Interaction of Rarefaction Waves of the Two-Dimensional …math0.bnu.edu.cn/~lijiequan/publications/ARMA...of two-dimensional Riemann problems, the expansion problem of a wedge of

Digital Object Identifier (DOI) 10.1007/s00205-008-0140-6Arch. Rational Mech. Anal. 193 (2009) 623–657

Interaction of Rarefaction Wavesof the Two-Dimensional Self-Similar Euler

Equations

Jiequan Li & Yuxi Zheng

Communicated by T.-P. Liu

Abstract

We construct classical self-similar solutions to the interaction of two arbitraryplanar rarefaction waves for the polytropic Euler equations in two space dimensions.The binary interaction represents a major type of interaction in the two-dimensionalRiemann problems, and includes in particular the classical problem of the expansionof a wedge of gas into vacuum. Based on the hodograph transformation, the methodemployed here involves the phase space analysis of a second-order equation andthe inversion back to (or development onto) the physical space.

1. Introduction

Consider the two-dimensional isentropic compressible Euler system⎧⎪⎨

⎪⎩

ρt + (ρu)x + (ρv)y = 0,

(ρu)t + (ρu2 + p)x + (ρuv)y = 0,

(ρv)t + (ρuv)x + (ρv2 + p)y = 0,

(1)

where ρ is the density, (u, v) is the velocity and p is the pressure given byp(ρ) = Kργ where K > 0 will be scaled to be one and γ > 1 is the gas constant.We are primarily interested in the so-called pseudo-steady case of (1); that is, thesolutions depend on the self-similar variables (ξ, η) = (x/t, y/t). The expansionof a wedge of gas into vacuum is such a case. Assuming the flow is irrotational,a classical hodograph transformation (see [22]) can be used to eliminate the twoself-similar variables, resulting in a partial differential equation of second order forthe speed of sound c in the velocity variables (u, v). It has been known that the diffi-culties of the procedure are that the transformation is degenerate for common wavessuch as the constant states and some other types of waves resembling the simplewaves of the steady Euler system, and the transformation of boundaries is difficult

624 Jiequan Li & Yuxi Zheng

to handle. In 2001, Li [14] carried out an analysis of the second order equationin the space (c, u, v), where he discovered a pair of variables resembling thewell-known Riemann invariants together with their invariant regions and estab-lished the existence of a solution to the expansion of a wedge of gas into vac-uum in the hodograph plane for wide ranges of the gas constant and the wedgeangle. Recently, in 2006, paper [16], in an attempt to establish the inversion of thehodograph mapping, clarified the concept of simple waves for (1). We show in thispaper that the hodograph transformation is non-degenerate precisely for non-simplewaves, and all the solutions constructed in [14] in the hodograph plane can now betransformed back to the self-similar plane. Thus we complete the procedural circleof construction of solutions.

We find that the circle of construction bears a very interesting similarity to theconstruction of centered rarefaction waves in the one-dimensional systems of con-servation laws. The self-similar variable(s) in both cases decouple from the phasespace(s), and the equations in the phase space(s) are solved first. The developmentof the solutions from the phase space onto the physical space(s) requires genuinenon-linearity in the one-dimensional case and a non-degeneracy condition of thehodograph transformation in the two-dimensional case.

The key ingredient of this paper is the simplification of the form of the equa-tions in the phase space brought about by the employment of the inclination anglesof characteristics, and the discovery of the precise forms of the equations for thesecond-order derivatives. The approach now presents itself as a method of signifi-cant potential for the study of the pseudo-steady Euler system in hyperbolic regions.It yields structure of solutions in addition to existence. We use this approach, forexample, to establish the Lipschitz continuity and monotonicity behavior of thevacuum boundary in the problem of a wedge of gas into vacuum and establishthe dependence of the location of the boundary on the wedge angle and gas con-stant. The approach is particularly suitable for studying two-dimensional Riemannproblems [27], since the apparent nature of the solutions of a Riemann problem ispiece-wise smooth.

The expansion problem of gas into vacuum has been a favorite for a longtime. The problem has been interpreted hydraulically as the collapse of a wedge-shaped dam containing water initially with a uniform velocity, see Levine [12].In Suchkov [25], a set of interesting explicit solutions were found. Mackie [20]proposed a scalar equation of second order for a potential function, studied theinterface of gas and vacuum by the method of unsteady Prandtl–Meyer expansionsand related it to the PSI approach in [22], from which Li [14] started with newmotivation from the success on the pressure gradient system [7]. In the contextof two-dimensional Riemann problems, the expansion problem of a wedge of gasinto vacuum is the interaction of two two-dimensional planar rarefaction waves.We see it as one of two possible interactions of continuous waves in the hyperbolicregion; this one expands without shocks but with a boundary degeneracy, whilethe other one forms shocks with a sonic boundary as well as a shock boundary.The method applies locally in both cases. A quick round-up of cases that involvehyperbolic regions of non-constant continuous waves [1–3,8,9,11,15,23,28] showthat the approach taken here has general applicability.

Interaction of Rarefaction Waves for Two-Dimensional Euler 625

Our main results are the simple form of the equations in the phase space (c, u, v)[system (73)], the existence of solutions of the expansion of a wedge of gas intovacuum (Theorem 7), and the detailed properties of the expansion (Theorems 8, 9).We provide some background information as well regarding the hodograph trans-formation and simple waves in Sections 2–5 for the convenience of non-expertreaders. Section 6 is for the phase space analysis, while Section 7 handles the gasexpansion problem. We point out that the main work of this paper is the estab-lishment of the invariance of various triangles and the validity of the inversion ofthe hodograph transformation with the associated trial-and-error process in findingthe precise forms for the second-order equations suitable for bootstrapping [seeformulae (82), (85)]. The difficulty of the inversion manifests itself in the fact thatthe characteristics in the phase space do not have a fixed convexity type althoughthe corresponding characteristics in the physical space do (in most cases), seeSections 7.5 and 7.7. We mention additionally that hodograph transforms havebeen used in various forms, see [4,21] and references therein.

Here is a list of our notations: ρ density, p pressure, (u, v) velocity, c = √γ p/ρ

speed of sound, i = c2/(γ − 1) enthalpy, γ gas constant, (ξ, η) = (x/t, y/t) theself-similar (or pseudo-steady) variables, ϕ pseudo-velocity potential, θ wedgehalf-angle, and

U = u − ξ, V = v − η, κ = (γ − 1)/2, m = (3 − γ )/(γ + 1).

α1 = 2m

3 + m + √(3 + m)2 + 4m

, tan n = √α1, tan m = √

m.

Λ+ = tan α, Λ− = tan β, ω = α − β

2, Λ± = U V ± c

√U 2 + V 2 − c2

U 2 − c2 .

∂± = ∂ξ +Λ±∂η, ∂± = ∂u + λ±∂v, ∂0 = ∂u, Λ±λ∓ = −1.

∂+ = (sin β,− cosβ) · (∂u, ∂v), ∂− = (sin α,− cosα) · (∂u, ∂v).

Letters C , C1 and C2 denote generic constants.

2. Primary system

Our primary system is system (1) in the self-similar variables (ξ, η)=(x/t, y/t):⎧⎨

⎩

(u − ξ)iξ + (v − η)iη + 2κ i (uξ + vη) = 0,(u − ξ)uξ + (v − η)uη + iξ = 0,(u − ξ)vξ + (v − η)vη + iη = 0.

(2)

We assume further that the flow is irrotational:

uη = vξ . (3)

Then, we insert the second and third equations of (2) into the first one to deducethe system,(2κ i − (u − ξ)2)uξ − (u − ξ)(v − η)(uη + vξ )+ (2κ i − (v − η)2)vη = 0,uη − vξ = 0,

(4)


supplemented by Bernoulli’s law

i + 1

2((u − ξ)2 + (v − η)2) = −ϕ, ϕξ = u − ξ, ϕη = v − η. (5)

We remark that the difference between the pseudo-steady flow (4) and thesteady case (32) (see Section 7) is that the latter is self-contained, since the soundspeed c can be expressed by a pointwise function of the velocity explicitly throughBernoulli’s law (33).

