Numerical Methods for Partial Differential Equations

Transcription

1 Numerical Methods for Partial Differetial Equatios Eric de Sturler Uiversity of Illiois at Urbaa-Champaig Cosider the liear first order hyperbolic equatio Øu Øu Øt + a(x, t) Øx = 0 I spite of its simple appearece it is ot always easy to solve accurately. Let s first review the exact solutio (agai). dx Alog the curve defied by dt = a du (x, t) we have dt = Øu Øt + Øu dx Øx dt = 0, ad so the solutio does ot chage. The solutio at ay poit i time is the defied by fidig where the charateristic curve goig through that poit crosses the boudary (i space or time) of the domai. Hece we ca start with a set of poits ad itegrate (trace) the characteristic curves from those poits /15/01

2 So we ca take a set of poits x0, x1, x2, x3,... ad itegrate the charateristic curves umerically with startig poit x(0) = xj. The for ay poit (x,t) o such a curve the solutio will be u(x,t) = u(xj, 0) = u 0 (x) the give iitial data (of the PDE). If a(x, t) is Lipschitz cotiuous i x ad cotiuous i t the characteristic curves caot cross ad the solutio is everywhere well-defied (i classical sese) give appropriate boudary ad iitial coditio. If a is costat the characteristic curves are straight parallel lies x at = k ad the solutio satisfies u(x, t) = u 0 (x at). If the equatio is oliear (quasi-liear) ad a depeds oly o u the characteristics are also straight lies (why?), but they will, i geeral, ot be parallel. The solutio satisfies u(x, t) = u 0 (x a(u)t) ad is costat alog the lie x a(u)t = k. As log as the characteristics do ot cross or fa out the solutio is well-defied. t u x /15/01

3 The characteristics play a essetial role i the aalysis of umerical schemes for hyperbolic equatios. I geeral we study so-called systems of coservatio laws: Øu Øt = Øf(u) Øx, where u is a vector of ukow fuctios ad f is a vector of so-called flux fuctios. Let u = [u 1 u 2 ] T ad f(u) = [f 1 (u) f 2 (u)] T the we ca write u t 1 u t 2 + Øf 1 Øf 1 Øu 1 Øu 2 Øf 2 Øf 2 Øu 1 Øu 2 Øu 1 Øx Øu 2 Øx = 0 u t 1 Øf 1 Øf 1 Øu 1 Øf 2 Øf 2 Øu 1 Øu 2 Øu 1 Øu Cosider Øx Øu = 0, ad write A(u) = Øf (Jacobia), the u 2 Øu t Øx the system ca be writte as ut = A(u)ux ad the characteristic speeds are give by the eigevalues of A(u). The system of equatios is called hyperbolic if A has all real eigevalues ad ad full set of idepedet eigevectors. I such a case the matrix A(u) is diagoalizable: A = S 1 S w SA = S where S is the matrix of left eigevectors. S ad are fuctios of u. This gives the characteristic ormal form Sut = Sux /15/01

4 If it is possible to defie a vector r(u) such that r t = Su t ad r x = Su x ( r is called a vector of Riema ivariats), the we ca write r t + r x = 0 where = (r) ad i geeral each compoet of will deped o all compoets of r. Hece, the characteristic curves will geerally ot be straight lies. For a system of more tha 2 equatios the Riema ivariats may ot exist. To solve this system usig the method of characteristics we have to itegrate first the characteristic curves ad the the solutio alog those curves. Especially i more spatial dimesios this becomes very complicated (although it is supposed to lead to beautiful geometric problems). Lookig at more pragmatic solutio methods we first cosider explicit methods. First we look at. ut + aux = 0 We use the (upwid) discretizatio U j +1 U j t + a U j U j 1 x = 0, which leads to U j +1 = U j a t x U j U j 1 t U j +1 = U j a t x U j U j 1 U +1 j = (1 )U j + U j 1, where = a t +1 x. So U j depeds o the two values U j ad U j 1. Each of these i tur depeds o two values at the time level 1, ad so o till we get to the iitial coditio. We see that depeds o the values U j 1, U j +1,, U j. U j +1 0,U j /15/01

