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14-point difference operator for the approximation of the first derivatives of a solution of Laplace’s equation in a rectangular parallelepiped

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Abstract

A 14-point difference operator is used to construct finite difference problems for the approxi- mation of the solution, and the first order derivatives of the Dirichlet problem for Laplace’s equations in a rectangular parallelepiped. The boundary functions φj on the faces Γj, j = 1, 2, …, 6 of the parallelepiped are supposed to have pth order derivatives satisfying the Hölder condition, i.e., φ j ε Cp,λ (Γj), 0 < λ < 1, where p = {4, 5}. On the edges, the boundary functions as a whole are continuous, and their second and fourth order derivatives satisfy the compatibility conditions which result from the Laplace equation. For the error uh - u of the approximate solution uh at each grid point (x1, x2, x3), |uh-u| ≤ cρ p-4 (x1, x2, x3)h⁴ is obtained, where u is the exact solution, ρ = ρ(x1, x2, x3) is the distance from the current grid point to the boundary of the parallelepiped, h is the grid step, and c is a constant independent of ρ and h. It is proved that when φj ε C p,λ, 0 < λ < 1, the proposed difference scheme for the approximation of the first derivative converges uniformly with order O(hp-1), p ε{4, 5}.
Filomat 32:3 (2018), 791–800
https://doi.org/10.2298/FIL1803791D
University of Niˇ
s, Serbia
Available at: http://www.pmf.ni.ac.rs/filomat
14-Point Dierence Operator for the Approximation of the First
Derivatives of a Solution of Laplace’s Equation in a Rectangular
Parallelepiped
aNear East University, Department of Mathematics, Nicosia, KKTC, Mersin 10, Turkey
Abstract. A 14-point dierence operator is used to construct ﬁnite dierence problems for the approxi-
mation of the solution, and the ﬁrst order derivatives of the Dirichlet problem for Laplace’s equations in a
rectangular parallelepiped. The boundary functions ϕjon the faces Γj,j=1,2, ..., 6 of the parallelepiped
are supposed to have pth order derivatives satisfying the H¨
older condition, i.e., ϕjCp (Γj), 0 < λ < 1,
where p={4,5}.On the edges, the boundary functions as a whole are continuous, and their second and
fourth order derivatives satisfy the compatibility conditions which result from the Laplace equation. For
the error uhuof the approximate solution uhat each grid point (x1,x2,x3),|uhu|cρp4(x1,x2,x3)h4is
obtained, where uis the exact solution, ρ=ρ(x1,x2,x3) is the distance from the current grid point to the
boundary of the parallelepiped, his the grid step, and cis a constant independent of ρand h. It is proved
that when ϕjCp,0<λ<1,the proposed dierence scheme for the approximation of the ﬁrst derivative
converges uniformly with order O(hp1),p∈ {4,5}.
1. Introduction
It is well known that the use of dierence operators with a low number of pattern and with the highest
order of accuracy for the approximate solution of dierential equations reduces the eective realization of
the obtained system of ﬁnite-dierence equations. Moreover, to enlarge the class of applied problems the
convergence of the dierence solutions are preferred to be investigated under the weakened assumptions on
the smoothness of the boundary conditions. All of these become more valuable in 3D problems, especially
the derivatives of the unknown solution are sought.
The application of derivatives arise in many applied problems such as problems in electrophysics in
which the ﬁrst derivatives of the potential function deﬁne the electrostatic ﬁeld [7], and in the fracture
problems where the ﬁrtst derivatives of the stress function deﬁne the components of the tangential stress
[8].
The investigation of approximate derivatives started in [9],where it was proved that the high order
dierence derivatives uniformly converge to the corresponding derivatives of the solution for the 2D
Laplace equation in any strictly interior subdomain, with the same order hwith which the dierence
solution converges on the given domain. The uniform convergence of the dierence derivatives over the
2010 Mathematics Subject Classiﬁcation. Primary 65M06; Secondary 65M12, 65M22
Keywords. Finite dierence method, approximation of the ﬁrst derivatives, error estimations, Laplace’s equation on parallelepiped
Received: 8 December 2016; 13 February 2017; Accepted: 19 March 2017
Communicated by Allaberen Ashyralyev
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 792
whole grid domain to the corresponding derivatives of the solution for the 2D Laplace equation with the
order O(h2) was proved in [14].In [5],for the ﬁrst and pure second derivatives of the solution of the 2D
Laplace equation special ﬁnite dierence problems were investigated. It was proved that the solution of
these problems converge to the exact derivatives with the order O(h4).
