Hydro code II. Simone Recchi

Transcription

1 Hydro code II Simoe Recchi

2 Literature E.F. Toro: Riema Solvers ad Numerical Methods for Fluid Dyamics, Spriger R.J. LeVeque: Fiite Volume Methods for Hyperbolic Problems, Cambridge Uiversity Press flash.uchicago.edu/site/flashcode irfu.cea.fr/projects/site_ramses/ramses.html

3 Visual outlie set up (grid based) Hydro code evolutio grid geeratio iitial values iput output storage solutio advacemet boudary values time stepper hydro solver source terms solutio update

4 set up (grid based) Hydro code evolutio grid geeratio iitial values iput output storage solutio advacemet boudary values time stepper hydro solver source terms solutio update

5 1 D grid geeratio Remider of uiform grid geeratio x 1/2 x=a x N+1/2 x=b C 1 C 0 C 1 C 2 C N 1 C N C N+1 C N+2 u 1 u 0 u 1 u 2 u N 1 u N+1 u N u N+2 The domai [a,b] has bee divided ito N equal cells C i. The scalar value u i is defied i the midpoit of C i (or it represets a average of u(x) i the cell C i ). C 1, C 0, C N+1, C N+2 : ghost (guard) cells

6 Variable grids x 1/2 x=a x i 3/2 x i 1/2 x i+1/2 x i+3/2 x i 1 x i x i+1 C 1 C 0 C 1 C 2 C i 1 u 1 u 0 u 1 u 2 u i 1 C i u i C i+1 u i+1 x i = x i 1/ 2 x i 1/ 2 All expressios ecoutered last week could be adapted by substitutig the cell size x with x i!

7 Variable grids All expressios ecoutered last week could be adapted by substitutig x with x i! For istace: average over cell u i 1 x i 1/2 x x u x, t dx i 1/2 i stadard discretizatio scheme for coserved laws time step determiatio ad so o u i 1 =u i t x i [ f i 1 /2 f i 1/ 2 ] t=b x i a v

8 Variable grids Notice that last week i the determiatio of the evolutio of desity with slope limiters =[ f v i 1/ 2 i 1/ 2 i 1 = i 3 t f i 1/ 2 f i 1 /2 i 1 v i 1/ r i 1/ 2 r i 1/ 2 i 1 2 v i 1/ 2 1 c ] S r 2 i 1/ 2 i 1 / 2 we have ot assumed that the grids are uiform!

9 Variable grids x i 3/2 x i 1/2 x i+1/2 x i+3/2 x i 1 x i x i+1 C 1 C 0 C 1 C 2 C i 1 u 1 u 0 u 1 u 2 u i 1 C i u i C i+1 u i+1 x i = x i 1/ 2 x i 1/ 2 Thigs to otice: 1. I reflectig boudaries mirrorig cells must be equal, e.g. x 0 = x 1, x 1 = x 2 2. Half idex cell sizes might be eeded (i.e. for the mometum advectio; see also below). They are give by: x i 1 /2 =x i x i 1 =0.5 x i 1/ 2 x i 3 / 2

10 set up (grid based) Hydro code evolutio grid geeratio iitial values iput output storage solutio advacemet boudary values time stepper hydro solver source terms solutio update

11 set up (grid based) Hydro code evolutio grid geeratio iitial values iput output storage solutio advacemet boudary values time stepper hydro solver source terms solutio update

12 Hydro solver outlie More o the Euler equatios Global solutios ad Riema solvers

13 Euler equatios t v =0 m m v P =0 t E t I these equatios we have eglected Exteral forces (e.g. gravity!) Viscosity! Possible source terms [ E P v ]=0

14 Navier Stokes equatios with sources t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė : Viscous stress tesor g: Gravity q: heat flux

15 Navier Stokes equatios with sources t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė : Viscous stress tesor g: Gravity q: heat flux It depeds o the divergece of v ad o the divergece of spatial derivatives of velocities. I 1 D it is proportioal to v xx!

