Stability of the Riemann solutions for a Chaplygin gas

J. Math. Anal. Appl. 409 (2014) 347–361

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Stability of the Riemann solutions for a Chaplygin gas Qu Aifang ∗ , Wang Zhen Wuhan Institute of Physics and Mathematics, Chinese Academy of Sciences, Wuhan 430071, China

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Article history: Received 17 December 2012 Available online 19 July 2013 Submitted by Kenji Nishihara Keywords: Stability Delta wave Chaplygin gas Riemann problem Wave interaction

abstract In this paper, we study the structural stability of solutions to the Riemann problem for one-dimensional isentropic Chaplygin gas. We perturb the Riemann initial data by taking three piecewise constant states and construct the global structure. By letting the perturbed parameter ε tend to zero, we prove that the Riemann solutions are stable under the local small perturbations of the Riemann initial data even when the initial perturbed density depending on the parameter but with no mass concentration limit. © 2013 Elsevier Inc. All rights reserved.

1. Introduction In this paper, we are concerned with the one-dimensional isentropic Chaplygin gas dynamics characterized by



ρt + (ρ u)x = 0, (ρ u)t + (ρ u2 + p)x = 0

(1.1)

with state equation p = − ρ1 , where ρ > 0 is the density, u is the velocity. One remarkable feature of this dynamics is that the characteristic fields are linearly degenerate. For the physical meaning of Chaplygin gas, we refer readers to [1–3,9]. Recently, there are many works on the existence of solution to the one-dimensional Chaplygin gas. The existence of classical solution to the Cauchy problem has been thoroughly studied in [8]. The research on the existence of solution including delta waves is still mainly concentrated on Riemann problems. For the isentropic case, Brenier found that it owns solution with concentration [2]. Guo et al. constructed the Riemann solution with delta wave [6]. Wang et al. proved the existence of the Riemann problem with delta initial data [13]. For the adiabatic case, Sheng et al. constructed the Riemann solution in [14]. The Riemann problem for one dimensional generalized Chaplygin gas dynamics is studied in [12], the characteristic fields of the dynamics in which are genuinely nonlinear. However, there are few works on the stability of the solution to one-dimensional isentropic Chaplygin gas, even on that of the Riemann solution. Consider the Riemann problem of (1.1) with initial data

 (ρl , ul ), x < 0 (ρ, u)(x, 0) = (ρr , ur ), x > 0, where ρl,r , ul,r are all given constants.

∗

Corresponding author. E-mail addresses: [email protected] (A. Qu), [email protected] (Z. Wang).

0022-247X/$ – see front matter © 2013 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jmaa.2013.07.018

(1.2)

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A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

In order to investigate the stability of Riemann solution for the Chaplygin gas, we consider the initial data consisting of three piecewise constant states as follows:

 (ρl , ul ), (ρ, u)(x, 0) = (ρm , um ), (ρr , ur ),

x < −ϵ −ϵ < x < ϵ x > ϵ,

(1.3)

where ε > 0 is arbitrarily small. If limε→0 ερm = 0, then by passing to the limit as ε → 0 in (1.3) we get the corresponding Riemann initial data (1.2), thus the initial data (1.3) can be viewed as a perturbation of the corresponding Riemann initial data (1.2). This method has been used by many authors to study the stabilities of Riemann solution [10,11] and even to construct the unique solution of Riemann solution [7,13]. For the Riemann problem of (1.1), a delta wave may be present in the solution. When constructing the global solution of (1.1) and (1.3), one needs to consider whether two adjacent waves intersect and interact with each other. It is not so easy to see whether two delta waves meet and how they interact with each other for the model considered here and thus needs some technical treatment. Since (1.1) and (1.2) are scaling invariant, people usually construct its self-similar solutions. By studying the limit structure of the solution of (1.1) and (1.3), we prove that the Riemann solutions of (1.1) and (1.2) are stable under this local small perturbation of the Riemann initial data by letting ε → 0. This paper is organized as follows. In Section 2 we recall and present some known results about the system (1.1) and its Riemann solution with initial data (1.2). In Section 3 we mainly discuss the interaction of the elementary waves, namely, shock waves, rarefaction waves and delta waves, for all the cases when the initial data are in the form of (1.3) with um and ρm being quantities independent of the parameter ε. By letting ε → 0 we get the limit solution and by comparing it with the corresponding Riemann problem of (1.1) and (1.2), we find that the Riemann solution is stable under this perturbation. In Section 4, we investigate the case that ρm is a function of ε satisfying limε→0 ρm = ∞ and limε→0 ερm = 0. By studying the limit structure of the perturbed solution we find that the Riemann problem is also stable under these local perturbation. A conclusion of this paper is given in Section 5. 2. Riemann problem for (1.1) and (1.2) The solution of (1.1) and (1.2) is referred to [5,6,13]. For the readers’ convenience we recall and present some facts as follows. The two eigenvalues of (1.1) are

λ1 = u − c and λ2 = u + c , where c = ρ1 is the sound speed. The corresponding eigenvectors are

⃗r1 = (ρ, −c )t and ⃗r2 = (ρ, c )t . System (1.1) is strictly hyperbolic for ρ > 0 and the fact ∇(ρ,u) λi · ⃗ ri ≡ 0, i = 1, 2 implies that λi (i = 1, 2) are linearly degenerate. Since (1.1) and (1.2) are invariant under scaling (x, t ) → (τ x, τ t ), we seek the self-similar solution x

ξ= .

(ρ, u)(x, t ) = (ρ, u)(ξ ),

t

Then (1.1) and (1.2) are reduced to the following boundary value problem of ordinary differential equations

 )ξ = 0 −ξ ρξ + (ρ u

−ξ (ρ u)ξ + ρ u − 2

(ρ, u)(−∞) = (ρl , ul )

1

ρ

 =0

(2.1)

ξ

(ρ, u)(+∞) = (ρr , ur ).

In addition to the solution (ρ, u) = const ., (2.1) has solution

ξ = u ± c = ul ± cl . For a bounded discontinuity at ξ = σ , the Rankine–Hugoniot conditions hold

 σ [ρ] = [ρ u], σ [ρ u] = [ρ u2 + p].

(2.2)

Hereafter, [ρ] = ρr − ρl , where ρl,r are the value of ρ on the left and right sides of the discontinuity, etc. It follows from (2.2) that

σ = u ± c = ul ± cl .

