\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 02, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE-2005/02\hfil Solution dependence \dots ]
{Solution dependence on problem parameters for initial-value
problems associated with the Stieltjes Sturm-Liouville equations}

\author[L. Battle\hfil EJDE-2005/02\hfilneg]
{Laurie Battle}

\address{Laurie Battle\hfill\break
Department of Mathematics and Computer Science \\
Campus Box 017 \\
Georgia College and State University \\
Milledgeville, GA, 31061, USA}
\email{laurie.battle@gcsu.edu}

\date{}
\thanks{Submitted September 10, 2004. Published January 2,2005.}
\subjclass[2000]{34A12, 34A30}
\keywords{Initial value problems; continuous dependence; linear systems}

\begin{abstract}
 We examine properties of solutions to a 2n-dimensional
 Stieltjes Sturm-Liouville initial-value problem.
 Existence and uniqueness of a solution has been previously
 proven, but we present a proof in order to establish properties
 of boundedness, bounded variation, and continuity.
 These properties are then used to prove that the solutions
 depend continuously on the coefficients and on the initial
 conditions under certain hypotheses. In a future paper,
 these results will be extended to eigenvalue problems, and we will
 examine dependence on the endpoints and boundary data in addition
 to the coefficients. We will find conditions under which the eigenvalues
 depend continuously and differentiably on these parameters.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theo}{Theorem}[section]
\newtheorem{lemma}[theo]{Lemma}
\newtheorem{cor}[theo]{Corollary}
\newtheorem{rem}[theo]{Remark}

\section{Introduction}\label{intro}

In this work, we examine properties of solutions of generalized
2n-dimensional Sturm-Liouville initial value problems of the form
\begin{equation}\label{slprob}
  \begin{gathered}
    dy=Ay\,dt+dP\,z \\
    dz=(dQ-\lambda dW)y+Dz\,dt
  \end{gathered}
\end{equation}
on an interval $[a,b]$.
Existence and uniqueness of a solution over the class of quasi-continuous
functions has already been established \cite{Hinton}, but we include part of
the proof in section \ref{IVP} to establish certain bounds and continuity
properties of the solution. In section \ref{seqIVP}, we determine conditions
under which the solution depends continuously on the coefficients.
This is accomplished by taking a sequence of initial value problems and
finding conditions under which the sequence of solutions converges to the
solution of the limit problem.

This work generalizes some earlier results. Kong and Zettl \cite{Kong1},
\cite{Kong3}, and Kong, Wu, and Zettl \cite{Kong1.1}, \cite{Kong2},
consider scalar, self-adjoint equations, whereas this work allows for Stieltjes
Hamiltonian systems. Thus difference equations are included in our formulation.
These authors take sequences of initial value problems to examine dependence
of the solution on the problem data, and they require $L^1$ convergence of the
coefficients. For example, in the second order equation $-(py')'+qy=\lambda wy$,
the approximate equation $-(p_1y')'+q_1y=\lambda w_1y$ is considered to be close
to the original equation if $\int_a^b |\frac{1}{p}-\frac{1}{p_1} |+\int_a^b
|q-q_1|+ \int_a^b |w-w_1|$ is small. We take the same approach of using
sequences of initial value problems but allow for more general modes of
convergence on the coefficients, with two sequences of coefficients converging
weakly in $L^1$, one sequence converging uniformly, and two sequences
converging pointwise. The $L^1$ convergence used in \cite{Kong1}-\cite{Kong3}
is a special case of our convergence conditions. In another related work,
Reid \cite{Reid} addresses this problem for Hamiltonian systems, but we relax
his hypotheses on the data and on the modes of convergence.
Knotts-Zides \cite{Knotts} extends Reid's results to more general conditions,
but her problem is only $2$-dimensional and requires $A=D=0$ in (\ref{slprob}).

The references mentioned above actually apply the results to eigenvalue problems
rather than initial value problems. Likewise, the results we find for initial
value problems will be extended to eigenvalue problems in a later paper.
We will find conditions under which not only the solutions, but also the
eigenvalues, depend continuously on the problem parameters. In addition,
conditions will be found under which the eigenvalues depend differentiably on
the problem data.

\section{Preliminaries}\label{pre}

Here we give a preliminary discussion of Stieltjes integrals and
some previous results by  Hinton \cite{Hinton}.

In this work, we take $N$ to be the ring of $2n \times 2n$ matrices and
we define the norm as follows: First we define the vector norm by
$\|\bar{x}\| := \sum_i |x_i|$ for $\bar{x}= (x_1, \dotsc, x_{2n})^T$.
Then define the norm on $N$ by $\|A\|:=\sum_{i,j} |a_{ij}|$ for
$A=\{a_{ij}\}$.
It follows that for $A$ in $N$ and $\bar{x}$ a $2n$-vector,
$\|A\bar{x}\|\le \|A\| \|\bar{x}\|$. Let $\mathbb{R}^m$ be the
space of $m$-vectors.

For some interval $[a,b]$ with $a<b$, a function
$F:[a,b]\to N$ is
said to be {\it quasi-continuous} if the left and right limits exist at
each interior point of $[a,b]$ and the appropriate one-sided limit
exists at each endpoint. A function $F:[a,b]\to N$ is said to
be of {\it bounded variation} if there exists a number $K$ such that
\[
  \sum_{i=1}^m \left|F(t_i)-F(t_{i-1})\right| \le K
\]
for any partition $a=t_0<t_1<\dotsc <t_m=b$. The greatest lower bound
of such constants $K$ is called the {\it total variation} of $F$ and is
denoted $\bigvee_a^b F$ or $\int_a^b \|dF(t)\|$. If a real function $F$ is
nondecreasing on an interval $[a,b]$, then $F$ is of bounded variation
with $\bigvee_a^b F = F(b)-F(a)$. A function of bounded variation
is also quasi-continuous. For $t\ge a$ and a function $F$ of bounded
variation on $[a,t]$, define the total variation function
$v_F(t) :=  \bigvee_a^t F$.
If $F$ is of bounded variation and is continuous, then $v_F$ is also
continuous \cite[Theorem 6.26]{Rudin}.

If $F$ is a function from $[a,b]$ to $N$, and if $G$ is a
function from $[a,b]$ to $N$ or to $\mathbb{R}^{2n}$, then the
{\it Cauchy-left integral}
  $(L)\int_a^b dF(x)G(x)$
denotes an element $Y$ of $N$ or $\mathbb{R}^{2n}$ with the
following property: for each
positive number $\epsilon$, there is a partition $s=\{s_i\}_0^n$
from $a$ to $b$ such that for any partition $t=\{t_i\}_0^m$ that is
a refinement of $s$, then
\[
  \big\|Y- \sum_{i=1}^m
         [F(t_i)-F(t_{i-1})]G(t_{i-1})\big\|
         < \epsilon
\]
The {\it Cauchy-right integral}
  $(R)\int_a^b dF(x)G(x)$
is the same except $G(t_{i-1})$ is replaced by $G(t_i)$.
The ordinary {\it Stieltjes integral}
  $\int_a^b dF(x)G(x)$
denotes an element $Y$ of $N$ or $\mathbb{R}^{2n}$ with the
following property: for each
positive number $\epsilon$, there is a partition $s=\{s_i\}_0^n$
from $a$ to $b$ such that for any partition $t=\{t_i\}_0^m$ that is
a refinement of $s$, then
\[
  \big\|Y- \sum_{i=1}^m
         [F(t_i)-F(t_{i-1})]G(t_i^*)\big\| < \epsilon,
\]
for all $t_i^*$ such that $t_i^*\in [t_{i-1},t_i]$.
We can define similarly the Cauchy-left integral
$(L)\int_a^b G(x)dF(x)$, the Cauchy-right integral $(R)\int_a^b
G(x)dF(x)$, and the Stieltjes integral $\int_a^b G(x)dF(x)$.
These integrals are known to be unique if they exist. The Cauchy integrals
are known to exist if $F$ is of bounded variation and $G$ is quasi-continuous.
If in addition either $F$ or $G$ is continuous, the Stieltjes
integral is known to exist. If the Stieltjes integral exists, then
the corresponding left-Cauchy
and right-Cauchy integrals exist and all three integrals are equal.

