\documentclass[reqno]{amsart} \usepackage[notref,notcite]{showkeys} %\usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2007(2007), No. 137, pp. 1--9.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2007 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2007/137\hfil On the solvability] {On the solvability of a fourth order operator differential equations} \author[A. Nouar\hfil EJDE-2007/137\hfilneg] {Ahmed Nouar} \address{D\'{e}partement des Sciences Fondamentales \\ Facult\'{e} des Sciences et des Sciences de L'Ingenieur\\ Universit\'{e} de Skikda\\ 21000 Skikda, Algeria} \email{hamdensk21@Yahoo.fr} \thanks{Submitted February 24, 2007. Published October 17, 2007.} \subjclass[2000]{47E05, 35K20} \keywords{Operator Pencil; operator differential equation; normal operator} \begin{abstract} We establish sufficient conditions for the solvability of a fourth order operator differential equation. These conditions are expressed in terms of operator coefficients. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}[theorem]{Definition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Preliminaries} In this paper we consider the problem \begin{gather} P( d/dt) u=\frac{d^{4}u}{dt^{4}}+A^{4}u+\sum_{j=0}^{4}A_j \frac{d^{4-j}u}{dt^{4-j}}=f, \label{e1} \\ u^{( s_{i}) }( 0) =0,i=0,1,\quad s_{i}\in \{0,1,2,3\} , \label{e2} \end{gather} where $f\in L_2 ( \mathbb{R}_+,H)$, and $A_0 ,A_1 ,A_2 ,A_{3},A_{4}$ and $A$ are unbounded linear operators in a separable Hilbert space $H$. By $L_2 ( \mathbb{R}_+,H)$ we denote the space of $H$-valued vector functions such that $\int_0 ^{+\infty }\| f( t) \|_{H}^{2}dt<+\infty$, endowed with the inner product $\langle f,g\rangle _{L_2 ( \mathbb{R}_+,H) }= \int_0 ^{+\infty }\langle f( t) ,g( t) \rangle _{H}dt,$ and $\mathbb{R}_+=[0,+\infty )$. Let $$P( \lambda ) =\lambda ^{4}I+A^{4}+\sum_{j=0}^4 \lambda ^{4-j}A_j , \label{e3}$$ be the operator pencil associated with the operator differential equation \eqref{e1}. Our aim is to establish sufficient conditions for the solvability of problem \eqref{e1}, \eqref{e2}, expressed in terms of the norms of the operator coefficients of the pencil (\ref{e3}). For the general theory of operator pencils see \cite{yak3, yak1}. In particular, unbounded operator pencils of second and fourth orders are treated in \cite{yak5, yak4}. Interest for such equations is justified not only by the relation which they have with the practical problems described by ordinary or partial differential equations but also by their important applications in many theoretical problems. Thus, the study of such equations is connected with the multicompleteness of the eigenvectors and the generalized eigenvectors of the corresponding operator