THE RIEMANN PROBLEM AND THE LIMIT SOLUTIONS AS MAGNETIC FIELD VANISHES TO MAGNETOGASDYNAMICS FOR GENERALIZED CHAPLYGIN GAS

. This paper is concerned with the Euler equations in the magnetogasdynamics for generalized Chaplygin gas. The global solutions to the Rie- mann problems of the Euler equations in the magnetogasdynamics for generalized Chaplygin gas are obtained constructively by using phase plane analysis method. The formation of delta shock wave is studied as magnetic ﬁeld vanishes. The limit behaviors of the Riemann solutions as magnetic ﬁeld vanishes are also obtained.


1.
Introduction. Magnetogasdynamics plays a very important role in engineering physics, astrophysics, nuclear science, plasma physics and many other areas [9]. It can be described mathematically by the Euler equations for the ideal magnetofluid. Since the full governing system for magnetogasdynamics is highly nonlinear and more complicated, it is necessary to study the various simplified models. One of the simplified model is the assumption that the flow wherein magnetic and velocity fields are orthogonal everywhere, such as in [5,10,11,12]. In the present paper, we consider the system of equations which governs the one dimensional unsteady simple flow of an isentropic, inviscid and perfectly conducting compressible fluid, subject to a transverse magnetic field, can be written as the following conservation laws    ρ t + (ρu) x = 0, where ρ ≥ 0, u, p, B = κρ and µ > 0 denote the density, the velocity, the pressure, the transverse magnetic field and the magnetic permeability, respectively. κ is a positive constant.
The system (1) with the equation of state is called the generalized Chaplygin gas magnetogasdynamic system, where s and α are constants. When α = 1, s = 1, it is called a Chaplygin gas, which was introduced by Chaplygin [3], Tsien [19] and von Karman [6] as a suitable mathematical approximation for calculating the lifting force on a wing of an airplane in aerodynamics.
The vanishing pressure limit method was investigated first by Li [8] in 2001, in which he obtained the limit of Riemann solutions of the isentropic Euler equations as pressure vanishes. For the magnetogasdynamic system, in [14], Shen studied the limits of Riemann solutions to the isentropic magnetogasdynamics for polytropic gas. In [15], Shao proved that the Riemann solutions of the isentropic Chaplygin gas magnetogasdynamics equations tends to a delta shock solution to transport equations as pressure and magnetic field vanish. For results on the vanishing pressure limit of Riemann solutions, the reader is referred to the papers ( [1,2,7,16,17,20,22]) and the references cited therein for more details.
In this paper, we are concerned with the Riemann problem for (1) and (2) with the following initial data where (u ± , ρ ± ) are two constant states. The organization of this paper is as follows. In Section 2, we solve the Riemann problem of (1) and (3) with (2) by characteristic analysis in phase plane. In Section 3, we recall the Riemann problem to the generalized Chaplygin gas dynamics system. In Section 4, we discuss the limit of Riemann solutions to (1), (2) and (3) as the magnetic field vanish, i.e. κ → 0.
2. Solutions to the Riemann problem (1) and (3) with (2). For any smooth solution, the system (1) can be written in the form where w = √ c 2 + b 2 is the magneto-acoustic speed, c = p (ρ) is the sound speed, and b 2 = B 2 (ρ)/(µρ) is the Alfven speed. The eigenvalues of system (1) are It follows that the system is strictly hyperbolic. The corresponding right-eigenvectors are By same computations we have where ∇ denotes the gradient with respect to (ρ, u). So, the first and second characteristic fields are genuinely nonlinear.
Since equations (1) and the Riemann data (3) are invariant under uniform stretching of coordinates where k is a constant. We consider self-similar solutions of (1) and (2) Then the Riemann problem is reduced to a boundary value problem of ordinary differential equations at infinity For smooth solutions, the equations (8) can be written as The general solution of (9) is constant states (u, ρ)(ξ) = Constant, (ρ > 0), and the singular solution consists of forward or backward rarefaction waves For a bounded discontinuous solution, the following Rankine-Hugoniot conditi- where [ρ] = ρ − ρ − , etc., and σ is the velocity of the discontinuity. For (11), we have two shock waves ← − S κ and − → S κ as follows The rarefaction waves  The curve − → S κ (u − , ρ − ) is monotonic increasing in the (u, ρ) plane. It has a straight line ρ = 0 as its asymptote. (See Fig.1) Proof. Differentiating the second equation of ← − R κ (u − , ρ − ) with respect to ρ, we get It is easy to see that u → +∞ when ρ → 0. Similarly, for the curve − → R κ (u − , ρ − ), we can obtain u ρ > 0, u ρρ < 0 and it has u → +∞ when ρ → +∞.
Differentiating the second equation of In addition, we obtain that u → −∞ when ρ → +∞. Similarly, for the curve − → S κ (u − , ρ − ), we can obtain u ρ > 0 and it has u → −∞ when ρ → 0 + . For any given (u − , ρ − ), the phase plane (ρ > 0) is divided into five regions. Then, by the analysis method in phase plane, according to the right state (u + , ρ + ) in the different region, we can construct the unique global Riemann solution connecting two constant states (u − , ρ − ) and (u + , ρ + ). As shown in Fig.1.
Theorem 2.2. For Riemann problem (1), (2) and (3), there exists a unique piecewise smooth solution, which consists of constant states, shocks and rarefaction waves. For any given (u + , ρ + ), the Riemann solutions are shown as follows: 3. Riemann problems of the Euler equations for generalized Chaplygin gas. The Riemann problems of the Euler equations modeling isentropic compressible fluids for Chaplygin gas were studied by D. Serre ([13]), Guo, Sheng and Zhang [4], Wang [21], etc. The Riemann problem to the Euler equations are with initial data (3) and the equation of state (2). For smooth solutions in the self-similar plane, we have the backward (forward) rarefaction wave curves For a bounded discontinuous solution, we get the backward (forward) shock wave curves From the above results, we see that the phase plane is divided into five regions (See Fig.2). In addition, we draw an L δ curve, which is determined as follows: For any given state (ρ + , u + ), the Riemann solutions are shown as follows we consider the delta shock wave solution.
The curves of elementary waves.

