Sheila Smitheman
Spectral Methods for Elliptic Equations in
Regular and Convex Irregular Polygons
Sheila Smitheman
University of Nottingham
Joint work with: A S Fokas (DAMTP, Cambridge)
ICMS, 31 May 2012
University of Nottingham
Overview
Sheila Smitheman
Overview
• Introduction to the problem
• Previous work
• Current work
• Future plans
ICMS, 31 May 2012
University of Nottingham
Introduction to the Problem
Sheila Smitheman
Introduction to the problem
Let D ⊂ R be a bounded domain with boundary ∂D. Suppose that
the differential form W is closed in D and that W involves
continuously differentiable functions. Then, according to Poincaré’s
lemma, the integral of W along ∂D vanishes.
Example Let u satisfy the Laplace equation
∂2u ∂2u
+ 2 = 0,
∂x2
∂y
(x, y) ∈ D ⊂ R2 .
Let
z = x + iy,
Then,
Z
e−iλz
∂D
z̄ = x − iy.
∂u
∂u
− iαλu dz + α dz̄ = 0,
(1 + α)
∂z
∂ z̄
λ ∈ C,
where α is an arbitrary constant.
ICMS, 31 May 2012
University of Nottingham
Introduction to the Problem
Sheila Smitheman
Example Let u satisfy the modified Helmholtz equation
∂2u ∂2u
2
+
−
4β
u = 0,
2
2
∂x
∂y
β > 0,
(x, y) ∈ D ⊂ R2 .
Then,
Z
z̄
∂u
∂u
β
+ iβλu dz −
+ u dz̄ = 0,
e−iβ (λz− λ )
∂z
∂ z̄
iλ
∂D
λ ∈ Cr{0}.
We concentrate on the case that D is the interior of the convex
polygon with corners {zj }n1 , indexed counter-clockwise, modulo n.
Let Sj denote the side (zj , zj+1 ). We parametrize this side with
respect to its midpoint, i.e. if z ∈ Sj ,
z(s) = mj + shj ,
ICMS, 31 May 2012
−π < s < π,
j = 1, . . . , n,
University of Nottingham
Introduction to the Problem
Sheila Smitheman
where
mj =
1
(zj+1 + zj ),
2
hj =
1
(zj+1 − zj ),
2
j = 1, . . . , n.
Let uj and unj denote u and the normal derivative of u on the side
Sj , i.e.
uj (s) = u(z(s), z̄(s)),
unj (s) =
∂u
(z(s), z̄(s)),
∂n
z(s) ∈ Sj ,
j = 1, . . . , n,
where ∂/∂n denotes the derivative in the outward normal direction
to the side Sj .
We assume that a Robin boundary condition is prescribed on each
side of the polygon, i.e.
unj (s) − γj uj (s) = gj (s),
j = 1, . . . , n,
where {γj }n1 are given complex constants and gj (s) are given
complex functions.
ICMS, 31 May 2012
University of Nottingham
Introduction to the Problem
Sheila Smitheman
In the particular case of γj = 0, the Robin boundary condition
becomes a Neumann boundary condition; similarly, as γj → ∞, it
becomes a Dirichlet boundary condition.
Proposition 1 Let u satisfy the Laplace equation in the polygonal
domain D with Robin boundary conditions. Then,
Z
n X
(γj + λhj )e−iλmj
j=1
π
e−iλhj s uj (s)ds
−π
+e−iλmj
ICMS, 31 May 2012
Z
π
e−iλhj s gj (s)ds
−π
= 0,
λ ∈ C.
University of Nottingham
Introduction to the Problem
Sheila Smitheman
Similarly, if u satisfies the modified Helmholtz equation in the
polygonal domain D with Robin boundary conditions, then, for
λ ∈ C r {0},
n X
γj + βλhj +
j=1
Z
βhj
λ
π
mj
−iβ λmj − λ
e
hj
−iβ λhj − λ s
e
−π
mj
−iβ λmj − λ
+e
ICMS, 31 May 2012
Z
π
hj
−iβ λhj − λ s
e
−π
uj (s)ds
gj (s)ds
)
= 0.
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Previous Work - S., Spence & Fokas
If q, q̃ satisfy qz z̄ = λq and q̃z z̄ = λq̃, then
∂
∂
(q̃qz − q q̃z ) −
(q q̃z̄ − q̃qz̄ ) = 0.
∂ z̄
∂z
The complex form of Green’s theorem gives
Z
[(q̃qz − q q̃z ) dz + (q q̃z̄ − q̃qz̄ ) dz̄] = 0.
∂ Ω̃
λ
λ
Separation of variables gives q̃ = e±(ikz+ ik z̄ ) , q̃ = e±(ikz̄+ ik z ) .
λ
−(ikz+ ik
z̄ )
Letting q̃ = e
gives the global relation
Z
λ
λ
−(ikz+ ik
z̄ )
q + qz̄ dz̄ = 0 ∀k ∈ C.
e
(qz + ikq) dz −
ik
∂ Ω̃
Replacing k by k̄ and taking the complex conjugate yields the
Schwarz conjugate global relation:
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Z
λ
ikz̄+ ik
z)
(
e
∂ Ω̃
Sheila Smitheman
λ
q − qz dz + (qz̄ − ikq) dz̄ = 0 ∀k ∈ C.
ik
Take Ω̃ to be a bounded, convex n-sided polygon with vertices
z1 , z2 , . . . , zn and sides S1 , S2 , . . . , Sn (indexed anticlockwise,
modulo n):
zj−1
HH
HH
HHu
zj+2
Ω̃
u
H
Sj+1
HH
u
HH
zj+1
Sj−1 HH
HH
Sj
u zj
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
On Sj , we
• parametrize z by
z = mj +shj , s ∈ [−π, π] ,
zj+1 − zj
hj :=
;
2π
zj + zj+1
,
mj :=
2
• reinterpret q as a function q (j) (s) of s ∈ [−π, π];
• have that
1 (j)
1 (j)
(j)
(j)
qz dz =
qs (s) + iqn (s) ds, qz̄ dz̄ =
qs (s) − iqn (s) ds,
2
2
(j)
(j)
where qs (s) and qn (s) are the derivatives of q tangential and
normal to Sj .
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
The global relations become
n
X
λ
(j)
mj −kmj ) d
i( λ
k
qn khj − hj
e
k
j=1
n X
λ
λ
i( λ
mj −kmj )d
(j)
k
=−
khj + hj e
q
khj − hj ∀k ∈ C
k
k
j=1
and
n
X
λ
λ
d
(j)
ei(kmj − k mj ) qn −khj + hj
k
j=1
n X
λ
λ
λ
(j) −kh + h
=−
khj + hj ei(kmj − k mj ) qd
∀k ∈ C,
j
j
k
k
j=1
where ĥ is the Fourier transform of h over (−π, π):
Z π
1
ĥ(k) :=
e−iks h(s)ds ∀k ∈ C.
2π −π
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Previous Work - S., Spence & Fokas
On Sj , taking lj = |Sj |,
1
π
qz(j) (s) =
qs(j) (s) + iqn(j) (s) = e−iαj qs(j) (s) + iqn(j) (s) .
2hj
lj
(j)
Matching the two representations of qz at zj , qz (π) and
(j−1)
qz
(−π), gives
π −iαj (j)
qs (π) + iqn(j) (π)
e
lj
π −iαj−1 (j−1)
(j−1)
qs
(−π) + iqn
=
e
(−π) .
lj−1
(j)
Taking real and imaginary parts and solving gives qn (π) and
(j−1)
qn
(−π).
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
(j)
(j)
(j)
For each j ∈ {1, 2, . . . , n}, we set Qn (s) = qn (s) − qn,⋆ (s),
(j)
(j)
(j)
where qn,⋆ (s) is linear on [−π, π] s.t. qn,⋆ (±π) = qn (±π).
(j)
Then Qn (±π) = 0.
Basis Functions
We associate with each side Sj of the polygon the same set of basis
functions {φr (s)}N
r=1 , with N even.
s+π
Sine functions: φr (s) := sin r
.
2

