\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 198, pp. 1--22.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/198\hfil Direct and inverse degenerate problems]
{Direct and inverse degenerate parabolic differential equations
 with multi-valued operators}

\author[A. Favini, A. Lorenzi, H. Tanabe \hfil EJDE-2015/198\hfilneg]
{Angelo Favini, Alfredo Lorenzi, Hiroki Tanabe}

\dedicatory{In memory of Alfredo Lorenzi}

\address{Angelo Favini \newline
Dipartimento di Matematica,
University of Bologna,
Piazza di Porta San Donato 5, 40126,
Bologna, Italy}
\email{angelo.favini@unibo.it}

\address{Alfredo Lorenzi \newline
Dipartimento di  Matematica ``F. Enriques'',
Universit\`a degli Studi di Milano
via Saldini 50, 20133 Milano, Italy}
\email{alfredo.lorenzi@unimi.it}

\address{Hiroki Tanabe \newline
Takarazuka, Hirai Sanso 12-13, 665-0817, Japan}
\email{h7tanabe@jttk.zaq.ne.jp}

\thanks{Submitted June 11, 2015. Published July 30, 2015.}
\subjclass[2010]{35R30,  34G10, 35K20, 35K50, 45N05, 45Q05}
\keywords{Identification problems; first-order equations and systems
\hfill\break\indent in Banach spaces;
linear parabolic integro-differential equations; existence and uniqueness}

\begin{abstract}
 Real interpolation spaces are used for solving some identification linear
 evolution problems in Banach spaces, under space regularity assumptions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

This note starts with the following {\it direct} problem in a Banach space $X$,
\begin{equation}
\begin{gathered}
\frac d{dt}y(t)+Ay(t)\ni f(t),\quad 0\leq t\leq T,\\
y(0)=y_0.
\end{gathered} \label{e1.1}
\end{equation}
Here $A$ is a possibly multivalued linear operator such that
\begin{equation}
\rho(A)\supset\Sigma_{\alpha}=\{\lambda\in\mathbb{C};\operatorname{Re}\lambda
\leq c(1+|\operatorname{Im}\lambda|)^{\alpha}\},
 \label{e1.2}
\end{equation}
and the following inequality holds for $\lambda\in\Sigma_{\alpha}$
\begin{equation}
\|(\lambda-A)^{-1}\|_{\mathcal{L}(X)}\leq c(1+|\lambda|)^{-\beta}, \label{e1.3}
\end{equation}
where $c$, $\alpha$ and $\beta$ are positive constants satisfying $\beta\leq\alpha\leq1$.
It is shown in the book by Favini and Yagi \cite{faviyagi} that $-A$ generates
a $C^{\infty}$-semigroup $e^{-tA}$, $0<t<\infty$. We are interested in extending
some of the results obtained in the paper by Favini, Lorenzi Tanabe \cite{falota},
where $A$ is assumed to be single-valued, to the case where $A$ is multi-valued.
In \cite{falota} supposing that $y_0\in D(A)$, $Ay_0\in{\tilde X}_A^{\theta}$,
$f\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$ or with
$(X,D(A))_{\theta,\infty}$ in place of ${\widetilde X}_A^{\theta}$,
the existence and uniqueness of a solution to \eqref{e1.1} with some regularity
 property is established, where $B([0,T];Y)$ stands for the set of all bounded
functions defined in $[0,T]$ with values in a Banach space $Y$.
In this paper we show analogous results in case $A$ is multivalued replacing
 $Ay_0\in{\tilde X}_A^{\theta}$ or $Ay_0\in(X,D(A))_{\theta,\infty}$ by
 $Ay_0\cap{\tilde X}_A^{\theta}\ne\emptyset$ or
 $Ay_0\cap(X,D(A))_{\theta,\infty}\ne\emptyset$.

We begin by modifying the definitions of the spaces $X_A^{\theta}$ and
${\widetilde X}_A^{\theta}$ in case where $A$ is not necessarily single valued
replacing $A(t+A)^{-1}$ and $Ae^{-tA}$ by $I-t(t+A)^{-1}$ and
$de^{-tA}/dt$ respectively and prove some preliminary results on these spaces.

The results proved for the {\it direct} problem are applied to the following
identification problem:
\begin{equation}
\begin{gathered}
{\frac d{dt}y(t)+Ay(t)\ni f(t)z+h(t)},\quad t\in[0,T],\\
y(0)=y_0,\\
\Phi[y(t)]=g(t),\quad t\in[0,T],
\end{gathered}  \label{e1.4}
\end{equation}
where $y\in C([0,T];X)$, $f\in C([0,T];\mathbb{C})$ are unknown,
$z\in X$, $h\in C([0,T];X)$, $g\in C([0,T];\mathbb{C})$ are given elements
and $\Phi\in X^*$. Under some regularity assumptions on $z,h$ and $g$
it is shown that a unique solution to problem \eqref{e1.4} exists.
An extension of this result to equations with several unknown scalar functions
\begin{equation}\label{e1.5}
\begin{gathered}
{\frac d{dt}y(t)+Ay(t)\ni \sum_{j=1}^nf_j(t)z_j+h(t)},\quad t\in[0,T],\\
y(0)=y_0,\\
\Phi_j[y(t)]=g_j(t),\quad j=1,\dots,n,\quad t\in[0,T],
\end{gathered}
\end{equation}
is also established.

The above results are applied to the following problems
\begin{equation}
\begin{gathered}
{\frac d{dt}Mu(t)+Lu(t)=f(t),\quad t\in[0,T]},\\
Mu(0)=Mu_0,
\end{gathered} \label{e1.6}
\end{equation}
and
\begin{equation}
\begin{gathered}
{\frac d{dt}Mu(t)+Lu(t)=\sum_{j=1}^nf_j(t)z_j+h(t),\quad t\in[0,T]},\\
Mu(0)=Mu_0,\\
\Phi_j[Mu(t)]=g_j(t),\quad j=1,\dots,n,\quad t\in[0,T],
\end{gathered} \label{e1.7}
\end{equation}
where $L$ and $M$ are linear closed operators such that
\begin{equation}
D(L)\subset D(M) \label{e1.8}
\end{equation}
and for $\lambda\in\Sigma_{\alpha}$ a bounded inverse of $\lambda M+L$ exists and
\begin{equation}
\|M(\lambda M+L)^{-1}\|_{\mathcal{L}(X)}\leq c(1+|\lambda|)^{-\beta}. \label{e1.9}
\end{equation}
Then $A=LM^{-1}$ satisfies \eqref{e1.2} and \eqref{e1.3}.
A solution to \eqref{e1.6} or \eqref{e1.7} is easily obtained from that
to \eqref{e1.1} or \eqref{e1.5} with $y_0=Mu_0$.

We refer to the monograph [5] for diverse problems concerning regular evolution
equations via Mathematical Physics. These identification problems were discussed
in the papers \cite{FLMT} and \cite{fata} by using at all different techniques.
 Our present approach allows us to weaken the assumptions and to improve the
regularity of solutions.

The plan of this paper is as follows.
In section 2 preliminary results on intermediate spaces are collected.
 Section 3 is devoted to the existence and uniqueness of a solution to the
direct problem \eqref{e1.1}. In Section 4 identification problem \eqref{e1.4}
is solved by transforming it to a simple Volterra integral equation for $f$.
In Section 5 the general identification problem \eqref{e1.4} is solved by
applying the Banach fixed point theorem. This result is extended to
equations \eqref{e1.5} with several unknown scalar functions in Section 6.
Finally, in Section 7 these results are applied to problems \eqref{e1.6}
and \eqref{e1.7}.

\section{Preliminaries}

Let $A$ be a possibly multivalued linear operator in the complex Banach space $X$
satisfying \eqref{e1.2} and \eqref{e1.3}. Then it is shown in
Chapter III of Favini and Yagi \cite{faviyagi} that $-A$ generates a
semigroup $e^{-tA}$, $0<t<\infty$, satisfying
\begin{equation}
\|\frac{d^i}{dt^i}e^{-tA}\|_{\mathcal{L}(X)}\leq C_0t^{(\beta-i-1)/\alpha},\quad
i=0,1,2,\quad C_0>0. \label{e2.1}
\end{equation}
The set $D(A)$ makes a Banach space with norm
\begin{equation}
\|u\|_{D(A)}=\inf_{\phi\in Au}\|\phi\|_X\quad\text{for }u\in D(A). \label{e2.2}
\end{equation}
It is known that
\begin{equation}\label{e2.3}
A^{-1}\frac d{dt}e^{-tA}=\frac d{dt}e^{-tA}A^{-1}
=-e^{-tA},\quad \lim_{t\to 0}e^{-tA}u=u\quad\text{for }u\in D(A).
\end{equation}
If $u\in D(A)$, then in view of \eqref{e2.3} and \eqref{e2.1} one has for
$\phi\in Au$
\[
\|\frac d{dt}e^{-tA}u\|_X=\|\frac d{dt}e^{-tA}A^{-1}\phi\|_X
=\|e^{-tA}\phi\|_X\leq C_0t^{(\beta-1)/\alpha}\|\phi\|_X.
\]
This implies, by \eqref{e2.2},
\[
\|\frac d{dt}e^{-tA}u\|_X\leq C_0t^{(\beta-1)/\alpha}\|u\|_{D(A)}.
\]
This inequality and the one obtained with the aid of an analogous argument imply
\begin{equation}\label{e2.4}
\|\frac d{dt}e^{-tA}\|_{\mathcal{L}(D(A),X)}\leq C_0t^{(\beta-1)/\alpha},\quad
\|\frac{d^2}{dt^2}e^{-tA}\|_{\mathcal{L}(D(A),X)}\leq C_0t^{(\beta-2)/\alpha}.
\end{equation}
\smallskip

