\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2002(2002), No. 58, pp. 1--13.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2002 Southwest Texas State University.} \vspace{1cm}} \begin{document} \title[\hfilneg EJDE--2002/58\hfil Boundary-value problems for the biharmonic equation] {Boundary-value problems for the biharmonic equation with a linear parameter} \author[Yakov Yakubov\hfil EJDE--2002/58\hfilneg] {Yakov Yakubov} \address{Yakov Yakubov \hfil\break Raymond and Beverly Sackler Faculty of Exact Sciences \hfil\break School of Mathematical Sciences, Tel-Aviv University \hfil\break Ramat-Aviv 69978, Israel} \email{yakubov@post.tau.ac.il} \date{} \thanks{Submitted April 25, 2002. Published June 18, 2002.} \subjclass[2000]{35J40} \keywords{Biharmonic equation, isomorphism, boundary-value problem} \begin{abstract} We consider two boundary-value problems for the equation $$\Delta^2 u(x,y)-\lambda \Delta u(x,y)=f(x,y)$$ with a linear parameter on a domain consisting of an infinite strip. These problems are not elliptic boundary-value problems with a parameter and therefore they are non-standard. We show that they are uniquely solvable in the corresponding Sobolev spaces and prove that their generalized resolvent decreases as $1/|\lambda|$ at infinity in $L_2(\mathbb{R}\times (0,1))$ and $W_2^1(\mathbb{R}\times (0,1))$. \end{abstract} \maketitle \newtheorem{theorem}{Theorem}[section] \numberwithin{equation}{section} \section{Formulation of the problem} The main objective of this paper is to find estimates for the generalized resolvent for the problem \begin{gather} L(\lambda,D_x,D_y)u:=\Delta^2 u(x,y)-\lambda \Delta u(x,y)=f(x,y), \quad (x,y)\in \Omega, \label{1.1}\\ u(x,0)=u(x,1)=\frac {\partial u(x,0)}{\partial y}=\frac {\partial u(x,1)}{\partial y}=0, \quad x\in \mathbb{R}, \label{1.2} \end{gather} where $D_x=\frac \partial{\partial x}$, $D_y=\frac \partial{\partial y}$, $\lambda\in \mathbb{C}$ and $\Omega :=(-\infty,\infty)\times[0,1]\subset \mathbb{R}^2$. Known results on this subject treat elliptic boundary-value problems with a parameter, mostly in bounded domains \cite{AN,AV,KM,KMR,Y,YY}. Note that (\ref{1.1})--(\ref{1.2}) is not an elliptic boundary-value problem with a parameter \cite[p. 98]{KMR} (one should add a term such as $\lambda^2u(x,y)$ to get ellipticity with a parameter). Moreover, $\Omega$ is an unbounded domain. This fact makes the problem non-standard and known results for boundary-value problems do not apply. We denote by $F_{x\to \sigma}$ the one-dimensional Fourier transform with respect to $x$, where $\sigma$ is the dual variable. Applying the operator $F_{x\to \sigma}$ to problem (\ref{1.1})--(\ref{1.2}) we obtain a boundary value problem for an ordinary differential equation of the fourth order with 2 parameters \begin{gather} L(\lambda,i\sigma,D_y)\widehat u := \big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+ \sigma^4+\lambda \sigma^2 \big] \widehat u(\sigma,y)=\widehat f(\sigma,y), \quad y \in [0,1], \label{1.3} \\ \widehat u(\sigma,0)=\widehat u(\sigma,1)=\frac {d\widehat u(\sigma,0)}{dy}= \frac {d\widehat u(\sigma,1)}{dy}=0, \label{1.4} \end{gather} where $\lambda\in \mathbb{C}$ and $\sigma \in \mathbb{R}$ are parameters, $\widehat u(\sigma,y):= (F_{x\to \sigma}u(x,y))(\sigma,y)$. To solve our main question, we start from the solvability of problem (\ref{1.3})--(\ref{1.4}) and get estimates of its solution depending on the parameters $\lambda$ and $\sigma$. \section{Isomorphism and coerciveness of the equation on the whole axis} Consider equation (\ref{1.3}) on the whole axis, i.e., $$L(\lambda,i\sigma,D_y)u:= \big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+\sigma^4+\lambda \sigma^2\big] u(y)=f(y), \quad y \in \mathbb{R}. \label{2.1}$$ \begin{theorem} \label{thm1} For all complex numbers $\lambda$ satisfying $|\arg\lambda| \le \pi-\varepsilon$, where $\varepsilon>0$ is arbitrary, and $\sigma\in \mathbb{R}$, the operator ${\mathbb{L}(\lambda,\sigma)}:$ $u\rightarrow {\mathbb{L}(\lambda,\sigma)}u:= L(\lambda,i\sigma,D_{y})u$ from $W^{4}_{q}(\mathbb{R})$ onto $L_{q}(\mathbb{R})$, where $q\in(1,\infty)$, is an isomorphism and for these $\lambda$ the following estimates hold for solutions of (\ref{2.1}) \begin{gather} \| u\|_{W_{q}^4(\mathbb{R})}+\sigma^2\| u\|_{W_{q}^2(\mathbb{R})} +\sigma^4\| u\|_{L_{q}(\mathbb{R})}\le C(\varepsilon)\| f\|_{L_{q}(\mathbb{R})}, \quad |\arg\lambda|\le \pi-\varepsilon, \quad \sigma \in \mathbb{R}, \label{2.2}\\ \| u\|_{W_{q}^2(\mathbb{R})}+\sigma^2\| u\|_{L_{q}(\mathbb{R})} \le \frac {C(\varepsilon)} {|\lambda|} \| f\|_{L_{q}(\mathbb{R})}, \quad |\arg\lambda|\le \pi-\varepsilon, \quad \sigma \in \mathbb{R}. \label{2.3} \end{gather} \end{theorem} \paragraph{Proof} The operator $\mathbb{L}(\lambda,\sigma)$ acts from $W_{q}^4(\mathbb{R})$ into $L_{q}(\mathbb{R})$ linearly and continuously. Let us prove that if $f \in L_{q}(\mathbb{R})$ then (\ref{2.1}) has a solution $u$ in $W_{q}^4(\mathbb{R})$ and for this solution estimates (\ref{2.2})-(\ref{2.3}) hold. With the substitution $$u''(y)-\sigma^2u(y)=v(y), \quad y \in \mathbb{R}, \label{2.4}$$ equation (\ref{2.1}) is reduced to $$v''(y)-(\sigma^2+\lambda)v(y)=f(y), \quad y \in \mathbb{R}. \label{2.5}$$ By a theorem in \cite[p. 109]{YY}, equation (\ref{2.5}), for $|\arg \lambda|\le \pi-\varepsilon$, $\sigma \in \mathbb{R}$, has a solution $v \in W_{q}^2(\mathbb{R})$ and $$\| v\|_{W_{q}^2(\mathbb{R})}+|\lambda+\sigma^2| \| v\|_{L_{q}(\mathbb{R})} \le C(\varepsilon) \| f\|_{L_{q}(\mathbb{R})}, \quad |\arg \lambda|\le \pi-\varepsilon, \ \sigma \in \mathbb{R}. \label{2.6}$$ Apply now the same theorem \cite[p. 109]{YY} to (\ref{2.4}). Then for $\sigma \in \mathbb{R}$, and for $v\in W_{q}^2(\mathbb{R})$, (\ref{2.4}) has a solution $u \in W_{q}^4(\mathbb{R})$ and \begin{gather} \| u\|_{W_{q}^4(\mathbb{R})}+\sigma^2\| u\|_{W_{q}^2(\mathbb{R})} +\sigma^4\| u\|_{L_{q}(\mathbb{R})}\le C(\| v\|_{W_{q}^2(\mathbb{R})} +\sigma^2 \| v\|_{L_{q}(\mathbb{R})}), \quad \sigma \in \mathbb{R}, \label{2.7}\\ \| u\|_{W_{q}^2(\mathbb{R})}+\sigma^2\| u\|_{L_{q}(\mathbb{R})} \le C\| v\|_{L_{q}(\mathbb{R})}, \quad \sigma \in \mathbb{R}. \label{2.8} \end{gather} Consequently, from (\ref{2.6}) and (\ref{2.7}) it follows that for $|\arg \lambda|\le \pi-\varepsilon$ and $\sigma \in \mathbb{R}$, (\ref{2.1}) has a solution $u \in W_{q}^4(\mathbb{R})$ and \begin{align*} \| u\|_{W_{q}^4(\mathbb{R})}+\sigma^2\| u\|_{W_{q}^2(\mathbb{R})} +\sigma^4\| u\|_{L_{q}(\mathbb{R})} \le & C(\varepsilon) (\| f\|_{L_{q}(\mathbb{R})}+\frac {\sigma^2} {|\lambda+\sigma^2|} \| f\|_{L_{q}(\mathbb{R})}) \\ \le& C(\varepsilon) \| f\|_{L_{q}(\mathbb{R})}, \quad |\arg \lambda|\le \pi-\varepsilon, \quad \sigma \in \mathbb{R}, \end{align*} i.e., estimate (2.2) has been proved. In the last inequality we have used that $$|\lambda+\sigma^2|\ge C(\varepsilon)(|\lambda|+\sigma^2),\quad |\arg\lambda|\le \pi-\varepsilon,\ \sigma\in\mathbb{R},$$ which can be easily checked. In what follows we will often use the fact. On the other hand, from (\ref{2.6}) and (\ref{2.8}), it follows that $$\| u\|_{W_{q}^2(\mathbb{R})}+\sigma^2\| u\|_{L_{q}(\mathbb{R})} \le \frac {C(\varepsilon)}{|\lambda+\sigma^2|} \| f\|_{L_{q}(\mathbb{R})} \le \frac {C(\varepsilon)}{|\lambda|} \| f\|_{L_{q}(\mathbb{R})},$$ with $|\arg \lambda|\le \pi-\varepsilon$ and $\sigma \in \mathbb{R}$; i.e., estimate (\ref{2.3}) holds.\quad\hfill\qed \section{Solvability of the boundary-value problem for the homogeneous equation} Consider a boundary-value problem for the ordinary differential equation of the fourth order on $[0,1]$ \begin{gather} L(\lambda,i\sigma,D_y)u:= \big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+\sigma^4+\lambda \sigma^2 \big] u(y)=0, \ \ y \in [0,1], \label{3.1}\\ u(0)=f_1, \quad u(1)=f_2, \quad \frac {d u(0)}{dy}=f_3, \quad \frac {d u(1)}{dy}=f_4, \label{3.2} \end{gather} where $\lambda\in \mathbb{C}$ and $\sigma \in \mathbb{R}$ are parameters, $f_\nu$ are complex numbers. Here (1.3) is homogeneous but the boundary conditions (1.4) are not. \begin{theorem} \label{thm2} For each $\varepsilon>0$ there exists $M>0$ such that for all complex numbers $\lambda$ satisfying $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$ and for all real numbers $\sigma\in\mathbb{R},\ \sigma\neq 0$ problem (\ref{3.1})--(\ref{3.2}) has a unique solution $u(y)$ that belongs to $C^{\infty}[0,1]$ and for this solution the following inequality holds for $n=0,1,2,\dots$ \begin{align} &\|u^{(n)}\|_{L_q(0,1)} \label{3.3} \\ &\le C(\varepsilon,K) \begin{cases}(|\lambda|^{\frac{n-1}2-\frac 1{2q}}+|\sigma|^{n-1})(|f_1|+|f_2|)\\ +(|\lambda|^{\frac{n-1}2-\frac 1{2q}}+\frac{|\sigma|^{n-1}} {|\lambda|^{1/2}})(|f_3|+|f_4|),& 0<|\sigma|\le K,\\ (\frac{(|\lambda|+\sigma^2)^{\frac{n+1}2-\frac 1{2q}}|\sigma|}{|\lambda|}+ \frac{(|\lambda|+\sigma^2)|\sigma|^{n-\frac 1 q}}{|\lambda|}) (|f_1|+|f_2|)\\ +(\frac{(|\lambda|+\sigma^2)^{\frac{n+1}2-\frac 1{2q}}}{|\lambda|} +\frac{(|\lambda|+\sigma^2)^{1/2}|\sigma|^{n-\frac 1 q}} {|\lambda|})(|f_3|+|f_4|),& |\sigma|> K. \end{cases} \nonumber \end{align} \end{theorem} \paragraph{Proof} Let us prove that for any complex numbers $f_{\nu}$, $\nu=1,\dots,4$, problem (\ref{3.1})--(\ref{3.2}) has a unique solution $u(y)$ in $C^{\infty}[0,1]$ and let us estimate this solution. A characteristic equation of (\ref{3.1}) has the form $$\omega^4-(2\sigma^2+\lambda)\omega^2+\sigma^4+\lambda \sigma^2=0 \label{3.4}$$ which has the following roots $$\omega_1=-|\sigma^2+\lambda|^{\frac 1{2}}\text{e}^{i\frac {\arg (\sigma^2+\lambda)}{2}}, \quad \omega_2=-|\sigma|,\quad \omega_3=|\sigma^2+\lambda|^{\frac 1{2}}\text{e}^{i\frac {\arg (\sigma^2+\lambda)} {2}},\quad \omega_4=|\sigma|.