s
s
 T
if r is odd,
r+1 π − T0 π
Chebyshev functions: φr (s) :=
 Tr+1 s − T1 s
if r is even,
π
π
constructed from the Chebyshev polynomials of the first kind
Tn (x) = cos(n cos−1 (x))
ICMS, 31 May 2012
∀n ∈ N ∪ {0}.
University of Nottingham
Previous Work - S., Spence & Fokas
sine basis functions
φr(s)
Sheila Smitheman
Chebyshev basis functions
φr(s)
1
2
0.5
1
0
0
−0.5
−1
r=1
r=2
r=3
r=4
−2
r=1
r=2
r=3
r=4
−1
0
2
s
−2
−2
0
2
s
(j)
For each j ∈ {1, 2, . . . , n}, we approximate the function Qn (s) by
a linear combination
(j)
Qn,N (s)
=
N
X
r=1
cr(j) φr(j) (s)
∀s ∈ [−π, π],
(j)
with the coefficients {cr }N
r=1 to be determined.
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Consider first global relation with λ ≥ 0 (Laplace, modified Helmholtz):
n
X
λ
λ
d
(j)
ei( k mj −kmj ) qn khj − hj
k
j=1
n X
λ
λ
λ
=−
khj + hj ei( k mj −kmj )d
q (j) khj − hj ∀k ∈ C.
k
k
j=1
d
(j)
λ
in the integrand of qn khj − k hj is
The term e
bounded as |k| → ∞ if, and only if, khj − λk hj ∈ R.
√2
−l±
l +4λ|hj |2
λ
Suppose khj − k hj = −l ∈ R. Then k = k± =
.
2hj
−i(khj − λ
k h j )s
For l > 0 we discard
k+ because k+ → 0 as l → ∞. So
√2
−l− l +4λ|hj |2
.
k = k− =
2hj
√
−l+ l2 +4λ|hj |2
For l < 0 we discard k− and take k = k+ =
.
2hj
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
So k = − hl̃j with ˜l =