\noindent\textbf{Definition.}
For $0<\theta<1$,
\begin{gather*}
 X_A^{\theta}=\{u\in X;\,|u|_{X_A^{\theta}}
=\sup_{0<t<\infty}t^{\theta}\|u-t(t+A)^{-1}u\|_X<\infty\},\\
 \|u\|_{X_A^{\theta}}=|u|_{X_A^{\theta}}+\|u\|_{X},
\\
 {\widetilde X}_A^{\theta}
=\big\{u\in X;|u|_{{\widetilde X}_A^{\theta}}
=\sup_{t>0}t^{(2-\beta-\theta)/\alpha}\|\frac d{dt}e^{-tA}u\|_X<\infty\big\},
\\
 \|u\|_{{\widetilde X}_A^{\theta}}=|u|_{{\widetilde X}_A^{\theta}}+\|u\|_{X}.
\end{gather*}
One of the definition of $(X,D(A))_{\theta,\infty}$ is
\begin{gather*}
\begin{aligned}
(X,D(A))_{\theta,\infty}=\Big\{&u=u_0(t)+u_1(t)\;\forall t\in(0,\infty);
 \sup_{0<t<\infty}\|t^{\theta}u_0(t)\|_X<\infty,\\
&\sup_{0<t<\infty}\|t^{\theta-1}u_1(t)\|_{D(A)}<\infty\Big\},
\end{aligned}\\
\begin{aligned}
&\|u\|_{(X,D(A))_{\theta,\infty}}\\
&=\inf_{u=u_0(t)+u_1(t)\;\forall t\in(0,\infty)}
\big\{\sup_{0<t<\infty}\|t^{\theta}u_0(t)\|_X
+\sup_{0<t<\infty}\|t^{\theta-1}u_1(t)\|_{D(A)}\big\}.
\end{aligned}
\end{gather*}

\begin{lemma}\label{lem2.1}
The following inclusion relation holds for $0<\theta<1$,
\[
X_A^{\theta}\subset(X,D(A))_{\theta,\infty}.
 \]
\end{lemma}

\begin{proof}
Suppose $u\in X_A^{\theta}$. Set
\[
u_0(t)=u-t(t+A)^{-1}u,\quad u_1(t)=t(t+A)^{-1}u.
\]
Then $u=u_0(t)+u_1(t)$ and
\[
\sup_{0<t<\infty}\|t^{\theta}u_0(t)\|_X
=\sup_{0<t<\infty}t^{\theta}\|u-t(t+A)^{-1}u\|_X<\infty.
\]
Since $(t+A)u_1(t)\ni tu$, one has $Au_1(t)\ni t(u-u_1(t))$, and hence
\[
\|u_1(t)\|_{D(A)}\leq t\|u-u_1(t)\|_X=t\|u-t(t+A)^{-1}u\|_X.
\]
Therefore,
\[
\sup_{0<t<\infty}\|t^{\theta-1}u_1(t)\|_{D(A)}
\leq\sup_{0<t<\infty}t^{\theta}\|u-(t+A)^{-1}u\|_X<\infty.
 \]
\end{proof}

\begin{lemma}\label{lem2.2}
The following inclusion relation holds for $0<\theta<1$,
\[
(X,D(A))_{\theta,\infty}\subset{\widetilde X}_A^{\theta}.
\]
\end{lemma}

\begin{proof}
Let $u\in(X,D(A))_{\theta,\infty}$. From \eqref{e2.1}, with $i=1$, and
the first inequality of \eqref{e2.4} it follows that
\begin{equation}
\|\frac d{dt}e^{-tA}u\|_X\leq C_0t^{(\beta-2+\theta)/\alpha}\|u\|_{(X,D(A))_{\theta,\infty}}.
 \label{e2.5}
\end{equation}
This readily implies $u\in {\widetilde X}_A^{\theta}$.
\end{proof}

\begin{lemma}\label{lem2.3}
Suppose $\alpha+\beta+\theta>2$. Then for $u\in(X,D(A))_{\theta,\infty}$, $e^{-tA}u\to  u$
as $t\to 0$. If $v\in D(A)$, the set $Av\cap(X,D(A))_{\theta,\infty}$ consists of at most a single element $ {-\lim_{t\to 0}\frac d{dt}e^{-tA}v}$.
\end{lemma}

\begin{proof}
Let $u\in D(A)$ and $\phi\in Au$. Then in view of \eqref{e2.3} one gets
\[
e^{-tA}u-u=\int_0^t\frac d{d\tau}e^{-\tau A}ud\tau
=\int_0^t\frac d{d\tau}e^{-\tau A}A^{-1}\phi d\tau
=-\int_0^te^{-\tau A}\phi d\tau.
\]
Hence, noting that the assumption implies $\alpha+\beta>1$, one deduces
\begin{align*}
\|e^{-tA}u-u\|_{X}
&=\|\int_0^te^{-\tau A}\phi d\tau\|_{X}\\
&\leq C_0\int_0^t\tau^{(\beta-1)/\alpha}\|\phi\|_{X}d\tau\\
&=C_0\frac{t^{(\beta-1)/\alpha+1}}{(\beta-1)/\alpha+1}\|\phi\|_{X}.
\end{align*}
This implies
\begin{equation}
\|e^{-tA}u-u\|_{X}\leq C_0\frac{t^{(\beta-1)/\alpha+1}}{(\beta-1)/\alpha+1}\|u\|_{D(A)}.
\label{e2.6}
\end{equation}
For $u\in X$,
\begin{equation}
\|e^{-tA}u-u\|_{X}\leq\|e^{-tA}u\|_{X}+\|u\|_{X}
\leq C_0t^{(\beta-1)/\alpha}\|u\|_{X}+\|u\|_{X}. \label{e2.7}
\end{equation}
Interpolating \eqref{e2.6} and \eqref{e2.7} yields that there exists a constant
$C$ such that for $u\in(X,D(A))_{\theta,\infty}$,
\begin{equation}
\|e^{-tA}u-u\|_{X}\leq Ct^{(\beta-1)/\alpha+\theta}\|u\|_{(X,D(A))_{\theta,\infty}},
\quad 0<t\leq1. \label{e2.8}
\end{equation}
Since
\[
\frac{\beta-1}{\alpha}+\theta-\frac{\alpha+\beta+\theta-2}{\alpha}=\frac{(1-\theta)(1-\alpha)}{\alpha}\geq0,
\]
one has
\[
\frac{\beta-1}{\alpha}+\theta\geq\frac{\alpha+\beta+\theta-2}{\alpha}>0.
\]
Hence the first assertion follows.
Suppose $v\in D(A)$ and $\phi\in Av\cap(X,D(A))_{\theta,\infty}$.
Then by the first assertion
\[
\frac d{dt}e^{-tA}v=\frac d{dt}e^{-tA}A^{-1}\phi=-e^{-tA}\phi\to -\phi
\]
as $t\to 0$.
\end{proof}

\begin{lemma}\label{lem2.4}
For $u\in X$, $t>0$, the following equality holds,
\[
 \int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau=(A+t)^{-1}u.
\]
\end{lemma}

\begin{proof}
With the aid of \eqref{e2.3} and integration by parts one deduces
\begin{align*}
\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau
&=-\int_0^{\infty}e^{-t\tau}\frac d{d\tau}e^{-\tau A}A^{-1}ud\tau\\
&=A^{-1}u-t\int_0^{\infty}e^{-t\tau}e^{-\tau A}A^{-1}ud\tau\\
&=A^{-1}\Big(u-t\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau\Big).
\end{align*}
Hence $\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau\in D(A)$ and
\[
A\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau\ni u
-t\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau.
\]
The assertion of the lemma readily follows.
\end{proof}

\begin{lemma}\label{lem2.5}
Suppose $\alpha+\beta+\theta>2$. If $u\in{\widetilde X}_A^{\theta}$, then
$e^{-tA}u$ converges as $t\to 0$. If the limit is equal to $u$:
$\lim_{t\to 0}e^{-tA}u=u$, then $u\in X_A^{(\alpha+\beta-2+\theta)/\alpha}$ and
\[
|u|_{X_A^{(\alpha+\beta-2+\theta)/\alpha}}\leq\Gamma((\alpha+\beta-2+\theta)/\alpha)
|u|_{{\widetilde X}_A^{\theta}}.
\]
\end{lemma}

\begin{proof}
For $0<s<t$, one has
\begin{align*}
\|e^{-tA}u-e^{-sA}u\|_X
&=\|\int_s^t\frac d{d\tau}e^{-\tau A}ud\tau\|_X
 \leq\int_s^t\tau^{(\beta-2+\theta)/\alpha}|u|_{{\widetilde X}_A^{\theta}}d\tau\\
&=\frac{t^{(\alpha+\beta-2+\theta)/\alpha}-s^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
 |u|_{{\widetilde X}_A^{\theta}}.
\end{align*}
Hence $e^{-tA}u$ converges as $t\to 0$. If $e^{-tA}u\to  u$,
then with the aid of integration by parts and Lemma \ref{lem2.4} one gets for $t>0$
\[
\int_0^{\infty}e^{-t\tau}\frac d{d\tau}e^{-\tau A}ud\tau
=-u+t\int_0^{\infty}e^{-t\tau}e^{-\tau A}ud\tau=-u+t(t+A)^{-1}u.
\]
Hence
\begin{align*}
\|u-t(t+A)^{-1}u\|_X
&=\|-\int_0^{\infty}e^{-t\tau}\frac d{d\tau}e^{-\tau A}ud\tau\|_X\\
&\leq\int_0^{\infty}e^{-t\tau}\tau^{(\beta-2+\theta)/\alpha}
 |u|_{{\widetilde X}_A^{\theta}}d\tau\\
&=t^{(2-\alpha-\beta-\theta)/\alpha}\Gamma((\alpha+\beta-2+\theta)/\alpha)|u|_{{\widetilde X}_A^{\theta}}.
\end{align*}
\end{proof}

\begin{lemma}\label{Lemma 2.6}
Let $\alpha+\beta+\theta>2$. If $v\in D(A)$ and $Av\cap{\widetilde X}_A^{\theta}\ne\emptyset$, then $ {\lim_{t\to 0}\frac d{dt}e^{-tA}v}$ exists.
\end{lemma}
\begin{proof}
Let $\phi\in Av\cap{\widetilde X}_A^{\theta}$. Then
\[ \lim_{t\to 0}\frac d{dt}e^{-tA}v=\lim_{t\to 0}\frac d{dt}e^{-tA}A^{-1}\phi=-\lim_{t\to 0}e^{-tA}\phi \]
exists by  Lemma \ref{lem2.5}.
\end{proof}


\begin{remark} \label{rmk2.1} \rm
The limit $\lim_{t\to 0}e^{-tA}u$ is not necessarily equal to $u$ if $A$
is multivalued. If $0\ne u\in A0$, then
\[
e^{-tA}u=-\frac d{dt}e^{-tA}A^{-1}u=-\frac d{dt}e^{-tA}0=0.
\]
Therefore $u\in{\widetilde X}_A^{\theta}$ for any $\theta\in(0,1)$ and
$\lim_{t\to 0}e^{-tA}u=0\ne u$. If $A$ is single valued, then
\[
A^{-1}\lim_{t\to 0}e^{-tA}u=\lim_{t\to 0}e^{-tA}A^{-1}u=A^{-1}u.
\]
Hence $\lim_{t\to 0}e^{-tA}u=u$ whenever the limit of the left hand side exists.
\end{remark}


\section{Main result concerning problem \eqref{e1.1}}

\begin{theorem} \label{thm3.1}
Suppose that $2\alpha+\beta+\theta>3,\;\,y_0\in D(A)$,
$Ay_0\cap(X,D(A))_{\theta,\infty}\ne\emptyset$ and
$f\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$.
Then problem \eqref{e1.1} admits a unique solution $y$ such that
\begin{equation}\label{e3.1}
\begin{gathered}
y\in C^1([0,T];X),\\
y'-f\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)\cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\end{gathered}
\end{equation}

