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

\AtBeginDocument{{\noindent\small {\em 
Electronic Journal of Differential Equations}, 
Vol. 2004(2004), No. 116, pp. 1--7.\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 2004 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2004/116\hfil Solvability of some integral equations]
{Solvability of some integral equations in Hilbert spaces}

\author[O. A. Ilhan\hfil EJDE-2004/116\hfilneg]
{Onur Alp Ilhan}

\address{Onur Alp Ilhan \hfill\break 
National University of Uzbekistan\\
Department of Mathematical Physics\\
Tashkent, Uzbekistan} 
\email{onuralp@sarkor.com, onuralpilhan@yahoo.com}

\thanks{Submitted August 25, 2004. Published October 5, 2004.}
\date{}
\subjclass[2000]{45A05} 
\keywords{Integral equations; compact operators; spectral theory}

\begin{abstract}
 We consider an integral equation of Fredholm and Volterra type
 with spectral parameter depending on time. Conditions of
 solvability are established when the initial value of the
 parameter coincides with an eigenvalue of Fredholm operator.
\end{abstract}
\maketitle


\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}


\section{introduction}

Let $H$ be a Hilbert space. We consider the integral equation
\begin{equation}
\int_{0}^{t}K(t,s)u(s)ds+Au(t)-\lambda(t)u(t)=f(t),\quad t>0, \label{e1}
\end{equation}
where $u:\mathbb{R}_{+}\to H$ is unknown, $A:H\to H$ is a linear
bounded self-adjoint operator, $K:Q\to\mathbb{R}$ is the kernel,
$f(t)$ is a given function, and $\lambda(t)$ is a function which
we may interpret as spectral parameter. Here
$\mathbb{R}_{+}=\{t\in R:t\geq0\}$ is the positive half-line, and
\begin{equation}
Q=\{(t,s)\in R^{2}:0\leq s\leq t<\infty\}. \label{e2}%
\end{equation}
The following equation is a typical example of this integral
equation:
\begin{equation}
\int_{0}^{t}K(t,s)P(x,s)ds-\int_{0}^{a}R(x,y)P(y,t)dy-\lambda(t)P(x,t)=f(x,t),
\label{e3}%
\end{equation}
where $0\leq x\leq a$ and $t>0$. We consider in Hilbert space
$H=L_{2}(0,a)$, where $a>0$. Equations of this type arise in the
theory of elasticity \cite{p1,s1}. The kernels $K(t,s)$ and
$R(x,y)$ are connected with some elastic creeping base and
$\lambda(t)$ is the given value which describes the elastic
properties of deformable body. We may reference also to work [4],
where the similar integral equation was considered. In case when
$\lambda(t)$ is constant, \eqref{e1} was considered in \cite{a1}.

\section{results}

We assume that $\lambda(t)$ is a continuous function for $t\geq0$.
We denote the range of $\lambda(s)$ on the interval $[0,t]$ by
\begin{equation}
\Lambda(t)=\{\lambda(s),\;0\leq s\leq t\}. \label{e4}%
\end{equation}
It is clear that $\Lambda(t)$ is a closed subset of $\mathbb{R}$.

We consider function $u:\mathbb{R}_{+}\to H$ in an abstract
setting, i. e., $u(t)\in H$ for $t\geq0$. Assuming that $u(t)$ is
bounded on each interval $[0\leq t\leq T]$, we denote
\[
\| u\|_{t}=\sup_{0\leq s\leq t}\| u(s)\|.
\]
We Assume also that the kernel $K(t,s)$ is continuous on the set
$Q$, as defined by \eqref{e2}. Denote
\begin{equation}
M(t)=\sup_{0\leq s\leq\tau\leq t} | K(\tau,s)|. \label{e5}
\end{equation}
It is clear that $M(t)$ is a monotonically increasing function.

We denote by $\sigma(A)$ the spectrum of a bounded self-adjoint
operator $A$. When parameter $\lambda$ is outside this spectrum,
i.e. $\lambda\notin \sigma(A)$, there exists the resolvent
$R_{\lambda}=(A-\lambda I)^{-1}$ of the operator $A$.

We consider the auxilary equation
\begin{equation}
\int_{0}^{t}K(t,s)R_{\lambda(s)} v(s)ds+v(t)=f(t),\quad t>0. \label{e6}
\end{equation}
Assuming that we have a solution $v(t)$ of this equation, by
putting $u(t)=R_{\lambda(t)}v(t)$ or $v(t)=(A-\lambda(t)I)u(t)$,
we obtain
\[
\int_{0}^{t}K(t,s)u(s)ds+(A-\lambda(t)I)u(t)=f(t),\quad t>0.
\]
This implies $u(t)$ is a solution of the equation \eqref{e1}.