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

349

Noting that although the characteristics λi , i = 1, 2 are linearly degenerate, the pressure p and thus the density ρ varies after crossing the waves. In view of this we distinguish the above two kinds of waves as in [4] by the variance of the density across the waves, i.e. R1 : u − ul = c − cl ,

c > cl ;

R2 : u − ul = −(c − cl ),

S1 : u − ul = c − cl ,

0 < c < cl ;

S2 : u − ul = cl − c ,

0 < c < cl ; c > cl .

Denote by J1 = R1 ∪ S1 = {(u, c )| u − c = ul − cl , c > 0} and J2 = R2 ∪ S2 = {(u, c )| u + c = ul + cl , c > 0}. For the Riemann problem of (1.1) and (1.2), if ul − cl < ur + cr , then (ur , cr ) can be connected to (ul , cl ) on the right by a J1 wave, an intermediate state (u1 , c1 ) and a J2 wave, where

 u + cr + ul − cl  u1 = r , 2 u + c − u + c r r l l c = . 1

(2.3)

2

Remark 2.1. For convenience, we also use (u, c ) to denote the corresponding state in the sequel without confusion since ρ = 1/c for 0 < c < ∞. If ul − cl > ur + cr , the overlap of the characteristics results in a δ -wave connecting the two states (ul , cl ) and (ur , cr ). The solution of (1.1) and (1.2) is

 (ρl , ul ), (ρ, u) = (w(t )δ(x − x(t )), uδ (t )), (ρr , ur ),

x < x(t ) x = x(t ) x > x(t ),

(2.4)

where x(t ), w(t )(≥ 0) and uδ (t ) satisfy the generalized R–H conditions

 dx(t )   = uδ ,     dt dw(t ) = uδ [ρ] − [ρ u],  dt      d(w uδ ) = u [ρ u] − [ρ u2 + p] δ

(2.5)

dt

and x(0) = 0, w(0) = 0, 1/ρ(x(t ), t ) = 0. Direct calculation gives

  w(t ) = ρl ρr ((ur −ul )2 − (cr − cl )2 )t , ρ u − ρl ul + ρl ρr ((ur − ul )2 − (cr − cl )2 )  uδ = r r , ρr − ρl

if ρr ̸= ρl ,

(2.6)

and

 w(t ) = (ρl ul − ρr ur )t , uδ =

1

2

(ur + ul ),

if ρr = ρl ,

(2.7)

t

while x(t ) = 0 uδ (τ )dτ = uδ t . Denote by S the line c = −u + ul − cl . In the phase (u, c )-plane, the upper half plane c > 0 is divided by J1 , J2 and S into five regions I, II, III, IV and V as depicted in Fig. 1. If (ur , cr ) ∈ I ∪ II ∪ III ∪ IV, the solution of the initial value problem (1.1) and (1.2) can be expressed briefly as

(ul , cl ) + J1 + (u1 , c1 ) + J2 + (ur , cr ),

(2.8)

where ‘‘+’’ means ‘‘followed by’’ as usual, and (u1 , c1 ) are given by (2.3). See Fig. 2. If (ur , cr ) ∈ J1 ∪ J2 , the solution to the Riemann problem contains only one wave (rarefaction wave or shock), which is a degenerate case with one wave having zero strength. As in [5], we will not pay much special attention to these degenerate cases in this paper.

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A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

Fig. 1. The upper half (u, c ) plane is divided into 5 regions.

Fig. 2. ur + cr > ul − cl .

Fig. 3. ur + cr ≤ ul − cl .

If (u, c ) ∈ V, the solution of (1.1) and (1.2) can be expressed briefly as

(ul , cl ) + δ S + (ur , cr ),

(2.9)

and on δ S we have

  (1/2(ul + ur )t , (ρl ul − ρr ur )t , 1/2(ul + ur )) ρ r ur − ρ l ul + ρr ur − ρl ul + ddtw  (x, w, uδ )(t ) = t , ρl ρr ((ur − ul )2 − (cr − cl )2 )t ,   ρr − ρl ρr − ρl

dw dt

ρl = ρr ,

 ,

ρl ̸= ρr .

(2.10)

See Fig. 3. Therefore, we have the following lemma. Lemma 2.2. For the Riemann problem of (1.1) and (1.2), if ur + cr > ul − cl , then (ur , cr ) can be connected to (ul , cl ) on the right by a J1 wave, an intermediate state (u1 , c1 ) given by (2.3) and a J2 wave. If ur + cr < ul − cl , then (ur , cr ) can be connected to (ul , cl ) on the right by a δ -wave given by (2.10). However, there is a special case left. If (ur , c r ) ∈ S, i.e., ur + c r = ul − cl , then we can neither directly connect (ul , cl ) and (ur , cr ) by Ji (i = 1, 2) waves nor by a δ -wave, since there is neither a reasonable intermediate state (c ̸= 0) to connect them nor any overlap of different families of characteristics. We get the solution by approximation. Note that S separates (n) (n) (n) (n) (n) (n) the region III and V, then for (ur , cr ) close enough to (ur , c r ) we have either (ur , cr ) ∈ III or (ur , cr ) ∈ V. (n)

(n)

(n)

(n)

If (ur , cr ) ∈ III and (ur , cr ) → (ur , c r ) as n → ∞, then the solution of (1.1) and (1.2) is

(u(rn) , cr(n) ) + S1 + (u(1n) , c1(n) ) + S2 + (ur , cr ),

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

351

Fig. 4. Case 3.1.

with (n)

(n)

u1 = (n)

c1

ur

2 (n)

=

+ cr(n) + ul − cl

ur

(n)

+ cr − ul + cl 2 (n)

.

(n)

As n → ∞, we have (u1 , c1 ) → (ul − cl , 0), while S1 and S2 coincide. (n)

(n)

(n)

(n)

If (ur , cr ) ∈ V and (ur , cr ) → (ur , c r ) as n → ∞, then the solution of (1.1) and (1.2) is

(u(rn) , cr(n) ) + D + (ur , cr ). As n → ∞, we have for ρ¯ r ̸= ρl

   (n)  w(n) (t ) = ρl ρr ((ul − u(n) )2 − (cl − c (n) )2 )t → 2t (n) ρr(n) u(rn) − ρl ul + dw dt (t ) 2 + ρ r (ul − cl − c r ) − ρl ul (n)   → = ul − cl .  uδ = (n) ρ r − ρl ρr − ρl (n) = c¯r , then  (n) (n) w (t ) = ρl (ul − ur )t → ρl (ul − u¯ r )t = 2t

If ρ¯ r = ρl , without loss of generality, we let cr (n)

