Well-Conditioned Collocation Schemes and New Triangular Spectral-Element Methods

Spectral methods Well-conditioned collocation New TSEM Further studies

Well-Conditioned Collocation Schemes and

New Triangular Spectral-Element Methods

Michael Daniel V. [email protected]

supervised by Li-Lian Wang

Nanyang Technological University

29 April 2014


Spectral methods

Background and historyPolynomial representation

Well-conditioned collocation

PreliminariesBirkhoff interpolation methodExtensions

New TSEM

PreliminariesNew transformImplementation

Further studies

CollocationTSEMFurther reading


Numerical solution of differential equations

Given f ∈ L2(Ω), g ∈ L2(∂Ω), find u such that

Lu = f in Ω differential operator,

Bu = g on ∂Ω boundary conditions.

f and g are given as data on predetermined points in thedomain Ω and on the boundary ∂Ω, respectively.

This allows for the determination of a numerical solution uN ,which can be computed in a few ways.


Lagrange interpolation

Given data (xi, yi), 0 ≤ i ≤ N , with pairwise distinct xi ∈ R,the Lagrange interpolation of the data is given as p ∈ PN

satisfyingp(xi) = yi for 0 ≤ i ≤ N,

computed by

p(x) =N∑

i=0

yiLi(x) where Li(x) =

0≤j≤N∏

i 6=j

x− xjxi − xj

.

Li is the Lagrange interpolation basis at the points xi,also as the nodal basis, as each function is 1 on one node and 0on the others, Li(xj) = δij .


Lagrange interpolation at quadrature points

If the nodes are Gauss-type quadrature points, with associatedweights ωi, then the Lagrange interpolation polynomials aregiven as

Li(x) =ωi

2

N∑

k=0

(2k + 1)Pk(xi)Pk(x)

for Legendre-Gauss-type points, and

Li(x) =ωi

π+

2ωi

π

N∑

k=1

Tk(xi)Tk(x)

for Chebyshev-Gauss-type points.


Collocation scheme

Given nodes ~xi, values f(~xi), g(~xi), 0 ≤ i ≤ N , find uN suchthat

LuN (~xi) = f(~xi) for each ~xi ∈ Ω;

BuN (~xi) = g(~xi) for each ~xi ∈ ∂Ω.

These equations form a linear system A~u = ~f , where theunknown is ~u = (uN (x0), . . . , uN (xN ))t.

If the components of the nodes are, depending on Ω and B,Gauss-Radau or Gauss-Lobatto points, the nodes are spectral

collocation points.


Second-order BVP with Lagrange interpolationGiven I = (−1, 1), r, s, f ∈ C(I) and u±, find u such that

−u′′ + ru′ + su = f in I; u(±1) = u±.

For nodes −1 = x0 < x1 < · · · < xN−1 < xN = 1, let

uN (x) =N∑

i=0

uiLi(x),

where Li is the nodal basis on xi. The collocation schemeis, for 0 < i < N ,

−N−1∑

k=1

ukL′′k(xi) + r(xi)

N−1∑

k=1

ukL′k(xi) + s(xi)ui

=f(xi) + u−(L′′0(xi)− r(xi)L

′0(xi)) + u+(L

′′N (xi)− r(xi)L

′N (xi)).


Linear system from Lagrange interpolation

When uN (x) =∑N

i=0 uiLi(x), with Li the nodal basis on xi:

(−D(2)in +ΛrD

(1)in +Λs)~u = ~f+u−(~d

(2)0 −Λr

~d(1)0 )+u+(~d

(2)N −Λr

~d(1)N ),

where

D(m)in = [L

(m)j (xi)]

N−1i,j=1,m = 1, 2 ~u = (u1, . . . , uN−1)

t,

Λφ = diag(φ(x1), · · · , φ(xN−1)), φ = r, s, ~f = (f(x1), . . . , f(xN−1))t,

~d(m)k = (L

(m)k (x1), . . . , L

(m)k (xN−1))

t, m = 1, 2, k = 0, N.

For spectral collocation points, D(m)in and ~d

(m)k , m = 1, 2,

k = 0, N , are computed accurately and efficiently.


Spectral collocation using Lagrange interpolationConsider

u′′(x)−(1+sinx)u′(x)+exu(x) = f(x), x ∈ (−1, 1); u(±1) = u±,

with f ∈ C1(I) and the exact solution u ∈ C3(I), given by

u(x) =

cosh(x+ 1)− x2/2− x, −1 ≤ x < 0,

cosh(x+ 1)− cosh(x)− x+ 1, 0 ≤ x ≤ 1.

101

102

103

10−14

10−12

10−10

10−8

10−6

10−4

10−2

N

BCOLLCOLPLCOL

101

102

103

10−14

10−12

10−10

10−8

10−6

10−4

10−2

N

BCOLLCOLPLCOL


Motivations and goals

Generate a collocation scheme that is

• Well-conditioned: condition number for Lagrangeinterpolation collocation for second-order BVP is O(N4)

• Stably, efficiently, accurately computed: as in Lagrangeinterpolation collocation

Previous methods use preconditioning or spectral integrationto generate systems with better condition numbers, but stilldependent on N


Well-conditioned collocation scheme

Generate a well-conditioned collocation scheme based on adifferent interpolation basis

• Uses integration on nodal functions to generate systemswith condition number independent of N

• Generates an optimal preconditioner—inverts thedifferential matrix of highest order

• Computed accurately, stably and efficiently—based onslowly-decaying coefficient matrices

The new interpolation basis has to be carefully verified andcomputed, as it does not always exist, and that modifications

may be needed to ensure the collocation scheme iswell-conditioned.