l+
l−
i( λ
k mj −kmj )
As l → −∞, e
√2
l +4λ|hj |2
2
√2
l +4λ|hj |2
2
Sheila Smitheman
if l > 0,
.
if l < 0.
→ 0, so unknowns become weakly coupled.
i( λ
k mj −kmj )
As l → ∞, e
→ ∞, so unknowns remain strongly coupled.
√2
l+ l +4λ|hj |2
.
So we take l > 0 and l̃ =
2
So k ∈ ˆlj := {k ∈ C : arg(k) = π − arg(hj )}.
Similarly, in the second global relation we take
p
2 + 4λ|h |2
˜l
l
+
l
j
∈ ˜lj , l > 0, ˜l =
,
k=−
2
hj
where ˜lj := {k ∈ C : arg(k) = arg(hj ) − π}.
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
l̂j Y
H
HH
HH
HH
H
th
j
HH
αj
αj HHu
αj 0
Sj
l̃j
Take αj := arg(hj ) for j ∈ {1, . . . , n}.
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
The global relations become: for p ∈ {1, 2, . . . , n} and l > 0,
N
n X
X
j=1 r=1
(j)
l̃ d
τp,j σp,j
φr (zl,p,j,1 )cr(j)
1
l̃
=
n
X
j=1
N
n X
X
j=1 r=1
=
n
X
j=1
where
(j)
l̃ d
(τp,j ) (σp,j ) φr (−zl,p,j,1 )cr(j)
1
l̃
(1)
(2)
d
(j)
l̃
(j) (z
zl,p,j,2 qd
τp,j σp,j
)
−
q
n,⋆ (zl,p,j,1 ) ,
l,p,j,1
1
l̃
d
(j)
(j) (−z
(τp,j ) (σp,j )l̃ zl,p,j,2 qd
)
−
q
n,⋆ (−zl,p,j,1 ) ;
l,p,j,1
1
l̃
zl,p,j,1
λ
hj
˜
+ hj hp ,
= −l
˜l
hp
−iλ(mj −mp )hp
τp,j = e
ICMS, 31 May 2012
,
zl,p,j,2
λ
hj
˜
+ hj hp ,
=l
˜l
hp
i
σp,j = e
mj −mp
hp
.
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Collocation Points
Real-valued boundary data q(z): we only use (1).
For each p ∈ {1, 2, . . . , n}, we take the real and imaginary parts of
1 3
N
N −1
(1) at the points l ∈ 2 , 2 , . . . , 2
and l ∈ 1, 2, . . . , 2
respectively.
This gives a system of nN equations for the unknown coefficients
(j)
n
{{cr }N
r=1 }j=1 ⊂ R.
Complex-valued boundary data q(z): we consider the sum and
1 3
N −1
and
difference of (1) and (2) at the points l ∈ 2 , 2 , . . . , 2
l ∈ 1, 2, . . . , N2 respectively.
This gives a system of nN equations for the unknown coefficients
(j)
n
{{cr }N
r=1 }j=1 ⊂ C.
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
real boundary data
complex boundary data
l̂j H
Yb r
HH
b r
HH
b r
HbHr
Hb H
r b
HH
r
0
r Real part of (1)
b Imaginary part of (1)
ICMS, 31 May 2012
Sheila Smitheman
ˆlj H
Yb r
HH
b r
HH
b r
HbHr
Hb H
r b
HH
r
r 0
b
r
b
r
b
r
b
r
b
r b
˜lj r (1)+(2) b (1)-(2)
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Numerical Results
100
We take λ = 100 and use the analytical solution q(z, z̄) = e11z+ 11 z̄
(j)
to generate complex valued boundary data {q (j) (s), qn (s)}nj=1 .
(j)
We compare the numerical approximation {qn,N (s)}nj=1 to the
(j)
exact data {qn (s)}nj=1 by taking 10001 evenly spaced points
−π = s1 < s2 < . . . < s10000 < s10001 = π
and calculating the error E∞
where
||qn ||∞ =
||qn − qn,N ||∞ =
ICMS, 31 May 2012
||qn − qn,N ||∞
:=
,
||qn ||∞
max
j∈{1, 2, ..., n}
max
j∈{1, 2, ..., n}
max
k∈{1, 2, ..., 10001}
max
k∈{1, 2, ..., 10001}
|qn(j) (sk )| ,
(j)
|qn(j) (sk ) − qn,N (sk )| .
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Regular Polygons
sine basis
Chebyshev basis
0
0
10
10
−2
∞
−4
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
10
−6
10
−8
10
relative error, E
relative error, E∞
10
0
10
1
2
10
10
basis functions per side, N
ICMS, 31 May 2012
3
10
−5
10
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
−10
10
0
10
1
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
sine basis
Chebyshev basis
1
3
10
triangle
square
pentagon
hexagon
octagon
condition number
condition number
10
0
2
10
triangle
square
pentagon
hexagon
octagon
1
10
0
10
1
2
10
10
basis functions per side, N
ICMS, 31 May 2012
10 0
10
1
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Irregular Polygons
sine basis
Chebyshev basis
0
0
10
10
−2
10
−2
∞
−4
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
10
−6
10
−8
10
relative error, E
relative error, E∞
10
0
10
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
−6
10
−8
10
1
2
10
10
basis functions per side, N
ICMS, 31 May 2012
−4
10
3
10
0
10
1
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
sine basis
condition number
3
10
triangle
square
pentagon
hexagon
octagon
condition number
2
10
Chebyshev basis
1
10
0
10
2
10
triangle
square
pentagon
hexagon
octagon
1
10
0
1
10
2
10
basis functions per side, N
ICMS, 31 May 2012
3
10
10 0
10
1
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
The Helmholtz Equation
Sheila Smitheman
λ < 0,
β :=
√
−λ
Modified Helmholtz equation: in the first global relation we took
p
2 + 4λ|h |2
˜l
l
+
l
j
.
∈ ˆlj , l > 0, ˜l =
k=−
hj
2
For l ≥ 2β|hj |, we continue to do this. When l = 2β|hj |, |k| = β.
For l < 2β|hj |, we choose k on the circle with centre 0, radius β.
Recall the first global relation:
n
X
λ
(j)
) qd
hj
kh
−
n
j
k
j=1
n X
λ
λ
λ
(j) kh − h
khj + hj ei( k mj −kmj ) qd
=−
∀k ∈ C.
j
j
k
k
j=1
ei(
λ
k mj −kmj
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
If k = |k|ei arg(k) = βeiφ and hj = |hj |ei arg(hj ) = |hj |eiαj , then
λ
khj − hj = 2β|hj | cos (φ + αj ) .
k
In order to satisfy khj − λk hj = −l, we take
π
l
−1
) − αj ∈
φ = cos (−
− αj , π − αj .
2β|hj |
2
| {z }
∈(−1, 0)
Similarly, in the second global relation we take k = βeiφ with
π
l
+ αj , π + αj .
) + αj ∈
φ = cos−1 (−
2β|hj |
2
| {z }
∈(−1, 0)
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
l̂j Y
HH
HH
H
Sheila Smitheman
HH
HH #
r
H π β
H
α
2
jH
αj
π0
2A
K Ar
"!
l̃j
Numerical Results
25
We take λ = −25 and use the analytical solution q(z, z̄) = e4z− 4 z̄
(j)
to generate complex valued boundary data {q (j) (s), qn (s)}nj=1 .
ICMS, 31 May 2012
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Regular Polygons, Sine Basis Functions
5
10
0
10
4
−2
10
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
−4
10
−6
10
−8
10
condition number
relative error, E∞
10
0
10
2
10
triangle
square
pentagon
hexagon
octagon
1
10
0
1
2
10
10
basis functions per side, N
ICMS, 31 May 2012
3
10
3
10
10
1
2
10
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Regular Polygons, Chebyshev Basis Functions
10
10
triangle
square
pentagon
hexagon
octagon
condition number
relative error, E
∞
0
10
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
−10
10
0
10
0
1
10
basis functions per side, N
ICMS, 31 May 2012
5
10
10
0
10
1
10
basis functions per side, N
University of Nottingham
Previous Work - S., Spence & Fokas
Sheila Smitheman
Summary
We
• Considered sine and Chebyshev basis functions.
• Determined collocation points for real and complex valued
boundary data which gave optimal algebraic convergence for
sine basis functions for Laplace and modified Helmholtz
equations, and spectral accuracy for Chebyshev basis functions
for these equations.
• Undertook preliminary studies on the Helmholtz equation. It
was found that both error control and the conditioning of the
system matrix were far worse than for the other linear, elliptic
equations; and that they deteriorated with increasing β.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Current Work - S. & Fokas
Laplace and modified Helmholtz equations (λ ≥ 0)
The sub-matrices A(p,j) ∈ CN,N , p, j ∈ {1, 2, . . . , n}, of the system
matrix A ∈ CnN,nN satisfy