Now we state some theorems that will be referenced in this paper
\cite{Hinton}. They are stated here without proof.

\begin{theo}\label{intbd}
  If $F$ is of bounded variation from $[a,b]$ to $N$, $G$ is
  quasi-continuous from $[a,b]$ to $N$ or to $\mathbb{R}^{2n}$, and
  $f$ and $g$ are real functions
  on $[a,b]$ such that for $x\le y$, $\bigvee_x^y F \le f(y)-f(x)$,
  and $g(x)=\|G(x)\|$, then
\[
  \big\| (L)\int_a^b dF(s)G(s)\big\|
  \le \big| (L)\int_a^b df(s)g(s)\big|
  \le \big( \bigvee_a^b f \big) \|g\|_{[a,b]},
\]
where $\|g\|_{[a,b]}:=\sup \{g(x): x\in [a,b]\}$.
\end{theo}

\begin{theo}\label{Gronwall}
  If $K\ge 0$, $h$ is a real nondecreasing function on $[a,b]$, and if
  $m$ is a real function on $[a,b]$ bounded above by some $T>0$ and
  such that for each $x$,
\[
  m(x)\le K+ (L)\int_a^x m(s)\,dh(s),
\]
then
\[
  m(x)\le Ke^{[h(x)-h(a)]}
\]
for each $x$.
\end{theo}

Let $\mathcal{F}$ be the set of all functions $F:[a,b]\times [a,b]
\to N$ such that
\begin{enumerate}
   \item $F(x,x)=I$ for all $x$,
   \item $F$ is quasi-continuous with respect to its first variable, and
   \item there is a real nondecreasing function $g$ on $[a,b]$ such
         that $g(a)=0$ and
    %{\footnotesize
     $$\|F(t,x)-F(t,y)\|\le \|g(x)-g(y)\|$$%}
     for all $t$, $x$, and $y$. Such a function $g$ is called a
     {\it super function} for $F$.
\end{enumerate}

In this paper, we consider a problem that involves a function $F(t)$,
which is not an element of $\mathcal{F}$ since it is a function of
only one variable. However, we can define $\tilde{F}(t,x):= I+F(x)-F(t)$,
which can be shown to be an element of $\mathcal{F}$.
This allows the following two theorems to apply to our case.

\begin{theo}\label{qcint}
  If $F\in \mathcal{F}$, $Q:[a,b]\to N$ is quasi-continuous,
  $X=L$ or $X=R$, and $P$ is defined on $[a,b]$ by
  \[
    P(t)=(X)\int_a^t d_sF(t,s) \,Q(s),
  \]
  then $P$ is quasi-continuous. Moreover, if $F$ is continuous
  with respect to its first variable, then
  $P$ is continuous.
\end{theo}

\begin{theo}\label{homog}
  Given $F\in \mathcal{F}$, there is a unique $M \in \mathcal{F}$ that is
  a solution of
  \[
    M(t,x)=I+ (L)\int_x^t d_sF(t,s) \,M(s,x)
  \]
  for all $t$ and $x$. Moreover, if $F$ is continuous with respect
  to its first variable, then so is $M$.
\end{theo}

\begin{theo}\label{inhomog}
  If $F\in \mathcal{F}$ and $G$ is a quasi-continuous function from
  $[a,b]$ to $N$ or to $\mathbb{R}^{2n}$, then there is a unique
  quasi-continuous function $Y$ on $[a,b]$ such that
  \[
    Y(t)=G(t)+ (L)\int_x^t d_sF(t,s) \,Y(s).
  \]
\end{theo}

Now we state two convergence theorems for Stieltjes integrals
\cite{Hilde}, followed by a convergence theorem for a sequence
of functions.

\begin{theo}[Helly's Integral Convergence Theorem]\label{Helly}
  Let $f$ be a continuous function on $[a,b]$ and let $\{g_n\}$ be a
  sequence of functions, uniformly of bounded variation on $[a,b]$, converging
  to a function $g$ at every point of $[a,b]$.
  Then
  \[
    \lim_{n\to \infty} \int_a^b f\,dg_n = \int_a^b f\,dg.
  \]
\end{theo}

\begin{theo}[Osgood's Theorem] \label{Osgood}
  Let $g$ be a function of bounded variation on $[a,b]$ and let
  $\{f_n\}$ be a sequence of functions which is uniformly bounded
  and converges pointwise to a function $f$ on $[a,b]$. If
  $\int_a^b f_n \,dg$ and $\int_a^b f \,dg$ exist, then
  \[
    \lim_{n\to \infty} \int_a^b f_n \,dg = \int_a^b f\,dg.
  \]
\end{theo}

\begin{theo}[Helly's Pointwise Convergence Theorem]\label{Helly2}
  If $f_n$ is a sequence of functions, uniformly of bounded variation
  on $[a,b]$ such that $f_n(a)$ is bounded in $n$, then there exists a
  subsequence $f_{n_m}$ and a function $f$ of bounded variation such
  that $\lim_{m\to \infty} f_{n_m} =f$ at every point of $[a,b]$.
\end{theo}

Here we introduce some notation that will be used throughout this
paper. Let $\mathcal{P}[a,b]$ denote the set of all partitions of the
interval $[a,b]$. Also, define, for $f$ integrable, $I_f(t) := \int_a^t f$.

\section{The Initial Value Problem}\label{IVP}

We consider the system of $2n$ equations
\begin{equation}\label{genprob}
  \begin{gathered}
    dy=Ay\,dt+dP\,z \\
    dz=(dQ-\lambda dW)y+Dz\,dt
  \end{gathered}
\end{equation}
which can be written as the Stieltjes integral equation
\begin{equation} \label{SIE}
    \begin{bmatrix}
      y(t,\lambda) \\
      z(t,\lambda)
    \end{bmatrix}
 =  \begin{bmatrix}
      y(a,\lambda) \\
      z(a,\lambda)
    \end{bmatrix}
 +\int_a^t
    \begin{bmatrix}
      A(s)ds & dP(s) \\
      dM(s,\lambda) & D(s)ds
    \end{bmatrix}
    \begin{bmatrix}
      y(s,\lambda) \\
      z(s,\lambda)
    \end{bmatrix}
\end{equation}
on $[a,b] \times K$, where K is a compact set in $\mathbb{C}$, and
$M(t,\lambda)=Q(t)- \lambda W(t)$.
Here, $y$ and $z$ are $n$-vectors and $A$, $P$, $Q$, $W$, and $D$ are $n
\times n$ real matrices. We require the following
conditions on the coefficients:
\begin{equation}\label{hyp}
\begin{aligned}
& A,D \in L_1([a,b]); \\
& P=P^T \text{ is continuous and nondecreasing with } P(a)=0; \\
&  Q=Q^T \text{ is of bounded variation on } [a,b]; \\
& W=W^T \text{ is nondecreasing with } W(a)=0,
\end{aligned}
\end{equation}

\begin{rem} \rm
\begin{enumerate}
     \item It is a consequence of a matrix function being of bounded variation that
       each component is of bounded variation.
     \item When we say a real symmetric matrix A is positive, we mean
       that all of the eigenvalues of A are positive. This is
       equivalent to the condition that $\langle A \bar{x},\bar{x} \rangle >0$
       for all $\bar{x}\ne \bar{0}$. The meaning of a matrix function
       $A(t)$ being nondecreasing follows accordingly, i.e., $A(t)$ is
       nondecreasing if $A(t_2)-A(t_1)$ is nonnegative for $t_2\ge t_1$.
     \item The symmetry condition on $Q$, $P$, and $W$ is needed later
       for self-adjoint problems, but it is not needed to prove
       existence of a solution.
\end{enumerate}
\end{rem}

In Reid's paper \cite{Reid}, the derivatives of $P$, $Q$, and $W$ are required
to be $L_1$ functions. We are allowing for more generality in the
coefficients that does not require a Banach space.

We define the following terms:
$$L :=
  \Big\|
      \begin{bmatrix}
        y(a,\lambda) \\
        z(a,\lambda)
      \end{bmatrix}
\Big\| \quad
F(t,\lambda):=
    \begin{bmatrix}
      \int_a^t A(s)\,ds  & P(t) \\
      M(t,\lambda) & \int_a^t D(s)\,ds
    \end{bmatrix}
$$
$$
f(t,\lambda):= \int_a^t \| A(s) \| \,ds + \int_a^t \| D(s) \| \,ds +
\bigvee_a^t P+ \bigvee_a^t M(\cdot,\lambda).
$$
Note that $\bigvee_a^t M(\cdot,\lambda)\le \bigvee_a^t Q +
|\lambda| \bigvee_a^t W$.

