\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 36, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2016 Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2016/36\hfil Boundary value problems] {Boundary value problems for fourth-order mixed type equation with fractional derivative} \author[A. S. Berdyshev, B. E. Eshmatov, B. J. Kadirkulov \hfil EJDE-2016/36\hfilneg] {Abdumauvlen S. Berdyshev, Bakhodir E. Eshmatov,\\ Bakhtiyor J. Kadirkulov} \address{Abdumauvlen S. Berdyshev \newline Kazakh National Pedagogical University named after Abai, Kazakhstan} \email{berdyshev@mail.ru} \address{Bakhodir E. Eshmatov \newline Karshi Engineering Economics Institute, Uzbekistan} \email{eshmatovb@mail.ru} \address{Bakhtiyor J. Kadirkulov \newline Tashkent State Institute of Oriental Studies, Uzbekistan} \email{kadirkulovbj@gmail.com} \thanks{Submitted October 1, 2015. Published January 26, 2016.} \subjclass[2010]{35M10} \keywords{Mixed type equation; direct and inverse problems; \hfill\break\indent Caputo fractional derivative; Mittag-Leffler function; separation of variables; Fourier series} \begin{abstract} In this work we study direct and inverse problems for fourth-order mixed type equations with the Caputo fractional derivative. Applying method of separation of variables we prove unique solvability of these problems. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{remark}[theorem]{Remark} \allowdisplaybreaks \section{Introduction and formulation of problems} It is known that the theory of boundary value problems for fractional order differential equations is one of the rapidly developing branches of the general theory of differential equations. Detailed information related with fractional calculus can be found in \cite{Kilbas}. Many problems in diffusion and dynamical processes, electrochemistry, biosciences, signal processing, system control theory lead to differential equations of fractional order. More information can be found in \cite{Jesus},\cite{Magin}. Mathematical modeling of many real life processes lead to problems of identifying coefficients or right-hand sides of differential equations, based on some known data of their solutions. These kinds of problems are called inverse problems. Inverse problems appear also in various fields such physics (inverse problems of quantum scattering theory), geophysics (inverse problems of magneto metrics, seismology, theory of potentials), biology, medicine, quality control of industrial products and etc. (see \cite{Kab,Shadan}). Similar problems studied in \cite{Furati,Kirane}. In this work we investigate direct and inverse problems for fourth-order mixed type equations. Let $\Omega =\{(x,t):00)$, ${{\Omega }^{-}}=\Omega \cap (t<0)$, where $p,q>0$ are real numbers. In $\Omega $ we consider the equations \begin{equation} \label{e1} \begin{gathered} \frac{{{\partial }^4}u}{\partial {{x}^4}}+{}_CD_{0t}^{\alpha }u =f(x,t), \quad t>0, \\ \frac{{{\partial }^4}u}{\partial {{x}^4}} +\frac{{{\partial }^2}u}{\partial {{t}^2}}=f(x,t),\quad t<0, \end{gathered} \end{equation} and \begin{equation} \label{e2} \begin{gathered} \frac{{{\partial }^4}u}{\partial {{x}^4}}+{}_CD_{0t}^{\alpha }u =f(x),\quad t>0, \\ \frac{{{\partial }^4}u}{\partial {{x}^4}}+\frac{{{\partial }^2}u} {\partial {{t}^2}}=f(x),\quad t<0, \end{gathered} \end{equation} where ${}_CD_{0t}^{\alpha }$ is the Caputo fractional operator of the order $\alpha \in (0,1] $ with respect to variable $t$ \cite[p. 92]{Kilbas}, \[ {}_CD_{0t}^{\alpha }u(x,t)=\frac{1}{\Gamma (1-\alpha )} \int_0^t{{{(t-\tau )}^{-\alpha }} \frac{\partial u(x,\tau )}{\partial \tau }d\tau ,\quad t>0}. \] We study the following problems. \smallskip \noindent\textbf{Direct problem.