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arXiv:1712.01313v1 [math.NA] 4 Dec 2017
Uniformly Convergent Difference Scheme for a Semilinear
Reaction-Diffusion Problem on a Shishikin 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
Shiskin mesh. Finally, we present four numerical experiments to confirm the theoretical results.
Key words: singularly perturbed, boundary value problem, numerical solution, difference scheme, nonlinear,
Shishkin mesh, layer–adapted mesh, ε–uniform convergent.
2010 Mathematics Subject Classification. 65L10, 65L11, 65L50.
1 Introduction
We consider the semilinear singularly perturbed problem
ε2y′′(x) = f(x, y) on (0,1) ,(1)
y(0) = 0, y(1) = 0,(2)
where εis a small positive parameter. We assume that the nonlinear function fis continuously differentiable,
i.e. for k≥2, f ∈Ck([0,1] ×R),and that it has a strictly positive derivative with respect to y
∂f
∂y =fy≥m > 0 on [0,1] ×R(m=const).(3)
The boundary value problem (1)–(2), under the condition (3), has a unique solution (see [15]). Numerical
treatment of the problem (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)–(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 [3,4,6–9,11,14] and performed numerical tests, in [5,12] 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 [13] we used
the difference schemes from [12] and calculated the values of the approximate solutions of the problem (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)–(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 [16], which is piecewise equidistant and consequently
simpler than the modified Shishkin mesh we have already used in our mentioned papers.
2 Difference schemes
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 =△di
γfi−1/2+△di+1
γfi/2,(4)
where di=β
tanh(βhi−1), ai=β
sinh(βhi−1),fi−1/2=fxi−1+xi
2,yi−1+yi
2and △di=di−ai.
From (4), we obtain next discrete problem
F y = (F y0, F y1,...,FyN)T,(5)
1corresponding author
1
where
Fy0=y0= 0,
Fyi=γ
△di+△di+1 ai+di
2yi−1−ai+di
2+ai+1 +di+1
2yi
+ai+1 +di+1
2yi+1 −△di
γfi−1/2−△di+1
γfi/2= 0, i = 1,2,...,N −1,(6)
FyN=yN= 0,
and y:= (y0, y1,...,yN)Tis the solution of the problem
F y = 0.(7)
Second difference scheme has the following form
(3ai+di+△di+1) ( ˜yi−1−˜yi)−(3ai+1 +di+1 +△di) ( ˜yi−˜yi+1 ) = ˜
fi−1+ 2 ˜
fi+˜
fi+1
γ(△di+△di+1),(8)
where di=β
tanh(βhi−1), ai=β
sinh(βhi−1),˜
fi=f(xi,˜yi) and △di=di−ai.
From (8), we obtain second discrete problem
G˜y= (G˜y0, G˜y1,...,G˜yN)T,(9)
where
G˜y0= ˜y0= 0 (10)
G˜yi=γ
△di+△di+1 (3ai+di+△di+1) ( ˜yi−1−˜yi)−(3ai+1 +di+1 +△di) (˜yi−˜yi+1) (11)
−˜
fi−1+ 2 ˜
fi+˜
fi+1
γ(△di+△di+1)#= 0, i = 1,...,N −1,(12)
G˜yN= ˜yN= 0,(13)
and ˜y= (˜y0,˜y1,...,˜yn)Tis the solution of the problem
G˜y= 0.(14)
3 Theoretical background
In this paper we use the maximum norm
kuk∞= max
06i6N|ui|,(15)
for any vector u= (u0, u1,...,un)T∈RN+1 and the corresponding matrix norm.
The next two theorems hold
Theorem 3.1. [12]The discrete problem (7)for γ≥fy,has the unique solution
y= (y0, y1, y2, . . . , y N−1, yN)T,with y0=yN= 0.Moreover, the following stability inequality holds
kw−vk∞61
mkF w −F vk∞,(16)
for any vectors v= (v0, v1,...,vN)T∈RN+1, w = (w0, w1,...,wN)T∈RN+1 .