Suppose we fix $\lambda$ and let
$Y(t):=
    \begin{bmatrix}
      y(t,\lambda) \\
      z(t,\lambda)
    \end{bmatrix}$.
Then (\ref{SIE}) can be written in the form
$Y(t)=Y(a)+ \int_a^t dF(s)\,Y(s)$. We now show that for fixed $\lambda$,
$\tilde{F}(t,x):= I+F(x,\lambda)-F(t,\lambda)$ is an element of the set
$\mathcal{F}$ which was defined in section \ref{pre}.
The first condition, $\tilde{F}(x,x)=I$ is clearly satisfied.
For the second condition, it suffices to show that $F(t,\lambda)$ is
quasi-continuous in its first variable. We know $P$ is continuous by
assumption. The remaining elements of $F$ are now shown to be of
bounded variation, which implies they are quasi-continuous. We know
$\int_a^t A$ and $\int_a^t D$ are of bounded
variation since $A$, $D\in L_1$, and $Q$ is of bounded variation by
assumption. $P$ and $W$ are of bounded variation because they are
nondecreasing.
The third condition is satisfied for $g(t):=f(t,x)$.
Since $\tilde{F}(t,x) \in \mathcal{F}$, Theorems \ref{qcint},
 \ref{homog}, and \ref{inhomog} apply to our problem, meaning that
equation \ref{genprob} has a unique quasi-continuous solution.
We repeat a proof of existence and uniqueness to establish certain uniform
bounds, to determine the continuity properties of the solution, and to
establish a Lipschitz condition with respect to a spectral parameter.

Define successive approximations for $k=1,2,3,\dots $ as follows:
Given initial conditions $y(a,\lambda)$ and $z(a,\lambda)$, let
\begin{gather}\label{SA1}
    \begin{bmatrix}
      y^{(0)}(t,\lambda) \\
      z^{(0)}(t,\lambda)
    \end{bmatrix}
 =  \begin{bmatrix}
      y(a,\lambda) \\
      z(a,\lambda)
    \end{bmatrix} \\
\label{SA}
    \begin{bmatrix}
      y^{(k)}(t,\lambda) \\
      z^{(k)}(t,\lambda)
    \end{bmatrix}
 = \begin{bmatrix}
      y^{(0)}(t,\lambda) \\
      z^{(0)}(t,\lambda)
    \end{bmatrix}
 +\int_a^t
    \begin{bmatrix}
      A(s)\,ds & dP(s) \\
      dM(s,\lambda) & D(s)\,ds
    \end{bmatrix}
    \begin{bmatrix}
      y^{(k-1)}(s,\lambda) \\
      z^{(k-1)}(s,\lambda)
    \end{bmatrix}
\end{gather}
First we examine some properties of the successive approximations.
The proof of the following lemma follows by a standard argument.

\begin{lemma}\label{SAprop1}
If successive approximations are defined as in \eqref{SA1},
\eqref{SA} subject to the conditions in \eqref{hyp}, then
$$
    \Big\|
      \begin{bmatrix}
        y^{(k)}(t,\lambda) \\
        z^{(k)}(t,\lambda)
      \end{bmatrix}
 -   \begin{bmatrix}
        y^{(k-1)}(t,\lambda) \\
        z^{(k-1)}(t,\lambda)
      \end{bmatrix}\Big\| \le L\dfrac{f(t,\lambda)^k}{k!}
$$
for $k=1,2,3,\dots$
\end{lemma}

\begin{lemma}\label{SAbounds}
If successive approximations are defined as in \eqref{SA1},
\eqref{SA} subject to the conditions in \eqref{hyp}, then
$    \Big\|
      \begin{bmatrix}
        y^{(k)}(t,\lambda) \\
        z^{(k)}(t,\lambda)
      \end{bmatrix}
\Big\|$ and
$\bigvee_a^t
      \begin{bmatrix}
        y^{(k)}(\cdot,\lambda) \\
        z^{(k)}(\cdot,\lambda)
      \end{bmatrix}$
are bounded independently of $k$ and $(t,\lambda)\in [a,b] \times K$.
\end{lemma}

\begin{proof}
   Using Lemma \ref{SAprop1},
    \begin{align*}
  \Big\|
      \begin{bmatrix}
        y^{(k)}(t,\lambda) \\
        z^{(k)}(t,\lambda)
      \end{bmatrix}
  \Big\|
&\le
  \Big\|
      \begin{bmatrix}
        y^{(0)}(t,\lambda) \\
        z^{(0)}(t,\lambda)
      \end{bmatrix}
    \Big\| +
\sum_{i=1}^k
  \Big\|
      \begin{bmatrix}
        y^{(i)}(t,\lambda) \\
        z^{(i)}(t,\lambda)
      \end{bmatrix}
 -  \begin{bmatrix}
        y^{(i-1)}(t,\lambda) \\
        z^{(i-1)}(t,\lambda)
      \end{bmatrix}
  \Big\| \\
&\le L+ L \sum_{i=1}^k \frac{f(t,\lambda)^i}{i!} \le Le^{f(t,\lambda)},
\end{align*}
Recall also that
\begin{equation}\label{indep}
  f(t,\lambda)\le \int_a^t\|A\| +\int_a^b\|D\|+\bigvee_a^b
  P+\bigvee_a^b Q +|\lambda| \bigvee_a^b W,
\end{equation}
which is bounded independently of $(t,\lambda)\in [a,b] \times K$.

We now turn to the bound on the total variaton.
Let $T=\{t_i\}_0^m \in \mathcal{P}[a,t]$.
Using Theorem \ref{intbd} and letting $C$ be a bound on
$\big\|
   \begin{bmatrix}
        y^{(k)}(\cdot,\lambda) \\
        z^{(k)}(\cdot,\lambda)
   \end{bmatrix}
    \big\|$,
\begin{align*}
\Big\|
\bigvee_a^t
      \begin{bmatrix}
        y^{(k)}(\cdot,\lambda) \\
        z^{(k)}(\cdot,\lambda)
      \end{bmatrix}
\big\|
&= \sup_{T\in P[a,t]} \sum_{i=1}^{m}
  \Big\|
      \begin{bmatrix}
        y^{(k)}(t_i) \\
        z^{(k)}(t_i)
      \end{bmatrix}
-  \begin{bmatrix}
        y^{(k)}(t_{i-1}) \\
        z^{(k)}(t_{i-1})
      \end{bmatrix}
  \Big\| \\
&= \sup_{T\in P[a,t]} \sum_{i=1}^{m}
  \Big\|
  \int_{t_{i-1}}^{t_i}
    \begin{bmatrix}
      A(s)\,ds & dP(s) \\
      dM(s,\lambda) & D(s)\,ds
    \end{bmatrix}
    \begin{bmatrix}
      y^{(k-1)}(s,\lambda) \\
      z^{(k-1)}(s,\lambda)
    \end{bmatrix}
  \Big\| \\
& \le \sup_{T\in P[a,t]} C \sum_{i=1}^{m}
  \int_{t_{i-1}}^{t_i} dv_F(s) \\
&=
C \int_a^t dv_F(s) \le C\int_a^t df(s,\lambda) = Cf(t,\lambda),
\end{align*}
since $\Delta v_F$ is bounded by $\Delta f$. Independence follows from
(\ref{indep}).
\end{proof}

Now we examine the continuity of solutions, starting by proving
that each successive approximation is quasi-continuous in the
first variable. This will be used in the subsequent theorem to prove
the solution is quasi-continuous.

\begin{lemma}\label{cont1}
 As a function of the first variable, $y^{(k)}(t,\lambda)$ is
 continuous and $z^{(k)}(t,\lambda)$ is quasi-continuous on
 $[a,b]$, for $k=0,1,2,\dotsc$. If $Q$ and $W$ are continuous, then
 $z^{(k)}(t,\lambda)$ is continuous on $[a,b]$, for $k=0,1,2,\dotsc$.
\end{lemma}

\begin{proof}
The quasi-continuity of $y^{(k)}(\cdot,\lambda)$ and
$z^{(k)}(\cdot,\lambda)$ follows from the fact
that they are of bounded variation (Lemma \ref{SAbounds}). Now we
prove that $y^{(k)}(\cdot,\lambda)$ is actually continuous.

