# Parallel Type Decomposition Scheme for Quasi-Linear Abstract Hyperbolic
Equation

Nana Dikhaminjia, Jemal Rogava and Mikheil Tsiklauri

###### Abstract

Cauchy problem for an abstract hyperbolic equation with the Lipschitz
continuous operator is considered in the Hilbert space. The operator
corresponding to the elliptic part of the equation is a sum of operators
$A_{1},\,A_{2},\,\ldots,\,A_{m}$. Each addend is a self-adjoint and positive
definite operator. A parallel type decomposition scheme for an approximate
solution of the stated problem is constructed. The main idea of the scheme is
that on each local interval classic difference problems are solved in parallel
(independently from each other) respectively with the operators
$A_{1},\,A_{2},\,\ldots,\,A_{m}$. The weighted average of the received
solutions is announced as an approximate solution at the right end of the
local interval. Convergence of the proposed scheme is proved and the
approximate solution error is estimated, as well as the error of the
difference analogue for the first-order derivative for the case when the
initial problem data satisfy the natural sufficient conditions for solution
existence.

Keywords and phrases: Decomposition scheme, Abstract hyperbolic equation,
Operator splitting, Cauchy problem, Parallel algorithm.
MSC 2010: 65M12, 65M15, 65M55, 49M27.

##  1 Introduction

First works dedicated to construction and investigation of decomposition
schemes were published in the fifties and sixties of the XX century (see G. A.
Baker, T. A. Oliphant [1], G. Birkhoff, R. S. Varga [2], G. Birkhoff, R. S.
Varga, D. Young [3], J. Douglas [7], J. Douglas, H. Rachford [8], E. G.
Diakonov [6], D. G. Gordeziani [9], N. N. Ianenko [14], [15], V. P. Ilin [16],
A. N. Konovalov [18], G. I. Marchuk, N. N. Ianenko [23], G. I. Marchuk, U. M.
Sultangazin [24], D. Peaceman, H. Rachford [28], A. A. Samarskii [34], [35]).
It may be said that the works of these authors become a basis for further
research on decomposition schemes.

Decomposition schemes in view of numerical calculation can be divided into two
groups: schemes of sequential account (see for example G. I. Marchuk [22], A.
A. Samarskii, P. N. Vabishchevich [36] ) and schemes of parallel account (D.
G. Gordeziani [9], [10], D. G. Gordeziani, H. V. Meladze [11], D. G.
Gordeziani, A. A. Samarskii [12], A. M. Kuzyk, V. L. Makarov [20]).

In view of numerical calculations, with the development of parallel
processing, obviously the parallel type decomposition schemes have clear
advantage. The present work discusses the construction and investigation of
parallel type decomposition scheme.

It is well-known that for error estimation of approximate solution of
evolution problem, usually the solution is required to be of the higher order
smooth than it is necessary following from natural conditions. In case of
decomposition schemes this detail gains more importance. Demand on increasing
the smoothness can be conditioned by the fact that the operators corresponding
to the problems obtained by splitting are non-commutative. Therefore, it is
important to build the decomposition schemes, whose numerical calculation and
methodology of error estimate of approximate solution does not require sharp
increase of the solution smoothness. These details are much more complicated
for the second order evolution equation compared to the first order one. One
of the reasons for this is that natural scheme for the first order evolution
equation is two-layer, and for the second order - three-layer. For the most
cases, in comparison with the two-layer schemes, investigation of the three-
layer schemes are related with certain difficulties. At the first sight, this
issue might be overcome: the second order evolution equation by introducing
the additional unknown can be deduced to the first order system. However, in
this case, if the operator in the initially given equation is self-adjoint, in
the obtained system there will be non self-adjoint matrix operator, that
significantly complicates solving of the corresponding discrete problem.

In the present work the decomposition scheme for the second order evolution
equation is proposed that does not require to increase smoothness of the
solution in view of numerical calculation. In addition, the methodology that
we use for error estimate of the approximate solution makes it possible to
find convergence order in the conditions with almost natural limitations.

For the investigation of the decomposition scheme we use polynomials of
certain class, which we call two-variable polynomial. These polynomials are
represented by means of second order classical Chebyshev polynomials.

We should note that several works are devoted to use of orthogonal polynomials
in approximate solution schemes for differential equations: V. L. Makarov
[21], A. G. Morris, T. S. Horner [25], V. A. Novikov, G. V. Demidov [26], V.
A. Rastrenin [29]. In the work [21] many aspects of using orthogonal
polynomials in the difference problems are presented quite widely.

Recent results related to the construction and investigation of decomposition
schemes for evolution equations are obtained by the following authors: S.
Blanes, F. Casas and M. Thalhammer [4], D. He, K. Pan and H. Hu [13], J. L.
Padgett and Q. Sheng [27], J. Zhao, R. Zhan and Y. Xu [38]. We also note the
work [5] in which a high-order accuracy decomposition scheme is considered for
an abstract hyperbolic equation.

##  2 Statement of the problem and decomposition scheme

Let us consider the Cauchy problem for abstract hyperbolic equation in the
Hilbert space $H$:

$\frac{d^{2}u(t)}{dt^{2}}+Au\left(t\right)+M\left(u\left(t\right)\right)=f\left(t\right),\quad
t\in\left[0,T\right],$ (2.1)
$u\left(0\right)=\varphi_{0},\quad\frac{du\left(0\right)}{dt}=\varphi_{1}.$
(2.2)

where $A$ is a self-adjoint ($A$ does not depend on $t$), positive definite
(generally unbounded) operator with the definition domain $D\left(A\right)$,
which is everywhere dense in $H$, i.e. $\overline{D\left(A\right)}=H,\quad
A=A^{\ast}$ and

$\left(Au,u\right)\geq\alpha\left\|u\right\|^{2},\quad\forall u\in
D\left(A\right),\quad\alpha=const>0,$

where by $\left\|\cdot\right\|$ and $\left(\cdot,\cdot\right)$ are defined
correspondingly the norm and scalar product in $H$; nonlinear operator
$M\left(\cdot\right)$ satisfies Liptschitz condition,

$\left\|M(u)-M(v)\right\|\leq a\left\|u-v\right\|,\quad\forall u,v\in H,$

$a=const>0$ ; $\varphi_{0}$ and $\varphi_{1}$ are given vectors from $H$;
$u\left(t\right)$ is a continuous, twice continuously differentiable, searched
function with values in $H$, and $f\left(t\right)$ is given continuous
function with values in $H.$

Similar to the linear case, $u\left(t\right)$ vector function with values in
$H$, defined on the interval $\left[0,T\right]$, is called a solution of the
problem (2.1), (2.2) if it satisfies the following conditions: (a)
$u\left(t\right)$ is twice continuously differentiable in the interval
$\left[0,T\right]$; (b) $u\left(t\right)\in D\left(A\right)$ for any $t$ from
$\left[0,T\right]$ , the function $\ Au\left(t\right)$ is continuous and
$M(u\left(t\right))$ is continuous; (c) $u\left(t\right)$ satisfies equation
(2.1) on the $\left[0,T\right]$ interval and the initial condition (2.2).

Remark 2.1. If $f\left(t\right)$ is continuously differentiable on
$\left[0,T\right]$ (or $f\left(t\right)\in D(A^{1/2})$ for any $t$ from
$\left[0,T\right]$ and the function $A^{1/2}f\left(t\right)$ is continuous),
$\varphi_{0}\in D(A)$ and $\varphi_{1}\in D(A^{1/2})$, then there exists only
solution $u\left(t\right)$ of the problem (2.1), (2.2) (without Lipschitz
continuous operator) that satisfies the condition: the function
$u^{\prime}\left(t\right)$ gets the values from $D(A^{1/2})$ and
$A^{1/2}u^{\prime}\left(t\right)$ is continuous on $\left[0,T\right]$ (see
[19], Theorem 1.5, p. 301).

Let

$A=\sum_{j=1}^{m}A_{j}\ ,\quad A_{j}=A_{j}^{\ast}\geq\alpha_{j}I\
,\quad\alpha_{j}=const>0\ .$ (2.3)

Then approximate solution of problem (2.1), (2.2) at the points
$t=t_{k+1}=(k+1)\tau$ , $k=1,\ldots,n-1$ , $\tau=T/n$ ($n>1$) is defined by
the following formula:

$v_{k+1}=\sum_{j=1}^{m}\eta_{j}y_{j,k+1}\ ,\quad\sum_{j=1}^{m}\eta_{j}=1,\quad
0<\eta_{j}<1,\quad$

where $y_{j,k+1}$ is a solution of the following difference problem:

$\eta_{j}\frac{y_{j,k+1}-2v_{k}+v_{k-1}}{\tau^{2}}+A_{j}y_{j,k+1}=\delta_{1,j}\left[f(t_{k})-M\left(v_{k}\right)\right]\
,$ (2.4)

$v_{0}=\varphi_{0}\ ,\quad v_{1}=\varphi_{0}+\tau\varphi_{1}\ ,$ (2.5)

where $j=1,\ldots,m$, $\delta_{1,j}$ is a Kronecker symbol.