(1,1)
A

 A(2,1)

 (3,1)
A=
 A

..

.

A(n,1)
ICMS, 31 May 2012
(1,2)
A
(2,2)
(1,3)
A
(2,3)
A
A
A(3,2)
..
.
A(3,3)
..
.
A(n,2)
A(n,3)
(1,n)
···
A
(2,n)
···
A
···
..
.
A(3,n)
..
.
···
A(n,n)










University of Nottingham
Current Work - S. & Fokas
with, for r ∈ {1, 2, . . . , N },
i
h
(p,j)
A
2l,r

1

l̃ c
l̃

φr (zl,p,j ) ,
σp,j
 2i Im τp,j
1
=

l̃ c
l̃

φr (zl,p,j ) ,
σp,j
 2 Re τp,j
where zl,p,j
Sheila Smitheman
l ∈ {1, 2, . . . , N/2},
l ∈ {1/2, 3/2, . . . , (N − 1)/2};
mj −mp
λ
h
i
j
= −˜l + hj hp , τp,j = e−iλ(mj −mp )hp , σp,j = e hp .
hp ˜l
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
A regular n-sided polygon has order n rotational symmetry:
hj+1 = hj e
2πi
n
.
λ
hj+1
+ hj+1 hp+1
So zl,p+1,j+1,1 = −˜l
˜l
hp+1
2πi
n
2πi
λ
hj e
− 2πi
˜
n
hj e
· hp e n
= −l
2πi +
˜l
hp e n
λ
hj
+ hj hp = zl,p,j .
= −˜l
˜l
hp
Similarly,
τp+1,j+1 = τp,j ,
σp+1,j+1 = σp,j .
Hence A(p+1,j+1) = A(p,j) for p, j ∈ {1, 2, . . . , n}.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Hence the system matrix is of semi-block circulant form


A1
A2 A3 · · · An−1
An


 An
A1 A2 · · · An−2 An−1 




N ×N

(A
∈
R
).
A =  An−1 An A1 · · · An−3 An−2 
i



..
..
..
..
..
..


