Lectures on Exact Solutions of Landau 1+1 Hydrodynamics Cheuk-Yin Wong Oak Ridge National Laboratory

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1 1 Dene Matter Summer School, Dubna, July 4, 015 Lecture on Exact Solution of Landau 1+1 Hydrodynamic Cheuk-Yin Wong Oak Ridge National Laboratory 1. Introduction. Exact analytical olution diplayed Analogou behavior in higher dimenion 3. Riemann olution 4. Khalatnikov olution 5. Comparion of Landau hydrodynamic with boot-invariant Hwa-Bjorken hydrodynamic

2 Exact 1+1 dimenional hydrodynamical olution 1. Help guide our intuition. Summarize important feature 3. Point out eential element 4. Can be ued to invetigate whether Hwa-Bjorken hydrodynamic i a limiting cae of Landau hydrodynamic

3 Content of Landau Hydrodynamic in z, t 3 1. A uniform energy denity of length z = l initially at ret. An equation of tate p = ɛ/3

4 Boundary condition 4 1. The flow velocity i zero at the center of he lab, v = 0 at z = 0.. The Khalatnikov olution need to match to the Riemann olution at the edge of the ytem.

5 What i the Riemann olution? 5 1. Riemann olution correpond to the ituation when the energy denity and the velocity can be expreed a a function of each other in which x and t do not explicitly appear.. Becaue of thi mutual dependence, the two independent degree of freedom i reduced to a ingle degree of freedom. 3. The Riemann olution give the uperpoition of (i) the propagation of the diturbance with the peed of ound and (ii) the propagation of the fluid element itelf with the flow velocity

6 What i the Khalatnikov olution? 6 1. Repreenting denity by ζ and velocity by y, ζ= 1 ( ) ( ɛ T 4 ln = ln ɛ 0 y= tanh 1 v It i a general explicit olution of (x = z l, t) a a function of (ζ, y).. It i neceary to invert (x(ζ, y), t(ζ, y)) to yield (ζ(x, t), y(x, t)) for the uual hydrodynamical equation. Inverion can be done only numerically. 3. The Khalatnikov olution cannot be applied to t/l < For t/l < 3, only the Riemann olution applie. T 0 )

7 Explicit form of the olution 7 Definition of variou quantitie ζ = 1 4 ln(ɛ/ɛ 0) = ln(t/t 0 ), ɛ/ɛ 0 = (T/T 0 ) 4 = e 4ζ, / 0 = (ɛ/ɛ 0 ) 3/4 = (T/T 0 ) 3 = e 3ζ (i) Riemann olution y= tanh 1 v, v= tanh y. y = ± ζ c. x t = tanh( ζ/c ) c 1 tanh( ζ/c ) c.

8 (ii) Khalatnikov olution ( t(ζ, y) = e ζ χ ( ζ x(ζ, y) = e ζ χ ζ χ(ζ, y) = l 3e ζ ζ y/ 3 ) χ coh y y inh y, ) χ inh y y coh y. e ζ I 0 [ ζ 1 3 y ] dζ, where I 0 (z) i the Beel function of purely imaginary argument, I α (z) = e iαπ/ J α (iz) 8

9 9 Riemann olution only for t/l < 3, ε / ε t / l = 0 t / l = 0.4 t / l = 0.8 t / l = 1. t / l = 1.6 (a) 0 3 y 1 (b) z / l Wong,Sen,Gerhard,Torrieri,Read,PRC90,064907( 14)

10 Solid line, Khalatinkov olution. Dahed line, Riemann olution. 10 ε / ε t / l = 3 3 (a) y t / l = (b) z / l Wong,Sen,Gerhard,Torrieri,Read,PRC90,064907( 14)

11 Solid line, Khalatinkov olution. Dahed line, Riemann olution. 0.4 τ / l =10 (a) 11 T / T τ / l = 10 (b) y / y y Wong,Sen,Gerhard,Torrieri,Read,PRC90,064907( 14)

