\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 92, pp. 1--30.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/92\hfil Time-dependent domains] {Time-dependent domains for nonlinear evolution operators and partial differential equations} \author[C.-Y. Lin\hfil EJDE-2011/92\hfilneg] {Chin-Yuan Lin} \dedicatory{Dedicated to Professor Jerome A. Goldstein on his 70th birthday} \address{Chin-Yuan Lin \newline Department of Mathematics \\ National Central University \\ Chung-Li 320, Taiwan} \email{cylin@math.ncu.edu.tw} \thanks{Submitted June 15, 2011. Published July 18, 2011.} \subjclass[2000]{47B44, 47H20, 35J25, 35K20} \keywords{Dissipative operators; evolution equations; \hfill\break\indent parabolic and elliptic equations} \begin{abstract} This article concerns the nonlinear evolution equation \begin{gather*} \frac{du(t)}{dt} \in A(t)u(t), \quad 0 \leq s < t < T, \\ u(s) = u_0 \end{gather*} in a real Banach space $X$, where the nonlinear, time-dependent, and multi-valued operator $A(t) : D(A(t)) \subset X \to X$ has a time-dependent domain $D(A(t))$. It will be shown that, under certain assumptions on $A(t)$, the equation has a strong solution. Illustrations are given of solving quasi-linear partial differential equations of parabolic type with time-dependent boundary conditions. Those partial differential equations are studied to a large extent. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction} \label{S:A} Let $(X, \|\cdot\|)$ be a real Banach space with the norm $\|\cdot\|$, and let $T > 0$, $\omega$ be two real constants. Consider the nonlinear evolution equation $$\label{E:A} \begin{gathered} \frac{du(t)}{dt} \in A(t)u(t), \quad 0 \le s < t < T, \\ u(s) = u_0, \end{gathered}$$ where $A(t) : D(A(t)) \subset X \to X$ is a nonlinear, time-dependent, and multi-valued operator. To solve \eqref{E:A}, Crandall and Pazy \cite{Cran} made the following hypotheses of (H1)--(H3) and the $t$-dependence hypothesis of either (H4) or (H5), for each $0 \le t \le T$. \begin{itemize} \item[(H1)] $A(t)$ is dissipative in the sense that $\|u - v\| \leq \|(u - v) - \lambda (g - h)\|$ for all $u, v \in D(A(t))$, $g \in (A(t) - \omega)u, h \in (A(t) - \omega)v$, and for all $\lambda > 0$. Equivalently, $\Re(\eta(g - h)) \leq 0$ for some $\eta \in G(u - v) \equiv \{\xi \in X^{*} : \|u - v\|^2 = \xi(u - v) = \|\xi\|_{X^{*}}^2 \}$, the duality map of $(u - v)$ \cite{Mi}. Here $(X^{*}, \|.\|_{X^{*}})$ is the dual space of $X$ and $\Re(z)$ is the real part of a complex number $z$. \item[(H2)] The range of $(I - \lambda A(t))$ contains the closure $\overline{D(A(t))}$ of $D(A(t))$ for small $0 < \lambda < \lambda_0$ with $\lambda_0\omega < 1$. \item[(H3)] $\overline{D(A(t))} = \overline{D}$ is independent of $t$. \item[(H4)] There are a continuous function $f : [0, T] \to X$ and a monotone increasing function $L : [0, \infty) \to [0, \infty)$, such that $\|J_{\lambda}(t)x - J_{\lambda}(\tau)x\| \leq \lambda \|f(t) - f(\tau)\| L(\|x\|)$ for $0 < \lambda < \lambda_0, 0 \leq t, \tau \leq T,$ and $x \in \overline{D}$. Here $J_{\lambda}(t)x \equiv (I - \lambda A(t))^{-1}$ exists for $x \in \overline{D}$ by (H1) and (H2). \item[(H5)] There is a continuous function $f : [0, T] \to X$, which is of bounded variation on $[0, T]$, and there is a monotone increasing function $L : [0, \infty) \to [0, \infty)$, such that $\|J_{\lambda}(t)x - J_{\lambda}(\tau)x\| \leq \lambda \|f(t) - f(\tau)\| L(\|x\|) (1 + |A(\tau)x|)$ for $0 < \lambda < \lambda_0, 0 \leq t, \tau \leq T,$ and $x \in \overline{D}$. Here $|A(\tau)x| \equiv \lim_{\lambda \to 0}\|\frac{(J_{\lambda}(\tau) - I)x}{\lambda}\|$ by (H1) and (H2), which can equal $\infty$ \cite{Crand,Cran}. \end{itemize} By defining the generalized domain $\hat{D}(A(t)) \equiv \{x \in \overline{D(A(t))} : |A(t)x| < \infty \}$ \cite{Crand,Wes}, they \cite{Cran} proved, among other things, that the limit $$\label{E:DvpdesA} U(t, s)x \equiv \lim_{n \to \infty} \prod_{i=1}^{n} J_{\frac{t - s}{n}}(s + i \frac{t - s}{n})x$$ exists for $x \in \overline{D}$ and that $U(t, s)u_0$ is a unique solution, in a generalized sense, to the equation \eqref{E:A} for $u_0 \in \overline{D}$. Because of the restriction in (H3) that $\overline{D(A(t))} = \overline{D}$ is independent of $t$, the boundary condition in the example in \cite{Cran} does not depend on time. In this paper, in order to enlarge the scope of applications, we will consider a different set of hypotheses, the dissipativity condition (H1), the range condition (H2'), and the time-regulating condition (HA) below. Here a similar set of hypotheses was considered in \cite{Lin2} but the results were not satisfactory. \begin{itemize} \item[(H2')] The range of $(I - \lambda A(t))$, denoted by $E$, is independent of $t$ and contains $\overline{D(A(t))}$ for all $t \in [0, T]$ and for small $0 < \lambda < \lambda_0$ with $\lambda_0 \omega < 1$. \item[(HA)] There is a continuous function $f : [0, T] \to \mathbb{R}$, of bounded variation, and there is a nonnegative function $L$ on $[ 0, \infty)$ with $L(s)$ bounded for bounded $s$, such that, for each $0 < \lambda < \lambda_0$, we have $\{J_{\lambda}(t)x - J_{\lambda}(\tau)y : 0 \leq t, \tau \leq T, x, y \in E \} = S_1(\lambda) \cup S_2(\lambda).$ Here $S_1(\lambda)$ denotes the set \begin{align*} &\big\{ J_{\lambda}(t)x - J_{\lambda}(\tau)y : 0 \leq t, \tau \leq T, x, y \in E, \\ &\quad \|J_{\lambda}(t)x -J_{\lambda}(\tau)y\| \leq L(\|J_{\lambda}(\tau)y\|)|t - \tau| \big\}, \end{align*} and $S_2(\lambda)$ denotes the set \begin{align*} &\big\{J_{\lambda}(t)x - J_{\lambda}(\tau)y : 0 \leq t, \tau \leq T, x, y \in E, \|J_{\lambda}(t)x - J_{\lambda}(\tau)y\|\\ & \leq (1 - \lambda \omega)^{-1}[ \|x - y\| + \lambda |f(t) - f(\tau)| L(\|J_{\lambda}(\tau)y\|)( 1 + \frac{\|(J_{\lambda}(\tau) - I)y\|}{\lambda})] \big\}. \end{align*} \end{itemize} We will show that the limit in \eqref{E:DvpdesA} for $x \in \overline{\hat{D}(A(s))} = \overline{D(A(s))}$ exists, and that this limit for $x = u_0 \in \hat{D}(A(s))$ is a strong solution to the equation \eqref{E:A}, if $A(t)$ satisfies additionally an embedding property in \cite{Li} of embeddedly quasi-demi-closedness. We then apply the abstract theory to quasi-linear, parabolic partial differential equations with boundary conditions depending on time $t$. We finally show that, in those applications, each quantity $J_{\frac{t - s}{n}}(s + i\frac{t - s}{n})h = [I - \frac{t - s}{n} A(s + i \frac{t - s}{n})]^{-1}h, \quad i = 1, 2, \ldots, n$ is the limit of a sequence where each term in the sequence is an explicit function $F(\phi)$ of the solution $\phi = \pounds_0^{-1}(h, \varphi)$ to the elliptic equation with $\varphi \equiv 0$: $$\label{E:TimeC} \begin{gathered} -\Delta v(y) = h, \quad y \in \Omega, \\ \frac{\partial v}{\partial \nu} + v = \varphi, \quad y \in \partial \Omega. \end{gathered}$$ Here for the dimension of the space variable $y$ equal to 2 or 3, the $\phi = \pounds_0^{-1}(h, 0)$ and the solution $\pounds_0^{-1}(h, \varphi)$ to \eqref{E:TimeC} can be computed numerically and efficiently by the boundary element methods \cite{Gau,Sch}. See Sections \ref{S:D} and \ref{S:E} for more details of these, including how $F(\phi)$ depends on $\phi$, and for other aspects of the treated partial differential equations. There are many related works, to cite a few, we mention \cite{Ba,Br,Bre,Bro,Cr,Cra,Cran,En,Eng,Gol,Lie,Hi,Kat,Kato,Li,Lin0,Lin1,Lin2,Lin4,Mi,Oh,Pa,Paz,Ro,Roy,We}, especially the \cite{Lin4} for the recent development on nonlinear evolution equations where the hypothesis (H2) is relaxed. The rest of this article will be organized as follows. Section \ref{S:VaryA} obtains some preliminary estimates, and Section \ref{S:B} deals with the main results, where the nonlinear operator $A(t)$ is equipped with time-dependent domain $D(A(t))$. The Appendix in Section \ref{S:VaryB} examines the difference equations theory in our papers \cite{Lin0,Lin1,Lin4}, whose results, together with those in Section \ref{S:VaryA}, will be used to prove the main results in Section \ref{S:B}. Section \ref{S:D} studies applications to linear or nonlinear partial differential equations of parabolic type, in which each corresponding elliptic solution $J_{\frac{t - s}{n}}(s + i \frac{t - s}{n})h$ will be derived theoretically. Finally, Section \ref{S:E} follows Section \ref{S:D} but derives each elliptic solution $J_{\frac{t - s}{n}}(s + i \frac{t - s}{n})h$ as the limit of a sequence where each term in the sequence is an explicit function of the solution $\phi$ to the elliptic equation \eqref{E:TimeC} with $\varphi \equiv 0$. In either Section \ref{S:D} or Section \ref{S:E}, other aspects of the treated partial differential equations are considered. \section{Some preliminary estimates} \label{S:VaryA} Within this section and the Sections \ref{S:B} and \ref{S:VaryB}, we can assume, without loss of generality, that $\omega \ge 0$ where $\omega$ is the $\omega$ in the hypothesis (H1). This is because the case $\omega < 0$ is the same as the case $\omega = 0$. This will be readily seen from the corresponding proofs. To prove the main results Theorems \ref{T:XA} and \ref{T:XB} in Section \ref{S:B}, we need to make two preparations. One preparation is this section, and the other is the Appendix in Section \ref{S:VaryB}. \begin{proposition} \label{P:A} Let $A(t)$ satisfy the dissipativity condition (H1), the range condition (H2') , and the time-regulating condition (HA), and let $u_0$ be in $D(A(s)) \subset E$ % ($\subset where$ 0 \le s \le T $. Let$ 0 < \epsilon < \lambda_0 $be so chosen that$ 0 < \epsilon \omega < 1 $, and let$ 0 \le t_i = s + i \epsilon \le T $where$ i \in \mathbb{N} $. Then $$\label{E:VaryA} \|u_i - u_0\| \le \eta^{i}L(\|u_0\|)(i\epsilon) + [\eta^{i - 1}b_1 + \eta^{i - 2}b_2 + \dots + \eta b_{i - 1} + b_i]$$ and $$\label{E:VaryB} \begin{split} \|\frac{u_i - u_{i-1}}{\epsilon}\| &\leq [(c_ic_{i - 1} \dots c_2)L(\|u_0\|) \quad \text{or (c_ic_{i - 1} \dots c_3) L(\|u_1\|) or \dots } \\ &\quad \text{or c_iL(\|u_{i - 2}\|) or L(\|u_{i - 1}\|) }] + [(c_ic_{i -1}\dots c_1)a_0 \\ &\quad + (c_ic_{i - 1} \dots c_2)d_1 + (c_ic_{i - 1}\dots c_3)d_2 + \dots + c_id_{i - 1} + d_i]. \end{split}$$ Here$u_i = \prod_{j = 1}^{i}J_{\epsilon}(t_{j})u_0$exists uniquely by the hypotheses (H1) and (H2'); \\$\eta = (1 - \epsilon \omega)^{-1} > 1$; \\$b_i = \eta \epsilon \|v_0\| + \eta \epsilon |f(t_i) - f(s)| L(\|u_0\|)(1 + \|v_0\|)$, where$ v_0 $is any element in$ A(s)u_0 $; \\$c_i= \eta [1 + L(\|u_{i - 1}\|)|f(t_i) - f(t_{i-1})|]$; \\$d_i = \eta L(\|u_{i - 1}\|)|f(t_i) - f(t_{i-1})|$; \\ the right sides of \eqref{E:VaryB} are interpreted as$[L(\|u_0\|)] + [c_1a_0 + d_1]$for$ i = 1 $; \\$[c_2L(\|u_0\|)$or$L(\|u_1\|)] + [c_2c_1a_0 + c_2d_1 + d_2] $for$ i = 2 $; \dots, and so on; and $a_0 = \|\frac{u_0 - u_{-1}}{\epsilon}\|,$ where$ u_{-1} $is defined by$u_0 - \epsilon v_0 = u_{-1}$, with$ v_0 $any element in$ A(s)u_0 $. \end{proposition} \begin{proof} We will use the method of mathematical induction. Two cases will be considered, and for each case, we divide the proof into two steps. \subsection*{Case 1} Here \eqref{E:VaryA} is considered \textbf{Step 1.