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139
Novi Sad J. Math.
Vol. 33, No. 2, 2003, 139–161
HIGH-ORDER METHODS FOR SEMILINEAR
SINGULARLY PERTURBED BOUNDARY VALUE
PROBLEMS1
Ivana Radeka2, Dragoslav Herceg2
Abstract. We considered finite difference methods of higher order for
semilinear singularly perturbed boundary value problems, consisted of
constructing difference schemes on nonuniform meshes. Construction of
schemes is presented and convergence uniform in perturbation parameter
for one method is shown on Bakhvalov’s type of mesh. Numerical exper-
iments demonstrated influence of different meshes on developed schemes.
AMS Mathematics Subject Classification (2000): 65L10
Key words and phrases: finite difference method, singularly perturbed
boundary value problem, high order, nonuniform mesh
1. Introduction
Our aim in this paper is to construct the difference schemes that have higher
order of convergence uniform in small parameter εcombined with appropriate
nonuniform mesh. In the paper [1] the idea was developed by Clavero, Gracia
and Lisbona for the linear problem on the Shishkin mesh. By including more
coefficients in the schemes we get one degree of freedom for their determination
when we obtain the expected order of convergence that will be used when trying
to provide stability and consistency of the method. We generalized the method
for the semilinear problem of the form:
Tεu=−εu00(x) + a(x)u0(x) + b(x, u(x)) = 0, x ∈(0,1),
Ru = (u(0), u(1)) = (0,0),
(1)
where 0 < ε << 1, a and bare the functions satisfying the following conditions
a(x)≥α > 0, x ∈(0,1)
0≤bu(x, u)≤G(x),(x, u)∈(0,1) ×R
(2)
a∈Ck([0,1]), b ∈Ck([0,1] ×R), k ∈N.(3)
1This paper was supported by the Ministry of Science, Technology and Development of
the Republic of Serbia under grant No 1840.
2Department of Mathematics and Informatics, Faculty of Science and Mathematics, Uni-
versity of Novi Sad, Trg D. Obradovi´ca 4, 21000 Novi Sad, Serbia and Montenegro
140 I. Radeka, D. Herceg
The condition (2) is the standard stability condition, which implies that both
(1) and the reduced problem, a(x)u0(x) + b(x, u(x)) = 0,have unique smooth
solutions uεand u0, respectively, and the conditions (2, 3) provides us with
useful bounds of the solution uεand its derivatives [16]:
¯¯¯u(i)
ε(x)¯¯¯≤M(1 + ε−iexp(−α1−x
ε)), x ∈(0,1),(4)
for all i∈ {0,1,2, ..., k + 1}.
In the following sections we develop the schemes of order two and three com-
bined with Bakhvalov’s type of meshes and for these methods prove convergence,
uniform in small parameter, with some restrictions that will be emphasized.
Numerical results in the last section confirmed the theoretical findings. The
accuracy obtained on Bakhvalov’s type of meshes is better than on Shishkin’s
mesh.
2. Difference schemes
For x∈(0,1), we define the operator
Th
εw(x) =
n
X
k=1
rkw(x+dkh) +
N1
X
k=n1
qkb(x+d0
kh, w(x+d0
kh)),(5)
where dk, k = 1,2, ..., N and d0
k, k =n1, ..., N1,1≤n1≤N1,are real numbers
such that x+dkh, x +d0
kh∈(0,1). The coefficients dk,and d0
kdiffer from each
other. The unknown coefficients rkand qkare determined so that Th
εw(x) = 0
for all w∈Ps[x],(the space of polynomials of degree not greater than s),
including the normalization condition PN1
k=n1qk= 1.
Let s≥2.Depending on the base of the polynomial space Ps[x],we get the
system of equations for the determination of the parameters rkand qk.If we
choose the base {1, x, x2, ..., xs}we get the following system of linear equations:
N
X
k=1
rk= 0
N
X
k=1
rk(x+dkh)−
N1
X
k=n1
qka(x+d0
kh) = 0
N
X
k=1
rk(x+dkh)2+
N1
X
k=n1
qk(2ε−2(x+d0
kh)a(x+d0
kh)) = 0(6)
....
N
X
k=1
rk(x+dkh)s+
N1
X
k=n1
qk(εs(s−1)(x+d0
kh)s−2−s(x+d0
kh)s−1a(x+d0
kh)) = 0
High-order methods for semilinear singularly . . . 141
N1
X
k=n1
qk= 1.
2.1. Scheme 2
For s= 2, N = 3, n1= 1, N1= 2 and d1=d0
1=−hi
h, d2=d0
2=
0, d3=hi+1
h,the system (6), expressed in the matrix form, using x=xiand
the notation a(xi) = ai, rj=rj(i) and qj=qj(i),as follows:
1 1 1 0 0
−hi0hi+1 −ai−1−ai
h2
i0h2
i+1 2hiai−10
0 0 0 1 1
r1(i)
r2(i)
r3(i)
q1(i)
q2(i)
=
0
0
−2ε
1
.(7)
The rank of the coefficient matrix is 4, so the system has one degree of freedom.
The value q1(i) will be chosen freely.
The solution of the system is
r1(i) = −2ε−q1(i)(2hi+hi+1)ai−1−q2(i)hi+1ai
hi(hi+hi+1),
r3(i) = −2ε+hiai−q1(i)hi(ai−1+ai)
hi+1(hi+hi+1 ),
r2(i) = −r1(i)−r2(i),
q2(i) = 1 −q1(i).
Because of the boundary conditions, it holds that w0=wn= 0,so we will
observe the discrete problem for wh:= (w1, w2, ..., wn−1)T∈Rn−1,using the
nodes xi, i = 0,1, ..., n :
F1wh:= r2(1)w1+r3(1)w2+q1(1)b(x0, w0) + q2(1)b(x1, w1)
Fiwh:= r1(i)wi−1+r2(i)wi+r3(i)wi+1 +q1(i)b(xi−1, wi−1)
+q2(i)b(xi, wi)i= 2,3, ..., n −2
Fn−1wh:= r1(n−1)wn−2+r2(n−1)wn−1
+q1(n−1)b(xi−1, wi−1) + q2(n−1)b(xn−1, wn−1).