12 Richke and Gyulay,NPA597,701(1996) 1

13 Richke and Gyulay,NPA597,701(1996) 13

14 Richke and Gyulay,NPA597,701(1996) 14

15 Riemann olution Method of finding the Rieman olution i from Landau and Lifhitz page 503 Fluid Mechnic 15 T 00 t T 10 t + + T 01 x = 0 T 11 x = 0 Becaue T ik depend on each other, we have T 00 t + T 01 x = 0 T 10 t + T 11 x = 0 T 00 T 10 = T 01 T 11 dt 00 dt 11 = dt 01 dt 10

16 16 T 00 = (ɛ + p)u 0 u 0 pg 00 = (ɛ + p)u 0 u 0 p = ɛu 0 u 0 + p(u 0 u 0 1) = ɛu 0 u 0 + pu 1 u 1 T 01 = T 10 = (ɛ + p)u 0 u 1 T 11 = (ɛ + p)u 1 u 1 pg 11 = (ɛ + p)u 1 u 1 + p = ɛu 1 u 1 + p(u 1 u 1 + 1) = ɛu 1 u 1 + pu 0 u 0 We ue p/ɛ = c, u 0 = coh y, u 1 = inh y, we can rewrite T µν and dt µν a T 00 = ɛ(u 0 u 0 + c u 1 u 1 ) = ɛ(coh y + c inh y) dt 00 = dɛ(coh y + c inh y) + ɛ coh y inh y(1 + c )dy T 01 = (ɛ + p)u 0 u 1 = dɛ(1 + c ) coh y inh y dt 01 = dɛ(1 + c ) coh y inh y + ɛ(1 + c )[coh y + inh y]dy T 11 = ɛu 1 u 1 + pu 0 u 0 = ɛ(inh y + c coh y) dt 11 = dɛ(inh y + c coh y) + ɛ coh y inh y(1 + c )dy

17 Contruct dt 00 dt 11 dt 01 dt 10, the dɛdɛ term i 17 dɛdɛ[(coh y + c inh y)(inh y + c coh y) (1 + c ) coh y inh y (1 + c ) coh y inh y] = dɛdɛ[(1 + c 4 ) coh y inh y + c (inh 4 y + coh 4 y) (1 + c + c 4 ) coh y inh y] = dɛdɛ[c (inh 4 y coh y inh y + coh 4 y)] = dɛdɛc (coh y inh y) = c (dɛ) Contruct dt 00 dt 11 dt 01 dt 10, the dydy term i dydy[ɛ coh y inh y(1 + c )ɛ coh y inh y(1 + c ) ɛ (1 + c ) (coh y + inh y) ] = dydyɛ (1 + c ) [4 coh y inh y (coh 4 y + coh y inh y + inh 4 )] = dydyɛ (1 + c ) [ (coh 4 y coh y inh y + inh 4 )] = dydyɛ (1 + c ) (coh y coh ) = dydyɛ (1 + c )

18 Contruct dt 00 dt 11 dt 01 dt 10, the dɛdy term i dɛdy{[(coh y + c inh y)ɛ coh y inh y(1 + c ) +ɛ coh y inh y(1 + c )(inh y + c coh y)] (1 + c ) coh y inh yɛ(1 + c )[coh y + inh y]} = dɛdy(1 + c ) coh y inh yɛ{(coh y + c inh y) +(inh y + c coh y) (1 + c )[coh y + inh y]} = dɛdy(1 + c ) coh y inh y ɛ{(1 + c ) coh y +(1 + c ) inh y (1 + c )[coh y + inh y]} = 0 Finally, we get We get dt 00 dt 11 dt 01 dt 10 = c (dɛ) ɛ (1 + c ) (dy) = 0 c ( dɛ ɛ ) = (1 + c ) (dy) dy = ± c dɛ (1 + c )ɛ = ± c dɛ ɛ + p 18