Here $B([0,T];Y)$ stands for the set of all bounded (not necessarily measurable)
 functions with values in $Y$ defined in $[0,T]$.
\end{theorem}

\begin{proof}
Note that $\alpha+\beta-2+\theta\geq2\alpha+\beta-3+\theta>0$. Set
\begin{equation}
\begin{gathered}
{y(t)=e^{-tA}y_0+\int_0^te^{-(t-s)A}f(s)ds},\\
 {y_1(t)=e^{-tA}y_0,\quad y_2(t)=\int_0^te^{-(t-s)A}f(s)ds}.
\end{gathered} \label{e3.2}
\end{equation}
In view of Lemma \ref{lem2.3}, $Ay_0\cap(X,D(A))_{\theta,\infty}$ consists of a
 single element $\phi= {-\lim_{t\to 0}\frac d{dt}e^{-tA}y_0}$. Hence
\begin{equation}
y_1'(t)=\frac d{dt}e^{-tA}y_0=\frac d{dt}e^{-tA}A^{-1}\phi=-e^{-tA}\phi. \label{e3.3}
\end{equation}
In view of \eqref{e2.5}
\begin{equation}
\|\frac d{dt}e^{-tA}\phi\|_X
\leq C_0t^{(\beta-2+\theta)/\alpha}\|\phi\|_{(X,D(A))_{\theta,\infty}}. \label{e3.4}
\end{equation}
By  of \eqref{e3.3} and Lemma \ref{lem2.3}
\begin{equation}
\lim_{t\to 0}y_1'(t)=-\phi. \label{e3.5}
\end{equation}
With the aid of \eqref{e3.4}, for $0\leq s<t\leq T$ one obtains
\begin{align*}
\|y_1'(t)-y_1'(s)\|_X
&=\|-e^{-tA}\phi+e^{-sA}\phi\|_X
 =\|\int_s^t\frac d{d\sigma}e^{-\sigma A}\phi \,d\sigma\|_X\\
&\leq C_0\int_s^t\sigma^{(\beta-2+\theta)/\alpha}\|\phi\|_{(X,D(A))_{\theta,\infty}}\,d\sigma\\
&\leq C_0\frac{(t-s)^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
 \|\phi\|_{(X,D(A))_{\theta,\infty}}.
\end{align*}
Therefore $y_1'\in C^{(\alpha+\beta-2+\theta)/\alpha}([0,T];X)$ and
\begin{equation}
|y_1'|_{C^{(\alpha+\beta-2+\theta)/\alpha}([0,T];X)}
\leq \frac{C_0}{(\alpha+\beta-2+\theta)/\alpha}\|\phi\|_{(X,D(A))_{\theta,\infty}}. \label{e3.6}
\end{equation}
For $s>0$, in view of \eqref{e3.3}, \eqref{e3.4} and $0<(2-\beta-\theta)/\alpha<1$,
one deduces
\begin{align*}
&s^{(2-\beta-\theta)/\alpha}\|\frac d{ds}e^{-sA}y_1'(t)\|_X\\
&=s^{(2-\beta-\theta)/\alpha}\|\frac d{ds}e^{-sA}e^{-tA}\phi\|_X\\
&=s^{(2-\beta-\theta)/\alpha}\|\frac{\partial}{\partial s}e^{-(s+t)A}\phi\|_X\\
&\leq C_0s^{(2-\beta-\theta)/\alpha}(s+t)^{(\beta-2+\theta)/\alpha}
 \|\phi\|_{(X,D(A))_{\theta,\infty}}\\
&=C_0\left(\frac s{s+t}\right)^{(2-\beta-\theta)/\alpha}
 \|\phi\|_{(X,D(A))_{\theta,\infty}}\leq C_0\|\phi\|_{(X,D(A))_{\theta,\infty}}.
\end{align*}
Hence $y_1'(t)\in {\widetilde X}_A^{\theta}$ and
\begin{equation}
|y_1'(t)|_{{\widetilde X}_A^{\theta}}\leq C_0\|\phi\|_{(X,D(A))_{\theta,\infty}}. \label{e3.7}
\end{equation}
Using \eqref{e3.3} one observes that
\[
e^{-\tau A}y_1'(t)=-e^{-\tau A}e^{-tA}\phi
=-e^{-(t+\tau)A}\phi\to -e^{-tA}\phi=y_1'(t)
\]
as $\tau\to 0$. Hence in view of Lemma \ref{lem2.5} $y_1'(t)\in X_A^{(\alpha+\beta-2+\theta)/\alpha}$ and
\begin{equation}
|y_1'(t)|_{X_A^{(\alpha+\beta-2+\theta)/\alpha}}\leq\Gamma((\alpha+\beta-2+\theta)/\alpha)|y_1'(t)|_{{\widetilde X}_A^{\theta}}. \label{e3.8}
\end{equation}
This inequality, \eqref{e3.7} and \eqref{e3.3} yield
\[
\|y_1'(t)\|_{X_A^{(\alpha+\beta-2+\theta)/\alpha}}
\leq C_0\Gamma((\alpha+\beta-2+\theta)/\alpha)\|\phi\|_{(X,D(A))_{\theta,\infty}}
+\|e^{-tA}\phi\|_{X}.
\]
Hence $y_1'\in B([0,T];X_A^{(\alpha+\beta-2+\theta)/\alpha})$ and
\begin{equation}
\begin{aligned}
&\|y_1'\|_{B([0,T];X_A^{(\alpha+\beta-2+\theta)/\alpha})}\\
&\leq C_0\Gamma((\alpha+\beta-2+\theta)/\alpha)
\|\phi\|_{(X,D(A))_{\theta,\infty}}
+\sup_{0\leq t\leq T}\|e^{-tA}\phi\|_{X}.
\end{aligned} \label{e3.9}
\end{equation}
The second term of the right hand side of \eqref{e3.9} is finite by \eqref{e3.3}
and \eqref{e3.5}. In view of \eqref{e2.5}
\begin{equation}
\begin{aligned}
\|\frac{\partial}{\partial t}e^{-(t-s)A}f(s)\|_X
&\leq C_0(t-s)^{(\beta-2+\theta)/\alpha}\|f(s)\|_{(X,D(A))_{\theta,\infty}}\\
&\leq C_0(t-s)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{aligned} \label{e3.10}
\end{equation}
Hence
\begin{equation}
\begin{aligned}
\|\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}f(s)ds\|_X
&\leq C_0\int_0^t(t-s)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}ds\\
&=C_0\frac{t^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{aligned} \label{e3.11}
\end{equation}
For $0\leq t<t'\leq T$ one has with the aid of the change of the order of
integration
\begin{align*}
&\int_t^{t'}\int_0^{\tau}\frac{\partial}{\partial\tau}e^{-(\tau-s)A}f(s)dsd\tau\\
&=\int_0^t\int_t^{t'}\frac{\partial}{\partial\tau}
 e^{-(\tau-s)A}f(s)d\tau ds+\int_t^{t'}\int_s^{t'}
 \frac{\partial}{\partial\tau}e^{-(\tau-s)A}f(s)d\tau ds.
\end{align*}
It is obvious that if $s<t<t'$, then
\[
\int_t^{t'}\frac{\partial}{\partial\tau}e^{-(\tau-s)A}f(s)d\tau
=e^{-(t'-s)A}f(s)-e^{-(t-s)A}f(s).
\]
Since $f(s)\in(X,D(A))_{\theta,\infty}$, in view of Lemma \ref{lem2.3},
 $e^{-(\tau-s)A}f(s)\to  f(s)$ as $\tau\to  s$. Hence
\begin{equation}
\int_s^{t'}\frac{\partial}{\partial\tau}e^{-(\tau-s)A}f(s)d\tau
=e^{-(t'-s)A}f(s)-f(s). \label{e3.12}
\end{equation}
Therefore
\begin{align*}
&\int_t^{t'}\int_0^{\tau}\frac{\partial}{\partial\tau}e^{-(\tau-s)A}f(s)dsd\tau\\
&=\int_0^t\big[e^{-(t'-s)A}f(s)-e^{-(t-s)A}f(s)\big]ds
 +\int_t^{t'}\big[e^{-(t'-s)A}f(s)-f(s)\big]ds\\
&=\int_0^{t'}e^{-(t'-s)A}f(s)ds-\int_0^te^{-(t-s)A}f(s)ds-\int_t^{t'}f(s)ds\\
&= y_2(t')-y_2(t)-\int_t^{t'}f(s)ds.
\end{align*}
This means that $y_2(t)$ is differentiable and
\begin{equation}
y_2'(t)=f(t)+\int_0^t\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\,d\sigma. \label{e3.13}
\end{equation}
With the aid of \eqref{e3.13}, for $0\leq s<t\leq T$, one has
\begin{equation}
\begin{aligned}
&(y_2'(t)-f(t))-(y_2'(s)-f(s))\\
&=\int_0^t\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\,d\sigma
 -\int_0^s\frac{\partial}{\partial s}e^{-(s-\sigma)A}f(\sigma)\,d\sigma \\
&=\int_0^s\Big[\frac{\partial}{\partial t}e^{-(t-\sigma)A}
 -\frac{\partial}{\partial s}e^{-(s-\sigma)A}\Big]f(\sigma)\,d\sigma
 +\int_s^t\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\,d\sigma \\
&=F_1(s,t)+F_2(s,t).
\end{aligned}\label{e3.14}
\end{equation}
Write $F_1(s,t)$ as
\begin{equation}
F_1(s,t)=\int_0^s\int_s^t\frac{\partial^2}{\partial\tau^2}e^{-(\tau-\sigma)A}d\tau
f(\sigma)\,d\sigma. \label{e3.15}
\end{equation}
Similarl to \eqref{e2.5} it can be shown that for
$u\in(X,D(A))_{\theta,\infty}$,
\[
\|\frac{d^2}{dt^2}e^{-tA}u\|_X\leq C_0t^{(\beta-3+\theta)/\alpha}
\|u\|_{(X,D(A))_{\theta,\infty}}.
\]
Hence
\begin{equation}
\begin{aligned}
\|\frac{\partial^2}{\partial\tau^2}e^{-(\tau-\sigma)A}f(\sigma)\|_X
&\leq C_0(\tau-\sigma)^{(\beta-3+\theta)/\alpha}\|f(\sigma)\|_{(X,D(A))_{\theta,\infty}}\\
&\leq C_0(\tau-\sigma)^{(\beta-3+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{aligned} \label{e3.16}
\end{equation}
Therefore
\begin{equation}
\begin{aligned}
&\|F_1(s,t)\|_X\\
&\leq C_0\int_0^s\int_s^t(\tau-\sigma)^{(\beta-3+\theta)/\alpha}
 \|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}d\tau \,d\sigma\\
&\leq \frac{C_0(t-s)^{(2\alpha+\beta-3+\theta)/\alpha}}{(3-\alpha-\beta-\theta)
 /\alpha\cdot(2\alpha+\beta-3+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{aligned}\label{e3.17}
\end{equation}
In view of \eqref{e3.10} one obtains
\begin{equation}
\begin{aligned}
\|F_2(s,t)\|_X
&\leq C_0\int_s^t(t-\sigma)^{(\beta-2+\theta)/\alpha}
 \|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}\,d\sigma\\
&\leq C_0\frac{(t-s)^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{aligned} \label{e3.18}
\end{equation}
From \eqref{e3.14}, \eqref{e3.17} and \eqref{e3.18} it follows that
\[
y_2'-f\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X).
\]
This and \eqref{e3.6} yield
\begin{equation}
 y'-f\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X). \label{e3.19}
\end{equation}
From \eqref{e3.13} one deduces that for $\tau>0$,
\begin{align*}
\frac d{d\tau}e^{-\tau A}(y_2'(t)-f(t))
&=\frac d{d\tau}e^{-\tau A}\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}f(s)ds\\
&=\int_0^t\frac d{d\tau}e^{-\tau A}\frac{\partial}{\partial t}e^{-(t-s)A}f(s)ds\\
&=\int_0^t\frac{\partial^2}{\partial t^2}e^{-(t-s+\tau)A}f(s)ds.
\end{align*}
With the aid of \eqref{e3.16} one obtains
\begin{align*}
&\|\frac d{d\tau}e^{-\tau A}(y_2'(t)-f(t))\|_X\\
&=\|\int_0^t\frac{\partial^2}{\partial t^2}e^{-(t-s+\tau)A}f(s)ds\|_X\\
&\leq C_0\int_0^t(t-s+\tau)^{(\beta-3+\theta)/\alpha}
 \|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}ds\\
&=C_0\frac{\tau^{(\alpha+\beta-3+\theta)/\alpha}-(t+\tau)^{(\alpha+\beta-3+\theta)/\alpha}}
{(3-\alpha-\beta+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}\\
&\leq\frac{C_0\tau^{(\alpha+\beta-3+\theta)/\alpha}}{(3-\alpha-\beta+\theta)/\alpha}
\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{align*}
This means that $y_2'(t)-f(t)\in {\widetilde X}_A^{\alpha+\theta-1}$ and
\begin{equation}\label{e3.20}
|y_2'(t)-f(t)|_{{\widetilde X}_A^{\alpha+\theta-1}}
\leq \frac{C_0}{(3-\alpha-\beta-\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{equation}
For $\tau>0$,
\begin{equation}\label{e3.21}
\begin{aligned}
e^{-\tau A}\big[y_2'(t)-f(t)\big]
&=e^{-\tau A}\int_0^t\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\,d\sigma\\
&=\int_0^t\frac{\partial}{\partial t}e^{-(t-\sigma+\tau)A}f(\sigma)\,d\sigma.
\end{aligned}
\end{equation}
As $\tau\to 0$ for $0<\sigma<t$,
\begin{equation}\label{e3.22}
\|\frac{\partial}{\partial t}e^{-(t-\sigma+\tau)A}f(\sigma)
-\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\|_{X}\to 0,
\end{equation}
in view of \eqref{e3.10},
\begin{equation}
\begin{aligned}
\|\frac{\partial}{\partial t}e^{-(t-\sigma+\tau)A}f(\sigma)\|_{X}
&\leq C_0(t-\sigma+\tau)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}\\
&\leq C_0(t-\sigma)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})},
\end{aligned} \label{e3.23}
\end{equation}
and
\begin{equation}\label{e3.24}
\int_0^t(t-\sigma)^{(\beta-2+\theta)/\alpha}\,d\sigma
=\frac{t^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}<\infty.
\end{equation}
It follows from \eqref{e3.21}, \eqref{e3.22}, \eqref{e3.23} and \eqref{e3.24}
 that as $\tau\to 0$,
\begin{equation}\label{e3.25}
e^{-\tau A}\big[y_2'(t)-f(t)\big]\to
\int_0^t\frac{\partial}{\partial t}e^{-(t-\sigma)A}f(\sigma)\,d\sigma=y_2'(t)-f(t).
\end{equation}
Therefore, noting $\alpha+\beta+(\alpha+\theta-1)=2\alpha+\beta+\theta-1>2$, $0<\alpha+\theta-1<1$
we can apply Lemma \ref{lem2.5} and \eqref{e3.20} to obtain that
$y_2'(t)-f(t)\in X_A^{(2\alpha+\beta-3+\theta)/\alpha}$ and
\begin{align*}
|y_2'(t)-f(t)|_{X_A^{(2\alpha+\beta-3+\theta)/\alpha}}
&\leq\Gamma((2\alpha+\beta-3+\theta)/\alpha)|y_2'(t)-f(t)|_{{\widetilde X}_A^{\alpha+\theta-1}} \\
&\leq\frac{C_0\Gamma((2\alpha+\beta-3+\theta)/\alpha)}{(3-\alpha-\beta-\theta)/\alpha}
\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{align*}
From this inequality and the boundedness of $\|y_2'(t)-f(t)\|_X$,
which follows from \eqref{e3.11}, and \eqref{e3.13} one concludes that
$y_2'-f\in B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha})$. This and \eqref{e3.9}
imply $y'-f\in B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha})$.
Finally we show that $y$ defined by \eqref{e3.2} satisfies \eqref{e1.1}. By virtue of \eqref{e3.13} one has
\[
 y'(t)=y_1'(t)+y_2'(t)=\frac d{dt}e^{-tA}y_0+f(t)
+\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}f(s)ds.
\]
Hence
\begin{align*}
A^{-1}y'(t)
&=A^{-1}\frac d{dt}e^{-tA}y_0+A^{-1}f(t)
 +\int_0^tA^{-1}\frac{\partial}{\partial t}e^{-(t-s)A}f(s)ds\\
&=-e^{-tA}y_0+A^{-1}f(t)-\int_0^te^{-(t-s)A}f(s)ds.
\end{align*}
Therefore, one obtains
\[
A^{-1}(y'(t)-f(t))=-e^{-tA}y_0-\int_0^te^{-(t-s)A}f(s)ds=-y(t).
\]
From this the assertion \eqref{e1.1} readily follows.
Thus the proof complete.
\end{proof}