(8)
The Jacobian matrix of the mapping F= (F1, F2, ..., Fn−1) is a tridiagonal
142 I. Radeka, D. Herceg
matrix of the form F0(wh) = tridiag{Ai
1, Ai
2, Ai
3},where for i= 1,2, ..., n −1,
Ai
1=−2ε−q1(i)(2hi+hi+1)ai−1−q2(i)hi+1 ai
hi(hi+hi+1)+q1(i)bu(xi−1, wi−1),
Ai
2=−(r1(i) + r2(i)) + q2(i)bu(xi, wi),
Ai
3=−2ε+hiai−q1(i)hi(ai−1+ai)
hi+1(hi+hi+1 ).
In order to show the stability of the method we will determine the coefficient
q1(i) so that the matrix F0(wh) becomes an M−matrix. We will prove the
following theorem:
Theorem 2.1. Let n0∈N, so that
Mka0k∞
n0
< α, 3M(ka0k∞+kGk∞)
n0
< α(9)
and the mesh Ih={xi;i= 0,1, ..., n}has the property
hi≤M
n0
, i = 1,2, ..., n.(10)
If we choose q1(i)for all i= 1,2, ...n −1,so that
0≤q1(i)≤1,(11)
and for ifor which stands that −2ε+hiai≥0,
q1(i) = ai
ai+ai−1
,(12)
then for all n≥n0the matrix F0(wh)is an M−matrix.
Proof. Using (11), it follows that 0 ≤q2(i)≤1, i = 1,2, ..., n. From the
conditions (9), (2) and (10) we have
−(2hi+hi+1)ai−1+hi(hi+hi+1 )bu(xi−1, wi−1)
≤hi
3M
n0
(ka0k∞+kGk∞)−hiai−1< hi(α−ai−1)<0,
so the coefficients Ai
1<0,for all i= 1,2, ..., n −1.Let i∈ {1,2, ..., n −1},
then for −2ε+hiai<0,we get Ai
3<0,and if the mentioned condition is not
satisfied, because of (12) we have
Ai
3=−2ε
hi+1(hi+hi+1 )<0.
High-order methods for semilinear singularly . . . 143
Then r1(i), r2(i)<0, i = 1,2, ..., n −1,and because of (2) it is true that
Ai
2>0,for all i= 1,2, ..., n −1.Hence, F0(wh) is an L−matrix.
If we introduce the vector v= (x1, x2, ..., xn−1)T,where xi∈Ih, i =
1,2, ..., n −1,we know that v > 0 and we will prove that (F0(wh)v)>0.
For i= 2,3, ..., n −2
Ai
1xi−1+Ai
2xi+Ai
3xi+1 =−hir1(i) + hi+1r3(i) + R,
with R=xi−1q1(i)bu(xi−1, wi−1) + xiq2(i)bu(xi, wi)≥0.
Because of (7) and (9) we have
−hir1(i) + hi+1r3(i) = q1(i)ai−1+q2(i)ai
=q1(i)(ai−hia0(θ)) + q2(i)ai
(13)
≥α−M
n0ka0k∞=c > 0.
So (F0(wh)v)i>0, i = 2,3, ..., n −2.Since x0= 0 and xn= 1, it follows that
A1
2x1+A1
3x2=A1
1x0+A1
2x1+A1
3x2,
An−1
1xn−2+An−1
2xn−1=An−1
1xn−2+An−1
2xn−1±r3(i)xn,
and using the fact that r3(i)<0 it leads to the conclusion
(F0(wh)v)1,(F0(wh)v)n−1>0.2
Nonzero components of the truncating error vector are
τi[uε] = 1
6(2ε(hi−hi+1) + hihi+1 ai
.−q1(i)hi(6ε+hi+1ai+ai−1(hi+hi+1 ))) u000
ε(xi)
−h3
i(2ε−(1 −q1(i))hi+1ai+q1(i)ai−1(2hi+hi+1 ))
24(hi+hi+1)u
IV
ε(θ1,0)
−h3
i+1(2ε+q1(i)hiai−1−(1 −q1(i))hiai)
24(hi+hi+1)u
IV
ε(θ3,0)
+1
6h3
iq1(i)ai−1u
IV
ε(θ1,1) + 1
2h2
iq1(i)ε u
IV
ε(θ1,2),
i= 1,2, ..., n −1,
where θ1,0, θ1,1, θ1,2∈(xi−1, xi) i θ3,0∈(xi, xi+1).
2.2. Scheme 3
Let s= 3, N = 3, n1= 1, N1= 3 and d1=d0
1=−hi
h, d2=d0
3= 0, d3=
hi+1
h, d0
2=−hi
2h.
We introduce {1, x(x+hi), x(x−hi+1), x3}the base of the space P3[x] for the
determination of the coefficients rj, qj, j = 1,2,3.If x=xi,we use the notation
144 I. Radeka, D. Herceg
xi−hi
2=xi−1/2, a(xi−1/2) = ai−1/2, w(xi−1/2) = wi−1/2.For w(x) = 1 we
get the equation
Th
εw(x) = r1(i) + r2(i) + r3(i) = 0,(14)
for w(x) = x(x+hi)
Th
εw(x) = r3(i)hi+1(hi+hi+1 ) + q1(i)hiai−1−q3(i)hiai+ 2ε= 0.