Fix $\lambda \in K$. By definition of successive approximations,
for $k=1,2,3,\dotsc$
\[
  y^{(k)}(t,\lambda)=y(a,\lambda)+\int_a^t A(s)\,y^{(k-1)}(s,\lambda)\,ds
  +\int_a^t  dP(s)\,z^{(k-1)}(s,\lambda).
\]
Let $g(t)= \int_a^tA(s)\,y^{(k-1)}(s,\lambda)\,ds$, and $h(t)=\int_a^t
dP(s)\,z^{(k-1)}(s,\lambda)$. To show that $g$ is continuous, note that
  $$
\|g(t_2)-g(t_1)\|=
  \Big\|\int_{t_1}^{t_2} A(s)\,y^{(k-1)}(s,\lambda)\,ds\Big\|
  \le C \int_{t_1}^{t_2}\|A(s)\|\,ds,
$$
where $C$ is a uniform bound on
$ \big\| \begin{bmatrix}
      y^{(k)}(t,\lambda) \\
      z^{(k)}(t,\lambda)
    \end{bmatrix}\big\| $.
Since $A \in L_1([a,b])$, $I_{\|A\|}(t) :=\int_a^t
\|A(s)\|\,ds$ is absolutely continuous, which implies $g$ is continuous.

Now for $h$, we have
\begin{align*}
  \|h(t_2)-h(t_1)\| &=\Big\|\int_{t_1}^{t_2}
  dP(s)\,z^{(k-1)}(s,\lambda)\Big\| \\
  &\le \int_{t_1}^{t_2} dv_P(s) \,\|z^{(k-1)}(s,\lambda)\| \\
  &\le C (v_P(t_2)-v_P(t_1)).
\end{align*}
Since $P$ is continuous, $v_P$ is also continuous; hence $h$ is continuous.
Now both $g$ and $h$ are continuous, so
$y^{(k)}(t,\lambda)$ is continuous as a function of $t$.

If $Q$ and $W$ are continuous, then $z^{(k)}(t,\lambda)$ can be shown to
be continuous using a similar argument.
\end{proof}

\begin{theo}\label{existence}
 The initial value problem \eqref{SIE} has a
unique solution in the space of quasi-continuous functions. This
solution is bounded in norm and in total variation independently of
$t$ and $\lambda$ on $[a,b] \times K$.
\end{theo}

\begin{proof}
We prove the existence of a solution by showing it is
the limit of the sequence of successive approximations.
By Lemma \ref{SAprop1}, we have for $p>k$,
\begin{align*}
  \Big\|
  \begin{bmatrix}
     y^{(p)}(t,\lambda) \\
     z^{(p)}(t,\lambda)
  \end{bmatrix}
 - \begin{bmatrix}
     y^{(k)}(t,\lambda) \\
     z^{(k)}(t,\lambda)
  \end{bmatrix} \Big\|
&\le \sum_{i=k+1}^p
  \Big\|
       \begin{bmatrix}
     y^{(i)}(t,\lambda) \\
     z^{(i)}(t,\lambda)
       \end{bmatrix}
-  \begin{bmatrix}
     y^{(i-1)}(t,\lambda) \\
     z^{(i-1)}(t,\lambda)
       \end{bmatrix}
 \Big\| \\
&\le L\sum_{i=k+1}^p \frac{f(t,\lambda)^i}{i!}
\le L\frac{f(t,\lambda)^{k+1}}{(k+1)!}e^{f(t,\lambda)} \\
&\le L\frac{f(b,\lambda)^{k+1}}{(k+1)!}e^{f(b,\lambda)}
\to 0 \quad \text{as } k\to \infty.
\end{align*}
Thus $\{\left[ y^{(k)}, z^{(k)}\right]\}$
is uniformly Cauchy in $[a,b]$.
   Taking the limit as $k\to \infty$ in equation (\ref{SA})
   and using the fact that the convergence of the successive
   approximations is uniform, we get that
   $$
 \begin{bmatrix}
     y(t,\lambda) \\
     z(t,\lambda)
 \end{bmatrix}
 := \lim_{k\to \infty}
       \begin{bmatrix}
     y^{(k)}(t,\lambda) \\
     z^{(k)}(t,\lambda)
       \end{bmatrix}
 $$
is a solution. We know that
each successive approximation is quasi-continuous in $t$ and uniformly
bounded in norm and in total variation by Lemma \ref{SAbounds}, so this
solution has these same properties.

The proof of uniqueness is a standard argument, using the Gronwall inequality
from Theorem \ref{Gronwall}.
\end{proof}

The first part of the following corollary follows from the fact that the
functions $y^{(k)}(t,\lambda)$ are continuous with respect to $t$ and that they
converge uniformly to $y(t,\lambda)$ as $k\to \infty$. The
second part follows from Lemma \ref{cont1} and the uniform convergence
of the successive approximations.

\begin{cor}\label{ycont}
\begin{enumerate}
  \item
  $y(t,\lambda)$ is continuous with respect to $t$.
  \item
    If $Q$ and $W$ are continuous on $[a,b]$, then
    $z(t,\lambda)$ is continuous for $t\in [a,b]$.
\end{enumerate}
\end{cor}

We conclude with some additional properties of solutions. First a
Lipschitz condition in $\lambda$, and then a property concerning
total variation of the solution.

\begin{lemma}\label{Lipschitz}
For any $\lambda_1$ and $\lambda_2$ in a compact set $K\subseteq \mathbb{C}$, we
have the Lipschitz condition
\[
\Big\|
    \begin{bmatrix}
      y(t,\lambda_2) \\
      z(t,\lambda_2)
    \end{bmatrix}
-  \begin{bmatrix}
      y(t,\lambda_1) \\
      z(t,\lambda_1)
    \end{bmatrix}
\Big\| \le c|\lambda_2- \lambda_1 |,
\]
where $c$ is a constant independent of $t\in [a,b]$ and
$\lambda_1, \lambda_2 \in K$.
\end{lemma}

\begin{proof}
Let
\begin{align*}
  \phi (t) &:=
  \Big\|
    \begin{bmatrix}
      y(t,\lambda_2) \\
      z(t,\lambda_2)
    \end{bmatrix}
-  \begin{bmatrix}
      y(t,\lambda_1) \\
      z(t,\lambda_1)
    \end{bmatrix}
\Big\| \\
&=\| y(t,\lambda_2)-y(t,\lambda_1)\|+\| z(t,\lambda_2)-
z(t,\lambda_1) \|.
\end{align*}
Using the equation $y(t,\lambda_i)=y(a,\lambda_i)+\int_a^t
A(s)\,y(s,\lambda_i)\,ds+\int_a^t dP(s)\,z(s,\lambda_i)$ for $i=1,2$ and
the fact that $y(a,\lambda_1)=y(a,\lambda_2)$, we have
\[
  y(t,\lambda_2)-y(t,\lambda_1)=
  \int_a^t A(s)\,[y(s,\lambda_2)-y(s,\lambda_1)]\,ds +\int_a^t dP(s)
  \,[z(s,\lambda_2)-z(s,\lambda_1)].
\]
Similarly, we have
\begin{align*}
&z(t,\lambda_2)-z(t,\lambda_1)\\
&=  \int_a^t dQ(s)\,[y(s,\lambda_2)-y(s,\lambda_1)] \\
&\quad -\int_a^t dW(s)\,[\lambda_2 y(s,\lambda_2)
  - \lambda_1\,y(s,\lambda_1)]
  +\int_a^t D(s)\,[z(s,\lambda_2)-z(s,\lambda_1)]\,ds.
\end{align*}
 By adding and subtracting the term
$\int_a^t dW(s) \,\lambda_1 \,y(s,\lambda_2)$, we get
\begin{align*}
&z(t,\lambda_2)-z(t,\lambda_1)\\
&=  \int_a^t dQ(s)\,[y(s,\lambda_2)-y(s,\lambda_1)]
  -\int_a^t dW(s)\,(\lambda_2 - \lambda_1)\,y(s,\lambda_2)\\
&-\int_a^t dW(s) \lambda_1 \,[y(s,\lambda_2)-y(s,\lambda_1)]
  +\int_a^t D(s)\,[z(s,\lambda_2)-z(s,\lambda_1)]\,ds.
\end{align*}
Let $C>0$ be a bound on  $
\big\| \begin{bmatrix}
      y(t,\lambda) \\
      z(t,\lambda)
    \end{bmatrix}
\big\|$, from Theorem \ref{existence}. Then
\begin{align*}
  \phi (t) &\le \int_a^t \|A(s)\| \,\| y(s,\lambda_2)-y(s,\lambda_1)\|\,ds
  + \int_a^t dv_P(s) \,\| z(s,\lambda_2)-z(s,\lambda_1)\| \\
  & \quad + \int_a^t dv_Q(s) \,\| y(s,\lambda_2)-y(s,\lambda_1)\|
  + |\lambda_2 - \lambda_1| \int_a^t dv_W(s)  \,\|y(s,\lambda_2)\| \\
  & \quad + |\lambda_1|\int_a^t dv_W(s)
  \,\| y(s,\lambda_2)-y(s,\lambda_1)\|
  + \int_a^t \|D(s)\| \,\|z(s,\lambda_2)-z(s,\lambda_1)\| \,ds \\
  &\le \int_a^t dv_A(s)\,\phi (s) + \int_a^t dv_P(s)\,\phi (s)
  + (L)\int_a^t dv_Q(s)\,\phi (s) + C \,| \lambda_2
  - \lambda_1|\,\bigvee_a^b W \\
  & \quad  + |\lambda_1|\,(L)\int_a^t dv_W(s) \,\phi (s) +
  \int_a^t dv_D(s)\,\phi (s).
\end{align*}
By Theorem \ref{Gronwall}, it follows that
$$
\phi (t) \le \big(C\,|\lambda_2 - \lambda_1|\,\bigvee_a^b W \big) \,e^{m(t)},
$$
where $m(t)=v_A(t)+v_P(t)+v_Q(t)+kv_W(t)+v_D(t)$,
where $k=\sup_{\lambda \in K} \lambda$.
Then $\phi (t) \le c\,|\lambda_2 - \lambda_1|$, where
$c=C\bigvee_a^b W\,e^{m(b)}$.
\end{proof}

