _T\) large enough (how large depends on \(\sup _{t\in [0,T]}C(E^s(v(t))\) for the first term of the right hand side of the above inequality to be negative for all \(t\in [0,T]\), we deduce that

$$\begin{aligned} E^s(u(t)) \le e^{\lambda _T t}E^s(u^0) +2 \int _0^t e^{\lambda _T (t-t')} E^s(f(t'))dt'. \end{aligned}$$

2.2 Proof of the theorem

Thanks to this energy estimate, we classically conclude (see e.g. [16]) the existence of a time

$$\begin{aligned} T^*=T^*(E^s(u^0))>0, \end{aligned}$$

and a unique solution \(u\in C([0,T^*];H^{s}(\mathbb {R}))\cap C^1([0,T^*];H^{s-3}(\mathbb {R}))\) to (1.1) as the limit of the iterative scheme

$$\begin{aligned} u_0=u^0,\quad \hbox { and }\quad \forall n\in \mathbb {N}, \quad \left\{ \begin{array}{l} {\mathcal L}(u^n,\partial )u^{n+1}=f,\\ u^{n+1}_{\vert _{t=0}}=u^0. \end{array}\right. \end{aligned}$$