Compressible Euler equations with second sound: asymptotics of discontinuous solutions. Beixiang Fang Reinhard Racke

Universität Konstanz Compressible Euler equations with second sound: asymptotics of discontinuous solutions Beixiang Fang Reinhard Racke Konstanzer Schriften in Mathematik Nr. 306, August 0 ISSN 430-3558 Fachbereich Mathematik und Statistik Universität Konstanz Fach D 97, 78457 Konstanz, Germany

Compressible Euler equations with second sound: asymptotics of discontinuous solutions Beixiang Fang # and Reinhard Racke* # Department of Mathematics, Shanghai Jiao Tong University Shanghai 0040, P.R.China; E-mail: bxfang@sjtu.edu.cn *Department of Mathematics and Statistics, University of Konstanz 78457 Konstanz, Germany; E-mail: reinhard.racke@uni-konstanz.de Abstract: We consider the compressible Euler equations in three space dimensions where heat conduction is modeled by Cattaneo s law instead of Fourier s law. For the arising purely hyperbolic system, the asymptotic behavior of discontinuous solutions to the linearized Cauchy problem is investigated. We give a description of the behavior as time tends to infinity and, in particular, as the relaxation parameter tends to zero. The latter corresponds to the singular limit and a formal convergence to the classical i.e. Fourier law for the heat flux - temperature relation Euler system. We recover a phenomenon observed for hyperbolic thermoelasticity, namely the dependence of the asymptotic behavior on the mean curvature of the initial surface of discontinuity; in addition, we observe a more complex behavior in general. AMS Subject Classification: 35B40, 35Q3 Keywords: discontinuous solutions, asymptotic behavior, curvature, hyperbolic heat conduction Introduction The compressible Euler system with Cattaneo s law for heat conduction consists of the equations t + div u = 0,. u t + div u u + I = 0,. E t + div E + pu + q = 0,.3 τq t + K e + q = 0,.4 where, u = u, u, u 3 T, E, p, q = q, q, q 3 T, e represent density, velocity, total energy, pressure, heat flux, and internal energy, respectively, cf. []. Functions depend on the time variable t R + and on the space variable x R 3 Cauchy problem. I is the identity matrix, K is a given positive definite matrix which we assume to equal I with > 0, and τ > 0 is the relaxation parameter. The relations p = γ e, E = u + e.5 are assumed to hold with a constant γ > polytropic gases. The case τ = 0 corresponds to Fourier s law of heat conduction, while τ > 0 represents Cattaneo s law.

We have initial conditions t = 0 = 0, ut = 0 = u 0, et = 0 = e 0, qt = 0 = q 0,.6 and we shall be interested in the asymptotic behavior of solutions to the corresponding linearized system cf...8 below with initial data that may have jumps on an initial surface σ given by σ = { x Ω 0 R 3 Φ 0 x = 0 }.7 as the level set of a C -function with Φ 0 : Ω 0 R 3 R 3,.8 Φ 0 on Ω 0,.9 defined on an open set Ω 0 R 3, which is a neighborhood of the surface σ. Assumption.9 is made without loss of generality, Φ 0 may be taken as the distance function Φ 0 x = dist x, σ, cp. [4]. The classical Euler system corresponding to τ = 0 has been widely investigated see for example [, 3] and the references therein. We are interested in describing the propagation of initial jumps in the data as time t tends to infinity, and, in particular, in giving information on the behavior of the propagating jumps as τ 0. That is, we shall describe the effect of the singular limit of the hyperbolic system to the usual hyperbolic-parabolic one. This kind of analysis has been carried out for the system of hyperbolic resp. hyperbolicparabolic thermoelasticity in [8] in one space dimension, and in [9] in three or two space dimensions for an extension to compressible Navier-Stokes equations cf. [6]. In the latter case, an interesting relation between the geometry of the initial surface of discontinuity σ and the asymptotic behavior was found saying that the behavior strongly depends on the mean curvature of σ. We shall demonstrate that, again, the mean curvature may determine the specific asymptotic behavior; but, in addition, we shall investigate the, in general, more complex behavior. The linearized system..8 is obviously more complicated. Indeed, the linearized equations of thermoelasticity in [9] can be considered as a special case of..8 with u 0, under the notation correspondence indicated in Section 3.. However, it turns out that the dominating terms describing the asymptotic behavior of the propagation of the initial jumps are basically the same as the ones for thermoelasticity in [9], and the mean curvature of the surface of initial discontinuities plays an important role. In using expansions with respect to the relaxation parameter τ, we shall obtain that the jumps evolving on the characteristic surfaces go to zero for the internal energy e or, equivalently, the temperature for polytropic gases as τ 0, a mirror of the singular limit and the approach of the hyperbolic-parabolic system. As in thermoelasticity [9], we notice that the decay of the jumps is faster for smaller heat conductive coefficient which is similar to a phenomenon observed for discontinuous solutions to the compressible Navier-Stokes equations by Hoff [5].

