\documentclass[reqno]{amsart} \usepackage{hyperref} \usepackage{amssymb} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2014 (2014), No. 108, pp. 1--18.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2014 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2014/108\hfil Well-posedness of boundary-value problems] {Well-posedness of boundary-value problems for partial differential equations of even order} \author[D. Amanov, A. Ashyralyev \hfil EJDE-2014/108\hfilneg] {Djumaklych Amanov, Allaberen Ashyralyev} % in alphabetical order \address{Djumaklych Amanov \newline Institute of Mathematics at National University of Uzbekistan, 29, Do`rmon yo`li street, Tashkent, Uzbekistan} \email{amanov\_d@rambler.ru} \address{Allaberen Ashyralyev \newline Department of Mathematics Fatih University, 34500, Buyukcekmece, Istanbul, Turkey and ITTU, Ashgabat, Turkmenistan } \email{aashyr@fatih.edu.tr} \thanks{Submitted October 15, 2013. Published April 15, 2014.} \subjclass[2000]{35G15} \keywords{Boundary value problem; well-posedness; a priori estimate; \hfill\break\indent regular and strong solutions; spectrum of problems} \begin{abstract} In this article, we establish the well-posedness of two boundary value problems for $2k$-th order partial differential equations. It is shown that the solvability of these problems depends on the evenness and oddness of the number $k$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction and formulation of the problems} There is a huge number of theoretical and applied works devoted to the study of partial differential equations of higher order. Solutions of some classical and nonclassical problems for specific partial differential equations of higher order can be found, for example, in the monographs by Egorov and Fyodorov \cite{e1}, Gazzola and Evans \cite{e2}, Grunau and Sweers \cite{g1}, Mizohata \cite{m1}, Peetre \cite{p1}, Polyanin \cite{p2}, Vragov \cite{v1}, and in the articles by Kirane and Qafsaoui \cite{k2} and Qafsaoui \cite{q1}. These problems are studied in various directions: qualitative properties of solutions, spectral problems, various statements of boundary value problems, and numerical investigations. The study of well-posedness of the Cauchy problem, Goursat problem and boundary value problem for partial differential equations and mixed type partial differential equations have been studied extensively in a large number of articles; see, for example, \cite{a1,a3,a4,d2,d3,d4,k3,k4,r1,s2,s4,t1,v1,z1} and the references therein. As examples, we present some of them. In \cite{a1,a3}, in the domain $\Omega =\{ (x,t) :00$ that depends only on numbers $k$ and $T$, and does not depend on the function $u(x,t)$ such that the following a priori estimate holds: \begin{equation} \| u\| _{W_2^{2k,2}(\Omega )} \leqslant C_1\| {Lu}\| _{L_2(\Omega )}, \label{e5} \end{equation} where \[ \| u\| _{W_2^{2k,2}(\Omega )}=\sum_{m=0}^{2k}{ \| {\frac{{\partial ^{m}u}}{{\partial x^{m}}}}\| _{L_2(\Omega )}^2+}\sum_{m=1}^2{\| {\frac{{\partial ^{m}u}}{{\partial t^{m}}}}\| _{L_2(\Omega )}^2+} \sum_{m=2}^{k+1} \| {\frac{{\partial ^{m}u}}{{\partial x^{m-1}\partial t}}}\| _{L_2(\Omega )}^2. \] \end{lemma} \begin{proof} Multiplying by $u(x,t)$ both sides of the equation \begin{equation} \frac{{\partial ^{2k}u}}{{\partial x^{2k}}}+\frac{{\partial ^2u}}{{ \partial t^2}}=Lu \label{e6} \end{equation} and taking the integral over the domain $\Omega $, we obtain \begin{equation} \int_0^{p}{\int_0^{T}{u( {\frac{{\partial ^{2k}u}}{{ \partial x^{2k}}}+\frac{{\partial ^2u}}{{\partial t^2}}}) }}\,dx\,dt =\int_0^{p}{\int_0^{T}u}Lu\,dt\,dx. \label{e7} \end{equation} Since $k$ is odd number, using equality \eqref{e7}, identities \begin{gather*} u\frac{{\partial ^2u}}{{\partial t^2}}=\frac{\partial }{{\partial t}} ( {u\frac{{\partial u}}{{\partial t}}}) -( {\frac{{\partial u }}{{\partial t}}}) ^2, \\ u\frac{{\partial ^{2k}u}}{{\partial x^{2k}}}=\sum_{m=0}^{k-1}{(-1)^{m} \frac{\partial }{{\partial x}}( {\frac{{\partial ^{m}u}}{{\partial x^{m} }}\frac{{\partial ^{2k-m-1}u}}{{\partial x^{2k-m-1}}}}) +(-1)^{k}( {\frac{{\partial ^{k}u}}{{\partial x^{k}}}}) ^2,} \end{gather*} and conditions \eqref{e2} and \eqref{e3}, we obain \[ \| {\frac{{\partial ^{k}u}}{{\partial x^{k}}}} \| _{L_2(\Omega )}^2+\| {\frac{ {\partial u}}{{\partial t}}}\| _{L_2(\Omega )}^2=\int_0^{p}{\int_0^{T}{( {-u}) Lu\,dt\,dx.