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 Properties 2 Consistency conditions Stability Properties Integration formula Pros and Cons 3 Examples 4 References 5
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 ) = u 0 ˙ (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:
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 ) = u 0 ˙ (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:
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 ) = u 0 ˙ (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)
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 ) = u 0 ˙ (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-di ff usion problem u ( t ) = ǫ∆u + f ( u , x , t ) , 0 < ǫ ≪ 1 ˙ (3)
Introduction Properties Pros and Cons Examples References Introduction Sti ff problems: standard explicit RK methods: easy application, restrictive time-step for stability
Introduction Properties Pros and Cons Examples References Introduction Sti ff 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 Sti ff 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 Y 0 = U n , (4) Y 1 = Y 0 + ˜ µ 1 τF 0 , (5) Y j = µ j Y j − 1 + ν j Y j − 2 + ( 1 − µ j − ν j ) Y 0 (6) + ˜ µ j τF j − 1 + ˜ γ j τF 0 ( 2 � j � s ) , (7) U n + 1 = Y s , n = 0, 1, . . . , (8) This can be rewritten in the standard RK form: j − 1 � Y j = U n + τ a jl F ( t n + c l τ , Y l ) , ( 0 � j � s ) (9) l = 0 where F j = F ( t n + c j τ , Y j ) , c j are defined by: c 0 = 0, (10) c 1 = ˜ µ 1 , (11) c j = µ j c j − 1 + ν j c j − 2 + ˜ µ j + ˜ γ j , ( 2 � j � s ) (12)
Introduction Properties Pros and Cons Examples References Outline Introduction 1 Properties 2 Consistency conditions Stability Properties Integration formula Pros and Cons 3 Examples 4 References 5
Introduction Properties Pros and Cons Examples References Consistency conditions Suppose U n = U ( t n ) , where U ( t ) , t � t n is a su ffi ciently smooth solution. All Y j satisfy an expansion Y j = U ( t n ) + c j τ ˙ U ( t n ) + X j τ 2 U ( 2 ) ( t n ) + O ( τ 3 ) (13) Substitute this into the RKC formula, we have X 0 = X 1 = 0, (14) X j = µ j X j − 1 + ν j X j − 2 + ˜ µ j c j − 1 ( 2 � j � s ) (15)
Introduction Properties Pros and Cons Examples References Consistency conditions consistent of order 1: if c s = 1. (16)
Introduction Properties Pros and Cons Examples References Consistency conditions consistent of order 1: if c s = 1. (16) consistent of order 2: if c 2 2 = 2˜ µ 2 c 1 , (17) c 2 3 = µ 3 c 2 2 + 2˜ µ 2 c 2 , (18) c 2 j = µ j c 2 j − 1 + ν j c 2 j − 2 + 2˜ µ j c j − 1 ( 4 � j � s ) (19)
Introduction Properties Pros and Cons Examples References Outline Introduction 1 Properties 2 Consistency conditions Stability Properties Integration formula Pros and Cons 3 Examples 4 References 5
Introduction Properties Pros and Cons Examples References Stability function Scalar test equation: ˙ U ( t ) = λU ( t ) (20) U n + 1 = P s ( z ) U n , z = τλ (21)
Introduction Properties Pros and Cons Examples References Stability function Scalar test equation: ˙ U ( t ) = λU ( t ) (20) U n + 1 = P s ( z ) U n , z = τλ (21) P s is defined recursively: P 0 ( z ) = 1, (22) P 1 ( z ) = 1 + ˜ µ 1 z , (23) P j ( z ) = ( 1 − µ j − ν j ) + ˜ γ j z + ( µ j + ˜ µ j z ) P j − 1 ( z ) + ν j P j − 2 ( z ) ( 2 � (24)