In [17] for the 3D Laplace equation the convergence of order O(h2) of the dierence derivatives to the
corresponding ﬁrst order derivatives of the exact solution is proved. It was assumed that ton the faces
the boundary functions have third derivatives satisfying the H¨
older condition. Furthermore, they are
continuous on the edges, and their second derivatives satisfy the compatibility condition that is implied
by the Laplace equation. Whereas in [16] when the boundary values on the faces of a parallelepiped are
supposed to have the fourth derivatives satisfying the H¨
older condition, the constructed dierence schemes
converge with order O(h2) to the ﬁrst and pure second derivatives of the exact solution. In [5] it is assumed
that the boundary functions on the faces have sixth order derivatives satisfying the H¨
older condition, and
the second and fourth order derivatives satisfy some compatibility conditions on the edges. Dierent
dierence schemes with the use of the 26-point dierence operator are constructed on a cubic grid with
mesh size h,to approximate the ﬁrst and pure second derivatives of the solution of the Dirichlet problem
with order O(h4).
In this paper O(hp1),p=4,5 order of approximation for the ﬁrst derivatives of the solution of 3D
Laplace’s equation is obtained under weaker assumptions on the smoothness of the boundary functions on
the faces of the parallelepiped than those used in [6]. Moreover, to construct the ﬁnite dierence problems
the dierence operator with a lower number of pattern is used.
Finally, the obtained theoretical results are supported by the illustration of numerical results.
2. Some Properties of a Solution of the Dirichlet Problem on a Rectangular Parallelepiped
Let R={(x1,x2,x3): 0 <xi<ai,i=1,2,3}be an open rectangular parallelepiped; Γj,j=1,2, ..., 6 be
its faces including the edges; Γjfor j=1,2,3 (j=4,5,6) belongs to the plane xj=0 ( xj3=aj3),and let
Γ = 6
j=1Γjbe the boundary of R;γµν = ΓµΓνbe the edges of the parallelepiped R.Ck(E) is the class of
functions that have continuous kth derivatives satisfying the H¨
older condition with an exponent λ(0,1).
Consider the boundary value problem
u=0 on R,u=ϕjon Γj,j=1,2, ..., 6 (1)
where 2
x2
1
+2
x2
2
+2
x2
3
, ϕjare given functions.
Assume that
ϕjCpΓj,0<λ<1,j=1,2, ..., 6,p{4,5}(2)
ϕµ=ϕνon γµν,(3)
2ϕµ
t2
µ
+2ϕν
t2
ν
+2ϕµ
t2
µν
=0 on γµν, (4)
4ϕµ
t4
µ
+4ϕµ
t2
µt2
µν
=4ϕν
t4
ν
+4ϕν
t2
νt2
νµ
on γµν,(5)
where 1 µ < ν 6, ν µ,3,tµν is an element in γµν ,and tµand tνis an element of the normal to γµν on
the face Γµand Γν,respectively.
The following Lemma follows from Theorem 2.1 in [12].
Lemma 2.1. Under conditions (2)(5),the solution u of the Dirichlet problem (1)belong to the H¨older class Cp(R),
0<λ<1,p{4,5}.
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 793
Lemma 2.2. Let ρ(x1,x2,x3)be the distance from the current point of the open parallelepiped R to its boundary and
let
lα1
x1+α2
x2+α3
x3, α2
1+α2
2+α2
3=1.
Then the next inequality holds
6u(x1,x2,x3)
l6
cρp+λ6(x1,x2,x3),(x1,x2,x3)R and p {4,5}(6)
where u is the solution of the problem (1),c is a constant independent of the direction of derivative
l.
Proof. Since uCp(R),p{4,5}(Lemma 2.1) the proof Lemma 2.2 follows with the use of Lemma 3 in [10]
(Chap.4, Sec.3)
3. Finite Dierence Problem
We introduce a cubic grid with a step h>0 deﬁned by the planes xi=0,h,2h, ..., i=1,2,3. It is assumed
that the edge lengths of Rand hare such that ai
h4 (i=1,2,3) are integers.
Let Dhbe the set of nodes of the grid constructed, Rh=RDh,Rh=RDh,Rk
hRhbe the set of nodes
of Rhlying at a distance of kh away from the boundary Γof R,and Γh= Γ Dh.