1

(ul + u(rn) ) →

1

(ul + u¯ r ) = ul − cl . 2 2 The limitation of the speed on the discontinuity is the same, while the density concentrates on it. We can define a wave connected (ul , cl ) and (ur , c r ) in this way. This wave is a δ -wave with weight w(t ) = 2t and speed uδ = ul − cl . Note that this is consistent with (2.10). Thus we can conclude that uδ

=

Lemma 2.3. For the Riemann problem of (1.1) and (1.2), if ur + cr ≤ ul − cl , then (ur , cr ) can be connected to (ul , cl ) by a delta wave giving by (2.10). 3. Solution of (1.1) and (1.3) and its limitation as ε → 0 In this part, we first construct the global solution of (1.1) and (1.3) with ρm , um being constant independent of ε and then get the limit solution by letting the parameter ε → 0. By comparing the structure of the limit solution of (1.1) and (1.3) with that of (1.1) and (1.2) we study the stability of the solution of (1.1) and (1.2). Denote by Vl = ul − cl , Vm+ = um + cm , Vm− = um − cm , Vr = ur + cr . According to the relative locations of Vi (i = l, m− , m+ , r ) on the u-axis there are six possibilities as follows.

(1) Vm+ ≤ Vl < Vr ; (2) Vl < Vr ≤ Vm− ; (3) Vl < min{Vm+ , Vr } and Vm− < Vr ; (4) Vm− < Vr ≤ Vl and Vm+ ≤ Vl ; (5) Vr ≤ Vm− < Vm+ ≤ Vl ; (6) Vr ≤ Vl < Vm+ and Vr ≤ Vm− . We proceed to construct the solution of each case and then get the corresponding structure of the limit solution as ε → 0. Case 3.1. Vm+ ≤ Vl < Vr . Since Vm+ ≤ Vl , then by Lemma 2.3 (ul , cl ) and (um , cm ) can be connected by a δ -wave δ S. In view of Vm− < Vm+ < Vr , the solution of (1.1) and (1.3) for small t can be expressed briefly by

(ul , cl ) + δ S + (um , cm ) + J1 + (u1 , c1 ) + J2 + (ur , cr ),  

(3.1)

where (u1 , c1 ) = ur +cr +2um −cm , ur +cr −2um +cm . See Fig. 4. The weight w(t ) and propagation speed uδ (t ) of δ S are given by (2.10) with the subscript r replaced by m. The speeds of J1 and J2 are λ1 = um − cm and λ2 = ur + cr respectively.

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A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

Next, we proceed to investigate whether δ S meets J1 in finite time and then interacts with it. Note that for ρm ̸= ρl ,

 uδ − λ 1 =

ρm ([u]2 ρl

− [c ]2 ) + um − ul ρm ρl

 =

ρm ([u]2 ρl

+ cm

−1

− [c ]2 ) + [u] − [c ] ρm ρl

(3.2)

−1

and

     ρm ρm ρm 2 2 2 ([u] − [c ] ) − ([u] − [c ]) = ([u] − [c ]) − 1 [u] + + 1 [c ] . (3.3) ρl ρl ρl On the one hand, we have [u] − [c ] = um − ul − (cm − cl ) = Vm− − Vl < 0 and [u] = um − ul < −(cm + cl ) < 0 for the case considered here. ρ On the other hand, if ρm < ρl , then ρm < 1, [c ] = cm − cl > 0. It follows from (3.3) that l

ρm ([u]2 − [c ]2 ) − ([u] − [c ])2 < 0. (3.4) ρl Combining it with (3.2), we have uδ − λ1 > 0 for ρm < ρl .     ρ ρ ρ If ρm > ρl , then ρm > 1, [c ] = cm − cl < 0. Since [u] − [c ] < 0 and ρm − 1 [u] + ρm + 1 [c ] < 0, then by (3.3) we l l l have

ρm ([u]2 − [c ]2 ) − ([u] − [c ])2 > 0, ρl which implies that uδ − λ1 > 0 for ρm > ρl . If ρm = ρl , then 1

uδ − λ 1 =

(um + ul ) − (um − cm ) =

2 Therefore, we have

(3.5)

1 2

(ul − um + 2cm ) > 0.

(3.6)

uδ > λ1 . This means that the delta wave issuing from (−ε, 0) intersects the J1 wave issuing from (ε, 0) at a point (x∗ , t ∗ ) in finite (u +u −c )ε time. The intersection point satisfies uδ t ∗ − ε = λ1 t ∗ + ε , which gives t ∗ = u −(u2ε −c ) , x∗ = uδ t ∗ − ε = u δ−(um −cm ) . At the m m m m δ δ ∗ time t = t , the weight and the speed are respectively

w(t ∗ ) =



ρl ρm ((um − ul )2 − (cm − cl )2 )t ∗ , (ρl ul − ρm um )t ,

if ρl ̸= ρm if ρl = ρm

and

 ρ u − ρl ul +   m m ∗ ρm − ρl u(t ) =   1 (u + u ),

dw dt

,

if ρl ̸= ρm

if ρl = ρm . m l 2 So far, we know that for t ∈ (0, t ∗ ) the solution of (1.1) and (1.3) is (3.1). For t > t ∗ we need to solve a new Riemann problem with delta initial data

(ρl , ul ), (ρ, u)(x, t ) = (w(t ∗ )δ(x − x(t )), u∗δ ), (ρ1 , u1 ), 

∗

x < x∗ x = x∗ x > x∗ ,

(3.7)

where u∗δ = uδ (t ∗ ). In order to determine the solution of (1.1) and (3.7) and according to the results of [13], we just need to compare u∗δ with ul − cl and u1 + c1 respectively. Similar to (3.2) we have for ρm ̸= ρl

 u∗δ

− (ul − cl ) =

ρl ([u]2 ρm

− [c ]2 ) + [u] − [c ] ρ

1 − ρl m

ρl ([u]2 − [c ]2 ) − ([u] − [c ])2 = ([u] − [c ]) ρm

.



(3.8)

    ρl ρl − 1 [u] + + 1 [c ] . ρm ρm

(3.9)

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

ρ

ρ

Now ρ l in (3.9) replaces ρm in (3.3). Note that [u] < −(cl + cm ) and −(cl + cm ) = m l



ρl − 1 [ u] + ρm 



353 ρl ρ m +1 ρ 1− ρ l m

[c ], then

 ρl  <1 > 0 for ρl ρm + 1 [c ] ρ  ρm < 0 for l > 1. ρm 

Combining it with (3.9) and then (3.8) yields that u∗δ < ul − cl

for ρm ̸= ρl .