We remark that our three-dimensional discussion immediately carries over to the one- and two-dimensional cases. Of course, for the one-dimensional case, the initial singularity only lies in the origin, so that there will be no mean curvature terms in the dominating terms describing the asymptotic behavior, which is similar as in thermoelasticity [8]. The paper is organized as follows: In Section we shall give the setting in a normalized form of a first-order system, and Section 3 presents the discussion of the asymptotics, the main result being summarized in Theorem 3.7. Linearization and decomposition We rewrite the equation for the conservation of energy.3, using.5, as u t u + u t + t e + u div u + u u+ div eu + p div u + u p + div q = 0. Using. and., we arrive at t e + div eu + p div u + div q = 0, or t e + u e + γ e div u + div q = 0. Now we linearize the equations..3 to the following system for the unknowns, ũ, ẽ, q, with now constant, u, e, q: t + u x + x ũ + u x + x ũ + u 3 x3 + x3 ũ 3 = 0,. t ũ + u x ũ + u x ũ + u 3 x3 ũ + γ e x + γ x ẽ = 0,. t ũ + u x ũ + u x ũ + u 3 x3 ũ + γ e x + γ x ẽ = 0,.3 t ũ 3 + u x ũ 3 + u x ũ 3 + u 3 x3 ũ 3 + γ e x 3 + γ x3 ẽ = 0,.4 t ẽ + u x ẽ + u x ẽ + u 3 x3 ẽ + γ e x ũ + x ũ + x3 ũ 3 + x q + x q + x3 q 3 = 0,.5 t q + τ x ẽ + τ q = 0,.6 t q + τ x ẽ + τ q = 0,.7 t q 3 + τ x 3 ẽ + τ q 3 = 0..8 We have the initial values t = 0 = 0, ũt = 0 = ũ 0, ẽt = 0 = ẽ 0, qt = 0 = q 0,.9 which may have jumps on the initial surface σ given by.8. 3

Introducing the vector V :=, ũ, ũ, ũ 3, ẽ, q, q, q 3 T we may rewrite equations..8 as the following first-order system t V + A x V + A x V + A 3 x3 V + A 0 V = 0,.0 where A := A := A 3 := u 0 0 0 0 0 0 γ e u 0 0 γ 0 0 0 0 0 u 0 0 0 0 0 0 0 0 u 0 0 0 0 0 γ e 0 0 u 0 0 0 0 0 0 τ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 u 0 0 0 0 0 0 0 u 0 0 0 0 0 0 γ e 0 u 0 γ 0 0 0 0 0 0 u 0 0 0 0 0 0 γ e 0 u 0 0 0 0 0 0 0 0 0 0 0 0 0 0 τ 0 0 0 0 0 0 0 0 0 0 0 u 3 0 0 0 0 0 0 0 u 3 0 0 0 0 0 0 0 0 u 3 0 0 0 0 0 γ e 0 0 u 3 γ 0 0 0 0 0 0 γ e u 3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 τ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 A 0 := 0 0 0 0 0 0 0 0 0 0 0 0 0 τ 0 0 0 0 0 0 0 0 τ 0 0 0 0 0 0 0 0 τ 4.,,,

Finally, we decompose the vectors ũ := ũ, ũ, ũ 3 T and q := q, q, q 3 T into their potential and solenoidal parts, ũ = ũ p + ũ s, q = q p + q s, with ũ p = q p = 0, div ũ s = div q s = 0. Correspondingly, the initial data ũ 0 and q 0 are also decomposed, ũ 0 = ũ 0,p + ũ 0,s, q = q 0,p + q 0,s. Since u is a constant vector, we have the identities u ũ p = u ũ p = 0, div u ũ s = 0, which imply a decomposition of the equations..8 into a system for, ũ p, ẽ, q p that has exactly the same structure as..8, and the following system for ũ s, q s, t ũ s + u ũ s = 0,. t q s + τ qs = 0.. As initial conditions we have t = 0 = 0, ũ p t = 0 = ũ 0,p, ẽt = 0 = ẽ 0, q p t = 0 = q 0,p,.3 and ũ s t = 0 = ũ 0,s, q s t = 0 = q 0,s..4 Since. is explicitly solvable, and since. is of a well-known type, cp. [, 7, 0], the propagation of the jumps of the solenoidal parts ũ s and q s can be easily obtained. Thus we only look at the system for, ũ p, ẽ, q p given in..8 with ũ, q replaced by ũ p, q p, with the initial data.3. Remark.. The decomposition will be used in the analysis of possible discontinuities for the eigenvalues λ = 0, ξ u. 3 Asymptotic behavior as t or τ 0 In order to obtain an appropriate representation of solutions V =, ũ p, ẽ, q p T to the first-order system.0 that allows a detailed description of the asymptotic behavior of the solutions as t or τ 0, we first determine, as in [9], the eigenvalues and eigenvectors of the associated matrix Âξ := ξ j A j, j= where ξ = ξ, ξ, ξ 3 T R 3. Then, expansions in the parameter τ will be given, followed by the investigation of the evolution of jumps across the corresponding characteristic surfaces. 5

3. The eigenvalues and their expansions in the parameter τ. We compute the characteristic polynomial of Âξ, P 3 λ, ξ; τ := detλi Âξ = detλi With ξu := ξ u, we directly have that λi ξ j A j = j= j= ξ j A j λ C. 3. λ ξu ξ ξ ξ 3 0 0 0 0 γ e ξ λ ξu 0 0 γ ξ 0 0 0 γ e ξ 0 λ ξu 0 γ ξ 0 0 0 γ e ξ 3 0 0 λ ξu γ ξ 3 0 0 0 0 γ eξ γ eξ γ eξ 3 λ ξu ξ ξ ξ 3 0 0 0 0 τ ξ λ 0 0 0 0 0 0 τ ξ 0 λ 0 0 0 0 0 τ ξ 3 0 0 λ. Let B := λ ξu ξ ξ ξ 3 0 γ e ξ λ ξu 0 0 γ ξ γ e ξ 0 λ ξu 0 γ ξ γ e ξ 3 0 0 λ ξu γ ξ 3 0 γ eξ γ eξ γ eξ 3 λ ξu and Then we have λ ξu ξ ξ ξ 3 γ B := e ξ λ ξu 0 0 γ e ξ 0 λ ξu 0 γ e ξ 3 0 0 λ ξu P 3 λ, ξ; τ = λ 3 det B λ τ ξ det B.. By a direct computation it follows that det B = λ ξu 3 {λ ξu γγ e ξ }, 3. and det B = λ ξu {λ ξu γ e ξ }, 3.3 6

