# Analysis of the Stability of the Riemann Problem for a Simplified Model in Magnetogasdynamics

Authors: Yujin Liu; Wenhua Sun
Journal: Advances in Mathematical Physics
 (2016)

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## Abstract

The generalized Riemann problem for a simplified model of one-dimensional ideal gas in magnetogasdynamics in a neighborhood of the origin(t>0) in the (x,t) plane is considered. According to the different cases of the corresponding Riemann solutions, we construct the perturbed solutions uniquely with the characteristic method. We find that, for some case, the contact discontinuity appears after perturbation while there is no contact discontinuity of the corresponding Riemann solution. For most cases, the Riemann solutions are stable and the perturbation can not affect the corresponding Riemann solutions. While, for some few cases, the forward (backward) rarefaction wave can be transformed into the forward (backward) shock wave which shows that the Riemann solutions are unstable under such local small perturbations of the Riemann initial data.


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## Body

## 1. Introduction

Magnetogasdynamics plays an important role when we study engineering physics and many other aspects (see [1–4] and the references cited therein) and is important for the hyperbolic theory. An inviscid and perfectly conducting compressible fluid, subject to a transverse magnetic field, is given by(1)ρt+divρu=0,ρut+divρu⊗u+pI-μrotH×H=0,ρE+12μH2t+divρuE+up-μu×H×H=0,Ht-rotu×H=0,divH=0,p=fρ,S,where ρ≥0, p,u,S,B≥0, μ, e, and E=e+u2/2 denote the density, pressure, velocity, specific entropy, transverse magnetic field, magnetic permeability, specific internal energy, and specific total energy, respectively. u=(u1,u2,u3) is the velocity of the fluid in the direction of (x1,x2,x3), H=(H1,H2,H3) is the magnetic field in the direction of (x1,x2,x3) and H=μB.Raja Sekhar and Sharma [5] studied the one-dimensional unsteady flow of an isentropic, inviscid, and perfectly conducting compressible fluid, subject to a transverse magnetic field:(2)ρt+ρux=0,ρut+ρu2+p+B22x=0,where B=kρ and k are positive constant [2, 5]. They constructed the Riemann solutions and discussed the interactions of the elementary waves.Shen [6] studied the Riemann problem for (2) further and found that the Riemann solutions converge to the corresponding Riemann solutions of the transport equations by letting both the pressure and the magnetic field vanish.Hu and Sheng [2] investigated the Riemann problem of the one-dimensional inviscid and perfectly conducting compressible fluid(3)τt-ux=0,ut+p+B22μx=0,E+B2τ2μt+pu+B2u2μx=0,with the initial data(4)τ,p,ux,0=τ±,p±,u±,±x>0,under the assumption B=kρ, where τ±, p±, u± are arbitrary constants and τ>0 is the specific volume. They obtained constructively the unique solution of the Riemann problem with the characteristic method.In [7], we removed the above assumption B=kρ and mainly consider the Riemann problem of the one-dimensional unsteady flow of an isentropic, inviscid, and perfectly conducting compressible fluid(5)ρt+ρux=0,ρut+ρu2+p+B22x=0,Bt+Bux=0,where p=Aργ holds for the polytropic gas, A is a positive constant, and γ is the adiabatic constant.Since the governing equations are nonlinear and complex, the results for magnetogasdynamics are less than the conventional gas dynamics while many results are similar to each other [1, 8]. However the contact discontinuity is very different from each other. Unlike the conventional gas dynamics, where the image of the contact discontinuity in the space (τ,p,u) is a straight line parallel to the τ-axis and the projection on the plane (p,u) is a point, the contact discontinuity is a plane curve in (τ,p,u) and the projection on (p,u) is a straight line parallel to the p-axis which causes the fact that the Riemann solutions are more complicated than that of the conventional gas dynamics.The generalized Riemann problem of the conventional gas dynamics is studied by many authors (see [9–14], etc.).In this paper, we mainly consider the generalized Riemann problem of (3) with the initial data(6)τ,p,ux,0=τ0±,p0±,u0±x,±x>0,where τ0±(x), p0±(x), and u0±(x) are arbitrary smooth functions satisfying (7)limx→0±τ0±,p0±,u0±x=τ±,p±,u±. Problem (3) and (6) may be regarded as a perturbation of the corresponding Riemann problem (3) and (4). Naturally, we wonder if the solutions of (3) and (6) are similar to the corresponding Riemann solutions of (3) and (4) in the neighborhood of the origin. According to the different cases of the corresponding Riemann solutions of (3) and (4), we construct the perturbed solutions with the characteristic method. We find that the contact discontinuity appears after perturbation while the corresponding Riemann solution has no contact discontinuity for some case. For most cases, the Riemann solutions are stable and the perturbation can not affect the corresponding Riemann solutions. While, for some few cases, the forward (backward) rarefaction wave can be transformed into the forward (backward) shock wave which shows that the Riemann solutions are unstable under such small perturbations of the Riemann initial data.The paper is organized as follows. Section2 gives the elementary waves and the Riemann solutions of (3) and (4) for our later discussions. According to different cases of the corresponding Riemann solutions of (3) and (4), in Section 3, we construct case by case the perturbed Riemann solutions of (3) and (6).