At first we assume that $\lambda(t)$ does not intersect the
spectrum of $A$, i.e. $\Lambda(t)\cap\sigma(A)=\emptyset$. Then
obviously the norms of resolvent $\| R_{\lambda}\|$ are bounded
for every $\lambda\in\Lambda(t)$.

Denote
\begin{equation}
B(t)=\sup_{\lambda\in\Lambda(t)}\Vert R_{\lambda}\Vert.\label{e7}%
\end{equation}
The following simple result shows that outside the spectrum of the
operator $A$ there is the solvability of the equation (1.1)
without any additional condition.

\begin{theorem} \label{thm1}
Let $\Lambda (t)\cap \sigma (A)=\emptyset $ for all $t\geq 0$.
Then there exists the unique solution $u(t)$ of equation
\eqref{e1} and
$$
\| u(t)\|\leq \| f\|_{t}B(t)\exp \{tM(t)B(t)\}..
$$
\end{theorem}


To prove this theorem it is sufficient to show that there exists
the unique solution of the auxiliary equation \eqref{e6}. This
fact is evident from the following lemma.

\begin{lemma} \label{lem1}
Let $v_{0}(t)=f(t)$ and
\begin{equation}
v_{k}(t)=-\int_{0}^{t}K(t,s)R_{\lambda (s)}v_{k-1}(s)ds,\quad
k=1,2,\dots \label{e8}
\end{equation}
Then
$$
\| v_{k}\|_{t}\leq \| f\|_{t}\frac{[M(t)B(t)]^{k}}{k!}t^{k}.
$$
\end{lemma}


\begin{proof}
We shall use induction. The estimate is trivial for $k=0$. Assume
that it is true for some $k$ and then prove it for $k+1$. It
follows from the equality
$v_{k+1}(t)=-\int_{0}^{t}K(t,s)R_{\lambda(s)}v_{k}(s)ds$ that
\begin{align*}
\| v_{k+1}(t)\|  &  \leq\int_{0}^{t}M(t)\| R_{\lambda(s)}\|\, \| f\|_{s}%
\frac{[M(s)B(s)]^{k}}{k!}s^{k}ds\\
&  \leq M(t)B(t)\| f\|_{t}\frac{[M(t)B(t)]^{k}}{k!}\int_{0}^{t}s^{k}ds\\
&  =\| f\|_{t}\frac{[M(t)B(t)]^{k+1}}{(k+1)!}t^{k+1}.
\end{align*}
\end{proof}

Using this lemma, it is easy to show that the function
\[
v(t)=\sum_{k=0}^{\infty}(-1)^{k+1}v_{k}(t)
\]
is the unique solution of the equation \eqref{e6} and it proves
Theorem \ref{thm1}.

The problem is more complicated when $\Lambda(t)$ has a common
point with spectrum of $A$ and that is the main idea of our
consideration. Suppose that $\lambda(0)$ coincides with one of the
isolated points of the spectrum of the operator $A=A^{\ast}$. We
suppose also that $\lambda(t)\notin\sigma(A)$ for all $t>0$. It
will be proved below that in this case the following quantity is
finite:
\begin{equation}
B_{1}(t)=\sup_{0<s\leq t} s\| R_{\lambda(s)}\| <\infty,\quad t>0. \label{e9}%
\end{equation}
Remember that the set $Q$ is defined by equality \eqref{e2}. We
introduce the class of kernels, which vanish at the point $(0,0)$
with some order $\alpha>0$.

We say that $K(t,s)\in M^{\alpha}(Q)$ if $K(t,s)\in C(Q)$,
$K(0,0)=0$ and
\begin{equation}
| K(t,s)| \leq K_{\alpha}t^{\alpha},\quad0<s<t, \label{e10}
\end{equation}
with some constant $K_{\alpha}>0$.