hence P 3 λ, ξ; τ = λ λ ξu { λλ ξu λ ξu γγ e ξ τ ξ λ ξu γ e ξ } := λ λ ξu Qλ, ξ; τ. 3.4 Remark 3.. In the one- and two-dimensional cases, we have similar expressions for the corresponding characteristic polynomials P d d =, : where u, ξ R d. P d λ, ξ; τ = λ d λ ξu d Qλ, ξ; τ, 3.5 It is clear that the polynomial P 3 λ, ξ; τ in 3.4 has eight roots λ j j =,,, 8 with λ = λ = 0, λ 3 = λ 4 = ξu, and λ 5, λ 6, λ 7, λ 8 being the roots to the equation Qλ, ξ; τ = 0. We observe that λ, λ, λ 3, λ 4 are independent of τ, while λ 5, λ 6, λ 7, λ 8 are not. So we turn to the expansion of eigenvalues λ 5, λ 6, λ 7, λ 8 in terms of the parameter τ as τ 0. Let F λ, τ := τqλ, ξ; τ = τλλ ξu{λ ξu γ γe} {λ ξu γ e}. By letting τ = 0 and solving the equation F λ, 0 = {λ ξu γ e} = 0, we obtain the solution λ 0 5,6 with λ 0 5 = ξu γ e. λ 0 6 = ξu + γ e. Note that if ξu = γ e, one of λ 5 and λ 6 is 0, which equals λ and λ. Since we have F λ λ, 0 = λ ξu, F λ λ0 5,6, 0 = ± γ e 0. By the implicit function theorem, we have λ 5,6 = λ 5,6 τ near λ 0 5,6, 0 from the equation F λ, τ = 0. Differentiating F λτ, τ = 0 with respect to τ, we get Thus, we obtain F λ λ τ + F τ = 0. λ 0 = F τ F λ λ0, 0 λ0, 0. 7

Since we obtain and Thus, We conclude F τ = λλ ξu{λ ξu γ γe} F τ λ0 5, 0 = γ e γ eξu γ e, F τ λ0 6, 0 = γ e γ eξu + γ e. λ 50 = γ eξu γ e, λ 60 = γ eξu + γ e. λ 5 τ = λ 0 5 + λ 50τ + Oτ = ξu γ e γ eξu γ eτ + Oτ, λ 6 τ = λ 0 6 + λ 60τ + Oτ = ξu + γ e γ eξu + γ eτ + Oτ. To expand λ 7 and λ 8, we define, with Λ := λ, GΛ, τ := F λ, τ λ4 = τ ξuλ{ ξuλ γ γeλ } Λ { ξuλ γ eλ }. Then G0, 0 = 0 and G G Λ 0, 0 = 0. Also, we have τ 0, 0 =, hence, using the implicit function theorem again, we get τ = τλ near 0, 0 from the equation GΛ, τ = 0, and we conclude G Λ + G τ τ = 0, hence τ 0 = 0, G Λ + G Λ τ τ + G τ τ + τ G τ Λ + G τ τ = 0. Since G Λ 0, 0 =, we have τ 0 =. Hence τλ = Λ + OΛ 3, and then that is, Λ = + OΛ, τ λ = τ + O τ, and we have λ 7 τ = τ + O, λ 8τ = τ + O. Summarizing, we obtain the following lemma on the eigenvalues of the matrix Âξ. 8

Lemma 3.. Let τ > 0 be sufficiently small. Then the characteristic polynomial P 3 λ, ξ; τ of Âξ has eight real roots, depending on τ, as below: λ τ = λ τ = 0, 3.6 λ 3 τ = λ 4 τ = ξu, 3.7 λ 5 τ = ξu γ e γ e ξu γ e τ + Oτ, 3.8 λ 6 τ = ξu + γ e γ e ξu + γ e τ + Oτ, 3.9 λ 7 τ = + O, 3.0 τ λ 8 τ = + O. 3. τ 3. The right and left eigenvectors Next, we compute the eigenvectors corresponding to the above eigenvalues of Âξ. First, we determine the right eigenvectors x = x,..., x 8 T satisfying the equation, assuming ξ =, Then There are two cases. Case. ξu = 0. λi ξ j A j x = 0. 3. j= P 3 λ, ξ; τ = λ 4 {λ λ γγ e τ λ γ e}. The characteristic polynomial and the matrix Âξ coincide with those in the thermoelastic situation in [9], in the sense of the following correspondencethe notations in the paper [9] are placed on the left side of, and the ones for our equations are placed on the right: div u p, up t ũp, θ ẽ, q p q p, α γ e, β γ, δ γ e, γ. Therefore, in order to focus on the new ingredients, we omit the right eigenvectors for this case. Case. ξu 0. Then we have the eigenvalues λ = λ = 0, λ 3 = λ 4 = ξu, and λ 5, λ 6, λ 7, λ 8, which are expressed by 3.8 3.. Case.. For the multiple eigenvalues λ = λ = 0, we have x 5 = 0 by the last three equations of 3., and the first four equations of 3. read ξux ξ x ξ x 3 ξ 3 x 4 = 0, γ eξ x ξux = 0, γ eξ x ξux 3 = 0, γ eξ 3 x ξux 4 = 0, 9

