1. Linear advection equations Let u R be a scalar unknown function of x R and t R + satisfying the Cauchy problem. u t + a(x, t)u x = 0, x R, t > 0,
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1 1 Liear advectio equatios Let u be a scalar ukow fuctio of x ad t + satisfyig the Cauchy problem u t + au x =, x, t >, 1 ux, = u x, where a is a give positive costat, ad u x is a kow fuctio To solve this problem, defie the characteristics ξt; x by We ca trivially fid that Let us calculate d dt u ξt; x, t, d dt ξt; x = a, ξ; x = x ξt; x = at + x d dt u ξt; x, t = u t + u d x dt ξt; x = u t ξ, t + aξ, tu x ξ, t = Hece uξt; x, t = uξ; x, = ux, = u x We ca use the solutio of ξ to write uat + x, t = u x If we set x = at + x, ie, x = x at, we get the solutio formula ux, t = u x at Thus 1 expresses that the iitial fuctio u is trasported with a costat velocity a The same reasoig works if ow a = ax, t where the map x ax, t is Lipschitz cotiuous for all t I this case let u = ux, t satisfy the Cauchy problem u t + ax, tu x =, x, t >, 2 ux, = u x First we observe that this equatio is ot coservative, the iterpretatio of ax, tu is ot the flux of u across a poit Now let ξt; x deote the uique solutio of the ordiary differetial equatio d 3 dt ξt; x = aξt; x, t, ξ; x = x By the chai rule we also ow fid that d dt u ξt; x, t = u t + u d x dt ξt; x = u t ξ, t + aξ, tu x ξ, t = Therefore uξt; x, t = u x I order to get a solutio formula, we must solve x = ξt; x i terms of x, or equivaletly let ζτ; x solve the backward characteristic equatio, 4 d ζτ; x = aζτ; x, t τ, dτ ζ; x = x The d uζτ; x, t τ =, dτ which meas that ux, t = uζ; x, t = uζt; x, = u ζt; x example with ax, t = x, u t + xu x =, ux, = u x The the characteristic equatio is with solutio d dt ξ = ξ, ξ = x, ξt; x = x e t 1 Let us study the simple
2 2 Solvig ξt; x = x i terms of x gives that x = xe t, ad thus ux, t = u xe t Let us look at aother example, x <, 5 ax = x x 1, 1 1 < x I this case the characteristics are straight lies ξt; x = x if x, ad ξt; x = x + t if x 1 If < x < 1 the characteristics are give by x e t t lx, ξt; x = 1 + t + lx, t > lx See Figure 1 for a picture of this I this case a is icreasig i x, therefore the characteristics are t x Figure 1 Characteristics i the x, t plae for 5 ot closer tha they were iitially Sice u is costat alog characteristics, this meas that max u xx, t max x x u x If a is decreasig, such a boud caot be foud, as the ext example shows Let ow 1 x <, 6 ax = 1 x x 1, 1 < x I this case the characteristics are give by x + t, t < x, 1 e t+x x <, t x ξt; x = 1 1 x e t x < 1, x 1 x See Figure 2 for a illustratio of these characteristics Let ow x be i the iterval, 1, ad assume that u is cotiuously differetiable Sice u is costat alog characteristics, also u, t is cotiuously differetiable for all t > By olle s theorem there is some x i the iterval 1 1 x e t, 1 such that u x x, t = u1, t u1 1 x e t, t 1 x e t = u 1 u x 1 x e t
