%\input amstex
\documentstyle {amsppt}
\input tmm.def

\magnification1200
%\hsize=348.0pt
%\pageheight{18.0truecm}

\define \Real {\Bbb R}
\define \A {\Cal A}
\define \Lin {\Cal L}
\define \M {\frak M}
\define \Om {\Omega}
\define \dOm {\partial\Omega}
\define \normA {\lnorm A\rnorm}
\define \lnorm#1\rnorm {\vphantom{#1}
              \mathop{\left\|\smash{#1}\right\|}\nolimits}
\define \lmod#1\rmod {\vphantom{#1}
              \mathop{\left|\smash{#1}\right|}\nolimits}
\define \proof {\pf{Proof}}
\define \endproof {\qed\endpf}
\define \bydef {\mathrel{\overset\text{def}\to{=}}}
\define \meas {\operatorname{meas}}
\define \({\bigl(}
\define \){\bigr)}
\redefine \kappa {\varkappa}

\NoBlackBoxes

\topmatter
\at Error bounds for Galerkin procedures \endat
\au N.~A.~Bobylev, Yu.~M.~Burman, S.~A.~Soloviev \endau
\shortau {N.~A.~Bobylev, Yu.~M.~Burman, S.~A.~Soloviev}
\endtopmatter

\document
\bh {1.\enspace A general way to obtain error bounds} Consider a Banach
space $E$ with the norm $\lnorm \cdot \rnorm$ and an operator $\A:E \to E$.
Let a sequence of finite-dimensional projectors $P_1,\,P_2,\, \dots,\,
P_n,\, \dots:E \to E$ converging strongly to the identity operator be given
and denote $E_n = P_nE$. The strong convergence $P_n \to I$ implies that
$$
\overline{\bigcup_{n} E_n} = E.
$$
Galerkin procedure is the following way for approximate solution of the
operator equation
$$
x = \A(x). \tag1.1
$$
This equation is replaced by the sequence of equations
$$
x = P_n\A(x) \tag1.2
$$
considered in $E_n$ (and being thus equations in finite dimensions) and
their solutions $x_n^*$ are treated as approximants to the solution $x_*$ of
equation ~\thetag{1.1}. The aim of this article is to obtain an a priori
error bound for Galerkin procedure, or a number $r_n$ such that the ball
$B(r_n, x_n^*)$ necessarily contains at least one solution of equation
~\thetag{1.1}. What follows is an abstract method to obtain such bounds,
and the succeeding sections contain some examples.

Let a real Banach space $E^{(1)}$ be imbedded completely continuously and
densely into $E$; denote the norm in $E^{(1)}$ as $\lnorm \cdot\rnorm_1$.
Let the operator $\A$ be continuously Frech\'et differentiable and such that
$\A(E) \subset E^{(1)}$ and for some $K > 0$
$$
\lnorm \A(x) \rnorm_1 \le K \qquad (x \in E). \tag1.3
$$
The set $\M = \{x \in E^{(1)}: \lnorm x\rnorm_1 \le 1\}$ is a compact in
$E$, therefore the sequence
$$
\alpha_n = \max_{\lnorm x\rnorm_1 \le 1} \lnorm (I - P_n)x\rnorm \tag1.4
$$
vanishes because $P_n \to I$ strongly as $n \to \infty$.

If a Galerkin approximant $x_n^*$ is already found then the error can be
estimated using the following

\tm {Theorem 1} With the operator $\A:E \to E$ being Lipschitz
$$
\lnorm \A(x) - \A(y) \rnorm \le \Lambda \lnorm x - y \rnorm \quad (x,y \in
E) \tag1.5
$$
where $\Lambda < 1$ and $x_n^*$ being a solution of equation \thetag{1.2},
equation \thetag{1.1} has a unique solution $x_*$ and the following estimate
holds:
$$
\lnorm x_* - x_n^*\rnorm \le \frac {\alpha_n K}{1 - \Lambda}. \tag1.6
$$
\endtm
\proof The fact that equation \thetag{1.1} has a unique solution $x_*$
follows from the contraction mapping principle. If $x_n^*$ is a Galerkin
approximant of $x_*$ (i.e. a solution of \thetag{1.2}), then it follows from
\thetag{1.3}--\thetag{1.5} that
$$
\allowdisplaybreaks\align
\lnorm x_* - x_n^*\rnorm &\le \lnorm \A(x_*) - \A(x_n)\rnorm + \lnorm
\A(x_n) - P_n\A(x_n)\rnorm\\
&\le \Lambda\lnorm x_* - x_n^*\rnorm + \alpha_n K,
\endalign
$$
which completes the proof.
\endproof

\rk {Remark} Theorem ~1 is an a posteriori estimate statement, it can say
nothing of whether equations \thetag{1.2} are solvable and their solutions
are unique. But this is actually true if the space $E$ is Hilbert and $P_n$
are orthoprojectors. Indeed, the norm of an orthoprojector is equal to $1$,
therefore under the conditions of the Theorem ~1 the operator $P_n\A$ is a
contraction mapping.
\endrk

\bh {2.\enspace Harmonic balance technique for seeking the periodic
solutions of weakly nonlinear systems} Consider a differential equation
$$
\dot x = Ax + f(t,x) \tag2.1
$$
where $x \in \Real^N,\ A$ is a constant matrix, and the function $f(t,x)$ is
$T$-periodic in $t$, and use Galerkin procedure for seeking the $T$-periodic
solutions of this equation. To do it, consider the space $E = L_2^N[0,T]$ of
square-summable functions $[0,T] \to \Real^N$ and the space $E^{(1)} =
H_N^k[0,T]$ of the functions $[0,T] \to \Real^N$ whose $k$-th derivatives
belong to $L_2^N$. The norms in $L_2^N$ and $H_N^k$ will be defined by the
formulae
$$
\allowdisplaybreaks\align
\lnorm x\rnorm_{L_2^N} &= \left( \int_0^T \lmod x(t)\rmod^2
\,dt\right)^{1/2}, \\
\lnorm x\rnorm_{H_N^k} &= \sum_{m=0}^{k} \left( \int_0^T \lmod x^{(m)}(t)
\rmod^2 \,dt \right)^{1/2}.
\endalign
$$
The space $H_N^k$ is imbedded into the space $L_2^N$ densely and completely
continuously.