} Find a function $u(x,t)$ such that: \begin{itemize} \item[(1)] $u$ is continuous in $\bar{\Omega }$, together with its derivatives appearing in the boundary conditions, \item[(2)] $u$ satisfies equation \eqref{e1} in ${{\Omega }^{+}}\cup {{\Omega }^{-}}$, \item[(3)] $u$ satisfies the boundary conditions \begin{gather} u(0,t)=u(1,t)=u_{xx}(0,t)=u_{xx}(1,t)=0,\quad -p\le t\le q, \label{e3} \\ u(x,-p)=0,\quad 0\le x\le 1,\label{e4} \end{gather} \item[(4)] $u$ satisfies the matching condition \begin{equation} {}_CD_{0t}^{\alpha }u(x,+0)=\frac{\partial u(x,-0)}{\partial t},\quad 00. \] For negative values of $t$, $-p\le t\le 0$, we set \begin{equation} {{d}_n}(t)=\int_0^{1}{u(x,t){{X}_n}(x)dx}. \label{e11} \end{equation} Differentiating twice with respect to $t$ and considering \eqref{e1}, \eqref{e3}, we obtain \[ {{{d}''}_n}(t)+\lambda _n^4{{d}_n}(t)=0, \] whose solution is \begin{equation} {{d}_n}(t)={{B}_n}\sin(\lambda _n^2t) +{{L}_n}\cos(\lambda _n^2t),\quad t\le 0. \label{e12} \end{equation} For determining the unknown coefficients of \eqref{e10} and \eqref{e12} we use the continuity property of the function in $\overline{\Omega }$ as well as the matching condition \eqref{e5}. We obtain \[ {{A}_n}={{L}_n},\quad {{B}_n}+\lambda _n^2{{A}_n}=0,\quad {{B}_n}\sin(\lambda _n^2p)-{{L}_n}\cos(\lambda _n^2p)=0. %\label{e13} \] This system has only the trivial solution, since its determinant ${{\Delta }_n}$ is not equal to zero because of condition \eqref{e7}. Hence, ${{c}_n}(t)={{d}_n}(t)=0$. Then from \eqref{e9} and \eqref{e11} it follows that \[ \int_0^{1}{u(x,t){{X}_n}(x)dx}\equiv 0,\quad t\in [-p;q],\quad n\in \mathbb{N}. \] Based on the completeness of system \eqref{e8} we conclude that $u(x,t)=0$ in $\overline{\Omega }$. The proof is complete. \end{proof} \begin{remark} \label{rmk1} \rm The set of numbers $p$ satisfying condition \eqref{e7} is not empty. For instance, if $p=\frac{1}{\pi }$, then ${{\Delta }_n}=\pm 1\ne 0$. \end{remark} Now we prove the existence of a solution for the direct problem. \begin{theorem} \label{thm2} Let $f(x,t)\in C_{x,t}^{4,0}( \overline{\Omega })$, $\frac{{{\partial }^{5}}f(x,t)}{\partial {{x}^{5}}}\in {{L}_2}(\Omega )$, \[ f(0,t)={{f}_{xx}}(0,t)={{f}_{xxxx}}(0,t)=0,\quad f(1,t)={{f}_{xx}}(1,t)={{f}_{xxxx}}(1,t)=0, \] and ${{\Delta }_n}\ne 0$ at $n\in \mathbb{N}$. Then the solution of direct problem exists. \end{theorem} \begin{proof} We use the method of separation of variables for $f\equiv 0$ in $\Omega$, we set $u(x,t)=X(x)T(t)\ne 0$. Substituting this into \eqref{e1}, taking boundary conditions \eqref{e3} into account with respect to $X(x)$ we get the spectral problem \begin{equation} {{X}^{IV}}(x)-{{\lambda }^4}X(x)=0,\quad X(0)=X(1)=X''(0)=X''(1)=0. \label{e14} \end{equation} This problem is selfadjoint and has complete system of eigenfunctions represented by \eqref{e8}. Let us set \[ u(x,t)=\begin{cases} {{u}^{+}}(x,t), & (x,t)\in {{\Omega }^{+}}, \\ {{u}^{-}}(x,t), & (x,t)\in {{\Omega }^{-}}. \end{cases} \] For the solution of the non-homogeneous equation \eqref{e1} we set \begin{gather} {{u}^{+}}(x,t)=\sum_{n=1}^{\infty }{u_n^{+}(t){{X}_n}(x)},\quad \text{in } {{\Omega }^{+}}, \label{e15} \\ {{u}^{-}}(x,t)=\sum_{n=1}^{\infty }{u_n^{-}(t){{X}_n}(x)},\quad \text{in } {{\Omega }^{-}}, \label{e16} \end{gather} where $u_n^{+}(t)$ and $u_n^{-}(t)$ are unknown functions. Solutions of \eqref{e15} and \eqref{e16} satisfy conditions \eqref{e3}. Let us expand $f(x,t)$ into the series \begin{equation} f(x,t)=\sum_{n=1}^{\infty }{{{f}_n}(t){{X}_n}(x)},\label{e17} \end{equation} where \[ {{f}_n}(t)=\int_0^{1}{f(x,t){{X}_n}(x)dx}.\label{e18} \] Substiting \eqref{e15}--\eqref{e17} in \eqref{e1} lead to the equations \begin{gather*} {}_CD_{0t}^{\alpha }u_n^{+}(t)+\lambda _n^4u_n^{+}(t)={{f}_n}(t),\quad t>0, \\ \frac{{{d}^2}}{d{{t}^2}}u_n^{-}(t)+\lambda _n^4u_n^{-}(t) ={{f}_n}(t),\quad t<0, \end{gather*} whose solutions are \cite[p. 231]{Kilbas} \begin{gather} u_n^{+}(t)={{A}_n}{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }}) +\int_0^t{{{(t-\tau )}^{\alpha -1}}{{E}_{\alpha ,\alpha }} (-\lambda _n^4{{(t-\tau )}^{\alpha }}){{f}_n}(\tau )d\tau }, \label{e19} \\ u_n^{-}(t)={{B}_n}\sin(\lambda _n^2t)+{{L}_n} \cos(\lambda _n^2t)+\frac{1}{\lambda _n^2}\int_{t}^{0}{{{f}_n}(\tau ) \sin(\lambda _n^2(\tau -t))d\tau }, \label{e20} \end{gather} where ${{A}_n},{{B}_n},{{L}_n}$ are unknown constants, ${{E}_{\alpha ,\beta }}(z)$ is the Mittag-Leffler type function \cite[p. 42]{Kilbas}, \[ {{E}_{\alpha ,\beta }}(z)=\sum_{n=0}^{\infty }{\frac{{{z}^{n}}}{\Gamma (\alpha n+\beta )}},\quad z,\beta \in \mathbb{C},\;\operatorname{Re}(\alpha )>0,\; {{E}_{\alpha ,1}}(z)={{E}_{\alpha }}(z). \] To determine the unknown constants of \eqref{e19} and \eqref{e20} we use the continuity of the looked for function in $\bar{\Omega }$ and condition \eqref{e5} of direct problem. We obtain \[ u_n^{+}(0)=u_n^{-}(0),\quad {}_CD_{0t}^{\alpha }u_n^{+}(0) =\frac{du_n^{-}(0)}{dt},\quad u_n^{-}(-p)=0. \] Then concerning the unknowns ${{A}_n},{{B}_n},{{L}_n}$, we obtain the system of equations \begin{gather*} {{A}_n}={{L}_n}, \\ {{f}_n}(0)-\lambda _n^4{{A}_n}=\lambda _n^2{{B}_n}, \\ {{B}_n}\sin(\lambda _n^2p)-{{L}_n}\cos(\lambda _n^2p) =\frac{1}{\lambda _n^2}\int_{-p}^{0}{{{f}_n}(\tau ) \sin(\lambda_n^2(\tau +p))d\tau }. \end{gather*} Since ${{\Delta }_n}\ne 0$ the coefficients ${{A}_n},{{B}_n},{{L}_n}$ can be uniquely determined. Substituting ${{A}_n},{{B}_n},{{L}_n}$ in \eqref{e19} and \eqref{e20} we find that \begin{gather} \begin{aligned} u_n^{+}(t)&=\frac{{{f}_n}(0)\sin(\lambda _n^2p)} {\lambda _n^2{{\Delta }_n}}{{E}_{\alpha }} (-\lambda _n^4{{t}^{\alpha }})+\int_0^t{{{(t-\tau )}^{\alpha -1}} {{E}_{\alpha ,\alpha }}(-\lambda _n^4{{(t-\tau )}^{\alpha }}){{f}_n} (\tau )d\tau } \\ &\quad -\frac{{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }})} {\lambda _n^2{{\Delta }_n}}\int_{-p}^{0}{{{f}_n}(\tau ) \sin(\lambda_n^2(\tau +p))d\tau }, \end{aligned} \label{e21}\\ u_n^{-}(t)=\frac{{{f}_n}(0)\sin(\lambda_n^2(t+p))} {\lambda _n^2{{\Delta }_n}}+\int_{-p}^{0}{{{K}_n}(t,\tau ) {{f}_n}(\tau )d\tau }, \label{e22} \end{gather} where \begin{equation} {{K}_n}(t,\tau )= \begin{cases} \frac{(\lambda _n^2\sin(\lambda _n^2t) -\cos(\lambda _n^2t))\sin(\lambda_n^2(\tau +p))} {\lambda _n^2{{\Delta }_n}},& -p\le \tau \le t, \\ \frac{(\lambda _n^2\sin(\lambda_n^2\tau) -\cos(\lambda _n^2\tau) )\sin(\lambda_n^2(t+p))} {\lambda _n^2{{\Delta }_n}}, & t\le \tau \le 0. \end{cases} \label{e23} \end{equation} Thus we find a formal solution of direct problem in ${{\Omega }^{+}}$ and ${{\Omega }^{-}}$, given by formulas \eqref{e15} and \eqref{e16}, respectively, where $u_n^{+}(t)$, $u_n^{-}(t)$ are defined by formulas \eqref{e21}, \eqref{e22}. We need to prove that this formal solution is a true solution. For this aim, we prove the convergence of series \eqref{e15}, \eqref{e16} and \begin{gather} \sum_{n=1}^{\infty }{\lambda _n^2u_n^{+}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{\lambda _n^2u_n^{-}(t){{X}_n}(x)}, \\ \sum_{n=1}^{\infty }{\lambda _n^4u_n^{+}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{{}_CD_{0t}^{\alpha }u_n^{+}(t){{X}_n}(x)}, \label{e24} \\ \sum_{n=1}^{\infty }{\lambda _n^4u_n^{-}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{\frac{{{d}^2}u_n^{-}(t)}{d{{t}^2}}{{X}_n}(x)}. \label{e25} \end{gather} First we prove convergence of series \eqref{e24} and \eqref{e25}. Convergence of other series can be done similarly. Let us prove convergence of the first series of \eqref{e24}. It can be majorized by \begin{equation} \sum_{n=1}^{\infty }{\lambda _n^4|u_n^{+}(t)|}. \label{e26} \end{equation} We have ${{\Delta }_n}\ne 0$, hence there exists $\delta >0$ such that $|{{\Delta }_n}|\ge \delta >0$. Further, we use the following properties of the Mittag-Leffler function ${{E}_{\alpha ,\beta }}(z)$: \begin{itemize} \item[(1)] at $\mu >0$, $\alpha ,\beta \in (0,1]$, $\alpha \le \beta$, ${{t}^{\alpha -1}}{{E}_{\alpha ,\beta }}(-\mu {{t}^{\alpha }})$ is completely monotone \cite{Kirane}, i.e. ${{(-1)}^{n}}{{[ {{t}^{\beta -1}}{{E}_{\alpha ,\beta }}(-\mu {{t}^{\alpha }}) ]}^{(n)}}\ge 0,\,n\in \mathbb{N}\cup \{0\}$; \item[(2)] at $\alpha \in (0,2)$, $\gamma \le |\operatorname{argz}|\le \pi$, $\beta \in R$, $\gamma \in ( \pi \alpha /2;\min \{\pi ;\pi \alpha \} )$ we have for \[ | {{E}_{\alpha ,\beta }}(z) |\le \frac{M}{1+|z|}, \] where $M$ is constant which does not depend on argument $z$ \cite[p. 136]{Dzhr}; \item[(3)] the following formulas are valid \cite[pp. 118, 120]{Dzhr}, \begin{equation} {{E}_{\alpha ,\mu }}(z)=\frac{1}{\Gamma (\mu )}+z{{E}_{\alpha ,\alpha +\mu }}(z),\quad \int_0^{z}{{{t}^{\mu -1}}{{E}_{\alpha ,\mu }} (\lambda {{t}^{\alpha }})dt}={{z}^{\mu }}{{E}_{\alpha ,\mu +1}} (\lambda {{z}^{\alpha }}). \label{e27} \end{equation} \end{itemize} Set \begin{gather*} {{I}_{1}}(t)=\frac{{{f}_n}(0)\sin(\lambda _n^2p)} {\lambda _n^2{{\Delta }_n}}{{E}_{\alpha }} (-\lambda _n^4{{t}^{\alpha }}),\\ {{I}_2}(t)=\int_0^t{{{(t-\tau )}^{\alpha -1}}{{E}_{\alpha ,\alpha }} (-\lambda _n^4{{(t-\tau )}^{\alpha }}){{f}_n}(\tau )d\tau }, \\ {{I}_{3}}(t)=\frac{{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }})} {\lambda _n^2{{\Delta }_n}}\int_{-p}^{0}{{{f}_n}(\tau ) \sin(\lambda_n^2(\tau +p))d\tau }. \end{gather*} We estimate the function ${{I}_{1}}(t)$: \[ |{{I}_{1}}(t)|=\Big| \frac{{{f}_n}(0)\sin(\lambda _n^2p)} {\lambda _n^2{{\Delta }_n}}{{E}_{\alpha ,1}} (-\lambda _n^4{{t}^{\alpha }}) \Big| \le \frac{M}{\lambda _n^2\delta }|{{f}_n}(0)|. \] Based on the condition imposed on $f(x,t)$ in the theorem, from \eqref{e18} we have \begin{equation} \begin{gathered} {{f}_n}(t)=\frac{1}{\lambda _n^4}\int_0^{1}{\frac{{{\partial }^4}f} {\partial {{x}^4}}\sqrt{2}\sin(\lambda_nx)\,dx} =\frac{1}{\lambda _n^4}{{f}_{n4}}(t), \\ |{{f}_n}(t)|\le \frac{1}{\lambda _n^4}N, N =\max_{[0;q]} |{{f}_{n4}}(t)|. \end{gathered} \label{e28} \end{equation} Then \begin{equation} \lambda _n^4|{{I}_{1}}(t)| \le \frac{MN}{\lambda _n^2\delta }=\frac{C}{\lambda _n^2}. \label{e29} \end{equation} Here and further, $C$ is a positive that may change from line to line. Considering \eqref{e28}, we have the estimate \[ | {{I}_2}(t)|\le \frac{N}{\lambda _n^4} \int_0^t{\big| {{(t-\tau )}^{\alpha -1}}{{E}_{\alpha ,\alpha }} (-\lambda _n^4{{(t-\tau )}^{\alpha }}) \big|d\tau }. \] From here, applying formula (27) we obtain \begin{equation} | {{I}_2}(t)|\le \frac{N}{\lambda _n^4} \left| 1-{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }}) \right| \le \frac{C}{\lambda _n^4}. \label{e30} \end{equation} Further, taking \eqref{e28} into account we have the estimate \begin{equation} | {{I}_{3}}(t)|\le \frac{M}{\lambda _n^2\delta } \int_{-p}^{0}{| {{f}_{n4}}(\tau )|d\tau }\le \frac{C}{\lambda _n^2}. \label{e31} \end{equation} From \eqref{e29}--\eqref{e31} it follows the convergence of series \eqref{e26}. Then based on Weierstrass' theorem \cite[p. 20]{Ilyin}, first series of \eqref{e24} converges absolutely and uniformly. Since \[ {}_CD_{0t}^{\alpha }u_n^{+}(t)={{f}_n}(t)-\lambda _n^4u_n^{+}(t), \] the absolute and uniform convergence of the second series of \eqref{e24} can be proved similarly. Now we prove the absolute and uniform convergence of the first series of \eqref{e25}. Let \[ {{J}_{1}}(t)=\frac{{{f}_k}(0)\sin(\lambda_n^2(t+p))} {\lambda _n^2{{\Delta }_n}},\quad {{J}_2}(t) =\int_{-p}^{0}{{{K}_n}(t,\tau ){{f}_n}(\tau )d\tau }. \] Considering \eqref{e28}, which is valid as well in the case $t\le 0$ , we obtain \begin{equation} \lambda _n^4| {{J}_{1}}(t) | =\lambda _n^4\big| \frac{{{f}_n}(0)\sin(\lambda_n^2(t+p))} {\lambda _n^2{{\Delta }_n}} \big| \le \frac{\lambda _n^2}{\delta }| {{f}_n}(0) | \le \frac{C}{\lambda _n^2}. \label{e32} \end{equation} Note that the function ${{K}_n}(t,\tau )$ defined by \eqref{e23} is bounded, i.e., $| {{K}_n}(t,\tau )|\le C,\,-p\le t,\tau \le 0$. From \eqref{e18} it follows that \[ {{f}_n}(t)=-\frac{1}{\lambda _n^{5}}\int_0^{1}{\frac{{{\partial }^{5}}f} {\partial {{x}^{5}}}\sqrt{2}\cos(\lambda_nx)dx}=-\frac{1}{\lambda _n^{5}}{{f}_{n5}}(t). \] Further, applying the inequality $2ab\le {{a}^2}+{{b}^2}$, and the Cauchy-Schwarz inequality, we have the estimate \begin{align*} \lambda _n^4| {{J}_2}(t) | &=\lambda _n^4\Big| \int_{-p}^{0}{{{K}_n}(t,\tau ){{f}_n}(\tau )d\tau } \Big| \\ &\le \frac{C}{{{\lambda }_n}}\int_{-p}^{0}{| {{f}_{n5}}(\tau ) |d\tau }\\ &\le C\Big[ \frac{1}{\lambda _n^2}+{{\Big( \int_{-p}^{0}{| {{f}_{n5}}(\tau )| d\tau } \Big)}^2} \Big] \\ &\le C\Big[ \frac{1}{\lambda _n^2}+\int_{-p}^{0}{{{| {{f}_{n5}}(\tau ) |}^2}d\tau } \Big] \\ &=C\big[ \frac{1}{\lambda _n^2}+\| {{f}_{n5}}(t) \||_{{{L}_2}(-p,0)}^2 \big]. \end{align*} Since \[ \sum_{n=1}^{\infty }{\frac{1}{\lambda _n^2}}=\frac{1}{6},\quad \sum_{n=1}^{\infty }{\| {{f}_{n5}}(t) \||_{{{L}_2}(-p,0)}^2} \le \| \frac{{{\partial }^{5}}f(x,t)}{\partial {{x}^{5}}} \||_{{{L}_2} (\Omega )}^2, \] from this and \eqref{e32} , it follows the absolute and uniform convergence of the series for $t\le 0$. The convergence of the second series can be