5 +1 0 We see that U j depeds o the values U j 1,U j, U j +1,, U j. P The solutio at (space-time) poit P depeds o the data i a triagle as idicated i the above figure. For a ihomogeeous equatio with rhs term h j the solutio at P depeds o data give at every grid poit i the triagle. Hece, we call this triagle the domai of depedece. I cotrast for the PDE the domai of depedece is just the characteristic curve up till (icludig) the iitial value o that curve. A ihomogeeous term will be itegrated over this curve. It turs out for the iteratio to be stable the domai of depedece of the umerical scheme (at each poit) must cotai the domai of depedece of the differetial equatio at that poit. This coditio is called the Courat-Friedrich-Lewy coditio or short CFL coditio. This coditio is ecessary but ot sufficiet. It is easy to see that if the coditio is ot satisfied the solutio may ot coverge, because we ca alter data alog the characteristic curve that is ot picked up by the umerical scheme (see also book p.88&89) /15/01

6 The CFL coditio also shows why we must use a upwid differece scheme (if we use a oe-sided scheme). Why? We see that the CLF coditio gives a time step costrait a t x [ 1 +1 We see that for a > 0 we eed to use U j 1 ad U j to approximate U j ad for a < 0 we eed to use U j+1 ad U j (for the explicit method). Takig both cases ito accout we get for the CFL coditio a t x [ 1 We might be tempted to take a cetral (first) differece to take both possibilities ito accout immediately. The cetral scheme gives U +1 j = U j a t 2 x U j+1 U j 1 If we satisfy x we satisfy the CFL coditio for both sigs of a. Now cosider the stability usig Fourier aalysis for costat a igorig boudary coditios. a t Substitute U j = (k)e ikj x ito the iteratio: (k)e ikj x a t 2 x ( (k)e ikj x e ik x (k)e ikj x e ik x ) = (k)e ikj x (1 a t 2 x ( e ik x e ik x )) = (k)e ikj x (1 a t 2 x 2i si k x ) = (k)e ikj x (1 a t x i si k x ) /15/01

7 So we get (k) = (1 a t x i si k x ) ad the scheme is ustable for (almost) ay k ad ay t. So the cetral scheme satisfies the CFL coditio but is evertheless ustable. Which shows that the CFL coditio is ecessary, but ot sufficiet. Aother scheme usig the same three poit o the old time level is the upwid scheme: U j +1 = U j a t x +xu j for a < 0 U j a t x xu j for a > 0 where a = a j. To study this scheme we oly eed to cosider oe case, say a > 0 U j +1 = U j a t x U j U j 1 +1 (k)e ikj x = (k)e ikj x a t x ( (k)e ikj x (k)e ikj x e ik x ) w +1 (k)e ikj x = (k)e ikj x (1 a t x ( 1 e ik x )) w (k) = (1 a t x ( 1 e ik x )) (k) 2 = (1 + e ik x )(1 + e ik x ) = (1 ) (1 ) e ik x (k) 2 = (1 ) (1 ) cos k x = + 2(1 ) (1 2 si 2 k 2 x ) (k) 2 = 1 4(1 ) si k x /15/01

8 I order for the scheme to be stable, we eed (k) 2 = 1 4(1 ) si 2 1 2k x [ 1 which meas (1 ) m 0 u 0 [ [ 1 ( 0 [ a t x [ 1) ad (1 ) [ 1 2 Now for 0 [ [ 1 we have max(1 ) = 1 4 for = 1 2. So the scheme is stable for 0 [ [ 1. We fid the essetially the same amplificatio factor for a < 0 where a is replaced by a. The scheme clearly satisfies the CFL coditio ad so i this case the CFL coditio gives the correct stability limits. Apart from stability we also eed to cosider the error behavior of this scheme. Agai we assume a > 0 ad that the grid satisfies the CFL cod. A 1- C P t B +1 The solutio at U j (P) equals the solutio at the poit where the characteristic crosses the lie t = t (C). We ca fid that value usig liear iterpolatio of the solutio betwee the poits ad. U j 1 This turs out to be exactly the value that the explicit first order upwid differece scheme gives. U j /15/01