So
r3(i) = −2ε−q1(i)hiai−1+q3(i)hiai
hi+1(hi+hi+1 ).(15)
For w(x) = x(x−hi+1)
Th
εw(x) = r1(i)hi(hi+hi+1) + q1(i)(2hi+hi+1 )ai−1(hi+hi+1)
+q2(i)(hi+hi+1)ai−1/2+q3(i)hi+1 ai+ 2ε= 0,
it follows that
r1(i)= −2ε−q1(i)(2hi+hi+1)ai−1−q2(i)(hi+hi+1)ai−1/2−q3(i)hi+1ai
hi(hi+hi+1).(16)
For w(x) = x3,we get the additional condition for the determination qj=
qj(i), j = 1,2,3:
Th
εw(x) = −h3
ir1(i) + h3
i+1r3(i)−3hi(q1(i)(2ε+hiai−1)
+q2(i)(ε+ 3h2
i
4ai−1/2)) = 0.
(17)
Applying the normalization condition we have
q1(i) + q2(i) + q3(i) = 1,(18)
The system for the determination of coefficients has five linearly independent
equations, so we can choose one unknown freely, let it be q2(i).
For wh:= (w1, w2, ..., wn−1)T∈Rn−1(w0=wn= 0) using the nodes
High-order methods for semilinear singularly . . . 145
xi, i = 0,1, ..., n, instead of a discrete problem of the form
F1wh:= r2(1)w1+r3(1)w2+q1(1)b(0,0) + q2(1)b(x1/2, w1/2)
+q3(1)b(x2, w2)
Fiwh:= r1(i)wi−1+r2(i)wi+r3(i)wi+1 +q1(i)b(xi−1, wi−1)
+q2(i)b(xi−1/2, wi−1/2) + q3(i)b(xi, wi)
i= 2,3, ..., n −2
Fn−1wh:= r1(n−1)wn−2+r2(n−1)wn−1
+q1(n−1)b(xn−1, wn−1) + q2(n−1)b(xn−1/2, wn−1/2)
+q3(n−1)b(1,0),
we will form another one, when we use the Taylor expansion
wi−1/2−hi+ 2hi+1
4(hi+hi+1)wi−1−hi+ 2hi+1
4hi+1
wi+h2
i
4hi+1(hi+hi+1 )wi+1=e
Ri(w),(19)
where
e
Ri(w) = 1
3! 1
8h2
i(hi+ 2hi+1)w000(xi) + h4
i(hi+ 2hi+1)
96(hi+hi+1)wI V (αi
1)
−h4
i
384 hi+1 wIV (αi
2) + h2
ihi+13
96(hi+hi+1)wI V (αi
3),
(20)
with αi
1∈(xi−1, xi), αi
2∈(xi−1/2, xi), αi
3∈(xi, xi+1).Let
ewi−1/2=hi+ 2hi+1
4(hi+hi+1)wi−1+hi+ 2hi+1
4hi+1
wi−h2
i
4hi+1(hi+hi+1 )wi+1,
then
b(xi−1/2, wi−1/2) = b(xi−1/2,ewi−1/2) + e
Ri(w)bu(xi−1/2,ewi−1/2)
+e
R2
i(w)
2buu(xi−1/2, θ0
i),
(21)
for θ0
i∈(ewi−1/2, wi−1/2).
146 I. Radeka, D. Herceg
Now, the discrete problem we are going to analyze has the following form
e
F1wh:= r2(1)w1+r3(1)w2+q1(1)b(0,0) + q2(1)b(x1/2,ew1/2)
+q3(1)b(x2, w2)
e
Fiwh:= r1(i)wi−1+r2(i)wi+r3(i)wi+1 +q1(i)b(xi−1, wi−1)
+q2(i)b(xi−1/2,ewi−1/2) + q3(i)b(xi, wi)
i= 2,3, ..., n −2
e
Fn−1wh:= r1(n−1)wn−2+r2(n−1)wn−1+q1(n−1)b(xn−1, wn−1)
+q2(n−1)b(xn−1/2,ewn−1/2) + q3(n−1)b(1,0),
(22)
where rj(i), qj(i), j = 1,2,3, i = 1,2,3, ..., n −1 are given by (14, 15, 16, 18),
and because of our approximation (19), instead of equation (17), we get
−h3
ir1(i) + h3
i+1r3(i)−3hi(q1(i)(2ε+hiai−1) + q2(i)(ε+ 3 h2
i
4ai−1/2))
−q2(i)1
8h2
i(hi+ 2hi+1)bu(xi−1/2,ewi−1/2) = 0.
The Jacobian matrix of the mapping e
F= ( e
F1,e
F2, ..., e
Fn−1) is a tridiagonal
matrix of the form e
F0(wh) = tridiag{Ai
1, Ai
2, Ai
3},where for i= 1,2, ..., n −1,
Ai
1=r1(i) + q1(i)bu(xi−1, wi−1) + q2(i)hi+ 2hi+1
4(hi+hi+1)bu(xi−1/2,ewi−1/2),
Ai
2=−(r1(i) + r3(i)) + q3(i)bu(xi, wi) + q2(i)hi+ 2hi+1
4hi+1 bu(xi−1/2,ewi−1/2),
Ai
3=r3(i)−q2(i)h2
i
4hi+1(hi+hi+1 )bu(xi−1/2,ewi−1/2).
We shall choose the coefficient q2(i) in the appropriate way so that the matrix
e
F0(wh) becomes an L−matrix. For that we will use the following lemma:
Lemma 2.1 Let n0∈N, so that for all n≥n0it stands that
3Mka0k∞
n0
< α(23)
and the mesh Ihis chosen so that hi+1 ≤hifor all i∈ {0,1, ...n −1}.In the
case when 2hikak∞<3εwe define
q1(i) = 2h3
i+ 2h3
i+1 −3q2(i)h2
i(hi+hi+1)
12h2
i(hi+hi+1)
(24)
High-order methods for semilinear singularly . . . 147
and
q2(i) = 2
3h2
i¡6ε(h2
i−hihi+1 −h2
i+1)−ai−1(h3
i+h3
i+1)+5aih2
ihi+1
+aih2
i+1(hi−hi+1 )¢/¡6ε−ai−1(hi+hi+1)−ai−1/2hi
+3aihi+1 +3
2h2
ibu(xi−1/2,ewi−1/2)´.
Then
−2
3< q2(i)<1,1−q2(i)−2q1(i) = 4
3−δ > 0,
where δ > 0is the constant independent of ε.
