Polynomial Particular Solutions for Poisson Problems. A.S. Muleshkov. Department of Mathematical Sciences. University of Nevada, Las Vegas, U.S.A.
Polynomial Particular Solutions for Poisson Problems A.S. Muleshkov Department of Mathematical Sciences University of Nevada, Las Vegas, U.S.A M.A. Golberg 517 Bianca Bay Street, Las Vegas, U.S.A. A. H.-D. Cheng Department of Civil Engineering University of Mississippi U.S.A. C.S. Chen Department of Mathematical Sciences University of Nevada, Las Vegas, U.S.A.
Polynomial Interpolation ¯ = [a, b] × [c, d] containing Ω in R2 and let f in (8) be Consider a rectangular region Ω ˆ extendable continuously to Ω. Then, it is well known that f can be approximated by a Chebyshev polynomial of the form qm,n (x, y) =
n m X X k=0 j=0
where
and
ajk Tj
µ
¶ µ ¶ ax − a − b 2y − c − d Tk b−a c−d
µ ¶ µ ¶ n m X X 1 πpj πqk cos , akj = f (xp, yq ) cos nm¯cj c¯k q=0 p=0 n m ¶ ³ pπ ´ a + b b−a cos xp = + ,0 ≤ p ≤ n 2 2 n ¶ µ ³ qπ ´ c + d d−c cos + yq = ,0 ≤ q ≤ n 2 2 n µ
(1)
(2)
(3) (4)
ˆ = [a, b] × [c, d] × [g, h] ⊇ Ω in R3 , then f can be approximated by Similarly, if Ω ql,m,n(x, y, z) µ ¶ µ ¶ µ ¶ l X n m X X 2y − c − d 2z − e − g ax − a − b Tk T = ajk Tj b − a c − d e−g j=0
(5) (6)
k=0 =0
where
ajk
µ ¶ µ ¶ µ ¶ ∫ X m X n X 1 πrj πsk πt = cos cos f (xr , ys, zt) cos . lmn¯cj c¯k c¯ r=0 s=0 t=0 l m n
(7)
and where (xr , ys, zt) are defined analogously to (3) - (4).
Here, Tn (x) = cos (n arccos x) is the Chebishev polynomial of degree n and c¯j = 1/2 , if j = 0 or j = n and c¯j = 0 otherwise. Similarly, c¯k is found. It can be shown that qn,m (x, y) and ql,m,n (x, y, z) converge spectrally to f , if f is sufficiently smooth.
The Spliting Method Let L be a linear partial differential operator, f - a given function and Ω - a bounded subset of Rd, d = 2, 3, with a boundary S. We consider the boundary value problem Lu(P ) = f (P ), Bu(P ) = g(P ),
P ∈ Ω, P ∈ ∂Ω,
(8) (9)
where B is a given boundary operator and g is a given boundary value. In this approach, we begin by defining a particular solution, up, which satisfies (8) but not necessarily (9). Then letting v = u − up. (10) v satisfies the homogeneous boundary value problem Lv(P ) = 0, P ∈ Ω, Bv(P ) = f (P ) − Bup,
P ∈ ∂Ω.
(11) (12)
Let {φk }k1 be a set of basis functions and approximate f by fˆ where fˆ =
N X
ak φk .
(13)
k=1
Then, an “approximate” particular solution, uˆp, is defined by uˆp =
N X
ak Φk
(14)
k=1
where {Φk }N 1 are obtained by analytically solving LΦk = φk .
(15)
By linearity, it follows that Lˆ up = fˆ and uˆp is used instead of up in subsequent calculations.
(16)
Evaluation of Particular Solutions The 2D Case A particular solution of ∆ψ = xny m,
m ≥ 0, n ≥ 0,
is given by [Cheng et al, 1994] [ n+2 2 ] X m!n!xm+2k y n−2k+2 k+1 (−1) , for m ≥ n, (m + 2k)!(n − 2k + 2)! ψ(x, y) = [k=1 m+2 ] 2 X m!n!xm−2k+2y n+2k k+1 (−1) for m < n, (m − 2k + 2)!(n + 2k)!
(17)
(18)
k=1
where [x] is the largest integer that is less than or equal to x. Chen et al. (2000) gave a different approach for deriving particular solutions when the right-hand side of (17) is a homogeneous polynomial of degree n. Theorem 1
A particular solution of ∆ψ =
n X k=0
Ak xn−k y k
(19)
is given by ψ=
n X
Pk xn−k+2y k
(20)
k=0
where
[ n−k 2 ]
Pk = The 3D Case
X (−1)m(k + 2m)!(n − k − 2m)! Ak+2m, for 1 ≤ k ≤ n. k!(n − k + 2)! m=0
(21)
A particular solution of ∆ψ = xl y mz n
(22)
can be obtained by inspection similar to the 2D case: n [m 2 ]+[ 2 ]+1
ψ=
X i=1
l+2i
x
min{i,[ m 2 ]+1}
X
j=max{1,i−[ n2 ]}
aij y m−2j+2z n−2i+2j
(23)
where −1 [(m − 2j + 4)(m − 2j + 3)ai−1,j−1 (l + 2i)(l + 2i − 1) + (n − 2i + 2j + 2)(n − 2i + 2j + 1)ai−1,j ] , −1 a11 = , ai0 = a0j = 0. (l + 2)(l + 1) aij =
(24) (25)
An alternative approach is to find a particular solution when the right-hand side is a homogeneous polynomial of degree n as we have shown in the 2D case. Let Qn(x, y, z) be an arbitrary homogeneous polynomial of power n > 0; i.e., Qn(x, y, z) =
n X
zk
k=0
For the equation
n−k X
Ak,mxn−k−my m.