3. Concept of hodograph transformation

We introduce briefly the well-known hodograph transformation. The originalform of a hodograph transformation is for a homogeneous quasi-linear systemof two first-order equations for two known variables (u, v) in two independentvariables (x, y). By regarding (x, y) as functions of (u, v) and assuming thatthe Jacobian does not vanish nor is infinity, one can re-write the system for theunknowns (x, y) in the variables (u, v), which is a linear system if the coefficientsof the original system do not depend on (x, y). See the book of Courant andFriedrichs [5]. Specifically, consider the system of two equations of the form,

(uv

)

x+ A(u, v; x, y)

(uv

)

y= 0, (6)

where the coefficient matrix A(u, v; x, y) is

A(u, v; x, y) =(

a11 a12a21 a22

)

. (7)

The two eigenvalues, denoted by Λ±, satisfies,

Λ2± − (a11 + a22)Λ± + |A| = 0. (8)

We introduce the hodograph transformation,

T : (x, y) → (u, v). (9)

Then (6) is reduced to the system(

yv−yu

)

+ A(u, v; x, y)

(−xvxu

)

= 0. (10)

Its eigenvalues, denoted by λ±, satisfy

a12λ2± − (a22 − a11)λ± − a21 = 0. (11)

Obviously, if the coefficient matrix A does not depend on (x, y), (10) becomesa linear system for the unknowns (x, y).

The following proposition establishes the invariance of characteristics underthe hodograph transformation.


Proposition 1. (Invariance of characteristics): A characteristic of (6) in the (x, y)plane is mapped into a characteristic of (10) in the (u, v) plane by the hodographtransform T .

Proof. Let y = y(x) be a characteristic with dydx = Λ±. Its image is v = v(u)

under the hodograph transform (9). Then we have

dy

dx= yu + yv · dv

du

xu + xv · dvdu

= Λ±, (12)

that is,dv

du= −Λ±xu − yu

Λ±xv − yv. (13)

Using (10) we find

(Λ±xu − yu)+ λ±(Λ±xv − yv) = 0. (14)

Therefore, we havedv

du= λ±( or λ∓), (15)

which completes the proof of this proposition. The idea of hodograph transformation does not obviously generalize to other

systems such as systems (4) of more than two simple equations or for inhomoge-neous systems.

For (4), we realize that the three variables (i, u, v) are functions of (ξ, η), so wecan still try to use (u, v) as the independent variables and regard (ξ, η) as functionsof (u, v) and ultimately regard i as a function of (u, v). In this way we may obtainan equation for i = i(u, v) in the plane (u, v) which eliminates (ξ, η). This wasdone in 1958 in a paper [22] by Pogodin et al. and has been referred to as the PSIapproach. The implementation is as follows. Let the hodograph transformation be

T : (ξ, η) → (u, v) (16)

for (2), reverse the roles of (ξ, η) and (u, v) and regard i as a function of (u, v).Then i as the function of u and v satisfies

(uξ vη − uηvξ )di = (iξ vη − iηvξ )du + (−iξuη + iηuξ )dv. (17)

We insert this into the law of momentum conservation of (2) and use the irrotation-ality condition (3) to obtain

ξ − u = iu,

η − v = iv.(18)

These interesting identities provide an explicit correspondence between the physicalplane and the hodograph plane provided that the transformation T is not degenerate.

Therefore, using (18), we convert (4) into a linear (in fact, linearly degenerate,see Section 6) system

(2κ i(u, v)− i2

u

)ηv + iuiv(ξv + ηu)+ (

2κ i − i2v

)ξu = 0,

ξv − ηu = 0.(19)


for the unknowns (ξ, η). The difficulty here is that i , as a function of u and v, cannotbe determined explicitly and point-wise. We will remark more later in contrast withthe steady case.

We continue to differentiate (18) with respect to u and v:

ξu = 1 + iuu, ξv = iuv,

ηu = iuv, ηv = 1 + ivv,(20)

and inserting these into the first equations of (19) to obtain,(2κ i − i2

u

)ivv + 2iuiviuv + (

2κ i − i2v

)iuu = i2

u + i2v − 4κ i. (21)

This is a very interesting second order partial differential equation for i alone. Sothe study of irrotational, pseudo-steady and isentropic fluid flow can proceed along(21).

We point out for the case γ = 1 that the dependent variable i = ln ρ, insteadof i = c2/(γ − 1), is used [13]. Then we can obtain a similar equation for i ,

(1 − i2

u

)ivv + 2iuiviuv + (

1 − i2v

)iuu = i2

u + i2v − 2. (22)

We will establish in the pseudo-steady case that the transform is not degenerate,that is,

JT (u, v; ξ, η) = ∂(u, v)

∂(ξ, η)= uξ vη − uηvξ = 0 (23)

in regions of non-simple waves, to be detailed later. In the direction from (u, v)plane to the (ξ, η) plane, it is more direct to compute

J−1T (u, v; ξ, η) = ξuηv − ξvηu = 0. (24)

Noting that (4) and (19) are all two-by-two systems, we use Proposition 1 toassert that the characteristics of (4) are mapped into the characteristics of (19) bythe hodograph transformation (16). Moreover, the eigenvalues of (4) are

Λ± = (u − ξ)(v − η)± c√(u − ξ)2 + (v − η)2 − c2

(u − ξ)2 − c2 , (25)

while the eigenvalues of (19) are

λ± =iuiv ± c

√(i2u + i2

v − c2)

c2 − i2v

. (26)

By using (18), it is easy to see that

λ± = − 1

Λ∓. (27)

Furthermore, there is a correspondence between Λ± and λ±. Indeed, let η = η(ξ)

be a characteristic curve in the (ξ, η) plane with dηdξ = Λ+ and be mapped onto a

curve v = v(u). Then, using (20) and (27), we have

Λ = dη

dξ= ηu + ηv

dvdu

ξu + ξvdvdu

, (28)


that is,

dv

du= −ξuΛ+ − ηu

ξvΛ+ − ηv= − (1 + iuu)Λ+ − iuv

iuvΛ+ − (1 + ivv)= − iuu + 1 + λ−iuv

iuv + λ−(ivv + 1). (29)

We rewrite (21) as

iuu + 1 + (λ− + λ+)iuv + λ−λ+(ivv + 1) = 0. (30)

Then we concludedv

du= λ+. (31)

Similarly, we obtain the correspondence between Λ− and λ−.

3.1. Steady Euler

The steady isentropic and ir-rotational Euler system of (1) has the form

(c2 − u2)ux − uv(uy + vx )+ (c2 − v2)vy = 0,uy − vx = 0,

(32)

where c is the sound speed, given by Bernoulli’s law

u2 + v2

2+ c2

γ − 1= k0

2, (33)

where k0 is a constant, see [5]. Using the hodograph transform from (x, y) to (u, v),we obtain a linear system,

(2κ i − u2)yv + uv(xv + yu)+ (2κ i − v2)xu = 0,xv − yu = 0.

(34)

The hodograph transform is valid in the region of non-simple waves. With the sameset of steps in deriving (18), we obtain

−u = iu, −v = iv. (35)

This can also be obtained formally from (18) by regarding the steady flow as thelimit of unsteady flow (1) in t → ∞. Comparing (35) with (18), we see that it ismuch more difficult to convert the hodograph plane of the steady case back intothe physical plane than the pseudo-steady case. However, system (34) has moreadvantage over (19) of the pseudo-steady case because i is expressed in an explicitform by Bernoulli’s law (33).


3.2. Similarity to one-dimensional problems

The current approach parallels the procedure that is used to find centeredrarefaction waves to genuinely non-linear strictly hyperbolic systems of conserva-tion laws in one space dimension. Recall for a one-dimensional system ut + f (u)x =0 of n equations, a centered rarefaction wave takes the form ξ = λk(u) for ak ∈ (1, n) and the state variable u satisfies the system of ordinary differential equa-tions ( f ′(u)− λk(u)I )uξ = 0, whose solutions are rarefaction wave curves in thephase space. The development (or inversion) of the phase space solutions onto theξ -axis requires the monotonicity of λk(u) along the vector field of the k-th righteigenvector rk ; that is, genuine non-linearity. For the self-similar two-dimensionalEuler system, we have a pair ξ = u+iu, η = v+iv from (18) in place of ξ = λk(u);and the second-order partial differential equation (21) in place of the ordinary dif-ferential system. For inversion to the physical space, we show that the JacobianJ−1

T of (24) does not vanish.