} Claim that \eqref{E:VaryA} is true for$ i = 1 $. This will follow from the arguments below. If$ (u_1 - u_0)\in S_1(\epsilon) $(defined in Section \ref{S:A}), then $\|u_1 - u_0\| = \|J_{\epsilon}(t_1)u_0 - J_{\epsilon}(s)(I - \epsilon A(s))u_0\| \leq L(\|u_0\|)|t_1 - s| \leq L(\|u_0\|)\epsilon,$ which is less than or equal to the right-hand side of \eqref{E:VaryA} with$ i = 1 $. On the other and, if$ (u_1 - u_0) \in S_2(\epsilon) $(defined in Section \ref{S:A}), then $\|u_1 - u_0\| \le \eta \|u_0 - u_0\| +\eta \epsilon \|v_0\| +\eta \epsilon |f(t_1) - f(s)|L(\|u_0\|)(1 + \|v_0\|),$ which is less than or equal to the right-hand side of \eqref{E:VaryA} with$ i = 1 $. Here$ v_0 $is any element in$ A(s)u_0 $. \textbf{Step 2.} By assuming that \eqref{E:VaryA} is true for$ i = i - 1 $, we shall show that it is also true for$ i = i $. If$ (u_i - u_0) \in S_1(\epsilon) $, then $\|u_i - u_0\| = \|J_{\epsilon}(t_i)u_{i - 1} - J_{\epsilon}(s)(I - \epsilon A(s))u_0\| \le L(\|u_0\|)|t_i - s| = L(\|u_0\|)(i\epsilon),$ which is less than or equal to the right side of \eqref{E:VaryA} with$ i = i $because of$ \eta^{i} > 1 $. On the other hand, if$ (u_i - u_0) \in S_2(\epsilon) $, then $\|u_i - u_0\| \le \eta \|u_{i - 1} - u_0\| + b_i$ where$\eta = (1 - \epsilon \omega)^{-1}and $b_i = \eta \epsilon \|v_0\| + \eta \epsilon|f(t_i) - f(s)| L(\|u_0\|)(1 + \|v_0\|).$ This recursive inequality, combined with the induction assumption, readily gives \begin{align*} \|u_i - u_0\| &\le \eta \{\eta^{i - 1}L(\|u_0\|)(i - 1)\epsilon + [\eta^{i - 2}b_1 + \eta^{i - 3}b_2 + \dots + \eta b_{i - 2} + b_{i - 1}]\} + b_i \\ &= \eta^{i}L(\|u_0\|)(i - 1)\epsilon + [\eta^{i - 1}b_1 + \eta^{i - 2}b_2 + \dots + \eta b_{i - 1} + b_i], \end{align*} which is less than or equal to the right-hand side of \eqref{E:VaryA} with i = i $because of$ (i - 1)\epsilon \le i\epsilon $. \subsection*{Case 2} Here \eqref{E:VaryB} is considered. \textbf{Step 1.} Claim that \eqref{E:VaryB} is true for$i = 1 $. This follows from the Step 1 in Case 1, because there it was shown that $\|u_1 - u_0\| \le L(\|u_0\|)\epsilon \quad \text{or b_1 },$ which, when divided by$ \epsilon $, is less than or equal to the right side of \eqref{E:VaryB} with$ i = 1 $. Here$ a_0 = \|v_0\| $, in which$ a_0 = (u_0 - u_{-1})/\epsilon $and$ u_{-1} \equiv u_0 - \epsilon v_0 $. \textbf{Step 2.} By assuming that \eqref{E:VaryB} is true for$ i = i - 1 $, we will show that it is also true for$ i = i $. If$ (u_i - u_{i - 1})\in S_1(\epsilon)$, then $\|u_i - u_{i - 1}\| \le L(\|u_{i - 1}\|) |t_i - t_{i - 1}| = L(\|u_{i - 1}\|) \epsilon.$ This, when divided by$ \epsilon $, has its right side less than or equal to one on the right-hand sides of \eqref{E:VaryB} with$ i = i $. If$ (u_i - u_{i-1}) \in S_2(\epsilon) , then \begin{align*} \|u_i - u_{i-1}\| &\leq (1 - \epsilon \omega)^{-1}[\|u_{i-1} - u_{i-2}\| \\ &\quad + \epsilon |f(t_i) - f(t_{i-1})| L(\|u_{i-1}\|)(1 + \frac{\|u_{i-1} - u_{i-2}\|}{\epsilon})]. \end{align*} By letting \begin{gather*} a_i = \frac{\|u_i - u_{i-1}\|} {\epsilon}, \\ c_i = (1 - \epsilon \omega)^{-1}[1 + L(\|u_{i - 1}\|)|f(t_i) - f(t_{i-1})|], \quad \text{and} \\ d_i = L(\|u_{i - 1}\|)(1 - \epsilon \omega)^{-1}|f(t_i) - f(t_{i-1})|, \end{gather*} it follows thata_i \leq c_ia_{i - 1} + d_i. Here notice that $u_0 - \epsilon v_0 = u_{-1}; \quad a_0 = \|\frac{u_0 - u_{-1}}{\epsilon}\| = \|v_0\|.$ The above inequality, combined with the induction assumption, readily gives \begin{align*} a_i &\leq c_i\big\{[(c_{i - 1}c_{i - 2} \dots c_2)L(\|u_0\|) \quad \text{or (c_{i - 1}c_{i - 2} \dots c_3) L(\|u_1\|)$or$ \dots $} \\ &\quad \text{or$ c_{i - 1}L(\|u_{i - 3}\|) $or$ L(\|u_{i - 2}\|) $}] + [(c_{i - 1}c_{i -2}\dots c_1)a_0 \\ &\quad + (c_{i - 1}c_{i - 2} \dots c_2)d_1 + (c_{i - 1}c_{i - 2}\dots c_3)d_2 + \dots \\ &\quad + c_{i - 1}d_{i - 2} + d_{i - 1}]\big\} + d_i \\ &\leq [(c_ic_{i - 1} \dots c_2)L(\|u_0\|) \quad \text{or$ (c_ic_{i - 1} \dots c_3) L(\|u_1\|) $or$ \dots $} \\ &\quad \text{or$ c_iL(\|u_{i - 2}\|) }] + [(c_ic_{i -1}\dots c_1)a_0 \\ &\quad + (c_ic_{i - 1} \dots c_2)d_1 + (c_ic_{i - 1}\dots c_3)d_2 + \dots + c_id_{i - 1} + d_i], \end{align*} each of which is less than or equal to one on the right sides of \eqref{E:VaryB} with i = i $. The induction proof is now complete. \end{proof} \begin{proposition} \label{P:VaryA} Under the assumptions of Proposition \ref{P:A}, the following are true if$ u_0 $is in$ \hat{D}(A(s)) = \{y \in \overline{D(A(s))}:|A(s)y| < \infty\} $: \begin{gather*} \|u_i - u_0\| \le K_1(1 - \epsilon \omega)^{-i}(2i + 1)\epsilon \le K_1e^{(T - s)\omega}(3)(T - s); \\ \|\frac{u_i - u_{i - 1}}{\epsilon}\| \le K_3; \end{gather*} where the constants$ K_1 $and$ K_3 $depend on the quantities: \begin{gather*} K_1 = K_1(L(\|u_0\|), (T - s), \omega, |A(s)u_0|, K_{B}); \\ K_2 = K_2(K_1, (T - s), \omega, \|u_0\|); \\ K_3 = K_3(L(K_2), (T - s), \omega, \|u_0\|, |A(s)u_0|, K_{B}); \\ K_{B} \text{ is the total variation of$ f $on$ [0, T] $}. \end{gather*} \end{proposition} \begin{proof} We divide the proof into two cases. \subsection*{Case 1} Here$ u_0 \in D(A(s)) $. It follows immediately from Proposition \ref{P:A} that \begin{gather*} \|u_i - u_0\| \le N_1(1 - \epsilon \omega)^{-i}(2i + 1)\epsilon \le N_1e^{(T - s)\omega}(3)(T - s); \\ \|\frac{u_i - u_{i - 1}}{\epsilon}\| \le N_3; \end{gather*} where the constants$ N_1 $and$ N_3 $depend on the quantities: \begin{gather*} N_1 = N_1(L(\|u_0\|), (T - s), \omega, \|v_0\|, K_{B}); \\ N_2 = N_2(N_1, (T - s), \omega, \|u_0\|); \\ N_3 = N_3(L(N_2), (T - s), \omega, \|u_0\|, \|v_0\|, K_{B}); \\ K_{B} \text{ is the total variation of$ f $on$ [0, T] $}. \end{gather*} We used here the estimate in \cite[Page 65]{Cran} $c_i\dots c_1 \leq e^{i\epsilon \omega} e^{e_i + \dots + e_1},$ where$ e_i = L(\|u_{i - 1}\|)|f(t_i) - f(t_{i-1})| $. \subsection*{Case 2} Here$ u_0 \in \hat{D}(A(s)) $. This involves two steps. \textbf{Step 1.} Let$ u_0^{\mu} = (I - \mu A(s)) ^{-1}u_0 $where$ \mu > 0 $, and let $u_i = \prod_{j = 1}^{i}J_{\epsilon}(t_{j})u_0; \quad u_i^{\mu} = \prod_{j = 1}^{i}J_{\epsilon}(t_{j})u_0^{\mu}.$ As in \cite[Lemma 3.2, Page 9]{Paz}, we have, by letting$ \mu \to 0 $, $u_0^{\mu} \to u_0;$ here notice that$ D(A(s)) $is dense in$ \hat{D}(A(s)) $. Also it is readily seen that $u_i^{\mu} = \prod_{k=1}^{i}(I - \epsilon A(t_{k})) ^{-1}u_0^{\mu} \to u_i = \prod_{k=1}^{i}(I - \epsilon A(t_{k}))^{-1} u_0$ as$ \mu \to 0 $, since$ (A(t) - \omega) $is dissipative for each$ 0 \leq t \leq T $. \textbf{Step 2.} Since$ u_0^{\mu} \in D(A(s)) $, Case 1 gives $$\label{E:TimeE} \begin{gathered} \|u_i^{\mu} - u_0^{\mu}\| \le N_1(L(\|u_0^{\mu}\|), (T - s), \omega, \|v_0^{\mu}\|, K_{B}) (1 - \epsilon \omega)^{-i}(2i + 1)\epsilon \\ \frac{\|u_i^{\mu} - u_{i-1}^{\mu}\|}{\epsilon} \leq N_3(L(N_2), (T - s), \omega, \|u_0^{\mu}\|, \|v_0^{\mu}\|, K_{B}), \end{gathered}$$ where $N_2 = N_2(N_1, (T - s), \omega, \|u_0^{\mu}\|),$ and$ v_0^{\mu} $is any element in$ A(s)(I - \mu A(s))^{-1}u_0 $. We can take $v_0^{\mu} = w_0^{\mu} \equiv \frac{(J_{\mu}(s) - I)u_0}{\mu},$ since$ w_0^{\mu} \in A(s)(I - \mu A(s))^{-1}u_0 $. On account of$ u_0 \in \hat{D}(A(s)) $, we have $\lim_{\mu \to 0}\|\frac{(J_{\mu}(s) - I)u_0}{\mu}\| = |A(s)u_0| < \infty.$ Thus, by letting$ \mu \to 0 $in \eqref{E:TimeE} and using Step 1, the results in the Proposition \ref{P:VaryA} follow. The proof is complete. \end{proof} \section{Main results} \label{S:B} Using the estimates in Section \ref{S:VaryA}, together with the difference equations theory, the following result will be shown in in Section \ref{S:VaryB}. \begin{proposition} \label{P:VaryB} Under the assumptions of Proposition \ref{P:XA}, the following inequality is true $a_{m, n} \le \begin{cases} L(K_2)|n\mu - m\lambda|, \quad &\text{if S_2(\mu) = \emptyset}; \\ c_{m, n} + s_{m, n} + d_{m, n} + f_{m, n} + g_{m, n}, \quad &\text{if S_1(\mu) = \emptyset }; \end{cases}$ where$ a_{m, n}, c_{m, n}, s_{m, n}, f_{m, n}, g_{m, n} $and$ L(K_2) $are defined in Proposition \ref{P:XA}. \end{proposition} In view of this and Proposition \ref{P:A}, we are led to the following claim. \begin{proposition} \label{P:XA} Let$ x \in \hat{D}(A(s)) $where$ 0 \le s \le T $, and let$ \lambda , \mu > 0 $,$ n, m \in \mathbb{N}, $be such that$ 0 \le (s +m \lambda), (s + n \mu) \le T $, and such that$ \lambda_0 > \lambda \geq \mu > 0 $for which$ \mu \omega, \lambda \omega < 1 $. If$ A(t) $satisfies the dissipativity condition (H1), the range condition (H2'), and the time-regulating condition (HA), then the inequality is true: $$\label{E:VaryC} a_{m, n} \le c_{m, n} + s_{m, n} + d_{m, n} + e_{m, n} + f_{m, n} + g_{m, n}.