3 t x Figure 2 The caracteristics for 6 We ca let x ad get that there must be some x i the iterval 1 e t, 1 such that u x x, t = u 1 u e t From this we see that the oly boud o the derivative that we ca hope for is of the type max u xx, t e t max x x u x 11 Numerics I If we preted that we do ot have the characteristics, ad still wat to kow the solutio, we ca try to approximate it by some umerical method To this ed we itroduce approximatios to the derivative ux ux x D ux =, x ux + x ux D + ux = ad x ux + x ux x D ux =, 2 x where x is a small positive umber Whe we deal with umerical approximatios, we shall always use the otatio u t to idicate a approximatio to u x, t for some iteger We also use the otatio x = x, x ±1/2 = x ± x 2 Now cosider the case where a = cost > As a semi discrete umerical scheme for 1 we propose to let u solve the ifiite system of ordiary differetial equatios 7 u t + ad u t =, u = u x We eed to defie a approximatio to ux, t for ay x ad t, we do this by liear iterpolatio, 8 u x x, t = u t + x x D u +1 t, for x [x, x +1 We wat to show that; a u x coverges to some fuctio u as x, ad b the limit u solves the equatio If u is cotiuously differetiable, we kow that a solutio to 1 exists ad we ca fid it by the method of characteristics Sice the equatio is liear, we ca study the error e x x, t = ux, t u x x, t I the calculatio that follows, we use the followig D + u D u = xd + D u ad D u +1 = D + u
4 4 Isertig the error ito the equatio, we fid for x x, x +1, Let ext f x be defied by so that t e x + a x e x = t u x a x u x = d dt [u t + x x D u +1 t] ad u +1 t = u t x x D u +1t ad + u t = ad u t ad + u t + a x x D D u +1 t = a xd + D u t + a x x D + D u t = a x x x D + D u t f x x, t = a x x x D + D u t for x [x, x +1, 9 e x t + a e x x = f x Usig the method of characteristics o this equatio gives 1 e x x, t = e x x at, + Hece we get the boud t fx at s, s ds 11 e x x, t sup e x x, + t f x L [,t] x Tryig to boud f x, ote first that f x x, t x D D + u t, so f x teds to zero with x if D D + u is bouded Writig w = D D + u ad applyig D D + to 7, we get w t + ad w =, w = D D + u x Now it s time to use a > To boud w observe that if w w 1, D w Hece if w t w 1 t the d dt w t = ad w t, similarly if for some t, w t w 1 t the w t This meas that if x u x if k D D + u k w t sup k D u k sup u x x Thus w is bouded if u is Lipschitz cotiuous It remais to study e x x, For x [x, x +1 e x x, = u x u x x x x u x +1 u x 2 x max x [x u x,x +1] The we have show the boud 12 u x x, t ux, t x 2 u L + t u L, for all x ad t > Strictly speakig, i order for this argumet to be valid, we must kow that equatio 9 oly has the solutio 1 This brigs us to aother topic
5 5 12 Etropy solutios I Without much extra effort, we ca geeralize slightly, ad we wat to esure that the equatio 13 u t + ax, tu x = fx, t oly has oe differetiable solutio If we let the characteristic curves be defied by 3, a solutio is give by verify this! u ξt; x, t = u x + t fξs; x, s ds I terms of the iverse characteristic ζ defied by 4 this formula reads verify this as well! ux, t = u ζt; x + t f ζτ; x, t τ dτ If u is differetiable, ad f is bouded, this formula gives a differetiable fuctio ux, t Now we ca tur to the uiqueess questio, sice 13 is liear, to show uiqueess meas to show that the equatio with f = ad u = oly has the zero solutio Therefore we cosider u t + ax, tu x = Now let ηu be a differetiable fuctio, ad multiply the above with η u to get = t ηu + a x ηu = ηu t + aηu x a x ηu Assume that η = ad ηu > for u, ad a x x, t < C for all x ad t If ηu, t is itegrable, the we ca itegrate this to get d ηux, t dx = a x x, tηux, t dx C ηux, t dx dt By Growall s iequality ηux, t dx e Ct ηu x dx If u =, the ηu = ad we must have ux, t = as well We have show that if u is a differetiable solutio such that ηu is itegrable for some differetiable fuctio η with η = ad ηu, ad a x is bouded the 13 has oly oe differetiable solutio The reaso for the ame etropy solutio is that ofte the model 13 with f is obtaied by the limit of a more physically realistic model, 14 u ε t + ax, tu ε x = εu ε xx as ε becomes small You ca thik of u ε as the temperature i a log rod movig with a speed a I this case ε is proportioal to the heat coductivity of the rod The equatio 14 has more regular solutios tha the iitial data u recall the parabolic regularity results If we multiply this equatio with a covex fuctio ηu ε we get η u ε t + aη u ε x = ε η u ε u ε x x εη u ε u ε x 2 The fuctio η is ofte called a etropy Usig that η u, we get that 15 η u ε t + aη u ε x a x η u ε ε η u ε u ε x x We ow defie a etropy solutio to be the limit of solutios to 14 as ε Formally, a etropy solutio to 13 should satisfy ηu t + aηu x a x ηu η ufx, t, for all covex fuctios η We shall see later that this is sufficiet to establish uiqueess eve if u is ot assumed to be differetiable
6 6 13 Numerics II Let us for the momet retur to the trasport equatio with a costat speed, viz, u t + au x =, where a > We wat to costruct a fully discrete scheme for this equatio ad the simplest is the explicit Euler scheme, 16 D t +u + ad u =,, ad u = u x Here D t + deotes the discrete forward time differece D+ut t ut + t ut =, t ad u should approximate ux, t, with t = t, We ca rewrite 16 as u +1 = u aλ u u 1, where λ = t/ x 1 Sice a is costat ad the equatio liear, we ca use Neuma stability aalysis Assume that the scheme produces approximatios which coverge to a solutio, i particular, this meas that u must be bouded idepedetly of x ad t We make the asatz the equatio is liear, so we might as well expad the solutio i a Fourier series that u = α e i x Isertig this ito the equatio for u +1 we get α +1 e i x = α e i x λa α e i x α e i 1 x so that If the scheme is stable, the α 1, we fid = α e i x 1 λa1 e i x, α = 1 λa1 cos x + i si x α 2 = 1 + 2λ 2 a 2 2λa1 + 1 λa cos x, which yields that α 1 if λa 1 If a = ax, t the scheme will be Neuma stable if λ max ax, t 1 x,t Cosider ow a scheme for the trasport equatio with a variable speed, 2, where ax, t >, with D t +u + a D u =, t+1 a = 1 ax, t dt t t We wish to establish the covergece of u To this ed, set e = u ux, t, where u is the uique solutio to 2 Isertig this ito the scheme we fid that D t +e + a D e = D t +ux, t + a D ux, t = 1 t t+1 = 1 x t = 1 x t =: t x u t x, t dt + a x t+1 x 1 t x t+1 x 1 x x 1 u x x, t dx ax, t u x x, t u x x, t dtdx x ax, t u xx z, t dz t x t u xt x, s ds dtdx t 1 If othig else is said, you ca safely assume that this is the defiitio of λ
7 7 Assumig ow that u xx ad u tx are bouded, choose M such that max u xx L, u tx L, a L } M The we get the boud M 2 x t x x 1 t+1 t x x + t t dtdx = M 2 x + t 2 Therefore the error will satisfy the followig iequality e +1 e 1 λa + λa e 1 + t M 2 }} 2 Υ x + t If a L λ < 1 this is a CFL-coditio, the Υ is a covex combiatio of e ad e 1, which is less tha or equal to max e, 1} e Takig the supremum over, first o the right, ad the of the left, we get } } sup e +1 sup e M 2 + t x + t 2 We also have that e +1 e 1 λa + λa e 1 t M 2 x + t, }} 2 Υ which implies that Now write ē = sup e Iductively, we