\begin{theo}\label{bv}
There is a nondecreasing function $q$ on $[a,b]$ such that
$\bigvee_{x_1}^{x_2} y(x,\lambda)\le q(x_2)-q(x_1)$ and
$\bigvee_{x_1}^{x_2} z(x,\lambda)\le q(x_2)-q(x_1)$ for $x_1<x_2$
and all $\lambda \in K$.
\end{theo}

\begin{proof}
Fix $\lambda \in K$. For any $T=\{t_i\}_0^m \in \mathcal{P}[x_1,x_2]$,
\begin{align*}
  \bigvee_{x_1}^{x_2} y(\cdot,\lambda)
  &= \sup_{T\in \mathcal{P}[x_1,x_2]}\sum_{i=1}^m
  \|y(t_i,\lambda)-y(t_{i-1},\lambda)\| \\
  &=  \sup_{T\in \mathcal{P}[x_1,x_2]}
  \sum_{i=1}^m \Big\|\int_{t_{i-1}}^{t_i}
  A(s)\,y(s,\lambda)\,ds + \int_{t_{i-1}}^{t_i} dP(s)\,z(s,\lambda)\Big\| \\
  &\le \sup_{T\in \mathcal{P}[x_1,x_2]}\sum_{i=1}^m \int_{t_{i-1}}^{t_i}
  \|A(s)\,y(s,\lambda)\,ds\| +
  \int_{t_{i-1}}^{t_i} \|dP(s)\,z(s,\lambda)\| \\
  &= \int_{x_1}^{x_2} \|A(s)\,y(s,\lambda)\,ds\| +
  \int_{x_1}^{x_2} \|dP(s)\,z(s,\lambda)\| \\
  &\le
  C\,\int_{x_1}^{x_2} \|A(s)\|\,ds+C\,\bigvee_{x_1}^{x_2} P \\
  &= q_1(x_2)-q_1(x_1),
\end{align*}
where $C$ is a bound
on $\big\|
     \begin{bmatrix}
      y(t,\lambda) \\
      z(t,\lambda)
    \end{bmatrix} \big\|$ and
  $q_1(x)=C\int_a^x \|A(s)\|\,ds + C\,\bigvee_a^x P$.
Note that $q_1$ is continuous since $P$ is continuous
\cite[Theorem 6.26]{Rudin}
and since $A\in L_1([a,b])$.
Now
\begin{align*}
  &\bigvee_{x_1}^{x_2} z(x,\lambda)\\
  &= \sup_{T\in \mathcal{P}[x_1,x_2]}\sum_{i=1}^m
  \|z(t_i,\lambda)-z(t_{i-1},\lambda)\| \\
  &=  \sup_{T\in \mathcal{P}[x_1,x_2]}\sum_{i=1}^m
  \Big\|\int_{t_{i-1}}^{t_i}
  (dQ(s)-\lambda \,dW(s))\,y(s,\lambda) +
  \int_{t_{i-1}}^{t_i} D(s)\,z(s,\lambda)\,ds \Big\| \\
  &\le  \sup_{T\in \mathcal{P}[x_1,x_2]}\sum_{i=1}^m \int_{t_{i-1}}^{t_i}
  \|(dQ(s)-\lambda \,dW(s)) \,y(s,\lambda)\| +
  \int_{t_{i-1}}^{t_i} \|D(s) \,z(s,\lambda)\,ds \| \\
  &\le C\,\int_{x_1}^{x_2} \|dQ(s)\| +C
  \,( \max_{\lambda \in K} |\lambda |)
  \int_{x_1}^{x_2} \|dW(s)\| +C\, \int_{x_1}^{x_2} \|D(s)\|\,ds \\
  &= q_2(x_2)-q_2(x_1),
\end{align*}
    where
  $q_2(x)=C\,\bigvee_a^x Q+ C
  \,( \max_{\lambda \in K} \|\lambda\| )
  \,\bigvee_a^x W+ C\,\int_a^x \|D(s)\|\,ds$.
Note that $q_2$ is nondecreasing but may have discontinuities due
to $Q$ and $W$.
Then $q(x):= q_1(x)+ q_2(x)$ satisfies the conclusion of
the theorem.
\end{proof}

\section{A Sequence of Initial Value Problems}\label{seqIVP}

In section \ref{IVP}, we examined the continuity of a solution to
the initial value problem as a function of $t$ and of $\lambda$. Now
we study continuity as a function of the problem coefficients.
To accomplish this,
we define a sequence of initial value problems satisfying the same
conditions as the problem in the previous section
(therefore the results from section \ref{IVP} apply to each problem in the
sequence). We will show that the sequence of solutions converges to the
solution of the limit problem.

The sequence of problems is defined as follows, for $n=1,2,3,...:$
\begin{equation}\label{genseqprob}
  \begin{gathered}
    dy_n=A_n\,y_n\,dt+dP_n\,z_n \\
    dz_n=(dQ_n-\lambda \,dW_n)\,y_n+D_n\,z_n\,dt
  \end{gathered}
\end{equation}
on $[a,b]\times K$, which written as a Stieltjes integral equation has the form
\[
    \begin{bmatrix}
      y_n(t,\lambda) \\
      z_n(t,\lambda)
    \end{bmatrix}
=   \begin{bmatrix}
      y_n(a,\lambda) \\
      z_n(a,\lambda)
    \end{bmatrix}
 +\int_a^t
    \begin{bmatrix}
      A_n(s)\,ds & dP_n(s) \\
      dM_n(s,\lambda) & D_n(s)\,ds
    \end{bmatrix}
    \begin{bmatrix}
      y_n(s,\lambda) \\
      z_n(s,\lambda)
    \end{bmatrix}
\]
where
\begin{equation}  \label{convtype}
\begin{gathered}
  A_n \to A \quad \text{weakly in } L_1([a,b]); \\
  D_n \to D \quad \text{weakly in } L_1([a,b]); \\
  P_n \to P \quad \text{uniformly in } [a,b]; \\
  Q_n \to Q \quad \text{pointwise in } [a,b]; \\
  W_n \to W \quad \text{pointwise in } [a,b]; \\
  y_n(a,\lambda)\to y(a,\lambda); \quad
  z_n(a,\lambda)\to z(a,\lambda); \\
  \bigvee_a^b Q_n \le \tilde{Q} , \quad
  \bigvee_a^b W_n \le \tilde{W} \quad\text{for all } n,
\end{gathered}
\end{equation}
for some constants $\tilde{Q}$ and $\tilde{W}$.
Note that $A_n, $ $P_n$,  $M_n=Q_n-\lambda W_n$,  and $D_n $
satisfy the same conditions as
$A$, $P$, $M$, and $D$, respectively, as stated in (\ref{hyp}).
Also note that we require $Q_n$ and $W_n$ to be uniformly of bounded
variation. The modes of convergence
required here are more inclusive than in previous work \cite{Reid},
which is appropriate due to the increased generality in the classes of
coefficients.