The 14-point dierence operator Son the grid is deﬁned as (see [19])
Su(x1,x2,x3)=1
56
8
6
X
p=1(1)
up+
14
X
q=7(3)
uq
,(x1,x2,x3)Rh,(7)
where P(m)is the sum extending over the nodes lying at a distance of m1/2haway from the point (x1,x2,x3)
and upand uqare the values of uat the corresponding nodes.
On the boundary Γof R, we deﬁne continuous on the entire boundary including the edges of R, the
function ϕas follows
ϕ=
ϕ1on Γ1
ϕjon Γj\ j1
i=1Γi!,j=2,3, ..., 6.(8)
Obviously,
ϕ=ϕjon Γj,j=1,2, ..., 6.
We consider the ﬁnite dierence problem approximating Dirichlet problem (1):
uh=Suhon Rh,uh=ϕon Γh,(9)
where Sis the dierence operator given by (7)and ϕis the function deﬁned by (8). By maximum principle,
the system (9)has a unique solution (see [11], Chap. 4).
In what follows and for simplicity, we denote by c,c1,c2, ... constants, which are independent of hand
the nearest factors, the identical notation will be used for various constants.
Consider two systems of grid equations
vh=Svh+1h,on Rh,vh=0 on Γh,(10)
vh=Svh+1h,on Rh,vh=0 on Γh,(11)
where 1hand 1hare given functions and 1h1hon Rh.
Lemma 3.1. The solutions vhand vhof systems (10)and (11)satisfy the inequality
vhvhon Rh.
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 794
The proof of Lemma 3.1 is similar to that of the comparison theorem in [11] (Chap.4, Sec.3).
Deﬁne
N(h)=min {a1,a2,a3}
2h,(12)
where [a]is the integer part of a.
Consider for a ﬁxed k,1kN(h) the systems of grid equations
vk
h=Svk
h+1k
hon Rk
h,vk
h=0 on Γh,(13)
where
1k
h=(1, ρ (x1,x2,x3)=kh,
0, ρ (x1,x2,x3),kh.
Lemma 3.2. The solution vk
hof the system (13)satisﬁes the inequality
vk
h(x1,x2,x3)Tk
h,1kN(h),(14)
where Tk
his deﬁned as
Tk
h=Tk
h(x1,x2,x3)=(5ρ
h,0ρ(x1,x2,x3)kh,
5k, ρ (x1,x2,x3)>kh.(15)
Proof. By the direct calculation of the expression STk
h,we obtain
Tk
hSTk
h(1, ρ (x1,x2,x3)=kh,
0, ρ (x1,x2,x3),kh,(16)
on Rh.On the basis of (13), inequalities (16) and taking the boundary condition Tk
h=0 on Γhinto account,
by Lemma 3.1, we get (14).
Let x0=(x10,x20 ,x30),be some point in Rh.By Taylor’s formula for the solution uof the problem (1)
around the point x0,we have
u(x1,x2,x3)=p5(x1,x2,x3;x0)+r5(x1,x2,x3;x0),(17)
where p5is ﬁfth-degree Taylor polynomial and r5is remainder.
Since uis a harmonic function and Sis linear, by taking into account that Sp5(x10 ,x20,x30;x0)=
u(x10,x20 ,x30) from (17) follows
Su(x10,x20 ,x30)=u(x10 ,x20,x30)+Sr5(x10 ,x20,x30 ;x0).(18)
Lemma 3.3. The following estimation holds
max
(x1,x2,x3)Rk
h
|Su u|c4
hp+λ
k6pλ,k=1,2, ..., N(h),p{4,5},(19)
where u is the solution of the Dirichlet problem (1),S is the dierence operator deﬁned by (7),and N(h)is given by
(12).
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 795
Proof. Let x0=(x10,x20,x30)be some point in R1
h,and let Pmnq =x10 +mh,x20 +nh,x30 +qh, where m,n,q=
0,±1,m2+n2+q2,0,be any point in the pattern of operator S.Then by using the integral form of the
remainder term of Taylor’s formula for each point Pmnq by virtue of Lemma 2.1 and Lemma 2.2, we obtain
r5x10 +mh,x20 +nh,x30 +qh;x0chp+λ,p∈ {4,5}.(20)
From the structure (7) of the operator Sfollows that its norm in the uniform metric is equal to one, then
using by (20), we have
|Sr5(x10,x20 ,x30;x0)|chp+λ,p∈ {4,5}.(21)
On the basis of (18) and (21) follows the inequality (19), for k=1.Let x0Rk
hbe an arbitrary point for
2kN(h),and let r5(x1,x2,x3;x0)be remainder term of the Taylor formula (17)in the Lagrange form.
