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ADV MATH
SCI JOURNAL
Advances in Mathematics: Scientic Journal
4
(2015), no.2, 139159
ISSN 1857-8365 UDC: 519.62
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR A
SEMILINEAR REACTION-DIFFUSION PROBLEM
SAMIR KARASULJI
1
, ENES DUVNJAKOVI AND HELENA ZARIN
Abstract.
In this work we consider the singularly perturbed one-dimensional semi-
linear reaction-diusion problem
"
2
y
00
(
x
) =
f
(
x; y
)
; x
2
(0
;
1)
; y
(0) = 0
; y
(1) = 0
;
where
f
is a nonlinear function. Here the second-order derivative is multiplied by a
small positive parameter and consequently, the solution of the problem has boundary
layers. A new dierence scheme is constructed on a modied Shishkin mesh with
O
(
N
)
points for this problem. We prove existence and uniqueness of a discrete so-
lution on such a mesh and show that it is accurate to the order of
N
2
ln
2
N
in
the discrete maximum norm. We present numerical results that verify this rate of
convergence.
1.
Introduction
We consider the semilinear singularly perturbed problem
(1.1)
"
2
y
00
(
x
) =
f
(
x; y
)
on
(0
;
1)
;
(1.2)
y
(0) = 0
; y
(1) = 0
;
where
0
< " <
1
:
We assume that the nonlinear function
f
is continuously dierentiable,
i.e. for
k
2
; f
2
C
k
([0
;
1]
R
)
;
and that it has a strictly positive derivative with respect
to
y
(1.3)
@f
@y
=
f
y
m >
0
on
[0
;
1]
R
(
m
=
const
)
:
A solution of
(1
:
1)
(1
:
2)
usually exhibits sharp boundary layers at the endpoints of
(0
;
1)
;
when the parameter
"
is near zero. When classical numerical methods are applied
to
(1
:
1)
(1
:
2)
, one does not obtain
"
uniform results on the entire interval
(0
;
1)
;
because
of which we shall use nonstandard discretization of
(1
:
1)
(1
:
2)
.
1
corresponding author
2010
Mathematics Subject Classication.
65L10, 65L11, 65L50.
Key words and phrases.
Singular perturbation, Nonlinear, Boundary layer, Shishkin mesh, Layer-
adapted mesh, Uniform convergence.
139
140 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
Many authors have considered the problem
(1
:
1)
(1
:
2)
under various hypotheses on
f;
see e.g. Herceg and Miloradovi¢ [7] , Uzelac and Surla [14] , Vulanovi¢ [15] , [16] , Sun
and Stynes [13] etc.
Uniformly convergent methods with respect to
"
for the problem
(1
:
1)
(1
:
2)
under
condition
(1
:
3)
have also been examined. Vulanovi¢ [15] applies a central dierence scheme
to the problem
(1
:
1)
(1
:
3)
and proves second-order uniform convergence on a specially
graded mesh of Bakhvalov type. D'Annunzio [3] uses a simple central dierence scheme on
a special locally quasi-equidistant mesh to solve a more general problem than
(1
:
1)
(1
:
3)
:
The last result was signicantly improved by Sun and Stynes in [13] using the mesh of
Shishkin type.
Our paper is devoted to the construction of approximations on a Shishkin-type mesh.
Our aim is to construct a dierence scheme with coecients which behave similar to the
solution of the starting problem. It is well-known that in modelling of the boundary
layer of the exact solution of the problem (1.1)
(1.3), it is used suitable exponential
functions with a perturbation parameter
"
. We intend to get a scheme for calculation
of the numerical solution with coecients which acting on the same or similar way as
mentioned exponential functions. A motivation for constructing this kind of dierence
scheme is getting as good numerical results as possible.
Discretization in this paper is based on the paper Boglaev [2]. Unlike our previous
work [5] where we only constructed a dierence scheme for the boundary value problem
(1
:
1)
(1
:
3)
and performed a numerical test, in this work we also prove the existence
and uniqueness of the numerical solution. Further, we show
"
uniform convergence of
the numerical solution to the exact solution on a suitable layer-adapted mesh. We also
verify the rate and order of convergence on a numerical example.
Remark 1.1.
Throughout the paper we denote by
C;
sometimes subscripted, a
generic positive constant that may take dierent values in dierent formulas, but is
always independent with respect to
N
and
"
.
2.
Construction of the nonlinear difference scheme
In this section we construct a dierence scheme which generates a system of nonlinear
equations and solving this system produces the values of the numerical solution at the
mesh points. The scheme will be constructed based on the results in solving the linear
boundary value problem and Green's function for a suitable dierential operator. The
method was rst introduced Boglaev [2].
Let us now consider the dierential equation
(1
:
1)
in an equivalent form
L
"
y
(
x
) :=
"
2
y
00
(
x
)
y
(
x
) =
(
x; y
(
x
))
on
[0
;
1]
;
where
(
x; y
) =
f
(
x; y
)
y;
and
m
is a chosen constant.
On an arbitrary grid
0 =
x
0
< x
1
< x
2
< ::: < x
N
= 1
;
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 141
consider the following boundary value problems
L
"
u
i
(
x
) := 0
on
(
x
i
; x
i
+1
)
;
u
i
(
x
i
) = 1
; u
i
(
x
i
+1
) = 0
;
(
i
= 0
;
1
; :::; N
1)
;
and
L
"
u
i
(
x
) := 0
on
(
x
i
; x
i
+1
)
;
u
i
(
x
i
) = 0
; u
i
(
x
i
+1
) = 1
;
(
i
= 0
;
1
; :::; N
1)
:
(2.1)
We denote the solutions of problems
(2
:
1)
by
u
I
i
(
x
)
; u
II
i
(
x
)
;
(
i
= 0
;
1
;
2
; :::; N
1)
;
respectively. Functions
u
I
i
and
u
II
i
are known from [11] , i.e.
(2.2)
u
I
i
(
x
) = sinh (
(
x
i
+1
x
))
sinh (
h
i
)
; u
II
i
(
x
) = sinh (
(
x
x
i
))
sinh (
h
i
)
; x
2
[
x
i
; x
i
+1
]
;
(
i
= 0
;
1
;
2
; :::; N
1)
;
where
=
p
"; h
i
=
x
i
+1
x
i
:
Consider a new boundary value problem
(2.3)
L
"
y
i
(
x
) =
(
x; y
i
(
x
))
on
(
x
i
; x
i
+1
)
;
y
i
(
x
i
) =
y
(
x
i
)
; y
i
(
x
i
+1
) =
y
(
x
i
+1
)
;
(
i
= 0
;
1
;
2
; :::; N
1)
:
It is clear that
y
i
(
x
)
y
(
x
)
on
[
x
i
; x
i
+1
]
;
(
i
= 0
;
1
;
2
; :::; N
1)
:
The solution of
(2
:
3)
is given by
y
i
(
x
) =
C
1
u
I
i
(
x
) +
C
2
u
II
i
(
x
) +
Z
x
i
+1
x
i
G
i
(
x; s
)
(
s; y
(
s
)) d
s; x
2
[
x
i
; x
i
+1
]
;
where
G
i
(
x; s
)
is the Green's function associated with the operator
L
"
on the interval
[
x
i
; x
i
+1
]
.
