AMath 574 February 11, 2011

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1 AMath 574 February 11, 2011 Today: Entropy conditions and functions Lax-Wendroff theorem Wednesday February 23: Nonlinear systems Reading: Chapter 13 R.J. LeVeque, University of Washington AMath 574, February 14, 2011

2 Entropy-violating numerical solutions Riemann problem for Burgers equation at t = 1 with u l = 1 and u r = 2: 2.5 Godunov with no entropy fix 2.5 Godunov with entropy fix High resolution with no entropy fix 2.5 High resolution with entropy fix R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.3]

3 Vanishing viscosity solution We want q(x, t) to be the limit as ɛ 0 of solution to q t + f(q) x = ɛq xx. This selects a unique weak solution: Shock if f (q l ) > f (q r ), Rarefaction if f (q l ) < f (q r ). R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

4 Vanishing viscosity solution We want q(x, t) to be the limit as ɛ 0 of solution to q t + f(q) x = ɛq xx. This selects a unique weak solution: Shock if f (q l ) > f (q r ), Rarefaction if f (q l ) < f (q r ). Lax Entropy Condition: A discontinuity propagating with speed s in the solution of a convex scalar conservation law is admissible only if f (q l ) > s > f (q r ), where s = (f(q r ) f(q l ))/(q r q l ). R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

5 Vanishing viscosity solution We want q(x, t) to be the limit as ɛ 0 of solution to q t + f(q) x = ɛq xx. This selects a unique weak solution: Shock if f (q l ) > f (q r ), Rarefaction if f (q l ) < f (q r ). Lax Entropy Condition: A discontinuity propagating with speed s in the solution of a convex scalar conservation law is admissible only if f (q l ) > s > f (q r ), where s = (f(q r ) f(q l ))/(q r q l ). Note: This means characteristics must approach shock from both sides as t advances, not move away from shock! R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

6 Approximate Riemann solvers For nonlinear problems, computing the exact solution to each Riemann problem may not be possible, or too expensive. Often the nonlinear problem q t + f(q) x = 0 is approximated by q t + A i 1/2 q x = 0, q l = Q i 1, q r = Q i for some choice of A i 1/2 f (q) based on data Q i 1, Q i. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

7 Approximate Riemann solvers For nonlinear problems, computing the exact solution to each Riemann problem may not be possible, or too expensive. Often the nonlinear problem q t + f(q) x = 0 is approximated by q t + A i 1/2 q x = 0, q l = Q i 1, q r = Q i for some choice of A i 1/2 f (q) based on data Q i 1, Q i. Solve linear system for α i 1/2 : Q i Q i 1 = p αp i 1/2 rp i 1/2. Waves W p i 1/2 = αp i 1/2 rp i 1/2 propagate with speeds sp i 1/2, r p i 1/2 are eigenvectors of A i 1/2, s p i 1/2 are eigenvalues of A i 1/2. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

8 Approximate Riemann solvers q t + Âi 1/2q x = 0, q l = Q i 1, q r = Q i Often Âi 1/2 = f (Q i 1/2 ) for some choice of Q i 1/2. In general Â i 1/2 = Â(q l, q r ). R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

9 Approximate Riemann solvers q t + Âi 1/2q x = 0, q l = Q i 1, q r = Q i Often Âi 1/2 = f (Q i 1/2 ) for some choice of Q i 1/2. In general Â i 1/2 = Â(q l, q r ). Roe conditions for consistency and conservation: Â(q l, q r ) f (q ) as q l, q r q, Â diagonalizable with real eigenvalues, For conservation in wave-propagation form, Â i 1/2 (Q i Q i 1 ) = f(q i ) f(q i 1 ). R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

10 Approximate Riemann solvers For a scalar problem, we can easily satisfy the Roe condition Â i 1/2 (Q i Q i 1 ) = f(q i ) f(q i 1 ). by choosing Â i 1/2 = f(q i) f(q i 1 ) Q i Q i 1. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

11 Approximate Riemann solvers For a scalar problem, we can easily satisfy the Roe condition Â i 1/2 (Q i Q i 1 ) = f(q i ) f(q i 1 ). by choosing Â i 1/2 = f(q i) f(q i 1 ) Q i Q i 1. Then r 1 i 1/2 = 1 and s1 i 1/2 = Âi 1/2 (scalar!). Note: This is the Rankine-Hugoniot shock speed. = shock waves are correct, rarefactions replaced by entropy-violating shocks. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

12 Weak solutions to Burgers equation u t + ( 1 2 u2) x = 0, u l = 1, u r = 2 Characteristic speed: u Rankine-Hugoniot speed: 1 2 (u l + u r ). Physically correct rarefaction wave solution: R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