Theorem 3.2. [5]The discrete problem (14)has a unique solution ˜yfor γ>fy. Also, for every u, v ∈RN+1
we have the following stabilizing inequality
ku−vk∞61
mkGu −Gvk∞.
In the following analysis we need the decomposition of the solution yof the problem (1)−(2) to the layer
component sand a regular component r, given in the following assertion.
2
Theorem 3.3. [19]The solution yto problem (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)6C, (17)
and s(j)(x)6Cε−je−x
ε√m+e−1−x
ε√m.(18)
4 Construction of the mesh
The solution of the problem (1)–(3) changes fast near the ends of our domain [0,1].Therefore, the mesh has
to be refined there. A Shishin mesh is used to resolve the layers. This mesh is piecewise equidistant and it’s
quite simple. It is constructed as follows (see [17]). 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, (19)
hi−1=2(1 −2λ)
Nfor i=i0+ 1,...,N −i0.(20)
If λ=1
4i.e. 1
462εln 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.(21)
From (19) and (20), we conclude that that the interval lengths satisfy
hi−1=8εln N
√mfor i= 1,...,i0, N −i0,...,N, (22)
and 1
N6hi−162
Nfor i=i0+ 1,...,N −i0.(23)
5 Uniform convergence
We will prove the theorem on uniform convergence of the difference schemes (4) and (8) 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√mappear. For these
functions we have that e−x
ε√m>e−1−x
ε√m,∀x∈[0,1/2] and e−x
ε√m6e−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−16hi,and in the second case hi−1>hi.
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 3which corresponds to xN/4−1,1/2and xN/4=λ.
3
Lemma 5.1. Assume that ε6C
N.In the part of the Shiskin mesh from Section 3, when xi, xi±1∈[xN/4,1/2],
we have the following estimate
|F yi|6C
N2, i =N/4,...,N/2−1.(24)
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 ε2y′′ =f(x, y), x ∈(0,1),we obtain
|F yi|6C1|ri−1−2ri+ri+1|+|si−1−2si+si+1|+ε2|y′′
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|=
r′′(ξ−
i)
2h2
i+r′′(ξ+
i)
2h2
i
6C2h2
i,
|si−1−2si+si+1|6C3
N2,
1
cosh(βhi)−162
(βhi)2=2ε2
γh2
i
6C4,
ε2|y′′
i−1|6C5ε2ε−2(e−xi−1
ε√m+e−1−xi−1
ε√m) + r′′
i−16C61
N2+ε2,
where ξ−
i∈(xi−1, xi) and ξ+
i∈(xi, xi+1).Finally, we have that
|F yi|6C
N2.(25)
Lemma 5.2. Assume that ε6C
N.In the part of the Shiskin mesh from Section 3, when xi=xN/4,we have
the following estimate
F yN/46C
N.(26)
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 (27)
Let us first estimate the expressions from (27) using the nonlinear terms. Due to Theorem 3.3, and the fact
that ε2y′′ =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
6C3ε2y′′(xN/4)6C4
N2.(28)
For the linear terms from (27), we have that
4
γ
cosh(βhi−1)−1
sinh(βhi−1)+cosh(βhi)−1
sinh(βhi)
(29)
·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(30)
=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 ).(31)
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)
6C5(|si−1−si|+|si−si+1|)6C6
N2.(32)
For the regular component r, due to cosh x−1
sinh x= tanh x
2and our assumption ε61/N, we get that
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 βhi
2
2(ri−ri+1) + 2(ri−1−ri)−2(ri−ri+1 )
6C7|ri−1−ri|+|ri−ri−1|+|ri−1−ri|+|ri−ri+1|
tanh(βhi)
6C8 εln N
N+1
N+
εln N
N+1
N
tanh(βhi)!6C
N.(33)
Now, collecting (28), (32) and (33), the statement of the lemma is therefore proven.