Birkhoff interpolation for second-order BVP

Given data (xi, y2i ), 0 < i < N , y0 and yN , with−1 = x0 < x1 < · · · < xN−1 < xN = 1, the Birkhoff

interpolation of the data is given as p ∈ PN satisfying

p(−1) = y0; p′′(xi) = y2i for 0 < i < N ; p(1) = yN

computed by p(x) = y0B0(x)+∑N−1

i=1 y2iBi(x)+ yNBN (x) where

B0(−1) = 1; B′′0 (xi) = 0, 0 < i < N ; B0(1) = 0;

Bj(−1) = 0; B′′j (xi) = δij , 0 < i < N ; Bj(1) = 0, 0 < j < N ;

BN (−1) = 0; B′′N (xi) = 0, 0 < i < N ; BN (1) = 1.

Bi is the Birkhoff interpolation basis at the points xi.


Birkhoff interpolation for second-order BVP

Given data (xi, y2i ), 0 < i < N , y0 and yN , with−1 = x0 < x1 < · · · < xN−1 < xN = 1, the Birkhoff

interpolation of the data is given as p ∈ PN satisfying

p(−1) = y0; p′′(xi) = y2i for 0 < i < N ; p(1) = yN

computed by p(x) = y0B0(x)+∑N−1

i=1 y2iBi(x)+ yNBN (x) where

B0(x) = (1− x)/2;

Bj(−1) = 0; B′′j (xi) = δij , 0 < i < N ; Bj(1) = 0, 0 < j < N ;

B0(x) = (1 + x)/2.

Bi is the Birkhoff interpolation basis at the points xi.


Stable anti-differentiation on orthogonal polynomials

Define the following antiderivatives:

∂(−1)x P0(x) =

1 + x

2; ∂(−1)

x Pk(x) =Pk+1(x)− Pk−1(x)

2k + 1, k > 0;

∂(−1)x T0(x) =

1 + x

2; ∂(−1)

x T1(x) =x2 − 1

2;

∂(−1)x Tk(x) =

Tk+1(x)

k + 1− Tk−1(x)

k − 1− 2(−1)k

k2 − 1, k > 1;

and ∂(−[m+1])x φ = ∂

(−1)x [∂

(−m)x φ]. Then

∫ x

−1φ(t) dt = ∂(−1)

x φ(x).

Note that ∂(−1)x Pk(−1) = ∂

(−1)x Tk(−1) = 0, k ≥ 0, but

∂(−1)x Pk(1) = 0 for k > 0 and ∂

(−1)x Tk(1) = 0, for odd k > 0.


Birkhoff interpolation for spectral collocation points

For Gauss-Lobatto quadrature points and weights (xi, ωi), theassociated Birkhoff interpolation polynomials Bi, 0 < i < N ,follow from

B′′i (x) =

ωi

2

N−2∑

k=0

(2k + 1)[Pk(xi)− PN−mN−k(xi)]Pk(x)

for Legendre-Gauss-Lobatto points, and

B′′i (x) =

ωi

π(1−TN−mN

(xi))+2ωi

π

N−2∑

k=1

[Tk(xi)−TN−mN−k(xi)]Tk(x)

for Chebyshev-Gauss-Lobatto points, where mj = j mod 2.



For Gauss-Lobatto quadrature points and weights (xi, ωi), theassociated Birkhoff interpolation polynomials Bi, 0 < i < N ,are given by

Bi(x) =ωi

2

N−2∑

k=0

(2k + 1)[Pk(xi)− PN−mN−k(xi)]∂

(−2)x Pk(x)

− (1 + x)ωi

4[1− PN−mN

(xi)]∂(−2)x P0(1)

− (1 + x)3ωi

4[xi − PN−mN−1

(xi)]∂(−2)x P1(1)

for Legendre-Gauss-Lobatto points, where mj = j mod 2.



For Gauss-Lobatto quadrature points and weights (xi, ωi), theassociated Birkhoff interpolation polynomials Bi, 0 < i < N ,are given by

Bi(x) =2ωi

π

N−2∑

k=1

[Tk(xi)− TN−mN−k(xi)]∂

(−2)x Tk(x)

− (1 + x)ωi

π

N−2∑

k=1

[Tk(xi)− TN−mN−k(xi)]∂

(−2)x Tk(1)

+ωi

2π(1− TN−mN

(xi))[2∂(−2)x (x)− ∂(−2)

x (1)(1 + x)]

for Chebyshev-Gauss-Lobatto points, where mj = j mod 2.


Collocation scheme using Birkhoff interpolationGiven I = (−1, 1), b, c, f ∈ C(I), γ > 0 and u±, find u such that

−u′′ + bu′ + cu = f in I; u(±1) = u±.

For nodes −1 = x0 < x1 < · · · < xN−1 < xN = 1, let

uN (x) =N∑

i=0

viBi(x),

where Bi is Birkhoff interpolation basis on xi. Thecollocation scheme is, for 0 < i < N ,

− vi + b(xi)

N−1∑

k=1

vkB′k(xi) + c(xi)

N−1∑

k=1

vkBk(xi)

=f(xi) +u−(b(xi)− c(xi)(1− xi))− u+(b(xi) + c(xi)(1 + xi))

2.


Linear system from Birkhoff interpolation

When uN (x) =∑N

i=0 viBi(x), with Bi the Birkhoffinterpolation basis on xi:

(−IN−1+ΛbB(1)in +ΛcB

(0)in )~v = ~f+

(u− − u+)Λb~1−Λc(u−~x− + u+~x+)

2,

where

B(m)in = [B

(m)j (xi)]

N−1i,j=1,m = 0, 1 ~v = (v1, . . . , vN−1)

t,

Λφ = diag(φ(x1), · · · , φ(xN−1)), φ = b, c, ~f = (f(x1), . . . , f(xN−1))t,

~1 = (1, . . . , 1)t, ~x± = ~1± (x1, . . . , xN−1)t.

For spectral collocation points, B(m)in , m = 0, 1, are computed

accurately and efficiently. Solution: ~u = u−~x− +B(0)in ~v+ u+~x+.