The function
G
i
(
x; s
)
in this case has the following form
G
i
(
x; s
) = 1
"
2
w
i
(
s
)
u
II
i
(
x
)
u
I
i
(
s
)
; x
i
x
s
x
i
+1
;
u
I
i
(
x
)
u
II
i
(
s
)
; x
i
s
x
x
i
+1
;
where
w
i
(
s
) =
u
II
i
(
s
)
u
I
i
0
(
s
)
u
I
i
(
s
)
u
II
i
0
(
s
)
=
sinh(
(
s
x
i
))
sinh(
h
i
)
sinh(
(
x
i
+1
s
))
sinh(
h
i
)
0
sinh(
(
x
i
+1
s
))
sinh(
h
i
)
sinh(
(
s
x
i
))
sinh(
h
i
)
0
=
sinh(
h
i
)
6
= 0
; s
2
[
x
i
; x
i
+1
]
;
because the solutions
u
I
i
and
u
II
i
are linearly independent.
From the boundary conditions in
(2
:
3)
;
we have that
C
1
=
y
(
x
i
) =:
y
i
; C
2
=
y
(
x
i
+1
) =:
y
i
+1
;
(
i
= 0
;
1
;
2
; : : : ; N
1)
:
Hence, the solution
y
i
(
x
)
of
(2
:
3)
on the interval
[
x
i
; x
i
+1
]
has the following form
(2.4)
y
i
(
x
) =
y
i
u
I
i
(
x
) +
y
i
+1
u
II
i
(
x
) +
Z
x
i
+1
x
i
G
i
(
x; s
)
(
s; y
(
s
)) d
s:
The boundary value problem
L
"
y
(
x
) :=
(
x; y
)
on
(0
;
1)
;
y
(0) =
y
(1) = 0
;
142 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
has a unique continuously dierentiable solution
y
2
C
k
+2
(0
;
1)
. Since
y
i
(
x
)
y
(
x
)
on
[
x
i
; x
i
+1
]
;
(
i
= 0
;
1
;
2
; :::; N
1)
we have that
(2.5)
y
0
i
(
x
i
) =
y
0
i
1
(
x
i
)
;
(
i
= 1
;
2
; :::; N
1)
:
Now, dierentiating
(2
:
4)
and using
(2
:
5)
;
we get
y
i
1
u
I
i
1
0
(
x
i
) +
y
i
h
u
II
i
1
0
(
x
i
)
u
I
i
0
(
x
i
)
i
+
y
i
+1
h
u
II
i
0
(
x
i
)
i
=
=
@
@x
"Z
x
i
+1
x
i
G
i
(
x; s
)
(
s; y
(
s
)) d
s
Z
x
i
x
i
1
G
i
1
(
x; s
)
(
s; y
(
s
)) d
s
#
x
=
x
i
:
(2.6)
Dene
a
i
:=
u
I
i
1
0
(
x
i
)
; c
i
:=
u
II
i
1
0
(
x
i
)
u
I
i
0
(
x
i
)
; b
i
:=
u
II
i
0
(
x
i
)
:
Using
(2
:
2)
we have that
a
i
=
sinh(
h
i
1
)
; b
i
=
sinh(
h
i
)
and
c
i
=
tanh(
h
i
1
)+
tanh(
h
i
)
:
Now
(2
:
6)
takes the following form
a
i
y
i
1
+
c
i
y
i
b
i
y
i
+1
=
Z
x
i
+1
x
i
@
@x
(
G
i
(
x; s
))
j
x
=
x
i
(
s; y
(
s
)) d
s
Z
x
i
x
i
1
@
@x
(
G
i
1
(
x; s
))
j
x
=
x
i
(
s; y
(
s
)) d
s:
(2.7)
After nding the derivatives on the right hand side
(2
:
7)
, we get that
(2.8)
a
i
y
i
1
c
i
y
i
+
b
i
y
i
+1
=1
"
2
"
x
i
R
x
i
1
u
II
i
1
(
s
)
(
s; y
(
s
)) d
s
+
x
i
+1
R
x
i
u
I
i
(
s
)
(
s; y
(
s
)) d
s
#
;
y
0
= 0
; y
N
= 0
;
(
i
= 1
;
2
; :::; N
1)
:
Clearly, we cannot explicitly compute the integrals in
(2
:
8)
in general. Therefore, we
approximate the function
(
x; y
(
x
))
on the interval
[
x
i
1
; x
i
]
by
i
1
=
x
i
1
+
x
i
2
;
y
i
1
+
y
i
2
;
where
y
i
are approximate values of the solution
y
of the problem
(1
:
1)
(1
:
2)
at points
x
i
:
Finally, from
(2
:
8)
we get the following dierence scheme
a
i
y
i
1
c
i
y
i
+
b
i
y
i
+1
=1
"
2
"
i
1
Z
x
i
x
i
1
u
II
i
1
(
s
) d
s
+
i
Z
x
i
+1
x
i
u
I
i
(
s
) d
s
#
;
(
i
= 1
;
2
; :::; N
1)
:
From
(2
:
2)
, we have
Z
x
i
x
i
1
u
II
i
1
(
s
)
ds
=1
cosh(
h
i
1
)
sinh(
h
i
1
)
1
1
sinh(
h
i
1
)
;
Z
x
i
+1
x
i
u
I
i
(
s
)
ds
=1
cosh(
h
i
)
sinh(
h
i
)
1
1
sinh(
h
i
)
:
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 143
Hence, our dierence scheme has the following form
a
i
y
i
1
c
i
y
i
+
b
i
y
i
+1
=1
i
1
(
d
i
a
i
) + 1
i
(
d
i
+1
a
i
+1
)
;
where
d
i
=
tanh(
h
i
1
)
:
After some computation, we get
(2.9)
a
i
+
d
i
2
y
i
1
a
i
+
d
i
2+
a
i
+1
+
d
i
+1
2
y
i
+
a
i
+1
+
d
i
+1
2
y
i
+1
=
4
d
i
f
i
1
+
4
d
i
+1
f
i
;
where
4
d
i
=
d
i
a
i
and
f
i
=
i
+
(
y
i
+
y
i
+1
)
=
2
,
(
i
= 1
;
2
; :::; N
1)
:
Using (2.9) let us introduce the discrete problem of the problem (1.1)
(1.3)
;
Fy
= ((
Fy
)
0
;
(
Fy
)
1
; : : : ;
(
Fy
)
N
)
T
= 0
;
where
(
Fy
)
0
:=
y
0
;
(
Fy
)
i
:=
a
i
+
d
i
2
y
i
1
a
i
+
d
i
2
+
a
i
+1
+
d
i
+1
2
y
i
+
a
i
+1
+
d
i
+1
2
y
i
+1
4
d
i
f
i
1
4
d
i
+1
f
i
;
(
i
= 1
;
2
; : : : ; N
1)
;
(
Fy
)
N
:=
y
N
;
and its equivalent normalized form
(2.10)
~
Fy
=
~
Fy
0
;
~
Fy
1
; : : : ;
~
Fy
N
T
= 0
;
where
(~
Fy
)
0
:=
y
0
;
(~
Fy
)
i
:=
4
d
i
+
4
d
i
+1
(
Fy
)
i
;
(
i
= 1
;
2
; : : : ; N
1)
;
(~
Fy
)
N
:=
y
N
:
Here we use the maximum norm
k
u
k
1
= max
0
6
i
6
N
j
u
i
j
;
for any vector
u
= (
u
0
; u
1
; : : : ; u
n
)
T
2
R
N
+1
and the corresponding matrix norm.
