Info. No lecture on Thursday in a week (March 17) PSet back tonight

Transcription

1 Lecture

2 Info No lecture on Thursday in a week (March 7) PSet back tonight

3 Nonlinear transport & conservation laws What if transport becomes nonlinear?

4 Remember: Nonlinear transport A first attempt at understanding what is going on comes from analyzing the characteristic lines Simplest example: lnviscid Burgers' equation We can use the implicit ansatz du dx u(x, t) =u(x ut, 0) Characteristics for initial conditions all meet in (x,t)=(,)! u(x, 0) = x Solution is not defined at (x,t)=(,) => so what do we do?

5 Nonlinear transport & conservation laws A more famous example of issues with characteristics is the Riemann problem: Take again Burger s equation: Riemann problem: How does solution evolve in time for initial conditions du dx u(x, 0) = A, x < 0, and u(x, 0) = B, x 0 t t A > B: shock A < B: fan

6 How to find shock wave or fan solution? Physically, we could add a bit of diffusion to avoid discontinuities, then let D -> 0 Not very feasible numerically see Lect. Note: There is an example in the book (analytical) for u t + uu x = u x x

7 Conservation laws Burger s equation creates discontinuities Mathematical description using integral form (=conservation law) In general, conservation laws read du dx u t + f(u, u x,...) x =0() d dt Z b a udx + f(u(b),u x (b),...) f(u(a),u x (a),...)=0 Special form above: Diffusion is also a conservation law

8 Weak solutions Integral form is not very handy, because we have to show it holds for all times and all intervals Instead: Use the weak form: Multiply PDE with a test function ɸ and integrate over time and space. Assuming ɸ has compact support (almost everywhere zero), integration by parts gives (for Burger s equation): u t + f x (u) =0, Z 0 Z [ tu + x f(u)]dxdt = Z Lecture (x, 0)u(x, 0)dx For Burgers eq: One can show (PSet 2) that the shock in the Riemann problem with initial condition u(x, 0) = ( u L x<0 u R x>0 with ul>ur ul t ur here: ul=0, ur =- is solved by: u(x, t) = ( u L u R x<st x>st with shock speed s = u L + u R 2 ul ur

9 RH jump condition The previous example was specific for Burger s equation. But there is a general condition that determines the shock speed. Rankine-Hugoniot (RH) jump condition: The shock speed s of a conservation law with flux f is s = f(u L) f(u R ) = [f] u L u R [u] Lecture u L, u R : value of u on the left & right of the shock Also valid for initial conditions that are not piecewise linear. Then u L, u R : value of u infinitesimally close to shock

10 Rarefaction waves How does the Riemann problem look like when there are no shocks? ul t ur Correct (weak) solution (for Burgers eq): Rarefaction wave: u(x, t) = 8 >< x t >: u L u R x<u L t u L t apple x apple u R t x>u R t ul ur A < B: fan ul ur Think of cars starting to drive at green light Unfortunately, a shock would also be a weak solution in this case!!! We need another criterium to decide for a shock vs. a fan

11 Entropy condition Note that characteristics must collide for a shock to occur, while they separate for fans. This can be formulated in the entropy condition: If f 0 (u L ) >s>f 0 (u R ) => shocks otherwise: fans with u(x,t) = x/t

12 A first numerical try Simple idea: Discretize flux, not PDE! du d du du Start from f (u) = 0 and not from - + f '(u) - = 0 at dx at dx E.g. upwind finite differences (assuming f > 0) Upwind (for f '>o) Uj,n+l - Uj,n + f (Uj+l,n)- f (Uj,n) = 0 At Ax The advantage of this approach (apart from allowing discont. u) is that the conservation law is automatically fulfilled: B Discrete conservation uj,n+l - uj, - FA,^) = 0 j=a Lecture

13 Numerical flux function In fact, we have just seen that any scheme of the form U j,n+ t U j,n = x [F (U j p,n,u j p+,n,...,u j+q,n ) F (U j p+,n,u j p+2,n,...,u j+q+,n )] is a conservative scheme (meaning total U inside a domain is determined only by flux F at boundaries) Of course we cannot freely combine values of U into a flux: Consistency The flux F (Uj+,,..., Uj-,) satisfies F (u,..., u) = f (u) Question: How to choose flux F in the best way