Second-order BVP with Birkhoff interpolation

Consider the example


with the exact solution u(x) = e(x2−1)/2. For Legendre spectral

collocation points,

NLagrange Birkhoff Preconditioned Lagrange

Cond.# Error iters Cond.# Error iters Cond.# Error iters

64 3.97e+05 3.82e-14 286 6.36 5.55e-16 10 2.86 1.67e-15 8

128 6.23e+06 4.42e-13 1251 6.46 1.11e-15 10 2.86 2.44e-15 8

256 9.91e+07 3.95e-13 6988 6.51 1.11e-15 11 2.86 2.55e-15 8

512 1.58e+09 1.02e-11 9457 6.54 1.89e-15 11 2.86 4.77e-15 8

1024 2.52e+10 6.58e-12 9697 6.55 3.44e-15 11 2.86 1.15e-14 9

The Lagrange linear system is preconditioned with

B(0)in = [D

(2)in ]−1. BiCGSTAB iteration is used (initial: ~0).


Second-order BVP with Birkhoff interpolation

Consider the example


with the exact solution u(x) = e(x2−1)/2. For Chebyshev

spectral collocation points,

NLagrange Birkhoff Preconditioned Lagrange

Cond.# Error iters Cond.# Error iters Cond.# Error iters

64 7.23e+05 8.38e-14 285 6.43 7.77e-16 10 2.86 1.44e-15 8

128 1.16e+07 2.87e-13 1304 6.50 7.77e-16 10 2.86 4.22e-15 8

256 1.85e+08 9.74e-13 5868 6.53 1.22e-15 11 2.86 6.55e-15 8

512 2.96e+09 4.51e-12 9987 6.55 1.78e-15 11 2.86 3.44e-15 8

1024 4.73e+10 1.27e-11 9938 6.56 3.77e-15 11 2.86 6.00e-15 9

The Lagrange linear system is preconditioned with

B(0)in = [D

(2)in ]−1. BiCGSTAB iteration is used (initial: ~0).


Second-order BVP with mixed boundary using

Birkhoff interpolation

Given a second-order BVP with the mixed-boundary conditions

u(−1)− u′(−1) = c−, u(1) + u′(1) = c+,

where c± are given. Compare condition numbers for Lagrangeinterpolation (LCOL) and Birkhoff interpolation (BCOL):

N−u′′ + u = f u′′ + u′ + u = f

Chebyshev Legendre Chebyshev LegendreBCOL LCOL BCOL LCOL BCOL LCOL BCOL LCOL

32 2.42 1.21e+05 2.45 6.66e+04 2.61 1.43e+05 2.61 7.87e+04

64 2.43 2.65e+06 2.45 1.41e+06 2.63 3.15e+06 2.63 1.68e+06

128 2.44 5.88e+07 2.45 3.09e+07 2.64 7.04e+07 2.64 3.70e+07

256 2.44 1.32e+09 2.45 6.88e+08 2.64 1.58e+09 2.64 8.26e+08

512 2.44 2.97e+10 2.44 1.54e+10 2.65 3.57e+10 2.65 1.86e+10

1024 2.44 6.71e+11 2.44 3.48e+11 2.65 8.08e+11 2.65 4.19e+11


First-order IVP with Birkhoff interpolationConsider the first-order IVP

u′(x)− sin(x)u(x) = f(x), x ∈ I = (−1, 1); u(−1) = u−,

with an oscillatory solution:

u(x) = 20 exp(− cos(x))

∫ x

−1exp(cos(t)) sin(500t2) dt.

−1 −0.5 0 0.5 1−0.5

0

0.5

1

1.5

2

x400 450 500 550 600 650 700 750

10−14

10−12

10−10

10−8

10−6

10−4

10−2

100

N

BCOLLCOL


Third-order BVP with Birkhoff interpolation

Consider the following problem: for x ∈ I = (−1, 1),

−u′′′(x) + r(x)u′′(x) + s(x)u′(x) + t(x)u(x) = f(x);

u(±1) = u±, u′(1) = u1,

where r, s, t and f are given continuous functions on I, and u−,u+ and u1 are given constants.

The condition numbers of the coefficient matrices for CGLpoints are tabulated.

N r ≡ s ≡ 0, t ≡ 1 r ≡ 0, s ≡ t ≡ 1 s ≡ 0, r ≡ t ≡ 1 r ≡ s ≡ t ≡ 1

128 1.16 1.56 2.22 1.80

256 1.16 1.56 2.22 1.80

512 1.16 1.56 2.23 1.80

1024 1.16 1.56 2.23 1.80


Third-order Korteweg-de VriesConsider the third-order Korteweg-de Vries (KdV) equation:

∂tu+ u∂xu+ ∂3xu = 0, x ∈ (−∞,∞), t > 0; u(x, 0) = u0(x),

with the exact soliton solution

u(x, t) = 12κ2sech2(κ(x− 4κ2t− x0)),

where κ and x0 are constants. Let τ be the time step size. Usethe Crank-Nicolson leap-frog scheme in time and the newcollocation method in space: find uk+1

N ∈ PN+1 such that for0 < j < N ,

uk+1N (Lxj)− uk−1

N (Lxj)

2τ+ ∂3x

(uk+1N + uk−1

N

2

)(Lxj)

=− ∂xukN (Lxj)u

kN (Lxj);

ukN (±L) = ∂xukN (L) = 0, k ≥ 0.


Third-order KdV results

Let κ = 0.3, x0 = −20, L = 50 and τ = 0.001.

80 90 100 110 120 130 140 150 16010

−7

10−6

10−5

10−4

10−3

10−2

10−1

N

t = 1t = 50

On left, the numerical evolution of the solution with t ≤ 50 andN = 160. On right, the maximum point-wise errors for variousN at t = 1, 50.


Fifth-order BVP with Birkhoff interpolationConsider the fifth-order problem:u(5)(x) + a(x)u′(x) + b(x)u(x) = f(x), x ∈ I = (−1, 1);

u(±1) = u′(±1) = u′′(1) = 0,

where a, b and f are given continuous functions on I.