hence {γ e ξu }x = 0. Case... If γ e ξu, then x = 0 and, since ξu 0, we have x = x 3 = x 4 = 0. This implies, by the fifth equation in 3., that ξ x 6 + ξ x 7 + ξ 3 x 8 = 0. Thus, denoting by Φ τ, Φ τ, ξ thinking of ξ = Φ 0 an orthonormal basis of R 3, we have the right eigenvectors for λ = λ = 0 as follows: r = 0 5, Φ T τ T, r = 0 5, Φ T τ T. Case... eigenvectors If γ e = ξu, then the multiplicity of the eigenvalue 0 is 3 and we have three r, r, and r = ξu, γ e ξt, 0, γ e ξ T T. Case.. Consider the eigenvalues λ 3 = λ 4 = ξu. Then the equations 3. imply ξ x + ξ x 3 + ξ 3 x 4 = 0, e ξx + ξx 5 = 0 e x + x 5 = 0, ξ x 6 + ξ x 7 + ξ 3 x 8 = 0, τ x 5 ξ + ξu x 6 x 7 x 8 = 0 τ ξ x 5 + ξuξ x 6 + ξ x 7 + ξ 3 x }{{} 8 = 0, =0 hence, we have x = x 5 = 0, and since ξu 0, x 6 = x 7 = x 8 = 0. Thus, the corresponding right eigenvectors are r 3 = 0, Φ T τ, 0 4 T, r 4 = 0, Φ T τ, 0 4 T. Case.3. Consider the last eigenvalues λ 5, λ 6, λ 7, λ 8 which satisfy the equation Qλ, ξ; τ = λλ ξu{λ ξu γγ e} τ {λ ξu γ e} = 0. The first four equations in 3. yield implying λ ξu x ξ x ξ x 3 ξ 3 x 4 = 0, γ ξ x + λ ξuξ x + ξ x 3 + ξ 3 x 4 γ ξ ξ 5 = 0, The last four equations yield {λ ξu γ e}x γ x 5 = 0. 3.3 γ e λ ξux + λ ξux 5 ξ x 6 + ξ x 7 + ξ 3 x 8 = 0, τ ξ x 5 + λξ x 6 + ξ x 7 + ξ 3 x 8 = 0, 0

implying γ e λ ξux + λ ξu τλ x 5 = 0. 3.4 The determinant Γ corresponding to the linear system 3.3, 3.4 is Γ = λ ξu{λ ξu γγ e} which equals 0 for the eigenvalues λ 5, λ 6, λ 7, λ 8. τλ {λ ξu γ e} = Qλ, ξ; τ, Letting x = γ, we have x 5 = {λ ξu γ e}, thus x = γ λ ξuξ, x 3 = γ λ ξuξ, x 4 = γ λ ξuξ 3, x 6 = λτ ξ x 5 = τλ {λ ξu γ e}ξ, x 7 = τλ {λ ξu γ e}ξ, x 8 = τλ {λ ξu γ e}ξ 3. Hence, the right eigenvectors for λ j j = 5, 6, 7, 8 are r j = γ, γ λ j ξuξ T, {λ j ξu γ e}, T {λ j ξu γ e}ξ T. τλ j One can verify that in case γ e = ξu, one of the above eigenvector coincides with r we obtained already. For the left eigenvectors y = y,..., y 8 satisfying y λi similar analysis and conclude as follows. Case. ξu = 0. ξ j A j = 0, we carry out a As for the right eigenvectors, we have in this case again the same situation as in [9]. Case. ξu 0. Case.. Consider the multiple eigenvalues λ = λ = 0. Case... If γ e ξu, the left eigenvectors are l = 0 5, Φ T τ, l = 0 5, Φ T τ. j= Case... If γ e = ξu, we have three left eigenvectors, l, l, and l = ξu, ξ T, 0, γ τ ξt. Case.. For the multiple eigenvalues λ 3 = λ 4 = ξu, the corresponding left eigenvectors are l 3 = 0, Φ T τ, 0 4, l4 = 0, Φ T τ, 0 4.

Case.3. Consider λ j j = 5, 6, 7, 8. The left eigenvectors are l j = γ e, λ j ξuξ T, γ e {λ j ξu γ e}, {λ j ξu γ e}ξ T. γ eλ j Again one can verify that in case γ e = ξu, one of the above left eigenvectors coincides with l already obtained previously. Summarizing, we obtain the following lemma on the eigenvectors for the matrix Âξ. Lemma 3.3. Suppose τ > 0 is sufficiently small such that the eigenvalues of Âξ are expressed as in Lemma 3.. Then the right eigenvectors of Âξ are r = 0 5, Φ T τ T, r = 0 5, Φ T τ T, r 3 = 0, Φ T τ, 0 4 T, r 4 = 0, Φ T τ, 0 4 T, r j = γ, γ λ j ξuξ T, {λ j ξu γ e}, T {λ j ξu γ e}ξ T, j = 5, 6, 7, 8, τλ j and the left eigenvectors are l = 0 5, Φ T τ, l = 0 5, Φ T τ, l 3 = 0, Φ T τ, 0 4, l 4 = 0, Φ T τ, 0 4, l j = C j γ e, λ j ξuξ T, γ e {λ j ξu γ e}, γ eλ j {λ j ξu γ e}ξ T, j = 5, 6, 7, 8, where C j j = 5, 6, 7, 8 are chosen constants such that the following normalization condition holds: l j r k = {, j = k, 0, j k, j, k = 5, 6, 7, 8. Moreover, by substituting the expansion of λ j j = 5, 6, 7, 8 in terms of τ, we obtain, via a

direct computation, that C 5 = C 6 = C 7 = C 8 = γ + Oτ, 3.5 e γ + Oτ, 3.6 e γ e τ + O τ, 3.7 γ e τ + O τ. 3.8 3.3 The evolution of the initial singularity With the help of the matrices of left resp. right eigenvectors with ξ = Φ 0, l. L := = Lx, R := [ r,..., r 8 ] = Rx, l 8 satisfying LR = I 8 8 and depending on x dependence on Φ 0, we transform the original differential equation.0 for V into an equation for W := LV, t W + LA j R xj W + LA j xj R + LA 0 R W = 0. j= j= Defining and an initial value à 0 := LA 0 R + LA j xj R j= W 0 := LV 0 = LV t=0, we may rewrite this as t W + LA j R xj W + Ã0W = 0, W t=0 = W 0. 3.9 j= In order to describe the evolution of jumps in the initial data, we need the evolution of the initial surface σ, along which jumps are present, σ Σ 0. This is given through the characteristic surfaces, Σ 0 := {t, x Φ 0 t, x := Φ 0 x = 0}, for the eigenvalues λ,, Σ := {t, x Φ t, x := ξu t + Φ 0 x = 0}, for the eigenvalues λ 3,4, Σ k := {t, x Φ k t, x := λ k t + Φ 0 x = 0}, for the eigenvalues λ k k = 5,..., 8. In case that ξu = 0, that is, λ, = λ 3,4, the right and left eigenvectors are the same as in the paper [9], therefore the argument for the evolution of the singularities is also the same. In order 3