pencil. This relation has been proved in \cite {Gas1} and permits to answer some questions arising while studying non self-adjoint differential operators. The notion of multicompleteness has been introduced by Keldysh in \cite{keld1, keld2} and many interesting results have been obtained during the last decades by Gasymov \cite{Gas1}, Yakubov \cite{yak3, yak2, yak1, yak4}, Markus \cite{markus}, Mirzoev \cite{mirz1, mirz2} and others. \section{Definitions} \begin{definition}\label{def1} \rm Let $A$ be a normal operator in a Hilbert space $H$ such that $\sigma ( A)$ is contained in the sector $S_{\varepsilon }=\{ \lambda : | \arg \lambda | <\varepsilon <\frac{\pi }{12}\} ,\quad \inf \mathop{\rm Re}\sigma ( A) =\sigma _0 >0.$ We assume also that the operators $B_j =A_j A^{-j}$, $j=\overline{0,4}$ are bounded. Under these conditions we will say that the pencil (\ref{e3}) belongs to the class $\Gamma$. \end{definition} It is clear that if the operator $A$ satisfies the conditions of the above definition it admits a polar decomposition in the form: $A=UC$ , where $C$ is a positively defined self-adjoint operator and $U$ is a unitary operator, therefore$\ C$ and $U$ commute and we have for any rational $k$ \begin{gather*} D( C^{k}) =D( A^{k}) , \\ \| A^{k}x\| _{H}=\| C^{k}x\| _{H},\quad x\in D( A^{k}) . \end{gather*} In $D( C^{k})$ we define the inner product $\langle x,y\rangle _{H_{k}}=\langle C^{k}x,C^{k}y\rangle _{H},\quad x,y\in D( A^{k}) .$ Then, we obtain a Hilbert space which we denote by $H_{k}$. By $D(\mathbb{R}_+,H)$ we denote the pre-Hilbertian space of the infinitely differentiable $H_{4}$-valued functions with a compact support in $\mathbb{R}_+$, relative to the inner product $\langle u,v\rangle _{W}=\langle u^{( 4) },v^{( 4) }\rangle _{L_2 ( \mathbb{R}_+,H) }+\langle C^{4}u,C^{4}v\rangle _{L_2 ( \mathbb{R}_+,H) }.$ By $W( \mathbb{R}_+,H)$ we denote the completion of $D(\mathbb{R}_+,H)$ with respect to the above inner product. We introduce also the following notation \begin{gather*} W(\mathbb{R}_+,H,s_0 ,s_1 ) =\{ u\in W( \mathbb{R}_+,H) ,u^{(s_0 ) }( 0) =u^{( s_1 ) }(0)=0\} , \\ M_{4,j}( s_0 ,s_1 ) =\sup_{u\in W(\mathbb{R}_+,H,s_0 ,s_1 )} \frac{\| C^{4-j}u^{( j) }\| _{L_2 }}{\| u\| _{W}}. \end{gather*} \begin{definition} \label{def2} \rm We say that the couple $( s_0 ,s_1 )$ belongs to the class $S$ if it takes one of these values: $(0,1), (0,2), (1,3), (2,3)$. \end{definition} \begin{definition} \label{def3} \rm We say that the problem \eqref{e1}, \eqref{e2} is regular if for any function $f\in L_2 ( \mathbb{R}_+,H)$ there exists a unique function $u\in W(\mathbb{R}_+,H,s_0 ,s_1 )$ which satisfies the equation \eqref{e1} a.e. and the conditions \eqref{e2}. We call such a function $u$ a regular solution of the problem \eqref{e1}, \eqref{e2}. \end{definition} \section{Main Results} We consider now the pencil operator $S_j (\lambda ,\gamma ,C)=\lambda ^{8}I+C^{8}-\gamma ( i\lambda ) ^{2j}C^{8-2j},$ defined in $H_{8}$ and where $\gamma$ is a real parameter. We use the notation $d_{4,j}=\begin{cases} 1,& j=0,4, \\ ( \frac{4}{j}) ^{j/4} ( \frac{4}{4-j}) ^{\frac{4-j}{4}}, & j=1,2,3. \end{cases}$ \begin{theorem} \label{thm1} For $\gamma \in ( 0,d_{4,j})$, $j\in \{0,\dots, 4\}$, the pencil operator $S_j (\lambda ,\gamma ,C)$ can be represented in the form $S_j (\lambda ,\gamma ,C)=S_j ^{-}(\lambda ,\gamma ,C)S_j ^{+}(\lambda ,\gamma ,C),$ where \begin{gather*} S_j ^{-}(\lambda ,\gamma ,C)=\sum_{i=0}^4 \alpha_{i,j}( \gamma ) \lambda ^{i}C^{4-i}, \\ S_j ^{+}(\lambda ,\gamma ,C)=S_j ^{-}(-\lambda ,\gamma ,C). \end{gather*} Moreover, the coefficients $\alpha _{i,j}$ are real and satisfy the following relations $$\begin{gathered} \alpha _{0,0}=\sqrt{1-\gamma }, \\ \alpha _{4,0}=1, \\ \alpha _{1,0}^{2}-2\alpha _{0,0}\alpha _{2,0}=0, \\ \alpha _{2,0}^{2}-2\alpha _{1,0}\alpha _{3,0}+2\alpha _{0,0}\alpha _{4,0}=0, \\ \alpha _{3,0}^{2}-2\alpha _{2,0}\alpha _{4,0}=0. \end{gathered} \label{e4}$$ and $$\begin{gathered} \alpha _{0,4}=1 \\ \alpha _{4,4}=\sqrt{1-\gamma } \\ \alpha _{1,4}^{2}-2\alpha _{0,4}\alpha _{2,4}=0 \\ \alpha _{2,4}^{2}-2\alpha _{1,4}\alpha _{3,4}+2\alpha _{0,4}\alpha _{4,4}=0 \\ \alpha _{3,4}^{2}-2\alpha _{2,4}\alpha _{4,4}=0, \end{gathered} \label{e5}$$ and for $j=1,2,3$ $$\begin{gathered} \alpha _{0,j}=\alpha _{4,j}=1, \\ \alpha _{1,j}^{2}-2\alpha _{2,j} =\begin{cases} -\gamma ,& j=1, \\ 0, &j\neq 1. \end{cases} \\ \alpha _{2,j}^{2}-2\alpha _{1,j}\alpha _{3,j}+2 =\begin{cases} -\gamma ,& j=2, \\ 0,& j\neq 2. \end{cases} \\ \alpha _{3,j}^{2}-2\alpha _{2,j} =\begin{cases} -\gamma ,& j=3, \\ 0,& j\neq 3. \end{cases} \end{gathered} \label{e6}$$ \end{theorem} \begin{proof} For $\gamma \in ( 0,d_{4,j})$, $j\in \{0,\dots 4\}$ the polynomial $Q_j (\lambda ,\gamma )=\lambda^{8}+1-\gamma (i\lambda )^{2j}$ has not purely imaginary roots, its roots are simple and symmetrically situated relatively to the real axis and the origin. So it can be represented in the form $$Q_j (\lambda ,\gamma )=S_j (\lambda ,\gamma )S_j (-\lambda ,\gamma ). \label{e7}$$ The roots of $S_j (\lambda ,\gamma )$ belong to the left half-plane and we can denote them by $\omega _1 ( \gamma )$, $\omega _2 (\gamma )$, $\overline{\omega _1 ( \gamma ) }$, $\overline{\omega _2 ( \gamma ) }$ where $\overline{\omega _1 (\gamma ) }$ and $\overline{\omega _2 ( \gamma ) }$ are the conjugate of $\omega _1 ( \gamma )$ and $\omega _2 (\gamma )$ respectively. In this case $S_j (\lambda ,\gamma )=( \lambda -\omega _1 ( \gamma ) ) ( \lambda -\omega _2 ( \gamma ) ) ( \lambda -\overline{\omega _1 ( \gamma ) }) ( \lambda - \overline{\omega _2 ( \gamma ) }) .