19 which i equation () of Landau and Lifhitz, page 503. We have now 19 dy = ± c dɛ (1 + c )ɛ = ± 1 c dɛ c (1 + c )ɛ = ± 1 c d ln(ɛ/ɛ 0 ) c (1 + c ) = ± 1 c d{ln[(ɛ/ɛ 0 ) c (1+c ) ]} = ± dζ c The Rieman olution i then y = ± ζ c

20 where we introduce c ζ = {ln[(ɛ/ɛ 0 )(1+c ) ]} = (1 + c ) c ζ = ln[(ɛ/ɛ 0 )] ɛ ɛ 0 = e we can how ( ɛ ɛ 0 ) c /(1+c ) = T T 0 (1+c ) c ζ ζ = ln( T T 0 ) c (1 + c ) ln[(ɛ/ɛ 0)] 0

21 1 The entrophy conervation equation, along u i We take projection along u i the baic hydro equation: Upon multiply by u i and ummed, we get i ik u i T x = 0 k T ik u i = 0 we hall ue ummed convention xk u i T ik x k = 0 {(ɛ + p)u i u k g ik p} u i = 0, x k u i [(ɛ + p)u k ui x + u i (ɛ + p)uk ik p iu ] u k x k i g x = 0, k (ɛ + p)u k ] u k p x k x = 0, k u k (ɛ + p) + (ɛ + p) uk p x k uk xk x = 0, k (ɛ + p) = T, d(ɛ + p) = T d i the entropy denity, u k T x k + T uk x k = 0, T (uk ) x k = 0, conervation of entropy (1) (u k ) x k = 0, conervation of entropy ()

22 The equation perpendicular to u i We take projection perpendicular to u i the baic hydro equation: T ik x k u iu k T kl x l = 0 (3) Upon multiply by u i and ummed, we get Let u expand out the term: ik u i T x k T kl ui u i u k = 0 x l u T kl k = 0 (4) x l ik u i T x k T ik x k = ui u k T kl x l = 0 x k[(ɛ + p)ui u k g ik p] u i u k x l[(ɛ + p)uk u l g kl p] = x k[(ɛ + p)ui u k g ik p] u i x l(ɛ + p)uk u l u k + u i (ɛ + p)u k u l x lu k + g kl u i u k x lp = [(ɛ + p)u k + u i x kui x k(ɛ + p)uk g ik x kp] ui x l(ɛ + p)uk u l u k + u i (ɛ + p)u k u l x lu k + g kl u i u k x lp = [(ɛ + p)u k g ik x kui x kp] + gkl u i u k xlp = 0 (5) Now, dp = dt, (ɛ + p) = T

23 So, we have For i = 1, we have u k u1 T x k 1 T = 0, u 0 u1 T x 0 [(ɛ + p)u k g ik x kui x kp] + gkl u i u k x lp = 0 [T u k g ik x kui x kt ] + gkl u i u k x lt = 0 T u k x ku i + u i u l xlt gik x kt = 0 u k ui T x k gik k T = 0 u k ui T x k i T = 0, a in Landau (4.5) (6) + u1 u1 T x 1 + 1T = 0, uing 1 = g 11 1 = 1 u 0 u1 T x 0 + (u1 T u 1 ) x 1 u 1 T u 1 x 1 + 1T = 0, uing 1 = g 11 1 = 1 u 0 u1 T x 0 + [T (u0 u 0 1)] x 1 u 1 T u 1 x 1 + 1T = 0, u 0 u1 T x 0 u 0 { u1 T x 0 u0 + u0 T x 1 T u0 + x } + T 1 + T u0 u0 x 1 u1 T u 1 x = 0, 1 (u 0 ) (u 1 ) ) = 0 x 1 3

24 4 the olution i u 1 T x 0 T u 1 = T 0 φ x 1 T u 0 = T 0 φ x 0 T u0 + = 0 x 1 dφ = T T 0 u 0 dx 0 + T T 0 u 1 dx 1 (7) Introduce χ, we have χ = φ + T u 0 x 0 T u 1 x 1 ; u 0 = coh y, u 1 = inh y (8) Then dχ = tu 0 d(t/t 0 ) + t(t/t 0 ) coh ydy xu 1 d(t/t 0 ) x(t/t 0 ) coh ydy = (tu 0 xu 1 )d(t/t 0 ) + (T/T 0 )(t inh y x coh y)dy dχ = (t coh x inh)d(t/t 0 ) + (T/T 0 )(t inh y x coh y)dy Therefore, χ = t coh y x inh y) (T/T 0 ) χ y = (T/T 0)(t inh y x coh y) (9)