obtained similarly. The proof is complete. \end{proof} \section{Inverse problem} For solving the inverse problem, we set \begin{gather} u(x,t)=\begin{cases} \sum_{n=1}^{\infty } u_n^{+}(t) X_n (x), &(x,t)\in \Omega ^{+} \\ \sum_{n=1}^{\infty } u_n^{-}(t) X_n (x), &(x,t)\in \Omega ^{-}, \end{cases} \label{e33} \\ f(x)=\sum_{n=1}^{\infty } f_n X_n(x), \label{e34} \end{gather} where $u_n^{+}(t),u_n^{-}(t)$ are unknown functions, $f_n$ is unknown coefficient. Substitution \eqref{e33} and \eqref{e34} in \eqref{e2}, leads to equations \begin{gather} {}_CD_{0t}^{\alpha }u_n^{+}(t)+\lambda _n^4u_n^{+}(t) ={{f}_n},\quad t>0, \label{e35} \\ \frac{{{d}^2}}{d{{t}^2}}u_n^{-}(t)+\lambda _n^4u_k^{-}(t) ={{f}_n},\quad t<0, \label{e36} \end{gather} whose solutions are \eqref{e19} and \eqref{e20}. From here, considering ${{f}_n}$ constant, we obtain \begin{gather} u_n^{+}(t)={{A}_n}{{E}_{\alpha }} (-\lambda _n^4{{t}^{\alpha }})+{{f}_n}{{t}^{\alpha }}{{E}_{\alpha ,\alpha +1}} (-\lambda _n^4{{t}^{\alpha }}), \label{e37} \\ u_n^{-}(t)={{B}_n}\sin(\lambda _n^2t)+{{L}_n} \cos(\lambda _n^2t)+\frac{{{f}_n}}{\lambda _n^4} (1-cos\lambda _n^2t), \label{e38} \end{gather} where ${{A}_n},{{B}_n},{{L}_n},\,{{f}_n}$ are unknown coefficients. To determine the unknown coefficients we use the continuity of the unkown function in $\bar{\Omega }$ and conditions \eqref{e5} and \eqref{e6}. We obtain \[ u_n^{+}(0)=u_n^{-}(0),\quad {{\,}_C}D_{0t}^{\alpha }u_n^{+}(0)=\frac{du_n^{-}(0)}{dt},\quad u_n^{+}(q)={{\varphi }_n},\,u_n^{-}(-p)={{\psi }_n}, \] where \begin{equation} {{\varphi }_n}=\int_0^{1} \varphi (x){{X}_n}(x)dx,\quad {{\psi }_n}=\int_0^{1}{\psi (x){{X}_n}(x)dx}. \label{e39} \end{equation} Then concerning the unknowns ${{A}_n},{{B}_n},{{L}_n},\,{{f}_n}$ we obtain system of equations \begin{equation} \begin{gathered} {{L}_n}={{A}_n}, \\ \lambda _n^4{{A}_n}+\lambda _n^2{{B}_n}-{{f}_n}=0, \\ {{A}_n}{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }}) +{{f}_n}{{q}^{\alpha }}{{E}_{\alpha ,\alpha +1}} (-\lambda _n^4{{q}^{\alpha }})={{\varphi }_n}, \\ \cos(\lambda _n^2p)\cdot {{A}_n} -\sin(\lambda _n^2p)\cdot {{B}_n}+\frac{1}{\lambda _n^4} (1-\cos(\lambda _n^2p)){{f}_n}={{\psi }_n}, \end{gathered} \label{e40} \end{equation} whose solutions are \begin{gather*} {{A}_n}={{\varphi }_n}+\frac{1-{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})} {{{\Delta }_{n*}}}({{\psi }_n}-{{\varphi }_n}),\quad {{B}_n}=\frac{\lambda _n^2}{{{\Delta }_{n*}}}({{\varphi }_n} -{{\psi }_n}),\\ {{L}_n}={{A}_n}, \quad {{f}_n}=\frac{\lambda _n^4{{E}_{\alpha }} (-\lambda _n^4{{q}^{\alpha }})}{{{\Delta }_{n*}}}({{\varphi }_n} -{{\psi }_n})+\lambda _n^4{{\varphi }_n}, \end{gather*} where \begin{equation} {{\Delta }_{n*}}=\lambda _n^2\sin(\lambda _n^2p) +\cos(\lambda _n^2p)-{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }}).\label{e41} \end{equation} Substituting of ${{A}_n},{{B}_n},{{L}_n}$ into \eqref{e37} and \eqref{e38}, we find \begin{gather} u_n^{+}(t)=\frac{({{\varphi }_n}-{{\psi }_n})}{{{\Delta }_{n*}}} \left( {{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }}) -{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }}) \right) +{{\varphi }_n}, \label{e42}\\ u_n^{-}(t)=\frac{\lambda _n^2\sin(\lambda _n^2t) -\cos(\lambda _n^2t)+{{E}_{\alpha }} (-\lambda _n^4{{q}^{\alpha }})}{{{\Delta }_{n*}}} ({{\varphi }_n}-{{\psi }_n})+{{\varphi }_n}.