Analogously we say that $f\in N^{\alpha}(\mathbb{R}_{+})$ if 
$f:\mathbb{R}_{+}\to H$ is continuous on the half-line $t\geq0$,
 $f(0)=0$ and
\begin{equation}
\| f\|_{(\alpha)}=\sup_{t>0} t^{-\alpha}\| f(t)\|<\infty. \label{e11}
\end{equation}


\begin{theorem} \label{thm2}
Let $K(t,s)\in M^{\alpha }(Q)$, $0<\alpha <1$. Suppose that
$\lambda (t)$ is continuously differentiable function on the
half-line $t\geq 0$. Let $\lambda(0)$  be an isolated point of the
spectrum $\sigma (A)$ and $\lambda (t)\notin \sigma (A)$ for all
$t>0$. If
\begin{equation}
\lambda'(0) \neq 0\,, \label{e12}
\end{equation}
then for an arbitrary function $f\in N^{\alpha }(\mathbb{R}_+)$
there exists a solution $u(t)$ of \eqref{e1} such that
\begin{equation}
\| u(t)\|\leq \| f\|_{(\alpha )}t^{\alpha }\exp [C(t)t^{\alpha }]
\label{e13}
\end{equation}
where
$C(t)=K_{\alpha }B_{1}(t)/\alpha$. %\hfill (14)
The function $B_{1}(t)$ and constant $K_{\alpha }$ are defined by
equalities \eqref{e9} and \eqref{e10} respectively.
\end{theorem}


As we will show in Example 1, condition \eqref{e12} is essential.
This condition means that spectral parameter $\lambda(t)$ has to
move away from the point $\lambda(0)$ with non-zero velocity.

First we prove some lemmas.

\begin{lemma} \label{lem2}
Suppose that $\lambda (t)$ is a continuously differentiable
function for $t\geq 0$. Let $\lambda(0)=\lambda _{0}$ be an
isolated point of the spectrum $\sigma (A)$ and $\lambda (t)\notin
\sigma (A)$  for all $t>0$. If condition \eqref{e12} holds for
every $T>0$, then
\begin{equation}
| \lambda (t)-\lambda _{0}| \geq c(T)t,\quad 0\leq t\leq T\,.
\label{e15}
\end{equation}
\end{lemma}


\begin{proof}
We divide the interval $[0,T]$ into two parts: $[0,\delta)$ and
$[\delta,T]$.

\noindent(1) According to the conditions of this lemma, for every
$\delta>0$ we have $\lambda(t)\notin\sigma(A)$ if $\delta\leq
t\leq T$. Hence $\lambda(t)-\lambda_{0}\neq0$ on this interval.
Then because of continuity,
\begin{equation}
| \lambda(t)-\lambda_{0}| \geq c_{1}(T,\delta)\geq c_{1}(T,\delta)\frac{t}%
{T},\quad\delta\leq t\leq T. \label{e16}%
\end{equation}
(2) Further, since $\lambda^{\prime}(0)\neq0$ then for small
enough $\delta>0$ we have $| \lambda^{\prime}(t)| \geq
c_{2}(\delta)>0$, $0\leq t<\delta$. Then using Lagrange formulae
we get
\begin{equation}
| \lambda(t)-\lambda_{0}| =| \lambda(t)-\lambda(0)| =|\lambda' (\tau)|
(t-0)\geq c_{2}(\delta)t,\quad0\leq t\leq\delta. \label{e17}%
\end{equation}
The required estimate \eqref{e15} obviously follows from
\eqref{e16} and \eqref{e17} with
$c(T)=\min\{\frac{1}{T}c_{1}(T,\delta),c_{2}(\delta)\}$.
\end{proof}

The following lemma proves the correctness of the definition
\eqref{e9}.

\begin{lemma} \label{lem3}
Under the conditions of Lemma \ref{lem2} for an arbitrary $T>0$,
\begin{equation}
t\| R_{\lambda (t)}\|\leq C(T)<\infty ,\quad 0<t\leq T\,.\label{e18}
\end{equation}
\end{lemma}