}} \] From this equality and the H\"{o}lder inequality it follows the estimate \[ \| {\frac{{\partial ^{k}u}}{{\partial x^{k}}}} \| _{L_2(\Omega )}^2+\| {\frac{ {\partial u}}{{\partial t}}}\| _{L_2(\Omega )}^2\leq \| u\| _{L_2(\Omega )} \text{ }\| {Lu}\| _{L_2(\Omega )}. \] Applying the inequality \[ ab\leqslant \frac{\varepsilon }{2}a^2+\frac{1}{{2\varepsilon }}b^2 \] for $a>0$, $b>0$ and $\varepsilon >0$, we obtain \begin{equation} \| {u_{t}}\| _{L_2(\Omega )}^2+\| {\frac{{\partial ^{k}u}}{{\partial x^{k}}}} \| _{L_2(\Omega )}^2\leqslant \frac{\varepsilon }{2 }\| u\| _{L_2(\Omega )}^2+ \frac{1}{{2\varepsilon }}\| {Lu}\| _{L_2(\Omega )}^2. \label{e8} \end{equation} Now, we obtain the estimate for $\| u\| _{L_2(\Omega )}^2$. Using condition \eqref{e3}, we obtain \[ u^2(x,t)=\int_0^{t}{\frac{\partial }{{\partial t}}}( { u^2(x,\tau )}) d\tau . \] Therefore, \[ u^2(x,t)\leqslant 2\int_0^{T}{| {u(x,t)u_{t}(x,t)}| dt.} \] Taking the integral over the domain $\Omega $ on both sides of the inequality and applying the H\"{o}lder inequality, we obtain \[ \| u\| _{L_2(\Omega)}^2\leqslant 2T\| u\| _{L_2(\Omega )}\| {u_{t}}\| _{L_2(\Omega )}. \] From that, it follows \[ \| u\| _{L_2(\Omega)}^2\leqslant 4T^2 \| {u_{t}}\| _{L_2(\Omega )}^2. \] The last estimate and inequality \eqref{e8} yield that \begin{equation} \| u\| _{L_2(\Omega )}^2\leqslant 4T^2( {\frac{\varepsilon }{2}\| u\| _{L_2(\Omega )}^2+\frac{1}{{2\varepsilon }}\| Lu\| _{L_2(\Omega )}^2}) . \label{e9} \end{equation} Adding the both sides of inequalities \eqref{e8} and \eqref{e9}, we obtain \[ \| u\| _{L_2(\Omega)}^2+\| {u_{t}}\| _{L_2(\Omega)}^2 +\| \frac{{\partial ^{k}u}}{{\partial x^{k}}}\| _{L_2(\Omega )}^2 \leqslant ( {4T^2+1}) ( \frac{\varepsilon }{2}\| u\| _{L_2(\Omega )}^2 +\frac{1}{{2\varepsilon }}\| {Lu}\| _{L_2(\Omega )}^2) \] for any $\varepsilon >0$. Choosing $\varepsilon =1/(4T^2+1)$, we can write \begin{equation} \| u\| _{L_2(\Omega)}^2+\| {u_{t}}\| _{L_2(\Omega )}^2+\| {\frac{{\partial {}^{k}u}}{{\partial x^{k}}}} \| _{L_2(\Omega )}^2\leqslant ( {4T^2+1}) ^2\| {Lu}\| _{L_2(\Omega )}^2. \label{e10a} \end{equation} Taking the square of both sides of \eqref{e6} and integrating the obtained equality over the domain $\Omega $, we obtain \begin{equation} \| {u_{tt}}\| _{L_2(\Omega)}^2+\| {\frac{{\partial ^{2k}u}}{{\partial x^{2k}}}} \| _{L_2(\Omega )}^2+2\int_{\Omega }{\frac{{ \partial ^2u}}{{\partial t^2}}\frac{{\partial ^{2k}u}}{{\partial x^{2k}}} \,dx\,dt}=\| {Lu}\| _{L_2(\Omega)}^2. \label{e11} \end{equation} We consider the integral \[ \int_{\Omega }{\frac{{\partial ^2u}}{{\partial t^2}}\frac{{ \partial ^{2k}u}}{{\partial x^{2k}}}\,dx\,dt.} \] We have that \begin{align*} \frac{\partial ^2u}{\partial t^2}\frac{\partial ^{2k}u}{\partial x^{2k}} &=\sum_{m=0}^{k-1}( {-1}) ^{m}\frac{\partial }{{\partial x}} \Big( {\frac{{\partial ^{2+m}u}}{{\partial t^2\partial x^{m}}}\frac{{ \partial ^{2k-m-1}u}}{{\partial x^{2k-m-1}}}}\Big)\\ &\quad +( {-1}) ^{k}\frac{\partial }{{\partial t}}\Big( {\frac{{ \partial ^{1+k}u}}{{\partial t\partial x^{k}}}\frac{{\partial ^{k}u}}{{ \partial x^{k}}}}\Big) +( {-1}) ^{k+1}\Big( {\frac{{\partial ^{k+1}u}}{{\partial t\partial x^{k}}}}\Big) ^2. \end{align*} If $m$ is odd, then $2k-m-1$ is even. Therefore, from \eqref{e2} it follows that $\frac{{\partial ^{2k-m-1}u}}{{\partial x^{2k-m-1}}}=0$ at $x=0$ and $x=p$. If $m$ is even, then from \eqref{e2} it follows that $\frac{{\partial ^{m+2}u}}{{\partial t^2\partial x^{m}}}=0$ at $x=0$ and $x=p$. Moreover, from \eqref{e3} it follows $\frac{{\partial ^{k}u}}{{\partial x^{k}}}=0$ at $t=0$ and $t=T$. Therefore, integrating both sides of the last equality over the domain $\Omega $, we obtain \[ 2\iint_{\Omega }{\frac{{\partial ^2u}}{{\partial t^2}}\frac{{ \partial ^{2k}u}}{{\partial x^{2k}}}\,dx\,dt}=2\| {\frac{{ \partial ^{k+1}u}}{{\partial x^{k}\partial t}}}\| _{L_2(\Omega )}^2. \] Using this equality and \eqref{e11}, we obtain \[ \| {u_{tt}}\| _{L_2(\Omega )}^2+2\| {\frac{{\partial ^{k+1}u}}{{\partial x^{k}\partial t}}}\| _{L_2(\Omega )}^2 +\| {\frac{{\partial ^{2k}u}}{{\partial x^{2k}}}}\| _{L_2(\Omega )}^2 =\| {Lu}\| _{L_2(\Omega )}^2. \] From that it follows \begin{equation} \| {u_{tt}}\| _{L_2(\Omega )}^2+\| {\frac{{\partial ^{k+1}u}}{{\partial x^{k}\partial t}}}\| _{L_2(\Omega )}^2 +\| {\frac{{\partial ^{2k}u}}{{\partial x^{2k}}}}\| _{L_2(\Omega )}^2 \leqslant \| {Lu} \| _{L_2(\Omega )}^2. \label{e12} \end{equation} Adding both sides of inequalities \eqref{e10a} and \eqref{e12}, we obtain \begin{equation} \begin{aligned} &\| u\| _{L_2(\Omega)}^2+\| {u_{t}}\| _{L{}_2(\Omega )}^2+\| {u_{tt}}\| _{L_2(\Omega )}^2+\| {\frac{{\partial ^{k+1}u}}{{\partial x^{k}\partial t}}}\|_{L_2(\Omega )}^2 +\| {\frac{{\partial ^{k}u}}{{\partial x^{k}}}} \| _{L_2(\Omega )}^2+\| {\frac{{\partial ^{2k}u}}{{\partial x^{2k}}}}\| _{L_2(\Omega )}^2\\ &\leqslant C_1\| {Lu}\| _{L_2(\Omega )}^2, \end{aligned} \label{e13} \end{equation} where $C_1=( {4T^2+1}) ^2+1$. To obtain estimates for the norm $\| {\frac{{\partial^{m}u}}{{\partial x^{m}}}}\| _{L_2(\Omega )}^2$, $m=1,2,\dots,2k-1$, we use the inequality \begin{equation} \| {\frac{{\partial ^{n}u}}{{\partial x^{n}}}} \| _{L_2(\Omega )}^2\leqslant \frac{1}{2} \| {\frac{{\partial ^{n-1}u}}{{\partial x^{n-1}}}} \| _{L_2(\Omega )}^2+\frac{1}{2} \| {\frac{{\partial ^{n+1}u}}{{\partial x^{n+1}}}}\| _{L_2(\Omega )}^2 \label{e14} \end{equation} that can easily be checked. Taking the sum of both sides of inequality \eqref{e14} with respect to $n$ from $1$ to $2k-1$, and using inequality \eqref{e13}, we obtain \begin{equation} \| {\frac{{\partial u}}{{\partial x}}}\| _{L_2(\Omega )}^2 +\| {\frac{{\partial ^{2k-1}u}}{{\partial x^{2k-1}}}}\| _{L_2(\Omega )}^2\leqslant C_1\| {Lu}\|_{L_2(\Omega )}^2. \label{e14a} \end{equation} In general, taking the sum of both sides of inequality \eqref{e14} with respect to $n$ from $m$ to $k-1$, and using inequality \eqref{e13}, we obtain \begin{equation} \| {\frac{{\partial ^{m}u}}{{\partial x^{m}}}} \| _{L_2(\Omega )}^2+\| {\frac{ {\partial ^{2k-m}u}}{{\partial x^{2k-m}}}}\| _{L_2(\Omega )}^2\leqslant C_1\| {Lu}\| _{L_2(\Omega )}^2. \label{e14b} \end{equation} Taking the sum of both sides of inequalities \eqref{e13} and \eqref{e14b} with respect to $m$ from $1$ to $k-1$, we obtain \begin{equation} \sum_{m=0}^2{\| {\frac{{\partial ^{m}u}}{{ \partial t^{m}}}}\| }_{L_2(\Omega )}^2+\sum_{m=1}^{2k}{\| {\frac{{\partial ^{m}u} }{{\partial x^{m}}}}\| }_{L_2(\Omega )}^2+\| {\frac{{\partial ^{k+1}u}}{{\partial x^{k}\partial t}}}\| _{L_2(\Omega )}^2\leqslant kC_1\| {Lu}\| _{L_2(\Omega)}^2. \label{e15} \end{equation} Next, we use the inequality \[ \| {\frac{{\partial ^{m}u}}{{\partial x^{m-1}\partial t}} }\| _{L_2(\Omega )}^2\leqslant | { | {\frac{{\partial ^2u}}{{\partial t^2}}}| } | _{L_2(\Omega )}\| {\frac{{\partial ^{2m-2}u }}{{\partial x^{2m-2}}}}\| _{L_2(\Omega )} \] that can easily be checked. Summing with respect to $m$ from $2$ to $k$ and using inequality \eqref{e15}, we obtain \begin{equation} \sum_{m=2}^{k}{\| {\frac{{\partial ^{m}u}}{{ \partial x^{m-1}\partial t}}}\| }_{L_2(\Omega )}^2\leqslant k(k-1)C_1\| {Lu}\| _{L_2(\Omega )}^2. \label{e16} \end{equation} Adding both sides of inequalities \eqref{e15} and \eqref{e16}, we obtain estimate \eqref{e5}. The proof is complete. \end{proof} \begin{corollary} \label{coro2.1} From estimate \eqref{e5} it follows that: \begin{itemize} \item[(i)] The regular solution of problem 1 is unique and continuously depends on $f(x,t)$. \item[(ii)] The inverse operator $L^{-1}$ exists and it is bounded. \item[(iii)] $\| {L^{-1}}\| \leqslant C_2$, $C_2=k^2[ {( {4T^2+1}) ^2+1}]$. \item[(iv)] $\ker(L)=\{0\}$. \item[(v)] The adjoint problem to problem 1 is well-posed. \end{itemize} \end{corollary} Second, we will study the regular solvability of problem 1. We seek a regular solution of problem 1 in the form of a Fourier series \begin{equation} u(x,t)=\sum_{n=1}^{\infty }{u_n(t)X_n(x)} \label{e17} \end{equation} expanded along the complete orthonormal system $X_n(x)=\sqrt{\frac{2}{p}}\sin \lambda _nx$ in $L_2(0,p)$, where $\lambda _n=\frac{{n\pi }}{p}$, $n\in N$. It is clear that $u(x,t)$ satisfies conditions \eqref{e2}. We expand the given function $f\in W_1(\Omega )$ in the form of a Fourier series along the functions $X_n(x)$, $n\in N$ \begin{equation} f(x,t)=\sum_{n=1}^{\infty }f_n(t)X_n(x), \label{e18} \end{equation} where \begin{equation} f_n(t)=\int_0^{P}f(x,t)X_n(x)dx. \label{e19} \end{equation} Substituting \eqref{e17} and \eqref{e18} into \eqref{e1}, we obtain \begin{equation} u_n''(t)-\lambda _n^{2k}u_n(t)=f_n(t),\quad 00. \label{e24} \end{equation} \begin{lemma} \label{lem2.2} If $f\in W_1(\Omega )$, then for any $t\in [ 0,T] $ the following estimate is valid: \begin{equation} | {u_n(t)}| \leqslant \frac{{C_0C_2}}{{\lambda _n^{2k+1+\alpha }}}. \label{e25} \end{equation} \end{lemma} \begin{proof} Integrating by parts with respect to $x$ in \eqref{e19}, we obtain \[ f_n(t)=\frac{1}{{\lambda _n}}f_n^{(1,0)}(t), \] where \[ f_n^{(1,0)}(t)=\int_0^{p}{\frac{{\partial f}}{{\partial x}}}\sqrt{ \frac{2}{p}}\cos \lambda _nx\,dx. \] Since $\frac{{\partial f}}{{\partial x}}\in {\rm Lip}_{\alpha }[ 0,p]$ is uniform with respect to $t$, then (see \cite{d2}) \[ | {f_n^{(1,0)}(t)}| \leqslant \frac{{C_2}}{{\lambda_n^{\alpha }}}. \] So, \begin{equation} | {f_n(t)}| \leqslant \frac{{C_2}}{{\lambda_n^{1+\alpha }}}. \label{e26} \end{equation} Now, we will estimate $| {u_n(t)}| $. Applying equality \eqref{e22} and estimates \eqref{e24} and \eqref{e26}, we obtain \begin{align*} | {u_n(t)}| &\leqslant \frac{{C_0C_2}}{{\lambda _n^{1+k+\alpha }}}\Big[ {\int_0^{t}{e^{-\lambda _n^{k}(t-\tau )}}d\tau +\int_{t}^{T}{e^{-\lambda _n^{k}(\tau -t)}d\tau }}\Big] \\ &\leqslant \frac{{C_0C_2}}{{\lambda _n^{2k+1+\alpha }}}\Big( { e^{-\lambda _n^{k}(T-t)}-e^{-\lambda _n^{k}t}}\Big) \\ &\leqslant \frac{{C_0C_2}}{{\lambda _n^{2k+1+\alpha }}}{e^{-\lambda _n^{k}(T-t)}} \leqslant \frac{{C_0C_2}}{{\lambda _n^{2k+1+\alpha }}}. \end{align*} The proof is complete. \end{proof} \begin{theorem} \label{thm2.1} If $f(x,t)\in W_1(\Omega )$, then there exists a regular solution of problem 1. \end{theorem} \begin{proof} We will prove uniform and absolute convergence of series \eqref{e17} and \begin{equation} \frac{{\partial ^{2k}u}}{{\partial x^{2k}}}=-\sum_{n=1}^{\infty }\lambda _n^{2k}u_n(t)X_n(x), \label{e27} \end{equation} \begin{equation} \frac{{\partial ^2u}}{{\partial t^2}}=\sum_{n=1}^{\infty }f_n(t)X_n(x)+\sum_{n=1}^{\infty }\lambda _n^{2k}u_n(t)X_n(x). \label{e28} \end{equation} From \eqref{e26} it follows that the series \eqref{e17} and the first series in \eqref{e28} uniformly and absolutely converge. In the same manner, from \eqref{e25} it follows that the second series in \eqref{e28} and the series \eqref{e27} uniformly and absolutely converge. Adding equality \eqref{e27} and \eqref{e28}, we note that solution \eqref{e17} satisfies equation \eqref{e1}. Solution \eqref{e17} satisfies boundary conditions \eqref{e2} owing to properties of the function $X_n(x)$ and conditions \eqref{e3} owing to properties of the kernel \eqref{e23a}. The proof is complete. \end{proof} Third, we will study an existence and uniqueness of the strong solution of problem 1. \begin{theorem} \label{thm2.2} For any $f\in L_2(\Omega )$ there exists a unique strong solution of problem 1 and it satisfies estimate \eqref{e5}, it continuously depends on $f(x,t)$ and it can be represented in the form of Fourier series \[ u(x,t)=\sum_{n=1}^{\infty }{u_n(t)X_n(x)} \] expanded in full orthonormal system $X_n(x)=\sqrt{\frac{2}{p}}\sin \lambda_nx$ in $L_2(0,p)$, where $\lambda _n=\frac{{n\pi }}{p}$, $n\in N$. \end{theorem} \begin{proof} It is possible to represent solution \eqref{e17} in the form \begin{equation} u(x,t)=\int_0^{T}{\int_0^{p}{K(x,t;\xi ,\tau )}}f(\xi,\tau )d\xi \partial \tau , \label{e29} \end{equation} where \begin{equation} K(x,t;\xi ,\tau )=-\sum_{n=1}^{\infty }{\frac{{X_n(x)X_n(\xi )}}{{ \lambda _n^{k}}}}K_n(t,\tau ). \label{e30} \end{equation} From estimate \eqref{e24} and $k\geqslant 2$ it follows that series \eqref{e30} converges uniformly and absolutely. Consequently \begin{equation} | {K(x,t;\xi ,\tau )}| \leqslant C_{3},\,C_{3}={\rm const}>0. \label{e31} \end{equation} From relations $C_0^{\infty }(\Omega )\subset W_1(\Omega )\subset L_2(\Omega )$ it follows that $W_1(\Omega )$ is dense in $L_2(\Omega)$. Then for any $f\in L_2(\Omega )$ there exists a sequence $\{ {f_{m}}\} \subset W_1(\Omega )$, $m=1,2,\dots$ such that $\| {f_{m}-f}\| _{L_2(\Omega )}\to 0$ as $m\to \infty $. Consequently, $\{ {f_{m}}\} $ is a Cauchy sequence in $L_2(\Omega )$. We denote by $u_{m}(x,t)\in V_1(\Omega )$ the solution of \eqref{e1} with the right side term $f_{m}(x,t)$. According to Theorem \ref{thm2.1} and