$$ Here \begin{gather*} a_{m, n} \equiv \|\prod_{i = 1}^{n} J_{\mu}(s + i \mu)x - \prod_{i = 1}^{m}J_{\lambda}(s + i \lambda)x\|; \\ \gamma \equiv (1 - \mu \omega)^{-1} > 1; \quad \alpha \equiv \frac{\mu}{\lambda}; \quad \beta \equiv 1 - \alpha; \\ c_{m, n} = 2K_1\gamma^{n}[(n\mu - m \lambda) + \sqrt{(n \mu - m \lambda)^2 + (n \mu)(\lambda - \mu)}]; \\ s_{m, n} = 2K_1\gamma^{n}(1 - \lambda \omega)^{-m} \sqrt{(n \mu - m \lambda)^2 + (n \mu)(\lambda - \mu)}; \\ d_{m, n} = [K_4\rho(\delta)\gamma^{n}(m \lambda)] + \{K_4\frac{\rho(T)}{\delta^2}\gamma^{n}[(m \lambda)(n \mu - m \lambda)^2 + (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2]\}; \\ e_{m, n} = L(K_2)\gamma^{n}\sqrt{(n \mu - m \lambda)^2 + (n \mu)(\lambda - \mu)}; \\ f_{m, n} = K_1[\gamma^{n}\mu + \gamma^{n}(1 - \lambda \omega)^{-m}\lambda]; \\ g_{m, n} = K_4\rho(|\lambda - \mu|)\gamma^{n}(m\lambda); \\ K_4 = \gamma L(K_2)(1 + K_3); \quad \delta > 0 \quad \text{is arbitrary}; \\ \rho(r) \equiv \sup\{|f(t) - f(\tau)| : 0 \leq t, \tau \leq T, |t - \tau| \leq r \} \end{gather*} where$\rho(r)$is the modulus of continuity of$ f $on$ [0, T] $; and$ K_1, K_2 $, and$ K_3 $are defined in Proposition \ref{P:VaryA}. \end{proposition} \begin{proof} We will use the method of mathematical induction and divide the proof into two steps. Step 2 will involve six cases. \textbf{Step 1.} \eqref{E:VaryC} is clearly true by Proposition \ref{P:VaryA}, if$ (m, n)= (0, n) $or$ (m, n) = (m, 0) $. \textbf{Step 2.} By assuming that \eqref{E:VaryC} is true for$ (m, n) = (m - 1, n - 1) $or$ (m, n) = (m, n - 1) $, we will show that it is also true for$ (m, n) = (m, n) . This is done by the arguments below. Using the nonlinear resolvent identity in \cite{Cr}, we have \begin{align*} a_{m, n} &= \|J_{u}(s + n \mu)\prod_{i = 1} ^{n - 1}J_{\mu}(s + i \mu)x\\ &\quad - J_{\mu}(s + m \lambda) [\alpha \prod_{i = 1}^{m - 1}J_{\lambda}( s + i \lambda)x + \beta \prod_{i = 1} ^{m}J_{\lambda}(s + i \lambda)x)]\|. \end{align*} Here \alpha = \mu/\lambda$and$ \beta = (\lambda - \mu)/\lambda$. Under the time-regulating condition (HA), it follows that, if the element inside the norm of the right side of the above equality is in$ S_1(\mu) $, then, by Proposition \ref{P:VaryA} with$ \epsilon = \mu $, $$\label{E:VaryMain2} a_{m, n} \le L(\|\prod_{i = 1}^{n}J_{\mu} (s + i \mu)x\|)|m \lambda - n \mu| \le L(K_2)|m \lambda - n \mu|,$$ which is less than or equal to the right-hand side of \eqref{E:VaryC} with$ (m, n) = (m, n) $, where$ \gamma ^{n} > 1 $. If that element instead lies in$ S_2(\mu) $, then, by Proposition \ref{P:VaryA} with$ \epsilon = \mu $, $$\label{E:VaryMain} \begin{split} a_{m, n} &\le \gamma (\alpha a_{m - 1, n - 1} +\beta a_{m, n - 1}) + \gamma \mu |f(s + m \lambda) - f(s + n \mu)|\\ &\quad\times L(\|\prod_{i = 1}^{n}J_{\mu}(s + i \mu)x\|)[1 + \|\frac{\prod_{i = 1}^{n}J_{\mu} (s + i \mu)x - \prod_{i = 1}^{n - 1} J_{\mu}(s + i \mu)x}{\mu}\|] \\ &\le [\gamma \alpha a_{m - 1, n - 1} + \gamma \beta a_{m, n - 1}] + K_4\mu \rho(|n \mu - m \lambda|), \end{split}$$ where$ K_4 = \gamma L(K_2)(1 + K_3) $and$ \rho(r) $is the modulus of continuity of$ f $on$ [0, T] $. From this, it follows that proving the relations is sufficient under the induction assumption: \begin{gather} \gamma \alpha p_{m - 1, n - 1} + \gamma \beta p_{m, n - 1} \le p_{m, n}; \label{E:VaryD} \\ \gamma \alpha q_{m - 1, n - 1} + \gamma \beta q_{m, n - 1} + K_4\mu \rho(|n \mu - m \lambda|) \le q_{m, n}; \label{E:VaryE} \end{gather} where$ q_{m, n} = d_{m, n} $, and$ p_{m, n} = c_{m, n} $or$ s_{m, n} $or$ e_{m, n} $or$ f_{m, n} $or$ g_{m, n} $. Now we consider five cases. \textbf{Case 1.} Here$ p_{m, n} = c_{m, n} . Under this case, \eqref{E:VaryD} is true because of the calculations, where $b_{m, n} = \sqrt{(n\mu - m\lambda)^2 + (n\mu)(\lambda - \mu)}$ was defined and the Schwartz inequality was used: \begin{gather*} \alpha[(n - 1)\mu - (m - 1)\lambda] + \beta[(n - 1)\mu - m\lambda] = (n \mu - m \lambda); \\ \begin{aligned} \alpha b_{m - 1, n - 1} + \beta b_{m, n - 1} &= \sqrt{\alpha}\sqrt{\alpha}b_{m - 1, n - 1} + \sqrt{\beta}\sqrt{\beta} b_{m, n - 1} \\ &\quad \le (\alpha + \beta)^{1/2}(\alpha b_{m - 1, n - 1}^2 + \beta b_{m, n - 1}^2)^{1/2} \\ &\quad \le \{(\alpha + \beta)(n\mu - m\lambda)^2 + 2(n\mu - m \lambda)[\alpha(\lambda - \mu) - \beta \mu] \\ &\quad + [\alpha(\lambda - \mu)^2 + \beta \mu^2] + (n - 1)\mu(\lambda - \mu)\}^{1/2} \\ &= b_{m, n}. \end{aligned} \end{gather*} Here $\alpha + \beta = 1; \quad \alpha(\lambda - \mu) - \beta \mu = 0; \quad \alpha(\lambda - \mu)^2 + \beta \mu^2 = \mu(\lambda - \mu).$ \textbf{Case 2.} Here p_{m, n} = s_{m, n} $. Under this case, \eqref{E:VaryD} is true, as is with the Case 1, by noting that $(1 - \lambda \omega)^{-(m - 1)} \le (1 - \lambda \omega)^{-m}.$ \textbf{Case 3.} Here$ q_{m, n} = d_{m, n} . Under this case, \eqref{E:VaryE} is true because of the calculations: \begin{align*} &\gamma \alpha d_{m - 1, n - 1} + \gamma \beta d_{m, n - 1} + K_4 \mu \rho(|n \mu - m \lambda|) \\ &\le \{\gamma \alpha[K_4\rho(\delta)\gamma^{n - 1}(m - 1)\lambda] + \gamma \beta[K_4 \rho(\delta)\gamma^{n - 1}(m \lambda)]\} \\ & \quad + \gamma \alpha \{K_4\frac{\rho(T)}{\delta^2} \gamma^{n - 1}[(m - 1)\lambda\left((n - 1)\mu - (m - 1)\lambda\right)^2 + (\lambda - \mu)\frac{(m - 1)m}{2}\lambda^2]\} \\ &\quad + \gamma \beta\{K_4\frac{\rho(T)}{\delta^2} \gamma^{n - 1}[(m \lambda)\left((n - 1)\mu - m\lambda\right)^2 + (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2]\} \\ &\quad + K_4\mu \rho(|n \mu - m \lambda|) \\ &= K_4\rho(\delta)\gamma^{n}[(\alpha + \beta)(m\lambda) - \alpha \lambda] \\ &\quad + K_4\frac{\rho(T)}{\delta^2} \gamma^{n}\{\alpha[(n\mu - m\lambda)^2 + 2(n \mu - m \lambda)(\lambda - \mu) + (\lambda - \mu)^2](m \lambda - \lambda) \\ &\quad + [\alpha (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2 - \alpha(\lambda - \mu)m\lambda^2] \\ &\quad + \beta[(n \mu - m \lambda)^2 - 2(n \mu - m \lambda)\mu + \mu^2](m \lambda) \\ &\quad +[\beta (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2]\} + K_4\mu \rho(|n\mu - m\lambda|) \\ &\quad \le K_4\rho(\delta)\gamma^{n}[(m\lambda) - \mu] + K_4\mu \rho(|n \mu - m \lambda|) \\ &\quad + K_4\frac{\rho(T)}{\delta^2}\gamma^{n}[(m\lambda)(n \mu - m \lambda)^2 + (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2 - \mu(n \mu - m \lambda)^2] \\ & \equiv r_{m, n}, \end{align*} where the negative terms [2(n\mu - m\lambda)(\lambda - \mu) + (\lambda - \mu)^2](-\lambda)$were dropped, $\alpha 2(n\mu - m\lambda)(\lambda - \mu) - \beta 2(n\mu - m\lambda)\mu = 0,$ and $[\alpha(\lambda - \mu)^2 + \beta \mu^2](m\lambda) = (m\lambda)\mu(\lambda - \mu),$ which cancelled $-\alpha(\lambda - \mu)m\lambda^2 = - (m\lambda)\mu(\lambda - \mu);$ it follows that$ r_{m, n} \le d_{m, n} , since \begin{align*} &K_4\mu \rho(|n\mu - m\lambda|) \\ & \le \begin{cases} K_4\mu \rho(\delta) \le K_4\mu\rho(\delta)\gamma^{n}, \quad &\text{if |n\mu - m\lambda| \le \delta }; \3pt] K_4\mu\rho(T)\frac{(n\mu - m\lambda)^2}{\delta^2} \le K_4\mu\rho(T)\gamma^{n}\frac{(n\mu - m\lambda)^2}{\delta^2}, \quad &\text{if |n\mu - m\lambda| > \delta }. \end{cases} \end{align*} \textbf{Case 4.} Here p_{m, n} = e_{m, n} . Under this case, \eqref{E:VaryD} is true, as is with the Case 1. \textbf{Case 5.} Here p_{m, n} = f_{m, n} . Under this case, \eqref{E:VaryD} is true because of the calculations: \begin{align*} \gamma \alpha f_{m - 1, n - 1} + \gamma \beta f_{m, n - 1} &= \gamma \alpha K_1[\gamma^{n - 1}\mu + \gamma^{n - 1} (1 - \lambda \omega)^{-(m - 1)}\lambda] \\ & \quad + \gamma \beta K_1[\gamma^{n - 1} \mu + \gamma^{n - 1}(1 - \lambda \omega)^{-m}\lambda] \\ & \le K_1[(\alpha + \beta)\gamma^{n}\mu + (\alpha + \beta)\gamma^{n}(1 - \lambda \omega)^{-m}\lambda], \\ &= f_{m, n}. \end{align*} \textbf{Case 6.} Here p_{m, n} = g_{m, n} . Under this case, \eqref{E:VaryD} is true because of the calculations: \begin{align*} \gamma \alpha g_{m - 1, n - 1} + \gamma \beta g_{m, n - 1} &\le K_4\gamma^{n}\rho(|\lambda - \mu|)\alpha(m - 1)\lambda + K_4\gamma^{n}\rho(|\lambda - \mu|)\beta(m \lambda) \\ & \le K_4\gamma^{n}\rho(|\lambda - \mu|)(\alpha + \beta)(m\lambda) \\ &= g_{m, n}. \end{align*} Now the proof is complete. \end{proof} Here is one of our two main results: \begin{theorem} \label{T:XA} If the nonlinear operator A(t) satisfies the dissipativity condition {\rm (H1)}, the range condition {\rm (H2')}, and the time-regulating condition {\rm (HA)} , then \[ U(s + t, s)u_0 \equiv \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t}{n}}(s + i \frac{t}{n})u_0 exists for u_0 \in \overline{\hat{D}(A(s))} = \overline{D(A(s))} $where$ s,t \ge 0 $and$ 0 \le (s + t) \le T $, and is the so-called a limit solution to the equation \eqref{E:A} . Furthermore, this limit$ U(s + t, s)u_0 $has the Lipschitz property $\|U( s + t, s)u_0 - U(s + \tau, s)u_0\| \le k |t - \tau|$ for$ 0 \le s + t, s + \tau \le T $and for$ u_0 \in \hat{D}(A(s)) $. \end{theorem} \begin{proof} For$ x \in \hat{D}(A(s)) $, it follows from Proposition \ref{P:XA}, by setting$ \mu = \frac{t}{n}, \lambda = \frac{t}{m}, $and$ \delta^2 = \sqrt{\lambda - \mu} $that, as$ n, m \to \infty $,$ a_{m, n} $converges to$ 0 $, uniformly for$ 0 \le (s + t) \le T $. Thus $\lim_{n \to \infty} \prod_{i = 1}^{n}J_{\frac{t}{n}}(s + i\frac{t}{n})x$ exists for$ x \in \hat{D}(A(s)) $. This limit also exits for$ x \in \overline{\hat{D}(A(s))} = \overline{D(A(s)))} $, on following the limiting arguments in Crandall-Pazy \cite{Cran}. On the other hand, setting$ \mu = \lambda = t/n$,$m = [\frac{t}{\mu}]$and setting$ \delta^2 = \sqrt{\lambda - \mu} , it follows that $$\label{E:TimeF} \lim_{n \to \infty}\prod_{i = 1}^{n}J_{\frac{t}{n}}(s + i \frac{t}{n})u_0 = \lim_{\mu \to 0}\prod_{i = 1}^{[\frac{t}{\mu}]}J_{\mu}(s + i\mu)u_0.