9 First ote that the characteristic speed is a. So the characteristic that +1 passes through x j crosses the lie t = t at the poit x j a t dx (the characteristic is defied by = a t dx = a dt) dt This gives a relative weight (i liear iterpolatio) to U j of 1 a t x ad to of. So liear iterpolatio leads to the update U j 1 a t x U +1 j = (1 a t ) x U j + a t x U j 1 which is exactly the update of the explicit scheme with upwid first order fiite differece. I additio, for small eough t all coefficiets are positive, ad i that case the right side coefficiets sum to 1 (left side coefficiet). So we satisfy a maximum priciple. Now cosider the trucatio error ( a > 0): T j = u j +1 u j t a u j u j 1 j x = [u t tu tt + + au x 1 2 xu xx + ] j T j = [ 1 2 tu tt 1 2 a xu xx] j + So the trucatio error is O( t + x). If a is costat we have u tt = a 2 u xx ad T j = 1( 2 1 )a xu xx + x ad so the choice t = a removes the lowest order compoets of the trucatio error. Note that this exactly the CFL coditio (limit o time step) /15/01

10 Proceedig i the usual way, we fid for the error e +1 j = (1 )e j + e j 1 tt j From which we ca derive, assumig 0 [ [ 1, E +1 = max j e +1 j [ E + tmax j T j ad with T j [ T we fid E [ tt [ t f T So if we assume T [ tm tt + a xm xx, we ca make the error arbitrarily small by takig small eough time steps ad grid size while satisfyig the requiremets for a maximum priciple. It is easy to see that our aalysis remais valid for variable a ad the equatio for the error remais valid with variable v. Moreover, we have to satisfy our stability requiremet at every poit. Also check that if a chages sig ad we adapt the local discretizatio (upwid) the equatio for the error remais valid (with = a j ). Clearly, if a ca chage sig we may have to adapt the boudary coditios. Mai poit: the aalysis remais valid. t x /15/01

11 However, if we allow the iitial coditio to cotai poits where the solutio or its derivatives are ot cotiuous we have a problem. The equatio is the ot well-defied. I the parabolic equatio we showed that after some ifiitesimal time the solutio will be differetiable (at least if the Fourier expasio coverges). This is obviously ot the case here. I geeral, discotiuities of the solutio or its derivatives (may) propagate through the solutio. Hece a classical solutio does ot exist ad we must look for a geeralized solutio. As we saw i a earlier example, for oliear equatios discotiuities may arise eve if the iitial coditio is cotiuous with cotiuous (higher) derivatives. Our stability aalysis does ot hold for such cases. For Fourier aalysis we agai assume that a > 0 ad costat. Sice we kow the exact solutio looks like u 0 (x at) we take for the Fourier aalysis u(x, t) = e i(kx+ t) The, u satisfies the differetial equatio exactly if = ka Note that the amplitude of u does ot chage (o dampig). The solutio at a give poit chages i a time step i phase by ak t We saw that the umerical solutio usig our first order upwid scheme has eigefuctios ad amplificatio factor e ikj x (k) = 1 (1 e ik x ) /15/01

12 We saw that the umerical solutio usig our first order upwid scheme has eigefuctios ad amplificatio factor. e ikj x (k) = 1 (1 e ik x ) For 0 [ [ 1 this scheme is dampig (i cotrast to exact solutio). For the chage i phase we fid arg = arcta Followig the book we fid (for small (k x)[1 1 6 ( 1 )(1 2 )(k x) 2 + ] sik x 1 + cos x k x) that the phase chage is Amplitude error: O((k x) 2 ) gives global error of O(k x) Phase error: ak t $ O([k x] 2 ), relative error O([k x] 2 ) Chage i amplitude: (k) 2 = 1 4(1 ) si 2 1 2k x sik x 1 + cos x Chage i phase: arg = arcta ad (for small k x) (k x)[1 1 6 ( 1 )(1 2 )(k x) 2 + ] We see that for = 1 ad = 1 2 we have o phase error, ad i additio for = 1 we also do t have a amplitude error. So for = 1 the umerical solutio is exact. Of course, this caot be achieved for variable a (for a fixed mesh) /15/01

13 I may cases the (first order) upwid scheme is ot sufficietly accurate; maily because of its dampig of high frequecies. This teds to smooth steep gradiets. We saw that the upwid scheme could be derived by assumig a straight (lie) characteristic with slope ad usig liear iterpolatio at the previous time level. So, let s cosider quadratic iterpolatio at the previous time level (ad everythig else the same). So, let s cosider quadratic iterpolatio at the previous time level (ad everythig else the same). Usig a Lagrage iterpolatio formula gives (1- ) x x P t A Q B C (1+ ) 2 u A + (1 2 )u B ( 1 ) 2 u C. This gives the iteratio U +1 j = ( 1+ ) 2 U j 1 + (1 2 )U j ( 1 ) 2 U j /15/01