the fid that e +1 } if if, the the above meas that } e M 2 t x + t 2 ē +1 ē + t M 2 x + t 2 ē ē M 2 + t 2 x + t = t M 2 x + t, 2 sice e = by defiitio Hece, the approximatios defied by 16 coverges to the uique solutio if u is twice differetiable with bouded secod derivatives We have see that if x ax, t is decreasig o some iterval, the best bouds for u xx ad u xt are likely to be of the form Ce Ct, which meas that the costat M is likely to be large if we wat to study the solutio for large or eve moderate times 14 Etropy solutios II Cosider the Cauchy problem u t + ax, tu x =, x, t >, 17 ux, = u x, where a is a cotiuously differetiable fuctio i this sectio ot assumed to be o-egative ecall that a etropy solutio is defied as the limit of the sigularly perturbed equatios 14 For ay positive ε, u ε satisfies 15, this meas that a limit should satisfy 18 ηuψ t + aηuψ x + a x ηuψ dxdt + ηu xψx, dx, for all test fuctios ψ C [,, ad for all covex η If ux, t is a fuctio i L 1 loc [, which satisfies 18 for all covex etropies η, the u is called a etropy solutio to 17 The poit of this is that we o loger require u to be differetiable, or eve cotiuous Therefore, showig that approximatios coverge to a etropy solutio should be much easier tha showig that the limit is a classical 2 solutio 2 I the sese that 17 should hold for all x ad t
8 8 We are goig to show that there is oly oe etropy solutio Agai, sice the equatio is liear, it suffices to show that u = i L 1 implies u, t = i L 1 To do this, we specify a particular test fuctio Set Ω = [,, ad let ϕ ε x, t = x2 Lt x 1+Lt ω ε x y dy, where w ε is a stadard mollifier, ad L is a costat such that L > a L Ω Note that if t < x 2 x 1 /2L, the ϕ ε, t is a approximatio to the characteristic fuctio of the iterval x 1 + Lt, x 2 Lt Next, choose some T < x 2 x 1 /2L, ad set h ε t = 1 t ω ε s T ds This is a approximatio to the characteristic fuctio of the iterval [, T Fially we choose the test fuctio ψ ε x, t = h ε tϕ ε x, t C Ω Isertig this i the etropy iequality 18 we get 19 ηuϕ ε h εt dxdt + h ε tηu Ω Ω t ϕ εx, t + ax, t x ϕ εx, t dxdt + a x ηuh ε ϕ ε dxdt + ηu ϕ ε x, dx We treat the secod itegral first, ad calculate Therefore t ϕ εx, t = L ω ε x x 2 + Lt + ω ε x x 1 Lt x ϕ εx, t = ω ε x x 2 + Lt ω ε x x 1 Lt t ϕ ε + a x ϕ ε = L + aω ε x x 2 + Lt + L aω ε x x 1 Lt a L ω ε x x 2 + Lt + ω ε x x 1 Lt, sice L is chose to be larger tha a Hece, if ηu, the the secod itegral i 19 is o-positive The first itegral i 19 reads ηuϕ ε x, tω ε t T dxdt = f ε tω ε t T dt, where Ω f ε t = Ω ϕ ε x, tηux, t dx If t u, t is cotiuous as a map from [, with values i L 1, the f is a cotiuous fuctio If f is cotiuous, the ftω ε t T dt ft, as ε I order to esure this, we defie a etropy solutio to have this property, see Defiitio 1 below We have that h ε tϕ ε x, t χ ΠT x, t, where Π T = x, t t T, x 1 + Lt x x 2 Lt} By sedig ε i 19 we the fid that 2 x2 x 1 ηux, dx + T x2 Lt x 1+Lt a x x, tηux, t dxdt x2 LT x 1+LT ηux, T dx,