Then Sr6(x10,x20 ,x30;x0)can be expressed linearly in terms of the 14 number of sixth derivatives of uat
some points on the open intervals connecting the points of pattern of the operator Swith the point x0.The
sum of the absolute values of the coecients multiplying the sixth derivatives does not exceed ch6which is
independent k0(2k0N(h))or the point x0Rk0
h.Using the estimation of the sixth derivatives by Lemma
2.2, for all k,2kN(h),we obtain
|Sr5(x10,x20 ,x30;x0)|c1
h6
(kh)6pλ=c1
hp+λ
k6pλ.(22)
By virtue of (18)and (22)follows the estimation (19).
Theorem 3.4. Assume that the boundary functions ϕjsatisfy conditions (2)(5). Then at each point (x1,x2,x3)Rh
|uhu|c0h4ρp4,p{4,5},(23)
where uhis the solution of the ﬁnite dierence problem (9), u is the exact solution of problem (1),and ρ=ρ(x1,x2,x3)
is the distance from the current point (x1,x2,x3)Rhto the boundary of the rectangular parallelepiped R.
Proof. Let εk
h,1kN(h),be a solution of the system
εk
h=Sεk
h+µk
hon Rh, εk
h=0 on Γh,(24)
where
µk
h=(Su uon Rk
h
0 on Rh\Rk
h..(25)
Let
εh=uhu on Rh.(26)
By (9)and (26)the error function εhsatisﬁes the system of equations
εh=Sεh+(Su u)on Rh, εh=0 on Γh.(27)
We represent a solution of the system (27)as follows
εh=
N(h)
X
k=1
εk
h,(28)
where N(h) deﬁned by (12), εk
h,1kN(h),is a solution of the system
εk
h=Sεk
h+σk
hon Rh, εk
h=0 on Γh,(29)
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 796
when
σk
h=(Su uon Rk
h
0 on Rh\Rk
h.(30)
Then on the basis of (28), (29), (30), Lemma 3.2 and Lemma 3.3,for the solution of (27), we have
|εh|
N(h)
X
k=1εk
h
N(h)
X
k=1
Tk
hmax
(x1,x2,x3)Rk
h
|Su u|c1h6
N(h)
X
k=1
Tk
h
(kh)6pλ
5c1hp+λ
ρ/h1
X
k=1
k
(k)6pλ+5c1h6
N(h)
X
k=ρ/h
ρ/h
(kh)6pλ
5c1hp+λ
ρ/h1
X
k=1
k5+p+λ+5c1hp1+λρ
N(h)
X
k=ρ/h
k6+p+λ
c2h4ρ4+p+λ+c3h4ρc4h4ρp4,p{4,5}.(31)
From (26)and (31),for any point (x1,x2,x3)Rh, we obtain
|uhu|=|εh|c0h4ρp4(x1,x2,x3),p∈ {4,5}.
4. Approximation of the First Derivative
4.1. Boundary Function is from C5
Let the boundary functions ϕj,j=1,2, ..., 6,in problem (1)on the faces Γjbe satisﬁed the conditions
ϕjC5Γj,0<λ<1,j=1,2, ..., 6,(32)
i.e., p=5 in (2).Let ube a solution of the problem (1)with the conditions (32)and (3)(5).
We put v=u
x1,and Φj=u
x1on Γj,j=1,2, ..., 6.It is obvious that the function vis a solution of the
following boundary value problem
v=0 on R,v= Φjon Γj,j=1,2, ..., 6,(33)
where uis a solution of the problem (1)for p=5.
We deﬁne an approximate solution of problem (33) as a solution of the following ﬁnite dierence problem
νh=Sνhon Rh, νh= Φjh (uh) on Γh
j,j=1,2, ..., 6,(34)
where uhis the solution of the problem (9), Φ1h(Φ4h) is the fourth order forward (backward) numerical
dierentiation operator (see [1], [2]) used in [6] with the 26-point dierence operator. On the nodes Γh
p,the
boundary values are deﬁned as Φph(uh)=∂ϕp
x1,p=2,3,5,6.
Theorem 4.1. The estimation is true
max
(x1,x2,x3)Rh
νhu
x1
ch4,(35)
where u is the solution of the problem (1), νhis the solution of the ﬁnite dierence problem (34).