Compare the generalized Lagrange interpolation p ∈ PN+3

satisfying, for u ∈ C5(I), u(±1) = u′(±1) = u′′(1) = 0,

p(yj) = u(yj), 0 < j < N ; p(±1) = p′(±1) = p′′(1) = 0,

where yjN−1j=1 are zeros of the Jacobi polynomial J

(3,2)N−1(x),

computed by p(x) =∑N−1

j=1 u(xj)Lj(x) where

Lj(x) =J(3,2)N−1(x)

(x− xj)∂xJ(3,2)N−1(xj)

(1− x)3(1 + x)2

(1− xj)3(1 + xj)2.


Fifth-order BVP resultsSolving by the three collocation schemesu(5)(x) + sin(10x)u′(x) + xu(x) = f(x), x ∈ I = (−1, 1);

u(±1) = u′(±1) = u′′(1) = 0, with solution u(x) = sin3(πx).

20 40 60 80 100 120 140 160 180 20010

−14

10−10

10−6

10−2

102

N

BCOLLCOLSCOL


Fifth-order Korteweg-de VriesConsider the fifth-order Korteweg-de Vries (KdV) equation:

∂tu+γu∂xu+ν∂3xu−µ∂5xu = 0, x ∈ (−∞,∞), t > 0; u(x, 0) = u0(x),

with the exact soliton solution

u(x, t) = η0+105ν2

169µγsech4

(√ν

52µ

[x−

(γη0 +

36ν2

169µ

)t− x0

]),

where γ, ν, µ, η0 and x0 are constants. Let τ be the time stepsize and ζj = Lxj . Use the Crank-Nicolson leap-frog scheme intime and the new collocation method in space: finduk+1N ∈ PN+3 such that for 0 < j < N ,

uk+1N (ζj)− uk−1

N (ζj)

2τ+ ν∂3x

(uk+1N + uk−1

N

2

)(ζj)− µ∂5x

(uk+1N + uk−1

N

2

)(ζj)

=− γ∂xukN (ζj)u

kN (ζj);

ukN (±L) = ∂xukN (±L) = ∂2xu

kN (L) = 0, k ≥ 0.


Fifth-order KdV results

Let µ = γ = 1, ν = 1.1, η0 = 0, x0 = −10, L = 50 andτ = 0.001.

50 60 70 80 90 100 110 120

10−8

10−6

10−4

10−2

N

t = 1t = 50t = 100


Two-dimensional BVP with partial diagonalizationConsider, as an example, the two-dimensional BVP:

∆u−γu = f in Ω = (−1, 1)2; u = 0 on ∂Ω,

where γ ≥ 0 and f ∈ C(Ω). The collocation scheme is: finduN (x, y) ∈ QN (Ω) := P2

N such that

(∆uN −γuN )(xi, yj) = f(xi, yj), 0 < i, j < N ; uN = 0 on ∂Ω,

where xi and yj are LGL points. Let

uN (x, y) =

N−1∑

k,l=1

uklBk(x)Bl(y),

and obtain the system:

UBtin +BinU − γBinUB

tin = F ,

where U = [ukl]0<k,l<N and F = [fkl]0<k,l<N .


Two-dimensional BVP with partial diagonalizationConsider, as an example, the two-dimensional BVP:

∆u−γu = f in Ω = (−1, 1)2; u = 0 on ∂Ω,

where γ ≥ 0 and f ∈ C(Ω). Consider the generalizedeigen-problem:

Bin~x = λ(IN−1 − γBin)~x.

Let Λ be the diagonal matrix of the eigenvalues, and E be thematrix whose columns are the corresponding eigenvectors. Then

BinE = (IN−1 − γBin)EΛ.

Set U = EV . Let ~vp be the transpose of pth row of V , andlikewise ~gp for G := E

−1(IN−1 − γBin)−1F . Solve the systems:

(Bin + λpIN−1)~vp = ~gp, p = 1, 2, . . . , N − 1.


Two-dimensional BVP resultsConsider ∆u = f in Ω, u = 0 on ∂Ω with the exact solution,

u(x, y) =

(sinh(x+ 1)− x− 1) cos(πy/2)exy, x < 0,

(sinh(x+ 1)− sinh(x)− 1− (sinh(2)− sinh(1)− 1)x3)

× cos(πy/2)exy, 0 ≤ x,

which is first-order differentiable in x and smooth in y. Fix Ny.

101

102

103

10−7

10−6

10−5

10−4

10−3

10−2

10−1

Nx

BCOLSGALslope: −2

101

102

103

10−7

10−6

10−5

10−4

10−3

10−2

10−1

Nx

BCOLslope: −2


Half-line BVP with Birkhoff interpolationConsider the following half-line problem:

−u′′(x) + a(x)u′(x) + b(x)u(x) = f(x), x ∈ (0,∞),

u(0) = u0, limx→∞

u(x) = 0,

where a, b and f are given continuous functions on the half-line,and u0 is a given constant.

Consider the Birkhoff-type interpolation p ∈ e−x/2PN satisfying,for u ∈ C2(0,∞), u→ 0 as x→ ∞,

p(0) = u(0); p′′(xj)− 14 p(xj) = u′′(xj)− 1

4u(xj), 1 ≤ j ≤ N.

Then p(x) = u(0)B0(x) +∑N

j=1

(u′′(xj)− 1

4u(xj))Bj(x), where

B0(0) = 1, B′′0 (xi)− 1

4B0(xi) = 0, 1 ≤ i ≤ N ;

Bj(0) = 0, B′′j (xi)− 1

4Bj(xi) = δij , 1 ≤ i, j ≤ N.


Half-line BVP resultsConsider the following half-line problem:−u′′(x) +√

xu′(x) + log(1 + x)u(x) = f(x), x ∈ (0,∞),

u(0) = u0, limx→∞

u(x) = 0,

with exact solution u(x) = (1 + x)−9/2.

