ON TWO-DIMENSIONAL RIEMANN PROBLEM FOR ...math0.bnu.edu.cn/~lijiequan/publications/pressure...of...

DISCRETE AND CONTINUOUSDYNAMICAL SYSTEMSVolume 4, Number 4, October 1998 pp. 609–634

ON TWO-DIMENSIONAL RIEMANN PROBLEM FORPRESSURE-GRADIENT EQUATIONS OF THE EULER SYSTEM

Peng Zhang

Beijing Information and Technology Institute, Beijing, 100101

Jiequan Li

Institute of Applied Mathematics, Academia Sinica, Beijing, 100080

Tong Zhang

Institute of Mathematics, Academia Sinica, Beijing, 100080

(Communicated by Fanghua Lin )

Abstract. We consider the two-dimensional Riemann problem for the pressure-gradient equations with four pieces of initial data, so restricted that only one ele-

mentary wave appears at each interface. This model comes from the flux-splitting of

the compressible Euler system. Lack of the velocity in the eigenvalues, the slip lineshave little influence on the structures of solutions. The flow exhibits the simpler

patterns than in the Euler system, which makes it possible to clarify the interaction

of waves in two dimensions. The present paper is devoted to analyzing the struc-tures of solutions and presenting numerical results to the two-dimensional Riemann

problem. Especially, we give the criterion of transition from the regular reflection tothe Mach reflection in the interaction of shocks.

1. Introduction. We are concerned with the pressure-gradient equations of thecompressible Euler system

ρt = 0,

(ρU)t +∇p = 0,

(ρE)t +∇ · (pU) = 0,

(1.1)

where ρ(t,X) ≥ 0 (X = (x, y)) is the density, U = (u, v) is the velocity, p ≥ 0 isthe pressure, E = e + |U |2/2 denotes the total energy per unit mass, e denotes theinternal energy given by e = p((γ−1)ρ)−1 for polytripic gases, γ > 1 is the adiabaticindex, ∇ designates the gradient operator with the space variable X. This systemcomes from the flux-splitting method in numerical analysis on the Euler system

ρt +∇ · (ρU) = 0,

(ρU)t +∇ · (ρU ⊗ U) +∇p = 0,

(ρE)t +∇ · (pU(E +p

ρ) = 0,

(1.2)

by separating the pressure from the inertia in the flux [AH, LC]. The pressure-gradient equations (1.1) are valid whenever the inertia effect is so small comparedto the pressure-gradient effect of the flow as to be negligible. As a matter of fact,assuming that the velocity U is small, we can obtain ρt from the first equation of

1991 Mathematics Subject Classification. 35L65, 58F40, 65M06, 76N10.Key words and phrases. pressure-gradient equations, two-dimensional Riemann problem, rar-

efaction wave, shock, slip line, regular reflection, Mach reflection, MmB scheme.The second author was supported by State Key Laboratory of Scientific and engineering com-

puting, Academia Sinica. The third author was supported by National Fundamental ResearchProgram of State Commission of Scien. and Tech. of China, and Academia Sinica.

609

610 PENG ZHANG, JIEQUAN LI AND TONG ZHANG

(1.2), and then the second equation can be derived from the conservation law ofmomentum of (1.2) because the quadratic terms ∇·(ρU⊗U) are much smaller thanthe linear terms (ρU)t or the pressure-gradient terms ∇p which are supposed to belarge, while the last equation can be found after dropping the cubic terms from theenergy equation of (1.2) due to the similar reasons. Thus, the pressure-gradientequations (1.1) have their own physical value.

The eigenvalues of this system in the direction (µ, ν) with µ2 + ν2 = 1 are

λ0 = 0, λ± = ±√

γ − 1γ

c, (1.3)

where c =√

γpρ−1 is the sonic speed. These eigenvalues are independent of thevelocity U , compared with those of the Euler system. We also point out that thevorticity of the flow is unchanged as the time passes. Therefore, this flow exhibitsmore simplicity.

Noting that the density ρ remains unchanged as time increases, we hope tosimplify (1.1) slightly but not to change the essential nonlinear structure of theflow. We always assume, from now on, that

ρ ≡ 1

instead of ρt = 0, and further take the transformation (γ − 1)−1p → p′ and (γ −1)−1X → X ′. Then (1.1) can be reduced to the system

Ut +∇p = 0,

Et +∇ · (pU) = 0,(1.4)

where E = p+|U |22

, and primes on X ′ and p′ are dropped just for simplicity. For a

smooth solution or in the region where the solution is smooth, (1.4) can be writtenas

Ut +∇p = 0,

pt + p∇ · U = 0.(1.5)

Therefore, we can obtain the wave equation

(pt

p)t = 4p, (1.6)

where 4 = ∂xx + ∂yy is the Laplace operator. This is very interesting and maybe one of the simplest second order quasi-linear hyperbolic equations awaiting forthe investigation. In the self-similar plane, (ξ, η)-plane ((ξ, η) = (x/t, y/t)), (1.6)or (1.4) is of mixed-type. Throughout the present paper we just pay our attentionto (1.4) rather than the original one (1.1).

In this paper, we consider the Riemann problem for (1.4). The initial data isconstant in each quadrant,

(p, u, v)(0, x, y) = (pi, ui, vi), i = 1, 2, 3, 4, (1.7)

which is so restricted that only one elementary wave, a rarefaction wave, a shock, ora slip line appears at each interface. We follow the same steps as in [ZZ] to analyzethe structure of solutions and present the numerical results with MmB scheme (localmaximum and minimun bounds preserving), which is applicable to a wide varietyof applications [WS]. This problem is discussed in the self-similar plane accordingto the different combinations of elementary waves at the interface of initial data.We classify twelve genuinely different configurations for the system (1.4). Each ofthem is analyzed with the method of characteristics, the boundaries of interactiondomains are clarified, and the corresponding numerical result is illustrated by thecontour plots of pressure and self-similar Mach number. A lot of problems never

2-D RIEMANN PROBLEM FOR PRESSURE-GRADIENT EQUATIONS 611

considered before are proposed. We explicitly give the criterion of transition fromthe regular reflection to the Mach reflection in the interaction of shocks. By thecomparision with the conjecture in [ZZ], we find that this system is very useful tounderstand the complicated flow patterns of the Euler system.

In Section 2, We discuss the resulting self-similar equations of (1.4) and classifythe combinations of elementary waves. We analyze the structure of solutions andpresent the numerical solutions in Section 3, which contains the description of waveinteraction. Our discussions are given in Section 4.

