modified nodal cubic spline collocation for poisson’s...

MODIFIED NODAL CUBIC SPLINE COLLOCATION FOR

POISSON’S EQUATION

ABEER ALI ABUSHAMA † AND BERNARD BIALECKI‡

Abstract. We present a new modified nodal cubic spline collocation scheme for solving theDirichlet problem for Poisson’s equation on the unit square. We prove existence and uniquenessof a solution of the scheme and show how the solution can be computed on an (N + 1) × (N + 1)uniform partition of the square with cost O(N2logN) using a direct fast Fourier transform method.Using two comparison functions, we derive an optimal fourth order error bound in the continuousmaximum norm. We compare our scheme with other modified nodal cubic spline collocation schemes,in particular, the one proposed by Houstis et al. in [8]. We believe that our paper gives the firstcorrect convergence analysis of a modified nodal cubic spline collocation for solving partial differentialequations.

Key words. nodal collocation, cubic splines, convergence analysis, interpolants

AMS subject classifications. 65N35, 65N12, 65N15, 65N22

1. Introduction. De Boor [7] proved that classical nodal cubic spline collocationfor solving two-point boundary value problems is only second–order accurate and nobetter. For two-point boundary value problems, Archer [2] and independently Danieland Swartz [6] developed a modified nodal cubic spline collocation (MNCSC) schemewhich is fourth order accurate. The approximate solution in this scheme satisfieshigher-order perturbations of the ordinary differential equation at the partition nodes.Based on the method of [2] and [6], Houstis et al. [8] derived a fourth order MNCSCscheme for solving elliptic boundary value problems on rectangles. For the Helmholtzequation, a direct fast Fourier transform (FFT) algorithm for solving this scheme wasproposed recently in [3].

In this paper, we consider the Dirichlet boundary value problem for Poisson’sequation

∆u = f in Ω, u = 0 on ∂Ω,(1.1)

where ∆ denotes the Laplacian, Ω = (0, 1)× (0, 1), and ∂Ω is the boundary of Ω. Letρx = xi

N+1i=0 be a uniform partition of [0, 1] in the x-direction such that xi = ih,

i = 0, . . . , N + 1, where h = 1/(N + 1). For the sake of simplicity, we assume that auniform partition ρy = yj

N+1i=0 of [0, 1] in the y-direction is such that yj = xj . Let

S3 be the space of cubic splines defined by

S3 = v ∈ C2[0, 1] : v|[xi−1,xi] ∈ P3, i = 1, . . . , N + 1,

where P3 denotes the set of all polynomials of degree ≤ 3, and let

SD = v ∈ S3 : v(0) = v(1) = 0.

Our MNCSC scheme for solving (1.1) is formulated as follows: Find uh ∈ SD ⊗ SD

such that

∆uh(xi, yj) −h2

6D2

xD2yuh(xi, yj) = f(xi, yj) −

h2

12∆f(xi, yj),

i, j = 0, . . . , N + 1.(1.2)

†Department of Mathematical and Computer Sciences, Colorado School of Mines, Golden, Col-orado 80401-1887, U.S.A. ([email protected])

‡Department of Mathematical and Computer Sciences, Colorado School of Mines, Golden, Col-orado 80401-1887, U.S.A. ([email protected])

1

The scheme (1.2) is motivated by the fourth order finite difference method for (1.1),see, for example, equation (7) in section 4.5 of [9]. Using uh = u = 0 on ∂Ω and (1.1),we see that (1.2) is equivalent to:

2D2xD2

yuh(xi, yj) = ∆f(xi, yj), i, j = 0, N + 1,(1.3)

D2xuh(xi, yj) −

h2

6D2


h2

12∆f(xi, yj),

i = 0, N + 1, j = 1, . . . , N,(1.4)

D2yuh(xi, yj) −

h2

6D2


h2

12∆f(xi, yj),

i = 1, . . . , N, j = 0, N + 1,(1.5)

∆uh(xi, yj) −h2

6D2


h2

12∆f(xi, yj), i, j = 1, . . . , N.(1.6)

The scheme (4.2)–(4.4) of [8] for (1.1) is: Find uh ∈ SD ⊗ SD satisfying (1.3) and

D2xuh(xi, yj) = f(xi, yj), i = 0, N + 1, j = 1, . . . , N,(1.7)

D2yuh(xi, yj) = f(xi, yj), i = 1, . . . , N, j = 0, N + 1,(1.8)

(Lx + Ly)uh(xi, yj) = f(xi, yj), i, j = 1, . . . , N,(1.9)

where, for i, j = 1, . . . , N ,

Lxv(xi, yj) =1

12

[

D2xv(xi−1, yj) + 10D2

xv(xi, yj) + D2xv(xi+1, yj)

]

,(1.10)

Lyv(xi, yj) =1

12

[

D2yv(xi, yj−1) + 10D2

yv(xi, yj) + D2yv(xi, yj+1)

]

.

Our scheme and that of [8] are identical at the corners of Ω. However, they aredifferent at the remaining partition nodes. While (1.4)–(1.6) involve perturbations ofboth the left- and right-hand sides, (1.9) involves a perturbation of the left-hand sideonly. Numerical results show that our scheme exhibits superconvergence phenomenawhile that of [8] does not.

An outline of this paper is as follows. We give preliminaries in section 2. Thematrix-vector form of our scheme, an existence and uniqueness proof of its solution,and a direct FFT algorithm for solving the scheme are presented in section 3. Insection 4, using two comparison functions, we derive a fourth order error bound inthe continuous maximum norm. In section 5, we give convergence analysis of thescheme in [4] that consists of (1.3)–(1.5) and (1.9). We also explain why convergenceanalysis of the scheme (1.3) and (1.7)–(1.9), given in [8], is incorrect. This is why, webelieve, our paper gives the first correct convergence analysis of MNCSC for solvingpartial differential equations. Section 6 includes numerical results obtained using ourscheme.

2

2. Preliminaries. We extend the uniform partition ρx = xiN+1i=0 outside of

[0, 1] using xi = ih, i = −3,−2,−1, N + 2, N + 3, N + 4, and introduce Ii = [xi−1, xi],

i = −2, . . . , N + 4. Let BmN+2m=−1 be the basis for S3 defined by

Bm(x) =

g1[(x − xm−2)/h], x ∈ Im−1,g2[(x − xm−1)/h], x ∈ Im,g2[(xm+1 − x)/h], x ∈ Im+1,g1[(xm+2 − x)/h], x ∈ Im+2,0, otherwise,

(2.1)

where

g1(x) = x3, g2(x) = 1 + 3x + 3x2 − 3x3.(2.2)

The basis functions are such that, for m = 0, . . . , N + 1,

Bm−1(xm) = 1, Bm(xm) = 4, Bm+1(xm) = 1,B′′

m−1(xm) = 6/h2, B′′m(xm) = −12/h2, B′′

m+1(xm) = 6/h2.(2.3)

Let BDmN+1

m=0 be the basis for SD defined by

BD0 = B0 − 4B−1, BD

1 = B1 − B−1,BD

m = Bm, m = 2, . . . , N − 1,BD

N = BN − BN+2, BDN+1 = BN+1 − 4BN+2.

(2.4)

It follows from (2.3) that

BD0 (x1) = 1, BD

1 (x1) = 4, BD1 (x2) = 1,

BDN (xN−1) = 1, BD

N (xN ) = 4, BDN+1(xN ) = 1,

(2.5)

[

BD0

]′′(x0) = −36/h2,

[

BD1

]′′(x0) = 0,

[

BD0

]′′(x1) = 6/h2,

[

BD1

]′′(x1) = −12/h2,

[

BD1

]′′(x2) = 6/h2,

[

BDN

]′′(xN−1) = 6/h2,

[

BDN

]′′(xN ) = −12/h2,

[

BDN+1

]′′(xN ) = 6/h2,

[

BDN

]′′(xN+1) = 0,

[

BDN+1

]′′(xN+1) = −36/h2.

(2.6)

It also follows from (2.5), (2.6), (2.4), and (2.3) that, for i = 1, . . . , N,

BDm(xi) −

h2

6[BD

m]′′(xi) =

6, m = i,0, m 6= i,

m = 0, . . . , N + 1.(2.7)

Throughout the paper, C denotes a generic positive constant that is independentof u and h.

