Runge Kutta Chebyshev Method for parabolic PDEs · Introduction PropertiesPros and...
Transcript of Runge Kutta Chebyshev Method for parabolic PDEs · Introduction PropertiesPros and...
Introduction Properties Pros and Cons Examples References
Runge Kutta Chebyshev Method forparabolic PDEs
Zheng ChenBrown UniversityApril 25th, 2012
Introduction Properties Pros and Cons Examples References
Overview
1 Introduction
2 PropertiesConsistency conditionsStability PropertiesIntegration formula
3 Pros and Cons
4 Examples
5 References
Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
u̇(t) = F(t,u(t)), 0 < t 6 T , u(0) = u0 (1)
which originate from spatial discretization of parabolic PDEs.Restrictions:
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from anormal matrix.
Example:
model heat equation
u̇(t) = ∆u (2)
reaction-diffusion problem
u̇(t) = ε∆u+ f(u, x, t), 0 < ε� 1 (3)
Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
u̇(t) = F(t,u(t)), 0 < t 6 T , u(0) = u0 (1)
which originate from spatial discretization of parabolic PDEs.Restrictions:
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Example:
model heat equation
u̇(t) = ∆u (2)
reaction-diffusion problem
u̇(t) = ε∆u+ f(u, x, t), 0 < ε� 1 (3)
Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
u̇(t) = F(t,u(t)), 0 < t 6 T , u(0) = u0 (1)
which originate from spatial discretization of parabolic PDEs.Restrictions:
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Example:model heat equation
u̇(t) = ∆u (2)
reaction-diffusion problem
u̇(t) = ε∆u+ f(u, x, t), 0 < ε� 1 (3)
Introduction Properties Pros and Cons Examples References
Introduction
initial value problem for the ODE systems:
u̇(t) = F(t,u(t)), 0 < t 6 T , u(0) = u0 (1)
which originate from spatial discretization of parabolic PDEs.Restrictions:
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Example:model heat equation
u̇(t) = ∆u (2)
reaction-diffusion problem
u̇(t) = ε∆u+ f(u, x, t), 0 < ε� 1 (3)
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 stableRKC methods:explicit, considerable time-step restrictionextended real stability interval with a length β ∝ s2
Introduction Properties Pros and Cons Examples References
Introduction
Stiff problems:standard explicit RK methods:easy application, restrictive time-step for stabilityimplicit RK methods:expensive to implement, unconditionally stable
RKC methods:explicit, considerable time-step restrictionextended real stability interval with a length β ∝ s2
Introduction Properties Pros and Cons Examples References
Introduction
Stiff problems:standard explicit RK methods:easy application, restrictive time-step for stabilityimplicit RK methods:expensive to implement, unconditionally stableRKC methods:explicit, considerable time-step restrictionextended real stability interval with a length β ∝ s2
Introduction Properties Pros and Cons Examples References
RKC formula
Y0 = Un, (4)Y1 = Y0 + µ̃1τF0, (5)Yj = µjYj−1 + νjYj−2 + (1 − µj − νj)Y0 (6)
+ µ̃jτFj−1 + γ̃jτF0 (2 6 j 6 s), (7)Un+1 = Ys, n = 0, 1, . . . , (8)
This can be rewritten in the standard RK form:
Yj = Un + τ
j−1∑l=0
ajlF(tn + clτ, Yl), (0 6 j 6 s) (9)
where Fj = F(tn + cjτ, Yj), cj are defined by:
c0 = 0, (10)c1 = µ̃1, (11)cj = µjcj−1 + νjcj−2 + µ̃j + γ̃j, (2 6 j 6 s) (12)
Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 PropertiesConsistency conditionsStability PropertiesIntegration 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 sufficientlysmooth solution. All Yj satisfy an expansion
Yj = U(tn) + cjτU̇(tn) + Xjτ2U(2)(tn) +O(τ
3) (13)
Substitute this into the RKC formula, we have
X0 = X1 = 0, (14)Xj = µjXj−1 + νjXj−2 + µ̃jcj−1 (2 6 j 6 s) (15)
Introduction Properties Pros and Cons Examples References
Consistency conditions
consistent of order 1: if
cs = 1. (16)
consistent of order 2: if
c22 = 2µ̃2c1, (17)
c23 = µ3c
22 + 2µ̃2c2, (18)
c2j = µjc
2j−1 + νjc
2j−2 + 2µ̃jcj−1 (4 6 j 6 s) (19)
Introduction Properties Pros and Cons Examples References
Consistency conditions
consistent of order 1: if
cs = 1. (16)
consistent of order 2: if
c22 = 2µ̃2c1, (17)
c23 = µ3c
22 + 2µ̃2c2, (18)
c2j = µjc
2j−1 + νjc
2j−2 + 2µ̃jcj−1 (4 6 j 6 s) (19)
Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 PropertiesConsistency conditionsStability PropertiesIntegration formula
3 Pros and Cons
4 Examples
5 References
Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:U̇(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)
Ps is defined recursively:
P0(z) = 1, (22)P1(z) = 1 + µ̃1z, (23)Pj(z) = (1 − µj − νj) + γ̃jz+ (µj + µ̃jz)Pj−1(z) + νjPj−2(z) (2 6 j 6 s)
(24)
for each stage, we have
Uj = Pj(z)Un, (0 6 j 6 s) (25)
Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:U̇(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)
Ps is defined recursively:
P0(z) = 1, (22)P1(z) = 1 + µ̃1z, (23)Pj(z) = (1 − µj − νj) + γ̃jz+ (µj + µ̃jz)Pj−1(z) + νjPj−2(z) (2 6 j 6 s)
(24)
for each stage, we have
Uj = Pj(z)Un, (0 6 j 6 s) (25)
Introduction Properties Pros and Cons Examples References
Stability function
Scalar test equation:U̇(t) = λU(t) (20)
Un+1 = Ps(z)Un, z = τλ (21)
Ps is defined recursively:
P0(z) = 1, (22)P1(z) = 1 + µ̃1z, (23)Pj(z) = (1 − µj − νj) + γ̃jz+ (µj + µ̃jz)Pj−1(z) + νjPj−2(z) (2 6 j 6 s)
(24)
for each stage, we have
Uj = Pj(z)Un, (0 6 j 6 s) (25)
Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates ecjz
for z→ 0 asPj = 1 + cjz+ Xjz
2 +O(z3). (26)
The choice of the stability function Pj(z) is the cental issue indeveloping the RKC methods.
Stability Region S = {z ∈ C : |Ps| 6 1}
Stability Boundary β(s) = max{−z : z 6 0, |Ps| 6 1}
Design rules:
β(s) is as large as possibleall coefficients must be known in analytic form
Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates ecjz
for z→ 0 asPj = 1 + cjz+ Xjz
2 +O(z3). (26)
The choice of the stability function Pj(z) is the cental issue indeveloping the RKC methods.
Stability Region S = {z ∈ C : |Ps| 6 1}Stability Boundary β(s) = max{−z : z 6 0, |Ps| 6 1}
Design rules:
β(s) is as large as possibleall coefficients must be known in analytic form
Introduction Properties Pros and Cons Examples References
Stability boundary
According to the consistency condition, Pj(z) approximates ecjz
for z→ 0 asPj = 1 + cjz+ Xjz
2 +O(z3). (26)
The choice of the stability function Pj(z) is the cental issue indeveloping the RKC methods.
Stability Region S = {z ∈ C : |Ps| 6 1}Stability Boundary β(s) = max{−z : z 6 0, |Ps| 6 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
Stability boundary
According to the consistency condition, Pj(z) approximates ecjz
for z→ 0 asPj = 1 + cjz+ Xjz
2 +O(z3). (26)
The choice of the stability function Pj(z) is the cental issue indeveloping the RKC methods.