2. Characteristics, discontinuities, elementary waves and classifications.In this section we present a preliminary analysis on (1.4) and give the classificationof combinations of the elementary waves at each interface of initial data.2.1 Characteristics. For smooth solutions, (1.4) can be reduced to the self-similarform

−ξuξ − ηuη + pξ = 0,

−ξvξ − ηvη + pη = 0,

−ξEξ − ηEη + (pu)ξ + (pv)η = 0.

(2.1)

This system can be simplified into a second order partial differential equation afteru and v have been eliminated

(p− ξ2)pξξ − 2ξηpξη + (p− η2)pηη +1p(ξpξ + ηpη)2 − 2(ξpξ + ηpη) = 0, (2.2)

which can also be derived from (1.6) by the same coordinate transformation and isnamed the transonic pressure-gradient equation of Euler system.

The initial data (1.7) becomes

limξ2+η2→∞

(u, v, p) = (ui, vi, pi), (ξ, η) in the ith quadrant. (2.3)

The eigenvalues of (2.1) or (2.2) are

λ0 =η

ξ, λ± =

ξη ±√

p(ξ2 + η2 − p)ξ2 − p

. (2.4)

λ0 is obviously linear, while λ± may be real or complex depending on whether ornot ξ2 + η2 > p.

Define characteristic curves Γj (j = 0,±) in the (ξ, η)-plane by

Γj :dη

dξ= λj ,

which are called pseudo-characteristics (characteristics for abbreviation) of (1.4) inthe supersonic domain for a given solution (u, v, p)(ξ, η). Each of these characteristiccurves has an end-point, which results from the geometry singularity of initial dataat the origin. Obviously, on the sonic circle ξ2 + η2 = c2 (c2 = p), Γ± are tangentto the sonic circle and perpendicular to Γ0. We orient each characteristic from theinfinity to its end-point.

Call that the vectors n1 and n2 form a left-hand system if the angle from n1 to n2

is less than π and greater than zero; otherwise, the vectors n1 and n2 form a right-hand system, as shown in Figure 2.1. Then we can give the following definition.


Right-hand system Left-hand systemFigure 2.1

Definition 1.1 A smooth solution is called a rarefaction wave, denoted by R, ifthe pressure decreases along a slip line. The rarefaction wave can be classified intotwo kinds, R±:

R = R+ if ∇ξηp and the direction of slip line form a left-hand system;R = R− if ∇ξηp and the direction of slip line form a right-hand system.

2.2 Discontinuity. Let η = η(ξ) be a smooth discontinuity of a bounded discon-

tinuous solution of (1.4) or (2.1) with the normal (ξσ − η,−σ, 1) (σ =dη

dξ). Then

the Rankine-Hugoniot relation should hold, i.e.,

(ξσ − η)[u]− σ[p] = 0,

(ξσ − η)[v] + [p] = 0,

(ξσ − η)[E]− σ[pu] + [pv] = 0,

(2.5)

where the quantity in the bracket is the jump across the discontinuity. We can findby solving these either a linear discontinuity

σ0 =η

ξ=

[v][u]

,

[p] = 0,(2.6)

or nonlinear discontinuities

dη

dξ= σ± = − [u]

[v]=

ξη ±√

p(ξ2 + η2 − p)ξ2 − p

,

[p]2 = p([u]2 + [v]2),(2.7)

where p =p + p1

2is the average of the pressures on the two sides of the discontinuity.

Compared with (2.4), the nonlinear discontinuities (2.7) can never be tangent tothe flow line or the λ± characteristic lines. We can also conclude that

[p] = ξ[u] + η[v],

[u] =ξp± η

√p(ξ2 + η2 − p)

p(ξ2 + η2)[p],

[v] =ηp∓ ξ

√p(ξ2 + η2 − p)

p(ξ2 + η2)[p]

(2.8)

corresponding to σ± respectively. This system has only two independent equations,but it gives an accurate relation between the states on the wave front and waveback of shocks. It is expected to be useful in solving the boundary value problemfor (2.1) with a shock as the boundary.

Denote c2 = p, and call ξ2 + η2 = c2 a Rankine-Hugoniot circle (R-H circle forshort) similar to the sonic circle ξ2 + η2 = c2. Therefore an R-H circle must be


located between two sonic circles C1 : ξ2 + η2 = c2 and C2 : ξ2 + η2 = c21 unless

either p or p1 vanishes. Just like the characteristics, a nonlinear discontinuity(2.7) is tangential to the corresponding R-H circle and perpendicular to a lineardiscontinuity there, and it has a tangent point on R-H circle as its end-point,due to the geometry singularity of initial data at the origin. We also orient thediscontinuity (2.6) from infinity to the origin and (2.7) from infinity to its end-point respectively.

Definition 2.2 A discontinuity is called a contact discontinuity or a slip line, de-noted by J , if it satisfies (2.6). A slip line can be classified into two kinds accordingto the sign of vorticity, that is

J± : curl(u, v) = ±∞. (2.9)

A discontinuity is called a shock, denoted by S, if it satisfies (2.7) and the pressurep increases across it along a slip line, that is, the pressure on the wave front islarger than on the wave back. The shock can be classified into two kinds:

S = S+ if ∇ξηp and the direction of slip line form a right-hand system;S = S− if ∇ξηp and the direction of slip line form a left-hand system.

2.3 Planar elementary waves. Consider bounded solutions of the form (u, v, p)(ξ)or (u, v, p)(η). We call these planar elementary waves (exterior waves). We discussthe following four cases as examples; the other cases can be treated similarly. Herewe denote (u1, v1, p1) and (u2, v2, p2) the states on the wave front and the waveback.

(i) Constant states: (u, v, p) = (u0, v0, p0) = const.(ii) Rarefaction waves:

R±(ξ) :

ξ =√

p, p2 < p1,

[u] = 2[√

p],[v] = 0,

η>< 0.

(2.10)

(iii) Shock waves:

S±(ξ) :

ξ =√

p, p2 > p1,

[u] =[p]√

p,

[v] = 0,

η>< 0.

(2.11)

(iv) Slip lines

J± :

ξ = σ0 = 0,

[p] = 0,

[u] = 0,

curl(u, v)|J = ±∞.

(2.12)

The last expression in (2.12) is also equivalent to v2 < v1 or v2 > v1 correspondingto the signs ”minus” or ”plus”.


ξη

Γ

Γ

Γ

Γ

ΓΓΓΓΓ

Γ

Γ

Γ

Γ Γ Γ Γ

+

+

+

−

−

−

0

000

0

00

− +

subsonic domain

C0

Figure 2.2

ξη

Γ

Γ

Γ

ΓΓ+

−

+

Γ−−

+2 1

C

C2

1

Γ−R

Γ+R +

12

12

−

subsonicdomain sonic stem

Figure 2.3

We can analyse the following facts about planar elementary waves with the samemethod as in [ZZ].