200 220 240 260 280 300 320 340 360

10−13

10−12

10−11

10−10

N

LCOLBCOL


Variational formulationConsider the Helmholtz equation: given f ∈ C(Ω), g ∈ C(∂ΩN),γ > 0 and ∂ΩD ∪ ∂ΩN = ∂Ω, find u such that

−∆u+γu = f in Ω; u = 0 on ∂ΩD;∂u

∂~n= 0 on ∂ΩN.

The variational or weak formulation of the Helmholtz equationis: Find u ∈ T ⊂ u ∈ H1(Ω) , u = 0 on ∂ΩD such that, forevery v ∈ T ,

∫

Ω∇u · ∇v d~x+ γ

∫

Ωuv d~x =

∫

Ωfv d~x+

∫

∂ΩN

gv d~s.

Selecting the test functions v to form a basis for T gives rise toa linear system.


Variational formulationConsider the Helmholtz equation: given f ∈ C(Ω), g ∈ C(∂ΩN),γ > 0 and ∂ΩD ∪ ∂ΩN = ∂Ω, find u such that

−∆u+γu = f in Ω; u = 0 on ∂ΩD;∂u

∂~n= 0 on ∂ΩN.

The variational or weak formulation of the Helmholtz equationis: Find u ∈ T ⊂ u ∈ H1(Ω) , u = 0 on ∂ΩD such that, forevery v ∈ T ,

B (∇u,∇v) = (∇u,∇v)Ω+γ (u, v)Ω = (f, v)Ω+〈g, v〉∂ΩN= G(v).

Selecting the test functions v to form a basis for T gives rise toa linear system.

Often, for spectral methods, Ω = (−1, 1)d, and T ⊂ PdN , the

space of tensorial polynomials of degree N in each component.Data gives interpolations IIN f ∈ Pd

N and IN g ∈ Pd−1N .


Linear system of variational formulationIf ψi, 0 ≤ i ≤ K, is a basis for u ∈ Pd

N , u = 0 on ∂ΩD, anduN (x) =

∑Ki=0 uiψi(x), for v = ψi, the discretized weak or

spectral-Galerkin formulation gives

BN (uN , ψi) =K∑

k=0

uk[(∇ψk,∇ψi)Ω + γ (ψk, ψi)Ω]

= (IIN f, ψi)Ω + 〈IN g, ψi〉N,∂ΩN= GN (ψi),

where the trace inner product is given by d− 1-dimensionalquadrature. The K + 1 equations gives the linear system

(S + γM)~u = ~f,

where S and M are the stiffness and mass matrices, resp.,

S = [(∇ψi,∇ψj)Ω]Ki,j=0, ~u = (u0, . . . , uK)t,

M = [(ψi, ψj)Ω]Ki,j=0,

~f = (GN (ψ0), . . . ,GN (ψK))t.


Spectral element method

Spectral element methods solve differential equations oversubdomains piecewise, in conjunction with some domain

decomposition method.


Spectral element method

Spectral element methods solve differential equations oversubdomains piecewise, in conjunction with some domaindecomposition method.

As in the finite-element method, let the domain Ω be asimplex.

Consider first the reference triangle

= (x, y), 0 < x, y, x+ y < 1

on the xy-plane. Herein, consider maps from the reference

square = (−1, 1)2 on the ξη-plane to , with the plan oftransforming the domain to to perform the operations.


Rectangle-triangle mapping: Duffy’s transform

Duffy’s transform uses the fol-lowing 7→ map:

x =(1 + ξ)(1− η)

4,

y =1 + η

2.

The inverse map for y < 1 is

ξ =2x

1− y− 1,

η = 2y − 1.(−1,−1) (1,−1)

(1, 1)(−1, 1)

Note that the entire line η = 1 is mapped to (0, 1) ∈ ∂.


Rectangle-triangle mapping: Duffy’s transform

Duffy’s transform uses the fol-lowing 7→ map:

x =(1 + ξ)(1− η)

4,

y =1 + η

2.

The inverse map for y < 1 is

ξ =2x

1− y− 1,

η = 2y − 1.(0, 0) (1, 0)

(0, 1)

Note that the entire line η = 1 is mapped to (0, 1) ∈ ∂.


Transformed gradient for Duffy’s transform

Using the → map, given u(x, y) ∈ H1(), determineu(ξ, η) = u(x, y).

For Duffy’s transform, the Jacobian is

J =1− η

8,

and the gradient on is transformed on to

∇u =2

1− η

(2∂ξu, (1 + ξ)∂ξu+ (1− η)∂ηu

),

which requires the consistency condition ∂ξu(ξ, 1) = 0 to bebuilt into the approximation space to obtain high-orderaccuracy, resulting in the reduction of dimension andmodification of the usual basis functions.


New triangular spectral-element methodDuffy’s transform generates clustering near one vertex and asingularity in the gradient that requires modifying basiselements, and interpolations cannot be generated by acorresponding nodal basis on , as one edge on ∂ is mappedto a vertex on ∂.


New triangular spectral-element methodDuffy’s transform generates clustering near one vertex and asingularity in the gradient that requires modifying basiselements, and interpolations cannot be generated by acorresponding nodal basis on , as one edge on ∂ is mappedto a vertex on ∂.

A new transform is used that introduces less clustering, andintroduces a singularity in the gradient that is analyticallyremovable in the inner product of the variational form, whichis also one-to-one, allowing for good interpolations generatedby a corresponding nodal basis on . The function space shouldallow for optimal projection error.

Removing the singularity has to be done carefully. In addition,the singularity induced by the transform is a hanging node

when used in combination with domain decomposition methods.


Rectangle-triangle mapping: new transformThe new transform uses thefollowing 7→ map:

x =(1 + ξ)(3− η)

8,

y =(3− ξ)(1 + η)

8.

The inverse map is

ξ = 1 + (x− y)− χ,

η = 1− (x− y)− χ, (−1,−1) (1,−1)

(1, 1)(−1, 1)

where

χ =√(x− y)2 + 4(1− x− y) =

2− ξ − η

2.