9 9 which implies that with f t = lim ε f ε t, ie, Growall s iequality the implies f T f f T f + a x L Ω f t = x2 Lt x 1+Lt T ηux, t dx f t dt, 1 + T a x L Ω exp a x L Ω T Choosig ηu = u p for 1 p <, ad sedig x 1 to ad x 2 to, we get 21 u, T Lp u Lp 1 + T a x L Ω e ax L Ω T 1/p, 1 p < Next, we ca let p to get 22 u, T L u L I order to formalize the precedig argumet, we defie Defiitio 1 A fuctio u = ux, t i C[, ; L 1 is called a etropy solutio to the problem u t + ax, tu x =, t >, x, ux, = u x, if for all covex fuctios ηu ad all o-egative test fuctios ϕ C Ω the iequality ηuψ t + aηuψ x + a x ηuψ dxdt + ηu xψx, dx, holds Theorem 1 Assume that a = ax, t is such that a x is essetially bouded, the the problem 2 has at most oe etropy solutio u = ux, t, ad the bouds 21 ad 22 hold emark From the proof of this theorem, we see that if we defie a etropy solutio to satisfy the etropy coditio oly for ηu = u, the we get uiqueess i C[,, L 1 15 Helly s theorem The total variatio of a fuctio o a iterval I is defied as u BV I = ux ux 1 sup all partitios The supremum is take over all fiite partitios x } N =1 of the iterval I, with x < x +1 The space of fuctios of bouded variatio is a Baach space whe equipped with the orm u BV I = u L I + u BV I Now we preset a useful compactess criterio It is ot stated i its optimal form, but what follows will prove very hady i the cotext of scalar coservatio laws Theorem 2 Helly s theorem Let u h } h> be a family of fuctios defied o [, T ] such that u h x, t u h x, s dx C t s + h, for all t, s T, ad sup u h, t BV C, t [,T ] for some costat C which is idepedet of h The there exists a sequece h } N, lim h =, such that u h, t coverges to a fuctio u, t This covergece is i Lip[, T ]; L 1
10 1 16 Numerics III We ow recosider the trasport equatio u t + ax, tu x =, t >, 23 ux, = u x, ad the correspodig differece scheme D t +u + a D u =, with a = 1 t t+1 t ax, t dt, u = 1 x x+1/2 x 1/2 u x dx, where as before, we assume that ax, t I order to have a approximatio defied for all x ad t we defie u x x, t = u for x, t = I := [x 1/2, x +1/2 [t, t +1, where t = t We wish to show that u x coverges to a etropy the oly oe! of 23 Now we do ot use the liearity, ad first prove that u x } x> has a coverget subsequece First we recall that the scheme ca be writte u +1 = 1 a λ u + a λu 1 We aim to use Helly s theorem to show the compactess, so we must estimate the total variatio of u x For t [t, t +1 this is give by u x, t BV = u u 1 We also have that u +1 u +1 1 = 1 a λu + a λu 1 1 a 1λu 1 a 1λu 2 = 1 a λu u 1 + a 1λu 1 u 2 By the CFL coditio a λ 1 for all ad, hece u +1 u a λ u u 1 + a 1 u 1 u 2 Therefore u a λ u u 1 + u +1 = = u u 1 u u 1 a a 1 u u 1 + u 1 u 2 a u u 1 Hece u x, t BV u x, BV u BV To show that u x has a uiformly i x bouded BV orm, we must also show that u x is uiformly bouded i L This is easy sice u +1 is a covex combiatio of u ad u 1, so ew maxima or miima are ot itroduced Therefore the secod coditio of Helly s theorem is assured
11 11 To show the first coditio, assume that s [t, t +1 ad that t is such that t s t The u x x, t u x x, s dx x u +1 u = x a λ u u 1 t a L u u 1 The if s [t, t +1 ad t [t +k, t +k+1 we have +k u x x, t u x x, s dx = t a L u BV m= x +k m= x +k m= u m+1 u m a m λ u m t a L u m u m 1 u m 1 k + 1 t a L u BV t s + t a L u BV Hece, also the first coditio i Helly s theorem is fulfilled ad we have the covergece of a subsequece u x u as x It remais to prove that u is the etropy solutio To do this, start by observig that u +1 = 1 a λu + a λu 1 1 a λu + a λu 1 = 1 a λ u + a λ u 1 This ca be rearraged as D+ t u + a D u, ad as 24 D+ t u + D a u u 1 D a By emark 14, to get uiqueess, it suffices to study the etropy formulatio as i Defiitio 1 oly for ηu = u The operators D, D + ad D+ t satisfy the followig summatio by parts formulae a D b = b D + a, if a ± = or b ± =, a D+b t = 1 t a b b D a t if a = or b = = =1 Let ϕ be a o-egative test fuctio i C Ω ad set ϕ = 1 IJ ϕx, t dxdt I We multiply 24 with t xϕ ad sum over ad Z, usig the summatio by parts formula above to get x t u D t ϕ + x t a u D+ ϕ u 1D a ϕ + x u ϕ =1 =
