\documentclass[reqno]{amsart} \usepackage{amssymb} \pdfoutput=1\relax\pdfpagewidth=8.26in\pdfpageheight=11.69in\pdfcompresslevel=9 \usepackage{hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 36, pp. 1--20.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/36\hfil An Lp-approach] {An Lp-approach for the study of degenerate parabolic equations} \author[R. Labbas, A. Medeghri, B.-K. Sadallah\hfil EJDE-2005/36\hfilneg] {Rabah Labbas, Ahmed Medeghri, Boubaker-Khaled Sadallah} % in alphabetical order \address{Rabah Labbas\hfill\break Laboratoire de Math\'{e}matiques\\ Universit\'{e} du Havre, U.F.R. Sciences et Techniques\\ b.p. 540, 76058 Le Havre, France} \email{labbas@univ-lehavre.fr} \address{Ahmed Medeghri\hfill\break Universit\'{e} de Mostaganem\\ b.p. 188, 27000, Mostaganem, Algeria} \email{medeghri@univ-mosta.dz} \address{Boubaker-Khaled Sadallah \hfill\break Ecole Normale Sup\'{e}rieure, D\'{e}partement\\ de Maths, 16050 Kouba, Alger, Algeria} \email{sadallah@wissal.dz} \date{} \thanks{Submitted July 16, 2004. Published March 29, 2005.} \subjclass[2000]{35K05, 35K65, 35K90} \keywords{Sum of linear operators, diffusion equation, \hfill\break\indent non rectangular domain, bounded imaginary powers of operators} \begin{abstract} We give regularity results for solutions of a parabolic equation in non-rectangular domains $U=\cup_{t\in ] 0,1[}\{ t\} \times I_{t}$ with $I_{t}=\{x:03/2$ by considering the following cases: (1) When $\varphi (t)=t^{\alpha }$, $\alpha >1/2$ with a regular right-hand side belonging to a subspace of $L^{p}(U)$ and under assumption $p>1+\alpha $. We use Labbas-Terreni results \cite{LabbasTerreni}. (2) When $\varphi (t)=t^{1/2}$ with a right-hand side taken only in $L^{p}(U)$. Our approach make use of the celebrated Dore-Venni results \cite{DoreVenni1}. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \section{Introduction} \subsection*{Statement of the Problem} We study the autonomous parabolic equation \begin{equation} \begin{gathered} D_{t}u(t,x)-D_{x}^{2}u(t,x)+\lambda m(t,x)u(t,x)=f(t,x),\quad (t,x)\in U \\ u\big|_{\partial U-\Gamma _{1}}=0, \end{gathered} \label{Equa1} \end{equation} with a positive spectral parameter $\lambda $ and some positive weight functions $m(\cdot)$ which will be specified below. The right-hand term $f$ is taken in $L^{p}( U) $ with \begin{equation} p>\frac{3}{2}. \label{hypo1} \end{equation} Here, $\Gamma _{1}=\{ 1\} \times ]0,1[$ and $U$ is a non-cylindrical domain of the form \[ U=\cup_{t\in ] 0,1[ }\{ t\} \times I_{t}, \] where \[ I_{t}=\{ x:00\text{ and }( x_{1}/\varphi _{1}( t) ,x_{2}/\varphi _{2}( t) ) \in D\} , \] where $D$ is some given cylindrical domain in $\mathbb{R}_{+}^{2}$ and $\varphi _{i}$, $i=1,2$ are similar to $\varphi $. The change of variables $( t,x) \mapsto (t,y)=( t,x/\varphi (t)) $ transforms $U$ into the rectangle $\Omega =]0,1[\times ]0,1[$. Putting $u(t,x)=w(t,y)$ and $f(t,x)=g(t,y)$, Problem \eqref{Equa1} is transformed, in $\Omega $ into the degenerate evolution problem \begin{equation} \begin{gathered} \begin{aligned} &\varphi (t)^{2}D_{t}w(t,y)-D_{y}^{2}w(t,y)-\varphi '(t)\varphi (t)yD_{y}w(t,y) +\lambda M(t,y)\varphi (t)^{2}w(t,y)\\ &=\varphi (t)^{2}g(t,y) \end{aligned}\\ w\big|_{ \partial \Omega -\Gamma _{1}}=0. \end{gathered} \label{Equa2} \end{equation} with $\Gamma _{1}=\{ 1\} \times ]0,1[$ and $M(t,y)=m(t,x)$. It is easy to see that $f\in L^{p}(U)$ if and only if $\varphi ^{-2+1/p}h\in L^{p}(\Omega )$ which implies that $h\in $ $L^{p}(\Omega )$, since \[ h=(\varphi ^{-2+1/p}h)\varphi ^{2-1/p} \] and $2-1/p>0$. Then the function $h=\varphi ^{2}g$ lies in the closed subspace of $L^{p}(\Omega )$ defined by \[ E_{\varphi }=\{ h\in L^{p}(0,1;L^{p}(0,1)):\text{ }\varphi ^{-2+1/p}h\in L^{p}(0,1;L^{p}(0,1))\} . \] This space is equipped with the norm \[ \Vert h\Vert _{E_{\varphi }}=\Vert \varphi ^{-2+1/p}h\Vert _{L^{p}(0,1;L^{p}(0,1))}. \] We can find in the Favini-Yagi book \cite{FaviniYagi} an important study of abstract problems of parabolic type with degenerate terms in the time derivative. In particular, these authors have given an alternative approach to the study of Equation (\ref{Equa2}), when it takes a particular following form \begin{gather*} t^{l}D_{t}w(t,y)-D_{y}^{2}w(t,y)=t^{l}g(t,y) \\ w\big|_{\partial \Omega -\Gamma _{1} }=0. \end{gather*} The authors of this article have considered the three cases $l>1$, $l=1$ and $l<1$, see \cite[p. 111.]{FaviniYagi}. They used the notion of multivalued linear operators and constructed fundamental solutions when the right-hand side has a H\"{o}lder regularity with respect to the time. In this work, we are especially interested in the question: What conditions the functions $\varphi $ and $m$ must verify in order that Problem (\ref{Equa1}) has a solution with optimal regularity, that is a solution $u$ belonging to the anisotropic Sobolev space \[ H_{p}^{1,2}( U) =\{ u\in L^{p}( U) :D_{t}u,D_{x}^{j}u\in L^{p}( U) ,\text{ }j=1,2\} ? \] Our approach is different from that used in the previous methods : it is based on the direct use of the sum theory of operators in Banach spaces. This is naturally suggested by Equation (\ref{Equa2}). We will prove that the answer to the previous question is positive in the following two cases. 1. When $f$ is regular, the function $t\mapsto \varphi '(t)\varphi (t)$ is H\"{o}lderian and the weight multiplier coefficient $m$ is function of the parabolic boundary in the sense that \[ m(t,x)=m(t)=( \varphi (t)) ^{-2}. \] It corresponds for instance to the model case $\varphi (t)=t^{\alpha }$, with $\alpha >1/2$ and $m(t)=t^{-2\alpha }$. The approach uses Labbas-Terreni results \cite{LabbasTerreni}. 