Theorem 2.1.
The discrete problem
(2
:
10)
for
f
y
;
has the unique solution
y
= (
y
0
; y
1
; y
2
; : : : ; y
N
1
; y
N
)
T
;
with
y
0
=
y
N
= 0
:
Moreover, the following stability
inequality holds
k
w
v
k
1
6
1
m
~
Fw
~
Fv
1
;
for any vectors
v
= (
v
0
; v
1
; : : : ; v
N
)
T
2
R
N
+1
; w
= (
w
0
; w
1
; : : : ; w
N
)
T
2
R
N
+1
:
144 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
Proof.
We use a technique from [7] and [16] , and the proof of existence of the solution of
~
F
(
y
) = 0
is based on the proof of the following relation:
~
F
0
(
y
)
1
1
C;
where
~
F
0
(
y
)
is the Fréchet derivative of
~
F
.
The Fréchet derivative
H
:= ~
F
0
(
y
)
of the operator dened in (2.10) is a tridiagonal
matrix. Let
H
= [
h
ij
]
:
The non-zero elements of this tridiagonal matrix are
h
0
;
0
=1
; h
N;N
= 1
;
h
i;i
=
4
d
i
+
4
d
i
+1
h
a
i
+
d
i
2
+
a
i
+1
+
d
i
+1
2
+
4
d
i
2
f
i
1
;yi
+
4
d
i
+1
2
f
i;yi
i
<
0
;
h
i;i
1
=
4
d
i
+
4
d
i
+1
h
a
i
+
4
d
i
2
1
f
i
1
;yi
1
i
>
0
;
h
i;i
+1
=
4
d
i
+
4
d
i
+1
h
a
i
+1
+
4
d
i
+1
2
1
f
i;yi
+1
i
>
0
;
where
f
j;y
k
=
@
@y
k
f
j
; j
2 f
i
1
; i
g
; k
2 f
i
1
; i; i
+ 1
g
:
Hence
H
is an
L
matrix.
Since
j
h
i;i
jj
h
i;i
1
j j
h
i
1
;i
j
=
4
d
i
+
4
d
i
+1
h
4
d
i
f
i
1
;yi
+
f
i
1
;yi
1
2
+
4
d
i
+1
f
i;yi
+
f
i;yi
+1
2
i
>
4
d
i
+
4
d
i
+1
h
4
d
i
m
+
m
2
+
4
d
i
+1
m
+
m
2
i
=
m;
the matrix
H
is also an
M
matrix and
(2.11)
H
1
1
1
m:
Now, by Hadamard`s Theorem (5.3.10 from [10] we can conclude that
~
F
dened in (2.10),
is a homeomorphism. Since
R
N
+1
is nonempty set, there exists a solution of the problem
(2.10) and regarding that 0 is the only image of F we come to the conclusion that
y
is
the only solution of the problem (2.10).
The second part of the proof is based on the part of the proof of Theorem 3 from [6] .
We have that
~
Fw
~
Fv
=
~
F
0
u
(
w
v
)
for some
u
= (
u
0
; u
1
; : : : ; u
N
)
T
2
R
N
+1
. Therefore
w
v
= ( ~
F
0
u
)
1
~
Fw
~
Fv
and nally because of (2.11) we have that
k
w
v
k
1
=
(~
F
0
u
)
1
(~
Fw
~
Fv
)
1
6
1
m
~
Fw
~
Fv
1
:
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 145
3.
Construction of the mesh
The solution changes rapidly near
x
= 0
and
x
= 1
:
Hence the mesh has to be
rened there. Various meshes have been proposed in the literature. The most frequently
analysed are the exponentially graded mesh of Bakhvalov [1] and piecewise uniform mesh
of Shishkin [12].
Here we shall use the smoothed Shishkin mesh from [8, 9] . Let
N
+ 1
be the number
of mesh points,
q
2
(0
;
1
=
2)
and
>
0
the mesh parameters. Dene the Shishkin-mesh
transition point by
:= min
"
p
m
ln
N; q
:
Let us chose
= 2
.
Remark 3.1.
For the sake of simplicity in representation, we assume that
= 2
"
(
p
m
)
1
ln
N
, as otherwise the problem can be analysed in the classical way.
We shall also assume that
qN
is an integer. This is easily achieved by choosing
q
= 1
=
4
and
N
divisible by 4 for example.
The mesh
4
:
x
0
< x
1
< ::: < x
N
is generated by
x
i
=
'
(
i=N
)
with the mesh
generating function
'
(
t
) :=
8
>
<
>
:
q
t t
2
[0
; q
]
;
p
(
t
q
)
3
+
q
t t
2
[
q;
1
=
2]
;
1
'
(1
t
)
t
2
[1
=
2
;
1]
;
where
p
is chosen so that
'
(1
=
2) = 1
=
2
;
i.e.
p
=
1
2
(1
q
)(
1
2
q
)
3
:
Note that
'
2
C
1
[0
;
1]
with
k
'
0
k
1
;
k
'
00
k
1
C:
Therefore the mesh sizes
h
i
=
x
i
+1
x
i
; i
= 0
;
1
;
2
; :::; N
1
satisfy (see [9] for details)
(3.1)
h
i
=
Z
(
i
+1)
=N
i=N
'
0
(
t
) d
t
CN
1
;
j
h
i
+1
h
i
j
=
Z
(
i
+1)
=N
i=N
Z
t
+1
=N
t
'
00
(
s
) d
s
C
N
2
:
4.
The error estimating of the nonlinear difference scheme
We will prove theorem on uniform convergence of the dierence scheme (2.9) on the
part of the mesh which corresponds to
[0
;
1
=
2]
;
while the proof on
[1
=
2
;
1]
can be analo-
gously derived.
