Initial-Boundary Value Problems for Nonlinear Dispersive Equations of Higher Orders Posed on Bounded Intervals with General Boundary Conditions

Abstract

The present article concerns general mixed problems for nonlinear dispersive equations of any odd-orders posed on bounded intervals. The results on existence, uniqueness and exponential decay of solutions are presented.

1. Introduction

In this work, we formulate mixed problems with general boundary conditions for the following dispersive equation:

$$u_t+\sum _{j=1}^l{(-1)}^{j+1}{D}_x^{2j+1}u+u{D}_{x}u=0,x\in (0,L);t>0,$$

(1)

where L is an arbitrary real positive number and $l\in \mathbb{N}$. We propose Equation (1) because it includes classical models such as the Korteweg-de Vries (KdV) equation, when $l=1$ [1,2,3,4] and the Kawahara equation, when $l=2$ [5,6,7,8]. Dispersive equations posed on bounded and unbounded intervals with the Dirichlet type boundary conditions were studied in [9,10,11,12,13,14,15,16,17,18,19]. It is known that the KdV and Kawahara equations were deduced on the whole real line, however, approximating the line either by bounded or unbounded intervals, one needs to consider initial-boundary value problems posed either on finite or semi-finite intervals [2,4,9,10,11,13,14,15,17,18,19,20,21,22].

Last years, publications on dispersive equations of higher orders appeared [14,16,23,24,25,26]. Usually, Dirichlet conditions such as $D_x^{i}u(t,0)=D_x^{i}u(t,L)=D_x^{l}u(t,L)=0$, $i=0,\dots ,l-1;$$t>0$ were imposed for Equation (1), see [25,26]. In [27], general mixed problems for linear multi-dimensional $(2b+1)$-hyperbolic equations were studied by means of functional analisys methods. In [28], we have studied boundary value problems for the following linear stationary dispersive equations on bounded intervals subject to general boundary conditions at the endpoints of intervals:

$$\lambda u+\sum _{j=1}^l{(-1)}^{j+1}{D}_x^{2j+1}u=f\left(x\right),x\in (0,L);l\in \mathbb{N},$$

(2)

where $\lambda$ > 0 and f is a given function. Equation (2) appears while solving Equation (1) making use of either the semigroup theory or semi-discrete approaches [13]. We formulate well posed initial-boundary value problems to Equation (1) imposing the same boundary conditions as for Equation (2) [28].

Our goal is to prove the existence, uniqueness of local and global regular solutions for the formulated problems as well as exponential decay for small initial data.

This article has the following structure: Section 2 contains notations and preliminaries. In Section 3, we formulate the initial-boundary value problems. In Section 4, we prove local existence and uniqueness of regular solutions as well as a “smoothing effect” of them similar to one established in [29] for the initial problem of the KdV equation. In Section 5, the global existence and uniqueness of regular solutions have been established for arbitrary initial data. In Section 6, the existence and uniqueness of small global regular solutions as well as their exponential decay have been established. Section 7 is a conclusion.

2. Notations and Auxiliary Facts

For $x\in (0,L)$, symbols $D^i=D_x^i=\frac{{\partial }^i}{\partial x^i},i\in \mathbb{N};D=D^1$ denote the partial derivatives of order i. By ${\parallel ·\parallel }_{\infty }$ we denote the norm in $L^{\infty }(0,L)$. In what follows, we denote by $(·,·)$ and $\parallel ·\parallel$ as the inner product and the norm in $L^2(0,L)$ and ${\parallel ·\parallel }_{H^m}$, $m\in \mathbb{N}$ stands for the norm in $L^2$-based Sobolev spaces [30].

Lemma 1

(See [26], Lemma 2.2). Let u belong to $H_0^1(0,L)$, then the following inequality holds:

$${\parallel u\parallel }_{\infty }\le \sqrt{2}{\parallel Du\parallel }^{\frac{1}{2}}{\parallel u\parallel }^{\frac{1}{2}}.$$

(3)

Lemma 2

(See [31], p. 125). Suppose u and $D^{m}u$, $m\in \mathbb{N}$ belong to $L^2(0,L)$. Then for the derivatives $D^{i}u$, $0\le i<m$, the following inequality holds:

$$\parallel D^{i}u\parallel \le C_1\parallel D^m{u\parallel }^{\frac{i}{m}}{\parallel u\parallel }^{1-\frac{i}{m}}+C_2\parallel u\parallel ,$$

(4)

where $C_1$, $C_2$ are constants depending only on L, m, i.

Lemma 3

(See [32]). Let u belong to $H_0^1(0,L)$, then

$$\parallel u\parallel \le \frac{L}{\pi }\parallel Du\parallel .$$

(5)

3. Formulation of the Problem

Consider the following evolution equation:

$$u_t+\sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}u+uDu=0,x\in (0,L);t>0$$

(6)

subject to initial data

$$\begin{array}{cc}\hfill & u(0,x)=u_0\left(x\right),x\in (0,L),\hfill \end{array}$$

(7)

where $u_0$ is a given function. In [28], formulation of boundary value problems for the stationary linear equation Equation (2) on the interval $(0,L)$ has been proposed. In the present work, we will use the same formulation for Equations (6) and (7):

$\mathbf{l}=\mathbf{1}$:

$$\begin{array}{c}\hfill u(t,0)=u(t,L)=Du(t,L)=0,t>0,\end{array}$$

(8)

$\mathbf{l}\ge \mathbf{2}$:

$$\begin{array}{cc}\hfill & u(t,0)=u(t,L)=0,t>0,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & D^{i}u(t,0)=\sum _{j=1}^l{a}_{ij}{D}^{j}u(t,0),i=l+1,\dots ,2l-1;t>0,\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill & D^{i}u(t,L)=\sum _{j=1}^{l-1}{b}_{ij}{D}^{j}u(t,L),i=l,\dots ,2l-1;t>0,\hfill \end{array}$$

(11)

where $a_{ij}$, $b_{ij}$ are real constants. Assumptions on the coefficients imply that the $L^2$-norm of the solutions of Equation (6) is decreasing. Multiplying Equation (6) by u and integrating over $(0,L)$, we get

$$\frac{1}{2}\frac{d}{dt}{\parallel u\parallel }^2\left(t\right)+\sum _{j=1}^l{(-1)}^{j+1}(D^{2j+1}u,u)\left(t\right)=0.$$

A way to obtain $\frac{d}{dt}{\parallel u\parallel }^2\left(t\right)\le 0,t>0$ is to choose $a_{ij}$, $b_{ij}$ such that ${\sum }_{j=1}^l{(-1)}^{j+1}(D^{2j+1}u,u)\left(t\right)\ge 0,t>0$. Making use of integration by parts, finite induction and Young’s inequality, we prove that the coefficients $a_{ij}$, $b_{ij}$ satisfy the following conditions, see [28]:

For $\mathbf{l}=\mathbf{2}$:

$$\begin{array}{cc}\hfill & B_1=b_{31}-\frac{1}{2}-\frac{b_{21}^2}{2}>0,A_1=-a_{31}+\frac{1}{2}-a_{32}^2>0,A_2=\frac{1}{4}.\hfill \end{array}$$

(12)

This implies that $b_{31}>\frac{1}{2}$, $a_{31}<\frac{1}{2}$, and $|a_{32}|,|b_{21}|$ should be sufficiently small or zero.

For $\mathbf{l}=\mathbf{3}$:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & B_1=b_{31}-b_{51}-\frac{1}{2}-b_{31}^2-\frac{1}{2}\left(\right|b_{32}|+|b_{52}|+|b_{41}\left|\right)>0,\hfill \\ & B_2=b_{42}-\frac{1}{2}-b_{32}^2-\frac{1}{2}\left(\right|b_{32}|+|b_{52}|+|b_{41}\left|\right)>0,\hfill \\ & A_1=a_{51}-\frac{1}{2}-\frac{1}{2}\left(\right|a_{52}|+|a_{41}|+|a_{53}\left|\right)>0,\hfill \\ & A_2=-a_{42}+\frac{1}{2}-\frac{1}{2}\left(\right|a_{52}|+|a_{41}|+|a_{43}\left|\right)>0,\hfill \\ & A_3=\frac{1}{4}-\frac{1}{2}\left(\right|a_{53}|+|a_{43}\left|\right)>0.\hfill \end{array}\end{array}$$

(13)

This implies that $b_{51}<-\frac{1}{2}$, $b_{42}>\frac{1}{2}$, $a_{51}>\frac{1}{2}$, $a_{42}<\frac{1}{2}$ and the remaining coefficients in Inequality (13) should be sufficiently small or zero.