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 797
Proof. Let
h=νhνon Rk,(36)
where ν=u
x1.From (34)and (36),we have
h=Sh+(Sνν)on Rh,
h= Φkh(uh)νon Γh
k,k=1,4, h=0 on Γh
p,p=2,3,5,6.
We put
h=1
h+2
h,(37)
where
1
h=S1
hon Rh,(38)
1
h= Φkh(uh)νon Γh
k,k=1,4, 1
h=0 on Γh
q,q=2,3,5,6; (39)
2
h=S2
h+(Sνν)on Rh, 2
h=0 on Γh
j,j=1,2, ..., 6.(40)
First, we estimate the dierence Φkh(uh)νon Γh
k,k=1,4 using the representation
Φkh(uh)ν=(Φkh (uh)Φkh(u))+(Φkh(u)v)(41)
Since Φkh(u),k=1,4 are the fourth order approximation of u/∂x1on Γkand by Lemma 2.1 the ﬁfth order
partial derivatives of the solution uare bounded in R, the dierence Φkh(u)vhas estimation (see [1], [2])
max
k=1,4max
(x1,x2,x3)Γk
k
|Φkh(u)ν|c1h4.(42)
To estimate Φkh(uh)Φkh (u),we take the fourth order forward formula (k=1),
Φ1h(uh)=1
12h[25ϕ1(x2,x3)+48uh(h,x2,x3)36uh(2h,x2,x3)(43)
+16uh(3h,x2,x3)3uh(4h,x2,x3)] on Γh
1.
Using the pointwise estimation (24) in Theorem 3.4, when p=5,and taking into account the values of the
distance function ρ(x1,x2,x3) in the formula (43), we have
|Φ1h(uh)Φ1h(u)|c2h4.(44)
The estimation (44) is true for the backward formula (k=4),also. On the basis of (41), (42), (44), by using
the maximum principle, for the solution of system (38),(39),we have
max
(x1,x2,x3)Rh1
hc3h4.(45)
The solution ε2
hof system (40)is the error function of the ﬁnite dierence solution for problem (33),when
the boundary functions Φj=u/∂x1,j=1,2, ..., 6,as follows from (2) (5) satisfy the conditions
ΦjC4Γj,0<λ<1,j=1,2, ..., 6,
Φµ= Φνon γµν,
2Φµ
t2
µ
+2Φν
t2
ν
+2Φµ
t2
µν
=0 on γµν
Then, on the basis of Theorem 4 in [19] for the error ε2
h, we have
max
(x1,x2,x3)Rh2
hc4h4.(46)
By virtue of (37),(45),and (46)follows the inequality (35).
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 798
4.2. Boundary Function From C4
Let the boundary functions ϕjC4Γj,0< λ < 1,j=1,2, ..., 6,in (1)(5),i.e., p=4 in (2),and let
v=u
x1and let Φj=u
x1on Γj,j=1,2, ..., 6,and consider the boundary value problem:
v=0 on R,v= Φjon Γj,j=1,2, ..., 6,(47)
where uis a solution of the boundary value problem (1).
We deﬁne the following third order numerical dierentiation operators Φνh, ν =1,4
Φ1h(uh)=1
6h[11ϕ1(x2,x3)+18uh(h,x2,x3)9uh(2h,x2,x3)
+2uh(3h,x2,x3)] on Γh
1,(48)
Φ4h(uh)=1
6h[11ϕ4(x2,x3)18uh(a1h,x2,x3)+9uh(a12h,x2,x3)
2uh(a13h,x2,x3)] on Γh
4,(49)
and we put
Φph(uh)=∂ϕp
x1
,on Γh
p,p=2,3,5,6 (50)
where uhis the solution of the ﬁnite dierence problem (9).
It is obvious that Φj,j=1,2, ..., 6,satisfy the conditions
ΦjC3Γj,0<λ<1,j=1,2, ..., 6,(51)
Φµ= Φνon γµν,(52)
2Φµ
t2
µ
+2Φν
t2
ν
+2Φµ
t2
µν
=0 on γµν.(53)
Let νhbe the solution of the following ﬁnite dierence problem
νh=Sνhon Rh, νh= Φjh on Γh
j,j=1,2, ..., 6,(54)
where Φjh,j=1,2, ..., 6,are deﬁned by (48)(50).
Theorem 4.2. Let the boundary function ϕjC4 (Γj),j=1, ..., 6.The estimation is true
max
(x1,x2,x3)Rh
νhu
x1
ch3,(55)
where u is the solution of the problem (1), νhis the solution of the ﬁnite dierence problem (54).
