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ADV MATH
SCI JOURNAL
723–38
printed version
electronic version
UNIFORMLY CONVERGENT DIFFERENCE SCHEME FOR A SEMILINEAR
REACTION-DIFFUSION PROBLEM ON A SHISHKIN MESH
SAMIR KARASULJI´
C1, ENES DUVNJAKOVI´
C, AND ELVIR MEMI´
C
ABSTRACT. In this paper we consider two difference schemes for numerical solving of a
one–dimensional singularly perturbed boundary value problem. We proved an ε–uniform
convergence for both difference schemes on a Shishkin mesh. Finally, we present four
numerical experiments to confirm the theoretical results.
1. INTRODUCTION
We consider the semilinear singularly perturbed problem
(1.1) ε2y00(x) = f(x, y)on (0,1) ,
(1.2) y(0) = 0, y(1) = 0,
where εis a small positive parameter. We assume that the nonlinear function fis contin-
uously differentiable, i.e. for k≥2, f ∈Ck([0,1] ×R),and that it has a strictly positive
derivative with respect to y
(1.3) ∂f
∂y =fy≥m > 0on [0,1] ×R(m=const).
The boundary value problem (1.1)–(1.2), under the condition (1.3), has a unique solution
(see [14]). Numerical treatment of the problem (1.1), has been considered by many
authors, under different condition on the function f, and made a significant contribution.
We are going to analyze two difference schemes for the problem (1.1)–(1.3). These
difference schemes were constructed using the method first introduced by Boglaev [1],
who constructed a difference scheme and showed convergence of order 1 on a modified
Bakhvalov mesh. In our previous papers using the method [1], we constructed new
difference schemes in [3, 4, 10, 6, 7, 8, 9, 13] and performed numerical tests, in [5, 11]
we constructed new difference schemes and we proved the theorems on the uniqueness
of the numerical solution and the ε–uniform convergence on the modified Shishkin mesh,
and again performed the numerical test. In [12] we used the difference schemes from
1corresponding author
2010 Mathematics Subject Classification. 65L10, 65L11, 65L50.
Key words and phrases. Singularly perturbed, boundary value problem, numerical solution, difference
scheme, nonlinear, Shishkin mesh, layer–adapted mesh, ε–uniform convergent.
23
24 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
[11] and calculated the values of the approximate solutions of the problem (1.1)–(1.3)
on the mesh points and then we constructed an approximate solution.
Since in the boundary layers, i.e. near x= 0 and x= 1,the solution of the problem
(1.1)–(1.3) changes rapidly, when parameter tends to zero, in order to get the ε–uniform
convergence, we have to use a layer-adapted mesh. In the present paper we are going to
use a Shishkin mesh [15], which is piecewise equidistant and consequently simpler than
the modified Shishkin mesh we have already used in our mentioned papers.
2. DIFFERENCE SCH EMES
For a given positive integer N, let it be an arbitrary mesh
0 = x0< x1<··· < xN−1< xN= 1,
with hi=xi−xi−1,for i= 1, . . . , N.
Our first difference scheme has the following form
ai+di
2yi−1−ai+di
2+ai+1 +di+1
2yi+ai+1 +di+1
2yi+1 =
=4di
γfi−1/2+4di+1
γfi/2,(2.1)
where di=β
tanh(βhi−1), ai=β
sinh(βhi−1), f i−1/2=fxi−1+xi
2,yi−1+yi
2and 4di=di−ai.
From (2.1), we obtain next discrete problem
F y = (F y0, F y1, . . . , F yN)T,
where
F y0=y0= 0,
F yi=γ
4di+4di+1 ai+di
2yi−1−ai+di
2+ai+1 +di+1
2yi
+ai+1 +di+1
2yi+1 −4di
γfi−1/2−4di+1
γfi/2= 0,
i= 1,2, . . . , N −1,
F yN=yN= 0,
and y:= (y0, y1, . . . , yN)Tis the solution of the problem
(2.2) F y = 0.
Second difference scheme has the following form
(2.3) (3ai+di+4di+1) ( ˜yi−1−˜yi)−(3ai+1 +di+1 +4di) ( ˜yi−˜yi+1 )
=˜
fi−1+ 2 ˜
fi+˜
fi+1
γ(4di+4di+1),
where di=β
tanh(βhi−1), ai=β
sinh(βhi−1),˜
fi=f(xi,˜yi)and 4di=di−ai.