Let $E_n \subset \L_2^N$ be the space of the trigonometric polynomials of
degree not higher than $n$ and let $P_n$ be the orthoprojector of the
Hilbert space $L_2^N$ to $E_n$, or the operator mapping the function $x(t)
\in L_2^N$ to the section of its Fourier series containing modes up to the
$n$-th degree:
$$
(P_nx)(t) = \frac{A_0}{2} + \sum_{m=1}^{n} \(A_m \cos \frac{2\pi mt}{T}
+ B_m \sin \frac{2\pi mt}{T} \)
$$
where
$$
A_m = \int_0^T x(t) \cos \frac{2\pi mt}{T} \,dt, \quad B_m = \int_0^T x(t)
\sin \frac{2\pi mt}{T} \,dt.
$$

Let the numbers $2\pi mi/T\;(m = 0,\pm 1, \pm 2, \dots)$ do not belong to
the spectrum of the matrix $A$. It is easy to see that all the $T$-periodic
solutions of equation ~\thetag{2.1}, and only they, satisfy the integral
equation
$$
x = \A(x) \tag2.2
$$
where
$$
\A(x)(t) = \int_0^T H(t-s;T) f(s,x(s))\,ds \tag2.3
$$
and $H(t;T)$ is a matrix-valued function $T$-periodic in $t$ and given for
$0 \le t < T$ by the equation
$$
H(t;T) = (I - e^{AT})^{-1} e^{At}. \tag2.4
$$

The harmonic balance approximant of a $T$-periodic solution of equation
~\thetag{2.1} is a trigonometric polynomial
$$
x_n(t) = \frac{a_0}{2} + \sum_{m=1}^{n} \(a_m \cos \frac{2\pi mt}{T} + b_m
\sin \frac{2\pi mt}{T} \)
$$
with the amplitudes $a_m,\, b_m$ chosen such that $x_n(t)$ satisfies the
equation
$$
\dot x_n = Ax_n + P_nf(t,x_n) \tag2.5
$$
where $P_n$ is the Fourier projector described above. Since the projectors
$P_n$ commute with the integration, equation ~\thetag{2.5} is equivalent to
$$
x_n = P_n\A(x_n),
$$
and so harmonic balance technique is a particular case of general Galerkin
procedure described above.

\tm {Lemma 1} For any $n$ and for any $x \in H_N^k$
$$
\lnorm (I - P_n)x\rnorm_{L_2^N} \le \fracwithdelims(){T}{2\pi(n+1)}^k \left(
\int_0^T \lmod x^{(k)}(t) \rmod^2 \,dt \right)^{1/2}. \tag2.6
$$
\endtm
\proof Without loss of generality the mean value of the function $x \in
H_N^k$ is zero, so its Fourier series is
$$
\sum_{m=1}^{\infty} \(A_m \cos \frac{2\pi mt}{T} + B_m \sin \frac{2\pi
mt}{T} \),
$$
therefore the Fourier series of the function $(I - P_n)x$ is
$$
\sum_{m = n+1}^{\infty} \(A_m \cos \frac{2\pi mt}{T} + B_m \sin \frac{2\pi
mt}{T} \).
$$
The derivative $x^{(k)}$ of the function $x$ belongs to $L_2^N$ and its
Fourier series is
$$
\sum_{m = n+1}^{\infty} \fracwithdelims(){d}{dt}^k  \(A_m \cos \frac{2\pi
mt}{T} + B_m \sin \frac{2\pi mt}{T} \).
$$
By virtue of the Parcevalle equation
$$
\allowdisplaybreaks\align
\lnorm (I - P_n)x\rnorm_{L_2^N}^2 &= \int_0^T \lmod (I - P_n)x(t)\rmod^2\,dt
\\
&=T/2 \sum_{m=n+1}^{\infty} (A_m^2 + B_m^2)\\
&\le \fracwithdelims(){T}{2\pi(n+1)}^{2k} T/2 \sum_{m = n+1}^{\infty}
\fracwithdelims() {2\pi m}{T}^{2k} (A_m^2 + B_m^2) \\
&\le \fracwithdelims(){T}{2\pi(n+1)}^{2k} T/2 \sum_{m = 1}^{\infty}
\fracwithdelims() {2\pi m}{T}^{2k} (A_m^2 + B_m^2) \\
&= \fracwithdelims(){T}{2\pi(n+1)}^{2k} \int_0^T \lmod x^{(k)}(t)\rmod^2\,dt
\qed
\endalign
$$
\endpf

\tm {Theorem 2} With the function $f(t,x)$ in equation \thetag{2.1} being
$T$-periodic in $t$ and satisfying the estimates:
$$
\allowdisplaybreaks\gather
\lmod f(t,x)\rmod \le C \quad (t \in [0,T],\; x \in \Real^N), \tag2.7\\
\lmod f(t,x) - f(t,y) \rmod \le D \lmod x - y \rmod \quad (t \in [0,T],\;
x,\,y \in \Real^N), \tag2.8
\endgather
$$
the matrix $A$ being such that
$$
\max_{t \in [-T,T]} \lnorm e^{At}\rnorm_{\Lin(\Real^N)} \le M \tag2.9
$$
and
$$
\lnorm (I - e^{AT})^{-1} \rnorm_{\Lin(\Real^N)} \le Q, \tag2.10
$$
and the following condition met:
$$
L = QMDT < 1, \tag2.11
$$
the $T$-periodic solution $x_*(t)$ of equation ~\thetag{2.1} exists and is
unique, its harmonic balance approximant $x_n^*$ for any $n$ also exists, is
unique and satisfies the inequality
$$
\lnorm x_* - x_n^*\rnorm_{L_2^N} \le \frac {T}{2\pi(n+1)}\frac {\(2TC^2(1 +
2\lnorm A\rnorm^2 M^2 Q^2 T)\)^{1/2}}{1 - QMDT}. \tag2.12
$$
\endtm
\proof Take in Theorem ~1 $E = L_2^N[0,T],\; E^{(1)} = \widetilde
{H_N^1}[0,T]$ where $\widetilde {H_N^1}[0,T]$ is a subspace of $H_N^1[0,T]$
consisting of functions with zero mean value in $[0,T]$. The norm in
$\widetilde {H_N^1}[0,T]$ is defined by the equation
$$
\lnorm x\rnorm_1 = \lnorm x\rnorm_{\widetilde {H_N^1}[0,T]} = \left(
\int_0^T \lmod \dot x(t)\rmod^2\,dt \right)^{1/2}.
$$
This norm is equivalent to the norm induced in $\widetilde {H_N^1}[0,T]$ by
the space $H_N^1[0,T]$. The norm in $L_2^N[0,T]$ will be denoted as
$\lnorm\cdot\rnorm_0$.