14 A first numerical try Upwind finite differences (assuming f > 0) Upwind (for f '>o) Uj,n+l - Uj,n + f (Uj+l,n)- f (Uj,n) = 0 At Ax Using F j,n = f(u j,n ), this can be written as U j,n+ t U j,n = x [F j+,n F j,n ] We can extend this scheme to capture both flux directions: ( f(u j,n ) f 0 (U j,n ) > 0 F j,n = f(u j,n ) f 0 (U j,n ) < 0 First-order method, but fails for certain initial conditions (see LeVeque. p. 34)

15 Lecture 8.086

16 Last time: Conservation laws Burger s equation creates discontinuities Mathematical description using integral form (=conservation law) In general, conservation laws read du dx u t + f(u, u x,...) x =0() d dt Z b a udx + f(u(b),u x (b),...) f(u(a),u x (a),...)=0 Special form above: Diffusion is also a conservation law

17 Last time: RH jump condition The previous example was specific for Burger s equation. But there is a general condition that determines the shock speed. Rankine-Hugoniot (RH) jump condition: The shock speed s of a conservation law with flux f is s = f(u L) f(u R ) = [f] u L u R [u] Lecture u L, u R : value of u on the left & right of the shock Also valid for initial conditions that are not piecewise linear. Then u L, u R : value of u infinitesimally close to shock

18 Last time: Conservation laws Simple idea: Discretize flux, not PDE! du d du du Start from f (u) = 0 and not from - + f '(u) - = 0 at dx at dx E.g. upwind finite differences (assuming f > 0) Upwind (for f '>o) Uj,n+l - Uj,n + f (Uj+l,n)- f (Uj,n) = 0 At Ax Using F j,n = f(u j,n ), this can be written as U j,n+ t U j,n = x [F j+,n F j,n ] Advantage: conservation law is automatically fulfilled! We can extend this scheme to capture both flux directions: F j,n = ( f(u j,n ) f 0 (U j,n ) > 0 f(u j,n ) f 0 (U j,n ) < 0

19 Lax-Friedrichs for nonlinear conservation laws Remember Lax-Friedrichs for linear -way wave eq: Lax- Friedrichs uj,n+l - 5(uj+l,n uj-,n) Uj+l,n - Uj-l,n = C At 2Ax The nonlinear, conservative formulation is: This can be written in the standard discrete conservative form (see before) U j,n+ t U j,n = x [F j+,n F j,n ] by introducing the LF flux Lax-Friedrichs flux Ax F~~ 5 = [f(u~) - 2at(U3+l - UJ) j+ + f (u3+l)]

20 Lax-Friedrichs for nonlinear conservation laws As in the linear case, Lax-Friedrichs is only first-order accurate. It captures both directions, but the central difference flux term creates extra diffusion! We could just try to use Lax-Wendroff (which was second order), but this creates oscillations Schemes that avoid oscillations are called total variation diminishing (TVD) methods Define total variation (TV): TV(u) = / dx TV(U) = x IUJ+I - U,I O" du A TVD method has the property TV(Un+l) < TV(Un)

21 Godunov s method Some problems so far: Upwind is good, but we need to know direction (i.e. velocity) of wave. Problem: System of equations! Lax-Friedrichs has no upwind direction problems, but high diffusion and thus low accuracy 2nd order methods (e.g. LW) are prone to oscillations. How can we find schemes (in particular numerical flux functions) that are high order, direction-independent and TVD? Godunov s method: A systematic way to construct TVD numerical flux functions (more details in Strang s book and in R.J. LeVeque, Numerical Methods for Conservation Laws) But: Godunov s theorem: Linear numerical schemes for solving partial differential equations (PDE s) that are TVD can be at most first-order accurate.

22 Flux-limiting methods Linear in Godunov s theorem means a scheme where the scheme itself (flux function, coefficients etc.) are constant, i.e do not change their form depending on the current solution. => Use nonlinear numerical scheme, e.g. flux-limiter methods Idea: Combine high-order flux F H (good for smooth part) with low-order flux F L (TVD, good for shock) F(U;j) = FL(U;j) + 4'(U;j) [FH(U;j) - FL(U;j)] Choose Φ close to 0 near shock, near away from shock

23 Flux-limiter for Lax-Wendroff Remember Lax-Wendroff for -way wave eq. u t +au x =0 U j,n+ t U j,n = a U j+,n U j,n 2 x Can write it like an upwind plus correction method: U j,n+ = U j,n r(u j,n U j,n ) + t U 2 a2 j+,n 2U j,n + U j,n x 2 2 r( r)(u j+,n 2U j,n + U j,n ) The correction term increases accuracy (next term in series expansion), but we have seen it leads to oscillations Lecture Corresponding flux: F (U, j) =au j + 2 a( r)(u j+ U j ) Then, LW can be written as U j,n+ t U j,n = x [F j+,n F j,n ] (F j+,n = F(U,j+) etc.)