12 12 Let the left had side of the above be called B x ad set A x = u x ϕ t + a u x ϕ x + a x u x ϕ dxdt + The we have We fid that Ω A x = B x + A x B x A x B x u ϕt D t ϕ dxdt A x B x = =1 I + u ϕt dxdt I + u a ϕ x D + ϕ dxdt, I + u D+ ϕ a a dxdt, I u u 1 ax ϕ dxdt, I a x u ϕ ϕ dxdt, I u 1 a x D a ϕ dxdt, I + u u ϕx, dx I + u ϕx, ϕ dx I u ϕx, dx To show that the limit u is a etropy solutio, we must show that the right had side of 25 ad all the terms vaish whe x becomes small This is dirty work, but someoe has to do it We start with the last term 33 Now u ϕx, ϕ dx = u ϕx, ϕx, dx = I I Next, 32 u u ϕx, dx ϕ L Ω u x u dx I I 1 ϕ L Ω u x u y dxdy I x I where x i ad y are i I, ad chose such that Usig this, we see that ϕ L Ω x u x u y, max u x u y = u x u y x,y I I I u u ϕx, dx x ϕ L Ω u BV
13 13 Next, 31 First observer that Therefore where I D a = D 1 t t+1 t ax D a 1 dxdt x t ax, t dt = 1 x t I I I x t ν ax x, t, a x x, t dxdt a x x, t a x y, s dydsdxdt ν ax x, t = max a x x, t a x y, s x y x, t s t } We must assume that a x is a cotiuous fuctio, the ν ax x, t will vaish as x teds to zero It follows that a x D a ϕ dxdt u L ν a x x, t ϕ dxdt, I u 1 We cotiue with 3, a x u ϕ ϕ dxdt ax L Ω u 1 L ϕx, t ϕy, s dydsdxdt I x t I I a x L Ω u L x x t ϕ x L Ω + t ϕ t L Ω Therefore, I a x u ϕ ϕ dxdt egardig 29, we have that u u 1 a x ϕ dxdt, I a x L Ω u L ϕ x L Ω + λ ϕ t L Ω x support of ϕ t +1 sup a x ϕ dxdt t I t+1 sup a x ϕ dxdt I t u BV sup a x ϕ dxdt I xt u BV a x L Ω ϕ L Ω, Ω u u 1 u u 1 where T is a costat such that suppϕ [, T We cotiue our labors, ad estimate 28 Note ow that sice ϕ has bouded support, all sums are fiite, the sum over is from J to J, with J x L/2, ad the sum over from to N, with N t T The, I u D+ ϕ a a dxdt ϕ x L Ω u L ϕ x L Ω u L t xν a x, t, LT ν a x, t 1,, 1 t I a a dxdt I t+1 t ax, t ax, s dsdxdt,
14 14 To estimate 27, we first ote that prove this yourself! ϕ x x, t D + ϕ ϕ xx L Ω x + ϕ xt L Ω t This meas that 27 u L a L Ω ϕ xx L Ω x + ϕ xt L Ω t T L, where T ad L are as before By ow you should be able to show that 26 tl u L ϕ t L Ω Now the ed is i sight, we estimate the right had side of 25 As before show this! we deduce that ϕt x, t D+ϕ t ϕxt L Ω x + ϕ tt L Ω t Fially!: right had of25 u L ϕ xt L Ω x + ϕ tt L Ω t T L To sum up, what we have show is that for ay test fuctio ϕx, t, Ω u ϕ t + a u ϕ x a x u dxdt + u ϕx, dx = lim A x x lim A x B x =, x if a x is locally cotiuous ad u BV Hece the scheme 23 produces a subseqece which coverges to the uique weak solutio Sice the limit is the uique etropy solutio, ay subsequece will produce a further subsequece which coverges to the same limit, ad thus the whole sequece coverges! If u is bouded, we have see that the scheme 23 coverges at a rate O x to the etropy solutio The sigificace of the above computatios is that we have the covergece for the time beig, without ay rate to the uique etropy solutio, eve if u is oly assumed to be i L 1 BV!