For $\mathbf{l}\ge \mathbf{4}$:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & B_i=\sum _{\begin{array}{c}k=1\\ 2k+i\ge l\end{array}}^{l-i}{(-1)}^{k+1}{b}_{2k+i,i}+(2-l)+\frac{(1-l)}{2}{b}_{li}^2-\frac{1}{2}\sum _{\begin{array}{c}j=1\\ j\ne i\end{array}}^{l-1}{\left(\sum _{\begin{array}{c}k=1\\ 2k+i\ge l\end{array}}^{l-i}\left|b_{2k+i,j}\right|\right)}^2>0,i=1,\dots ,l-1,\hfill \\ \hfill & A_i=\sum _{\begin{array}{c}k=1\\ 2k+i\ge l+1\end{array}}^{l-i}{(-1)}^k{a}_{2k+i,i}+(5-2l)-\frac{1}{2}\sum _{\begin{array}{c}j=1\\ j\ne i\end{array}}^{l-1}{\left(\sum _{\begin{array}{c}k=1\\ 2k+i\ge l+1\end{array}}^{l-i}\left|a_{2k+i,j}\right|\right)}^2-\frac{1}{2}\sum _{\begin{array}{c}k=1\\ 2k+i\ge l+1\end{array}}^{l-i}\left|a_{2k+i,l}\right|>0,i=1,\dots ,l-1,\hfill \\ & A_l=\frac{1}{4}-\frac{1}{2}\sum _{i=1}^{l-1}\left(\sum _{\begin{array}{c}k=1\\ 2k+i\ge l+1\end{array}}^{l-i}\left|a_{2k+i,l}\right|\right)>0.\hfill \end{array}\end{array}$$

(14)

It follows that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & b_{l+1,l-1}>l-2,b_{l+j,l-j}>\frac{1}{2}{\left(\sum _{m=1}^{\frac{j-1}{2}}\left|b_{l+2m-1,l-2m+1}\right|\right)}^2+l-2,\underset{\left(j\mathrm{odd}\right)}{\underbrace{j=3,\dots ,l-1}},\hfill \\ \hfill & b_{l+2,l-2}<2-l,b_{l+j,l-j}<-\frac{1}{2}{\left(\sum _{m=1}^{\frac{j}{2}-1}\left|b_{l+2m,l-2m}\right|\right)}^2+2-l,\underset{\left(j\mathrm{even}\right)}{\underbrace{j=4,\dots ,l-1}},\hfill \\ \hfill & a_{l+1,l-1}<5-2l,a_{l+j,l-j}<-\frac{1}{2}{\left(\sum _{m=1}^{\frac{j-1}{2}}\left|a_{l+2m-1,l-2m+1}\right|\right)}^2+5-2l,\underset{\left(j\mathrm{odd}\right)}{\underbrace{j=3,\dots ,l-1}},\hfill \\ \hfill & a_{l+2,l-2}>2l-5,a_{l+j,l-j}>\frac{1}{2}{\left(\sum _{m=1}^{\frac{j}{2}-1}\left|a_{l+2m,l-2m}\right|\right)}^2+2l-5,\underset{\left(j\mathrm{even}\right)}{\underbrace{j=4,\dots ,l-1}}\hfill \end{array}\end{array}$$

(15)

and the remaining coefficients of the Inequality (14) should be sufficiently small or zero.

Assuming these coefficients equal to zero in Inequalities (12)–(14), we get the following boundary conditions for all $l\in \mathbb{N}$, [28]:

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & u(t,0)=u(t,L)=D^{l}u(t,L)=0,t>0,\hfill \\ \hfill & D^{l+j}u(t,0)=a_{l+j,l-j}{D}^{l-j}u(t,0),j=1,\dots ,l-1;t>0,\hfill \\ \hfill & D^{l+j}u(t,L)=b_{l+j,l-j}{D}^{l-j}u(t,L),j=1,\dots ,l-1;t>0\hfill \end{array}\end{array}$$

(16)

with $b_{31}>\frac{1}{2}$, $a_{31}<\frac{1}{2}$ for $l=2$; $b_{51}<-\frac{1}{2}$, $b_{42}>\frac{1}{2}$, $a_{51}>\frac{1}{2}$, $a_{42}<\frac{1}{2}$ for $l=3$ and Inequality (15) for $l\ge 4$.

Remark 1

(See [28], Remark 1). We call (10) and (11) general boundary conditions because they follow from a more general form:

$$\begin{array}{cc}\hfill & \sum _{i=1}^{2l-1}{\alpha }_{ki}{D}^{i}u(t,0)=0,k=1,\dots ,l-1;t>0,\hfill \\ \hfill & \sum _{i=1}^{2l-1}{\beta }_{ki}{D}^{i}u(t,L)=0,k=1,\dots ,l;t>0,\hfill \end{array}$$

where ${\alpha }_{ki},{\beta }_{ki}$ are real numbers.

Remark 2.

In this work, we will study the case $l\ge 2$. For the case $l=1$ see [26].

4. Local Regular Solutions

Let T be a real positive number and $Q_T=(0,T)\times (0,L)$. Consider the linear evolution equation

$$u_t+\sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}u=g(t,x)\mathrm{in}{Q}_T$$

(17)

subject to initial-boundary conditions Equations (7) and (16), with the coefficients satisfying $b_{31}>\frac{1}{2}$, $a_{31}<\frac{1}{2}$ for $l=2$; $b_{51}<-\frac{1}{2}$, $b_{42}>\frac{1}{2}$, $a_{51}>\frac{1}{2}$, $a_{42}<\frac{1}{2}$ for $l=3$ and Inequality (15) for $l\ge 4$, where g is a given function. Define the linear operator in $L^2(0,L)$:

$$Au\equiv \sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}u;$$

$$D\left(A\right)\equiv \{u\in H^{2l+1}(0,L):u\mathrm{satisfies} \mathrm{the} \mathrm{boundary} \mathrm{conditions}Equation\left(16\right)\}.$$

Theorem 1

(See [28], Theorem 4.1). Let $f\in L^2(0,L)$. Then for all $\lambda >0$ a stationary equation: $\lambda u+Au=f\left(x\right),x\in (0,L)$ subject to boundary conditions Equation (16) (omitting t) admits a unique regular solution $u=u\left(x\right)\in H^{2l+1}(0,L)$ satisfying

$${\parallel u\parallel }_{H^{2l+1}}\le C\parallel f\parallel$$

(18)

where C is a constant depending on L, l, λ, $a_{l+j,l-j}$, $b_{l+j,l-j},j=1,\dots ,l-1$.

Theorem 2.

Let $T>0$, $u_0\in D\left(A\right)$ and $g\in H^1(0,T;L^2(0,L))$ be given. Then, problem Equations (7), (16) and (17) has a unique solution $u=u(t,x)$:

$$u\in C([0,T];D\left(A\right))\cap C^1([0,T];L^2(0,L)).$$

Proof. 

Due to Theorem 1, the operator $\lambda I+A$ is surjective for all $\lambda >0$. On the other hand, by in [28], (33), we obtain

$$(Au,u)\ge \sum _{i=1}^{l-1}{B}_i{\left(D^{i}u\left(L\right)\right)}^2+\sum _{i=1}^l{A}_i{\left(D^{i}u\left(0\right)\right)}^2\ge 0\mathrm{for} \mathrm{all}u\in D\left(A\right).$$

(19)

By the semigroup theory, the result is proven. (See [33], Lemma 2.2.3 and Corollary 2.4.2) □

Theorem 3.