24 Flux-limiter for Lax-Wendroff Lax-Wendroff flux (prev. slide): F (U, j) =au j + 2 a( r)(u j+ U j ) This flux is an upwind flux (au j ) corrected with a high-order flux! We replace F(U,j) by F (U, j) =au j + 2 a( r)(u j+ U j ) j How should we choose? j Naturally, it should depend on the quantity It can be shown: No oscillations for j = U j U j U j+ U j close to for U smooth far from for shock in U 0 apple ( ) apple 2 and 0 apple ( ) apple 2 for all Θ For nonlinear equations: replace speed a by the local speed (see LeVeque, p. 8)

25 Flux-limiter for Lax-Wendroff LW-flux with flux-limiter: F (U, j) =au j + 2 a( r)(u j+ U j ) j Choices for ( ) φ(θ) 2t LW is not total vari tion diminis k, Θ

26 Finite volume method Another interpretation of the discrete conservation law: Finite volume method Directly approximates the integral form: - (j+i)~x Cell average Uj,n approximates uj,, = - / u(x, nat) dx AX (,-+)AX Change in I cell average '' U(X, (j++)~x (3++)Ax dt (J-;)*x ) dx + [f(uj++,n) - i(uj-+,n)] = 0 Integrate in time from n to n+ to obtain the finite volume conservation law /' U(X, (j-+)ax LA, tn+~) (n+l)at dx - l-;)ax (3+4)Ax U(X, tn) dx + (n+l)at f (uj+;,t) dt- LA, f (u,-;, t) dt = 0

27 Finite volume method - /(j+i)~x Using uj,, = - u(x, nat) dx the integral form (j++)~x tn+~) /' U(X, (j-+)ax LA, (n+l)at can be written as t (ū j,n+ AX (,-+)AX dx - l-;)ax (3+4)Ax U(X, tn) dx + (n+l)at f (uj+;,t) dt- LA, f (u,-;, t) dt = 0 ū j,n )+ t x Z (n+) n t t f(u j+ 2,t)dt t x Z (n+) n t t f(u j,t)dt =0 2 Using the time-averaged flux f j+ 2,n = t Z (n+) n t t f(u j+ 2,t)dt this becomes t (ū j,n+ ū j,n )+ x fj+ 2,n fj 2,n =0

28 Finite volume method Treat ū j,n as independent variables U j,n, approximate average fluxes by F j+ : t (U j,n+ U j,n )+ x F j+ 2,n F j 2,n =0 2,n Easiest choice for numerical flux F: Assume f is constant over one time step, ie. F j+ 2,n = f(u j+ 2,n t ) Note: We need to interpolate U to half-grid points! Note: Other choices of F can lead to implicit schemes

29 Why finite volumes? Grid can be arbitrary, conservation law always fulfilled y j+ NW N NE y j WW W nw w n P ne e n e E y EE y j- sw s se y SW x S SE j i x x i- x i x i+

30 AREPO finite volume code (Volker Springel) Example

31 Navier-Stokes / Flow problems Tricky equation, tricky numerics Incompressible Navier-Stokes equation: du -+(u-v)u=-vp+-nu+ dt Re divu=v=u=o f momentum balance continuity/incompr. equation In this dimensionless form, the only physical parameter is the Reynolds number Re inertial forces (a velocity U) (a length L) Reynoldsnumber Re= % viscous forces (kinematic viscosity v) - Pressure p is a Lagrange multiplier, adjusted such that the incompressibility condition is fulfilled.

32 Example: 2D flow in a lid-driven cavity Consider 2D box, with no-slip boundary conditions at three boundaries: = [0,l x ] [0,l y ]. The domain is In 2D, we can write the NS equations in component form as u t + p x = (u 2 ) x (uv) y + Re (u xx + u yy ) v t + p y = (uv) x (v 2 ) y + Re (v xx + v yy ) Lecture u x + v y = 0 The boundary conditions will be u(x, l y )=u N (x) u(x, 0) = u S (x) u(0,y)=0 u(l x,y)=0 v(x, l y ) = 0 v(x, 0) = 0 v(0,y)=v W (y) v(l x,y)=v E (y)