$$ Now, to show the Lipschitz property, \eqref{E:TimeF} and Crandall-Pazy \cite[Page 71]{Cran} will be used. From Proposition \ref{P:VaryA}, it is derived that \begin{gather*} \begin{aligned} \|u_{n} - u_{m}\| &\le \|u_{n} - u_{n - 1}\| + \|u_{n - 1} - u_{n - 2}\| + \dots + \|u_{m + 1} - u_{m}\| \\ &\le K_3 \mu (n - m) \quad \text{for } x \in \hat{D}(A(s)), \end{aligned}\\ u_{n} = \prod_{i = 1}^{n}J_{\mu}(s + i \mu)x, \quad u_{m} = \prod_{i = 1}^{m}J_{\mu}(s + i \mu)x, \end{gather*} where n = [t/\mu]$,$m = [\tau/\mu]$,$t > \tau $and$ 0 < \mu < \lambda_0 $. The proof is completed by making$ \mu \to 0 $and using \eqref{E:TimeF}. \end{proof} Now discretize \eqref{E:A} as $$\label{E:TimeB} \begin{gathered} u_i - \epsilon A(t_i)u_i \ni u_{i-1}, \\ u_i \in D(A(t_i)), \end{gathered}$$ where$ n \in \mathbb{N} $is large, and$ \epsilon $is such that$ s \le t_i = s + i \epsilon \le T $for each$ i = 1, 2, \ldots, n $. Here notice that, for$ u_0 \in E $,$ u_i $exists uniquely by the hypotheses (H1) and (H2'). Let$ u_0 \in \hat{D}(A(s)) $, and construct the Rothe functions \cite{Ka,Ro}. Let \begin{gather*} \chi^{n}(s) = u_0, \quad C^{n}(s) = A(s), \\ \chi^{n}(t) = u_i, \quad C^{n}(t) = A(t_i) \quad \text{for } t \in (t_{i-1}, t_i], \end{gather*} and let \begin{gather*} u^{n}(s) = u_0, \\ u^{n}(t) = u_{i-1} + (u_i - u_{i-1}) \frac{t - t_{i-1}}{\epsilon} \quad \text{for } t \in (t_{i-1}, t_i] \subset [s, T]. \end{gather*} Since$ \|\frac{u_i - u_{i-1}}{\epsilon}\| \leq K_3 $for$ u_0 \in \hat {D}(A(s)) $by Proposition \ref{P:A}, it follows that, for$ u_0 \in \hat{D}(A(s)) $, $$\label{E:B} \begin{gathered} \lim_{n \to \infty}\sup_{t \in [0, T]}\|u^{n}(t) - \chi^{n}(t)\| = 0, \\ \|u^{n}(t) - u^{n}(\tau)\| \leq K_3|t - \tau|, \end{gathered}$$ where$ t, \tau \in (t_{i-1}, t_i] $, and that, for$ u_0 \in \hat{D}A(s)) $, $$\label{E:C} \begin{gathered} \frac{du^{n}(t)}{dt} \in C^{n}(t)\chi^{n}(t), \\ u^{n}(s) = u_0, \end{gathered}$$ where$ t \in (t_{i-1}, t_i] $. Here the last equation has values in$ B([s, T]; X) $, which is the real Banach space of all bounded functions from$ [s, T] $to$X$. \begin{proposition} \label{P:B} If$ A(t) $satisfies the assumptions in Theorem \ref{T:XA}, then $\lim_{n \to \infty}u^{n}(t) = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t - s}{n}}(s + i \frac{t}{n})u_0$ uniformly for finite$ 0 \le (s +t) \le T $and for$ u_0 \in \hat{D}(A(s)) $. \end{proposition} \begin{proof} The asserted uniform convergence will be proved by using the Ascoli-Arzela Theorem \cite{Roy}. Pointwise convergence will be proved first. For each$ t \in [s, T) $, we have$ t \in [t_i, t_{i+1}) $for some$ i $, and so$ i = [\frac{t - s}{\epsilon}] $, the greatest integer that is less than or equal to$ \frac{t - s}{\epsilon} $. That$ u_i $converges is because, for each above$ t , \label{E:TimeG} \begin{aligned} \lim_{\epsilon \to 0}u_i &= \lim_{\epsilon \to 0}\prod_{k=1}^{i}(I - \epsilon A(t_{k}))^{-1}u_0 \\ &= \lim_{n \to \infty} \prod_{k=1}^{n}[I - \frac{t - s}{n}A(s + k\frac{t - s}{n})]^{-1}u_0 \end{aligned} by \eqref{E:TimeF}, which has the right side convergent by Theorem \ref{T:XA}. Since $\|\frac{u_i - u_{i - 1}}{\epsilon}\| \le K_3$ for u_0 \in \hat{D}(A(s)) $, we see from the definition of$ u^{n}(t) $that $\lim_{n \to \infty}u^{n}(t) = \lim_{\epsilon \to 0}u_i = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t - s}{n}}(s + i\frac{t - s}{n})u_0$ for each$ t $. On the other hand, due to $\|\frac{u_i - u_{i - 1}}{\epsilon}\| \le K_3$ again, we see that$ u^{n}(t) $is equi-continuous in$ C([s, T]; X) $, the real Banach space of all continuous functions from$ [s, T] $to$X$. Thus it follows from the Ascoli-Arzela theorem \cite{Roy} that, for$ u_0 \in \hat{D}(A(s)) $, some subsequence of$ u^{n}(t) $(and then itself) converges uniformly to some $u(t) = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t - s}{n}}(s + i \frac{t - s}{n})u_0 \in C([s, T]; X).$ This completes the proof. \end{proof} Now consider a strong solution. Let$ (Y, \|\cdot\|_{Y}) $be a real Banach space, into which the real Banach space$ (X, \|\cdot\|) $is continuously embedded. Assume additionally that$ A(t) $satisfies the embedding property of embeddedly quasi-demi-closedness: \begin{itemize} \item[(HB)] If$ t_{n} \in [0, T] \to t $, if$ x_{n} \in D(A(t_{n})) \to x $, and if$ \|y_{n}\| \leq k $for some$ y_{n} \in A(t_{n})x_{n} $, then$ \eta(A(t)x) $exists and $|\eta(y_{n_{l}}) - z| \to 0$ for some subsequence$ y_{n_{l}} $of$ y_{n} $, for some$ z \in \eta(A(t)x) $, and for each$ \eta \in Y^{*} \subset X^{*} $, the real dual space of$ Y $. \end{itemize} Here is the other main result. \begin{theorem} \label{T:XB} Let$ A(t) $satisfy the dissipativity condition {\rm (H1)}, the range condition {\rm (H2')}, the time-regulating condition {\rm (HA)}, and the embedding property {\rm (HB)}. Then equation \eqref{E:A}, for$ u_0 \in \hat{D}(A(s)) $, has a strong solution $u(t) = \lim_{n \to \infty} \prod_{i = 1}^{n}J_{\frac{t - s}{n}}(s + i \frac{t}{n})u_0$ in$ Y $, in the sense that \begin{gather*} \frac{d}{dt}u(t) \in A(t)u(t) \quad \text{in$ Y $for almost every$ t \in (0, T) $}; \\ u(s) = u_0. \end{gather*} The solution is unique if$ Y \equiv X $. Furthermore, $\|u(t) - u(\tau)\|_{X} \le K_3|t - \tau|$ for$ 0 \le s \le t$,$\tau \le T $, a result from Theorem \ref{T:XA}. \end{theorem} The results in the above theorem follow from Theorem \ref{T:XA} and the proof in \cite[page 364]{Li}, \cite[pages 262-263]{Lin2}. \begin{remark} \label{T:XC} \rm The results in Sections \ref{S:VaryA} and \ref{S:B} are still true if the range condition (H2') is replaced by the weaker condition (H2'') below, provided that the initial conditions$ u_0 \in \hat{D}(A(s)) (\supset D(A(s)) ) $and$ u_0 \in \overline{\hat{D}(A(s))} = \overline{D(A(s))} (\supset D(A(s)) )$are changed to the condition$ u_0 \in D(A(s)) $. This is readily seen from the corresponding proofs. Here \begin{itemize} \item[(H2'')] The range of$ (I - \lambda A(t)) $, denoted by$ E $, is independent$ t $and contains$ D(A(t)) $for all$ t \in [0, T] $and for small$ 0 < \lambda < \lambda_0 $with$ \lambda_0\omega < 1 $. \end{itemize} \end{remark} \section{Applications to partial differential equations (I)} \label{S:D} Within this section,$ K $will denote a constant that can vary with different occasions. Now we make the following assumptions: \begin{itemize} \item[(A1)]$ \Omega $is a bounded smooth domain in$ \mathbb{R}^{n}, n \geq 2, $and$ \partial \Omega $is the boundary of$ \Omega $. \item[(A2)]$ \nu(x) $is the unit outer normal to$ x \in \partial \Omega $, and$ \mu $is a real number such that$ 0 < \mu < 1 $. \item[(A3)]$ \alpha(x, t, p) \in C^2(\overline \Omega \times \mathbb{R}^{n}) $is true for each$ t \in [0, T] $, and is continuous in all its arguments. Furthermore,$ \alpha(x, t, p) \geq \delta_0 > 0 $is true for all$ x, z $, and all$ t \in [0, T] $, and for some constant$ \delta_0 > 0 $. \item[(A4)]$ g(x, t, z, p) \in C^2(\overline \Omega \times \mathbb{R} \times \mathbb{R}^{n}) $is true for each$ t \in [0, T] $, is continuous in all its arguments, and is monotone non-increasing in$ z $for each$ t , x $, and$ p $. \item[(A5)]$ \frac{g(x,t, z, p)}{\alpha(x, t, p)} $is of at most linear growth in$ p $, that is , $| \frac{g(x, t, z, p)}{\alpha(x, t, p)} | \leq M(x, t, z)(1 + |p|)$ for some continuous function$ M $and for all$ t \in [0, T] $when$ |p| $is large enough. \item[(A6)]$ \beta(x, t, z) \in C^{3}(\Omega \times \mathbb{R}) $is true for each$ t \in [0, T] $, is continuous in all its arguments, and is strictly monotone increasing in$ z $so that$ \beta_{z} \geq \delta_0 > 0 $for the constant$ \delta_0 > 0 $in (A3). \item[(A7)] \begin{gather*} |\alpha(x, t, p) - \alpha(x, \tau, p)| \leq |\zeta(t) - \zeta(\tau)| N_1(x, |p|), \\ |g(x, t, z, p) - g(x, \tau, z, p)| \leq |\zeta(t) - \zeta(\tau)|N_2(x, |z|, |p|), \\ |\beta(x, t, z) - \beta(x, \tau, z)| \leq |t - \tau|N_3(x, |z|) \end{gather*} are true for some continuous positive functions$ N_1, N_2, N_3 $and for some continuous function$ \zeta $of bounded variation. \end{itemize} Define the$t$-dependent nonlinear operator$ A(t) : D(A(t)) \subset C(\overline \Omega) \to C(\overline \Omega) $by \begin{gather*} D(A(t)) = \{ u \in C^{2 + \mu}(\overline \Omega) : \frac{\partial u}{\partial \nu} + \beta(x, t, u) = 0 \quad \text{on$ \partial \Omega $} \} \quad \text{and} \\ A(t)u = \alpha(x, t, Du) \Delta u + g(x, t, u, Du) \quad \text{for$ u \in D(A(t)) $}. \end{gather*} \begin{example} \label{T:XD} \rm Consider the equation $$\label{E:XA} \begin{gathered} \frac{\partial}{\partial t}u(x, t) = \alpha(x, t, Du)\Delta u + g(x, t, u, Du), \quad (x, t) \in \Omega \times (0, T), \\ \frac{\partial}{\partial \nu}u + \beta(x, t, u) = 0 , \quad x \in \partial \Omega, \\ u(x, 0) = u_0, \end{gathered}$$ for$ u_0 \in D(A(0)) $. The above equation has a strong solution $u(t) = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t}{n}}(i \frac{t}{n})u_0$ in$ L^2(\Omega) $with $\frac{\partial}{\partial \nu}u(t) + \beta(x, t, u(t)) = 0, \quad x \in \partial \Omega,$ and the solution$ u(t) $satisfies the property $\sup_{t \in [0, T]}\|u(t)\|_{C^{1 + \mu} (\overline{\Omega})} \le K$ for some constant$ K $. \end{example} \begin{proof} It was shown in \cite[Pages 264-268]{Lin2} that$ A(t) $satisfies the dissipativity condition (H1), the range condition (H2'') with$ E = C^{\mu}(\overline{\Omega}) $for any$ 0 < \mu < 1 $, and satisfies the time-regulating condition (HA) and the embedding property (HB). Here the third line on \cite[Page 268]{Lin2}: $\times [\|N_2(z, \|v\|_{\infty}, \|Dv\|_{\infty})\|_{\infty} + \frac{\|N_1(z, \|Dv\|_{\infty})\|_{\infty}}{\delta_1} \|A(\tau)v\|_{\infty})]$ should have$ \|A(\tau)v\|_{\infty} $replaced by $[\|A(\tau)v\|_{\infty} + \|g(z, \tau, v, Dv)\|_{\infty}].