$$ Then for $|\arg \lambda|\le \pi-\varepsilon$, $\sigma \in \mathbb{R}$, $\sigma\ne 0$, the general solution of (\ref{3.1}) has the form $$u(y)=C_1 \text{e}^{\omega_1 y}+ C_2 \text{e}^{\omega_2 y} + C_3 \text{e}^{\omega_3(y-1)} + C_4 \text{e}^{\omega_4(y-1)}. \label{3.5}$$ Substituting (\ref{3.5}) into (\ref{3.2}), we obtain a system for finding $C_i$, $i=1,\dots,4$, $$\begin{gathered} C_1+ C_2 + C_3 \text{e}^{-\omega_3}+ C_4 \text{e}^{-\omega_4}=f_1,\\ C_1\text{e}^{\omega_1}+ C_2\text{e}^{\omega_2} + C_3 + C_4 =f_2,\\ C_1\omega_1+ C_2\omega_2 + C_3\omega_3 \text{e}^{-\omega_3}+ C_4\omega_4 \text{e}^{-\omega_4}=f_3,\\ C_1 \omega_1\text{e}^{\omega_1}+ C_2 \omega_2\text{e}^{\omega_2} + C_3\omega_3 + C_4\omega_4 =f_4.\\ \end{gathered} \label{3.6}$$ Since for $|\arg \lambda|\le \pi-\varepsilon$, $\sigma \in \mathbb{R}$ we have $\frac {\pi}{2}+\frac \varepsilon 2 <\arg\omega_1<\frac {3\pi}2-\frac \varepsilon 2$ and $|\arg\omega_3|<\frac {\pi}{2}-\frac \varepsilon 2$, then $\mathop{\rm Re}\omega_1<-\delta(\varepsilon)(|\lambda|^{\frac 12}+ |\sigma|)$ and $-\mathop{\rm Re}\omega_3<-\delta(\varepsilon)(|\lambda|^{\frac 12}+|\sigma|)$, where $\delta(\varepsilon)>0$. The determinant of system (\ref{3.6}) is $$D(\lambda,\sigma)=\begin{vmatrix} 1 & 1 &\text{e}^{-\omega_3} &\text{e}^{-\omega_4}\\ \text{e}^{\omega_1} &\text{e}^{\omega_2} &1 &1\\ \omega_1 &\omega_2 &\omega_3 \text{e}^{-\omega_3} &\omega_4 \text{e}^{-\omega_4}\\ \omega_1\text{e}^{\omega_1} & \omega_2\text{e}^{\omega_2} &\omega_3 &\omega_4 \end{vmatrix}.$$ Calculating this determinant and taking into account that $\omega_3=-\omega_1$ and $\omega_4=-\omega_2$, we obtain $$D(\lambda,\sigma)=\omega_2\Big[(\omega_1^2+\omega_2^2)(1-\text{e}^{2\omega_1})\frac {1-\text{e}^{2\omega_2}} {\omega_2} -2\omega_1(1+\text{e}^{2(\omega_1+\omega_2)}+\text{e}^{2\omega_2}+\text{e}^{2\omega_1}- 4\text{e}^{\omega_1+\omega_2})\Big].$$ Let $0<|\sigma|\le K$. Because of $\lim\limits_{\sigma\to 0}\frac{1-\text{e}^{2\omega_2}}{\omega_2} =-2\neq 0$, one can choose $M$ such a big that for all $|\arg\lambda|\le\pi-\varepsilon$, $|\lambda|\ge M$ the following true ($\omega_1^2=\omega_2^2+\lambda=\sigma^2+\lambda$) $$|(\omega_1^2+\omega_2^2)(1-\text{e}^{2\omega_1})\frac {1-\text{e}^{2\omega_2}}{\omega_2}|\ge C(K,\varepsilon)(|\lambda|+\sigma^2)$$ and $$|2\omega_1(1+\text{e}^{2(\omega_1+\omega_2)}+\text{e}^{2\omega_2}+\text{e}^{2\omega_1}- 4\text{e}^{\omega_1+\omega_2})|\le\frac{C(K,\varepsilon)}2(|\lambda|+\sigma^2).$$ Then $|D(\lambda,\sigma)|\ge\frac{C(K,\varepsilon)}2 |\sigma|(|\lambda|+\sigma^2)$. In the case of $|\sigma|> K$, we write the determinant as \begin{align*} D(\lambda,\sigma)=&[\omega_2(1-\text{e}^{\omega_1})-\omega_1(1-\text{e}^{\omega_2})] [\omega_2(1+\text{e}^{\omega_1})-\omega_1(1+\text{e}^{\omega_2})] \\ &+(\omega_2-\omega_1)^2 \text{e}^{2(\omega_1+\omega_2)}+R(\lambda,\sigma), \end{align*} where $|R(\lambda,\sigma)|\le C(|\lambda|^{1/2}+|\sigma|)/\text{e}^{C(\varepsilon)|\sigma|}$. Using that $\omega_1^2=\omega_2^2+\lambda$ we have \begin{align*} D(\lambda,\sigma)=&\frac{-\lambda(1-\text{e}^{\omega_2})^2+2\omega_2^2\text{e}^{\omega_2}-\omega_2^2\text{e}^ {2\omega_2}- 2\omega_2^2\text{e}^{\omega_1}+\omega_2^2\text{e}^{2\omega_1}}{\omega_2(1-\text{e}^{\omega_1})+ \omega_1(1-\text{e}^{\omega_2})}\\ &\times\frac{-\lambda(1+\text{e}^{\omega_2})^2-2\omega_2^2\text{e}^{\omega_2}-\omega_2^2\text{e}^ {2\omega_2} +2\omega_2^2\text{e}^{\omega_1}+\omega_2^2\text{e}^{2\omega_1}}{\omega_2(1+\text{e}^{\omega_1})+ \omega_1(1+\text{e}^{\omega_2})}\\ &+\frac{\lambda^2}{(\omega_2+\omega_1)^2}\text{e}^{2(\omega_1+\omega_2)}+R(\lambda,\sigma). \end{align*} We have $|\omega_2+\omega_1|^2=(\mathop{\rm Re}(\omega_2+\omega_1))^2+(\mathop{\rm Im}(\omega_2+\omega_1))^2= (-|\sigma|+\mathop{\rm Re}\omega_1)^2+(\mathop{\rm Im}\omega_1)^2=\sigma^2+|\omega_1|^2-2|\sigma|\mathop{\rm Re}\omega_1\ge\sigma^2+|\lambda+\sigma^2|\ge C(\varepsilon)(|\lambda|+\sigma^2)$, because $\mathop{\rm Re}\omega_1<0$. Therefore, for all $|\arg\lambda|\le\pi-\varepsilon$, $|\lambda|\ge M$ and $|\sigma|>K$, \begin{align*} |D(\lambda,\sigma)|&\ge C(K)\frac{|\lambda|^2}{|\lambda|+\sigma^2}-C(\varepsilon)\frac{|\lambda|^2}{(|\lambda|+\sigma^2)\text{e}^ {C(\varepsilon)(|\lambda|^{1/2}+|\sigma|)}}-C\frac{|\lambda|^{1/2}+|\sigma|}{\text{e}^{C(\varepsilon)|\sigma|}}\\ &\ge \frac{C(K)}2 \frac{|\lambda|^2}{|\lambda|+\sigma^2}. \end{align*} Hence, for $|\arg\lambda|\le\pi-\varepsilon$, $|\lambda|\ge M$ and $\sigma\in\mathbb{R}$, $\sigma\neq 0$, system (\ref{3.6}) has a unique solution \begin{gather*} C_1=\frac { \begin{vmatrix} f_1 & 1 &\text{e}^{-\omega_3} &\text{e}^{-\omega_4}\\ f_2 &\text{e}^{\omega_2} &1 &1\\ f_3 &\omega_2 &\omega_3 \text{e}^{-\omega_3} &\omega_4 \text{e}^{-\omega_4}\\ f_4 & \omega_2\text{e}^{\omega_2} &\omega_3 &\omega_4 \end{vmatrix}} {D(\lambda,\sigma)}, \quad C_2=\frac {\begin{vmatrix} 1 & f_1 &\text{e}^{-\omega_3} &\text{e}^{-\omega_4}\\ \text{e}^{\omega_1} &f_2 &1 &1\\ \omega_1 &f_3 &\omega_3 \text{e}^{-\omega_3} &\omega_4 \text{e}^{-\omega_4}\\ \omega_1\text{e}^{\omega_1} &f_4 &\omega_3 &\omega_4 \end{vmatrix}} {D(\lambda,\sigma)}, \\ C_3=\frac {\begin{vmatrix} 1 & 1 &f_1 &\text{e}^{-\omega_4}\\\text{e}^{\omega_1} &\text{e}^{\omega_2} &f_2 &1\\ \omega_1 &\omega_2 &f_3 &\omega_4 \text{e}^{-\omega_4}\\ \omega_1\text{e}^{\omega_1} & \omega_2\text{e}^{\omega_2} &f_4 &\omega_4 \end{vmatrix}} {D(\lambda,\sigma)}, \quad C_4=\frac {\begin{vmatrix} 1 & 1 &\text{e}^{-\omega_3} &f_1\\ \text{e}^{\omega_1} &\text{e}^{\omega_2} &1 &f_2\\ \omega_1 &\omega_2 &\omega_3 \text{e}^{-\omega_3} &f_3\\ \omega_1\text{e}^{\omega_1} & \omega_2\text{e}^{\omega_2} &\omega_3 &f_4 \end{vmatrix}} {D(\lambda,\sigma)}, \end{gather*} where $$|D(\lambda,\sigma)|\ge C(\varepsilon,K)\begin{cases} |\sigma|(|\lambda|+\sigma^2),\quad 0<|\sigma|\le K,\\ \frac{|\lambda|^2}{|\lambda|+\sigma^2},\quad |\sigma|>K.