Let $u_0\in D\left(A\right)$. Then there exists a real $T_0\in (0,T)$ such that Equations (6), (7) and (16) has a unique regular solution $u=u(t,x)$:

$$\begin{array}{cc}& u\in L^{\infty }(0,T_0;H^{2l+1}(0,L))\cap L^2(0,T_0;H^{(2l+1)+l}(0,L));\hfill \\ & u_t\in L^{\infty }(0,T_0;L^2(0,L))\cap L^2(0,T_0;H^l(0,L)).\hfill \end{array}$$

Proof. 

Define $g=-vDv$; $v,v_t\in L^{\infty }(0,T;L^2(0,L))\cap L^2(0,T;H^l(0,L))$. Making use of Inequality (3), one can see that $g\in H^1(0,T;L^2(0,L))$, then by Theorem 2, we can define an operator P related to Equations (7), (16) and (17) such that $v\mapsto u=Pv$. Define the Banach space

$$E=\{v(t,x):v,v_t\in L^{\infty }(0,T;L^2(0,L))\cap L^2(0,T;H^l(0,L));v(0,·)\equiv u_0\}$$

with the norm

$${\parallel v\parallel }_E^2=\underset{t\in (0,T)}{ess\; sup}{\{\parallel v\parallel }^2\left(t\right)+\parallel v_t{\parallel }^2\left(t\right)\}+{\int }_0^T\sum _{j=1}^l\left[\parallel D^j{v\parallel }^2\left(t\right)+{\parallel D^j{v}_t\parallel }^2\left(t\right)\right]dt$$

and consider $1<R<+\infty$ such that

$$(1+L)(1+l)\left(2\parallel u_0{\parallel }_{H^l}^4+{\parallel u_0\parallel }_{H^{2l+1}}^2\right)\frac{M_2}{M_1}\le R^2$$

(20)

and a ball $B_R={\{v\in E:\parallel v\parallel }_E^2\le 8R^2\}$. Here, $M_1=\underset{i\in \{1,\dots ,l-1\}}{min}\{B_i,A_i,A_l\}$ and $M_2$ is the maximum among the coefficients of the derivatives ${\left(D^{l}u(t,0)\right)}^2$, ${\left(D^{i}u(t,0)\right)}^2$, ${\left(D^{i}u(t,L)\right)}^2$, $i=1,\dots ,l-1;t>0$ (see Inequality (19) and [28], p. 389).

Remark 3.

Note that by Inequalities (12)–(14), $A_l=\frac{1}{4}$ for all $l\ge 2$, therefore $M_1\le \frac{1}{4}$. On the other hand, $1\le M_2<+\infty$ for all $l\ge 2$. This provides that $\frac{M_1}{M_2}\le \frac{1}{4}$ for all $l\ge 2$.

Lemma 4.

There is a real $T_{*}>0$ such that $P\left(B_R\right)\subset B_R$.

Proof. 

Let $v\in B_R$, then due to Inequality (20), we get

$${\parallel Dv\parallel }^2\left(t\right)\le {\parallel D{u}_0\parallel }^2+{\int }_0^T\left[{\parallel Dv\parallel }^2\left(s\right)+{\parallel D{v}_s\parallel }^2\left(s\right)\right]ds\le \left(\frac{M_1}{M_2}+8\right)R^2\le 9R^2,t\in (0,T).$$

(21)

Making use of Inequalities (3) and (21), we find

$${\parallel v\parallel }_{\infty }^2\left(t\right)\le \left(34\right)R^2,\parallel v_t{\parallel }_{\infty }^2\left(t\right)\le 2[8R^2+\parallel D{v}_t{\parallel }^2\left(t\right)],t\in (0,T).$$

(22)

Estimate 1.

Multiplying Equation (17) by $2u$, integrating over $(0,L)$ and making use of Inequalities (21) and (22), we obtain

$$\begin{array}{cc}\hfill & \frac{d}{dt}{\parallel u\parallel }^2\left(t\right)+2M_{1}U(t,0,L)\le -2(vDv,u)\left(t\right)\le {\parallel vDv\parallel }^2\left(t\right)+{\parallel u\parallel }^2\left(t\right)\hfill \\ & \le {\parallel v\parallel }_{\infty }^2{\left(t\right)\parallel Dv\parallel }^2\left(t\right)+{\parallel u\parallel }^2\left(t\right)\le C_1+{\parallel u\parallel }^2\left(t\right),t\in (0,T),\hfill \end{array}$$

(23)

where $U(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^{i}u(t,L)\right)}^2+{\left(D^{i}u(t,0)\right)}^2\right]+{\left(D^{l}u(t,0)\right)}^2$ and $C_1=9\left(34\right)R^4$. By the Gronwall Lemma and Inequality (20),

$${\parallel u\parallel }^2\left(t\right)\le e^T\left(\frac{R^2}{2}+C_{1}T\right),t\in (0,T).$$

For $T_1=min\left\{ln2,\frac{2M_1{R}^2}{C_1{M}_2},\frac{2M_1{R}^2}{C_1{M}_{2}L}\right\}$, we find

$${\parallel u\parallel }^2\left(t\right)\le 2R^2,t\in (0,T_1).$$

(24)

Substituting Inequality (24) into Inequality (23), integrating the result over $(0,T_1)$ and making use of Inequality (20), we get

$$\begin{array}{cc}\hfill & {\int }_0^{T_1}U(t,0,L)dt\le \frac{1}{2M_1}\left[3\frac{M_1}{M_2}{R}^2+\frac{M_1}{M_2}{R}^2\right]=\frac{2R^2}{M_2}.\hfill \end{array}$$

(25)

Estimate 2.

Multiplying Equation (17) by $2xu$ and making use of Inequalities (21), (22) and (24), we obtain

$$\begin{array}{cc}\hfill & \frac{d}{dt}(x,u^2)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^{i}u(t,L)\right)}^2-2M_{2}U(t,0,L)+\sum _{j=1}^l(2j+1){\parallel D^{j}u\parallel }^2\left(t\right)\hfill \\ & \le L\left[{\parallel vDv\parallel }^2\left(t\right)+{\parallel u\parallel }^2\left(t\right)\right]\le L(C_1+2R^2),t\in (0,T_1).\hfill \end{array}$$

(26)

Integrating Inequality (26) over $(0,T_1)$ and making use of Inequalities (20) and (25), we conclude

$$\begin{array}{c}\hfill {\int }_0^{T_1}\sum _{j=1}^l{\parallel D^{j}u\parallel }^2\left(t\right)dt\le 2R^2.\end{array}$$

(27)

Estimate 3.

Differentiating Equation (17) with respect to t, multiplying the result by $2u_t$ and making use of Inequalities (21) and (22), one gets for an arbitrary $\epsilon >0$

$$\begin{array}{cc}\hfill & \frac{d}{dt}\parallel u_t{\parallel }^2\left(t\right)+2M_1{U}_t(t,0,L)\le \epsilon \left(\parallel v_t{Dv\parallel }^2\left(t\right)+{\parallel vD{v}_t\parallel }^2\left(t\right)\right)+\frac{2}{\epsilon }{\parallel u_t\parallel }^2\left(t\right)\hfill \\ & \le \epsilon C_2\left(1+\parallel D{v}_t{\parallel }^2\left(t\right)\right)+\frac{2}{\epsilon }{\parallel u_t\parallel }^2\left(t\right),t\in (0,T),\hfill \end{array}$$

where $U_t(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^i{u}_t(t,L)\right)}^2+{\left(D^i{u}_t(t,0)\right)}^2\right]+{\left(D^l{u}_t(t,0)\right)}^2$ and $C_2=C_2\left(R\right)$ is a fixed positive constant. Taking $\epsilon =\frac{M_1}{\left(16\right)M_2{C}_2}$, we reduce it to the inequality

$$\frac{d}{dt}\parallel u_t{\parallel }^2\left(t\right)+2M_1{U}_t(t,0,L)\le \frac{\left(32\right)M_2{C}_2}{M_1}{\parallel u_t\parallel }^2\left(t\right)+\frac{M_1}{\left(16\right)M_2}\left(1+\parallel D{v}_t{\parallel }^2\left(t\right)\right),t\in (0,T).$$

(28)