$ Hence Remark \ref{T:XC} and Theorems \ref{T:XA} and \ref{T:XB} are applicable. It remains to prove that$ u(t) $satisfies the mentioned property and the middle equation in \eqref{E:XA} in$ C(\overline{\Omega}) $. This basically follows from \cite[pages 264-268]{Lin2}. To this end, the$ u_i $in \eqref{E:TimeB} will be used. Since$ A(t) $satisfies (H1), (H2''), and (HA), it follows from Proposition \ref{P:VaryA} and Remark \ref{T:XC} that $\|\frac{u_i - u_{i - 1}}{\epsilon}\| = \|A(t_i)u_i\|_{\infty} \le K_3 \quad \text{and} \quad \|u_i\|_{\infty} \leq K_2.$ Thus, from linear$ L^{p} $elliptic theory \cite{Tr,Gi2}, it follows that$\|u_i\|_{W^{2, p}} \le K$for some constant$ K $, whence $$\label{E:TimeLA} \|u_i\|_{C^{1 + \eta}} \le K$$ for any$ 0 < \eta < 1 $by the Sobolev embedding theorem \cite{Gi2}. This, together with the interpolation inequality \cite{Gi2} and the Ascoli-Arzela theorem \cite{Gi2, Roy}, implies that a convergent subsequence of$ u_i $converges in$ C^{1 + \mu}(\overline{\Omega}) $for any$ 0 < \lambda < \eta < 1 $. Therefore, on account of \eqref{E:TimeG} and Proposition \ref{P:B}, $\sup_{t \in [0, T]}\|u(t)\|_{C^{1 + \mu}} \le K$ results for$ u_0 \in D(A(0)) $, and$ u(t) $satisfies the middle equation in \eqref{E:XA} in$ C(\overline{\Omega}) $. The proof is complete \end{proof} Consider the linear equation $$\label{E:ZA} \begin{gathered} \frac{\partial u(x,t)}{\partial t} = \sum_{i, j = 1}^{n}a_{ij}(x, t)D_{ij}u(x,t) + \sum_{i = 1}^{n}b_i(x, t)D_iu(x, t) + c(x, t)u(x, t) \\ \quad \text{for (x, t) \in \Omega \times (0, T) }, \\ \frac{\partial}{\partial \nu}u + \beta(x, t) u = 0, \quad x \in \partial \Omega, \\ u(x, 0) = u_0, \end{gathered}$$ in which the following are assumed. Let$ a_{ij}(x, t) = a_{ji}(x, t) $, and let $\lambda_{\rm min} |\xi|^2 \leq \sum_{i, j}^{n}a_{ij}(x, t)\xi_i\xi_{j} \le \lambda_{\rm max} |\xi|^2$ for some positive constants$ \lambda_{\rm min}$,$\lambda_{\rm max} $, for all$ \xi \in \mathbb{R}^{n} $, and for all$ x, t $. Let $a_{ij}(x, t),\; b_i(x, t),\; c(x, t) \in C^{\mu}(\overline{\Omega})$ uniformly for all$ t $, be continuous in all their arguments, and be of bounded variation in$ t $uniformly for$ x $. Let$c(x, t) \le 0$for all$ x, t $, $\beta(x, t) \in C^{1 + \mu}(\overline{\Omega}), \quad 0 < \mu < 1$ for all$ t $, and$ \beta(x, t) \ge \delta > 0 $for some constant$ \delta > 0 $. Finally, let$ \beta(x, t) $and$ c(x, t) $be continuous in all its arguments, and let$ \beta(x, t) $be Lipschitz continuous in$ t $uniformly for$ x $. \begin{example} \label{T:ZA} \rm If$ \sum_{i, j}a_{ij}(x, t)D_{ij}u(x, t) = a_0(x, t)\Delta u(x, t) $for some$ a_0(x, t) $, then the equation \eqref{E:ZA}, for$ u_0 \in D(A(0)) $, has a strong solution $u(t) = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t}{n}}(i \frac{t}{n})u_0$ in$ L^2(\Omega) $with $\frac{\partial}{\partial \nu}u(t) + \beta(x, t)u(t) = 0, \quad x \in \partial \Omega,$ and$ u(t) $satisfies the property $\sup_{t \in [0, T]}\|u(t)\|_{C^{1 + \mu}(\overline{\Omega})} \le K.$ \end{example} \begin{proof} Linear elliptic equation theory \cite[Pages 128-130]{Gi2} shows that the corresponding operator$ A(t) $satisfies the range condition (H2'') with$ E = C^{\mu}(\overline{\Omega}) $. The arguments in \cite[Pages 267-268]{Lin2} shows that$ A(t) $satisfies the dissipativity condition (H1), the time-regulating condition (HA), and the embedding property (HB). The proof is complete, after applying Remark \ref{T:XC}, Theorems \ref{T:XA} and \ref{T:XB}, and the proof for Theorem \ref{T:XD}. \end{proof} \begin{example} \label{T:ZB} \rm Suppose that $a_{ij}(x),\; b_i(x),\; c(x) \in C^{1 + \mu}(\overline{\Omega}),\; \beta(x) \in C^{2 + \mu}(\overline{\Omega})$ are independent of$ t $, where$ 0 < \mu < 1 $. Then equation \eqref{E:ZA} has a unique classical solution $u(t) = \lim_{n \to \infty}\prod_{i = 1}^{n} J_{\frac{t}{n}}(i \frac{t}{n})u_0 = \lim_{n \to \infty}(I - \frac{t}{n}A)^{-n}u_0$ for$u_0 \in D(A) $with$ Au_0 \in D(A) $, and the solution has the properties that$ \frac{du(t)}{dt} $is Lipschitz continuous in$ t $, and that $\|\frac{du}{dt}\|_{C^{1 + \mu}(\overline{\Omega})} \le K.$ Furthermore,$ \frac{d}{dt}u $is differentiable in$ t $and$ \frac{d^2}{dt^2}u(t) $is Lipschitz continuous in$ t $, if$ u_0 $is in$ D(A^{3}) $such that$ A^{3}u_0 \in D(A) $. More regularity of$ \frac{du}{dt} $in$ t $can be obtained iteratively. \end{example} \begin{remark} \label{rmk2} \rm In order for$ u_0 $to be in$ D(A^2) $, more smoothness assumptions should be imposed on the coefficient functions$ a_{ij}(x), b_i(x), c(x) $and$ \beta(x) $. \end{remark} \begin{proof} Here observe that the operator$ A $is not closed, and so \cite[Theorem 1 Page 363]{Li} does not apply directly. The$ u_i $in \eqref{E:TimeB} will be used, and$ u_0 \in D(A) $with$ Au_0 \in D(A) $will be assumed for a moment. It follows that $Au_i = \frac{u_i - u_{i - 1}}{\epsilon} = (I - \epsilon A)^{-i}(Au_0),$ and hence, by \eqref{E:TimeLA} which is for the proof of Theorem \ref{T:XD}, $\|Au_i\|_{C^{1 + \eta}(\overline{\Omega})} = \|(I - \epsilon A)^{-i}(Au_0)\|_{C^{1 + \eta}(\overline{\Omega})} \le K$ for$ Au_0 \in D(A) $and for any$ 0 < \eta < 1 $. This implies $\|u_i\|_{C^{3 + \eta}(\overline{\Omega})} \le K$ by the Schauder global estimate with more smoothness in the linear elliptic theory \cite{Gi2}. Consequently, on using the interpolation inequality \cite{Gi2} and the Ascoli-Arzela theorem \cite{Gi2,Roy}, we have $Au_i \to Au(t) = U(t)(Au_0)$ through some subsequence with respect to the topology in$ C^{1 + \lambda}(\overline{\Omega}) $for any$ 0 < \lambda < \eta < 1 $. Here $U(t)u_0 \equiv \lim_{n \to \infty} (I - \frac{t}{n}A)^{-n}u_0.$ The rest follows from \cite[Page 363]{Li}, where the Lipschitz property in Theorem \ref{T:XA} and Remark \ref{T:XC} will be used. \end{proof} Now consider the linear equation with the space dimension$ 1 $: $$\label{E:MA} \begin{gathered} \frac{\partial u}{\partial t} = a(x, t)u_{xx} + b(x, t)u_{x} + c(x, t)u, \quad (x, t) \in (0, 1) \times (0, T), \\ u'(j, t) = (-1)^{j}\beta_{j}(j, t)u(j, t), \quad j = 0, 1, \\ u(x, 0) = u_0(x). \end{gathered}$$ Here we assume that$ a, b, c $are jointly continuous in$ x \in [0, 1]$,$t \in [0, T] $, and are of bounded variation in$ t $uniformly for all$ x $, that$ c(x, t) \le 0 $and$ a(x, t) \ge \delta_0 $for some constant$ \delta_0 > 0 $, and finally that$ \beta_{j} \ge \delta_0 > 0, j = 0, 1 $are jointly continuous in$ x, t $, and are Lipschitz continuous in$ t $, uniformly over$ x $. Let$ A(t) $:$ D(A(t)) \subset C[0, 1] \to C[0, 1] $be the operator defined by \begin{gather*} A(t)u \equiv a(x, t)u'' + b(x, t)u' + c(c, t)u \quad \text{for$ u \in D(A(t)) $where} \\ D(A(t)) \equiv \{v\in C^2[0, 1]: v'(j) = (-1)^{j} \beta_{j}(j, t)v(j), j = 0, 1 \}. \end{gather*} Following \cite{Li} and the proof for the previous case of higher space dimensions, and applying linear ordinary differential equation theory \cite{Co, LinA} and Theorem \ref{T:XB}, the next example is readily proven. Here the range condition (H2') is satisfied with$ E = C[0, 1] \supset \overline{D(A(t))} $for all$ t $. \begin{example} \label{T:MA} \rm Equation \eqref{E:MA} has a strong solution $u(t) = \lim_{n \to \infty} (I - \frac{t}{n}A)^{-n}u_0$ in$ L^2(0, 1) $for$ u_0 \in \hat{D}(A(0)) $, and$ u(t) $satisfies the middle equation in \eqref{E:MA} and the Lipschitz property $\|u(t) -u(\tau)\|_{\infty} \le k|t - \tau|$ for$ u_0 \in \hat{D}(A(0)) $and for$ 0 \le t, \tau \le T $. \end{example} In the case that$ a, b, c, \beta_{j}$, for$j = 0, 1$, are independent of$ t $, the Theorem 1 in \cite[Page 363]{Li}, together with the Lipschitz property in the Theorem \ref{T:XA} in this paper, will readily deliver the following example. Here it is to be observed that the corresponding operator$ A $is closed. \begin{example} \label{T:MB} \rm If the coefficient functions$ a, b, c, \beta_{j}, j = 0, 1 $are independent of$ t $, then the equation \eqref{E:MA} has a unique classical solution $u(t) = \lim_{n \to \infty}(I - \frac{t}{n}A)^{-n}u_0$ for$ u_0 \in D(A) $with$ Au_0 \in \overline{D(A)} $. This$ u(t) $has this property that the function$ \frac{du}{dt} $is continuous in$ t $. Furthermore,$ u(t) $is Lipschitz continuous in$ t $for$ u_0 \in \hat{D}(A) $, and$ \frac{du}{dt} $is Lipschitz continuous in$ t $for$ u_0 \in D(A) $with$ Au_0 \in \hat{D}(A) $, and is differentiable in$ t $for$ u_0 \in D(A^2) $with$ A^2u_0 \in \overline{D(A)} $. More regularity of$ \frac{du}{dt} $can be obtained iteratively. \end{example} \begin{remark}\label{rmk3} \rm In order for$ u_0 $to be in$ D(A^2) $, more smoothness assumptions should be imposed on the coefficient functions$ a(x), b(x), c(x) $, and$ \beta_{j}, j = 0, 1 $. \end{remark} \section{Applications to partial differential equations (II)} \label{S:E} In this section, it will be further shown that, for each concrete$ A(t) $in Section \ref{S:D}, the corresponding quantity $J_{\frac{t}{n}}(i\frac{t}{n})h = [I - \frac{t}{n}A(i\frac{t}{n})]^{-1}h, \quad i = 1, 2, \ldots, n$ is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $. We start with the case of linear$ A(t) $and consider the parabolic equation \eqref{E:ZA}. \begin{proposition} \label{T:NA} For$ h \in C^{\mu}(\overline{\Omega}) $, the solution$ u $to the equation $$\label{E:TimeXA} [I - \epsilon A(t)]u = h$$ where$ 0 \le t \le T $and$ \epsilon > 0 $, is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $. Here$ A(t) $is the linear operator corresponding to the parabolic equation \eqref{E:ZA}. \end{proposition} \begin{proof} The linear operator$ A(t): D(A(t)) \subset C(\overline{\Omega}) \to C(\overline{\Omega}) $is defined by \begin{gather*} A(t)u \equiv \sum_{i, j}a_{ij}(x, t)D_{ij} u + \sum_ib_i(x, t)D_iu + c(x, t)u \\ \text{for$ u \in D(A(t)) \equiv \{u \in C^{2 + \mu}(\overline{\Omega}): \frac{\partial u}{\partial \nu} + \beta(x, t)u = 0 \quad $on$ \partial \Omega \} $}. \end{gather*} Solvability of \eqref{E:TimeXA} follows from \cite[Pages 128-130]{Gi2}, where the method of continuity \cite[Page 75]{Gi2} is used. By writing out