Namely, in the analysis of the value of the error the functions
e
x
"
p
m
and
e
1
x
"
p
m
appear. For these functions we have that
e
x
"
p
m
>
e
1
x
"
p
m
;
8
x
2
[0
;
1
=
2]
and
e
x
"
p
m
6
e
1
x
"
p
m
;
8
x
2
[1
=
2
;
1]
. In the boundary layer in the neighbourhood of
x
= 0
, we have
that
e
x
"
p
m
>> e
1
x
"
p
m
, while in the boundary layer in the neighbourhood of
x
= 1
we
have that
e
x
"
p
m
<< e
1
x
"
p
m
:
Based on the above, it is enough to prove the theorem
on the part of the mesh which corresponds to
[0
;
1
=
2]
with the exclusion of the function
e
1
x
"
p
m
, or on
[1
=
2
;
1]
but with the exclusion of the function
e
x
"
p
m
. Note that we need
to take care of the fact that in the rst case
h
i
1
6
h
i
;
and in the second case
h
i
1
>
h
i
.
146 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
The proof uses the decomposition of the solution
y
of the problem
(1
:
1)
(1
:
2)
to the
layer component
s
and a regular component
r
, given in the following assertion.
Theorem 4.1.
[15]
The solution
y
to problem
(1
:
1)
(1
:
2)
can be represented in the
following way:
y
=
r
+
s;
where for
j
= 0
;
1
; :::; k
+ 2
and
x
2
[0
;
1]
we have that
(4.1)
r
(
j
)
(
x
)
6
C;
and
s
(
j
)
(
x
)
6
C"
j
e
x
"
p
m
+
e
1
x
"
p
m
:
On equidistant part of the mesh, that is for
x
i
; x
i
1
2
[0
;
]
and
i
= 1
;
2
; : : : ; N=
4
1
,
we will use Taylor expansions for the function
y
y
i
1
y
i
=
y
0
i
h
i
1
+
y
00
i
2
h
2
i
1
y
000
i
6
h
3
i
1
+
y
(
iv
)
(
i
)
24
h
4
i
1
;
y
i
y
i
+1
=
y
0
i
h
i
y
00
i
2
h
2
i
y
000
i
6
h
3
i
y
(
iv
)
(
+
i
)
24
h
4
i
;
(4.2)
while for function
f
we will use Taylor expansions
f
i
1
=
f
x
i
1
+
x
i
2
;
y
i
1
+
y
i
2
=
"
2
y
00
i
1
2
"
2
y
000
i
h
i
1
+1
2
f
y
(
x
i
; y
i
)
y
0
i
h
i
1
+
y
00
i
2
h
2
i
1
y
000
i
6
h
3
i
1
+
y
(
iv
)
(
i
)
24
h
4
i
1
+1
8
f
yy
(
i
;
i
)(
y
i
y
i
1
)
2
+1
8
f
xx
(
i
;
i
)
h
2
i
1
+1
4
f
xy
(
i
;
i
)(
y
i
y
i
1
)
h
i
1
;
f
i
=
f
x
i
+
x
i
+1
2
;
y
i
+
y
i
+1
2
=
"
2
y
00
i
+1
2
"
2
y
000
i
h
i
+1
2
f
y
(
x
i
; y
i
)
y
0
i
h
i
+
y
00
i
2
h
2
i
+
y
000
i
6
h
3
i
+
y
(
iv
)
(
+
i
)
24
h
4
i
+1
8
f
yy
(
+
i
;
+
i
)(
y
i
+1
y
i
)
2
+1
8
f
xx
(
+
i
;
+
i
)
h
2
i
+1
4
f
xy
(
+
i
;
+
i
)(
y
i
+1
y
i
)
h
i
;
(4.3)
where
y
i
=
y
(
x
i
)
;
i
2
((
x
i
1
+
x
i
)
=
2
; x
i
)
;
i
2
(
x
i
1
; x
i
)
;
+
i
2
(
x
i
;
(
x
i
+
x
i
+1
)
=
2)
;
+
i
2
(
x
i
; x
i
+1
)
;
i
2
((
y
i
1
+
y
i
)
=
2
; y
i
)
and
+
i
2
(
y
i
;
(
y
i
+
y
i
+1
)
=
2)
:
For
x
i
; x
i
1
2
x
N=
4
1
;
[
[
;
1
=
2]
, i.e.
i
=
N=
4
; : : : ; N=
2
1
, we will use Taylor expansions for the
function
y
y
i
1
y
i
=
y
0
i
h
i
1
+
y
00
(
i
)
2
h
2
i
1
;
i
2
(
x
i
1
; x
i
)
;
y
i
y
i
+1
=
y
0
i
h
i
y
00
(
+
i
)
2
h
2
i
;
+
i
2
(
x
i
; x
i
+1
)
:
(4.4)
Let us start with the following three lemmas that will be further used in the proof
of the uniform convergence on the part of the mesh from
x
3 which corresponds to
x
N=
4
1
;
1
=
2
; x
N=
4
=
:
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 147
Lemma 4.1.
Assume that
"
6
C
N
:
In the part of the modied Shishkin mesh from
x
3 when
x
i
; x
i
1
2
x
N=
4
1
;
[
[
;
1
=
2]
, we have the following estimate:
cosh(
h
i
1
)
1
sinh(
h
i
1
)
f
i
1
cosh(
h
i
)
1
sinh(
h
i
)
f
i
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
C
N
2
;
(
i
=
N=
4
; :::; N=
2
1)
:
Proof of the lemma is given in Appendix 7.1.
Lemma 4.2.
Assume that
"
6
C
N
:
In the part of the modied Shishkin mesh from
x
3 when
x
i
; x
i
1
2
x
N=
4
1
;
[
[
;
1
=
2]
, we have the following estimate
cosh(
h
i
1
)
1
2 sinh(
h
i
1
)
(
y
i
1
y
i
)
cosh(
h
i
)
1
2 sinh(
h
i
)
(
y
i
y
i
+1
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
C
N
2
;
(
i
=
N=
4
; :::; N=
2
1)
:
Proof of the lemma is given in Appendix 7.2.
Lemma 4.3.
Assume that
"
6
C
N
:
In the part of the modied Shishkin mesh from
x
3 when
x
i
; x
i
1
2
x
N=
4
1
;
[
[
;
1
=
2]
, we have the following estimate
1
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
y
i
1
y
i
sinh(
h
i
1
)
y
i
y
i
+1
sinh(
h
i
)
6
C
N
2
; i
=
N=
4
; :::; N=
2
1
:
Proof of the lemma is given in Appendix 7.3.
The proof of the theorem on
"
uniform convergence is based on the relation
k
y
y
k
1
6
C
~
Fy
~
Fy
1
:
Since
~
Fy
= 0
;
it only remains to estimate
~
Fy
1
:
Now we can state the main theorem
on
"
uniform convergence of our dierence scheme and the specially chosen layer-adapted
mesh.
Theorem 4.2.