.
.
.
.
.
.


A2
A3
A4
···
An
A1
Furthermore, A has additional structure corresponding to clockwise
and clockwise rotation being related by complex conjugation.
Consider A2 = A(1,2) and An = A(1,n) .
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
For r ∈ {1, 2, . . . , N },
A{2,n} 2l,r

1

l̃
l̃
cr (zl,1,{2,n} ) , l ∈ {1, 2, . . . , N/2},

σ1,{2,n}
φ
2i Im τ1,{2,n}
1
=

l̃
l̃
cr (zl,1,{2,n} ) , l ∈ {1/2, 3/2, . . . , (N − 1)/2};

σ1,{2,n}
φ
2 Re τ1,{2,n}
where
zl,1,{2,n}
h{2,n}
λ
˜
+ h{2,n} h1 ,
= −l
˜l
h1
−iλ(m{2,n} −m1 )h1
τ1,{2,n} = e
Since h2 = h1 e
2πi
n
and hn = h1 e−
zl,1,2
ICMS, 31 May 2012
,
2πi
n
i
σ1,{2,n} = e
m{2,n} −m1
h1
.
,
2πi
λ
2 − 2πi
˜
n
= −le
+ |h1 | e n = zl,1,n .
˜l
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
For both sine and Chebyshev basis functions, φr (·) is even for r
odd and odd for r even. Hence
Z π
Z −π
−s↔s
cr (z̄) =
2π φ
e−iz̄s φr (s)ds = −
eiz̄s φr (−s)ds
−π
π

Z π
 2π φ
cr (z), r odd,
=
e−izs φr (−s)ds =
 −2π φ
cr (z), r even,
−π
giving that
cr (zl,1,n ) = φ
cr (zl,1,2 ) =
φ
Since m2 = m1 e
2πi
n


cr (zl,1,2 ),
φ
 −φ
cr (zl,1,2 ),
− 2πi
n
and mn = m1 e
r odd,
r even.
,
2πi
−
n
−iλ e
−1 m1 h1
τ1,2 = e−iλ(m2 −m1 )h1 = e
2πi
−
i
|z1 |2 sin( 2π
−iλ e n −1 · 2π
N )
=e
ICMS, 31 May 2012
λ
=e
2πi
e− n
|z1 |2
−1 · 2π
sin( 2π
N )
= τ1,n ;
University of Nottingham
Current Work - S. & Fokas
and, similarly, σ1,2 = σ1,n .


1
l̃
l̃
cr (zl,1,n ) =
So τ1,n
σ1,n φ

and hence
[An ]2l,r
Sheila Smitheman
1
l̃
l̃ φ
cr (zl,1,2 ),
τ1,2 σ1,2
1
l̃
l̃ φ
cr (zl,1,2 ),
−τ1,2 σ1,2
r odd,
r even,

1

l̃ c
l̃

φr (zl,1,n ) , l ∈ {1, 2, . . . , N/2},
σ1,n
2i Im τ1,n
1
=

l̃ c
l̃

φr (zl,1,n ) , l ∈ {1/2, 3/2, . . . , (N − 1)/2},
σ1,n
2 Re τ1,n


− [A2 ]2l,r , l ∈ {1, 2, . . . , N/2}, r odd,




 [A ]
2 2l,r , l ∈ {1, 2, . . . , N/2}, r even,
=

[A2 ]2l,r , l ∈ {1/2, 3/2, . . . , (N − 1)/2}, r odd,




− [A ]
, l ∈ {1/2, 3/2, . . . , (N − 1)/2}, r even,
2 2l,r
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
i.e.
[An ]i1 ,i2 = (−1)(i1 +i2 ) [A2 ]i1 ,i2 ,
i1 , i2 ∈ {1, 2, . . . , N }.
This is equivalent to

An = DA2 D,




D := 




Similarly, An−1 = DA3 D, . . ..
ICMS, 31 May 2012

1
−1
..
.
1
−1




 ∈ RN ×N .




University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
In summary, for the Laplace or modified Helmholtz equation in a
regular n-sided polygon, the system matrix is semi-block circulant


A1
A2 A3 · · · An−1
An


 An
A1 A2 · · · An−2 An−1 





A =  An−1 An A1 · · · An−3 An−2 



..
..
..
..
..
..


.
.
.
.
.
.