14 The scheme, U +1 j = (1+ ) 2 U j 1 + (1 2 )U j (1 ) Lax-Wedroff scheme. We ca also write this as U j +1 = U j 0x U j x 2 U j Fourier aalysis the leads to (k) = 1 i si k x 2 2 si k x This gives 2 = (1 2 ) si k x arg = arcta 2 U j+1 sik x si k x l [1 1 6 ( 1 2 ) 2 + ], is called the 2 = (1 2 ) si 4 1 2k x arg = arcta si k x si k x l [1 1 6 ( 1 2 ) 2 + ] Compared with the upwid scheme we see that Lax-Wedroff has much smaller dampig si 4 (k x/2) (especially for small k x) has same order of phase error (but larger o average) The phase error is always egative, so the scheme lags i time /15/01

15 For use with variable a we rederive the scheme usig Taylor approx. u(x, t + t) = u(x, t) + tu t t2 u tt t3 u ttt + Next we covert time derivatives ito space derivatives usig the PDE: u t = au x This gives u tt = (au x ) t = a t u x au xt = a t u x a(u t ) x = a t u x a( au x ) x u tt = a t u x + a(au x ) x We substitute ito the Taylor approximatio... u(x, t + t) = u(x, t) a tu x t2 [ a t u x + a(au x ) x ] + O( t 3 ) Discretizatio the gives U +1 j = U j a j t 0xU j x t2 (a t ) 0xU j j x + a j x a j Oe simplificatio is to replace the (a t ) j term: a j t 0xU j x 1( 2 a t ) j t 2 0xU j x xu j x 2 = t a j + 1( 2 a t ) j t 0xU j x +1/2 l ta 0xU j j x We see from U +1 j = (1+ ) 2 U j 1 + (1 2 )U j (1 ) 2 U j+1 that the scheme does ot satisfy a maximum priciple, sice ot all coefficiets are positive. (they do form a covex combiatio). Hece, we may expect the scheme to have oscillatios /15/01

16 Coservatio laws take the form u t + Ø Øx f (u) where u is a fuctio of x ad t ad f is a fuctio of u oly. We have Ø Øx f (u) = f u u x, which we have already studied with f u h a(u) We ca write Lax-Wedroff for this equatio: u t = [f(u)] x u tt = f xt = f tx = [f u u t ] x = [f u f x ] x = [af x ] x u t = [f(u)] x u tt = f xt = f tx = [f u u t ] x = [f u f x ] x = [af x ] x This gives for Lax-Wedroff (usig previous Taylor approximatio): U j +1 = U j t x 0x f U j ( t x )2 x a U j x f U j Typically, i x a U j we replace U j!1/2 by 2 U j + U j!1. 1 Furthermore, substitutig for F j f U j ad A j!1/2 for a U j!1/2 we get U j +1 = U j t 2 x 1 A t j+1/2 x +x F j A t j 1/2 x x F j /15/01

17 The coservative form of Burger s (iviscid) equatio ut + uux = 0 is give by ut + ( 1 2u 2 ) x = 0 which gives f(u) = 1 2u 2 As log as o shocks develop, i.e. o characteristics cross or fa out, the solutio is easily obtaied from the characteristics: dx dt = u alog which the (iitial) solutio propagates (is costat). The solutio is the give by. u(x, t) = u 0 (x tu(x, t)) After the characteristics cross the solutio becomes formally multivalued. I certai coditios this eve makes sese physically, such as breakig waves (o the beach), but i may others, such as pressure i a gas, it does ot. I the latter case we eed to pick oe solutio from the set of solutios. Typically oly oe solutio is physically relevat. I the quasiliear case it is kow there is oly oe physically relevat solutio. I geeral we refer to some selected physical solutio as a weak or geeralized solutio. There are several ways to derive this solutio. Oe way is the vaishig viscosity solutio. First we cosider the solutio with small viscosity: u : ut + uux = uxx. The we take the limit for the viscosity ( ) goig to zero: lim d0 u /15/01