Rectangle-triangle mapping: new transformThe new transform uses thefollowing 7→ map:

x =(1 + ξ)(3− η)

8,

y =(3− ξ)(1 + η)

8.

The inverse map is

ξ = 1 + (x− y)− χ,

η = 1− (x− y)− χ, (0, 0) (1, 0)

(0, 1)

( 1

2,

1

2)

where

χ =√(x− y)2 + 4(1− x− y) =

2− ξ − η

2.


Transformed gradientUsing the → map, given u(x, y) ∈ H1(), determineu(ξ, η) = u(x, y).

For the new transform, the Jacobian is

J =2− ξ − η

16=χ

8,

and the gradient on is transformed on to

∇u =1

χ

(2(∇ · u) + ∇⊺u, 2(∇ · u)− ∇⊺u

),

where

∇u = (∂ξu, ∂ηu) and ∇⊺u = (1− ξ)∂ξu− (1− η)∂ηu.

Originally, the consistency condition ∇ · u(1, 1) = 0 was builtinto the approximation space. This singularity can be

removed, however; observe that∫∫

χ−1 dξ dη = 8 ln 2.


Function space

Using the → map of the new transform, givenu(ξ, η) ∈ P2

N = QN (), determine u(x, y) = u(ξ, η). Then

u(x, y) = p(x, y) + χ(x, y)q(x, y)

∈ YN () = PN ()⊕ χPN−1(),

where p ∈ PN () has total degree N , and q ∈ PN−1().

This transformation is bijective: u ∈ YN () is mapped tou ∈ QN (), using the → inverse map of the newtransform.


Nodal and modal basisLet ζj, 0 ≤ j ≤ N , be the Legendre-Gauss-Lobatto points,and let Lj be the Lagrange interpolation basis on ζj. Thenodal basis of YN () on nodes

(xij , yij) =

((1 + ζi)(3− ζj)

8,(3− ζi)(1 + ζj)

8

)

is Ψij, 0 ≤ i, j ≤ N , where

Ψij(x, y) = Li(1 + (x− y)− χ)Lj(1− (x− y)− χ).

Consider the C0-modal basis on (−1, 1):

φ0(ζ) =1− ζ

2, φN (ζ) =

1 + ζ

2, φi(ζ) =

i(Pi−1(ζ)− Pi+1(ζ))

2(2i+ 1),

where 0 < i < N and Pi are the Legendre polynomials. Themodal basis of YN () is Ψij, 0 ≤ i, j ≤ N , where

Ψij(x, y) = φi(1 + (x− y)− χ)φj(1− (x− y)− χ).


Projection error

Consider the projection ΠN : L2() → YN (),

(ΠNu− u, v) = 0, for all v ∈ YN ().

Theorem

For any u ∈ Hr(), with r ≥ 0,

‖ΠNu− u‖ ≤ cN−r|u|r,,

where c is a positive constant independent of N and u.


Projection error

Consider the projection Π1N : H1() → YN (),

(∇(Π1

Nu− u),∇v) +

(Π1

Nu− u, v) = 0, for all v ∈ YN ().

Theorem


‖Π1Nu− u‖µ, ≤ cNµ−r|u|r,, µ = 0, 1,



Projection error

Consider the projectionΠ1,0

N : H10 () → Y 0

N () = YN () ∩H10 (),

(∇(Π1,0

N u− u),∇v)

= 0, for all v ∈ Y 0N ().

Theorem

For any u ∈ H10 () ∩Hr(), with r ≥ 1,

‖Π1,0N u− u‖µ, ≤ cNµ−r|u|r,, µ = 0, 1,



Interpolation error

Let ζj, 0 ≤ j ≤ N , be the Legendre-Gauss-Lobatto points,

and Ψij be the nodal basis of YN (). Given any u ∈ C(),define the interpolant of u by

(IIN u)(x, y) =N∑

i,j=0

u

((1 + ζi)(3− ζj)

8,(3− ζi)(1 + ζj)

8

)Ψij(x, y)

∈ YN ().

Theorem


‖IIN u− u‖µ, ≤ cN−r(|u|r, + |u|r−1,),



Interpolation error

Let ζj, 0 ≤ j ≤ N , be the Legendre-Gauss-Lobatto points,

and Ψij be the nodal basis of YN (). Given any u ∈ C(),define the interpolant of u by

(IIN u)(x, y) =N∑

i,j=0

u

((1 + ζi)(3− ζj)

8,(3− ζi)(1 + ζj)

8

)Ψij(x, y)

∈ YN ().

Theorem

For any u ∈ H2(),

‖IIN u−u‖µ, ≤ cN−2(|u|2,+‖(∂y−∂x)2u‖χ−1,+‖∇·u‖χ−1,),



Computing the mass matrix

Let ψij, 0 ≤ i, j ≤ N , be a basis of YN (), and

φij(ξ, η) =N∑

m,n=0

pmnij Pm(ξ)Pn(η) = ψij(x, y).

Then M = P′MP , where P = [pmn

ij ], 0 ≤ i, j,m, n ≤ N and

M is a pentadiagonal matrix whose entries are

1

16

∫∫

Pm(ξ)Pm′(ξ)Pn(η)Pn′(η)(2− ξ − η) dξ dη,

where 0 ≤ m,n,m′, n′ ≤ N .


Computing the stiffness matrix

Let ψij, 0 ≤ i, j ≤ N , be a basis of YN (), andφij(ξ, η) = ψij(x, y). Then S = S1 + S2, where

S1 =

[(∇ · φij , ∇ · φi′j′

)χ−1,

]N

i,j,i′,j′=0

,

S2 =1

4

[(∇⊺φij , ∇⊺φi′j′

)χ−1,

]N

i,j,i′,j′=0

.

Each entry is a computable combination of

apq =

∫∫

Pp(ξ)Pq(η)

χdξ dη, 0 ≤ p, q ≤ 2N.