25 The invere olution i χ t = T 0 T coh y T 0 T χ x = T 0 T inh y T 0 T χ y inh y χ y coh y 5

26 The Khalatnikov Equation for χ We can get an equation for χ. We have the equation along u i, the entropy conervation equation Note that Note that Entropy equation i We have therefore u 0 x 0 u 0 x = (u0, x) 0 (t, x) u 1 x = (u1, t) 1 (t, x) (u 0, x) (t, x) + u1 x 1 = 0 = = = = { (t, x) (u 0, x) (T, y) (t, x) u 0 (t) (x) (t) u 0 (t) 0 (u 1 ) (t) (t) (t) (u 1 (t) u 0 (x) (x) (x) u 0 (x) (x) (x) (u 1 ) (x) (t) (x) (u 1 ) (x) 1 0 (u1, t) (t, x) = (u0 ) (t) = ) (u1 (x) = 0 } (u1, t) (t, x) } { (u 0, x) (T, y) (u1, t) (T, y) = 0 = 0 6

27 Let u et T 0 a a unit for T and add thi unit factor to T back later on thi lead to coh y (T ) (x) (T ) { coh y (T ) which i Landau (4.13). coh y (y) (x) (y) (x) (y) (u 0, x) (T, y) (u1, t) (T, y) = 0 ( coh y, x) ( inh y, t) (T, y) (T, y) inh y (T ) (t) (T ) inh y (T ) inh y (y) (t) (y) (t) (y) = 0 } coh y (y) = 0 (x) (T ) + inh y (y) T { (x coh y t inh y) (x inh y t coh y)} (y) coh y (y) (x) (T ) + inh y (y) (t) (T ) = 0 T { (x coh y t inh y) (x inh y t coh y)} (y) inh y (x) (T ) + coh y (t) (T ) = 0 T { (x coh y t inh y) (x inh y t coh y)} (y) (x inh y t coh y) = 0, T (t) (T ) = 0 7

28 We introduce the tranformation χ t = T 0 T coh y T 0 T χ x = T 0 T inh y T 0 T χ y inh y χ y coh y 8 Hydro equation become (x coh y t inh y) = T 0 T χ y (coh y inh y) = T 0 χ T y χ (x inh y t coh y) = T 0 T (inh y coh χ y) = T 0 T T { T 0 χ T y + T χ 0 T } + T χ 0 T = 0 T 0 T { T 0 χ T y + T χ 0 T } + T 0 χ T = 0 { T 0 T T T = 1 T T = T c T = 1 T c p = 1 T p ɛ ɛ = c χ y + T χ 0 T } + T 0T c χ T = 0

29 9 We get We get finally { T 0 T ζ = ln T/T 0, (T/T 0 ) = e ζ dζ = T 0dT 1 = dt T T 0 T χ T = χ ζ ζ T = χ 1 ζ T χ T = ζ T ζ [ χ T ] = 1 T = 1 T [ χ ζ 1 χ y + T 0 ζ [ χ T ] T χ 1 T ζ T ζ ] = χ 1 ζ T χ ζ χ ζ Thi i jut Landau (4.15a), Khalatnikov equation. 1 T 1 T } + T 0T c ( χ 1 ζ T χ 1 ζ T ) = 0 χ y χ ζ χ c ζ + χ c ζ = 0 χ y χ c ζ + (c 1) χ ζ = 0