18 A alterative way to derive/defie a weak solutio is a solutio that satisfies the equivalet itegral form of the same physical problem. Also ote that at ay time to the left ad to the right of the shock the solutio is well-defied ad at the shock the solutio jumps. So all we eed to kow is the speed with which the shock propagates. It turs out (as we will see) that the shock moves with the average speed across the shock: S = f(ur ) f(u L ) u R u L i u). which is ofte deoted as S = [f] [u] (jump i f divided by jump This coditio for the shock speed is called the Rakie-Hugoiot coditio. Let s cosider the itegral form of the advectio equatio: the chage over a time iterval of the desity or cocetratio u iside some volume is give by the total iflow (per volume) over that time iterval: G u(t + t)dg = G u(t)dg t ØG f(u) $ da d dt G u (t)dg = ØG f(u) $ da Applyig Gauss divergece theorem to the right side gives d dt G u(t)dg = G $ f(u)dg w G u t + $ f(u)dg = 0 I oe dimesio this becomes G u t + f(u) x dg = /15/01

19 This itegral G ut + f(u) x dg = 0 reduces to our differetial equatio if we assume the relatio holds at every poit (arbitrarily small volume) at all times. However, eve if this is ot the case, the itegral equatio still holds. Writig it (more geeral) as G u(t + t)dg = G u(t)dg t t+ t [ ØG f(u) $ da] requires either differetiability i time or i space; just that the itegrals exists. Ay solutio u(x, t) that satisfies this itegral equatio is said to be a weak solutio if the differetial equatio. Now we use the itegral equatio to derive the shock speed. We apply the itegral equatio to a small iterval aroud the shock. We assume the shock is at y(t) where a < y < b. The oedimesioal form of d b dt a u(t)dx = f(u a ) f(u b ) Let I = b a udx = y a udx + b y udx, the we get d dt G u (t)dg = ØG f(u) $ da d dt I = y a u t dx + u L dy dt + b y u t dx u R dy dt = y a f x dx + u L s + b y f x dx u R s d dt I = f (u a ) f(u L ) f(u b ) + f(u R ) + u L s u R s = f(u a ) f(u b ) f(u L ) + f(u R ) + u L s u R s = 0 g f(u R ) f(u L ) = s[u R u L ] is /15/01

20 A alterative derivatio that does t require brigig the d/dt uder the itegral is as follows. We assume a iterval x 1 (t) < y(t) < x 2 (t), which is sufficietly small that we assume u costat o each side of the shock. d x 2 x dt x 1 udx + 2 x1 [f(u)] x dx = 0 w d x 2 dt x 1 udx + f(u(x 2, t)) f(u(x 1, t)) = 0 d x 2 dt x 1 udx = d y x dt x1 udx + 2 y udx = d y dt x1 u 1 x dx + 2 y u 2 dx d ( dt y x 1 )u 1 + d ( dt x 2 y)u 2 = [u 1 u 2 ] dy dt = [u 1 u 2 ]s Fially, this gives (agai) [u 1 u 2 ]s + f(u 2 ) f(u 1 ) = 0 w s = f (u 2 ) f(u 1 ) u 2 u 1 = [ f(u)] [u] Burger s equatio ca be simulated usig Lax-Wedroff for the coservative form U j +1 = U j t 2 x 1 A t j+1/2 x +x F j A t j 1/2 x x F j We ca write the upwid scheme i a aalogous form U j +1 = U j 1 2 t x 1 sga j+1/2 +x F j sg a j 1/2 x F j where a j!1/2 h!xf j!x U j I practice, oe sometimes switches adaptively betwee the two schemes to get the best of both (less dampig ad less oscillatios), followig the work by va Leer /15/01

21 Lax-Wedroff is particularly useful for systems of equatios. The mai reaso is that the stability coditio ivolves oly the magitude of the velocity, ot the sig. Cosider agai ut + [f(u)] x = 0, but with u a vector of ukow fuctios This gives ut = fx ad utt = ftx = (Aut) x = (Afx) x where A is the matrix Øf Øu = Lax-Wedroff the becomes Øf i Øu j ij U j +1 = U j ( t x ) 0x f U j ( t x )2 x A U j x f U j which ca be simplified as before. For the special case that A is costat ad f(u) = Au the scheme becomes U j +1 = U j ( t x ) A 0x U j ( t x )2 A 2 x 2 U j ad we ca apply Fourier aalysis. We substitute (where must be a eigevector of ) U j = e ikj x Û Û A This is a solutio if [ I (I i( t x ) si k xa 2( t x )2 si k xa2 )]Û = 0 For this Û must be a eigevector; let be the correspodig eigevalue The (usig = t/ x) = 1 i sik x 2 2 si k x which is exactly what we get for sigle equatio /15/01