By the Gronwall Lemma,

$$\begin{array}{c}\hfill \parallel u_t{\parallel }^2\left(t\right)\le e^{\frac{\left(32\right)M_2{C}_2}{M_1}T}\left[\parallel u_t{\parallel }^2\left(0\right)+\frac{M_1}{\left(16\right)M_2}{\int }_0^T\left(1+\parallel D{v}_t{\parallel }^2\left(t\right)\right)dt\right].\end{array}$$

Due to Inequalities (3) and (20),

$$\parallel u_t{\parallel }^2\left(0\right)\le (1+l)\left(\parallel u_{0}D{u}_0{\parallel }^2+{\parallel u_0\parallel }_{H^{2l+1}}^2\right)\le (1+l)\left(2\parallel u_0{\parallel }_{H^l}^4+{\parallel u_0\parallel }_{H^{2l+1}}^2\right)\le \frac{M_1}{(1+L)M_2}{R}^2\le \frac{R^2}{4}.$$

Choosing $T_2=min\left\{\frac{M_{1}ln2}{\left(32\right)M_2{C}_2},\frac{M_1^2}{\left(64\right)M_2^2{C}_2},\frac{3M_1}{8\left(16\right)M_2{C}_2{L}^2}\right\}$, we find

$$\parallel u_t{\parallel }^2\left(t\right)\le 2R^2,t\in (0,T_2).$$

(29)

Substituting Inequality (29) into Inequality (28), integrating the result over $(0,T_2)$ and making use of Inequality (20), we get

$$\begin{array}{cc}\hfill & {\int }_0^{T_2}{U}_t(t,0,L)dt\le \frac{1}{2M_1}\left[\frac{M_1}{M_2}{R}^2+2\frac{M_1}{M_2}{R}^2+\frac{M_1}{M_2}{R}^2\right]=\frac{2R^2}{M_2}.\hfill \end{array}$$

(30)

Estimate 4.

Differentiating Equation (17) with respect to t, multiplying the result by $2x{u}_t$ and making use of Inequalities (21), (22) and (29), we obtain

$$\begin{array}{cc}\hfill & \frac{d}{dt}(x,u_t^2)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^i{u}_t(t,L)\right)}^2-2M_2{U}_t(t,0,L)+\sum _{j=1}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)\hfill \\ & \le \epsilon L\left(\parallel v_t{Dv\parallel }^2\left(t\right)+{\parallel vD{v}_t\parallel }^2\left(t\right)\right)+\frac{2L}{\epsilon }{\parallel u_t\parallel }^2\left(t\right)\le \epsilon L{C}_2\left(1+\parallel D{v}_t{\parallel }^2\left(t\right)\right)+\frac{4L}{\epsilon }{R}^2,t\in (0,T_2).\hfill \end{array}$$

Taking $\epsilon =\frac{M_1}{\left(16\right)M_2{C}_{2}L}$, integrating over $(0,T_2)$, and making use of Inequalities (20) and (30), we find

$$\begin{array}{c}\hfill {\int }_0^{T_2}\sum _{j=1}^l{\parallel D^j{u}_t\parallel }^2\left(t\right)dt\le 2R^2.\end{array}$$

(31)

For $T_{*}=min\{T_1,T_2\}$, it follows from Inequalities (24), (27), (29) and (31) that $Pv=u\in B_R$. This completes the proof of Lemma 4. □

Lemma 5.

There is a real $T_{\star }>0$ such that the mapping P is a contraction in $B_R$.

Proof. 

For $v_1,v_2\in B_R$, denote $u_i=P{v}_i,i=1,2$, $w=v_1-v_2$, $z=u_1-u_2$ and $Z(t,0,L)\equiv {\sum }_{i=1}^{l-1}[{\left(D^{i}z(t,L)\right)}^2+{\left(D^{i}z(t,0)\right)}^2]+{\left(D^{l}z(t,0)\right)}^2$. Then z satisfies the equation

$$z_t+\sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}z=-v_{1}Dw-wD{v}_{2}in{Q}_T,$$

(32)

boundary conditions Equation (16) and initial data $z(0,·)\equiv 0$.

Similar arguments used in the proof of Lemma 4 show that ${\parallel z\parallel }_E\le \frac{1}{2}{\parallel w\parallel }_E$. Therefore, P is a contraction in $B_R$.  □

According to Lemmas 4 and 5 and the Banach Fixed Point Theorem with $T_0=min\{T_{*},T_{\star }\}$, problem Equations (6), (7) and (16) has a unique generalized solution $u=u(t,x)$:

$$\begin{array}{cc}\hfill & u\in L^{\infty }(0,T_0;H^l(0,L));\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & u_t\in L^{\infty }(0,T_0;L^2(0,L))\cap L^2(0,T_0;H^l(0,L)).\hfill \end{array}$$

(34)

Rewrite Equation (6) in the form

$$u+\sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}u=u-u_t-uDu:=F(t,x).$$

(35)

Due to Relations (33) and (34), it follows that $uDu\in H^1(0,T_0;L^2(0,L))$, hence $F\in L^{\infty }(0,T_0;L^2(0,L))$. Making use of Inequality (18), we get

$$u\in L^{\infty }(0,T_0;H^{2l+1}(0,L)).$$

(36)

Acting as in [26], Lemma 4.3, we find

$$u\in L^2(0,T_0;H^{(2l+1)+l}(0,L)).$$

(37)

Combining Relations (34), (36) and (37), we complete the proof of Theorem 3. □

Remark 4.

The local result presented in Theorem 3 can be obtained under the following boundary conditions:

$$\begin{array}{cc}\hfill & D^{i}u(t,0)=\sum _{j=0}^l{a}_{ij}{D}^{j}u(t,0),i=l+1,\dots ,2l;t>0,\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill & D^{i}u(t,L)=\sum _{j=0}^{l-1}{b}_{ij}{D}^{j}u(t,L),i=l,\dots ,2l;t>0\hfill \end{array}$$

(39)

instead of Equations (8)–(11) (see [28], Remark 3). We also call Equations (38) and (39) general boundary conditions because they follow from a more general form:

$$\begin{array}{cc}\hfill & \sum _{i=0}^{2l}{\alpha }_{ki}{D}^{i}u(t,0)=0,k=1,\dots ,l;t>0\hfill \\ \hfill & \sum _{i=0}^{2l}{\beta }_{ki}{D}^{i}u(t,L)=0,k=1,\dots ,l+1;t>0,\hfill \end{array}$$

where ${\alpha }_{ki},{\beta }_{ki}$ are real numbers (see Remark 1).

5. Global Regular Solutions

Theorem 4.

Let $u_0\in H^{2l+1}(0,L)$ satisfying Equation (16). Then for all $T>0$, problem Equations (6), (7) and (16) has a unique regular solution $u=u(t,x)$:

$$\begin{array}{cc}& u\in L^{\infty }(0,T;H^{2l+1}(0,L))\cap L^2(0,T;H^{(2l+1)+l}(0,L));\hfill \\ & u_t\in L^{\infty }(0,T;L^2(0,L))\cap L^2(0,T;H^l(0,L)).\hfill \end{array}$$

Proof. 

We will obtain a priori estimates independent of $t\in (0,T)$.

Estimate 1.

Multiplying Equation (6) by $2u$, we obtain

$$2(u_t,u)\left(t\right)+2M_{1}U(t,0,L)\le 0,$$

(40)

where $U(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^{i}u(t,L)\right)}^2+{\left(D^{i}u(t,0)\right)}^2\right]+{\left(D^{l}u(t,0)\right)}^2$ and $M_1=\underset{i\in \{1,\dots ,l-1\}}{min}\{B_i,A_i,A_l\}$.

Consequently,

$$\parallel u\parallel \left(t\right)\le \parallel u_0\parallel ,t\in (0,T).$$

(41)

Estimate 2.