## 2. Solutions of the Riemann Problem (3) and (4)

First, we give briefly the elementary waves and the Riemann solutions for system (3) and list some notations which are used in subsequent sections. We refer readers to [2] for details.There are three eigenvalues of (3) which are λ-=-(p-epBBτ/μ+eτ/ep)1/2, λ0=0, and λ+=(p-epBBτ/μ+eτ/ep)1/2. If ep>0 and eτ+p>0, they are real and distinct; thus (3) is a strictly hyperbolic system. It is easily shown that the characteristic fields λ± are genuinely nonlinear and the characteristic field λ0 is linearly degenerate.For the polytropic gas, the forward or backward rarefaction wave in the(τ,p,u) space consisting of all the states which can be connected to the state Q0(τ0,p0,u0) on the right is given by(8)R⇄:pτγ=p0τ0γ,u=u0±∫p0pγpτ+B2τ/μγpdp.The contact discontinuity can be expressed as(9)J:u=p+B22μ=σ=0, and it is easy to see that J is a curve with u=Const. in the (τ,p,u) space and the projection on the (p,u) plane is a straight line parallel to the p-axis.The forward or backward shock wave in the(τ,p,u) space consisting of all the states which can be connected to the state Q0(τ0,p0,u0) on the right is given by(10)
S
⇄
:
p
+
θ
2
p
0
+
θ
2
3
B
2
2
μ
+
B
0
2
2
μ
τ
=
p
0
+
θ
2
p
+
θ
2
3
B
0
2
2
μ
+
B
2
2
μ
τ
0
,
u
=
u
0
±
p
+
B
2
2
μ
-
p
0
-
B
0
2
2
μ
-
τ
-
τ
0
p
+
B
2
/
2
μ
-
p
0
-
B
0
2
/
2
μ
1
/
2
,where θ2=γ-1/γ+1 and B0=k/τ0.For convenience and conciseness, denote the projection ofR⇄(S⇄) on the (τ,p) plane and (p,u) plane by Ru(Su) and R⇄τ(S⇄τ), respectively. Denote the contact discontinuity J by J< when pl<pr, τl<τr and J> when pl>pr, τl>τr.In order to construct the Riemann problem of (3) and (4), we denote W←-τ(Q-τ)=R←-τ(Q-τ)∪S←-τ(Q-τ) and W→+τ(Q+τ)=R→+τ(Q+τ)∪S→+τ(Q+τ), where Q-τ and Q+τ are, respectively, the projections of Q- and Q+ on the plane (p,u). Draw W←-τ(Q-τ) from Q-τ and W→+τ(Q+τ) from Q+τ in the plane (p,u), respectively. According to the properties of W←-τ(Q-τ) and W→+τ(Q+τ), they intersect with each other at most once. Therefore, there are five cases: W←-τ(Q-τ)∩W→+τ(Q+τ)=(R←-τ(Q-τ)∩R→+τ(Q+τ)) or (S←-τ(Q-τ)∩R→+τ(Q+τ)) or (R←-τ(Q-τ)∩S→+τ(Q+τ)) or (S←-τ(Q-τ)∩S→+τ(Q+τ)) or ⌀.For the last case, we easily know there is a vacuum solution. In what follows, we just need to consider the first case since the other cases can be studied similarly.SupposeW←-τ(Q-τ)∩W→+τ(Q+τ)=R←-τ(Q-τ)∩R→+τ(Q+τ)={Q∗τ} (see Figure 1); we know there exists (p∗,u∗) satisfying(11)u∗=u-+∫p-p∗γpτ+kB/μγpdp,pτγ=p-τ-γ,u∗=u+-∫p+p∗γpτ+kB/μγpdp,pτγ=p+τ+γ.Figure 1
