Post on 04-Jun-2020
Contour Integral Methods for PDEsPractical computation with conformal maps and double exponential quadrature
JAC Weideman & N HaleStellenbosch University, South Africa
Contour Integral Methods for PDEsBackground: The Trapezoidal Rule
Trapezoidal rule for integrals on R:∫∞
−∞
f(x) dx ≈ h∞∑
k=−∞
f(kh), h > 0
Convergence can be superfast:
Example:∫∞
−∞
e−x2
√1 + x2
dx = 1.5241093 . . .
Re z
Im z
ia
−ia
h Abs. Err e−2π/h
0.4 5.5 · 10−7 1.5 · 10−7
0.2 5.7 · 10−14 2.4 · 10−14
0.1 9.0 · 10−27 5.2 · 10−27
Theorem (Martensen 1986): If the in-tegrand is analytic in a strip of half-width a about R then the discretizationerror is
DE = O(e−2πa/h)
JAC Weideman - Stellenbosch University Slide 1/22
Contour Integral Methods for PDEsBackground: The Trapezoidal Rule
When the sum is truncated: h∞∑
k=−∞
f(kh) ≈ hn∑
k=−n
f(kh)
This introduces a second error, the truncation error. If integrand decays rapidlywe estimate this as the magnitude of the first omitted term
Example:∫∞
−∞
e−x2
√1 + x2
dx
⇒ TE = O(e−(nh)2)
An estimate for the optimal step lengthfollows from balancing the truncationand discretization errors
2πh
= (nh)2⇒ h =
(2π)1/3
n2/3 0 5 10 15 20 25 30 35 40 45 5010
−16
10−14
10−12
10−10
10−8
10−6
10−4
10−2
n
Abs
olut
e E
rror
Actual Error
e−(2πn)2/3
JAC Weideman - Stellenbosch University Slide 2/22
Contour Integral Methods for PDEsIntroduction
Let x ∈ [0,1], t ∈ [0,∞), u ∈ Rm×1, A ∈ Rm×m, with σ(A) ∈ R−
Semi-discrete Parabolic PDE:
∂u∂t
= Au, t > 0
Initial condition:
u(0) = f
Contour integral:
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Semi-discrete Elliptic PDE:
∂2u∂x2 + Au = 0, 0 < x < 1
Boundary conditions:
u(0) = 0, u(1) = f
Contour integral:
u(x) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
E(x; z) = sin(x√
z)/ sin(√
z)
JAC Weideman - Stellenbosch University Slide 3/22
Contour Integral Methods for PDEsIntroduction
Let x ∈ [0,1], t ∈ [0,∞), u ∈ Rm×1, A ∈ Rm×m, with σ(A) ∈ R−
Semi-discrete Parabolic PDE:
∂u∂t
= Au, t > 0
Initial condition:
u(0) = f
Contour integral:
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Semi-discrete Elliptic PDE:
∂2u∂x2 + Au = 0, 0 < x < 1
Boundary conditions:
u(0) = 0, u(1) = f
Contour integral:
u(x) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
E(x; z) = sin(x√
z)/ sin(√
z)
JAC Weideman - Stellenbosch University Slide 3/22
Contour Integral Methods for PDEsIntroduction
Let x ∈ [0,1], t ∈ [0,∞), u ∈ Rm×1, A ∈ Rm×m, with σ(A) ∈ R−
Semi-discrete Parabolic PDE:
∂u∂t
= Au, t > 0
Initial condition:
u(0) = f
Contour integral:
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Semi-discrete Elliptic PDE:
∂2u∂x2 + Au = 0, 0 < x < 1
Boundary conditions:
u(0) = 0, u(1) = f
Contour integral:
u(x) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
E(x; z) = sin(x√
z)/ sin(√
z)
JAC Weideman - Stellenbosch University Slide 3/22
Contour Integral Methods for PDEsIntroduction
References for the Parabolic case:Talbot [1979]Sheen, Sloan & Thomee [2000]Gavrilyuk & Makarov [2001]Lopez-Fernandez & Palencia[2004]W & Trefethen [2007]etc
References for the Elliptic case:Gavrilyuk, Makarov & Vasylyk[2006, 2011]
Gavrilyuk et al prove very generaltheorems in an abstract Banachspace
Our aim is practical: optimalcontours and step sizes, attentionto the numerical linear algebra, etc
JAC Weideman - Stellenbosch University Slide 4/22
Contour Integral Methods for PDEsIntroduction
Numerical approach to contour integrals:conformal mapping + trapezoidal rule ⇒ double exponential quadratureDomain of analyticity of each integrand?Both integrals singular when z = λ < 0, for any λ ∈ σ(A). When choosing acontour we therefore exclude R− (or (−∞,−`2], if dominant λ is known)
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Because the exponential function isentire, there are no other singularities.In floating-point arithmetic, however,ezt is “practically singular” for Re(z)sufficiently large
u(x) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
By contrast
E(x; z) = sin(x√
z)/ sin(√
z)
has isolated poles at z = π2,4π2,9π2,etc. We exclude [π2,∞)
JAC Weideman - Stellenbosch University Slide 5/22
Contour Integral Methods for PDEsIntroduction
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Re z
Im z
Γ
Red: “Floating-pt singularity” of ezt
u(x) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
Re z
Im z
Γ
Blue: σ(A) Red: Poles of E(x; z)
JAC Weideman - Stellenbosch University Slide 6/22
Contour Integral Methods for PDEsConformal Mapping for Elliptic Case
Until further notice we focus on the Elliptic case.