\begin{theorem}  \label{thm3.2}
Suppose that $\alpha+\beta>3/2$, $\theta>2(2-\alpha-\beta)$, $y_0\in D(A)$,
$Ay_0\cap{\widetilde X}_A^{\theta}\ne\emptyset$,
$f\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$ and
$\lim_{\tau\to 0}e^{-\tau A}f(t)=f(t)$ for every $t\in[0,T]$.
Then problem \eqref{e1.1} admits a unique solution $y$ such that
$y\in C^1([0,T];X)$ and
\begin{equation}\label{e3.26}
y'-f\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)
\cap B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\end{equation}
\end{theorem}

\begin{proof}
Define functions $y_1$ and $y_2$ by \eqref{e3.2}.
Let $\phi\in Ay_0\cap{\widetilde X}_A^{\theta}$. Since
\[
\alpha+\beta-2+\theta>\alpha+\beta-2+2(2-\alpha-\beta)=2-\alpha-\beta\geq0,
\]
we can show as for \eqref{e3.6} and \eqref{e3.7} that
\begin{equation}
\begin{gathered}
|y_1'|_{C^{(\alpha+\beta-2+\theta)/\alpha}([0,T];X)}
\leq \frac{|\phi|_{{\widetilde X}_A^{\theta}}}{(\alpha+\beta-2+\theta)/\alpha}, \label{e3.27}\\
|y_1'(t)|_{{\widetilde X}_A^{\theta}}\leq|\phi|_{{\widetilde X}_A^{\theta}},
\end{gathered}
\end{equation}
and furthermore \eqref{e3.8} holds. Consequently the following statements
are obtained as \eqref{e3.9} was
\begin{gather}
y_1'\in B([0,T];X_A^{(\alpha+\beta-2+\theta)/\alpha}), \label{e3.28}\\
\|y_1'\|_{B([0,T];X_A^{(\alpha+\beta-2+\theta)/\alpha})}
\leq \Gamma((\alpha+\beta-2+\theta)/\alpha)|\phi|_{{\widetilde X}_A^{\theta}}
+\sup_{0\leq t\leq T}\|e^{-tA}\phi\|_{X}. \label{e3.29}
\end{gather}
The second term in the right-hand side of \eqref{e3.29} is finite since
$\lim_{t\to 0}e^{-tA}\phi$ exists by Lemma \ref{lem2.5}.