Elementary waves in the(u,p) plane.Denote(12)f1p1≔u-+∫p1p-γpτ+kBτ/μγpdp,pτγ=p-τ-γ,p1∈0,p-,f2p2≔u+-∫p2p+γpτ+kBτ/μγpdp,pτγ=p-τ-γ,p2∈0,p+,g1p1≔p1+B12τ12μ,g2p2≔p2+B22τ22μ,where τ1 satisfies(13)p1+θ2p-+θ23B12τ12μ+B-22μτ1=p-+θ2p1+θ23B-2τ12μ+B122μτ-,p1>p-,p1τ1γ=p-τ-γ,p1≤p-and τ2 satisfies(14)p2+θ2p++θ23B22τ22μ+B+22μτ2=p++θ2p2+θ23B+2τ22μ+B222μτ+,p2>p+,p2τ2γ=p+τ+γ,p2≤p+.Denote(15)h1p1≔u--p1+B12τ12μ-p--B-22μτ--τ1,h2p2≔u++p2+B22τ22μ-p+-B+22μτ+-τ2, where τ1 satisfies the first equation of (13) and τ2 satisfies the first equation of (14).Let(16)f1p1=f2p2,g1p1=g2p2,(17)f1p1=h2p2,g1p1=g2p2,(18)h1p1=f2p2,g1p1=g2p2.In [2], the authors proved that only one of the above three equations, (16), (17), and (18), is solvable and the solution is unique which implies that there exists a unique contact discontinuity J joining the two states which are located on R⇄ and S⇆, respectively.Next we list case by case the Riemann solutions of (3) and (4) and refer readers to [2] for details.Case 1 (p-τ-γ=p+τ+γ).
It holds thatg1(p∗)=g2(p∗) and p1=p2=p∗ is the solution of (16). Thus, the Riemann solution is R←+R→; here the symbol “+” means “followed by.” We notice that for this case there is no contact discontinuity.Case 2 (p-τ-γ<p+τ+γ).Subcase 2.1  (
f
1
(
0
)
≤
u
+
(see Figure2)). It is obvious that there exists a point p^1∈(p∗,p+) such that f1(0)=f2(p^1) and g1(0)<g2(p^1). It follows that the Riemann solution is R←+J<+R→.Figure 2
f
1
(
0
)
≤
u
+
=
f
2
(
p
+
).Subcase 2.2  (
f
1
(
0
)
>
u
+
(see Figure3)). Since there exists a point p^2∈(0,p∗) such that f1(p^2)=u+, we divide it into two subcases.Figure 3
f
1
(
0
)
>
u
+
=
f
2
(
p
+
).Subcase 2.2.1. If g1(p^2)≤g2(p+), due to h2′(p2)>0 and h2(p+)=f2(p+)<f1(0), considering that the curve h2(p2) has no asymptote, then there exists a point p^∗∈(p+,+∞) such that h2(p^∗)=f1(0). From the smoothness of the curve and g2(p^∗)>g2(0), there exists (p¯1,p¯2)∈(0,p^2)×(p+,p^∗) such that f1(p¯1)=h2(p¯2) and g1(p¯1)=g2(p¯2). Therefore, (p¯1,p¯2) is the point which we seek for and the Riemann solution is R←+J<+R→.Subcase 2.2.2. If g1(p^2)>g2(p+), from the continuity we know that the Riemann solution is R←+J<+S→.Case 3 (p-τ-γ>p+τ+γ).Subcase 3.1  (
u
-
≤
f
2
(
0
)
(see Figure4)). It is obvious that there exists a point p^3∈(p∗,p-) such that f1(p^3)=f2(0) and g1(p^3)>g2(0). And we get the Riemann solution R←+J>+R→.Figure 4
f
1
p
-
=
u
-
≤
f
2
(
0
).Subcase 3.2  (
u
-
>
f
2
(
0
)
(see Figure5)). Since there exists a point p^4∈(0,p∗) such that f2(p^4)=u-, we divide it into two subcases.Figure 5
f
1
p
-
=
u
-
>
f
2
(
0
).Subcase 3.2.1. If g1(p-)≥g2(p^4), the Riemann solution is R←+J>+R→.Subcase 3.2.2. If g1(p-)<g2(p^4), the Riemann solution is S←+J>+R→.From the above discussions, we have the result as follows [2].Theorem 1.
There exists uniquely piecewise smooth solution to the Riemann problem (3) and (4).