Introduction Properties Pros and Cons Examples References Stability function Scalar test equation: ˙ U ( t ) = λU ( t ) (20) U n + 1 = P s ( z ) U n , z = τλ (21) P s is defined recursively: P 0 ( z ) = 1, (22) P 1 ( z ) = 1 + ˜ µ 1 z , (23) P j ( z ) = ( 1 − µ j − ν j ) + ˜ γ j z + ( µ j + ˜ µ j z ) P j − 1 ( z ) + ν j P j − 2 ( z ) ( 2 � (24) for each stage, we have U j = P j ( z ) U n , ( 0 � j � s ) (25)
Introduction Properties Pros and Cons Examples References Stability boundary According to the consistency condition, P j ( z ) approximates e c j z for z → 0 as P j = 1 + c j z + X j z 2 + O ( z 3 ) . (26) The choice of the stability function P j ( z ) is the cental issue in developing the RKC methods. Stability Region S = { z ∈ C : | P s | � 1 } Design rules:
Introduction Properties Pros and Cons Examples References Stability boundary According to the consistency condition, P j ( z ) approximates e c j z for z → 0 as P j = 1 + c j z + X j z 2 + O ( z 3 ) . (26) The choice of the stability function P j ( z ) is the cental issue in developing the RKC methods. Stability Region S = { z ∈ C : | P s | � 1 } Stability Boundary β ( s ) = max { − z : z � 0, | P s | � 1 } Design rules:
Introduction Properties Pros and Cons Examples References Stability boundary According to the consistency condition, P j ( z ) approximates e c j z for z → 0 as P j = 1 + c j z + X j z 2 + O ( z 3 ) . (26) The choice of the stability function P j ( z ) is the cental issue in developing the RKC methods. Stability Region S = { z ∈ C : | P s | � 1 } Stability Boundary β ( s ) = max { − z : z � 0, | P s | � 1 } Design rules: β ( s ) is as large as possible
Introduction Properties Pros and Cons Examples References Stability boundary According to the consistency condition, P j ( z ) approximates e c j z for z → 0 as P j = 1 + c j z + X j z 2 + O ( z 3 ) . (26) The choice of the stability function P j ( z ) is the cental issue in developing the RKC methods. Stability Region S = { z ∈ C : | P s | � 1 } Stability Boundary β ( s ) = max { − z : z � 0, | P s | � 1 } Design rules: β ( s ) is as large as possible all coe ffi cients must be known in analytic form
Introduction Properties Pros and Cons Examples References Outline Introduction 1 Properties 2 Consistency conditions Stability Properties Integration formula Pros and Cons 3 Examples 4 References 5
Introduction Properties Pros and Cons Examples References shifted Chebyshev polynomials Chebyshev polynomial of the first kind T s ( x ) = cos ( sarccosx ) , − 1 � x � 1 (27) Eg: for the 1st order consistent polys, the shifted Chebyshev poly P s ( z ) = T s ( 1 + z s 2 ) , − β ( s ) � z � 0 (28) yields the largest value: β ( s ) = 2 s 2 . From the three-terms recursion formula for Chebyshev polynomials, we get: P 0 ( z ) = 1, P 1 ( z ) = 1 + z s 2 , P j ( z ) = 2 ( 1 + z s 2 ) P j − 1 ( z )− P j − 2 ( z ) , j � 2, (29) which gives the analytical form of the integration coe ff s µ 1 = 1 /s 2 , µ j = 2, ˜ µ j = 2 /s 2 , ν j = − 1, ˜ ˜ γ j = 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 P j ( z ) = a j + b j T j ( w 0 + w 1 z ) , 0 � j � s (31) RKC1: ( w 0 ) , w 0 = 1 + ǫ s 2 , w 1 = T s ( w 0 ) a j = 0, b j = T − 1 s ( w 0 ) , ( 0 � j � s ) j T ′ (32) Therefore, β ( s ) ≃ ( w 0 + 1 ) T ′ s ( w 0 ) ≃ ( 2 − 4 ǫ 3 ) s 2 , ǫ → 0 (33) T s ( w 0 ) choose ǫ = 0.05, then β ( s ) = 1.90 s 2 .
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