Proof. Using conditions (2) and 2hikak∞<3εit follows that
3h2
i(6ε−ai−1(hi+hi+1)−ai−1/2hi+3aihi+1 +3
2h2
ibu(xi−1/2,ewi−1/2)) >0,
(25)
We can prove that q2(i)<1 from the fact that q2(i)<1 if and only if
2(6ε(h2
i−hihi+1 −h2
i+1)−ai−1(h3
i+h3
i+1)+5aih2
ihi+1
+aih2
i+1(hi−hi+1 )) −3h2
i(6ε−ai−1(hi+hi+1)−ai−1/2hi
+3aihi+1 +3
2h2
ibu(xi−1/2,ewi−1/2)) <0.
Because of (25) we can prove that q2(i)>−2
3,using the fact that q2(i)>−2
3,
if and only if
6ε(h2
i−hihi+1 −h2
i+1)−ai−1(h3
i+h3
i+1) +5aih2
ihi+1 +aih2
i+1(hi−hi+1 )
+h2
i(6ε−ai−1(hi+hi+1)−ai−1/2hi+3aihi+1 +3
2h2
ibu(xi−1/2,ewi−1/2))
>0.
The form (24) leads to
1−q2(i)−2q1(i) = −1
2q2(i) + 2h2
i+hihi+1 −h2
i+1
3h2
i
,
and using the bounds for the coefficients q1(i) and q2(i) we have
−1
2q2(i) + 2h2
i+hihi+1 −h2
i+1
3h2
i
<1
3+2h2
i+hihi+1
3h2
i≤4
3
and
−1
2q2(i) + 2h2
i+hihi+1 −h2
i+1
3h2
i
>−1
2+2
3+hihi+1 −h2
i+1
3h2
i≥0.
148 I. Radeka, D. Herceg
So, there exists a constant δ > 0 such that the following stands:
1−q2(i)−2q1(i) = 4
3−δ > 0.
2
Theorem 2.2. Let i∈ {1,2, ...n −1}.If 2hikak∞<3ε, q1(i)and q2(i)be
defined as in the previous lemma, and for ifor which it holds that 2hikak∞≥
3ε, the coefficients are given in the form
q1(i) = ai−q2(i)(ai+hi
4bu(xi−1/2,ewi−1/2))
ai+ai−1
(26)
and
q2(i) = µ2ε(hi+1
hi−1)(ai+ai−1) + ai(6ε+hiai−1)¶/
µ(−3ε+hiai−1/2
4−h2
ibu(x1/2,ewi−1/2)
8)(ai+ai−1)
+(ai+hibu(x1/2,ewi−1/2)
4)(6ε+hiai−1))¶.
Let n0∈Nbe the number for which the following conditions are satisfied
max{γ, 1}M(4 ka0k∞+ 3 kGk∞)
n0
<min{1, δ}α,(27)
2γM (ka0k∞+kGk∞)kak∞
n0
< α2,(28)
and
γM 2(ka0k2
∞+ 4 kGk2
∞)
n2
0
< α2,(29)
let δbe determined in the previous lemma, and γ= max{|qj(i)|;j= 1,2,3}.
Then for n≥n0it follows that F0(wh)is an L−matrix.
Proof. Let i∈ {1,2, ...n −1}and 2hikak∞<3ε, from the previous lemma and
(27) it follows that
−2ε−q1(i)hiai−1+q3(i)hiai−q2(i)h2
i
4bu(xi−1/2,ewi−1/2)
≤ −2ε+hi(4
3−δ)kak∞+hi1
3min{1, δ}α
≤ −2ε+hi4
3kak∞<0,
High-order methods for semilinear singularly . . . 149
that is Ai
3<0.We have
−2ε−q1(i)(2hi+hi+1)ai−1−q2(i)(hi+hi+1 )ai−1/2−q3(i)hi+1ai
+q1(i)hi(hi+hi+1)bu(xi−1, wi−1)
+q2(i)hi(hi+ 2hi+1)
4bu(xi−1/2,ewi−1/2)
=−2ε−hi+1ai(1 −2q1(i)−q2(i)) −2q1(i)hiai−q2(i)hiai±hiai
+q1(i)a0(η1)hi(2hi+hi+1) + hi
2a0(η2)q2(i)(hi+hi+1)
+q1(i)hi(hi+hi+1)bu(xi−1, wi−1)
+q2(i)hi(hi+ 2hi+1)
4bu(xi−1/2,ewi−1/2)
<−2ε+4
3hiai−hiai+hika0k∞γ4M
n0
+hikGk∞γ3M
n0
<−2ε+4
3hiai−hiai+hiα < 0,
so Ai
1<0.It only remains to show that Ai
2>0,which follows from
q3(i)ai(hi−hi+1)−q2(i)hi+1 ai−1/2−q1(i)(hi+hi+1)ai−1
−q3(i)hihi+1bu(xi, wi)−q2(i)hi(hi+2hi+1 )
4bu(xi−1/2,ewi−1/2)
<2ε.
If 2hikak∞≥3ε, then because of (26) we have
Ai
3=−2ε
hi+1(hi+hi+1 )<0.