## 3. Solutions of the Generalized Riemann Problem (3) and (6)

In what follows, we investigate the solutions for the discontinuous initial value problem (3) with (6) in a neighborhood of the origin (t>0) on the (x,t) plane. From the results in [12], the classical solution (τl,pl,ul)(x,t)((τr,pr,ur)(x,t)) can be defined in a strip domain Dl(Dr) for a local time. The right boundary of Dl has characteristic OA:x=λ-t, and the left boundary of Dr has characteristic OB:x=λ+t (see Figure 6).Figure 6
The discussed region.According to the different cases of the corresponding Riemann solutions of (3) and (4), we construct the solutions case by case for (3) with (6). For simplicity, we only consider some interesting phenomena. For the other cases, a similar discussion to that of the generalized Riemann problem (3) and (6) can be carried out, and we omit them. For simplicity, we use the same symbols after perturbation since there is not any confusion.Case 1 (whenp-τ-γ=p+τ+γ, the corresponding Riemann solution is R←+R→).
After perturbation, we obtain two subcases which are thatp-τ-γ<p+τ+γ or p-τ-γ>p+τ+γ. In what follows, we discuss our problem in two subcases.Subcase 1.1  (
p
-
τ
-
γ
<
p
+
τ
+
γ
).Subcase 1.1.1. If f1(0)≤u+ or f1(0)>u+ and g1(p^2)≤g2(p+) after perturbation, we obtain that the perturbed Riemann solution is R←+J<+R→. For this subcase, the perturbation can not affect the corresponding Riemann solution which shows that the corresponding Riemann solution is stable.Subcase 1.1.2. If f1(0)>u+ and g1(p^2)>g2(p+) after perturbation, we obtain that the perturbed Riemann solution is R←+J<+S→. For this subcase, the perturbation can affect the corresponding Riemann solution which shows that the corresponding Riemann solution is unstable.Subcase 1.2  (
p
-
τ
-
γ
>
p
+
τ
+
γ
).Subcase 1.2.1. If u-≤f2(0) or u->f2(0) and g1(p-)≥g2(p^4) after perturbation, it follows that the perturbed Riemann solution is R←+J>+R→.Subcase 1.2.2. If u->f2(0) and g1(p-)<g2(p^4) after perturbation, it yields that the perturbed Riemann solution is S←+J>+R→.Theorem 2.
For Case 1 (see Figure7), although there is no contact discontinuity of the corresponding Riemann solution, the contact discontinuity appears after perturbation. Furthermore, for Subcases 1.1.1 and 1.2.1, we observe that the corresponding Riemann solution is stable, while for Subcases 1.1.2 and 1.2.2, the backward (forward) rarefaction wave of the corresponding Riemann solution can be transformed into the backward (forward) shock wave after perturbation which shows that the structure of the Riemann solution is unstable.Figure 7
Solutions in Case 1.
(a)
The Riemann solution
(b)
The perturbed solution in the neighborhood of the origin (t>0)Case 2 (p-τ-γ<p+τ+γ).
After perturbation, it still holds thatp-τ-γ<p+τ+γ. In what follows, we discuss in detail our perturbed Riemann problem for this case.Subcase 2.1. If f1(0)≤u+, the corresponding Riemann solution is R←+J<+R→. After perturbation, we obtain that f1(0)≤u+ or f1(0)>u+; we construct the perturbed Riemann solutions as follows.Subcase 2.1.1. When f1(0)≤u+ or f1(0)>u+ and g1(p^2)≤g2(p+) after perturbation, we know the perturbed Riemann solution is still R←+J<+R→.Subcase 2.1.2. When f1(0)>u+ and g1(p^2)>g2(p+) after perturbation, the perturbed Riemann solution is R←+J<+S→.Subcase 2.2. If f1(0)>u+ and g1(p^2)≤g2(p+), the corresponding Riemann solution is R←+J<+R→. After perturbation, we have f1(0)>u+, g1(p^2)≤g2(p+), or g1(p^2)>g2(p+), we construct the perturbed Riemann solutions as follows.Subcase 2.2.1. When f1(0)>u+ and g1(p^2)≤g2(p+) after perturbation, we obtain that the perturbed Riemann solution is still R←+J<+R→.Subcase 2.2.2. When f1(0)>u+ and g1(p^2)>g2(p+) after perturbation, the perturbed Riemann solution is R←+J<+S→.Subcase 2.3. If f1(0)>u+ and g1(p^2)>g2(p+), the corresponding Riemann solution is R←+J<+S→. After perturbation, we obtain that f1(0)>u+ and g1(p^2)>g2(p+) still hold and it follows that the perturbed Riemann solution is R←+J<+S→.Theorem 3.