Lemma 2.1. BDm

N+1

m=0 of (2.4) satisfy maxx∈[0,1]

|BDm(x)| ≤ C,m = 0, . . . , N + 1.

Proof. For each fixed m = −1, . . . , N + 2, using Ii = [xi−1, xi] and xi = ih, we have

0 ≤ (x − xm−2)/h ≤ 1, x ∈ Im−1, 0 ≤ (x − xm−1)/h ≤ 1, x ∈ Im,0 ≤ (xm+1 − x)/h ≤ 1, x ∈ Im+1, 0 ≤ (xm+2 − x)/h ≤ 1, x ∈ Im+2.

(2.8)

Equations (2.2) and (2.8) give

|g1[(x − xm−2)/h]| ≤ 1, x ∈ Im−1, |g2[(x − xm−1)/h]| ≤ 7, x ∈ Im,|g2[(xm+1 − x)/h]| ≤ 7, x ∈ Im+1, |g1[(xm+2 − x)/h]| ≤ 1, x ∈ Im+2.

(2.9)

3

Using (2.1) and (2.9), we see that maxx∈[0,1]

|Bm(x)| ≤ C,m = −1, . . . , N + 2. Hence

the required inequality follows from (2.4) which implies that each BDm is a linear

combination of at most two of the functions BnN+2n=−1. 2

For BDm

N+1

m=0 of (2.4), we introduce N × N matrices A and B defined by

A = (ai,m)Ni,m=1, ai,m = [BD

m]′′(xi), B = (bj,n)Nj,n=1, bj,n = BD

n (yj).(2.10)

It follows from (2.4), (2.3), (2.5), and (2.6) that

A = 6h−2T, B = T + 6I,(2.11)

where I is the identity matrix and the N × N matrix T is given by

T =

−2 11 −2 1

. . .. . .

. . .

1 −2 11 −2

.(2.12)

Lemma 2.2. If B[u1, . . . , uN ]T = [v1, . . . , vN ]T , where B is defined in (2.10),then max

1≤i≤N|ui| ≤ C max

1≤i≤N|vi|.

Proof. It follows from (2.10), (2.11), and (2.12) that |bi,i| −∑

i6=j

|bi,j | ≥ 2, i = 1, . . . , N .

Hence the required result follows, for example, from the discussion on page 21 in [1].2

In what follows, [u1,1, . . . , uN,N ]T is the short notation for

[u1,1, . . . , u1,N , u2,1, . . . , u2,N , . . . , uN,1, . . . , uN,N ]T .

Lemma 2.3. If u = [u1,1, . . . , uN,N ]T and v = [v1,1, . . . , vN,N ]T are such that(B ⊗ B)u = v, where B is defined in (2.10), then max

1≤i,j≤N|ui,j | ≤ C max

1≤i,j≤N|vi,j |.

Proof. Since B ⊗ B = (B ⊗ I)(I ⊗ B), we have

v = (B ⊗ I)w, w = (I ⊗ B)u.(2.13)

Using ( 2.13) and Lemma 2.2, we obtain

max1≤i,j≤N

|ui,j | ≤ C max1≤i,j≤N

|wi,j |, max1≤i,j≤N

|wi,j | ≤ C max1≤i,j≤N

|vi,j |,

which imply the required inequality. 2

It is well known (see Theorem 4.5.2 of [10]) that for T of (2.12), we have

QTQ = Λ, QQ = I,(2.14)

where the N × N matrices Λ and Q are given by

Λ = diag(λi)Ni=1, λi = −4sin2 iπ

2(N + 1),(2.15)

Q = (qi,j)Ni,j=1, qi,j =

(

2

N + 1

)1/2

sinijπ

N + 1.(2.16)

4

Lemma 2.4. If v = [v1,1, . . . , vN,N ]T and w = [w1,1, . . . , wN,N ]T are such that

[

T

h2⊗ I + I ⊗

T

h2+

h2

6

(

T

h2⊗

T

h2

)]

v = w,(2.17)

where T is the matrix defined in (2.12), then max1≤i,j≤N

v2i,j ≤ Ch2

N∑

i=1

N∑

j=1

w2i,j.

Proof. The matrix in (2.17) arises in the fourth order finite difference method for (1.1).Hence the desired result follows, for example, from the last unnumbered equation onpage 296 in [9]. 2

Finally, we observe that the matrix-vector form of

φi,j =

N∑

m=1

c(1)i,m

N∑

n=1

c(2)j,nψm,n, i, j = 1, . . . , N,(2.18)

is

φ = (C1 ⊗ C2)ψ,(2.19)

where C1 =(

c(1)i,m

)N

i,m=1, C2 =

(

c(2)j,n

)N

j,n=1, and

φ = [φ1,1, . . . , φN,N ]T , ψ = [ψ1,1, . . . , , ψN,N ]T .

3. Matrix-Vector Form of Scheme. Since dim(SD ⊗ SD) = (N + 2)2, thescheme (1.3)–(1.6) involves (N +2)2 equations in (N +2)2 unknowns. Using the basis

BDm

N+1

m=0 of (2.4) for the space SD, we have

uh(x, y) =

N+1∑

m=0

N+1∑

n=0

um,nBDm(x)BD

n (y).(3.1)

Substituting (3.1) into (1.3), we obtain

2

N+1∑

m=0

N+1∑

n=0

um,n[BDm]′′(xi)[B

Dn ]′′(yj) = ∆f(xi, yj), i, j = 0, N + 1.(3.2)

Using (2.6), we conclude that (3.2) gives

ui,j =h4

2592∆f(xi, yj), i, j = 0, N + 1.(3.3)


N+1∑

m=0

N+1∑

n=0

um,n[BDm]′′(xi)

(

BDn (yj) −

h2

6[BD

n ]′′(yj)

)

= f(xi, yj) −h2

12∆f(xi, yj), i = 0, N + 1, j = 1, . . . , N.

(3.4)

Using (2.6) and (2.7), we see that (3.4) gives

ui,j = −h2

216f(xi, yj) +

h4

2592∆f(xi, yj), i = 0, N + 1, j = 1, . . . , N.(3.5)

5

Using (3.5) and symmetry with respect to x and y, we conclude that (1.5) gives

ui,j = −h2

216f(xi, yj) +

h4

2592∆f(xi, yj), i = 1, . . . , N, j = 0, N + 1.(3.6)


N+1∑

m=0

N+1∑

n=0

um,n

(

[BDm]′′(xi)B

Dn (yj) +

[

BDm(xi) −

h2

6[BD

m]′′(xi)

]

[BDn ]′′(yj)

)

= f(xi, yj) −h2

12∆f(xi, yj), i, j = 1, . . . , N.

(3.7)

Moving the terms involving um,nN+1n=0 , m = 0, N + 1, um,n

Nm=1, n = 0, N + 1, to

the right-hand side of (3.7), we get

N∑

m=1

N∑

n=1

um,n

(

[BDm]′′(xi)B

Dn (yj) +

[

BDm(xi) −

h2

6[BD

m]′′(xi)

]

[BDn ]′′(yj)

)

= pi,j , i, j = 1, . . . , N,

(3.8)

where

pi,j = f(xi, yj) −h2

12∆f(xi, yj)

−∑

m=0,N+1

N+1∑

n=0

um,n

(

[BDm]′′(xi)B

Dn (yj) +

[

BDm(xi) −

h2

6[BD

m]′′(xi)

]

[BDn ]′′(yj)

)

−

N∑

m=1

∑

n=0,N+1

um,n

(

[BDm]′′(xi)B

Dn (yj) +

[

BDm(xi) −

h2

6[BD

m]′′(xi)

]

[BDn ]′′(yj)

)

.

Using (2.18)–(2.19), we write (3.8) as[

A ⊗ B +

(

B −h2

6A

)

⊗ A

]

u = p,(3.9)

where u = [u1,1, . . . , uN,N ]T , p = [p1,1, . . . , pN,N ]T , and A, B are defined in (2.10).Using (2.11), we see that

A ⊗ B +

(

B −h2

6A

)

⊗ A =6

h2[6T ⊗ I + (6I + T ) ⊗ T ] ,

and hence the system (3.9) simplifies to

6h−2 [6T ⊗ I + (6I + T ) ⊗ T ]u = p.(3.10)

We are now ready to prove existence and uniqueness of uh in SD ⊗ SD that satisfies(1.3)–(1.6).