to concentrate on the new ingredients, we omit the analysis of this part and refer the reader to [9] for details. Hereafter, we always assume that ξu 0. We also assume that ξu γ e to avoid that λ 5 or λ 6 equals 0. By [f] Σj we denote the jump of f along Σ j, i.e. the difference of the values of the function f on both sides of the surface Σ j, j = 0,, 5, 6, 7, 8. Then an analogous argument to the one in [9] will lead us to the next two results. Lemma 3.4. Let W be a bounded piecewise smooth solution to the Cauchy problem 3.9. Then W, are continuous on Σ 8 k=5 Σ k, W3,4 are continuous on Σ 0 8 k=5 Σ k, and W i i = 5, 6, 7, 8 are continuous on Σ 0 Σ 8 k=5,k i Σ k. That is, [W i ] Σk = 0, i =,, k =, or i = 3, 4, k = 0. [W i ] Σk = 0, i =,, 3, 4, k = 5, 6, 7, 8. [W i ] Σk = 0, i = 5, 6, 7, 8, k = 0,. [W i ] Σk = 0, i, k = 5, 6, 7, 8, i k. This means that, for any i =,..., 8, the singularity of W i can only propagate along the characteristic surface corresponding to λ i. Lemma 3.5. W, are continuous on Σ 0, so are W 3,4 on Σ. That is, [W i ] Σ0 = 0, i =,, [W i ] Σ = 0, i = 3, 4. Remark 3.6. To obtain this lemma, we exploit the property that ũ p q p 0, which comes from the decomposition ũ = ũ p + ũ s and q = q p + q s. Therefore, we present the proof. Proof. By definition, we have W = q p Φ τ, W = q p Φ τ, W 3 = ũ p Φ τ, W 4 = ũ p Φ τ, where Φ τ, Φ τ and Φ 0 are unit vectors perpendicular to each other. We only prove the continuity of W, across Σ 0 : [W i ] Σ0 = 0, i =,. the continuity of W 3,4 across Σ can be analogously verified. We have, via the decomposition q = q p + q s, that q p 0, in the sense of distribution, and also classically away from Σ 0. Therefore, for a bounded domain G R + R 3 with G Σ 0 and any ϕ Cc Ω 3, we have q p ϕ = 0, G 4

which implies [ q p ] Σ0 Φ 0 ϕ = 0. G Σ 0 Hence, since ϕ is arbitrary, we obtain that [ q p ] Σ0 Φ 0 = 0, which yields that [W i ] Σ0 = 0, i =,. In view of the Lemmas 3.4 and 3.5, the remaining problem is to discuss the propagation of the jumps of W k on Σ k for k = 5, 6, 7, 8. The k-th equation k = 5, 6, 7, 8 in 3.9 can be written as 8 8 t W k + LA j R km xj W m = Ã0 km W m. Since λ k LA j R xj Φ 0 j= m= j= m= kk = 0, we have that the operator t + 3 LA j R kk xj is tangential to Σ k = { λ k t + Φ 0 x = 0}. Analogously, since all entries in the k-th line of the matrix λ k I 3 j= LA j R xj Φ 0 = λ k I Λ vanish, we obtain that, when m k, the vector j= 0, LA R km, LA R km, LA 3 R km is orthogonal to the normal direction λ k, Φ 0 T of Σ k, thus the operator LA j R km xj is also tangential to Σ k when m k. Therefore, by applying Lemma 3.4 and Lemma 3.5, we obtain that the jumps [W k ] Σk k = 5, 6, 7, 8 are governed by the the following transport equations: t + LA j R kk xj + Ã0 kk [W k ] Σk = 0, 3.0 j= j= with initial conditions: [W k ] Σk t=0 = [Wk 0] {Φ 0 x=0}. Thus, to determine the behavior of [W k ] Σk, it is essential to study LA 0 R + 3 LA j xj R. j= We first look at the part LA 0 R. For k = 5,..., 8, we have Ã0 kk with Ã0 = LA 0 R kk = τ 8 l kj r jk = τ C k {λ k ξu γ e}, γ eλ k τλ k j=6 5

Thus, by 3.8 3. and 3.5 3.8, we have LA 0 R 55 = γ e + Oτ γ eτ λ 5 {λ 5 ξu + γ eλ 5 ξu γ e} = + Oτ γ e γ eτ λ 5 and similarly Also and similarly { γ eξu γ eτ γ e + Oτ = γ e + Oτ, LA 0 R 77 = LA 0 R 66 = γ e + Oτ. γ eτ + O τ γ eτ λ 7 { } τ + O γ e = τ + O τ = τ + O τ, { τ + O τ + γ e LA 0 R 88 = τ + O τ. Now we turn to the second part LA j xj R for k = 5, 6, 7, 8. Direct computations follow that j= kk } } lk A x r k = l k + u l k + γ el k5 x r k + u l k3 x r 3k + u l k4 x r 4k + γ l k + u l k5 + τ l k6 x r 5k + l k5 x r 6k, lk A x r k = u l k x r k + l k + u l k3 + γ el k5 x r 3k + u l k4 x r 4k + γ l k3 + u l k5 + k τ l k7 x r 5k + l k5 x r 7k, lk A 3 x3 r k = u 3 l k x3 r k + u 3 l k3 x3 r 3k + l k, u 3 l k4 + γ el k5 x3 r 4k + γ l k4 + u 3 l k5 + τ l k8 x3 r 5k + l k5 x3 r 8k. 6