From (2.3), we obtain second discrete problem
(2.4) G˜y= (G˜y0, G˜y1, . . . , G˜yN)T,
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 25
where
G˜y0= ˜y0= 0
G˜yi=γ
4di+4di+1 (3ai+di+4di+1) ( ˜yi−1−˜yi)−(3ai+1 +di+1 +4di) ( ˜yi−˜yi+1 )
−˜
fi−1+ 2 ˜
fi+˜
fi+1
γ(4di+4di+1)#= 0, i = 1, . . . , N −1,
(2.5)
G˜yN= ˜yN= 0,
and ˜y= (˜y0,˜y1,...,˜yn)Tis the solution of the problem
(2.6) G˜y= 0.
3. THEORETICAL BAC KGROUND
In this paper we use the maximum norm
kuk∞= max
0≤i≤N|ui|,
for any vector u= (u0, u1, . . . , un)T∈RN+1 and the corresponding matrix norm.
The next two theorems hold:
Theorem 3.1. [11] The discrete problem (2.2) for γ≥fy,has the unique solution
y= (y0, y1, y2, . . . , y N−1, yN)T,with y0=yN= 0.Moreover, the following stability in-
equality holds
kw−vk∞≤1
mkF w −F vk∞,
for any vectors v= (v0, v1, . . . , vN)T∈RN+1, w = (w0, w1, . . . , wN)T∈RN+1.
Theorem 3.2. [5] The discrete problem (2.6) has a unique solution ˜yfor γ≥fy. Also, for
every u, v ∈RN+1 we have the following stabilizing inequality
ku−vk∞≤1
mkGu −Gvk∞.
In the following analysis we need the decomposition of the solution yof the problem
(1.1) −(1.2) to the layer component sand a regular component r, given in the following
assertion.
Theorem 3.3. [18] The solution yto problem (1.1) −(1.2) can be represented in the
following way:
y=r+s,
where for j= 0,1, ..., k + 2 and x∈[0,1] we have that
r(j)(x)≤C,
and s(j)(x)≤Cε−je−x
ε√m+e−1−x
ε√m.
26 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
4. CONSTRUCTIO N OF THE ME SH
The solution of the problem (1.1)–(1.3) changes fast near the ends of our domain [0,1].
Therefore, the mesh has to be refined there. A Shishkin mesh is used to resolve the layers.
This mesh is piecewise equidistant and it’s quite simple. It is constructed as follows (see
[16]). For given a positive integer N, where Nis divisible by 4, we divide the interval
[0,1] into three subintervals
[0, λ],[λ, 1−λ],[1 −λ, 1].
We use equidistant meshes on each of these subintervals, with 1 + N
4points in each of
[0, λ]and [1 −λ, 1],and 1 + N
2points in [1 −λ, 1].We define the parameter λby
λ= min 1
4,2εln N
√m,
which depends on Nand ε. The basic idea here is to use a fine mesh to resolve the part
of the boundary layers. More precisely, we have
0 = x0< x1< . . . < xi0< . . . < xN−i0< . . . < xN−1< xN= 1,
with i0=N/4, xi0=λ, xN−i0= 1 −λ, and
hi−1=4λ
Nfor i= 1, . . . , i0, N −i0, . . . , N ,(4.1)
hi−1=2(1 −2λ)
Nfor i=i0+ 1, . . . , N −i0.(4.2)
If λ=1
4i.e. 1
4≤2εln N
N,then 1
Nis very small relative to ε. This is unlike in practice, and
in this case the method can be analyzed using standard techniques. Hence, we assume
that
λ=2εln N
√m.
From (4.1) and (4.2), we conclude that that the interval lengths satisfy
hi−1=8εln N
√mfor i= 1, . . . , i0, N −i0, . . . , N ,
and 1
N≤hi−1≤2
Nfor i=i0+ 1, . . . , N −i0.
5. UNIFORM CONVE RGENCE
We will prove the theorem on uniform convergence of the difference schemes (2.1) and
(2.3) on the part of the mesh which corresponds to [0,1/2],while the proof on [1/2,1]
can be analogously derived.
Namely, in the analysis of the value of the error the functions e−x
ε√mand e−1−x
ε√m
appear. For these functions we have that e−x
ε√m≥e−1−x
ε√m,∀x∈[0,1/2] and e−x
ε√m≤
e−1−x
ε√m,∀x∈[1/2,1]. In the boundary layer in the neighbourhood of x= 0, we have
that e−x
ε√m>> e−1−x
ε√m, while in the boundary layer in the neighbourhood of x= 1 we
have that e−x
ε√m<< e−1−x
ε√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
ε√m, or on [1/2,1] but with the exclusion of the function e−x
ε√m. Note that we need
to take care of the fact that in the first case hi−1≤hi,and in the second case hi−1≥hi.