The dierence scheme
(2
:
9)
on the mesh from Section 3 is uniformly
convergent with respect to
";
and
max
0
i
N
j
y
(
x
i
)
y
i
j
C
ln
2
N
N
2
;
where
y
(
x
)
is the solution of the problem
(1
:
1)
(1
:
2)
; y
is the corresponding numer-
ical solution of
(2
:
9)
, and
C >
0
is a constant independent with respect to
N
and
"
.
Proof.
Suppose rst that
x
i
2
[0
;
]
; i
= 1
; : : : ; N=
4
1
:
On this part of the mesh, we
have that
h
i
1
=
h
i
C
"
ln
N
N
:
The scheme
(2
:
9)
for the function
y
can be represented as
(4.5)
(
Fy
)
i
=
cosh(
h
i
)+1
2
[
y
i
1
y
i
(
y
i
y
i
+1
)]
cosh(
h
i
)
1
(
f
i
1
+
f
i
)
;
(
i
= 1
; :::; N=
4
1)
:
Now, putting
(4
:
2)
;
(4
:
3)
and
cosh(
h
i
) = 1 + (
h
i
)
2
2+
O
ln
4
N=N
4
;
148 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
into
(4
:
5)
, we get that
(
Fy
)
i
=
2
h
2
i
4+
O
(ln
4
N=N
4
)
y
00
i
h
2
i
+
1 +
2
h
2
i
4+
O
(ln
4
N=N
4
)
y
(
iv
)
(
i
)
h
4
i
+
y
(
iv
)
(
+
i
)
h
4
i
24
O
(ln
4
N=N
4
)
2
"
2
y
00
i
+1
2
f
y
(
x
i
; y
i
)
y
00
i
h
2
i
+
y
(
iv
)
(
i
)
h
4
i
+
y
(
iv
)
(
+
i
)
h
4
i
24
+
SD
i
2
h
2
i
2
1
2
f
y
(
x
i
; y
i
)
y
00
i
h
2
i
+
y
(
iv
)
(
i
)
h
4
i
+
y
(
iv
)
(
+
i
)
h
4
i
24
+
SD
i
;
where
SD
i
=1
8
f
yy
(
i
;
i
)(
y
i
y
i
1
)
2
+1
8
f
xx
(
i
;
i
)
h
2
i
+1
4
f
xy
(
i
;
i
)(
y
i
y
i
1
)
h
i
+1
8
f
yy
(
+
i
;
+
i
)(
y
i
+1
y
i
)
2
+1
8
f
xx
(
+
i
;
+
i
)
h
2
i
+1
4
f
xy
(
+
i
;
+
i
)(
y
i
+1
y
i
)
h
i
:
For the sake of normalization, dividing (4.5) with
cosh(
h
i
)
1
we get
(4.6)
2
j
(~
Fy
)
i
j
=
(
Fy
)
i
cosh(
h
i
)
1
C
ln
2
N
N
2
:
Now, suppose that
x
i
2
[
x
N=
4
1
;
]
[
[
;
1
=
2]
for
i
=
N=
4
; : : : ; N=
2
1
. The scheme
(2
:
9)
for the function
y
can be written as
(
Fy
)
i
=cosh(
h
i
1
)
1
2 sinh(
h
i
1
)(
y
i
1
y
i
)
cosh(
h
i
)
1
2 sinh(
h
i
)(
y
i
y
i
+1
)
+
y
i
1
y
i
sinh(
h
i
1
)
y
i
y
i
+1
sinh(
h
i
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
f
i
1
cosh(
h
i
)
1
sinh(
h
i
)
f
i
;
(
i
=
N=
4
; :::; N=
2
1)
:
From the inequality
(
Fy
)
i
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
1
cosh(
h
i
1
)
1
sinh(
h
i
1
)+cosh(
h
i
)
1
sinh(
h
i
)
cosh(
h
i
1
)
1
2 sinh(
h
i
1
)
(
y
i
1
y
i
)
cosh(
h
i
)
1
2 sinh(
h
i
)
(
y
i
y
i
+1
)
+
y
i
1
y
i
sinh(
h
i
1
)
y
i
y
i
+1
sinh(
h
i
)
+
cosh(
h
i
1
)
1
sinh(
h
i
1
)
f
i
1
cosh(
h
i
)
1
sinh(
h
i
)
f
i
;
based on Lemma 1Lemma 3, we have that
j
(~
Fy
)
i
j
=
(
Fy
)
i
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
C
N
2
:
(4.7)
According to
(4
:
6)
and
(4
:
7)
, the proof is complete.
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 149
5.
The numerical results
In this section we present numerical results to conrm the uniform accuracy of the
scheme
(2
:
9)
:
Consider the following problem from [14]
(5.1)
"
2
y
00
= (1 +
y
)(1 + (1 +
y
)
2
)
on
(0
;
1)
;
(5.2)
y
(0) =
y
(1) = 0
;
whose exact solution is unknown. The nonlinear system of equations is solved by Newton's
method with initial guess
y
0
=
1
that represents the reduced solution. The value
of the constant
= 4
has been chosen so that the condition
f
y
(
x; y
)
;
8
(
x; y
)
2
[0
;
1]
[
y
L
; y
U
]
[0
;
1]
R
is fullled, where
y
L
and
y
U
are lower and upper solutions
of the test problem
(5
:
1)
(5
:
2)
and their values are
y
L
=
1
and
y
U
= 0
:
Because of
the fact that the exact solution is unknown, we dene the computed error
E
N
and the
computed rate of convergence Ord in the usual way (double-mesh method, see [4, 13, 14]
E
N
= max
0
i
N
~
y
2
N
(
x
i
)
y
N
(
x
i
)
;
Ord
=ln
E
N
ln
E
2
N
ln
2
k
k
+1
;
where
N
= 2
k
; k
= 6
;
7
; :::;
11
; y
N
(
x
i
)
is the numerical solution on a mesh with
N
subin-
tervals, and
~
y
2
N
(
x
i
)
is the numerical solution on a mesh with
2
N
subintervals and the
transition point altered slightly to
~
"
= min
n
1
4
;
2
"
p
m
ln
N
2
o
:
0.2
0.4
0.6
0.8
1.0
x
-1.0
-0.8
-0.6
-0.4
-0.2
y
0.2
0.4
0.6
0.8
1.0
x
-1.0
-0.8
-0.6
-0.4
-0.2
y
Figure 1.
Graphics of approximate solutions for values
"
= 2
3
; "
= 2
10
:
In Figure 1 we display the computed solution of (5.1)(5.2) for two values of the pa-
rameter
"
. For dierent values of the perturbation parameter
"
, in Table 1 we present
the results of numerical experiments that clearly conrm the robustness of the method
as well as that theoretical and experimental results match.