If ρm = ρl , then u∗δ − (ul − cl ) = 1/2(um + ul ) − (ul − cl ) = 1/2(Vm+ − Vl ) < 0. Thus, we also have u∗δ < ul − cl . By the conclusion of [13, Case 3.2], we know that the solution of (1.1) and (3.7) satisfies

 (ρ , u ),   l l (w1 (t )δ(x − x(t )), uδ1 (t )), (ρ, u)(x, t ) = ¯ u¯ )(x, t ),  (ρ, (ρ1 , u1 ),

x < x(t ), x = x(t ), x(t ) < x < (u1 + c1 )(t − t ∗ ) + x∗ , x > (u1 + c1 )(t − t ∗ ) + x∗

(3.10)

where (ρ, ¯ u¯ )(x, t ) = (ρ∗ , u∗ )(t¯) along (u1 + c1 )t − x = (u1 + c1 )t¯ − x(t¯) for t ≥ t ∗ and (ρ∗ , u∗ )(t ) = (2(u1 + c1 − uδ1 (t ))−1 , 1/2(u1 + c1 + uδ1 (t ))) is the right state of the δ -shock wave δ S1 satisfying (2.5) with initial data (x, w, uδ )(t ∗ ) = (x∗ , w(t ∗ ), u∗δ ). For t ≥ t ∗ we have w1 (t ) = w( ˜ t − t ∗ ) with w( ˜ t ) satisfying dw( ˜ t) dt

= −1 − ρl uδ1 (t + t ∗ ) + ρl ul

(3.11)

and



w ˜ 2+

dw ˜



dt

= w( ˜ 0)(ρl ul + 1 − ρl ud e∗ ).

(3.12)

It follows from (3.12) that if w( ˜ 0) = 0, then w( ˜ t ) ≡ 0 or w( ˜ t ) = −2t for t > 0, and if w( ˜ 0) ̸= 0, then w( ˜ t ) is determined by

w ˜ + A0 ln(A0 − w) ˜ −w ˜ 0 − A0 ln(A0 − w ˜ 0 ) = −2t , where A0 = speed

1 2

(3.13)

w ˜ 0 (ρl ul + 1 − ρl u∗δ ) and w ˜ 0 = w( ˜ 0). Here w ˜ 0 = w(t ∗ ) > 0 for t ∗ > 0. Thus w1 (t ) is given by (3.13). The

uδ1 (t ) = ul + cl −

2A0

ρl w1 (t )

(3.14)

and x(t ) = x∗ +



t t∗

uδ1 (τ )dτ .

(3.15)

One can refer to [13] for more details. We point it out here that w ˜ 0 and thus A0 depend on ε since they are functions of t ∗ . The fact limt →∞ uδ1 (t ) = ul − cl < u1 + c1 = ur + cr implies that δ S1 will never overtake those J2 waves. The structure of the solution is shown in Fig. 4. Let Ωm = {(x, t ) | uδ t − ε < x < (um − cm )t + ε, 0 ≤ t ≤ t ∗ }. As ε → 0, we can see that (x∗ , t ∗ ), (−ε, 0) and (ε, 0) go to (0, 0). The region Ωm shrinks to the origin point with no mass concentration. Since limε→0 w ˜ 0 = 0, then limε→0 A0 (ε) = 0. Inserting ˜ 0 = 0 into (3.12) yields that w(t ) ≡ 0 since  w  w(t ) ≥ 0. It follows from (3.11) that uδ (t ) = ul − cl . Here (ρ∗ , u∗ )(t ) = 2(ur + cr − (ul − cl ))−1 , 21 (ur + cr + ul − cl ) is constant. The limit solution is consistent with the corresponding Riemann solution (2.8), thus the Riemann problem is stable with respect to the perturbation in this case. Case 3.2. Vl < Vr ≤ Vm− . In this case, for sufficiently small t, the solution of (1.1) and (1.3) can be expressed briefly as

(ul , cl ) + J1 + (u1 , c1 ) + J2 + (um , cm ) + δ S + (ur , cr ),  u +c +u −c u +c −u +c  where (u1 , c1 ) = m m2 l l , m m2 l l , λ1 = ul − cl , λ2 = um + cm . The weight w(t ) and propagation speed uδ (t ) of δ S are given by(2.10) with the subscript  l replaced by m. Similar analysis as in Case 3.1 gives the intersection point of J1 and (um +cm +u+δ)ε 2ε ∗ ∗ δ S, (x , t ) = , um +cm −uδ , and in addition, uδ > ur + cr . Then we have u1 − c1 (= ul − cl ) < ur + cr < uδ . um +cm −uδ

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A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

Fig. 5. Case 3.2.

For t > t ∗ , the solution of (1.1) and (1.3) satisfies

 (ρ , u ),   1 1 (ρ, ¯ u¯ )(x, t ), (ρ, u)(x, t ) = (w   1 (t )δ(x − x(t )), uδ1 (t )), (ρr , ur ),

x < (ul − cl )t , (ul − cl )(t − t ∗ ) + x∗ < x < x(t ), x = x(t ), x > x(t )

(3.16)

where (ρ, ¯ u¯ )(x, t ) = (ρ∗ , u∗ )(t¯) along (u1 − c1 )t − x = (u1 − c1 )t¯ − x(t¯) for t ≥ t ∗ and (ρ∗ , u∗ )(t ) = (2(uδ1 (t ) − (u1 − c1 ))−1 , 1/2(u1 − c1 + uδ1 (t ))) is the left state of the δ -shock wave δ S satisfying (2.5) with initial data (x, w, uδ )(t ∗ ) = (x∗ , w(t ∗ ), u∗δ ). For t ≥ t ∗ we have w1 (t ) = w( ˜ t − t ∗ ) with w( ˜ t ) satisfying dw( ˜ t) dt

= −1 + ρr uδ1 (t + t ∗ ) − ρr ur ,

t >0

and determined by (3.13) with A0 replaced by A1 = uδ1 (t ) = ur − cr +

(3.17) 1 2

w ˜ 0 (ρr u∗δ − ρr ur + 1), while

2A1

(3.18)

ρr w1 (t )

and x(t ) = x∗ +



t t∗

uδ1 (τ )dτ .