2. When $f$ $\ $is taken only in $L^{p}( U) $, $\varphi (t)=\sqrt{% t}$ and $m(t,x)=t^{-1}$. Here, we use a celebrated Dore-Venni Theorem given in \cite{DoreVenni1}. This work is also an extension of the Hilbertian case ($p=2$) studied in Sadallah \cite{Sadallah}. The author has considered the cases $\lambda =0$, $m=0$. \subsection*{Assumptions and main results} \subsubsection*{First case} Let $\varphi (t)=t^{\alpha }$, with $\alpha >1/2$, $m(t)=t^{-2\alpha }$ and assume that \begin{equation} p-1>\alpha . \label{hypothese1} \end{equation} For $\sigma \in ]0,1[$, we introduce the following subspace of $L^{p}( U) $ (with a slight abuse) \begin{align*} &L_{\varphi ^{2\sigma }}^{p}( 0,1;W_{\varphi }^{2\sigma ,p})\\ & =\{ f\in L^{p}( U) :\int_0^1 \varphi (t)^{2\sigma p} \int_0^{\varphi(t)} \int_0^{\varphi(t)} \frac{\vert f(t,x)-f(t,x')\vert ^{p}}{\vert x-x'\vert ^{2\sigma p+1}}\, dx\,dx'\,dt<\infty \} . \end{align*} The main result regarding this case is the following. \begin{theorem} \label{thm1} Assume \eqref{hypothese1}. Let $\sigma \in ] 0,1[ $ such that $0<\sigma <1/2p$. Then, there exists $\lambda ^{\ast }$ such that $\forall \lambda \geqslant \lambda ^{\ast }$ and for all $f\in L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p}) $, Problem \eqref{Equa1} has a unique solution $u\in H_{p}^{1,2}(U)$ fulfilling the regularity properties: $u$, $D_{t}u$, $D_{x}u$ and $D_{x}^{2}u$ belong to $L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p}) $. \end{theorem} \subsubsection*{Second case} For $\varphi (t)=\sqrt{t}$ and $m(t)=t^{-1}$, we will prove the following statement. \begin{theorem} \label{thm2} For $f\in L^{p}( U) $ and $\lambda \geqslant 1/2p$, Problem \eqref{Equa1} has a unique solution $u\in H_{p}^{1,2}(U)$. \end{theorem} Note that in the second case, we obtain maximal results for any $\lambda\geqslant 1/2p$. \section{On the sum of linear operators} \subsection*{Definitions} Let $\Lambda $ be a closed linear operator in a complex Banach space $E$. Then, $\Lambda $ is said sectorial if \begin{itemize} \item[(i)] $D(\Lambda )$ and $\mathop{\rm Im}(\Lambda )$ are dense in $E$, \item[(ii)] $\ker(\Lambda )=\{ 0\} $, \item[(iii)] $]-\infty ,0[\subset \rho (\Lambda )$ and there exists a constant $M\geqslant 1$ such that \[ \Vert t( \Lambda +tI) ^{-1}\Vert \leqslant M \] for all $t>0$. \end{itemize} We recall that for a sectorial operator $\Lambda $, the complex powers $\Lambda ^{z}$, $z\in \mathbb{C}$, are well defined but not necessarily bounded, see Komatsu \cite{Komatsu}. When $\Lambda ^{is}\in L(E)$ for any $s\in \mathbb{R}$ and \[ \sup_{s\in [ -1,1]}\Vert \Lambda ^{is}\Vert<\infty , \] we say that $\Lambda \in Bip(E)$, see Pr\"{u}ss-Sohr \cite{PrussSohr}. In general, the knowledge of $\Lambda ^{is}$ is important for the determination of the domain of $\Lambda ^{z}$, see Triebel \cite{Triebel}. Consider two closed linear operators $A$ and $B$ with domains $D(A)$ and $D(B)$ in $E$. Their sum is defined by \[ Sv=Av+Bv, \quad v\in D(S)=D(A)\cap D(B). \] \subsection*{First approach} Let us assume that $A$ and $B$ satisfy the following assumptions: \begin{itemize} \item[(LT1)] There exists $r$, $C_{A}$, $C_{B}>0$, $\epsilon _{A}$, $\epsilon _{B}>0$ such that \\ (i) $\rho (-A)\supset \sum_{\epsilon _{A}}=\{ z:\vert z\vert \geq r,\;\vert Arg(z)\vert <\epsilon _{A}\}$ \\ and for all $z\in \sum_{\epsilon _{A}}$, $\Vert( A+zI) ^{-1}\Vert _{L(E)}\leqslant C_{A}/\vert z\vert$, \\ (ii) $\rho (-B)\supset \sum_{\epsilon _{B}}=\{ z:\vert z\vert \geq r,\; \vert Arg(z)\vert <\epsilon _{B}\}$ \\ and for all $z\in \sum_{\epsilon _{B}}$, $\Vert ( B+zI) ^{-1}\Vert _{L(E)}\leqslant C_{B}/\vert z\vert$, \\ (iii) $\epsilon _{A}+\epsilon _{B}>\pi$, \end{itemize} where $\rho (-A)$ and $\rho (-B)$ denote the resolvent sets of ($-A)$ and ($-B)$ respectively. We suppose that there exist $C>0$, $\lambda _{0}>0$ (with $\lambda _{0}\in \rho (-A)$), $\tau $ and $\rho $ such that \begin{itemize} \item[(LT2)] (i) For all $\lambda \in \rho (-A)$ and all $\mu \in \rho (-B)$, \[ \Vert (A+\lambda _{0})( A+\lambda I) ^{-1}[ (A+\lambda _{0})^{-1};( B+\mu I) ^{-1}] \Vert _{L(E)} \leqslant \frac{C}{\vert \lambda \vert ^{1-\tau }.\vert \mu \vert ^{1+\rho }}, \] (ii) $0\leqslant \tau <\rho \leqslant 2$. \end{itemize} For any $\sigma \in ]$ $0,1[$ and $1\leqslant p\leqslant +\infty $, let us introduce the real Banach interpolation spaces $D_{A}(\sigma ,p)$ between $% D(A)$ and $E$ (or $D_{B}(\sigma ,p)$ between $D(B)$ and $E$) which are characterized by \[ D_{A}(\sigma ,p)=\{ \xi \in E:t\mapsto \Vert t^{\sigma }A(A+tI)^{-1}\xi \Vert _{E}\in L_{\ast }^{p}\} , \] where $L_{\ast }^{p}$ denotes the space of $p$-integrable functions on $% (0,+\infty )$ with the measure $dt/t$. For $p=+\infty $, \[ D_{A}(\sigma ,\infty )=\big\{ \xi \in E:\underset{t>0}{\text{ }\sup }% \Vert t^{\sigma }A(A+tI)^{-1}\xi \Vert _{E}<\infty \big\} . \] Then the main result proved in Labbas-Terreni \cite{LabbasTerreni} is \begin{theorem} \label{thmLT} Under Assumptions (LT1) and (LT2), there exists $\lambda ^{\ast }$ such that $\forall \lambda $ $\geqslant \lambda ^{\ast }$ and $\forall h\in D_{A}(\sigma ,p)$, Equation $Aw+Bw+\lambda w=h$, has a unique solution $w\in D(A)\cap D(B)$ with the regularities $Aw,Bw$ $\in D_{A}(\theta ,p)$ and $Aw\in D_{B}(\theta ,p)$ for any $\theta $ verifying $% \theta \leqslant \min (\sigma ,\rho -\tau )$. \end{theorem} \subsection*{Second approach} Here, we assume that $E$ is an U.M.D complex Banach space (Unconditional Martingales Differences Property). It means that for all $p\in ]1,\infty [ $, the Hilbert transform is bounded on $L^{p}( \mathbb{R}% ,E) $, (see Burckholder \cite{Burkholder}). In concrete terms, any space built on $L^{p}$ is U.M.D if $p$ belongs to $(1,\infty )$. We suppose that $A$ and $B$ satisfy \begin{itemize} \item[(DV1)] (i) $D(A)$ and $D(B)$ are dense in $E$ \\ (ii) $\rho (A)\supset ]-\infty ,0]$ and there exists $M_{A}\geqslant 1$ so that for all $\lambda \geqslant 0$, $\Vert (A+\lambda I\text{ })^{-1}\Vert _{L(E)}\leqslant M_{A}/(1+\lambda )$ \\ (iii) $\rho (B)\supset ]-\infty ,0]$ and there exists $M_{B}\geqslant 1$ such that for all $\lambda \geqslant 0$, $\Vert (B+\lambda I)^{-1}\Vert _{L(E)}\leqslant M_{B}/(1+\lambda )$. \item[(DV2)] There exist $K>0$, $\theta _{A}$, $\theta _{B}\in [ 0,\pi [$ such that \\ (i) $A^{is}\in L(E)$ and for all $s\in \mathbb{R}$, $\Vert A^{is}\Vert \leq Ke^{\vert s\vert \theta _{A}}$, \\ (ii) $B^{is}\in L(E)$ and for all $s\in \mathbb{R}$, $\Vert B^{is}\Vert \leq Ke^{\vert s\vert \theta _{B}}$, \\ (iii) $\theta _{A}+\theta _{B}<\pi$. \item[(DV3)] For all $\xi \in \rho (-A)$ and all $\eta \in \rho (-B)$, \[ ( A+\xi I) ^{-1}( B+\eta I) ^{-1}=( B+\eta I) ^{-1}( A+\xi I) ^{-1}. \] \end{itemize} In general, Hypothesis (DV2) is hard to check in concrete cases. We shall use the following Dore-Venni Theorem (see \cite{DoreVenni1}). \begin{theorem} \label{thm4} Assume (DV1), (DV2), (DV3). Then $A+B$ is closed, invertible and $( A+B) ^{-1}\in L(E)$. \end{theorem} \section{Proof of Theorem \ref{thm1}} Let $\varphi ( t) =t^{\alpha }$ with $\alpha >1/2$, $m(t)=t^{-2\alpha }$ and assume \begin{equation} p>1+\alpha . \label{HypoTheorem1} \end{equation} Set $X=L^{p}(0,1)$ and $w(t)=w(t,.)