\end{cases}$$ Calculating these determinants one can obtain that $$|C_{1,3}|\le C(\varepsilon,K)\begin{cases}\frac{|\lambda|}{(|\lambda|+\sigma^2)^{\frac 3 2}} [|f_1|+|f_2|+(|f_3|+|f_4|)\frac{|\lambda|+\sigma^2}{|\lambda|}],& 0<|\sigma|\le K,\\ \frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda|}[(|f_1|+ |f_2|)|\sigma|+|f_3|+|f_4|],& |\sigma|> K,\end{cases}$$ and $$|C_{2,4}|\le C(\varepsilon,K)\begin{cases}\frac{|\lambda|}{|\sigma|(|\lambda|+\sigma^2)^{\frac 3 2}} [(|f_1|+|f_2|)(|\lambda|+\sigma^2)^{1/2}+|f_3|+|f_4|], & 0<|\sigma|\le K,\\ \frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda|}[(|f_1|+ |f_2|)(|\lambda|+\sigma^2)^{1/2}+|f_3|+|f_4|], & |\sigma|> K,\end{cases}$$ or $$|C_{1,3}|\le C(\varepsilon,K)\begin{cases}\frac 1{|\lambda|^{1/2}} [|f_1|+|f_2|+|f_3|+|f_4|],& 0<|\sigma|\le K,\\ \frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda|}[(|f_1|+ |f_2|)|\sigma|+|f_3|+|f_4|],& |\sigma|> K,\end{cases}$$ and $$|C_{2,4}|\le C(\varepsilon,K)\begin{cases}\frac 1{|\sigma|} [|f_1|+|f_2|+\frac 1{|\lambda|^{1/2}}(|f_3|+|f_4|)],& 0<|\sigma|\le K,\\ \frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda|}[(|f_1|+ |f_2|)(|\lambda|+\sigma^2)^{1/2}+|f_3|+|f_4|], & |\sigma|> K.\end{cases}$$ From (\ref{3.5}) for $n=0,1,2,\dots$, we have \begin{align*} \|u^{(n)}&\|_{L_q(0,1)}\\ \le& |C_1||\omega_1|^n\|\text{e}^{\omega_1\cdot}\|_{L_q(0,1)}+ |C_2||\omega_2|^n\|\text{e}^{\omega_2\cdot}\|_{L_q(0,1)}\\ &+|C_3||\omega_3|^n\|\text{e}^{\omega_3(\cdot-1)}\|_{L_q(0,1)}+ |C_4||\omega_4|^n\|\text{e}^{\omega_4(\cdot-1)}\|_{L_q(0,1)}\\ \le & C(\varepsilon)[(|C_1|+|C_3|)(|\lambda|+\sigma^2)^{\frac n 2 -\frac 1{2q}}+(|C_2|+|C_4|)|\sigma|^n \big(\frac{1-\text{e}^{-|\sigma|q}}{|\sigma|}\big)^{1/q}]\\ \le& C(\varepsilon,K) \begin{cases}(|\lambda|^{\frac{n-1}2-\frac 1{2q}}+|\sigma|^{n-1})(|f_1|+|f_2|)\\ +(|\lambda|^{\frac{n-1}2-\frac 1{2q}}+\frac{|\sigma|^{n-1}} {|\lambda|^{1/2}})(|f_3|+|f_4|), & 0<|\sigma|\le K,\\[2pt] (\frac{(|\lambda|+\sigma^2)^{\frac{n+1}2-\frac 1{2q}}|\sigma|}{|\lambda|}+ \frac{(|\lambda|+\sigma^2)|\sigma|^{n-\frac 1 q}}{|\lambda|}) (|f_1|+|f_2|) \\ +(\frac{(|\lambda|+\sigma^2)^{\frac{n+1}2-\frac 1{2q}}}{|\lambda|} +\frac{(|\lambda|+\sigma^2)^{1/2}|\sigma|^{n-\frac 1 q}} {|\lambda|})(|f_3|+|f_4|), & |\sigma|> K. \end{cases} \end{align*} \quad\hfill\qed From (\ref{3.3}), in particular, follows $$\|u(\cdot)\|_{L_q(0,1)}\le C(\varepsilon,K) \begin{cases}\frac 1{|\sigma|}(|f_1|+|f_2|)+\frac 1{|\sigma||\lambda|^ {\frac 1 2}}(|f_3|+|f_4|), &0<|\sigma|\le K,\\ \frac{|\lambda|+\sigma^2}{|\lambda||\sigma|^{1/q}}(|f_1|+|f_2|)\\ +\frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda||\sigma|^{1/q}}(|f_3|+|f_4|), & |\sigma|>K,\end{cases}\label{3.7}$$ and $$\|u'(\cdot)\|_{L_q(0,1)}\le C(\varepsilon,K)\begin{cases} |f_1|+|f_2|+\frac 1{|\lambda|^ {\frac 1 {2q}}}(|f_3|+|f_4|), & 0<|\sigma|\le K,\\ \frac{(|\lambda|+\sigma^2)|\sigma|^{1-\frac 1 q}}{|\lambda|}(|f_1|+|f_2|)\\ +\frac{(|\lambda|+\sigma^2)^{1-\frac 1 {2q}}}{|\lambda|}(|f_3|+|f_4|), &|\sigma|>K. \end{cases}\label{3.8}$$ \section{Isomorphism of the boundary value problem with a linear parameter and estimates for its solution} Now we consider the main problem (\ref{1.1})--(\ref{1.2}). \begin{theorem} \label{thm3} For each $\varepsilon>0$ there exists $M>0$ such that for all complex numbers $\lambda$ satisfying $|\arg\lambda|\le\pi-\varepsilon$, $|\lambda|\ge M$, the operator $\mathbb{L}(\lambda): u\to \mathbb{L}(\lambda) u:= L(\lambda,D_x,D_y)u$ from $W^{s+4}_2(\mathbb{R}\times(0,1)$, $u(x,0)=u(x,1)=u'_y(x,0)=u'_y(x,1))$ onto $W^s_2(\mathbb{R}\times(0,1))$, where $s\ge 0$, is an isomorphism and for these $\lambda$ the following estimates hold $$\|u\|_{W^k_2(\mathbb{R}\times(0,1))}\le C(\varepsilon)\frac 1{|\lambda|}\|f\|_ {W^k_2(\mathbb{R}\times(0,1))},\quad k=0,1,$$ where $u(x,y)$ is a solution of (\ref{1.1})--(\ref{1.2}). \end{theorem} \paragraph{Proof} Fix $\lambda$ such that $|\arg\lambda|\le\pi-\varepsilon,\ |\lambda|\ge M$. In this case problem (\ref{1.1})--(\ref{1.2}) becomes an elliptic boundary value problem with constant coefficients (for the definition see, e.g., \cite{NP}). Then, the required isomorphism follows from \cite[Theorem 1.1, p.44]{NP}. Indeed, to check the