It follows from formulae \thetag{2.3}, \thetag{2.4}, \thetag{2.7},
\thetag{2.9}, and \thetag{2.10} that for any $x \in L_2^N$ and for any $t
\in [0,T]$
$$
\lmod \A(x)(t)\rmod \le QMCT,
$$
consequently,
$$
\lnorm \A(x)\rnorm_0^2 \le Q^2 M^2 C^2 T^3.
$$
Since $\frac {d}{dt} \A(x)(t) = A\A(x)(t) + f(t,x(t))$, there holds the
inequality
$$
\allowdisplaybreaks\align
\int_0^T \bigl| \frac {d}{dt} \A(x)(t) \bigr|^2\,dt &\le 2\lnorm A\rnorm^2
\lnorm \A(x)\rnorm_0^2 + 2TC^2 \\
&\le 2TC^2 (1 + \lnorm A\rnorm^2 M^2 Q^2 T^2) \bydef K^2. \tag2.14
\endalign
$$

The norm $\lnorm (I - P_n)x\rnorm_0$ does not change if a constant is added
to the function $x$, therefore it can be assumed that $\A(x) \in \widetilde
{H_N^1}$ and $\lnorm \A(x)\rnorm_1 \le K$ where $K$ is defined by formula
~\thetag{2.14}. Thus Lemma ~1 (with $k = 1$) yields that
$$
\alpha_n = \frac {T}{2\pi(n+1)}
$$
(all the notations fit those introduced in Section ~1).

It follows from formulae \thetag{2.3}, \thetag{2.4}, and
\thetag{2.8}--\thetag{2.10} that for any $x,\,y \in L_2^N$ and for any $t
\in [0,T]$
$$
\lmod \A(x)(t) - \A(y)(t)\rmod \le QMD \int_0^T \lmod x(s) - y(s)\rmod\, ds
\le QMD\sqrt{T} \lnorm x - y\rnorm_0.
$$
Consequently,
$$
\lnorm \A(x) - \A(y)\rnorm_0 \le QMDT \lnorm x - y\rnorm_0 \bydef \Lambda
\lnorm x - y\rnorm_0
$$

It was assumed that $\Lambda < 1$. The projectors $P_n$ are orthoprojectors
in the Hilbert space $L_2^N$, therefore the Remark after Theorem ~1
completes the proof.
\endproof

Theorem ~2 gives rather pessimistic estimate of convergence rate for of
harmonic balance approximants to the $T$-periodic solution, it is $O(1/n)$
as $n \to \infty$. This estimate can however be improved for systems of
special kind such as the systems of automatic control.

Forced oscillations of the system of automatic control are $T$-periodic
solutions of the differential equation
$$
L\(\frac{d}{dt}\) x = M\(\frac{d}{dt}\) f(t,x) \tag2.22
$$
where $x \in \Real,\ f(t,x)$ is a $T$-periodic in $t$ function (feedback)
and $L(p)$ and $M(p)$ are polynomials
$$
\allowdisplaybreaks\align
L(p) &= p^l + a_1 p^{l-1} + \dots + a_l, \\
M(p) &= b_0 p^m + b_1 p^{m-1} + \dots + b_m
\endalign
$$
with $l > m$. Harmonic balance approximants of the $T$-periodic solutions of
equation \thetag{2.23} are $T$-periodic trigonometrical polynomials
$$
x_n(t) = \frac{a_0}{2} + \sum_{k=0}^{n} \(a_k \cos \frac{2\pi kt}{T} + b_k
\sin \frac{2\pi kt}{T} \)
$$
satisfying the equation
$$
L\(\frac{d}{dt}\) x_n = P_n M\(\frac{d}{dt}\) f(t,x_n).
$$

One can readily see that the $T$-periodic solutions of equation
\thetag{2.22} (and only they) are solutions of the integral equation
$$
x = \Cal B(x) \bydef \int_0^T h(t-s;T)f(s,x(s))\,ds; \tag2.23
$$
here $h(t;T) = \(c,H(t;T)\gamma\)$ where $H(t;T)$ is given by equation
\thetag{2.4} with
$$
A = \pmatrix
0 &1 &0 &\hdots &0\\
0 &0 &1 &\hdots &0\\
\vdots &\vdots &\vdots &\ddots &\vdots \\
0 &0 &0 &\hdots &1\\
-a_l &-a_{l-1} &-a_{l-2} &\hdots &-a_1
\endpmatrix, \tag2.24
$$
and
$$
c = \pmatrix
1\\ 0\\ \vdots\\ 0\\ 0
\endpmatrix,
\quad
\gamma = \pmatrix\format\l\\
\gamma_1\\ \gamma_2\\ \vdots\\ \gamma_{l-1}\\ \gamma_l
\endpmatrix
$$
where
$$
\aligned
&\gamma_1 = \gamma_2 = \dots = \gamma_{l-m-1} = 0,\\
&\gamma_{l-m} = b_0,\\
&\gamma_{l-m+1} + \gamma_{l-m} a_1 = b_1,\\
&\hdotsfor 1\\
&\gamma_l + \gamma_{l-1} a_1 + \dots + \gamma_{l-m} a_m = b_m.
\endaligned \tag2.25
$$

To formulate Theorem ~3 note that the eigenvalues of the matrix $A$ are the
roots of the polynomial $L(p)$, so if the numbers $2\pi mi/T\;(m = 0, \pm 1,
\pm 2, \dots)$ are supposed not to be roots of the polynomial $L(p)$ then
the matrix $I - e^{AT}$ is invertible, and the numbers $M$ and $Q$ can be
defined by formulae \thetag{2.9} and \thetag{2.10}, respectively, with the
matrix $A$ given by equation \thetag{2.24}.

\tm {Theorem 3} With the function $f(t,x)$ in equation \thetag{2.22} being
$T$-periodic in $t$ and satisfying the estimates:
$$
\allowdisplaybreaks\alignat2
\lmod \gamma \rmod \lmod f(t,x)\rmod &\le C \quad &&(t \in [0,T],\; x \in
\Real), \tag2.26\\
\lmod \gamma \rmod \lmod f(t,x) - f(t,y)\rmod &\le D \lmod x - y\rmod \quad
&&(t \in [0,T],\; x,\,y \in \Real) \tag2.27\\
\endalignat
$$
where $\gamma$ is given by formula ~\thetag{2.25}, the numbers $2\pi
mi/T\;(m = 0, \pm 1, \pm 2, \dots)$ not being roots of the polynomial
$L(p)$, and $QMDT < 1$, it is true that equation \thetag{2.23} has a unique
$T$-periodic solution $x_*(t)$, for any $n$ there exists its unique harmonic
balance approximant $x_n^*$, and the following estimate takes place:
$$
\lnorm x_* - x_n^*\rnorm_0 \le \fracwithdelims(){T}{2\pi(n+1)}^{l-m}\frac
{\(2TC^2(1 + 2\lnorm A\rnorm^2 M^2 Q^2 T)\)^{1/2}}{1 - QMDT}. \tag2.28
$$
\endtm
\proof Define the operator $\A:L_2^l[0,T] \to L_2^l[0,T]$ by the formula
$$
\A(z)(t) = \int_0^T H(t-s;T) f(s,\(c,z(s)\))\,ds
$$
where the function $H(t;T)$ is defined by formula \thetag{2.4} with the
matrix $A$ given by equation \thetag{2.24}. Then for any $z \in L_2^l$
$$
\Cal B(\(c,z\)) = \(c, \A(z)\).
$$
For an arbitrary $x \in L_2$ take $z \in L_2^l$ such that $x = \(c,z\)$.
Then, as suggests formula \thetag{2.24},
$$
\fracwithdelims(){d}{dt}^{l-m} \Cal B(x)(t) = \(c,
\fracwithdelims(){d}{dt}^{l-m} \A(z)(t)\) = \(c_*, \frac{d}{dt} \A(z)(t)\)
\tag2.29
$$
where the $(l-m)$-th component of the vector $c_*$ is equal to $1$ and all
the rest are zeros. Therefore
$$
\allowdisplaybreaks\align
\int_0^T \bigl| \fracwithdelims(){d}{dt}^{l-m} \Cal B(x)(t) \bigr|^2\, dt
&\le \int_0^T \bigl| \frac{d}{dt} \A(z)(t) \bigr|^2\, dt\\
&\le 2TC^2 (1 + \lnorm A\rnorm^2 M^2 Q^2 T^2) \bydef K^2,
\endalign
$$
the latter inequality is proved as in Theorem ~2. Therefore it follows from
Lemma ~1 (with $k = l-m$) that
$$
\lnorm (I - P_n)\Cal B(x)\rnorm_0 \le \fracwithdelims(){T}{2\pi(n+1)}^{l-m}
K.
$$
The proof is completed as that of Theorem ~2.
\endproof