$, then \eqref{Equa2} is equivalent to the abstract degenerate Cauchy problem \begin{equation} \begin{gathered} t^{2\alpha }w'(t)+L(t)w(t)+\lambda w(t)=t^{2\alpha }g(t),\quad t\in (0,1), \\ w(0)=0, \end{gathered} \label{Equa3} \end{equation} where the family $( L(t)) _{t\in [ 0,1]}$ is defined by \begin{equation} \begin{gathered} D(L(t))=\{ \psi \in W^{2,p}(0,1): \psi (0)=\psi (1)=0\} \\ ( [ L(t)]\psi ) (y)=-\psi ''(y)-\alpha t^{2\alpha -1}y\psi '(y)\text{ for a.e. }t\in (0,1). \end{gathered} \label{formule4} \end{equation} Observe that \begin{equation} \overline{D(L(t))}=X. \end{equation} Then we must solve \begin{equation} \begin{gathered} t^{2\alpha }w'(t)+L(t)w(t)+\lambda w(t)=h(t),\quad t\in (0,1) \\ w(0)=0, \end{gathered} \label{Equa5} \end{equation} where $h$ belongs to the space \[ E_{1}=\big\{ h\in L^{p}(0,1;X): t^{-2\alpha +\alpha /p}h\in L^{p}(0,1;X)\big\} . \] Equation (\ref{Equa5}) can be written in the form \[ Bw+Aw+\lambda w=h, \] where \begin{equation} \begin{gathered} D(A)=\{ w\in E_{1}:t^{-2\alpha +\alpha /p}w\in L^{p}(0,1;W^{2,p}(0,1)\cap W_{0}^{1,p}(0,1))\} \\ ( Aw) (t)=L(t)w(t),\quad t\in (0,1), \end{gathered} \label{Formule6} \end{equation} and \begin{equation} \begin{gathered} D(B)=\{ w\in E_{1}:t^{\alpha /p}w'\in L^{p}(0,1;X)\text{ and } w(0)=0\} \\ (Bw)(t)=t^{2\alpha }w'(t),\quad t\in (0,1). \end{gathered} \label{Formule7} \end{equation} Note that the trace $w( 0) $ is well defined in $D(B)$. In fact, we have \[ t^{\alpha /p}w\in L^{p}(0,1;X), \quad t^{\alpha /p}w'\in L^{p}(0,1;X), \] and $\alpha /p+1/p<1$ in virtue of (\ref{HypoTheorem1}). Then $w$ is continuous on $[ 0,1] $, (see \cite[Lemma, p. 42]{Triebel}). Now, to apply Theorem (\ref{thmLT}), we will verify Assumptions $(LT1)$ and $(LT2)$. \begin{proposition} \label{prop5} The operators $A$ and $B$ are linear closed with dense domains in $E_{1}$. Moreover, they satisfy Assumption $(LT1)$. \end{proposition} The properties of operator $B$ are based on the solvability of the spectral equation $Bw+zw=h$. Fix some positive $\mu _{0}$ and let $z$ such that $\mathop{\rm Re}(z)\geqslant \mu _{0}$. Then the problem \begin{gather*} t^{2\alpha }w'(t)+zw(t)=h(t) \\ w(0)=0, \end{gather*} admits the solution \begin{align*} w(t) &=\big( ( B+zI) ^{-1}h\big) (t) \\ &=\exp \big( \frac{z}{(2\alpha -1)t^{2\alpha -1}}\big) \int_{0}^{t}\sigma ^{-2\alpha }h(\sigma )\exp \big( \frac{-z}{(2\alpha -1)\sigma ^{2\alpha -1}}\big) d\sigma \end{align*} Let us check that this formula is well defined on $[0,1]$ and gives $w(0)=0$. Set $\mu =z/(2\alpha -1)$, then \begin{align*} \Vert w(t)\Vert &\leqslant \exp \big( \frac{\mathop{\rm Re}\mu }{t^{2\alpha -1}}\big) \int_{0}^{t}\Vert \sigma ^{-2\alpha +\alpha /p}h(\sigma )\Vert \sigma ^{-\alpha /p}\exp \big( \frac{-\mathop{\rm Re}\mu }{\sigma ^{2\alpha -1}} \big) d\sigma \\ &\leqslant \Big( \int_{0}^{t}\Vert \sigma ^{-2\alpha +\alpha /p}h(\sigma )\Vert ^{p}d\sigma \Big) ^{1/p}\Big( \int_{0}^{t}\sigma ^{-\alpha q/p}d\sigma \Big) ^{1/q} \\ &\leqslant \big( \frac{1}{1-\alpha q/p}\big) ^{1/q}t^{1/q-\alpha /p}\Vert h\Vert _{E} \end{align*} where $1/p+1/q=1$. Hence $w(t)$ is defined and $w(0)=0$ since \[ 1/q-\alpha /p=1-1/p-\alpha /p>0 \] means $p>1+\alpha $. On the other hand we can write \begin{equation} \label{formulresolvante} \begin{aligned} t^{-2\alpha +\alpha /p}( ( B+zI) ^{-1}h) (t) &= t^{-2\alpha +\alpha /p}w(t)\\ &=\int_{0}^{1}\sigma ^{-2\alpha +\alpha /p}h(\sigma )K_{\mu }(t,\sigma )d\sigma , \end{aligned} \end{equation} where \begin{equation} K_{\mu }(t,\sigma )=\begin{cases} \frac{1}{t^{2\alpha -\alpha /p}\sigma ^{\alpha /p}}\exp \mu (t^{-2\alpha +1}-\sigma ^{-2\alpha +1})& \text{if }t>\sigma \\ 0 & \text{if }t<\sigma . \end{cases} \label{formulenoyau} \end{equation} Therefore, \begin{align*} \int_{0}^{1}\vert K_{\mu }(t,\sigma )\vert d\sigma &=\frac{1}{t^{2\alpha -\alpha /p}}\exp ( \mathop{\rm Re}(\mu ).t^{-2\alpha +1}) \int_{0}^{t}\frac{\exp ( -\mathop{\rm Re}(\mu ).\sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}}d\sigma \\ &\leqslant \frac{1}{2\alpha -1}\exp ( \mathop{\rm Re}\mu .t^{-2\alpha +1}) \int_{t^{-2\alpha +1}}^{+\infty }\exp ( -\mathop{\rm Re}(\mu ).s) ds \\ &\leqslant \frac{1}{\mathop{\rm Re}(z)}, \end{align*} and \begin{equation} \max_{t\in [ 0,1]}\int_{0}^{1}\vert K_{\mu }(t,\sigma )\vert d\sigma \leqslant \frac{1}{\mathop{\rm Re}(z)}. \label{formule8} \end{equation} Furthermore, one has \begin{align*} &\int_{0}^{1}\vert K_{\mu }(t,\sigma )\vert dt \\ &=\frac{\exp ( -\mathop{\rm Re}\mu .\sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}}\int_{\sigma }^{1}\frac{1}{t^{2\alpha -\alpha /p}}\exp ( \mathop{\rm Re}\mu .t^{-2\alpha +1}) dt \\ &=\frac{1}{2\alpha -1}\frac{\exp ( -\mathop{\rm Re}(\mu ).\sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}}\int_{1}^{\sigma ^{-2\alpha +1}} \frac{1}{s^{\frac{\alpha }{p(2\alpha -1)}}}\exp ( \mathop{\rm Re}\mu .s) ds, \end{align*} and \begin{align*} &\int_{1}^{\sigma ^{-2\alpha +1}}\frac{1}{s^{\frac{\alpha }{p(2\alpha -1)}}}% \exp ( \mathop{\rm Re}\mu s) ds \\ &=\int_{1}^{\frac{1+\sigma ^{-2\alpha +1}}{2}}\frac{1}{s^{\frac{\alpha }{% p(2\alpha -1)}}}\exp ( \mathop{\rm Re}\mu s) ds+\int_{\frac{1+\sigma ^{-2\alpha +1}}{2}}^{\sigma ^{-2\alpha +1}}\frac{1}{s^{\frac{\alpha }{% p(2\alpha -1)}}}\exp ( \mathop{\rm Re}\mu s) ds \\ &\leqslant \int_{1}^{\frac{1+\sigma ^{-2\alpha +1}}{2}}\exp ( \mathop{\rm Re}% \mu s) ds+\frac{1}{( \frac{1+\sigma ^{-2\alpha +1}}{2}) ^{% \frac{\alpha }{p(2\alpha -1)}}}\int_{\frac{1+\sigma ^{-2\alpha +1}}{2}% }^{\sigma ^{-2\alpha +1}}\exp ( \mathop{\rm Re}\mu s) ds \\ &=I_{1}+I_{2}. \end{align*} Then \begin{align*} I_{1} &\leqslant \frac{1}{\mathop{\rm Re}\mu }\left[ \exp ( \mathop{\rm Re}\mu \frac{(1+\sigma ^{-2\alpha +1})}{2}) -\exp ( \mathop{\rm Re}\mu ) % \right] \\ &\leqslant \frac{1}{\mathop{\rm Re}\mu }\exp ( \mathop{\rm Re}\mu \frac{(1+\sigma ^{-2\alpha +1})}{2}) , \end{align*} and \begin{align*} &\frac{1}{2\alpha -1}\frac{\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}}I_{1} \\ &\leqslant \frac{1}{\mathop{\rm Re}(z)}\frac{\exp ( -\mathop{\rm Re}\mu (\sigma ^{-2\alpha +1}-1)/2) }{\sigma ^{\alpha /p}} \\ &\leqslant \frac{1}{\mathop{\rm Re}(z)}\frac{\exp ( -\mu _{0}(\sigma ^{-2\alpha +1}-1)/2) }{\sigma ^{\alpha /p}} \\ &\leqslant \frac{C_{1}(\alpha ,p)}{\mathop{\rm Re}(z)}, \end{align*} since the function \[ \sigma \mapsto \frac{\exp ( -\mu _{0}(\sigma ^{-2\alpha +1}-1)/2) }{\sigma ^{\alpha /p}} \] is continuous on $[0,1]$. Moreover \begin{align*} &\frac{1}{2\alpha -1}\frac{\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}}I_{2} \\ &\leqslant \frac{1}{2\alpha -1}\frac{\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) }{\sigma ^{\alpha /p}( \frac{1+\sigma ^{-2\alpha +1}}{2}) ^{\frac{\alpha }{p(2\alpha -1)}}}\int_{\frac{1+\sigma ^{-2\alpha +1}}{2}}^{\sigma ^{-2\alpha +1}}\exp ( \mathop{\rm Re}\mu s) ds \\ &\leqslant \frac{C_{2}(\alpha ,p)}{\mathop{\rm Re}(z).\sigma ^{\alpha /p}( \frac{1+\sigma ^{-2\alpha +1}}{2}) ^{\frac{\alpha }{p(2\alpha -1)}}} \\ &\leqslant \frac{C_{3}(\alpha ,p)}{\mathop{\rm Re}(z)} \end{align*} in virtue of the fact that \[ \underset{\sigma \to 0}{\lim }\frac{1}{\sigma ^{\alpha /p}( 1+\sigma ^{-2\alpha +1}) ^{\frac{\alpha }{p(2\alpha -1)}}}=1. \] Consequently, there exists some constant $C(\alpha ,p)>0$ such that \begin{equation} \underset{\sigma \in [ 0,1]}{\max }\int_{0}^{1}\vert K_{\mu }(t,\sigma )\vert dt\leqslant \frac{C(\alpha ,p)}{\mathop{\rm Re}(z)}. \label{formule9} \end{equation} Now, using Schur interpolation Lemma together with (\ref{formule8}) and (\ref{formule9}), we obtain \[ \Vert t^{-2\alpha +\alpha /p}w\Vert _{L^{p}(0,1;X)}\leqslant \frac{C(\alpha ,p)}{\mathop{\rm Re}(z)}\Vert t^{-2\alpha +\alpha /p}h\Vert _{L^{p}(0,1;X)}, \] from which it follows \[ \Vert (B+zI)^{-1}\Vert _{L(E_{1})}\leqslant \frac{C(\alpha ,p)}{% \mathop{\rm Re}(z)}. \] Thus, we can take $\epsilon _{B}=\pi /2-\epsilon _{0}$ (for each $\epsilon _{0}\in ]0,\pi /2[$). The study of the spectral properties of the operator $A$ are based on those of operators $L(t)$. For each $t$, we write \[ ( L(t)\Psi ) =L_{0}\Psi +P(t)\Psi \ \] with \begin{equation} \begin{gathered} D(L_{0})=\{ \Psi \in W_{p}^{2}(0,1):\Psi (0)=\Psi (1)=0\} \\ L_{0}\Psi =-\Psi '', \end{gathered}\label{formule10bis} \end{equation} and \begin{equation} \begin{gathered} D(P(t))=W^{1,p}(0,1) \\ P(t)\Psi =-\alpha t^{2\alpha -1}y\Psi '=-b(t)y\Psi '. \end{gathered}\label{formule11} \end{equation} It is easy to see that the operator $L_{0}$ is sectorial. Moreover, thanks to H\"{o}lder inequality, for $\psi \in D(L_{0})\subset D(P(t))$ we have \begin{align*} \Vert P(t)\Psi \Vert _{L^{p}( 0,1) } &=\Big(\int_0^1 \Big\vert -b(t)y\Psi '(y)\vert ^{p}dy \Big) ^{1/p} \\ &=\Big( \int_0^1 \vert -b(t)y\Big(\int_{0}^{y}( s\Psi ''(s)) ds+\int_{y}^{1}( 1-s) \Psi ''(s)ds\Big) \Big\vert ^{p}dy\Big) ^{1/p} \\ &\leqslant \Big( \int_0^1 \Big\vert -b(t)y\int_{0}^{y}( s\Psi ''(s)) ds\Big\vert ^{p}dy\Big) ^{1/p} \\ &\quad+\Big( \int_0^1 \Big\vert -b(t)y\int_{y}^{1}( 1-s) \Psi ''(s)ds \Big\vert ^{p}dy\Big) ^{1/p} \\ &\leqslant b(t)\big( C_{1}(p)\Vert \Psi ''\Vert _{_{L^{p}( 0,1) }}+C_{2}(p)\Vert \Psi ''\Vert _{_{L^{p}( 0,1) }}\big) \\ &\leqslant C(p)\Vert \Psi \Vert _{D(L_{0})}. \end{align*} On the other hand, let us set \begin{align*} m(t):L^{p}( 0,1) &\to L^{p}( 0,1) \\ \psi &\mapsto [ m(t)\psi ](y)=-b(t)y\psi (y), \\ i:W^{1,p}( 0,1) &\to L^{p}( 0,1) \\ \psi &\mapsto \psi , \\ d:W^{2,p}( 0,1) &\to W^{1,p}( 0,1) \\ \psi &\mapsto d(\psi )=\psi '. \end{align*} Then $P(t)=m(t)\circ i\circ d$. Thus, $P(t)$ is compact from $D(L_{0})$ into $E_{1}$ since $i$ is compact and $d$, $m(t)$ are continuous. So for any $t\in [ 0,1]$, the operator $L(t)$ is sectorial (see Lunardi \cite[Proposition 2.4.3, p.65]{Lunardi}) and consequently there exists some $r_{0}>0$ such that \[ \rho (-L(t))\supset \Sigma _{\pi -\epsilon _{1}}=\{ z: \vert z\vert \geqslant r_{0},\;\vert \arg z\vert <\pi -\epsilon _{1}\} \] where $\epsilon _{1}\in ]0,\pi /2[$. Now, for $h\in E_{1}$ and $\lambda \in $ $\Sigma _{\pi -\epsilon _{1}}$ the spectral equation \[ Aw+zw=h \] is equivalent to \[ L(t)w(t)+zw(t)=h(t),\quad t\in (0,1) \] which admits a unique solution, \[ w(t)=( L(t)+z) ^{-1}h(t). \] Hence \[ \Vert w(t)\Vert _{L^{p}(0,1)}\leqslant \frac{K}{\vert z\vert }\Vert h(t)\Vert _{L^{p}(0,1)} \] and \[ \Vert w\Vert _{E_{1}}=\Big( \int_{0}^{1}\Vert t^{-2\alpha +\alpha /p}w(t)\Vert ^{p}dt\Big) ^{1/p} \leqslant \frac{K}{\vert z\vert }\Vert h\Vert _{E_{1}}. \] Finally $A$ and $B$ satisfy $(LT1)$. \begin{proposition} \label{prop1} Operators $A$ and $B$ satisfy Hypothesis (LT2). \end{proposition} In our case, the domains $D(L(t))=D(L(0))$ are constant. Let us verify the Sobolevskii's estimate: There exists $M>0$ such that for all $t,\sigma \in [ 0,1]$, \begin{equation} \label{S} \Vert [ (L(t)L(\sigma )^{-1}-I)] \Vert _{L(X)}\leqslant M\vert t-\sigma \vert ^{\rho }\,, \end{equation} where $\rho =\min (1,2\alpha -1)$. For $g\in X=L^{p}(0,1)$, the equation $\psi =L(\sigma )^{-1}g$ is equivalent to \begin{gather*} ( [ L(\sigma )]\psi ) (y)=-D_{y}^{2}\psi (y)-\varphi (\sigma )\varphi '(\sigma )yD_{y}\psi (y)=g(y), \\ \psi (0)=\psi (1)=0, \end{gather*} and \begin{equation} \label{formule12} [ (L(t)-L(\sigma ))L(\sigma )^{-1}g] (y) =( \varphi (\sigma )\varphi '(\sigma )-\varphi (t)\varphi '(t)) yD_{y}\psi (y). \end{equation} Then, we get \begin{align*} \Vert [ (L(t)-L(\sigma ))L(\sigma )^{-1}g] \Vert _{X} &\leqslant \alpha \vert t^{2\alpha -1}-\sigma ^{2\alpha -1}\vert \Vert y\psi '\Vert _{X} \\ &\leqslant M_{1}\vert t-\sigma \vert ^{\min (1,2\alpha -1)}\Vert \psi ''\Vert _{X} \\ &\leqslant M_{2}\vert t-\sigma \vert ^{\min (1,2\alpha -1)}\Vert \psi \Vert _{W^{2,p}(0,1)} \\ &\leqslant M\vert t-\sigma \vert ^{\min (1,2\alpha -1)}\Vert g\Vert _{L^{p}(0,1)}. \end{align*} To prove (LT2), it is sufficient to estimate \[ \Vert A( A+\lambda I) ^{-1}[ A^{-1};( B+zI)^{-1}] \Vert _{L(E_{1})} \] where $\lambda \in \rho (-A)$ ,$z\in \rho (-B)$. Let $h\in E_{1}$. Then \begin{align*} \Delta &=( t^{-2\alpha +\alpha /p}A( A+\lambda I) ^{-1} [A^{-1};( B+zI) ^{-1}] h) (t) \\ &=t^{-2\alpha +\alpha /p}( A( A+\lambda I) ^{-1}(A^{-1}( B+zI) ^{-1} -( B+zI) ^{-1}A^{-1})h) (t)\\ &= t^{-2\alpha +\alpha /p}L(t)(L(t)+\lambda I)^{-1} \{ L(t)^{-1}\big( ( B+zI) ^{-1}h\big) (t)\\ &\quad -\big( (B+zI) ^{-1}A^{-1}h\big) (t)\} . \end{align*} Using the representation given in (\ref{formulresolvante}) and the kernel defined in (\ref{formulenoyau}) where we have put $\mu =z/(2\alpha -1)$, we obtain \begin{align*} \Delta &=L(t)(L(t)+\lambda I)^{-1}\int_{0}^{1}\sigma ^{-2\alpha +\alpha /p}K_{\mu }(t,\sigma )( L(t)^{-1}-L(\sigma )^{-1}) h(\sigma )d\sigma \\ &=\int_{0}^{1}\sigma ^{-2\alpha +\alpha /p}K_{\mu }(t,\sigma )L(t)(L(t)+\lambda I)^{-1}( L(t)^{-1}-L(\sigma )^{-1}) h(\sigma )d\sigma \\ &=\int_{0}^{1}\sigma ^{-2\alpha +\alpha /p}K_{\mu }(t,\sigma )(L(t)+\lambda I)^{-1}( I-L(t)L(\sigma )^{-1}) h(\sigma )d\sigma \end{align*} since the domains $D(A(t))$ are constant. Also \[ \Vert \Delta \Vert _{X}=\frac{K}{\vert \lambda \vert }% \int_{0}^{1}\vert K_{\mu }(t,\sigma )\vert \vert t-\sigma \vert ^{\rho }\sigma ^{-2\alpha +\alpha /p}\Vert h(\sigma )\Vert _{X}d\sigma , \] with $\rho =\min (1,2\alpha -1)$. Recall that \begin{align*} &\int_{0}^{1}K_{\mu }(t,\sigma )\vert t-\sigma \vert ^{\rho }d\sigma \\ &=\frac{1}{t^{2\alpha -\alpha /p}}\exp ( \mathop{\rm Re}\mu t^{-2\alpha +1}) \int_{0}^{t}\sigma ^{-\alpha /p}( t-\sigma ) ^{\rho }\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \,. \end{align*} Then by H\"{o}lder inequality, one has \begin{align*} &\int_{0}^{t}\sigma ^{-\alpha /p}( t-\sigma ) ^{\rho }\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \\ &\leqslant \Big( \int_{0}^{t}\sigma ^{-\alpha /p}\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \Big) ^{1-\rho }\\ &\quad \times \Big( \int_{0}^{t}\sigma ^{-\alpha /p}( t-\sigma ) \exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \Big) ^{\rho } \end{align*} and \begin{align*} J_{1} &=\Big( \int_{0}^{t}\sigma ^{2\alpha -\alpha /p}\sigma ^{-2\alpha }\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \Big)^{1-\rho } \\ &\leqslant \frac{( t^{2\alpha -\alpha /p}) ^{1-\rho }}{( 2\alpha -1) ^{1-\rho }}\frac{1}{( \mathop{\rm Re}\mu ) ^{1-\rho }} ( \exp ( -\mathop{\rm Re}\mu t^{-2\alpha +1}) ) ^{1-\rho }\, \end{align*} \begin{align*} J_{2} &=\Big( \int_{0}^{t}\sigma ^{-\alpha /p}( t-\sigma ) \exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \Big) ^{\rho } \\ &=\Big( \int_{0}^{t}\sigma ^{2\alpha -\alpha /p}( t-\sigma ) \sigma ^{-2\alpha }\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \Big) ^{\rho } \\ &\leqslant \frac{( t^{2\alpha -\alpha /p}) ^{\rho }}{( 2\alpha -1) ^{\rho }}\frac{1}{( \mathop{\rm Re}\mu ) ^{\rho }}% \Big( \int_{0}^{t}( t-\sigma ) \varkappa '(\sigma )d\sigma \Big) ^{\rho } \end{align*} where $\varkappa (\sigma )=\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) $. Using an integration by parts, we obtain% \begin{align*} \int_{0}^{t}( t-\sigma ) \varkappa '(\sigma )d\sigma &=\int_{0}^{t}\exp ( -\mathop{\rm Re}\mu \sigma ^{-2\alpha +1}) d\sigma \\ &=\int_{0}^{t}\sigma ^{2\alpha }\sigma ^{-2\alpha }\exp ( -\mathop{\rm Re} \mu \sigma ^{-2\alpha +1}) d\sigma \\ &\leqslant \frac{t^{2\alpha }}{( 2\alpha -1) }\frac{1}{\mathop{\rm Re} \mu }\exp ( -\mathop{\rm Re}\mu t^{-2\alpha +1}) \end{align*} from which we deduce that \[ J_{2}\leqslant \frac{( t^{2\alpha -\alpha /p}) ^{\rho }}{( 2\alpha -1) ^{\rho }}\frac{1}{( \mathop{\rm Re}\mu ) ^{\rho }}% \frac{( t^{2\alpha }) ^{\rho }}{( 2\alpha -1) ^{\rho }}% \frac{1}{( \mathop{\rm Re}\mu ) ^{\rho }}( \exp ( -\mathop{\rm Re}% \mu t^{-2\alpha +1}) ) ^{\rho }. \] Finally we have \begin{align*} &\int_{0}^{1}K_{\mu }(t,\sigma )\vert t-\sigma \vert ^{\rho}d\sigma \\ &\leqslant \frac{\exp ( \mathop{\rm Re}\mu t^{-2\alpha +1}) } {t^{2\alpha -\alpha /p}}\frac{( t^{2\alpha -\alpha /p}) ^{1-\rho } }{( 2\alpha -1) ^{1-\rho }}\frac{( \exp ( -\mathop{\rm Re}\mu t^{-2\alpha +1}) ) ^{1-\rho }}{( \mathop{\rm Re}\mu ) ^{1-\rho }}\\ &\quad \times \frac{( t^{2\alpha -\alpha /p}) ^{\rho }}{( 2\alpha -1) ^{\rho }}\frac{1}{( \mathop{\rm Re}\mu ) ^{\rho }}\frac{ ( t^{2\alpha }) ^{\rho }}{( 2\alpha -1) ^{\rho }}\frac{ ( \exp ( -\mathop{\rm Re}\mu t^{-2\alpha +1}) ) ^{\rho }}{ ( \mathop{\rm Re}\mu ) ^{\rho }} \\ &\leqslant \frac{( t^{2\alpha }) ^{\rho }}{( 2\alpha -1) ^{1+\rho }}\frac{1}{( \mathop{\rm Re}\mu ) ^{1+\rho }}, \end{align*} and \[ \underset{t}{\max }\int_{0}^{1}K_{\mu }(t,\sigma )\vert t-\sigma \vert ^{\rho }d\sigma \leqslant \frac{C}{( \mathop{\rm Re}\mu ) ^{1+\rho }}. \] Similarly one has \[ \underset{\sigma \in [ 0,1]}{\max }\int_{0}^{1}K_{\mu }(t,\sigma )\vert t-\sigma \vert ^{\rho }dt\leqslant \frac{C}{( \mathop{\rm Re% }\mu ) ^{1+\rho }}. \] In virtue of Schur's lemma, we conclude that \[ \Vert A( A+\lambda I) ^{-1}[ A^{-1};( B+zI) ^{-1}] \Vert _{L(E_{1})} \leqslant \frac{C}{\vert \lambda \vert ( \mathop{\rm Re}\mu ) ^{1+\rho }}=\frac{C}{\vert \lambda \vert ( \mathop{\rm Re} z) ^{1+\rho }} \] which implies \[ \Vert A( A+\lambda I) ^{-1}[ A^{-1};( B+zI) ^{-1}] \Vert _{L(E_{1})}\leqslant \frac{C}{\vert \lambda \vert \vert z\vert ^{1+\rho }} \] for any $\lambda \in \rho (-A)$ and any $\mu $ belonging to a suitable sectorial curve. Then (LT2) is verified with $\tau =0$ and $\rho =\min(1,2\alpha -1)$. For more details concerning the commutator used in (LT2), see Labbas-Terreni \cite{LabbasTerreni}. Applying Theorem \ref{thmLT}, we deduce the following statement. \begin{proposition} \label{prop7} There exists $\lambda ^{\ast }$ such that for all $\lambda \geqslant \lambda ^{\ast }$ and $h\in D_{A}(\sigma ,p)$ Problem \eqref{Equa5} has a unique solution $w\in D(A)\cap D(B)$ such that for all $\theta \leqslant 1/2p$ \begin{itemize} \item[(i)] $L(.)w\in D_{A}(\theta ,p)$ \item[(ii)] $\varphi (t)w'\in D_{A}(\theta ,p)$ \item[(iii)] $L(.)w\in D_{B}(\theta ,p)$. \end{itemize} \end{proposition} Observe that we have a similar result when $h\in D_{B}(\sigma,p))$. To make precise the time and the space regularity of $w$ we need to specify the space $D_{A}(\sigma ,p)$. One has \[ D_{A}(\sigma ,p) =\begin{cases} \big\{ w\in E_{1}:t^{-2\alpha +\alpha /p}w\in L^{p}(0,1;W^{2\sigma ,p}(0,1)),\\ w(t,0)=w(t,1)=0\big\} &\text{if }2\sigma >1/p, \\[4pt] \{ w\in E:t^{-2\alpha +\alpha /p}w\in L^{p}(0,1;W^{2\sigma ,p}(0,1))\} &\text{if }2\sigma <1/p. \end{cases} \] Indeed we know that \[ D_{A}(\sigma ,p)=\{ w\in E_{1}:\Vert \xi ^{1-\sigma }Ae^{-\xi A}w\Vert _{E_{1}}\in L_{\ast }^{p}\} , \] because $(-A)$ is a generator of the analytic semigroup $\{ e^{-\xi A}\} $, $\xi \geqslant 0$. Now, $w\in D_{A}(\sigma ,p)$ implies \[ \Vert \xi ^{1-\sigma }Ae^{-\xi A}w\Vert _{E_{1}}\in L_{\ast }^{p} \] or equivalently \begin{align*} &\int_{0}^{\infty }\Vert \xi ^{1-\sigma }Ae^{-\xi A}w\Vert _{E_{1}}^{p}\frac{d\xi }{\xi } \\ &=\int_{0}^{\infty }\Vert t^{-2\alpha +\alpha /p}\xi ^{1-\sigma }Ae^{-\xi A}w\Vert _{_{L^{p}( 0,1;L^{p}(0,1)) }}^{p}\frac{d\xi }{\xi } \\ &=\int_{0}^{\infty }\Big( \int_{0}^{1}\Vert t^{-2\alpha +\alpha /p}\xi ^{1-\sigma }( Ae^{-\xi A}w) ( t) \Vert _{_{L^{p}( 0,1) }}^{p}dt\Big) \frac{d\xi }{\xi } <+\infty . \end{align*}% Since \[ ( Ae^{-\xi A}w) (t)=L(t)_{\text{ }}e^{\xi L(t)}( w(t)), \] by Fubini's Theorem, we obtain \begin{align*} &\int_{0}^{\infty }\Vert \xi ^{1-\sigma }Ae^{-\xi A}w\Vert _{E_{1}}^{p}\frac{d\xi }{\xi } \\ &=\int_{0}^{\infty }\Big( \int_{0}^{1}\Vert t^{-2\alpha +\alpha /p}\xi ^{1-\sigma }L(t)_{\text{ }}e^{\xi L(t)}( w(t)) \Vert _{_{L^{p}( 0,1) }}^{p}dt\Big) \frac{d\xi }{\xi } \\ &=\int_{0}^{1}\Vert t^{-2\alpha +\alpha /p}\Vert ^{p}\Big( \int_{0}^{\infty }\Vert \xi ^{1-\sigma }L(t)e^{\xi L(t)}( w(t)) \Vert _{_{L^{p}( 0,1) }}^{p}\frac{d\xi }{\xi }% \Big) dt<+\infty , \end{align*} which means that, for almost every $t$, the function $y\mapsto t^{-2\alpha +\alpha /p}w(t)(y)$ is in $D_{L(t)}(\sigma ,p)$. It is well known that this last constant space is \begin{align*} D_{L(t)}(\sigma ,p) &=D_{L(0)}(\sigma ,p)=( W^{2,p}(0,1)\cap W_{0}^{1,p}(0,1);L^{p}( 0,1) ) _{1-\sigma ,p} \\ &=\begin{cases} \{ w\in W^{2\sigma ,p}(0,1)\text{, }w(0)=w(1)=0\} & \text{if } 2\sigma >1/p \\ W^{2\sigma ,p}(0,1) & \text{if }2\sigma <1/p. \end{cases} \end{align*} Let $\sigma $ be a fixed positive number satisfying $\sigma <1/2p$. From the above proposition, we have the following statement. \begin{proposition} \label{prop8} For all $h$ such that $t^{-2\alpha +\alpha /p}h\in L^{p}(0,1;W^{2\sigma ,p}(0,1))$, Problem \eqref{Equa5} admits a unique solution $w$ fulfilling the regularity properties: \begin{itemize} \item[(i)] $w\in L^{p}(\Omega )$, $t^{-2\alpha +\alpha /p}w\in L^{p}(\Omega )$, $w(0)=0$ \item[(ii)] $t^{-2\alpha +\alpha /p}D_{y}^{2}w\in L^{p}(\Omega )$ \item[(iii)] $t^{\alpha /p}D_{t}w\in L^{p}(\Omega )$, \item[(iv)] $t^{-2\alpha +\alpha /p}D_{y}^{2}w\in L^{p}(0,1;W^{2\sigma,p}(0,1))$ \item[(v)] $t^{\alpha /p}D_{t}w\in L^{p}(0,1;W^{2\sigma ,p}(0,1))$. \end{itemize} \end{proposition} Let us recall that $h(t,y)=t^{2\alpha }g(t,y)$, $g(t,y)=f(t,x)$ and $% w(t,y)=u(t,x)$ where $(t,y)=(t,x/t^{\alpha })$. Then \begin{align*} &\int_0^1 \Vert t^{-2\alpha +\alpha /p}h(t,.)\Vert _{W^{2\sigma }(0,1)}^{p}dt \\ &=\int_0^1 t^{\alpha -2\alpha p}\int_0^1 \int_0^1 \frac{\vert h(t,y)-h(t,y')\vert ^{p}}{\vert y-y'\vert ^{2\sigma p+1}}dy\,dy'\,dt \\ &=\int_0^1 t^{2\sigma \alpha p}\int_0^{t^{\alpha }} \int_0^{t^{\alpha }} \frac{\vert f(t,x)-f(t,x')\vert ^{p}}{\vert x-x'\vert ^{2\sigma p+1}}dx\,dx'\,dt, \end{align*} from which we deduce that $\varphi ^{-2+1/p}h\in L^{p}(0,1;W^{2\sigma ,p}(0,1))$ signifies that \[ \int_0^1 \varphi (t)^{2\sigma p}\underset{0}{\overset{% \varphi (t)}{\int }}\int_0^{\varphi(t)} \frac{% \vert f(t,x)-f(t,x')\vert ^{p}}{\vert x-x'\vert ^{2\sigma p+1}}dxdx'dt<\infty . \] We denote this condition by \[ f\in L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p}). \] Similarly, we can prove the following equivalences. \begin{proposition} \label{prop9} \quad \begin{itemize} \item[(i)] $t^{-2\alpha +\alpha /p}w\in L^{p}(0,1;L^{p}(0,1))$ if and only if $u\in L^{p}(U)$ \item[(ii)] $t^{-2\alpha +\alpha /p}D_{y}^{2}w\in L^{p}(0,1;W^{2\sigma ,p}(0,1))$ if and only if $D_{x}^{2}u\in L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p})$ \item[(iii)] $t^{\alpha /p}D_{t}w\in L^{p}(0,1;W^{2\sigma ,p}(0,1))$ if and only if $D_{t}u\in L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p})$. \end{itemize} \end{proposition} Then Theorem \ref{thm1} is a consequence of the previous propositions. Note that the regularity of the solution in this result is maximal, when $f\in L_{\varphi ^{2\sigma }}^{p}(0,1;W_{\varphi }^{2\sigma ,p})$. \section{Proof of Theorem \ref{thm2}} Assume that $\varphi (t)=\sqrt{t}$, $m=t^{-1}$ and $\lambda \geqslant 1/2p$. Recall that $u(t,x)=w(t,y)$. Then problem (\ref{Equa1}) is equivalent to the following problem in $X=L^{p}(0,1)$ \begin{equation} \begin{gathered} tw'(t)+Lw(t)+\lambda w(t)=tg(t)\quad t\in (0,1), \\ w(0)=0, \end{gathered} \label{Equa13} \end{equation} where \begin{equation} \begin{gathered} D(L)=\{ \psi \in W^{2,p}(0,1)\text{ }:\psi (0)=\psi (1)=0\} \\ ( L\psi ) (y)=-\psi ''(y)-\frac{1}{2}y\psi '(y). \end{gathered} \label{formule14} \end{equation} Observe that $\overline{D(L)}=X$. It is sufficient to consider the equation \begin{equation} \begin{gathered} tw'(t)+Lw(t)+\frac{1}{2p}w(t)=h_{1}(t),\quad