(i) The solution is a constant state (u, v, p) = (u0, v0, p0). Its sonic curve is thecircle

C0 : ξ2 + η2 = p0.

This flow is subsonic inside the circle and supersonic outside the circle. The slipline is a ray through the origin. Each of the characteristics Γ± is a ray frominfinity to the sonic circle tangentially. The clockwise ray corresponds to Γ− andthe counterclockwise ray corresponds to Γ+, as illustrated in Figure 2.1.

ξη Γ

2 1

−

+Γ

C1

subsonicdomain

ΓΓ−+

−S12

C12

+

S12−

C2

Figure 2.4

(ii) R+12(ξ) (η > 0) and R−12(ξ) (η < 0) connecting two constant states (u2, v2, p2)

and (u1, v1, p1), as drawn in Figure 2.2. Γ0 are half-ray lines with the origin as thevertex. Γ± are

Γ+ :

ξ =

√p, if η > 0,

(ξ − k)2 + η2 = k2, if η < 0,Γ− :

(ξ − k)2 + η2 = k2, if η > 0,

ξ =√

p, if η < 0.

where k > c2 is an arbitrary constant.


The sonic curve of R+12(ξ) (η > 0) and R−12(ξ) (η < 0) is

η = 0, (√

p2 ≤ ξ ≤ √p1),

which is called the sonic stem.

(iii) S+21(ξ) (η > 0) and S−21(ξ) (η < 0) connecting two constant states (u2, v2, p2)

and (u1, v1, p1), as drawn in Fig.2.3. S±21(ξ) is tangent to the R-H circle

C12 : ξ2 + η2 =p1 + p2

2.

Since c1 < σ12 < c2, the sonic circle C1 and the part of the sonic circle C2 areimaginary. And on two sides of a point on S±21(ξ) in the supersonic domain, thereare three nonlinear characteristic lines are incoming while the left is outgoing, asshown in Figure 2.3.

(iv) J±21 connecting two constant states (u2, v2, p2) and (u1, v1, p1).

J±21 :

ξ = 0,

p1 = p2,

u1 = u2,

(2.13)

where the sign ”plus” or ”minus” is equivalent to that v2 < v1 or v2 > v1. Sincethe pressures on two sides are equal, the sonic circles C1 and C2 are the same andthe halves are imaginary. The two constant states are cut off to shift with differentvelocities.

2.4 Classification. Under the restriction that the initial data (2.1) is so chosenthat only a shock wave, a rarefaction wave or a contact discontinuity appears at eachinterface, there exist 12 genuinely different combinations of exterior waves exceptthree trivial cases, since all other combinations can be transformed into these casesby coordinate rotation and/or reflection transformations.

4R : R+12R

+23R

−34R

−41, R+

12R−23R

+34R

−41;

4S : S−12S+23S

−34S

+14, S−12S

−23S

+34S

+41;

2R + 2S : S−12R−23S

−34R

−14;

2J + 2R : R+12J

+23J

−34R

−14, R−12J

+23J

−34R

+14;

2J + 2S : S−12J−23J

+34S

+14, S+

12J−23J

+34S

−14;

2J + R + S : R+12J

+23J

+34S

+14, R−12J

+23J

+34S

−14, J12R

−23J34S

−14.

3. The analysis on the structures of solutions and numerical results. Inthis section, we will analyze the structure of solutions to the Riemann problem withthe generalized characteristic method, and then present the numerical solutions.The following abbreviations are often used,

Φij = 2(√

pi −√pj), Ψij =pi − pj√pi + pj

2

, (3.1)

where ij ∈ 12, 23, 34, 41. These abbreviations satisfy

Φij = −Φji, Ψij = −Ψji. (3.2)


ξη

C1

subsonicdomain

subsonicdomain2 1

C2 C

J 12J−12+

2

C112

v <v v >v2 1 2 1

Figure 2.5

3.1 The interaction of four rarefaction waves. There are two cases for theinteraction of four rarefaction waves and each of them has two subcases.

Configuration A R+12R

+23R

−34R

−14.

The four constant states must satisfy the following system (A):

R+12 : u1 − u2 = Φ12, v1 = v2, p1 > p2,

R+23 : v2 − v3 = Φ23, u2 = u3, p2 > p3,

R−34 : u4 − u3 = Φ43, v4 = v3, p4 > p3,

R−41 : v4 − v1 = Φ41, u1 = u4, p1 > p4.

(3.3)

Therefore,√

p1 +√

p3 =√

p2 +√

p4. For any fixed p1, u1, v1, p2, p3, we can obtainp4, ui, vi (i = 2, 3, 4) from the first column and the second column of (3.3).

ξη

Γ−

R12+

R

R

R

C

C1C2

Γ−

Γ+

Γ

Γ−

+

+

−

Γ

Γ+

−

23

34

41

3

4

21

3H

G

F

E

A

BC

D

P

−

Γ−

Γ+ Γ+

Figure 3.1

R+12(ξ) and R−14(η) from the infinity meet together at P before they reach their

own sonic circles. Then they will interact with each other. So, the part of theboundary of the interaction region should be extension of characteristic lines Γ−and Γ+ from P. Γ− penetrates R+

12(ξ) and ends at A firstly and then goes straightuntil it intersects R−23(η) at C before R−23(η) arrives at the corresponding sonic stem.This characteristic curve Γ− continues to pass through R−23 and ends at E, and goesstraight again until it is tangent to the sonic circle C3 at G. By the discussion of


the last section, we kown that AP and CE are circular arcs. AC is tangent to AP

and CE at A and C respectively, and EG is tangent to CE at E. The equivalentis true for Γ+ from P . We illustrate these in Figure 3.1.

This case has two subcases depending on whether EG and HF are tangent to C3

or not before they intersect each other. For simplicity, we just consider the case forp2 = p4. Then the solution is axially symmetric ξ = η. So the former occurs if andonly if 0 <

ηG

ξG≤ 1 or vice versa for the latter. This is because EG is perpendicular

to the radius of C3 through G. After a routine calculation, we arrive at

ηG

ξG=

x(x−√2x− 1)− (2− x)(x− 1)2

(x− 1)√

(2− x)[2x(x−√2x− 1)− (2−)(x− 1)2], (3.4)

where x =p1

p3. It can be shown that there is a unique x0(

.= 1.84494) such thatηG

ξG> 1 if and only if 2 > x > x0. Thus we obtain

Theorem 3.1 The characteristic curves Γ− and Γ+ from P are tangent to thesonic circle C3 before they interact each other if and only if 2 > x > x0(

.= 1.884494).