condition there, one should prove that there is no eigenvalue $\mu$ on the imaginary axis of the spectral problem \begin{gather*} u^{(4)}(y)+(2\mu^2-\lambda)u''(y)+(\mu^4-\lambda\mu^2)u(y)=0,\ \quad y\in (0,1),\\ u(0)=u(1)=u'(0)=u'(1)=0. \end{gather*} Let us prove this by contradiction. If there is such $\mu$ then $\mu^2\le 0$ and there is an eigenfunction $u$, $u\not\equiv 0$, i.e., $\int\limits^1_0 |u(y)|^2dy> 0$. Moreover, $\int\limits^1_0|u'(y)|^2dy> 0$, otherwise $u(y)$ would be constant and, taking into account $u(0)=0$, $u(y)\equiv 0$. Multiply the first equation of the above spectral problem by $\overline{u(y)}$ and integrate by parts on $(0,1)$. Then, using boundary conditions, we get $$\int^1_0|u''(y)|^2dy+(\lambda-2\mu^2)\int^1_0|u'(y)|^2dy+\mu^2(\mu^2-\lambda) \int^1_0|u(y)|^2dy=0,$$ i.e., \begin{gather*} \int\limits^1_0|u''(y)|^2dy+(\mathop{\rm Re}\lambda-2\mu^2)\int\limits^1_0|u'(y)|^2dy+ \mu^2(\mu^2-\mathop{\rm Re}\lambda)\int\limits^1_0|u(y)|^2dy=0,\\ \mathop{\rm Im}\lambda\int\limits^1_0|u'(y)|^2dy-\mu^2\mathop{\rm Im}\lambda\int\limits^1_0|u(y)|^2dy =0.\end{gather*} But from the second equation follows that $\mathop{\rm Im}\lambda=0$. Then $\mathop{\rm Re}\lambda\ge M>0$ and this contradicts to the first equation. Now prove estimates of the theorem. First, consider a solution of problem (\ref{1.3})--(\ref{1.4}). We find the solution of problem (\ref{1.3})--(\ref{1.4}) in the form $\widehat u=u_1+u_2$, where $u_1$ is a restriction on [0,1] of a solution $\tilde{u}_1$ of the equation $$\big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+\sigma^4+\lambda \sigma^2 \big] \tilde u_1(\sigma,y)=\tilde f(\sigma,y), \quad y \in\mathbb{R},$$ where $\tilde f\in L_q(\mathbb{R})$ is an extension of $\widehat f\in L_q(0,1)$ such that the extension operator $\widehat f\to\tilde f: L_q(0,1)\to L_q(\mathbb{R})$ is bounded \cite[p.314]{T}. Then for $u_1$, from Theorem 2.1, estimates (\ref{2.2})-(\ref{2.3}) hold, i.e., $$\begin{gathered} \| u_1(\sigma,\cdot)\|_{W_{q}^4(0,1)}+\sigma^2\| u_1(\sigma,\cdot)\|_{W_{q}^2 (0,1)}+\sigma^4\| u_1(\sigma,\cdot)\|_{L_{q}(0,1)} \le C(\varepsilon)\|\widehat f(\sigma,\cdot)\|_{L_{q}(0,1)},\\ \| u_1(\sigma,\cdot)\|_{W_{q}^2(0,1)} +\sigma^2\| u_1(\sigma,\cdot)\|_{L_{q}(0,1)} \le \frac {C(\varepsilon)} {|\lambda|} \|\widehat f(\sigma,\cdot)\|_{L_{q}(0,1)}, \end{gathered}\label{4.1}$$ where $|\arg \lambda|\le \pi-\varepsilon, \ \sigma \in \mathbb{R}$. The second summand $u_2$ in the form of $\widehat u$ is, by virtue of Theorem 3.1, a unique solution of the problem $$\begin{gathered} \big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+\sigma^4+\lambda \sigma^2 \big] u_2(\sigma,y)=0, \quad y \in [0,1],\\ u_2(\sigma,0)=-u_1(\sigma,0), \quad u_2(\sigma,1)=-u_1(\sigma,1),\\ \frac {du_2(\sigma,0)}{dy}=-\frac {du_1(\sigma,0)}{dy}, \quad \frac {d u_2(\sigma,1)}{dy}=-\frac {d u_1(\sigma,1)}{dy}, \end{gathered}\label{4.2}$$ where $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$, and $\sigma\in\mathbb{R},\ \sigma\neq 0$. Then from (\ref{3.7}) for the solution $u_2$ of problem (\ref{4.2}) we have $$\|u_2(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon,K) \begin{cases}\frac 1{|\sigma|}(|u_1(\sigma,0)|+|u_1(\sigma,1)|)\\ +\frac 1{|\sigma||\lambda|^{1/2}}(|u'_1(\sigma,0)| +|u'_1(\sigma,1)|),& 0<|\sigma|\le K,\\[3pt] \frac{|\lambda|+\sigma^2}{|\lambda||\sigma|^{1/q}}(|u_1(\sigma,0)|+|u_1(\sigma,1)|)\\ +\frac{(|\lambda|+\sigma^2)^{1/2}}{|\lambda||\sigma|^{1/q}}(|u'_1(\sigma,0)| +|u'_1(\sigma,1)|), & |\sigma|>K,\end{cases}\label{4.3}$$ and from (\ref{3.8}), $$\big\|\frac{du_2(\sigma,\cdot)}{dy}\big\|_{L_q(0,1)}\le C(\varepsilon,K) \begin{cases} |u_1(\sigma,0)| +|u_1(\sigma,1)|\\ +\frac 1{|\lambda|^{\frac 1 {2q}}}(|u'_1(\sigma,0)| +|u'_1(\sigma,1)|), &0<|\sigma|\le K,\\[3pt] \frac{(|\lambda|+\sigma^2)|\sigma|^{1-\frac 1 q}}{|\lambda|}(|u_1(\sigma,0)|+|u_1(\sigma,1)|)\\ +\frac{(|\lambda|+\sigma^2)^{1-\frac 1 {2q}}}{|\lambda|}(|u'_1(\sigma,0)|+|u'_1(\sigma,1)|), &|\sigma|>K.\end{cases}\label{4.4}$$ From \cite[Ch.3, \S 10, Theorem 10.4]{BIN} it follows that $$\|u^{(j)}\|_{C[0,1]}\le C (h^{1-\gamma}\|u^{(\ell)}\|_{L_q(0,1)}+ h^{-\gamma}\|u\|_{L_q(0,1)}),$$ where $j<\ell,\ 0 K$, \begin{align*} \|u_2(\sigma,\cdot)\|_{L_q(0,1)} \le& C(\varepsilon,K)\frac{(|\lambda|+\sigma^2)^{1/2} |\sigma|^{1-\frac 1 q}}{|\lambda||\sigma|}[(|u_1(\sigma,0)|+ |u_1(\sigma,1)|)(|\lambda|+\sigma^2)^{1/2}\\ &+|u'_1(\sigma,0)|+|u'_1(\sigma,1)|]\\ \le &C(\varepsilon,K)\frac{|\sigma|^{1-\frac 1 q}}{|\lambda|^{1/2}} [(|u_1(\sigma,0)| +|u_1(\sigma,1)|)|\lambda|^{1/2}\\ &+(|u_1(\sigma,0)|+|u_1(\sigma,1)|)|\sigma|+|u'_1(\sigma,0)|+|u'_1(\sigma,1)|]\\ \le &C(\varepsilon,K)[(|u_1(\sigma,0)|+|u_1(\sigma,1)|)|\sigma|^{1-\frac 1 q}\\ &+(|u_1(\sigma,0)|+|u_1(\sigma,1)|)\frac{|\sigma|^2}{|\lambda|^{1/2}} +(|u'_1(\sigma,0)|+|u'_1(\sigma,1)|)\frac{|\sigma|^{1-\frac 1 q}}{|\lambda|^{1/2}}]\\ \le& C(\varepsilon,K)[\mu_1^{-2+\frac 1 