\bh {3.\enspace The Dirichlet problem for weakly nonlinear elliptic
equations}\ Let $\Om \subset \Real^N$ be a bounded domain with sufficiently
smooth boundary $\dOm$ and let
$$
L = \sum_{i,j = 1}^{N} a_{ij}(x)\frac{\partial^2}{\partial x_i \partial x_j}
+ \sum_{i = 1}^{N} a_i(x) \frac{\partial}{\partial x_i} + a(x) \tag3.1
$$
be a 2nd order strictly elliptic operator in $\Om$. Denote $H_k$ the Hilbert
space of the functions $u: \Om \to \Real$ having generalized derivatives up
to the $k$-th order, with the norm
$$
\lnorm u\rnorm_k = \sum_{m=0}^{k} \left( \int_{\Om} \sum\Sb \alpha_1,
\dots, \alpha_N \ge 0\\ \alpha_1 + \dots + \alpha_N = m \endSb \biggl| \frac
{\partial^m u(x)}{\partial x_1^{\alpha_1} \dots \partial x_N^{\alpha_N}}
\biggr|^2 \,dx \right)^{1/2}. \tag3.2
$$
The corresponding scalar products will be referred to as $\(\cdot,
\cdot\)_k$. Obviously
$$
\lnorm u\rnorm_0 \le \lnorm u\rnorm_1 \le \dots \tag3.3
$$
and therefore
$$
H_0 \supset H_1 \supset \dots.
$$
It can be proved (see e.g. \cite{1}) that all these imbeddings are dense and
completely continuous. Assume also that the coefficiets $a_{ij}(x),\,
a_i(x)$, and $a(x)$ are such that the operator $L$ is symmerical with
respect to the scalar product $\(\cdot,\cdot\)_0$.

Since the operator $L$ is strictly elliptic, there exist such positive
constants $c_1$ and $c_2$ that for any $u \in H_1$
$$
c_1\lnorm u\rnorm_1^2 - c_2\lnorm u\rnorm_0^2 \le \(Lu,u\)_0 \tag3.4
$$
(the quadratic form $\(Lu,u\)_0$ can be extended to the whole $H_1$ though
the operator $L$ is not defined in this space). This inequality can be
proved by integration by parts; see \cite{1} for more detail. It can be also
proved (see also \cite{1}) that the opreator $L$ is a bounded operator from
$H_2$ to $H_0$; denote its norm as $\kappa$:
$$
\lnorm Lu\rnorm_0 \le \kappa \lnorm u\rnorm_2. \tag3.5
$$

The eigenvalues of the symmetric strictly elliptic operator $L$ are real and
form a monotone sequence:
$$
\lambda_1 < \lambda_2  < \dots < \lambda_n < \dots.
$$
It follows from ~\thetag{3.4} that $\lambda_1 > -c_2$. Each eigenspace $E_i
\subset H_0$ related to the eigevnalue $\lambda_i$ is finite-dimensional,
and $E_i$ for different $i$ are orthogonal with respect to the scalar
product of $H_0$. Let functions $u_{i1}(x),\, \dots,\, u_{id_i}(x)$ form an
$H_0$-orthonormed basis of $E_i$, then $\frak A = \{u_{ik}: i = 1,2,\dots,\,
1 \le k \le d_i\}$ is an orthonormed basis of the whole $H_0$ (for proof see
e.g. \cite{1}). Denote the orthoprojector of $H_0$ to the linear hull of the
spaces $E_1,\, \dots,\, E_n$ as $P_n$, or
$$
(P_nu)(x) = \sum_{m=1}^{n} \sum_{k=1}^{d_m} A_{mk} u_{mk}(x)
$$
where
$$
A_{mk} = \int_\Om u_{mk}(x)u(x)\,dx.
$$

It will be assumed that zero is not an eigenvalue of the operator $L$;
denote
$$
\lambda_* = \min_{i} \lmod \lambda_i\rmod. \tag3.6
$$

\tm {Lemma 2} For any $n$ and for any $u(\cdot) \in H_2$
$$
\lnorm (I - P_n)u\rnorm_0 \le \frac{c_2 + \kappa}{\sqrt{c_1(\lambda_{n+1} +
c_2)}} \lnorm u\rnorm_2.
$$
\endtm
\proof Let $\frak A$ be the basis in $H_0$ described above. Since $H_2
\subset H_0$, there exists a development
$$
u = \sum_{m=1}^{\infty} \sum_{k=1}^{d_m} A_{mk} u_{mk}.
$$
Therefore
$$
(I - P_n)u = \sum_{m=n+1}^{\infty} \sum_{k=1}^{d_m} A_{mk} u_{mk}.
$$
Since the projectors $P_n$ commute with the operator $L$, it follows from
\thetag{3.4} and \thetag{3.5} that
$$
\allowdisplaybreaks\align
\lnorm (I - P_n)u\rnorm_1^2 &\le \frac{1}{c_1} \((L + c_2)u, (I -
P_n)u\)_0\\
&= \frac{1}{c_1} \sum_{m=n+1}^{\infty} \sum_{k=1}^{d_m} (\lambda_m + c_2)
A_{mk}^2 \\
&\le \frac{1}{c_1(\lambda_{n+1} + c_2)} \sum_{m=n+1}^{\infty}
\sum_{k=1}^{d_m} (\lambda_m + c_2)^2 A_{mk}^2 \\
&= \frac{1}{c_1(\lambda_{n+1} + c_2)} \lnorm (L + c_2)u\rnorm_0^2 \le
\frac{(c_2 + \kappa)^2}{c_1(\lambda_{n+1} + c_2)} \lnorm u\rnorm_2^2. \qed
\endalign
$$
\endpf