JIANJUN CHEN AND WANCHENG SHENG
for any φ ∈ C ∞ 0 (R × R + ). With this definition, the delta shock wave solutions can be written as For the delta shock wave solutions, we have the following generalized Rankine-Hugoniot conditions: with initial data: x(0) = 0, w(0) = 0. To guarantee uniqueness, the discontinuity must satisfy the δ−entropy condition and when ρ − = ρ + ,  Fig.3).
Then we take Thus, by Lemma 4.6, when 0 < κ < κ 3 , the Riemann solution consists of a backward shock wave ← − S κ and a forward shock wave Then we take Thus, when 0 < κ < κ 4 , the Riemann solution consists of a backward shock wave ← − S κ and a forward shock wave − → S κ . Denoteκ 0 = min{κ 3 , κ 4 } > 0, we get the result.
(2) lim Proof. In view of (37) and Lemma 4.9, we get Similarly, we get So we obtain (1). Using the Rankine-Hugoniot conditions (11) for ← − S κ and − → S κ and Lemma 3.1, we In addition as ρ − = ρ + and Proof. Using the Rankine-Hugoniot conditions (11), we have and According to the first equation in (44), (45) and Lemma 4.10, we obtain According to the second equations in (44) and (45) and Lemma 4.10, we obtain Therefore, we get The proof is completed.
Theorem 4.12. Let 0 < s < s 0 and (u + , ρ + ) ∈ V(u − , ρ − ). For a fixed κ > 0, the Riemann solution of (1) and (3) with (2) is that constructed in Section 2. Then ρ and ρu converge to the Riemann solution of (13) and (3) in the sense of distributions as κ → 0. The limit functions ρ and ρu are the sums of a step function and a δ-measure with weights and respectively, which form a delta shock wave solution.
Similarly, we have Then we get Then, we obtain that for any test function φ ∈ C 1 0 (−∞, +∞). As done previously for (47), we obtain that By definition, the last term on the right-hand side of (52) equals to Then we get the result.