Thus, to construct approximate solution $v_{k+1}$ for problem (2.1), (2.2) at
the point $t_{k+1}$ , it is necessary to solve $m$ problems independent from
each other. Therefore, scheme (2.4) can be called parallel type decomposition
scheme. These kind schemes for the first time were discussed in the work by D.
Gordeziani (see [9], [10]). Parallel type decomposition schemes also are
considered in the works: D. Gordezianis, A. Samarski [12], D. G. Gordeziani,
H. V. Meladze [11], A. M. Kuzyk, V. L. Makarov [20]. Specifically the scheme
(2.4) (without Lipschitz-continuous operator) is given in [31].

##  3 Representation of the approximate solution error by means of Chebyshev
polynomial and the main theorem

Let the problem (2.1), (2.2) has a solution.

Then the equation (2.1) at the point $t=t_{k+1}$ can be written as

$\frac{u(t_{k+1})-2u(t_{k})+u(t_{k-1})}{\tau^{2}}+Au(t_{k+1})=g_{k}\ ,$ (3.1)

where

$\displaystyle g_{k}$ $\displaystyle=$
$\displaystyle\widetilde{f}(t_{k})+A\left[u(t_{k+1})-u(t_{k})\right]$
$\displaystyle+\tau^{-2}\int_{t_{k}}^{t_{k+1}}(t_{k+1}-t)\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{k}\right)\right]dt$

$+\tau^{-2}\int_{t_{k-1}}^{t_{k}}(t-t_{k-1})\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{k}\right)\right]dt\
.$

and where $\widetilde{f}(t)=f(t)-M(u(t))$.

From (3.1) it follows that

$u(t_{k+1})-2Lu(t_{k})+Lu(t_{k-1})=\tau^{2}Lg_{k}\ ,$ (3.2)

where $k=1,\ldots,n-1$,

$L=\left(I+\tau^{2}A\right)^{-1}.$

From (2.4) we have

$y_{j,k+1}-2S_{j}v_{k}+S_{j}v_{k-1}=\tau^{2}\eta_{j}^{-1}\delta_{1,j}S_{j}\left[f(t_{k})-M\left(v_{k}\right)\right]\
,$ (3.3)

where $j=1,\ldots,m$ ,

$S_{j}=\left(I+\tau^{2}\eta_{j}^{-1}A_{j}\right)^{-1}.$

If we multiply both sides of equality (3.3) on $\eta_{j}$ and summarize, we
get

$v_{k+1}-2Sv_{k}+Sv_{k-1}=\tau^{2}\psi_{k}\ ,$ (3.4)

where $k=1,\ldots,n-1$,

$\displaystyle S$ $\displaystyle=$ $\displaystyle\sum_{j=1}^{m}\eta_{j}S_{j}\
,\quad\psi_{k}=\sum_{j=1}^{m}\delta_{1,j}S_{j}\left[f(t_{k})-M\left(v_{k}\right)\right]$
$\displaystyle=$ $\displaystyle
S_{1}\left[f(t_{k})-M\left(v_{k}\right)\right]\ .$

If we take (3.4) from (3.2), we get

$z_{k+1}-2Sz_{k}+Sz_{k-1}=r_{k}\ ,$ (3.5)

where $z_{k}=v_{k}-u(t_{k})$,

$\displaystyle r_{k}$ $\displaystyle=$ $\displaystyle
r_{0,k}+r_{1,k}-L\left(\tau^{2}r_{2,k}+r_{3,k}\right)+r_{4,k}\ ,$
$\displaystyle r_{0,k}$ $\displaystyle=$
$\displaystyle\left(S-L\right)u(t_{k})\ ,\quad
r_{1,k}=\left(S-L\right)\left[u(t_{k})-u(t_{k-1})\right]\ ,$ $\displaystyle
r_{2,k}$ $\displaystyle=$ $\displaystyle A\left[u(t_{k+1})-u(t_{k})\right]\
,\quad r_{4,k}=\tau^{2}\left[\psi_{k}-L\widetilde{f}(t_{k})\right]\ ,$

$\displaystyle r_{3,k}$ $\displaystyle=$
$\displaystyle\int_{t_{k}}^{t_{k+1}}(t_{k+1}-t)\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{k}\right)\right]dt$
$\displaystyle+\int_{t_{k-1}}^{t_{k}}(t-t_{k-1})\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{k}\right)\right]dt\
.$

To present solution of equation (3.5) in explicit form, we need certain class
polynomials, which we call two-variable Chebyshev polynomials. These
polynomials are defined by the following recurrent relation:

$\displaystyle\widetilde{U}_{k+1}(x,y)$ $\displaystyle=$ $\displaystyle
x\widetilde{U}_{k}(x,y)-y\widetilde{U}_{k-1}(x,y),\quad k=1,2,...\quad,$ (3.6)
$\displaystyle\widetilde{U}_{1}(x,y)$ $\displaystyle=$ $\displaystyle
x,\quad\widetilde{U}_{0}(x,y)=1.$

Notice that the works [33], [32] are devoted to research of three-layer semi-
discrete schemes by means of Chebyshev polynomials.

We call $\widetilde{U}_{k}(x,y)$ two-variable Chebyshev polynomials as
$U_{k}(x)=\widetilde{U}_{k}\left(2x,1\right)$ represents second order
Chebyshev polynomials.

The following formula can be easily obtained

$\widetilde{U}_{k}(x,y)=\sqrt{y^{k}}\widetilde{U}_{k}(\xi,1),\quad\xi=\frac{x}{\sqrt{y}},\quad
y>0\ ,$ (3.7)

that relates $\widetilde{U}_{k}(x,y)$ to $U_{k}(x)$.

Now we can explicitly write solution of equation (3.5) by means of polynomials
$\widetilde{U}_{k}(x,y)$. Using induction we get

$z_{k+1}=\widetilde{U}_{k}\left(2S,S\right)z_{1}-S\widetilde{U}_{k-1}\left(2S,S\right)z_{0}+\sum_{i=1}^{k}\widetilde{U}_{k-i}\left(2S,S\right)r_{i}\
.$ (3.8)

Obviously, as $S_{j}$ (j$=1,\ldots,m$) are self-adjoint, non-negative, bounded
operators, then operator $S$ also will be self-adjoint non-negative and
bounded. Hence, as known (see, e. g., [30], Chapter VI), there exists the only
square root $S^{1/2}$ . Taking this into account and using formula (3.7) , we
obtain

$\widetilde{U}_{k}\left(2S,S\right)=B^{k}\widetilde{U}_{k}\left(2B,I\right)=B^{k}U_{k}(B)\
,\quad B=S^{1/2}.$

Due to this equality, (3.8) will be

$z_{k+1}=B^{k}U_{k}(B)z_{1}-B^{k+1}U_{k-1}(B)z_{0}+\sum_{i=1}^{k}B^{k-i}U_{k-i}(B)r_{i}\
.$ (3.9)

Formula (3.9) is a main relation, by using of which the following theorem is
proved.

Theorem 3.1. If the problem (2.1), (2.2) has a solution, then the estimate is
valid for approximate solution error:

$\left\|z_{k+1}\right\|\leq\exp\left(ct_{k-1}\right)\left(\gamma_{0}\left\|\frac{\Delta
z_{0}}{\tau}\right\|+\gamma_{1}\left\|z_{0}\right\|+\Theta_{k}(\tau)\right)\
,$ (3.10)

where $z_{k}=v_{k}-u(t_{k})$, $\Delta z_{0}=z_{1}-z_{0}$ , $c=\nu^{-1/2}a$ ,
$\nu=\min\limits_{1\leq j\leq m}\left(\alpha_{j}\right)$,
$\gamma_{0}=\nu^{-1/2}+c\tau^{2}$, $\gamma_{1}=1+c\tau$ ,