\label{e43} \end{gather} Further, from \eqref{e33}, \eqref{e34}, \eqref{e42}, \eqref{e43} we obtain formal solution of the inverse problem, which is given by formulas \begin{gather} u(x,t)=\sum_{n=1}^{\infty }{\Big( \frac{{{\varphi }_n}-{{\psi }_n}} {{{\Delta }_{n*}}}\left( {{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }}) -{{E}_{\alpha }}(-\lambda _n^4{{t}^{\alpha }}) \right) +{{\varphi }_n} \Big)X_n(x),\quad t\ge 0}, \label{e44} \\ \begin{gathered} u(x,t)=\sum_{n=1}^{\infty }{\Big( \frac{\lambda _n^2\sin(\lambda _n^2t) -\cos(\lambda _n^2t)+{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})} {{{\Delta }_{n*}}}({{\varphi }_n}-{{\psi }_n})+{{\varphi }_n} \Big)X_n(x)},\\ \text{for } t\le 0, \end{gathered}\label{e45} \\ f(x)=\sum_{n=1}^{\infty }{\Big( \frac{\lambda _n^4{{E}_{\alpha }} (-\lambda _n^4{{q}^{\alpha }})}{{{\Delta }_{n*}}}({{\varphi }_n} -{{\psi }_n})+\lambda _n^4{{\varphi }_n} \Big){{X}_n}(x)}. \label{e46} \end{gather} \begin{theorem} \label{thm3} Let $\varphi (x)\in {{C}^{6}}[0,1]$, ${{\varphi }^{(7)}}(x)\in {{L}_2}(0,1)$, \[ \frac{{{d}^{2k}}\varphi (x)}{d{{x}^{2k}}}\big| _{x=0} =\frac{{{d}^{2k}}\varphi (x)}{d{{x}^{2k}}}\big| _{x=1} =0,\quad k=\overline{0,3}, \] $\psi (x)\in {{C}^{6}}[0,1]$, ${{\psi }^{(7)}}(x)\in {{L}_2}(0,1)$, \[ \frac{{{d}^{2k}}\psi (x)}{d{{x}^{2k}}}\big| _{x=0} =\frac{{{d}^{2k}}\psi (x)}{d{{x}^{2k}}}\big| _{x=1} =0,\quad k=\overline{0,3}, \] and ${{\Delta }_{n*}}\ne 0$ at $n \in \mathbb{N}$. Then there exists the unique solution for the inverse problem. \end{theorem} \begin{proof} Since ${{\Delta }_{n*}}\ne 0$, the uniqueness of the solution follows from \eqref{e44}--\eqref{e46}, and from the completeness of \eqref{e8}. Therefore, we prove only the existence of the solution. Solutions \eqref{e44}--\eqref{e46} satisfy \eqref{e2} and conditions \eqref{e3}, \eqref{e5}, \eqref{e6}. We need to prove that this formal solution is a true solution. For this aim we prove the convergence of series \eqref{e44}--\eqref{e46} and \begin{gather} \sum_{n=1}^{\infty }{\lambda _n^2u_n^{+}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{\lambda _n^2u_n^{-}(t){{X}_n}(x)}, \\ \sum_{n=1}^{\infty }{\lambda _n^4u_n^{+}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{{}_CD_{0t}^{\alpha }u_n^{+}(t){{X}_n}(x)}, \\ \sum_{n=1}^{\infty }{\lambda _n^4u_n^{-}(t){{X}_n}(x)},\quad \sum_{n=1}^{\infty }{\frac{{{d}^2}u_n^{-}(t)}{d{{t}^2}}{{X}_n}(x)}.\label{e47} \end{gather} Let us prove convergence of the series of \eqref{e47}. We have ${{\Delta }_{n*}}\ne 0$, hence there exists $\delta >0$ such that $|{{\Delta }_{n*}}|\ge \delta >0$. From here, taking properties of Mittag-Leffler function we get \[ |u_n^{-}(t)|=\Big| \frac{\lambda _n^2 \sin(\lambda _n^2t)-\cos(\lambda _n^2t) +{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})}{{{\Delta }_{n*}}}({{\varphi }_n} -{{\psi }_n})+{{\varphi }_n} \Big| \le C\lambda _n^2( |{{\varphi }_n}|+|{{\psi }_n}| ). \] Based on the conditions imposed on $\varphi (x)$ and $\psi (x)$ in theorem \ref{thm3}, from \eqref{e39} we obtain \begin{gather} {{\varphi }_n}=-\frac{1}{\lambda _n^{7}}\int_0^{1} {\frac{{{d}^{7}}\varphi }{\partial {{x}^{7}}}\sqrt{2}\cos{{\lambda }_n}xdx} =-\frac{1}{\lambda _n^{7}}{{\varphi }_{n7}},\label{e48} \\ {{\psi }_n}=-\frac{1}{\lambda _n^{7}}\int_0^{1} {\frac{{{d}^{7}}\psi }{\partial {{x}^{7}}}\sqrt{2}\cos{{\lambda }_n}x\,dx} =-\frac{1}{\lambda _n^{7}}{{\psi }_{n7}}.