Theorem 5.1. The discrete problem (7)on the mesh from Section 3is uniformly convergent with respect to ε
and
max
i|yi−yi|6C
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},
(34)
where yis the solution of the problem (1)–(3),yis the corresponding solution of (7)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 [12, Theorem 4.2]. It is hold that
|F yi|6Cln2N
N2, i = 0,1,...,N/4−1.(35)
Now, suppose that xi, xi±i∈[xN/4+1, xN/2−1].Based on Lemma 5.1, we have that
|F yi|6C
N2.(36)
In the case i=N/4,now based on Lemma 5.2, we have that
5
F yN/46C
N.(37)
Finally, the proof in the case i=N/2 is trivial, because the mesh on this part is equidistant and the influence
of the layer component is negligible. Therefore
F yN/26C
N2.(38)
Using inequalities (35), (36), (37) and (38), we complete the proof of the theorem.
Let us show the ε–uniform convergence of second difference scheme, i.e (8).
Lemma 5.3. Assume that ε6C
N.In the part of the Shiskin mesh from Section 3, when xi, xi±1∈[xN/4,1/2],
we have the following estimate
|Gyi|6C
N2, i =N/4,...,N/2−1.(39)
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 )
−ε2y′′
i−1−y′′
i+y′′
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(y′′
i−1−2y′′
i+y′′
i+1)·cosh(βhi)−1
γ
=γ(yi−1−2yi+yi+1)−2γ(yi−1−2yi+yi+1)
cosh(βhi)−1−ε2(y′′
i−1−y′′
i+y′′
i+1).(40)
Using Theorem 3.3, Taylor expansion, assumption ε61
Nand the properties of the mesh from Section 3, let us
estimate the expressions from (40). We get that
|yi−1−2yi+yi+1|6C1(|ri−1−ri+ri+1|+|si−1−2si+si+1 |)
6C2 r′′(ξ+
i) + r′′(ξ−
i)
2h2
i+e−xi−1
ε√m!6C3
N2,(41)
1
cosh(βhi)−162
(βhi)2=2ε2
γh2
i
6C3,(42)
ε2y′′
i−1−y′′
i+y′′
i+16ε2r′′
i−1−r′′
i+r′′
i+1+ε2s′′
i−1−s′′
i+s′′
i+1
6C4ε2 1 + e−xi−1
ε√m
ε2!6C5
N2,(43)
where ξ−
i∈(xi−1, xi), ξ+
i∈(xi, xi+1).
Now using (40), (41),(42) and (43), we obtain (39).
Lemma 5.4. Assume that ε6C
N.In the part of the Shiskin mesh from Section 3, when xi=xN/4,we have
the following estimate
GyN/46C
N.(44)
Proof. Using (12), let us write Gyiin the following form
Gyi=γ
△di+△di+1
[(4ai+△di+△di+1)(yi−1−yi)−(4ai+1 +△di+△di+1)(yi−yi+1 )]
−(fi−1+ 2fi+fi+1)
=4γ
△di+△di+1
[ai(yi−1−yi)−ai+1(yi−yi+1)] + γ(yi−1−2yi+yi+1)−(fi−1+ 2fi+fi+1).(45)
6
In a similar way, as in the previously lemmas, we can get
|yi−1−2yi+yi+1|6|si−1−2si+si+1|+|ri−1−2ri+ri+1 |6C11
N2+1
N,(46)
|fi−1+ 2fi+fi+1|6C2
N2.(47)
Using the identity c osh x−1
sinh = tanh x
2and Theorem 3.3, we have that
4γ
△di+△di+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 |.(48)
Due to Theorem 3.3 and assumption ε6C
N,hold the next inequalities
γ
tanh βhi−1
2+ tanh β hi
2
64γ
tanh βhi
2
6C1,(49)
1
sinh(βhi−1)|si−1−si|61
βhi−1|si−1−si|6C2·1
ln N
N·1
N2=C2
Nln N,(50)
1
sinh(βhi)|si−si+1 |61
βhi|si−si+1 |6C3
N2,(51)
1
sinh(βhi−1)|ri−1−ri|61
βhi−1|ri−1−ri|61
ln N
N·C4
εln N
N=C4ε, (52)
1
sinh(βhi)|ri−ri+1 |61
βhi|ri−ri+1 |6C5
N.(53)
Now, using (45), (46), (47), (48), (49), (50), (51), (52) and (53), we obtain (44).