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 27
Let us start with the following two lemmas that will be further used in the proof of
the first uniform convergence theorem on the part of the mesh from Section 3 which
corresponds to xN/4−1,1/2and xN/4=λ.
Lemma 5.1. Assume that ε≤C
N.In the part of the Shishkin mesh from Section 3, when
xi, xi±1∈[xN/4,1/2],we have the following estimate
|F yi| ≤ C
N2, i =N/4, . . . , N/2−1.
Proof. On this part of the mesh holds hi−1=hi,so we have that
F yi=γ
2(cosh(βhi)−1)
·(1 + cosh(βhi)(yi−1−2yi+yi+1 )−cosh(βhi)−1
γ(fi−1/2+fi/2)
=γ
2yi−1−2yi+yi+1 −fi−1/2+fi/2
γ−γ
cosh(βhi)−1(yi−1−2yi+yi+1 ).
Because of Theorem 3.3, and the fact that ε2y00 =f(x, y), x ∈(0,1),we obtain
|F yi| ≤ C1|ri−1−2ri+ri+1|+|si−1−2si+si+1 |+ε2|y00
i−1|
+1
cosh(βhi)−1(|ri−1−2ri+ri+1 |+|si−1−2si+si+1|).
Again, due to Theorem 3.3 and Taylor expansion, the following inequalities hold
|ri−1−2ri+ri+1|=
r00(ξ−
i)
2h2
i+r00(ξ+
i)
2h2
i≤C2h2
i,
|si−1−2si+si+1| ≤ C3
N2,
1
cosh(βhi)−1≤2
(βhi)2=2ε2
γh2
i≤C4,
ε2|y00
i−1| ≤ C5ε2ε−2(e−xi−1
ε√m+e−1−xi−1
ε√m) + r00
i−1≤C61
N2+ε2,
where ξ−
i∈(xi−1, xi)and ξ+
i∈(xi, xi+1).Finally, we have that
|F yi| ≤ C
N2.
Lemma 5.2. Assume that ε≤C
N.In the part of the Shishkin mesh from Section 3, when
xi=xN/4,we have the following estimate
F yN/4≤C
N.
28 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
Proof. Let us estimate
F yN/4
∞,consider F yiin the following form
F yi=γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)
·1 + cosh(βhi−1)
2 sinh(βhi−1)yi−1−1 + cosh(βhi−1)
2 sinh(βhi−1)+1 + cosh(βhi)
2 sinh(βhi)yi
+1 + cosh(βhi)
2 sinh(βhi)yi+1 −cosh(βhi−1)−1
γsinh(βhi−1)fi−1/2−cosh(βhi)−1
γsinh(βhi)fi/2, i =N/4
(5.1)
Let us first estimate the expressions from (5.1) using the nonlinear terms. Due to
Theorem 3.3, and the fact that ε2y00 =f(x, y), x ∈(0,1),we have that
γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)−cosh(βhi−1)−1
γsinh(βhi−1)fi−1/2−cosh(βhi)−1
γsinh(βhi)fi/2
≤C3ε2y00(xN/4)≤C4
N2.(5.2)
For the linear terms from (5.1), we have that
γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)
·1 + cosh(βhi−1)
2 sinh(βhi−1)yi−1−1 + cosh(βhi−1)
2 sinh(βhi−1)+1 + cosh(βhi)
2 sinh(βhi)yi+1 + cosh(βhi)
2 sinh(βhi)yi+1
=γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)1 + cosh(βhi−1)
2 sinh(βhi−1)(yi−1−yi)−1 + cosh(βhi)
2 sinh(βhi)(yi−yi+1 ).
According Theorem 3.3, for the layer component s, we have that
γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)
1 + cosh(βhi−1)
2 sinh(βhi−1)(si−1−si)−1 + cosh(βhi)
2 sinh(βhi)(si−si+1 )
≤C5(|si−1−si|+|si−si+1|)≤C6
N2.(5.3)
For the regular component r, due to cosh x−1
sinh x= tanh x
2and our assumption ε≤1/N,
we get that
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 29
γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)
1 + cosh(βhi−1)
2 sinh(βhi−1)(ri−1−ri)−1 + cosh(βhi)
2 sinh(βhi)(ri−ri+1 )
=γ
tanh βhi−1
2+ tanh β hi
2
·
tanh βhi−1
2
2(ri−1−ri) + tanh β hi
2
2(ri−ri+1) + 2(ri−1−ri)−2(ri−ri+1 )
≤C7|ri−1−ri|+|ri−ri−1|+|ri−1−ri|+|ri−ri+1|
tanh(βhi)
≤C8 εln N
N+1
N+
εln N
N+1
N
tanh(βhi)!≤C
N.