Computing the stiffness matrix

Let ψij, 0 ≤ i, j ≤ N , be a basis of YN (), andφij(ξ, η) = ψij(x, y). Then S = S1 + S2, where

S1 =

∫∫

(∇ · φij)(∇ · φi′j′)χ

dξ dη

N

i,j,i′,j′=0

,

S2 =1

4

∫∫

(∇⊺φij)(∇⊺φi′j′)

χdξ dη

N

i,j,i′,j′=0

.

Each entry is a computable combination of

apq =

∫∫

Pp(ξ)Pq(η)

χdξ dη, 0 ≤ p, q ≤ 2N.


Removing the singularity in the stiffness matrix

1. Compute a0q,0 ≤ q ≤ 4N . q

p

Use

a0q =

∫ 1

−1Pq(η) ln

3− η

2︸︷︷︸by quadrature

+

∫ 1

−1Pq(η) ln

2

1− η︸︷︷︸2 if q=0, else 2/q(q+1)

.



1. Compute a0q,0 ≤ q ≤ 4N .

2. Compute a1q,1 ≤ q ≤ 4N − 1.

q

p

Use

a1q = 2a0q −(q + 1)a0,q+1 + qa0,q−1

2q + 1.




2. Compute a1q,1 ≤ q ≤ 4N − 1.

3. For p = 2, 3, . . . , 2N ,

compute apq,p ≤ q ≤ 4N − p.

q

p

Use

apq = ap−2,q +2p− 1

2q + 1(ap−1,q+1 − ap−1,q−1).




2. Compute a1q,1 ≤ q ≤ 4N − 1.

3. For p = 2, 3, . . . , 2N ,


q

p

Use

apq = ap−2,q +2p− 1

2q + 1(ap−1,q+1 − ap−1,q−1).




2. Compute a1q,1 ≤ q ≤ 4N − 1.

3. For p = 2, 3, . . . , 2N ,


q

p

Use

apq = ap−2,q +2p− 1

2q + 1(ap−1,q+1 − ap−1,q−1).




2. Compute a1q,1 ≤ q ≤ 4N − 1.

3. For p = 2, 3, . . . , 2N ,compute apq,p ≤ q ≤ 4N − p.

4. For 0 = q < p = 2N , set

apq = aqp.

q

p


Numerical resultsConsider the elliptic equation:

−∆u+u = f in ; u|Γ1= 0;

∂u

∂~n

∣∣∣Γ2

= g,

where Γ1 is the edges x = 0 and y = 0, Γ2 is the hypotenuse of, and with the exact solution:

u(x, y) = ex+y−1 sin(3xy

(y −

√32 x+

√34

)).

For comparison, consider

−∆u+u = f in S = (0, 1/√2)2; u|Γ′

1= 0;

∂u

∂~n

∣∣∣Γ′

2

= g,

where Γ′1 is the edges x = 0 and y = 0 and Γ′

2 is the edgesx = 1/

√2 and y = 1/

√2, with exact solution

u(x, y) = exp(−(

1√2− x)(

1√2− y))

sin(3xy

(y −

√32 x+

√34

)).


Numerical results

5 10 15 20 25

10−15

10−10

10−5

100

N

erro

r

L2 error, rectangle

L∞ error, rectangle

L2 error, triangle

L∞ error, triangle

modal basis

5 10 15 20 25

10−15

10−10

10−5

100

Ner

ror

L2 error, rectangle


L2 error, triangle


nodal basis


Numerical results

Consider the elliptic equation:

−∆u+u = f in ; u|Γ1= 0;

∂u

∂~n

∣∣∣Γ2

= g,

where Γ1 is the edges x = 0 and y = 0, Γ2 is the hypotenuse of, with the finite regularity exact solution:

u(x, y) = (1− x− y)52 (exy − 1) ∈ H3−ǫ()

The counterpart on the square S takes the form:

u(x, y) =

(1√2− x

) 52(

1√2− y

) 52

(exy − 1), ∀ (x, y) ∈ S.


Numerical results

0.6 0.8 1 1.2 1.4 1.6−8

−7

−6

−5

−4

−3

−2

log10

(N)

log 10

(err

or)

L2 error, rectangle


L2 error, triangle


modal basis

0.6 0.8 1 1.2 1.4 1.6−8

−7

−6

−5

−4

−3

−2

log10

(N)lo

g 10(e

rror

)

L2 error, rectangle


L2 error, triangle


nodal basis


Arbitrary triangleFor a triangle any, with vertices counterclockwise at (x1, y1),(x2, y2) and (x3, y3), the invertible map → any is

(x, y) = (x1, y1)(1− ξ)(1− η)

4+ (x2, y2)

(1 + ξ)(3− η)

8

+ (x3, y3)(3− ξ)(1 + η)

8.

Using this map to determine u(ξ, η) = u(x, y), the mass matrixis determined by

(u, v)any=F

8(u, v)χ, ,

where

F = (x2 − x1)(y3 − y1)− (x3 − x1)(y2 − y1) 6= 0.


Arbitrary triangle

For a triangle any, with vertices counterclockwise at (x1, y1),(x2, y2) and (x3, y3), the invertible map → any is

(x, y) = (x1, y1)(1− ξ)(1− η)

4+ (x2, y2)

(1 + ξ)(3− η)

8

+ (x3, y3)(3− ξ)(1 + η)

8.

Using this map to determine u(ξ, η) = u(x, y), the stiffnessmatrix is determined by

(∇u,∇v)any=

A

2F

(∇ · u, ∇ · v

)χ−1,

+C

8F

(∇⊺u, ∇⊺v

)χ−1,

− B

4F

[(∇ · u, ∇⊺v

)χ−1,

+(∇⊺u, ∇ · v

)χ−1,

].

where A, B and C are determined from xi, yi, 1 ≤ i ≤ 3.