Multiplying Equation (6) by $2xu$, we get

$$2(u_t,xu)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^{i}u(t,L)\right)}^2-2M_{2}U(t,0,L)+\sum _{j=1}^l(2j+1){\parallel D^{j}u\parallel }^2\left(t\right)+2(uDu,xu)\left(t\right)\le 0,$$

(42)

where $M_2$ is calculated in [28], p. 389. Making use of Inequalities (3) and (41), we estimate

$$\begin{array}{cc}\hfill & 2(uDu,xu)\left(t\right)=-\frac{2}{3}(u,u^2)\left(t\right)\ge -\frac{2\sqrt{2}}{3}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right){\parallel u\parallel }^{\frac{1}{2}}\left(t\right){\parallel u\parallel }^2\left(t\right)\hfill \\ \hfill & \ge -\frac{2\sqrt{2}}{3}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\parallel u_0{\parallel }^{\frac{5}{2}}\ge -{\parallel Du\parallel }^2\left(t\right)-C{\parallel u_0\parallel }^{\frac{10}{3}},\hfill \end{array}$$

(43)

where C is a positive constant. On the other hand, due to Inequality (40) and the fact that $\frac{M_2}{M_1}\ge 1$ for all $l\ge 2$, we get

$$\begin{array}{c}\hfill -2M_{2}U(t,0,L)\ge \frac{2M_2}{M_1}(u_t,u)\left(t\right)\ge 2(u_t,u)\left(t\right).\end{array}$$

(44)

Substituting Inequalities (43) and (44) into Inequality (42), we find

$$\begin{array}{c}\hfill \frac{d}{dt}(1+x,u^2)\left(t\right)+{2\parallel Du\parallel }^2\left(t\right)+\sum _{j=2}^{l-1}(2j+1)\parallel D^j{u\parallel }^2\left(t\right)\le C{\parallel u_0\parallel }^{\frac{10}{3}}.\end{array}$$

(45)

After integration of Inequality (45) over $(0,T)$, we conclude

$${\int }_0^T\sum _{j=1}^l\parallel D^j{u\parallel }^2\left(t\right)dt\le C{\parallel u_0\parallel }^2$$

(46)

where $C=C(T,L,l,\parallel u_0\parallel )$ is a positive constant.

Estimate 3.

Differentiate Equation (6) with respect to t, multiply the result by $2u_t$ to obtain

$$\begin{array}{c}\hfill 2(u_{tt},u_t)\left(t\right)+2M_1{U}_t(t,0,L)+2(D\left[u{u}_t\right],u_t)\left(t\right)\le 0,\end{array}$$

(47)

where $U_t(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^i{u}_t(t,L)\right)}^2+{\left(D^i{u}_t(t,0)\right)}^2\right]+{\left(D^l{u}_t(t,0)\right)}^2$. Making use of Inequalities (3) and (41), we estimate for an arbitrary $\epsilon >0$

$$\begin{array}{cc}\hfill & 2(D\left[u{u}_t\right],u_t)\left(t\right)=-2(u{u}_t,D{u}_t)\left(t\right)\ge -\frac{1}{\epsilon }{\left(\right|u|}^2,|u_t{|}^2)\left(t\right)-\epsilon {\parallel D{u}_t\parallel }^2\left(t\right)\hfill \\ \hfill & \ge -\frac{2}{\epsilon }\parallel Du\parallel \left(t\right)\parallel u\parallel \left(t\right)\parallel u_t{\parallel }^2\left(t\right)-\epsilon \parallel D{u}_t{\parallel }^2\left(t\right)\ge -\left(\frac{1}{{\epsilon }^2}\parallel u_0{\parallel }^2+{\parallel Du\parallel }^2\left(t\right)\right)\parallel u_t{\parallel }^2\left(t\right)-\epsilon {\parallel D{u}_t\parallel }^2\left(t\right).\hfill \end{array}$$

(48)

Substituting Inequality (48) into Inequality (47), we find

$$U_t(t,0,L)\le -\frac{1}{M_1}(u_{tt},u_t)\left(t\right)+\frac{1}{2M_1}\left(\frac{1}{{\epsilon }^2}\parallel u_0{\parallel }^2+{\parallel Du\parallel }^2\left(t\right)\right)\parallel u_t{\parallel }^2\left(t\right)+\frac{\epsilon }{2M_1}{\parallel D{u}_t\parallel }^2\left(t\right).$$

(49)

Estimate 4.

Differentiate Equation (6) with respect to t, multiply the result by $2x{u}_t$ and integrate over $(0,L)$. The result reads

$$\begin{array}{cc}\hfill & 2(u_{tt},x{u}_t)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^i{u}_t(t,L)\right)}^2-2M_2{U}_t(t,0,L)\hfill \\ & +\sum _{j=1}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)+2(D\left[u{u}_t\right],x{u}_t)\left(t\right)\le 0.\hfill \end{array}$$

(50)

Making use of Inequalities (3) and (41), we estimate

$$\begin{array}{cc}\hfill & 2(D\left[u{u}_t\right],x{u}_t)\left(t\right)=-2(u{u}_t,u_t+xD{u}_t)\left(t\right)\ge -2\left(\right|u|,u_t^2)\left(t\right)-L^2(u^2,u_t^2)\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right)\hfill \\ & \ge -2\sqrt{2}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\parallel u_0{\parallel }^{\frac{1}{2}}\parallel u_t{\parallel }^2\left(t\right)-2L^2\parallel Du\parallel \left(t\right)\parallel u_0\parallel \parallel u_t{\parallel }^2\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right)\hfill \\ & \ge -C\left(1+{\parallel Du\parallel }^2\left(t\right)\right)\parallel u_t{\parallel }^2\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right)\hfill \end{array}$$

(51)

for some positive constant $C=C(L,\parallel u_0\parallel )$. On the other hand, taking into account Inequality (49) with $\epsilon =\frac{M_1}{M_2}$ and exploiting the relation $\frac{M_2}{M_1}\ge 1$ for all $l\ge 2$, we obtain

$$-2M_2{U}_t(t,0,L)\ge 2(u_{tt},u_t)\left(t\right)-C\left(1+{\parallel Du\parallel }^2\left(t\right)\right)\parallel u_t{\parallel }^2\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right)$$

(52)

for some positive constant $C=C(M_1,M_2,\parallel u_0\parallel )$. Substituting Inequalities (51) and (52) into Inequality (50), we get

$$\frac{d}{dt}(1+x,u_t^2)\left(t\right)+\parallel D{u}_t{\parallel }^2\left(t\right)+\sum _{j=2}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)\le C\left(1+{\parallel Du\parallel }^2\left(t\right)\right)(1+x,u_t^2)\left(t\right).$$

(53)

Due to Inequality (46), $1+{\parallel Du\parallel }^2\left(t\right)\in L^1(0,T)$, whence by the Gronwall Lemma,

$$\parallel u_t{\parallel }^2\left(t\right)\le (1+x,u_t^2)\left(t\right)\le C\left(\parallel u_0{\parallel }_{H^l}^4+{\parallel u_0\parallel }_{H^{2l+1}}^2\right).$$

(54)

Substituting Inequality (54) into Inequality (53) and integrating over $(0,T)$, we find

$${\int }_0^T\sum _{j=1}^l{\parallel D^j{u}_t\parallel }^2\left(t\right)dt\le C\left(\parallel u_0{\parallel }_{H^l}^4+{\parallel u_0\parallel }_{H^{2l+1}}^2\right)$$

(55)

with a positive constant $C=C(T,L,l,a_{l+j,l-j},b_{l+j,l-j},\parallel u_0\parallel ),j=1,\dots ,l-1$.

Estimates Inequalities (41), (46), (54) and (55) allow us to extend the local solution ensured by Theorem 3 to all $T>0$ and to prove the existence of a generalized solution $u=u(t,x)$:

$$u\in L^{\infty }(0,T;H^l(0,L));u_t\in L^{\infty }(0,T;L^2(0,L))\cap L^2(0,T;H^l(0,L)).$$

(56)

Acting as by the proof of Theorem 3 and making use of Relation (56), we get

$$u\in L^{\infty }(0,T;H^{2l+1}(0,L))\cap L^2(0,T;H^{(2l+1)+l}(0,L)).$$

The existence part of Theorem 4 is proved.

Lemma 6.

A regular solution of Equations (6), (7) and (16) is uniquelly defined.

Proof. 