16 Navier Stokes equatios with sources : Viscous stress tesor g: Gravity q: heat flux It depeds o the divergece of v ad o the divergece of spatial derivatives of velocities. I 1 D it is proportioal to v xx! Remember the viscous advectio equatio u t au x = u xx to prevet wave breakig. Itroducig such terms is a cheap way to cosider viscous effects (remember the artificial viscosity of the exercise!) = mi 0, C 1 v r v r

17 Navier Stokes equatios with sources Remember the viscous advectio equatio u t au x = u xx to prevet wave breakig. Itroducig such terms is a cheap way to cosider viscous effects (remember the artificial viscosity of the exercise!) Remember also the modified equatio of the first order upwid scheme u t au x = 1 2 a x 1 c u xx= um u xx it is agai a viscous term! This shows that some form of artificial viscosity is always preset i umerical schemes (i particular firstorder oes). Sometimes this is ot eough ad some extra artificial viscosity must be added. If the built i artificial viscosity is eough to prevet oscillatios, a extra artificial viscosity term is ot eeded.

18 Navier Stokes equatios with sources t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė : Viscous stress tesor g: Gravity q: heat flux g might deped o some (fixed) exteral field (e.g. a fixed dark matter halo) > costat vector! It might be also self gravity of the gas. I this case the (elliptic!) Poisso equatio must be solved! > Complicated 2 =4 G

19 Navier Stokes equatios with sources g might deped o some (fixed) exteral field (e.g. a fixed dark matter halo) > costat vector! It might be also self gravity of the gas. I this case the (elliptic!) Poisso equatio must be solved! > Complicated 2 =4 G i 1 D the calculatio is much simpler sice oe has: g=g M R R 2 g is a vector quatity (defied at the itercell boudary i a staggered grid); M is the mass withi R The self gravity ca be thus very easily implemeted i the 1 D Superova explosio code. However it will appear evidet that gravity is (almost) egligible i this problem.

20 Navier Stokes equatios with sources t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė : Viscous stress tesor g: Gravity q: heat flux Fourier's law relates the heat flux to the temperature gradiet, i.e. q= T (parabolic problem!) The thermal coductivity ca be costat or variable (see exercise) > Solved through Crak Nicolso method

21 Serial vs. global solutio t v =0 m m v P =0 t t [ P v ]=0 This system ca be solved i two ways: serially: three scalar advectio equatios are solved separately globally; the sigle vector advectio equatio W t [ F W ] x = 0, W t A W W x = 0 where F represets the fluxes of the ivolved variables, is solved (bars are omitted hereafter).

22 Serial vs. global solutio W t [ F W ] x =0, W = m, F W = m m u P u P To obtai the global solutio of this vector equatio, we eed some tool to uderstad the global behaviour of the thermodyamic variables. This tool is offered by the solutio of the Riema Problem (RP): W t [ F W ] x =0 W x, 0 =W 0 x = W L W R x 0 x 0 Two (costat!) thermodyamic states i cotact.

23 Riema Problem Replace t=0 with t ; x=0 with x i 1/2 (chage of referece frame), W L with W i 1, W R with W i ad you will immediately realize why is the Riema problem so relevat to the solutio of the Euler equatios. t t 1 W t [ F W ] x =0 W x, t =W 0 x = W i 1 x x i 1 / 2 W i x x i 1 /2 1 /2 t W i 1 W i Costat states i cotact! t x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x

24 Riema Problem W t [ F W ] x =0 W x, 0 =W 0 x = W L W R x 0 x 0 Remember Sod shock problem (Nigel's lecture), m, = 1, 0, 2.5 x , 0, 0.25 x 0.5 Simple RP (iitial velocities are zero).

