JOURNAL OF COMPUTATIONAL AND APPLIED MATHEMATICS
ELSEVIER
Journal of Computational and Applied Mathematics 85 (1997) 1-10
An accurate solution of the Poisson equation by the Chebyshev-Tau method H.I. S i y y a m a'*, M.I. S y a m b a Department of Mathematical Science, Jordan University of Science and Technology, Irbid, Jordan b Department of Mathematics, Yarmouk University, Irbid, Jordan Received 30 October 1996; received in revised form 21 April 1997
Abstract A new Tau method is presented for the two-dimensionalPoisson equation. Comparisonof the results for the test problem u(x, y ) = sin(47~x)sin(4ny) with those computed by Haidvogel and Zang, using the matrix diagonalizationmethod, and Dang-Vu and Delcarte, using the Chebyshevcollocation method, indicates that our method would be more accurate.
Keywords: Poisson equation; Chebyshevpolynomials; Tau method AMS classification." 65
1. Introduction Haidvogel and Zang [3] developed a matrix diagonalization method for the solution of the twodimensional Poisson equation. This method is efficient but requires a preprocessing calculation of the eigenvalues and eigenvectors which limits the accuracy of the solution to that of the preprocessing calculations, especially at large N values. Dang-Vu and Delcarte [2] developed a Chebyshev collocation method for solving the same problem. Their method has the same accuracy as the matrix diagonalization method when N is small and it is more accurate when N is large. In this paper we present a new alternative method for solving
Au(x,y)=Uxx +Uyy= f(x,y),
x, y E ( - 1 , 1 ) ,
u ( + l , y ) = u ( x , + l )=O. which is more accurate than the above two methods. * Corresponding author. E,mail:
[email protected]. 0377-0427/97/$17.00 (~) 1997 Elsevier Science B.V. All rights reserved PH S0377-0427(97)00104-0
(1.1)
H.I. Siyyam, M.I. Syam/ Journal of Computational and Applied Mathematics 85 (1997) 1-10
Let us describe the utility of our method. Consider the following boundary-value problem: z(+l)=0,
z"(x)=g(x),
-l 1 and if n # m .
~"'~=
Other properties of Chebyshev polynomials include the recursion relation Tn+l(X)
= 2xTn(x)
-
T._l(x)
(2.2)
for n t> 1 and the endpoint relation T,(±I) = (+1)".
(2.3)
Suppose that f ( x ) E C2[-1, 1] and f ' " ( x ) is a piecewise continuous function on [-1, 1]. Then for d2 D2 f ( x ) = -~x2f ( x ) , we have that
D2 f ( x ) = ~ .1-(2)T.(x) n=O converges uniformly on [-1, 1] where f(2): L Cn
~ p(p2 p=n+2 p+n even
_
nZ)fp,
(2.4)
and
2 f l f ( x ) Tn(x) dx. fn = rcc----~, I vii - x 2 The coefficients also satisfy the recursion relation
c n - l J n¢(q) -1
¢'(q) =2nf(n q-l),
-- Jn+l
For more details, see [4, 1].
n >>.1, q -~" 1,2.
(2.5)
4
H.1. Siyyam, M.I. Syam/Journal of Computational and Applied Mathematics 85 (1997) 1-10
3. Chebyshev-Tau method for solving two-dimensional Poisson equation The basic topics of this section involve the Chebyshev-Tau method to discretize a class of linear boundary-value problems of the form of Problem (1.1). To explain the total procedure both analytic and numerical results are presented. Referring to the boundary-value problem (1.1), approximate u and f in terms of Chebyshev polynomials as N
uN(x, y) = ~
ak(x) Tk(y),
k=0
and N
fN(x,y) = Z
bk(x)Tk(y).
k=0
For the approximate solution uN, the residual is given by
RN(uN ) = AuN(x, y) -- fN(x, y).
(3.1)
Thus, the residual can be written as N
RN(uu) = ~ [a~2)(x)+ a~'(x) -- bk(x)]T~(y),
(3.2)
k=0
where a~2) is given in (2.4) and a~'(x) is the second derivative of ak(x) with respect to x. As in a typical Galerkin scheme, we generate ( N - 1) second-order ordinary differential equations by orthogonalizing the residual with respect to the basis functions Tk(y) (RN, Tk(y)) =
f
l RN
Tk(y)
dy = O,
for k = 0 : N - 2 .