(3.19)

The structure of the solution of (1.1) and (1.3) is shown in Fig. 5. Letting ε → 0 and similar to Case 3.1, we get the limit solution, which is the corresponding Riemann solution (2.8) of (1.1) and (1.2). Thus the Riemann problem is stable with respect to the perturbation in this case. Case 3.3. Vl < min{Vm+ , Vr } and Vm− < Vr . In this case, for sufficiently small t, the solution of (1.1) and (1.3) can be expressed briefly as

(ul , cl ) + J1 + (u1 , c1 ) + J2 + (um , cm ) + J1′ + (u2 , c2 ) + J2′ + (ur , cr ),  u +c +u −c u +c −u +c    where (u1 , c1 ) = m m2 l l , m m2 l l , (u2 , c2 ) = ur +cr +2um −cm , ur +cr −2um +cm , λ1 = ul − cl , λ2 = um + cm , λ′1 = um − cm , λ′2 = ur + cr .   λ2 ε

− ε, cεm . In view of ul − cl = u1 − c1 < ur − cr = u2 + c2 , the interaction results in two new waves J1′′ and J2′′ with λ′′1 = ul − cl , λ′′2 = ur + cr . The  ur +cr +ul −cl ur +cr −ul +cl  intermediate state between them is (u3 , c3 ) = , . The structure of the solution is shown in Fig. 5, 2 2   where (x∗ , t ∗ ) = ucm ε , cε , Ωm = {(x, t ) | (um + cm )t − ε < x < (um − cm )t + ε, 0 ≤ t ≤ t ∗ }. m m As ε → 0, (x∗ , t ∗ ), (−ε, 0), (ε, 0) all go to (0, 0) and Ωm shrinks to the origin point with no mass concentration. The limit situation is a J1 wave with speed λ1 = ul − cl plus a J2 wave with speed λ2 = ur + cr , which is exactly the corresponding Since λ′1 < λ2 , J2 meets J1′ in finite time and the intersection point is

cm

Riemann solution. Thus, the Riemann problem is stable under the perturbation in this case. Case 3.4. Vm− < Vr ≤ Vl and Vm+ ≤ Vl . In this case, for sufficiently small t, the solution of (1.1) and (1.2) can be expressed briefly as

(ul , cl ) + δ S + (um , cm ) + J1 + (u1 , c1 ) + J2 + (ur , cr ), where the weight w(t ) and propagation speed uδ (t ) of δ S are   given by (2.10) with the subscript r replaced by m, λ1 = um − cm , λ2 = ur + cr , (u1 , c1 ) = ur +cr +2um −cm , ur +cr −2um +cm . Similar to the analysis in Case 3.1 we know that δ S meets J1 in finite time. The intersection point is (x∗ , t ∗ ) =



(uδ +um −cm )ε , uδ −(u2mε +cm ) uδ −(um −cm )



.

In order to get the solution in the part t > t ∗ , we need to solve the Riemann problem (1.1) and (3.7). Since Vm+ < Vl and due to the analysis in Case 3.1 we have um − cm < uδ (t ∗ ) < ul − cl .

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

355

In view that uδ (t ∗ ) is independent with u1 + c1 (= ur + cr ) and um − cm < ur + cr < ul − cl , it has two possibilities: one is uδ (t ∗ ) ≥ ur + cr , the other is uδ (t ∗ ) < ur + cr . If uδ (t ∗ ) ≥ ur + cr , then u1 + c1 ≤ uδ (t ∗ ) ≤ ul − cl . By the conclusion of [13, Case 3.3], the interaction of δ S and J1 results in a delta wave δ S1 with

 ∗ ∗  w u (t − t ∗ ) − 12 [ρ u2 − c ](t − t ∗ )2    + x∗ , [ρ = 0]     w∗ − [ρ u]t   x1 ( t ) =  ∗ ∗ ∗      −w + [ρ u](t − t ) + w1 (t − t ) + x∗ , [ρ ̸= 0] (3.20) [ρ]    w1 (t ) = {ρ1 ρl ([u]2 − [c ]2 )(t − t ∗ )2 + 2w∗ (u∗ [ρ] − [ρ u])(t − t ∗ ) + w∗ 2 }1/2     ∗ ∗ 2 ∗  u (t ) = w u + [ρ u]x1 (t ) − [ρ u − c ](t − t ) δ1 w1 (t ) and u1 + c1 < limt →∞ uδ1 (t ) < ul − cl . Thus δ S1 will meet the J2 wave issuing from (ε, 0) at a point (x+ , t + ) since u1 + c1 = ur + cr . (x+ , t + ) is determined by x1 (t ) = (ur + cr )t + ε . At the time t = t + , a new Riemann problem is formed with delta initial data, which satisfy ur + cr < uδ1 (t + ) < ul − cl . Similar to δ S1 , a new delta wave δ S2 resulted from the new Riemann problem is determined. The structure of the global solution is shown in Fig. 7. If uδ (t ∗ ) < ur + cr , then uδ (t ∗ ) < u1 + c1 < ul − cl . By the conclusion of [13, Case 3.4], the interaction of δ S and J1 results in a delta wave δ S1 adjacent to a family of J2 waves and there exists a unique t ∗∗ > t ∗ such that uδ1 (t ∗∗ ) = u1 + c1 . When t > t ∗∗ , it will overtake and finally penetrates all of the J2 waves at a point (x+ , t + ). Since ur + cr < uδ1 (t + ) < ul − cl , a new delta wave δ S2 arises from (x+ , t + ) and meets the J2 wave issuing from (ε, 0) at a point (x++ , t ++ ). It also holds that ur + cr < uδ2 (t ++ ) < ul − cl . Then a new delta wave δ S3 arises from (x++ , t ++ ). Here we construct the solution without giving the explicit expression of the solutions since they are all obtained by analyzing the existence of solutions to corresponding ordinary differential equations. One can refer the related details to [13]. The structure of the solution is shown in Fig. 8. Next, we go on to investigate the limit solution of this case. If uδ (t ∗ ) ≥ ur + cr , then limε→0 (x+ , t + ) = (0, 0). In fact, if limε→0 uδ (t ∗ ) = ur + cr , then due to the definition of t + , we have t + = 0 for (x∗ , t ∗ ) = (0, 0). If limε→0 uδ (t ∗ ) > ur + cr , then t+ =

(ur + cr )t ∗ − x∗ → 0 as ε → 0. uδ (t + − t ∗ ) − (ur + cr )

(3.21)

Noting that limε→0 c1 = ur +c2r −um > 0, there is no concentration formed due to the shrink of the region with state ① (see Fig. 7) as ε → 0. In addition, we have w(0) = 0. Therefore, as ε → 0, the limit solution consists of a delta wave as shown in Fig. 3. If uδ (t ∗ ) < ur + cr , then limε→0 (x∗∗ , t ∗∗ ) = (0, 0) and w(0) = 0. By (3.13) we have t + = 0 for w(0) = 0, uδ = ur + cr . Thus, t ++ = 0. Therefore, as ε → 0, we have