fully how the method of continuity is used, it will be seen that the solution$ u $is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $. To this end, set \begin{gather*} U_1 = C^{2 + \mu}(\overline{\Omega}), \quad U_2 = C^{\mu}(\overline{\Omega}) \times C^{1 + \mu}(\partial \Omega), \\ L_{\tau}u = \tau [u - \epsilon A(t)u] + (1 - \tau)(- \Delta u) \quad \text{in} \quad \Omega, \\ N_{\tau}u = \tau[\frac{\partial u}{\partial \nu} + \beta(x, t)u] + (1 - \tau) (\frac{\partial u}{\partial \nu} + u) \quad \text{on} \quad \partial \Omega, \end{gather*} where$ 0 \le \tau \le 1 $. Define the linear operator$ \pounds_{\tau}: U_1 \to U_2 $by $\pounds_{\tau}u = (L_{\tau}u, N_{\tau}u)$ for$ u \in U_1 $, and assume that$ \pounds_{s} $is onto for some$ s \in [0, 1] . It follows from \cite[Pages 128-130]{Gi2} that \begin{align} \label{E:PA} \|u\|_{U_1} \le C \|\pounds_{\tau}u\|_{U_2}, \end{align} where the constant C $is independent of$ \tau $. This implies that$ \pounds_{s} $is one to one, and so$ \pounds_{s}^{-1} $exists. By making use of$ \pounds_{s}^{-1} $, the equation, for$ w_0 \in U_2 $given, $\pounds_{\tau}u = w_0 %(h, 0)$ is equivalent to the equation $u = \pounds_{s}^{-1}w_0 + (\tau - s)\pounds_{s}^{-1}( \pounds_0 - \pounds_1)u,$ from which a linear map$S: U_1 \to U_1$, $Su = S_{s}u \equiv \pounds_{s}^{-1}w_0 + (\tau - s)\pounds_{s}^{-1}( \pounds_0 - \pounds_1)u$ is defined. The unique fixed point$ u $of$ S = S_{s}$will be related to the solution of \eqref{E:TimeXA}. By choosing$ \tau \in [0, 1] such that \begin{align} \label{E:PB} |s - \tau| < \delta \equiv [C( \|\pounds_0\|_{U_1 \to U_2} + \|\pounds_1\|_{U_1 \to U_2})]^{-1}, \end{align} it follows that S = S_{s} $is a strict contraction map. Therefore$ S $has a unique fixed point$ w $, and the$ w $can be represented by $\lim_{n \to \infty}S^{n}0 = \lim_{n \to \infty}(S_{s})^{n}0$ because of$ 0 \in U_1 $. Thus$ \pounds_{\tau} $is onto for$ |\tau - s| < \delta $. It follows that, by dividing$ [0, 1] $into subintervals of length less than$ \delta $and repeating the above arguments in a finite number of times,$ \pounds_{\tau} $becomes onto for all$ \tau \in [0, 1] $, provided that it is onto for some$ \tau \in [0, 1] $. Since$ \pounds_0 $is onto by the potential theory \cite[Page 130]{Gi2}, we have that$ \pounds_1 $is also onto. Therefore, for$ w_0 = (h, 0) $, the equation $\pounds_1u = w_0$ has a unique solution$ u $, and the$ u $is the seeked solution to \eqref{E:TimeXA}. Here it is to be observed that$ \phi \equiv \pounds_0^{-1}(h, 0) $is the unique solution$ \pounds_0^{-1}(h, \varphi) $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $: \begin{gather*} -\Delta v = h, \quad x \in \Omega, \\ \frac{\partial v}{\partial \nu} + v(x) = 0 \quad \text{on} \quad \partial \Omega, \end{gather*} and that \begin{gather*} S0 = S_00 = \pounds_0^{-1}(h, 0), \\ S^20 = (S_0)^20 = \pounds_0^{-1}(h, 0) + \pounds_0^{-1}[|\tau -0| (\pounds_0 - \pounds_1)\pounds_0^{-1}(h, 0)], \\ \dots. \end{gather*} The proof is complete. \end{proof} \begin{remark} \label{rmk4}\rm$\bullet$The solution$ u $is eventually represented by %this integral formula $u(x) = \pounds_0^{-1}H((h, 0)), %\int_{\Omega}G(x, y)h(y)(y) \, dy$ where$ H((h, 0)) $is a convergent series in which each term is basically obtained by, repeatedly, applying the linear operator$ (\pounds_0 - \pounds_1)\pounds_0^{-1} $to$ (h, 0) $for a certain number of times.$\bullet$The quantity$ \pounds_0^{-1}(h, \varphi) $, for each$ (h, \varphi) \in U_2 $given, can be computed numerically and efficiently by the boundary element methods \cite{Gau,Sch}, if the dimension of the space variable$ x $equals$ 2 $or$ 3 $.$\bullet$The constant$ C $above in \eqref{E:PA} and \eqref{E:PB} depends on$ n, \mu, \lambda_{\rm min}, \Omega $, and on the coefficient functions$ a_{ij}(x, t), b_i(x, t), c(x, t), \beta(x, t) $, and is not known explicitly \cite{Gi2}. % \cite[First Edition, Page 134]{Gi2}), Therefore, the corresponding$ \delta $cannot be determined in advance, and so, when dealing with the elliptic equation \eqref{E:TimeXA} in Proposition \ref{T:NA} numerically, it is more possible, by choosing$ \tau \in [0, 1] $such that$ |s - \tau| $is smaller, that the sequence$ S^{n}0 $will converge, for which$|s - \tau| < \delta$occurs. \end{remark} Next, we extend the above techniques to the case of nonlinear$ A(t) $, and consider the nonlinear parabolic equation \eqref{E:XA}; more work is required in this case. \begin{proposition} \label{T:OA} For$ h \in C^{\mu}(\overline{\Omega}) $, the solution$ u $to the equation \eqref{E:TimeXA} $[I - \epsilon A(t)]u = h$ where$ 0 \le t \le T $and$ \epsilon > 0 $, is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $. Here$ A(t) $is the nonlinear operator corresponding to the parabolic equation \eqref{E:XA}, and$ \beta(x, t, 0) \equiv 0 $is assumed additionally. \end{proposition} \begin{proof} The nonlinear operator$ A(t): D(A(t)) \subset C(\overline{\Omega}) \to C(\overline{\Omega}) $is defined by \begin{gather*} D(A(t)) = \{u \in C^{2 + \mu}(\overline{\Omega}): \frac{\partial u}{\partial \nu} + \beta(x, t, u) = 0 \quad \text{on} \quad \partial \Omega\}, \\ A(t)u = \alpha(x, t, Du)\Delta u + g(x, t, u, Du), \quad u \in D(A(t)). \end{gather*} Equation \eqref{E:TimeXA} with the nonlinear$ A(t) $has been solved in \cite{Lin2}, but here the proof will be based on the contraction mapping theorem as in the proof of Proposition \ref{T:NA}. To this end, set \begin{gather*} U_1 = C^{2 + \mu}(\overline{\Omega}), \\ U_2 = C^{\mu}(\overline{\Omega}) \times C^{1 + \mu}(\partial \Omega), \\ L_{\tau}u = \tau[u - \epsilon A(t)u] + (1 - \tau)(u - \Delta u), \quad x \in \Omega, \\ N_{\tau}u = \tau[\frac{\partial u}{\partial \nu} + \beta(x, t, u)] + (1 - \tau)( \frac{\partial u}{\partial \nu} + u) \quad \text{on } \partial \Omega, \end{gather*} where$ 0 \le \tau \le 1 $. Define the nonlinear operator$ \pounds_{\tau}: U_1 \to U_2 $by $\pounds_{\tau}u = (L_{\tau}u, N_{\tau}u)$ for$ u \in U_1 $, and assume that$ \pounds_{s} $is onto for some$ s \in [0, 1]$. As in proving that$ A(t) $satisfies the dissipativity (H1) where the maximum principle was used,$ \pounds_{s} $is one to one, and so$ \pounds_{s}^{-1} $exists. By making use of$ \pounds_{s}^{-1} $, the equation, for$ w_0 \in U_2 $given,$\pounds_{\tau}u = w_0$is equivalent to the equation $u = \pounds_{s}^{-1}[w_0 + (\tau - s)(\pounds_0 - \pounds_1)u],$ from which a nonlinear map \begin{gather*} S: U_1 \to U_1, \\ Su = S_{s}u \equiv \pounds_{s}^{-1}[w_0 + (\tau - s)(\pounds_0 - \pounds_1)u] \quad \text{for$ u \in U_1 $} \end{gather*} is defined. The unique fixed point of$ S = S_{s} $will be related to the solution of \eqref{E:TimeXA} with nonlinear$ A(t) $. By restricting$ S = S_{s} $to the closed ball of the Banach space$ U_1 $, $B_{s, r, w_0} \equiv \{u \in U_1: \|u - \pounds_{s}^{-1}w_0\|_{C^{2 + \mu}} \le r > 0\},$ and choosing small enough$ |\tau - s| $, we will show that$ S = S_{s} $leaves$ B_{s, r, w_0} $invariant. This will be done by the following steps 1 to 4. \textbf{Step 1.} It follows as in \cite[Pages 265-266]{Lin2} that, for$ \pounds_{\tau} v = (f, \chi) $, $$\label{E:Time1A} \begin{gathered} \|v\|_{\infty} \le k_{\{\|f\|_{\infty}, \|\chi\|_{C(\partial \Omega)}\}}, \\ \|Dv\|_{C^{\mu}} \le k_{\{\|v\|_{\infty}\}}\|Dv\|_{\infty} + k_{\{\|v\|_{\infty}, \|f\|_{\infty}, \|\chi\|_{C(\partial \Omega)}\}}, \\ \|v\|_{C^{1 + \mu}} \le k_{\{\|\chi\|_{C(\partial \Omega)}, \|f\|_{\infty}\}}, \\ \|v\|_{C^{2 + \mu}} \le K \|\pounds_{\tau}v\|_{U_2} = K \|\pounds_{\tau}v\|_{C^{\mu}(\overline{\Omega}) \times C^{1 + \mu}(\partial \Omega)} \end{gathered}$$ where$ k_{\{\|f\|_{\infty}\}} $is a constant depending on$ \|f\|_{\infty} $, and similar meaning is defined for other constants$ k $'s; further,$ K $is independent of$ \tau $, but depends on$ n, \delta_0, \mu, \Omega $, and on the coefficient functions$ \alpha(x, t, Dv), g(x, t, v, Dv), \beta(x, t, v) $, which have incorporated the dependence of$ v, Dv $into$ \|\pounds_{\tau}v\|_{U_2} $. \textbf{Step 2.} It is readily seen that, for$ v \in C^{2 + \mu}(\overline{\Omega}) $with$ \|v\|_{C^{2 + \mu}} \le R > 0 $, we have $$\label{E:Time2A} \|\pounds_{\tau}v\|_{U_2} \le k_{\{R\}}\|v\|_{C^{2 + \mu}},$$ where$ k_{\{R\}} $is independent of$ \tau . \textbf{Step 3.} It will be shown that, if $\|u\|_{C^{2 + \mu}} \le R, \quad \|v\|_{C^{2 + \mu}} \le R > 0,$ then $$\label{E:Time3A} \|\pounds_{\tau}u - \pounds_{\tau}v\|_{U_2} \le k_{\{R\}}\|u -v\|_{C^{2 + \mu}}.$$ It will be also shown that, if $\pounds_{\tau}u = (f, \chi_1), \quad \pounds_{\tau}v = (w, \chi_2),$ then \label{E:Time3B} \begin{aligned} \|u - v\|_{C^{2 + \mu}} &\le k_{\{\|\pounds_{\tau}u\|_{U_2}, \|\pounds_{\tau}v\|_{U_2}\}} [\|f - w\|_{C^{\mu}} + \|\chi_1 - \chi_2\|_{C^{1 + \mu}}] \\ &= k_{\{\|\pounds_{\tau}u\|_{U_2}, \|\pounds_{\tau}v\|_{U_2}\}}\|\pounds_{\tau}u - \pounds_{\tau}v\|_{U_2}. \end{aligned} Here K_{\{R\}} $and$ K_{\{\|\pounds_{\tau}u\|_{U_2}, \|\pounds_{\tau}v\|_{U_2}\}} $are independent of$ \tau . Using the mean value theorem, we have that \begin{gather*} \begin{aligned} f - w &= L_{\tau}u - L_{\tau}v \\ &= (u - v) - (1 - \tau)\Delta (u - v) - \tau \epsilon [\alpha \Delta(u - v) \\ &\quad + \alpha_{p}(x,t,p_1)(Du - Dv)\Delta v + g_{p}(x, t, u, p_2)(Du - Dv) \\ &\quad + g_{z}(x, t, z_1, Dv)(u - v)], \quad x \in \Omega, \end{aligned} \\ \frac{\partial (u - v)}{\partial \nu} + [\beta(x, t, u) - \beta(x, t, v)] = \chi_1 - \chi_2 \quad \text{on } \partial \Omega , \end{gather*} were p_1, p_2 $are some functions between$ Du $and$ Dv $, and$ z_1 $is some function between$ u $and$ v $. It follows as in \eqref{E:Time2A} that $\|\pounds_{\tau}u - \pounds_{\tau}v\|_{U_2} \le k_{\{R\}}\|u - v\|_{C^{2 + \mu}},$ which is the desired estimate. On the other hand, the maximum principle yields $\|u - v\|_{\infty} \le k_{\{\|f - w\|_{\infty}, \|\chi_1 - \chi_2\|_{\infty}\}}$ and \eqref{E:Time1A} yields $\|u\|_{C^{2 + \mu}} \le K \|\pounds_{\tau}u\|_{U_2}, \quad \|v||_{C^{2 + \mu}} \le K \|\pounds_{\tau}v\|_{U_2}.