$\displaystyle\Theta_{k}(\tau)$ $\displaystyle=$
$\displaystyle\tau^{2}\sum_{i=1}^{k}\left[c_{1}\left\|Au(t_{i})\right\|+c_{3}\left\|\widetilde{f}(t_{i})\right\|\right]$
$\displaystyle+\tau\sum_{i=1}^{k}\left[c_{1}J_{i}(t_{i-1},\
A^{1/2}u)+c_{2}J_{i}(t_{i+1},\ A^{1/2}u)\right]$
$\displaystyle+c_{2}\sum_{i=1}^{k}\int_{t_{i-1}}^{t_{i+1}}\left[J_{i}\left(t,\
A^{1/2}u\right)+J_{i}\left(t,\ A^{-1/2}\widetilde{f}\right)\right]dt\ ,$

and where

$\displaystyle J_{i}\left(t,u\right)$ $\displaystyle=$
$\displaystyle\left\|u(t_{i})-u(t)\right\|,\quad\widetilde{f}(t)=f(t)-M(u(t)),$
$\displaystyle c_{1}$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{m}\eta_{j}^{-3/2}\left(\eta_{j}^{-1}a_{j}+1\right),\quad
a_{j}=\left\|A_{j}A^{-1}\right\|<\infty\ ,$ $\displaystyle c_{2}$
$\displaystyle=$ $\displaystyle m+c_{0}\ ,\quad
c_{0}=\sum_{j=1}^{m}\eta_{j}^{-1/2}a_{j}\ ,\quad
c_{3}=\eta_{1}^{-1/2}+m+c_{0}\ .$

Result 2.2. If the functions $f(t)$ and $A^{1/2}$

$u(t)$ on $[0,\ T]$ satisfies Holder condition by the index $\lambda$
($0<\lambda\leq 1$), then

$\left\|u(t_{k})-v_{k}\right\|\leq c\tau^{\lambda},\quad c=const>0.$

##  4 Auxiliary lemmas and remarks

Remark 4.1. As $A_{j}A^{-1}$ ($j=1,\ldots,m$) are closed operators (it can be
proved easily), therefore due to Closed Graph Theorem they are bounded, i.e.
$a_{j}=\left\|A_{j}A^{-1}\right\|<\infty$ .

Lemma 4.2. For any $j$ ($j=1,\ldots,m$) $D(A^{1/2})\subset D(A_{j}^{1/2})$ and

$\left\|A_{j}^{1/2}u\right\|\leq\left\|A^{1/2}u\right\|,\quad\forall u\in
D(A^{1/2}).$ (4.1)

Proof: According to condition (2.3) we have:

$\left(A_{j}u,u\right)\leq\left(Au,u\right),\quad\forall u\in D(A)\subset
D(A_{j}).$

From here it follows

$\left\|A_{j}^{1/2}u\right\|\leq\left\|A^{1/2}u\right\|,\quad\forall u\in
D(A)\subset D(A_{j}).$ (4.2)

It is known that $D(A)$ is a ball for $A^{1/2}$(see [17], p. 354). This means:
for each $u\in D(A^{1/2})$ there exists a sequence $u_{n}\in D(A)$ such that
$u_{n}\rightarrow u$ and $A^{1/2}u_{n}\rightarrow A^{1/2}u$. From here,
according to (4.2) it follows that $A_{j}^{1/2}u_{n}$ is a Cauchy Sequence.
Obviously it will be convergent due to completeness of $H$. Since
$A_{j}^{1/2}$ is closed ( $A_{j}$ is given as self-adjoint and positive
definite), therefore $u\in D(A_{j}^{1/2})$ and $A_{j}^{1/2}u_{n}\rightarrow
A_{j}^{1/2}u$. Thus $D(A^{1/2})\subset D(A_{j}^{1/2})$, and in addition, the
inequality

$\left\|A_{j}^{1/2}u_{n}\right\|\leq\left\|A^{1/2}u_{n}\right\|,\quad u_{n}\in
D(A)\subset D(A_{j})$

gives (4.1).

Remark 4.3. If self-adjoint positive definite operators $A$ and $B$ are such
that $D(A)\subset D(B)$ and $B\leq A$
($\left(Bu,u\right)\leq\left(Au,u\right)$ , $\forall u\in D(A)$ ), then
$A^{-1}\leq B^{-1}$ .

Indeed, let’s $u=B^{-1}f$ and $\ v=A^{-1}f$ , $f\in H$. The relation is true
(see the proof of Theorem VI.2.21 from [17]):

$\displaystyle\left(A^{-1}f,\ f\right)^{2}$ $\displaystyle=$
$\displaystyle\left(v,\ Bu\right)^{2}=\left(B^{1/2}v,\ B^{1/2}u\right)^{2}$
$\displaystyle\leq$
$\displaystyle\left\|B^{1/2}v\right\|^{2}\left\|B^{1/2}u\right\|^{2}=\left(Bv,v\right)\left(Bu,u\right)$
$\displaystyle\leq$ $\displaystyle\left(Av,v\right)\left(Bu,u\right)=\left(f,\
A^{-1}f\right)\left(f,\ B^{-1}f\right)$ $\displaystyle=$
$\displaystyle\left(A^{-1}f,\ f\right)\left(B^{-1}f,\ f\right).$

After reduction we get

$\left(A^{-1}f,\ f\right)\leq\left(B^{-1}f,\ f\right),$

i. e. $A^{-1}\leq B^{-1}$.

Remark 4.4 The formula is valid

$S-L=\tau^{2}\sum_{j=1}^{m}\eta_{j}^{-1}\left(I-S_{j}\right)\left(\eta_{j}^{-1}A_{j}A^{-1}-I\right)AL\
.$ (4.3)

Indeed, as

$I-S=\sum_{j=1}^{m}\eta_{j}\left(I-S_{j}\right)=\tau^{2}\sum_{j=1}^{m}A_{j}S_{j}\
,$

therefore

$\displaystyle S-L$ $\displaystyle=$
$\displaystyle\left[S\left(I+\tau^{2}A\right)-I\right]L=\left[\left(S-I\right)+\tau^{2}AS\right]L$
$\displaystyle=$
$\displaystyle\tau^{2}\sum_{j=1}^{m}\left(\eta_{j}S_{j}A-A_{j}S_{j}\right)L\
.$

From here, taking the equality into account

$\displaystyle\eta_{j}S_{j}A-A_{j}S_{j}$ $\displaystyle=$
$\displaystyle\eta_{j}S_{j}A-S_{j}A_{j}=S_{j}\left(\eta_{j}A-A_{j}\right)$
$\displaystyle=$
$\displaystyle\left(S_{j}-I\right)\left(\eta_{j}A-A_{j}\right)+\left(\eta_{j}A-A_{j}\right)\
,$

we get (4.3) .

We can consider the closeness of operators $S$ and $L$ also by the following
formulas:

$\displaystyle S_{j}$ $\displaystyle=$
$\displaystyle\tau^{4}\eta_{j}^{-2}A_{j}^{2}S_{j}-\tau^{2}\eta_{j}^{-1}A_{j}+I\
,$ $\displaystyle S$ $\displaystyle=$
$\displaystyle\tau^{4}\sum_{j=1}^{m}\eta_{j}^{-1}A_{j}^{2}S_{j}-\tau^{2}A+I\
,$ $\displaystyle L$ $\displaystyle=$
$\displaystyle\tau^{4}A^{2}L-\tau^{2}A+I\ .$

Lemma 4.5. The inequalities are valid:

$\displaystyle\left\|\left(I-S\right)^{-1/2}ALf\right\|$ (4.4)
$\displaystyle\leq$
$\displaystyle\tau^{-1}\left(m\left\|A^{1/2}Lf\right\|+c_{0}\left\|A^{1/2}L^{1/2}f\right\|\right),\quad
f\in H,$

$\left\|\left(I-S\right)^{-1/2}\left(S-L\right)f\right\|\leq\tau^{2}c_{1}\left\|ALf\right\|,\quad
f\in H,$ (4.5)

$\left\|\left(I-S\right)^{-1/2}LAu\right\|\leq\tau^{-1}c_{2}\left\|A^{1/2}u\right\|,\quad
u\in D(A),$ (4.6)

$\left\|\left(I-S\right)^{-1/2}Lf\right\|\leq\tau^{-1}c_{2}\left\|A^{-1/2}f\right\|,\quad
f\in H.$ (4.7)

Proof: Let us prove the inequality (4.4) . As

$I-S\geq\eta_{j}\left(I-S_{j}\right)=\tau^{2}A_{j}S_{j}>0,$

therefore (see remark 4.3)

$\displaystyle\left(I-S\right)^{-1}$ $\displaystyle\leq$
$\displaystyle\eta_{j}^{-1}\left(I-S_{j}\right)^{-1}=\tau^{-2}\left(I+\tau^{2}\eta_{j}^{-1}A_{j}\right)A_{j}^{-1}$
(4.8) $\displaystyle=$
$\displaystyle\tau^{-2}A_{j}^{-1}+\eta_{j}^{-1}I\leq\left(\tau^{-1}A_{j}^{-s}+\eta_{j}^{-s}I\right)^{2},\quad
s=\frac{1}{2}\ .$