(5.4)
Now, collecting (5.2), (5.3) and (5.4), the statement of the lemma is therefore proven.
Theorem 5.1. The discrete problem (2.2) on the mesh from Section 3is uniformly conver-
gent with respect to εand
max
i|yi−yi| ≤ C
ln2N
N2, i ∈ {0,1, . . . , N/4−1}
1
N2, i ∈ {N/4+1,...,3N/4−1}
1
N, i ∈ {N/4,3N/4}
ln2N
N2, i ∈ {3N/4+1, . . . , N },
where yis the solution of the problem (1.1)–(1.3),yis the corresponding solution of (2.2)
and C > 0is a constant independent of Nand ε.
Proof. We are going to divide the proof of this theorem in four parts.
Suppose first that xi, xi±1∈[0, λ], i = 1, . . . , N/4.The proof for this part of the mesh
has already been done in [11, Theorem 4.2]. It is hold that
(5.5) |F yi| ≤ Cln2N
N2, i = 0,1, . . . , N/4−1.
Now, suppose that xi, xi±i∈[xN/4+1, xN/2−1].Based on Lemma 5.1, we have that
(5.6) |F yi| ≤ C
N2.
In the case i=N/4,now based on Lemma 5.2, we have that
(5.7) F yN/4≤C
N.
30 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
Finally, the proof in the case i=N/2is trivial, because the mesh on this part is
equidistant and the influence of the layer component is negligible. Therefore
(5.8) F yN/2≤C
N2.
Using inequalities (5.5), (5.6), (5.7) and (5.8), we complete the proof of the theorem.
Let us show the ε–uniform convergence of second difference scheme, i.e (2.3).
Lemma 5.3. Assume that ε≤C
N.In the part of the Shishkin mesh from Section 3, when
xi, xi±1∈[xN/4,1/2],we have the following estimate
(5.9) |Gyi| ≤ C
N2, i =N/4, . . . , N/2−1.
Proof. Let us rewrite G˜yiin the following form
G˜yi=γ
2cosh(βhi)−1
sinh(βhi)2(cosh(βhi) + 1)
sinh(βhi)(yi−1−yi)−2(cosh(βhi) + 1)
sinh(βhi)(yi−yi+1 )
−ε2y00
i−1−y00
i+y00
i+1
γ·2(cosh(βhi)−1)
sinh(βhi)
=γ
cosh(βhi)−1[(cosh(βhi)−1)(yi−1−2yi+yi+1)−2(yi−1−2yi+yi+1 )
−ε2(y00
i−1−2y00
i+y00
i+1)·cosh(βhi)−1
γ
=γ(yi−1−2yi+yi+1)−2γ(yi−1−2yi+yi+1 )
cosh(βhi)−1−ε2(y00
i−1−y00
i+y00
i+1).(5.10)
Using Theorem 3.3, Taylor expansion, assumption ε≤1
Nand the properties of the mesh
from Section 3, let us estimate the expressions from (5.10). We get that
|yi−1−2yi+yi+1| ≤ C1(|ri−1−ri+ri+1 |+|si−1−2si+si+1|)
≤C2 r00(ξ+
i) + r00(ξ−
i)
2h2
i+e−xi−1
ε√m!≤C3
N2,(5.11)
1
cosh(βhi)−1≤2
(βhi)2=2ε2
γh2
i≤C3,(5.12)
ε2y00
i−1−y00
i+y00
i+1≤ε2r00
i−1−r00
i+r00
i+1+ε2s00
i−1−s00
i+s00
i+1
≤C4ε2 1 + e−xi−1
ε√m
ε2!≤C5
N2,(5.13)
where ξ−
i∈(xi−1, xi), ξ+
i∈(xi, xi+1).
Now using (5.10), (5.11),(5.12) and (5.13), we obtain (5.9).
Lemma 5.4. Assume that ε≤C
N.In the part of the Shishkin mesh from Section 3, when
xi=xN/4,we have the following estimate
(5.14) GyN/4≤C
N.