Corollary \ref{coro2.1}, there exists a unique solution of the form \begin{equation} u_{m}(x,t)=\int_0^{T}\int_0^{p}K(x,t;\xi ,\tau )f_{m}(\xi ,\tau )\,d\xi\,d\tau ,\quad m=1,2,\dots \label{e32} \end{equation} of the equation \begin{equation} Lu_{m}(x,t)=f_{m}(x.t),\quad m=1,2,\dots. \label{e33} \end{equation} Since $u_{m}(x,t)\in V_1(\Omega )$, according to \eqref{e5} we have \[ \| {u_{m}-u_{l}}\|_{W_2^{2k,2}}\leqslant C_1\| {Lu_{m}-Lu_{l}} \| _{L_2(\Omega )} =C_1\| {f_{m}-f_{l}}\| _{L_2(\Omega )}\to 0 \] as $m\to \infty $, $l\to \infty $, that $\{ {u_{m}}\} $ is a Cauchy sequence in $W_2^{2k,2}(\Omega )$. According to completeness of the $W_2^{2k,2}(\Omega )$, there exists a unique limit $u(x,t)=\lim_{m\to \infty }u_{m}(x,t)\in W_2^{2k,2}(\Omega )$. Passing to limit in \eqref{e33} as $m\to \infty $, we obtain \begin{equation} \overline{L}u=\lim_{m\to \infty }Lu_{m}(x,t)= \mathop{\lim }_{m\to \infty }f_{m}(x,t)=f,\,\,\,\,\,\,f\in L_2(\Omega ); \label{e34} \end{equation} i.e., \[ \| {Lu_{m}-f}\| _{L_2(\Omega)}\to 0\quad \text{as }m\to \infty . \] Consequently, according to Definition 2.2, the function $u(x,t)\in W_2^{2k,2}(\Omega )$ is the strong solution of problem 1. Hence, it follows that the domain $D(\overline{L})$ of definition of operator $\overline{L}$ consists of all strong solutions of problem 1 and $R(\overline{L})=L_2(\Omega )$. So, $u_{m}(x,t)\in V_1(\Omega )$ from \eqref{e5} we have that \[ (u_{m},u_{m})_{W_2^{2k,2}(\Omega )}\leqslant C_1^2( {Lu_{m},Lu_{m}}) _{L_2(\Omega )}. \] Passing to the limit in the above inequality as $m\to \infty $, we obtain \begin{equation} \| u\|_{W_2^{2k,2}}^2\leqslant C_1^2\| {\overline{L}u} \| _{L_2(\Omega )}^2, \label{e35} \end{equation} where $u\in W_2^{2k,2}(\Omega )$, $\overline{L}u=f\in L_2(\Omega )$. We conclude that estimate \eqref{e5} is also true for the strong solution $u(x,t)$. Passing to limit in \eqref{e33} as $m\to \infty $, we obtain \[ u(x,t)=\int_0^{T}{\int_0^{P}{K(x,t;\xi ,\tau )}}f(\xi,\tau )d\xi \partial \tau , \] where $u(x,t)\in W_2^{2k,2}(\Omega )$, $f\in L_2(\Omega )$. From estimate \eqref{e35} it follows that the strong solution of problem 1 is unique and it continuously depends on $f(x,t)$. The proof of Theorem \ref{thm2.2} is complete. \end{proof} \begin{corollary} \label{coro2.2} The strong solution $u(x,t)$ is almost everywhere solution in $\Omega $. This follows from \eqref{e34} and $u(x,t)\in W_2^{2k,2}(\Omega )$. \end{corollary} Fourth, we will study the spectrum of problem 1. \begin{definition} \label{def2.4} The spectrum of a problem is the set of eigenvalues of the operator of the problem. \end{definition} According to Definition 2.4, the spectrum of problem 1 is the set of eigenvalues of operator $\overline{L}$. \begin{theorem} \label{thm2.3} The spectrum of problem 1 consists of real eigenvalues of finite multiplicity of the operator $\overline{L}$. \end{theorem} \begin{proof} From \eqref{e5} and \eqref{e29} we conclude that it is defined a bounded symmetric operator $L^{-1}$ on $W_1(\Omega )$ which is the inverse of the operator $L$ and acts from $W_1(\Omega )$ to $V_1(\Omega )$ by the rule \[ (L^{-1}f)(x,t)=\int_0^{T}{\int_0^{p}{K(x,t;\xi ,\tau )}} f(\xi ,\tau )\,d\xi\,d\tau . \] It can be extended to whole space $L_2(\Omega )$. This extension will be denoted by $\overline{L^{-1}}$, the closure of $L^{-1}$, $D(\overline{L^{-1})}=L_2(\Omega )$. The operator $\overline{L^{-1}}$ is symmetric, bounded and defined on the whole space $L_2(\Omega )$, so it is self-adjoint. It follows from \eqref{e31} that $K(x,t;\xi ,\tau ))\in L_2( {\Omega \times \Omega }) $; therefore $\overline{L^{-1}}$ is a compact operator in $L_2(\Omega )$. Then the spectrum of the operator $\overline{L^{-1}}$ is discrete and consists of real eigenvalues of finite multiplicity. The relation between eigenvalues of the operator $\overline{L^{-1}}$ and $\overline{L}$ is as follows of monograph \cite{d3}: if $\mu _n\neq 0$ is an eigenvalue of the operator $\overline{L^{-1}}$, then $\mu _n^{-1}$ is eigenvalue of the operator $\overline{L}$. The proof is complete. \end{proof} \section{Well-posedness of problem 1 when $k$ is even} Let $k$ be even number. In the same manner, we seek a regular solution of problem 1 in the form of Fourier series \eqref{e17} and we obtain