For the remainder of this section we fix $\lambda$ and write
$ \begin{bmatrix}
      y(t) \\
      z(t)
    \end{bmatrix}
=  \begin{bmatrix}
      y(t,\lambda) \\
      z(t,\lambda)
    \end{bmatrix}$,
understanding that there is still a dependence on $\lambda$.
We define the following terms for use in the proof:
\begin{gather*}
  L= \Big\|
    \begin{bmatrix}
      y(a) \\
      z(a)
    \end{bmatrix}
  \Big\|; \quad
  L_n=
  \Big\| \begin{bmatrix}
      y_n(a) \\
      z_n(a)
    \end{bmatrix}
  \Big\|;\\
f(t)=\int_a^t \| A(s) \| \,ds + \int_a^t \| D(s) \| \,ds +
  \bigvee_a^t P+ \bigvee_a^t M;\\
f_n(t)=\int_a^t \| A_n(s) \| \,ds + \int_a^t \| D_n(s) \| \,ds +
  \bigvee_a^t P_n+ \bigvee_a^t M_n.
\end{gather*}
Recall that in Lemma \ref{SAbounds} we proved the existence
of a uniform bound on all successive approximations for one
initial value problem. Now we extend that uniform bound to all
successive approximations for all of the initial value problems
in the sequence.

\begin{lemma}\label{SAseqbound} Let  $ \begin{bmatrix}
      y_n^{(k)}(t) \\
      z_n^{(k)}(t)
    \end{bmatrix}$,
 $k=1,2,3,\dots$ be the $kth$ successive approximation for the
  $nth$ problem in the sequence \eqref{genseqprob}
  for $n=1,2,3,\dots$, defined in the same manner as for
  problem \eqref{genprob}. Then under assumptions \eqref{convtype},
  $\big\|\begin{bmatrix}
      y_n^{(k)}(t) \\
      z_n^{(k)}(t)
    \end{bmatrix}
  \big\|$ and
  $\bigvee_a^b
    \begin{bmatrix}
      y_n^{(k)} \\
      z_n^{(k)}
    \end{bmatrix}$ are bounded independently of $n$, $k$, and $t$.
\end{lemma}

\begin{proof}
By the proof of Lemma \ref{SAbounds}, we have
\[
\Big\|
      \begin{bmatrix}
        y_n^{(k)}(t) \\
        z_n^{(k)}(t)
      \end{bmatrix}
    \Big\| \le L_ne^{f_n(b)}; \quad
\bigvee_a^t
      \begin{bmatrix}
        y_n^{(k)} \\
        z_n^{(k)}
      \end{bmatrix}
 \le L_ne^{f_n(b)}f_n(b).
\]
Therefore, the proof will be complete when we find a uniform bound on
$\{L_n\}_1^{\infty}$ and $\{f_n(b)\}_1^{\infty}$.
The sequence $\{L_n\}_1^{\infty}$ is bounded because
of the assumption that $L_n\to L$.

For $f_n(b)= \int_a^b \| A_n(s) \| \,ds + \int_a^b
\| D_n(s) \| \,ds + \bigvee_a^b P_n+ \bigvee_a^b M_n$,
first observe that the weak convergence of $\{A_n\}_1^{\infty}$
and $\{D_n\}_1^{\infty}$
guarantees a uniform bound on $\int_a^b \|A_n(s)\|ds$ and
$\int_a^b \|D_n(s)\|ds$ for all $n$ \cite[p.430]{Reid}. If $P(t)=
\{P_{ij}\}$, then for $T=\{t_i\}_1^m \in \mathcal{P}[a,b]$,
\begin{align*}
  \bigvee_a^b P &= \sup_{T\in \mathcal{P}[a,b]} \sum_{k=1}^m
  \|P(t_k)-P(t_{k-1})\| \\
  &=\sup_{T\in \mathcal{P}[a,b]}\sum_{i,j=1}^n \sum_{k=1}^m
  |P_{ij}(t_k)-P_{ij}(t_{k-1})|.
\end{align*}
Since $P$ is symmetric and nondecreasing, together with the fact that
$P(t)\ge P(a)=0$, any component of $P$ is
bounded by the square root of the product of the corresponding
diagonal entries (which are nonnegative),
i.e., $|P_{ij}|\le \sqrt{P_{ii}P_{jj}}$. Then using the fact that
$\sqrt{ab}\le \frac{1}{2}(a+b)$ for any positive numbers $a$ and $b$,
we have $|P_{ij}|\le \frac{1}{2}(P_{ii}+P_{jj})$. Then continuing
from above, we have
\begin{align*}
  \bigvee_a^b P
  &\le \frac{1}{2} \sum_{i,j=1}^n \sup_{T\in \mathcal{P}[a,b]} \sum_{k=1}^m
  |P_{ii}(t_k)-P_{ii}(t_{k-1})|+|P_{jj}(t_k)-P_{jj}(t_{k-1})| \\
   &= \frac{1}{2} \sum_{i,j=1}^n (\bigvee_a^b P_{ii} +
  \bigvee_a^b P_{jj}).
\end{align*}

Now define $\hat{P}(t):=
\sum_{i=1}^n P_{ii}(t)$. Since the diagonal entries of $P$
are nondecreasing and have initial value zero, we have
$\sum_{i=1}^n \bigvee_a^b P_{ii}=\bigvee_a^b \hat{P}$. Therefore,
\[
  \bigvee_a^b P \le
  \frac{1}{2} \sum_{i,j=1}^n \Big(\bigvee_a^b P_{ii}
  + \bigvee_a^b P_{jj}\Big)
  =\frac{1}{2} \Big(n\bigvee_a^b \hat{P}+n\bigvee_a^b \hat{P}\Big) \\
  =n\bigvee_a^b \hat{P} = n\hat{P}(b),
\]
with the last equality holding because $\hat{P}$ is continuous and
nondecreasing with $\hat{P}(a)=0$.
Similarly,
\[
  \bigvee_a^b P_j \le n\hat{P_j}(b).
\]
Since $\hat{P_j}(b)\to \hat{P}(b)$, there exists a bound on
$\{\hat{P_j}(b)\}_1^{\infty}$. Hence there exists a bound
on $\bigvee_a^b P_j$ for all $j$.

Recall we have a uniform bound on $\bigvee_a^b M_n$ for all $n$ by
assumption. Therefore, $\{f_n(b)\}_{n=1}^{\infty}$ is bounded and
the proof is completed.
\end{proof}

In preparation for showing the convergence of the sequence of
solutions, we now show the same is true for each set of successive
approximations.

\begin{lemma}\label{SAconv}
Under the assumptions in \eqref{convtype},
  for each $k=0,1,2,\dots$.
  \begin{enumerate}
  \item $y_n^{(k)}(t)\to y^{(k)}(t)$ uniformly on
    $[a,b]$ as $n\to \infty$, and
  \item $z_n^{(k)}(t)\to z^{(k)}(t)$ pointwise on
    $[a,b]$ as $n\to \infty$.
  \end{enumerate}
\end{lemma}

\begin{proof}
The proof is by induction on $k$. First, note that
\begin{gather*}
  \|y^{(0)}(t)-y_n^{(0)}(t)\| = \|y(a)-y_n(a)\|\to 0, \\
  \|z^{(0)}(t)-z_n^{(0)}(t)\| = \|z(a)-z_n(a)\|\to 0
\end{gather*}
by assumption. Assume the lemma to be true for some $k$, and now
consider $y$ and $z$ separately to show the lemma is true for
$k+1$.