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 31
Proof. Using (2.5), let us write Gyiin the following form
Gyi=γ
4di+4di+1
[(4ai+4di+4di+1)(yi−1−yi)−(4ai+1 +4di+4di+1 )(yi−yi+1)]
(5.15)
−(fi−1+ 2fi+fi+1)
=4γ
4di+4di+1
[ai(yi−1−yi)−ai+1(yi−yi+1 )]
+γ(yi−1−2yi+yi+1)−(fi−1+ 2fi+fi+1 ).
In a similar way, as in the previously lemmas, we can get
|yi−1−2yi+yi+1| ≤ |si−1−2si+si+1 |+|ri−1−2ri+ri+1| ≤ C11
N2+1
N,
(5.16)
|fi−1+ 2fi+fi+1| ≤ C2
N2.
(5.17)
Using the identity cosh x−1
sinh = tanh x
2and Theorem 3.3, we have that
4γ
4di+4di+1
[ai(yi−1−yi)−ai+1(yi−yi+1 )]
=4γ
tanh βhi−1
2+ tanh β hi
21
sinh(βhi−1)|si−1−si| − 1
sinh(βhi)|si−si+1 |
+1
sinh(βhi−1)|ri−1−ri| − 1
sinh(βhi)|ri−ri+1 |.(5.18)
Due to Theorem 3.3 and assumption ε≤C
N,hold the next inequalities
γ
tanh βhi−1
2+ tanh β hi
2≤4γ
tanh βhi
2≤C1,(5.19)
1
sinh(βhi−1)|si−1−si| ≤ 1
βhi−1|si−1−si| ≤ C2·1
ln N
N·1
N2=C2
Nln N,(5.20)
1
sinh(βhi)|si−si+1 | ≤ 1
βhi|si−si+1 | ≤ C3
N2,(5.21)
1
sinh(βhi−1)|ri−1−ri| ≤ 1
βhi−1|ri−1−ri| ≤ 1
ln N
N·C4
εln N
N=C4ε,(5.22)
1
sinh(βhi)|ri−ri+1 | ≤ 1
βhi|ri−ri+1 | ≤ C5
N.(5.23)
Now, using (5.15), (5.16), (5.17), (5.18), (5.19), (5.20), (5.21), (5.22) and (5.23), we
obtain (5.14).
32 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
Theorem 5.2. The discrete problem (2.4) on the mesh from Section 3is uniformly conver-
gent with respect to εand
max
i|yi−˜yi| ≤ C
ln2N
N2, i ∈ {0,1, . . . , N/4−1}
1
N2, i ∈ {N/4+1,...,3N/4−1}
1
N, i ∈ {N/4,3N/4}
ln2N
N2, i ∈ {3N/4+1, . . . , N },
where yis the solution of the problem (1.1)–(1.3),˜yis the corresponding solution of (2.6)
and C > 0is a constant independent of Nand ε.
Proof. Again, let us divide the proof on four parts.
Suppose first that xi, xi±1∈[0, λ], i = 1, . . . , N/4.The proof for this part of the mesh has
already been done in [5, Theorem 4.4]. It is proved that
(5.24) |Gyi| ≤ Cln2N
N2, i = 0,1, . . . , N/4−1.
Secondly, suppose that xi, xi±1∈[xN/4+1 , xN/2−1].Due to Lemma 5.3, we have that
(5.25) |Gyi| ≤ C
N2.
In the case i=N/4,based on Lemma 5.4, we have the following estimate
(5.26) GyN/4≤C
N.
At the end, in the case i=N/2,the proof is trivial, because of the properties of the mesh
and the layer component. Hence, it is true that
(5.27) |Gyi| ≤ C
N2.
Using (5.24), (5.25), (5.26) and (5.27), we complete the statement of the theorem.
6. NUMERICAL EXPER IMENTS
In this section we present numerical results to confirm the uniform accuracy of the
discrete problems (2.2) and (2.6). Both discrete problems will be checked on two dif-
ferent examples. First one is the linear boundary value problem, whose exact solution is
known. Second example is the nonlinear boundary value problem whose exact solution
is unknown.
