Numerical Solutions to Partial Differential Equations Zhiping Li - PowerPoint PPT Presentation

Numerical Solutions to Partial Differential Equations Zhiping Li LMAM and School of Mathematical Sciences Peking University

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme What is a Modified Equation of a Difference Scheme? 1 Let h , τ be the spatial and temporal step sizes. 2 Let U m +1 = B − 1 [ B 0 U m + F m ] be a difference scheme. 1 3 Let { U m j } m ≥ 0 , j ∈ J , be a solution to the scheme. 4 Let P = P h ,τ be a parameterized differential operator. 5 Let X h ,τ = { ˜ U smooth : ˜ U m = U m j , ∀ m ≥ 0 , j ∈ J } . j 6 If P ˜ U = 0, for some ˜ U ∈ X h ,τ , ∀ h , τ , then the differential equation Pu = 0 is called a modified equation of the difference scheme U m +1 = B − 1 [ B 0 U m + F m ]. 1 U = O ( τ q + h q ), for some 7 The q th order modified equation: P ˜ ˜ U ∈ X h ,τ , ∀ h , τ . 2 / 29

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme How to Derive the Modified Equation — an Example Such P is not unique. We want P = D + H x , with Du = 0 the original equation, H x a higher order partial differential operator with respect to x . 1 1D advection equation: u t + au x = 0, a > 0. U m +1 − U m U m j − U m 2 Upwind scheme: j j + a j − 1 = 0. τ h 3 Let ˜ U be smooth and ˜ U m = U m j . j 4 Taylor expand ˜ U at ( x j , t m ) U t + 1 U tt + 1 � m � 2 τ 2 ˜ 6 τ 3 ˜ U m +1 ˜ U + τ ˜ ˜ = U ttt + · · · j , j U x + 1 U xx − 1 � m 2 h 2 ˜ 6 h 3 ˜ � ˜ U − h ˜ ˜ U m j − 1 = U xxx + · · · j , 3 / 29

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme How to Derive the Modified Equation — an Example 5 Hence, ˜ − ˜ ˜ j − ˜ U m +1 U m U m U m j j j − 1 0 = + a τ h j +1 j +1 � m � m � m � � � τ 2 ˜ U ttt + ah 2 ˜ U t + a ˜ ˜ τ ˜ U tt − ah ˜ j + O ( τ 3 + h 3 ) . = U x U xx U xxx 2 6 U t + a ˜ ˜ U x = 0 , the first order modified equation. (original one) 6 � � U t + a ˜ ˜ ah ˜ U xx − τ ˜ U x = 1 : the second order. 7 U tt 2 � � � ah 2 ˜ U xxx + τ 2 ˜ � U t + a ˜ ˜ ah ˜ U xx − τ ˜ U x = 1 − 1 , the 3rd. 8 U tt U ttt 2 6 9 But the latter two are not in the preferred form. 4 / 29

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme How to Derive the Modified Equation — an Example (continue) 10 By � � � � � τ 2 ˜ U ttt + ah 2 ˜ � = O ( τ 3 + h 3 ) U t + a ˜ ˜ τ ˜ U tt − ah ˜ + 1 + 1 U x U xx U xxx 2 6 U xx + 1 � � + O ( τ 2 + h 2 ) , U xt = − a ˜ ˜ ah ˜ U xxx − τ ˜ ⇒ U xtt 2 U xt + 1 � � + O ( τ 2 + h 2 ) U tt = − a ˜ ˜ ah ˜ U xxt − τ ˜ ⇒ U ttt 2 U xx − 1 = a 2 ˜ � � a 2 h ˜ U xxx − ah ˜ U xxt − a τ ˜ U xtt + τ ˜ + O ( τ 2 + h 2 ) . U ttt 2 U x = 1 U xx + O ( τ 2 + h 2 ) . U t + a ˜ ˜ 2 ah (1 − ν )˜ ⇒ 5 / 29

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme How to Derive the Modified Equation — an Example (continue) 11 Hence, ˜ U t + a ˜ 2 ah (1 − ν )˜ U x = 1 U xx is the 2nd order modified equation. Similarly, we have the 3rd order modified equation: U x = 1 U xx − 1 U t + a ˜ ˜ 2 ah (1 − ν )˜ 6 ah 2 (1 − ν )(1 − 2 ν )˜ U xxx . 6 / 29

Derive the Modified Equation by Difference Operator Calculus 1 Express a difference operator by a series of differential operators (Taylor expansion). For example, △ + t = e τ∂ t − 1. 2 Formally inverting the expression, a differential operator can then be expressed by a power series of a difference operator. 3 For example, ∂ t = τ − 1 ln (1 + τ D + t ), where D + t := τ − 1 △ + t . + t + τ 2 + t − τ 3 This yields ∂ t = D + t − τ 2 D 2 3 D 3 4 D 4 + t + · · · . = ( � ∞ 4 For a difference scheme D + t U m = A x U m k =0 α k ∂ k x ) U m j , j j substitute D + t by � ∞ k =0 α k ∂ k x in the series expression of ∂ t , and collect the terms with the same powers of ∂ x , we are led ∞ to the modified equation � � � ˜ β k ∂ k ∂ t − U = 0 . x k =0