For Subcases 2.1.1, 2.2.1, and 2.3 (see Figure8), the corresponding Riemann solution remains unchanged after perturbation which shows that the corresponding Riemann solution is stable and the perturbation can not affect the corresponding Riemann solution. While, for Subcases 2.1.2 and 2.2.2, the forward rarefaction wave of the corresponding Riemann solution can be transformed into the forward shock wave after perturbation which shows that the corresponding Riemann solution is unstable.Figure 8
Solutions in Case 2.
(a)
A perturbed solution in the neighborhood of the origin (t>0)
(b)
A perturbed solution in the neighborhood of the origin (t>0)Case 3 (p-τ-γ>p+τ+γ).
After perturbation, it still holds thatp-τ-γ>p+τ+γ.Subcase 3.1. If u-≤f2(0), the corresponding Riemann solution is R←+J>+R→. After perturbation, we obtain that u-≤f2(0) or u->f2(0); we construct the perturbed Riemann solutions as follows.Subcase 3.1.1. When u-≤f2(0) or u->f2(0) and g1(p-)≥g2(p^4) after perturbation, we know the perturbed Riemann solution is still R←+J>+R→.Subcase 3.1.2. When u->f2(0) and g1(p-)<g2(p^4) after perturbation, the perturbed Riemann solution is S←+J>+R→.Subcase 3.2. If u->f2(0) and g1(p-)≥g2(p^4), the corresponding Riemann solution is R←+J>+R→. After perturbation, we have u->f2(0), g1(p-)≥g2(p^4), or g1(p-)<g2(p^4); we construct the perturbed Riemann solutions as follows.Subcase 3.2.1. When u->f2(0) and g1(p-)≥g2(p^4) after perturbation, we obtain that the perturbed Riemann solution is still R←+J>+R→.Subcase 3.2.2. When u->f2(0) and g1(p-)<g2(p^4) after perturbation, the perturbed Riemann solution is S←+J>+R→.Subcase 3.3. If u->f2(0) and g1(p-)<g2(p^4), the corresponding Riemann solution is S←+J>+R→. After perturbation, we find that u->f2(0) and g1(p-)<g2(p^4) still hold and it follows that the perturbed Riemann solution is S←+J>+R→.Theorem 4.
For Subcases 3.1.1, 3.2.1, and 3.3 (see Figure9), the corresponding Riemann solution remains unchanged after perturbation which shows that the corresponding Riemann solution is stable. While, for Subcases 3.1.2 and 3.2.2, the backward rarefaction wave of the corresponding Riemann solution can be transformed into the backward shock wave after perturbation which shows that the corresponding Riemann solution is unstable.Figure 9
Solutions in Case 3.
(a)
A perturbed solution in the neighborhood of the origin (t>0)
(b)
A perturbed solution in the neighborhood of the origin (t>0)Based on the above discussions, we summarize our main result as follows.Theorem 5.
There exists uniquely solution for the generalized Riemann problem (3) and (6) in a neighborhood of the origin (t>0) in the (x,t) plane. The contact discontinuity may appear after perturbation although there is no contact discontinuity in the corresponding Riemann solution. Furthermore, for most cases, the perturbed Riemann solutions of (3) and (6) are stable. While, for some few cases, the perturbed Riemann solutions of (3) and (6) are unstable under such local small perturbations of the Riemann data.



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