\noindent\textbf{(1)} By definition of successive approximations,
 \begin{align*}
   y^{(k+1)}(t) -y_n^{(k+1)}(t) &= y(a)+\int_a^t [A(s)\,y^{(k)}(s)\,ds +
   dP(s)\,z^{(k)}(s)]\\
 &\quad - y_n(a)-\int_a^t [A_n(s)\,y_n^{(k)}(s)\,ds +
   dP_n(s)\,z_n^{(k)}(s)].
\end{align*}
Then
\begin{equation}\label{terms}
\begin{aligned}
&y^{(k+1)}(t) -y_n^{(k+1)}(t)\\
&= [y(a)-y_n(a)] +   \int_a^t [dP(s)-dP_n(s)]\,z^{(k)}(s) \\
&\quad -\int_a^t dP_n(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]
   +\int_a^t [A(s)-A_n(s)]\,y^{(k)}(s)\,ds \\
&\quad - \int_a^t A_n(s)\,[y_n^{(k)}(s)-y^{(k)}(s)]\,ds.
 \end{aligned}
\end{equation}
  We will show that each of these terms converges to zero uniformly in $[a,b]$.
First, $\|y(a)-y_n(a)\|\to 0$ by assumption.

For the second term in \eqref{terms}, we use integration by parts and
the fact that $P(a)=P_n(a)=0$ to write
\begin{equation}\label{second}
\begin{aligned}
&\int_a^t [dP(s)-dP_n(s)]\,z^{(k)}(s)\\
&=  -\int_a^t [P(s)-P_n(s)]\,dz^{(k)}(s) +[P(t)-P_n(t)]\,z^{(k)}(t).
\end{aligned}
\end{equation}
Since $P$ and $P_n$ are continuous and therefore bounded on $[a,b]$,
\begin{align*}
&\Big\|\int_a^t [dP(s)-dP_n(s)]\,z^{(k)}(s)\Big\|\\
&\le     \Big(\max_{t\in [a,b]}\|P(t)-P_n(t)\|\Big) \,
\bigvee_a^b z^{(k)} +    \|P(t)-P_n(t)\| \,\|z^{(k)}(t)\|.
\end{align*}
This converges to zero as $n\to\infty$ since $\bigvee_a^b
z^{(k)}$ and $\|z^{(k)}(t)\|$ are bounded (by Lemma \ref{SAbounds})
and since $P_n\to P$ uniformly by assumption.

Rewrite the third term in (\ref{terms}) as
\begin{equation}\label{third}
\begin{aligned}
&\int_a^t dP_n(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\\
&= \int_a^t [dP_n(s)-dP(s)]\,[z_n^{(k)}(s)-z^{(k)}(s)]
  +\int_a^t dP(s)\,[z_n^{(k)}(s)-z^{(k)}(s)].
\end{aligned}
\end{equation}
Now treat the first of the two integrals on the right-hand side of
(\ref{third}) the same way as in (\ref{second}). Performing
integration by parts and taking the norm, we get
\begin{align*}
&\Big\|\int_a^t [dP_n(s)-dP(s)]
  \,[z_n^{(k)}(s)-z^{(k)}(s)]\Big\|\\
& \le
  \Big(\max_{t\in [a,b]}\|P_n(t)-P(t)\| \Big) \bigvee_a^b [z_n^{(k)}-z^{(k)}]
  +\|P_n(t)-P(t)\|\,\|z_n^{(k)}(t)-z^{(k)}(t)\|,
\end{align*}
which converges uniformly
to zero for the same reasons as in (\ref{second}).

The pointwise convergence of the second integral on the right-hand
side of (\ref{third}) follows from Theorem \ref{Osgood}.
This theorem  applies since $P$ is of bounded variation, since
$[z_n^{(k)}(s)-z^{(k)}(s)]$ is uniformly bounded, and since
$z_n^{(k)}(t)\to z^{(k)}(t)$ pointwise by induction
hypothesis. Now we use the Ascoli-Arzela Theorem to show the
convergence is actually uniform and not only pointwise. The sequence
$\{\int_a^t dP(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\}_{n=1}^{\infty}$ is
uniformly bounded on $[a,b]$ since
\[
  \Big\|\int_a^t dP(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\Big\|
\le  \Big(\sup_{t\in [a,b]}\|z_n^{(k)}(t)-z^{(k)}(t)\|\Big)\,\bigvee_a^b P.
\]
This term is finite since $z^{(k)}$ and $z_n^{(k)}$ are
quasi-continuous by Lemma \ref{cont1} and are uniformly bounded by Lemma \ref{SAseqbound}.
To show the sequence is equicontinuous, fix $n$ and $k$ and let
$g(t)=\int_a^t dP(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]$. Now
\begin{align*}
\|g(t_2)-g(t_1)\| &=
\Big\|\int_{t_1}^{t_2} dP(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\Big\| \\
&\le \Big(\sup_{t\in [a,b]}\|z_n^{(k)}(t)-z^{(k)}(t)\|\Big)
\bigvee_{t_1}^{t_2} P\\
&= \Big(\sup_{t\in [a,b]}\|z_n^{(k)}(t)-z^{(k)}(t)\|\Big)
(v_P(t_2)-v_P(t_1))
\end{align*}
Now $P$ being continuous implies that $v_P$ is continuous, and therefore
$g$ is also continuous by this last inequality. Since
$\|z_n^{(k)}(t)-z^{(k)}(t)\|$ is bounded independently of $n$, $k$,
and $t$ by Lemma \ref{SAseqbound}, the sequence is equicontinuous
on $[a,b]$. Then by the
Ascoli-Arzela Theorem, the sequence has a uniformly
convergent subsequence. Since the sequence converges pointwise
and has a uniformly
convergent subsequence, the sequence itself converges uniformly.

We rewrite the fourth term on the right-hand side of (\ref{terms})
using the the function $I_A(t)=\int_a^t A(s)ds$.
Then exactly as done in (\ref{second}), integrate by
parts and take the norm to get
\begin{equation}\label{fifth}
\begin{aligned}
&\Big\|\int_a^t [A(s)-A_n(s)]\,y^{(k)}(s)\,ds\Big\|\\
& =
\Big\|\int_a^t [dI_A(s)-dI_{A_n}(s)]\,y^{(k)}(s)\,ds\Big\| \\
&\le \Big(\sup_{t\in [a,b]} \|I_A(s)-I_{A_n}(s)\|\Big)\,
\bigvee_a^b y^{(k)}
+\|I_A(t)-I_{A_n}(t)\|\,\|y^{(k)}(t)\|
\end{aligned}
\end{equation}
which converges uniformly to zero since $\bigvee_a^b y^{(k)}$ and
$\|y^{(k)}(t)\|$ are bounded, and since $A_n\to A$ weakly
in $L_1([a,b])$ implies that $I_{A_n}\to I_A$ uniformly
\cite[p.430]{Reid}.

For the last term in (\ref{terms}), we have
\[
\Big\|\int_a^t A_n(s)\,[y_n^{(k)}(s)-y^{(k)}(s)]\,ds\Big\|
 \le   \Big(\sup_{t\in [a,b]}\|y_n^{(k)}(t)-y^{(k)}(t)\|\Big)
  \int_a^t \|A_n(s)\|\,ds,
\]
Now $\int_a^t \|A_n(s)\|\,ds$ is uniformly bounded since
$A_n\to A$ weakly in $L_1([a,b])$ \cite[p.430]{Reid}, and
$y_n^{(k)}(t)\to y^{(k)}(t)$ uniformly by induction hypothesis.
Therefore the term on the right-hand side of (\ref{fifth}) converges
uniformly to zero.

We have shown that for $y^{(k+1)}(t)-y_n^{(k+1)}(t)$ written as
the sum of five terms, each term converges to zero uniformly on
$[a,b]$, completing the inductive step.

\noindent\textbf{(2)}
Similarly to part 1, we write
\begin{align*}
z^{(k+1)}(t)-z_n^{(k+1)}(t) &= z(a)+\int_a^t [dM(s)y^{(k)}(s)
     +D(s)\,z^{(k)}(s)\,ds] \\
&\quad   -z_n(a)-\int_a^t [dM_n(s)\,y_n^{(k)}(s)
     +D_n(s)\,z_n^{(k)}(s)\,ds].
\end{align*}
Then
\begin{equation}\label{zterms}
\begin{aligned}
& z^{(k+1)}(t)-z_n^{(k+1)}(t)\\
&= [z(a)-z_n(a)]
   +\int_a^t[dM(s)-dM_n(s)]\,y^{(k)}(s) \\
&\quad -\int_a^t dM_n(s)\,[y_n^{(k)}(s)-y^{(k)}(s)]
   +\int_a^t [D(s)-D_n(s)]\,z^{(k)}(s)\,ds \\
&\quad -\int_a^t D_n(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\,ds.
 \end{aligned}
\end{equation}
  We will show that each of these terms converges to zero
  pointwise in $[a,b]$.
First, $\|z(a)-z_n(a)\|\to 0$ by assumption.