For the problems from our examples whose exact solution is known, we calculate EN
as
EN= max
0≤i≤Ny(xi)−yN(xi)or EN= max
0≤i≤Ny(xi)−˜yN(xi),
for the problems, whose exact solution is unknown, we calculate EN, as
(6.1) EN= max
0≤i≤Ny2N
S(xi)−yN(xi)or EN= max
0≤i≤N˜y2N
S(xi)−˜yN(xi),
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 33
the rate of convergence Ord we calculate in the usual way
Ord = ln EN−ln E2N
ln 2k
k+1
where N= 2k, k = 6,7,...,11, yN(xi),˜yN(xi)are the values of the numerical solutions
on a mesh with N+ 1 mesh points, and y2N
S(xi),˜y2N
S(xi)are the values of the numerical
solutions on a mesh with 2N+ 1 mesh points and the transition points altered slightly to
λS= min n2
4,2ε
√mln N
2o.
Remark 6.1. In a case when the exact solution is unknown we use the double mesh method,
see [2, 16, 17] for details.
Example 1. Consider the following problem
2y00 =y+ 1 −2ε2+x(x−1) for x∈(0,1), y(0) = y(1) = 0.
The exact solution of this problem is given by y(x) = e−x
+e−1−x
1 + e−1
−x(x−1) −1.The
nonlinear system was solved using the initial condition y0=−0.5and the value of the
constant γ= 1.
N EnOrd EnOrd EnOrd
268.1585e−04 2.00 2.8932e−03 2.02 2.5827e−02 2.05
272.7762e−04 2.00 9.7397e−04 2.01 8.5547e−03 1.96
289.0650e−05 2.00 3.1625e−04 2.00 2.8566e−03 1.99
293.5410e−05 2.00 1.2353e−04 2.00 1.2111e−03 2.00
210 1.5738e−05 2.00 5.4904e−05 2.00 4.9827e−04 2.00
211 7.7116e−06 −2.6903e−05 −2.4415e−04 −
ε2−32−52−10
N EnOrd EnOrd EnOrd
263.9901e−02 2.04 3.9901e−02 2.04 3.9901e−02 2.04
271.3288e−02 1.93 1.3288e−02 1.93 1.3288e−02 1.93
284.5122e−03 1.99 4.5122e−03 1.99 4.5122e−03 1.99
291.7709e−03 1.98 1.7709e−03 1.98 1.7709e−03 1.98
210 7.9347e−04 1.98 7.9347e−04 1.98 7.9347e−04 1.98
211 3.9158e−04 −3.9158e−04 −3.9158e−04 −
ε2−15 2−25 2−30
N EnOrd EnOrd EnOrd
263.9901e−02 2.04 4.0243e−02 2.02 4.0248e−02 2.02
271.3288e−02 1.93 1.3581e−02 1.92 1.3582e−02 1.92
284.5122e−03 1.99 4.6375e−03 1.97 4.6381e−03 1.97
291.7709e−03 1.98 1.8372e−03 1.98 1.8375e−03 1.98
210 1.7709e−03 1.98 8.2321e−04 1.98 8.2331e−04 1.98
211 3.9158e−04 −4.0626e−04 −4.0631e−04 −
ε2−35 2−40 2−45
TABLE 1. Errors ENand convergence rates Ord for approximate solu-
tions from Example 1.
Example 2. Consider the following problem
ε2y00 =y3+y−2for (0,1), y(0) = y(1) = 0,(6.2)
whose exact solution is unknown. The nonlinear system was solved using the initial
condition y0= 1,that represents the reduced solution. The value of the constant γ= 4
has been chosen so that the condition γ≥fy(x, y),∀(x, y)∈[0,1] ×[yL, yU]⊂[0,1] ×R
is fulfilled, where yLand yUare lower and upper solutions, respectively, of the problem
34 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
(6.2). Because of the fact that the exact solution is unknown, we are going to calculate
Enusing (6.1).
N EnOrd EnOrd EnOrd
267.1345e−04 2.02 3.7134e−03 2.01 1.5182e−02 2.09
272.4017e−04 2.01 1.2564e−04 2.01 4.9236e−03 1.96
287.7985e−05 2.00 3.1655e−04 2.00 1.6403e−03 2.09
293.0463e−05 2.00 1.2959e−04 2.00 5.1903e−04 2.00
210 1.3539e−05 2.00 3.9986e−05 2.00 1.6001e−04 2.00
211 6.6341e−06 −1.2096e−05 −4.8389e−05 −
ε2−32−52−10
N EnOrd EnOrd EnOrd
261.5181e−02 2.09 1.5181e−02 2.09 1.5181e−02 2.09
274.9236e−03 1.96 4.9236e−03 1.96 4.9236e−03 1.96
281.6403e−03 2.00 1.6403e−03 2.00 1.6403e−03 2.00
295.1903e−04 2.00 5.1903e−04 2.00 5.1903e−04 2.00
210 1.6001e−04 2.00 1.6001e−04 2.00 1.6001e−04 2.00
211 4.8389e−05 −4.8389e−05 −4.8389e−05 −
ε2−15 2−25 2−30
N EnOrd EnOrd EnOrd
261.5181e−02 2.09 1.5184e−02 2.09 1.5795e−02 2.09
274.9236e−03 1.96 4.9221e−03 1.96 5.1202e−03 1.96
281.6403e−03 2.00 1.6436e−03 1.99 1.7097e−03 1.99
295.1903e−04 2.00 6.4509e−04 2.00 6.7102e−04 2.00
210 1.6002e−04 2.00 2.8669e−04 2.00 2.9823e−04 2.00
211 4.8390e−05 −1.4048e−04 −1.4613e−04 −
ε2−35 2−40 2−45
TABLE 2. Errors ENand convergence rates Ord for approximate solu-
tions from Example 2.
