Runge Kutta Chebyshev Method for parabolic ... - Brown University

Introduction Properties Pros and Cons Examples References
Runge Kutta Chebyshev Method for
parabolic PDEs
Zheng Chen
Brown University
April 25th, 2012

Introduction Properties Pros and Cons Examples References
Overview
1 Introduction
2 Properties
Consistency conditions
Stability Properties
Integration formula
3 Pros and Cons
4 Examples
5 References

Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
˙u(t) = F(t, u(t)), 0 < t � T, u(0) = u0
which originate from spatial discretization of parabolic PDEs.
Restrictions:
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane
Example:
(1)

Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
˙u(t) = F(t, u(t)), 0 < t � T, u(0) = u0
which originate from spatial discretization of parabolic PDEs.
Restrictions:
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from a
normal matrix.
Example:
(1)

Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
˙u(t) = F(t, u(t)), 0 < t � T, u(0) = u0
which originate from spatial discretization of parabolic PDEs.
Restrictions:
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from a
normal matrix.
Example:
model heat equation
(1)
˙u(t) = ∆u (2)

Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
˙u(t) = F(t, u(t)), 0 < t � T, u(0) = u0
which originate from spatial discretization of parabolic PDEs.
Restrictions:
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from a
normal matrix.
Example:
model heat equation
reaction-diffusion problem
(1)
˙u(t) = ∆u (2)
˙u(t) = ɛ∆u + f(u, x, t), 0 < ɛ ≪ 1 (3)

Introduction Properties Pros and Cons Examples References
Introduction
Stiff problems:
standard explicit RK methods:
easy application, restrictive time-step for stability

Introduction Properties Pros and Cons Examples References
Introduction
Stiff problems:
standard explicit RK methods:
easy application, restrictive time-step for stability
implicit RK methods:
expensive to implement, unconditionally stable

Introduction Properties Pros and Cons Examples References
Introduction
Stiff problems:
standard explicit RK methods:
easy application, restrictive time-step for stability
implicit RK methods:
expensive to implement, unconditionally stable
RKC methods:
explicit, considerable time-step restriction
extended real stability interval with a length β ∝ s 2

Introduction Properties Pros and Cons Examples References
RKC formula
Y0 = Un, (4)
Y1 = Y0 + ˜µ1τF0, (5)
Yj = µjYj−1 + νjYj−2 + (1 − µj − νj)Y0
(6)
+ ˜µjτFj−1 + ˜γjτF0 (2 � j � s), (7)
Un+1 = Ys, n = 0, 1, . . . , (8)
This can be rewritten in the standard RK form:
�j−1
Yj = Un + τ
l=0
ajlF(tn + clτ, Yl), (0 � j � s) (9)
where Fj = F(tn + cjτ, Yj), cj are defined by:
c0 = 0, (10)
c1 = ˜µ1, (11)
cj = µjcj−1 + νjcj−2 + ˜µj + ˜γj, (2 � j � s) (12)

Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 Properties
Consistency conditions
Stability Properties
Integration formula
3 Pros and Cons
4 Examples
5 References

Introduction Properties Pros and Cons Examples References
Consistency conditions
Suppose Un = U(tn), where U(t), t � tn is a sufficiently
smooth solution. All Yj satisfy an expansion
Yj = U(tn) + cjτ ˙U(tn) + Xjτ 2 U (2) (tn) + O(τ 3 ) (13)
Substitute this into the RKC formula, we have
X0 = X1 = 0, (14)
Xj = µjXj−1 + νjXj−2 + ˜µjcj−1 (2 � j � s) (15)

Introduction Properties Pros and Cons Examples References
Consistency conditions
consistent of order 1: if
cs = 1. (16)

Introduction Properties Pros and Cons Examples References
Consistency conditions
consistent of order 1: if
consistent of order 2: if
cs = 1. (16)
c 2 2 = 2 ˜µ2c1, (17)
c 2 3 = µ3c 2 2 + 2 ˜µ2c2, (18)
c 2 j = µjc 2 j−1 + νjc 2 j−2 + 2 ˜µjcj−1 (4 � j � s) (19)

Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 Properties
Consistency conditions
Stability Properties
Integration formula
3 Pros and Cons
4 Examples
5 References

Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:
˙U(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)

Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:
Ps is defined recursively:
˙U(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)
P0(z) = 1, (22)
P1(z) = 1 + ˜µ1z, (23)
Pj(z) = (1 − µj − νj) + ˜γjz + (µj + ˜µjz)Pj−1(z) + νjPj−2(z) (2 �
(24)

Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:
Ps is defined recursively:
˙U(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)
P0(z) = 1, (22)
P1(z) = 1 + ˜µ1z, (23)
Pj(z) = (1 − µj − νj) + ˜γjz + (µj + ˜µjz)Pj−1(z) + νjPj−2(z) (2 �
(24)
for each stage, we have
Uj = Pj(z)Un, (0 � j � s) (25)

Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates e cjz
for z → 0 as
Pj = 1 + cjz + Xjz 2 + O(z 3 ). (26)
The choice of the stability function Pj(z) is the cental issue in
developing the RKC methods.
Stability Region S = {z ∈ C : |Ps| � 1}
Design rules:

Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates e cjz
for z → 0 as
Pj = 1 + cjz + Xjz 2 + O(z 3 ). (26)
The choice of the stability function Pj(z) is the cental issue in
developing the RKC methods.
Stability Region S = {z ∈ C : |Ps| � 1}
Stability Boundary β(s) = max{−z : z � 0, |Ps| � 1}
Design rules:

Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates e cjz
for z → 0 as
Pj = 1 + cjz + Xjz 2 + O(z 3 ). (26)
The choice of the stability function Pj(z) is the cental issue in
developing the RKC methods.
Stability Region S = {z ∈ C : |Ps| � 1}
Stability Boundary β(s) = max{−z : z � 0, |Ps| � 1}
Design rules:
β(s) is as large as possible

Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates e cjz
for z → 0 as
Pj = 1 + cjz + Xjz 2 + O(z 3 ). (26)
The choice of the stability function Pj(z) is the cental issue in
developing the RKC methods.
Stability Region S = {z ∈ C : |Ps| � 1}
Stability Boundary β(s) = max{−z : z � 0, |Ps| � 1}
Design rules:
β(s) is as large as possible
all coefficients must be known in analytic form

Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 Properties
Consistency conditions
Stability Properties
Integration formula
3 Pros and Cons
4 Examples
5 References

Introduction Properties Pros and Cons Examples References
shifted Chebyshev polynomials
Chebyshev polynomial of the first kind
Ts(x) = cos(sarccosx), −1 � x � 1 (27)
Eg: for the 1st order consistent polys, the shifted Chebyshev
poly
Ps(z) = Ts(1 + z
), −β(s) � z � 0
s2 (28)
yields the largest value: β(s) = 2s 2 . From the three-terms
recursion formula for Chebyshev polynomials, we get:
P0(z) = 1, P1(z) = 1+ z
s2 , Pj(z) = 2(1+ z
s2 )Pj−1(z)−Pj−2(z), j � 2,
(29)
which gives the analytical form of the integration coeffs
˜µ1 = 1/s 2 , µj = 2, ˜µj = 2/s 2 , νj = −1, ˜γj = 0, 0 � j � s (30)

Introduction Properties Pros and Cons Examples References
1st order case: RKC1
For 1st and 2nd order RKC, we have this general form
RKC1:
Pj(z) = aj + bjTj(w0 + w1z), 0 � j � s (31)
aj = 0, bj = T −1
j (w0), w0 = 1 + ɛ
Therefore,
β(s) � (w0 + 1)T ′ s(w0)
Ts(w0)
choose ɛ = 0.05, then β(s) = 1.90s 2 .
s2 , w1 = Ts(w0)
T ′ , (0 � j � s)
s(w0)
(32)
� (2 − 4ɛ
3 )s2 , ɛ → 0 (33)

Introduction Properties Pros and Cons Examples References
1st order case: RKC1
Then compare these with the recursive definition of Pj, we get,
˜µ1 = w1
, (34)
w0
µj = 2w0
˜µj = 2w1
bj
bj−1
bj
, νj = − bj
bj−1
Ts(w0)T ′
j
cj = (w0)
T ′ s(w0)Tj(w0) � j2 /s 2
bj−2
, (35)
, ˜γj = 0, (2 � j � s) (36)
(37)

Introduction Properties Pros and Cons Examples References
2nd order case: RKC2
T ′′
j (w0)
(T ′
j
s2 , w1 = T ′ s(w0)
T ′′
aj = 1 − bjTj(w0), bj =
w0 = 1 + ɛ
(w0)) 2 , (2 � j � s) (38)
, (39)
s (w0)
a0 = 1 − b0, a1 = 1 − b1w0, b0 = b1 = b2 (40)
β(s) � (w0 + 1)T ′′
s (w0)
T ′ �
s(w0)
2
3 (s2 − 1)(1 − 2
ɛ), ɛ → 0 (41)
15
From the pictures,we can tell ɛ = 2
13 is a suitable choice.