  1 1   (ul + ur )t , (ρl ul − ρr ur )t , (ul + ur ) ,   2 2   (x, w, uδ )(t ) =  2 − [c ]2 ) [ρ u] + dw(t )  [ρ u ] t + w( t ) ρ ρ ([ u ] l r dt   , , ,  [ρ] [ρ] [ρ]

if ρl = ρr if ρl ̸= ρr ,

which is exactly the corresponding Riemann solution (2.10) of (1.1) and (1.2). Thus the Riemann problem is stable with respect to the perturbation in this case. Similar to the analysis in Case 1, the limit situation is a δ wave, which is exactly the corresponding Riemann solution. Case 3.5. Vr ≤ Vm− < Vm+ ≤ Vl . In this case, for sufficiently small t, the solution of (1.1) and (1.2) can be expressed briefly as

(ul , cl ) + δ S1 + (um , cm ) + δ S2 + (ur , cr ). The speeds of these two delta waves are respectively

1   (ul + um ),

if ρl = ρm ,  2 2   ρm um − ρl ul + ρl ρm ((ul − um ) − (cl − cm ) ) , if ρl ̸= ρm ρm − ρl 1  if ρr = ρm ,  (ur + um ), 2  uδ 2 = 2 2   ρr ur − ρm um + ρr ρm ((ur − um ) − (cr − cm ) ) , if ρr ̸= ρm . ρm − ρr uδ 1 =

2

(3.22)

(3.23)

356

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

The weights of them are respectively

 (ρl ul − ρm um )t , if ρl = ρm , w1 (t ) =  2 2 ρl ρm ((ul − um ) − (cl − cm ) )t , if ρl ̸= ρm  (ρm um − ρr ur )t , if ρr = ρm , w2 (t ) =  ρr ρm ((ur − um )2 − (cr − cm )2 )t , if ρr ̸= ρm .

(3.24)

(3.25)

In order to see whether δ S1 meets δ S2 in finite time, we need to compare uδ1 and uδ2 . If ρl = ρm = ρr , then uδ1 − uδ2 = 21 (ul − ur ) > 0, i.e. uδ1 > uδ2 .

√

If ρl = ρm and ρm ̸= ρr , then uδ1 = 12 (ul + um ) > um + cm , uδ2 = as in Case 3.1 gives that uδ2 < um + cm . Thus, we have uδ1 > uδ2 . ρm um −ρ u +

+

ρ ρm ((u −um )2 −(cl −cm )2 )

, uδ 2 =

1 2

Similar analysis

(ur + um ) < um − cm . Also, similar analysis

ρl ρm ((ul − um )2 − (cl − cm )2 ) ρm − ρl  um (ρm − ρl ) + ρl (um − ul ) + ρl ρm ((ul − um )2 − (cl − cm )2 ) = ρm − ρl  ρl (um − ul ) + ρl ρm ((ul − um )2 − (cl − cm )2 ) = um + ρ ρ (c − cm )  √ m √l l cm (um − ul ) + cm cl − cm + cm (ul − um )2 − (cl − cm )2 = um + cl − cm  ρr ur − ρm um + ρr ρm ((ur − um )2 − (cr − cm )2 ) uδ 2 = ρr − ρm ρr (ur − um ) + ρr ρm ((ur − um )2 − (cr − cm )2 ) = um + ρ ρm (cm − cr )  √r√ cm (ur − um ) + cm cr − cm + cm (ur − um )2 − (cr − cm )2 . = um − cr − cm Denote by A = ul − um , B = cl − cm , C = ur − um , D = cr − cm , then √ √  √ √ cm B + cm A2 − B2 − cm A = uδ1 (um , cm , A, B), uδ 1 = um + B  √ √ √ √ cm D + cm C 2 − D2 + cm C = uδ1 (um , cm , C , D). uδ 2 = um − uδ 1 =

ρr ρm ((ur −um )2 −(cr −cm )2 ) . ρm −ρr

√

l l l l If ρl ̸= ρm and ρm = ρr , then uδ1 = ρm −ρl as in Case 3.1 gives that uδ1 > um − cm . Thus, we have uδ1 > uδ2 . Next, we focus on the case ρl ̸= ρm and ρm ̸= ρr . Note that

ρm um − ρl ul +

ρr ur −ρm um ρm −ρr



D

(3.26)

(3.27)

Note that A = ul − um > cm + cl > 0 in view of Vm+ < Vl , B = cl − cm > −cm , C = ur − um < −(cm + cr ) < 0 in view of Vr < Vm− , D = cr − cm > −cm . Straightforward calculation gives

 √  √ A B + cm − cm √ √ (uδ1 )A (um , cm , A, B) = cm B A2 − B2 2 A (B + cm ) − cm (A2 − B2 )  =  √ √ √ B A B + cm A2 − B2 + cm (A2 − B2 ) 1

1

=

A B + cm

1

√

cm

A2 + Bcm

√

√

A2 − B2 +

√

cm (A2 − B2 )

>0

 √  √ C D + cm + cm √ D C 2 − D2 C 2 (C + cm ) − cm (C 2 − D2 )  =  √ √ √ D C D + cm C 2 − D2 + cm (C 2 − D2 )

(uδ2 )C (um , cm , C , D) = −

=

1

√ C

D + cm

√

C 2 + Dcm

C 2 − D2 +

√

cm (C 2 − D2 )

> 0,

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

357

which yield (uδ1 )A (um , cm , A, B) > 0, (uδ2 )C (um , cm , C , D) > 0. Thus, we have uδ1 (um , cm , A, B) > uδ1 (um , cm , cm + cl , B) 

√

√

cm

B + cm



(cm + cl )2 − B2 −

= um +

√



cm (cm + cl )

B

uδ2 (um , cm , C , D) < uδ2 (um , cm , −(cm + cr ), D) 

√

cm

√

D + cm



(cm + cr )2 − D2 +

= um −

√

cm (cm + cr )



D

.