22 Sice = 1 i sik x 2 2 si k x (with = t/ x ) must hold for every eigevalue we have [ 1 g t x [ 1 for all Hece, the scheme is stable if t x [ 1 where is the spectral radius of A, the maximum absolute eigevalue. This geeralizes the cocept of stability to systems of equatios. Note that sice stability for Lax-Wedroff depeds oly o the magitude of the eigevalues, ot o their sig, there is o eed to chage the form of the update with a chage i sig. Chagig the update depedig o the sig would be hard for systems of equatios. We would eed to compute the eigevalues ad eigevectors ad decompose the solutio alog the eigevectors, ad use forward or backward differeces based o the sig of the eigevalues. Lax-Wedroff also has a two-step variat that avoids the computatio of the Jacobia matrix : A U +1/2 j+1/2 = 1 2 U j + U j+1 1( t ) 2 x f U j+1 f U j U +1 j = U j ( t ) x f U +1/2 j+1/2 +1/2 f U j 1/2 It is based o predictig the values at x +1/2 +1/2 j+1/2,t j+1/2 cetral scheme for updatig the ext time level. ad the usig a /15/01

23 Aother popular scheme for u t + au x = 0 is the box scheme P t U +1/2 +1/2 j +U j+1 2 t + a x U +1 j+1/2+u j+1/2 2 x = 0 (for costat a) If ot costat it should be evaluated at +1/2 a a j+1/2. With some work we fid that the scheme is secod order accurate i space ad i time. Whe applied to u t + [f(u)] x = 0 the scheme is writte as t U +1/2 +1/2 j +U j+1 2 t + x F +1 j+1/2+f j+1/2 2 x = 0 where F j h f U j. Although the scheme is implicit, i simple cases it becomes explicit if boudary value is kow. U +1 j+1 = U +1/2 j 1 + j+1/2 1 +1/2 1 j+1/2 U j+1 U+1 j, where +1/2 = a j+1/2. j+1/2 +1/2 t x Assumig a > 0 ad the value of U 0 (t) +1 give, for U 1 all values o the +1 right are kow ad we ca update U 1 explicitly. After that we ca +1 update U 2 explicitly, ad so o. Likewise if a < 0 o the right boudary. For oliear equatios i coservative form the update is slightly more complicated as a oliear equatio has to be solved. For systems of equatios the method becomes truly implicit /15/01

24 Sice ot all coefficiets i the box scheme are positive, it does ot satisfy the maximum priciple. However, it is easy to see that as log as the scheme is applied i the right directio (give ) the CFL coditio is always satisfied. a From Fourier aalysis we fid: (k) = cos 1 2 k x i si 1 2 k x cos 1 2 k x+i si 1 2 k x So (k) = 1 ad arg (k) = 2 arcta ta 1 2 k x l ( 1 2 ) 2 + So we see that the scheme has o dampig ad is absolutely stable as log as we take the discretizatio i the right directio. The box method has a phase advace error for ad a phase lag > 1 error for. [ 1 Sice the scheme does ot satisfy the maximum priciple it also causes oscillatios. Sice the scheme is ot dampig, these oscillatios become mode severe tha for Lax-Wedroff. If these oscillatios become too severe they ca be damped usig a average with. U +1 + (1 )U > /15/01

25 The last scheme we discuss (believe me there are may more) is leap-frog. For the oliear advectio equatio i coservative form u t + [f(u)] x the leap-frog scheme is as follows U j +1 U j 1 2 t + f U j+1 f U j 1 2 x = 0 For u t + au x = 0 it becomes U +1 j = U 1 j a t x U j+1 U j 1 Hece, the method is explicit, but eeds additioal work to start. Assumig U 0 is give, we eed a alterative way to compute U 1. We ca use Lax-Wedroff or ay other coveiet oe-step scheme. After this the leap-frog scheme ca update the further time levels explicitly By ispectio we see that the CFL coditio requires [ 1. I the case f = au (costat a) Fourier aalysis leads to the followig equatio for (k) i si k x = 0 This has the two solutios 1,2 = i si k x! [1 2 si 2 k x] 1/ /15/01