Introduction Properties Pros and Cons Examples References
2nd order case: RKC2
Then compare these with the recursive definition of Pj, we get,
˜µ1 = b1w1, (42)
µj = 2w0
˜µj = 2w1
c1 = c2
bj
bj−1
bj
bj−1
T ′
c2
�
(w0)
2
, νj = − bj
bj−2
, (43)
, ˜γj = −(1 − bj−1Tj−1(w0)) ˜µj, (2 � j � s)
4 , cj = T ′ s(w0)T ′′
j (w0)
T ′′
s (w0)T ′
j (w0) � j2 − 1
s2 − 1
(44)
(2 � j � s)
(45)

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Cons: Restrictions on the problem

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Quadratic increase of β(s) with the number of stages
Cons: Restrictions on the problem

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Quadratic increase of β(s) with the number of stages
Requires at most seven vectors of storage
Cons: Restrictions on the problem

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Quadratic increase of β(s) with the number of stages
Requires at most seven vectors of storage
no particular difficulties for vectorization and/or
parallelization
Cons: Restrictions on the problem

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Quadratic increase of β(s) with the number of stages
Requires at most seven vectors of storage
no particular difficulties for vectorization and/or
parallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane

Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:
Explicit and designed for modestly stiff problems
Quadratic increase of β(s) with the number of stages
Requires at most seven vectors of storage
no particular difficulties for vectorization and/or
parallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in a
narrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from a
normal matrix.

Introduction Properties Pros and Cons Examples References
Eg 1: linear heat conduction problem
ut = ∆u + f(x, y, z, t), 0 < x, y, z < 1, 0 � t � 0.7 (46)
u(x, y, z, t) = tanh(5(x + 2y + 1.5z − 0.5 − t)) (47)
Take uniform grid with h = 0.025, then 39 3 = 59319 equations:
Table: Results for RKC and BDF
Tol maxError #steps #F-evals CPU(s)
RKC BDF RKC BDF RKC BDF RKC BDF
1.E-1 8.9E-3 9.9E-1 6 7 402 46 186 35
1.E-2 1.7E-3 8.3E-2 15 16 729 160 338 122
1.E-3 3.7E-4 1.0E-2 27 34 786 237 366 185
1.E-4 3.9E-5 1.2E-3 57 70 1087 474 507 371
1.E-5 4.3E-6 1.3E-5 129 112 1682 984 787 770
1.E-6 6.5E-7 1.9E-5 262 168 2445 1151 1149 913

Introduction Properties Pros and Cons Examples References
Eg 2: combustion problem
ct = ∆c−Dce −δ/T , LTt = ∆T+αDce −δ/T , 0 < x, y, z < 1, 0 � t � 0.3
(48)
L = 0.9, α = 1, δ = 20, D = Re δ /αδ, with R = 5
The grid spacing is h = 1/(N + 0.5), with N = 40,
2 × 40 3 = 128000 equations
Tol maxError #steps #F-evals CPU(s)
RKC BDF RKC BDF RKC BDF RKC BDF
1.E-4 5.4E-1 8.7E-1 51 33 525 285 420 412
1.E-5 1.8E-1 7.6E-1 124 91 781 659 630 957
1.E-6 3.9E-2 1.2E-1 270 201 1270 1141 1030 1702
1.E-7 8.7E-3 1.2E-3 581 286 2147 1548 1758 2376
The low accuracy is expected from the local instability of the
problem.

Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,
Convergence properties of the Runge-Kutta-Chebyshev
method, Numer. Math. 57, 157-178, 1990.

Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,
Convergence properties of the Runge-Kutta-Chebyshev
method, Numer. Math. 57, 157-178, 1990.
B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: An
explicit solver for parabolic PDEs, J. Comp. Appl. Math.
88, 315-326, 1997.

Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,
Convergence properties of the Runge-Kutta-Chebyshev
method, Numer. Math. 57, 157-178, 1990.
B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: An
explicit solver for parabolic PDEs, J. Comp. Appl. Math.
88, 315-326, 1997.
J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKC
time-stepping for Advection-Diffusion-Reaction Problems,
J. Comput. Phys. 201, 61-79, 2004.

Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,
Convergence properties of the Runge-Kutta-Chebyshev
method, Numer. Math. 57, 157-178, 1990.
B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: An
explicit solver for parabolic PDEs, J. Comp. Appl. Math.
88, 315-326, 1997.
J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKC
time-stepping for Advection-Diffusion-Reaction Problems,
J. Comput. Phys. 201, 61-79, 2004.
J.G. Verwer and B.P. Sommeijer, An Implicit-Explicit
Runge-Kutta-Chebyshev Scheme for Diffusion-Reaction
Equations, SIAM J. Scientific Computing 25, 1824-1835,
2004.

Introduction Properties Pros and Cons Examples References
Thank you!