Consider the following problem:
$$
\align
&Lu = f(x,u,\nabla u), \tag3.7\\
&u\bigl|_{\dOm} = 0. \tag3.8
\endalign
$$
The problem \thetag{3.7}--\thetag{3.8} can be represented in the form of the
operator equation
$$
u = \A(u) \tag3.9
$$
where
$$
\A(u)(x) \bydef L^{-1} f(x,u(x),\nabla u(x)).
$$
and $L^{-1}$ represents the operator inverse to $L$ in the space
$\widetilde{H^1} \subset H^1$ of the functions with zero trace in the
boundary $\dOm$ of the domain $\Om$. Since the operator $L$ is strictly
elliptic and zero is not its eigenvalue, it is a bounded operator from $H_0$
to $H_2$; denote its norm as $\nu$:
$$
\lnorm L^{-1}u\rnorm_2 \le \nu \lnorm u\rnorm_0.
$$
Since $L$ is $H_0$-symmetrical, the operator $L^{-1}$ also satisfies the
bounds
$$
\allowdisplaybreaks\gather
\lnorm L^{-1}u\rnorm_0 \le \frac{1}{\lambda_*}\lnorm u\rnorm_0,\\
\(L^{-1}u,u\)_0 \le \frac{1}{\lambda_*} \lnorm u\rnorm_0^2.
\endgather
$$

A version of Galerkin procedure will be used for approximate solution of the
problem \thetag{3.7}--\thetag{3.8}. A solution of the equation
$$
u_n = P_n\A(u_n) \qquad (u_n \in E_n \bydef P_n H_0) \tag3.10
$$
will be used as an approximant $u_n^*$ of the solution of the problem.

\tm {Theorem 4} With the function $f$ satisfying the estimates:
$$
\allowdisplaybreaks\gather
\lmod f(x,u,p)\rmod \le C \qquad (x,\,u \in \Real,\; p \in \Real^N),
\tag3.11\\
\lmod f(x,u_1,p_1) - f(x,u_2,p_2)\rmod \le D (\lmod u_1 - u_2\rmod + \lmod
p_1 - p_2\rmod), \tag3.12\\
(x,\,u_1,\,u_2 \in \Real,\; p_1,\,p_2 \in \Real^N)
\endgather
$$
and the following inequality holding:
$$
\Lambda \bydef 2D^2 \( + \frac {1}{c_1 \lambda_*} + \frac{1}{\lambda_*^2} +
\frac {c_2}{c_1 \lambda_*^2} \) < 1
$$
equation \thetag{3.9} has a unique solution $u_*$, equation \thetag{3.10}
has a unique solution $u_n^*$, and the following estimate is true:
$$
\lnorm u_* - u_n^*\rnorm_1^2 \le \frac{(c_2 + \kappa)^2 \nu^2 C^2 \meas\Om}
{c_1(\lambda_{n+1} + c_2)(1 - \Lambda)^2}.
$$
\endtm

To prove the Theorem, we need the following

\tm {Lemma 3} For any $u \in H_0$
$$
\lnorm L^{-1}u\rnorm_1^2 \le \( \frac {1}{c_1 \lambda_*} + \frac {c_2}{c_1
\lambda_*^2} \) \lnorm u\rnorm_0^2.
$$
\endtm
\proof Applying inequality \thetag{3.4} and formula \thetag{3.6} we have
$$
\allowdisplaybreaks\align
\lnorm L^{-1}u\rnorm_1^2 &\le \frac{1}{c_1} \((L + c_2)L^{-1}u, L^{-1}u\)_0
\\
&= \frac{1}{c_1} \(L^{-1}u,u\)_0 + \frac{c_2}{c_1} \(L^{-1}u,L^{-1}u\)_0 \le
\( \frac {1}{c_1 \lambda_*} + \frac {c_2}{c_1 \lambda_*^2} \) \lnorm
u\rnorm_0^2. \qed
\endalign
$$
\endpf

Proceed now to the proof of the Theorem.

\proof Prove first that Galerkin approximating equations \thetag{3.10}
have unique solutions with all $n$. The function $u_n \in P_n H_0$ is a
solution of \thetag{3.10} if and only if the function $v_n = Lu_n \in P_n
H_0$ is a solution of the equation
$$
v_n = P_n f(x, L^{-1}v_n(x), \nabla L^{-1}v_n(x)) \bydef P_n \Cal B(v_n).
\tag3.13
$$

It follows from the condition \thetag{3.12} that
$$
\lmod \Cal B(v_1) - \Cal B(v_2)\rmod \le D (\lmod L^{-1}(v_1 - v_2)\rmod +
\lmod \nabla L^{-1} (v_1 - v_2)\rmod),
$$
and therefore by virtue of Lemma ~3
$$
\allowdisplaybreaks\align
\lnorm \Cal B(v_1) - \Cal B(v_2)\rnorm_0^2 &\le 2 D^2 (\lnorm L^{-1}(v_1 -
v_2)\rnorm_0^2 + \lnorm L^{-1}(v_1 - v_2)\rnorm_1^2) \\
&\le \frac {2 D^2}{\lambda_*^2} \lnorm v_1 - v_2\rnorm_0^2 + 2 D^2 \( \frac
{1}{c_1 \lambda_*} + \frac {c_2}{c_1 \lambda_*^2} \) \lnorm v_1 -
v_2\rnorm_0^2 \\
&= \Lambda \lnorm v_1 - v_2\rnorm_0^2.
\endalign
$$
Since the projectors $P_n$ are orthoprojectors in the space $H_0$, it is
true that $\lnorm P_n\rnorm_{\Lin(H_0)} = 1$, and the fact that equation
\thetag{3.13} (and therefore equation \thetag{3.10}) is uniquely solvable
follows from the contraction mapping principle.