Thus LA j xj R kk = l k A x r k + l k A x r k + l k A 3 x3 r k j= = l k + γ el k5 x r k + x r 3k + x3 r 4k + l k5 x r 6k + x r 7k + x3 r 8k + u l k x r k + l k3 x r 3k + l k4 x r 4k + u l k x r k + l k3 x r 3k + l k4 x r 4k + u 3 l k x3 r k + l k3 x3 r 3k + l k4 x3 r 4k + γ l k x r 5k + l k3 x r 5k + l k4 x3 r 5k + l k5 u x r 5k + u x r 5k + u 3 x3 r 5k + τ l k6 x r 5k + l k7 x r 5k + l k8 x3 r 5k. To get more details on the expressions involving derivatives above, we recall that, for k, j = 5, 6, 7, 8, we have and r k r 3k r 4k r 6k r 7k r 8k = γ λ k Φ 0 u Φ 0, r 5k = {λ k Φ 0 u γ e}, = lj = C j γ e, λ j ξuξ T, We consider first k = 5: k τλ k {λ k Φ 0 u γ e} Φ 0. γ e {λ j ξu γ e}, {λ j ξu γ e}ξ T. γ eλ j λ 5 Φ 0 u = γ e γ e Φ 0 u γ eτ + Oτ, and, for i, j =,, 3, { γ xi r j+5 = xi γ e γ e Φ 0 u } γ eτ + Oτ xj Φ 0 = γ γ e xi x j Φ 0 + Oτ, for i =,, 3, xi r 55 = xi { γ e γ e Φ 0 u γ eτ + Oτ γ e } = γ eγ e xi Φ 0 uτ + Oτ = Oτ. 7

moreover, for i, j =,, 3, { xi r j+55 = xi λ5 Φ 0 u γ e } xj Φ 0 τλ 5 = Thus, with the expression 3.5, τλ 5 { λ 5 Φ 0 u γ e} xi x j Φ 0 x i λ 5 τλ { λ 5 Φ 0 u γ e} xj Φ 0 5 λ 5 Φ 0 u xi λ 5 + xi Φ 0 u xj Φ 0 τλ 5 = γ eγ e xi x j Φ 0 + Oτ γ eγ xi Φ 0 u e Φ 0 u γ e x j Φ 0 + Oτ + γ eγ xi Φ 0 u e Φ 0 u γ e x j Φ 0 + Oτ = γ eγ e xi x j Φ 0 + Oτ. l k + γ el k5 x r k + x r 3k + x3 r 4k =C 5 γ e + {λ5 Φ 0 u γ e} γ γ e Φ 0 + Oτ γ e = Φ 0 + Oτ, l k5 x r 6k + x r 7k + x3 r 8k =C 5 γ e {λ 5 Φ 0 u γ e} γ eγ e Φ 0 + Oτ =Oτ, u j l k xj r k + l k3 xj r 3k + l k4 xj r 4k j= =C 5 λ 5 Φ 0 u γ = Φ0 T D Φ 0 u + Oτ =Oτ, 3 γ e xi Φ 0 u x x i Φ 0 + u x x i Φ 0 + u 3 x3 x i Φ 0 + Oτ i= where the last equality holds because Φ 0 and Φ 0 T D Φ 0 = Φ 0 T = 0. 8

Finally, γ l k x r 5k + l k3 x r 5k + l k4 x3 r 5k = C 5 γ λ 5 Φ 0 u Φ 0 Oτ = Oτ, l k5 u x r 5k + u x r 5k + u 3 x3 r 5k = C 5 γ e {λ 5 Φ 0 u γ e}uoτ = Oτ, τ l k6 x r 5k + l k7 x r 5k + l k8 x3 r 5k = C 5 {λ 5 Φ 0 u γ e} Φ 0 Oτ τ γ eλ 5 = Oτ. Hence, concluding the above computation, we obtain, for k = 5, that j= LA j xj R 55 = γ e Φ 0 + Oτ. 3. For k = 6, with the expression 3.6, we obtain similarly that, for i, j =,, 3, λ 6 Φ 0 u = γ e γ e Φ 0 u + γ eτ + Oτ, xi r j+6 = γ γ e xi x j Φ 0 + Oτ, xi r 56 = γ e γ e xi Φ 0 uτ + Oτ, xi r j+56 = γ γ e xi x j Φ 0 + Oτ and j= LA j xj R 66 = γ e Φ 0 + Oτ. 3. 9

For k = 7, we have, for i, j =,, 3, λ 7 = τ + O, { } γ xi r j+7 = xi λ 7 Φ 0 u xj Φ 0 γ = xi { τ x j Φ 0 + O τ } γ = xi x τ j Φ 0 + O τ, { xi r 57 = xi λ7 Φ 0 u γ e } { } = xi + O τ, τ = O τ { xi r 5+j7 = xi λ7 Φ 0 u γ e } xj Φ 0 τλ 7 Thus, with the expression 3.7, { } τ = xi + O τ + O τ xj Φ 0 τ τ = 3 xi x τ j Φ 0 + O τ. l k + γ el k5 x r k + x r 3k + x3 r 4k =C 7 γ e + {λ7 Φ 0 u γ e} = γ e τ Φ0 + O τ, l k5 x r 6k + x r 7k + x3 r 8k =C 7 = γ e {λ 7 Φ 0 u γ e} Φ 0 + O τ, τ u j l k xj r k + l k3 xj r 3k + l k4 xj r 4k j= =C 7 λ 7 Φ 0 u =Oτ. i= γ γ τ Φ0 + O τ 3 Φ 0 + O τ τ xi Φ 0 u x xi Φ 0 + O τ τ 0