The rarefaction waves R+12 and R−41 interact at P to penetrate each other. The

interaction domain is bounded by the characteristics Γ− and Γ+ from P (and apart of C3 for the first subcase). So, we need to consider the Goursat problem forthe system (2.1) with Γ− and Γ+ as the Cauchy support. The following theoremshows that this problem has a unique continuous supersonic rarefactive solution ina neighborhood of P .

Theorem 3.2 The Goursat problem for (2.1) with PA and PB as the Cauchysupport has a unique smooth supersonic solution in a neighborhood of P .

To prove this theorem, we consider this problem in the polar coordinates (r, θ)with r =

√ξ2 + η2 and tan θ =

η

ξ. Let

(U

V

)=

(cos θ sin θ

− sin θ cos θ

)(u

v

).

Then (2.1) can be written as

AWr + BWθ + c = 0 (3.5)

where

A =

−r 0 10 −r 0p 0 −r

, B =

0 0 0

0 01r

0 −p

r0

,

and

c =(

0, 0,pU

r

)T

, w = (U, V, p)T.

Here and in the following the superscript T represents the transposition of vectors.The eigenvalues of (3.5) are

µ0 = 0, µ± = ± 1r√

m− 1,


where m =r2

pis the square of Mach number. Then the characteristic curves are

defined bydθ

dr= µi, i = 0,±,

and the left characteristic vectors l0 and l± associated with µ0 and µ±, respectively,are

(1, 0, 0), (−1,±√m− 1,−r

p).

Thus, we can write (3.4) as the standard form,

(−r, 0, 1)(∂

∂r+ µ0

∂

∂θ)

U

V

p

= 0,

(0,−r,±√m− 1)(∂

∂r+ µ±

∂

∂θ)

U

V

p

= µ±rU

∆= b±.

(3.6)

Introduce the notationd

dir=

∂

∂r+ µi

∂

∂θ, i = 0,±,

l0 = (−r, 0, 1), l± = (0,−r,±√m− 1),(3.7)

whered

dirrepresents the directional derivative along the ith characteristic curve.

So we express (3.6) as

l0 · dw

d0r= 0,

l± · dw

d±r= b±.

(3.8)

Now we begin to consider the interaction of R+12 and R−14 in a neighborhood of

P . In the polar coordinate system–the (r, θ)-plane, the circular arcs PA and PBcan be expressed as

PA : r = 2 cos θ, (π/4 ≤ θ ≤ θA),PB : r = 2 sin θ, (θB ≤ θ ≤ π/4),

(3.9)

where p1 is assumed to be 1, θA and θB are the polar angles of A and B. Thus,the Cauchy data of this Goursat problem are

W |dPA= (2 cos 2θ cos θ,−2 cos 2θ sin θ, 4 cos4 θ)T ∆= w−,

W |dPB= (−2 cos 2θ sin θ,−2 cos 2θ cos θ, 4 sin4 θ)T ∆= w+.

(3.10)

To guarantee that the Goursat problem has a unique smooth solution, we canfurther check the following compatibility conditions are satisfied.

(1) w−(P ) = w+(P ). This is obvious from (3.9).(2)

l− · dw−

d−r

∣∣dPA= µ−rU |dPA

,

l+ · dw+

d+r

∣∣dPB= µ+rU |dPB

.

These show that the Cauchy data on PA and PB are compatible with the equations(3.6).


(3)

1µ+ − µ0

l0 · dw+

d+r|P =

1µ− − µ0

l0 · dw−

d−r|P (3.11)

This equation shows that at P , the Cauchy data on PB is compatible with that onPA. As a matter of fact, from (3.7), we have

l0 · dw

d−r= l0 · ( dw

d−r− dw

d0r) = l0 · (µ− − µ0)

∂w

∂θ,

l0 · dw

d+r= l0 · ( dw

d+r− dw

d0r) = l0 · (µ+ − µ0)

∂w

∂θ.

(3.12)

Noticing that P = (√

2, π/4) in the (r, θ)-plane and then substituting (3.6) intothese equations, we can check that (3.11) holds.

By now, Theorem 3.2 follows by the theorems in the paper [WW]. ¤The numerical solution in Figure 3.2 illustrates that there is a subsonic region

inside the interaction domain of four exterior rarefaction waves and the supersonicrarefactive solution extends continuously towards the sonic curve. The paper [Z]showed that a positive H1

loc solution exists provided that the sonic curve belongsto C2,α. An open problem is how we can extend the local rarefactive solution ina neighborhood of P to the sonic curve and determine the regularity of the soniccurve. An other open problem is whether the pressure vanishes in a subset of theinteraction domain. The initial data for Figure 3.2 is p1 = 0.525, p2 = 0.2252,p3 = 0.0155, λx = λy = 0.2 with time steps n = 540.

20 40 60 80 100 120 140 160 180 200

20

40

60

80

100

120

140

160

180

200

Configuration A

Pressure contour curves

10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

0.5

1

1.5 2

2 2

2.5

2.5

2.5 3

3.5

Configuration A

Pseudo-Mach number contour curves

Figure 3.2.

Configuration B R+12R

−23R

+34R

−41. For this case, we have the system (B)

R+12 : u1 − u2 = Φ12, v1 − v2 = 0, p1 > p2,

R−23 : v2 − v3 = −Φ23, u2 − u3 = 0, p2 < p3,

R+34 : u3 − u4 = −Φ34, v3 − v4 = 0, p3 > p4,

R−14 : v4 − v1 = Φ41, u1 − u4 = 0, p4 < p1,

(3.13)

from which, we obtain p1 = p3 > p2 = p4. It follows that u1 = u4 > u2 = u3,v1 = v2 > v3 = v4 and u1 − u2 = v1 − v4. Hence, for any fixed p1 > p2, u1, v1, wecan get p3 and p4, ui, vi (i = 2, 3, 4) from the first column and the second columnof (3.13). This solution is symmetric with respect to ξ − η = 0 and ξ + η = 0.