25 Riema Problem, m, = 1, 0, 2.5 x , 0, 0.25 x 0.5 Simple RP (iitial velocities are zero). Nedwich's solutio.

26 Riema Problem New itermediate states have bee produced! The solutio of the RP is self similar! I.e. it develops alog rays x x itf t t itf =costat

27 RP ad Goduov method How ca we exploit the solutio of the RP? We solve the RP at each iterface ad average the ew state at each cell! t t 1 W i 1 1 /2 t t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x (remember that each state ca be defied as cell average, e.g.) W 1 i 1 x i 1/ 2 x x W x, t 1 dx i 1/ 2 i

28 RP ad Goduov method t t 1 W i 1 1 /2 t t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x W 1 i 1 x i 1/ 2 x x W x, t 1 dx i 1/ 2 i Problems: This itegral is ofte difficult to calculate We do ot kow what to do if rays itersect, i.e. Courat coditio is: a v t x 1 2

29 RP ad Goduov method t t 1 R i 1 /2 R i 1 /2 1 /2 t t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x To simplify we ca cosider oly the solutio of the RP at the iterfaces! (We idicated it with R i 1/2 ). At this poit we cosider the (Goduov) flux with state W equal to these itermediate states, i.e.: W i 1 =W i t God x i [ F i 1 /2 F i 1 / 2 ], F i 1 /2 =F i 1 / 2 = F R i 1/ 2 Courat umbers must be (as usual) smaller tha 1 ad ot of 0.5!

30 RP ad Goduov method t t 1 R i 1 /2 R i 1 /2 1 /2 t t u i 1 u i u i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x Example: advectio equatio u t au x =0 with a>0 (remember: advected quatities remai ualtered!). Clearly: R i 1 /2 =u i 1 u 1 i =u i t > Upwid scheme!, R i 1 / 2 =u i x i [ F u i 1 F u i ]=u i c u i 1 u i

31 RP ad Goduov method t t 1 R i 1 /2 R i 1 /2 1 /2 t t u i 1 u i u i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x Example: advectio equatio u t au x =0 with a<0. R i 1 /2 =u i, R i 1/ 2 =u i 1 u 1 i =u i t x i [ F u i F u i 1 u 1 i =u i c u i u i 1 ]=u i c u i u i 1 c u i 1 u i

32 RP ad Goduov method t t 1 R i 1 /2 R i 1 /2 1 /2 t t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x Example: cotiuity equatio t v x =0 states W i = i, v i T with piecewise costat This is ot a simple extesio of the previous example: R i 1/2 ad R i+1/2 represet ew states! Two waves are emaated from the iterfaces, oe with velocity v i (v i+1 ) ad oe with velocity 0!

33 RP ad Goduov method Two waves are emaated from the iterfaces, oe with velocity v i (v i+1 ) ad oe with velocity 0! Thik at coveyor belts with differet speeds t v i v i+1 v i v i+1 v i v i+1 v i v i+1

34 RP ad Goduov method Two waves are emaated from the iterface, oe with velocity v i (v i+1 ) (shock!) ad oe with velocity 0 (cotact!) A ew state has bee created! t cotact shock v i v i+1

35 RP ad Goduov method t R i 1 /2 W i 1/ 2 R i 1 /2 W i 1/ 2 t 1 1 /2 t t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x How ca we solve the RP (i.e. calculate the itermediate states W*) the? > Coservatio of desity flux at the iterface! i v i = i 1 /2 v i 1, i 1/ 2 = v i i v i 1 i 1 = i t x i [ f i 1/ 2 f i 1/ 2 ]= i t x i i 1 v i 1 i v i

36 RP ad WAF methods Secod way of exploitig the solutio of the RP: we average the fluxes at half timestep! (Weighted average flux WAF methods) t t 1 1 /2 t WAF F i 1 / 2 WAF F i 1 / 2 t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x

37 RP ad WAF methods t t 1 1 /2 t WAF F i 1 / 2 WAF F i 1 / 2 t W i 1 W i W i 1 x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x i 1 x i 3 / 2 x (remember that we wished to defie fluxes at half iteger time idices staggered grid!) The WAF flux ca be defied as: F WAF i 1 / 2 1 x i x x i 1 i 1 /2 F [W x, t 1 / 2 ] dx where W(x,t +1/2 ) is the solutio of the RP evaluated after half a temporal timestep.