Proof. The proof of Theorem 4.2 is similar to that of Theorem 4.1, with the following dierences in estimation
for the errors 1
hand 2
hin (37) : (i) putting p=4 in Lemma 2.1 and Theorem 3.4, and taking into account
that the formulae (48) and (49) are the third order, the estimation
max
(x1,x2,x3)Rh1
hc5h3.
is proved. (ii) on the basis of (51)-(53) and Theorem 2 in [19], we obtain
max
(x1,x2,x3)Rh1
hc6h3.
Remark 4.3. We have investigated the method of high order approximations of the ﬁrst derivative u/∂x1.
The same results are obtained for the derivatives u/∂xl,l=2,3 analogously, by using the same order
forward and backward formulae in appropriate faces of the parallelepiped.
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 799
5. Numerical Results
Example 5.1. Let R={(x1,x2,x3): 0 <xi<1,i=1,2,3},and let Γj,j=1, ..., 6 be its faces. We consider the
following problem:
u=0 on R,u=ϕ(x1,x2,x3)on Γj,j=1, ..., 6,(56)
where
ϕ(x1,x2,x3)=x31
22
x2
1+x2
2
2
+x2
1+x2
2(5+1
30 )
2.cos 5+1
30.arctan x2
x1 (57)
is the exact solution of this problem,which is in C5,1/30.
We solve the system (9) and (34) to ﬁnd the approximate solution uhfor uand approximate ﬁrst derivative
vhfor u
x1respectively.
In Tables 1 and 2 the maximum errors are given. Table 1 shows that the convergence order more than 4
which is corresponds to the product ρin Theorem 3.4. Table 2 justiﬁed estimation (35) in Theorem 4.1, i.e.,
the fourth order convergence.
hkuhukRhEm
u
237,5172E09 32,13
242,3396E10 32,66
257,1637E12 32,74
262,1883E13 32,75
276,6827E15
Table1 Errors for the solution in maximum norm
hkvhvkRhEm
v
234.5436E03 13,40
243.3909E04 14,76
252.2975E05 15,40
261.4922E06 15,70
279.5053E08
Table 2 Errors for the ﬁrst derivative in maximum norm with the fourth-order formulae
Example 5.2. Let ube a solution of problem (56) when the boundary function ϕis chosen from C4,1
30 as
ϕ(x1,x2,x3)=x31
22
x2
1+x2
2
2
+x2
1+x2
2(4+1
30 )
2.cos 4+1
30.arctan x2
x1.
Table 3 and 4 give the fourth order convergence when boundary function is from C4,1
30 for both, solution
and ﬁrst derivative which are the numerical justiﬁcation of Theorem 3.4 and 4.2 respectively.
hkuhukRhEm
u
233,4801E08 16,20
242,1486E09 16,36
251,3135E10 16,37
268,0228E12 16,37
274,8998E13
A.A. Dosiyev, H. Sarikaya /Filomat 32:3 (2018), 791–800 800
Table 3 Errors for the solution in maximum norm
hkvhvkRhEm
v
238,5126E03 6,49
241,3161E03 7,29
251,8065E04 7,66
262,3598E05 7,83
273,0144E06
Table 4 Errors for the ﬁrst derivative in maximum norm with the third-order formulae
In Tables 1-4 we have used the following notations:
kUhUkRh=max
Rh
|UhU|and Em
U=kUU2mkRh
kUU2(m+1)kRh
where Ube the exact solution of the continuous
problem, and Uhbe its approximate values on Rh.
6. Conclusion
Three dierent schemes with the 14-point dierence operator are constructed on a cubic grid with
mesh size h, whose solutions separately approximate the solution of the Dirichlet problem for 3D Laplace’s
equation with the order O(h4ρp4), p∈ {4,5},where ρ=ρ(x1,x2,x3)is the distance from the current point
(x1,x2,x3)Rhto the boundary of the rectangular parallelepiped Rand its ﬁrst derivatives with the orders
O(hp1).
The obtained results can be used to highly approximate the derivatives of the solution of 3D Laplace’s
boundary value problems on a prism with an arbitrary polygonal base and on polyhedra by developing
the combined or composite grid methods [13, 15]. For the 2D case see [3, 4, 18, 20].
References
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equation on the rectangular parallelepiped, Russ. J. Numer. Anal. Math Model. 19 (2004) 269–278
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