Example 3. Consider the following problem
2y00 =y+ 1 −2ε2+x(x−1) for x∈(0,1), y(0) = y(1) = 0.
The exact solution of this problem is given by y(x) = e−x
+e−1−x
1 + e−1
−x(x−1) −1.The
nonlinear system was solved using the initial condition y0=−0.5and the value of the
constant γ= 1.
Example 4. Consider the following problem
ε2y00 =y3+y−2for (0,1), y(0) = y(1) = 0,
whose exact solution is unknown. The nonlinear system was solved using the initial
condition y0= 1,that represents the reduced solution. The value of the constant γ= 4
has been chosen so that the condition γ≥fy(x, y),∀(x, y)∈[0,1] ×[yL, yU]⊂[0,1] ×R
is fulfilled, where yLand yUare lower and upper solutions, respectively, of the problem
(6.2). Because of the fact that the exact solution is unknown, we are going to calculate
Enusing (6.1).
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 35
N EnOrd EnOrd EnOrd
269.0262e−04 2.05 4.4799e−03 2.03 3.9479e−02 2.01
272.8729e−04 1.91 1.4999e−03 1.92 1.3362e−03 1.93
289.8102e−05 1.95 5.1221e−04 1.95 4.5373e−03 1.96
293.9049e−05 1.99 2.0484e−04 1.99 1.8060e−03 1.97
210 1.7496e−05 1.99 9.1409e−05 1.99 8.1249e−04 1.97
211 8.6345e−06 −4.4951e−05 −4.0241e−04 −
ε2−32−52−10
N EnOrd EnOrd EnOrd
263.9479e−02 2.01 3.9479e−02 2.01 3.9479e−02 2.01
271.3362e−03 1.93 1.3362e−03 1.93 1.3362e−03 1.93
284.5373e−03 1.96 4.5373e−03 1.96 4.5373e−03 1.96
291.8060e−03 1.97 1.8060e−03 1.97 1.8060e−03 1.97
210 8.1249e−04 1.97 8.1249e−04 1.97 8.1249e−04 1.97
211 4.0241e−04 −4.0241e−04 −4.0241e−04 −
ε2−15 2−25 2−30
N EnOrd EnOrd EnOrd
263.9479e−02 2.01 3.9483e−02 2.01 3.9485e−02 2.01
271.3362e−03 1.93 1.3363e−03 1.93 1.3364e−03 1.93
284.5373e−03 1.96 4.5377e−03 1.95 4.5378e−03 1.95
291.8060e−03 1.97 1.8147e−03 1.97 1.8180e−03 1.97
210 8.1249e−04 1.97 8.1641e−04 1.97 8.1645e−04 1.97
211 4.0241e−04 −4.0434e−04 −4.0436e−04 −
ε2−35 2−40 2−45
TABLE 3. Errors ENand convergence rates Ord for approximate solu-
tions from Example 3.