q}|\sigma|^{1-\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\mu_1^{1/q}|\sigma|^{1-\frac 1 q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)} +\mu_2^{-2+\frac 1 q}\frac{|\sigma|^{2}}{|\lambda|^{1/2}}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\mu_2^{1/q}\frac{|\sigma|^{2}}{|\lambda|^{1/2}}\|u_1(\sigma,\cdot)\|_{L_q(0,1)} +\mu_3^{-1+\frac 1 q}\frac{|\sigma|^{1-\frac 1 q}}{|\lambda|^{1/2}}\|u_1''(\sigma,\cdot) \|_{L_q(0,1)}\\ &+\mu_3^{1+\frac 1 q}\frac{|\sigma|^{1-\frac 1 q}}{|\lambda|^{1/2}} \|u_1(\sigma,\cdot)\|_{L_q(0,1)}]. \end{align*} Choose $\mu_1=|\sigma|^{1+q},\ \mu_2=|\lambda|^{\frac q 2},\ \mu_3^{1+\frac 1 q}= |\lambda|^{1/2}|\sigma|^{1+\frac 1 q}$, where $|\sigma|> K\gg 1$ and $|\lambda|\ge M\gg 1$. Then, by virtue of (\ref{4.1}), we obtain \begin{aligned} \|u_2(\sigma,\cdot)\|_{L_q(0,1)}\le &C(\varepsilon,K)[\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+\sigma^2 \|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+|\lambda|^{-q}\sigma^2\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \sigma^2\|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+|\lambda|^{-\frac q{q+1}}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+\sigma^2\|u_1(\sigma,\cdot) \|_{L_q(0,1)}]\\ \le &C(\varepsilon,K)\frac 1{|\lambda|}\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)},\quad |\sigma|>K. \label{4.6} \end{aligned} On the other hand, for $0<|\sigma|\le K$, from (\ref{4.3}) and (\ref{4.5}) we have ($|\lambda|\ge M$) \begin{align*} \|u_2(\sigma,\cdot)\|_{L_q(0,1)} \le& C(\varepsilon,K)[\frac 1{|\sigma|}(|u_1(\sigma,0)|+|u_1(\sigma,1)|) \\ &+\frac 1{|\sigma||\lambda|^{1/2}}(|u'_1(\sigma,0)|+|u'_1(\sigma,1)|)]\\ \le& C(\varepsilon,K)\frac 1{|\sigma|}[\mu_1^{-2+\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \mu_1^{1/q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\mu_2^{-1+\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \mu_2^{1+\frac 1 q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)}]. \end{align*} Choose $\mu_1=\mu_2=R\gg 1$. It implies, by virtue of (\ref{4.1}), that $$\|u_2(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon,K)\frac 1{|\sigma||\lambda|}\|\widehat f(\sigma,\cdot)\| _{L_q(0,1)},\quad 0<|\sigma|\le K.$$ The last inequality with (\ref{4.6}) taking into account gives us the following estimate for a solution of (\ref{4.2}) for $\sigma\in\mathbb{R}$, $\sigma\neq 0$ $$\|u_2(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon)\frac 1{|\lambda|}\big(\|\widehat f(\sigma,\cdot)\| _{L_q(0,1)}+\frac 1{|\sigma|}\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)}\big). \label{4.7}$$ Then from (\ref{4.1}) and (\ref{4.7}) for a solution $\widehat u(\sigma,y)=u_1(\sigma,y)+u_2(\sigma,y)$ of problem (\ref{1.3})--(\ref{1.4}) we have $$\|\widehat u(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon)\frac 1{|\lambda|} \big(\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)}+\frac 1{|\sigma|}\| \widehat f(\sigma,\cdot)\|_{L_q(0,1)}\big),$$ where $\sigma\in\mathbb{R}$, $\sigma\neq 0$, $|\arg\lambda|\le\pi-\varepsilon$, $|\lambda|\ge M$. Multiplying the last inequality on $|\sigma|$ we obtain $$\|\widehat{u'_x}(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon)\frac 1{|\lambda|}(\|\widehat{f'_x}(\sigma,\cdot)\| _{L_q(0,1)}+\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)}),\label{4.8}$$ where $\sigma\in\mathbb{R},\ \sigma\neq 0,\ |\arg\lambda|\le\pi-\varepsilon,\ |\lambda|\ge M$. From (\ref{4.4}) and (\ref{4.5}) we have for $0<|\sigma|\le K$ and $|\lambda|\ge M$ \begin{align*} \big\|\frac{du_2(\sigma,\cdot)}{dy}\big\|_{L_q(0,1)}\le& C(\varepsilon,K)[|u_1(\sigma,0)|+|u_1(\sigma,1)|+ |u'_1(\sigma,0)|+|u'_1(\sigma,1)|]\\ \le &C(\varepsilon,K)[\mu_1^{-2+\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \mu_1^{1/q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\mu_2^{-1+\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \mu_2^{1+\frac 1 q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)}]. \end{align*} Choose $\mu_1=\mu_2=R\gg 1$. Then (\ref{4.1}) implies $$\big\|\frac{du_2(\sigma,\cdot)}{dy}\big\|_{L_q(0,1)} \le C(\varepsilon,K)\frac 1{|\lambda|}\|\widehat f(\sigma,\cdot)\| _{L_q(0,1)},\quad 0<|\sigma|\le K.\label{4.9}$$ On the other hand, from (\ref{4.4}) and (\ref{4.5}), we have for $|\sigma|> K$ \begin{align*} \big\|\frac{du_2(\sigma,\cdot)}{dy}\big\|_{L_q(0,1)} \le& C(\varepsilon,K)[|\sigma|^{1-\frac 1 q} (|u_1(\sigma,0)|+|u_1(\sigma,1)|)\\ &+\frac{|\sigma|^{3-\frac 1 q}}{|\lambda|}(|u_1(\sigma,0)|+|u_1(\sigma,1)|) +|\lambda|^{-\frac 1{2q}}(|u'_1(\sigma,0)|+|u'_1(\sigma,1)|)\\ &+\frac{|\sigma|^{2-\frac 1 q}}{|\lambda|} (|u'_1(\sigma,0)|+|u'_1(\sigma,1)|)]\\ \le &C(\varepsilon,K)[\mu_1^{-2+\frac 1 q}|\sigma|^{1-\frac 1 q}\|u_1''(\sigma,\cdot)\| _{L_q(0,1)}\\ &+\mu_1^{1/q}|\sigma|^{1-\frac 1 q}\|u_1(\sigma,\cdot)\|_{L_q(0,1)} +\mu_2^{-2+\frac 1 q}\frac{|\sigma|^{3-\frac 1 q}}{|\lambda|}\|u_1''(\sigma,\cdot)\|_ {L_q(0,1)}\\ &+\mu_2^{1/q}\frac{|\sigma|^{3-\frac 1 