In the present case the following inequality holds instead of \eqref{e3.10},
\begin{equation}
\begin{aligned}
\|\frac{\partial}{\partial t}e^{-(t-s)A}f(s)\|_X
&\leq(t-s)^{(\beta-2+\theta)/\alpha}|f(s)|_{{\widetilde X}_A^{\theta}}\\
&\leq(t-s)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{aligned} \label{e3.30}
\end{equation}
Noting that \eqref{e3.12} holds since $e^{-(\tau-s)A}f(s)\to  f(s)$ as
$\tau\to  s$ by assumption, it is possible to show that
$y_2(t)=\int_0^te^{-(t-s)A}f(s)ds$ is differentiable and \eqref{e3.13} holds.
 Let $F_1(s,t)$ and $F_2(s,t)$ be functions defined by \eqref{e3.14}.
We have for $u\in{\widetilde X}_A^{\theta}$,
\begin{align*}
\|\frac{d^2}{dt^2}e^{-t A}u\|_X
&=4\|\big(\frac d{dt}e^{-(t/2)A}\big)^2u\|_X
 \leq4\|\frac d{dt}e^{-(t/2)A}\|_{\mathcal{L}(X)}\|\frac d{dt}e^{-(t/2)A}u\|_X \\
&\leq 4\frac{C_0}2\big(\frac t2\big)^{(\beta-2)/\alpha}
 \frac12\left(\frac t2\right)^{(\beta-2+\theta)/\alpha}|u|_{{\widetilde X}_A^{\theta}}\\
&=C_1t^{(2\beta-4+\theta)/\alpha}|u|_{{\widetilde X}_A^{\theta}},
\end{align*}
where $C_1=C_02^{(4-2\beta-\theta)/\alpha}$. Hence
\begin{equation}
\begin{aligned}
\|\frac{d^2}{d\tau^2}e^{-(\tau-\sigma)A}f(\sigma)\|_X
&\leq C_1(\tau-\sigma)^{(2\beta-4+\theta)/\alpha}|f(\sigma)|_{{\widetilde X}_A^{\theta}}\\
&\leq C_1(\tau-\sigma)^{(2\beta-4+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{aligned} \label{e3.31}
\end{equation}
With the aid of \eqref{e3.15} and \eqref{e3.31} we obtain
\begin{equation}
\begin{aligned}
&\|F_1(s,t)\|_X\\
&\leq C_1\int_0^s\int_s^t(\tau-\sigma)^{(2\beta-4+\theta)/\alpha}
 \|f\|_{B([0,T];{\widetilde X}_A^{\theta})}d\tau \,d\sigma\\
&\leq \frac{C_1\alpha^2}{(4-\alpha-2\beta-\theta)(2\alpha+2\beta-4+\theta)}
 (t-s)^{(2\alpha+2\beta-4+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{aligned} \label{e3.32}
\end{equation}
Using \eqref{e3.30}, one obtains
\begin{equation}
\begin{aligned}
\|F_2(s,t)\|_X
&\leq\int_s^t(t-\sigma)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}
 \,d\sigma\\
&=\frac{(t-s)^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
 \|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{aligned} \label{e3.33}
\end{equation}
Inequalities \eqref{e3.32} and \eqref{e3.33} imply that
$y_2'-f\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)$.
This and \eqref{e3.27} yield $y'-f\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)$.
With the aid of \eqref{e3.31} one observes that
\begin{align*}
\|\frac d{d\tau}e^{-\tau A}(y_2'(t)-f(t))\|_X
&=\|\int_0^t\frac{\partial^2}{\partial t^2}e^{-(t-s+\tau)A}f(s)ds\|_X\\
&\leq C_1\int_0^t(t-s+\tau)^{(2\beta-4+\theta)/\alpha}\|f\|_{B([0,T];
 {\widetilde X}_A^{\theta})}ds\\
&=C_1\frac{\tau^{(\alpha+2\beta-4+\theta)/\alpha}-(t+\tau)^{(\alpha+2\beta-4+\theta)/\alpha}}
 {(4-\alpha-2\beta+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}\\
&\leq\frac{C_1\tau^{(\alpha+2\beta-4+\theta)/\alpha}}{(4-\alpha-2\beta+\theta)/\alpha}
 \|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{align*}
This means that $y_2'(t)-f(t)\in {\widetilde X}_A^{\alpha+\beta+\theta-2}$ and
\begin{equation}\label{e3.34}
|y_2'(t)-f(t)|_{{\widetilde X}_A^{\alpha+\beta+\theta-2}}
\leq \frac{C_1}{(4-\alpha-2\beta+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{equation}
From \eqref{e3.21}, \eqref{e3.22}, \eqref{e3.24} and the inequality
\begin{align*}
\|\frac{\partial}{\partial t}e^{-(t-\sigma+\tau)A}f(\sigma)\|_{X}
&\leq(t-\sigma+\tau)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})}\\
&\leq C_1(t-\sigma)^{(\beta-2+\theta)/\alpha}\|f\|_{B([0,T];{\widetilde X}_A^{\theta})},
\end{align*}
which follows from \eqref{e3.30}. Then \eqref{e3.25} holds.
Therefore noting $\alpha+\beta+(\alpha+\beta+\theta-2)=2\alpha+2\beta+\theta-2>2$ we may apply
Lemma \ref{lem2.5} and \eqref{e3.34} to obtain that
$y_2'(t)-f(t)\in X_A^{(2\alpha+2\beta-4+\theta)/\alpha}$ and
\begin{align*}
|y_2'(t)-f(t)|_{X_A^{(2\alpha+2\beta-4+\theta)/\alpha}}
&\leq\Gamma((2\alpha+2\beta-4+\theta)/\alpha)|y_2'(t)-f(t)|_{{\widetilde X}_A^{\alpha+\beta+\theta-2}} \\
&\leq\frac{C_1\Gamma((2\alpha+2\beta-4+\theta)/\alpha)}{(4-\alpha-2\beta+\theta)/\alpha}
 \|f\|_{B([0,T];{\widetilde X}_A^{\theta})}.
\end{align*}
From this inequality, \eqref{e3.13} and \eqref{e3.30} one concludes that
\[
y_2'-f\in B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\]
This and \eqref{e3.28} imply $y'-f\in B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha})$.
The proof that $y$ defined by \eqref{e3.2} satisfies \eqref{e1.1} is the same
 as that in the proof of Theorem \ref{thm3.1}.
\end{proof}

\begin{remark} \label{rmk3.1} \rm
Suppose that the assumptions of Theorem \ref{thm3.1} are satisfied, and
let $k\in{\mathbb R}$. Consider the  problem
\begin{equation}\label{e3.35}
\begin{gathered}
\frac d{dt}y(t)+Ay(t)+ky(t)\ni f(t),\quad 0\leq t\leq T,\\
y(0)=y_0.
\end{gathered}
\end{equation}
The operator $-(A+k)$ generates a differentiable semigroup
$e^{-t(A+k)}=e^{-kt}e^{-tA}$.
Then define $C_0'=C_0e^{\max\{k,0\}T}$, so that for $0<t\leq T$ one gets
\[
\|\frac{d^i}{dt^i}e^{-t(A+k)}\|_{\mathcal{L}(X)}
\leq C_0't^{(\beta-i-1)/\alpha},\quad i=0,1,2.
\]
The unique solution to \eqref{e3.35} is given by
\[
y(t)=e^{-t(A+k)}y_0+\int_0^te^{-(t-s)(A+k)}f(s)ds.
\]
Since $e^{kt}y(t)$ is a solution to \eqref{e1.1} with $f(t)$ replaced by
$e^{kt}f(t)$, $y$ satisfies
$y\in C^1([0,T];X)$ and
\[
y'+ky-f\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)
\cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\]
Furthermore, the following results hold just as \eqref{e3.13} and \eqref{e3.11}:
\begin{gather*}
\frac d{dt}\int_0^te^{-(t-s)(A+k)}f(s)ds
=f(t)+\int_0^t\frac{\partial}{\partial t}e^{-(t-s)(A+k)}f(s)ds,\\
\|\int_0^t\frac{\partial}{\partial t}e^{-(t-s)(A+k)}f(s)ds\|_X
\leq \frac{C_0't^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
\|f\|_{B([0,T];(X,D(A))_{\theta,\infty})}.
\end{gather*}
An analogous remark holds also for Theorem \ref{thm3.2}.
\end{remark}

\section{An identification problem for the differential equation}

In this section we consider problem \eqref{e1.4}:
\begin{gather}
\frac d{dt}y(t)+Ay(t)\ni f(t)z+h(t),\quad t\in[0,T], \label{e4.1}\\
y(0)=y_0, \label{e4.2}\\
\Phi[y(t)]=g(t),\quad t\in[0,T], \label{e4.3}
\end{gather}

\begin{theorem} \label{thm4.1}
Suppose that $2\alpha+\beta+\theta>3$, $y_0\in D(A)$,
$Ay_0\cap(X,D(A))_{\theta,\infty}\ne\emptyset$, $z\in (X,D(A))_{\theta,\infty}$,
$h\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$,
$g\in C^1([0,T];\mathbb{C})$, $\Phi\in X^*$, $\Phi[y_0]=g(0)$, $\Phi[z]\ne0$.
Then problem \eqref{e4.1}-\eqref{e4.3} admits a unique solution $(y,f)$ such that
\begin{equation}\label{e4.4}
\begin{gathered}
y\in C^1([0,T];X),\quad f\in C([0,T];\mathbb{C}),\\
y'-f(\cdot)z-h\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)
\cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
Supposing that $f\in C([0,T];\mathbb{C})$ is known, define a function $y$ by
\begin{equation}
y(t)=e^{-tA}y_0+\int_0^tf(s)e^{-(t-s)A}zds+\int_0^te^{-(t-s)A}h(s)ds. \label{e4.5}
\end{equation}
Since $\alpha+\beta+\theta-2\geq2\alpha+\beta+\theta-3>0$, by Lemma \ref{lem2.3} and \eqref{e2.5}
the following statements hold:
\begin{equation}
e^{-tA}z\to  z\text{ as }t\to 0,\quad
\|\frac d{dt}e^{-tA}z\|_X\leq C_0t^{(\beta-2+\theta)/\alpha}\|z\|_{(X,D(A))_{\theta,\infty}}. \label{e4.6}
\end{equation}
Hence $\int_0^tf(s)e^{-(t-s)A}zds$ is differentiable and
\begin{equation}
\frac d{dt}\int_0^tf(s)e^{-(t-s)A}zds
=f(t)z+\int_0^tf(s)\frac{\partial}{\partial t}e^{-(t-s)A}zds. \label{e4.7}
\end{equation}
According to the proof of Theorem \ref{thm3.1} (cf. \eqref{e3.13} and \eqref{e3.11})
$\int_0^te^{-(t-s)A}h(s)ds$ is differentiable and
\begin{gather}
\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds
=h(t)+\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}h(s)ds, \label{e4.8}\\
\|\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}h(s)ds\|_X
\leq C_0\frac{t^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
\|h\|_{B([0,T];(X,D(A))_{\theta,\infty})}. \label{e4.9}
\end{gather}
Hence $y$ is differentiable and
\begin{equation}
\frac d{dt}y(t)
=\frac d{dt}e^{-tA}y_0+f(t)z+\int_0^tf(s)
\frac{\partial}{\partial t}e^{-(t-s)A}zds
+\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds. \label{e4.10}
\end{equation}
Assuming that $y(t)$ satisfies \eqref{e4.3} one deduces from \eqref{e4.10} the following identity
\begin{equation}
\begin{aligned}
g'(t)&=\Phi\Big[\frac d{dt}e^{-tA}y_0\Big]+f(t)\Phi[z]
   +\int_0^tf(s)\Phi\Big[\frac{\partial}{\partial t}e^{-(t-s)A}z\Big]ds\\
&\quad +\Phi\Big[\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds\Big].
\end{aligned}\label{e4.11}
\end{equation}
Rewriting \eqref{e4.11} one obtains the following integral equation to
be satisfied by $f$:
\begin{equation}
\begin{aligned}
&f(t)+\chi\int_0^tf(s)\Phi\Big[\frac{\partial}{\partial t}e^{-(t-s)A}z\Big]ds\\
&=\chi g'(t)-\chi\Phi\Big[\frac d{dt}e^{-tA}y_0\Big]
-\chi\Phi\Big[\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds\Big],
\end{aligned} \label{e4.12}
\end{equation}
where $\chi=\Phi[z]^{-1}$. Set
\begin{equation}\label{e4.13}
\begin{gathered}
\kappa(t)=\chi\Phi\Big[ {\frac d{dt}}e^{-tA}z\Big],\\
 {\psi(t)=\chi g'(t)-\chi\Phi\Big[\frac d{dt}e^{-tA}y_0\Big]
-\chi\Phi\Big[\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds\Big]}.
\end{gathered}
\end{equation}
Then \eqref{e4.12} is rewritten as
\[
f(t)+\int_0^t\kappa(t-s)f(s)ds=\psi(t),
\]
or briefly
\begin{equation}
f+\kappa*f=\psi. \label{e4.14}
\end{equation}
By  \eqref{e4.6} one has
\[
|\kappa(t)|\leq C_0|\chi|\|\Phi\|t^{(\beta-2+\theta)/\alpha}
\|z\|_{(X,D(A))_{\theta,\infty}}.
\]
In view of \eqref{e3.5}, \eqref{e4.8} and \eqref{e4.9} one observes
that $\psi\in C([0,T];\mathbb{C})$. Let $r$ be the solution to the integral equation
\[
\kappa+r+r*\kappa=0.
\]
This equation is solved by successive approximations, and the solution $r$ satisfies
\[
\kappa+r+\kappa*r=0,\quad |r(t)|\leq C_2t^{(\beta-2+\theta)/\alpha},\quad C_2>0.
\]
The integral equation \eqref{e4.14} admits a unique solution
$f\in C([0,T];\mathbb{C})$ given by
$f=\psi+r*\psi$,
or
\begin{equation}
f(t)=\psi(t)+\int_0^tr(t-s)\psi(s)ds. \label{e4.15}
\end{equation}
It is easy to verify that if we define $y$ by \eqref{e4.5} with $f$
given by \eqref{e4.15}, the pair $(y,f)$ satisfies \eqref{e4.1} and \eqref{e4.2}.
From \eqref{e4.10} and \eqref{e4.11} it follows that
\[
g'(t)=\Phi\Big[\frac d{dt}y(t)\Big]=\frac d{dt}\Phi[y(t)].
\]
From this equality and the compatibility condition $\Phi[y_0]=g(0)$
\eqref{e4.3} follows. Since
$f(\cdot)z+h\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$,
the assertion \eqref{e4.4} follows from Theorem \ref{thm3.1}.
\end{proof}