\label{e49} \end{gather} From here it follows that given series can be majorized by \[ \sum_{n=1}^{\infty }{\frac{1}{{{\lambda }_n}} \left( |{{\varphi }_{n7}}|+|{{\psi }_{n7}}| \right)}. \] Since \begin{gather*} \frac{1}{{{\lambda }_n}}\left( |{{\varphi }_{n7}}|+|{{\psi }_{n7}}| \right) \le \frac{1}{2}\Big( \frac{1}{\lambda _n^2}+2|{{\varphi }_{n7}}{{|}^2} +2|{{\psi }_{n7}}{{|}^2} \Big), \quad \sum_{n=1}^{\infty }{\frac{1}{\lambda _n^2}}=\frac{1}{6},\\ \sum_{n=1}^{\infty }{|{{\varphi }_{n7}}{{|}^2}} \le \| {{\varphi }^{(7)}}(x) \||_{{{L}_2}(0,1)}^2,\quad \sum_{n=1}^{\infty }{|{{\psi }_{n7}}{{|}^2}} \le \| {{\psi }^{(7)}}(x) \||_{{{L}_2}(0,1)}^2, \end{gather*} it follows the absolute and uniform convergence of the first series of \eqref{e47}. Now we consider the second series of \eqref{e47}. Since \begin{gather*} \frac{{{d}^2}u_n^{-}(t)}{d{{t}^2}} =\frac{{{\varphi }_n}-{{\psi }_n}}{{{\Delta }_{n*}}} (\lambda _n^{6}\sin(\lambda _n^2t)-\lambda _n^4\cos(\lambda _n^2t)), \\ \big| \frac{{{d}^2}u_n^{-}(t)}{d{{t}^2}} \big| \le C\cdot \lambda _n^{6}\left( |{{\varphi }_n}|+|{{\psi }_n}| \right), \end{gather*} from this \eqref{e48} and \eqref{e49}, it follows the absolute and uniform convergence of the second series of \eqref{e47}. This completes the proof. \end{proof} \begin{remark} \label{rmk2} \rm Set of numbers $p$, satisfying condition ${{\Delta }_{n*}}\ne 0$ is not empty. For instance, if $p=2/\pi$, then ${{\Delta }_{n*}}=1-{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})$. Since ${{\lambda }_n}\ne 0$, $ n\in \mathbb{N}$, $0<{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})<1$, it follows that ${{\Delta }_{n*}}\ne 0$. \end{remark} \begin{remark} \label{rmk3} \rm If condition ${{\Delta }_{n*}}\ne 0$ is not valid for $n=k$ and for some $p$ and $q$, i.e. \[ {{\Delta }_{k*}}=\lambda _k^2\sin(\lambda _k^2p) +\cos(\lambda _k^2p)-{{E}_{\alpha }}(-\lambda _k^4{{q}^{\alpha }})=0, \] homogeneous system \eqref{e40} has nontrivial solution (${{\varphi }_n}={{\psi }_n}=0$). Then homogeneous inverse problem also has nontrivial solution. For instance, when \[ {{A}_k}={{L}_k}=1,\quad {{B}_k}=\frac{\lambda _k^2}{{{E}_{\alpha }} (-\lambda _k^4{{q}^{\alpha }})-1},\quad {{f}_k}=\frac{\lambda _k^4{{E}_{\alpha }} (-\lambda _k^4{{q}^{\alpha }})}{{{E}_{\alpha }} (-\lambda _k^4{{q}^{\alpha }})-1} \], the functions \begin{gather*} u(x,t)=\begin{cases} \Big( \frac{{{E}_{\alpha }}(-\lambda _k^4{{q}^{\alpha }}) -{{E}_{\alpha }}(-\lambda _k^4{{t}^{\alpha }})}{{{E}_{\alpha }} (-\lambda _n^4{{q}^{\alpha }})-1} \Big) \sqrt{2}\sin(\lambda _k^2x), &t>0, \\[4pt] \Big( \frac{\lambda _k^2\sin(\lambda_k^2t) -\cos(\lambda_k^2t)-{{E}_{\alpha }}(-\lambda _k^4{{q}^{\alpha }})} {{{E}_{\alpha }}(-\lambda _n^4{{q}^{\alpha }})-1} \Big) \sqrt{2} \sin(\lambda _k^2x), &t<0, \end{cases} \\ f(x)=\frac{\lambda _k^4{{E}_{\alpha }}(-\lambda _k^4{{q}^{\alpha }})} {{{E}_{\alpha }}(-\lambda _k^4{{q}^{\alpha }})-1} \sqrt{2}\sin(\lambda _k^2x) \end{gather*} form a solution of the homogeneous problem. \end{remark} \subsection*{Acknowledgements} This work was partially supported by Grant No 3293/GF4 from the Ministry of education and science of the Republic of Kazakhstan. \begin{thebibliography}{00} \bibitem{Dzhr} M. 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