In this case q1(i) is of the form (26), so using the condition (28) it follows that
−2ε−q1(i)(2hi+hi+1)ai−1−q2(i)(hi+hi+1 )ai−1/2−q3(i)hi+1ai
+q1(i)hi(hi+hi+1)bu(xi−1, wi−1)
+q2(i)hi(hi+ 2hi+1)
4bu(xi−1/2,ewi−1/2)
≤ −2ε−2(hi+hi+1)ai−1ai+hi+1 a2
i
ai+ai−1
+q2(i)
ai+ai−1
(ai(ai−1−ai−1/2)
≤ −2ε+1
ai+ai−1¡−2(hi+hi+1)α2−hi+1 α2
+γhi(hi+hi+1 )kak∞ka0k∞+γ(3h2
i+ 4hihi+1)kak∞kGk∞¢
150 I. Radeka, D. Herceg
=−2ε+1
ai+ai−1µhi(−2α2+γM
n0kak∞ka0k∞)
+3γM
n0kak∞kGk∞+hi+1(−3α2+γM
n0kak∞ka0k∞
+4γM
n0kak∞kGk∞)¶
<0
and Ai
1<0.It can be shown that Ai
2>0,using the fact that
−q3(i)ai(hi−hi+1) + q2(i)ai−1/2hi+1 +q1(i)ai−1(hi+hi+1 )
+q3(i)hihi+1bu(xi, wi) + q2(i)hi(hi+2hi+1 )
4bu(xi−1/2,ewi−1/2)≥0.
considering that q1(i) is of the form (26). Hence, F0(wh) is an L−matrix. 2
For the coefficients qj(i), j = 1,2,3 we do not have the nonnegativity prop-
erty so we shall prove that e
F0(wh) is an M−matrix only for the case when
b(x, u) is the linear function in u, that is for
b(x, u) = e
b(x)u−f(x),(30)
with e
band fare functions smooth enough. From (2) it follows that e
b(x)≥0,
for x∈(0,1).
Theorem 2.3. Let all conditions from the previous theorem using the function
b(x, u)be of the form (30), for all i∈ {1,3
2,2, ..., n −1}
e
b(xi)3
2
M
n0
γ°
°
°e
b0°
°
°∞≤e
b(xi)2,(31)
then for n≥n0the matrix F0(wh)is an M−matrix.
Proof. Using the vector v= (x1, x2, ..., xn−1)T,with xi∈Ih, i = 1,2, ..., n −1,
we know that v > 0 and we can show that (F0(wh)v)>0, so the theorem
holds.2
The truncating error is
τi[uε] = 1
48((4q1(i)ai−1−q2(i)ai−1/2)h3
i
−2(q1(i)ai−1−q3(i)ai)(h2
i+1hi−hi+1 h2
i)
+4εhi+1(hi−hi+1 ) + 2εh2
i(3q2(i) + 12q1(i)−2)) uIV
ε(xi)
−bu(xi−1/2,euε(xi−1/2)) µh4
i(hi+ 2hi+1)
96(hi+hi+1)uI V
ε(αi
1)
High-order methods for semilinear singularly . . . 151
−h4
i
384 hi+1
uIV
ε(αi
2) + h2
ihi+13
96(hi+hi+1)uI V
ε(αi
3)¶
−e
R2
i(uε)
2buu(xi−1/2, θ0
i) + h4
i
120(hi+hi+1)(2ε+q3(i)aihi+1
+q2(i)ai−1/2(hi+hi+1) + q1(i)ai−1(2hi+hi+1 )) u
V
ε(θ1,0)
−h4
i+1(2ε+q1(i)hiai−1−q3(i)hiai)
120(hi+hi+1)u
V
ε(θ3,0)
−1
24h4
iq1(i)ai−1u
V
ε(θ1,1)−1
384h4
iq2(i)ai−1/2u
V
ε(θ2,1)
−1
6q1(i)εh3
iu
V
ε(θ1,2)−1
48q2(i)εh3
iu
V
ε(θ2,2),
i= 1,2, ..., n −1,
where θ1,0, θ1,1, θ1,2, αi
1∈(xi−1, xi), θ2,1, θ2,2, αi
2∈(xi−1/2, xi), θ3,0, αi
3∈
(xi, xi+1) i θ0
i∈(euε(xi−1/2), uε(xi−1/2)), and e
Riis given with (20).
3. Meshes
In order to obtain a good approximation for the exact solution of the problem
(1) we use the nonuniform meshes that are dense in the neighborhood of the
point x= 1,where the boundary layer appears. We considered two types of
meshes, Bakhvalov’s and Shiskin’s. Because of getting better numerical results
when applying Bakhvalov’s type of meshes, we are going to prove the uniform
convergence of the method obtained on a mesh of type, constructed by Vulanovi´c
([16], [?]). The mesh, further on called H-mesh, is generated by the function
λ(t) = (λ1(t) = λ0
2(τ)t, t ∈[0, τ ]
λ2(t) = 1 −Aε(1−t)
q−(1−t), t ∈[τ, 1]
(32)
with
τ= 1 −q−pAqε(1 −q+Aε)
1 + Aε ,
and the constants Aand qsatisfy
q∈(0,1), A ∈(0, q/ε),(33)
so that the transition point has the property τ∈(1 −q, 1).The mesh points are
xi=λµi
n¶, i = 0,1, . . . , n.
The Shiskin mesh we use in numerical experiments has a generating function of
the form
λ(t) = ½λ1(t) = 2(1 −τ)t, t ∈[0,0.5]
λ2(t) = 1 −τ+ 2τ(t−0.5), t ∈[0.5,1] ,
152 I. Radeka, D. Herceg
with the transition point τ= min{0.5, εα ln n}.We have to emphasize the fol-
lowing property of the nodes of the H-mesh where hi=xi−xi−1
Lemma 3.1 For i∈ {1,2, ..., n −1},it holds true that hi≥hi+1 and hi≤M1
n.
4. Convergence
4.1. Scheme 2 and H-mesh
Lemma 4.1 For the discrete problem (8) applied on an H-mesh when a, b are
functions smooth enough, then for n≥n0,and i∈ {1,2, ..., n −1}for which
holds −2ε+hiai<0,the coefficient q1(i)is of the form
q1(i) = hi−hi+1
3hi
,(34)
otherwise is of the form (12). If the constants of the mesh satisfy (33) and
additionally q > 3
n, then
|τi[uε]| ≤
M(h2
i+1
hiexp(−α1−xi+1
ε)),i+1
n≤τ,
M(h2
i+1
εn−2), otherwise.