26 The product of the roots equals 1. Hece, for the scheme to be stable, we must require the roots to be complex (so that both have modulus 1) The roots are complex for all k iff 1 2 si 2 k x m 0. Hece [ 1. So Fourier aalysis gives the same coditio as the CFL umber. Note that i the stable case there is o dampig. However, sice there are two solutios for we have oe solutio mode that is spurious. The correct solutio mode correspods to the positive root T = i sik x + [1 2 si 2 k x] 1/2 arg T = arcsi( si k x) l ( 1 2 ) 2 + For the spurious solutio mode we have S l 1[1 + i ] This gives a mode that oscillates ad moves i the wrog directio. I some cases this mode has to be filtered out /15/01

27 The real advatage of the leap-frog method is for (certai) coupled first order equatios: u t + av x = 0 v t + au x = 0 Note that this system of first order equatios is equivalet to the secod order wave equatio (costat ): a u tt + a 2 u xx = 0 It turs out we ca implemet the leap-frog method very efficietly i a so-called staggered scheme. The staggered scheme looks as follows +1-1 V U j-1 j j+1 j-1 j j+1 U j +1 U j t V +3/2 j+1/2 V+1/2 j+1/2 + a V j+1/2 +1/2 V+1/2 j 1/2 x = 0 +1 U j +1 t + a U j+1 x = /15/01

28 This (staggered) discretizatio ca be writte as tu + xv = 0 ad tv + xu = 0 where = a t x ad the two equatios are evaluated at grid poits that are offset by half a cell. I order to apply Fourier aalysis we apply the aalysis to a vector cotaiig both a U compoet ad a V compoet. U j, V j+1/2 = e ikj x (Û, Vˆ) This satisfies the equatios if 1 2i si 1 2 k x Û 2i si 1 2 k x 1 Vˆ = 0 1 2i si 1 2 k x Û for o-zero 2i si 1 2 k x 1 Vˆ = 0 Û Vˆ this meas det 1 2i si 1 2 k x 2i si 1 2 k x 1 = 0 w si k x = 2 2(1 2 2 si k x ) + 1 = 0! = (1 2 2 si k x )! 2 si 1 2 k x si k x! = (1 2 2 s 2 )! 2 s s 2 It is easy to see that for 2 s 2 > 1 oe of the roots will have magitude larger tha 1. So, we must have 2 s 2 [ /15/01

29 ! = (1 2 2 s 2 )! 2 s s 2 For 2 s 2 = 1 we have! = 1. For 2 s 2 < 1 we have! = (1 2 2 s 2 )! 2i s 1 2 s 2, ad 2 = (1 2 2 s 2 ) s 2 (1 2 s 2 ) = 1 For the argumets of the two solutios we have arg! =! arcsi(2 s[1 2 s 2 ] 1/2 ) l! [ ( 1 2 ) 2 + ] So for 2 [ 1 ( s 2 [ 1) the scheme is ot dampig. Note that = 1 correspods to the CFL coditio. The two modes travel i opposite directios with equal speeds. The accuracy of this scheme is better tha the box scheme. The aalysis of amplitude ad phase errors is detailed further i sectio 4.9. Please study carefully! /15/01

30 Some of the umerical schemes discussed require more boudary coditios to start tha give for the PDE. Hece, we must fid meaigful boudary coditios i other ways. This holds, for example, for both the Lax-Wedroff scheme ad the ustaggered leap-frog scheme. Oe possibility is to use the characteristic curves that move ito the domai to defie a boudary coditio cosistet with the physics. Aother possibility is to use coservatio priciples to derive a boudary coditio that gives the same coservatio property for the differece scheme as for the PDE. We also must assess the stability ad accuracy resultig from our choices. Some relatively easy choices are to use schemes that do ot eed iformatio from both sides for the last poit (o the side where o boudary coditio is give) such as the upwid scheme. Aother possibility is to set the first or higher derivatives of uspecified variables to zero o the boudary. To maitai some coservative scheme we could proceed as follows Assume u(0, t) = 0 ad 1 u(x, t)dx = 1 0 u t dx = 1 0 f x dx d dt f x dx = f(u(0, t)) f(u(1, t)) For Lax-Wedroff we could base the boudary value o x j U +1 +1/2 j U j = t[f(u 1/2 ) +1/2 f(u J 1/2 )] /15/01