Theorem 3.1. There exists unique uh in SD ⊗ SD satisfying (1.3)–(1.6).Proof. Since the number of equations in (1.3)–(1.6) is equal to the number of un-knowns, we assume that the right-hand side in (1.3)–(1.6) is zero, and show thatuh = 0 is the only solution of the resulting scheme. Using (3.1), (3.3), (3.5), and(3.6), we have

um,n = 0, m = 0, N + 1, n = 0, . . . , N + 1, m = 1, . . . , N, n = 0, N + 1.(3.11)

6

Clearly

6h−2[6T ⊗ I + (6I + T ) ⊗ T ] = 36

[

T

h2⊗ I + I ⊗

T

h2+

h2

6

(

T

h2⊗

T

h2

)]

.(3.12)

Hence it follows from (3.10) with p replaced by 0, (3.12), and Lemma 2.4 that

um,n = 0, m, n = 1, . . . , N.(3.13)

Equations (3.1), (3.11), and (3.13) give uh = 0. 2

Using Q of (2.16), we see that (3.10) is equivalent to

6h−2(Q ⊗ I) [6T ⊗ I + (6I + T ) ⊗ T ] (Q ⊗ I)(Q−1 ⊗ I)u = (Q ⊗ I)p.(3.14)

Introducing u′ = (Q−1 ⊗ I)u and p′ = (Q ⊗ I)p, and using (3.14) and (2.14), weobtain

6h−2 [6Λ ⊗ I + (6I + Λ) ⊗ T ]u′ = p′,(3.15)

where Λ is defined in (2.15). The system (3.15) reduces to the N independent systems

6h−2 [6λiI + (6 + λi)T ]u′i = p′

i, i = 1, . . . , N,(3.16)

where u′i = [u′

i,1, . . . , u′i,N ]T , p′

i = [p′i,1, . . . , p′i,N ]T , i = 1, . . . , N .

We have the following algorithm for solving (3.10):Step 1. Compute p′ = (Q ⊗ I)p.Step 2. Solve the N systems in (3.16).Step 3. Compute u = (Q ⊗ I)u′.Since the entries of Q in (2.16) are given in terms of sines, steps 1 and 3 are performedeach using FFTs at a cost O(N2 log N). In step 2, the systems are tridiagonal, sothis step is performed at a cost O(N2). Thus the total cost of the algorithm isO(N2 log N).

4. Convergence Analysis. In what follows, C(u) denotes a generic positiveconstant that is independent of h, but depends on u.

Our goal is to show that if u in C6(Ω) and uh in SD ⊗ SD are the solutions of(1.1) and (1.3)–(1.6), respectively, then

‖u − uh‖C(Ω) ≤ C(u)h4,(4.1)

where ‖g‖C(Ω) = maxx∈Ω

|g(x)| for g in C(Ω).

To prove (4.1), for u in C4(Ω), we introduce two comparison functions, the splineinterpolants S and Z in SD ⊗ SD of u defined respectively by

D2xD2

yS(xi, yj) = D2xD2

yu(xi, yj), i, j = 0, N + 1,(4.2)

D2xS(xi, yj) −

h2

6D2

xD2yS(xi, yj) = D2

xu(xi, yj) −h2

12D4

xu(xi, yj)

−h2

6D2

xD2yu(xi, yj), i = 0, N + 1, j = 1, . . . , N,

(4.3)

7

D2yS(xi, yj) −

h2

6D2

xD2yS(xi, yj) = D2

yu(xi, yj) −h2

12D4

yu(xi, yj)

−h2

6D2

xD2yu(xi, yj), i = 1, . . . , N, j = 0, N + 1,

(4.4)

S(xi, yj) = u(xi, yj), i, j = 1, . . . , N,(4.5)

and

D2xD2

yZ(xi, yj) = D2xD2

yu(xi, yj), i, j = 0, N + 1,(4.6)

D2xZ(xi, yj) = D2

xu(xi, yj), i = 0, N + 1, j = 1, . . . , N,(4.7)

D2yZ(xi, yj) = D2

yu(xi, yj), i = 1, . . . , N, j = 0, N + 1,(4.8)

Z(xi, yj) = u(xi, yj), i, j = 1, . . . , N.(4.9)

It follows from (1.1) that

f = D2xu + D2

yu, ∆f = D4xu + D4

yu + 2D2xD2

yu.(4.10)

Hence, using u = 0 on ∂Ω, we see that (1.3)–(1.5) reduce, respectively, to

D2xD2

yuh(xi, yj) = D2xD2

yu(xi, yj), i, j = 0, N + 1,(4.11)

D2xuh(xi, yj) −

h2

6D2

xD2yuh(xi, yj) = D2

xu(xi, yj) −h2

12D4

xu(xi, yj)

−h2

6D2

xD2yu(xi, yj), i = 0, N + 1, j = 1, . . . , N,

(4.12)

D2yuh(xi, yj) −

h2

6D2

xD2yuh(xi, yj) = D2

yu(xi, yj) −h2

12D4

yu(xi, yj)

−h2

6D2

xD2yu(xi, yj), i = 1, . . . , N, j = 0, N + 1.

(4.13)

Comparing (4.11)–(4.13) and (4.2)–(4.4), we see that uh and S are defined in thesame way for i = 0, N + 1, j = 0, . . . , N + 1, and i = 1, . . . , N , j = 0, N + 1. On theother hand, (4.6)–(4.8) are a simplified, tensor product, version of (4.2)–(4.4).

The triangle inequality gives

‖u − uh‖C(Ω) ≤ ‖u − Z‖C(Ω) + ‖Z − S‖C(Ω) + ‖S − uh‖C(Ω).(4.14)

In what follows, we bound the three terms on the right-hand side of (4.14).

8

4.1. Bounding ‖u − Z‖C(Ω). We need the following results.

Lemma 4.1. Let the interpolant Ixv in S3 of v in C2[0, 1] be defined by

(Ixv)′′(xi) = v′′(xi), i = 0, N + 1, Ixv(xi) = v(xi), i = 0, . . . , N + 1.(4.15)

Then

maxx∈[0,1]

|v(x) − Ixv(x)| ≤ C maxx∈[0,1]

|v′′(x)|h2.(4.16)

If v ∈ C4[0, 1], then

maxx∈[0,1]

|v(x) − Ixv(x)| ≤ C maxx∈[0,1]

|v(4)(x)|h4.(4.17)

Proof. First we prove (4.16). Using the discussion on page 404 in [5], we have

Ixv(x) = v(xi) + Bi(x − xi) + Ci(x − xi)2 + Di(x − xi)

3, x ∈ [xi, xi+1],(4.18)

for i = 0, . . . , N , where

Bi = −h

6ri+1 −

h

3ri +

1

h[v(xi+1) − v(xi)] , Ci =

ri

2, Di =

1

6h(ri+1 − ri),(4.19)

and ri = (Ixv)′′(xi). Equations (4.18) and (4.19) give

Ixv(x) − v(x) = Ai(x)−h

6ri+1(x − xi) −

h

3ri(x − xi) +

ri

2(x − xi)

2

+1

6h(ri+1 − ri)(x − xi)

3, x ∈ [xi, xi+1],(4.20)

where

Ai(x) = v(xi) − v(x) +v(xi+1) − v(xi)

h(x − xi), x ∈ [xi, xi+1].(4.21)

Using (4.20) and the triangle inequality, we obtain, for x ∈ [xi, xi+1],

|Ixv(x) − v(x)| ≤ |Ai(x)| + h2

(

|ri| +1

3|ri+1|

)

≤ |Ai(x)| +4

3h2 max

0≤i≤N+1|ri|.(4.22)

We introduce

E = (ei,j)N+1i,j=0 =

11 4 1

. . .. . .

. . .

1 4 11

,r = [r0, . . . , rN+1]

T ,p = [p0, . . . , pN+1]

T ,

where

pi =

v′′(xi), i = 0, N + 1,h−2[v(xi−1) − 2v(xi) + v(xi+1)], i = 1, . . . , N.