A2
A3
A4
···
An
A1
A has additional structure corresponding to clockwise and
clockwise rotation being related by complex conjugation:
An = DA2 D, An−1 = DA3 D, . . . , D = diag(1, −1, . . . , 1, −1) ∈ RN ×N .
These results have been proved for the Laplace equation in a less
intuitive way by Saridakis, Sifalakis & Papadopoulou (J. Comput.
Appl. Math. 236 2515–2528 (2012)).
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
What does this mean for matrix inversion?
Saridakis, Sifalakis & Papadopoulou (2012)
1. converted the linear system for the Laplace equation into an
equivalent one in which the system matrix has block diagonal
form;
2. solved this new system using both direct and indirect methods,
speeding up relevant matrix-vector multiplications using the
FFT.
They never found the inverse matrix explicitly, and their method
was O(N 3 n + N 2 nlogn).
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Instead, we use the following result
Propag. 45(10), 1565–1567 (1997))
circulant matrix

B1
B2 B3

 Bn
B1 B2


−1
A =
 Bn−1 Bn B1

..
..
..

.
.
.

B2
B3 B4
of Vescovo (IEEE T. Antenn.
for the inverse of a semi-block
···
Bn−1
Bn
···
Bn−2
Bn−1
···
..
.
Bn−3
..
.
Bn−2
..
.
···
Bn
B1
where
n
1 X j−1 (k) −1
αk [A ] ,
Bj = 2
n
k=1
αk = e
2π(k−1)
i
n





,




n
,
A(k)
1X
Aj (αk )j−1 .
=
n j=1
Furthermore,
Bn = DB2 D, Bn−1 = DB3 D, . . . , D = diag(1, −1, . . . , 1, −1) ∈ RN ×N .
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
So, once we have found A1 , . . . , A⌈ n+1 ⌉ , we have to
2
1. construct the N × N matrices A⌈ n+1 ⌉+1 , . . . , An - O(N 2 n)
2
2. construct the N × N matrices A(1) , . . . , A(n) - O(N 2 n2 )
3. invert the matrices A(1) , . . . , A(n) - O(N 3 n)
4. construct the N × N matrices B1 , . . . , B⌈ n+1 ⌉ - O(N 2 n2 )
2
5. construct the N × N matrices B⌈ n+1 ⌉+1 , . . . , Bn - O(N 2 n)
2
We use Matlab’s backslash command for the third step.
Overall, our simple implementation of the above is
O(N 3 n + N 2 n + N 2 n2 ) ≈ O(N 3 n).
Furthermore, the inverse matrix is found explicitly - so it can be
stored and re-used.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Matrix inversion times in seconds for sine basis functions:
SSP ≡ values quoted in Saridakis, Sifalakis, Papadopoulou (2012).
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
What about Chebyshev basis functions?
As we’ve already seen, these can offer spectral - rather than
algebraic - convergence. But we previously found that...
• computing their Fourier transforms efficiently was difficult;
• conditioning of the resulting system matrix was poor,
especially for the Helmholtz equation. It deteriorated with
√
both increasing N and increasing β = −λ.
The second issue is particularly unwelcome because as β increases,
the solutions become more oscillatory and so larger values of N
must be chosen to achieve “reasonable” error control.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Regular Polygons, Chebyshev Basis Functions
10
10
triangle
square
pentagon
hexagon
octagon
condition number
relative error, E
∞
0
10
triangle
square
pentagon
hexagon
octagon
order 1
order 2
order 3
−10
10
0
10
0
1
10
basis functions per side, N
ICMS, 31 May 2012
5
10
10
0
10
1
10
basis functions per side, N
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Our Solution
Recall that we use the Chebyshev functions

s
s
 T
if r is odd,
r+1 π − T0 π
φr (s) :=
 Tr+1 s − T1 s
if r is even,
π
π
constructed from the Chebyshev polynomials of the first kind
Tn (x) = cos(n cos−1 (x))
ICMS, 31 May 2012
∀n ∈ N.
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
S. & Fokas (2012) For k ∈ C r {0},
c1 (k) =
φ
4
1
sin(kπ)],
[cos(kπ)
−
2
2
k π
kπ
and for r = 2, 3, . . .,

h
i(r+1)2 i
1

− kπ
cos(kπ) − kπ sin(kπ)


kπ

n
oi



c
[
[

+2(r + 1) φ
, r odd,
r−1 (k) + φr−3 (k) + . . . + φ2 (k)


cr (k) =
φ

h


r(r+2)
i

−
sin(kπ)

kπ
kπ


n
oi



c
[
[
+2(r + 1) φ
, r even.
r−1 (k) + φr−3 (k) + . . . + φ1 (k)
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Recall that we wish to evaluate A1 , . . . , A⌈ n+1 ⌉ .
2
For r ∈ {1, 2, . . . , N },