15 15 17 Systems of equatios Now we geeralize, ad let u : + be a solutio of the liear system u t + Au x =, x, t >, 34 ux, = u x, where A is a matrix with real ad differet eigevalues λ i } i=1 We order these such that λ 1 < λ 2 < < λ The matrix A will also have liearly idepedet right eigevectors r 1,, r, such that Ar i = λ i r i Similarly, it has idepedet left eigevectors l 1,, l For k m, l k ad r m are orthogoal, sice λ m r m l k = Ar m l k = r m l k A = λ k r m l k Let l 1 L =, = r 1 r l Normalize the eigevectors so that l k r i = δ i, ie, L = I The λ 1 LA = λ We ca multiply 34 with L from the left to get Lu t + LAu x =, ad itroducig w by u = w, we fid that λ 1 35 w t + w x = λ This is decoupled equatios, oe for each compoet of w, These have the solutios w i t + λ w i i =, for i = 1,, x w i x, t = w i x λ i t, Example The liear wave equatio Now cosider the liear wave equatio; α : + is a solutio of α tt c 2 α xx =, x, t >, αx, = α x, α t x, = β x, where c is a positive costat Defiig implies that The matrix u 1 t u c2 2 x = u 2 t u 1 x = u = αt α x, c 2 or u t + u 1 x = c 2 A = 1
16 16 has eigevalues ad eigevectors Thus λ 1 = c, r 1 = = c 1, λ 2 = c, r 2 = c c, L = 1 = c Hece we fid that w1 = 1 u1 + cu 2 w 2 2c u 1 + cu 2 Writig the solutio i terms of α x ad α t, we fid that Therefore, c 1 1 c 1 c α t x, t + cα x x, t = β x + ct + cα x + ct α t x, t + cα x x, t = β x ct + cα x ct α x x, t = 1 2 α x + ct + α x ct + 1 2c β x + ct β x ct, α t x, t = 1 2 β x + ct + β x ct + c 2 α x + ct α x ct To fid α, we ca itegrate the last equatio i t, αx, t = 1 2 α x + ct + α x ct + 1 2c x+ct x ct β y dy, after a chage of variables i the itegral ivolvig β This is the famous D Alembert formula for the solutio of the wave equatio 171 The iema problem eturig to the geeral case, we ote that the solutio to 34 u = ux, t ca be writte as ux, t = [l i u x λ i t] r i i=1 w i x, tr i = i Now we shall look at a type of iitial value problem where u is give by two costat values, viz, u l x <, 36 u x = u r x, where u l ad u r are two costats vectors This type of iitial value problem is called a iema problem, amed after a Germa mathematicia called iema For a sigle equatio = 1, the weak solutio to this iema problem reads u l x < λ 1 t, ux, t = u x λ 1 t = u r x λ 1 t Note that u is ot cotiuous, evertheless it is the uique etropy solutio to 36 check this! For two equatios = 2, we write 2 2 u l = [l i u l ] r i u r = [l i u r ] r i i=1 To help you stay awake, the otatio is deliberately cofusig A subscript r or l refers to right or left A symbol ot subscript r i or l i here i = 1 or i = 2 refers to the left ad right eigevectors of A Usig the that we ca fid the solutio of each compoet separately, [l 1 u l ] x < λ 1 t, [l 2 u l ] x < λ 2 t, [l 1 ux, t] = [l 2 ux, t] = [l 1 u r ] x λ 1 t, [l 2 u r ] x λ 2 t i=1