Finally, and γ l k x r 5k + l k3 x r 5k + l k4 x3 r 5k = C 7 γ λ 7 Φ 0 u Φ 0 O τ = Oτ, l k5 u x r 5k + u x r 5k + u 3 x3 r 5k = C 7 γ e {λ 7 Φ 0 u γ e}uo τ = O τ, τ l k6 x r 5k + l k7 x r 5k + l k8 x3 r 5k = C 7 {λ 7 Φ 0 u γ e} Φ 0 O τ γ eλ 7 = O. Hence, concluding the above computation, we obtain, for k = 7, that LA j xj R 77 = j= τ Φ 0 + O τ. 3.3 τ For k = 8, with the expression 3.8, we obtain similarly that, for i, j =,, 3, λ 8 = τ + O, γ xi r j+8 = xi x τ j Φ 0 + O τ, xi r 58 = O τ, xi r 5+j8 = 3 xi x τ j Φ 0 + O τ LA j xj R 88 = j= Φ 0 + O τ. 3.4 τ Summarizing, we have obtained the asymptotic expansions γ e Ã0 kk = γ e Φ 0 + Oτ, for k = 5, 6, τ Φ 0 + O τ, for k = 7, 8. τ For any x 0 σ = {x R 3 Φ 0 x = 0}, let t t, x k t; 0, x 0 = t, x k t; 0, x0, x k t; 0, x0, x k 3 t; 0, x0 3.5

describe the characteristic line of the operator t + 3 j= LA jr kk xj passing through 0, x 0, which lies on the characteristic surface Σ k for k = 5, 6, 7, 8: x k j t; 0, x 0 = LA j R kk t, x, x, x 3, t x k j 0; 0, x 0 = x 0 j, j =,, 3, and define [W k ] Σk t := [W k ] Σk t, x k t; 0, x 0. For k = 5, 6, 7, 8, let D k t; τ = t that is, in view of 3.5, γ γ e D 5 t; τ = et γ γ e D 6 t; τ = et + D 7 t; τ = τ t t τ D 8 t; τ = τ t + τ 0 0 t 0 Ã0 kk x k s; 0, x 0 ds, t 0 t 0 Φ 0 x 5 t; 0, x 0 + Oτ ds, 3.6 Φ 0 x 6 t; 0, x 0 + Oτ ds, 3.7 Φ 0 x 7 t; 0, x 0 + O τ Then we conclude from 3.0 and 3.5 that, for k = 5, 6, 7, 8, { [W k ] Σk t = [Wk 0 ] σ exp t 0 ds, 3.8 Φ 0 x 8 t; 0, x 0 + O τ ds. 3.9 } Ã0 kk xs; 0, x 0 ds = [Wk 0 ] σ e Dkt;τ. 3.30 Remembering the relation V = RW, we compute V =, ũ p, ẽ, q p T as = γ 8 W j, j=5 ũ p = Φ τ W 3 + Φ τ W 4 + Φ 0 ẽ = 8 j=5 8 j=5 {λ j Φ 0 u γ e}w j, q p = Φ τ W + Φ τ W + Φ 0 8 j=5 γ λ j Φ 0 uw j, τλ j {λ j Φ 0 u γ e}w j.

Therefore, in view of Lemma 3.4, we have [ ] Σk = γ [W k ] Σk, for k = 5, 6, 7, 8, [ũ p ] Σk = [ẽ] Σk = Φ 0 γ Φ 0 γ γ e + Oτ[W k ] Σk, for k = 5, 6, τ + O τ[wk ] Σk, for k = 7, 8, ± γ e γ e Φ 0 u γ e τ + Oτ[W k ] Σk, for k = 5, 6 τ + O τ[w k ] Σk, for k = 7, 8, [ q p ] Σk = { Φ 0 ± γ γ e + Oτ[W k ] Σk for k = 5, 6, Φ 0 3 τ + O τ[w k ] Σk for k = 7, 8. Since W = LV, we have Thus we compute, for k = 5, 6: W k = = = and for k = 7, 8: W k = [W 0 k ] 0 = [LV 0 ] 0, W k = l k + j= γ + Oτ e l kj ũ j + l k5 ẽ + 8 l kj q j. j=8 { γ e + λ k Φ 0 u Φ 0 ũ p + γ e {λ k Φ 0 u γ e}ẽ + {λ k Φ 0 u γ e} Φ 0 q } p γ eλ k γ γ e γ e Φ0 ũ p + Oτ + Oτẽ + Oτ Φ 0 q p, Φ 0 ũ p + Oτ, γ γ e { γ eτ + O τ γ e + λ k Φ 0 u Φ 0 ũ p + γ e {λ k Φ 0 u γ e}ẽ + {λ k Φ 0 u γ e} Φ 0 q } p γ eλ k τ γ eτ 3 + O τ Φ 0 ũ p + O τ = γ e + τ = ẽ + O τ τ ẽ τ 3 = τ ẽ + O τ. τ 3 Φ 0 q p + O τ, Φ 0 γ eũ p + q p + γ e τ + O τ 3

