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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 Introduction 1

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Introduction Properties Pros and Cons Examples References

Runge Kutta Chebyshev Method for parabolic PDEs

Zheng Chen Brown University April 25th, 2012

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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

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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 (1) 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:

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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 (1) 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:

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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 (1) 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 ˙ u(t) = ∆u (2)

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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 (1) 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 ˙ u(t) = ∆u (2) reaction-diffusion problem ˙ u(t) = ǫ∆u + f(u, x, t), 0 < ǫ ≪ 1 (3)

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Introduction Properties Pros and Cons Examples References

Introduction

Stiff problems: standard explicit RK methods: easy application, restrictive time-step for stability

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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

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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 ∝ s2

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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: Yj = Un + τ

j−1

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)

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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

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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) + Xjτ2U(2)(tn) + O(τ3) (13) Substitute this into the RKC formula, we have X0 = X1 = 0, (14) Xj = µjXj−1 + νjXj−2 + ˜ µjcj−1 (2 j s) (15)

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Introduction Properties Pros and Cons Examples References

Consistency conditions

consistent of order 1: if cs = 1. (16)

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Introduction Properties Pros and Cons Examples References

Consistency conditions

consistent of order 1: if cs = 1. (16) consistent of order 2: if c2

2 = 2˜

µ2c1, (17) c2

3 = µ3c2 2 + 2˜

µ2c2, (18) c2

j = µjc2 j−1 + νjc2 j−2 + 2˜

µjcj−1 (4 j s) (19)

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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

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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)

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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) Ps is defined recursively: 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)

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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) Ps is defined recursively: 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)

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Introduction Properties Pros and Cons Examples References

Stability boundary

According to the consistency condition, Pj(z) approximates ecjz for z → 0 as Pj = 1 + cjz + Xjz2 + O(z3). (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:

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Introduction Properties Pros and Cons Examples References

Stability boundary

According to the consistency condition, Pj(z) approximates ecjz for z → 0 as Pj = 1 + cjz + Xjz2 + O(z3). (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:

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Introduction Properties Pros and Cons Examples References

Stability boundary

According to the consistency condition, Pj(z) approximates ecjz for z → 0 as Pj = 1 + cjz + Xjz2 + O(z3). (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

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Introduction Properties Pros and Cons Examples References

Stability boundary

According to the consistency condition, Pj(z) approximates ecjz for z → 0 as Pj = 1 + cjz + Xjz2 + O(z3). (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

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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

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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 s2 ), −β(s) z 0 (28) yields the largest value: β(s) = 2s2. 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/s2, µj = 2, ˜ µj = 2/s2, νj = −1, ˜ γj = 0, 0 j s (30)

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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 j s (31) RKC1: aj = 0, bj = T −1

j

(w0), w0 = 1 + ǫ s2 , w1 = Ts(w0) T ′

s(w0), (0 j s)

(32) Therefore, β(s) ≃ (w0 + 1)T ′

s(w0)

Ts(w0) ≃ (2 − 4ǫ 3 )s2, ǫ → 0 (33) choose ǫ = 0.05, then β(s) = 1.90s2.

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Introduction Properties Pros and Cons Examples References

1st order case: RKC1

Then compare these with the recursive definition of Pj, we get, ˜ µ1 = w1 w0 , (34) µj = 2w0 bj bj−1 , νj = − bj bj−2 , (35) ˜ µj = 2w1 bj bj−1 , ˜ γj = 0, (2 j s) (36) cj = Ts(w0)T ′

j (w0)

T ′

s(w0)Tj(w0) ≃ j2/s2

(37)

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Introduction Properties Pros and Cons Examples References

2nd order case: RKC2

aj = 1 − bjTj(w0), bj = T ′′

j (w0)

(T ′

j (w0))2 ,

(2 j s) (38) w0 = 1 + ǫ 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)

≃ 2 3(s2 − 1)(1 − 2 15ǫ), ǫ → 0 (41) From the pictures,we can tell ǫ = 2

13 is a suitable choice.

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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 bj bj−1 , νj = − bj bj−2 , (43) ˜ µj = 2w1 bj bj−1 , ˜ γj = −(1 − bj−1Tj−1(w0))˜ µj, (2 j s) (44) c1 = c2 T ′

2(w0) ≃ c2

4 , cj = T ′

s(w0)T ′′ j (w0)

T ′′

s (w0)T ′ j (w0) ≃ j2 − 1

s2 − 1 (2 j s) (45)

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Introduction Properties Pros and Cons Examples References

Pros and Cons

Pros: Explicit and designed for modestly stiff problems Cons: Restrictions on the problem

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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

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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

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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

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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

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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.

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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 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 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

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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 = Reδ/αδ, with R = 5 The 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 33 525 285 420 412 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.

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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.

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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.

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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.

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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.

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