Thus we have

$\displaystyle\left\|\left(I-S\right)^{-s}f\right\|$ $\displaystyle\leq$
$\displaystyle\left\|\left(\tau^{-1}A_{j}^{-s}+\eta_{j}^{-s}I\right)f\right\|$
$\displaystyle\leq$
$\displaystyle\tau^{-1}\left\|A_{j}^{-s}f\right\|+\eta_{j}^{-s}\left\|f\right\|,\quad
f\in H.$

Using this inequality we get (below everywhere $s=1/2$):

$\displaystyle\left\|\left(I-S\right)^{-s}ALf\right\|$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-s}\sum_{j=1}^{m}A_{j}Lf\right\|$ (4.9)
$\displaystyle\leq$
$\displaystyle\tau^{-1}\sum_{j=1}^{m}\left(\left\|A_{j}^{s}Lf\right\|+\tau\eta_{j}^{-s}\left\|A_{j}Lf\right\|\right).$

According to lemma 4.2 we have

$\left\|A_{j}^{s}Lf\right\|\leq\left\|A^{s}Lf\right\|.$ (4.10)

According to remark 4.1 we have

$\left\|A_{j}Lf\right\|\leq\left\|A_{j}A^{-1}\right\|\left\|ALf\right\|=a_{j}\left\|ALf\right\|.$
(4.11)

From (4.9), taking into account (4.10) and (4.11), we obtain

$\left\|\left(I-S\right)^{-s}ALf\right\|\leq
m\tau^{-1}\left\|A^{s}Lf\right\|+c_{0}\left\|ALf\right\|.$

Hence, taking into account inequality

$\left\|ALf\right\|=\left\|A^{s}L^{s}\left(A^{s}L^{s}f\right)\right\|\leq\left\|A^{s}L^{s}\right\|\left\|A^{s}L^{s}f\right\|\leq\tau^{-1}\left\|A^{s}L^{s}f\right\|$

(4.4) is obtained.

Let us prove inequality (4.5). From (4.8) we have

$\left\|\left(I-S\right)^{-1/2}h\right\|\leq\eta_{j}^{-1/2}\left\|\left(I-S_{j}\right)^{-1/2}h\right\|,\quad
h\in H.$ (4.12)

If in (4.12) we substitute $h=\left(I-S_{j}\right)f$ , we get

$\left\|\left(I-S\right)^{-1/2}\left(I-S_{j}\right)f\right\|\leq\eta_{j}^{-1/2}\left\|\left(I-S_{j}\right)^{1/2}f\right\|\leq\eta_{j}^{-1/2}\left\|f\right\|.$
(4.13)

From this inequality and (4.3) it follows

$\displaystyle\left\|\left(I-S\right)^{-1/2}\left(S-L\right)f\right\|$
$\displaystyle\leq$
$\displaystyle\tau^{2}\sum_{j=1}^{m}\eta_{j}^{-1}\left\|\left(I-S\right)^{-1/2}\left(I-S_{j}\right)\left(\eta_{j}^{-1}A_{j}A^{-1}-I\right)ALf\right\|$
$\displaystyle\leq$
$\displaystyle\tau^{2}\sum_{j=1}^{m}\eta_{j}^{-3/2}\left\|\left(\eta_{j}^{-1}A_{j}A^{-1}-I\right)ALf\right\|\leq\tau^{2}c_{1}\left\|ALf\right\|\
.$

Let us prove the inequality (4.6) . Obviously, operators $A$ and $L$ are
commutative on $D(A)$. Then due to (4.4), the relation is valid:

$\displaystyle\left\|\left(I-S\right)^{-1/2}LAu\right\|$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}ALu\right\|$ $\displaystyle\leq$
$\displaystyle\tau^{-1}\left(m\left\|A^{1/2}Lu\right\|+c_{0}\left\|A^{1/2}L^{1/2}u\right\|\right)$
$\displaystyle\leq$
$\displaystyle\tau^{-1}\left(m\left\|LA^{1/2}u\right\|+c_{0}\left\|L^{1/2}A^{1/2}u\right\|\right)$
$\displaystyle\leq$ $\displaystyle\tau^{-1}c_{2}\left\|A^{1/2}u\right\|.$

Obviously inequality (4.7) is a result of (4.6).

Lemma 4.6. For operator polynomials $U_{k}(B)$ the following estimate is
valid:

$\left\|BU_{k}(B)\right\|\leq\left(\tau\sqrt{\nu}\right)^{-1},\quad\nu=\min_{1\leq
j\leq m}\left(\alpha_{j}\right)\ ,$ (4.14)

$\left\|U_{k}(B)\left(I-B^{2}\right)^{1/2}\right\|\leq 1\ ,$ (4.15)

$\left\|U_{k}(B)-BU_{k-1}(B)\right\|\leq 1,$ (4.16)

$\left\|BU_{k}(B)-U_{k-1}(B)\right\|\leq 1.$ (4.17)

Proof: As it is well-known, for Chebyshev second-order polynomials the
estimate is valid (see, e. g., [37]):

$\left|U_{k}(x)\right|\leq\frac{1}{\sqrt{1-x^{2}}}\ ,\quad x\in]-1,1[\ ,$
(4.18)

that follows from the well-known formula

$U_{k}\left(x\right)=\frac{\sin\left(\left(k+1\right)\arccos
x\right)}{\sqrt{1-x^{2}}},\quad x\in]-1,1[\ .$

From this formula, by means of simple calculations, we also obtain the
estimate

$\left|U_{k}(x)-U_{k-1}(x)\right|\leq\sqrt{\frac{2}{1+x}}\ ,\quad x\in]-1,1[\
.$ (4.19)

Indeed we have

$\displaystyle\left|U_{k}(x)-U_{k-1}(x)\right|$ $\displaystyle=$
$\displaystyle\frac{2}{\sqrt{1-x^{2}}}\left|\cos(k+\frac{\theta}{2})\sin\frac{\theta}{2}\right|$
$\displaystyle\leq$
$\displaystyle\frac{2}{\sqrt{1-x^{2}}}\sqrt{\frac{1-\cos\theta}{2}}$
$\displaystyle=$ $\displaystyle\sqrt{\frac{2}{1+x}}\ ,\quad x\in]-1,1[\ ,$

where $\theta=\arccos x$ .

The following estimation is valid:

$\left|U_{k}(x)-xU_{k-1}(x)\right|\leq\left|\cos((k+1)\theta)\right|\leq 1.$
(4.20)

Obviously, similarly we get the estimate

$\left|xU_{k}(x)-U_{k-1}(x)\right|\leq 1.$ (4.21)

For the estimation of norm of polynomial operators $U_{k}(B)$ , we need to
estimate the spectrum of operator $B=S^{1/2}$, that obviously can be reduced
to the estimate of spectrum of operator $S$ .

Let us first estimate spectrum of operator
$S_{j}=\left(I+\tau^{2}\eta_{j}^{-1}A_{j}\right)^{-1}$ . We obviously have

$\left(I+\tau^{2}\eta_{j}^{-1}A_{j}\right)\geq\left(1+\tau^{2}\eta_{j}^{-1}\alpha_{j}\right)I>0\
.$

From here, due to remark 4.3, it follows

$0<S_{j}\leq\left(1+\tau^{2}\eta_{j}^{-1}\alpha_{j}\right)^{-1}I\leq\left(1+\tau^{2}\nu\right)^{-1}I\
.$ (4.22)

If we take into account representation of $S$, then according to (4.22), we
have

$0<S\leq\sum_{j=1}^{m}\eta_{j}\left(1+\tau^{2}\nu\right)^{-1}I=\left(1+\tau^{2}\nu\right)^{-1}I\
.$

This means that

$Sp\left(S\right)\subset[0,\ \left(1+\tau^{2}\nu\right)^{-1}]$

From here it follows that (due to well-known theorem on spectral mapping)

$Sp\left(B\right)\subset[0,\ \left(1+\tau^{2}\nu\right)^{-1/2}]\ .$ (4.23)

Further we can easily show estimates of (4.14)-(4.17).

Let us show estimate (4.14). As is known, norm of operator-function, when the
argument represents self-adjoint bounded operator, is equal to the $C$-norm of
the corresponding scalar function on the spectrum (see, e. g., [30], Chapter
VII). Due to this result we have

$\left\|BU_{k}(B)\right\|\leq\max_{x\in Sp(B)}\left|xU_{k}(x)\right|\ .$

From here, by the inequality (4.18) and relation (4.23), we get

$\left\|BU_{k}(B)\right\|\leq\max_{x\in
Sp(B)}\frac{x}{\sqrt{1-x^{2}}}\leq\frac{1}{\tau\sqrt{\nu}}\ .$

Similarly is obtained (4.15). Indeed we have

$\displaystyle\left\|U_{k}(B)\left(I-B^{2}\right)^{1/2}\right\|$
$\displaystyle\leq$ $\displaystyle\max_{x\in
Sp(B)}\left|U_{k}(x)\sqrt{1-x^{2}}\right|$ $\displaystyle\leq$
$\displaystyle\max_{x\in
Sp(B)}\left(\frac{1}{\sqrt{1-x^{2}}}\sqrt{1-x^{2}}\right)=1\ .$

From the inequalities (4.20) and (4.21) respectively follows the estimates
(4.16) and (4.17) .