4. Simple waves

4.1. Concept of simple waves

Now we recall some facts about simple waves. Simple waves were system-atically studied, for example in [10], for hyperbolic systems in two independentvariables,

ut + A(u)ux = 0, (36)

where u = (u1, . . . , un), the n × n matrix A(u) has real and distinct eigenvalues

λ1 < · · · < λn for all u under consideration. They are defined as a special familyof solutions of the form

u = U (φ(x, t)). (37)

The function φ = φ(x, t) is scalar. Substituting (37) into (36) yields

U ′(φ)φt + A(U (φ))U ′(φ)φx = 0, (38)

which implies that −φt/φx is an eigenvalue of A(U (φ)) and U ′(φ) is the associatedeigenvector. This concludes that in the (x, t) plane a simple wave is associated witha kind of characteristic field, say, λk , and spans a domain in which characteristicsof the k-kind are straight along which the solution is constant.

The property of simple waves can be analyzed using Riemann invariants.A Riemann invariant is a scalar function w = w(x, t) satisfying the followingcondition,

rk · grad w = 0, (39)

for all values of u, where rk is the k-th right eigenvector of A. Using the Riemanninvariants, it can be shown that a state in a domain adjacent to a domain of constantstate is always a simple wave.

In general, system (36) cannot be diagonalized because it does not have a fullcoordinate system of Riemann invariants [6]. Note that in (36) the coefficient matrix


A depends on u only. Once A depends on x and t as well as u, the treatment in[10] and [6] breaks down. For example, we are unable to use the same techniquesto show that it is a simple wave to be adjacent to a constant state.

4.2. Simple waves for pseudo-steady Euler equations

We introduce in a traditional manner a simple wave for (4) as a solution (u, v) =(u, v)(ξ, η) that is constant along the level set l : l(ξ, η) = C for some functionl(ξ, η), where C is constant. That is, this solution has the form,

(u, v)(ξ, η) = (F,G)(l(ξ, η)). (40)

Inserting this into (4) gives((2κ i − U 2)lξ − U V lη, −U V lξ + (2κ i − V 2)lη

lη −lξ

) (F ′G ′

)

= 0. (41)

Here we use U := u −ξ, V := v−η for short. It turns out that (F ′,G ′) = (0, 0) orthere exists a singular solution for which the coefficient matrix becomes singular.The former just gives a trivial constant solution. But for the latter, l(ξ, η) satisfies

(2κ i − U 2)l2ξ − 2U V lξ lη + (2κi − V 2)l2

η = 0; (42)

that is,

− lξlη

= U V ± 2κ i(U 2 + V 2 − 2κ i)1/2

U 2 − 2κ i=: Λ±, (43)

which implies that the level curves l(ξ, η) = C are characteristic lines, and

F ′ +Λ±G ′ = 0 (44)

holds along each characteristic line l(ξ, η) = C locally at least.In a recent paper by Li et al. [16], the pseudo-steady full Euler is shown to

have a characteristic decomposition. Let us quote several identities from that paper.First, the flow will be isentropic and irrotational adjacent to a constant state. Thenthe pseudo-characteristics are defined as

dη

dξ= U V ± c

√U 2 + V 2 − c2

U 2 − c2 ≡ Λ±. (45)

Here c is the speed of sound c2 = γ p/ρ. RegardingΛ± as simple straight functionsof the three independent variables (U, V, c2), we have

∂UΛ = Λ(UΛ− V )/Θ, ∂VΛ = (V − UΛ)/Θ, ∂c2Λ = −(1 +Λ2)/(2Θ)(46)

where Θ := Λ(c2 − U 2)+ U V . Then we further obtain

∂±u +Λ∓∂±v = 0, (47)

∂±c2 = −2κ(U∂±u + V ∂±v

), (48)

∂±Λ± = [∂UΛ± −Λ−1∓ ∂VΛ± − 2κ(U − V/Λ∓)∂c2Λ±

]∂±u, (49)


where ∂± = ∂ξ + Λ±∂η. We keep ∂± for later use in the hodograph plane. Thus,if one of the quantities (u, v, c2) is a constant along Λ−, so are the remaining twoand Λ−. The same is true for the plus family Λ+. Hence we have

Proposition 2. (Section 4, [16]) For the irrotational and isentropic pseudo-steadyflow (2) or (4), we have the following characteristic decomposition

∂+∂−u = h∂−u, ∂−∂+u = g∂+u, (50)

where h = h(u, v, c) and g = g(u, v, c) are some functions. Similar decompo-sitions hold for v, c2 and Λ±. We further conclude that simple waves are wavessuch that one family of characteristic curves are straight along which the physicalquantities (u, v, c2) are constant.

5. Convertibility

We are now ready to discuss the non-degeneracy of hodograph transformation(16).

Theorem 1. (sufficient and necessary condition) Let the ir-rotational, isentropicand pseudo-steady fluid flow (2) be smooth at a point (ξ, η) = (ξ0, η0). Then theJacobian JT (u, v; ξ, η) of the hodograph transformation (16) vanishes in a neigh-borhood of the point if the flow is a simple wave in the neighborhood. Conversely,if the Jacobian JT (u, v; ξ, η) vanishes in a neighborhood of the point, then the flowis a simple wave in the neighborhood.

Proof. Assume first that c2 − V 2 = 0 at (ξ, η) = (ξ0, η0). We compute

JT (u, v; ξ, η) = uξ vη − uηvξ

= − 1

c2 − V 2 · [((c2 − U 2)uξ − 2U V uη

)uξ

] − u2η (51)

= − 1

c2 − V 2 · [(c2 − U 2)u2

ξ − 2U V uξuη + (c2 − V 2)u2η

].

Therefore, the degeneracy of the transformation implies

(c2 − U 2)u2ξ − 2U V uξuη + (c2 − V 2)u2

η = 0. (52)

It follows that−uξ

uη= Λ±. (53)

That is,uξ +Λ+uη = 0, or uξ +Λ−uη = 0, (54)

at (ξ0, η0). For the former, we deduce that ∂+u = 0 along the wholeΛ−-characteristic line through (ξ0, η0) in view of (50) in Proposition 2, and sodo ∂+v and ∂+c. Therefore, we conclude that the wave is a simple wave associatedwith Λ+.


Conversely, if a point (ξ, η) = (ξ0, η0) is in the region of a simple wave, thenEquation (54) holds either for the plus or minus families. From there we go up thederivation to find that the Jacobian vanishes in the same neighborhood.

The case that c2 − (v− η)2 = 0 is a special planar simple wave. Therefore, theconclusion follows naturally.

We comment that the Jacobian JT (u, v; ξ, η) can be factorized as

JT (u, v; ξ, η) = − 1

Λ−Λ+∂+u · ∂−u = −∂+v · ∂−v. (55)

6. Phase space system of equations

In this section, we use the inclination angles of characteristics as useful variablesto rewrite (21) in the hodograph plane. We proceed as follows. We first transformthe second-order equation (21) into a first-order system of equations as in [14].Introduce

X = iu, Y = iv. (56)

Then we deduce a 3 × 3 system of first-order equations,

⎡

⎣c2 − Y 2 XY 0

0 1 00 0 1

⎤

⎦

⎡

⎣XYi

⎤

⎦

u

+⎡

⎣XY c2 − X2 0−1 0 00 0 0

⎤

⎦

⎡

⎣XYi

⎤

⎦

v

=⎡

⎣X2 + Y 2 − 2c2

0X

⎤

⎦ , (57)

where c2 = 2κ i . This system is equivalent to (21) for C1 solutions if the givendatum for Y is compatible with the datum for iv . The characteristic equation is

(c2 − Y 2)λ2 − 2XYλ+ c2 − X2 = 0 (58)

besides the trivial factor λ. This system has three eigenvalues

λ0 = 0,dv

du= λ± = XY ± √

c2(X2 + Y 2 − c2)

c2 − Y 2 = c2 − X2

XY ∓ √c2(X2 + Y 2 − c2)

,(59)

from which we deduce that (57) is hyperbolic if X2 + Y 2 − c2 > 0 provided thati > 0 and c2 − Y 2 = 0 (or c2 − X2 = 0). If c2 − Y 2 = 0 or c2 − X2 = 0,we have planar rarefaction waves in the neighborhood. The three associated lefteigenvectors with (59) are

l0 = (0, 0, 1), l∓ = (1, λ±, 0). (60)


We multiply (57) by the left eigen matrix M = (l+, l−, l0) (here and onwardthe superscript means transpose) from the left-hand side to obtain

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

Xu + λ−Yu + λ+(Xv + λ−Yv) = X2 + Y 2 − 2c2

c2 − Y 2 ,

Xu + λ+Yu + λ−(Xv + λ+Yv) = X2 + Y 2 − 2c2

c2 − Y 2 ,

iu = X.