ξη

AP Γ−

R 41−

Γ−

Γ+R12

+Γ+

B

C2

C1

ξ=ηξ=−η

D

E

Γ

Γ

−

+

Figure 3.3

As shown in Figure 3.3, the two forward rarefaction waves R+12 and R−41 coming

from infinity interact at P . Then Γ− (Γ+) from P penetrates R+12 (R−41) firstly to

enter into the constant state (p2, u2, v2) ((p4, u4, v4)) and are tangent to the soniccircle C2 at C(E). Symmetrically, R−23(η) and R+

34(ξ) interact from Q. Γ− (Γ+)penetrates R+

34 (R−23) to enter into (p4, u4, v4) ((p2, u2, v2)) and are tangent to C2

at F (G).Since the symmetric axes ξ = η and ξ = −η are just slip lines, which form

compressive corners, the numerical solution shows that there are two symmetricshock waves separating the subsonic domain from the supersonic domains. Withthe same arguments, we can prove that the solution in a neighborhood of P (Q) israrefactive. The problem is to solve the mixed-type equation (2.2) with the Cauchydata on Γ− and Γ+. The initial data for Figure 3.4 are p1 = 4.2, p2 = 1.4, u1 = 2,v1 = 0.2677, λx = λy = 0.1 with the time steps n = 620.

3.2 The interaction of four shocks. There are two kinds of configurations ofsolutions.Configuration C S−12S

+23S

−34S

+14. For this case, the four constant states satisfy

the system (C)

S−12 : u1 − u2 = −Ψ12, v1 = v2, p1 > p2,

S+23 : v2 − v3 = Ψ23, u2 = u3, p2 < p3,

S−34 : u3 − u4 = Ψ34, v3 = v4, p3 > p4,

S+41 : v4 − v1 = −Ψ41, u2 = u3, p4 < p1.

(3.14)

Similar to Configuration B, we can prove that p1 = p3 > p2 = p4. This results in

u1 = u4 < u2 = u3, v1 = v2 < v3 = v4, u1 − u2 = v1 − v4.


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10

20

30

40

50

60

70

80

90

100

Configuration B


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10

20

30

40

50

60

70

80

90

100

0.5

1

1

1

1.5

1.5

1.5

2

2 2

2

2

2

2.5

2.5

3

3

3.5

3.5

Configuration B


Figure 3.4.

Therefore, the solution is symmetric to ξ = η and ξ = −η. For any fixed p2 < p1,u1, v1, we can get ui, vi (i = 2, 3, 4) from the first column and the second columnof (3.14).

Let us analyze how the four shocks from infinity interact to match together.Obviously, if S−12 and S+

23 meet at the point P = (ξ0, η0) before they reach the soniccircles C1 and C2, the collision of these two shocks results in the formation of tworeflection shocks S+

15 and S−35, which separate the constant states from the state(p5, u5, v5), as shown in Figure 3.5. From (2.8), we have

u5 − u1 =ξ0p15 + η0

√p15(ξ2

0 + η20 − p15)

p15(ξ20 + η2

0)(p5 − p1),

u5 − u3 =ξ0p35 − η0

√p35(ξ2

0 + η20 − p35)

p35(ξ20 + η2

0)(p5 − p3),

(3.15)

where pij =pi + pj

2. Eliminating u5 in (3.15) and noting that p1 = p3 and ξ0 =

−η0 = −√p12, we get

u3 − u1 =2η0

√p15(ξ2

0 + η20 − p15)

p15(ξ20 + η2

0)(p5 − p1),

which is equivalent to√

p1 + p2(u3 − u1)x = 2√

2√

x(p1 + p2 − x)(x− p1), (3.16)

where x = p15. We get by solving (3.16) that

x± =5p1 + p2 −

√−7p2

1 + 10p1p2 + p22

4.

Therefore,

p5 =3p1 + p2 −

√−7p2

1 + 10p1p2 + p22

2.

It follows that p5 is real if and only if p2/p1 ≥ 4√

2−5. Obviously, if p2/p1 ≥ 4√

2−5,

ξ20 + η2

0 −p5 + p1

2=

14[3p2 − p1 +

√−7p2

1 + 10p1p2 + p22] > 0.

Since

ξ20 + η2

0 − p5 =12[p2 − p1 +

√−7p2

2 + 10p1p2 + p22],


we see that ξ20 +η2

0−p5 > 0 if and only if p2/p1 >23. Besides, it can be proved that

p5 − p1 =12[p1 + p2 −

√−7p2

1 + 10p1p2 + p22] > 0,

if p2/p1 ≥ 4√

2−5. Based on the above discussion, we obtain the following theorem.

Theorem 3.3 S−12 and S+23 interact at the point P if and only if

1 > p2/p1 ≥ 4√

2− 5.

The state (u5, v5, p5) at the point P is supersonic if 1 > p2/p1 > 2/3; it is sonic ifp2/p1 = 2/3; and it is subsonic if 2/3 > p2/p1 ≥ 4

√2− 5.

In view of symmetry, the symmetric axis ξ = −η can be regarded as a rigid wall.Therefore, S+

15 should be a regular reflection shock of S−12. We restate Theorem 3.4as follows.

ξη

C2

S12−

34S−

S

41S

23

C1ξ=η

ξ=−η

+

+

1

2

34

5

S15

35S −

+

Figure 3.5

Theorem 3.3′ For the shock S−12, the regular reflection shock occurs if and onlyif 1 > p2/p1 ≥ 4

√2 − 5. The wave back of S+

15 is supersonic if 1 > p2/p1 ≥ 2/3;sonic if p2/p1 = 2/3; and subsonic if 2/3 > p2/p1 ≥ 4

√2− 5.

Thus, we can consider this configuration by two subcases according to the criticalvalue 4

√2− 5.

Subcase C1 1 > p2/p1 > 4√

2 − 5. S−12 and S+23 meet at P to form regular

reflection shocks with the state (p5, u5, v5) in between. If 1 > p2/p1 > 2/3, thestate (p5, u5, v5) is supersonic and S+

15 is tangent to the R-H circles C15 at P . If2/3 > p2/p1 > 4

√2−5, (p5, u5, v5) is subsonic or sonic, so S+

15 separates (p1, u1, v1)from the subsonic domain, goes straight until it reaches C5, then bends towardsC5 clockwise, and becomes weaker and weaker. Finally, it ends at the intersectionpoint of ξ = η and C1∗, where p∗ is the state on the wave back. The equivalentis true for S−34 and S+

41. We illustrate this subcase in Figure 3.6. The problem isto solve a free boundary value problem for (2.2) with the boundary consisting ofreflection shocks and the parts of sonic circle for the former, or the reflection shocksfor the latter. In the system of polar coordinates, the boundary conditions can beconsidered as follows. Letting r = r(θ) be a smooth discontinuity, we write (2.7)


and (2.8) in the form,

dr

dθ= −r

[v][u]

=±r

√r2 − p√p

,

[p] = r[u],[p]2 = p([u]2 + [v]2).