(4.23)

It follows from the discussion on pages 400 and 401 in [5] that Er = p. Since

|ei,i| −∑

i6=j

|ei,j | ≥ 1, i, j = 0, . . . , N + 1, the discussion on page 21 in [1] implies that

max0≤i≤N+1

|ri| ≤ C max0≤i≤N+1

|pi|.(4.24)

9

Using Taylor’s theorem, we obtain

|v(xi−1) − 2v(xi) + v(xi+1)| ≤ Ch2 maxx∈[0,1]

|v′′(x)|, i = 1, . . . , N.(4.25)

It follows from (4.24), (4.23) and (4.25), that

max0≤i≤N+1

|ri| ≤ C maxx∈[0,1]

|v′′(x)|.(4.26)

Using Taylors’ theorem to expand v(x), x ∈ [xi, xi+1], around xi, we have

v(x) = v(xi) + (x − xi)v′(xi) +

(x − xi)2

2v′′(ξi,x), xi ≤ ξi,x ≤ x.(4.27)

Using (4.21), (4.27), and the triangle inequality, we obtain, for x ∈ [xi, xi+1],

|Ai(x)| =

∣

∣

∣

∣

(x − xi)2

2v′′(ξi,x) −

h

2(x − xi)v

′′(ξi,xi+1)

∣

∣

∣

∣

≤ h2 maxx∈[0,1]

|v′′(x)|.(4.28)

Inequality (4.16) follows from (4.22), (4.26), and (4.28).A proof of (4.17) is given in the proof of Theorem 2.3.4 in [1]. 2

Lemma 4.2. If u ∈ C4(Ω), Z in SD ⊗SD is defined by (4.6)–(4.9), and Ixu andIyu are defined in (4.15), then for (x, y) in Ω, we have

Z(x, y) = Ix(Iyu)(x, y), D2x(Iyu)(x, y) = Iy(D2

xu)(x, y).(4.29)

Proof. Let CiN+3i=0 be the basis for S3 such that

Ci(xj) = δij , i, j = 0, . . . , N + 1,

C ′′i (xj) = 0, i = 0, . . . , N + 1, j = 0, N + 1,

CN+2(xj) = CN+3(xj) = 0, j = 0, . . . , N + 1,

C ′′N+2(x0) = C ′′

N+3(xN+1) = 1, C ′′N+2(xN+1) = C ′′

N+3(x0) = 0,

(4.30)

where δij is the Kronecker delta. Using (4.15) and (4.30), we have for (x, y) ∈ Ω,

Ix(Iyu)(x, y)

= Ix

N+1∑

j=0

u(x, yj)Cj(y) + D2yu(x, y0)CN+2(y) + D2

yu(x, yN+1)CN+3(y)

=

N+1∑

i=0

N+1∑

j=0

u(xi, yj)Cj(y) + D2yu(xi, y0)CN+2(y) + D2

yu(xi, yN+1)CN+3(y)

Ci(x)

+

N+1∑

j=0

D2xu(x0, yj)Cj(y) + D2

xD2yu(x0, y0)CN+2(y)

+D2xD2

yu(x0, yN+1)CN+3(y)]

CN+2(x) +

N+1∑

j=0

D2xu(xN+1, yj)Cj(y)

+D2xD2

yu(xN+1, y0)CN+2(y) + D2xD2

yu(xN+1, yN+1)CN+3(y)]

CN+3(x).

10

Since u = 0 on ∂Ω, all terms involving C0(x), CN+1(x), C0(y), CN+1(y) drop outwhich implies that Ix(Iyu) ∈ SD ⊗ SD. Using (4.30), we verify that Ix(Iyu) satisfies(4.6)–(4.9), that is, (4.6)–(4.9) hold with Ix(Iyu) in place of Z. Hence, the uniquenessof the interpolant Z implies the first equation in (4.29). To prove the second equationin (4.29), we use (4.15) and (4.30) to see that for (x, y) ∈ Ω,

Iy(D2xu)(x, y)

=

N+1∑

j=0

D2xu(x, yj)Cj(y) + D2

xD2yu(x, y0)CN+2(y) + D2

xD2yu(x, yN+1)CN+3(y)

= D2x

N+1∑

j=0

u(x, yj)Cj(y) + D2yu(x, y0)CN+2(y) + D2

yu(x, yN+1)CN+3(y)

= D2x(Iyu)(x, y). 2

Theorem 4.1. If u ∈ C4(Ω) and Z in SD ⊗ SD is defined by (4.6)–(4.9), then‖u − Z‖C(Ω) ≤ C(u)h4.

Proof. Using (4.29) and the triangle inequality, we have

‖u − Z‖C(Ω) ≤ ‖u − Ixu‖C(Ω) + ‖Ix(u − Iyu) − (u − Iyu)‖C(Ω)

+ ‖u − Iyu‖C(Ω).(4.31)

For any fixed y in [0, 1], Ixu(·, y) is the cubic spline interpolant of u(·, y). Using this,symmetry with respect to x and y, and (4.17), we have

‖u − Ixu‖C(Ω) ≤ C(u)h4, ‖u − Iyu‖C(Ω) ≤ C(u)h4.(4.32)

For any fixed y in [0, 1], Ix(u−Iyu)(·, y) is the cubic spline interpolant of (u−Iyu)(·, y).Hence it follows from (4.16) that

‖Ix(u − Iyu) − (u − Iyu)‖C(Ω) ≤ C‖D2x(u − Iyu)‖C(Ω)h

2.(4.33)

Using (4.29) and(4.16), we obtain

‖D2x(u − Iyu)‖C(Ω) = ‖D2

xu − Iy(D2xu)‖C(Ω) ≤ C‖D2

xD2yu‖C(Ω)h

2.(4.34)

Combining (4.33) and (4.34), we have

‖Ix(u − Iyu) − (u − Iyu)‖C(Ω) ≤ C(u)h4.(4.35)

The desired inequality now follows from (4.31), (4.32), and (4.35). 2

4.2. Bounding ‖Z − S‖C(Ω). We start by proving the following lemma.

Lemma 4.3. If u ∈ C4(Ω) and

S(x, y) =

N+1∑

m=0

N+1∑

n=0

sm,nBDm(x)BD

n (y), Z(x, y) =

N+1∑

m=0

N+1∑

n=0

zm,nBDm(x)BD

n (y),(4.36)

are defined by (4.2)–(4.5) and (4.6)–(4.9), respectively, then

|sm,n − zm,n| ≤ C(u)h4, m, n = 0, . . . , N + 1.

11

Proof. Using (4.2), (4.6), and following the derivation of (3.3) from (1.3), we obtain

sm,n = zm,n, m, n = 0, N + 1.(4.37)

Next we prove the required inequality for m = 0, n = 1, . . . , N . Using (4.7), we have

D2x(S − Z)(x0, yj) = D2

xS(x0, yj) − D2xu(x0, yj), j = 1, . . . , N.(4.38)

It follows from (4.36), (4.37), and (2.6) that

D2x(S − Z)(x0, yj) = −36h−2

N∑

n=1

(s0,n − z0,n)BDn (yj), j = 1, . . . , N.(4.39)

Using (4.36), (2.6), (2.4), (2.3), and (2.5), we obtain, for j = 1, . . . , N ,

D2xS(x0, yj) = −36h−2

N+1∑

n=0

s0,nBDn (yj) = −36h−2(s0,j−1 + 4s0,j + s0,j+1).(4.40)

Substituting (4.39) and (4.40) into (4.38), and multiplying through by −h2/36, wehave

N∑

n=1

(s0,n − z0,n)BDn (yj) = s0,j−1 + 4s0,j + s0,j+1 +

h2

36D2

xu(x0, yj)(4.41)

for j = 1, . . . , N . Using (4.2), (4.3), and following the derivations of (3.3) from (1.3)and (3.5) from (1.4), we obtain

s0,j =h4

1296D2

xD2yu(x0, yj), j = 0, N + 1,(4.42)

and

s0,j = −h2

216

[

D2xu(x0, yj) −

h2

12D4

xu(x0, yj) −h2

6D2

xD2yu(x0, yj)

]