##  5 Error estimate of approximate solution

In this section we continue to prove Theorem 3.1, that concerns error estimate
of approximate solution.

Let us rewrite formula (3.9) in the following form

$\displaystyle z_{k+1}$ $\displaystyle=$ $\displaystyle\tau
B^{k}U_{k}(B)\frac{\Delta
z_{0}}{\tau}+B^{k}\left(U_{k}(B)-BU_{k-1}(B)\right)z_{0}$ (5.1)
$\displaystyle+\sum_{i=1}^{k}B^{k-i}U_{k-i}(B)r_{i}\ ,$

where $\Delta z_{0}=z_{1}-z_{0}$ .

Let us note that $\Delta z_{0}/\tau$ is an error of the difference analog of
the first order derivative of approximate solution at the point $t=0$ ,

$\frac{\Delta z_{0}}{\tau}=\frac{\Delta u_{0}}{\tau}-\frac{\Delta u(0)}{\tau}\
,\quad\Delta u(0)=u(\tau)-u(0)\ .$

If we move to norms in (5.1), we obtain

$\displaystyle\left\|z_{k+1}\right\|$ $\displaystyle\leq$
$\displaystyle\tau\left\|B^{k-1}\right\|\left\|BU_{k}(B)\right\|\left\|\frac{\Delta
z_{0}}{\tau}\right\|$
$\displaystyle+\left\|B^{k}\right\|\left\|U_{k}(B)-BU_{k-1}(B)\right\|\left\|z_{0}\right\|$
$\displaystyle+\sum_{i=1}^{k}\left\|B^{k-i}\right\|\left\|U_{k-i}(B)\left(I-B^{2}\right)^{1/2}\right\|\left\|\left(I-B^{2}\right)^{-1/2}r_{i}\right\|\
.$

From here, taking into account estimates (4.14), (4.15) and (4.16)
($\left\|B\right\|\leq 1$), we have

$\left\|z_{k+1}\right\|\leq\nu^{-1/2}\left\|\frac{\Delta
z_{0}}{\tau}\right\|+\left\|z_{0}\right\|+\sum_{i=1}^{k}\left\|\left(I-B^{2}\right)^{-1/2}r_{i}\right\|\
.$ (5.2)

Obviously we have

$\displaystyle\left\|\left(I-B^{2}\right)^{-1/2}r_{i}\right\|$
$\displaystyle=$ $\displaystyle\left\|\left(I-S\right)^{-1/2}r_{i}\right\|$
(5.3) $\displaystyle\leq$
$\displaystyle\lambda_{0,i}+\lambda_{1,i}+\tau^{2}\lambda_{2,i}+\lambda_{3,i}+\lambda_{4,i}\
,$

where

$\displaystyle\lambda_{s,i}$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}r_{s,i}\right\|\ ,\quad s=0,\ 1,\
4,$ $\displaystyle\lambda_{s,i}$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}Lr_{s,i}\right\|\ ,\quad s=2,\ 3\
.$

Using (4.5), we get:

$\displaystyle\lambda_{0,i}$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}\left(S-L\right)u(t_{i})\right\|$
(5.4) $\displaystyle\leq$
$\displaystyle\tau^{2}c_{1}\left\|ALu(t_{i})\right\|\leq\tau^{2}c_{1}\left\|Au(t_{i})\right\|,$

$\displaystyle\lambda_{1,i}$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}\left(S-L\right)\left[u(t_{i})-u(t_{i-1})\right]\right\|$
(5.5) $\displaystyle\leq$
$\displaystyle\tau^{2}c_{1}\left\|AL\left[u(t_{i})-u(t_{i-1})\right]\right\|$
$\displaystyle\leq$ $\displaystyle\tau c_{1}\left\|\tau
A^{1/2}L\right\|\left\|A^{1/2}\left[u(t_{i})-u(t_{i-1})\right]\right\|$
$\displaystyle\leq$ $\displaystyle\tau
c_{1}\left\|A^{1/2}\left[u(t_{i})-u(t_{i-1})\right]\right\|.$

Using (4.6) and (4.7), respectively we get:

$\displaystyle\tau^{2}\lambda_{2,i}$ $\displaystyle=$
$\displaystyle\tau^{2}\left\|\left(I-S\right)^{-1/2}LA\left[u(t_{i+1})-u(t_{i})\right]\right\|$
(5.6) $\displaystyle\leq$ $\displaystyle\tau
c_{2}\left\|A^{1/2}\left[u(t_{i+1})-u(t_{i})\right]\right\|,$

$\lambda_{3,i}=\left\|\left(I-S\right)^{-1/2}Lr_{3,i}\right\|\leq\tau^{-1}c_{2}\left\|A^{-1/2}r_{3,i}\right\|$

$\displaystyle\leq$
$\displaystyle\tau^{-1}c_{2}\left\|\int_{t_{i}}^{t_{i+1}}(t_{i+1}-t)A^{-1/2}\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right]dt\right\|$
$\displaystyle+\tau^{-1}c_{2}\left\|\int_{t_{i-1}}^{t_{i}}(t-t_{i-1})A^{-1/2}\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right]dt\right\|$

$\leq
c_{2}\int_{t_{i-1}}^{t_{i+1}}\left\|A^{-1/2}\left[u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right]\right\|dt\
.$

From here, taking into account equation (2.1) , we have

$\displaystyle\lambda_{3,i}$ $\displaystyle\leq$ $\displaystyle
c_{2}\int_{t_{i-1}}^{t_{i+1}}\left\|A^{1/2}\left[u\left(t\right)-u\left(t_{i}\right)\right]\right\|dt$
(5.7)
$\displaystyle+c_{2}\int_{t_{i-1}}^{t_{i+1}}\left\|A^{-1/2}\left[\widetilde{f}(t)-\widetilde{f}(t_{i})\right]\right\|dt\
.$

Let us estimate $\lambda_{4,i}$ . Let’s rewrite $r_{4,i}$ as

$\displaystyle r_{4,i}$ $\displaystyle=$
$\displaystyle\tau^{2}\left[\psi_{i}-L\widetilde{f}(t_{i})\right]=\tau^{2}S_{1}\left[f(t_{i})-M\left(v_{i}\right)\right]-\tau^{2}L\widetilde{f}(t_{i})$
(5.8) $\displaystyle=$
$\displaystyle\tau^{2}S_{1}\left[f(t_{i})-M(u(t_{i}))\right]+\tau^{2}S_{1}\left[M(u(t_{i}))-M\left(v_{i}\right)\right]-\tau^{2}L\widetilde{f}(t_{i})$
$\displaystyle=$
$\displaystyle\tau^{2}S_{1}\left[M(u(t_{i}))-M\left(v_{i}\right)\right]+\tau^{2}S_{1}\widetilde{f}(t_{i})-\tau^{2}L\widetilde{f}(t_{i})$
$\displaystyle=$
$\displaystyle\tau^{2}S_{1}\left[M(u(t_{i}))-M\left(v_{i}\right)\right]+\tau^{2}(S_{1}-I)\widetilde{f}(t_{i})+\tau^{2}(I-L)\widetilde{f}(t_{i})$
$\displaystyle=$
$\displaystyle\tau^{2}S_{1}\left[M(u(t_{i}))-M\left(v_{i}\right)\right]+\tau^{2}(S_{1}-I)\widetilde{f}(t_{i})+\tau^{4}AL\widetilde{f}(t_{i})\
.$

Hence we have

$\displaystyle\lambda_{4,i}$ $\displaystyle=$
$\displaystyle\left\|\left(I-S\right)^{-1/2}r_{4,i}\right\|$ (5.9)
$\displaystyle\leq$
$\displaystyle\tau^{2}\left\|\left(I-S\right)^{-1/2}S_{1}\left[M(u(t_{i}))-M\left(v_{i}\right)\right]\right\|$
$\displaystyle+\tau^{2}\left\|\left(I-S\right)^{-1/2}\left(I-S_{1}\right)\widetilde{f}(t_{i})\right\|$
$\displaystyle+\tau^{4}\left\|\left(I-S\right)^{-1/2}AL\widetilde{f}(t_{i})\right\|.$

If we use inequality (4.12), we get:

$\displaystyle\left\|\left(I-S\right)^{-1/2}S_{1}h\right\|$
$\displaystyle\leq$
$\displaystyle\eta_{1}^{-1/2}\left\|\left(I-S_{1}\right)^{-1/2}S_{1}h\right\|$
(5.10) $\displaystyle=$
$\displaystyle\eta_{1}^{-1/2}\left\|\left(\tau^{2}\eta_{1}^{-1}A_{1}S_{1}\right)^{-1/2}S_{1}h\right\|$
$\displaystyle=$
$\displaystyle\tau^{-1}\left\|A_{1}^{-1/2}S_{1}^{1/2}h\right\|\leq\left(\tau\sqrt{\nu}\right)^{-1}\left\|h\right\|,$

$\left\|\left(I-S\right)^{-1/2}\left(I-S_{1}\right)h\right\|\leq\eta_{1}^{-1/2}\left\|\left(I-S_{1}\right)^{1/2}h\right\|\leq\eta_{1}^{-1/2}\left\|h\right\|.$
(5.11)

Taking into account that$\left\|\tau A^{1/2}L^{1/2}\right\|\leq 1$, then from
(4.4) we get

$\displaystyle\tau^{2}\left\|\left(I-S\right)^{-1/2}ALf\right\|$ (5.12)
$\displaystyle\leq$
$\displaystyle\tau\left(m\left\|A^{1/2}Lf\right\|+c_{0}\left\|A^{1/2}L^{1/2}f\right\|\right)\leq(m+c_{0})\left\|f\right\|.$

From (5.9), taking into account estimates (5.10), (5.11) and (5.12) , we get

$\lambda_{4,i}\leq\tau\nu^{-1/2}\left\|M(u(t_{i}))-M\left(v_{i}\right)\right\|+\tau^{2}c_{3}\left\|\widetilde{f}(t_{i})\right\|.$
(5.13)

where $c_{3}=\eta_{1}^{-1/2}+m+c_{0}$ .

Since nonlinear operator $M$ satisfies Lipschietz condition, therefore from
(5.13) we have

$\lambda_{4,i}\leq\tau\nu^{-1/2}a\left\|u(t_{i})-v_{i}\right\|+\tau^{2}c_{3}\left\|\widetilde{f}(t_{i})\right\|,$
(5.14)

where $a$ is a Lipchietz constant.

If in (5.3) we substitute (5.4), (5.5), (5.6), (5.7) and (5.14), we get

$\displaystyle\left\|\left(I-B^{2}\right)^{-1/2}r_{i}\right\|$ (5.15)
$\displaystyle\leq$
$\displaystyle\tau\nu^{-1/2}a\left\|z_{i}\right\|+\tau^{2}c_{1}\left\|Au(t_{i})\right\|+\tau
c_{1}J_{i}(t_{i-1},\ A^{1/2}u)$ $\displaystyle+\tau c_{2}J_{i}(t_{i+1},\
A^{1/2}u)+\tau^{2}c_{3}\left\|\widetilde{f}(t_{i})\right\|$
$\displaystyle+c_{2}\int_{t_{i-1}}^{t_{i+1}}\left[J_{i}\left(t,\
A^{1/2}u\right)+J_{i}\left(t,\ A^{-1/2}\widetilde{f}\right)\right]dt\ ,$

where

$J_{i}\left(t,u\right)=\left\|u(t_{i})-u(t)\right\|.$

From (5.2), taking into account (5.15),we get

$\delta_{k+1}\leq c\tau\delta_{k}+\lambda_{k}\ ,$ (5.16)

where $\delta_{k}=\left\|z_{k}\right\|$ , $c=\nu^{-1/2}a$ ,

$\lambda_{k}=\nu^{-1/2}\left\|\frac{\Delta
z_{0}}{\tau}\right\|+\left\|z_{0}\right\|+\Theta_{k}(\tau)\ .$

From (5.16), according to discrete analog of Gronwell’s lemma, we have

$\delta_{k+1}\leq\exp\left(ct_{k-1}\right)\left(c\tau\delta_{1}+\lambda_{k}\right)\
.$

Obviously from here, taking into account inequality

$\delta_{1}=\left\|z_{1}\right\|\leq\tau\left\|\frac{\Delta
z_{0}}{\tau}\right\|+\left\|z_{0}\right\|$

We get the estimate (3.10) .

##  6  Error estimate for the difference analog of the first order derivative
of approximate solution

In this section, on the basis of the results obtained in the previous section,
we obtain the a priori estimates for error of the difference analog of the
first order derivative of the decomposition (2.4) scheme solution.

Theorem 6.1. If the problem (2.1), (2.2) has a solution and $\varphi_{0}\in
D(A)$, then the estimate is valid:

$\left\|\frac{\Delta z_{k}}{\tau}\right\|\leq\left\|\frac{\Delta
z_{0}}{\tau}\right\|+\left(\tau
t_{k}\right)^{-1/2}\left\|z_{0}\right\|+\widetilde{\Theta}_{k}(\tau)\ ,$ (6.1)

where $\Delta z_{k}=z_{k+1}-z_{k}$.

$\displaystyle\widetilde{\Theta}_{k}(\tau)$ $\displaystyle=$
$\displaystyle\tau\sum_{i=1}^{k}\left[c_{5}J_{i}\left(t_{i-1},u^{\prime\prime}\right)+J_{i}\left(t_{i+1},u^{\prime\prime}\right)+ac_{6}J_{i}\left(t_{i-1},u\right)\right]$
$\displaystyle+\tau\sum_{i=1}^{k}\left[c_{7}J_{i}\left(t_{i-1},f\right)+J_{i}\left(t_{i-1},f\right)\right]$
$\displaystyle+\sum_{i=1}^{k}\left(\int_{t_{i-1}}^{t_{i+1}}J_{i}\left(t,u^{\prime\prime}\right)dt+\tau
a\left\|z_{i}\right\|\right)$
$\displaystyle+\tau\left[c_{1}\left\|A\varphi_{0}\right\|+c_{6}\left(\left\|f(0)\right\|+\left\|M(\varphi_{0})\right\|\right)\right]\
,$

and where $c_{5}=c_{4}+c_{1}$ , $\ c_{6}=\eta_{1}^{-1/2}+m+c_{0}$ ,
$c_{7}=c_{4}+c_{1}+c_{6}$ ,

$c_{4}=\sum_{j=1}^{m}\eta_{j}^{-1}\left(\eta_{j}^{-1}a_{j}+1\right)\ ,\quad
J_{i}\left(t,u\right)=\left\|u(t_{i})-u(t)\right\|\ .$

Proof: Due to formula (3.9) we have

$z_{k+1}-z_{k}=\left(R_{k}-R_{k-1}\right)z_{1}-\left(R_{k-1}-R_{k-2}\right)B^{2}z_{0}+\Phi_{k}\
,$ (6.2)

where $R_{k}=B^{k}U_{k}(B)$ ,

$\Phi_{k}=\sum_{i=1}^{k}R_{k-i}r_{i}-\sum_{i=1}^{k-1}R_{k-1-i}r_{i}\ .$

Let us rewrite (6.2) as

$\displaystyle\frac{\Delta z_{k}}{\tau}$ $\displaystyle=$
$\displaystyle\left(R_{k}-R_{k-1}\right)\frac{\Delta z_{0}}{\tau}$ (6.3)
$\displaystyle+\tau^{-1}\left[R_{k}+B^{2}R_{k-2}-\left(I+B^{2}\right)R_{k-1}\right]z_{0}+\tau^{-1}\Phi_{k}\
,$

where $\Delta z_{k}=z_{k+1}-z_{k}$ .