Theorem 5.2. The discrete problem (9)on the mesh from Section 3is uniformly convergent with respect to ε
and
max
i|yi−˜yi|6C
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},
(54)
where yis the solution of the problem (1)–(3),˜yis the corresponding solution of (14)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
|Gyi|6Cln2N
N2, i = 0,1,...,N/4−1.(55)
Secondly, suppose that xi, xi±1∈[xN/4+1, xN/2−1].Due to Lemma 5.3, we have that
|Gyi|6C
N2.(56)
In the case i=N/4,based on Lemma 5.4, we have the following estimate
GyN/46C
N.(57)
7
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
|Gyi|6C
N2.(58)
Using (55), (56), (57) and (58), we complete the statement of the theorem.
6 Numerical experiments
In this section we present numerical results to confirm the uniform accuracy of the discrete problems (7) and
(14). Both discrete problems will be checked on two different 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 ENas
EN= max
06i6Ny(xi)−yN(xi)or EN= max
06i6Ny(xi)−˜yN(xi),(59)
for the problems, whose exact solution is unknown, we calculate EN, as
EN= max
06i6Ny2N
S(xi)−yN(xi)or EN= max
06i6N˜y2N
S(xi)−˜yN(xi),(60)
the rate of convergence Ord we calculate in the usual way
Ord = ln EN−ln E2N
ln 2k
k+1
(61)
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 [2,17,18] for
details.
Example 6.1. Consider the following problem
ǫ2y′′ =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
ǫ+e−1−x
ǫ
1 + e−1
ǫ−x(x−1) −1.The nonlinear system was
solved using the initial condition y0=−0.5 and the value of the constant γ= 1.
Example 6.2. Consider the following problem
ε2y′′ =y3+y−2 for (0,1), y(0) = y(1) = 0,(62)
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] ×Ris fulfilled, where yLand yUare lower and upper solutions,
respectively, of the problem (62). Because of the fact that the exact solution is unknown, we are going to
calculate Enusing (60).
Example 6.3. Consider the following problem
ǫ2y′′ =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
ǫ+e−1−x
ǫ
1 + e−1
ǫ−x(x−1) −1.The nonlinear system was
solved using the initial condition y0=−0.5 and the value of the constant γ= 1.
8
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 solutions from Example 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 solutions from Example 6.2.
Example 6.4. Consider the following problem
ε2y′′ =y3+y−2 for (0,1), y(0) = y(1) = 0,(63)
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] ×Ris fulfilled, where yLand yUare lower and upper solutions,
respectively, of the problem (62). Because of the fact that the exact solution is unknown, we are going to
calculate Enusing (60).
9
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 solutions from Example 6.3.
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 solutions from Example 6.4.
10
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
6.1 and Example 6.3
11
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 6.2
and Example 6.4
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Department of Mathematics
Faculty of sciences and Mathematics,
University of Tuzla
Univerzitetska 4, 75000 Tuzla,
Bosnia and Herzegovina
E-mail address: samir.karasuljic@untz.ba
Department of Mathematics
Faculty of sciences and Mathematics,
University of Tuzla
Univerzitetska 4, 75000 Tuzla,
Bosnia and Herzegovina
E-mail address: enes.duvnjakovic@untz.ba
Department of Mathematics
Faculty of sciences and Mathematics,
University of Tuzla
Univerzitetska 4, 75000 Tuzla,
Bosnia and Herzegovina
E-mail address: memic 91 elvir@hotmail.com
13