,
This leads to the elementwise equation
a~2)(x) + a~'(x) = bk(x).
(3.3)
Since
ak(x ) = rka~2)_2(x) + ska~2)(x) + wka~2(x ) for k : 2 : N , SO
ak :
rk(bk-2
" 2) + sk(bk -- ak") + wk(bk+2 -- ak+ " 2) - ak_ for k = 2 :N,
where Ck-2
rk-4k(k_
--ek+2
l)'
Sk--2(k2_l)'
ek+4
w k - 4 k ( k + l-)'
and ek = 1 for k N.
H.I. Siyyam, M.I. Syaml Journal of Computational and Applied Mathematics 85 (1997) 1-10
5
For simplicity, let us assume that N is an even positive integer. Let D 2 -- d2/dx 2 be the differential operator. Since N
UN(X, + 1 ) = 0 = ~
(+1)kak(x),
k=0 SO
ao(x) + a2(x) + . . - + aN(x) = 0
and a l ( x ) + a3(x) + - - . + aN-l(X) = O. Thus, we have the following two systems
( Ae + DZge )ae = Re
(3.4)
( Ao + D2 Bo )ao = Ro,
(3.5)
and
where ae -- (ao, a 2 , . . . , aN) r, ao = (al, a 3 , . . . , aN-I )T -0 -1
0
0
0
0
...
0
0
0
O
1
...
1
1
r2
S2
W2
0
...
0
0
0
0
1
...
0
0
0
I"4
$4
W4
...
0
0
0
0
Ae--
Be
0
0
...
1
0
0
0
0
0
...
I'N_ 4
SN_ 4
WN_ 4
0
0
0
...
0
1
0
0
0
0
...
0
I"N_ 2
SN_ 2
0
0
0
0
0
...
0
0
rN
0
-0
0
0
0
...
0
0
0
O
1
1
...
1
1"
r3
s3
w3
0
...
0
0
0
0
0
1
...
0
0
0
r5
ss
w5
...
0
0
0
0
..°
Ao=
BO
0
0
...
1
0
0
0
0
0
...
rN-5
SN--5 WN--5 0
0
0
...
0
1
0
0
0
0
...
0
rN_ 3
SN_ 3
0
0
0
0
0
...
0
0
rN_~
0
H.L Siyyam, M.I. Syam/Journal of Computational and Applied Mathematics 85 (1997) 1-10 0 r2bo + s2b2 + w2b4
rab2 + s4b4 + w4b6 re=
rN-4bN-6 + SN-4bN-4 -[- WN--4bN--2 ru-zbu-4 + SN--zbN--2 rNbN-2 0 r3bl + s3b3 + w3b5 rsb3 + s5b5 + wsb7 re=
rN-sbN--7 + SN--sbN--7 + WN-sbN--7 rN--3bN--5 + SN--3bN--3 rN-lbN-3 Since {Tk(y), k=O, 1 , . . . , N } is linearly independent over • and uN(+l, y) -- 0, we see that ae(+l)=0
and
ao(+l)=0.
(3.6)
From Eqs. (3.4)-(3.6), we see that the two systems are similar. For this reason, we will discuss the solution of the following system:
ae(+l)=O.
(Ae + OZBe)ae =Re,
(3.7)
Multiply both sides of the differential equation (3.7) by 1
Ae 1 =
-1
-1
...
-1
-1
0
1
0
...
0
0
0
0
1
...
0
0
0
0
0
...
1
0
0
0
0
...
0
1
to get
(I + D2 C, )a, = qre,
ae(+l) =0.
(3.8)
H.I. Siyyam, M.I. SyamlJournal of Computationaland Applied Mathematics 85 (1997) 1-10
7
Let qe be the (N/2 + 1 ) x 1 vector which is defined as follows: the ith component of q, is the Taylor expansion of the ith component of qr~ about x = 0 of degree N/2 + 1 for all 1