\begin{theorem} \label{thm4.2}
Suppose that $\alpha+\beta>3/2$, $2(2-\alpha-\beta)<\theta<1$,
$y_0\in D(A)$, $Ay_0\cap{\widetilde X}_A^{\theta}\ne\emptyset$,
$z\in{\widetilde X}_A^{\theta}$, $\lim_{t\to 0}e^{-tA}z=z$,
$g\in C^1([0,T];\mathbb{C})$, $\Phi\in X^*$, $\Phi[y_0]=g(0)$,
$\Phi[z]\ne0$, $h\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$,
$\lim_{\tau\to 0}e^{-\tau A}h(t)=h(t)$ for every $t\in[0,T]$.
Then problem \eqref{e4.1}-\eqref{e4.3} admits a unique solution $y$ such that
$y\in C^1([0,T];X)$ and
\[
y'-f\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)\cap B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\]
\end{theorem}

\begin{proof}
Supposing that $f\in C([0,T];X)$ is known, define the function $y$ by \eqref{e4.5}.
Let $\phi\in Ay_0\cap{\widetilde X}_A^{\theta}$. Then in view of Lemma \ref{lem2.5},
\[
\frac d{dt}e^{-tA}y_0=\frac d{dt}e^{-tA}A^{-1}\phi=-e^{-tA}\phi
\]
converges as $t\to 0$. Hence
\begin{equation}\label{e4.16}
\text{the function $t\mapsto\frac d{dt}e^{-tA}y_0$ belongs to }C([0,T];X).
\end{equation}
By the definition of ${\widetilde X}_A^{\theta}$ one has
\begin{equation}
\|\frac d{dt}e^{-tA}z\|_X\leq t^{(\beta-2+\theta)/\alpha}|z|_{{\widetilde X}_A^{\theta}}.
\label{e4.17}
\end{equation}
Hence using the assumption $\lim_{t\to 0}e^{-tA}z=z$ one observes that
$\int_0^tf(s)e^{-(t-s)A}zds$ is differentiable and \eqref{e4.7} holds.
According to the proof of Theorem \ref{thm3.2} (cf. \eqref{e3.30})
$\int_0^te^{-(t-s)A}h(s)ds$ is differentiable, and equality \eqref{e4.8} and the inequality
\begin{equation}\label{e4.18}
\|\int_0^t\frac{\partial}{\partial t}e^{-(t-s)A}h(s)ds\|_X
\leq \frac{t^{(\alpha+\beta-2+\theta)/\alpha}}{(\alpha+\beta-2+\theta)/\alpha}
\|h\|_{B(0,T;{\widetilde X}_A^{\theta})}
\end{equation}
holds.
Hence $y$ is differentiable and \eqref{e4.10} holds. Assuming that $y(t)$
satisfies \eqref{e4.3} one deduces \eqref{e4.14} from \eqref{e4.10} as in the
proof of Theorem \ref{thm4.1}, where $\kappa$ and $\psi$ are functions defined by
\eqref{e4.13}. By virtue of \eqref{e4.17} one has
\[
|\kappa(t)|\leq |\chi|\|\Phi\|_{X^*}t^{(\beta-2+\theta)/\alpha}
|z|_{{\widetilde X}_A^{\theta}}.
\]
In view of \eqref{e4.16}, \eqref{e4.8} and \eqref{e4.18} one observes
$\psi\in C([0,T];\mathbb{C})$. The remaining part of the proof is the
same as that of Theorem \ref{thm4.1}.
\end{proof}

\section{Equations with several unknown scalar functions}

In this section we consider the problem consisting of recovering several
unknown scalar functions $f_1,\dots,f_n$ and a vector function $y$ such that
\begin{gather}
\frac d{dt}y(t)+Ay(t)\ni \sum_{j=1}^nf_j(t)z_j+h(t),\quad t\in[0,T], \label{e5.1}\\
 y(0)=y_0, \label{e5.2}\\
\Phi_j[y(t)]=g_j(t),\quad j=1,\dots,n,\quad t\in[0,T]. \label{e5.3}
\end{gather}

\begin{theorem} \label{thm5.1}
Suppose $2\alpha+\beta+\theta>3$, $y_0\in D(A)$,
$Ay_0\cap(X,D(A))_{\theta,\infty}\ne\emptyset$,
$z_j\in (X,D(A))_{\theta,\infty}$, $j=1,\dots,n$,
$h\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$,
$g_j\in C^1([0,T];\mathbb{C})$, $\Phi_j\in X^*$,
$\Phi_j[y_0]=g_j(0),\,j=1,\dots,n$, and
\begin{equation}
\det \begin{pmatrix}\Phi_1[z_1] & \dots &\Phi_1[z_n]\\
\dots& & \dots \\
\Phi_n[z_1]&\dots&\Phi_n[z_n]\end{pmatrix} \ne0.
 \label{e5.4}
\end{equation}
Then, problem \eqref{e5.1}--\eqref{e5.3} admits a unique solution
$(y,f_1,\dots,f_n)$ such that
\begin{equation}
\begin{gathered}
y\in C^1([0,T];X),\quad f_1,\dots,f_n\in C([0,T];\mathbb{C}), \\
y'-\sum_{j=1}^nf_j(\cdot)z_j-h\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)
 \cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\end{gathered} \label{e5.5}
\end{equation}
\end{theorem}

The proof is performed in parallel to the proof of Theorem \ref{thm4.1}.
If $f_1,\dots,f_n\in C([0,T];\mathbb{C})$ are known, $y$ is given by
\[
 y(t)=e^{-tA}y_0+\int_0^t\sum_{j=1}^nf_j(s)e^{-(t-s)A}z_jds
+\int_0^te^{-(t-s)A}h(s)ds.
\]
Just as the proof of \eqref{e4.10} one deduces from this equality
\begin{equation}
\begin{aligned}
y'(t)&=\frac d{dt}e^{-tA}y_0+\sum_{j=1}^nf_j(t)z_j\\
&\quad +\int_0^t\sum_{j=1}^nf_j(s)D_te^{-(t-s)A}z_jds
+\frac d{dt}\int_0^te^{-(t-s)A}h(s)ds.
\end{aligned}\label{e5.6}
\end{equation}
It follows from \eqref{e5.3} and \eqref{e5.6} that
\begin{align*}
g_i'(t)&=\Phi_i[y'(t)] \\
&=\Phi_i\big[D_te^{-tA}y_0\big]+\sum_{j=1}^nf_j(t)\Phi_i[z_j]
 +\int_0^t\sum_{j=1}^nf_j(s)\Phi_i[D_te^{-(t-s)A}z_j]ds \\
&\quad +\Phi_i\Big[D_t\int_0^te^{-(t-s)A}h(s)ds\Big].
\end{align*}
This is rewritten as
\begin{equation}
\begin{aligned}
&\begin{pmatrix}
\Phi_1[z_1] & \dots &\Phi_1[z_n]\\
\dots && \dots\\
\Phi_n[z_1]&\dots& \Phi_n[z_n]
\end{pmatrix}
\begin{pmatrix} f_1(t)\\ \dots \\ f_n(t)\end{pmatrix} \\
&=\begin{pmatrix}
g_1'(t)-\Phi_1[D_te^{-tA}y_0]-\Phi_1\big[D_t \int_0^te^{-(t-s)A}h(s)ds\big]\\
\dots\\
g_n'(t)-\Phi_n[D_te^{-tA}y_0]-\Phi_n\big[D_t \int_0^te^{-(t-s)A}h(s)ds\big]
\end{pmatrix} \\
&\quad \int_0^t \begin{pmatrix}
\Phi_1[D_te^{-(t-s)A}z_1]&\dots&\Phi_1[D_te^{-(t-s)A}z_n]\\
\dots& & \dots \\
\Phi_n[D_te^{-(t-s)A}z_1]&\dots &\Phi_n[D_te^{-(t-s)A}z_n]
\end{pmatrix}
\begin{pmatrix} f_1(s)\\ \dots\\ f_n(s)\end{pmatrix}\,ds.
\end{aligned}\label{e5.7}
\end{equation}
Set
\[
\mathcal{A}=\begin{pmatrix}
\Phi_1[z_1]&\dots&\Phi_1[z_n]\\
\dots&&\dots \\
\Phi_n[z_1]&\dots&\Phi_n[z_n]
\end{pmatrix}.
\]
Then by assumption \eqref{e5.4}, $\mathcal{A}^{-1}$ exists. Set
\begin{gather*}
\Phi(t)=\mathcal{A}^{-1}\begin{pmatrix}
g_1'(t)-\Phi_1[D_te^{-tA}y_0]
 -\Phi_1\big[D_t \int_0^te^{-(t-s)A}h(s)ds\big]\\
\dots\\
g_n'(t)-\Phi_n[D_te^{-tA}y_0]-\Phi_n
\big[D_t \int_0^te^{-(t-s)A}h(s)ds\big]\end{pmatrix},\\
K(t)=\mathcal{A}^{-1}\begin{pmatrix}
\Phi_1[D_te^{-tA}z_1]&\dots& \Phi_1[D_te^{-tA}z_n]\\
\dots&& \dots \\
\Phi_n[D_te^{-tA}z_1]&\dots&\Phi_n[D_te^{-tA}z_n]
\end{pmatrix},\quad
F(t)=\begin{pmatrix} f_1(t)\\
\dots\\
f_n(t) \end{pmatrix}.
\end{gather*}
Tt follows from \eqref{e5.7} that
\[
F(t)=\Phi(t)-\int_0^tK(t-s)F(s)\,ds.
\]
Since
\[
\|K(t)\|_{\mathcal{L}(\mathbb{C}^n)}\leq Ct^{(\beta-2+\theta)/\alpha},
\]
the remaining part of the proof is the same as that of Theorem \ref{thm4.1}.
Analogously the following theorem is established.