150 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
N E
n
Ord
E
n
Ord
E
n
Ord
2
6
4
:
5606
e
04 2
:
42 3
:
1000
e
03 1
:
99 7
:
9175
e
03 1
:
99
2
7
1
:
2375
e
04 2
:
04 1
:
0597
e
03 2
:
00 2
:
7093
e
03 2
:
00
2
8
3
:
9496
e
05 2
:
01 3
:
4620
e
04 2
:
00 8
:
8552
e
04 2
:
00
2
9
1
:
2398
e
05 2
:
00 1
:
0956
e
04 2
:
00 2
:
8030
e
04 2
:
00
2
10
3
:
8199
e
06 2
:
00 3
:
3815
e
05 2
:
00 8
:
6565
e
05 2
:
00
2
11
1
:
1562
e
06 2
:
00 1
:
0229
e
05 2
:
00 2
:
6187
e
05 2
:
00
2
12
3
:
4400
e
07 2
:
00 3
:
0434
e
06 2
:
00 7
:
7911
e
06 2
:
00
2
13
1
:
0093
e
07
8
:
9294
e
07
2
:
2860
e
06
"
2
3
2
5
2
10
N E
n
Ord
E
n
Ord
E
n
Ord
2
6
7
:
9175
e
03 1
:
99 7
:
9175
e
03 1
:
99 7
:
9175
e
03 1
:
99
2
7
2
:
7093
e
03 2
:
00 2
:
7093
e
03 1
:
99 2
:
7093
e
03 1
:
99
2
8
8
:
8552
e
04 2
:
00 8
:
8552
e
04 2
:
00 8
:
8552
e
04 2
:
00
2
9
2
:
8030
e
04 2
:
00 2
:
8030
e
04 2
:
00 2
:
8030
e
04 2
:
00
2
10
8
:
6565
e
05 2
:
00 8
:
6565
e
05 2
:
00 8
:
6565
e
05 2
:
00
2
11
2
:
6187
e
05 2
:
00 2
:
6187
e
05 2
:
00 2
:
6187
e
05 2
:
00
2
12
7
:
7911
e
06 2
:
00 7
:
7911
e
06 2
:
00 7
:
7911
e
06 2
:
00
2
13
2
:
2860
e
06
2
:
2860
e
06
2
:
2860
e
06
"
2
15
2
25
2
30
N E
n
Ord
E
n
Ord
E
n
Ord
2
6
7
:
9175
e
03 1
:
99 7
:
9176
e
03 1
:
99 7
:
9175
e
03 1
:
98
2
7
2
:
7093
e
03 1
:
99 2
:
7095
e
03 2
:
00 2
:
7223
e
03 2
:
01
2
8
8
:
8552
e
04 2
:
00 8
:
8553
e
04 2
:
00 8
:
5552
e
04 2
:
00
2
9
2
:
8030
e
04 2
:
00 2
:
8030
e
04 2
:
00 2
:
8030
e
04 2
:
00
2
10
8
:
6565
e
05 2
:
00 8
:
6565
e
05 2
:
00 8
:
6571
e
05 1
:
99
2
11
2
:
6187
e
05 2
:
00 2
:
6189
e
05 2
:
00 2
:
6329
e
05 1
:
99
2
12
7
:
7911
e
06 2
:
00 7
:
7915
e
06 2
:
00 7
:
8633
e
06 1
:
99
2
13
2
:
2860
e
06
2
:
2864
e
06
2
:
3213
e
06
"
2
35
2
40
2
45
Table 1.
Errors
E
N
and convergence rates Ord for approximate solutions.
6.
Discussion
In this paper we present a discretization of a one-dimensional semilinear reaction-
diusion problem, with suitable assumptions that ensure the existence and uniqueness of
the continuous problem. We prove the existence and uniqueness of the numerical solution,
the
"
uniform convergence using a suitable layer-adaptive mesh and nally we perform a
numerical experiment which agrees with theoretical results.
The presented method should be expandable to discretization of higher dimensional
boundary value problems without major problems. Namely, hyperbolic functions appear
in the dierence scheme coecients, so one has to choose the discretization in which
these hyperbolic functions remain functions of one variable, which is not dicult to
do. In this case we could separate the terms in which the same variables appear and
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 151
the analysis would be reduced to the one presented in this paper. In the case of the
discretization in which hyperbolic functions of several variables appear, the analysis would
be more dicult. Clearly, the above discussion is only related to suitable Shishkin-type
meshes. Using Bakhvalov-type meshes the analysis of one-dimensional boundary value
problems becomes substantially more dicult, as is the analysis of the dicretization of
higher dimensional boundary value problems.
7.
Appendix
7.1.
Proof of the Lemma 4.1.
Proof.
Due to
"
2
y
00
(
x
i
) =
f
(
x
i
; y
(
x
i
))
;
Theorem 4.1 for both components
r
and
s
in part
of the mesh
x
N=
4
1
;
[
[
;
1
=
2]
and the assumption
"
6
C
N
;
we have that
cosh(
h
i
1
)
1
sinh(
h
i
1
)
f
i
1
cosh(
h
i
)
1
sinh(
h
i
)
f
i
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
cosh(
h
i
1
)
1
sinh(
h
i
1
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
j
f
i
1
j
+
cosh(
h
i
)
1
sinh(
h
i
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
j
f
i
j
6
j
f
i
1
j
+
j
f
i
j
6
C
N
2
:
7.2.
Proof of the Lemma 4.2.
Proof.
Let us use again the decomposition from Theorem 4.1 and assumption
"
6
C
N
.
For the layer component
s
we have that
cosh(
h
i
1
)
1
2 sinh(
h
i
1
)
(
s
i
1
s
i
)
cosh(
h
i
)
1
2 sinh(
h
i
)
(
s
i
s
i
+1
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
6
2(
j
s
i
1
s
i
j
+
j
s
i
s
i
+1
j
)
:
Using (4.2),
e
x
"
p
m
>
e
1
x
"
p
m
;
8
x
2
[0
;
1
=
2]
;
the monotonicity of the function
e
x
"
p
m
and putting
x
N=
4
1
=
2
"
ln
N
p
m
N=
4
1
N=
4
into
e
x
"
p
m
, we get that
j
s
i
1
s
i
j
+
j
s
i
s
i
+1
j
6
4
j
s
i
1
j
6
C
1
e
2 ln
N
N
4
N
=
C
1
e
2 ln
N
e
8 ln
N
N
6
C
1
N
2
1 + 8 ln
N
N
+
6
C
N
2
:
(7.1)
For the regular component
r
, due to
cosh
x
1
sinh
x
= tanh
x
2
;
we have that
cosh(
h
i
1
)
1
2 sinh(
h
i
1
)
(
r
i
1
r
i
)
cosh(
h
i
)
1
2 sinh(
h
i
)
(
r
i
r
i
+1
)
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
=
2
1
tanh
h
i
1
2
+ tanh
h
i
2
tanh
h
i
1
2
(
r
i
1
r
i
)
tanh
h
i
2
(
r
i
r
i
+1
)
:
152 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
Using Taylor expansions
(4
:
4)
for
r
i
1
and
r
i
+1
;
we get that
2
1
tanh
h
i
1
2
+ tanh
h
i
2
tanh
h
i
1
2
(
r
i
1
r
i
)
tanh
h
i
2
(
r
i
r
i
+1
)
=
2
1
tanh
hi
1
2
+tanh
hi
2
tanh
h
i
1
2
r
0
i
h
i
1
+
r
00
(
i
)
2
h
2
i
1
+ tanh
h
i
2
r
0
i
h
i
+
r
00
(
+
i
)
2
h
2
i
6
C
h
2
i
1
+
h
2
i
+
j
r
0
i
j
tanh
hi
1
2
+tanh
hi
2
tanh
h
i
1
2
h
i
1
tanh
h
i
2
h
i
:
(7.2)
Because of the monotonicity of the function
tanh
x
and
(4
:
1)
, we get the following estimate
j
r
0
i
j
tanh
h
i
1
2
+ tanh
h
i
2
tanh
h
i
1
2
h
i
1
tanh
h
i
2
h
i
6
C
tanh
h
i
1
2
tanh
h
i
2
h
i
1
h
i
:
Now, let us use the second inequality from
(3
:
1)
for
h
i
h
i
1
in order to get
tanh
h
i
1
2
tanh
h
i
2
h
i
1
h
i
=
tanh
h
i
1
2
tanh
h
i
2
h
i
1
h
i
1
+
C
N
2
=
tanh
h
i
1
2
tanh
h
i
2
1
h
i
1
+
C
N
2
=tanh
h
i
2
tanh
h
i
1
2
tanh
h
i
2
h
i
1
+
C
N
2
=
e
hi
1
e
hi
+1
e
hi
1
1
e
hi
1
+1
e
hi
1
e
hi
+1
h
i
1
+
C
N
2
= 2
e
h
i
e
h
i
1
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
= 2
(
h
i
h
i
1
) +
2
(
h
2
i
h
2
i
1
)
2!