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10

20

30

40

50

60

70

80

90

100

Configuration C1


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10

20

30

40

50

60

70

80

90

100

0.5

1

1.5

2

2

2

2

2.5

2.5

2.5

2.5 3

3

Configuration C1


Figure 3.6.

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10

20

30

40

50

60

70

80

90

100

Configuration C2


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10

20

30

40

50

60

70

80

90

100

0.5

1

1

1.5

2

2

2

2

2.5

2.5

3

3

3.5

3.5

Configuration C2


Figure 3.7.

Then by some calculations, one can obtain that on r = r(θ) there holds

A(r, p, p)pr + B(r, p, p)pθ = 0,

where

A = −(

r2 − p

p+

r2 − p

p− r2(p− p)

2p2 r2 dr

dθ

), B = − 2

r2

(dr

dθ

)3

+r2(p− p)

2p2 .

Therefore, this problem is the free boundary problem for (2.2) with the direc-tional derivative as the boundary value, which is something like the third Dirichletboundary-value problem but is more difficult. It still remains open. We takethe initial data for the numerical solution as p1 = 1.5, p2 = 1.2, u1 = −0.1291,v1 = −0.1291, λx = λy = 0.1 with time steps n = 840.


Subcase C2 p2/p1 < 4√

2 − 5. For this subcase, the Mach reflection shock willoccur. There is a triple configuration of shocks on S−12. The Mach stem reachesthe rigid wall vertically, and the reflection shock separates (p1, u1, v1) from thesubsonic domain. The initial data for the numerical solution (see ConfigurationC2) is p1 = 2.5, p2 = 0.8, u1 = 0, v1 = 0, λx = λy = 0.05 with the time stepsn = 1100.

ξη

S41+

3

C1

S

S34

23

ξ=ηS12

−

C C12

C23

C2

−

+

Figure 3.8

10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

Configuration D


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10

20

30

40

50

60

70

80

90

100

1

2

3

3 4

4

4 5

6

7

8

Configuration D


Figure 3.9.Configuration D S−12S

−23S

+34S

+14. For this case, we have the system (D)

S−12 : u1 − u2 = −Ψ12, v1 − v2 = 0, p1 > p2,

S−23 : v2 − v3 = −Ψ23, u2 − u3 = 0, p2 > p3,

S+34 : u4 − u3 = −Ψ34, v4 − v3 = 0, p4 > p3,

S+14 : v1 − v4 = −Ψ14, u1 − u4 = 0, p1 > p4.

(3.17)

Based on this system, we can verify that

p2 = p4, (p1

p2+ 1)(

p3

p2+ 1) = 4. (3.18)

Therefore, u1 = u4 < u2 = u3, v1 = v2 < v3 = v4, u1 − u2 = v1 − v4. The solutionis symmetric to ξ = η. For any fixed p2 < p1, u1, v1, we can obtain p2 and p4 from(3.18), ui, vi from the first and the second columns of (3.17).


Just as in Configuration C, ξ = η can be viewed as a rigid wall. S−23 meetsξ = η before it reaches the sonic circle C2. If 1 > p3/p2 > 4

√2− 5, then a regular

reflection shock occurs. Otherwise, a Mach reflection shock appears. The reflectionshock bends clockwise. S−12 begins to bend counterclockwise with decreasing wavestrength after it meets the sonic circle C1. This shock matches the reflection shockon its R-H circle, where the minimum of the pressure is taken. The initial datafor the numerical solution is p2 = 1.2, p1 = 2.5, λx = λy = 0.05 with time stepsn = 1100.

ηξ

Γ

Γ

Γ

Γ

−

−

S

S

CC

C

12

R

R −

−

−

−

23

34

41

12

2

1

−

−

Figure 3.10

3.3 The interaction of two rarefaction waves and two shock wavesConfiguration E S−12R

−23S

−34R

−14. The four constant states satisfy the system (E)

S−12 : u1 − u2 = −Ψ12, v1 − v2 = 0, p1 > p2,

R−23 : v2 − v3 = −Φ23, u2 − u3 = 0, p2 < p3,

S−34 : u3 − u4 = Ψ34, v3 − v4 = 0, p3 > p4,

R−41 : v4 − v1 = Φ41, u4 − u1 = 0, p4 < p1.

(3.19)

From this system, we obtain p1 = p3 > p2 = p4. So, u1 = u4 < u2 = u3, v1 =v2 > v3 = v4, u1 − u2 = v4 − v1. For any fixed p2 < p1, u1, v1, we can obtain p3,p4, and ui, vi (i = 2, 3, 4) from the first and the second column from (3.19).

Since the solution is rotationally symmetric with the rotation angle α = π/2, itsuffices to consider R−41 and S−34. S−34 first meets the sonic circle C3 and then bendsinward to separate (p4, u4, v4) from the subsonic domain with decreasing strength.This shock matches the weak shock resulting from R−41 at its R-H circle C4∗, wherep∗ is the pressure on the wave back. The numerical results show that R−41 extends tointeract with S−12. There is a subsonic domain, being bounded by the shocks. Theinitial data we choose is: p1 = 2.525, p2 = 1.4, u1 = 0, v = 0.8116, λx = λy = 0.1with time steps n = 480.

3.4 The interaction of two rarefaction waves and two slip lines. We dividethis into two cases, Configurations F and G. The initial data distribution satisfiesp2 = p3 = p4 and u1 = u3 = u4 > u2 and v1 = v2 = v3 > v4. For any fixed p2, p1,u1, v1, we can easily find ui, vi (i = 2, 3, 4) from (2.10) and (2.12). The solution issymmetric to ξ = η.


10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

Configuration E


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10

20

30

40

50

60

70

80

90

100

0.5 1

1 1

1.5

2

2.5

3

3

3

3

3.5

3.5

3.5

3.5

4

4

4.5

4.5

Configuration E


Figure 3.11.

ηξ

Γ

Γ

C

C

R −41

2

1

RΓ

12

−

+++Γ

J

J23

34+

+−

Γ

Γ+

−

Figure 3.12

Configuration F R+12J

+23J

−34R

−14. The four constant states satisfy the system (F)

R+12 : u1 − u2 = Φ12, v1 − v2 = 0, p1 > p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J−34 : u3 = u4, v3 > v4, p3 = p4,

R−14 : v4 − v1 = Φ41, u4 − u1 = 0, p4 < p1.

(3.20)

So, p1 > p2 = p3 = p4. R+12 and R−41 meet at P and interact with each other.