$ Thus, it follows from the Schauder global estimate \cite{Gi2} that $\|u - v\|_{C^{2 + \mu}} \le k_{\{\|\pounds_{\tau}u\|_{U_2}, \|\pounds_{\tau}\|_{U_2}\}} \|\pounds_{\tau}u - \pounds_{\tau}v\|_{U_2},$ which is the other desired estimate. \textbf{Step 4.} Consequently, for$ u \in B_{s, r, w_0} , we have that, by \eqref{E:Time1A}, $$\label{E:Time4A} \|u\|_{C^{2 + \mu}} \le r + \|\pounds_{s}^{-1}w_0\|_{C^{2 + \mu}} \le r + K\|w_0\|_{U_2} \equiv R_{\{r, \|w_0\|_{U_2}\}},$$ and that \begin{align*} & \|Su - \pounds_{s}^{-1}w_0\|_{C^{2 + \mu}} \\ &\le k_{\{\|w_0\|_{U_2}, \|w_0 + (\tau - s) (\pounds_0 - \pounds_1)u\|_{U_2}\}} \|(\tau - s)(\pounds_0 - \pounds_1)u\|_{U_2} \quad \text{by \eqref{E:Time3B}} \\ &\le |\tau - s|k_{\{\|w_0\|_{U_2}, R_{\{r, \|w_0\|_{U_2}\}}\}} \quad \text{by \eqref{E:Time2A} and \eqref{E:Time4A}}. \end{align*} Here the constant k_{\{\|w_0\|_{U_2}, R_{\{r, \|w_0\|_{U_2}\}}\}} $when$ w_0 $given and$ r $chosen, is independent of$ \tau $and$ s $. Hence, by choosing some sufficiently small$ \delta_1 > 0 $, there results $S = S_{s}: B_{s, r, w_0} \subset U_1 \to B_{s, r, w_0} \subset U_1$ for$ |\tau -s| < \delta_1 $; that is,$ B_{s, r, w_0} $is left invariant by$ S = S_{s}$. Next, it will be shown that, for small$ |\tau - s| $,$ S = S_{s} $is a strict contraction on$ B_{s, r, w_0} $, from which$ S = S_{s} $has a unique fixed point. Because, for$ u, v \in B_{s, r, w_0} $, $\|u\|_{C^{2 + \mu}} \le R_{\{r, \|w_0\|_{U_2}\}}, \quad \|v\|_{C^{2 + \mu}} \le R_{\{r, \|w_0\|_{U_2}\}} \quad \text{by \eqref{E:Time4A}},$ it follows that, by \eqref{E:Time2A}, $$\label{E:Time5A} \begin{gathered} \|w_0 + (\tau - s)(\pounds_0 - \pounds_1)u\|_{U_2} \le k_{\{\|w_0\|_{U_2}, R_{\{r, \|w_0\|_{U_2}\}}\}}, \\ \|w_0 + (\tau - s)(\pounds_0 - \pounds_1)v\|_{U_2} \le k_{\{\|w_0\|_{U_2}, R_{\{r, \|w_0\|_{U_2}\}}\}}, \end{gathered}$$ and that, by \eqref{E:Time3A}, $$\label{E:Time5B} \|(\tau - s)[(\pounds_0 - \pounds_1)u - (\pounds_0 - \pounds_1)v]\|_{U_2} \le |\tau - s|k_{\{R_{\{r, \|w_0\|_{U_2}\}}\}}\|u - v\|_{C^{2 + \mu}}.$$ Therefore, on account of \eqref{E:Time3B}, \eqref{E:Time5A}, and \eqref{E:Time5B}, we obtain $\|Su - Sv\|_{C^{2 + \mu}} \\ \le |\tau - s|k_{\{R_{\{r, \|w_0\|_{U_2}\}}, \|w_0\|_{U_2}\}} k_{\{ R_{\{r, \|w_0\|_{U_2}\}}\}} \|u - v\|_{C^{2 + \mu}}.$ Here the constant$ k_{\{R_{\{r, \|w_0\|_{U_2}\}}, \|w_0\|_{U_2}\}} k_{\{R_{\{r, \|w_0\|_{U_2}\}}\}} $when$ w_0 $given and$ r $chosen, is independent of$ \tau $and$ s $. Hence, by choosing some sufficiently small$ \delta_2 > 0 $, it follows that $S = S_{s}: B_{s, r, w_0} \to B_{s, r, w_0}$ ia a strict contraction for $|\tau - s| < \delta_2 \le \delta_1.$ Furthermore, the unique fixed point$ w $of$ S = S_{s} $can be represented by $\lim_{n \to \infty}S^{n}0 = \lim_{n \to \infty}(S_{s})^{n}0$ if$ \beta(x, t, 0) \equiv 0 $and if$ r= r_{\{K\|w_0\|_{U_2}\}} $is chosen such that $$\label{E:TimeQA} r = r_{\{K\|w_0\|_{U_2}\}} \ge K\|w_0\|_{U_2} \ge \|\pounds_{s}^{-1}w_0\|_{C^{2 + \mu}}$$ (by \eqref{E:Time4A}); this is because$ 0 $belongs to$ B_{s, r, w_0} $in this case. Thus$ \pounds_{\tau} $is onto for$ |\tau - s|< \delta_2$. It follows that, by dividing$ [0, 1] $into subintervals of length less than$ \delta_2 $and repeating the above arguments in a finite number of times,$ \pounds_{\tau} $becomes onto for all$ \tau \in [0, 1] $, provided that it is onto for some$ \tau \in [0, 1]$. Since$ \pounds_0 $is onto by linear elliptic theory \cite{Gi2}, we have that$ \pounds_1 $is also onto. Therefore, the equation, for$ w_0 = (h, 0) $, $\pounds_1u = w_0$ has a unique solution$ u $, and the$ u $is the sought solution to \eqref{E:TimeXA}. Here it is to be observed that$ \psi \equiv \pounds_0^{-1}(h, 0) $is the unique solution to the elliptic equation \begin{gather*} v -\Delta v = h, \quad x \in \Omega, \\ \frac{\partial v}{\partial \nu} + v(x) = 0 \quad \text{on} \quad \partial \Omega, \end{gather*} and that, by Proposition \ref{T:NA},$ \psi $is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the elliptic equation \eqref{E:TimeC} with$ \varphi \equiv 0 $. It is also to be observed that \begin{gather*} S0 = S_00 = \pounds_0^{-1}(h, 0), \\ S^20 = (S_0)^20 = \pounds_0^{-1}[(h, 0) + |\tau -0| (\pounds_0 - \pounds_1)\pounds_0^{-1}(h, 0)], \\ \dots, \end{gather*} where$ (\pounds_0 - \pounds_1)\pounds_0^{-1} $is a nonlinear operator. The proof is complete. \end{proof} \begin{remark} \label{rmk5} \rm The constants$ k_{\{R_{\{r, \|w_0\|_{U_2}\}}\}} $and$ k_{\{R_{\{r, \|w_0\|_{U_2}\}}, \|w_0\|_{U_2}\}} k_{\{ R_{\{r, \|w_0\|_{U_2}\}}\}} $, when$ w_0 $is given and when$ r $is chosen and conditioned by \eqref{E:TimeQA}, is not known explicitly, and so the corresponding$ \delta_2 $cannot be determined in advance. Hence, when dealing with the elliptic equation \eqref{E:TimeXA} in Proposition \ref{T:OA} numerically, it is more possible, by choosing$ \tau \in [0, 1] $such that$ |\tau - s| $is smaller, that the sequence$ S^{n}0 $will converge, for which$|\tau - s| < \delta_2 \le \delta_1$occurs. \end{remark} Finally, what will be considered is the linear equation \eqref{E:MA} of space dimension$1$. \begin{proposition} \label{T:PA} For$ h \in C[0, 1] $, the solution$ u $to the equation \eqref{E:TimeXA} $[I - \epsilon A(t)]u = h$ where$ 0 \le t \le T $and$ \epsilon > 0 $, is the limit of a sequence where each term in the sequence is an explicit function of the solution$ \phi $to the ordinary differential equation $$\label{E:TimeYA} \begin{gathered} v - v'' = h \quad x \in (0, 1), \\ v'(j) = (-1)^{j}v(j), \quad j = 0, 1. \end{gathered}$$ Here$ A(t) $is the linear operator corresponding to the parabolic equation \eqref{E:MA}. \end{proposition} \begin{proof} The linear operator$ A(t): D(A(t)) \subset C[0, 1] \to C[0, 1] $is defined by \begin{gather*} A(t)u \equiv a(x, t)u'' + b(x, t)u' + c(x, t)u \quad \text{for$ u \in D(A(t)) $where} \\ D(A(t)) \equiv \{v\in C^2[0, 1]: v'(j) = (-1)^{j} \beta_{j}(j, t)v(j), \quad j = 0, 1 \}. \end{gather*} The contraction mapping theorem in the proof of Proposition \ref{T:NA} will be used in order to solve the equation \eqref{E:TimeXA}. To this end, set, for$ 0 \le \tau \le 1 , \begin{gather*} U_1 = C^2[0, 1], \quad U_2 = C[0, 1] \times {\mathbb R}^2, \\ L_{\tau}u = \tau[u - \epsilon A(t)u] + (1 - \tau)(u - u''), \\ \begin{aligned} N_{\tau}u &= \Big(\tau [u'(0) - \beta_0(0, t)u(0)] + (1 - \tau)[u'(0) - u(0)], \\ &\quad \tau [u'(1) + \beta_1(1, t)u(1)] + (1 - \tau)[u'(1) + u(1)]\Big). \end{aligned} \end{gather*} Define the linear operator\pounds_{\tau} : U_1 \to U_2 $by $\pounds_{\tau}u = (L_{\tau}u, N_{\tau}u)$ for$ u \in U_1 $, and assume that$ \pounds_{s} $is onto for some$ s \in [0, 1] $. The following will be readily derived.$\bullet$For$ u \in C^2[0, 1] $, we have $$\label{E:TimeWB} \|\pounds_{\tau}u\|_{U_2} = \|\pounds_{\tau}u\|_{C[0, 1] \times {\mathbb R}^2} \le k_{\{a, b, c, \beta_0, \beta_1\}}\|u\|_{C^2},$$ where$ k_{\{a, b, c, \beta_0, \beta_1\}} $is independent of$ \tau $, and can be computed, depending on the given$ a(x, t), b(x, t), c(x, t), \beta_0(0, t) $, and$ \beta_1(1, t) $.$\bullet$For$ \pounds_{\tau}u = (h, (r, s)) $, the maximum principle shows $\|u\|_{\infty} \le \|h\|_{\infty} + |\frac{r}{\beta_0(0, t)}| + |\frac{s}{\beta_1(1, t)}|.$ This, together with the known interpolation inequality \cite[Page 65]{Gol} or \cite[Pages 7-8]{Mi} $\|u'\|_{\infty} \le \frac{2}{\lambda} \|u\|_{\infty} + \frac{\lambda}{2}\|u''\|_{\infty}$ for any$ \lambda > 0 $, applied to$ \pounds_{\tau}u = (h, (r, s)) $, it follows that, by choosing small enough$ \lambda = \lambda_1 $, $$\label{E:TimeWA} \|u\|_{C^2} \le k_{\{\lambda_1, a, b, c, \beta_0, \beta_1\}}(\|h\|_{\infty} + |r| + |s|) = k_{\{\lambda_1, a, b, c, \beta_0, \beta_1\}}\|\pounds_{\tau} u\|_{U_2},$$ where$ k_{\{\lambda_1, a, b, c, \beta_0, \beta_1\}} $is independent of$ \tau $and can be computed explicitly. \medskip On account of the estimate \eqref{E:TimeWA},$ \pounds_{s} $is one to one, and so$ \pounds_{s}^{-1} $exists. Thus, making use of$ \pounds_{s}^{-1} $, the equation, for$ w_0 \in U_2 $given,$\pounds_{\tau}u = w_0 $is equivalent to the equation $u = \pounds_{s}^{-1}w_0 + (\tau - s)\pounds_{s}^{-1}( \pounds_0 - \pounds_1)u,$ from which a linear map \begin{gather*} S: U_1 = C^2[0, 1] \to U_1 = C^2[0, 1], \\ Su = S_{s}u \equiv \pounds_{s}^{-1}w_0 + (\tau - s)\pounds_{s}^{-1}(\pounds_0 - \pounds_1)u, \quad u \in U_1 \end{gather*} is defined. Because of \eqref{E:TimeWA} and \eqref{E:TimeWB}, it follows that this$ S $is a strict contraction if $|\tau - s| < \delta = [k_{\{\lambda_1, a, b, c, \beta_0, \beta_1\}}2k_{\{a, b, c, \beta_0, \beta_1\}}]^{-1}.$ The rest of the proof will be the same as that for Proposition \ref{T:NA}, in which the equation, for$ w_0 = (h, (0, 0)) $, $\pounds_1u = w_0$ has a unique solution$ u $, and the$ u $is the sought solution. \end{proof} \begin{remark} \label{rmk6} \rm$\bullet$The$ \delta = [k_{\{\lambda_1, a, b, c, \beta_0, \beta_1\}}2k_{\{a, b, c, \beta_0, \beta_1\}}]^{-1} $in the above proof of Proposition \ref{T:PA} can be computed explicitly.$\bullet$The quantity$ \pounds_0^{-1}(h, (0, 0)) $is represented by the integral $\pounds_0^{-1}(h, (0, 0)) = \int_0^{1}g_0(x, y)h(y) \, dy,$ where$ g_0(x, y) $is the Green function associated with the boundary value problem \begin{gather*} u - u'' = h \quad \text{in} \quad (0, 1), \\ u'(j) = (-1)^{j}u(j), \quad j = 0, 1. \end{gather*} This$ g_0(x, y) $is known explicitly by a standard formula.