q}}{|\lambda|}\|u_1(\sigma,\cdot)\|_{L_q(0,1)} +\mu_3^{-1+\frac 1 q}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\mu_3^{1+\frac 1 q} \|u_1(\sigma,\cdot)\|_{L_q(0,1)} +\mu_4^{-1+\frac 1 q}\frac{|\sigma|^{2-\frac 1 q}}{|\lambda|}\|u_1''(\sigma,\cdot)\|_ {L_q(0,1)}\\ &+\mu_4^{1+\frac 1 q}\frac{|\sigma|^{2-\frac 1 q}}{|\lambda|}\|u_1(\sigma,\cdot)\|_{L_q(0,1)}]. \end{align*} Choose $\mu_1=\mu_2=|\sigma|^{1+q}$, $\mu_3=R\gg 1$, $\mu_4=|\sigma|^{\frac{2q+1}{q+1}}$, where $|\sigma|> K\gg 1$. Then, by (\ref{4.1}), \begin{aligned} \big\|\frac{du_2(\sigma,\cdot)}{dy}\big\|_{L_q(0,1)} \le& C(\varepsilon,K)[\|u_1''(\sigma,\cdot) \|_{L_q(0,1)}+\sigma^2\|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\frac 1{|\lambda|}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+ \frac 1{|\lambda|}\sigma^4\|u_1(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}+\|u_1(\sigma,\cdot)\|_{L_q(0,1)}+ \frac {|\sigma|}{|\lambda|}\|u_1''(\sigma,\cdot)\|_{L_q(0,1)}\\ &+\frac 1{|\lambda|}\sigma^4 \|u_1(\sigma,\cdot)\|_{L_q(0,1)}]\\ \le& C(\varepsilon,K)\frac 1{|\lambda|}\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)}, \quad |\sigma|> K.\label{4.10} \end{aligned} Therefore, for a solution $\widehat u(\sigma,y)=u_1(\sigma,y)+u_2(\sigma,y)$ of problem (\ref{1.3})--(\ref{1.4}) from (\ref{4.1}), (\ref{4.9}), (\ref{4.10}), and $\widehat{u'_y}(\sigma,y)=\widehat{u}'_y(\sigma,y)$ we have $$\|\widehat{u'_y}(\sigma,\cdot)\|_{L_q(0,1)}=\|\widehat{u}'_y(\sigma,\cdot)\|_ {L_q(0,1)}\le C(\varepsilon)\frac 1{|\lambda|}\|\widehat f(\sigma,\cdot)\| _{L_q(0,1)},\label{4.11}$$ where $\sigma\in\mathbb{R}$, $\sigma\neq 0$, $|\arg\lambda|\le \pi-\varepsilon$, $|\lambda|\ge M$. Consider now a solution $u(x,y)$ of the main problem (\ref{1.1})--(\ref{1.2}). Since $u(x,0)=u(x,1)=0$, for all $x\in\mathbb{R}$, $$\begin{gathered} \|u\|_{L_q(\mathbb{R}\times (0,1))}\le C \|u'_y\|_{L_q(\mathbb{R}\times (0,1))},\\ \|u\|_{W^1_q(\mathbb{R}\times (0,1))}\le C(\|u'_y\|_{L_q(\mathbb{R}\times (0,1))}+ \|u'_x\|_{L_q(\mathbb{R}\times (0,1))}). \end{gathered}\label{4.12}$$ Finally, taking $q=2$, from (\ref{4.8}), (\ref{4.11}), (\ref{4.12}) and the Parseval equality, for a solution $u(x,y)$ of problem (\ref{1.1})--(\ref{1.2}) we have $$\|u\|_{L_2(\mathbb{R}\times (0,1))}\le C \|u'_y\|_{L_2(\mathbb{R}\times (0,1))}= C\|\widehat{u'_y}\|_{L_2(\mathbb{R}\times (0,1))} \le C(\varepsilon)\frac 1{|\lambda|}\|f\|_{L_2(\mathbb{R}\times (0,1))},$$ and \begin{align*} \|u\|_{W^1_2(\mathbb{R}\times (0,1))}&\le C(\|\widehat{u'_y}\|_{L_2(\mathbb{R}\times (0,1))} +\|\widehat{u'_x}\|_{L_2(\mathbb{R}\times (0,1))})\\ &\le C(\varepsilon)\frac 1{|\lambda|}(\|f'_x\|_{L_2(\mathbb{R}\times (0,1))}+ \|f\|_{L_2(\mathbb{R}\times (0,1))})\\ &\le C(\varepsilon)\frac 1{|\lambda|}\|f\|_{W^1_2(\mathbb{R}\times (0,1))}, \end{align*} where $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$. \qed \section{ A boundary value problem for the biharmonic equation with the second order derivative in boundary conditions} Consider now the following problem in the strip $\Omega:= (-\infty, \infty)\times [0,1]\subset\mathbb{R}^2$, \begin{gather} L(\lambda,D_x,D_y)u:=\Delta^2 u(x,y)-\lambda \Delta u(x,y)=f(x,y), \quad (x,y)\in \Omega, \label{5.1} \\ \begin{gathered} L_1u:= u(x,0)=0,\quad L_2u:= u(x,1)=0\\ L_3u:= \frac {\partial^2 u(x,0)}{\partial y^2}=0,\quad L_4u:= \frac {\partial^2 u(x,1)}{\partial y^2}=0, \quad x\in \mathbb{R}.\end{gathered} \label{5.2} \end{gather} Applying the Fourier transform operator $F_{x\to \sigma}$ to (\ref{5.1})--(\ref{5.2}), we obtain a boundary-value problem for an ordinary differential equation of the fourth order with 2 parameters \begin{gather} L(\lambda,i\sigma,D_y)\widehat u := \big[\frac {d^4}{dy^4}-(2\sigma^2 +\lambda)\frac {d^2}{dy^2}+ \sigma^4+\lambda \sigma^2 \big] \widehat u(\sigma,y) =\widehat f(\sigma,y), \quad y \in [0,1], \label{5.3}\\ \widehat u(\sigma,0)=\widehat u(\sigma,1)=\frac {d^2\widehat u(\sigma,0)}{dy^2}= \frac {d^2\widehat u(\sigma,1)}{dy^2}=0, \label{5.4} \end{gather} where $\lambda\in \mathbb{C}$ and $\sigma \in \mathbb{R}$ are parameters, $\widehat u(\sigma,y):= (F_{x\to \sigma}u(x,y))(\sigma,y)$. These two problems, (\ref{5.1})--(\ref{5.2}) and (\ref{5.3})--(\ref{5.4}), are much more easier to handle than (\ref{1.1})--(\ref{1.2}) and (\ref{1.3})--(\ref{1.4}), respectively. Moreover, we can get here a more complete result. \begin{theorem} \label{thm4} For each $\varepsilon>0$ there exists $M>0$ such that for all complex numbers $\lambda$ satisfying $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$, the operator ${\mathbb{L}(\lambda)}:$ $u\rightarrow {\mathbb{L}(\lambda)}u:= L(\lambda,D_x,D_{y})u$ from $W^{4}_{2}(\mathbb{R}\times(0,1),L_\nu u=0,\nu=1,\dots,4)$ onto $L_{2}(\mathbb{R}\times(0,1))$ is an isomorphism and for these $\lambda$ the following estimates hold $$\|u\|_{W^{4}_{2}(\mathbb{R}\times(0,1))}\le C(\varepsilon)\|f\| _{L_{2}(\mathbb{R}\times(0,1))},\label{5.5}$$ and $$\|u\|_{W^{2}_{2}(\mathbb{R}\times(0,1))}\le