Stability Region S = {z ∈ C : |Ps| 6 1}Stability Boundary β(s) = max{−z : z 6 0, |Ps| 6 1}
Design rules:β(s) is as large as possibleall coefficients must be known in analytic form
Introduction Properties Pros and Cons Examples References
Outline
1 Introduction
2 PropertiesConsistency conditionsStability PropertiesIntegration 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 6 x 6 1 (27)
Eg: for the 1st order consistent polys, the shifted Chebyshevpoly
Ps(z) = Ts(1 +z
s2 ),−β(s) 6 z 6 0 (28)
yields the largest value: β(s) = 2s2. From the three-termsrecursion 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
µ̃1 = 1/s2,µj = 2, µ̃j = 2/s2,νj = −1, γ̃j = 0, 0 6 j 6 s (30)
Introduction Properties Pros and Cons Examples References
1st order case: RKC1
For 1st and 2nd order RKC, we have this general form
Pj(z) = aj + bjTj(w0 +w1z), 0 6 j 6 s (31)
RKC1:
aj = 0,bj = T−1j (w0),w0 = 1 +
ε
s2 ,w1 =Ts(w0)
T ′s(w0)
, (0 6 j 6 s)
(32)Therefore,
β(s) ' (w0 + 1)T ′s(w0)
Ts(w0)' (2 −
4ε3)s2, ε→ 0 (33)
choose ε = 0.05, then β(s) = 1.90s2.
Introduction Properties Pros and Cons Examples References
1st order case: RKC1
Then compare these with the recursive definition of Pj, we get,
µ̃1 =w1
w0, (34)
µj = 2w0bj
bj−1, νj = −
bj
bj−2, (35)
µ̃j = 2w1bj
bj−1, γ̃j = 0, (2 6 j 6 s) (36)
cj =Ts(w0)T
′j (w0)
T ′s(w0)Tj(w0)
' j2/s2 (37)
Introduction Properties Pros and Cons Examples References
2nd order case: RKC2
aj = 1 − bjTj(w0), bj =T ′′j (w0)
(T ′j (w0))2 , (2 6 j 6 s) (38)
w0 = 1 +ε
s2 , w1 =T ′s(w0)
T ′′s (w0)
, (39)
a0 = 1 − b0, a1 = 1 − b1w0, b0 = b1 = b2 (40)
β(s) ' (w0 + 1)T ′′s (w0)
T ′s(w0)
' 23(s2 − 1)(1 −
215ε), ε→ 0 (41)
From the pictures,we can tell ε = 213 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,
µ̃1 = b1w1, (42)
µj = 2w0bj
bj−1, νj = −
bj
bj−2, (43)
µ̃j = 2w1bj
bj−1, γ̃j = −(1 − bj−1Tj−1(w0))µ̃j, (2 6 j 6 s)
(44)
c1 =c2
T ′2(w0)
' c2
4, cj =
T ′s(w0)T
′′j (w0)
T ′′s (w0)T
′j (w0)
' j2 − 1s2 − 1
(2 6 j 6 s)
(45)
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 stagesRequires at most seven vectors of storageno particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:Explicit and designed for modestly stiff problemsQuadratic increase of β(s) with the number of stages
Requires at most seven vectors of storageno particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:Explicit and designed for modestly stiff problemsQuadratic increase of β(s) with the number of stagesRequires at most seven vectors of storage
no particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:Explicit and designed for modestly stiff problemsQuadratic increase of β(s) with the number of stagesRequires at most seven vectors of storageno particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problem
The eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal matrix.
Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:Explicit and designed for modestly stiff problemsQuadratic increase of β(s) with the number of stagesRequires at most seven vectors of storageno particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problemThe eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex plane
Jacobian matrix should not deviate too much from anormal matrix.