Therefore,

√ uδ1 − uδ2

√

B + cm

>

cm



(cm + cl )2 − B2 B

+

D + cm

 (cm + cr )2 − D2 D

−

√

cm



cm + cl B

+

cm + cr



D

(B + cm )((cm + cl ) − B ) − cm (cm + cl ) √   √ B B + cm (cm + cl )2 − B2 + cm (cm + cl ) 2

=

√

+

= √

2

2

(D + cm )((cm + cr )2 − D2 ) − cm (cm + cr )2 √   √ D D + cm (cm + cr )2 − D2 + cm (cm + cr ) (cm + cr )2 − D(D + cm ) (cm + cl )2 − B(B + cm )   +√ . √ √ B + cm (cm + cl )2 − B2 + cm (cm + cl ) D + cm (cm + cr )2 − D2 + cm (cm + cr )

Since B = cl − cm , D = cr − cm , we deduce that uδ1 > uδ2 . Summarizing, we have uδ1 > uδ2 . Therefore, δ S1 meets δ S2 in finite time. The intersection point is (x∗ , t ∗ ) = δ1 +uδ2 )ε u δ −u δ

 (u

1

2

, uδ

2ε −uδ2 1



. The weight of the delta wave at that point is the sum of the two delta waves’ weights. However, the

velocity of the wave at that point needs to be determined. It seems reasonable to assume that the velocity of the gas particles at any point (x, t ) takes one value. Under this assumption, the conservation law of momentum yields

w1 (t ∗ )uδ1 (t ∗ ) + w2 (t ∗ )uδ2 (t ∗ ) = (w1 (t ∗ ) + w2 (t ∗ ))uδ3 (t ∗ ),

(3.28)

which implies that uδ3 (t ∗ ) =

w1 (t ∗ )uδ1 (t ∗ ) + w2 (t ∗ )uδ2 (t ∗ ) . w1 (t ∗ ) + w2 (t ∗ )

(3.29)

Let u∗δ3 := uδ3 (t ∗ ) and w3∗ := w3 (t ∗ ) = w1 (t ∗ ) + w2 (t ∗ ). Then it follows that uδ2 (t ∗ ) < u∗δ3 < uδ1 (t ∗ ). Now at the point (x∗ , t ∗ ) a new Riemann problem is formed with delta initial data

 (ρl , ul ), (ρ, u)(x, t ∗ ) = (w3∗ δ(x − x(t )), u∗δ3 ), (ρ , u ), r r

x < x∗ x = x∗ x > x∗ .

(3.30)

Similar analysis as in Cases 3.1 and 3.2 respectively deduces that uδ1 < ul − cl and uδ2 > ur + cr . Therefore, we have ur + cr < u∗δ3 < ul − cl .

(3.31)

By the conclusion of [13, Case 3.3], the solution of (1.1) and (3.30) is

 ∗ ∗ w3 uδ3 (t − t ∗ ) − 12 [ρ u2 − c ](t − t ∗ )2   + x∗ ,  ∗ ∗) − [ρ u ]( t − t w 3 x( t ) = ∗ ∗    −w3 + [ρ u](t − t ) + w3 (t ) + x∗ , [ρ]

if [ρ] = 0 (3.32) if [ρ] ̸= 0,

w3 (t ) = {ρl ρr ([u]2 − [c ]2 )(t − t ∗ )2 + 2w0 (˜uδ [ρ] − [ρ u])(t − t ∗ ) + w02 }1/2

(3.33)

and uδ3 (t ) =

(w1 + w2 )˜uδ + [ρ u]x(t ) − [ρ u2 − c ]t . w(t )

The structure of the solution is shown in Fig. 9.

(3.34)

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A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

As ε → 0, we have (x∗ , t ∗ ) → (0, 0), w3∗ → 0. Inserting x∗ = t ∗ = w ∗ = 0 into (3.32)–(3.34), we get the limit solution

  1 1   ( u + u ) t , (ρ u − ρ u ) t , ( u + u ) , r l l r r l r   2 l 2    (x, w, uδ )(t ) = 2 − [c ]2 ) [ρ u] + dw(t )  [ρ u ] t + w( t ) ρ ρ ([ u ] l r dt   , , ,  [ρ] [ρ] [ρ]

ifρl = ρr (3.35) if ρl ̸= ρr ,

This is exactly the corresponding Riemann solution of (1.1) and (1.2). Thus the Riemann problem is stable with respect to the perturbation in this case. Case 3.6. Vr ≤ Vl < Vm+ and Vr ≤ Vm− . In this case, for sufficiently small t, the solution of (1.1) and (1.2) can be expressed briefly as

(ul , cl ) + J1 + (u1 , c1 ) + J2 + (um , cm ) + δ S + (ur , cr ). Similar to the analysis as that in Case 4, we know that the limit situation is a δ wave, which is exactly the corresponding Riemann solution. Thus the Riemann solution is stable under the perturbation. 4. Perturbation depending on parameter ε In this part we study the solution of (1.1) and (1.3) under the assumption that um is constant, while ρm depends on ε satisfying lim ρm = lim ρm (ε) = +∞,

ε→0

ε→0

lim ρm ε = 0.

(4.1)

ε→0

According to the relations among Vl , Vr and um , the Riemann problem is divided into 9 cases:

(1) um < Vl < Vr ; (4) Vl < um < Vr ; (7) Vr < um < Vl ;

(2) Vl < Vr < um ; (5) um < Vr < Vl ; (8) Vr < Vl < um ;

(3) Vl = um < Vr or Vl < um = Vr ; (6) Vr = um < Vl ; (9) Vr < um = Vl .

In the sequel, we construct the solution of each case for sufficiently small positive ε with emphasis on the limit structure of the perturbed solution by passing ε → 0. Case 4.1. um < Vl < Vr . For fixed sufficiently small positive ε and due to the assumption (4.1), we have Vm+ < Vl < Vr , which corresponds to Case 3.1 in Section 3. Then the solution of (1.1) and (1.3) in this case is determined. See Fig. 4. Next, we investigate the limitation of the perturbed solution as ε → 0. By (3.2) we have

     ρm − ρρml ε ε ρρml − 1 ρl ε  =  . =  ρm uδ − λ1 ([u]2 − [c ]2 ) + [u] − [c ] [u]2 − [c ]2 + ρρml ([u] − [c ]) ρl Then it follows from the assumption (4.1) that lim (t ∗ , x∗ ) = lim

ε→0

ε→0



2ε uδ − λ 1

,

(uδ + λ1 )ε uδ − λ1

ε uδ −λ1



(4.2)

→ 0 as ε → 0. Thus,

= (0, 0).

In addition, 2 ρl ρm ε([u]2 − [c ]2 )



w(t ∗ ) =



ρl ρm ([u]2 − [c ]2 )t ∗ =



ρm ε ρl

−



ρl ε ρm



  [u]2 − [c ]2 + ρρml ([u] − [c ])

→ 0 as ε → 0.