$\bullet$As before, we have \begin{gather*} S0 = S_00 = \pounds_0^{-1}(h, (0, 0)), \\ S^20 = S_0^20 = \pounds_0^{-1}(h, (0, 0)) + \pounds_0^{-1}[|\tau - 0|( \pounds_0 - \pounds_1)\pounds_0^{-1}(h, (0, 0))], \\ \dots. \end{gather*} \end{remark} \section{Appendix} \label{S:VaryB} In this section, the Proposition \ref{P:VaryB} in Section \ref{S:B} will be proved, using the theory of difference equations. We now introduce its basic theory \cite{Mic}. Let $\{b_{n}\} = \{b_{n}\}_{n \in \{0\}\cup {\mathbb N}} = \{ b_{n} \}_{n = 0}^{\infty}$ be a sequence of real numbers. For such a sequence$ \{b_{n}\} $, we further extend it by defining $b_{n} = 0 \quad \text{if n = -1, -2, \ldots. }.$ The set of all such sequences$ \{b_{n}\} $'s will be denoted by$ S $. Thus, if$ \{a_{n}\} \in S $, then$0 = a_{-1} = a_{-2} = \dots$. Define a right shift operator$ E : S \to S $by $E\{b_{n}\} = \{b_{n + 1}\} \quad \text{for \{b_{n}\} \in S }.$ For$ c \in {\mathbb R} $and$ c \ne 0 $, define the operator$ (E - c)^{*} : S \to S $by $(E - c)^{*}\{b_{n}\} = \{c^{n}\sum_{i = 0}^{n - 1} \frac{b_i}{c^{i + 1}}\}$ for$ \{b_{n}\} \in S $. Here the first term on the right side of the equality, corresponding to$ n = 0 $, is zero. Define, for$ \{b_{n}\} \in S $, \begin{gather*} (E - c)^{i *}\{b_{n}\} = [(E - c)^{*}]^{i}\{b_{n}\}, \quad i = 1, 2, \ldots; \\ (E - c)^{0}\{b_{n}\} = \{b_{n}\}. \end{gather*} It follows that$ (E - c)^{*} $acts approximately as the inverse of$ (E - c) $in this sense $(E - c)^{*}(E - c)\{b_{n}\} = \{b_{n} - c^{n}b_0\}.$ Next we extend the above definitions to doubly indexed sequences. For a doubly indexed sequence$ \{\rho_{m, n}\} = \{\rho_{m, n}\}_{m, n = 0}^{\infty} $of real numbers, let $E_1\{\rho_{m, n} \} = \{\rho_{m + 1, n} \}; \quad E_2\{\rho_{m, n} \} = \{\rho_{m, n + 1} \}.$ Thus,$ E_1 $and$ E_2 are the right shift operators, which acts on the first index and the second index, respectively. It is easy to see that $E_1E_2 \{\rho_{m, n}\} = E_2E_1 \{\rho_{m, n}\}\,.$ Before we prove the Proposition \ref{P:VaryB}, we need the following four lemmas, which are proved in \cite{Lin1,Lin0,Lin1,Lin4}, respectively. \begin{lemma} \label{L:VaryA} If \eqref{E:VaryMain} is true, then \label{E:VaryMain3} \begin{aligned} \{a_{m, n}\} &\leq (\alpha \gamma (E_2 - \beta \gamma)^{*})^{m} \{a_{0, n}\} + \sum_{i = 0}^{m - 1}(\gamma \alpha(E_2 - \gamma \beta)^{*})^{i} \{(\gamma \beta)^{n}a_{m - i, 0}\} \\ &\quad + \sum_{j = 1}^{m}(\gamma \alpha)^{j - 1}((E_2 - \gamma \beta)^{*})^{j} \{r_{m + 1 - j, n + 1}\}, \end{aligned} where r_{m, n} = K_4\mu \rho(|n\mu - m\lambda|) $. \end{lemma} \begin{lemma} \label{L:VaryB} The following equality holds: $((E_2 - \beta \gamma)^{*})^{m}\{n \gamma^{n}\} = \{\frac{n \gamma^{n}}{\alpha^{m}} \frac{1}{\gamma^{m}} - \frac{m \gamma^{n}}{\alpha^{m + 1}}\frac{1}{\gamma^{m}} + \Big(\sum_{i = 0}^{m - 1} \binom{n}{i}\frac{\beta^{n - i}}{\alpha^{m + 1 - i}} (m - i)\frac{1}{\gamma^{m}}\Big) \gamma^{n}\}.$ Here$ \gamma, \alpha $and$ \beta $are defined in Proposition \ref{P:XA}. \end{lemma} \begin{lemma} \label{L:VaryC} The following equality holds: $((E - \beta \gamma)^{*})^{j}\{\gamma^{n} \} = \big\{\Big(\frac{1}{\alpha^{j}} - \frac{1}{\alpha^{j}}\sum_{i = 0}^{j - 1}\binom{n}{i} \beta^{n - i}\alpha^{i}\Big)\gamma^{n - j}\big\} = \big\{\Big(\frac{1}{\alpha^{j}} \sum_{i = j}^{n}\beta^{n - i}\alpha^{i}\Big)\gamma^{n - j}\big\}$ for$ j \in \mathbb{N} $. Here$ \gamma, \alpha $and$ \beta are defined in Proposition \ref{P:XA} \end{lemma} \begin{lemma} \label{L:VaryD} The following equality holds: \begin{align*} (E - \beta \gamma)^{m *}\{n^2\gamma^{n}\} &= \gamma^{n - m}\{\frac{n^2}{\alpha^{m}} - \frac{(2m)n}{\alpha^{m + 1}} + (\frac{m(m - 1)}{\alpha^{m + 2}} + \frac{m(1 + \beta)}{\alpha^{m + 2}}) \\ &\quad - \sum_{j = 0}^{m - 1}\big(\frac{(m - j)(m - j - 1)} {\alpha^{m - j + 2}} + \frac{(m - j)(1 + \beta)}{\alpha^{m - j + 2}}\big)\binom{n}{j} \beta^{n - j}\}. \end{align*} Here \gamma, \alpha $, and$ \beta $are defined in Proposition \ref{P:XA}. \end{lemma} \begin{proof}[Proof of Proposition \ref{P:VaryB}] If$ S_2(\mu) = \emptyset $, then \eqref{E:VaryMain2} is true, and so $a_{m, n} \le L(K_2)|n\mu - m\lambda|.$ If$ S_1(\mu) = \emptyset $, then \eqref{E:VaryMain} is true, and so the inequality \eqref{E:VaryMain3} follows by Lemma \ref{L:VaryA}. Since, by Proposition \ref{P:VaryA}, \begin{gather*} a_{0, n} \le K_1\gamma^{n}(2n + 1)\mu; \\ a_{m - i, 0} \le K_1(1 - \lambda \omega)^{-m}[2(m - i) + 1]\lambda; \end{gather*} it follows from Lemma \ref{L:VaryC} and from the Proposition 3 and its proof of \cite[Pages 115-116]{Lin0} that the first two terms of the right side of the inequality \eqref{E:VaryMain3} is less than or equal to $c_{m, n} + s_{m, n} + f_{m, n}.$ We finally estimate the third term, denoted by$ \{t_{m, n}\} $, of the right-hand side of \eqref{E:VaryMain3}. Observe that, using the subadditivity of$ \rho , we have \begin{align*} \{t_{m, n}\} &\leq \sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *}K_4 \mu \{\rho(|\lambda - \mu|) + \rho(|n \mu -m \lambda + j \lambda|)\} \\ &\leq \sum_{j = 1}^{m}(\gamma \alpha)^{j - 1}(E_2 - \gamma \beta)^{ j *}K_4 \mu \{\gamma^{n}\rho(|\lambda - \mu|) + \gamma^{n}\rho(|n \mu - (m - j)\lambda|)\} \\ &\equiv \{u_{m, n}\} + \{v_{m, n}\}, \end{align*} where \gamma = ( 1 - \mu \omega)^{-1} > 1 . It follows from Lemma \ref{L:VaryC} that \begin{align*} \{u_{m, n}\} &\le \{K_4\mu\gamma^{n}\rho(|\lambda - \mu|)\sum_{j = 1}^{m} \alpha^{j - 1}\frac{1}{\alpha^{j}}\sum_{i = 1}^{n}\binom{n}{i}\beta^{n - i} \alpha^{i}\} \\ &\le \{K_4\gamma^{n}\rho(|\lambda - \mu|)\mu \frac{1}{\alpha}m\} = \{K_4\rho(|\lambda - \mu|)\gamma^{n}(m\lambda)\}. \end{align*} To estimate \{v_{m, n}\} $, as in Crandall-Pazy \cite[page 68]{Cran}, let$ \delta > 0 $be given and write $\{v_{m, n}\} = \{I^{(1)}_{m, n}\} + \{I^{(2)}_{m, n}\},$ where$ \{I^{(1)}_{m, n}\} $is the sum over indices with$ |n \mu - (m - j)\lambda| < \delta $, and$ \{I^{(2)}_{m, n}\} $is the sum over indices with$ |n \mu - (m - j)\lambda| \geq \delta . As a consequence of Lemma \ref{L:VaryC}, we have \begin{align*} \{I^{(1)}_{m, n}\} &\leq \{K_4 \mu \gamma^{n}\rho(\delta)\sum_{j = 1} ^{m}\alpha^{j - 1}\frac{1}{\alpha^{j}}\sum_{i = j}^{n}\binom{n}{i} \beta^{n - i}\alpha^{i}\} \\ &\leq \{K_4 \rho(\delta) \mu \gamma^{n}m \frac{1}{\alpha}\} = \{K_4 \rho(\delta)\gamma^{n}m \lambda \}. \end{align*} On the other hand, \begin{align*} \{I^{(2)}_{m, n}\} &\leq K_4\mu \rho(T)\sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *}\{\gamma^{n}\} \\ &\leq K_4 \mu \rho(T)\sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *} \{\gamma^{n}\frac{[n \mu - (m - j)\lambda]^2}{\delta^2}\}, \end{align*} which will be less than or equal to $\{K_4\frac{\rho(T)}{\delta^2}\gamma^{n}[(m\lambda) (n\mu -m\lambda)^2 + (\lambda - \mu)\frac{m(m + 1)}{2}\lambda^2]\}$ and so the proof is complete. This is because of the calculations, where Lemmas \ref{L:VaryB}, \ref{L:VaryC}, and \ref{L:VaryD} were used: \begin{gather*} [n\mu - (m - j)\lambda]^2 = n^2\mu^2 - 2(n\mu)(m - j)\lambda + (m - j)^2\lambda^2; \\ \begin{aligned} &\sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *}\{\gamma^{n}n^2\}\mu^2 \\ & = \gamma^{n - 1}\sum_{j = 1}^{m}\alpha^{j - 1} \{\frac{n^2}{\alpha^{j}} - \frac{2jn}{\alpha^{j + 1}} +[\frac{j(j - 1)}{\alpha^{j + 2}} + \frac{j(1 + \beta)}{\alpha^{j + 2}}] \\ &\quad - \sum_{i = 0}^{j - 1}[\frac{(j - i)(j - i - 1)}{ \alpha^{j - i + 2}} + \frac{(j - i)(1 + \beta)}{\alpha^{j - i + 2}}] \binom{n}{i}\beta^{n - i}\}\mu^2 \\ & \le \gamma^{n}\sum_{j = 1}^{m}\{\frac{n^2}{\alpha} - \frac{2jn}{\alpha^2} + [\frac{j(j - 1)}{\alpha^{3}} + \frac{j(1 + \beta)}{\alpha^{3}}]\}\mu^2, \end{aligned} \end{gather*} where the negative terms associated with \sum_{i = 0}^{j - 1} were dropped; \begin{align*} &\sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *}\{\gamma^{n}n\}[2\mu(m - j)\lambda](-1) \\ &= \sum_{j = 1}^{m}(\gamma \alpha)^{j - 1}\{\gamma^{n - j}[ \frac{n}{\alpha^{j}} - \frac{j}{\alpha^{j + 1}} \\ &\quad + \sum_{i = 0}^{j - 1}\binom{n}{i} \beta^{n - i}\alpha^{i - j - 1}(j - i)]\}[2\mu(m - j)\lambda](-1) \\ &\le \sum_{j = 1}^{m}\gamma^{n}\{\frac{n}{\alpha} - \frac{j}{\alpha^2}\}[2\mu(m - j)\lambda](-1), \\ & = \sum_{j = 1}^{m}\gamma^{n}\alpha^{-1} \{- 2(n\mu)(m\lambda) + j[2n\mu \lambda + \frac{2\mu}{\alpha}(m\lambda)] - j^2(\frac{2\mu \lambda}{\alpha})\}; \end{align*} where the negative terms associated with \sum_{i = 0}^{j - 1}were dropped; \begin{align*} &\sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} (E_2 - \gamma \beta)^{j *}\{\gamma^{n}\}(m - j)^2\lambda^2 \\ &= \sum_{j = 1}^{m}(\gamma \alpha)^{j - 1} \{\gamma^{n - j}[\frac{1}{\alpha^{j}} - \frac{1}{\alpha^{j}} \sum_{i = 0}^{j - 1}\binom{n}{i}\beta^{n - i}\alpha^{i}]\} (m - j)^2\lambda^2 \\ & \le \sum_{j = 1}^{m}\gamma^{n}\alpha^{-1}(m^2 - 2mj + j^2)\lambda^2, \end{align*} where the negative terms associated with \sum_{i = 0}^{j - 1} $were dropped. Adding up the right sides of the above three inequalities and grouping them as a polynomial in$ j $of degree two, we have the following: The term involving$ j^{0} = 1 $has the factor $\mu \frac{1}{\alpha}\sum_{j = 1}^{m}[n^2\mu^2 - 2 (n\mu)(m\lambda)+(m\lambda)^2] = (m\lambda)(n\mu - m\lambda)^2;$ the term involving$ j^2 $has the factor $\frac{\mu^2}{\alpha^{3}} - \frac{2\mu \lambda}{\alpha^2} + \frac{\lambda^2}{\alpha} = 0;$ the term involving$ j $has two parts, one of which has the factor $\frac{2n\mu \lambda}{\alpha} + \frac{2\mu m \lambda}{\alpha^2} - \frac{2m\lambda^2}{\alpha} - \frac{2n\mu^2}{\alpha^2} = 0,$ and the other of which has the factor $\mu \sum_{j = 1}^{m}(\frac{1 + \beta}{\alpha^{3}} - \frac{1}{\alpha^{3}})j\mu^2 = (\lambda - \mu)\frac{m(m + 1)}{2} \lambda^2.$ The proof is complete. \end{proof} \begin{remark} \label{rmk7} \rm The results in Proposition \ref{P:VaryB} are true for$ n, m \ge 0 $, but a similar result in the \cite[Proposition 4, page 236]{Lin1} has the restriction$ n\mu - m\lambda \ge 0 $which is not suitable for a mathematical induction proof. \end{remark} \subsection*{Acknowledgments} The author wishes to thank very much Professor Jerome A. 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