Let $u_1$ and $u_2$ be two distinct regular solutions of Equations (6), (7) and (16), then the difference $w=u_1-u_2$ satisfies the equation

$$w_t+\sum _{j=1}^l{(-1)}^{j+1}{D}^{2j+1}w+\frac{1}{2}D[u_1^2-u_2^2]=0,$$

(57)

boundary conditions Equation (16) and initial data $w(0,·)\equiv 0$.

Estimate 5.

Multiplying Equation (57) by $2w$, we find

$$\begin{array}{c}\hfill 2(w_t,w)\left(t\right)+2M_{1}W(t,0,L)+(D[u_1^2-u_2^2],w)\left(t\right)\le 0,\end{array}$$

(58)

where $W(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^{i}w(t,L)\right)}^2+{\left(D^{i}w(t,0)\right)}^2\right]+{\left(D^{l}w(t,0)\right)}^2$. Since $u_1,u_2$ are regular solutions of Equations (6), (7) and (16), then

$$\parallel u_i{\parallel }_{H^l}\left(t\right)\le C<+\infty ,i=1,2\mathrm{for} \mathrm{a}.\mathrm{e}.t\in (0,T),$$

(59)

where $C=C(L,l,a_{l+j,l-j},b_{l+j,l-j}),j=1,\dots ,1-1$. Making use of Inequalities (3) and (59), we estimate for an arbitrary $\epsilon >0$

$$\begin{array}{cc}\hfill & (D[u_1^2-u_2^2],w)\left(t\right)\ge -\left(\right|u_1+u_2\left|\right|w|,|Dw\left|\right)\left(t\right)\ge -{\parallel u_1+u_2\parallel }_{\infty }\left(t\right)\left(\frac{1}{2\epsilon }{\parallel w\parallel }^2\left(t\right)+\frac{\epsilon }{2}{\parallel Dw\parallel }^2\left(t\right)\right)\hfill \\ \hfill & \ge -2\sqrt{2}C\left(\frac{1}{2\epsilon }{\parallel w\parallel }^2\left(t\right)+\frac{\epsilon }{2}{\parallel Dw\parallel }^2\left(t\right)\right).\hfill \end{array}$$

(60)

Substituting Inequality (60) into Inequality (58), we obtain

$$\begin{array}{c}\hfill W(t,0,L)\le -\frac{1}{M_1}(w_t,w)\left(t\right)+\frac{\sqrt{2}C}{M_1}\left(\frac{1}{2\epsilon }{\parallel w\parallel }^2\left(t\right)+\frac{\epsilon }{2}{\parallel Dw\parallel }^2\left(t\right)\right).\end{array}$$

(61)

Estimate 6.

Multiplying Equation (57) by $2xw$, we get

$$\begin{array}{cc}\hfill & 2(w_t,x{w}_t)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^{i}w(t,L)\right)}^2-2M_{2}W(t,0,L)\hfill \\ & +\sum _{j=1}^l(2j+1){\parallel D^{j}w\parallel }^2\left(t\right)+2(D[u_1^2-u_2^2],xw)\left(t\right)\le 0.\hfill \end{array}$$

(62)

Making use of Inequalities (3) and (59), we estimate for an arbitrary $\epsilon >0$

$$\begin{array}{cc}\hfill & 2(D[u_1^2-u_2^2],xw)\left(t\right)\ge -\left(\right|u_1+u_2\left|\right|w|,|w|+x\left|Dw\right|)\left(t\right)\hfill \\ & \ge -\parallel u_1+u_2{\parallel }_{\infty }\left(t\right)\left({\parallel w\parallel }^2\left(t\right)+\frac{1}{2\epsilon }(x,w^2)\left(t\right)+\frac{\epsilon L}{2}{\parallel Dw\parallel }^2\left(t\right)\right)\hfill \\ & \ge -2\sqrt{2}C\left[\left(1+\frac{1}{2\epsilon }\right)(1+x,w^2)\left(t\right)+\frac{\epsilon L}{2}{\parallel Dw\parallel }^2\left(t\right)\right].\hfill \end{array}$$

(63)

Substituting Inequalities (61) and (63) into Inequality (62), we reduce it to the inequality

$$\begin{array}{cc}\hfill & \frac{d}{dt}(1+x,w^2)\left(t\right)+\left[3-\epsilon \left(\frac{\sqrt{2}C{M}_2}{M_1}+\sqrt{2}CL\right)\right]{\parallel Dw\parallel }^2\left(t\right)+\sum _{j=2}^l(2j+1){\parallel D^{j}w\parallel }^2\left(t\right)\hfill \\ & \le \left[\frac{\sqrt{2}C{M}_2}{M_1\epsilon }+2\sqrt{2}C\left(1+\frac{1}{2\epsilon }\right)\right](1+x,w^2)\left(t\right).\hfill \end{array}$$

Taking $\epsilon >0$ such that $3-\epsilon \left(\frac{\sqrt{2}C{M}_2}{M_1}+\sqrt{2}CL\right)>0$ and applying the Gronwall Lemma, we obtain $\parallel w\parallel \left(t\right)\equiv 0$, $t\in (0,T)$. This completes the proof of Lemma 6. □

Uniqueness part of Theorem 4 is thereby proved. □

6. Exponential Decay of Small Regular Solutions

Theorem 5.

Let $u_0\in H^{2l+1}(0,L)$ satisfy Equation (16) and

$$\parallel u_0\parallel <min\{m_1,m_2\},$$

(64)

where

$$\begin{array}{cc}\hfill & m_1={\left(\frac{{\pi }^2}{C_0{L}^2}\right)}^{3/4},m_2={\left[\frac{{\pi }^2}{L^2}{\left(2\sqrt{2}{K}_0(1+C_1{K}_0)\right)}^{-1}\right]}^2,\hfill \\ \hfill & K_0={\left((1+L)(1+l)\left(2\parallel u_0{\parallel }_{H^l}^4+{\parallel u_0\parallel }_{H^{2l+1}}^2\right)+(2+L+C_0\parallel u_0{\parallel }^{4/3})\parallel u_0{\parallel }^2\right)}^{1/3},\hfill \\ & C_0=2^{-2/3}·3^{-1/3},C_1=\left(\frac{2L^2}{\sqrt{2}}+\frac{2M_2^2}{\sqrt{2}{M}_1^2}\right){\parallel u_0\parallel }^{1/2}.\hfill \end{array}$$

Then Equations (6), (7) and (16) has a unique global regular solution $u=u(t,x)$:

$$\begin{array}{cc}\hfill & u\in L^{\infty }((0,+\infty );H^{2l+1}(0,L))\cap L^2((0,+\infty );H^{(2l+1)+l}(0,L));\hfill \\ & u_t\in L^{\infty }((0,+\infty );L^2(0,L))\cap L^2((0,+\infty );H^l(0,L))\hfill \end{array}$$

satisfying the inequalities:

$$\begin{array}{c}\hfill {\parallel u\parallel }^2\left(t\right)\le C{e}^{-\theta t},\parallel u_t{\parallel }^2\left(t\right)\le C{e}^{-\theta t},{\parallel u\parallel }_{H^{2l+1}}^2\left(t\right)\le C{e}^{-\theta t},\end{array}$$

where $\theta ={\pi }^2/(1+L)L^2$.

Proof. 

We need global in t a priori estimates of local solutions in order to prolong them for all $t>0$.

Estimate 1.

Estimates Inequalities (40) and (41) are valid in our case:

$$U(t,0,L)\le -\frac{(u_t,u)}{M_1}\left(t\right),\parallel u\parallel \left(t\right)\le \parallel u_0\parallel t>0,$$

(65)

where $U(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^{i}u(t,L)\right)}^2+{\left(D^{i}u(t,0)\right)}^2\right]+{\left(D^{l}u(t,0)\right)}^2$ and $M_1=\underset{i\in \{1,\dots ,l-1\}}{min}\{B_i,A_i,A_l\}$.

Estimate 2.