(61)

Introduce the inclination angles α, β (−π/2 < α, β < π/2) of Λ+ andΛ−-characteristics by

tan α = Λ+, tan β = Λ−. (62)

Note that, see (27),

Λ+ = − 1

λ−, Λ− = − 1

λ+; (63)

and denoteA := tan(α/2), B := tan(β/2). (64)

This explains the Riemann invariant introduced in [14]. Then we find that X , Y arerelated with A, B through the following identities,

A = X − √X2 + Y 2 − c2

c − Y,

B = − X − √X2 + Y 2 − c2

c + Y,

(65)

or

X = c(1 − AB)

A − B, Y = c(A + B)

A − B. (66)

In terms of α, β, we have

X = ccos α+β

2

sinω, Y = c

sin α+β2

sinω, (67)

forω := (α − β)/2. (68)

We observe that the variables α, β are Riemann invariants for (61). In fact, we canwrite (61) as

∂+α = 1 + γ

4c· sin(α − β)

sin β· [m − tan2 ω],

∂−β = 1 + γ

4c· sin(α − β)

sin α· [m − tan2 ω],

∂0c = κcos α+β

2

sinω,

(69)

where we use the notations of directional derivatives,

∂+ = ∂

∂u+ λ+

∂

∂v, ∂− = ∂

∂u+ λ−

∂

∂v, ∂0 = ∂

∂u, (70)


and keep the letter m for

m = 1 − κ

1 + κ= 3 − γ

1 + γ. (71)

We further introduce the normalized directional derivatives along characteris-tics,

∂+ = (sin β,− cos β) · (∂u, ∂v), ∂− = (sin α,− cosα) · (∂u, ∂v). (72)

They are coordinate-free. Using them, we write (69) as,

∂+α = 1 + γ

4c· sin(α − β) · [m − tan2 ω] =: G(α, β, c),

∂−β = 1 + γ

4c· sin(α − β) · [m − tan2 ω] ≡ G(α, β, c),

∂0c = κcos α+β

2

sinω.

(73)

In particular, we note that

∂+c = −κ, ∂−c = κ. (74)

This highlights that the first two equations of (73) are entirely decoupled from thethird c-equation. In addition, each of the first two equations of (73) is actually adecomposition of the second-order equation (21) for c.

Note from (62) and (63) that λ+ = − cot β and λ− = − cot α. The system(69) is linearly degenerate in the sense of Lax [10]. For the particular case thattan((α − β)/2) = m for 1 < γ < 3, the first two equations become homogeneousequations

αu + λ+αv = 0,βu + λ−βv = 0,

(75)

which always have a unique global continuous solution provided that the corre-sponding initial and/or boundary data have a uniform bound in C1 norm (compare[18]). In fact, the explicit solutions of Suchkov [25] in the expansion problem ofa wedge of gas into a vacuum is such a case, see Remark 1 in Section 7.

The mapping (X,Y ) → (α, β) is bijective as long as system (61) is hyperbolic.We summarize the above as follows, which is similar to [14]:

Theorem 2. The two-dimensional pseudo-steady, irrotational, isentropic flow (21)can be transformed into a linearly degenerate system of first-order partial dif-ferential equations (69) or (73) provided that the transform (X,Y ) → (α, β) isinvertible, that is, system (61) is hyperbolic.

Regarding ∂−α and ∂+β, we have second-order equations although we areunable to obtain explicit expressions for them like (73). By direct computations,we obtain

Lemma 1. (commutator relation of ∂±) For any quantity I = I (u, v), there holds

∂−∂+ I − ∂+∂− I = ∂−λ+ − ∂+λ−λ− − λ+

(∂− I − ∂+ I ). (76)


Lemma 2. (commutator relation of ∂±) For any quantity I = I (u, v), there holds,

∂−∂+ I − ∂+∂− I = tanω(∂− I + ∂+ I )∂+α, (77)

where ∂+α is given in (73). Noting ∂+α = ∂−β in (73), we can also use ∂−β in(77).

Using these commutator relations, we easily derive:

Theorem 3. Assume that the solution of (73) (α, β) ∈ C2. Then we have

∂+∂−α + W ∂−α = Q(ω, c),−∂−∂+β + W ∂+β = Q(ω, c),

(78)

where W (ω, c) and Q(ω, c) are

W (ω, c) := 1 + γ

4c[(m − tan2 ω)(3 tan2 ω − 1) cos2 ω + 2 tan2 ω],

Q(ω, c) := (1 + γ )2

16c2 sin(2ω)(m − tan2 ω)(3 tan2 ω − 1).(79)

Proof. The proof is simple. Recall from (74) that

∂+c = −κ, ∂−c = κ. (80)

Then we apply the commutator relation to obtain [setting I = α in (77)]

∂+∂−α = ∂−∂+α + tanω(∂−α + ∂+α)∂−β. (81)

Using the expressions of ∂+α and ∂−β in (73), we compute directly to yield theresult in (78) and the proof of Theorem 3 is complete.

We can prove the next two theorems with straight computation.


∂+∂−(α + β)+ W ∂−(α + β) = a(ω, c)∂+(α + β)

−∂−∂+(α + β)+ W ∂+(α + β) = a(ω, c)∂−(α + β),(82)

where

a(ω, c) := γ + 1

4c[2 tan2 ω − (m − tan2 ω) cos(2ω)]

= γ + 1

4ccos2 ω(tan2 ω + α2)(tan2 ω − α1),

(83)

where

α2 := 1

2

[3 + m +

√(3 + m)2 + 4m

], α1 := 2m

3 + m + √(3 + m)2 + 4m

.

(84)



(∂+ + W )(Z − ∂−α) = γ + 1

4c(tan2 ω + 1)(Z − ∂+β)

(−∂− + W )(Z − ∂+β) = γ + 1

4c(tan2 ω + 1)(Z − ∂−α),

(85)

where

Z := γ + 1

2ctanω. (86)

We need Theorem 4 for lower bound and Theorem 5 for upper bound of thederivatives ∂−α and ∂+β.

7. The gas expansion problem

We now use the hodograph transformation and the decomposition of the previ-ous section to study the expansion of a wedge of gas into vacuum. The problem wasstudied earlier in [12,14,20,25], and especially by Li in [13,14], but the solutionof Li is in the hodograph plane for 1 γ < 3, and the behavior of the vacuumboundary was left open. We continue the effort of Li and prove that the solutionin the hodograph plane can be transformed back to the physical self-similar planefor all γ > 1 and the vacuum boundary is a Lipschitz continuous curve which ismonotone in the upper and lower parts of the wedge, respectively. We also deter-mine explicitly the relative location of the vacuum boundary with respect to thevertical position of the explicit solution of Suchkov [25]. Moreover, we can drawa clear picture of the distribution of characteristics. For notational simplicity in thissection, we use m, m0, and n defined by

tan2 m = m, m0 = 1/√

m, tan2 n = α1 (87)

for 1 < γ < 3; and m ≡ 0, n ≡ 0 for γ 3. Note that n < m for 1 γ < 3.

7.1. The planar rarefaction waves

First we prepare our planar rarefaction waves. Assume that the initial data for(1) is

(ρ, u, v)(x, y, 0) =(ρ1, 0, 0), for n1x + n2 y > 0,vacuum, for n1x + n2 y < 0,

(88)

where n21 + n2

2 = 1, and ρ1 is a constant. The solution of (1) and (88) takes theform, see [15],

(ρ, u, v)(x, y, t) =⎧⎨

⎩

(ρ1, 0, 0), ζ > 1,(ρ, u, v)(ζ ), −1/κ ζ 1,vacuum, ζ < −1/κ,

(89)


where ζ = n1ξ + n2η, (ξ, η) = (x/t, y/t), and the solution (c, u, v) has beennormalized so that c1 = 1. The rarefaction wave solution (ρ, u, v)(ζ ) satisfies

ζ = n1u + n2v + c,n1

κc − u = n1

κ,

n2

κc − v = n2

κ. (90)

Note that this rarefaction wave corresponds to a segment in the hodograph plane,n2u − n1v = 0, −n1/κ u 0.