38 RP ad WAF methods F i 1 / 2 WAF t 1 /2 t 2 1 F F 3 i 1 / 2 i 1 / 2 F i 1 / 2 4 F i 1 / 2 t W i 1 W i x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x If N waves emaate from a RP, the WAF flux ca be approximated as: F WAF i 1 / 2 1 x i x x F [W x, t 1 / 2 N 1 k i 1 ] dx = k k =1 F i 1/ 2 i 1 /2 F (k) : itermediate fluxes (fluxes of itermediate states); k : weights

39 RP ad WAF methods W i 1 /2 WAF t 1 /2 t 2 1 W W 3 i 1 /2 i 1 /2 W i 1 /2 4 W i 1 /2 t W i 1 W i x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x Equivalet approach: recostruct a average WAF state W WAF ad the calculate its flux W WAF i 1 /2 1 x i x xi W x, t 1/ 2 dx 1 = k=1 i 1/ 2 F WAF i 1/ 2 = F W i 1/ 2 WAF N 1 k k W i 1 / 2

40 RP ad WAF methods t 1 /2 t 2 1 W W 3 i 1 /2 i 1 /2 W i 1 /2 4 W i 1 /2 A 0 A 1 A 2 A 3 A t W i 1 W i x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x Weights k are easy to calculate if the slopes k (propagatio speeds) are kow. They are proportioal to the legth of the segmets A k 1 A k, i.e.: k = A k 1 A k x i 1/ 2

41 RP ad WAF methods Weights k are easy to calculate if the slopes k (propagatio speeds) are kow (removed idex from x for simplicity). It is: k = A k 1 A k x k = 1 2 c i 1/ 2 Oe ca thus obtai: k 1 k ; c i 1 / 2 = t k x, c0 = 1, c N 1 =1 c i 1 /2 F WAF N 1 k i 1 / 2 = k k =1 F i 1 / 2 = 1 2 F i 1 F i 1 2 N k=1 k k c i 1/ 2 F i 1/ 2 k k F i =F W i, F i 1 = F W i 1, F i 1 /2 =F 1 k i 1 / 2 F i 1/ 2 (otice that F i 1/2 (1) =F i 1, F i 1/2 (N+1) =F i ) Alteratively: W WAF i 1 /2 = 1 2 W i 1 W i 1 2 k =1 k k 1 W i 1 /2 =W k i 1/ 2 W i 1/ 2 N k k c i 1 / 2 W i 1 / 2, F WAF WAF i 1/ 2 = F W i 1 /2

42 RP ad WAF methods t 1 /2 t 1 u i 1/ 2 2 u i 1/ 2 A 0 A 1 A 2 1 =a t u i 1 u i x i 3 / 2 x i 1 x i 1 /2 x i x i 1 /2 x Example: advectio equatio k = A k 1 A k x, 1 = c, 2 = c f WAF i 1/ 2 = 1 2 [ au i 1 1 c au i 1 c ] This is the Lax Wedroff flux!

43 Flux limited WAF methods k = A k 1 A k x, 1 = c, 2 = c f WAF i 1/ 2 = 1 2 [ au i 1 1 c au i 1 c ] This is the Lax Wedroff flux! How ca we costruct flux limited (i.e. o oscillatig) WAF methods? We remid the costructio of scalar flux limited methods: we realized that the Lax Wedroff method ca be obtaied from a piecewise liear data recostructio with slopes give by the dowwid slopes. We the modified the data jumps S through a flux limiter fuctio i order to make the iitial data mootoe. f i 1/ 2 =a u i 1 a u i 1 2 a 1 c S i 1 / 2 S i 1 /2 = i 1 / 2 u i 1 /2

44 Flux limited WAF methods f i 1/ 2 =a u i 1 a u i 1 2 a 1 c S i 1 / 2 S i 1 /2 = i 1 / 2 u i 1 /2 sice: F WAF i 1 / 2 = 1 2 F i 1 F i 1 2 k=1 N k c i 1/ 2 k F i 1 /2 by aalogy with the scalar case we ca assume that the (mootoicity preservig) WAF flux is: WAF = 1 2 F i 1 F i 1 2 k=1 F i 1 / 2 S i 1/2 (k) : limited versio of the k th jump, i.e. k S i 1 /2 N k k [1 1 c ] k i 1 / 2 S i 1/ 2 k k = i 1 / 2 W i 1/ 2