C(\varepsilon)\frac 1{|\lambda|}\|f\| _{L_{2}(\mathbb{R}\times(0,1))},\label{5.6}$$ for a solution $u(x,y)$ of problem (\ref{5.1})--(\ref{5.2}). \end{theorem} \paragraph{Proof} The required isomorphism follows from \cite[Theorem 1.1, p.44]{NP} (the proof is done as in the proof of Theorem \ref{thm3}). To get estimates (\ref{5.5}) and (\ref{5.6}), first consider a solution $\widehat u(\sigma,y)$ of problem (\ref{5.3})--(\ref{5.4}). Substituting $v(\sigma,y):=\frac{d^2\widehat u(\sigma,y)}{dy^2}-\sigma^2 \widehat u(\sigma,y)$, one can consider, instead of (\ref{5.3})--(\ref{5.4}), the two problems $$\begin{gathered} \frac{d^2\widehat u(\sigma,y)}{dy^2}-\sigma^2\widehat u(\sigma,y)=v(\sigma,y),\ y\in [0,1],\\ \widehat u(\sigma,0)=\widehat u(\sigma,1)=0,\end{gathered} \label{5.7}$$ and $$\begin{gathered} \frac{d^2 v(\sigma,y)}{dy^2}-(\sigma^2+\lambda)v(\sigma,y)=\widehat f(\sigma,y),\ y\in [0,1],\\ v(\sigma,0)=v(\sigma,1)=0.\end{gathered} \label{5.8}$$ From a theorem in \cite[p. 110]{YY} for problem (\ref{5.7}) for each fixed $\sigma$, such that $|\sigma|>K$ an isomorphism from $W_q^4(0,1)$ onto $W_q^2(0,1)$ follows and for a solution $\widehat u(\sigma,y)$ of problem (\ref{5.7}) the following estimate holds $$\sum^4_{k=0}|\sigma|^{4-k}\|\widehat u(\sigma,\cdot)\|_{W^k_q(0,1)}\le C(\|v(\sigma,\cdot)\|_{W^2_q(0,1)}+|\sigma|^2\|v(\sigma,\cdot)\|_{L_q(0,1)}),\quad |\sigma|>K.\label{5.9}$$ For $|\sigma|\le K$ one can easily obtain that for a solution of (\ref{5.7}), $$\|\widehat u(\sigma,\cdot)\|_{W^4_q(0,1)}\le C\|v(\sigma,\cdot)\|_{W^2_q(0,1)},\ |\sigma|\le K.\label{5.10}$$ From the same theorem \cite[p. 110]{YY} for problem (\ref{5.8}) for each fixed $\sigma\in\mathbb{R}$ an isomorphism from $W_q^2(0,1)$ onto $L_q(0,1)$ follows and for a solution $v(\sigma,y)$ of problem (\ref{5.8}) the following estimate holds $$\|v(\sigma,\cdot)\|_{W_q^2(0,1)}+|\lambda+\sigma^2|\|v(\sigma,\cdot)\|_{L_q(0,1)}\le C(\varepsilon)\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)},\label{5.11}$$ where $|\arg\lambda|\le \pi-\varepsilon$, $|\lambda|\ge M$. We have $|\lambda+\sigma^2|\ge C(\varepsilon)(|\lambda|+\sigma^2)$ (see section 2). Then, from (\ref{5.9}), (\ref{5.10}), and (\ref{5.11}) for problem (\ref{5.3})--(\ref{5.4}) an isomorphism from $W_q^4((0,1),L_\nu \widehat u=0,\nu=1,\dots,4)$ onto $L_q(0,1)$ follows for $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$ for each fixed $\sigma\in \mathbb{R}$. Moreover, for a solution $\widehat u(\sigma,y)$ of problem (\ref{5.3})--(\ref{5.4}) the following estimate holds $$\sum^4_{k=0}|\sigma|^{4-k}\|\widehat u(\sigma,\cdot)\|_{W^k_q(0,1)}\le C(\varepsilon)\|\widehat f(\sigma,\cdot)\|_{L_q(0,1)},\label{5.12}$$ where $\sigma\in\mathbb{R}$, $|\arg\lambda|\le \pi-\varepsilon,\ |\lambda|\ge M$. Taking now $q=2$ and using the Parseval equality to (\ref{5.12}) we obtain for a solution $u(x,y)$ of problem (\ref{5.1})--(\ref{5.2}) estimate (\ref{5.5}). From (\ref{5.5}) and equation (\ref{5.1}) follows $$|\lambda|\|\Delta u\|_{L_{2}(\mathbb{R}\times(0,1))}\le C(\varepsilon)\|f\| _{L_{2}(\mathbb{R}\times(0,1))}.\label{5.13}$$ On the other hand, from \cite[Theorem 1.1, p.44]{NP} it follows that $$\|u\|_{W^2_{2}(\mathbb{R}\times(0,1))}\le C\|\Delta u\|_{L_{2} (\mathbb{R}\times(0,1))}. \label{5.14}$$ From (\ref{5.13}) and (\ref{5.14}) we obtain estimate (\ref{5.6}).\qed \begin{thebibliography}{00} \bibitem{AN} Agmon, S. and Nirenberg, L., Properties of solutions of ordinary differential equations in Banach spaces, {\it Comm. Pure Appl. Math.}, {\bf 16} (1963), 121--239. % \bibitem{AV} Agranovich, M. S. and Vishik, M. I., Elliptic problems with a parameter and parabolic problems of general type, {\it Uspekhi Mat. Nauk}, {\bf 19}, 3 (1964), 53--161 (Russian; English translation in {\it Russian Math. Surveys}, {\bf 19}, 3 (1964), 53--159). % \bibitem{BIN} Besov, O. V., Ilin, V. P. and Nikolskii, S. M., {\it Integral Representations of Functions and Embedding Theorems,} Halsted Press, New York, v.I, 1978. % \bibitem{KM} Kozlov, V. A. and Maz'ya, V. G., {\it Differential Equations with Operator Coefficients}, Springer, 1999. % \bibitem{KMR} Kozlov, V. A., Maz'ya, V. G. and Rossmann, J., {\it Elliptic Boundary Value Problems in Domains with Point Singularities}, AMS, Math. Surv. and Monogr., v.52, 1997. % \bibitem{NP} Nazarov, S. A. and Plamenevskii, B. A., {\it Elliptic Problems in Domains with Piecewise Smooth Boundaries}, Walter de Gruyter, Berlin, 1994. % \bibitem{T} Triebel, H., {\it Interpolation Theory. Function Spaces. Differential Operators}, North-Holland, Amsterdam, 1978. % \bibitem{Y} Yakubov, S., {\it Completeness of Root Functions of Regular Differential Operators,} Longman, Scientific and Technical, New York, 1994. % \bibitem{YY} Yakubov, S. and Yakubov, Ya., {\it Differential-Operator Equations. Ordinary and Partial Differential Equations}, Chapman and Hall/CRC, Boca Raton, 2000. % \end{thebibliography} \end{document}