By the simple transformation we get

$\displaystyle R_{k}+B^{2}R_{k-2}-\left(I+B^{2}\right)R_{k-1}$
$\displaystyle=$ $\displaystyle
B^{k-1}\left[B\left(U_{k}+U_{k-2}\right)-\left(I+B^{2}\right)U_{k-1}\right]$
$\displaystyle=$ $\displaystyle-B^{k-1}\left(I-B^{2}\right)U_{k-1}\ .\ $

From here, taking into account (4.15), we have

$\displaystyle\left\|R_{k}+B^{2}R_{k-2}-\left(I+B^{2}\right)R_{k-1}\right\|$
(6.4) $\displaystyle\leq$
$\displaystyle\left\|B^{k-1}\left(I-B^{2}\right)^{1/2}\right\|\left\|U_{k-1}\left(I-B^{2}\right)^{1/2}\right\|\leq\left\|B^{k-1}\left(I-B^{2}\right)^{1/2}\right\|$
$\displaystyle\leq$ $\displaystyle\max_{0\leq x\leq
1}\left[x^{k-1}\left(1-x^{2}\right)^{1/2}\right]\leq\frac{1}{\sqrt{k}}\ .$

Obviously, according to (4.17), we have

$\displaystyle\left\|R_{k}-R_{k-1}\right\|$ $\displaystyle=$
$\displaystyle\left\|B^{k-1}\left(BU_{k}(B)-U_{k-1}(B)\right)\right\|$ (6.5)
$\displaystyle\leq$
$\displaystyle\left\|B^{k-1}\right\|\left\|BU_{k}(B)-U_{k-1}(B)\right\|\leq
1.$

We can give to $\Phi_{k}$ the following form

$\displaystyle\Phi_{k}$ $\displaystyle=$
$\displaystyle\sum_{i=1}^{k}R_{k-i}\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)-\sum_{i=1}^{k-1}R_{k-1-i}\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)$
$\displaystyle+\sum_{i=1}^{k}R_{k-i}\left(r_{0,i}+\tau^{2}\zeta_{i}\right)-\sum_{i=1}^{k-1}R_{k-1-i}\left(r_{0,i}+\tau^{2}\zeta_{i}\right)$

$\displaystyle=$
$\displaystyle\sum_{i=1}^{k}\left(R_{k-i}-R_{k-1-i}\right)\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)$
$\displaystyle+\sum_{i=1}^{k}R_{k-i}\left(r_{0,i}+\tau^{2}\zeta_{i}\right)-\sum_{i=2}^{k}R_{k-i}\left(r_{0,i-1}+\tau^{2}\zeta_{i-1}\right)$

$\displaystyle=$
$\displaystyle\sum_{i=1}^{k}\left(R_{k-i}-R_{k-1-i}\right)\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)$
(6.6)
$\displaystyle+\sum_{i=1}^{k}R_{k-i}\left[\left(r_{0,i}-r_{0,i-1}\right)+\tau^{2}\left(\zeta_{i}-\zeta_{i-1}\right)\right]$
$\displaystyle+R_{k-1}\left(r_{0,0}+\tau^{2}\zeta_{0}\right)\ ,$

where $R_{-1}=0$ ,

$\zeta_{i}=(S_{1}-I)\widetilde{f}(t_{i})+\tau^{2}AL\widetilde{f}(t_{i})\ .$

Taking into account (6.5), we have

$\displaystyle\left\|\left(R_{k-i}-R_{k-1-i}\right)\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)\right\|$
(6.7) $\displaystyle\leq$
$\displaystyle\left(\left\|r_{1,i}\right\|+\tau^{2}\left\|Lr_{2,i}\right\|+\left\|Lr_{3,i}\right\|+\left\|r_{4,i}-\tau^{2}\zeta_{i}\right\|\right)\
.$

By formula (4.3), we get

$\displaystyle\left\|r_{1,i}\right\|$ $\displaystyle=$
$\displaystyle\left\|\left(S-L\right)\left[u(t_{i})-u(t_{i-1})\right]\right\|$
(6.8) $\displaystyle=$
$\displaystyle\tau^{2}\left\|\sum_{j=1}^{m}\eta_{j}^{-1}\left(I-S_{j}\right)\left(\eta_{j}^{-1}A_{j}A^{-1}-I\right)AL\left[u(t_{i})-u(t_{i-1})\right]\right\|$
$\displaystyle\leq$
$\displaystyle\tau^{2}c_{4}\left\|A\left[u(t_{i})-u(t_{i-1})\right]\right\|,$

Obviously, for $Lr_{2,i}$ we have

$\left\|Lr_{2,i}\right\|\leq\left\|A\left[u(t_{i+1})-u(t_{i})\right]\right\|.$
(6.9)

From inequalities (6.8) and (6.9) , with account of equation (2.1), we get

$\displaystyle\left\|r_{1,i}\right\|+\tau^{2}\left\|Lr_{2,i}\right\|$ (6.10)
$\displaystyle\leq$
$\displaystyle\tau^{2}c_{4}\left(\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+\left\|f(t_{i})-f(t_{i-1})\right\|\right)$
$\displaystyle+\tau^{2}\left(\left\|u^{\prime\prime}\left(t_{i+1}\right)-u^{\prime\prime}\left(t_{i}\right)\right\|+\left\|f(t_{i+1})-f(t_{i})\right\|\right)\
.$

It is also obvious, that from (5.8) it follows

$\left\|r_{4,i}-\tau^{2}\zeta_{i}\right\|\leq\tau^{2}\left\|S_{1}\left[M\left(u(t_{i})\right)-M\left(v_{i}\right)\right]\right\|\leq
a\tau^{2}\left\|u(t_{i})-v_{i}\right\|.$ (6.11)

From the representation of $r_{3,i}$ it follows that

$\displaystyle\left\|Lr_{3,i}\right\|$ $\displaystyle\leq$
$\displaystyle\int_{t_{i}}^{t_{i+1}}(t_{i+1}-t)\left\|u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right\|dt$
(6.12)
$\displaystyle+\int_{t_{i-1}}^{t_{i}}(t-t_{i-1})\left\|u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right\|dt$
$\displaystyle\leq$
$\displaystyle\tau\int_{t_{i-1}}^{t_{i+1}}\left\|u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right\|dt\
.$

If we substitute inequalities (6.10), (6.11) and (6.12) in (6.7), we get

$\displaystyle\left\|\left(R_{k-i}-R_{k-1-i}\right)\left(r_{i}-r_{0,i}-\tau^{2}\zeta_{i}\right)\right\|$
(6.13) $\displaystyle\leq$
$\displaystyle\tau^{2}c_{4}\left(\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+\left\|f(t_{i})-f(t_{i-1})\right\|\right)$
$\displaystyle+\tau^{2}\left(\left\|u^{\prime\prime}\left(t_{i+1}\right)-u^{\prime\prime}\left(t_{i}\right)\right\|+\left\|f(t_{i+1})-f(t_{i})\right\|\right)$
$\displaystyle+\tau\int_{t_{i-1}}^{t_{i+1}}\left\|u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right\|dt+ac_{4}\tau^{2}\left\|u(t_{i})-v_{i}\right\|\
.$

According to estimate (4.15), we have

$\displaystyle\left\|R_{k-i}\left(r_{0,i}-r_{0,i-1}\right)\right\|$ (6.14)
$\displaystyle=$
$\displaystyle\left\|R_{k-i}\left(I-B^{2}\right)^{1/2}\left(I-B^{2}\right)^{-1/2}\left(r_{0,i}-r_{0,i-1}\right)\right\|$
$\displaystyle\leq$
$\displaystyle\left\|U_{k-i}\left(I-B^{2}\right)^{1/2}\right\|\left\|\left(I-S\right)^{-1/2}\left(r_{0,i}-r_{0,i-1}\right)\right\|$
$\displaystyle\leq$
$\displaystyle\left\|\left(I-S\right)^{-1/2}\left(r_{0,i}-r_{0,i-1}\right)\right\|.$

Similarly we get

$\displaystyle\left\|R_{k-i}\left(\zeta_{i}-\zeta_{i-1}\right)\right\|$ (6.15)
$\displaystyle\leq$
$\displaystyle\left\|\left(I-S\right)^{-1/2}\left(\zeta_{i}-\zeta_{i-1}\right)\right\|$
$\displaystyle\leq$
$\displaystyle\left\|\left(I-S\right)^{-1/2}(S_{1}-I)\left(\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right)\right\|$
$\displaystyle+\tau^{2}\left\|\left(I-S\right)^{-1/2}AL\left(\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right)\right\|.$

If we substitute in (6.14) representation of $r_{0,i}$ ($r_{0,i}=$
$\left(S-L\right)u(t_{i})$ ) and take into account estimate (4.5), we get

$\left\|R_{k-i}\left(r_{0,i}-r_{0,i-1}\right)\right\|\leq\tau^{2}c_{1}\left\|A\left[u(t_{i})-u(t_{i-1})\right]\right\|\
.$ (6.16)

Similarly we have

$\left\|R_{k-1}r_{0,0}\right\|\leq\tau^{2}c_{1}\left\|Au(0)\right\|\
=\tau^{2}c_{1}\left\|A\varphi_{0}\right\|\ ,\quad\varphi_{0}\in D(A).$ (6.17)

From (6.16), taking into account equation (2.1), we have

$\displaystyle\left\|R_{k-i}\left(r_{0,i}-r_{0,i-1}\right)\right\|$ (6.18)
$\displaystyle\leq$
$\displaystyle\tau^{2}c_{1}\left(\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+\left\|f(t_{i})-f(t_{i-1})\right\|\right)\
.$

If in (6.15) we take into account estimates (4.13), (4.4) and
$\tau\left\|A^{1/2}L^{1/2}h\right\|\leq h$, we get

$\displaystyle\left\|R_{k-i}\left(\zeta_{i}-\zeta_{i-1}\right)\right\|$ (6.19)
$\displaystyle\leq$
$\displaystyle\eta_{1}^{-1/2}\left\|\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right\|+\tau
m\left\|A^{1/2}L\left(\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right)\right\|$
$\displaystyle+\tau
c_{0}\left\|A^{1/2}L^{1/2}\left(\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right)\right\|\leq
c_{6}\left\|\widetilde{f}(t_{i})-\widetilde{f}(t_{i-1})\right\|$
$\displaystyle\leq$ $\displaystyle
c_{6}\left(\left\|f(t_{i})-f(t_{i-1})\right\|+a\left\|u(t_{i})-u(t_{i-1})\right\|\right)\
,$

where $c_{6}=\eta_{1}^{-1/2}+m+c_{0}$ .