13 Weak solutions to Burgers equation u t + ( 1 2 u2) x = 0, u l = 1, u r = 2 Characteristic speed: u Rankine-Hugoniot speed: 1 2 (u l + u r ). Entropy violating weak solution: R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

14 Weak solutions to Burgers equation u t + ( 1 2 u2) x = 0, u l = 1, u r = 2 Characteristic speed: u Rankine-Hugoniot speed: 1 2 (u l + u r ). Another Entropy violating weak solution: R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

15 Transonic rarefactions Sonic point: u s = 0 for Burgers since f (0) = 0. Consider Riemann problem data u l = 0.5 < 0 < u r = 1.5. In this case wave should spread in both directions: R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

16 Transonic rarefactions Entropy-violating approximate Riemann solution: s = 1 2 (u l + u r ) = 0.5. Wave goes only to right, no update to cell average on left. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

17 Transonic rarefactions If u l = u r then Rankine-Hugoniot speed is 0: Similar solution will be observed with Godunov s method if entropy-violating approximate Riemann solver used. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

18 Entropy-violating numerical solutions Riemann problem for Burgers equation at t = 1 with u l = 1 and u r = 2: 2.5 Godunov with no entropy fix 2.5 Godunov with entropy fix High resolution with no entropy fix 2.5 High resolution with entropy fix R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.3]

19 Approximate Riemann solver Q n+1 i = Q n i t [ A + Q x i 1/2 + A ] Q i+1/2. For scalar advection m = 1, only one wave. W i 1/2 = Q i 1/2 = Q i Q i 1 and s i 1/2 = u, A Q i 1/2 = s i 1/2 W i 1/2, A + Q i 1/2 = s + i 1/2 W i 1/2. For scalar nonlinear: Use same formulas with W i 1/2 = Q i 1/2 and s i 1/2 = F i 1/2 / Q i 1/2. Need to modify these by an entropy fix in the trans-sonic rarefaction case. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.3]

20 Entropy fix Q n+1 i Revert to the formulas = Q n i t [ A + Q x i 1/2 + A ] Q i+1/2. A Q i 1/2 = f(q s ) f(q i 1 ) A + Q i 1/2 = f(q i ) f(q s ) if f (Q i 1 ) < 0 < f (Q i ). left-going fluctuation right-going fluctuation High-resolution method: still define wave W and speed s by W i 1/2 = Q i Q i 1, { (f(qi ) f(q s i 1/2 = i 1 ))/(Q i Q i 1 ) if Q i 1 Q i f (Q i ) if Q i 1 = Q i. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.3]

21 Godunov flux for scalar problem The Godunov flux function for the case f (q) > 0 is F n i 1/2 = = f(q i 1 ) if Q i 1 > q s and s > 0 f(q i ) if Q i < q s and s < 0 f(q s ) if Q i 1 < q s < Q i. min f(q) Q i 1 q Q i if Q i 1 Q i max f(q) if Q i Q i 1, Q i q Q i 1 Here s = f(q i) f(q i 1 ) Q i Q i 1 is the Rankine-Hugoniot shock speed. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.1]

22 Entropy-violating numerical solutions Riemann problem for Burgers equation with q l = 1 and q r = 2: 2.5 Godunov with no entropy fix 2.5 Godunov with entropy fix High resolution with no entropy fix 2.5 High resolution with entropy fix R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 12.3]

23 Entropy (admissibility) conditions We generally require additional conditions on a weak solution to a conservation law, to pick out the unique solution that is physically relevant. In gas dynamics: entropy is constant along particle paths for smooth solutions, entropy can only increase as a particle goes through a shock. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

24 Entropy (admissibility) conditions We generally require additional conditions on a weak solution to a conservation law, to pick out the unique solution that is physically relevant. In gas dynamics: entropy is constant along particle paths for smooth solutions, entropy can only increase as a particle goes through a shock. Entropy functions: Function of q that behaves like physical entropy for the conservation law being studied. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

25 Entropy (admissibility) conditions We generally require additional conditions on a weak solution to a conservation law, to pick out the unique solution that is physically relevant. In gas dynamics: entropy is constant along particle paths for smooth solutions, entropy can only increase as a particle goes through a shock. Entropy functions: Function of q that behaves like physical entropy for the conservation law being studied. NOTE: Mathematical entropy functions generally chosen to decrease for admissible solutions, increase for entropy-violating solutions. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

26 Entropy functions A scalar-valued function η : lr m lr if the Hessian matrix η (q) with (i, j) element η ij(q) = 2 η q i q j is positive definite for all q, i.e., satisfies is a convex function of q v T η (q)v > 0 for all q, v lr m. Scalar case: reduces to η (q) > 0. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