the equation \[ u_n''+\lambda _n^{2k}u_n(t)=f_n(t),\quad 00. \] It is difficult to show the numbers $p$ and $T$ satisfying the last condition. Therefore, we give an other variant of solution of problem 1 in the case of even $k$. In the case of even $k$, solvability of problem 1 depends on domain geometry. We consider the following spectral problem. \subsection*{Problem 3} Find the solution $u(x,t)$ of the equation \begin{equation} \frac{{\partial ^{2k}u}}{{\partial x^{2k}}}+\frac{{\partial ^2u}}{{ \partial t^2}}=\lambda u \label{e36} \end{equation} satisfying \eqref{e2} and \eqref{e3}, where $\lambda $ is spectral parameter. This problem has the following eigenvalues \[ \lambda _{nm}=( {\frac{{n\pi }}{p}}) ^{2k}-( {\frac{{n\pi }}{ T}}) ^2 \] and eigenfunctions \[ u_{nm}(x,t)=\frac{2}{\sqrt{pT}}\sin \frac{{n\pi }}{p}x\sin \frac{{n\pi }}{T} t,\quad m,n=1,2,\dots \] and they form a complete orthonormal system in $L_2(\Omega )$. We expand the functions $u(x,t)$ and $f(x,t)$ into the Fourier series in functions $u_{nm}(x,t)$, \begin{gather} u(x,t)=\sum_{n,m=1}^{\infty }{\alpha _{nm}u_{nm}(x,t)}, \label{e37}\\ f(x,t)=\sum_{n,m}^{\infty }{f_{nm}u_{nm}(x,t)}, \label{e38} \end{gather} where \begin{equation} f_{nm}=\int_0^{P}\int_0^{T}{f(x,t)u_{nm}(x,t)}\,dt\,dx \label{e39} \end{equation} and $\alpha _{nm}$ are unknown Fourier coefficients of the function $u(x,t)$. Substituting \eqref{e37} and \eqref{e38} into \eqref{e1}, we obtain the solution of problem 1 in the form \begin{equation} u(x,t)=\sum_{n,m=1}^{\infty }{\frac{{f_{nm}}}{{\lambda _{nm}}} u_{nm}(x,t)}. \label{e40} \end{equation} Suppose that $m=n^{k}$, $T=p^{k}/ \pi ^{k-1}$. Then $\lambda _{nm}=\lambda _{nn^{k}}=0$ and problem 1 has infinitely many of linearly independent solutions in the form \[ u_{nn^{k}}(x,t)=\sin \big(\frac{{n\pi }}{p}x \big) \sin\big( \frac{{n^{k}\pi ^{k}}}{{p^{k}}}t\big), \quad n=1,2,\dots \] at $f(x,t)=0$. Number $\lambda _{nn^{k}}=0$ is an eigenvalue of infinite multiplicity of the spectral problem \eqref{e36}, \eqref{e2}, \eqref{e3}. In this case, the solution of problem 1 exists, if the following orthogonality conditions hold \[ \int_0^{p} \int_0^{T} f(x,t)\sin \big(\frac{{n\pi }}{p}x\big) \sin \big(\frac{{n^{k}\pi ^{k}}}{p^{k}}t\big) \,dt\,dx=0,\quad n=1,2,\dots. \] Now, suppose that $p$ and $T$ such that $\frac{{P^{k}}}{{T\pi ^{k-1}}}$ is irrational algebraic number of second degree (see \cite{s3}). Then according to Liouville's theorem \cite{s3} there exists number $\varepsilon _0>0$ such that \[ \big| {\frac{{p^{k}}}{{T\pi ^{k-1}}}-\frac{{n^{k}}}{m}}\big| \geqslant \frac{{\varepsilon _0}}{{m^2}}. \] In this case \begin{equation} \lambda _{nm}\geqslant \frac{{\pi ^{k+1}}}{{Tp^{k}}}\varepsilon _0. \label{e41} \end{equation} \begin{theorem} \label{thm3.1} Let $f\in L_2(\Omega )$ and number $\frac{{P^{k}}}{{T\pi ^{k-1}}}$ be irrational algebraic number of second degree. Then there exists the solution of problem 1, it belongs to $L_2(\Omega )$, and satisfies estimate \begin{equation} \| u\| _{L_2(\Omega )}\leqslant C\| f\| _{L_2(\Omega )}, \label{e42} \end{equation} where $C=\frac{{Tp^{k}}}{{\pi ^{k+1}\varepsilon _0}}$ and it continuously depends on $f(x,t)$. \end{theorem} \begin{proof} Since $f\in L_2(\Omega )$. Then for $f$, the Parseval equality is true \[ \sum_{n,m=1}^{\infty }{| {f_{nm}}| ^2=\| f\| }_{L_2(\Omega )}^2. \] Owing to it, for any $S>0$ and natural number $N$, we have \begin{align*} \| u{_{N+S,N+S}-u_{NN}}\|_{L_2(\Omega )}^2 &=\sum_{n=N+1}^{N+S}{\sum_{m=N+1}^{N+S}{\frac{{| {f_{nm}}| ^2}}{{\lambda _{nm}^2}}}}\\ &\leqslant C\sum_{n=N+1}^{N+S}{\sum_{m=N+1}^{N+S}{| {f_{nm}}| }}^2\to 0 \quad \text{as } N\to \infty. \end{align*} Consequently, the series \eqref{e40} converges in $L_2(\Omega )$ and $u(x,t)\in L_2(\Omega )$. It is easy to show that $\|u\| _{L_2(\Omega )}\leqslant C\| f\| _{L_2(\Omega )}$. From that it follows continuously dependence of the solution of problem 1 on $f(x,t)$. The proof is complete. \end{proof} Now, we study the regular solvability of the problem 1 for $k$ even. We denote \begin{align*} W_2(\Omega )=\Big\{& f: f\in C_{x,t}^{2k,2}(\overline{\Omega }),\, \frac{{ \partial ^{2k+2}f}}{{\partial x^{k+1}\partial t}}\in L_2(\Omega ),\, \frac{{\partial ^{4}f}}{{\partial x\partial t^{3}}}\in L_2(\Omega ),\\ & \frac{{\partial ^{2l}f(0,t)}}{{\partial x^{2l}}}=\frac{{\partial ^{2l}f(p,t)}}{{\partial x^{2l}}}=0,\,l=0,1,2,\dots,k;\\ &\frac{{\partial^{2s}f(x,0)}}{{\partial t^{2s}}} =\frac{{\partial ^{2s}f(x,T)}}{{\partial t^{2s}}}=0,\,s=0,1\Big\} . \end{align*} \begin{theorem} \label{thm3.2} If $P^{k}/(T\pi ^{k-1})$ is irrational algebraic number of second degree and $f\in W_2(\Omega )$, then there exists regular solution of problem 1 and satisfies estimate \eqref{e42}. \end{theorem} \begin{proof} We will prove uniform and absolute convergence of series \eqref{e37} and that \begin{gather} \frac{{\partial ^{2k}u}}{{\partial x^{2k}}}=\sum_{n,m=1}^{\infty }{ \frac{{f_{nm}}}{{\lambda _{nm}}}( {\frac{{n\pi }}{p}}) } ^{2k}u_{nm}(x,t), \label{e43} \\ \frac{{\partial ^2u}}{{\partial t^2}}=-\sum_{n,m=1}^{\infty }{ \frac{{f_{nm}}}{{\lambda _{nm}}}( {\frac{{m\pi }}{T}}) } ^2u_{nm}(x,t). \label{e44} \end{gather} The series \begin{equation} \sum_{n,m=1}^{\infty }{n^{2k}}| {f_{nm}}| \label{e45} \end{equation} is majorant for series \eqref{e43} and the series \begin{equation} \sum_{n,m=1}^{\infty }{m^2}| {f_{nm}}| \label{e46} \end{equation} is majorant for series \eqref{e44} . Integrating \eqref{e39} by parts $2k+1$ times with respect to $x$ and one time with respect to $t$, we obtain \begin{gather} |f_{nm}|=\frac{{Tp^{2k+1}}}{{\pi ^{2k+2}}}\frac{1}{{mn^{2k+1}}} |f_{nm}^{(2k+1,1)}|, \label{e47} \\ f_{nm}=-\frac{{pT^{3}}}{{\pi ^{4}nm^{3}}}f_{nm}^{(1,3)}, \label{e48} \end{gather} where \begin{gather*} f_{nm}^{(2k+1,1)}=\int_0^{p} \int_0^{T} \frac{{\partial^{2k+2}f}}{{\partial x^{2k+1}\partial t}}\frac{2}{\sqrt{pT}} \cos \big(\frac{{n\pi}}{p}x\big)\cos \big(\frac{{m\pi }}{T}t\big)\,dt\,dx, \\ f_{nm}^{(1,3)}=\int_0^{P}{\int_0^{T}{\frac{{\partial ^{4}f} }{{\partial x\partial t^{3}}}\frac{2}{\sqrt{PT}}\cos\big( \frac{{n\pi }}{P}x\big) \cos\big(\frac{{m\pi }}{T}t\big)\,dt\,dx}}. \end{gather*} Using \eqref{e47}, the H\"{o}lder inequality and the Bessel inequality from \eqref{e45}, we obtain \[ \sum_{n,m=1}^{\infty }{n^{2k}}| {f_{nm}}| \leqslant \frac{{Tp^{2k+1}}}{{6\pi ^{2k}}}\| {\frac{{ \partial ^{2k+2}f}}{{\partial x^{2k+1}\partial t}}}\| _{L_2(\Omega )}. \] Consequently, series\ \eqref{e43} uniformly and absolutely converges. In \ref{e46} using \eqref{e48}, the H\"{o}lder inequality and the Bessel inequality, we obtain \[ \sum_{n,m=1}^{\infty }{m^2}| {f_{nm}}| \leqslant \frac{{pT^{3}}}{{6\pi ^2}}\| {\frac{{\partial ^{4}f}}{{ \partial x\partial t^{3}}}}\| _{L_2(\Omega )}. \] Hence, absolutely and uniformly convergence of series\ \eqref{e44} follows. Adding equality \eqref{e43} and \eqref{e44}, we note that solution \eqref{e37} satisfies equation \eqref{e1}. Solution \eqref{e37} satisfies boundary conditions \eqref{e2} and \eqref{e3} owing to properties of functions $ u_{nm}(x,t)$. From $W_2(\Omega )\subset L_2(\Omega )$ for $f\in W_2(\Omega )$ it follows estimate (42). The proof is complete. \end{proof} \begin{corollary} \label{coro3.1} In the case even $k$, solvability of problem 1 depends on domain geometry. A minor change of numbers $p$ and $T$ can lead to an ill-posed problem. \end{corollary} \section{Well-posedness of problem 2} First, we will study the regular solvability of problem 2. Let $k$ be odd number, then problem 2 is not correct. Indeed, if we seek the solution of problem 2 in the form \eqref{e17}, we obtain \[ u_n(t)=\frac{1}{{\lambda _n^{k}}}\int_0^{t}{f_n(\tau )} sh\lambda _n^{k}(t-\tau )d\tau \] and $|u_n(t)|\to \infty $ as $n\to \infty $. Therefore, we will consider the case when $k$ is even. We denote \begin{gather*} V_2(\Omega )=\{U: U\in C_{x,t}^{2k-2,1}(\overline{\Omega })\cap C_{x,t}^{2k,2}(\Omega );\text{ \eqref{e2} and \eqref{e4} are satisfied}\}\\ \begin{aligned} W_{3}(\Omega )=\Big\{& f: f\in C_{x,t}^{k,0}(\overline{\Omega }),\, \frac{{\partial ^{k+1}f}}{{\partial x^{k+1}}}\in L_2(\Omega );\, \frac{{\partial ^{2m}f}}{{\partial x^{2m}}}=0\\ & \text{at $x=0$ and $x=p$, $m=0,1,\dots,\frac{{k-2}}{2}$}\Big\}. \end{aligned} \end{gather*} We define the operator $A$ mapping $V_2 (\Omega )$ into $C(\Omega )$ by the rule $Au = Lu$. Let $\overline A$ be the closure of $A$ in $L_2 (\Omega )$. Note that the definition of regular solution $u(x,t)\in V_2(\Omega )$ of problem 2 is same as Definition 2.1. Moreover, the definition