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Similarly, for the energy, u 3 = 12 ρv 2 + ρe and q3 = (u 3 + p)v : ∂t 1 2 v + e + ∂ y (P(τ, e)v) = 0. 2 The conservation of mass gives nothing new since it was already used to construct the change of variables. With u 1 = ρ and q1 = ρv, we only obtain the trivial equation 1t + 0 y = 0. To complete the system of equations for the unknowns (τ, v, e) we have to involve a trivial conservation law. For example with u 4 ≡ 1 and q4 ≡ 0, we obtain ∂t τ = ∂ y v. We note that in lagrangian variables the perfect gas law is written P = (γ − 1)e/τ .

When u 0 is also of class C 1 , a classical solution is of class C 1 for t ≥ 0. To avoid the related phenomena of propagation with infinite speed, we suppose in addition that u 0 is bounded on R. The linear case First of all let us examine the case in which f is given by the formula f (u) = cu, c being a constant. Then d (u(x + ct, t)) = (u t + cu x )(x + ct, t) = 0. dt Thus, t → u(x + ct, t) is a constant, with value u 0 (x). 1). For all initial data, there therefore exists one and only one solution which has the same regularity.

9) along , is an entropy solution. Finally we obtain the following result. 7 Let u be a piecewise C 1 function in Q, whose discontinuities are carried by the union of Lipschitz curves, pairwise disjoint. 10) on . 10) in detail. We can fix a point of and suppose that u + = u − . Let I be the interval with extremities u − and u + . In choosing k outside of I , we obtain successively two inequalities which, together, express the Rankine–Hugoniot condition. We thus have [ f (u)] dX = . dt [u] Finally we take k ∈ I , that is to say k = au − + (1 − a)u + , a ∈ [0, 1].

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