In particular, when we consider the rarefaction wave propagates in thex-direction, that is, (n1, n2) = (1, 0), this wave can be expressed as

x/t = u + c, c = κu + 1, v ≡ 0, −1/κ u 0. (91)

That is, in the hodograph (u, v) plane, this rarefaction wave is mapped onto asegment v ≡ 0, −1/κ u 0, on which we have

i = 1

2κ(κu + 1)2, iu = κu + 1, iuu = κ. (92)

When we consider the expansion problem of a wedge of gas in the next subsection,we need to know not only the derivatives of i with respect to u in (92), but alsothe derivatives with respect to v, on the segment v ≡ 0, −1/κ u 0. For thispurpose, we insert (92) into (21) to obtain

(i2v

)

u − κ + 1

κu + 1i2v = −(κ + 1)(κu + 1). (93)

Solving this equation in terms of i2v yields,

i2v =

⎧⎪⎨

⎪⎩

(κu + 1)2[

1

m+

(

C2 − 1

m

)

(κu + 1)1−κκ

]

, for γ = 3,

(1 + u)2[C2 − 2 ln(1 + u)], for γ = 3,

(94)

where C is an integral constant. This was obtained in [12].

7.2. A wedge of gas

We place the wedge symmetrically with respect to the x-axis and the sharpcorner at the origin, as in Fig. 1a. This problem is then formulated mathematicallyas seeking the solution of (1) with the initial data,

(i, u, v)(t = 0, x, y) =(i0, u0, v0), −θ < δ < θ,

(0, u, v), otherwise,(95)

where i0 > 0, u0 and v0 are constant, (u, v) is the velocity of the wave front, notbeing specified in the state of vacuum, δ = arctan y/x is the polar angle, and θ isthe half-angle of the wedge restricted between 0 and π/2. This can be consideredas a two-dimensional Riemann problem for (1) with two pieces of initial data (95).As we will see below, this problem is actually the interaction of two whole planarrarefaction waves, see Fig. 1b. We note that the solution we construct is valid forany portions of (95) as the solutions are hyperbolic.


(a) (b)

, ,, ,

Fig. 1. The expansion of a wedge of gas

The gas away from the sharp corner expands into the vacuum as planarrarefaction waves R1 and R2 of the form (i, u, v)(t, x, y) = (i, u, v)(ζ ) (ζ =(n1x + n2 y)/t) where (n1, n2) is the propagation direction of waves. We assumethat initially the gas is at rest, that is, (u0, v0) = (0, 0). Otherwise, we replace(u, v) by (u − u0, v − v0) and (ξ, η) by (ξ − u0, η − v0) in the following com-putations [see also (2)]. We further assume that the initial sound speed is unitsince the transformation (u, v, c, ξ, η) → c0(u, v, c, ξ, η) with c0 > 0 can makeall variables dimensionless. Then the rarefaction waves R1, R2 emitting from theinitial discontinuities l1, l2 are expressed in (90) with (n1, n2) = (sin θ,− cos θ)and (n1, n2) = (sin θ, cos θ), respectively. These two waves begin to interact atP = (1/ sin θ, 0) in the (ξ, η) plane due to the presence of the sharp corner and awave interaction region, called the wave interaction region D, is formed to separatefrom the planar rarefaction waves by k1, k2,

k1 : (1 − κ2)ξ21 − (κη1 + 1)2 = 2(1−κ)/κC(κη1 + 1)(κ+1)/κ

for ξ1 > 0,−1 η1 1/κ,k2 : (1 − κ2)ξ2

2 − (κη2 + 1)2 = 2(1−κ)/κC(κη2 + 1)(κ+1)/κ

for ξ2 > 0,−1/κ η2 1,

(96)

where k1 and k2 are two characteristics from P , associated with the non-lineareigenvalues of system (2), see [15,27], and the constant C is

C = (γ + 1)

[1√

γ (γ + 1)

](γ+1)/(γ−1)

× [(3 − γ )(γ )−(γ+1)/(2(γ−1)) + (γ + 1)(γ )(γ−3)/(2(γ−1))

],

(97)

and ξ1 = ξ cos θ + η sin θ,η1 = −ξ sin θ + η cos θ,

ξ2 = ξ cos θ − η sin θ,η2 = ξ sin θ + η cos θ.

(98)

So, the wave interaction region D is bounded by k1, k2 and the interface of gas withvacuum, connecting D and E , see Fig. 1b. The solution outside D consists of theconstant state (i0, u0, v0), the vacuum and the planar rarefaction waves R1 and R2.Problem A. Find a solution of (2) inside the wave interaction region D, subjectto the boundary values on k1 and k2, which are determined continuously from therarefaction waves R1 and R2.


This problem is a Goursat-type problem for (2) since k1 and k2 are characteristics.Our strategy to solve this problem is to use the hodograph transform, solve the asso-ciated problem in the hodograph plane and show that the hodograph transformationis invertible.

Note that initial data (95) is irrotational, we conclude that the flow is alwaysirrotational provided that it is continuous. So the irrotationality condition (3) holdsand all results about the hodograph transformation can be used to treat this problem.Then Problem A can be converted into a problem in the hodograph plane.

For this purpose, we need to map the wave interaction region D in the (ξ, η)plane into a region Ω in the (u, v) plane. Notice that the mapping of the planarrarefaction waves R1 and R2 into (u, v) plane are exactly two segments

H1 : u cos θ + v sin θ = 0, (− sin θ/κ u 0) andH2 : u cos θ − v sin θ = 0, (− sin θ/κ u 0).

(99)

The boundary values of c on H1, H2, are

c|H1 = 1 + κv′ =: c10, c|H2 = 1 + κv′′ =: c2

0, (100)

where v′ = u sin θ − v cos θ and v′′ = u sin θ + v cos θ . Obviously,

0 c10, c2

0 1. (101)

Thus the wave interaction region Ω is bounded by H1, H2 and the interface ofvacuum connecting D and E in the hodograph (u, v)-plane, see Fig. 2. We defineΩmore precisely to contain the boundaries H1 and H2, but not the vacuum boundaryc = 0.

Boundary conditions. We need to derive the necessary boundary conditionson H1 and H2, respectively. This can be done simply by using coordinate transfor-mations for (92) and (94). Indeed, denote temporarily [compare (94)]

Γ (u,C) :=⎧⎨

⎩

[1m + (

C2 − 1m

)(κu + 1)

1−κκ

] 12, forγ = 3,

[C2 − 2 ln(1 + u)] 12 , forγ = 3.

(102)

Fig. 2. Wave interaction region in the hodograph plane


Then we have

iu = (1 − κv′)Γ (v′,C1) cos θ + sin θ

,

iv = (1 − κv′)(Γ (v′,C1) sin θ − cos θ

,

on H1, (103)

andiu = (1 + κv′′)

Γ (v′′,C2) cos θ + sin θ

,

iv = (1 + κv′′)−Γ (v′′,C2) sin θ + cos θ

,

on H2, (104)

where C1 and C2 are two constants. Applying the compatibility condition that iu ,iv are continuous at (u, v) = (0, 0), we obtain

C1 = −C2 = cot θ. (105)

Thus, we obtain the boundary conditions as in (103) and (104).In order to evaluate the boundary values of α, β, we substitute (100), (103),

(104) into (65) to deduce

A|H1 = sin θ

1 + cos θ= tan(θ/2),

B|H1 = −−Γ (v′, cot θ) sin θ + (1 + cos θ)

Γ (v′, cot θ)(1 + cos θ)+ sin θ=: B1,

A|H2 = −Γ (v′′, cot θ) sin θ + (1 + cos θ)

Γ (v′′, cot θ)(1 + cos θ)+ sin θ=: A2,

B|H2 = −1 + cos θ

sin θ= cot(θ/2).

(106)

Thus the boundary values for α, β on H1 and H2 are

α|H1 = θ, β|H1 = 2 arctan(−B1),

α|H2 = 2 arctan(A2), β|H2 = −θ. (107)

The boundary values of c on H1 and H2 are given in (100). Now Problem Abecomes:

Problem B. Find a solution (α, β, c) of (69) with boundary values (107) and (100),in the wave interaction region Ω in the hodograph plane.