Introduction Properties Pros and Cons Examples References
Pros and Cons
Pros:Explicit and designed for modestly stiff problemsQuadratic increase of β(s) with the number of stagesRequires at most seven vectors of storageno particular difficulties for vectorization and/orparallelization
Cons: Restrictions on the problemThe eigenvalues of the Jacobian matrix should lie in anarrow strip along the negative axis of the complex planeJacobian matrix should not deviate too much from anormal 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 6 t 6 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 393 = 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 46 186 351.E-2 1.7E-3 8.3E-2 15 16 729 160 338 1221.E-3 3.7E-4 1.0E-2 27 34 786 237 366 1851.E-4 3.9E-5 1.2E-3 57 70 1087 474 507 3711.E-5 4.3E-6 1.3E-5 129 112 1682 984 787 7701.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 6 t 6 0.3
(48)L = 0.9,α = 1, δ = 20,D = Reδ/αδ, with R = 5The grid spacing is h = 1/(N+ 0.5), with N = 40,2× 403 = 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 4121.E-5 1.8E-1 7.6E-1 124 91 781 659 630 9571.E-6 3.9E-2 1.2E-1 270 201 1270 1141 1030 17021.E-7 8.7E-3 1.2E-3 581 286 2147 1548 1758 2376
The low accuracy is expected from the local instability of theproblem.
Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,Convergence properties of the Runge-Kutta-Chebyshevmethod, Numer. Math. 57, 157-178, 1990.
B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: Anexplicit solver for parabolic PDEs, J. Comp. Appl. Math.88, 315-326, 1997.J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKCtime-stepping for Advection-Diffusion-Reaction Problems,J. Comput. Phys. 201, 61-79, 2004.J.G. Verwer and B.P. Sommeijer, An Implicit-ExplicitRunge-Kutta-Chebyshev Scheme for Diffusion-ReactionEquations, SIAM J. Scientific Computing 25, 1824-1835,2004.
Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,Convergence properties of the Runge-Kutta-Chebyshevmethod, Numer. Math. 57, 157-178, 1990.B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: Anexplicit solver for parabolic PDEs, J. Comp. Appl. Math.88, 315-326, 1997.
J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKCtime-stepping for Advection-Diffusion-Reaction Problems,J. Comput. Phys. 201, 61-79, 2004.J.G. Verwer and B.P. Sommeijer, An Implicit-ExplicitRunge-Kutta-Chebyshev Scheme for Diffusion-ReactionEquations, SIAM J. Scientific Computing 25, 1824-1835,2004.
Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,Convergence properties of the Runge-Kutta-Chebyshevmethod, Numer. Math. 57, 157-178, 1990.B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: Anexplicit solver for parabolic PDEs, J. Comp. Appl. Math.88, 315-326, 1997.J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKCtime-stepping for Advection-Diffusion-Reaction Problems,J. Comput. Phys. 201, 61-79, 2004.
J.G. Verwer and B.P. Sommeijer, An Implicit-ExplicitRunge-Kutta-Chebyshev Scheme for Diffusion-ReactionEquations, SIAM J. Scientific Computing 25, 1824-1835,2004.
Introduction Properties Pros and Cons Examples References
References
J.G. Verwer, W.H. Hundsdorfer and B.P. Sommeijer,Convergence properties of the Runge-Kutta-Chebyshevmethod, Numer. Math. 57, 157-178, 1990.B.P. Sommeijer, L.F. Shampine and J.G. Verwer, RKC: Anexplicit solver for parabolic PDEs, J. Comp. Appl. Math.88, 315-326, 1997.J.G. Verwer, B.P. Sommeijer and W. Hundsdorfer, RKCtime-stepping for Advection-Diffusion-Reaction Problems,J. Comput. Phys. 201, 61-79, 2004.J.G. Verwer and B.P. Sommeijer, An Implicit-ExplicitRunge-Kutta-Chebyshev Scheme for Diffusion-ReactionEquations, SIAM J. Scientific Computing 25, 1824-1835,2004.
Introduction Properties Pros and Cons Examples References
Thank you!