30 Khalatnikov olution 30 Firt, we need to eliminate the firt order term by the tranformation χ(ζ, y) = f(ζ)z(ζ, y), χ ζ = f (ζ)z + f(ζ)z (ζ, y) χ ζ = f (ζ)z + f (ζ)z + f(ζ)z (ζ, y) We chooe f according to c (f ) + (c 1)f = 0 f f = c 1 c f(ζ) = e c 1 c ζ 1 c = e c ζ f (ζ) = 1 c f c f (ζ) = ( 1 c ) f c χ(ζ, y) = e 1 c c ζ Z(ζ, y)

31 31 The Khaltnikov equation become, prime i for d/dζ χ(ζ, y) = e 1 c c ζ Z(ζ, y) χ y χ c ζ + (c 1) χ ζ = 0 f Z y c (f Z + f Z + fz ) + (c 1)(f Z + fz ) = 0 f Z y + Z [ c f] + Z [ c f + (c 1)f] + Z[c f + (c 1)f ] = 0 f Z y + Z [ c f] + Z[ c f + (c 1)f ] = 0 Z y c Z + Z [ c f + (c 1)f ] = 0 f Z y c Z + Z[ c ( 1 c c Z y c Z + Z[ ( (1 c ) ) + ( 1 c )(c c 1)] = 0 4c ) + ( (1 c ) )] = 0 c Z y c Z + Z[( (1 c ) )] = 0 4c Z y Z c ζ + Z[((1 c ) 4c )] = 0 (10) We write it a [ Z ζ Z (1 c c y ) )] Z = 0 (11) c Thi i the equation we would like to olve. We make a change of variable. We could ue +ζ or ζ below, but becaue

32 3 our ζ i alway negative, o we chooe to ue ζ below, a argued by Beuf et al. [? ] ( ) ζ α = ζ + c y, β = ζ c y, (1) ζ = 1 (α + β), c y = 1 (α β), (13) ζ = α ζ α + β ζ β = α β dαdβ = dζc dy ( ) ( = ζ α ) β c y = α c y α + β c y ( ) ( = c y α ) β ( c y ) = ( α β ) β = α β ( α ) = 4 β α β (14) We olve thi equation by uing green function (Beuf et al (008)) [ ] [ ] ζ (1 c Z ) Z = 0 c y c [ ] 4 (1 c Z(α, β) ) Z(α, β) = 0 α β c [ ] (1 c Z(α, β) ) Z(α, β) = 0 (15) α β G(α, β) α β 4c [ ] (1 c ) G(α, β) = δ(α)δ(β) (16) 4c

33 33 Conider the following and we would like to chooe a to make it work. G(α, β) = Θ(α)Θ(β)I 0 (a αβ) α G(α, β) = δ(α)θ(β)i 0 (a αβ) + Θ(α)Θ(β)I 0a 1 α 1/ β, ( where I (z) = di 0 (z)/dz, ) = δ(α)θ(β)i 0 (a αβ)) + Θ(α)Θ(β)I 0a( 1 ) β α αβ G(α, β) = δ(α)δ(β)i 0 (a αβ) + δ(α)θ(β)i 0a 1 β 1/ α + Θ(α)δ(β)I 0a( 1 ) β α +Θ(α)Θ(β)I 0 a( 1 β α ) α a(1 ) β + Θ(α)Θ(β)I 0a( 1 1 ) 1 α β 1/ = δ(α)δ(β)i 0 (0) + Θ(α)Θ(β)I 0 a( 1 β α ) α a(1 ) β + Θ(α)Θ(β)I 0a( ) αβ = δ(α)δ(β) + Θ(α)Θ(β)I 0 ( a 4 ) + Θ(α)Θ(β)I a 4 1 αβ (17) So, we have [ ] (1 c αβ G(α, β) ) G(α, β) 4c = δ(α)δ(β) + a 4 Θ(α)Θ(β)I 0 + a [ ] 1 (1 c 4 αβ Θ(α)Θ(β)I (z) ) G(α, β) 4c = δ(α)δ(β) + a 4 Θ(α)Θ(β)I 0 + a [ ] 1 (1 c 4 αβ Θ(α)Θ(β)I (z) ) Θ(α)Θ(β)I(z) Beel equation for I 0 (z) i (Abramowitz page ) 4c z d I ν dz + z di ν dz (z + ν )I ν = 0 (18)