\begin{proof}
We use the well known estimate (see, e.g. \cite{w1})
\begin{equation}
\| R_{\lambda}(A)\|\leq1/\mathop{\rm dist}\big(\lambda,\sigma
(A)\big) \label{e19}
\end{equation}
which is fulfilled for any self-adjoint operator $A$. We denote
the distance from the point $\lambda$ to the spectrum $\sigma(A)$
of operator $A$ by
\[
\mathop{\rm dist}\big(\lambda,\sigma(A)\big)
=\inf_{\mu\in\sigma(A)}|\lambda-\mu| .
\]
It follows from estimate \eqref{e19} that in order to get
\eqref{e18} it is enough to prove the inequality
\begin{equation}
\mathop{\rm dist}\big(\lambda,\sigma(A)\big) \geq Ct,\quad
C>0,\quad0<t\leq T.
\label{e20}%
\end{equation}
Denote
\begin{equation}
d(t)=\mathop{\rm dist}\big(\lambda,\sigma(A)\big) . \label{e21}
\end{equation}
On each interval $[0,T]$, obviously,
\begin{equation}
|\lambda(t)-\lambda_{0}| \leq C_{0}(T),\quad0\leq t\leq T. \label{e22}
\end{equation}
Since $\lambda(0)=\lambda_{0}$ is an isolated point of
$\sigma(A)$, the distance from point $\lambda_{0}$ to other points
of $\sigma(A)$ is positive:
\begin{equation}
\rho= \mathop{\rm
dist}\big(\lambda(0),\sigma(A)\backslash\lambda_{0}\big)>0.
\label{e23}%
\end{equation}
(1) According to conditions of lemma $d(0)=\mathop{\rm
dist}\big(\lambda(0),\sigma(A)\big)=0$. Then it is clear that for
small enough $\delta>0$ $d(t)\leq\rho/2$, $0\leq t\leq\delta$.

Hence for these $t$ the nearest point of $\sigma(A)$ to
$\lambda(t)$ is
\begin{equation}
\lambda_{0}: d(t)=| \lambda(t)-\lambda_{0}|\quad\mbox{if }0\leq
t\leq\delta.
\label{e24}%
\end{equation}
(2) Further, on compact interval $[\delta,T]$ and according to
condition $\lambda(t)\notin\sigma(A)$, we have $d(t)>0$ and hence
\[
d(t)\geq d_{0}(T,\delta)>0\quad\mbox{for }\delta\leq t\leq T.
%\label{e25}
\]
Taking into consideration this inequality and \eqref{e22}, we
obtain
\begin{equation}
d(t)\geq d_{0}(T,\delta)\geq\frac{d_{0}(T,\delta)}{C_{0}(T)}|
\lambda (t)-\lambda_{0}| =C_{1}| \lambda(t)-\lambda_{0}|
,\quad\delta\leq t\leq T.
\label{e26}%
\end{equation}
(3) From \eqref{e24} and \eqref{e26} it follows that
\begin{equation}
d(t)\geq C_{2}(T)| \lambda(t)-\lambda_{0}| ,\quad0\leq t\leq T, \label{e27}%
\end{equation}
In this case the required estimate \eqref{e20} follows from
\eqref{e27}, definition \eqref{e21} and Lemma \ref{lem2}.
\end{proof}

\begin{lemma} \label{lem4}
Let $v_{k}(t)$ be as defined by \eqref{e8}  and $v_{0}(t)=f(t)$.
If all of conditions of  Theorem \ref{thm2} are fulfilled then
\begin{equation}
\| v_{k}\|_{t}\leq \| f\|_{(\alpha )}[C(t)]^{k}\frac{t^{\alpha
(k+1)}}{k!}, \label{e28}
\end{equation}
where $C(T)=K_{\alpha }B_{1}(t)/\alpha $.
\end{lemma}


\begin{proof}
The estimate \eqref{e28} is obviously true for $k=0$. Assuming
that it is true for some $k$ we get
\[
\| v_{k+1}(t)\|\leq K_{\alpha}t^{\alpha}| |
f\|_{(\alpha)}\int_{0}^{t}\|
R_{\lambda(s)}\|[C(s)]^{k}\frac{s^{\alpha(k+1)}}{k!}ds.
\]
Further we apply estimate \eqref{e18},
\begin{align*}
\| v_{k+1}(t)\|  &  \leq\| f\|_{(\alpha)}K_{\alpha}B_{1}(T)[C(t)]^{k}%
t^{\alpha}\frac{1}{k!} \int_{0}^{t}s^{\alpha(k+1)-1}ds\\
&  =\| f\|_{(\alpha)}[C(t)]^{k+1}\frac{t^{\alpha(k+2)}}{(k+1)!}.
\end{align*}

\end{proof}

Using this lemma we complete the proof of Theorem \ref{thm2} as in
the proof of Theorem \ref{thm1}.

As mentioned above, the condition \eqref{e12} in Theorem
\ref{thm2} is important as confirmed in the following example.