1

l̃ c
l̃

φr (zl,1,j ) ,
σ1,j
 2i Im τ1,j
1
[Aj ]2l,r =

l̃ c
l̃

φr (zl,1,j ) ,
σ1,j
 2 Re τ1,j
where zl,p,j
l ∈ {1, 2, . . . , N/2},
l ∈ {1/2, 3/2, . . . , (N − 1)/2};
mj −mp
λ
h
i
j
= −˜l + hj hp , τp,j = e−iλ(mj −mp )hp , σp,j = e hp .
hp ˜l
For each j and l, our result can be used to sequentially calculate
[Aj ]2l,1 , [Aj ]2l,2 , . . . , [Aj ]2l,N without any need for recursive
function calls - all we need to do is store the values of
c1 (zl,1,j ), φ
c2 (zl,1,j ), . . . , φc
φ
N (zl,1,j ) as they are calculated.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Fornberg & Flyer (Proc. R. Soc. A 467(2134), 2983–3003 (2011))
“The k-values we use should be well separated from each other, so
that all the relations that are obtained by substituting them into
[the global relations] will be relatively independent of each other.
On the other hand, k-values of large magnitude will make the
integrals strongly dominated by a very narrow range of z-values.
Numerical experiments suggest it to be a good strategy to
distribute the k-values somewhat randomly within a domain
|k| ≤ R, where R has to be appropriately chosen.
Direct use of a random number generator is a bad strategy since, by
the nature of randomness, dense clusters will occasionally form...
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Several options are available for creating scattered numbers that
lack any apparent regularities, but which nevertheless will “avoid”
each other in a systematic manner.
Halton nodes have this feature, and are particularly easy to
generate.
For example, the routine haltonset in Matlabs Statistics Toolbox or
haltonseq from Matlab Central can be used to generate node sets in
a d-dimensional unit cube.
We use d = 2, and then ignore those falling outside the inscribed
circle to the unit square.
Translation and scaling will then provide good node sets satisfying
|k| ≤ R”.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
2D: the first 1000 Halton nodes:
1
0.75
0.5
0.25
0
0
ICMS, 31 May 2012
0.25
0.5
0.75
1
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
But... for regular polygons we lose the symmetry we got from
choosing points on rays.
So we consider instead the 1D Halton nodes
0
ICMS, 31 May 2012
0.25
0.5
0.75
1
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Complex boundary data:
previous approach
l̂j H
Yb r
HH
b r
HH
b r
HbHr
Hb H
r b
Hr
H
r 0
b
r
b
r
b
r
b
r
b
r
b
l̃j
ICMS, 31 May 2012
new approach
^
l
j
0
~
lj
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Summary
We have derived and implemented an elegant and essentially
analytic scheme for approximating the Dirichlet boundary data
(rather than, as is usually the case, its tangential derivative) from
given Robin boundary data for the three basic linear, elliptic PDEs
formulated in the interior of a regular polygon.
Our approach appears to yield both spectral accuracy and a well
conditioned system matrix whose structure reflects the rotational
symmetry of the domain, allowing its inverse to be computed
efficiently.
The only non-analytic part is the determination of the Fourier
(j) (z
transforms of the known boundary data on each side, qd
l,p,j,1 ),
if this cannot be achieved analytically.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
We have made many assumptions in our work:
• that there exists a unique solution to the global relation
- A C L Ashton
• that q being a solution of this global relation is necessary and
sufficient for it to be a solution of the original PDE
- A C L Ashton
• that q has sufficient regularity to approximate the functions
(j)
(j)
(j)
Qn (s) = qn (s) − qn,⋆ (s), j ∈ {1, . . . , n}, by sequences of
linear combinations
(j)
Qn,N (s) =
N
X
r=1
ICMS, 31 May 2012
cr(j) φr(j) (s)
∀s ∈ [−π, π],
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
(j)
with the coefficients {cr }N
r=1 satisfying, for p ∈ {1, 2, . . . , n},
N
n X
X
j=1 r=1
(j)
l̃ d
φr (zl,p,j,1 )cr(j)
τp,j σp,j
1
l̃
n
X
=
j=1
N
n X
X
j=1 r=1
=
n
X
j=1
where
d
(j)
(τp,j ) (σp,j )l̃ φr (−zl,p,j,1 )cr(j)
1
l̃
d
(j)
(j) (−z
(τp,j ) (σp,j )l̃ zl,p,j,2 qd
)
−
q
n,⋆ (−zl,p,j,1 ) ;
l,p,j,1
1
l̃
zl,p,j,1
hj
λ
˜
= −l
+ hj hp ,
˜l
hp
−iλ(mj −mp )hp
τp,j = e
ICMS, 31 May 2012
d
(j)
l̃
(j) (z
τp,j σp,j
zl,p,j,2 qd
)
−
q
(zl,p,j,1 ) ,
n,⋆
l,p,j,1
1
l̃
,
zl,p,j,2
hj
λ
˜
=l
+ hj hp ,
˜l
hp
i
σp,j = e
mj −mp
hp
;
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
(the sum and difference of these equations being taken at the
1 3
N
N −1
points l ∈ 2 , 2 , . . . , 2
and l ∈ 1, 2, . . . , 2
respectively) - A C L Ashton, that the system matrix which
results is invertible, and that
(j)
max Qn,N − Qn(j) → 0 as N → ∞
j∈{1, ..., n}
for some appropriate norm k · k.
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
The Fourier Transform of the Known Boundary data
(j) , j ∈ {1, 2, . . . , n}, cannot be found analytically.
Suppose that qd
Analogous to the (assumed) Chebyshev approximations
(j)
Qn,N (·) =
N
X
r=1
(2)
(N )
cr(j) φr(j) (·) ∈ span{φ(1)
,
φ
,
.
.
.
,
φ
r
r
r },
j ∈ {1, . . . , n}
(j)
of the unknown boundary functions {Qn }nj=1 , the known
boundary functions {Q(j) }N
j=1 with
(j)
Q(j) (s) = q (j) (s) − q⋆ (s),
(j)
(j)
qn,⋆ (s) linear s.t. qn,⋆ (±π) = qn(j) (±π),
j ∈ {1, . . . , n}
(j)
+1
can be approximated by functions {QN }nj=1 ⊂ span{{Tj }N
j=1 }
(C W Clenshaw & A R Curtis, Numer. Math. 2, 197–205 (1960)):
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
(j)
QN (s)
Sheila Smitheman
=
N
+1
X
′′
ar T r
r=0
where
s
π
,
πtr
πt
2
Q(j) π cos
cos
ar =
N + 1 t=0
N +1
N +1
N
2 X′′ (j)
πtr
πt
=
cos
,
π cos
Q
N + 1 t=1
N +1
N +1
N
+1
X
′′
where
′′
means first and last terms in the sum are divided by two.
(Recall that the Chebyshev functions:

 T
s
r+1 π − T0
φr (s) :=
 Tr+1 s − T1
π
ICMS, 31 May 2012
s
π
s
π
if r is odd,
if r is even,
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
were constructed from the Chebyshev polynomials of the first kind
Tn (x) = cos(n cos−1 (x))
∀n ∈ N ∪ {0}.)
(j)
Then QN is the Lagrangian interpolation polynomial of Q(j) such
that
πt
πt
(j)
= Q(j) π cos
, t = 0, 1, . . . N +1.
QN π cos
N +1
N +1
(3)
It follows that
(j)
0 = Q(j) (π) = QN (π) =
N
+1
X
′′
ar Tr (1) =
r=0
(j)
0 = Q(j) (−π) = QN (−π) =
N
+1
X
′′
r=0
ICMS, 31 May 2012
ar Tr (−1) =
N
+1
X
′′
ar ,
r=0
N
+1
X
′′
(−1)r ar .
r=0
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Hence
a0 + 2a1 = −2
where
′
N
+1
X
′
ar
&
a0 − 2a1 = −2
r=2
N
+1
X
′
(−1)r ar ,
r=2
means the last term is halved, giving that (recall N is even)
"N +1
#
N
N
+1
2
X
X
X
′
′
a2k ,
a0 = −
ar +
(−1)r ar = −2
r=2
a1 = −
ICMS, 31 May 2012
"N +1
1 X
′
2
r=2
ar +
r=2
N
+1
X
′
r=2
k=1
#
(−1)r+1 ar = −
N
2
+1
X
′
a2k−1 .
k=2
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Hence
(j)
QN (s)
=
N
+1
X
′′
ar T r
r=0
N
2
s
π
+1
s
π NX
s
1
′
= a0 T 0
+ a1 T 1
+
ar T r
2
π
s
π
r=2
N
2 +1
h
s
h
s
s i X
s i
X
′
=
a2k−1 T2k−1
a2k T2k
− T0
+
− T1
π
π
π
π
k=1
k=2
N
2
N
2
=
X
a2k φ2k−1 (s) +
k=1
where
ar =
2
N +1
+1
X
′
a2k−1 φ2k−2 (s) =
′′
t=1
Q(j) π cos
′
ar+1 φr (s),
r=1
k=2
N
X
N
X
πt
N +1
cos
πtr
N +1
.
(j)
Convergence results for QN → Q(j) as N → ∞ are standard, and
d
(j)
d
(j) .
these give convergence for QN → Q
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
There is a modified version of the FFT for efficiently computing the
ar ’s (W M Gentleman, Comm. ACM 15, 343–346 (1972)).
(j)
The Fourier transforms of the functions {QN }nj=1 follow from
d
(j)
QN (k) =
N
X
′
r=1
cr (k),
ar+1 φ
cr ’s calculated as discussed previously.
with the φ
ICMS, 31 May 2012
University of Nottingham
Current Work - S. & Fokas
Sheila Smitheman
Future work
(1) Derivation of a suitable choice of collocation points for the
Helmholtz equation for regular polygons; extension to irregular
polygons.
(2) Proof of convergence of the spectral collocation scheme
developed in (1).
(3) Determination of a robust and efficient method for inverting
the system matrix for irregular polygons (the relevant matrix
for regular polygons can be inverted analytically).
(4) Computation of the solution inside the domain.
(5) Investigation of the exterior problem for a polygon.
(6) Investigation of solutions with corner singularities.
ICMS, 31 May 2012
University of Nottingham
References
Sheila Smitheman
References
A S Fokas, N Flyer, S A Smitheman and E A Spence, A
Semi-Analytical Numerical Method for Solving Evolution and
Elliptic Partial Differential Equations, J. Comput. Appl. Math.
227(1), 59–74 (2009) (Invited Paper).
S A Smitheman, E A Spence and A S Fokas, A Spectral Collocation
Method for the Laplace and Modified Helmholtz Equations in a
Convex Polygon, IMA J. Numer. Anal. 30(4), 1185–1205 (2010).
S A Smitheman and A S Fokas, A Semi-Analytical Spectral
Method for Elliptic Equations in Regular Polygons, in preparation.
ICMS, 31 May 2012
University of Nottingham