In order to solve Problem B, we estimate the boundary values (107) and (100).

Lemma 3. (boundary data estimate) For the boundary data (107) on the boundariesHi , i = 1, 2, we have the following estimates:

(i) If θ < m, there holds

2θ (α − β)|Hi 2m. (108)

(ii) If θ > m, there holds

2m (α − β)|Hi 2θ. (109)


Proof. For the first case, that is, θ < m, by noting 0 1 + κv′, 1 + κv′′ 1, weestimate to get

tan(θ/2) B|H1 −m0 tan(θ/2)+ 1

m0 + tan(θ/2)=: mθ , (110)

where m0 = 1/√

m. It is easy to check that

tan(θ/2 + arctan mθ ) = √m = tan m. (111)

Therefore,

2θ (α − β)|H1 2m. (112)

Similarly, we can prove the second inequality on H2 in (108).For the second case where θ > m, the proof is also similar if 1 < γ < 3. If

γ 3, it is evident that

− tan(θ/2) A|H2 , B|H1 tan(θ/2). (113)

Then the proof is complete. The local existence of solutions at the origin (u, v) = (0, 0) follows routinely

from the idea [19, Chap. 2] or [26]. We need only to check the compatibilitycondition to this problem, that is,

1

λ+

[

l0 · ∂+K − κ cosα + β

2sin−1 ω

]

= 1

λ−

[

l0 · ∂−K − κ cosα + β

2sin−1 ω

]

(114)at (u, v) = (0, 0), where K = (α, β, c) and l0 = (0, 0, 1). That is, we need tocheck if there holds

1

λ+

[

∂+c − κ cosα + β

2sin−1 ω

]

= 1

λ−

[

∂−c − κ cosα + β

2sin−1 ω

]

. (115)

This is obviously true by using (74). Hence we have

Lemma 4. (local existence) There is a δ > 0 such that the C1-solution of (69) and(100), (107) exists uniquely in the region Ω = (u, v) ∈ Ω;−δ < u < 0, whereδ depends only on the C0 and C1 norms of α, β on the boundaries H1 and H2.

We do not give the proof. For details, see [19, Chap. 2] or [26].Next we will extend the local solution to the whole region Ω . Therefore, some

a priori estimates on the C0 and C1 norms of α, β and i , are needed. The norm ofi comes from the norms of α and β, see the third equation of (69). Therefore, weneed only the estimate on α and β. Recall that the derivation of (69) is based on thestrict hyperbolicity of the flow, i > 0. These will be achieved when we estimatethe C0 norms of α and β, see Section 7.3. The main existence theorem is stated asfollows. Let l be the interface of the gas with the vacuum.


Theorem 6. (global existence in the hodograph plane) There exists a solution(α, β, i) ∈ C1 to the boundary value problem (69) with boundary values (100)and (107) (Problem B) inΩ . The vacuum interface l exists and is Lipschitz contin-uous.

We prove this theorem by two steps. We estimate the solution itself in Section 7.3and then proceed with estimates on the gradients in Section 7.4. The proof ofTheorem 6 is also given in Section 7.4.

After we solve Problem B, we show the inversion of hodograph transforma-tion in Section 7.5, which establishes the existence of the gas expansion problem,Problem A.

Theorem 7. (global existence in the physical plane) There exists a solution(c, u, v) ∈ C1 of (2) for the gas expansion problem (Problem A) in the waveinteraction region D in the self-similar (ξ, η)-plane for all γ 1 and all wedgehalf-angle θ ∈ (0, π/2).

7.3. The maximum norm estimate on (α, β, c)

We estimate the solution (α, β, c) itself, that is, the C0 norm of α, β and c. Weadopt the method of invariant regions [24]. The notations m and n are given in thebeginning of this section.

Lemma 5. Suppose that there exists a C1 solution (α(u, v), β(u, v), c(u, v)) toproblem (69), (100) and (107) in Ω . Then the C0-norms of α and β have uniformbounds:

(i) If θ n, then θ α 2m − θ and −2m + θ β −θ ;(ii) If n < θ < m, then 2θ α − β 2m, θ α, and β −θ ;

(iii) If θ > m, then 2m α − β 2θ, α θ, β −θ .

Proof. Case (i). Consider using system (73). We construct a square, shown inFig. 3a, with left and upper sides denoted by L1 and L2, respectively. By Lemma 3,we know that the data belong to this square. Note that L1 corresponds to H1 andL2 to H2. We have

G(α, β, c) > 0, on L1, L2. (116)

We have the opposite on the other two sides. Note that the vector (sin β,− cosβ)on H1 points toward the interior of Ω , and the vector (sin α,− cosα) on H2 pointtowards outside of Ω , see Fig. 2. Thus, such a square bounded by L1 and L2 isinvariant. [The square exists and remains in the fourth quadrant and is invariantfor all the cases θ 2m, but we have smaller invariant regions (that is, invarianttriangles) for all θ > n.]

Case (ii). We show in this case that the upper triangle in Fig. 3a is invariant. LetL denote the diagonal line α − β = 2m. We show that the solution remains aboveL .

Consider using Equation (82) to establish ∂±(α+β) 0. We note that ∂+α > 0and ∂−β > 0 before the solution hits L , so α θ and β −θ , thus ω > n and


(a) (b) (c)

Fig. 3. Invariant regions

a(ω, c) > 0 for the solution, recalling that n is such that a = 0 at θ = n. Inaddition, there holds ∂+(α + β) > 0 on H2 and ∂−(α + β) > 0 on H1. A simplebootstrapping argument on Equation (82) implies that both ∂±(α + β) > 0 insidethe domain.

Thus, we find

∂+(α − β) = −∂+(α + β)+ 2∂+α 2∂+α = γ + 1

2csin(2ω)(m − tan2 ω)

(117)or

c∂+ψ −(γ + 1)(tan2 ω)ψ −(3 − γ )ψ (118)

forψ := m − tan2 ω. (119)

Inequality (118) implies that ψ > 0 in the domain c > 0. Thus the upper triangleis invariant.

Case (iii). First let γ ∈ [1, 3) so that θ > m > 0. We show that the lowertriangle is invariant, see Fig. 3b. The proof is similar. Again L denotes the lineα − β = 2m, which is below the line α − β = 0. Then we have

G(α, β, c) < 0, on L1, L2 (120)

as in the case for the invariant square. We need to show that the solution does notgo above the line L .

Consider using Equation (82) to establish ∂±(α+β) 0. We note that ∂+α < 0and ∂−β < 0 before the solution hits L , so α θ and β −θ , thus ω < θ andω > m > n and a(ω, c) > 0 for the solution before the solution hits L . In addition,there hold ∂+(α+β) < 0 on H2 and ∂−(α+β) < 0 on H1. A simple bootstrappingargument on Equation (82) implies that both ∂±(α + β) < 0 inside the domain.

Thus, we find

∂+(α − β) = −∂+(α + β)+ 2∂+α 2∂+α = γ + 1

2csin(2ω)(m − tan2 ω)

(121)or

c∂+ψ −(γ + 1)(tan2 ω)ψ −(γ + 1)(tan2 θ)ψ (122)


forψ := m − tan2 ω. (123)

Inequality (122) implies that ψ < 0 in the domain c > 0. Thus the upper triangleis invariant.

Lastly, we consider γ 3 and m = 0. We show that the lower triangle isinvariant and so α − β > 0, see Fig. 3c. Let L denote the line α − β = 0. Similarto the previous case, we need to show that the solution does not cross L . We havesimilarly obtained ∂±(α + β) < 0 inside the domain. Thus, we find

∂+(α−β) = −∂+(α+β)+2∂+α 2∂+α = γ + 1

2csin(2ω)(m − tan2 ω) (124)

orc∂+(α − β) −(γ + 1) sinω cosω(tan2 ω − m), (125)

orc∂+ sinω −(γ + 1) sinω(tan2 θ + |m|). (126)

Inequality (126) implies that α − β > 0 in the domain c > 0. Thus the uppertriangle is invariant. Corollary 1. For solutions (α, β, c) of (73), (107) and (100), we have:

(i) If n < θ < m, then G(α, β, c) > 0 and ∂+α > 0, ∂−β > 0 for all (u, v) ∈ Ω .(ii) If θ > m, then G(α, β, c) < 0 and ∂+α < 0, ∂−β < 0 for all (u, v) ∈ Ω .