Analogously to (6.19), the inequality is true

$\left\|R_{k-1}\zeta_{0}\right\|\leq
c_{6}\left\|\widetilde{f}(t_{0})\right\|\leq
c_{6}\left(\left\|f(0)\right\|+\left\|M(\varphi_{0})\right\|\right)\ .$ (6.20)

Obviously, from inequalities (6.18) and (6.19) it follows

$\displaystyle\left\|R_{k-i}\left[\left(r_{0,i}-r_{0,i-1}\right)+\tau^{2}\left(\zeta_{i}-\zeta_{i-1}\right)\right]\right\|$
(6.21) $\displaystyle\leq$
$\displaystyle\tau^{2}\left(c_{1}\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+ac_{6}\left\|u(t_{i})-u(t_{i-1})\right\|\right)$
$\displaystyle+\tau^{2}(c_{1}+c_{6})\left\|f(t_{i})-f(t_{i-1})\right\|\ .$

If in (6.6) we move to norms and take into account inequalities (6.13),
(6.21), (6.17) and (6.20), we get

$\displaystyle\left\|\Phi_{k}\right\|$ $\displaystyle\leq$
$\displaystyle\tau^{2}c_{4}\sum_{i=1}^{k}\left(\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+\left\|f(t_{i})-f(t_{i-1})\right\|\right)$
$\displaystyle+\tau^{2}\sum_{i=1}^{k}\left(\left\|u^{\prime\prime}\left(t_{i+1}\right)-u^{\prime\prime}\left(t_{i}\right)\right\|+\left\|f(t_{i+1})-f(t_{i})\right\|\right)$
$\displaystyle+\sum_{i=1}^{k}\left(\tau\int_{t_{i-1}}^{t_{i+1}}\left\|u^{\prime\prime}\left(t\right)-u^{\prime\prime}\left(t_{i}\right)\right\|dt+a\tau^{2}\left\|z_{i}\right\|\right)$

$\displaystyle+\tau^{2}\sum_{i=1}^{k}\left(c_{1}\left\|u^{\prime\prime}\left(t_{i}\right)-u^{\prime\prime}\left(t_{i-1}\right)\right\|+ac_{6}\left\|u(t_{i})-u(t_{i-1})\right\|\right)$
$\displaystyle+\tau^{2}(c_{1}+c_{6})\sum_{i=1}^{k}\left\|f(t_{i})-f(t_{i-1})\right\|$
$\displaystyle+\tau^{2}c_{1}\left\|A\varphi_{0}\right\|+\tau^{2}c_{6}\left(\left\|f(0)\right\|+\left\|M(\varphi_{0})\right\|\right)\
.$

Or the same

$\displaystyle\left\|\Phi_{k}\right\|$ $\displaystyle\leq$
$\displaystyle\tau^{2}\sum_{i=1}^{k}\left[c_{5}J_{i}\left(t_{i-1},u^{\prime\prime}\right)+J_{i}\left(t_{i+1},u^{\prime\prime}\right)+ac_{6}J_{i}\left(t_{i-1},u\right)\right]$
(6.22)
$\displaystyle+\tau^{2}\sum_{i=1}^{k}\left[c_{7}J_{i}\left(t_{i-1},f\right)+J_{i}\left(t_{i-1},f\right)\right]$
$\displaystyle+\tau\sum_{i=1}^{k}\left(\int_{t_{i-1}}^{t_{i+1}}J_{i}\left(t,u^{\prime\prime}\right)dt+\tau
a\left\|z_{i}\right\|\right)$
$\displaystyle+\tau^{2}\left[c_{1}\left\|A\varphi_{0}\right\|+c_{6}\left(\left\|f(0)\right\|+\left\|M(\varphi_{0})\right\|\right)\right]\
,$

where $c_{7}=c_{4}+c_{1}+c_{6}$ .

From (6.3), taking into account (6.4), (6.5) and (6.22), it follows (6.1).

Result 2. If functions $f(t)$ and $u^{\prime\prime}\left(t\right)$ on the
interval $[0,\ T]$ satisfies Holder condition with the index $\lambda$
($0<\lambda\leq 1$) , then

$\left\|u^{\prime}\left(t_{k}\right)-\frac{v_{k+1}-v_{k}}{\tau}\right\|\leq
c\tau^{\lambda},\quad c=const>0.$ (6.23)

With account of (6.1), estimate (6.23) follows from the following equality

$\displaystyle u^{\prime}\left(t_{k}\right)-\frac{v_{k+1}-v_{k}}{\tau}$
$\displaystyle=$
$\displaystyle\tau^{-1}\int_{t_{k}}^{t_{k+1}}\left[u^{\prime}\left(t_{k}\right)-u^{\prime}\left(t\right)\right]dt-\frac{z_{k+1}-z_{k}}{\tau}$

##  7 Numerical Results

The calculations were performed for the following problem.

$\displaystyle\frac{\partial^{2}u}{\partial t^{2}}-\Delta u=sin(u)+f(x,y,t),\
\ (x,y,t)\in\Omega\times(0,T),$ $\displaystyle u(x,y,0)=\varphi_{0}(x,y),\ \ \
u^{\prime}(x,y,0)=\varphi_{1}(x,y),$ $\displaystyle\Omega=(0,1)\times(0,1),$
$\displaystyle u_{\partial\Omega\times[0,T]}=0.$

For this problem two different test case were calculated.

Test 1.

$\displaystyle
u\left(x,y,t\right)=t^{7/2}\sin\left(2{\pi}{x}\right)\cos\left(2{\pi}{x}\right),$
$\displaystyle(x,y,t)\in[0,1]\times[0,1]\times[0,1],$
$\displaystyle\tau=h_{x}=h_{y}=0.05.$

Fig 1 shows exact solution, approximate solution and error. This test example
is interesting as fourth derivative of the solution is discontinuous. As we
see from Fig. 1 the numerical algorithm was able to resolve this problem with
good accuracy. Maximum error equals to $0.0046$.

Figure 1: Exact and approximate solutions of Test 1.

Test 2.

$\displaystyle
u\left(x,y,t\right)=t^{7/2}\sin\left(10{\pi}{x}\right)\cos\left(10{\pi}{x}\right),$
$\displaystyle(x,y,t)\in[0,1]\times[0,1]\times[0,1],$
$\displaystyle\tau=h_{x}=h_{y}=0.01.$

Similarly, Fig 2 shows exact solution, approximate solution and error. This
test example is interesting as fourth derivative of the solution is
discontinuous and also has high oscillations, in the interval
$\left(0,1\right)$ it changes the sign ten times. As we see from Fig. 2 the
numerical algorithm was able to detect these oscillations and recovered
original function with good accuracy. Maximum error for this case equals to
$0.0074$.

Figure 2: Exact and approximate solutions of Test 2.

On Fig. 3, there is given a dependence of the logarithm of the relative error
of the approximated solution on the logarithm of number of division by special
variable. Aim of this figure is to find the convergence rate of the method by
means of the numerical experiment. If the method is second order of accuracy,
then, the curve of the function (logarithm of the solution error) should
approach to the line, the tangent of which equals two. On Fig. 3 it is clearly
seen that, the curve approaches the line, the tangent of which equals to two,
and this verifies the theoretical result proved in the article.

Figure 3: Dependence of logarithm of relative error on logarithm of number of
divisions for Test 2.

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## Authors’ addresses:

Nana Dikhaminjia

School of Business, Technology and Education, Ilia State University (ISU),
Kakutsa Cholokashvili Ave 3/5, Tbilisi 0162, Georgia.
E-mail<EMAIL_ADDRESS>
Jemal Rogava

Faculty of Exact and Natural Sciences, Ivane Javakhishvili Tbilisi State
University (TSU), Ilia Vekua Institute of Applied Mathematics (VIAM), 2
University St., Tbilisi 0186, Georgia.
E-mail<EMAIL_ADDRESS>
Mikheil Tsiklauri

Missouri University of Science and Technology, Electromagnetic Compatibility
Laboratory, 4000 Enterprise Drive, Rolla, MO 65409, USA.
E-mail<EMAIL_ADDRESS><EMAIL_ADDRESS>