of strong solution of problem 2 is same as Definition 2.2, but in this case $\{u_n\} \subset V_2(\Omega )$. We seek a regular solution of problem 2 in the form \eqref{e17}. Then, for unknown functions $u_n(t)$ we have \[ u_n''+\lambda _n^{2k}u_n(t)=f_n(t),00$ that depends only on numbers $T$ and $k$, and does not depend on the function $u(x,t)$ such that \begin{equation} \| u\| _{W_2^{(k,1)}(\Omega )}\leqslant C\| {Lu}\| _{L_2(\Omega )}, \label{e58} \end{equation} where \[ \| u\| _{W_2^{k,1}(\Omega )}^2=\sum_{m=0}^{k}{ \| {\frac{{\partial ^{m}u}}{{\partial x^{m}}}}\| }_{L_2(\Omega )}^2 +\| {\frac{{\partial u}}{{\partial t}}} \| _{L_2(\Omega )}^2. \] \end{lemma} \begin{proof} Multiplying by $\frac{{\partial u}}{{\partial t}}$ both sides of the identity \eqref{e6} and integrating it over the domain $\Omega _{\tau }=\{ (x,t):00. \label{e63} \end{equation} The method of proof of Theorem \ref{thm2.2} enables us to establish the following fact. \begin{theorem} \label{thm4.2} For any $f\in L_2(\Omega )$ there exists unique strong solution of problem 2 and it satisfies estimate \eqref{e58}, it continuously depends on $f(x,t)$ and it can be represented in the form \eqref{e61}. \end{theorem} Fourth, we will study the spectrum of problem 2. \begin{definition} \label{def4.1} \rm We say that a problem has the Volterra property (see \cite{k1}) if the inverse operator of the problem has the Volterra property. \end{definition} \begin{theorem} \label{thm4.3} If the number $k$ is even, then the spectrum of the problem 2 is empty. \end{theorem} \begin{proof} From \eqref{e58}, \eqref{e61} and \eqref{e63} we conclude that it is defined operator $A^{-1}$ on $W_{3}(\Omega )$ which is inverse of the operator $A$ and acts from $W_{3}(\Omega )$ to $V_2(\Omega )$ by the rule \begin{equation} ( {A^{-1}f}) (x,t)=\int_0^{p}{\int_0^{t}{ K(x,t;\xi ,\tau )}}f(\xi ,\tau )\,d\tau\,d\xi . \label{e64} \end{equation} It follows from \eqref{e63} that $K(x,t;\xi ,\tau )\in L_2(\Omega \times \Omega )$, therefore $A^{-1}$ is a compact operator in $L_2(\Omega )$. As $ D(A^{-1})\equiv W_{3}(\Omega )$ is dense in $L_2(\Omega )$, the operator $ A^{-1}$ can be extended to whole space $L_2(\Omega )$. This extension, we denote it by $\overline{A^{-1}}$, the closure of $A^{-1}$, $D(\overline{ A^{-1}})=L_2(\Omega )$. $\overline{A^{-1}}$ is a compact operator in $L_2(\Omega )$. Now we show that $\sigma (\overline{A^{-1}})=\{0\} $, where $\sigma (\overline{A^{-1}})$ is the spectrum of the operator $\overline{A^{-1}}$ (see \cite{s1}). For this purpose we calculate the spectral radius of the operator $\overline{A^{-1}}$ (see \cite{s1}). \[ r(\overline{A^{-1}})=\lim_{n\to \infty }( { \| {\overline{A^{-n}}}\| }) ^{1/n}. \] It is easy to show that \[ ( {\overline{A^{-n}}f}) (x,t)=\int_0^{p}{ \int_0^{t}{K_n(x,t;\xi ,\tau )}}f(\xi ,\tau )\,d\tau\,d\xi , \] where \begin{gather*} K_n(x,t;\xi ,\tau )=\int_0^{p}{\int_0^{T}{K(x,t;\xi',\tau ')K_{n-1} (\xi ',\tau ';\xi ,\tau )}}d\tau 'd\xi ',\quad n=2,3,\ldots , \\ K_1(x,t;\xi ,\tau )=K(x,t;\xi ,\tau ), \\ | {K_n(x,t;\xi ,\tau )}| \leqslant \frac{{C^{n}p^{n-1}(t-\tau )^{n-1}}}{{(n-1)!}}. \end{gather*} By direct computation, we obtain \[ \| {\overline{A^{-n}}}\| \leqslant \frac{{(CpT)^{n}}}{{n!}}. \] Using Stirling formula (see \cite{i1}) for $n!$ we convince that $r(\overline{A^{-1}})=\{0\} $. Since the spectrum of $\overline{A^{-1}}$ lays in the circle $| \lambda | \leqslant r(\overline{A^{-1}})$, the spectrum consists of one point a zero. Consequently, operator $\overline{A^{-1}}$ has the Volterra property. The relation between eigenvalues of the operator $\overline{A^{-1}}$ and $\overline{A}$ is as follows \cite{d1}: If $\mu _n\neq 0$ is eigenvalue of the operator $\overline{A^{-1}}$, then $\mu _n^{-1}$ is eigenvalue of operator $\overline{A}$. Since zero is not eigenvalue of the operator $\overline{A^{-1}}$, then the spectrum of the problem 2 is empty set. The proof of Theorem \ref{thm4.3} is complete. \end{proof} \begin{corollary} \label{coro4.2} For even $k$ the problem 2 has the Volterra property. \end{corollary} Note that problem 2 is not self-adjoint. \subsection*{Problem $2^{\ast}$} Obtain the solution $v(x,t)$ of \eqref{e1} in the domain $\Omega $ satisfying conditions \eqref{e2} and $v(x,T)=0$, $v_{t}(x,T)=0$, $0