45 Flux limited WAF methods WAF = 1 2 F i 1 F i 1 2 k=1 F i 1 / 2 N k k [1 1 c ] k i 1 / 2 S i 1/ 2 Alteratively: W WAF i 1 /2 = 1 2 W i 1 W i 1 2 k =1 WAF WAF = F W i 1/ 2 F i 1/ 2 N k sig c i 1/ 2 k k [1 1 c i 1 /2 ]S i 1/ 2 S i 1/2 (k) : still limited versio of the k th jump, i.e. k S i 1 /2 k k = i 1 / 2 W i 1/ 2

46 Flux limited WAF methods S i 1/2 (k) : still limited versio of the k th jump, i.e. k S i 1 /2 k k = i 1 / 2 W i 1/ 2 For scalar problems we had: i 1/ 2 = u I 1/ 2, I = u i 1/ 2 i 1 a 0 i 1 a 0 : ratio of the upwid jump to the local jump. I vector problems is the ratio of wave jumps, but wave jumps are vectors! How ca we defie their divisio? (Geerally they are ot parallel!)

47 Flux limited WAF methods : ratio of the upwid jump to the local jump. I vector problems is the ratio of wave jumps, but wave jumps are vectors! How ca we defie their divisio? 1. Oe selected sigle quatity which is kow to chage across waves (for istace desity) is chose to calculate the jumps, i.e. k i 1/ 2 = k I 1 / 2 i 1 /2, I= i 1 c i 1/ 2 0 k k i 1 c i 1/ The vectors are projected oto a preferetial directio (for istace the oe of the local jump), therefore oe has: k i 1/ 2 = W k I 1/ 2 k W i 1/ 2 k k W i 1/ 2, I = i 1 c k i 1/ 2 0 k k W i 1 / 2 i 1 c i 1/ 2 0

48 Flux limited WAF methods k i 1/ 2 k i 1/ 2 = k I 1 / 2 i 1 /2 = W k I 1/ 2 k W i 1/ 2, I= i 1 c i 1/ 2 0 k k i 1 c i 1/ 2 0 Limiter fuctios are the same as i the scalar problem, for istace: k k W i 1/ 2, I = i 1 c k i 1/ 2 0 k k W i 1 / 2 i 1 c i 1/ 2 0 Va Leer limiter: mimod = 1 =max [0, mi 1, ] superbee =max [0, mi 1, 2, mi 2, ]

49 Flux limited WAF methods Thigs to otice: 1. W ca be the vector of coserved quatities W=(,m, ) T, but it ca be also the vector of primitive quatities W=(,v,P) T! 2. To calculate the timestep to advace the solutio as usual the Courat umber must be smaller tha 1. We must however take ito accout the speeds of all the waves, emaatig from all the iterfaces, take the fastest at the time level (S max () ) ad calculate the timestep accordig to geerally S max () is well approximated by S max t=b x S max =max i [ v i a i ] but fastest waves are always possible. Commo practice: start with a very low b, icrease it i the first few timesteps util reachig 0.4 or 0.5

50 Flux limited WAF methods: summary Calculate the timestep x t=b max i [ v i a i ] assumig a small value of b durig the first timesteps For each pair of data (W i 1,W i ) solve the RP. Calculate itermediate (k) (k) (k) states W i 1/2, fluxes F i 1/2, jumps W i 1/2 ad wave speeds k Calculate Courat umbers related to each wave: c i 1 / 2 k = t k x