Proof. If we denote the exact solution of the problem (1) by uεthen we observe
the truncating error given earlier for Scheme 2. Let i∈ {1,2, ..., n −1}.In the
case when −2ε+hiai<0 and −2ε+hiai≥0 we have different forms of the
coefficient q1(i).In both cases we have
|τi[uε]| ≤ M(h2
i¯¯¯u000
ε(xi)¯¯¯+ max{ε, hi}(h2
i¯¯¯¯u
IV
ε(θ1,0)¯¯¯¯
+h3
i+1
hi+hi+1 ¯¯¯¯u
IV
ε(θ3,0)¯¯¯¯) + h3
i¯¯¯¯u
IV
ε(θ1,1)¯¯¯¯+εh2
i¯¯¯¯u
IV
ε(θ1,2)¯¯¯¯..
Using (4) it follows that
|τi[uε]| ≤ M(h2
i(1 + ε−3exp(−α1−xi
ε))(35)
+ max{ε, hi}(h2
i(1 + ε−4exp(−α1−xi
ε))
+h3
i+1
hi+hi+1 ¯¯¯¯u
IV
ε(θ3,0)¯¯¯¯) + h3
i(1 + ε−4exp(−α1−xi
ε))
+εh2
i(1 + ε−4exp(−α1−xi
ε)).
So,we will consider two cases:
High-order methods for semilinear singularly . . . 153
1. Let i+1
n≤τThen hi=hi+1 and max{ε, hi}=hi,because of the condition
ε < 1
n. If we use the integral form of the error in the Taylor expansion of
the function u
IV
ε(x), it follows that
u
IV
ε(θ3,0) = 4
h4
i+1 Zxi+1
xi
(xi+1 −s)3uIV
ε(s)ds.
Using (4) we have
¯¯uIV
ε(θ3,0)¯¯≤4
h4
i+1 Zxi+1
xi
(xi+1 −s)3M(1 + ε−4exp(−α1−s
ε))ds
≤M+M
ε4h4
i+1 Zxi+1
xi
(xi+1 −s)3exp(−α1−s
ε)ds,
that is
|τi[uε]| ≤ M(h2
i+ (h2
iε−3+h3
iε−4+h3
iε−4+h2
iε−3) exp(−α1−xi
ε)
+1
hi
exp(−α1−xi+1
ε)).
For s≥0 it holds true that skexp(−s)≤M1, k ∈N,using xi=xi+1 −hi+1
it follows
hk−1
i
εkexp(−α1−xi
ε) = 1
hi
exp(−α1−xi+1
ε)hk
i
εkexp(−αhi
ε)
≤M1
1
hi
exp(−α1−xi+1
ε).
So,
|τi[uε]| ≤ M(h2
i+1
hi
exp(−α1−xi+1
ε))
2. Let τ < i+1
n.Then
|τi[uε]| ≤ M(h2
i+ (h2
iε−3+h2
imax{ε, hi}ε−4+h3
iε−4
+h2
iε−3) exp(−α1−xi
ε)) + h2
iε−3exp(−α1−xi+1
ε)).
Now, we have two possibilities
(a) 1 −q+3
n<i+1
n.Then
h2
i
ε2exp(−α1−xi+1
ε) = h2
i
ε2exp Ã−αAq 1−i+1
n
q−1 + i+1
n!
≤1
n2ÃAq
(q−1 + i−1
n)2!2
exp Ã−αAq(1 −i+1
n)
q−1 + i+1
n!≤M1n−2.
154 I. Radeka, D. Herceg
Because of
h3
i
ε3exp(−α1−xi
ε)≤h3
i
ε3exp(−α1−xi+1
ε)≤M1n−2,
we have the statement.
(b) τ < i+1
n<1−q+3
n.Then
h2
i
ε2exp(−α1−xi+1
ε)
≤1
n2³Aq
(q−1+τ)2´2exp ³−α1−λ2(1−q+3
n)
ε´≤Mn−2,
follows from
µAq
(q−1 + τ)2¶2
exp(−Mn)≤µAq
(q−1 + τ)2¶2
M1n−k≤M,
for k∈Nbig enough.
So the theorem is proven.2
If we define the functions for i= 1,2, ..., n −1 and t > 0
φi(t) =
n
Y
j=i+1
1
1 + thj
ε
with φn(t) = 1,we can prove the following lemma:
Lemma 4.2 Let
0< α0< α, t ≤α0/2and b(x, u)≥0(36)
for (x, u)∈(0,1) ×Runder the conditions of Theorem 2.1. Then
Th
εφi(t)≥C(t)
ε+thi
φi(t).
We shall use also the result of the following lemma:
Lemma 4.3 For t∈(0, α0)and i= 0,1, ..., n it holds true that
exp(−α1−xi
ε)) ≤φi(t)
Using the results from the previous section and the previous lemma we obtain
the main conclusion:
High-order methods for semilinear singularly . . . 155
Theorem 4.1. If we denote the solution of the discrete problem (8) by w∗
applied on an H-mesh and if uh
εis the discrete exact solution of the problem
(1), then for the functions aand bsmooth enough, under the conditions (2),
(36), and also under conditions of Theorem 2.1, and previous lemmas, then for
n≤1/√ε, it follows that
°
°uh
ε−w∗°
°∞≤Ch2.
Proof. Let whand vhbe the mesh functions. Using the results of Theorem 2.1,
we have that F0(wh+s(vh−wh)) is M−matrix. For some s∈(0,1),from
wh
0≥uh
0, wh
n≥uh
ni T h
εwh
i≥Th
εuh
i, i = 1,2, ..., n −1,
it follows that
wh−uh=F0(wh+s(vh−wh))−1(F(wh)−F(uh)) ≥0,
so the operator Th
εsatisfies the discrete maximum principle. Defining the barrier
function
ψi(t) = Cµ(1 + xi)n−2+ (1 + thi+1
ε)φi(t)¶
for i= 0,1, ...n, and hn+1 some positive number, we conclude that
ψ0(t)±(uh
ε−w∗)0≥0ψn(t)±(uh
ε−w∗)n≥0,
and for i= 1,2, ...n −1 using previous lemmas it follows that
Th
ε(ψi(t)±(uh
ε−w∗)i)
≥Th
ε(C((1 + xi)n−2+φi+1(t))) − |τi[uε]|
≥Cµ(q1(i)ai−1+q2(i)ai)n−2+C(t)
ε+thi+1
φi+1(t)¶− |τi[uε]|
> C1n−2+C2
ε+thi
φi+1(t)− |τi[uε]| ≥ 0.
Using the discrete maximum principle for the observed operator, we have
¯¯(uh
ε−w∗)i¯¯≤ψi(t).(37)
Let k∈ {0,1, ..., n −1}be the number that
k+ 1
n≤τ < k+ 2
n.(38)
We will show that for all i≤k+ 1 is satisfied φi+1 (t)≤Mn−2.It stands that
φi+1(t)≤
n
Y
j=k+3
1
1 + thj
ε
156 I. Radeka, D. Herceg
Because of (38) we have
k+ 3 ≤2 + n−nq−pAqε(1 −q+Aε)
1 + Aε =tn.
Using
hj
ε=Anq
(j−1 + n(q−1))(j+n(q−1))
≥Anq
(−1
2+j+n(q−1))2,
and n√ε≤1 it follows that
φi+1(t)≤
n
Y
j=k+3
1
1 + thj
ε≤
n
Y
j=btnc
1
1 + tAnq
(−1
2+j+n(q−1))2
≤
n−btnc
Y
j=1
1
1 + 4Anqt(1+Aε)2
(1+A(1+2nq)ε+2j(1+Aε)+2n√Aqε(1−q+Aε))2
≤
2
Y
j=1
1
4Anqt(1+Aε)2
(1+A2q√ε+4+5Aε+2√Aq(1−q+Aε))2
≤(5 + 2Aq + 5A+ 2pAq(1 −q+A))2
4Anqt ≤M n−2.
So, for all i≤k+ 1 from (37) and the previous conclusions it follows that
¯¯(uh
ε−w∗)i¯¯≤Mn−2.
Now, we define a new barrier function
ϕi(t) = C¡(1 + xi)n−2+n−2φi(t)¢
for i=k+ 1, k + 2, ...n. Then
ϕk+1(t)±(uh
ε−w∗)k+1 ≥0, ϕn(t)±(uh
ε−w∗)n≥0,
and
Th
ε(ϕi(t)±(uh
ε−w∗)i)
≥Th
ε(C((1 + xi)n−2+n−2φi(t))) − |τi[uε]|
> C1n−2+n−2C2
ε+thi
φi(t)− |τi[uε]| ≥ 0.
So, ¯¯(uh
ε−w∗)i¯¯≤Mn−2
for i∈ {0,1, ..., n}and the theorem is proven.2
High-order methods for semilinear singularly . . . 157
4.2. Scheme 3 and H-mesh
In a similar way as in the previous theorem, we can get the following con-
clusion:
Theorem 4.2. If we denote the solution of the discrete problem (22) by w∗
applied on an H-mesh and if uh
εis the discrete exact solution of the problem (1),
then for the functions aand bsmooth enough, under the conditions (2), (36),
and also under the conditions of Theorems 2.2 and 2.3, then for n≤1/√ε, it
follows that °
°uh
ε−w∗°
°∞≤Ch3.
5. Numerical results
The obtained theoretical results are confirmed by numerical experiments.
Exact solutions of the tested examples are known, so the error is measured by
En=°
°uh
ε−w∗°
°∞,where w∗is the solution of the discrete problem, whereas
uh
ε= (uε(x0), ..., uε(xn))T,for uεexact solution of the observed problem. The
order of convergence is calculated with
Ordn=ln En−ln E2n
ln 2 .
The approximations, obtained from (8) and 22 applied on an H-mesh and
Shiskin (S) mesh are tested for the different values of εand n. The results con-
firmed the order of convergence of the methods, but the error Enwas smaller for
H-mesh, which is a consequence of the greater number of nodes in the boundary
layer. Newton’s method is used for solving the nonlinear system of equations
F(wh) = 0 with the initial approximation w0= (u0(x0), ..., u0(xn))T, u0as the
solution of the reduced problem. The stop criterion applied is
max ©°
°wk−wk−1°
°∞,°
°F(wk)°
°∞ª<10−3.
Some of the tested problems are:
Example 1
−εu00 + (1 + x(1 −x)) u0=f(x), u (0) = u(1) = 0,(39)
where f(x) is the function for which
uε(x) = 1−e−(1−x)/ε
1−e−1/ε −cos π
2x,
is the exact solution.
Example 2
−εu00 +u0+u2+u=f(x), u (0) = u(1) = 0,(40)
where f(x) is the function for which
158 I. Radeka, D. Herceg
uε(x) = 1−e−x/ε
e1/ε −1+x
is the exact solution.
Table 1:Example 1 (Scheme 2 and H-mesh with A= 7 and q= 0.5)
n
ε64 128 256 512 1024 2048
2−44.60(-5) 1.10(-5) 2.68(-6) 6.61(-7) 1.64 (-7) 4.09(-8) En
2.07 2.03 2.02 2.01 2.00 Ordn
2−66.35(-5) 1.54(-5) 3.80(-6) 9.44(-7) 2.35(-7) 5.87(-8) En
2.04 2.02 2.01 2.00 2.00 Ordn
2−81.31(-4) 4.64(-5) 8.54(-6) 2.12(-6) 5.28(-7) 1.32(-7) En
1.50 2.44 2.01 2.00 2.00 Ordn
2−10 1.41(-4) 3.98(-5) 1.24(-5) 4.39(-6) 1.27(-6) 2.22(-7) En
1.83 1.68 1.50 1.79 2.52 Ordn
2−12 1.65(-4) 4.18(-5) 1.11(-5) 3.17(-6) 9.90(-7) 3.47(-7) En
1.98 1.91 1.81 1.68 1.51 Ordn
2−14 1.81(-4) 4.54(-5) 1.14(-5) 2.95(-6) 7.92(-7) 2.26(-7) En
2.00 1.99 1.95 1.90 1.81 Ordn
2−16 1.89(-4) 4.77(-5) 1.19(-5) 2.99(-6) 7.61(-7) 1.97(-7) En
1.98 2.00 1.99 1.98 1.95 Ordn
2−18 1.94(-4) 4.86(-5) 1.22(-5) 3.06(-6) 7.66(-7) 1.93(-7) En
2.00 1.99 2.00 2.00 1.99 Ordn
2−20 1.97(-4) 4.93(-5) 1.24(-5) 3.10(-6) 7.74(-7) 1.94(-7) En
2.00 2.00 2.00 2.00 2.00 Ordn
Table 2: Example 1 (Scheme 2 and S-mesh with σ0= 4)
n
ε64 128 256 512 1024 2048
2−44.55(-4) 1.08(-4) 2.65(-6) 6.54(-7) 1.62(-7) 4.05(-8) En
2.07 2.03 2.02 2.01 2.00 Ordn
2−61.28(-4) 2.58(-4) 5.11(-6) 9.9(-7) 1.88(-7) 3.49(-8) En
1.99 2.01 2.00 2.00 2.00 Ordn
2−82.6(-4) 8.83(-4) 3.16(-4) 4.3(-6) 1.03(-6) 2.47(-7) En
1.56 1.48 2.88 2.06 2.06 Ordn
2−10 2.19(-4) 6.18(-4) 1.91(-4) 6.6(-6) 2.56(-6) 3.51(-7) En
1.82 1.70 1.53 1.37 2.86 Ordn
2−12 2.06(-4) 5.32(-4) 1.43(-4) 4.05(-6) 1.26(-6) 4.37(-7) En
1.95 1.90 1.82 1.69 1.52 Ordn
2−14 2.02(-4) 5.1(-4) 1.3(-4) 3.37(-6) 9.04(-7) 2.57(-7) En
1.99 1.97 1.95 1.890 1.81 Ordn
2−16 2.01(-4) 5.04(-4) 1.27(-4) 3.2(-6) 8.15(-7) 2.11(-7) En
2.00 1.99 1.98 1.97 1.94 Ordn
2−18 2.01(-4) 5.03(-4) 1.26(-4) 3.15(-6) 7.92(-7) 2.(-7) En
2.00 2.00 2.00 1.99 1.98 Ordn
2−20 2.01(-4) 5.02(-4) 1.26(-4) 3.14(-6) 7.86(-7) 1.97(-7) En
2.00 2.00 2.00 2.00 2.00 Ordn
High-order methods for semilinear singularly . . . 159
Table 3: Example 3 (Scheme 2 and H-mesh with A= 4 and q= 0.8)
n
ε64 128 256 512 1024 2048
2−47.44(-5) 1.65(-5) 3.87(-6) 9.38(-7) 2.31(-7) 5.73(-8) En
2.18 2.09 2.04 2.02 2.01 Ordn
2−61.36(-4) 2.95(-5) 6.84(-6) 1.64(-6) 4.03(-7) 9.97(-8) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−81.50(-4) 3.26(-5) 7.57(-6) 1.82(-6) 4.46(-7) 1.10(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−10 1.54(-4) 3.35(-5) 7.79(-6) 1.87(-6) 4.59(-7) 1.14(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−12 1.55(-4) 3.38(-5) 7.85(-6) 1.89(-6) 4.63(-7) 1.15(-7) En
2.20 2.11 2.06 2.03 2.00 Ordn
2−14 1.55(-4) 3.39(-5) 7.86(-6) 1.89(-6) 4.64(-7) 1.15(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−16 1.55(-4) 3.39(-5) 7.86(-6) 1.89(-6) 4.64(-7) 1.15(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−18 1.55(-4) 3.39(-5) 7.86(-6) 1.89(-6) 4.64(-7) 1.15(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
2−20 1.55(-4) 3.39(-5) 7.86(-6) 1.89(-6) 4.64(-7) 1.15(-7) En
2.20 2.11 2.06 2.03 2.01 Ordn
Table 4: Example 1 (Scheme 3 and H-mesh with A= 2 and q= 0.8)
n
ε32 64 128 256 512 1024
2−43.56(-5) 6.30(-6) 1.40(-6) 2.11(-7) 2.86(-8) 3.71(-9) En
2.50 2.17 2.73 2.88 2.95 Ordn
2−62.97(-3) 1.29(-4) 1.11(-5) 1.82(-7) 3.27(-8) 4.71(-9) En
4.52 3.54 5.94 2.47 2.80 Ordn
2−83.27(-3) 1.30(-4) 4.75(-5) 7.35(-6) 2.47(-7) 4.66(-9) En
4.66 4.77 2.69 1.57 5.73 Ordn
2−10 3.36(-3) 1.31(-4) 3.55(-6) 2.46(-7) 5.04(-8) 1.46(-8) En
4.68 5.20 3.85 2.29 1.79 Ordn
2−12 3.38(-3) 1.31(-4) 3.21(-6) 1.39(-7) 1.67(-8) 3.32(-9) En
4.69 5.35 4.53 3.06 2.33 Ordn
2−14 7.44(-3) 3.08(-4) 9.23(-6) 3.09(-7) 1.77(-8) 1.65(-9) En
4.70 5.39 4.82 3.64 3.01 Ordn
2−16 3.39(-3) 1.31(-4) 3.09(-6) 1.02(-7) 6.77(-9) 5.64(-10) En
3.64 5.40 4.91 3.92 3.59 Ordn
2−18 3.39(-3) 1.31(-4) 3.08(-6) 1.00(-7) 6.25(-9) 4.38(-10) En
3.79 5.40 4.93 4.01 3.84 Ordn
2−20 3.39(-3) 1.31(-4) 3.08(-6) 1.00(-7) 6.15(-9) 4.43(-10) En
3.93 5.40 4.94 4.02 3.80 Ordn
160 I. Radeka, D. Herceg
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Received by the editors April 22, 2003