z =12π2
(1 + i sinh w
), |Im w | <
12π
Re z
Im z
π20
Re w
Im w
i2π
− i2π
JAC Weideman - Stellenbosch University Slide 7/22
Contour Integral Methods for PDEsQuadrature Sum
Contour of integration:
Γ : z =12π2
(1 + i sinhθ
), −∞ < θ < ∞
12πi
∫Γ
E(x; z)((zI − A)−1f
)dz =
12πi
∫∞
−∞
E(x; z(θ))((z(θ)I − A)−1f
)z′(θ) dθ
Introduce grid with spacing h
θk = (k + 12)h, k = 0,±1,±2, . . .
The approximation is then
u(x) ≈h
2πi
∞∑k=−∞
E(x; z(θk ))((
z(θk )I − A)−1
f)z′(θk )
=hπ
Im
n−1∑k=0
E(x; z(θk ))((
A − z(θk )I)−1
f)z′(θk )
JAC Weideman - Stellenbosch University Slide 8/22
Contour Integral Methods for PDEsPractical Implementation of Quadrature Sum
For each node of the quadrature sum a shifted linear system has to be solved:
u(x) =hπ
Im
n−1∑k=0
E(x; z(θk ))z′(θk )vk
, (A − z(θk )I
)︸ ︷︷ ︸
complex shift
vk = f
Since A is typically large, we wish to minimize number of resolvent solves.Can do this by(a) choosing a step size h for optimal accuracy(b) using efficient numerical algorithms for solving the shifted linear systems
In this talk we focus mainly on (a). Considerations for (b) includedirect vs iterativeserial vs parallelSylvester or Lyapunov solvers when A is in Kronecker product form
JAC Weideman - Stellenbosch University Slide 9/22
Contour Integral Methods for PDEsFeatures of The Method
∂2u∂x2 + Au = 0, 0 < x < 1
u(0) = 0, u(1) = fu(x) =
hπ
Im
n−1∑k=0
E(x; z(θk ))z′(θk )vk
Key points to note:
Solution of resolvent systems (A − z(θk )I)vk = f can be implemented inparallel, e.g., parfor in MATLABIf original problem is posed in Rd+1, subproblems in Rd are solvedSolution is computed at a fixed x “slice”, not in the entire domainAccuracy is lost when x → 1, which is the edge where the BC is prescribedOther types of BCs can also be handled, e.g., u(0) = g,u(1) = f . AlsoNeumann and mixedNeed to pick step size h carefully, otherwise convergence is slow as n→∞
JAC Weideman - Stellenbosch University Slide 10/22
Contour Integral Methods for PDEsDiscretization Error Estimate for the Trapezoidal/Midpoint Rule
If integrand is analytic in strip ofhalf-width a
DE = O(e−2πa/h)
In this case a = 12π
DE = O(e−π2/h)
Geometric convergence as h → 0,but only if infinite series can besummed exactly
Re w
Im w
i2π
− i2π
In practice the infinite series is truncated and a second error, namely theTruncation Error, is incurred
JAC Weideman - Stellenbosch University Slide 11/22
Contour Integral Methods for PDEsTruncation Error Estimate
Estimate truncation error as first omitted term in series
u(x) =hπ
Im
n−1∑k=0
E(x; z(θk ))z′(θk )vk
, E(x; z) =sin(x
√z)
sin(√
z)
⇒ TE = O(e−(1−x) Im√
z(θn))
Now, as n→∞ with h fixed
z(θn) ∼i2π2 sinhθn ∼
i4π2 eθn ⇒
√z(θn) ∼
eπi/4
2πeθn/2
Because θn ∼ nh it follows that
TE = O(e−(1−x)πenh/2/(2√
2))
Note: Double exponential decay in integrand⇒ double exponential quadratureJAC Weideman - Stellenbosch University Slide 12/22
Contour Integral Methods for PDEsEstimate for Optimal Step Size h
Recall estimates for discretization and truncation errors, respectively,
DE = O(e−π2/h), TE = O(e−(1−x)πenh/2/(2
√2))
Set exponential orders equal
π2
h=
1 − x
2√
2πe
w︷︸︸︷hn/2 (
w ew = t ⇒ w = W(t))
︸ ︷︷ ︸Solve for optimal step size h in terms of the Lambert W -function
h∗ =2n
W( √
2πn1 − x
)The corresponding estimate for the total error, as n→∞, is
Total Error = O(e−π
2/h∗)
Next we check the validity of these estimates on a toy problemJAC Weideman - Stellenbosch University Slide 13/22
Contour Integral Methods for PDEsModel Problems