27 Entropy functions Entropy function: η : lr m lr Entropy flux: ψ : lr m lr chosen so that η(q) is convex and: η(q) is conserved wherever the solution is smooth, η(q) t + ψ(q) x = 0. Entropy decreases across an admissible shock wave. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

28 Entropy functions Entropy function: η : lr m lr Entropy flux: ψ : lr m lr chosen so that η(q) is convex and: η(q) is conserved wherever the solution is smooth, η(q) t + ψ(q) x = 0. Entropy decreases across an admissible shock wave. Weak form: x2 x 1 η(q(x, t 2 )) dx + x2 x 1 t2 t 1 η(q(x, t 1 )) dx with equality where solution is smooth. t2 ψ(q(x 1, t)) dt t 1 ψ(q(x 2, t)) dt R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

29 Entropy functions How to find η and ψ satisfying this? η(q) t + ψ(q) x = 0 For smooth solutions gives η (q)q t + ψ (q)q x = 0. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

30 Entropy functions How to find η and ψ satisfying this? η(q) t + ψ(q) x = 0 For smooth solutions gives η (q)q t + ψ (q)q x = 0. Since q t = f (q)q x this is satisfied provided ψ (q) = η (q)f (q) R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

31 Entropy functions How to find η and ψ satisfying this? η(q) t + ψ(q) x = 0 For smooth solutions gives η (q)q t + ψ (q)q x = 0. Since q t = f (q)q x this is satisfied provided ψ (q) = η (q)f (q) Scalar: Can choose any convex η(q) and integrate. Example: Burgers equation, f (u) = u and take η(u) = u 2. Then ψ (u) = 2u 2 = Entropy function: ψ(u) = 2 3 u3. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

32 Weak solutions and entropy functions The conservation laws ( ) 1 u t + 2 u2 = 0 and both have the same quasilinear form x u t + uu x = 0 ( u 2 ) t + ( 2 3 u3 ) x = 0 but have different weak solutions, different shock speeds! R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

33 Weak solutions and entropy functions The conservation laws ( ) 1 u t + 2 u2 = 0 and both have the same quasilinear form x u t + uu x = 0 ( u 2 ) t + ( 2 3 u3 ) x = 0 but have different weak solutions, different shock speeds! Entropy function: η(u) = u 2. A correct Burgers shock at speed s = 1 2 (u l + u r ) will have total mass of η(u) decreasing. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

34 Entropy functions x2 x 1 η(q(x, t 2 )) dx + x2 x 1 t2 t 1 η(q(x, t 1 )) dx t2 ψ(q(x 1, t)) dt ψ(q(x 2, t)) dt t 1 comes from considering the vanishing viscosity solution: q ɛ t + f(q ɛ ) x = ɛq ɛ xx R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

35 Entropy functions x2 x 1 η(q(x, t 2 )) dx + x2 x 1 t2 t 1 η(q(x, t 1 )) dx t2 ψ(q(x 1, t)) dt ψ(q(x 2, t)) dt t 1 comes from considering the vanishing viscosity solution: Multiply by η (q ɛ ) to obtain: q ɛ t + f(q ɛ ) x = ɛq ɛ xx η(q ɛ ) t + ψ(q ɛ ) x = ɛη (q ɛ )q ɛ xx. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

36 Entropy functions x2 x 1 η(q(x, t 2 )) dx + x2 x 1 t2 t 1 η(q(x, t 1 )) dx t2 ψ(q(x 1, t)) dt ψ(q(x 2, t)) dt t 1 comes from considering the vanishing viscosity solution: qt ɛ + f(q ɛ ) x = ɛqxx ɛ Multiply by η (q ɛ ) to obtain: η(q ɛ ) t + ψ(q ɛ ) x = ɛη (q ɛ )qxx. ɛ Manipulate further to get η(q ɛ ) t + ψ(q ɛ ) x = ɛ ( η (q ɛ )qx ɛ ) x ɛη (q ɛ ) (qx) ɛ 2. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

37 Entropy functions Smooth solution to viscous equation satisfies η(q ɛ ) t + ψ(q ɛ ) x = ɛ ( η (q ɛ )qx ɛ ) x ɛη (q ɛ ) (qx) ɛ 2. Integrating over rectangle [x 1, x 2 ] [t 1, t 2 ] gives x2 x 1 η(q ɛ (x, t 2 )) dx = + ɛ ɛ ( t2 t 1 t2 t 1 t2 x2 t 1 x2 x 1 ψ(q ɛ (x 2, t)) dt η(q ɛ (x, t 1 )) dx t2 t 1 ) ψ(q ɛ (x 1, t)) dt [ η (q ɛ (x 2, t)) q ɛ x(x 2, t) η (q ɛ (x 1, t)) q ɛ x(x 1, t) ] dt x 1 η (q ɛ ) (q ɛ x) 2 dx dt. Let ɛ 0 to get result: Term on third line goes to 0, Term of fourth line is always 0. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