(4.43)

for j = 1, . . . , N . Since u = 0 on ∂Ω, (4.42) is the same as (4.43) with j = 0, N + 1.This observation and (4.43) imply that for j = 1, . . . , N , we have

s0,j±1 = −h2

216

[

D2xu(x0, yj±1) −

h2

12D4

xu(x0, yj±1) −h2

6D2

xD2yu(x0, yj±1)

]

.(4.44)


D2xu(x0, yj±1) = D2

xu(x0, yj) ± hD2xDyu(x0, yj) +

h2

2D2

xD2yu(x0, ξ

±j ),(4.45)

where yj−1 ≤ ξ−j ≤ yj , yj ≤ ξ+j ≤ yj+1. Using (4.44), (4.43), and (4.45), we obtain

∣

∣

∣

∣

s0,j−1 + 4s0,j + s0,j+1 +h2

36D2

xu(x0, yj)

∣

∣

∣

∣

≤ C(u)h4, j = 1, . . . , N.(4.46)

It follows from (4.46) that (4.41) is a system in s0,n − z0,nNn=1 with the matrix B

defined in(2.10) and with each entry on the right-hand side bounded in absolute valueby C(u)h4. Hence, Lemma 2.2 implies

max1≤n≤N

|s0,n − z0,n| ≤ C(u)h4.(4.47)

12

Using (4.47) and symmetry with respect to x and y, we also have

max1≤n≤N

|sN+1,n − zN+1,n| ≤ C(u)h4,

max1≤m≤N

|sm,n − zm,n| ≤ C(u)h4, n = 0, N + 1.(4.48)

Finally we prove the required inequality for m,n = 1, . . . , N. Using (4.5) and (4.9), wehave (S − Z)(xi, yj) = 0, i, j = 1, . . . , N , which, by (4.36) and (4.37), can be writtenas

N∑

m=1

N∑

n=1

(sm,n − zm,n)BDm(xi)B

Dn (yj) = di,j , i, j = 1, . . . N,(4.49)

where

di,j =

∑

m=0,N+1

N∑

n=1

+

N∑

m=1

∑

n=0,N+1

(zm,n − sm,n)BDm(xi)B

Dn (yj).

Since for any fixed i, j, each of the above double sums reduces to at most three terms,using the triangle inequality, (4.47), (4.48), and Lemma 2.1, we obtain

|di,j | ≤ C(u)h4, i, j = 1, . . . , N.(4.50)

It follows from (2.18)–(2.19) that (4.49) is a system in zm,n − sm,nNm,n=1 with the

matrix B ⊗B, where B is defined in (2.10). Hence, for m,n = 1, . . . , N, the requiredinequality follows from (4.50) and Lemma 2.3. 2

Theorem 4.2. If u ∈ C4(Ω) and S, Z in SD ⊗ SD are defined by (4.2)–(4.5)and (4.6)–(4.9), respectively, then ‖Z − S‖C(Ω) ≤ C(u)h4.

Proof. Since Z − S is continuous on Ω, there is (x∗, y∗) in Ω such that

‖Z − S‖C(Ω) = |(Z − S)(x∗, y∗)| .

Hence, (4.36) and the triangle inequality imply

‖Z − S‖C(Ω) ≤

N+1∑

m=0

N+1∑

n=0

|sm,n − zm,n||BDm(x∗)||B

Dn (y∗)|.

Since the above double sum reduces to at most nine terms, the required inequalityfollows from Lemmas 4.3 and 2.1. 2

4.3. Bounding ‖S−uh‖C(Ω) and ‖u−uh‖C(Ω) . We need the following results.

Lemma 4.4. If u ∈ C6(Ω) and S in SD ⊗ SD is defined by (4.2)–(4.5), then fori = 0, N + 1, j = 1, . . . , N ,

∣

∣D2xD2

yS(xi, yj) − D2xD2

yu(xi, yj)∣

∣ ≤ C(u)h2,(4.51)

∣

∣

∣

∣

D2xS(xi, yj) − D2

xu(xi, yj) +h2

12D4

xu(xi, yj)

∣

∣

∣

∣

≤ C(u)h4.(4.52)

13

Proof. We prove (4.51) for i = 0; for i = N + 1, (4.51) follows by symmetry withrespect to x. Using (4.36), we obtain

D2xD2

yS(x0, yj) =N+1∑

m=0

N+1∑

n=0

sm,n

[

BDm

]′′(x0)

[

BDn

]′′(yj), j = 1, . . . , N,

and hence (2.4), (2.3), and (2.6) imply

D2xD2

yS(x0, yj) = −216h−4(s0,j−1 − 2s0,j + s0,j+1), j = 1, . . . , N.(4.53)

Equations (4.53), (4.43), and (4.44) give, for j = 1, . . . , N,

D2xD2

yS(x0, yj) − D2xD2

yu(x0, yj) = −D2xD2

yu(x0, yj)+h−2

[

D2xu(x0, yj−1) − 2D2

xu(x0, yj) + D2xu(x0, yj+1)

]

−1

12

[

D4xu(x0, yj−1) − 2D4


]

−1

6

[

D2xD2

yu(x0, yj−1) − 2D2xD2

yu(x0, yj) + D2xD2

yu(x0, yj+1)]

.

(4.54)


D2xu(x0, yj±1) = D2


h2

2D2

xD2yu(x0, yj)

±h3

3!D2

xD3yu(x0, yj) +

h4

4!D2

xD4yu(x0, ξ

±j ),

(4.55)

D4xu(x0, yj±1) = D4


h2

2D4

xD2yu(x0, η

±j ),(4.56)

D2xD2

yu(x0, yj±1) = D2xD2

yu(x0, yj) ± hD2xD3

yu(x0, yj) +h2

2D2

xD4yu(x0, κ

±j ),(4.57)

where yj−1 ≤ ξ−j , η−j , κ−

j ≤ yj , yj ≤ ξ+j , η+

j , κ+j ≤ yj+1. Equations (4.55)–(4.57) give

∣

∣h−2[

D2xu(x0, yj−1) − 2D2


]

− D2xD2

yu(x0, yj)∣

∣ ≤ C(u)h2,∣

∣D4xu(x0, yj−1) − 2D4


∣

∣ ≤ C(u)h2,∣

∣D2xD2

yu(x0, yj−1) − 2D2xD2

yu(x0, yj) + D2xD2

yu(x0, yj+1)∣

∣ ≤ C(u)h2,

and hence (4.51) for i = 0 follows from (4.54) and the triangle inequality. Using (4.3)and (4.51), we obtain (4.52). 2

Lemma 4.5. If u ∈ C6(Ω) and S in SD ⊗ SD is defined by (4.2)–(4.5), then, fori, j = 1, . . . , N , we have

∣

∣

∣

∣

D2xS(xi, yj) − D2

xu(xi, yj) +h2

12D4

xu(xi, yj)

∣

∣

∣

∣

≤ C(u)h4,(4.58)

∣

∣

∣

∣

D2yS(xi, yj) − D2

yu(xi, yj) +h2

12D4

yu(xi, yj)

∣

∣

∣

∣

≤ C(u)h4,(4.59)

|D2xD2

yS(xi, yj) − D2xD2

yu(xi, yj)| ≤ C(u)h2.(4.60)

14

Proof. First we prove (4.58). For i = 0, . . . , N + 1, j = 1, . . . , N , we introduce

di,j = D2xS(xi, yj) −

[

D2xu(xi, yj) −

h2

12D4

xu(xi, yj)

]

.(4.61)

Then

di−1,j + 4di,j + di+1,j = φi,j − ψi,j , i, j = 1, . . . , N,(4.62)

where

φi,j = D2xS(xi−1, yj) + 4D2

xS(xi, yj) + D2xS(xi+1, yj)

−6

[

D2xu(xi, yj) +

h2

12D4

xu(xi, yj)

]

,(4.63)

ψi,j = D2xu(xi−1, yj) −

h2

12D4

xu(xi−1, yj) + 4

[

D2xu(xi, yj) −

h2

12D4

xu(xi, yj)

]

+D2xu(xi+1, yj) −

h2

12D4

xu(xi+1, yj) − 6

[

D2xu(xi, yj) +

h2

12D4

xu(xi, yj)

]

= D2xu(xi−1, yj) − 2D2

xu(xi, yj) + D2xu(xi+1, yj)

−h2

12

[

D4xu(xi−1, yj) + 10D4

xu(xi, yj) + D4xu(xi+1, yj)

]

.