51 Flux limited WAF methods: summary Calculate the wave jump ratios accordig to oe of the followig formulae: k i 1/ 2 k i 1/ 2 Calculate limited jumps Va Leer limiter: mimod = k I 1 / 2 i 1 /2 = W k I 1/ 2 k W i 1/ 2 k S i 1 /2 k, I= i 1 c i 1/ 2 0 k k i 1 c i 1/ 2 0 k W i 1/ 2, I = i 1 c k i 1/ 2 0 k k W i 1 / 2 i 1 c i 1/ 2 0 k k = i 1 / 2 W i 1/ 2 = 1 =max [0, mi 1, ] superbee =max [0, mi 1, 2, mi 2, ]

52 Flux limited WAF methods: summary Calculate the itercell flux accordig to or: WAF = 1 2 F i 1 F i 1 2 k=1 F i 1 / 2 W WAF i 1 /2 = 1 2 W i 1 W i 1 2 k =1 WAF WAF = F W i 1/ 2 F i 1/ 2 Update the solutio Update the boudaries Go to the ext time level N N k k [1 1 c ] k i 1 / 2 S i 1/ 2 k sig c i 1/ 2 W i 1 =W i t x i [ F i 1 /2 F i 1 / 2 ] k k [1 1 c i 1 /2 ]S i 1/ 2

53 Solutio of the RP i pills The RP solutio is self similar! (It develops alog rays of costat x/t) The Euler equatio produces three waves, propagatig at speeds 1 = v a, 2 =v, 3 =v a The cetral wave is always a cotact! The other two waves ca be shocks or rarefactio waves

54 Solutio of the RP i pills Across a cotact oly the desity chages! Velocity ad pressure remai costat (remember the coveyor belts). P ad v are thus also called Geeralized Riema ivariats of the cotact wave. t cotact shock v i v i+1

55 Solutio of the RP i pills Shocks are zero legth discotiuities. Across them, the fluxes of mass, mometum ad eergy are coserved. This leads to the Rakie Hugoiot jump shock coditios! (M: Mach umber) P 2 = 2 P 1 1 M v 2 = 1 v M = 1 1 2/ M 2 Rarefactio waves are regios of liear variatios of the thermodyamic quatities. The Geeralized Riema ivariats of rarefactio waves are: v 2a / 1, v 2a / 1, across 1 =v a across 3 =v a

56 Solutio of the RP i pills RP solvers ca be either exact (obtaied through a iterative procedure) or approximate (usually obtaied through a liearizatio of the Euler equatios). I approximate Riema solvers the rarefactio wave is usually approximated as a sigle state. Approximate structure of the Riema solutio (W=(,m, )T ): t S L S S R W L W L W R W R W L W R remember that oly the desity chages across the cotact! x

57 Solutio of the RP i pills The HLLC approximate Riema solver: P =max [0,0.5 P L P R 0.5 v R v L a ], =0.5 L R, a=0.5 a L a R S L = v L a L q L, S R = v R a R q R, q L, R = 1 P P [ L, R 2 1 P 1 1 ]1/ P 2 P P L, R L, R S = P R P L L v L S L v L R v R S R v R L S L v L R S R v R W L, R = R S L, R v L, R L, S L, R S 1 S [ L, R S v S L, R L, R ] P L, R L, R S L, R v L, R F L, R = F L, R S L, R W L, R W L, R

58 WAF methods last remarks 1. I spherical symmetry, as i the case of scalar problems, the Euler equatios ca be solved as: W i 1 =W i 3 t F i 1 / 2 F i 1 / r i 1 / 2 r i 1 / 2 ad the WAF fluxes aalyzed so far must be multiplied by r 2, i.e. F WAF i 1 / 2 F WAF 2 i 1 / 2 r i 1 /2 alteratively, oe ca solve the plaar Euler equatio with (geometric) source terms W t [ F W ] x =S W W = m, F W = m m v P u P, S W = 2 r v v 2 v P

59 WAF methods last remarks 2. I global solutios of the Euler equatio it is atural to defie scalar ad vector quatities (like m) at the grid ceter sice here is where the state W is defied > it is difficult to defie staggered grid as we did for serial solutios. However, this is ot importat because we shifted the defiitio of velocities i order to calculate fluxes at iterfaces! This is what we did also with WAF methods. t t 1,, P 1 /2 t m, v, t x i 1 /2 x i x i 1 /2 x