+
3
(
h
3
i
h
3
i
1
)
3!
+
:::
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
= 2
+
1
X
n
=1
n
(
h
n
i
h
n
i
1
)
n
!
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
:
(7.3)
From the identity
(7.4)
a
n
b
n
= (
a
b
)(
a
n
1
+
a
n
2
b
+
:::
+
ab
n
2
+
b
n
1
)
; n
2
N
;
and
h
i
1
6
h
i
,
i
= 1
;
2
; : : : ; N=
2
1
, we get the majorization
h
n
i
h
n
i
1
= (
h
i
h
i
1
)(
h
n
1
i
+
h
n
2
i
h
i
1
+
:::
+
h
i
h
n
2
i
1
+
h
n
1
i
1
)
6
n
(
h
i
h
i
1
)
h
n
1
i
:
(7.5)
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 153
Moreover, for the last relation from
(7
:
3)
, due to
(7
:
5)
;
we have
2
+
1
X
n
=1
n
(
h
n
i
h
n
i
1
)
n
!
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
6
2
(
h
i
h
i
1
)
+
1
X
n
=0
(
h
i
)
n
n
!
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
= 2
(
h
i
h
i
1
)
e
h
i
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
= 2
(
h
i
h
i
1
)(
e
h
i
1 + 1)
(
e
h
i
1)(
e
h
i
1
+ 1)
h
i
1
+
C
N
2
6
2(
h
i
h
i
1
)
e
hi
1
e
hi
1
h
i
1
e
hi
1
+1
+ 2(
h
i
h
i
1
)
h
i
1
(
e
hi
1)(
e
hi
1
+1)
+
C
N
2
6
C
h
i
h
i
1
+1
N
2
:
(7.6)
Now, collecting
(3
:
1)
;
(7
:
1)
;
(7
:
2)
;
(7
:
3)
and
(7
:
6)
, the statement of the lemma is therefore
proven.
7.3.
Proof of the Lemma 4.3.
Proof.
We are using again the decomposition from Theorem 4.1 and expansions
(4
:
4)
.
For the regular component
r
, we have that
1
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
r
i
1
r
i
sinh(
h
i
1
)
r
i
r
i
+1
sinh(
h
i
)
6
cosh(
h
i
)
1
sinh(
h
i
)
r
i
1
r
i
sinh(
h
i
1
)
r
i
r
i
+1
sinh(
h
i
)
=
sinh(
h
i
)(
r
i
1
r
i
)
sinh(
h
i
1
)(
r
i
r
i
+1
)
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
0
i
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
i
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
i
1
!
(cosh(
h
i
)
1) sinh(
h
i
1
)
+
r
00
(
+
i
)
2
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
2
i
+
r
00
(
i
)
2
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
2
i
1
(cosh(
h
i
)
1) sinh(
h
i
1
)
:
(7.7)
Let us rst estimate the expressions from
(7
:
7)
using the rst derivatives. Now we
have that
154 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
r
0
i
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
i
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
i
1
!
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
0
i
h
i
1
+
3
h
3
i
1
3!
+
:::
h
i
h
i
+
3
h
3
i
3!
+
:::
h
i
1
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
0
i
h
i
1
h
i
+
1
X
n
=1
2
n
(
h
2
n
i
h
2
n
i
1
)
(2
n
+ 1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)! sinh(
h
i
1
)
:
(7.8)
The identity
(7
:
4)
yields
2
n
(
h
2
n
i
h
2
n
i
1
)
(2
n
+ 1)! =
2
n
(
h
2
i
h
2
i
1
)(
h
2(
n
1)
i
+
h
2(
n
2)
i
h
2
i
1
+
:::
+
h
2
i
h
2(
n
2)
i
1
+
h
2(
n
1)
i
1
)
(2
n
+ 1)!
6
2
n
(
h
2
i
h
2
i
1
)
nh
2(
n
1)
i
(2
n
+ 1)!
<
2
n
(
h
2
i
h
2
i
1
)
h
2(
n
1)
i
(2
n
)!
;
8
n
2
N
:
(7.9)
If we use the last expression from
(7
:
9)
into
(7
:
8)
, together with
(4
:
1)
, we get
r
0
i
h
i
1
h
i
+
1
X
n
=1
2
n
(
h
2
n
i
h
2
n
i
1
)
(2
n
+ 1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)! sinh(
h
i
1
)
6
r
0
i
h
i
1
h
i
+
1
X
n
=1
2
n
(
h
2
i
h
2
i
1
)
h
2(
n
1)
i
(2
n
)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)! sinh(
h
i
1
)
6
2
r
0
i
h
i
1
sinh(
h
i
1
)
h
2
i
(
h
i
h
i
1
)
+
1
X
n
=1
2
n
h
2(
n
1)
i
(2
n
)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
6
C
(
h
i
h
i
1
)
:
(7.10)
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 155
For the terms from
(7
:
7)
with second derivatives we have
r
00
(
+
i
)
2
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
2
i
+
r
00
(
i
)
2
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
2
i
1
(cosh(
h
i
)
1) sinh(
h
i
1
)
6
r
00
(
+
i
)
2
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
2
i
(cosh(
h
i
)
1) sinh(
h
i
1
)
+
r
00
(
i
)
2
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
2
i
1
(cosh(
h
i
)
1) sinh(
h
i
1
)
:
(7.11)
Again, considering
(4
:
1)
, for the rst summand from the last expression from
(7
:
11)
;
we
have that
r
00
(
+
i
)
2
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
h
2
i
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
00
(
+
i
)
2sinh(
h
i
1
)
h
2
i
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
00
(
+
i
)
h
2
i
2
cosh(
h
i
)
1
6
r
00
(
+
i
)
h
2
i
2
h
2
i
6
C"
2
;
(7.12)
while the second summand can be estimated using
r
00
(
i
)
2
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
h
2
i
1
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
r
00
(
i
)
2
h
i
+
1
X
n
=0
(
h
i
)
2
n
(2
n
+1)!
h
2
i
1
+
1
X
n
=0
(
h
i
)
2
n
(2
n
)!