Multiply Equation (6) by $2xu$ and integrate over $(0,L)$ to obtain

$$2(u_t,xu)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^{i}u(t,L)\right)}^2-2M_{2}U(t,0,L)+\sum _{j=1}^l(2j+1){\parallel D^{j}u\parallel }^2\left(t\right)+2(uDu,xu)\left(t\right)\le 0,$$

(66)

where $M_2$ is calculated in [28], p. 389. Making use of Inequalities (3) and (65), we estimate

$$\begin{array}{cc}\hfill & 2(uDu,xu)\left(t\right)=-\frac{2}{3}(u,u^2)\left(t\right)\ge -\frac{2\sqrt{2}}{3}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right){\parallel u\parallel }^{\frac{1}{2}}\left(t\right){\parallel u\parallel }^2\left(t\right)\hfill \\ \hfill & \ge -{\parallel Du\parallel }^2\left(t\right)-C_0\parallel u_0{\parallel }^{\frac{4}{3}}{\parallel u\parallel }^2\left(t\right),\hfill \end{array}$$

(67)

where $C_0=2^{-2/3}·3^{-1/3}$. Substituting Inequalities (65) and (67) into Inequality (66) and using Equation (5), we get

$$\begin{array}{cc}\hfill & \frac{d}{dt}(1+x,u^2)\left(t\right)+{\parallel Du\parallel }^2\left(t\right)+\sum _{j=2}^l(2j+1){\parallel D^{j}u\parallel }^2\left(t\right)+\frac{1}{1+L}\left(\frac{{\pi }^2}{L^2}-C_0{\parallel u_0\parallel }^{\frac{4}{3}}\right)(1+x,u^2)\left(t\right)\le 0.\hfill \end{array}$$

Due to Inequalities (5) and (64),

$$\frac{d}{dt}(1+x,u^2)\left(t\right)+\theta (1+x,u^2)\left(t\right)\le 0,$$

where $\theta ={\pi }^2/(1+L)L^2$. Consequently

$${\parallel u\parallel }^2\left(t\right)\le C{e}^{-\theta t}.$$

(68)

Estimate 3.

By Inequalities (3) and (65), Inequality (66) becomes

$$\begin{array}{cc}\hfill & 2L\sum _{i=1}^{l-1}{B}_i{\left(D^{i}u(t,L)\right)}^2+\sum _{j=1}^l(2j+1){\parallel D^{j}u\parallel }^2\left(t\right)\le -2(u_t,(1+x)u)\left(t\right)-2(uDu,xu)\left(t\right)\hfill \\ \hfill & \le 2(1+L)\parallel u_t\parallel \left(t\right)\parallel u\parallel \left(t\right)+\frac{2\sqrt{2}}{3}{\parallel Du\parallel }^{\frac{1}{2}}{\left(t\right)\parallel u\parallel }^{\frac{1}{2}}\left(t\right){\parallel u\parallel }^2\left(t\right)\hfill \\ \hfill & \le {\parallel Du\parallel }^2\left(t\right)+(1+L)\parallel u_t{\parallel }^2\left(t\right)+(1+L+C_0\parallel u_0{\parallel }^{\frac{4}{3}}{)\parallel u\parallel }^2\left(t\right).\hfill \end{array}$$

Thus

$${\parallel u\parallel }_{H^l}^2\left(t\right)\le (1+L)\parallel u_t{\parallel }^2\left(t\right)+(2+L+C_0\parallel u_0{\parallel }^{\frac{4}{3}}{)\parallel u\parallel }^2\left(t\right).$$

(69)

Estimate 4.

Differentiate Equation (6) with respect to t, multiply the result by $2u_t$ and integrate over $(0,L)$ to obtain

$$\begin{array}{c}\hfill 2(u_{tt},u_t)\left(t\right)+2M_1{U}_t(t,0,L)+2(D\left[u{u}_t\right],u_t)\left(t\right)\le 0,\end{array}$$

(70)

where $U_t(t,0,L)\equiv {\sum }_{i=1}^{l-1}\left[{\left(D^i{u}_t(t,L)\right)}^2+{\left(D^i{u}_t(t,0)\right)}^2\right]+{\left(D^l{u}_t(t,0)\right)}^2$. Repeating arguments used to prove Inequality (48), we estimate for an arbitrary $\epsilon >0$

$$\begin{array}{c}\hfill 2(D\left[u{u}_t\right],u_t)\left(t\right)\ge -\frac{2}{\epsilon }\parallel Du\parallel \left(t\right)\parallel u\parallel \left(t\right)\parallel u_t{\parallel }^2\left(t\right)-\epsilon {\parallel D{u}_t\parallel }^2\left(t\right).\end{array}$$

(71)

Substituting Inequality (71) into Inequality (70) and using Inequality (65), we find

$$U_t(t,0,L)\le -\frac{1}{M_1}(u_{tt},u_t)\left(t\right)+\frac{1}{\epsilon M_1}\parallel u_0\parallel \parallel Du\parallel \left(t\right)\parallel u_t{\parallel }^2\left(t\right)+\frac{\epsilon }{2M_1}{\parallel D{u}_t\parallel }^2\left(t\right).$$

(72)

Estimate 5.

Differentiate Equation (6) with respect to t, multiply the result by $2x{u}_t$ to obtain

$$\begin{array}{cc}\hfill & 2(u_{tt},x{u}_t)\left(t\right)+2L\sum _{i=1}^{l-1}{B}_i{\left(D^i{u}_t(t,L)\right)}^2-\underset{I_1}{\underbrace{2M_2{U}_t(t,0,L)}}\hfill \\ & +\sum _{j=1}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)+\underset{I_2}{\underbrace{2(D\left[u{u}_t\right],x{u}_t)\left(t\right)}}\le 0.\hfill \end{array}$$

(73)

Taking into account Inequality (72) with $\epsilon =\frac{M_1}{2M_2}$ and exploiting the relation $\frac{M_2}{M_1}\ge 1$ for all $l\ge 2$, we obtain

$$\begin{array}{cc}\hfill & I_1\ge 2(u_{tt},u_t)\left(t\right)-\frac{4M_2^2}{M_1^2}\parallel u_0\parallel \parallel Du\parallel \left(t\right)\parallel u_t{\parallel }^2\left(t\right)-\frac{1}{2}{\parallel D{u}_t\parallel }^2\left(t\right).\hfill \end{array}$$

On the other hand, repeating arguments used to prove Inequality (51), we estimate

$$\begin{array}{cc}\hfill & I_2\ge -2\sqrt{2}\parallel u_0{\parallel }^{\frac{1}{2}}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\left[1+\frac{2L^2}{\sqrt{2}}\parallel u_0{\parallel }^{\frac{1}{2}}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\right]\parallel u_t{\parallel }^2\left(t\right)-\frac{1}{2}{\parallel D{u}_t\parallel }^2\left(t\right).\hfill \end{array}$$

Making use of Inequalities (65) and (69), we find

$$\begin{array}{cc}\hfill & I_1+I_2\ge 2(u_{tt},u_t)\left(t\right)-2\sqrt{2}\parallel u_0{\parallel }^{\frac{1}{2}}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\left[1+\left(\frac{2L^2}{\sqrt{2}}+\frac{2M_2^2}{\sqrt{2}{M}_1^2}\right)\parallel u_0{\parallel }^{\frac{1}{2}}{\parallel Du\parallel }^{\frac{1}{2}}\left(t\right)\right]\parallel u_t{\parallel }^2\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right)\hfill \\ & \ge 2(u_{tt},u_t)\left(t\right)-2\sqrt{2}\parallel u_0{\parallel }^{\frac{1}{2}}{\left((1+L)\parallel u_t{\parallel }^2\left(t\right)+(2+L+C_0\parallel u_0{\parallel }^{\frac{4}{3}})\parallel u_0{\parallel }^2\right)}^{\frac{1}{4}}[1+\left(\frac{2L^2}{\sqrt{2}}+\frac{2M_2^2}{\sqrt{2}{M}_1^2}\right){\parallel u_0\parallel }^{\frac{1}{2}}\hfill \\ & \times {\left((1+L)\parallel u_t{\parallel }^2\left(t\right)+(2+L+C_0\parallel u_0{\parallel }^{\frac{4}{3}})\parallel u_0{\parallel }^2\right)}^{\frac{1}{4}}]\parallel u_t{\parallel }^2\left(t\right)-{\parallel D{u}_t\parallel }^2\left(t\right).\hfill \end{array}$$