$ This proves that if we write $S_j (\lambda ,\gamma )=\sum_{i=0}^{4}\alpha _{i,j}\lambda ^{i}$, then the coefficients $\alpha _{i,j}$ are real. The relations (\ref{e4}), (\ref{e5}) and (\ref{e6}) can be obtained by using the Viete theorem and identifying the coefficients in the left and right parts of the relation (\ref{e7}). Let $E_{\sigma }$ be the spectral decomposition of the operator $C$. We have \begin{align*} S_j (\lambda ,\gamma ,C) &=\lambda ^{8}I+C^{8}-\gamma ( i\lambda )^{2j}C^{8-2j} \\ &=\int_{\sigma _0 }^{\infty }(\lambda ^{8}+\sigma ^{8}-\gamma ( i\lambda ) ^{2j}\sigma ^{8-2j})dE_{\sigma }\\ &=\int_{\sigma_0 }^{\infty }\sigma ^{8}(\frac{\lambda ^{8}}{\sigma ^{8}} +1-\gamma \frac{( i\lambda ) ^{2j}}{\sigma ^{2j}})dE_{\sigma } \\ &=\int_{\sigma _0 }^{\infty }\sigma ^{8}Q_j (\frac{\lambda }{\sigma } ,\gamma )dE_{\sigma }\\ &=\int_{\sigma _0 }^{\infty }\sigma ^{8}S_j (\frac{\lambda }{\sigma }, \gamma )S_j (-\frac{\lambda }{\sigma },\gamma )dE_{\sigma } \\ &=\int_{\sigma _0 }^{\infty }\sigma ^{4}S_j (\frac{\lambda }{\sigma } ,\gamma )dE_{\sigma }\int_{\sigma _0 }^{\infty }\sigma ^{4}S_j (- \frac{\lambda }{\sigma },\gamma )dE_{\sigma } \\ &=(\sum_{i=0}^{4}\alpha _{i,j}\lambda ^{i}C^{4-i})(\sum_{i=0}^{4}\alpha _{i,j}(-\lambda )^{i}C^{4-i}). \end{align*} Then taking $S_j ^{-}(\lambda ,\gamma ,C)=\sum_{i=0}^{4}\alpha _{i,j}\lambda ^{i}C^{4-i},$ we obtain $S_j (\lambda ,\gamma ,C)=S_j ^{-}(\lambda ,\gamma ,C)S_j ^{+}(\lambda ,\gamma ,C)$. \end{proof} \begin{theorem} \label{thm2} Let $\gamma \in ( 0,d_{4,j})$ and $u\in W(\mathbb{R}_+,H)$. Then for $j\in \{ 0,\dots 4\}$, $\| S_j ^{-}( d/dt,\gamma ,C) u\| _{L_2 }^{2}=\| u\| _{W}^{2}-\gamma \| C^{4-j}u^{( j) }\| _{L_2 }^{2}-\langle G_j \widetilde{f},\widetilde{f}\rangle _{H^{4}},$ where \begin{gather*} \widetilde{f}=\big( C^{7/2}u(0),C^{5/2}u'(0),C^{3/2}u^{\prime \prime }(0),C^{1/2}u^{^{\prime \prime \prime }}(0)\big) \in H\oplus H\oplus H\oplus H=H^{4}, \\ G_j =\begin{bmatrix} \alpha _1 \alpha _0 & \alpha _2 \alpha _0 & \alpha _{3}\alpha _0 & \alpha _{4}\alpha _0 \\ \alpha _2 \alpha _0 & \alpha _2 \alpha _1 -\alpha _{3}\alpha _0 & \alpha _{3}\alpha _1 -\alpha _{4}\alpha _0 & \alpha _{4}\alpha _1 \\ \alpha _{3}\alpha _0 & \alpha _{3}\alpha _1 -\alpha _{4}\alpha _0 & \alpha _{3}\alpha _2 -\alpha _{4}\alpha _1 & \alpha _{4}\alpha _2 \\ \alpha _{4}\alpha _0 & \alpha _{4}\alpha _1 & \alpha _{4}\alpha _2 & \alpha _{4}\alpha _{3} \end{bmatrix}, \end{gather*} and $\alpha _{i}\equiv \alpha _{i,j}(\gamma )$. \end{theorem} \begin{proof} By a theorem of the intermediate derivatives \cite[pp.19-25]{Lion}, $C^{4-j}u^{( j) }(\cdot) \in L_2 (\mathbb{R}_+,H)$ and $u^{( j) }( 0) \in D( C^{4-j-\frac{1}{2}})$. We have \begin{align*} \| S_j ^{-}( d/dt,\gamma ,C) u\| _{L_2 (\mathbb{R}_+,H) }^{2} &=\| \sum_{i=0}^{4}\alpha _{i,j}C^{4-i}u^{(i)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}\\ &= \sum_{i=0}^{4}\alpha _{i,j}^{2}\| C^{4-i}u^{(i)}\| _{L_2 ( \mathbb{R}_+,H)}^{2}\\ &\quad +2\sum_{i=1}^{4}\sum_{s=0}^{i-1}\alpha _{i,j}\alpha _{s,j} \mathop{\rm Re}\langle C^{4-i}u^{(i)},C^{4-s}u^{(s)}\rangle _{L_2 ( \mathbb{R}_+,H) }. \end{align*} Calculating the expressions $\langle C^{4-i}u^{(i)},C^{4-s}u^{(s)}\rangle _{L_2 ( \mathbb{R}_+,H) }$ by using successive integrations by parts and taking into account that $C$ is a selfadjoint operator we obtain \begin{align*} &\| S_j ^{-}( d/dt,\gamma ,C) u\| _{L_2 (\mathbb{R}_+,H) }^{2}\\ &=\alpha _{0,j}^{2}\| C^{4}u\|_{L_2 ( \mathbb{R}_+,H) }^{2} +\alpha _{4,j}^{2}\|u^{(4)}\| _{L_2 ( \mathbb{R}_+,H) }^{2} + (\alpha _{3,j}^{2}-2\alpha _{2,j}\alpha _{4,j})\| Cu^{(3)}\| _{L_2 ( \mathbb{R}_+,H) }^{2} \\ &\quad +(\alpha_{2,j}^{2}-2\alpha _{1,j}\alpha _{3,j} +2\alpha _{0,j}\alpha_{4,j})\| C^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}\\ &\quad +(\alpha _{1,j}^{2}-2\alpha _{2,j}\alpha _{0,j})\| C^{3}u^{(1)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}-\langle G_j \widetilde{f},\widetilde{f}\rangle _{H^{4}}. \end{align*} Then using the relations (\ref{e4}), (\ref{e5}) and (\ref{e6}) we obtain $\| S_j ^{-}( d/dt,\gamma ,C) u\| _{L_2 ( \mathbb{R}_+,H) }^{2}=\| u\| _{W}^{2}-\gamma \| C^{4-j}u^{(j)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}-\langle G_j \widetilde{f},\widetilde{f}\rangle _{H^{4}}.$ \end{proof} \begin{remark} \label{rmk1} \rm Theorem \ref{thm2} shows that $M_{4,j}( s_0 ,s_1 )$ can be written with respect to $S_j ^{-}( d/dt,\gamma,C)$ and $G_j$. So, according to Theorem \ref{thm1} and the definition of $G_j$, it can be written with respect to $\gamma$ and $\alpha _{i,j}$. In fact, the pencil $S_j ( \lambda ,\gamma ,C)$ from which we have obtained the pencil $S_j ^{-}( d/dt,\gamma ,C)$ has been especially defined to this end. \end{remark} Let $G_j (s_0 ,s_1 ,\gamma )$ be the matrix obtained from the matrix $G_j$ by deleting the lines $s_0 +1$ and $s_1 +1$ and the columns $s_0 +1$ and $s_1 +1$. \begin{theorem}[\cite{mirz3}] \label{thm3} Let $j\in \{ 0,\dots 4\}$, then $M_{4,j}( s_0 ,s_1 ) =\begin{cases} d_{4,j}^{-1/2}, &\text{if }\det G_j (s_0 ,s_1 ,\gamma ) \neq 0, \text{ for any } \gamma \in ( 0,d_{4,j}) \\ \mu _{4,j}^{-1/2}( s_0 ,s_1 ), & \text{otherwise,} \end{cases}$ where $\mu _{4,j}( s_0 ,s_1 )$ is the smallest positive root from $( 0,d_{4,j})$ of the equation $\det G_j ( s_0 ,s_1 ,\gamma ) =0$ with respect to $\gamma$. \end{theorem} \begin{remark} \label{rmk2} \rm Theorem \ref{thm3} implies that to calculate $M_{4,j}(s_0 ,s_1 )$, one must determine the matrix $G_j (s_0 ,s_1 ,\gamma )$ from the matrix $G_j$ obtained in Theorem \ref{thm2} and resolve the equation $\det G_j ( s_0 ,s_1 ,\gamma )=0$ with respect to $\gamma$ by using the relations (\ref{e4}), (\ref{e5}) and (\ref{e6}) from Theorem \ref{thm1}. \end{remark} \begin{lemma}\label{lemme1} Let the operator $A$ be as in Definition \ref{def1} and $u\in W(\mathbb{R}_+,H,s_0 ,s_1 )$ with $(s_0 ,s_1 )\in S$, then $\| u\| _{W}\leq (1-2\sin 2\varepsilon )^{-1/2} \| P_0 u\| _{L_2 ( \mathbb{R}_+,H) },$ where $P_0 u=\frac{d^{4}u}{dt^{4}}+A^{4}u.$ \end{lemma} \begin{proof} If $u\in W(\mathbb{R}_+,H,s_0 ,s_1 )$, we have $\| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }^{2}=\| A^{4}u\| _{L_2 ( \mathbb{R}_+,H) }^{2}+ \| u^{(4)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}+2\mathop{\rm Re} \langle A^{4}u,u^{(4)}\rangle _{L_2 ( \mathbb{R}_+,H) }.$ Using integration by parts, we have \begin{align*} &\langle A^{4}u,u^{(4)}\rangle _{L_2 (\mathbb{R}_+,H) }\\ &=\langle A^{\frac{7}{2}}u,{A^{\ast}}^{\frac{1}{2}}u^{(3)}\rangle \big| _0 ^{\infty } -\langle A^{\frac{5}{2}}u',{A^{\ast}}^{\frac{3}{2}}u^{(2)}\rangle \big| _0 ^{\infty} +\langle A^{2}u^{( 2) },A^{\ast 2}u^{(2) }\rangle _{L_2 ( \mathbb{R}_+,H)}. \end{align*} So, for $( s_0 ,s_1 ) \in S$, we obtain $\| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }^{2}=\| u\| _{W}^{2}+2\mathop{\rm Re}\langle A^{2}u^{( 2) },A^{\ast 2}u^{( 2) }\rangle _{L_2 ( \mathbb{R}_+,H) }.$ We have \begin{align*} &\langle A^{2}u^{( 2) },A^{\ast 2}u^{( 2) }\rangle _{L_2 ( \mathbb{R}_+,H) }\\ &=\langle A^{2}u^{(2) },(A^{\ast 2}A^{-2}-I)A^{2}u^{( 2) }\rangle _{L_2 ( \mathbb{R}_+,H) } +\| A^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}\\ &\geq \| A^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H)}^{2}-\| A^{\ast 2}A^{-2}-I\| \| A^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}. \end{align*} Now, we have also \begin{align*} \| A^{\ast 2}A^{-2}-I\| &=\| \int_{\sigma(A)} \big( \frac{\overline{\sigma }^{2}}{\sigma ^{2}}-1\big) dE_{\sigma } \| \\ &=\| \int_{\sigma ( A) } \frac{\overline{ \sigma }^{2}-\sigma ^{2}}{\sigma ^{2}}dE_{\sigma }\| \\ &= \| \int_{\sigma ( A) }\big( \frac{ e^{-2i\theta }-e^{2i\theta }}{e^{2i\theta }}\big) dE_{\sigma }\| \end{align*} where $\theta =\arg \sigma$. The spectrum $\sigma ( A)$ of the operator $A$ is contained in the sector $S_{\varepsilon }$ defined by Definition \ref{def1} and this means that $\big| \frac{e^{-2i\theta }-e^{2i\theta }}{e^{2i\theta }}\big| \leq 2\sin 2\varepsilon ,$ this proves that $\langle A^{2}u^{( 2) },A^{\ast 2}u^{( 2) }\rangle _{L_2 ( \mathbb{R}_+,H) } \geq ( 1-2\sin 2\varepsilon ) \| A^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}.