(4.64)

Since S(·, yj) ∈ S3, (2.1.7) in [1], (4.5), and S = u = 0 on ∂Ω, imply that

D2xS(xi−1, yj) + 4D2

xS(xi, yj) + D2xS(xi+1, yj)

= 6h−2 [u(xi−1, yj) − 2u(xi, yj) + u(xi+1, yj)] , i, j = 1, . . . , N.(4.65)


u(xi±1, yj) = u(xi, yj) ± hDxu(xi, yj) +h2

2D2

xu(xi, yj)±h3

3!D3

xu(xi, yj)

+h4

4!D4

xu(xi, yj)±h5

5!D5

xu(xi, yj) +h6

6!D6

xu(ξ±i , yj),

where xi−1 ≤ ξ−i ≤ xi, xi ≤ ξ+i ≤ xi+1, and hence

∣

∣h−2 [u(xi−1, yj) − 2u(xi, yj) + u(xi+1, yj)]

−

[

D2xu(xi, yj) +

h2

12D4

xu(xi, yj)

]∣

∣

∣

∣

≤ C(u)h4, i, j = 1, . . . , N.(4.66)

Using (4.63), (4.65), and (4.66), we obtain

|φi,j | ≤ C(u)h4, i, j = 1, . . . , N.(4.67)


D2xu(xi±1, yj) = D2

xu(xi, yj) ± hD3xu(xi, yj) +

h2

2D4

xu(xi, yj)

±h3

3!D5

xu(xi, yj) +h4

4!D6

xu(ξ±i , yj),

D4xu(xi±1, yj) = D4

xu(xi, yj) ± hD5xu(xi, yj) +

h2

2D6

xu(η±i , yj),

15

where xi−1 ≤ ξ−i , η−i ≤ xi, xi ≤ ξ+

i , η+i ≤ xi+1, and hence (4.64) gives

|ψi,j | ≤ C(u)h4, i, j = 1, . . . , N.(4.68)

Using (4.61) and (4.52), we have

|di,j | ≤ C(u)h4, i = 0, N + 1, j = 1, . . . , N.(4.69)

It follows from (4.67)–(4.69) that moving d0,j and dN+1,j to the right-hand side of

(4.62), we obtain, for each j = 1, . . . , N , a system in di,jNi=1 with the matrix B of

(2.10)–(2.12), and with each entry on the right-hand side bounded in absolute valueby C(u)h4. Hence (4.58) follows from (4.61) and Lemma 2.2, and (4.59) follows from(4.58) by symmetry with respect to x and y.

Next we prove (4.60). Since S(x, ·) ∈ S3 for x ∈ [0, 1], (2.1.7) in [1] gives

D2yS(x, yj−1) + 4D2

yS(x, yj) + D2yS(x, yj+1)

= 6h−2 [S(x, yj−1) − 2S(x, yj) + S(x, yj+1)] , j = 1, . . . , N, x ∈ [0, 1].(4.70)

Differentiating (4.70) twice with respect to x, we obtain, for j = 1, . . . , N , x ∈ [0, 1],

D2xD2

yS(x, yj−1) + 4D2xD2

yS(x, yj) + D2xD2

yS(x, yj+1)= 6h−2

[

D2xS(x, yj−1) − 2D2

xS(x, yj) + D2xS(x, yj+1)

]

.(4.71)

Using (4.71) with x = xi−1, xi, xi+1, we obtain, for i, j = 1, . . . , N ,

D2xD2

yS(xi−1, yj−1) + 4D2xD2

yS(xi−1, yj) + D2xD2

yS(xi−1, yj+1)= 6h−2

[

D2xS(xi−1, yj−1) − 2D2

xS(xi−1, yj) + D2xS(xi−1, yj+1)

]

,(4.72)

D2xD2

yS(xi, yj−1) + 4D2xD2

yS(xi, yj) + D2xD2

yS(xi, yj+1)= 6h−2

[

D2xS(xi, yj−1) − 2D2

xS(xi, yj) + D2xS(xi, yj+1)

]

,(4.73)

D2xD2

yS(xi+1, yj−1) + 4D2xD2

yS(xi+1, yj) + D2xD2

yS(xi+1, yj+1)= 6h−2

[

D2xS(xi+1, yj−1) − 2D2

xS(xi+1, yj) + D2xS(xi+1, yj+1)

]

.(4.74)

Adding (4.72), (4.74) and (4.73) multiplied through by 4, and using (4.65) and S =u = 0 on ∂Ω, we obtain

D2xD2

yS(xi−1, yj−1) + 4D2xD2

yS(xi−1, yj) + D2xD2

yS(xi−1, yj+1)+4D2

xD2yS(xi, yj−1) + 16D2

xD2yS(xi, yj) + 4D2

xD2yS(xi, yj+1)

+D2xD2

yS(xi+1, yj−1) + 4D2xD2

yS(xi+1, yj) + D2xD2

yS(xi+1, yj+1)= 36h−4αi,j , i, j = 1, . . . , N,

(4.75)

where

αi,j = u(xi−1, yj−1) − 2u(xi, yj−1) + u(xi+1, yj−1)−2u(xi−1, yj) + 4u(xi, yj) − 2u(xi+1, yj)+u(xi−1, yj+1) − 2u(xi, yj+1) + u(xi+1, yj+1).

(4.76)

Using (4.76) and the discussion on pages 290–292 in [9], we have

∣

∣h−4αi,j − D2xD2

yu(xi, yj)∣

∣ ≤ C(u)h2, i, j = 1, . . . , N.(4.77)

16

Equation (4.75) is equivalent to

D2xD2

y(S − u)(xi−1, yj−1) + 4D2xD2

y(S − u)(xi−1, yj)+D2

xD2y(S − u)(xi−1, yj+1) + 4D2

xD2y(S − u)(xi, yj−1)

+16D2xD2

y(S − u)(xi, yj) + 4D2xD2

y(S − u)(xi, yj+1)+D2

xD2y(S − u)(xi+1, yj−1) + 4D2

xD2y(S − u)(xi+1, yj)

+D2xD2

y(S − u)(xi+1, yj+1) = 36h−4αi,j − βi,j , i, j = 1, . . . , N,

(4.78)

where

βi,j = D2xD2

yu(xi−1, yj−1) + 4D2xD2

yu(xi−1, yj) + D2xD2

yu(xi−1, yj+1)+4D2

xD2yu(xi, yj−1) + 16D2

xD2yu(xi, yj) + 4D2

xD2yu(xi, yj+1)

+D2xD2

yu(xi+1, yj−1) + 4D2xD2

yu(xi+1, yj) + D2xD2

yu(xi+1, yj+1).(4.79)


D2xD2

yu(xi−1, yj±1) = D2xD2

yu(xi, yj) − hD3xD2

yu(xi, yj) ± hD2xD3

yDyu(xi, yj) + ε±i,j ,

D2xD2

yu(xi+1, yj±1) = D2xD2

yu(xi, yj) + hD3xD2


yDyu(xi, yj) + σ±i,j ,

D2xD2

yu(xi, yj±1) = D2xD2


yu(xi, yj) + µ±i,j ,

D2xD2

yu(xi±1, yj) = D2xD2


yu(xi, yj) + ν±i,j .

where∣

∣ε±i,j∣

∣,∣

∣σ±i,j

∣

∣,∣

∣µ±i,j

∣

∣,∣

∣ν±i,j

∣

∣ ≤ C(u)h2, i, j = 1, . . . , N , and hence (4.79) gives

∣

∣βi,j − 36D2xD2

yu(xi, yj)∣

∣ ≤ C(u)h2, i, j = 1, . . . , N.(4.80)

It follows from (4.77) and (4.80) that the right-hand side of (4.78) is bounded inabsolute value by C(u)h2. Using (4.2) and moving terms involving