Hence, [W 0 k ] σ = [ 0 ] σ γ [ Φ 0 ũ p 0 σ ] + Oτ, k = 5, 6. γ e τ [ẽ 0] σ + O τ, k = 7, 8. Thus we have [ ] Σk = γ [W k ] Σk = γ [W 0 k ] σ e D kt;τ = [ 0 ] σ [ Φ 0 ũ p 0 σ ] + Oτ e Dkt;τ, k = 5, 6, γ e τ [ẽ 0] σ + O τ e D kt;τ, k = 7, 8. [ũ p ] Σk = = Φ 0 γ Φ 0 γ γ e + Oτ[W k ] Σk, k = 5, 6, τ + O τ [Wk ] Σk, k = 7, 8, γ e Φ 0 [ 0 ] σ + [ Φ 0 ũ p 0 σ ] + Oτ e Dkt;τ, γ e k = 5, 6, Φ 0 γ τ [ẽ 0 ] σ + O τ e D k t;τ, k = 7, 8, [ẽ] Σk = = { ± γ e γ e Φ 0 u γ e τ + Oτ[W k ] Σk, k = 5, 6, τ + O τ [Wk ] Σk, k = 7, 8, Oτ e D kt;τ, k = 5, 6, [ẽ 0] σ + O τ e D k t;τ, k = 7, 8. [ q p ] Σk = = { Φ 0 ± γ γ e + Oτ[W k ] Σk, k = 5, 6, Φ 0 3 τ + O τ [W k ] Σk, k = 7, 8, Φ 0 γ γ e ±[ 0 ] σ k = 5, 6, Φ 0 [ẽ 0 ] σ + O τ e D k t;τ, τ k = 7, 8. [ Φ 0 ũ p 0 ] σ + Oτ e D kt;τ γ e 4

Summarizing we obtain the following characteristic behavior of the evolving jumps along Σ k. Theorem 3.7. Suppose that the initial data of V 0 = 0, ũ p 0, ẽ 0, q p 0 T for the system.0 may have jumps on σ = { Φ 0 x = 0 } with Φ 0 x =. Then the jumps will propagate along the characteristic surfaces Σ k k = 5, 6, 7, 8 and the propagation is given by [ ] Σk = [ 0 ] σ [ Φ 0 ũ p 0 σ ] + Oτ e Dkt;τ, k = 5, 6, γ e Oτ e D kt;τ, k = 7, 8. [ũ p ] Σk = γ e Φ 0 [ 0 ] σ + [ Φ 0 ũ p 0 σ ] + Oτ e Dkt;τ, γ e k = 5, 6, Φ 0 O τ e D kt;τ, k = 7, 8. [ẽ] Σk = [ q p ] Σk = Oτ e Dkt;τ, k = 5, 6, [ẽ 0] σ + O τ e D k t;τ, k = 7, 8. Φ 0 γ γ e ±[ 0 ] σ [ Φ 0 ũ p 0 ] σ + Oτ e Dkt;τ, γ e k = 5, 6, Φ 0 [ẽ 0 ] σ + O τ e D k t;τ, k = 7, 8. τ where D k t; τ with k = 5, 6, 7, 8 is given as in 3.6 3.9. That is, as t or τ 0, the propagation of the jumps of V =, ũ p, ẽ, q p T depends on the parameters of the coefficients of the equations.0 and the mean curvature of the initial surface σ: H = Φ0. On the characteristic surfaces Σ 5 and Σ 6, as τ 0, the jumps of, ũ p, q p will remain while the jumps of ẽ will vanish of order Oτ, which shows a smoothing effect in the system.0 when τ 0.. On the characteristic surfaces Σ 5 or Σ 6 resp., as t, the jumps of, ũ p, q p will decay exponentially as long as γ e γ e Φ 0 or γ e + γ e Φ 0 resp. are positive. 3. On the characteristic surfaces Σ 7 and Σ 8, the jumps of V will decay exponentially as τ 0 or t for a fixed small τ > 0. 5

Remark 3.8. The key terms describing the asymptotical behaviors in Theorem 3.7 are similar to the ones for the equations of thermoelasticity with second sound in [9]. Hence, it would be possible to add nonhomogeneous terms or semilinear terms on the right-hand side of the equations..8 and carry out a similar analysis as in [9]. Acknowledgments: The authors were supported by the Chinese-German NSF/DFGproject Partial Differential Equations and Applications in Geometry and Physics, DFG-GZ: 446 CHV 3/67/0-. Beixiang Fang is also supported in part by the NNSF of China under Grant No. 080096 and 0300. References [] Beals, M.: Propagation and Interaction of Singularities in Nonlinear Hyperbolic Problems. Birkhäuser, Boston 989. [] Courant, R. and Friedrichs, K.O.: Supersonic flows and shock waves. Springer-Verlag, New York 976. [3] Dafermos, C.M.: Hyperbolic conservation laws in continuum physics. Grundlehren math. Wiss. 35. Springer-Verlag, Berlin 00. [4] Gilbarg, D. and Trudinger, N.S.: Elliptic Partial Differential Equations of Second Order, Grundlehren math. Wiss. 4, Springer- Verlag, Berlin 983. [5] Hoff, D.: Dynamics of singularity surfaces for compressible, viscous flows in two space dimensions. Commun. Pure Appl. Math. 55 00, 365 407. [6] Li, Z.: Asymptotic behavior of discontinuous solutions to two kind hyperbolic-parabolic coupled equations. Shanghai Jiao Tong University, Ph.D. thesis 00. [7] Métivier, G.: Propagation, interaction and reflection of discontinuities progressing waves for semilinear hyperbolic systems. Amer. J. Math. 989, 39 87. [8] Racke, R. and Wang, Y.-G.: Asymptotic behavior of discontinuous solutions to thermoelastic systems with second sound. J. Anal. Appl. Z. Anal. Anwendungen 4 005, 7 35. [9] Racke, R. and Wang, Y.-G.: Asymptotic behavior of discontinous solutions in 3-d thermoelasticity with second sound. Quart. Appl. Math. 66 008, 707 74. [0] Rauch, J. and Reed, M.: Discontinous progressing waves for semilinear systems. Commun. PDE 0 985, 033 075. [] Sideris, T.C., Thomases, B. and Wang, D.: Long time behavior of solutions to the 3D compressible Euler equations with damping. Commun. Partial Differential Equations 8 003, 795 86. 6