Unstructured TSEM with LDG-H

To use this TSEM on an unstructured mesh, the hybridized

local discontinuous Galerkin method is used.

• DG methods enjoy a large degree of flexibility,non-conformity and locality. In particular, DG methods

can handle hanging nodes in meshes, while providing ascheme to handle the coupling on the mesh. Having thehanging node in a predictable position allows for efficientcomputation.

• LDG-H makes use of auxillary functions, which renders theelliptic problem into a system of first-order differentialequations. For those inner products, the rectangle-triangle

map does not induce a singularity.

• LDG-H generates a global system whose degrees of

freedom are only those on the interior edges.


Results for unstructured TSEMConsider the model problem

−∆u+u = f, in Ω = [0, 1]2; u = 0 on ∂Ω,

with the highly-oscillating exact solution

u(x, y) = sin(10πx) cos(10πy).

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

x

y

5× 5

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

x

y

15× 15

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

xy

25× 25


Results for unstructured TSEM

2 4 6 8 10 12 1410

−12

10−10

10−8

10−6

10−4

10−2

100

102

Polynomial order

Ave

rage

ele

men

t−w

ise

H1 err

or

5 × 5, τ = 15 × 5, τ = 100015 × 15, τ = 115 × 15, τ = 100025 × 25, τ = 125 × 25, τ = 1000

10−1

10−8

10−6

10−4

10−2

h

Ave

rage

ele

men

t−w

ise

L2 e

rror

P = 3P = 4P = 5


Results for structured TSEM versus perturbationConsider the model problem

−∆u+u = f, in Ω = [0, 1]2; u = 0 on ∂Ω,

with the highly-oscillating exact solutionu(x, y) = sin(10πx) cos(10πy). Ω is triangulated into twomeshes of varying coarseness, denoted 5× 5 and 10× 10, eithermaintaining the regular underlying mesh or perturbing slightly

on the internal vertices.

5 10 15 20 25

10−12

10−14

10−10

10−8

10−6

10−4

10−2

100

N

aver

age

elem

ent−

wis

e L

2 err

or

5 10 15 20 25

10−9

10−11

10−7

10−5

10−3

10−1

101

N

aver

age

elem

ent−

wis

e H

1 err

or


Unstructured TSEM with perturbed solutionConsider the model problem

−∆u+u = f, in Ω = [0, 1]2; u = 0 on ∂Ω,

with the low-oscillation exact solution with slight

highly-oscillating perturbation

u(x, y) = sin(2πx) cos(2πy) + ǫ cos(20πx) sin(20πy).

2 4 6 8 10 12 14 16 18 20 2210

−12

10−10

10−8

10−6

10−4

10−2

100

N

aver

age

elem

ent−

wis

e L

2 err

or

ε = 0, τ = 1ε = 0, τ = 1000ε = 1e−8, τ = 1ε = 1e−8, τ = 1000ε = 1e−4, τ = 1ε = 1e−4, τ = 1000

2 4 6 8 10 12 14 16 18 20 2210

−10

10−8

10−6

10−4

10−2

100

102

N

aver

age

elem

ent−

wis

e H

1 err

or

ε = 0, τ = 1ε = 0, τ = 1000ε = 1e−8, τ = 1ε = 1e−8, τ = 1000ε = 1e−4, τ = 1ε = 1e−4, τ = 1000


Unstructured TSEM with perturbed traceConsider the model problem

−∆u+u = f, in Ω = [0, 1]2; u = 0 on ∂Ω,

with the low-oscillation exact solution

u(x, y) = sin(2πx) cos(2πy).

The perturbation ǫ cos(20πx) sin(20πy) is introduced after

solving the global system, before applying the local solvers.

2 4 6 8 10 12 14 16 18 20 2210

−12

10−10

10−8

10−6

10−4

10−2

100

N

aver

age

elem

ent−

wis

e L

2 err

or

ε = 0, τ = 1

ε = 0, τ = 1000

ε = 1e−8, τ = 1

ε = 1e−8, τ = 1000

ε = 1e−4, τ = 1

ε = 1e−4, τ = 1000

2 4 6 8 10 12 14 16 18 20 2210

−10

10−8

10−6

10−4

10−2

100

102

N

aver

age

elem

ent−

wis

e H

1 err

or

ε = 0, τ = 1

ε = 0, τ = 1000

ε = 1e−8, τ = 1

ε = 1e−8, τ = 1000

ε = 1e−4, τ = 1

ε = 1e−4, τ = 1000


Well-conditioned collocation

Research for the method in the first part, which produceswell-conditioned collocation schemes, three directions areworthy of further investigation.

• Investigate the notion for well-conditionedpolynomial-based collocation methods for other situations,e.g., the spline collocation, radial basis functions and somenon-polynomial bases.

• Extension of the well-conditioned collocation approach tomultiple dimensions.

• Obtain the optimal error estimates for the Birkhoffinterpolations.


Tetrahedral spectral elements

The new TSEM on unstructured meshes based on the DGformulation is worthy of deep investigation. Furtherdevelopment can be taken in the following directions:

• Apply the TSEM to more challenging problems such as theStokes equations and the Navier-Stokes equations.

• Develop a three-dimensional unstructured tetrahedralTSEM.

• Prove global convergence of the unstructured TSEM.


Further reading

Robert Kirby, Spencer Sherwin and Bernardo Cockburn. ToCG or to HDG: a comparative study. Journal of ScientificComputing, vol. 51 (1), 183–212, 2012

Michael Daniel Samson, Li-Lian Wang and Huiyuan Li. Anew triangular spectral element method I:

implementation and analysis on a triangle. NumericalAlgorithms, vol. 64 (3), 519–547, 2013

Li-Lian Wang, Michael Daniel Samson and Xiaodan Zhao.A well-conditioned collocation method using a

pseudospectral integration matrix. Accepted to SIAMJournal on Scientific Computing, 2014

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Well-Conditioned Collocation Schemes and New Triangular Spectral-Element Methods

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