\subsection*{Example}

Consider equation
\begin{equation}
\int_{0}^{t}tu(s,x)ds+3x\int_{0}^{1}yu(t,y)dy-(1+t^{2})u(t,x)=tx,\label{e29}%
\end{equation}
which is of the same type as equation \eqref{e1}. In this case
$K(t,s)=t$, $R(x,y)=-3xy$, $f(t,x)=tx$, $\lambda(t)=1+t^{2}$.
Consider the operator
\[
Au(x)=3x\int_{0}^{1}yu(y)dy.
\]
and look for an eigenvalue $\lambda$ and eigenfunction $v$ such
that $Av=\lambda v$. Obviously,
\[
R(A)=\{v\in L_{2}[0,1]:v(x)=Cx,\;C\in R\}
\]
and in as much as $v\in R(A)$ we get
\[
(A-\lambda)v(x)=(A-\lambda)Cx=3Cx\int_{0}^{1}y^{2}dy-\lambda
Cx=Cx(1-\lambda )=0.
\]
This implies $\lambda=1$ is an eigenvalue and hence it is isolated
point of spectrum of compact operator $A$. Then all of conditions
of Theorem \ref{thm2} are fulfilled except for \eqref{e12},
because $\lambda(0)=1\in\sigma(A)$, but $\lambda^{\prime}(0)=0$.

Assume that there exists a solution $u(t,x)$ of \eqref{e29}.
Denote
\[
g(t,x)=u(t,x)-a(t)x,\quad{with}\quad a(t)=3\int_{0}^{1}xu(t,x)dx.
\]
Obviously, the function $f(x)=x$ is orthogonal to $g(t,x)$ in the
Hilbert space $L_{2}[0,1]$:
\[
\int_{0}^{1}xg(t,x)dx=\int_{0}^{1}xu(t,x)dx-a(t)\int_{0}^{1}x^{2}%
dx=a(t)/3-a(t)/3=0.
\]
According to the definition $u(t,x)=g(t,x)+a(t)x$, and putting
this representation in \eqref{e29}, we get
\begin{align*}
t\int_{0}^{t}g(s,x)ds+3x\int_{0}^{1}yg(t,y)dy-(1+t^{2})g(t,x) &  \\
+xt\int_{0}^{t}a(s)ds+3xa(t)\int_{0}^{1}y^{2}dy-(1+t^{2})a(t)x &
=tx.
\end{align*}
Note that because of orthogonality of the functions $x$ and
$g(t,x)$ this equality is equivalent to the following two
equalities:
\begin{gather}
t\int_{0}^{t}g(s,x)ds-(1+t^{2})g(t,x)=0,\label{e30}\\
\int_{0}^{t}a(s)ds-ta(t)=1-\frac{3}{t}\int_{0}^{1}yg(t,y)dy\,.\label{e31}%
\end{gather}
After differentiating \eqref{e30}, we get
\[
\int_{0}^{t}g(s,x)ds+tg(t,x)-2tg(t,x)-(1+t^{2})g_{t}(t,x)=0.
\]
Hence, $g_{t}(0,x)=0$. Further, we may rewrite \eqref{e30} as
\[
\int_{0}^{t}g(s,x)ds-(t+1/t)g(t,x)=0
%\label{e32}
\]
and after differentiating this equation, we get
\[
g(t,x)-g(t,x)+\frac{1}{t^{2}}g(t,x)-(t+1/t)g_{t}=0,
\]
or
\[
t(t^{2}+1)g_{t}(t,x)-g(t,x)=0.
\]
Now we find the solution of this ordinary differential equation
\[
g(t,x)=C(x)\frac{t}{\sqrt{t^{2}+1}}.
\]
Then
\[
g_{t}(t,x)=\frac{C(x)}{(t^{2}+1)^{3/2}}%
\]
and because of the condition $g_{t}(0,x)=0$, we have
$g_{t}(0,x)=C(x)=0$ Consequently, $g(t,x)\equiv0$  Then equation
\eqref{e31} takes the form
\[
\int_{0}^{t}a(s)ds-ta(t)=1,
\]
and when we put $t=0$ we obtain $0=1$. This contradiction proves
that there is no solution to \eqref{e29} because
$\lambda^{\prime}(0)=0$.

\subsection*{Acknowledgements}

I am grateful to Professor Sh. Alimov for his valuable assistance.
I would like to express my thanks to reviewer from the Electron.
J. Diff. Eqns. for his/her helpful remarks.

\begin{thebibliography}{9}                                                                                                %


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\end{thebibliography}


\end{document}