17 17 Combiig this we see that [l 1 u l ] r 1 + [l 2 u l ] r 2 = u l x < λ 1 t, ux, t = [l 1 ux, t] r 1 + [l 2 ux, t] r 2 = [l 1 u r ] r 1 + [l 2 u l ] r 2 =: u m, tλ 1 < x tλ 2, [l 1 u r ] r 1 + [l 2 u r ] r 2 = u r x λ 1 t, We see that i r 1, r 2 coordiates, u l = wl z l, u m = wr z l, ad u r = wr z r We ca also view the solutio i phase space We see that for ay u l ad u r, ux, t = u l for x < λ 1 t ad ux, t = u r for x λ 2 t I the middle ux, t = u m for λ 1 t x < λ 2 t The middle value u m is o the itersectio of the lie through u l parallel to r 1 ad the lie through u r parallel to r 2 See Figure 3 u r t r 1 r 2 u m u l u l u r x u m Figure 3 The solutio of the iema problem Left: i x, t space, right: i phase space Now we ca fid the solutio to the iema problem for ay, u l x < λ 1 t, ux, t = u i λ i t x < λ i+1 t, i = 1,, 1, u r λ t x, where u i = i [l u r ] r + =1 =i+1 [l u l ] r Observe that this solutio ca also be viewed i phase space as the path from u = u l to u = u r obtaied by goig from u i 1 to u i o a lie parallel to r i for i = 1,, This viewpoit will be importat whe we cosider oliear equatios 172 Numerics for liear systems with costat coefficiets If λ i >, the we kow that the scheme D t +w i, + λ i D w i, =, will produce a sequece of fuctios w i, x } which coverges to the uique etropy solutio of Similarly, if λ i <, the scheme w i t + λ w i i x = D t +w i, + λ i D + w i, =
18 18 will give a coverget sequece Both of these schemes will be coverget oly if t x λ i I eigevector coordiates, the resultig scheme for u reads 37 D t +w + Λ + D w + Λ D + w =, where λ 1 λ 1 Λ =, ad Λ + = λ λ Multiplyig by L from the left, ad usig that u = w yields 38 D t +u + A + D u + A D + u =, where A ± = LΛ ± If the CFL-coditio t x max λ i = max λ 1, λ } i holds, the the scheme 38 will produce a coverget sequece The limit u will be the uique etropy solutio to 39 w t + Λw x =, where w = Lu ad Λ = Λ + + Λ The meaig of a etropy solutio to the equatio writte i characteristic variables is that for some covex fuctio ˆηw, the etropy solutio should satisfy ˆηw t + Λˆηw x, i the weak sese A etropy solutio to 34 is the limit if such a limit exists of the parabolic reqularizatio u ε t + Au ε x = εu ε xx as ε To check if we have a covex etropy η :, we take the ier product of the above with ηu ε to get η u ε t + η u ε Au ε x ε η u ε u ε x x, by the covexity of η We wat to write the secod term o the left as the x derivative of some fuctio qu ε Usig q u ε x = q u ε u ε x, we see that if this is so, the q = η a i for = 1,, u u i i This is equatios for 2 ukows η ad q Thus we caot expect ay solutio if > 2 However we have a solutio if 2 q = 2 q, u k u u u k or 2 η a ik = 2 η a i for 1, k u i i u u i i u k Oe importat example which satisfies this is ηu = u 2 /2, sice η uiu k = Hece, a etropy usig the etropy ηu = u /2 solutio satisfies This meas that u 2 t + qu x weakly u, t L 2 u L 2, ad thus there is at most oe etropy solutio to 34
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