34 34 We chooe z = (1 c ) αβ c a = (1 c ) c z = a αβ (19) The lat three term in (18) give [ αβ + a 4 Θ(α)Θ(β)I 0 + a 4 G(α, β) = Θ(α)Θ(β)I 0 ( (1 c ) αβ) (0) c 1 αβ Θ(α)Θ(β)I (z) [ ] (1 c ) Θ(α)Θ(β)I(z)] = z 4 Θ(α)Θ(β)I 0 + z 4 Θ(α)Θ(β)I 0(z) z 4 Θ(α)Θ(β)I 0(z) (1) which give zero if I 0 i the olution. So, we have Then [ ] (1 c αβ G(α, β) ) G(α, β) = δ(α)δ(β) () with A general olution i 4c 4c G(α, β) = Θ(α)Θ(β)I 0 ( (1 c ) αβ) (3) c Z(α, β) = χ(α, β) = e 1 c c ζ Z(α, β) dα dβ G(α α, β β )F (α, β ) (4) [ ] (1 c αβ Z(α, β) ) Z(α, β) = 4c dα dβ δ(α α )δ(β β )F (α, β ) = F (α, β)

35 35 a required. We change variable: χ(α, β) = e 1 c c ζ dα dβ G(α α, β β )F (α, β ) (5) dαdβ = dζc dy δ(α)δ(β)dαdβ = δ(ζ)δ(c y)dζc dy = 1 δ(ζ)δ(c y) δ(α)δ(β)dαdβ = dζc dy δ(α)δ(β) = δ(ζ)δ(c y) Let u write the Green function for G(ζ, y): Then [ ] (1 c αβ G(α, β) ) G(α, β) = δ(α)δ(β) = δ(ζ)δ(c y) 4c (6) (7) Eq. (16) give { ( ) ( ) } [ ] 1 (1 c G(α, β) ) G(α, β) = δ(ζ)δ(c y) 4 ζ c y 4c { ( ) ( ) } [ ] (1 c G(α, β) 4 ) G(α, β) = δ(ζ)δ(c y) ζ c y 4c (8) χ(ζ, y) = e 1 c c ζ dζ c dy G(ζ ζ, y y ) F (η, y ) (9)

36 36 We introduce and chooe then we have G(α, β) = Θ(α)Θ(β)I 0 ( (1 c ) αβ) c = Θ(α)Θ(β)I 0 ( (1 c ) ζ c c y ) = Θ( ζ + c y)θ( ζ c y)i 0 ( (1 c ) ζ c y ) χ(ζ, y) = e 1 c c ζ χ(ζ, y) = e 1 c c ζ c = G(ζ, y) (30) dζ c dy Θ[ (ζ ζ ) + c (y y )]Θ[ (ζ ζ ) c (y y )] I 0 ( (1 c ) (ζ ζ ) c c (y y ) ) F (ζ, y ) (ζ ζ ) = ζ, ζ = ζ + ζ, dζ = dζ (31) F (ζ, y ) = f(ζ )δ(c y ) (3) dζ Θ[ (ζ ζ ) + c y]θ[ (ζ ζ ) c y]i 0 ( (1 c ) (ζ ζ ) c c y ) f(ζ ) We conider only the region of y > 0. We have from the tep function Θ[ (ζ ζ ) + c (y y )]Θ[ (ζ ζ ) c (y y )], (ζ ζ ) + c y > 0, (ζ ζ ) > c y (ζ ζ ) c y > 0, (ζ ζ ) > c y > c y, ζ ζ = ζ > c y We conider only the region of y > 0, becaue there i complete ymmetry in y. Then Θ[ ζ + c y]θ[ ζ c y]= Θ[ ζ c y + c y]θ[ ζ c y] = Θ[ ζ c y] becaue c y > 0, (33) (34)