The inequality
$$
\lnorm \A(u)\rnorm_2 \le \kappa C \sqrt{\meas\Om} \tag3.14
$$
follows from \thetag{3.5} and \thetag{3.11}. Inequality \thetag{3.12} and
Lemma ~3 imply that for any $u_1,\,u_2 \in H_1$
$$
\allowdisplaybreaks\align
&\lnorm \A(u_1) - \A(u_2)\rnorm_1^2 \\
&\le \( \frac {1}{c_1 \lambda_*} + \frac {c_2}{c_1 \lambda_*^2} \)
\int_{\Om} \lmod f(x,u_1(x),\nabla u_1(x)) - f(x,u_2(x),\nabla
u_2(x))\rmod^2\, dx\\
&\le 2D^2\( \frac {1}{c_1 \lambda_*} + \frac {c_2}{c_1 \lambda_*^2} \)
\(\int_{\Om} \lmod u_1(x) - u_2(x) \rmod^2\, dx + \int_{\Om} \lmod \nabla
u_1(x) - \nabla u_2(x) \rmod^2\, dx\) \\
&\le 2D^2\( \frac {1}{c_1 \lambda_*} + \frac {c_2}{c_1 \lambda_*^2} \)
\lnorm u_1 - u_2\rnorm_1^2 \le \Lambda \lnorm u_1 - u_2\rnorm_1^2.
\endalign
$$
Estimate \thetag{3.14}, Lemma ~2, and Theorem ~1 complete the proof.
\endproof

The estimation process in Theorem ~3 may be greatly simplified if $L$ is a
Laplace-Belitrami operator. Let a Riemannian metric $G$ be given in a domain
$\Om \subset \Real^N$. Let us introduce Hilbert spaces $H_k$ as before with
the norms given by the formula \thetag{3.2} where $\(\cdot, \cdot\)$ under
the integral sign means scalar product in $\Om$ relative to the Riemannian
metric $G$. Let $\Delta$ be Laplace--Beltrami operator corresponding to $G$,
therefore
$$
\((1 - \Delta)u,v\)_k = \(u,v\)_{k+1} \tag3.15
$$
and
$$
\lnorm (1 - \Delta)u\rnorm_k = \lnorm u\rnorm_{k+2} \tag3.16
$$
with any $k$ and $u,\,v$ such that the appropriate norms exist. Consider the
Dirichlet problem
$$
\align
&\Delta u = f(x,u,\nabla u), \tag3.17\\
&u\bigl|_{\dOm} = 0 \tag3.18
\endalign
$$
(where $\nabla u$ represents the gradient of the function $u(x)$ relative to
the Riemannian metric $G$) and try to find its approximate solutions using a
method analogous to that used for the problem \thetag{3.7}--\thetag{3.8}.
Let
$$
\lambda_* \bydef \lambda_1 < \lambda_2 < \dots
$$
be eigenvalues of the operator $\Delta$, $E_n$ be corresponding eigenspaces,
$P_n$ be $H_0$-orthoprojectors to the linear hull of $E_1,\, \dots,\, E_n$.
Then the Galerkin approximants $u_n^*$ of the solution of the problem
\thetag{3.17}--\thetag{3.18} are the solutions of equations
$$
u_n = P_n \Delta^{-1}f(x,u_n,\nabla u_n) \tag3.20
$$
where $\Delta^{-1}$ is an operator inverse to $\Delta$ with the boundary
condition \thetag{3.18}. Since $-\Delta$ is a positive operator with respect
to the scalat product $\(\cdot, \cdot\)_0$, all $\lambda_i$ are positive.

Equalities \thetag{3.15} and \thetag{3.16} imply

\tm {Lemma 4} The eigenspaces $E_n$ for different $n$ are orthogonal in
$H_1$ and the projectors $P_n: H_1 \to H_1$ are orthoprojectors.
\endtm

The following estimate is a corollary of Lemma ~4:

\tm {Lemma 5} With any $u \in H_1$ and with any $n$
$$
\lnorm (I - P_n)u\rnorm_1^2 \le -\frac{1}{\lambda_{n+1}} \(\Delta u, u\)_1.
$$
\endtm
\proof Using Lemma ~4 choose an $H_1$-orthonormed basis $u_{n1},\, \dots,\,
u_{nd_n}$ in each $E_n$, therefore $\frak A = \{ u_{nk} : n = 1,2,\dots,\; k
= 1,2,\dots,d_n\}$ is an orthonormed basis in the whole $H_1$. If
$$
u = \sum_{m=1}^{\infty} \sum_{k=1}^{d_m} A_{mk} u_{mk},
$$
then
$$
\allowdisplaybreaks\align
\lnorm (I - P_n)u\rnorm_1^2 &= \sum_{m=n+1}^{\infty} \sum_{k=1}^{d_m}
A_{mk}^2 \le \frac{1}{\lambda_{n+1}}\sum_{m=n+1}^{\infty} \sum_{k=1}^{d_m}
\lambda_m A_{mk}^2 \\
&\le \frac{1}{\lambda_{n+1}}\sum_{m=1}^{\infty} \sum_{k=1}^{d_m} \lambda_m
A_{mk}^2 = -\frac{1}{\lambda_{n+1}} \(\Delta u, u\)_1. \qed
\endalign
$$
\endpf

In the sequel $\meas\Om$ means the measure of $\Om$ with respect to the
Riemannian metric $G$.

\tm {Theorem 5} With the function $f$ satisfying the estimates
$$
\allowdisplaybreaks\gather
\lmod f(x,u,p)\rmod \le C \qquad (x,\,u \in \Real,\; p \in \Real^N),
\tag3.20\\
\lmod f(x,u_1,p_1) - f(x,u_2,p_2)\rmod \le D (\lmod u_1 - u_2\rmod + \lmod
p_1 - p_2\rmod), \tag3.21\\
(x,\,u_1,\,u_2 \in \Real,\; p_1,\,p_2 \in \Real^N)
\endgather
$$
and
$$
\Lambda \bydef \frac {\sqrt{2(\lambda_* + 1)}}{\lambda_*} D < 1,
$$
Dirichlet problem \thetag{3.17}--\thetag{3.18} has a unique solution $u_*$,
equation \thetag{3.20} has a unique solution $u_n^*$, and the estimate
$$
\lnorm u_* - u_n^*\rnorm_1 \le \frac{C}{\lambda_{n+1}} \frac {\sqrt{\(1 +
\frac {1}{\lambda_*}\) \meas\Om}}{1 - \Lambda}
$$
holds.
\endtm
\proof Denote
$$
\A(u) = \Delta^{-1} f(x,u,\nabla u).
$$
It follows from equation \thetag{3.15} that with any $u \in H_0$
$$
\lnorm \Delta^{-1} u\rnorm_1^2 = \lnorm \Delta^{-1} u\rnorm_0^2 -
\(\Delta^{-1} u, u\)_0 \le \frac {\lambda_* + 1}{\lambda_*^2} \lnorm
u\rnorm_0^2,
$$
and therefore
$$
\allowdisplaybreaks\align
\lnorm \A(u_1) - \A(u_2)\rnorm_1^2 &\le \frac {\lambda_* + 1}{\lambda_*^2}
\int_{\Om} \lmod f(x,u_1(x),\nabla u_1(x)) - f(x,u_2(x),\nabla
u_2(x))\rmod^2\, dx\\
&\le \frac {2(\lambda_* + 1)}{\lambda_*^2} D^2 \lnorm u_1 - u_2\rnorm_1^2 =
\Lambda^2 \lnorm u_1 - u_2\rnorm_1^2.
\endalign
$$
Since $P_n$ are $H_1$-orthoprojectors, $\lnorm P_n\rnorm_{\Lin(H_1)} = 1$,
and the same estimate holds for all the operators $P_n\A$. Consequently, the
existence and uniqueness of $u_*$ and $u_n^*$ follows from the contraction
mapping principle.