t\in (0,1), \\ w(0)=0, \end{gathered} \label{Equa15} \end{equation} where $h_{1}$ is in the space \[ E_{2}=\{ h\in L^{p}(0,1;L^{p}(0,1)):t^{-1+1/2p}h\in L^{p}(0,1;L^{p}(0,1))\} . \] Then, we can write $Bw+Aw=h_{1}$, where \begin{gather*} D(A)=\{ w\in E_{2}:t^{-1+\frac{1}{2p}}w\in L^{p}(0,1;W^{2,p}(0,1)\cap W_{0}^{1,p}(0,1))\} \\ ( Aw) (t)=Lw(t),\quad t\in (0,1), \end{gather*} and \begin{equation} \begin{gathered} D(B)=\{ w\in E_{2}:t^{\frac{1}{2p}}w'\in L^{p}(0,1;X)\} \\ (Bw)(t)=t\,w'(t)+\frac{1}{2p}w(t),\quad t\in (0,1). \end{gathered} \label{formule17} \end{equation} Since $p>3/2$ then $1/2p+1/p<1$ and $w(0)$ is well defined in $D(B)$. We will see below that necessarily $w(0)=0$. Observe that the condition (DV3) is satisfied. In order to apply Theorem \ref{thm4} we need to verify (DV1), (DV2). \begin{proposition} \label{prop11} Operators $A$ and $B$ are linear and closed with dense domains in $E_{2}$. Moreover, they satisfy (DV1) and (DV2). \end{proposition} Regarding the operator $B$, the proof is based on the solvability of the spectral equation \[ Bw+zw=h_{1}, \] where $h_{1}\in E_{2\text{ }}$and $z\geqslant 0$. Then \[ t\,w'(t)+\frac{1}{2p}w(t)+zw(t)=h_{1}(t) \] admits the following solution \[ w(t)=t^{-z-1/2p}\int_{0}^{t}s^{z-1+1/2p}h_{1}(s)ds. \] Observe that $w(0)=0$ since \begin{align*} \Vert t^{-z-1/2p}\int_{0}^{t}s^{z+1/2p-1}h_{1}(s)ds\Vert _{X} &\leqslant \int_{0}^{t}s^{-1/2p}s^{-1+1/2p}\Vert h_{1}(s)\Vert ds \\ &\leqslant \Vert h_{1}\Vert _{E}\Big( \int_{0}^{t}s^{-q/2p}ds\Big) ^{1/q} \\ &\leqslant \frac{t^{\frac{2p-3}{2p-2}}}{( 1-q/2p) ^{1/q}}% \Vert h_{1}\Vert _{E_{2}}, \end{align*} and $p>3/2$ (recall that $1/p+1/q=1)$. On the other hand, we have \[ \Vert (zI\text{ }+B)^{-1}h_{1}\Vert _{E_{2}}^{p}=\Vert w\Vert _{E_{2}}^{p}=\int_{0}^{1}\Vert t^{^{-1+1/2p}}w(t)\Vert _{X}^{p}dt, \] and \[ t^{^{-1+1/2p}}w(t) =\int_{0}^{t}t^{^{-z-1}}s^{z-1+1/2p}h(s)ds =\int_{0}^{1}K_{z}(t,s)s^{^{-1+1/2p}}h(s)ds, \] where \[ K_{z}(t,s)=\begin{cases} t^{^{-z-1}}s^{z} &\text{if }st. \end{cases} \] So \[ \sup_{t\in [ 0,1]} \int_{0}^{1}\vert K_{z}(t,s)\vert ds \leqslant \sup_{t\in [ 0,1]} \int_{0}^{t}t^{^{-z-1}}s^{z}ds \leqslant \frac{1}{z+1}, \] and for all $z>0$, \begin{align*} \sup_{s\in [ 0,1]}\int_{0}^{1}\vert K_{z}(t,s)\vert dt \leqslant \sup_{s\in [ 0,1]}\int_{s}^{1}t^{^{-z-1}}s^{z}dt \leqslant \frac{1}{z}. \end{align*} Then, there exists a constant $C(p)$ such that \[ \Vert (B+zI)^{-1}\Vert _{L(E_{2})}\leqslant \frac{C(p)}{z} \] for any large $z$. Moreover, we can see that \[ (B^{-1}h_{1})(t)=t^{-1/2p}\int_{0}^{t}s^{-1+1/2p}h(s)ds, \] and \[ \Vert ( B^{-1}h_{1}) (t)\Vert _{X}\leqslant t^{1-1/2p}\Vert h_{1}\Vert _{E_{2}}, \] which implies \[ \Vert B^{-1}h_{1}\Vert _{E_{2}}\leqslant C(p)\Vert h_{1}\Vert _{E_{2}}. \] So $B^{-1}\in L(E_{2})$ and $ii)$ of $(DV1)$ is verified. Now, to prove that $B\in Bip(E_{2})$, we set \[ Bw=B_{0}w+\frac{1}{2p}w, \] with \begin{gather*} D(B_{0}) =\{ w\in E:t^{\frac{1}{2p}}w'\in L^{p}(0,1;X),\,w(0)=0\} =D(B) \\ (B_{0}w)(t) =t\,w'(t),\quad t\in (0,1). \end{gather*} Consequently, it is sufficient to show that $B_{0}\in Bip(E_{2})$ (see \cite[Theorem 3]{PrussSohr}). Due to the Dore-Venni similar techniques used in \cite{DoreVenni2}, we have, for $t\in (0,1)$ and $\mathop{\rm Re}(z)\in ]0,1[$ \begin{align*} ( B_{0}^{-z}w) (t) &=\frac{1}{\Gamma ( z) \Gamma ( 1-z) } \int_{0}^{\infty }\lambda ^{-z}\big( ( \lambda +B_{0}) ^{-1}w\big) (t)d\lambda \\ &= \frac{1}{\Gamma ( z) \Gamma ( 1-z) } \int_{0}^{\infty }\lambda ^{-z}t^{-\lambda }\int_{0}^{t}s^{\lambda -1}w(s)dsd\lambda \\ &= \frac{1}{\Gamma ( z) \Gamma ( 1-z) } \int_{0}^{\infty }\lambda ^{-z}\int_{0}^{t}( \frac{s}{t}) ^{\lambda }w(s)\frac{ds}{s}d\lambda \\ &= \frac{1}{\Gamma ( z) \Gamma ( 1-z) } \int_{0}^{\infty }\lambda ^{-z}\int_{0}^{t}\exp \big( \lambda \ln ( \frac{s}{t}) \big) w(s)\frac{ds}{s}d\lambda \\ &= \frac{1}{\Gamma ( z) \Gamma ( 1-z) } \int_{0}^{t}w(s)\big( \ln ( \frac{t}{s}) \big) ^{z-1}\Big( \int_{0}^{\infty }\mu ^{-z}e^{-\mu }d\mu \Big) \frac{ds}{s} \\ &= \frac{1}{\Gamma ( z) }\int_{0}^{t}( \ln \frac{t}{s}) ^{z-1}w(s)\frac{ds}{s}. \end{align*} which implies that, for $\tau <0,$% \[ ( B_{0}^{-z}w) (e^{\tau })=\frac{1}{\Gamma ( z) } \int_{-\infty }^{\tau }( \tau -\sigma ) ^{z-1}w(e^{\sigma })d\sigma . \] Let us set $\omega =-1+\frac{1}{2p}$, and \begin{gather*} \psi ( \sigma ) =\frac{1}{\Gamma ( z) }( \max \{ \sigma ,0\} ) ^{z-1}e^{\sigma (\omega +1/p)} \quad \text{for }\sigma \in \mathbb{R}, \\ \phi (\sigma )=\begin{cases} e^{\sigma (\omega +1/p)}w(e^{\sigma }) &\text{for }\sigma \in (-\infty ,0), \\ 0&\text{ for }\sigma \in (0,\infty ). \end{cases} \end{gather*} Then it is easy to verify that \[ ( \psi \ast \phi ) ( \tau ) =e^{\tau (\omega +1/p)}( B_{0}^{-z}w) ( e^{\tau }) , \] for any $\tau \in \mathbb{R}_{-}$. Then we have \begin{align*} \Vert \phi \Vert _{L^{p}( (-\infty ,+\infty );X) } &= \Big( \int_{-\infty }^{0}\Vert e^{\sigma (\omega +1/p)}w(e^{\sigma })\Vert _{X}^{p}d\sigma \Big) ^{1/p} \\ &=\Big( \int_{-\infty }^{0}e^{\sigma (\omega p+1)}\Vert w(e^{\sigma })% \text{ }\Vert _{X}^{p}d\sigma \Big) ^{1/p} \\ &=\Big( \int_{0}^{1}\Vert t^{\omega }w(t)\Vert _{X}^{p}dt\Big) ^{1/p} \\ &=\Vert w\Vert _{E_{2}}. \end{align*} Since $\psi \in L^{1}( \mathbb{R}) $, its Fourier transform $% F(\psi )$ is \begin{align*} F(\psi )(\xi ) &=\frac{1}{\sqrt{2\pi }\Gamma ( z) }% \int_{0}^{\infty }e^{-i\sigma \xi }\sigma ^{z-1}e^{\sigma ( \omega +1/p) }d\sigma \\ &=\frac{1}{\sqrt{2\pi }\Gamma ( z) }\int_{0}^{\infty }e^{-\sigma ( i\xi -\omega -1/p) }\sigma ^{z-1}d\sigma \\ &=\frac{1}{\sqrt{2\pi }}\big( -\frac{1}{p}-\omega +i\xi \big) ^{-z}. \end{align*} This last equality is obtained by using the curve \[ \gamma _{\epsilon ,R}=\gamma _{\epsilon ,R}^{1}\cup \gamma _{R}^{2}\cup \gamma _{\epsilon ,R}^{3}\cup \gamma _{\epsilon ,R}^{4}, \] with \begin{gather*} \gamma _{\epsilon ,R}^{1} :[\epsilon ,R]\longrightarrow \mathbb{R}; \sigma \mapsto \sigma , \\ \gamma _{R}^{2} :[0,\theta ]\longrightarrow \mathbb{R}; \sigma \mapsto \gamma (\sigma )=R\,e^{i\sigma }, \\ \gamma _{\epsilon ,R}^{3} :[R,\epsilon ]\longrightarrow \mathbb{R}; \sigma \mapsto \sigma (-\frac{1}{p}-\omega +i\xi ), \\ \gamma _{\epsilon ,R}^{4} :[\theta ,0]\longrightarrow \mathbb{R};% \sigma \mapsto \gamma (\sigma )=\epsilon e^{i\sigma }, \end{gather*} where $\theta =\arg (-\frac{1}{p}-\omega +i\xi )$. It follows that \begin{align*} 0&=\lim_{\epsilon \to 0,R\to \infty } \int_{\gamma _{\epsilon ,R}}e^{-u( i\xi -\omega -1/p) }u^{z-1}du \\ &=\int_{0}^{\infty }e^{-\sigma ( i\xi -\omega -1/p) }\sigma ^{z-1}d\sigma -\Gamma ( z) \big( -\frac{1}{p}-\omega +i\xi \big) ^{-z}. \end{align*} Therefore, \begin{gather*} \frac{d}{d\xi }F(\psi )( \xi ) =-iz( -1/p-\omega +i\xi ) ^{-1}F(\psi )( \xi ) \\ \frac{d^{2}}{d\xi ^{2}}F(\psi )( \xi ) =-z( z+1) ( -1/p-\omega +i\xi ) ^{-2} F(\psi )( \xi ) \end{gather*} Now, let us put $z=\eta -ir$ with $(\eta ,r)\in ] 0,1[ \times \mathbb{R}$ and \[ \widetilde{( \psi \ast \phi ) }(\tau )= \begin{cases} ( \psi \ast \phi ) ( \tau ) &\text{for }\tau \in (-\infty ,0) \\ 0 & \text{for }\tau \in (0,+\infty ). \end{cases} \] Then, the extended Mikhlin's multiplier theorem gives \begin{align*} &\Vert ( B_{0}^{-z}w) \Vert _{E_{2}} \\ &=\Vert \widetilde{( \psi \ast \phi ) }\Vert _{L^{p}( (-\infty ,+\infty );X) } \\ &\leqslant C\max_{0\leqslant j\leqslant 2} \Big( \sup_{\xi\in \mathbb{R}} \big\vert \xi ^{j}\frac{d^{j}}{d\sigma ^{j}}F(\psi )( \xi ) \big\vert \Big) \Vert \phi \Vert _{L^{p}( X) } \\ &\leqslant C( 1+\vert z\vert ^{2}) \sup_{\xi \in \mathbb{R}}\Big[ \big( ( \omega +\frac{1}{p}) ^{2}+\xi ^{2}\big) ^{-\mathop{\rm Re}z/2}e^{( \mathop{\rm Im}z\arg ( -\omega -\frac{1}{p}+i\xi ) ) } \Big] \Vert w\Vert _{E_{2}} \\ &\leqslant C( 1+\eta ^{2}+r^{2}) e^{\frac{\pi }{2}\vert r\vert }\Vert w\Vert _{E_{2}}. \end{align*} since $-\omega -1/p=1-1/2p-1/p>0$. We then deduce the following proposition. \begin{proposition} \label{prop12} For any $r\in $ $\mathbb{R}$, $B_{0}^{ir}$ is a bounded operator in $E_{2}$. Moreover $r\to B_{0}^{ir}$ is a strongly continuous group, and there exists a constant $C>0$ such that \[ \Vert B_{0}^{ir}\Vert _{L(E_{2})}\leqslant C(1+r^{2})e^{\frac{\pi }{2}\vert r\vert }, \] \end{proposition} See A.9 in the appendix of Dore-Venni \cite{DoreVenni1}. Consequently, condition (ii) of $(DV2)$ is satisfied with $\theta _{B}=\pi /2+\varepsilon $ for any $\varepsilon >0$. Now, we are concerned with the operator $A$ and its realization defined by \[ ( L\Psi ) =L_{0}\Psi +P\Psi \ \] where $L_{0}$ and $P$ are defined as in (14) and (15) by \begin{gather*} D(L_{0})=\{ \Psi \in W_{p}^{2}(0,1):\Psi (0)=\Psi (1)=0\}, \quad L_{0}\Psi =-\Psi ''; \\ D(P)=W^{1,p}(0,1), \quad P\Psi =-\frac{1}{2}y\Psi '. \end{gather*} It is easy to see that the operator $L_{0}$ is sectorial. Moreover, thanks to H\"{o}lder inequality, for $\psi \in D(L_{0})\subset D(P)$, we have \[ \Vert P\Psi \Vert _{L^{p}( 0,1) }\leqslant C(p)\Vert \Psi \Vert _{D(L_{0})}. \] And the operator $P=m\circ i\circ d$ is compact from $D(L_{0})$ into $E_{2}$ while $L$ is sectorial. Furthermore, there exists some $r_{0}>0$ such that \[ \rho (-L)\supset \Sigma _{\pi -\epsilon _{1}}=\{ \lambda :\text{ }% \vert \lambda \vert \geqslant r_{0}\text{ },\vert \arg \lambda \vert <\pi -\epsilon _{1}\} , \] where $\epsilon _{1}\in ]0,\pi /2[$. Observe that the classical second order differential equations theory with Dirichlet conditions gives, for all $\delta >0$, \begin{gather*} \rho (-(L+\delta I))\supset \Sigma _{\pi -\epsilon _{1}}=\{ \lambda :\vert \arg \lambda \vert <\pi -\epsilon _{1}\} \\ \forall \lambda \in \Sigma _{\pi -\epsilon _{1}},\Vert (L_{\delta }+\lambda I\text{ })^{-1}\Vert _{L(E_{2})}\leqslant M_{\delta }/(1+\vert \lambda \vert ), \end{gather*} where $L_{\delta }=L+\delta I$. Consequently, $L_{\delta }$ is a positive operator. According to Labbas-Moussaoui \cite{LabbasMoussaoui}, $( L_{0}+\delta I) ^{is}$ forms a strongly continuous group such that for any $\gamma_{1}>0$, there exists $M_{0}>0$ satisfying \[ \Vert ( L_{0}+\delta I) ^{is}\Vert _{L(E_{2})}\leqslant M_{0}e^{\gamma _{1}\vert s\vert }, \quad \forall s\in \mathbb{R}\,. \] On the other hand, for any $\eta \in ]0,1[$, $( L_{0}+\delta I)^{\eta }$ is defined and its domain $D( ( L_{0}+\delta I)^{\eta }) $ coincides with the complex interpolation space \[ [ L^{p}(0,1),D( L_{0}+\delta I)] _{\eta } \] which is contained in $W^{1,p}(0,1)=D(P)$ when $\eta $ is near to $1$, see Triebel \cite{Triebel}. Then, using Dore-Venni \cite{DoreVenni3}, the operator $( L_{\delta }) ^{is}$ is bounded for all $s\in \mathbb{R% }$ and \[ \Vert ( L_{\delta }) ^{is}\Vert _{L(E_{2})}\leqslant M_{\delta }'e^{(\varepsilon +\gamma _{1})\vert s\vert },\quad \forall s\in \mathbb{R}\; \forall \varepsilon >0\,. \] So, we have a same result for the operator $A+\delta I$, for any $\delta >0$. Now, applying the same techniques used in the proofs of Theorem A1 and Lemma A2, pages 89-90 in \cite{GigaSohr}, we obtain, as $\delta \to 0$, \[ \theta _{A}=\theta _{L_{\delta }}=\varepsilon +\gamma _{1}\in ]0,\pi /2[. \] The condition $\theta _{A}+\theta _{B}<\pi $ is verified and, finally, $A$ and $B$ satisfy (DV1), (DV2) and (DV3). Then, Theorem \ref{thm4} leads to \begin{proposition} \label{prop13} Given $h_{1}\in E_{2}$, Problem \eqref{Equa13} has a unique solution $w$ satisfying \begin{gather*} w\in E_{2},w(0)=0, \\ t^{-1+\frac{1}{2p}}w\in L^{p}(0,1;W^{2,p}(0,1)\cap W_{0}^{1,p}(0,1)), \\ t^{\frac{1}{2p}}w'\in L^{p}(0,1;L^{p}(0,1)). \end{gather*} \end{proposition} In the triangle $U$, using the changes $h_{1}(t,y)=t.g(t,y)$, $g(t,y)=f(t,x)$ with $f\in L^{p}(U)$, we then have $h_{1}\in E_{2}$. Consequently, from this last Proposition, we have \begin{enumerate} \item $w\in E_{2}$, that is $t^{-1}u\in L^p(U)$ \item $t^{-1+1/2p}D_{x}^{2}w\in L^{p}(0,1;L^{p}(0,1))$ that is $D_{x}^{2}u\in L^{p}(U)$. \end{enumerate} Thus, Equation \eqref{Equa1} implies $D_{t}u=D_{x}^{2}u-\lambda t^{-1}u+f\in L^{p}(U)$. This completes the proof of Theorem \ref{thm2}. 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