Laplace in a square [0,1] × [0,1]
∂2u∂x2 +
∂2u∂y2︸︷︷︸A u
= 0
Boundary conditions:
u(x ,0) = u(x ,1) = 0, 0 < x < 1
u(0, y) = 0, u(1, y) = 1, 0 < y < 1 1
0.5
01
0.5
1
0.5
00
xy
Will also considersame problem on the rectangle [0,1] × [0,10], which is equivalent to the lastof the famous Ten Digit problems of Nick Trefethen (2002)same problem in the box [0,1] × [0,1] × [0,10]
JAC Weideman - Stellenbosch University Slide 14/22
Contour Integral Methods for PDEsVerification of Optimal Step-Size Formula
Optimal formula for h∗ is shown as dashed curve in the (n,h) parameter plane:experiment (left) and theory (right). (Error computed at x = 0.5)
Above the dashed curve the discretization error dominates O(e−π2/h)
Below it, the truncation error dominates O(e−(1−x)πenh/2/(2√
2))
JAC Weideman - Stellenbosch University Slide 15/22
Contour Integral Methods for PDEsAsymptotic Behavior of Error
The principal (real) branch of the W -function satisfies W(t) ∼ log t , t →∞Therefore, for fixed x and n→∞
h∗ =2n
W( √
2πn1 − x
)∼
2n
log( n1 − x
)Total Error = O
(e−π
2/h∗)
= O
(e−
12π
2n/ log(n/(1−x)))
“Almost” geometric convergence,typical of double exponential quadrature
What if x ≈ 1? Can modify contour andchoice of h to obtain O
(e−c
√n)
convergence(Gavrilyuk et al, 2006, 2011)
JAC Weideman - Stellenbosch University Slide 16/22
Contour Integral Methods for PDEsFine Tuning: Using Spectral Information
If dominant eigenvalue of A is known, one can replace R− by (−∞,−`2], i.e.,
z =12
(π2− `2) +
i2
(π2 + `2) sinhθ, −∞ < θ < ∞
Number of resolvent solves0 5 10 15 20 25 30
Rel
ativ
e er
ror
10 -10
10 -5
10 0
Rectangle: [0,1]x[0,10]. x = 0.5
Without spectral infoWith spectral info
JAC Weideman - Stellenbosch University Slide 17/22
Contour Integral Methods for PDEsA 3D Problem
Laplace in a box [0,1] × [0,1] × [0,10]
∂2u∂x2 +
∂2u∂y2 +
∂2u∂z2︸ ︷︷ ︸
A u
= 0
Shifted systems (A −zI)v = f solved byApproximating 2D Laplacian A byKronecker product of Chebyshevdifferentiation matrices D (20 × 20)This leads to a Sylvester systemthat can be solved efficiently bylyap in MATLAB
Timing and accuracy results coming up
Boundary conditions:One square edge u = 1, all others 0This is the 3D version of the last of theTrefethen 10 Digit problems:
JAC Weideman - Stellenbosch University Slide 18/22
Contour Integral Methods for PDEsA 3D Problem: numerical results
Error computed at midpoint of[0,1] × [0,1] × [0,10]
ParticleInBox_opt
Number of terms in Trap rule 15
Elapsed time is 0.006097 seconds.
ans =
7.298817657065122e-107.298817657048526e-10
[Dell Precision M400 laptop]
Trefethen & W, SIAM Review 2014
JAC Weideman - Stellenbosch University Slide 19/22
Contour Integral Methods for PDEsParabolic vs Elliptic: convergence rates & stability
Parabolic:∂u∂t
= Au, t > 0
u(t) =1
2πi
∫Γ
ezt((zI − A)−1f
)dz
Elliptic:∂2u∂x2 + Au = 0, 0 < x < 1
u(t) =1
2πi
∫Γ
E(x; z)((zI − A)−1f
)dz
JAC Weideman - Stellenbosch University Slide 20/22
Contour Integral Methods for PDEsParabolic vs Elliptic: sensitivity to parameters
Parabolic:∂u∂t
= Au, t > 0
n
h
2 4 6 8 10 12 14 16 18 20
0.05
0.1
0.15
0.2
0.25
−14
−12
−10
−8
−6
−4
−2
Elliptic:∂2u∂x2 + Au = 0, 0 < x < 1
n
h
5 10 15 20 25 30
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
−14
−12
−10
−8
−6
−4
−2
JAC Weideman - Stellenbosch University Slide 21/22
Contour Integral Methods for PDEsFurther Reading
To see details for Parabolic PDEs: To see proofs:
JAC Weideman - Stellenbosch University Slide 22/22