1
!
sinh(
h
i
1
)
6
r
00
(
i
)
2
h
i
h
2
i
1
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
+
r
00
(
i
)
2
h
2
i
1
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
6
r
00
(
i
)
h
i
h
2
i
1
(
h
i
)
2
h
i
1
+
r
00
(
i
)
2
h
2
i
1
h
i
h
i
1
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
6
C
(
"
2
+
h
i
1
h
i
)
:
(7.13)
For the layer component
s
, rst we have that
156 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
1
cosh(
h
i
1
)
1
sinh(
h
i
1
)
+
cosh(
h
i
)
1
sinh(
h
i
)
s
i
1
s
i
sinh(
h
i
1
)
s
i
s
i
+1
sinh(
h
i
)
6
sinh(
h
i
)(
s
i
1
s
i
)
sinh(
h
i
1
)(
s
i
s
i
+1
)
(cosh(
h
i
)
1) sinh(
h
i
1
)
=
+
1
X
n
=0
(
h
i
)
2
n
+1
(2
n
+1)!
(
s
i
1
s
i
)
+
1
X
n
=0
(
h
i
1
)
2
n
+1
(2
n
+1)!
(
s
i
s
i
+1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
6
h
i
(
s
i
1
s
i
)
h
i
1
(
s
i
s
i
+1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
+
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
(
s
i
1
s
i
)
h
i
1
+
1
X
n
=1
(
h
i
1
)
2
n
(2
n
+1)!
(
s
i
s
i
+1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
:
(7.14)
The rst summand can be bounded with
h
i
(
s
i
1
s
i
)
h
i
1
(
s
i
s
i
+1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
6
h
i
h
s
0
i
h
i
1
+
s
00
(
i
)
2
h
2
i
1
i
h
i
1
h
s
0
i
h
i
+
s
00
(
+
i
)
2
h
2
i
i
2
h
2
i
2
h
i
1
6
"
2
s
00
(
i
)
h
i
1
h
i
+
s
00
(
+
i
)
6
C"
2
e
i
"
p
m
"
2
h
i
1
h
i
+
e
+
i
"
p
m
"
2
!
6
C
N
2
:
(7.15)
For the second summand in
(7
:
14)
we get
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR ... 157
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
(
s
i
1
s
i
)
h
i
1
+
1
X
n
=1
(
h
i
1
)
2
n
(2
n
+1)!
(
s
i
s
i
+1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
6
h
i
sinh(
h
i
1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
j
s
i
1
s
i
j
+
h
i
1
h
i
1
+
1
X
n
=1
(
h
i
1
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
j
s
i
s
i
+1
j
and
+
1
X
n
=1
(
h
i
1
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
j
s
i
s
i
+1
j
6
C
N
2
:
(7.16)
In the expression
(7.17)
h
i
sinh(
h
i
1
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
j
s
i
1
s
i
j
;
there is a ratio
h
i
sinh(
h
i
1
)
:
Though inequality
h
i
sinh(
h
i
1
)
6
h
i
h
i
1
holds true, the quotient
h
i
h
i
1
is not bounded for
x
i
=
and
"
!
0
:
This is why we are going to estimate the
expression
(7
:
17)
separately on the transition part and on the nonequidistant part of the
mesh.
In the case
i
=
N=
4
, we can write
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+ 1)!(
s
i
1
s
i
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)! sinh(
h
i
1
)
=
sinh(
h
i
)
h
i
cosh(
h
i
)
1
s
i
1
s
i
sinh(
h
i
1
)
;
158 S. KARASULJI, E.DUVNJAKOVI AND H. ZARIN
since
P
+
1
n
=1
x
2
n
(2
n
)!
= cosh
x
1
and
P
+
1
n
=1
x
2
n
+1
(2
n
+1)!
= sinh
x
x;
8
x
2
R
:
The function
r
(
x
) =
sinh
x
x
cosh
x
1
takes values from the interval
(0
;
1)
when
x >
0
. Thus
(7.18)
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
(
s
i
1
s
i
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
=
sinh(
h
i
)
h
i
cosh(
h
i
)
1
s
i
1
s
i
sinh(
h
i
1
)
6
j
s
i
1
s
i
j
sinh(
h
i
1
)
6
C
ln
N
N
3
ln
N
N
=
C
N
2
:
When
i
=
N=
4 + 1
; : : : ; N=
2
1
, we can use
P
+
1
n
=1
x
2
n
(2
n
+1)!
P
+
1
n
=1
x
2
n
(2
n
)!
=
sinh
x
x
x
(cosh
x
1)
=
p
(
x
)
and
0
< p
(
x
)
<
1
3
for
x >
0
:
Therefore
h
i
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+1)!
(
s
i
1
s
i
)
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
sinh(
h
i
1
)
6
h
i
h
i
1
+
1
X
n
=1
(
h
i
)
2
n
(2
n
+ 1)!
+
1
X
n
=1
(
h
i
)
2
n
(2
n
)!
j
s
i
1
s
i
j
6
C
N
2
:
(7.19)
Using
(3
:
1)
;
(7
:
10)
;
(7
:
12)
;
(7
:
13)
;
(7
:
15)
;
(7
:
16)
;
(7
:
18)
and
(7
:
19)
completes the proof of
the lemma.
Acknowledgements.
Helena Zarin supported by the Ministry of Education and Science of the Republic of
Serbia under grant 174030.
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:
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:
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:
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,
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:
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:
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:
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:
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:
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:
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:
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Department of Mathematics
Faculty of sciences, University of Tuzla
Univerzitetska 4, 75000, Bosnia and Herzegovina
E-mail address
:
samir.karasuljic@untz.ba
Department of Mathematics
Faculty of sciences, University of Tuzla
Univerzitetska 4, 75000, Bosnia and Herzegovina
E-mail address
:
enes.duvnjakovic@untz.ba
Department of Mathematics and Informatics
Faculty of sciences, University of Novi Sad
Trg Dositeja Obradovi¢a 4, 21000 Novi Sad, Serbia
E-mail address
:
helena.zarin@dmi.uns.ac.rs