Then, it is true that
$$
-\(\Delta \A(u), \A(u)\)_1 \le \biggl( 1 + \frac{1}{\lambda_*} \biggr)
\int_{\Om} \lmod f(x,u(x),\nabla u(x))\rmod^2\, dx \le \biggl( 1 +
\frac{1}{\lambda_*} \biggr) C^2 \meas\Om
$$
and by virtue of Lemma ~5
$$
\lnorm (I - P_n)\A(u)\rnorm_1 \le \frac{C}{\lambda_{n+1}} \sqrt{\(1 + \frac
{1}{\lambda_*}\) \meas\Om}.
$$
The proof is completed in the usual way.
\endproof

\bh {4.\enspace A generalization: Galerkin perturbed procedure}\ It is
sometimes convenient to use for approximate solution of the operator
equation \thetag{1.1} the following technique called Galerkin perturbed
procedure: this equation is replaced by the sequence of equations
$$
x_n = \A_n(x_n) \qquad (x_n \in E_n) \tag4.1
$$
where the operators $\A_n:E_n \to E_n$ are close in some sense to Galerkin
approximating operators $P_n\A$ but not equal to them. An analog of Theorem
~1 can be proved in this case also. Let a Banach space $E^{(1)}$ be imbedded
densely and completely continuously to the space $E$.

\tm {Theorem 6} The following conditions met:
\roster
\item the space $E^{(1)}$ contains all the subspaces $E_n$ and all the
operators $\A_n$ are bounded in $\lnorm\cdot\rnorm_1$-norm by one constant:
$$
\lnorm \A_n(x_n)\rnorm_1 \le K \qquad (x_n \in E_n);
$$
\item the inequality holds:
$$
\sup\Sb \lnorm x_n\rnorm_1 \le K\\ x_n \in E_n\endSb \lnorm \A(x_n) -
\A_n(x_n)\rnorm \le \beta_n;
$$
\item the operators $\A$ and $\A_n$ are Lipschitz:
$$
\allowdisplaybreaks\align
\lnorm A(x) - \A(y)\rnorm &\le \Lambda_1\lnorm x - y\rnorm \quad (x,\,y \in
E),\\
\lnorm A_n(x_n) - \A_n(y_n)\rnorm &\le \Lambda_2^{(n)}\lnorm x_n - y_n\rnorm
\quad (x_n,\,y_n \in E_n),
\endalign
$$
and $\Lambda_1,\,\Lambda_2^{(n)} < 1$ with all $n$,
\endroster
equations \thetag{1.1} and \thetag{4.1} have unique solutions $x_*$ and
$x_n^*$, respectively, and there holds the estimate
$$
\lnorm x_* - x_n^*\rnorm \le \frac {\beta_n}{1 - \Lambda_1}.
$$
\endtm

The proof is analog to that of Theorem ~1.

Take up again the differential equation
$$
\dot x = Ax + f(t,x) \tag4.2
$$
where $x \in \Real^N,\ A$ is a constant matrix, and the function $f(t,x)$ is
$T$-periodic in $t$, and use Theorem ~6 for approximating the $T$-periodic
solution of this equation by means of the finite differences technique. To
do this, assume $E = C_N$ and $E^{(1)} = W_{\infty, N}^1$, or the space of
Lipschitz functions $[0,T] \to \Real^N$ with the norm

$$
\lnorm x\rnorm_1 = \sup_{0 \le t_1,t_2 \le T} \frac {\lmod x(t_1) -
x(t_2)\rmod} {\lmod t_1 - t_2\rmod}.
$$
Let $E_n$ be the space of continuous functions $x:[0,T] \to \Real^N$ linear
in each segment $[kT/n,\, (k+1)T/n],\;(k = 0,\dots,n-1)$, and $P_n$ be the
projector $C_N \to E_n$ such that
$$
(P_n x)(kT/n) = x(kT/n)
$$
for any $k = 0,\dots,n$ and any $x(\cdot) \in C_N$.

If the matrix $A$ has no eigenvalues of the form $2\pi ki/T,\, k = 0, \pm1,
\dots$, then the $T$-periodic solutions of equation \thetag{4.2} (and only
they) are solutions of the operator equation
$$
x = \A(x) \tag4.3
$$
where the operator $\A: C_N \to C_N$ is given by formula \thetag{2.3}.
Assume now that $1$ is not an eigenvalue of the operator $I - (I +
\frac{AT}{n})^n$ with all $n$. For the Galerkin approximant $x_n \in E_n$
take a $T$-periodic function satisfying for any $k = 0, \dots, n-1$ the
equation
$$
\frac {x_n^{(k+1)} - x_n^k}{T/n} = A x_n^{(k)} + f(kT/n,x_n^{(k)}),
$$
where $x_n^{(k)}$ stands for $x_n(kT/n)$. This means that $x_n$ is a
solution of the equation
$$
x_n^{(k)} = \sum_{s=0}^{n-1} \frac {T}{n} h_n^{(k-s)} f(sT/n, x_n^{(s)})
\bydef \A_n(x_n) \tag4.4
$$
where
$$
h_n^{(s)} = (I - (I + \frac{AT}{n})^n)^{-1} (I + \frac{AT}{n})^{s}
$$
for $s = 0, \dots, n-1$ and $h_n^{(s \pm n)} = h_n^{(s)}$ for any $s$.

\tm {Theorem 7} The following conditions met:
\roster
\item matrices $I - e^{AT}$ and $I - (I + AT/n)^n$ are invertible with
$$
\allowdisplaybreaks\align
\lnorm (I - e^{AT})^{-1}\rnorm &\le Q, \tag4.5\\
\lnorm (I - (I + AT/n)^n)^{-1}\rnorm &\le Q_n; \tag4.6
\endalign
$$
\item matrix $A$ satisfies the estimates
$$
\allowdisplaybreaks\align
\max_{t \in [0,T]} \lnorm e^{At}\rnorm &\le M, \tag4.7\\
\max_{k = 0,\dots,n} \lnorm (I + AT/n)^k\rnorm &\le M_n; \tag4.8
\endalign
$$
\item function $f(t,x)$ is continuous in all its variables, $T$-periodic in
$t$, and satisfies the estimates
$$
\allowdisplaybreaks\gather
\lmod f(t,x)\rmod \le C, \tag4.9\\
\lmod f(t_1,x_1) - f(t_2,x_2) \rmod \le D_1 \lmod t_1 - t_2\rmod + D_2 \lmod
x_1 - x_2 \rmod \tag4.10\\
(t,\,t_1,\,t_2 \in [0,T],\; x,\,x_1,\,x_2 \in \Real^N);
\endgather
$$
\item the inequalities
$$
\allowdisplaybreaks\align
\Lambda_1 &\bydef QMD_2T < 1\\
\Lambda_2^{(n)} &\bydef Q_n M_n D_2T < 1
\endalign
$$
hold,
\endroster
equations \thetag{4.3} and \thetag{4.4} have unique solutions $x_*$ and
$x_n^*$, respectively, and there holds the estimate
$$
\lnorm x_* - x_n^*\rnorm \le \frac{QT^2}{n(1 - \Lambda_1)} (M(D_1 + D_2C(1
+ Q_n M_n \normA T)) + \normA e^{\normA T}(C + Q_nM_n))
$$
\endtm

To prove Theorem ~7, we need an elementary inequality.