Proof. They follow trivially from the invariant triangles. Remark 1. If the angle of the wedge θ and the adiabatic index γ are related by

tan2 θ = 3 − γ

γ + 1, (127)

for 1 < γ < 3, that is, θ = m, then boundary value (107) becomes constant(α, β)|Hj = (θ,−θ), j = 1, 2. In this case the invariant region shrinks to a point(θ,−θ) on the line α − β = 2m. Note that the source terms of (73) vanish on theboundaries H1, H2. We can use (75) to get an explicit solution,

c = 1 + κ

sin θu, (128)

where − sin θ/κ u 0. We further use (18) to get an explicit solution for theoriginal gas expansion problem,

c = 1 + κ(ξ sin θ − 1)

κ + sin2 θ,

u = sin θ(ξ sin θ − 1)

κ + sin2 θ,

v = η.

(129)

This solution was first observed in [25].


Remark 2. In the proof of Lemma 5, we observe that

cos((α + β)/2)

sinω> δ (130)

for some constant δ > 0. It follows from the third equation of (73) that

c < 1 + δu (131)

for u < 0 and thus c vanishes at u > −1/δ (assuming γ > 1). Therefore, thereexists a curve u = u(v) such that c(u(v), v) = 0 where u = u(v) is well-definedin the (u, v) plane. This is the interface of gas and vacuum.

Corollary 2. For the gas expansion problem, the mappings (X,Y ) → (α, β) and(X,Y ) → (A, B) are all bijective in the whole region Ω .

Proof. It suffices to check the non-degeneracy of the Jacobian, say from (X,Y ) →(α, β),

J (X,Y ;α, β) = − 1

sin2 ω· cot ω. (132)

In view of Lemma 5, we obtain the conclusion. Corollary 2 shows that we can convert system (57) into system (69) and, there-

fore, use system (69) or (73) to discuss Problem B in the hodograph plane.

7.4. Gradient estimates and the proof of Theorem 6

In order to establish the existence of smooth solutions in the whole wave inter-action region Ω , we need to establish gradient estimates for system (69) or (73).Due to the degeneracy of interface l, we cut off a sufficient thin strip between theinterface l and the level set of c = ε, ε > 0. The remaining sub-domain is denotedby Ωε, in which c > ε. We first show that there is a unique solution on Ωε. Thenwe extend the solution to Ω by using the argument of the arbitrariness of ε > 0.

Lemma 6. (gradient estimate) Consider system (69) or (73) with boundary values(107) and (100). Assume that there is a C1 solution (α, β) inΩε, then the C1 normof α and β has a uniform bound, which only depends on the C0 and C1 norms ofboundary values (107). That is, there is a constant C > 0, depending only on theboundary data (107) and (100), but not on ε, such that

‖(α, β)‖C1(Ωε) C/ε2, (133)

where ‖ · ‖C1(Ωε)represents the C1-norm.

Proof. We use (78) to integrate ∂−α and ∂+β along λ+ and λ−-characteristics,respectively. Noting (74), we know that the integral path has a limited length. Alsowe note that Q has a uniform bound C/ε2 in Ωε. Then we deduce that ∂−α and∂+β are uniformly bounded in Ωε,

|∂−α| < C/ε2, |∂+β| < C/ε2. (134)


On the other hand, since G has a bound C/ε inΩε [see (73)], so are ∂+α and ∂−β,

|∂+α| < C/ε, |∂−β| < C/ε. (135)

Hence using the identities,

∂u = − sin−1(2ω)(cosα ∂+ −cosβ ∂−), ∂v = − sin−1(2ω)(sin α ∂+ −sin β ∂−),(136)

and using the hyperbolicity α = β inΩε, we conclude that ∂uα, ∂vα, ∂uβ and ∂vβare uniformly bounded in Ωε, as expressed in (133). Lemma 7. (modulus estimate) Assume that the solution (α, β) ∈ C2(Ωε). Thenwe have the following modulus estimate,

‖(α, β)‖C1,1(Ωε)< C/ε2, (137)

where ‖·‖C1,1(Ωε)represents the C1,1-norm, and C1,1(Ωε) is the space of functions

whose C1-derivatives are Lipschitz continuous.

Proof. We mainly follow [6, Lemma 3.6, p. 291] to obtain the estimate (137) byusing (78). The verification process is specified as follows. For any point (u, v)insideΩε, we denote v = v(u; u, v) the λ+-characteristic curve from (u, v). Thenwe draw the characteristics v = v(u; u1, v1) and v = v(u; u2, v2) from two points(u1, v1) and (u2, v2), u1 u2, and they intersect H1 at two points (u1, v1) and(u2, v2) (compare Fig. 2). We want to use (78) to show the Lipschitz continuity of∂−α. The same is true for ∂+β. Denote

Θ(u, v(u; u, v)) = exp

(∫ u

uW (q, v(q; u, v)) dq

)

Q(u, v(u; u, v)). (138)

Recall that on H1, ∂−α(u, v) ≡ 0. Then we obtain by integrating (78) along theλ+-characteristics,

∂−α(u1, v1) =∫ u1

u1

Θ(u, v(u; u1, v1)) du,

∂−α(u2, v2) =∫ u2

u2

Θ(u, v(u; u2, v2)) du.

(139)

Therefore, we proceed to obtain

|∂−α(u1, v1)− ∂−α(u2, v2)|

=∣∣∣∣

∫ u1

u1

Θ(u, v(u; u1, v1)) du −∫ u2

u2

Θ(u, v(u; u2, v2)) du

∣∣∣∣

∣∣∣∣

∫ u1

u2

[Θ(u, v(u; u1, v1))−Θ(u, v(u; u2, v2))] du

∣∣∣∣

+∣∣∣∣

∫ u1

u2

Θ(u, v(u; u1, v1)) du

∣∣∣∣ +

∣∣∣∣

∫ u1

u2

Θ(u, v(u; u2, v2)) du

∣∣∣∣

=: T1 + T2 + T3.

(140)


Obviously, since |Θ(u, v(u; u2, v2))| C/ε2 for some constant C , we have

T3 C |u1 − u2|. (141)

To estimate T1, we use the definition ofv = v(u; u, v): dv(u;u,v)du = λ+(u, v(u; u, v))

and obtaind

du

∂v(u; u, v)

∂v= ∂λ+

∂v(u, v(u; u, v)) · ∂v(u; u, v)

∂v. (142)

Integration along v = v(u; u, v) yields,

∂v(u; u, v)

∂v= exp

(∫ u

u

∂λ+∂v

(q, v(q; u, v)) dq

)

. (143)

Recall again that λ+ = − cot β and apply Lemma 6. Then we deduce:

∣∣∣∣∂v(u; u, v)

∂v

∣∣∣∣ C/ε2. (144)

Noting that∂v(u; u, v)

∂v= −λ+(u, v(u; u, v)) · ∂v(u; u, v)

∂ u, (145)

we obtain ∣∣∣∣∂v(u; u, v)

∂ u

∣∣∣∣ C/ε2. (146)

Therefore, we have

T1 ∣∣∣∣

∫ u1

u2

[∂Θ

∂ u(u1 − u2)+ ∂Θ

∂v(v1 − v2)

]

du

∣∣∣∣ C/ε2(|u1 − u2| + |v1 − v2|),

(147)where we use (144) and (146) as well as the property of W and Q.

It remains to estimate T2. By the definition of v = v(u; u, v), we can show, bythe Gronwall inequality, that

|u1 − u2| C(|u1 − u2| + |v1 − v2|). (148)

So we have

T2 C/ε2(|u1 − u2| + |v1 − v2|). (149)

Hence we conclude that

|∂−α(u1, v1)− ∂−α(u2, v2)| C/ε2(|u1 − u2| + |v1 − v2|) (150)

for some C independent of ε.Besides, we can use (73) to obtain the Lipschitzian property of ∂+α and ∂−β

directly. Thus we complete the proof.


Proof of Theorem 6. With the classical technique in [18] or [7], we obtain theglobal solution in Ωε by the extension from the local solution.

In view of Lemma 4, we obtain a local solution (α, β, c) in Ωδ = (u, v) ∈Ωε;−δ < u < 0. We take a level set of c, denoted by Υc, in Ωδ . On this curve,(α, β, c) is known from the local solution and (α, β) ∈ C1(Υc) in view of Lemma 7.Then our problem becomes to find a solution of (69) in the remaining region, subjectto the data on H1, H2 and Υc.