60 set up (grid based) Hydro code evolutio grid geeratio iitial values iput output storage solutio advacemet boudary values time stepper hydro solver source terms solutio update

61 Source terms t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė If you are solvig the equatios serially, you ca defie source terms where the correspodet quatities are defied (grid ceters for desity ad eergy, itercell boudaries for mometum). If you are solvig globally, the Euler equatio with source terms look like that: W t [ F W ] x =S W

62 Source terms W t [ F W ] x =S W Basic strategy: time splittig method! We first solve the pure advectio hyperbolic problem: W t [ F W ] x =0 W the we solve the system of Ordiary differetial equatios d W dt =S W usig the solutio of the pure advectio problem as iitial coditio. Ay of the methods leared i the first two discretizatio lectures could be theoretically adopted to solve this system

63 Source terms Example: scalar advectio equatio with source term u t au x = u the aalytical solutio is: u x, t =u 0 x at e t where u 0 represets the iitial coditios. The solutio of the pure advectio problem is: u t au x =0, u x, t =u 0 x at If we ow solve the ODE u'= u usig as iitial coditios the solutio of the pure advectio problem we obtai: u x, t = u x, t e t =u 0 x at e t

64 Source terms W t [ F W ] x =S W W t [ F W ] x =0 W i 1 =W i t x [ F i 1/ 2 F i 1/ 2 ] At this poit the system of ODEs d W dt =S W ca be solved, for istace with the Euler method W i 1 = W i 1 t S t, W i 1

65 W t [ F W ] x =0 Source terms W i 1 =W i t x [ F i 1/ 2 F i 1/ 2 ] W i 1 = W i 1 t S t, W i 1 t v =0 m m v P =0 t t v = P v Oce the pure advectio equatios have bee solved, the coolig ca be thus dealt with i this way: 1 i = 1 i t i i 1 m H c V i m H 2 T

66 Source terms W t [ F W ] x =0 W i 1 =W i t x [ F i 1/ 2 F i 1/ 2 ] W i 1 = W i 1 t S t, W i 1 This is sufficiet for first order (e.g. Goduov) methods. For (almost) secod order methods like the WAF methods it is more balaced to look for a secod order ODE solver. Two possibilities: 1. Heu method or similar schemes W i 1 = W i 1 t S t, W i 1 S [t 1, W i 1 t S t, W i 1 ] 2

67 Source terms 1. Heu method or similar schemes W 1 i = W 1 i t S t, W 1 i S [t 1, W 1 i t S t, W 1 i ] 2 2. A secod order accurate splittig scheme i which first the ODE for the source term is solved advacig the solutio by half a timestep (with a first order method) the the solutio is advected, the the solutio of the ODE is advaced by the remaiig half of the timestep. This ca be represeted by the sequece W i 1 =S t / 2 A t S t / 2 W i S: operator solutio of the ODE; A: operator solutio of the advectio

68 Source terms W i 1 =S t / 2 A t S t / 2 W i I order to update the eergy accordig to the coolig fuctio, we thus eed the followig 3 steps: 1. 1 /2 i = i t i 2 m H 2 i c V i 2. i 1 = i 1 / 2 t x [ f i 1/ 2 f i 1 / 2 ] 3. 1 i = 1 i t 1 i 2 m H 2 1 i 1 c V i

69 Source terms ad time steppig t v = m m v P = g ṁ t E t [ E P v ]= v q v g Ė Each source term itroduces a ew typical timescale (i additio to the Courat timescale), amely: t =, t m= ṁ m, t E t c = Ė E All these timescales must be take ito accout i the time steppig routie!

70 Source terms ad time steppig t =, t m= ṁ m, t E t c = Ė E All these timescales must be take ito accout i the time steppig routie! I order ot to be boud by these timesteps, a implicit method (for istace backward Euler) could be employed (remember: implicit methods are ucoditioally stable!). I the case of the coolig fuctio: 1 i = 1 i t i i 1 m H c V i

71 Thak you for the attetio!