\begin{theorem} \label{thm5.2}
Suppose $2(2-\alpha-\beta)<\theta<1$, $y_0\in D(A)$,
$Ay_0\cap{\widetilde X}_A^{\theta}\ne\emptyset$,
$z_j\in {\widetilde X}_A^{\theta}$,  $\lim_{t\to 0}e^{-tA}z_j=z_j$,
$j=1,\dots,n$, $h\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$,
$\lim_{\tau\to 0}e^{-\tau A}h(t)=h(t)$ for every $t\in[0,T]$,
$g_j\in C^1([0,T];\mathbb{C}),\,\Phi_j\in X^*$,
$\Phi_j[y_0]=g_j(0),\,j=1,\dots,n$, and
\[
\det \begin{pmatrix}
\Phi_1[z_1]&\dots&\Phi_1[z_n]\\
\dots&&\dots\\
\Phi_n[z_1]&\dots&\Phi_n[z_n]
\end{pmatrix}\ne0.
\]
Then, problem \eqref{e5.1}-\eqref{e5.3} admits a unique solution
$(y,f_1,\dots,f_n)$ such that
\begin{gather*}
y\in C^1([0,T];X),\quad f_1,\dots,f_n\in C([0,T];\mathbb{C}),\\
y'-\sum_{j=1}^nf_j(\cdot)z_j-h\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)
\cap B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\end{gather*}
\end{theorem}

\section{Problems \eqref{e1.6} and \eqref{e1.7}}

Let $L$ and $M$ be two linear closed operators satisfying \eqref{e1.8}
and \eqref{e1.9}. Set $A=LM^{-1}$. Namely
\begin{equation}\label{e6.1}
\begin{gathered}
D(A)=MD(L)=\{Mu: u\in D(L)\},\\
Ay=\{Lu: y=Mu,\, u\in D(L)\}\text{ for } y\in D(A).
\end{gathered}
\end{equation}
It is shown in  Favini and  Yagi \cite{faviyagi} that $A$ satisfies
\eqref{e1.2} and \eqref{e1.3}. The graph-norm of $D(A)$ is defined by
\[
\|y\|_{D(A)}=\inf\{\|Lu\|_X: y=Mu,\ u\in D(L)\}\quad{\rm for}\quad y\in D(A).
\]
Consider the problem
\begin{gather}
\frac d{dt}Mu(t)+Lu(t)=f(t),\quad t\in[0,T], \label{e6.2}\\
Mu(0)=Mu_0. \label{e6.3}
\end{gather}

\begin{theorem} \label{thm6.1}
Suppose that $2\alpha+\beta+\theta>3$, $u_0\in D(L)$,
$Lu_0\in(X,D(A))_{\theta,\infty}$,
$f\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$.
Then problem \eqref{e6.2}--\eqref{e6.3} admits a unique solution $u$ such that
\begin{equation}\label{e6.4}
\begin{gathered}
Mu\in C^1([0,T];X),\\
Lu\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)\cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
Set $y_0=Mu_0$. Then $y_0\in D(A)$, and $Ay_0\cap(X,D(A))_{\theta,\infty}$
is not empty, since it contains $Lu_0$. In view of Theorem \ref{thm3.1} there
exists a solution $y$ to \eqref{e1.1}, $A$ being defined by \eqref{e6.1}.
Set $u(t)=L^{-1}(f(t)-y'(t))$. Then
\begin{equation}\label{e6.5}
Lu(t)=f(t)-y'(t)\in Ay(t)=LM^{-1}y(t).
\end{equation}
Since $L$ is bijective, it follows from \eqref{e6.5} that $u(t)\in M^{-1}y(t)$,
or $y(t)=Mu(t)$. From the first equation of \eqref{e6.5} equation \eqref{e6.2} follows.
It is obvious that $u$ satisfies \eqref{e6.3}.
\end{proof}

Analogously, using Theorem \ref{thm3.2} instead of Theorem \ref{thm3.1}, the following theorem
is  obtained.

\begin{theorem} \label{thm6.2}
Suppose that $\theta>2(2-\alpha-\beta)$, $u_0\in D(L)$,
$Lu_0\in{\widetilde X}_A^{\theta}$,
$f\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$,
$\lim_{\tau\to 0}e^{-\tau A}f(t)=f(t)$ for every $t\in[0,T]$.
Then problem \eqref{e6.2}--\eqref{e6.3} admits a unique solution $u$ such that
\begin{equation}\label{e6.6}
\begin{gathered}
Mu\in C^1([0,T];X),\\
Lu\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)\cap B([0,T];X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\end{gathered}
\end{equation}
\end{theorem}

Next, consider the  problem
\begin{gather}
\frac d{dt}Mu(t)+Lu(t)=\sum_{j=1}^nf_j(t)z_j+h(t),\quad t\in[0,T], \label{e6.7}\\
Mu(0)=Mu_0, \label{e6.8}\\
\Phi_j[Mu(t)]=g_j(t),\quad j=1,\dots,n,\quad t\in[0,T]. \label{e6.9}
\end{gather}

\begin{theorem} \label{thm6.3}
Suppose $2\alpha+\beta+\theta>3$, $u_0\in D(L)$, $Lu_0\in(X,D(A))_{\theta,\infty}$,
$z_j\in (X,D(A))_{\theta,\infty}$, $j=1,\dots,n$,
$h\in C([0,T];X)\cap B([0,T];(X,D(A))_{\theta,\infty})$, $\Phi_j\in X^*$,
$g_j\in C^1([0,T;\mathbb{C})$,
$\Phi_j[Mu_0]=g_j(0)$, $j=1,\dots,n$, and \eqref{e5.4} holds.
Then problem \eqref{e6.7}--\eqref{e6.9} admits a unique solution
$(u,f_1,\dots,f_n)$ such that
\begin{equation}
\begin{gathered}
Mu\in C^1([0,T];X),\quad f_1,\dots,f_n\in C([0,T];\mathbb{C}),
\\
Lu\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X)\cap B([0,T];
X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\end{gathered} \label{e6.10}
\end{equation}
\end{theorem}

\begin{proof}
Let $(y,f_1,\dots,f_n)$ be a solution to problem \eqref{e5.1}-\eqref{e5.3}
with $A$ defined by \eqref{e6.1} and define a function $u$ by
\[
u(t)=L^{-1}\Big[\sum_{j=1}^nf_j(t)z_j+h(t)-y'(t)\Big],\quad t\in [0,T]. \]
Then in view of \eqref{e5.1}
\begin{equation}\label{e6.11}
Lu(t)=\sum_{j=1}^nf_j(t)z_j+h(t)-y'(t)\in Ay(t)=LM^{-1}y(t).
\end{equation}
Since $L$ is injective, one gets $u(t)\in M^{-1}y(t)$. This implies
\begin{equation}\label{e6.12}
Mu(t)=y(t).
\end{equation}
The first equation in \eqref{e6.11} and \eqref{e6.12} imply \eqref{e6.7}.
It is obvious that \eqref{e6.8} and \eqref{e6.9} hold. From the first equation
in \eqref{e6.11} and the second equation in \eqref{e5.5} it follows that
\[
 Lu=\sum_{j=1}^nf_j(\cdot)z_j+h-y'\in C^{(2\alpha+\beta-3
+\theta)/\alpha}([0,T];X)\cap B([0,T];X_A^{(2\alpha+\beta-3+\theta)/\alpha}).
\]
Thus the regularity property \eqref{e6.10} is proved.
\end{proof}

The following theorem is analogously established.

\begin{theorem} \label{thm6.4}
Suppose $\theta>2(2-\alpha-\beta)$, $u_0\in D(L)$,
$Lu_0\in{\widetilde X}_A^{\theta}$, $z_j\in {\widetilde X}_A^{\theta}$,
$\lim_{t\to 0}e^{-tA}z_j=z_j$, $j=1,\dots,n$,
$h\in C([0,T];X)\cap B([0,T];{\widetilde X}_A^{\theta})$,
$\lim_{\tau\to 0}e^{-\tau A}h(t)\\ =h(t)$ for every $t\in[0,T]$,
$g_j\in C^1([0,T];\mathbb{C})$, $\Phi_j\in X^*$,
$\Phi_j[Mu_0]=g_j(0)$, $j=1,\dots,n$, and \eqref{e5.4} holds.
Then, problem \eqref{e6.7}--\eqref{e6.9} admits a unique solution
$(u,f_1,\dots,f_n)$ such that
\begin{gather*}
Mu\in C^1([0,T];X),\quad f_1,\dots,f_n\in C([0,T];\mathbb{C}),\\
Lu\in C^{(2\alpha+2\beta-4+\theta)/\alpha}([0,T];X)\cap B([0,T];
X_A^{(2\alpha+2\beta-4+\theta)/\alpha}).
\end{gather*}
\end{theorem}

\begin{remark} \label{rmk6.1}\rm
When $L$ is the realization in $L^2(\Omega)$ of a second order
strongly elliptic linear differential operator $\mathcal{L}$ with the
Dirichlet boundary condition in a bounded domain $\Omega$ and $M$ is
the multiplication operator by a function belonging to $L^\infty(\Omega)$
one has $\alpha=1$, $\beta=1/2$. Hence the assumption $2\alpha+\beta+\theta>3$ of
Theorems \ref{thm6.1} and \ref{thm6.3} is not satisfied for $\theta=1/2$, and the
assumption $\theta>2(2-\alpha-\beta)$ of Theorems \ref{thm6.2} and \ref{thm6.4}
 is not satisfied for
$\theta\in(0,1)$. A treatment of this case is given in \cite{tanabe}.
Furthermore, owing to the inclusion relations
$D({\widetilde A})\subset L^2(\Omega)_A^{1/2}\subset(L^2(\Omega),D(A))_{1/2,\infty}$
the assumptions are described by using a clearer space $D({\widetilde A})$ than
${\widetilde L^2(\Omega)}_A^{1/2}$, $(L^2(\Omega),D(A))_{1/2,\infty}$
in \cite{tanabe}, where ${\widetilde A}={\widetilde L}M^{-1}$ and ${\widetilde L}$
is the realization of $\mathcal{L}$ in $H^{-1}(\Omega)=H_0^1(\Omega)^*$.
\end{remark}