D2xD2

y(S − u)(xi, yj), i = 0, N + 1, j = 1, . . . , N, i = 1, . . . , N, j = 0, N + 1,

to the right-hand side of (4.78), we obtain a system in D2xD2

y(S − u)(xi, yj)N

i,j=1

with the matrix B ⊗ B, where B is given in (2.10)–(2.12). By (4.51) and symmetrywith respect to x and y, each entry on the right-hand side in this system is boundedin absolute value by C(u)h2. Therefore, (4.60) follows from Lemma 2.3. 2

Lemma 4.6. If u ∈ C6(Ω) and

uh(x, y) =

N+1∑

m=0

N+1∑

n=0

um,nBDm(x)BD

n (y), S(x, y) =

N+1∑

m=0

N+1∑

n=0

sm,nBDm(x)BD

n (y),(4.81)

are defined by (1.3)–(1.6) and (4.2)–(4.5), respectively, then

max1≤m,n≤N

|sm,n − um,n| ≤ C(u)h4, m, n = 1, . . . , N.(4.82)

Proof. Using (4.2)–(4.4), (4.11)–(4.13), and following the derivations of (3.3) from(1.3), (3.5) from (1.4), and (3.6) from (1.5), we conclude that

sm,n = um,n, m = 0, N + 1, n = 0, . . . , N + 1, m = 1, . . . , N, n = 0, N + 1.(4.83)

We define wi,jNi,j=1 by

∆(S − uh)(xi, yj) −h2

6D2

xD2y(S − uh)(xi, yj) = wi,j , i, j = 1, . . . , N.(4.84)

17

Using (4.84), (1.6), and (4.10), we obtain

wi,j = D2xS(xi, yj) + D2

yS(xi, yj) −h2

6D2

xD2yS(xi, yj)

−D2xu(xi, yj) − D2

yu(xi, yj) +h2

12

[

D4xu(xi, yj) + D4

yu(xi, yj) + 2D2xD2

yu(xi, yj)]

,

and hence (4.58)–(4.60) and the triangle inequality imply that

|wi,j | ≤ C(u)h4, i, j = 1, . . . , N.(4.85)

Introducing v = [s1,1 − u1,1, . . . , sN,N − uN,N ]T , w = [w1,1, . . . , wN,N ]T , using (4.84),(4.81), (4.83), and following the derivation of (3.10) from (1.6), we obtain

6h−2 [6T ⊗ I + (6I + T ) ⊗ T ]v = w.(4.86)

Since h = 1/(N +1), (4.85) gives h2

N∑

i,j=1

w2i,j ≤ C2(u)h8 and hence (4.82) follows from

(4.86), (3.12), and Lemma 2.4. 2

Theorem 4.3. If u ∈ C6(Ω) and uh and S in SD⊗SD are defined by (1.3)–(1.6)and (4.2)–(4.5), respectively, then ‖S − uh‖C(Ω) ≤ C(u)h4.

Proof. Since uh − S is continuous on Ω, there is (x∗, y∗) in Ω such that

‖uh − S‖C(Ω) = |(S − uh)(x∗, y∗)|.

Hence, (3.1), (4.36), (4.83), and the triangle inequality give

‖uh − S‖C(Ω) ≤N

∑

m=1

N∑

n=1

|sm,n − um,n||BDm(x∗)||B

Dn (y∗)|.

Since the above double sum reduces to at the most nine terms, the desired resultfollows from Lemmas 4.6 and 2.1. 2

Theorem 4.4. If u in C6(Ω) and uh in SD ⊗ SD are the solutions of (1.1) and(1.3)–(1.6), respectively, then ‖u − uh‖C(Ω) ≤ C(u)h4.

Proof. The required inequality follows from (4.14) and Theorems 4.1, 4.2, 4.3. 2

5. Other Schemes. Consider the scheme for solving (1.1) formulated as follows:Find uh ∈ SD ⊗ SD satisfying (1.3)–(1.5) and (1.9). This scheme is essentially thesame as the scheme (4.1)–(4.3) in [4], except that (1.5) is replaced in [4] with

1

12[13D2

yuh(xi, yj) − 2D2yuh(xi, yj+1) + D2

yuh(xi, yj+2)] = f(xi, yj), j = 0,

1

12[D2

yuh(xi, yj−2) − 2D2yuh(xi, yj−1) + 13D2

yuh(xi, yj)] = f(xi, yj), j = N + 1,

where i = 1, . . . , N . It follows from (3.1), the discussion in section 3, and (2.9) of [4]that the the matrix-vector form of (1.3)–(1.5) and (1.9) is

(A ⊗ B + B ⊗ A)u = p,(5.1)

18

where u = [u1,1, . . . , uN,N ]T , p = [p1,1, . . . , pN,N ]T ,

pi,j = f(xi, yj)−∑

m=0,N+1

N+1∑

n=0

um,n

[

LxBDm(xi)B

Dn (yj) + BD

m(xi)LyBDm(yj)

]

−

N∑

m=1

∑

n=0,N+1

um,n

[

LxBDm(xi)B

Dn (yj) + BD

m(xi)LyBDn (yj)

]

,

ui,jN+1j=0 , i = 0, N + 1, ui,j

Ni=1, j = 0, N + 1, are given in (3.3), (3.5), (3.6),

A =1

2h2(T 2 + 12T ), B = T + 6I,(5.2)

and T is defined in (2.12).Lemma 5.1. Assume A, B are as in (5.2) and v = [v1,1, . . . , vN,N ]T , w =

[w1,1, . . . , wN,N ]T are such that (A ⊗ B + B ⊗ A)v = w. Then

max1≤i,j≤N

v2i,j ≤ Ch2

N∑

i=1

N∑

j=1

w2i,j .

Proof. It follows from (5.2) that

A ⊗ B + B ⊗ A =1

2h2(T 2 ⊗ T ) +

3

h2(T 2 ⊗ I) +

6

h2(T ⊗ T ) +

36

h2(T ⊗ I)

+1

2h2(T ⊗ T 2) +

3

h2(I ⊗ T 2) +

6

h2(T ⊗ T ) +

36

h2(I ⊗ T ) = 36[r(T ) + s(T )],

(5.3)

where for an N × N matrix P ,

r(P ) = h−2(P ⊗ I + I ⊗ P ),(5.4)

s(P ) =1

3h2(P ⊗ P ) +

1

72h2

(

P 2 ⊗ P + P ⊗ P 2)

+1

12h2

(

P 2 ⊗ I + I ⊗ P 2)

.(5.5)

First, we will show that

([r(T ) + s(T )]z, z) ≤2

9(r(T )z, z), z ∈ RN2

,(5.6)

where (·, ·) is the standard inner product in RN2

. It follows from (2.14) and QT = Qfor Q of (2.16) that

(r(T )z, z) = ([Q ⊗ Q]r(Λ)[Q ⊗ Q]z, z) = (r(Λ)[Q ⊗ Q]z, [Q ⊗ Q]z),

(s(T )z, z) = ([Q ⊗ Q]s(Λ)[Q ⊗ Q]z, z) = (s(Λ)[Q ⊗ Q]z, [Q ⊗ Q]z),

where Λ is given in (2.15). Hence (5.6) is equivalent to

([r(Λ) + s(Λ)]z, z) ≤2

9(r(Λ)z, z), z ∈ RN2

,

which, by (5.4), (5.5), and (2.15), is in turn equivalent to

g(λi, λj) ≤ 0, i, j = 1, . . . , N,(5.7)

19

where

g(x, y) =7

9(x + y) +

1

3xy +

1

72(x2y + xy2) +

1

12(x2 + y2).

It follows from (2.15) that −4 ≤ λi ≤ 0, i = 1, . . . , N . Hence, (5.7) follows from

g(x, y) ≤ 0, x, y ∈ [−4, 0],

which is established using elementary calculus.The matrices r(T ) and s(T ) are symmetric, r(T )s(T ) = s(T )r(T ), and −r(T ) is

positive definite. Hence, (5.6) and 6) on page 135 in [9] imply that

‖r(T )z‖2 ≤9

2‖[r(T ) + s(T )]z‖2, z ∈ RN2

.(5.8)

where ‖ · ‖2 is the two vector norm. It is known (see, for example, the embeddingtheorem on page 281 in [9]) that

max1≤i,j≤N

z2i,j ≤

1

4h2‖r(T )z‖2

2, z = [z1,1, . . . , zN,N ]T ∈ RN2

.(5.9)

Hence the desired result follows from (5.9), (5.8), and (5.3). 2

Theorem 5.1. If u ∈ C6(Ω) and uh and S are defined by (1.3)–(1.5) and (1.9),and (4.2)–(4.5), respectively, then ‖S − uh‖C(Ω) ≤ C(u)h4.