\tm {Lemma 6} For any $t > 0$ and any $n = 0, 1, \dots$ the following
inequality hold:
$$
e^t - (1 + t/n)^n < te^t/n, \tag4.11
$$
\endtm

Inequality \thetag{4.11} follows from the estimate
$$
(1 + t/n)^n < e^t < (1 + t/n)^{n+1} \tag4.12
$$
which is easily proved by induction on $n$.

Proceed now to the proof of the Theorem.

\proof Let $x,\,y \in C_N$. Then by virtue of formulae \thetag{2.3},
\thetag{4.5}, \thetag{4.7}, and \thetag{4.10}
$$
\lnorm \A(x) - \A(y)\rnorm \le QMD_2T \lnorm x - y\rnorm = \Lambda_1 \lnorm
x - y\rnorm. \tag4.13
$$
The inequality
$$
\lnorm \A_n(x_n) - \A_n(y_n)\rnorm \le \Lambda_2^{(n)} \lnorm x_n -
y_n\rnorm \qquad (x_n,\, y_n \in E_n) \tag4.14
$$
can be proved in the same way. Since $\Lambda_1,\, \Lambda_2^{(n)} < 1$, the
operators $\A$ and $\A_n$ are contraction mappings and therefore their fixed
points $x_*$ and $x_n^*$ exist and are unique.

It follows from \thetag{4.6}, \thetag{4.8}, and \thetag{4.9} that
$$
\lnorm \A_n(x_n)\rnorm \le Q_n M_n TC.
$$
Evidently, for a function $x_n \in E_n$ the following formula holds:
$$
\lnorm x_n\rnorm_1 = \max_{k=0,\dots,n-1} \frac{\lmod x_n^{(k+1)} -
x_n^{(k)}\rmod} {T/n}.
$$
Since
$$
\frac{1}{T/n} (\A_n(x_n)^{(k+1)} - \A_n(x_n)^{(k)}) = A\A_n(x_n)^{(k)} +
f(kT/n,x_n^{(k)}),
$$
it is true that
$$
\lnorm \A_n(x_n)\rnorm_1 \le C(1 + Q_n M_n T\normA). \tag4.15
$$

Let now $x_n^* = \A_n(x_n^*)$ and $x_n^*(kT/n) = x^{(k)}$. It follows from
\thetag{4.5}, \thetag{4.8}, and \thetag{4.9} that
$$
\allowdisplaybreaks\align
\|\A(x_n^*) &- \A_n(x_n^*)\| \\
&\le Q\biggl| \int_0^T e^{A(T-s)} f(s,x_n^*(s))\,ds - \frac{T}{n}
\sum_{k=0}^{n-1} e^{\frac{n-k}{n}AT} f(kT/n, x^{(k)}) \biggr| \\
&+ QTC \max_{k=0,\dots,n} \lnorm e^{\frac{k}{n} AT} - (I + AT/n)^k\rnorm \\
&+ M_n TC \lnorm (I - e^{AT})^{-1} - (I - (I + AT/n)^n)^{-1}\rnorm. \tag4.16
\endalign
$$
Denote three terms of the latter sum as $I_1,\, I_2$, and $I_3$,
respectively, and estimate them.

Evidently, with any function $x \in W_{\infty, N}^1$ the following
estimate holds:
$$
\biggl| \int_0^T x(s)\,ds - \frac{T}{n} \sum_{k=0}^{n-1} x(kT/n) \biggr| \le
T^2 \lnorm x\rnorm_1/n.
$$
Using this inequaltiy and formulae \thetag{4.10} and \thetag{4.15}, we
obtain for the first term in \thetag{4.16}
$$
\allowdisplaybreaks\align
\lmod I_1\rmod &\le \frac{QMT^2}{n} \lnorm f(s,x_n^*(s))\rnorm_1 \le
\frac{QMT^2}{n} (D_1 + D_2 \lnorm x_n^*\rnorm_1) \\
&\le \frac{QMT^2}{n} (D_1 + D_2 C(1 + Q_n M_n \normA T)). \tag4.17
\endalign
$$

Then, it follows from Lemma ~6 that with any $k = 0, \dots, n$
$$
\allowdisplaybreaks\align
\lnorm e^{\frac{k}{n} AT} - (I + AT/n)^k\rnorm &= \biggl\| \sum_{s=0}^{k}
\binom{k}{s} \frac{A^s T^s}{n^s} - \sum_{s=0}^{\infty} \frac{1}{s!}
\frac{k^s A^s T^s}{n^s} \biggr\| \\
&\le \sum_{s=0}^{k} \biggl( \frac{k^s}{s!} - \binom{k}{s} \biggr)
\frac{T^s}{n^s} \normA^s + \sum_{s=k+1}^{\infty} \frac{k^sT^s}{n^s s!}
\normA^s\\
&= e^{k\normA T/n} - (1 + \normA T/n)^k\\
&\le e^{k\normA T/n} \frac{\normA T}{n} \le \frac{\normA T}{n} e^{\normA T},
\endalign
$$
and therefore
$$
\lmod I_2\rmod \le QT^2 C\normA e^{\normA T}/n. \tag4.18
$$

Since for any operators $P$ and $Q$ the inequality
$$
\lnorm P^{-1} - Q^{-1}\rnorm \le \lnorm P^{-1}\rnorm \lnorm Q^{-1}\rnorm
\lnorm P - Q\rnorm
$$
holds, an inequality
$$
\lnorm (I - e^{AT})^{-1} - (I - (I + AT/n)^n)^{-1}\rnorm \le QQ_n \normA T
e^{\normA T}/n
$$
follows from \thetag{4.18}. Consequently,
$$
\lmod I_3\rmod \le QQ_n M_n T^2 C\normA e^{\normA T}/n. \tag4.19
$$

The proof is completed by summing inequalities \thetag{4.17}--\thetag{4.19}
and applying Theorem ~6.
\endproof

\enddocument