More on Consistency, Stability and Convergence Modified Equation Analysis Modified Equation of a Difference Scheme Derive Modified Equation by Difference Operator Calculus — an Example 1 Advection-diffusion equation: u t + au x = cu xx , x ∈ R , t > 0. U m +1 − U m U m j +1 − U m U m j +1 − 2 U m j + U m 2 Explicit scheme: j j − 1 j − 1 j + a = c . τ 2 h h 2 3 By Taylor series expansions of △ 0 x U m and δ 2 x U m j , we have j � � ∂ x + 1 � � x + 1 �� D + t ˜ ˜ 6 h 2 ∂ 3 ∂ 2 12 h 2 ∂ 4 U = − a x + · · · + c x + · · · U . 4 The modified equation obtained: � ˜ ah 2 − 6 ac τ + 2 a 3 τ 2 � ˜ U x = 1 U xx − 1 U t + a ˜ ˜ 2 c − a 2 τ � � U xxx 2 6 ch 2 − 2 a 2 τ h 2 − 6 c 2 τ + 12 a 2 c τ 2 − 3 a 4 τ 3 � ˜ + 1 � U xxxx + · · · . 12 8 / 29

More on Consistency, Stability and Convergence Modified Equation Analysis Dissipation and Dispersion of Modified Equations What is the use of a Modified Equation 1 Difference solutions approximate higher order modified equation with higher order of accuracy. 2 Well-posedness of the modified equations provides useful information on the stability of the scheme. 3 Amplitude and phase errors of the modified equations on the Fourier mode solutions provide the corresponding information for the scheme. 4 Convergence rate of the solution of the modified equation to the solution of the original equation also provides the corresponding information for the scheme. 5 In particular, the dissipation and dispersion of the solutions of the modified equations can be very useful. 9 / 29

Dissipation and Dispersion Terms of the Modified Equation 1 Fourier mode e i ( kx + ω t ) ⇒ modified eqn. ˜ U t = � ∞ x ˜ m =0 a m ∂ m U . x e i ( kx + ω t ) = ( i k ) m e i ( kx + ω t ) ⇒ dispersion relation: 2 Notice ∂ m ∞ ∞ ( − 1) m − 1 a 2 m − 1 k 2 m − 1 − i � � ( − 1) m a 2 m k 2 m . ω ( k ) = m =1 m =0 3 Denote ω ( k ) = ω 0 ( k ) + i ω 1 ( k ), where ∞ ∞ � ( − 1) m − 1 a 2 m − 1 k 2 m − 1 , ω 1 ( k ) := − � ( − 1) m a 2 m k 2 m . ω 0 ( k ) := m =1 m =0 4 The Fourier mode solution e i ( kx + ω ( k ) t ) = e − ω 1 ( k ) t e i ( kx + ω 0 ( k ) t ) . 5 Even order spatial derivative terms change the amplitude. 6 Odd order spatial derivative terms change the phase speed. 7 Even and odd order terms are called dissipation and dispersion terms of the modified equations respectively.

More on Consistency, Stability and Convergence Modified Equation Analysis Dissipation and Dispersion of Modified Equations Dissipation and Dispersion of Modified Equation — an Example 1 Consider third order modified equation of the upwind scheme for the advection equation with a > 0 as an example: U x = 1 U xx − 1 U t + a ˜ ˜ 2 ah (1 − ν )˜ 6 ah 2 (1 − ν )(1 − 2 ν )˜ U xxx . 2 We have here a 0 = 0, a 1 = − a , a 2 = 1 2 ah (1 − ν ), a 3 = − 1 6 ah 2 (1 − ν )(1 − 2 ν ), a m = 0, m ≥ 4. 3 Thus, we have ω 0 ( k ) = − ak + 1 − ω 1 ( k ) = − 1 6 a (1 − ν )(1 − 2 ν ) k 3 h 2 , 2 a (1 − ν ) k 2 h . 4 If CFL condition is not satisfied ⇒ − ω 1 ( k ) > 0 ⇒ unstable. 5 For kh ≪ 1 ⇒ relative phase error O ( k 2 h 2 ). 11 / 29

Dissipation and Dispersion of Modified Equation — another Example 1 Consider the Lax-Wendroff scheme of the advection equation: U m +1 − U m U m j +1 − U m U m j +1 − 2 U m j + U m = 1 j j j − 1 j − 1 2 a 2 τ + a . h 2 τ 2 h 2 The modified equation (compare with (4.5.16) and (4) on p.8 of this slides) : U x = − 1 U xxx − 1 U t + a ˜ ˜ 6 ah 2 (1 − ν 2 )˜ 8 ah 3 ν (1 − ν 2 )˜ U xxxx + · · · . 3 For kh ≪ 1, dispersion and dissipation components of ω ( k ): � 1 − 1 � ω 0 ( k ) ≈ a 1 k − a 3 k 3 = − ak 6(1 − ν 2 ) k 2 h 2 , − ω 1 ( k ) ≈ a 0 − a 2 k 2 + a 4 k 4 = − 1 8 a ν (1 − ν 2 ) k 4 h 3 . 4 ν 2 > 1 ⇒ − ω 1 ( k ) > 0 ⇒ unstable. 5 For kh ≪ 1 ⇒ phase lag, relative phase error O ( k 2 h 2 ).

More on Consistency, Stability and Convergence Modified Equation Analysis Dissipation and Dispersion of Modified Equations Necessary Stability Conditions Given by the Modified Equation m =0 ( − 1) m a 2 m k 2 m > 0 ⇒ the scheme is unstable. 1 − ω 1 = � ∞ 2 In the case of a 0 = 0, a finite difference scheme is generally unstable if a 2 < 0, or a 2 = 0 but a 4 > 0. 3 The case when a 0 = 0, a 2 > 0, a 4 > 0 is more complicated. For kh ≪ 1, Fourier mode solutions are stable, for kh big, say kh = π , they can be unstable, in particular, high frequency modes are unstable when a 2 m = 0, ∀ m > 2. Remark: In fact, for high frequency modes, − ω 1 = � ∞ m =0 ( − 1) m a 2 m k 2 m does not necessarily make sense, since it may not converge in general. 13 / 29