The pointwise convergence to zero of the second term on the right-hand
side of (\ref{zterms}),
$\int_a^t[dM(s)-dM_n(s)]\,y^{(k)}(s)$, follows from Theorem \ref{Helly}.

For the third term on the right-hand side of (\ref{zterms}), we have
\[
  \Big\|\int_a^t
  dM_n(s)\,[y_n^{(k)}(s)-y^{(k)}(s)]\Big\| \le
  \Big(\sup_{t\in [a,b]} \|y_n^{(k)}(t)-y^{(k)}(t)\|\Big)
  \Big(\sup_n \bigvee_a^b M_n\Big),
\]
which converges uniformly to zero since $\bigvee_a^b M_n$ is
bounded independently of $n$ and since $y_n^{(k)}(t)\to
y^{(k)}(t)$ uniformly by inductive hypothesis.

We rewrite the fourth term on the right-hand side of (\ref{zterms})
using the function
$I_D(t)=\int_a^t D(s) ds$
and the same method of integrating by parts, getting
\begin{align*}
& \Big\|\int_a^t [D(s)-D_n(s)] \,z^{(k)}(s) \,ds \Big\|\\
& =
  \Big\|\int_a^t [dI_D(s)-dI_{D_n}(s)] \,z^{(k)}(s) \Big\| \\
&\le \Big(\sup_{t\in [a,b]} \|I_D(s)-I_{D_n}(s)\|\Big)\,\bigvee_a^b z^{(k)}
  +\|I_D(t)-I_{D_n}(t)\|\,\|z^{(k)}(t)\|.
\end{align*}
The right-hand side converges uniformly to zero since $\bigvee_a^b z^{(k)}$ and
$\|z^{(k)}(t)\|$ are bounded and since $D_n\to D$ weakly
in $L_1([a,b])$ implies that $I_{D_n}\to I_D$ uniformly.

We write the last term in (\ref{zterms}) as
\begin{align*}
&\int_a^t D_n(s)\,[z_n^{(k)}(s)-z^{(k)}(s)]\,ds \\
&=
  \int_a^t [dI_{D_n}(s)-dI_D(s)]\,[z_n^{(k)}(s)-z^{(k)}(s)]
  + \int_a^t dI_D(s)\,[z_n^{(k)}(s)-z^{(k)}(s)].
\end{align*}
The first of these two integrals can be shown to converge uniformly
to zero using integration by parts as seen before. Theorem
\ref{Osgood} can be used to show pointwise convergence of the
second integral. This
theorem applies because $D\in L_1([a,b])$ implies that $I_D$ is of
bounded variation, and because $z_n^{(k)}(t) \to z^{(k)}(t)$
pointwise by induction hypothesis. The convergence can be shown to be
uniform using the Ascoli-Arzela Theorem as in part 1, i.e., as in the
last term of (\ref{third}).

Having shown that $z^{(k+1)}(t)-z_n^{(k+1)}(t)$ can be written as
the sum of five terms, one converging
pointwise and four converging uniformly in $[a,b]$, the inductive
step is complete.
\end{proof}

\begin{cor}\label{zunif}
The convergence $\|z^{(k)}(t)-z_n^{(k)}(t)\|\to 0$ as
$n\to \infty$
is uniform on $[a,b]$ if $M_n\to M$ uniformly as $n\to \infty$
on $[a,b]$, for $k=1,2,\dots$.
\end{cor}
\begin{proof}
In the proof of Lemma \ref{SAconv} (2),
$z^{(k+1)}(t)-z_n^{(k+1)}(t)$ was written as the sum of five
terms, all of which converge uniformly to zero except for the term
$\int_a^t[dM(s)-dM_n(s)]y^{(k)}(s)$, which converges pointwise.
Under the assumption that $M_n\to M$ uniformly, we now show
that this term converges uniformly. Using integration by parts in the
same manner as used in the proof of the previous Lemma,
\begin{align*}
&\Big\|\int_a^t[dM(s)-dM_n(s)]\,y^{(k)}(s)\Big\| \\
&\le
  \Big(\sup_{t\in [a,b]} \| M(t)-M_n(t)\| \Big) \bigvee_a^b y^{(k)}
  +\| M(t)-M_n(t)\rVert\,\lVert y^{(k)(t)}\|,
\end{align*}
which converges uniformly to zero since $\bigvee_a^b y^{(k)}$ and
$\|y^{(k)}\|$ are bounded and since $M_n\to M$ uniformly.
\end{proof}

\begin{theo}\label{conv}
Let   $ \begin{bmatrix}
      y(t) \\
      z(t)
    \end{bmatrix}$
    be the solution of \eqref{genprob} and
  $\begin{bmatrix}
      y_n(t) \\
      z_n(t)
    \end{bmatrix}$
 be the solution of \eqref{genseqprob}, and assume
 \eqref{convtype} holds. Then as $n\to \infty$,
  \begin{enumerate}
    \item $y_n(t)\to y(t)$ uniformly in $[a,b]$, and
    \item  $z_n(t)\to z(t)$ pointwise in $[a,b]$.
  \end{enumerate}
\end{theo}

\begin{proof} Assume (\ref{convtype}) holds.

\noindent\textbf{(1)} We write
\[
  \|y(t)-y_n(t)\|\le \|y(t)-y^{(k)}(t)\|+
      \|y^{(k)}(t)-y_n^{(k)}(t)\| + \|y_n^{(k)}(t)- y_n(t)\|.
\]
As in the proof of Thoerem \ref{existence}, using Lemma
\ref{SAprop1}, we have
\[
  \Big\|
      \begin{bmatrix}
        y_n^{(k)}(t) \\
        z_n^{(k)}(t)
      \end{bmatrix}
 -  \begin{bmatrix}
        y_n(t) \\
        z_n(t)
      \end{bmatrix}
 \Big\| \le L_n \sum_{i=k+1}^{\infty} \dfrac{f_n(b)^i}{i!},
\]
  which tends to zero  as $k\to \infty$. Since the $L_n$ and
  $f_n(b)$ are bounded independent of $n$, this convergence is
  independent of $n$, and it is
  clearly independent of $t$ and $k$ as well. So given any $\epsilon
  >0$, there exists a $k_0>0$ such that if $k\ge k_0$, then
  $$\|y_n^{(k)}(t) - y_n(t) \|<\frac{\epsilon}{3}$$
  for all $n=1,2,3,\dots$. and $t\in [a,b]$.

  Also, by the convergence of successive approximations to the solution
  as shown in the proof of Theorem \ref{existence}, we can assume for
  the same $k_0$ that $k\ge k_0$ implies
  $$
  \|y(t)-y^{(k)}(t)\|<\frac{\epsilon}{3}
  $$
  for all $n=1,2,3,\dotsc$
  and $t\in [a,b]$. By Lemma \ref{SAconv}, we can assume that for
  the same $k_0$ that $k\ge k_0$ implies
  $$\|y_n^{(k)}(t) - y^{(k)}(t) \|<\frac{\epsilon}{3}.$$
  Therefore, for $k\ge k_0$,
  $\|y(t)-y_n(t)\|<\epsilon$
  for $n=1,2,3,\dots$ and $t\in [a,b]$.

\noindent\textbf{(2)}  Write
\[
  \|z(t)-z_n(t)\|\le \|z(t)-z^{(k)}(t)\|+
      \|z^{(k)}(t)-z_n^{(k)}(t)\| + \|z_n^{(k)}(t)- z_n(t)\|,
\]
      and in a similar manner obtain
      $\|z(t)-z_n(t)\|< \epsilon$,
      but this time the convergence of the middle term is pointwise rather than
      uniform (see Lemma \ref{SAconv}).
\end{proof}

The next corollary follows directly from Corollary \ref{zunif} and
Theorem \ref{SAconv} (2).

\begin{cor}
The convergence $z_n(t) \to z(t)$ is uniform if
$M_n \to M$ is uniform.
\end{cor}

Theorem \ref{conv} gives convergence of solutions to initial value problems
under weak conditions of convergence of the coefficients. In a second work,
we will show how this gives the continuous and differentiable dependence of
eigenvalues and eigenfunctions on the data--coefficients, boundary conditions,
and endpoints.



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\end{document}