Substituting $I_1+I_2$ into Inequality (73), we get

$$\begin{array}{cc}\hfill & \frac{d}{dt}(1+x,u_t^2)\left(t\right)+\parallel D{u}_t{\parallel }^2\left(t\right)+\sum _{j=2}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)\hfill \\ & +\parallel D{u}_t{\parallel }^2\left(t\right)-2\sqrt{2}\parallel u_0{\parallel }^{\frac{1}{2}}K\left(t\right)\left(1+C_{1}K\left(t\right)\right){\parallel u_t\parallel }^2\left(t\right)\le 0.\hfill \end{array}$$

(74)

Here

$$K\left(t\right)={\left((1+L)\parallel u_t{\parallel }^2\left(t\right)+(2+L+C_0\parallel u_0{\parallel }^{4/3})\parallel u_0{\parallel }^2\right)}^{1/3},C_1=\left(\frac{2L^2}{\sqrt{2}}+\frac{2M_2^2}{\sqrt{2}{M}_1^2}\right){\parallel u_0\parallel }^{\frac{1}{2}}.$$

Using Inequality (5), this inequality can be rewritten as

$$\begin{array}{cc}\hfill & \frac{d}{dt}(1+x,u_t^2)\left(t\right)+\parallel D{u}_t{\parallel }^2\left(t\right)+\sum _{j=2}^l(2j+1){\parallel D^j{u}_t\parallel }^2\left(t\right)\hfill \\ & +\frac{1}{1+L}\left[\frac{{\pi }^2}{L^2}-2\sqrt{2}{\parallel u_0\parallel }^{\frac{1}{2}}K\left(t\right)\left(1+C_{1}K\left(t\right)\right)\right](1+x,u_t^2)\left(t\right)\le 0.\hfill \end{array}$$

(75)

Taking into account Inequality (64), the fact that $K\left(0\right)\le K_0$ and standard arguments, see [14], we reduce it to the form

$$\begin{array}{c}\hfill \frac{d}{dt}(1+x,u_t^2)\left(t\right)+\theta (1+x,u_t^2)\left(t\right)\le 0,\end{array}$$

where $\theta ={\pi }^2/(1+L)L^2$. This implies

$$\parallel u_t{\parallel }^2\left(t\right)\le (1+x,u_t^2)\left(t\right)\le C{e}^{-\theta t}.$$

(76)

Returning to Inequality (74) with Inequality (76) and integrating over $(0,+\infty )$, we obtain

$$u_t\in L^2((0,+\infty );H^l(0,L)).$$

Finally, substituting Inequalities (68) and (76) into Inequality (69), we find

$${\parallel u\parallel }_{H^l}^2\left(t\right)\le C{e}^{-\theta t}.$$

(77)

Estimate 6.

(Regularity) Rewrite Equation (6) in the form

$${(-1)}^{l+1}{D}^{2l+1}u=-u_t-\sum _{j=1}^{l-1}{(-1)}^{j+1}{D}^{2j+1}u-uDu.$$

We estimate

$$\parallel D^{2l+1}u\parallel \left(t\right)\le \parallel u_t\parallel \left(t\right)+\sum _{2j\le l-1}\parallel D^{2j+1}u\parallel \left(t\right)+\sum _{l-1<2j<2l}\parallel D^{2j+1}u\parallel \left(t\right)+\parallel uDu\parallel \left(t\right).$$

(78)

For $l=2$, we have ${\sum }_{2j\le l-1}\parallel D^{2j+1}u\parallel \left(t\right)=0$ and for $l\ge 3$, due to Inequality (77),

$$\sum _{2j\le l-1}\parallel D^{2j+1}{u\parallel \left(t\right)\le l\parallel u\parallel }_{H^l}\left(t\right)\le C{e}^{-\frac{\theta }{2}t}.$$

(79)

Making use of Inequalities (3) and (77), we obtain

$$\begin{array}{c}\hfill \parallel uDu\parallel \left(t\right)\le \sqrt{2}{\parallel u\parallel }_{H^l}^2\left(t\right)\le C{e}^{-\theta t}.\end{array}$$

(80)

On the other hand, Inequality (4) implies

$$\parallel D^{2j+1}u\parallel \left(t\right)\le C_1^j\parallel D^{2l+1}{u\parallel }^{{\alpha }^j}{\left(t\right)\parallel u\parallel }^{1-{\alpha }^j}\left(t\right)+C_2^j\parallel u\parallel \left(t\right)(l-1<2j<2l),$$

where ${\alpha }^j=\frac{2j+1}{2l+1}$ and $C_1^j$, $C_2^j$ are constants depending on L, l. Making use of the Young inequality with an arbitrary $\epsilon >0$, we get

$$\parallel D^{2j+1}u\parallel \left(t\right)\le \epsilon \parallel D^{2l+1}u\parallel \left(t\right)+(C^j\left(\epsilon \right)+C_2^j)\parallel u\parallel \left(t\right).$$

Summing over $l-1<2j<2l$ and taking into account Inequality (68), we find

$$\begin{array}{c}\hfill \sum _{l-1<2j<2l}\parallel D^{2j+1}u\parallel \left(t\right)\le l\epsilon \parallel D^{2l+1}u\parallel \left(t\right)+C\left(\epsilon \right)e^{-\frac{\theta }{2}t}.\end{array}$$

(81)

Substituting Inequalities (76), (79), (80) and (81) into Inequality (78) and taking $\epsilon =\frac{1}{2l}$, we obtain

$$\parallel D^{2l+1}u\parallel \left(t\right)\le C\left(3e^{-\frac{\theta }{2}t}+e^{-\theta t}\right)\le C{e}^{-\frac{\theta }{2}t}.$$

(82)

Again by Inequality (4), for all $i=l+1,\dots ,2l$, there are constants $C_1^i$, $C_2^i$ depending only on L, l such that

$$\parallel D^{i}u\parallel \left(t\right)\le C_1^i\parallel D^{2l+1}{u\parallel }^{{\alpha }^i}{\left(t\right)\parallel u\parallel }^{1-{\alpha }^i}\left(t\right)+C_2^i\parallel u\parallel \left(t\right)\mathrm{with}{\alpha }^i=\frac{i}{2l+1}.$$

By the Young inequality and Inequalities (68) and (82), we get

$$\parallel D^{i}u\parallel \left(t\right)\le C{e}^{-\frac{\theta }{2}t},i=l+1,\dots ,2l.$$

(83)

According to Inequalities (77), (82) and (83), we conclude that

$${\parallel u\parallel }_{H^{2l+1}}^2\left(t\right)\le C{e}^{-\theta t}$$

(84)

with $\theta ={\pi }^2/(1+L)L^2$. Repeating the arguments that appears in the proof of Lemma 4.3 in [26], and taking into account Inequality (84), we establish a “smoothing effect”:

$$u\in L^2((0,+\infty );H^{(2l+1)+l}(0,L)).$$

Similar arguments used in the proof of Lemma 6 with Inequality (77) instead of Inequality (59), show the uniqueness of the solution. The proof of Theorem 5 is complete. □

7. Conclusions

Making use of the formulation of a linear stationary version of Equation (1) in [28], we prove in Theorem 3 local existence and uniqueness of regular solutions. In Theorem 4, we prove global in $t\in (0,T)$ existence and uniqueness of regular solution for arbitrary smooth initial data and arbitrary $T>0$. In Theorem 5, we prove global in $t\in (0,+\infty )$ existence and uniqueness of regular solutions as well as their exponential decay of $\parallel u\parallel \left(t\right)$, $\parallel u_t\parallel \left(t\right)$ and ${\parallel u\parallel }_{H^{2l+1}}\left(t\right)$ for small initial data. A smoothing effect has been established: if $u_0\in H^{2l+1}(0,L)$, then $u\in L^2((0,+\infty );H^{(2l+1)+l}(0,L))$. Our results can be used for constructing of numerical schemes while studying various models of initial-boundary value problems for higher-order dispersive equations.