(4.3)

The intermediate state (u1 , c1 ) → , . Since > 0, there is no mass concentration formed as 2 the region with state ① shrinks. By the analysis in Case 3.1 we have w(t ) ≡ 0 and uδ = ul − cl . The limit solution consists of two contact discontinuities and the structure is just that of the corresponding Riemann problem of (1.1) and (1.2). Thus the Riemann problem is stable under the perturbation.

 u r + cr + u m

ur +cr −um 2



ur +cr −um 2

Case 4.2. Vl < Vr < um . This case is similar to Case 4.1 and the Riemann solution is stable under the perturbation. Case 4.3. Vl = um < Vr or Vl < um = Vr . If Vl = um < Vr , then we have Vl < min{Vm+ , Vr } and Vv− < Vr for fixed small positive ε . It corresponds to Case 3.3 in Section 3, the solution in which has been determined. See Fig. 6.

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

359

Fig. 6. Case 3.3.

Fig. 7. Case 3.4 (a).

Fig. 8. Case 3.4 (b).

Fig. 9. Case 3.5.

Note that limε→0 (u1 , c1 ) = limε→0 ul − cl +

   , c2m = (ul −cl , 0), limε→0 (u2 , c2 ) = limε→0 ur +cr +2um −cm , ur +cr −2um +cm and ur + cr − um > 0. As ε → 0, we want to see whether the region with state ① will shrink into a 

=

 u r + cr + u m 2

,

 ur +cr −um 2

cm 2

discontinuity with mass concentration on it. Note that

(x∗ − (ul − cl )t ∗ + ε)ρm =



um ε cm

− (ul − cl )

ε cm

 + ε ρm = ερm → 0,

as ε → 0

(4.4)

which implies that there is no concentration formed as the region with state ① shrinks. Thus the Riemann solution is stable under this perturbation.

360

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

Similarly, for Vl < um = Vr , the Riemann solution is stable under this perturbation. Case 4.4. Vl < um < Vr . For fixed small positive ε , we have Vl < Vm− < Vm+ < Vr . It corresponds to Case 3.3 in Section 3. In  u −c +u u +c −u  this case the structure of the solution of (1.1) and (1.3) is shown in Fig. 6. Since (u1 , c1 ) → l 2l m , m 2l l , (u2 , c2 ) →

m l l > 0, ur +c2r −um > 0, there is no mass concentration formed in the limit solution as ε → 0. 2 Similar analysis as that in Case 3.3 yields that the limit solution is the same as that of (1.1) and (1.2), and thus the Riemann problem is stable under this perturbation.

 u r + cr + u m 2

 , ur +c2r −um and

u +c −u

Case 4.5. um < Vr < Vl . For fixed small positive ε , we have Vm+ < Vl < Vr , which corresponds to Case 3.4 in Section 3. The solution is determined as shown in Figs. 7 and 8. We have limε→0 (x∗ , t ∗ ) = (0, 0). Similar analysis as that in Case 3.4 yields that the limit solution is the same as that of (1.1) and (1.2), and thus the Riemann problem is stable under this perturbation. Case 4.6. Vr = um < Vl . For fixed small positive ε , we have Vm+ < Vl and Vm− < Vr < Vl , which corresponds to Case 3.4 in Section 3. The solution is determined as shown in Figs. 7 and 8. We go on to see the limitation of the solution as ε → 0. Similar to the analysis in Case 4.5, we have limε→0 (x∗ , t ∗ ) = limε→0 (x∗∗ , t ∗∗ ) = (0, 0) in the case shown in Fig. 7 and limε→0 (x∗∗ , t ∗∗ ) = limε→0 (x++ , t ++ ) = (0, 0) in the case shown in Fig. 8. In view of limε→0 ρ1 = limε→0 ρm = ∞, a concentration may form due to the shrink of the region with state ①. Since 2ερm



ρm ρl

ρm ((ur + cr )t ∗ − x∗ ) = −ερm +  [ u] 2 − [ c ] 2 +

− 



ρl ρm

ρl ([u] ρm

 → 0,

as ε → 0,

(4.5)

− [c ])

similar to Case 4.3, there is no mass concentration formed due to the shrink of the region with state ①. Similar to the discussion in Case 4.5 we know that the Riemann solution is stable under this perturbation. Case 4.7. Vr < um < Vl . For fixed small positive ε , we have Vr < Vm− < Vm+ < Vl , which corresponds to Case 3.5 in Section 3. The solution is determined as shown in Fig. 7. Since

ε uδ1 − uδ2

ε(ρm − ρl )(ρr − ρm )     (ρr − ρm ) ρm um − ρl ul + − [c ]2 ) − (ρm − ρl ) ρr ur − ρm um + ρρml ([u]2 − [c ]2 )    ρr ε 1 − ρρml − 1 ρm      =    ρl ρl ρl ρr ρr ρr 2 2 2 − [c ]2 ) − 1 u − u + ([ u ] − [ c ] ) − 1 − u − u + ([ u ] m l r m ρm ρm ρm ρm ρm ρm    ρl ρr ρm ε 1 − ρm −1 ρm     =   √ um (ρr + ρl ) + ρl ul − ρr ur − ρm ρl ([u]2 − [c ]2 ) + ρr ([u]2 − [c ]2 ) + O ρ1m

=





ρl ([u]2 ρm

→ 0 as ε → 0. Thus, limε→0 (t ∗ , x∗ ) = limε→0

(4.6)

 (u

δ1 +uδ2 )ε u δ −u δ 1 2

, uδ

lim w3∗ = lim (w1 (t ∗ ) + w2 (t ∗ )) = 0.

ε→0

ε→0

1

2ε −u δ

 2

= (0, 0). On the other hand, (4.7)

The analysis about the limit solution in Case 3.5 is also valid here. The limit solution is given by (3.35). In this case the solution is stable under the perturbation. Case 4.8. Vr < Vl < um . For fixed small positive ε , we have Vr < Vl < Vm+ , Vr < Vm− , which corresponds to Case 3.6 in Section 3. The solution has been determined. As ε → 0, the limit solution can be get in the same way as that in Case 4.5. It is stable under this perturbation. Case 4.9. Vr < um = Vl . For fixed small positive ε , we have Vr < Vl < Vm+ , which corresponds to Case 3.6 in Section 3. The solution has been determined. Since (u1 , c1 ) = (ul − cl , cm ), we have limε→0 ρ1 = limε→0 ρm = ∞. Similar to the analysis in Case 4.6, we know that under the perturbation (1.3), the Riemann solution is stable in this case. 5. Conclusion From the above analysis we can see that under the perturbation (1.3) the solution of (1.1) and (1.2) is stable if limε→0 ερm = 0.

A. Qu, Z. Wang / J. Math. Anal. Appl. 409 (2014) 347–361

361

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