Γ− (Γ+) from P penetrates R+12 (R−41) and are finally tangent to C2. J+

23 and J−34first reach C3 and then enter into the subsonic domain. Therefore the interactionregion is bounded by Γ− and Γ+ and the part of C3. In view of Theorem 3.2, thereexists a unique a rarefactive smooth solution in the neighborhood of P . R+

12 andR−41 penetrate each other. It seems from the numerical solutions that the part ofthe solution in the subsonic domain is compressive. We choose the initial data asp1 = 2.525, p2 = 1.4, u1 = 0, v1 = −0.8116, λx = λy = 0.2 with time steps n = 400.


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10

20

30

40

50

60

70

80

90

100

Configuration F


10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

0.5

1

1.5

1.5

2

2

2.5

2.5 2.5

Configuration F


Figure 3.13.

Configuration G R−12J+23J

−34R

+14. The four constant states satisfy (G)

R−12 : u1 − u2 = −Φ12, v1 − v2 = 0, p1 < p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J−34 : u3 = u4, v3 > v4, p3 = p4,

R+14 : v4 − v1 = −Φ41, u4 − u1 = 0, p4 > p1.

(3.21)

So, p1 < p2 = p3 = p4. R−12 and R+41 reach their singularities. On the upper-right

part, there exists a weak shock separating (p1, u1, v1) from the subsonic domain. Wewill discuss this configuration carefully in the next section. Here we take the initialdata for the numerical solution as p1 = 2.525, p2 = 3.4, u1 = 0, v1 = −0.5098,λx = λy = 0.1 with time steps n = 520.

ηξ

J

J23

34

+C1

C2

R 12−

R 41+

−

Γ Γ

Γ+

Γ+

−−

Figure 3.14

3.5 The interaction of two shocks and two slip lines. We also have twocases for this interaction, Configurations H and I. The four constant states satisfyp2 = p3 = p4, u1 = u3 = u4 < u2 and v1 = v2 = v3 < v4. For any fixed p2, p1,u1 and v1, we can find ui, vi (i = 2, 3, 4) from (2.11) and (2.12). The solution issymmetric to ξ = η.


10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

Configuration G


10 20 30 40 50 60 70 80 90 100

10

20

30

40

50

60

70

80

90

100

0.5

1

1.5

2

2.5

2.5

2.5 2.5

2.5

2.5

3

Configuration G


Figure 3.15.

Configuration H S−12J−23J

+34S

+14. For this case, we have the system (H)

S−12 : u1 − u2 = −Ψ12, v1 − v2 = 0, p1 > p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J−34 : u3 = u4, v3 > v4, p3 = p4,

S+41 : v1 − v4 = −Ψ14, u4 − u1 = 0, p4 < p1.

(3.22)

So, p1 > p2 = p3 = p4. S−12 and S+41 bends towards the upper right part after it

reaches the sonic circle C1. Then they becomes weaker and weaker until they matchtogether at their R-H circle, as shown in Figure 3.16. These shocks bound a ball-like subsonic domain from below. The solution in the subsonic domain should besmooth and increasing along the radial direction. The initial data for the numericalsolution is p1 = 2.525, p2 = 1.4, u1 = 0, v1 = 0.8031, λx = λy = 0.1 with the timesteps n = 540.

ηξ

J

J23

34

+C

C

−

C12

2

1

S −12

S41+

Figure 3.16


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10

20

30

40

50

60

70

80

90

100

Configuration H


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10

20

30

40

50

60

70

80

90

100

0.5

1 1.5

2

2.5

2.5

3

3

3.5

3.5

3.5

4

4

4

Configuration H


Figure 3.17.

Configuration I S+12J

−23J

+34S

−14. We have the system (I)

S+12 : u1 − u2 = Ψ12, v1 − v2 = 0, p1 < p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J−34 : u3 = u4, v3 > v4, p3 = p4,

S−41 : v1 − v4 = Ψ41, u4 − u1 = 0, p4 > p1,

(3.23)

which gives p1 < p2 = p3 = p4. If p1/p2 ≥ 4√

2 − 5, S+12 reaches ξ = η and a

regular reflection shock S2∗ occurs with the state (p∗, u∗, v∗) on the wave back.The wave back of this shock at P is supersonic if p1/p2 > 2/3; sonic if p1/p2 = 2/3and subsonic if 2/3 > p1/p2 > 4

√2 − 5. If p1/p2 ≤ 4

√2 − 5, the Mach reflection

shock occurs in the same way as in Configuration D, see Figure 3.18. The reflectionshock propagates with decreasing wave strength and finally vanishes at its R-Hcircle. Symmetrically, the equivalent is true for the reflection shock from S−41.These reflection shocks bound a subsonic domain, which is like a bubble. Theinitial data for the numerical solution is p1 = 2.525, p2 = 4.4, u1 = 0, v1 = 1.0076,λx = λy = 0.1 with time steps n = 530.

ηξ

J

J23

34

+C

−

C21

C12

S12

S41

+

−ξ=η

S

S

+

−

Figure 3.18


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20

30

40

50

60

70

80

90

100

Configuration I


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10

20

30

40

50

60

70

80

90

100

0.5

1

1.5

2

2.5

2.5

2.5

2.5

3

3

3

3

3.5 4

Configuration I


Figure 3.19.

3.6 The interaction of a rarefaction wave, a shock and two slip lines.There are three cases for this interaction. The two involve two neighboring sliplines and the left involves two non-neighboring slip lines. Compared with the casesin Subsection 3.5, the solutions are no longer symmetric.Configuration J R+

12J+23J

+34S

+41. The four constant states satisfy the system (J)

R+12 : u1 − u2 = Φ12, v1 = v2, p1 > p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J+34 : u3 = u4, v3 < v4, p3 = p4,

S+41 : v4 − v1 = −Ψ41, u4 = u1, p4 < p1.

(3.24)

ηξ

J

J23

34

+C12

2C

C1

S41+

+

ΓΓ ++R 12

+

Figure 3.20

So, p1 > p2 = p3 = p4, u1 = u3 = u4 < u1, v1 = v2 = v3 < v4. The shockwave S+

41 bends clockwise with decreasing wave strength after it reaches the soniccircle C1. On the wave back of the rarefaction wave R+

12 is a weak shock, whichpropagates and finally vanishes on its R-H circle. Thus, the shocks and the part ofC1 bound a subsonic domain. The initial data for the numerical solution is p1 = 2.5,p2 = 1, u1 = 0, v1 = 0, λx = λy = 0.1 with time steps n = 620.