\section{Problems for systems}

Let us consider the following inverse problem: Recover
$y_i,f_{ij}$, $i=1,\ldots,n$, $j=1,\ldots,N$, such that
\begin{gather}
\begin{gathered}
y_1'=A_1y_1+B_{11}y_1+\dots+B_{1n}y_n+f_{11}(t)z_1+\dots+f_{1N}(t)z_N+h_1(t),\\
\dots \\
y_n'=A_ny_n+B_{n1}y_1+\dots+B_{nn}y_n+f_{n1}(t)z_1+\dots+f_{nN}(t)z_N+h_n(t),
\end{gathered} \label{e7.1}
\\
y_1(0)=y_{10},\ldots,y_n(0)=y_{n0},\\
\Phi_j[y_i(t)]=g_{ji}(t),\quad i=1,\ldots,n; j=1,\ldots,N. \label{e7.3}
\end{gather}
Assume  $0<\beta\leq\alpha\leq1$ and $2\alpha+\beta+\theta>3$. It is also assumed
that for $i,j=1,\dots,n$, $A_i$ and $B_{ij}$  satisfy
\begin{gather*}
\|(\lambda-A_i)^{-1}\|_{\mathcal{L}(X)}
\leq \frac C{(1+|\lambda|)^{\beta}}\quad\text{for }\lambda\in\Sigma_{\alpha}
=\{\lambda in\mathbb{C};\operatorname{Re}\lambda\geq-c(1+|\operatorname{Im}\lambda|)^{\alpha}\},\\
B_{ij}\in\mathcal{L}(D(A_j),X_{A_i}^\theta).
\end{gather*}
Also assume that $y_{i0}\in D(A_i)$, $A_iy_{i0}\in(X,D(A_i))_{\theta,\infty}$,
$z_j\in\cap_{k=1}^n(X,D(A_k))_{\theta,\infty}$,
$h_i\in C([0,T];X)\cap B([0,T];(X,D(A_i))_{\theta,\infty}$,
$g_{ji}\in C^1([0,T];\mathbb{C})$, $\Phi_j\in X^*$,
$\Phi_j[y_{i0}]=g_{ji}(0)$, $i=1,\ldots,n$, $j=1,\ldots,N$.
Set
\[
y=\begin{pmatrix}
y_1\\ \dots\\ y_n\end{pmatrix},\quad
A=\begin{pmatrix} A_1&0&\dots&0\\
\dots&&&\dots \\
0&0&\dots&A_n\end{pmatrix},\quad
B=\begin{pmatrix}
B_{11}&\dots&B_{1n}\\
\dots &&\dots\\
B_{n1}&\dots&B_{nn}
\end{pmatrix}.
\]
Then $A$ and $A+B$ generate infinitely differentiable semigroups in $X^n$.
The system \eqref{e7.1} is written as
\begin{align*}
y'&=(A+B)y+f_{11}(t) \begin{pmatrix}
z_1\\0\\ \dots\\ 0\end{pmatrix}
+f_{12}(t)\begin{pmatrix}z_2\\ 0 \\ \dots\\ 0\end{pmatrix}
+\dots+f_{1N}(t)\begin{pmatrix}z_N\\ 0\\ \dots\\ 0\end{pmatrix}\\
&\quad +f_{21}(t)\begin{pmatrix}0\\ z_1\\ \dots\\ 0\end{pmatrix}
+f_{22}(t)\begin{pmatrix}0\\ z_2\\ \dots\\ 0\end{pmatrix}
 +\dots+f_{2N}(t)\begin{pmatrix}0\\ z_N\\ \dots \\ 0\end{pmatrix}+\dots\\
&\quad +f_{n1}(t)\begin{pmatrix}0\\ 0\\ \dots \\ z_1\end{pmatrix}
+f_{n2}(t)\begin{pmatrix}0\\ 0\\ \dots \\ z_2\end{pmatrix}+\dots
+f_{nN}(t)\begin{pmatrix}0\\ 0\\ \dots\\ z_N \end{pmatrix}
+\begin{pmatrix}h_1(t)\\ h_2(t)\\ \dots\\h_n(t)\end{pmatrix}.
\end{align*}
Theorem \ref{thm6.3} applies provided that
\[
\det \begin{pmatrix}
\Phi_1[z_1]& \dots &\Phi_1[z_N]\\
\dots&&\dots\\
\Phi_N[z_1]&\dots&\Phi_N[z_N]
\end{pmatrix}\ne 0.
\]
Indeed, the further information reduces to the $N$ linear systems
in the unknowns $f_{i1},\ldots,f_{iN},i=1,\ldots,n$, whose determinant
is just the indicated above. Identification problem \eqref{e7.1}--\eqref{e7.3}
admits a unique solution
$y=(y_1,\ldots,y_n)^t$, $f_{ij}$, $i=1,\ldots,n$, $j=1,\ldots,N$ such that
\begin{gather*}
y\in C^1([0,T];X^n),\quad f_{ij}\in C^1([0,T];\mathbb{C}),\quad
 i=1,\ldots,n,\; j=1,\ldots,N,\\
\begin{aligned}
&(A+B)y\in C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X^n)\cap
 B([0,T];X_{A+B}^{(2\alpha+\beta-3+\theta)/\alpha})\\
& \subset C^{(2\alpha+\beta-3+\theta)/\alpha}([0,T];X^n)
 \cap \big[B([0,T];(X,D(A_1))_{(2\alpha+\beta-3+\theta)/\alpha,\infty})\\
&\times\dots\times B([0,T];(X,D(A_n))_{(2\alpha+\beta-3+\theta)/\alpha,\infty})\big].
\end{aligned}
\end{gather*}

\subsection*{Example}
Let us consider the inverse problem
\begin{gather}
\frac d{dt}(A+1)y+Ay=\int_0^tk(t-s)Ay(s)ds+f(t)z+h(t),\quad0\leq t\leq T, \label{e7.4}\\
(A+1)y(0)=(A+1)y_0, \label{e7.5}\\
\Phi[(A+1)y(t)]=\Phi[g(t)],\quad0\leq t\leq T. \label{e7.6}
\end{gather}
We suppose that $-1$ is a simple pole for the resolvent of $A$,
i.e. $(A+1+\lambda)^{-1}$ exists for $0<|\lambda|\leq\varepsilon$ and
\[
\|(A+1+\lambda)^{-1}\|_{\mathcal{L}(X)}\leq\frac C{|\lambda|}.
\]
A change of variable $y(t)=e^{\kappa t}x(t)$ transforms equation \eqref{e7.4} into
\begin{align*}
&\frac d{dt}(A+1)x(t)+\kappa(A+1)x(t)+Ax(t)\\
&=\int_0^tk(t-s)e^{-\kappa(t-s)}Ax(s)ds+f_1(t)z+e^{-\kappa t}h(t),
\end{align*}
where $f_1(t)=e^{-\kappa t}f(t)$. Now
\begin{align*}
&\lambda(A+1)+\kappa(A+1)+A=(\lambda+\kappa+1)A+\lambda+\kappa\\
&=(\lambda+\kappa+1)\Big(A+\frac{\lambda+\kappa}{\lambda+\kappa+1}\Big)
=(\lambda+\kappa+1)\Big(A+1-\frac1{\lambda+\kappa+1}\Big).
\end{align*}
Hence if $0<|\lambda+\kappa+1|^{-1}<\varepsilon$, i.e.
$|\lambda+\kappa+1|>\varepsilon^{-1}$, $\lambda(A+1)+\kappa(A+1)+A$
has a bounded inverse. Take $\kappa$ so large that $\kappa+1>\varepsilon^{-1}$.
 Then $(A+1- {\frac1{\lambda+\kappa+1}})^{-1}\in\mathcal{L}(X)$ exists for
$\lambda\in \mathbb{C}\setminus S(-1-\kappa,\varepsilon^{-1})$ and
\begin{align*}
&(A+1)(\lambda(A+1)+\kappa(A+1)+A)^{-1}\\
& =(A+1)(\lambda+\kappa+1)^{-1}\big(A+1-\frac1{\lambda+\kappa+1}\big)^{-1}\\
& =(\lambda+\kappa+1)^{-1}
\big\{1+\frac1{\lambda+\kappa+1}\big(A+1-\frac1{\lambda+\kappa+1}\big)^{-1}\big\},
\end{align*}
so that
\[
\|(A+1)(\lambda(A+1)+\kappa(A+1)+A)^{-1}\|_{\mathcal{L}(X)}
\leq C|\lambda+\kappa+1|^{-1}
\]
for $|\lambda+\kappa+1|>\varepsilon^{-1}$. Hence the previous results
(See also  Favini and Tanabe \cite{fata}) apply for $\alpha=\beta=1$.

However, this pole case allows a better treatment. First of all the change
of variable $y=e^{-t}x$ transforms the given problem \eqref{e7.4}--\eqref{e7.6}
into
\begin{gather}
\frac d{dt}(A+1)x(t)-x(t)=\int_0^tk_1(t-s)Ax(s)ds+f_1(t)z+h_1(t), \label{e7.7}\\
(A+1)x(0)=(A+1)y_0, \label{e7.8}\\
\Phi[(A+1)x(t)]=g_1(t). \label{e7.9}
\end{gather}
where $k_1(t)=e^tk(t)$, $f_1(t)=e^tf(t)$,
$h_1(t)=e^th(t)$, $g_1(t)=e^tg(t)$. If $-1$ is a simple pole for
$(A-\lambda)^{-1}$, so that
\[
X=N(A+1)\oplus R(A+1),
\]
and $P$ denotes the projection onto $N(A+1)$, problem \eqref{e7.7}--\eqref{e7.9}
reduces to
\begin{gather}
\begin{aligned}
& \frac d{dt}(A+1)(1-P)x(t)-(1-P)x(t)\\
&=\int_0^tk_1(t-s)A(1-P)x(s)ds+f_1(t)(1-P)z+(1-P)h_1(t),
\end{aligned} \label{e7.10}\\
(A+1)(1-P)x(0)=(A+1)(1-P)y_0, \label{e7.11}\\
\Phi[(A+1)(1-P)x(t)]=g_1(t), \label{e7.12}\\
\begin{aligned}
-Px(t)&=\int_0^tk_1(t-s)P(A+1-1)x(s)ds+f_1(t)Pz+Ph_1(t) \\
&=-\int_0^tk_1(t-s)Px(s)ds+f_1(t)Pz+Ph_1(t).
\end{aligned} \label{e7.13}
\end{gather}
Since the restriction of $A+1$ to $R(A+1)$ is boundedly invertible,
the change of variable $(A+1)(1-P)x(t)=\xi(t)$ transforms \eqref{e7.10}--\eqref{e7.12}
into
\begin{gather}
\frac d{dt}\xi(t)-R\xi(t)
=\int_0^tk_1(t-s)[1-R]\xi(s)ds+f_1(t)(1-P)z+(1-P)h_1(t), \label{e7.14}
\\
\xi(0)=(A+1)(1-P)y_0, \label{e7.15}\\
\Phi[\xi(t)]=g_1(t), \label{e7.16}
\end{gather}
where $R$ indicates the inverse of the restriction of $A+1$ to $R(A+1)$.
Therefore, if $k$ is continuous in $[0,T]$, $\Phi[(1-P)z]\ne0$,
$h\in C([0,T];X)$, $(1-P)y_0\in D(A)$, $g\in C^1([0,T];\mathbb{C})$,
 problem \eqref{e7.14}--\eqref{e7.16} admits a unique strict solution
$(\xi,f_1)$. Hence, we have a unique strict solution $((1-P)x,f_1)$
to \eqref{e7.10}--\eqref{e7.12}. Since $f_1$ is now known,
we only remain to solve integral equation \eqref{e7.13}, that is uniquely solvable.
Notice that this improves the preceding result, since condition
 $\Phi[z]\ne0$ is replaced by the weaker condition $\Phi[(1-P)z]\ne0$.


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\end{document}