Proof. Following the proof of Lemma 4.6, we define wi,jNi,j=1 by

(Lx + Ly)(S − uh)(xi, yj) = wi,j , i, j = 1, . . . , N.(5.10)

Using (5.10), (1.9), (1.1), we obtain

wi,j = LxS(xi, yj) − D2xu(xi, yj) + LyS(xi, yj) − D2

yu(xi, yj).

Equations (1.10), (4.58), and (4.52) give, for i, j = 1, . . . , N ,

LxS(xi, yj) − D2xu(xi, yj) = D2

xS(xi, yj) − D2xu(xi, yj)

+1

12[D2

xS(xi−1, yj) − 2D2xS(xi, yj) + D2

xS(xi+1, yj)]

= −h2

12D4

xu(xi, yj) +1

12[D2

xu(xi−1, yj) − 2D2xu(xi, yj) + D2

xu(xi+1, yj)]

−h2

144[D4

xu(xi−1, yj) − 2D4xu(xi, yj) + D4

xu(xi+1, yj)] + εi,j ,

where |εi,j | ≤ C(u)h4, i, j = 1, . . . , N . Hence Taylor’s theorem and similar considera-tions for LyS(xi, yj) − D2

yu(xi, yj) show that (4.85) holds. It follows from (4.81) and(4.83) that the matrix-vector form of (5.10) is

(A ⊗ B + B ⊗ A)v = w,

where A, B are as in (5.2), v = [s1,1−u1,1, . . . , sN,N −uN,N ]T , w = [w1,1, . . . , wN,N ]T .Hence Lemma 5.1 implies (4.82) and the desired result follows from the proof ofTheorem 4.3. 2

Theorem 5.2. If u in C6(Ω) and uh in SD ⊗ SD are the solutions of (1.1), and(1.3)–(1.5) and (1.9), respectively, then

‖u − uh‖C(Ω) ≤ C(u)h4.(5.11)

20

Proof. The required inequality follows from (4.14) and Theorems 4.1, 4.2, 5.1. 2

It is claimed in Theorem 4.1 of [8] that for the scheme (1.3) and (1.7)–(1.9), onehas (5.11) provided that u ∈ C6(Ω). The proof of this claim in [8] is based on using Zdefined in (4.6)–(4.9) as a comparison function. It is claimed, for example, in Lemma2.1 of [8] that Z has properties (4.58) and (4.59), that is, (4.58) and (4.59) hold withZ in place of S. Unfortunately, numerical examples indicate that such property doesnot hold even in one dimensional case. Specifically, for u(x) = x(x−1)ex and Z ∈ SD

such that

Z(xi) = u(xi), i = 1, . . . , N, Z ′′(xi) = u′′(xi), i = 0, N + 1,

we only have

max1≤i≤N

∣

∣

∣

∣

Z ′′(xi) − u′′(xi) +h2

12u(4)(xi)

∣

∣

∣

∣

= Ch2

and not better. It should be noted that the convergence analysis of [6] for two-pointboundary value problems involves the comparison function S ∈ SD defined by

S(xi) = u(xi), i = 1, . . . , N, S′′(xi) = u′′(xi) −h2

12u(4)(xi), i = 0, N + 1,

which, in part, was motivation for the definition (4.2)–(4.5). The convergence analysisof the scheme (4.2)–(4.4) in [8] remains an open problem. We believe that such analysismay require proving stability not only with respect to the right-hand side but alsowith respect to the boundary conditions.

6. Numerical Results. We used scheme (1.3)–(1.6) and algorithm of section 3to solve a test problem (1.1). The computations were carried out in double precision.We determined the nodal and global errors using the formulas

‖w‖h = max0≤i,j≤N+1

|w(xi, yj)|, ‖w‖C(Ω) ≈ max0≤i,j≤501

|w(ti, tj)|,

where ti = i/501, i = 1, . . . , 501. Convergence rates were determined using theformula

rate =log(eN/2/eN )

log[(N + 1)/(N/2 + 1)],

where eN is the error corresponding to the partition ρx × ρy.We took f in (1.1) corresponding to the exact solution

u(x, y) = 3exy(x2 − x)(y2 − y).

We see from the results in Tables 1 and 2 that the scheme (1.3)–(1.6) produces fourthorder accuracy for u in both the discrete and the continuous maximum norms. Wealso observe superconvergence phenomena since the derivative approximations at thepartition nodes are of order four.

REFERENCES

[1] J. H. Ahlberg, E. N. Nilson, and J. L. Walsh, The Theory of Splines and Their Applications,Academic Press, New York, 1967.

21

Table 1

Nodal errors and convergence rates for u, ux, uy, and uxy

‖u − uh‖h ‖(u − uh)x‖h ‖(u − uh)y‖h ‖(u − uh)xy‖h

N Error Rate Error Rate Error Rate Error Rate4 7.305–05 9.219–04 9.219–04 1.673–028 6.574–06 4.097 8.789–05 3.998 8.789–05 3.998 2.134–03 3.504

16 5.036–07 4.040 6.715–06 4.044 6.715–06 4.044 2.158–04 3.60332 3.585–08 3.983 4.712–07 4.005 4.712–07 4.005 1.881–05 3.67864 2.380–09 4.001 3.129–08 4.001 3.129–08 4.001 1.497–06 3.734

128 1.534–10 4.000 2.017–09 4.000 2.017–09 4.000 1.127–07 3.774

Table 2

Global errors and convergence rates for u, ux, uy, and uxy

‖u − uh‖C(Ω) ‖(u − uh)x‖C(Ω) ‖(u − uh)y‖C(Ω) ‖(u − uh)xy‖C(Ω)

N Error Rate Error Rate Error Rate Error Rate4 8.630–05 1.125–03 1.125–03 1.654–028 8.606–06 3.922 1.186–04 3.827 1.186–0 3.827 2.169–03 3.456

16 6.776–07 3.996 1.232–05 3.561 1.232–05 3.561 2.593–04 3.33932 4.763–08 4.003 1.267–06 3.429 1.267–06 3.429 2.997–05 3.25364 3.157–09 4.003 1.308–07 3.350 1.308–07 3.350 3.308–06 3.251

128 2.038–10 3.998 1.467–08 3.192 1.467–08 3.192 3.823–07 3.148

[2] D. Archer, An O(h4) cubic spline collocation method for quasilinear parabolic equations, SIAMJ. Number. Anal., 14 (1977), 620–637.

[3] B. Bialecki, G. Fairweather, and A. Karageorghis, Matrix decomposition algorithms for modified

spline collocation for Helmholtz problems, SIAM J. Sci. Comput., 24 (2003), 1733–1753.[4] B. Bialecki, G. Fairweather, and A. Karageorghis, Optimal superconvergent one step nodal

cubic spline collocation methods, SIAM J. Sci. Comput., 27 (2005), 575–598.[5] W. Cheney, and D. Kincaid, Numerical Mathematics and Computing, Brooks Cole, California,

1999.[6] J. W. Daniel and B. K. Swartz, Extrapolated collocation for two–point boundary–value problems

using cubic splines, J. Inst. Math Appl., 16 (1975), 161–174.[7] C. de Boor, The Method of Projections as Applied to the Numerical Solution of Two Point

Boundary Value Problems Using Cubic Splines, Ph.D. thesis, University of Michigan, AnnArbor, Michigan, 1966.

[8] E. N. Houstis, E. A. Vavalis, and J. R. Rice, Convergence of O(h4) cubic spline collocation

methods for elliptic partial differential equations, SIAM J. Numer. Anal., 25 (1988), 54–74.[9] A. A. Samarski, The Theory of Difference Schemes, Marcel Dekker, Inc., New York, Basel,

2001.[10] C. Van Loan, Computational Frameworks for the Fast Fourier Transform, SIAM, Philadelphia,

1992.

22

modified nodal cubic spline collocation for poisson’s...

Documents