N EnOrd EnOrd EnOrd
268.8623e−04 2.09 3.4567e−03 2.11 1.1656e−02 2.10
272.8728e−05 1.92 1.1085e−03 1.93 3.7537e−03 1.91
289.8102e−05 1.96 3.7643e−04 1.95 1.2923e−03 1.98
293.9049e−05 1.98 1.5054e−04 1.98 4.1404e−04 1.99
210 1.7496e−05 1.98 6.7451e−05 1.99 1.2855e−04 2.00
211 8.6345e−06 −3.3169e−05 −3.8914e−05 −
ε2−32−52−10
N EnOrd EnOrd EnOrd
261.1656e−02 2.10 1.1656e−02 2.10 1.1656e−02 2.10
273.7537e−03 1.91 3.7537e−03 1.91 3.7537e−03 1.91
281.2923e−03 1.98 1.2923e−03 1.98 1.2923e−03 1.98
294.1404e−04 1.99 4.1404e−04 1.99 4.1404e−04 1.99
210 1.2855e−04 2.00 1.2855e−04 2.00 1.2855e−04 2.00
211 3.8914e−05 −3.8914e−05 −3.8914e−05 −
ε2−15 2−25 2−30
N EnOrd EnOrd EnOrd
261.1656e−02 2.10 1.1656e−02 2.10 1.1656e−02 2.10
273.7537e−03 1.91 3.7537e−03 1.91 3.7537e−03 1.91
281.2923e−03 1.98 1.2923e−03 1.98 1.2923e−03 1.98
294.1404e−04 1.99 4.1404e−04 1.99 4.1404e−04 1.99
210 1.2855e−04 2.00 1.2855e−04 2.00 1.2855e−04 2.00
211 3.8914e−05 −3.8914e−05 −3.8914e−05 −
ε2−35 2−40 2−45
TABLE 4. Errors ENand convergence rates Ord for approximate solu-
tions from Example 4.
7. CONCLUSION
In this paper we proved ε–uniform convergence on two differential schemes. In the
parts of the domain where lies boundary layers we showed ε–convergence of the order
O(ln2N−2),while out of the layers the order of the ε–convergence are of order O(N−2)
and in the transition points λand 1−λare of order O(N−1).At the end of the paper we
perform numerical experiments which agrees with theoretical results.
36 S. KARASULJI´
C, E. DUVNJAKOVI´
C, AND E. MEMI´
C
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
numerical solution ǫ=2-3
exact solution= ǫ=2-3
(A)ε= 2−3, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
numerical solution ǫ=2-5
exact solution= ǫ=2-5
(B)ε= 2−5, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
numerical solution ǫ=2-7
exact solution= ǫ=2-7
(C)ε= 2−7, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
numerical solution ǫ=2-9
exact solution= ǫ=2-9
(D)ε= 2−9, N = 32
FIGURE 1. Graphics of the numerical and exact solutions for N= 32 and
ε= 2−3,2−5,2−7,2−9for Example 1 and Example 3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
numerical solution ǫ=2-3, N=32
numerical solution= ǫ=2 -3, N=64
(A)ε= 2−3, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
numerical solution ǫ=2-5, N=32
numerical solution= ǫ=2 -5, N=64
(B)ε= 2−5, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
numerical solution ǫ=2-7, N=32
numerical solution= ǫ=2 -7, N=64
(C)ε= 2−7, N = 32
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
0
0.2
0.4
0.6
0.8
1
numerical solution ǫ=2-9, N=32
numerical solution= ǫ=2 -9, N=64
(D)ε= 2−9, N = 32
FIGURE 2. Graphics of the numerical and solutions for N= 32,64 and
ε= 2−3,2−5,2−7,2−9for Example 2 and Example 4
8. ACKNOWLEDGM ENTS
This paper is the part of Project "Numeriˇ
cko rješavanje kvazilinearnog singularno–
perturbacionog jednodimenzionalnog rubnog problema". The paper has emanated from
UNIFORMLY CONVERGENT DIFFERENCE SCHEME . . . 37
research conducted with the partial financial support of Ministry of education and sci-
ences of Federation of Bosnia and Herzegovina and University of Tuzla under grant 01/2-
3995-V/17 of 18.12.2017.
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38 S. KARASULJI´
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C
DEPART ME NT O F MATHE MATI CS
FACULTY OF S CI EN CE S AND MATHEMATICS ,
UNIVER SI TY O F TUZ LA, UNIVERZ IT ET SKA 4, 75000 TUZLA,, BOSN IA A ND HE RZ EGOVI NA
E-mail address:
DEPART ME NT O F MATHE MATI CS
FACULTY OF S CI EN CE S AND MATHEMATICS ,
UNIVER SI TY O F TUZ LA, UNIVERZ IT ET SKA 4, 75000 TUZLA,, BOSN IA A ND HE RZ EGOVI NA
E-mail address:
DEPART ME NT O F MATHE MATI CS
FACULTY OF S CI EN CE S AND MATHEMATICS ,
UNIVER SI TY O F TUZ LA, UNIVERZ IT ET SKA 4, 75000 TUZLA,, BOSN IA A ND HE RZ EGOVI NA
E-mail address:
