Survey on the numerical methods for ODE’s
using the sequence of successive approximations
Alexandru Mihai Bica
Department of Mathematics, University of Oradea,
Str. Armatei Romane no. 5, 410087 Oradea, Romania
[email protected], [email protected]
Abstract
Here we present the error estimations in the numerical methods for Cauchy problems corresponding to ODE’s, which use the sequence of suc-cessive approximations and quadrature rules such as trapezoidal rule and perturbed trapezoidal rule.
AMS Mathematics Subject Classification: 65L05, 34K28.
Keywords and phrases: method of successive approximations, trapezoidal quadrature rule, perturbed trapezoidal quadrature rule.
1
Introduction
There are many results concerning to the numerical methods for ODE’s. The properties of the sequence of successive approximations and his application to the numerical methods for ODE’s was obtained in [1], [5], [17], [20], [21], [22] and [6].Some of the results in the numerical methods for ODE’s obtained in the last 30 years can be found in [4], [9]-[16], [18], [23], [24], [26]-[30]. Here, we continue the work from [6] and [7].
Let a, b >0, x0, y0 ∈ R and f : [x0−a, x0+a]×[y0−b, y0+b] → R a
continuous function which verify the Lipschitz condition in the second argument. We consider the initial value problem :
y0 =f(x, y)
y(x0) =y0
(1)
having in the spaceC1[x
0−h, x0+h] an unique solution, which can be obtained by the successive approximations method starting from any y∈C1[x
0−h, x0+h] ( in particular, we can start from y0 ), where h = min(a,Mb ) and M ∈ R+ is such that|f(x, y)| ≤M,∀(x, y)∈[x0−a, x0+a]×[y0−b, y0+b].
The Lipschitz condition in the second argument is: ∃L >0 such that
The Cauchy’s problem (1) is equivalent with the following Volterra integral equation,
y(x) =y0+
x
Z
x0
f(s, y(s))ds. (3)
Applying the Banach’s fixed point principle to the integral equation (3) we obtain the sequence of successive approximations,
y0(x) =y0, ∀x∈[x0, x0+h] (4)
ym(x) =y0+
x
Z
x0
f(s, ym−1(s))ds, ∀x∈[x0, x0+h],∀m∈N∗.
In [20], at pages 84-95, D.V.Ionescu prove the uniform convergence onI = [x0−h, x0+h],of the sequence (ym)m∈Ngiven in (4). Using the sequence (4)
and a uniform partition of [x0, x0+h], D.V.Ionescu give a numerical method to approximate the solution of (1). In this method he apply the classical trape-zoidal quadrature rule ( for functions with continuous second order derivatives ) in [22], to compute the integrals from (4) on the knots of the uniform partition. In [14] is used the sequence of successive approximations and the classical quadrature trapezoidal rule for functions with continuous second order deriva-tives to approximate the solutions of Fredholm and Volterra integral equations.
In [21], using the quadrature formula of K.Petr ( see [25] ),
b
Z
a
f(x)dx= b−a
2 ·[f(a) +f(b)]−
(b−a)2 12 ·[f
0(b)−f0(a)] +R(f) (5)
with
|R(f)| ≤ (b−a)
5
720 · fIV
(6)
D.V.Ionescu obtains a numerical method for the problem (1) in which the remainder estimation is obtained by the method from [20] and [22] and by the inequality (6). The inequality (6) holds for functions f ∈ C4(D), where
D = [x0 −a, x0 +a]×[y0−b, y0 +b]. Unfortunately, the fourth derivative
d4f(t,x(t))
dt4 have 13 terms, and therefore the remainder containing more terms, increase. This is the reason for which we consider that we can stop to the third derivative in the estimation of the remainder, trying to use an adequate quadrature formula.
In [6] and [7], we have used a recent remainder estimation in the formula (5) obtained by N.S.Barnett and S.S.Dragomir in [3] and .[2] We sumarize the results from [2] and [3] in the following inequality,
b
Z
a
f(x)dx−b−a
2 [f(a) +f(b)] +
(b−a)2 12 [f
0(b)−f0(a)]
≤
(b−a)4 160 · kf
000k, if f ∈C3[a, b]
(b−a)5
720 L, if f
000∈Lip[a, b]
(b−a)5 720 ·
fIV
, if f ∈C4[a, b],
(7)
whereLis the Lipschitz constant of the third derivativef000 and
Lip[a, b] ={g∈C[a, b] :gis Lpischitzian}.
Using a uniform partition of the interval [a, b],
∆ :a=x0< x1< ... < xn−1< xn =b, xi=a+
i(b−a)
n ,∀i= 0, n,
in [2], N.S.Barnett and S.S.Dragomir obtain the following quadrature rule,
b
Z
a
f(x)dx= b−a
2n [f(a)+2
n−1 X
i=1
f(xi)+f(b)]−
(b−a)2 12n2 [f
0(b)−f0(a)]+R
n(f) (8)
with the remainder estimation,
|Rn(f)| ≤
(b−a)4 160n3 · kf
000k. (9)
Iff000 is Lipschitzian with the constant L, then the remainder estimation, ac-cording to [3], is,
|Rn(f)| ≤
(b−a)5L
720n4 , (10)
and if f ∈ C4[a, b], then, according to [3] and [21], the following inequality, which is used in [21], holds:
|Rn(f)| ≤
(b−a)5 720n4 ·
fIV
. (11)
In [6] and [7] we have obtained the error estimation in the numerical method for ODE’s which use the sequence of successive approximations and the quadra-ture rule (8) with the remainder estimation (9) and (10), respectively.
On the other hand, here we use also the trapezoidal quadrature rule obtained in [13] and [8]:
b
Z
a
f(x)dx=b−a
2n ·[f(a) + 2
n−1 X
i=1
f(a+i(b−a)
with the remainder estimation,
|Rn(f)| ≤
(b−a)2L
4n , if f ∈Lip[a, b] ( in [13] )
(b−a)2·kf0k
4n , if f ∈C
1[a, b] ( in [13] )
(b−a)3L0
12n2 , if f0∈Lip[a, b] ( in [8] ) (b−a)3kf00k
12n2 , if f ∈C 2[a, b],
(13)
whereLis the Lipschitz constant off andL0 is the Lipschitz constant off0.
2
Numerical method using the trapezoidal
quadrature rule
Let ∆ an uniform partition of [x0, x0+h],
∆ :x0< x1< ... < xn−1< xn=x0+h, xi=x0+
ih
n,∀i= 1, n.
On the knotsxi, i= 1, n, the sequence of successive approximations (4) is,
y0(xi) =y0, ∀i= 0, n.
ym(xi) =y0+
xi R
x0
f(s, ym−1(s))ds, ∀i= 0, n,∀m∈N∗. (14)
We define the functions,
Fm: [x0, x0+h]→R, Fm(x) =f(x, ym(x)), ∀x∈[x0, x0+h],∀m∈N.
To compute the integrals from (14) we use the quadrature rule (12)-(13), and obtain:
ym(xi) =y0+
xi Z
x0
Fm−1(s)ds=y0+
h
2n[Fm−1(x0) + 2
i−1 X
j=1
Fm−1(xj)+
+Fm−1(xi)] +Rm,i(f) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, ym−1(xj))+
and the remainder estimation is,
|Rm,i(f)| ≤
h2
4n·L, ∀i= 1, n,∀m∈N
∗,
h2
4n ·
F
0
m−1
, ∀i= 1, n,∀m∈N ∗
h3 12n2 ·L
0, ∀i= 1, n,∀m∈
N∗,
h3
12n2 · Fm00−1
, ∀i= 1, n,∀m∈N∗
(16)
whereL0 is an upper bound of the Lipschitz constants of the functions Fm0 −1, m∈N∗ ( andL is an upper bound of the Lipschitz constants of the functions
Fm−1, m∈N∗ ).
The relations (15) and (16) lead to the following algorithm :
ym(x0) =y0, ∀m∈N and y0(xi) =y0, ∀i= 0, n,
y1(xi) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y0(xj)) +f(xi, y0(xi))]+
+R1,i(f) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y0) +f(xi, y0)]+
+R1,i(f) =y1(xi) +R1,i(f), ∀i= 1, n (17)
y2(xi) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y1(xj) +R1,j(f)) +f(xi, y1(xi)+
+R1,i(f))] +R2,i(f) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y1(xj))+
+f(xi, y1(xi))] +R2,i(f) =y2(xi) +R2,i(f), ∀i= 1, n. (18)
By induction, form∈N,m≥3, we obtain :
ym(xi) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, ym−1(xj) +Rm−1,j(f))+
+f(xi, ym−1(xi) +Rm−1,j(f))] +Rm,i(f) =
=y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, ym−1(xj)) +f(xi, ym−1(xi))]+
Letγ≥0 such that,
|f(t, u)−f(x, u)| ≤γ|t−x|, ∀t, x∈[x0−a, x0+a],∀u∈[y0−b, y0+b]. (20) Iff ∈C1([x
0−a, x0+a]×[y0−b, y0+b]),since
Fm0 (x) =
∂f
∂x(x, ym(x)) + ∂f
∂y(x, ym(x))·y
0
m(x) =
= ∂f
∂x(x, ym(x)) + ∂f
∂y(x, ym(x))·f(x, ym−1(x)),
follows that
Fm0 −1
≤M1(M+ 1), ∀m∈N∗.
where,
M1= max(
∂f ∂x
,
∂f ∂y
).
Theorem 1 If f ∈ C([x0−a, x0+a]×[y0−b, y0+b]) and hL < 1, in the
conditions (2) and (20) then the functions Fm, m∈ N, are Lipschitzian with
the Lipschitz constant γ+LM, and the sequence (ym(xi))m∈N∗ given by (19)
approximate the solution of the Cauchy problem (1) on the knots ti, i= 1, n,
with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+(
1 1−hL)·
h2
4n(γ+LM),∀i= 1, n, m∈N
∗.
(21)
Proof. We have successively, by induction,
|F0(t1)−F0(t2)| ≤γ|t1−t2|, ∀t1, t2∈[x0, x0+h],
|Fm(t1)−Fm(t2)| ≤γ|t1−t2|+L|ym(t1)−ym(t2)| ≤γ|t1−t2|+
+L
t2 Z
t1
|f(s, ym−1(s))|ds≤(γ+LM)|t1−t2|, ∀t1, t2∈[x0, x0+h], ∀m∈N∗.
From the Banach’s fixed point principle we have the following estimation
|y∗(x)−ym(x)| ≤
(Lh)m
1−Lh· ky0−y1kC , ∀x∈[x0, x0+h],∀m∈N
∗, (22)
in the approximation of the solution of (1) by the sequence of successive ap-proximations, where
kukC= max{|u(t)|:t∈[a, b]}, ∀u∈C[a, b].
We observe thatky0−y1kC≤ |y0|+M h. From the relations (19) we have
ym(xi)−ym(xi) =
Rm,i(f)
Moreover,
R2,i(f)
≤ |R2,i(f)|+
h
2n(2L
i−1 X
j=1
|R1,j(f)|+L|R1,i(f)|)≤
≤ |R2,i(f)|+Lh·
h2
4n(γ+LM)≤(1 +hL)· h2
4n(γ+LM), ∀i= 1, n. (23)
and
Rm,i(f)
≤ |Rm,i(f)|+
h
2n[L
Rm−1,i(f)) + 2L
i−1 X
j=1
Rm−1,j(f) ]≤
≤ h
4M000 160n3 +
hL
2n(
Rm−1,i(f)) + 2
i−1 X
j=1
Rm−1,j(f) )≤
≤[1 +hL+...+ (hL)m−1]· h
2
4n(γ+LM), ∀i= 1, n, ∀m∈N
∗. (24) IfhL <1 then
Rm,i(f) ≤(
1 1−hL)·
h2
4n(γ+LM), ∀i= 1, n, ∀m∈N
∗. (25)
Since y
∗(x
i)−ym(xi)
≤ |y
∗(x
i)−ym(xi)|+
ym(xi)−ym(xi)
, ∀i= 1, n, m∈N ∗,
from (22) and (25) follows the inequality (21).
Theorem 2 Iff ∈C1([x
0−a, x0+a]×[y0−b, y0+b]) andhL <1, then the
sequence (ym(xi))m∈N∗ given by (19) approximate the solution of the Cauchy problem (1) on the knotsti, i= 1, n, with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+(
1 1−hL)·
h2
4nM1(M+1),∀i= 1, n, m∈N
∗. (26)
Proof. Since Fm0 −1
≤M1(M+ 1), ∀m∈N∗,
by the inequality (16) in an analogous manner as in the proof of Theorem 1, we obtain
Rm,i(f) ≤(
1 1−hL)·
h2
4nM1(M + 1), ∀i= 1, n, ∀m∈N
∗
If f ∈C1([x0−a, x0+a]×[y0−b, y0+b]) we can consider the Lipschitz conditions for the first partial derivatives :
∂f
∂x(t1, u)− ∂f ∂x(t2, u)
≤L10|t1−t2|, ∀t1, t2∈[x0, x0+h],∀u∈[y0−b, y0+b] (27)
∂f
∂x(t, u1)− ∂f ∂x(t, u2)
≤L01|u1−u2|, ∀t∈[x0, x0+h],∀u1, u2∈[y0−b, y0+b] (28)
∂f
∂y(t1, u)− ∂f ∂y(t2, u)
≤L12|t1−t2|, ∀t1, t2∈[x0, x0+h],∀u∈[y0−b, y0+b] (29)
∂f
∂y(t, u1)− ∂f ∂y(t, u2)
≤L21|u1−u2|, ∀t∈[x0, x0+h],∀u1, u2∈[y0−b, y0+b] (30) whereL10≥0, L01≥0, L12≥0, L21≥0.
Theorem 3 If f ∈ C1([x
0−a, x0+a]×[y0−b, y0+b]) and Lh < 1, in the
conditions (27)-(30) then the functionsFm0 , m∈N, are Lipschitzian with the
Lipschitz constant
L0=L10+M(L01+L12) +M2L21+M1(γ+LM),
and the sequence (ym(xi))m∈N∗ given by (19) approximate the solution of the Cauchy problem (1) on the knotsti, i= 1, n, with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h) + (
1 1−hL)·
L0h3
12n2, ∀i= 1, n,∀m∈N ∗.
(31)
Proof. Firstly, we obtain
|F00(t1)−F00(t2)| ≤(L10+M L01)|t1−t2|, ∀t1, t2∈[x0, x0+h] and by induction follows for anyt1, t2∈[x0, x0+h] and m∈N∗ that,
|Fm0 (t1)−Fm0 (t2)| ≤
∂f
∂x(t1, ym(t1))− ∂f
∂x(t2, ym(t2))
+
+
∂f
∂y(t1, ym(t1))− ∂f
∂y(t2, ym(t2))
· |f(t1, ym−1(t1))|+
∂f
∂y(t2, ym(t2))
·
· |f(t1, ym−1(t1))−f(t2, ym−1(t2))| ≤L10|t1−t2|+L01|ym(t1)−ym(t2)|+ +M(L12|t1−t2|+L21|ym(t1)−ym(t2)|) +M1(γ+LM)|t1−t2| ≤
From inequality (16), in a similar way as in the Theorem 1, we obtain the estimation (31).
Iff ∈C2([x0−a, x0+a]×[y0−b, y0+b]) we have,
Fm00(x) =∂ 2f
∂x2(x, ym(x)) + 2
∂2f
∂x∂y(x, ym(x))·f(x, ym−1(x))+
+∂ 2f
∂y2(x, ym(x))·[f(x, ym−1(x))] 2+∂f
∂y(x, ym(x))·[ ∂f
∂x(x, ym−1(x))+
+∂f
∂y(x, ym−1(x))·f(x, ym−2(x))]
and therefore
|Fm00(x)| ≤M2(M + 1)2+M12(M+ 1) =M
00, ∀x∈[x
0, x0+h], (32) where
M2= max{
∂2f
∂x2
,
∂2f
∂x∂y
,
∂2f
∂y2
}.
Theorem 4 If f ∈C1([x
0−a, x0+a]×[y0−b, y0+b]) andLh <1 then the
sequence (ym(xi))m∈N∗ given by (19) approximate the solution of the Cauchy problem (1) on the knotsti, i= 1, n, with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+(
1 1−hL)·
M00h3
12n2 , ∀i= 1, n,∀m∈N ∗.
(33)
Proof. Using the inequality (32) in a similar way as in the Theorem 1, we obtain the estimation (33).
3
Numerical methods using the perturbed
trape-zoidal quadrature rule
Using the perturbed trapezoidal quadrature rule (8), with the remainder esti-mation (9) and the method of successive approxiesti-mations we have obtained in [6] the numerical method :
ym(xi) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, ym−1(xj)) +f(xi, ym−1(xi))]−
− h
2
12n2[
∂f
∂x(xi, ym−1(xi)) + ∂f
∂y(xi, ym−1(xi))·f(xi, ym−2(xi))−
−∂f
∂x(x0, y0)− ∂f
∂y(x0, y0)·f(x0, y0)] +Rm,i(f), ∀i= 1, n,∀m∈N
with the remainder estimation,
|Rm,i(f)| ≤
h4 160n3 ·
Fm000−1
, ∀i= 1, n,∀m∈N∗. (35) After elementary calculus we have,
Fm000−1(x) =∂ 3f
∂x3(x, ym(x)) + 3
∂3f
∂x2∂y(x, ym(x))·y 0
m(x) + 3
∂3f
∂x∂y2(x, ym(x))·
·[y0m(x)]2+∂ 3f
∂y3(x, ym(x))·[y 0
m(x)]
3+ 3 ∂2f
∂x∂y(x, ym(x))·y
00
m(x)+
3∂ 2f
∂y2(x, ym(x))·y 0
m(x)·y
00
m(x)+
∂f
∂y(x, ym(x))·y
000
m(x), ∀x∈[x0, x0+h], ∀m∈N.
and,
ym00(x) = ∂f
∂x(x, ym−1(x)) + ∂f
∂y(x, ym−1(x))·y
0
m−1(x),
y000m(x) =
∂2f
∂x2(x, ym−1(x)) + 2
∂2f
∂x∂y(x, ym−1(x))·y
0
m−1(x) +
∂2f
∂y2(x, ym−1(x))·
·[ym0 −1(x)]2+∂f
∂y(x, ym−1(x))·y
00
m−1(x), ∀x∈[x0, x0+h],∀m∈N∗.
y00(x) =y000(x) =y0000(x) = 0, ∀x∈[x0, x0+h]. For
∂αf
∂xα1∂yα2
= max{
∂αf(x, y)
∂xα1∂yα2
:x∈[x0, x0+h], y∈[y0−b, y0+b]}
and|α| ≤3, α1+α2=|α|, α1, α2∈ {0,1,2,3}, let
M1= max{ ∂f ∂x , ∂f ∂y },
M2= max{
∂2f ∂x2 ,
∂2f ∂x∂y ,
∂2f ∂y2 }
M3= max{
∂3f ∂x3 ,
∂3f ∂x2∂y
,
∂3f ∂x∂y2 ,
∂3f ∂y3 }. So,
|Fm000(x)| ≤M3+ 3M M3+ 3M3M2+M3M3+ 3M1M2(M+ 1)+ +3M M1M2(M+ 1) +M1[M2+ 2M M2+M2M2+M12(M+ 1) =M3(M+ 1)3+
and therefore, the inequality (35) became:
|Rm,i(f)| ≤
h4M000
160n3, ∀i= 1, n,∀m∈N
∗. (36)
The method (34) lead to the algorithm,
ym(x0) =y0, ∀m∈Nandy0(xi) =y0, ∀i= 0, n,
y1(xi) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y0(xj)) +f(xi, y0(xi))]−
− h
2
12n2[
∂f
∂x(xi, y0(xi))− ∂f
∂x(x0, y0)] +R1,i(f) =y0+ h
2n[f(x0, y0)+
+2
i−1 X
j=1
f(xj, y0) +f(xi, y0)]−
h2 12n2[
∂f
∂x(xi, y0)− ∂f
∂x(x0, y0)]+
+R1,i(f) =y1(xi) +R1,i(f), ∀i= 1, n
y2(xi) =y0+
h
2n[f(x0, y0)+2
i−1 X
j=1
f(xj, y1(xj)+R1,j(f))+f(xi, y1(xi)+R1,i(f))]
− h
2
12n2[
∂f
∂x(xi, y1(xi) +R1,i(f)) + ∂f
∂y(xi, y1(xi) +R1,i(f))·f(xi, y0)− ∂f
∂x(x0, y0)
−∂f
∂y(x0, y0)·f(x0, y0)] +R2,i(f) =y0+ h
2n[f(x0, y0) + 2
i−1 X
j=1
f(xj, y1(xj))+
+f(xi, y1(xi))]−
h2 12n2[
∂f
∂x(xi, y1(xi)) + ∂f
∂y(xi, y1(xi))·f(xi, y0)− ∂f
∂x(x0, y0)
−∂f
∂y(x0, y0)·f(x0, y0)] +R2,i(f) =y2(xi) +R2,i(f), ∀i= 1, n
By induction, form∈N,m≥3, we have obtained :
ym(xi) =y0+
h
2n[f(x0, y0)+2
i−1 X
j=1
f(xj, ym−1(xj)+Rm−1,j(f))+f(xi, ym−1(xi)+
+Rm−1,j(f))]−
h2 12n2[
∂f
∂x(xi, ym−1(xi) +Rm−1,j(f)) + ∂f
∂y(xi, ym−1(xi)+
+Rm−1,j(f))·f(xi, ym−2(xi) +Rm−2,j(f))−
∂f
∂x(x0, y0)− ∂f
∂y(x0, y0)·f(x0, y0)]
+Rm,i(f) =y0+
h
2n[f(x0, y0) + 2
i−1 X
j=1
− h
2
12n2[
∂f
∂x(xi, ym−1(xi)) + ∂f
∂y(xi, ym−1(xi))·f(xi, ym−2(xi))− ∂f
∂x(x0, y0)−
−∂f
∂y(x0, y0)·f(x0, y0)]+Rm,i(f) =ym(xi)+Rm,i(f), ∀i= 1, n, ∀m≥3. (37)
For the remainder estimation we have obtained ( for m ∈ N, m ≥ 2 ) the
inequality:
Rm,i(f) ≤
1−[hL+12hn22(M2(M+ 1) +M1L)]m 1−[hL+12hn22(M2(M + 1) +M1L)]
· h
4M000
160n3 ,∀i= 1, n, (38) which lead to the following result :
Theorem 5 ( in [6] ) : IfLh <11
12 andf ∈C 3([x
0−a, x0+a]×[y0−b, y0+b])
then, for n ∈ N, n > hp2 M
2(M + 1) +M1L, the sequence (ym(xi))m∈N given in (37) approximates on the knotsxi ,i= 1, n, the solutiony∗ of the Cauchy’s
problem (1) with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+
+ h
4M000 160n3(1−[hL+ h2
12n2(M2(M + 1) +M1L)])
, ∀i= 1, n,∀m∈N, m≥2.
(39) With the perturbed trapezoidal quadrature rule (8) having the remainder estimation (10), in [7] we have obtained for the numerical method given in (34) and (37) the error estimation :
Rm,i(f) ≤
h5L000
720n4(1−[hL+ h2
12n2(M2(M+ 1) +M1L)])
, ∀i= 1, n,∀m≥2,
(40) where L000 ≥ 0 is the Lipschitz constant of the functionsFm000, m ∈ N, in the
Lipschitz conditions of the third order derivatives
∂3f
∂x3,
∂3f
∂x2∂y,
∂3f
∂x∂y2,
∂3f
∂y3
in each argument.
In this case we have obtained in [7] the following result :
Theorem 6 ( in [7] ) : Iff ∈C3([x
0−a, x0+a]×[y0−b, y0+b]), Lh < 1112
and the third order partial derivatives
∂3f ∂x3,
∂3f ∂x2∂y,
∂3f ∂x∂y2,
are Lipschitzian in each argument then the sequence(ym(xi))m∈Napproximates the solution of the initial value problem (1) on the knotsxi ,i= 1, n, and for
n∈N, n > hp2 M2(M+ 1) +M1L, with the error estimation :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+
+ h
5L000 720n4(1−[hL+ h2
12n2(M2(M + 1) +M1L)])
, ∀i= 1, n,∀m∈N, m≥2.
(41) Iff ∈C4([x
0−a, x0+a]×[y0−b, y0+b]),then for
M4= max{
∂4f
∂xi∂yj
:i, j= 0,4, i+j= 4}
and after elementary calculus we obtain the fourth derivative Fm(IV)(x) which
has 13 terms,
Fm(IV)(x) = ∂ 4f
∂x4(x, ym(x)) + 4
∂4f
∂x3∂y(x, ym(x))·y 0
m(x) +
∂4f
∂x4(x, ym(x))·
·[ym0 (x)]4+ 4 ∂ 4f
∂x∂y3(x, ym(x))·[y 0
m(x)]
3+ 3 ∂4f
∂x3∂y(x, ym(x))·[y 0
m(x)]
2+
3 ∂ 4f
∂x2∂y2(x, ym(x))·[y 0
m(x)]
2+ 6 ∂3f
∂x2∂y(x, ym(x))·y 00
m(x) + 12
∂3f
∂x∂y2(x, ym(x))·
·ym0 (x)·y00m(x) + 6∂ 3f
∂y3(x, ym(x))·[y 0
m(x)]
2·y00
m(x) + 4
∂2f
∂x∂y(x, ym(x))·y
000
m(x)+
3∂ 2f
∂y2(x, ym(x))·[y 00
m(x)]
2+4∂ 2f
∂x2(x, ym(x))·y 0
m(x)·y
000
m(x)+
∂f
∂y(x, ym(x))·y
IV m (x)
and the following estimations holds
F
(IV)
m (x)
≤M4(M+1) 4+6M
1M3(M+1)2+3M2M12(M+1) 2+4M2
2(M+1) 3+
+4M2M12(M + 1) 2+M
1M000 =MIV, ∀x∈[x0, x0+h], ∀m∈N.
Then, for the remainder from (34) we have the estimation,
|Rm,i(f)| ≤
h5MIV
720n4 , ∀i= 1, n,∀m∈N ∗.
Theorem 7 ( in [6] ) :If f ∈ C4([x0−a, x0 +a]×[y0−b, y0+b]), in the
conditions of Theorem 5, the error estimation (28) became :
y
∗(x
i)−ym(xi)
≤
(Lh)m
1−Lh(|y0|+M h)+
+ h
5MIV
720n4(1−[hL+ h2
12n2(M2(M + 1) +M1L)])
, ∀i= 1, n,∀m∈N, m≥2,
(42)
∀i= 1, n, ∀m∈N, m≥2.
Remark 8 The error estimations given in the Theorems 1-7 present the adapt-ability of the method of successive approximations, corresponding to the proper-ties of the functionf.
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