(26)
m=0
∆ψ = Qn(x, y, z),
(27)
we are looking for a particular solution in the form ψ=
n+2 X k=0
zk
n−k+2 X
Pk,mxn−k−my m.
(28)
m=0
Substituting (28) into (27) and comparing the coefficients of ∆ψ with Qn, we obtain the
following system of equations: Ak,m = (n − k − m + 2)(n − k − m + 1)Pk,m + (m + 2)(m + 1)Pk,m+2 +(k + 2)(k + 1)Pk+2,m, k ≥ 0, m ≥ 0, m + k ≤ n.
(29)
The above system contains (n + 1)(n + 2)/2 equations with (n + 3)(n + 4)/2 unknowns. This means that there are 2n + 5 free parameters which can be set to zero due to the fact that the particular solution is not unique. The unknowns Pk,n in (29) can be partitioned into [(n + 4) /2] subsets as follows: S[ n+2 ] = {Pk,m : k + m = n + 2 or m + m = n + 1} 2
S[ n2 ] = {Pk,m : k + m = n or m + m = n − 1}
S[ n−2 ] = {Pk,m : k + m = n − 2 or m + m = n − 3} 2 ... If n is odd, then S0 = {P0,0, P0,1, P1,0} . If n is even, then S0 = {P0,0} . Since S[ n+2 ] contains 2 2n+5 elements, all of these elements can be set to zero. In (29), we observe that Sj , 0 ≤ j ≤ [ n2 ], can be expressed in terms of the elements of Sj+1. We summarize the above stated procedure into the following algorithm.
Algorithm for evaluating Pk,m (k ≥ 0, m ≥ 0, k + m = l ≤ n + 2) : INPUT n (degree of the homogeneous polynomial) INPUT Ak,m, k = 0, · · · , n; m = 0, · · · , n. Step 1 For k = 0, · · · , n + 2 set Pk,n−k+2 = 0 Step 2 For k = 0, · · · , n + 1 set Pk,n−k+1 = 0 Step 3 For l = n, · · · , 0, For k = 0, · · · , l set m = l − k and Pk,m =
Ak,m − (m + 2)(m + 1)Pk,m+2 − (k + 2)(k + 1)Pk+2,m (n − l + 2)(n − l + 1)
OUTPUT For k = 0, · · · , n + 2; m = 0, · · · , n − k + 2, Print Pk,m. Remark: The above algorithm does not work for degree zero. For convenience, in the case of a particular solution of degree zero, Q0 = A0,0, we use the particular solution A0,0x2/2.
Our algorithm for finding polynomial particular solutions for Poisson’s equation consists of the following steps: 1. approximate f by using (1) - (2); 2. expand the interpolants in monomial form using a symbolic language such as Mathematica; 3. use (20) - (21) or the algorithm given in this section to obtain particular solutions for homogeneous polynomials and then add the results. Finally, to solve the homogeneous Laplace equation, we use the Method of Fundamental Solutions (MFS).
Numerical Example Consider the 2D Poisson problem ∆u(x, y) = f (x, y), u(x, y) = g(x, y),
(x, y) ∈ Ω, (x, y) ∈ ∂Ω,
where Ω = [1, 2]2 and 7πx 3πy 5πy 7π2 7πx 751π 2 πx πx sin sin sin sin + cos cos f (x, y) = − 144 6 4 4 4 12 6 4 2 5πy 15π πx 7πx 3πy 5πy 3πy sin + sin sin cos cos , × sin 4 4 8 6 4 4 4 7πx 3πy 5πy πx sin sin sin g(x, y) = sin , 6 4 4 4 The exact solution is 7πx 3πy 5πy πx sin sin sin . u(x, y) = sin 6 4 4 4
(30) (31)
20 0 -20 1
-40 2
1.8 x
1.2
12
1.8
1.6 y
Forcing term
0.5 0 -0.5 1
1.2
1.4 y 1.6
1.8
2
2
1.8
x 1.4 1.6
Exact Solution
1.2
1
To approximate the forcing term using Chebyshev interpolation, we choose m = n in (1) since the solution domain is a square. We note that the particular solution is not unique. In this example, the approximate particular solution up(x, y) was obtained by taking n = 7. The MFS was applied to evaluate the homogeneous solution. In the MFS, we choose 40 uniformly distributed collocation points on the boundary. The same number of source points on the fictitious boundary, a circle with radius 2 and center (1.5, 1.5), were also chosen. The approximate solutions uˆ are evaluated at six randomly chosen points. The numerical results in Table 1 are in excellent agreement with the exact solution. x 1.2 1.2 1.3 1.5 1.4 1.6 Table 1:
y 1.3 1.4 1.5 1.8 1.4 1.7
ku − uˆk∞ 5.605 × 10−5 1.408 × 10−6 1.552 × 10−4 2.558 × 10−4 1.052 × 10−4 1.673 × 10−4
The errors of u − uˆ at six points.