$ Hence $\| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }^{2}\geq \| u\| _{W}^{2}+2(1-2\sin 2\varepsilon )\| A^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2},$ and so $\| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }^{2}\geq \| u\| _{W}^{2}+2\| C^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}-4\sin 2\varepsilon \| C^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}.$ Since \begin{align*} \| C^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2} &= \int_0^{\infty } \langle C^{2}u^{( 2) },C^{2}u^{( 2) }\rangle dt \\ &= \langle C^{\frac{5}{2}}u',C^{\frac{3}{2}}u^{( 2) }\rangle \big|_0 ^{\infty } -\langle C^{\frac{7}{2}}u,C^{\frac{1}{2}}u^{( 3) }\rangle \big| _0 ^{\infty } +\int_0^{\infty } \langle C^{4}u,u^{( 4) }\rangle dt \\ &= \int_0^{\infty } \langle C^{4}u,u^{(4) }\rangle dt \\ &\leq \| C^{4}u\| _{L_2 ( \mathbb{R}_+,H) }\| u^{( 4) }\| _{L_2 ( \mathbb{R}_+,H) } \\ &\leq \frac{1}{2}\big( \| C^{4}u\| _{L_2 ( \mathbb{R}_+,H) }^{2}+\| u^{( 4) }\|_{L_2 ( \mathbb{R}_+,H) }^{2}\big) \\ &= \frac{1}{2}\| u\| _{W}^{2} \end{align*} we conclude that \begin{align*} \| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }^{2} &\geq (1-2\sin 2\varepsilon )\| u\| _{W}^{2}+ 2\| C^{2}u^{(2)}\| _{L_2 ( \mathbb{R}_+,H) }^{2}\\ &\geq (1-2\sin 2\varepsilon )\| u\| _{W}^{2}, \end{align*} and finally, we have $\| u\| _{W}\leq (1-2\sin 2\varepsilon )^{-\frac{1}{2} }\| P_0 u\| _{L_2 ( \mathbb{R}_+,H) }.$ \end{proof} \begin{lemma} \label{lemme2} Let the operator $A$ be as in Definition \ref{def1}, then for $(s_0 ,s_1 )\in S$, the operator $P_0$ is a bijection between $W(\mathbb{R}_+,H,s_0 ,s_1 )$ and $L_2 ( \mathbb{R}_+,H)$. \end{lemma} \begin{proof} Consider the problem \begin{gather*} P_0 u=g, \\ u^{(s_0 )}(0)=u^{(s_1 )}(0)=0, \end{gather*} where $g\in L_2 ( \mathbb{R}_+,H)$ and $(s_0 ,s_1 )\in S$. The unique solution of this problem is given by (see \cite{Gas1}) $u(t)=V_1 (t)\Psi _0 +U_1 (t)\Psi _1 +\int_0 ^{\infty }G(t,x)g(x)dx,$ where $G(t,x)$ is the Green function of the problem \begin{gather*} P_0 u=g, \\ u(0)=u^{'}(0)=0, \end{gather*} and $V_1 (t)$ and $U_1 (t)$ are analytic semigroups generated by the operators $w_1 A$ and $w_2 A$ respectively, where $w_1$and $w_2$ are the roots on the left half plane of the equation $w^{4}=-1$. The elements $\Psi _0$and $\Psi _1$can be obtained by solving the system \begin{gather*} V_1 ^{(s_0 )}(t)\Psi _0 +U_1 ^{(s_0 )}(t)\Psi _1 +\int_0 ^{\infty }G^{(s_0 )}(t,x)g(x)dx=0, \\ V_1 ^{(s_1 )}(t)\Psi _0 +U_1 ^{(s_1 )}(t)\Psi _1 +\int_0 ^{\infty }G^{(s_1 )}(t,x)g(x)dx=0. \end{gather*} \end{proof} \begin{theorem} Let pencil \eqref{e3} belongs to the class $\Gamma$ and the couple $(s_0 ,s_1 )$ be in the class $S$ and $\sum_{j=0}^4 \beta _j \|A_{4-j}A^{j-4}\| <1,$ where $\beta _j =M_{4,j}( s_0 ,s_1 ) ( 1-2\sin 2\varepsilon) ^{-1/2}$. Then, problem \eqref{e1}, \eqref{e2} admits a unique regular solution. \end{theorem} \begin{proof} According to the Lemma \ref{lemme2} we can look for the solution of \eqref{e1}, \eqref{e2} in the form $u=$ $P_0 ^{-1}v$, where $v\in L_2 (\mathbb{R}_{+},H)$. In this case, problem \eqref{e1}, \eqref{e2} takes the form $$(I+P_1 P_0 ^{-1})v=f, \label{e8}$$ where $v\in L_2 (\mathbb{R}_{+},H)$, \begin{gather*} P_0 =\frac{d^{4}}{dt^{4}}+A^{4},\\ D( P_0 ) =W( \mathbb{R}_+,H,s_0 ,s_1 ) ,\\ P_1 =\sum_{j=0}^{4}A_{4-j}\frac{d^{j}}{dt^{j}}. \end{gather*} A sufficient condition for the existence of a unique solution of (\ref{e8}) is that $\| P_1 P_0 ^{-1}\| <1.$ Let $h\in L_2 (\mathbb{R}_{+},H)$, we have \begin{align*} \| P_1 P_0 ^{-1}h\| _{L_2 (\mathbb{R}_{+},H)} &=\| \sum_{j=0}^{4}A_{4-j}\frac{d^{j}}{dt^{j}} (P_0 ^{-1}h)\| _{L_2 (\mathbb{R}_{+},H)} \\ &=\| \sum_{j=0}^{4}A_{4-j}A^{j-4}A^{4-j}\frac{d^{j}}{dt^{j}} (P_0 ^{-1}h)\| _{L_2 (\mathbb{R}_{+},H)} \\ & \leq \sum_{j=0}^{4}\| A_{4-j}A^{j-4}\| \| A^{4-j}\frac{d^{j}}{dt^{j}}(P_0 ^{-1}h)\| _{L_2 (\mathbb{R}_{+},H)} \\ &=\sum_{j=0}^{4}\| A_{4-j}A^{j-4}\| \| C^{4-j} \frac{d^{j}}{dt^{j}}(P_0 ^{-1}h)\| _{L_2 (\mathbb{R}_{+},H)}. \end{align*} Since $P_0 ^{-1}h\in W(\mathbb{R}_{+},H,s_0 ,s_1 )$, $\| C^{4-j}\frac{d^{j}}{dt^{j}}(P_0 ^{-1}h)\| _{L_2 (\mathbb{R} _{+},H)}\leq M_{4,j}( s_0 ,s_1 ) \| P_0 ^{-1}h\| _{W}.$ Moreover, according to Lemma \ref{lemme1}, we have $\| P_0 ^{-1}h\| _{W}\leq (1-2\sin 2\varepsilon )^{-\frac{1 }{2}}\| h\| _{_{L_2 (\mathbb{R}_{+},H)}}.$ In this case $\| P_1 P_0 ^{-1}h\| _{L_2 (\mathbb{R}_{+},H)} \leq \sum_{j=0}^{4}\| A_{4-j}A^{j-4}\| M_{4,j}( s_0 ,s_1 ) (1-2\sin 2\varepsilon )^{-\frac{1}{2} }\| h\| _{_{L_2 (\mathbb{R}_{+},H)}}.$ Then $\| P_1 P_0 ^{-1}\| \leq \sum_{j=0}^{4}M_{4,j}( s_0 ,s_1 ) (1-2\sin 2\varepsilon )^{-1/2}\| A_{4-j}A^{j-4}\| .$ In view of the corresponding condition of the theorem, the above inequality implies $\| P_1 P_0 ^{-1}\| <1.$ \end{proof} \begin{remark} \label{rmk3} \rm The value of $M_{4,j}( s_0 ,s_1 )$ obtained in Theorem \ref{thm3} gives us the best estimation for the norm of the operator $P_1 P_0 ^{-1}$ in the sense that the coefficients $\beta _j$ are optimal in the inequality $\sum_{j=0}^4 \beta _j \|A_{4-j}A^{j-4}\| <1.$ \end{remark} \subsection*{Acknowledgements} We would like to thank the anonymous referees for their constructive and helpful comments which improved the quality of this paper. \begin{thebibliography}{00} \bibitem{Gas1} M. 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