38 Entropy functions Weak form of entropy condition: 0 [ φt η(q) + φ x ψ(q) ] dx dt + φ(x, 0)η(q(x, 0)) dx 0 for all φ C0 1 (lr lr) with φ(x, t) 0 for all x, t. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

39 Entropy functions Weak form of entropy condition: 0 [ φt η(q) + φ x ψ(q) ] dx dt + φ(x, 0)η(q(x, 0)) dx 0 for all φ C0 1 (lr lr) with φ(x, t) 0 for all x, t. Informally we may write η(q) t + ψ(q) x 0. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec. 11.4]

40 Lax-Wendroff Theorem Suppose the method is conservative and consistent with q t + f(q) x = 0, F i 1/2 = F(Q i 1, Q i ) with F( q, q) = f( q) and Lipschitz continuity of F. If a sequence of discrete approximations converge to a function q(x, t) as the grid is refined, then this function is a weak solution of the conservation law. Note: Does not guarantee a sequence converges (need stability). Two sequences might converge to different weak solutions. Also need to satisfy an entropy condition. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

41 Sketch of proof of Lax-Wendroff Theorem Multiply the conservative numerical method by Φ n i to obtain Q n+1 i Φ n i Q n+1 i = Q n i t x (F n i+1/2 F n i 1/2 ) = Φ n i Q n i t x Φn i (F n i+1/2 F n i 1/2 ). R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

42 Sketch of proof of Lax-Wendroff Theorem Multiply the conservative numerical method by Φ n i to obtain Q n+1 i Φ n i Q n+1 i = Q n i t x (F n i+1/2 F n i 1/2 ) = Φ n i Q n i t x Φn i (F n i+1/2 F n i 1/2 ). This is true for all values of i and n on each grid. Now sum over all i and n 0 to obtain n=0 i= Φ n i (Q n+1 i Q n i ) = t x n=0 i= Φ n i (F n i+1/2 F n i 1/2 ). Use summation by parts to transfer differences to Φ terms. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

43 Sketch of proof of Lax-Wendroff Theorem Obtain analog of weak form of conservation law: x t [ n=1 i= + ( Φ n i Φ n 1 i t n=0 i= ) Q n i ( Φ n i+1 Φ n i x ) ] Fi 1/2 n = x i= Φ 0 i Q 0 i. Consider on a sequence of grids with x, t 0. Show that any limiting function must satisfy weak form of conservation law. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

44 Analog of Lax-Wendroff proof for entropy Show that the numerical flux function F leads to a numerical entropy flux Ψ such that the following discrete entropy inequality holds: η(q n+1 i ) η(q n i ) t [ ] Ψ n i+1/2 x Ψn i 1/2. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

45 Analog of Lax-Wendroff proof for entropy Show that the numerical flux function F leads to a numerical entropy flux Ψ such that the following discrete entropy inequality holds: η(q n+1 i ) η(q n i ) t [ ] Ψ n i+1/2 x Ψn i 1/2. Then multiply by test function Φ n i, sum and use summation by parts to get discrete form of integral form of entropy condition. = If numerical approximations converge to some function, then the limiting function satisfies the entropy condition. R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

46 Entropy consistency of Godunov s method For Godunov s method, F (Q i 1, Q i ) = f(q i 1/2 ) where Q i 1/2 is the constant value along x i 1/2 in the Riemann solution. Let Ψ n i 1/2 = ψ(q i 1/2 ) Discrete entropy inequality follows from Jensen s inequality: The value of η evaluated at the average value of q n is less than or equal to the average value of η( q n ), i.e., ( ) 1 xi+1/2 η q n (x, t n+1 ) dx 1 xi+1/2 η ( q n (x, t n+1 ) ) dx. x x i 1/2 x x i 1/2 R.J. LeVeque, University of Washington AMath 574, February 14, 2011 [FVMHP Sec ]

Notes: Outline. Shock formation. Notes: Notes: Shocks in traffic flow

Notes: Outline. Shock formation. Notes: Notes: Shocks in traffic flow Outline Scalar nonlinear conservation laws Traffic flow Shocks and rarefaction waves Burgers equation Rankine-Hugoniot conditions Importance of conservation form Weak solutions Reading: Chapter, 2 R.J.