37 χ(ζ, y) = e 1 c c ζ dζ Θ[ (ζ ζ ) + c y]θ[ (ζ ζ ) c y] 37 = e 1 c c ζ = e 1 c c ζ = e 1 c c ζ I 0 ( (1 c ) (ζ ζ ) c c y ) f(ζ ) dζ Θ[ (ζ ζ ) c y]i 0 ( (1 c ) (ζ ζ ) c c y ) f(ζ ) dζ Θ[(ζ ζ) c y]i 0 ( (1 c ) (ζ ζ ) c y ) f(ζ ) c y dζ I 0 ( (1 c ) c c (ζ ) c y ) f(ζ + ζ) We need to look for f(ζ + ζ) to atify the boundary condition. the Khalatnikov olution mut atify two boundary condition (35)

38 Boundary condition (1) It mut match to the Rieman olution when ζ = c y, a given by (1). Thi occur when χ(ζ, y) i no longer a function of ζ and y, the cae when there are two degree of frrdom. Intead, χ i a contant. Note that beaue x and t are related to χ by derivative, χ i free up to a contant, and we can chooe thi contant to be zero, when we match to the Riemann olution. 38 In order to atify thi boundary condition, we mut have (i) χ(ζ, y) = 0 at Rieman matching (36) which give zero when χ(ζ, y) = e 1 c ζ c ζ c y dζ I 0 ( (1 c ) (ζ c ) c y ) f(ζ + ζ) (37) ζ = c y or ζ = c y (38) () When y = 0, whatever i ɛ(ζ), the matter i at ret, y = 0 and that occur at x = l. The midpoint i the point of reflection and no motion. Eq.(??) give (ii) at y = 0, we mut have (39) x = e ζ ( χ ζ ) χ inh y y coh y, α = y ζ χ χ x = e coh y = e ζ y y = l χ y = e ζ l y=0 (40)

39 39 We get from (37) that χ(ζ, y) y = y=0 y e 1 c ( y=0 = e 1 c ζ c ζ c y c ζ ( c )I 0 ( (1 c ) dζ I 0 ( (1 c ) (ζ c ) c y ) f(ζ + ζ) (ζ ) c y ) f(ζ + ζ) The term proportional to y inide the bracket i zero at y = 0. So, we are left with ( χ(ζ, y) y = e 1 c c ζ ( c )I 0 ( (1 c ) (ζ ) c y ) y=0 We have therefore f(ζ + ζ) c c = e 1 c c ζ ( c ) f(ζ + ζ) We put thi back into χ, and we get from (37), χ(ζ, y) = e 1 c ζ c ζ f(ζ + ζ) 1 c = e( c +1)(ζ+ζ 1 c ) = e( c +1)ζ e ( 1 c c +1)ζ ) c c c y = e 1 c ζ c ζ = l c e ζ = l c e ζ c y ζ c y ζ c y ζ =c, and y=0 + y integral... ζ =c y, and y=0 ) f(ζ + ζ) ) (41) = e ζ l (4) = e ζ le 1 c c ζ /c (43) dζ I 0 ( (1 c ) (ζ c ) c y ) f(ζ + ζ) dζ I 0 ( (1 c ) c dζ I 0 ( (1 c ) c dζ e 1+c c ζ I 0 ( (1 c ) (ζ ) c y )( 1) le( 1 c c +1)ζ e 1+c c ζ (ζ ) c y )e 1+c c ζ, change ζ ζ c 1 c = le( c +1)ζ e 1+c c ζ (44) c c (ζ ) c y ), the Khalatnikov olution (45)

40 τ / l =10 (a) T / T τ / l = 10 (b) y / y y Wong,Sen,Gerhard,Torrieri,Read,PRC90,064907( 14)

41 Concluion Exact olution of 1+1 dimenional hydrodynamic give a compact form of the olution with the Landau initial condition. Landau hydrodynamic remain different from Hwa-Bjorken hydrodynamic even at very late time 3. Hwa-Bjorken hydrodynmic mut obtain the flow condition not from hydrodynamic but from other type of initial condition.