31 A importat way of choosig ad aalyzig boudary coditios is to cosider what happes to a wave travellig out of the domai. Care must be take ot to create spurious solutios reflectig back ito the domai. A detailed aalysis for oe particular example is give i the book (pp ). Cosider the equatios u t + av x = 0 v t + au x = 0 with boudary coditio v = 0 at x = 0 ad oe additioal bc (to be specified). Exact solutio: waves movig to the right: u R = e ik(x at) ad v R = e ik(x at) waves movig to the left: u L = e ik(x+at) ad v L = e ik(x+at) We cosider solvig these equatios usig the ustaggered leap-frog method. We substitute for possible solutios: U j = j Û ad V j = j Vˆ (ote these are compoets of a vector) We substitute this ito the differece scheme U j +1 U j 1 + a t x V j+1 V j +1 V j 1 + a t x U j+1 V j 1 = 0 U j 1 = 0 Substitutig our solutio above gives ad ( 2 1) ( 2 1) ( 2 1) ( 2 1) Û Vˆ = /15/01

32 ( 2 1) ( 2 1) ( 2 1) ( 2 1) Û Vˆ = 0 For a otrivial solutio we eed det ( 2 1) ( 2 1) ( 2 1) ( 2 1) = 0 g ( 2 1) ( 2 1) 2 = 0 We are iterested i spatial solutios (to defie boudary coditios) so we cosider solutio give some. We fid =! ( ) 1/2!1/2 where = ( 2 1). Cosider ow the solutio for some low frequecy mode We take l e ika t l 1 + ika t; this gives l 2ik x. For small eough we get from =! ( ) 1/2!1/2 that l!(1 )!1/2 This gives (1 ) 1/2 l e ik x which correspods to a right movig wave ad (1 ) 1/2 l e ik x which correspods to a left movig wave. The two waves correspodig to (1 )!1/2 are the spurious modes from the leap-frog method. For each we deote the roots as RT, LT, RS, LS. From the approximate solutios for above we see that RS = RT, ad LS = LT = 1 RT /15/01

33 For the eigevalues from ( 2 1) ( 2 1) ( 2 1) ( 2 1) Û we get for each solutio. Vˆ = 0 Û Vˆ = 1 2 We ca costruct solutios with various spatial behavior ad a give time variatio (from choice of ) from these four solutios. Now suppose that the true left movig wave has uit amplitude ad there is o cotamiatio with a spurious mode (additioal bc). Solutio: U j = V j LT j Û LT VˆLT j + RT Û RT VˆRT + Û RS VˆRS We must pick the boudary coditios such that ad take desired values. Clearly we wat = 0 (o spurious mode) ad = 1 to match the exact solutio. It turs out we ca do this by takig U 0 U 1 = 0 ad V 0 + V 1 = 0 The coditio o V approximates the coditio v = 0 o the left. The first correspods to a zero first derivative. Now we substitute these ito our solutio ad use the relatios for. We use the ormalizatio Û LT = Û RT = Û RS = 1. The first boudary coditio gives: (1 LT ) + (1 RT ) + (1 RS ) = 0 which becomes (relatios for ) 1 (1 RT ) + (1 RT ) + (1 + RT ) = /15/01

34 From the secod boudary coditio, V 0 + V 1 = 0, we get 1 (1 + RT )VˆLT + (1 + RT )VˆRT + (1 RT )VˆRS = 0 Usig the relatio (previously give) for the eigevectors we get (after some further algebraic maipulatio --see book--) 1 (1 + RT ) + (1 + RT ) (1 RT ) = 0 Together with 1 (1 RT ) + (1 RT ) + (1 + RT ) = 0 1 this gives the solutio: = RT ad = 0 For higher dimesioal (spatial) problems several of the schemes ca be exteded relatively easily; however, the stability properties may chage. May specialized schemes exist that approximate upwid schemes. Please study sectio 4.11 carefully /15/01