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10

20

30

40

50

60

70

80

90

100

Configuration J


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10

20

30

40

50

60

70

80

90

100

0.5

1 1.5

2

2.5

2.5 3

3.5

3.5

3.5

4

4 4

4.5

4.5 4.5

Configuration J


Figure 3.21.

ηξ

J

J23

34

+C12

+

R 12

C1

Γ Γ

C2

S 41−

−

−−

Figure 3.22

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10

20

30

40

50

60

70

80

90

100

Configuration K


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10

20

30

40

50

60

70

80

90

100

0.5

1

1.5

1.5

1.5

2

2

2.5

Configuration K


Figure 3.23.


Configuration K R−12J+23J

+34S

−41. The four constant states satisfy the system (K)

R−12 : u1 − u2 = −Φ12, v1 = v2, p1 < p2,

J+23 : u2 < u3, v2 = v3, p2 = p3,

J+34 : u3 = u4, v3 < v4, p3 = p4,

S−41 : v4 − v1 = Ψ41, u4 = u1, p4 > p1,

(3.25)

which results in p1 < p2 = p3 = p4, u1 = u3 = u4 < u1,v1 = v2 = v3 < v4.As in Configuration J, on the wave back of R−12 is a shock. which matches theextension of S−41 on the R-H circle. We take the initial data for the numerical resultas p1 = 2.525, p2 = 6.2, u1 = 0, v1 = 0, λx = λy = 0.1 with the time steps n = 640.

ηξ

J34

C12

C1 C2

S 41−

J12

R 23−

Γ−

Γ −

Figure 3.24

Configuration L J12R−23J34S

−41. The four constant states satisfy the system (L):

J12 : u1 = u2, p1 = p2,

R−23 : v2 − v3 = −Φ23, v2 = v3, p2 < p3,

J34 : u3 = u4, p3 = p4,

S−41 : v4 − v1 = Ψ41, u2 = u3, p4 > p2.

(3.26)

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10

20

30

40

50

60

70

80

90

100

Configuration L


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10

20

30

40

50

60

70

80

90

100

Configuration L


Figure 3.25.


So, we havep1 = p2 < p3 = p4, u1 = u2 = u3 = u4.

The shock S−41 stops at the R-H circle, where it matches the shock from R−23. Thesubsonic domain is bounded by the part of sonic circle from the lower-right andthe shocks. The initial data we choose is p1 = 2.525, p3 = 7.2, u1 = 0, v1 = 0,v2 = −0.5, v3 = −2.6885, v4 = 2.1201, λx = λy = 0.1 with time steps n = 560.

4. Discussions. Since the structures of solutions to two-dimensional Riemannproblem for gas dynamics are conjectured in [ZZ], few analytic results are availableto prove the complicated flowfield patterns. The present work just attempt tosimplify the Euler system so that it is possible to establish analytic theories on theinteraction of elementary waves in two dimensions. The results show that the sliplines have little influence on the structures of solutions so that the interaction ofrarefaction waves and shocks can be studied thoroughly. The present paper alsogives lots of flowfield patterns similar to those in [ZZ], but is much simpler. Thecriterion of transition from the regular reflection shock to the Mach reflection shockis presented and expected to be useful in the understanding of the oblique shockreflection.

In the self-similar plane, the pressure-gradient equations are of mixed-type. How-ever, in many flowfield patterns presented here, the subsonic domain is bounded byshock waves and the parts of sonic circles, which proposes the free boundary-valueproblem for (2.1) with shocks and parts of sonic circles as the boundary. Owing to(2.8), we can solve this problem for (2.2) provided that the solution is smooth in thesubsonic domain. So far, there is no general theory on the boundary-value problemfor high-order partial different equations. The elegant second-order equation (2.2)may becomes a touchstone. By the observation on the numerical results, we findthe maximum principle is taken in the subsonic domain.

ηξ

J

J23

34

+

C2

R 12−

R 41+

−

Γ Γ

Γ+

Γ+

−−

P

Q

EC1

F

G

H

Γ

Γ

+

−

Figure 4.1

Our purpose to solve two dimensional Riemann problem is just to study howelementary waves interact. Oblique shock reflection is the most famous examples.In present paper, we classify elementary waves R, S, and J into two kinds, re-spectively. The numerical solutions shows that the interaction of rarefaction wavesthemselves, or rarefaction waves and other types of elementary waves may result inthe occurrence of shocks, see Configurations A, B, F, E, G, J, K, L. These phenom-


ena are completely different from those in one dimension. We can further checkthat all these are compatible with the maximum principle aforementioned, whichcan be shown by taking Configuration G as an example, see Figure 4.1. Drawthe characteristics Γ+ and Γ− from P and Q, respectively. They are tangent tothe sonic circle C1 at F and H. Then a quarter of C2, the circular arc FH, Γ+

and Γ− bound a domain Ω, outside of which the solution is determined by thefour constant states (ui, vi, pi) (i = 1, 2, 3, 4), and consists of four constant states,rarefaction waves R−12 and R+

41 and two slip lines J+23 and J−34. Assume the solution

consists of rarefaction waves inside Ω. Then the solution takes its minimum at theorigin because all slip lines point towards there, which contradicts the maximumprinciple.

References[AH] R.K.Agarwal and D.W.Halt, A modified CUSP scheme in wave/particle split form for un-

structed grid Euler flow, Frontiers of Computational Fluid Dynamics, David A. Caughey andMohamed M Hafez, 1994:155-163.

[LC] Y. Li and Y. Cao, Second order ”large particle” difference method, Sciences in China, inChinese, A. 8. 1985.

[WW] R. H. Wang and Z. Q. Wu, On mixed initial boundary value problem for quasilinear hyper-

bolic system of partial differential equations in two independent variables, J. Jilin University,2(1963), 459-502.

[WY] H. M. Wu and S. L. Yang, MmB- A new class of accurate high resolution schemes for conser-

vation laws in two dimensions, IMPACT of Computing in Sciences and Engineering, 1(1989),217-289.

[Z] Y. X. Zheng, Existence of solutions to the transonic pressure gradient equations of the com-

pressible Euler equations in elliptic regions, to appear in Comm.in Partial Diff. Equat.

[ZZ] T. Zhang and Y. X. Zheng, Conjecture on the structure of solution of the Riemann problem

for two-dimensional gas dynamics systems, SIAM J. Math. Anal. Vol. 21, No. 3, pp. 593-630,

May, 1995.

Received for publication December 1997.Email addressPeng Zhang: pzhanght.rol.cn.netJiequan Li: ljqamath6.amt.ac.cnTong Zhang: tzhangmath03.math.ac.cn

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