Denote the slope of Υc by s0,

s0 := dv

du= −cu

cv= − cot

α + β

2. (151)

Then we have

1

s0− 1

λ−= sinω

cos α+β2 cosα

> 0,1

s0− 1

λ+= − sinω

cos α+β2 cosβ

< 0. (152)

This shows that the level setΥc is not a characteristic and λ±-characteristics alwayspoints toward the right hand side of Υc. Thus, we follow the proof of Lemma 4.1 in[7, p. 294], using Lemmas 6 and 7, to finish the proof of the existence of solutionsin Ωε.

Owing to the arbitrariness of width ε > 0, we use the contradiction argumentto show that the C1 solution (α, β, c) can be extend to the whole region Ω .

The discussion of vacuum boundary is left in Section 7.7.2.

7.5. Inversion

We now establish the global one-to-one inversion of the hodograph transform.Consider the hodograph transformation T : (ξ, η) → (u, v). The mapping (18)defines a domain via ξ = u + iu, η = v + iv. We need to show that the JacobianJ−1

T (u, v; ξ, η) in (24) does not vanish:

J−1T (u, v; ξ, η) = ξuηv − ξvηu = (1 + iuu)(1 + ivv)− i2

uv = 0. (153)

We calculate, on the one hand, multiplying (21) with (1 + iuu),(2κi − i2

u

)i2uv + 2iuiviuv(1 + iuu)+ (

2κi − i2v

)(1 + iuu)

2

= (2κi − i2u )

[i2uv − (1 + iuu)(1 + ivv)

].

(154)

On the other hand, from (26) and (58) we have(2κi − i2

u

)i2uv + 2iuiviuv(1 + iuu)+ (

2κi − i2v

)(1 + iuu)

2

= (2κi − i2

v

)(∂+iu + 1)(∂−iu + 1). (155)

Then we obtain

J−1T (u, v; ξ, η) = − (∂+ X + 1)(∂− X + 1)

λ−λ+= − (∂+ X + sin β)(∂− X + sin α)

cosα cosβ,

(156)by using the definition of ∂±, see (72). This is parallel to (55). Therefore, in orderto show that J−1

T (u, v; ξ, η) does not vanish, it is equivalent to prove that:


Lemma 8. The non-degeneracy of the Jacobian J−1T (u, v; ξ, η) is equivalent to

∂+ X + sin β = 0 and ∂− X + sin α = 0. (157)

Recall the expression of X in terms of α, β in (67). Then we compute

∂+ X + sin β = −κ cos α+β2

sinω− 1 + κ

2cot ω cosβ[m − tan2 ω]

+ sin β + c

2

cosα

sin2 ω∂+β,

∂− X + sin α = κcos α+β

2

sinω+ 1 + κ

2cot ω cosα[m − tan2 ω]

+ sin α − c

2

cosβ

sin2 ω∂−α.

(158)

They are easily simplified to be

∂+ X + sin β = − 1 + κ

sin(α − β)cosα + c

2

cosα

sin2 ω∂+β = c

2

cosα

sin2 ω[∂+β − Z ],

∂− X + sin α = 1 + κ

sin(α − β)cosβ − c

2

cosβ

sin2 ω∂−α = − c

2

cosβ

sin2 ω[∂−α − Z ],

(159)where Z = (1 + γ ) tanω/(2c), as denoted in (86) before. Note that on the bound-aries H1 and H2, the values ∂+β and ∂−α are, respectively,

∂+β|H2 ≡ 0, ∂−α|H1 ≡ 0. (160)

Therefore, (157) follows from the following Lemma.

Lemma 9. There holds∂+β < Z , ∂−α < Z (161)

in the region Ω , see Fig. 2.

Proof. By identities (85), the boundary condition on Z , and (160) we obtain theinequalities (161). More precisely, we note that both inequalities hold at the origin,and thus they hold in a neighborhood of the origin. Let c = ε ∈ (0, 1) be the firstlevel curve on which at least one of the two strict inequalities becomes equality.Note that the level curves of c are transversal to the characteristics. We integrateidentities (85) in the domain ε < c < 1 along characteristics to yield strict inequal-ities (161), resulting in a contradiction. Thus, both strict inequalities must holdup to c = 0.

While the non-vanishing Jacobian guarantees local one-to-one, we need globalone-to-one, which is guaranteed by the monotonicity of ξ andη along characteristicsfor the cases θ ∈ (0, 2m). In fact, ∂±ξ = 1+∂± X = 0, or±∂±ξ < 0 more precisely,following the above lemma. From (20), (30) we have ∂+ξ = −λ−∂+η, ∂−ξ =−λ+∂−η. Thus ξ and η have the same monotonicity along a plus characteristic


curve since −λ− > 0, but opposite monotonicity along a minus characteristicssince −λ+ < 0 for the cases θ ∈ (0, 2m) for which α > 0 and β < 0. For anytwo points in the interaction zone in the (u, v) plane, there exist two characteristiccurves connecting the two points. Either ξ or η is monotone along the connectingpath. Thus, no two points from the (u, v) domain maps to one point in the (ξ, η)plane when θ ∈ (0, 2m).

For θ 2m, the angles α and β may become negative or positive, respectively,thus the monotonicity of η ceases along characteristics. However, the variable ξremains monotone decreasing along both characteristics regardless of the signs ofα or β because of ±∂±ξ < 0. See Fig. 4b where we have indicated three points a,b1, and b2. If b1 is located above the plus characteristic curve passing a, the variableξ is monotone decreasing from a to b1 because it is so along both characteristiccurves and the minus characteristic curve is oriented toward the vacuum, see Fig. 4a.Let b2 be located below the plus characteristic curve passing a, then we use thefact from next subsections [that the two characteristic curves in the (ξ, η) plane areconvex and concave, respectively] so that points a and b2 will not give the samevalue of η. (We have drawn two tangent-lines at the intersection point to show thata and b2 indeed have different η values) Thus no two different points on the (u, v)plane map to a single point of the (ξ, η) plane.

We have, therefore, established the global one-to-one property.

7.6. Proof of Theorem 7

The above estimates are sufficient for the proof of Theorem 7. For completeness,we sum it as follows. First we use the hodograph transformation (16) to convertProblem A into Problem B. Since the region D in Fig. 1b is a wave interactionregion, the Jacobian JT (u, v; ξ, η) does not vanish in view of Theorem 1, so thehodograph transformation (16) is valid. Then we solve Problem B in Theorem 6.In Section 7.5, we have shown that the hodograph transformation is invertible inthe entire domain of interaction, using properties to be established in the nextSection 7.7. Thus the proof of Theorem 7 is complete, once we finish Section 7.7.

(a) (b) ,,

Fig. 4. Global one-to-one between (ξ, η) and (u, v)


7.7. Properties of the solutions

7.7.1. Convexity of characteristics in the physical plane Now we discuss theconvexity ofΛ±-characteristics in the mixed wave region D, in the (ξ, η) plane. Itis a rather simple way to look at this from the correspondence between the (ξ, η)plane and the (u, v) plane.

Consider the hodograph transformation T of (16). We note, by using the chainrule, that,

∂u + λ+∂v =(∂ξ

∂u+ λ+

∂ξ

∂v

)

∂ξ +(∂η

∂u+ λ+

∂η

∂v

)

∂η. (162)

We rewrite (19) as

∂ξ

∂u+ λ+

∂ξ

∂v= −λ−

(∂η

∂u+ λ+

∂η

∂v

)

. (163)

Using (18), we have∂ξ

∂u+ λ+

∂ξ

∂v= ∂+ X + 1. (164)

Thus, we derive a differential relation from (162), by noting Λ+ = −1/λ−,

∂+ = (∂+ X + sin β)(∂ξ +Λ+∂η). (165)

Similarly, we have∂− = (∂− X + sin α)(∂ξ +Λ−∂η). (166)

Acting (165) on Λ+ and (166) on Λ− as well as using the definition of α, β (thatis, Λ+ = tan α, Λ− = tan β), we obtain

(∂ξ +Λ+∂η)Λ+ = (1 + tan2 α) · (∂+ X + sin β)−1 · ∂+α,(∂ξ +Λ−∂η)Λ− = (1 + tan2 β) · (∂− X + sin α)−1 · ∂−β. (167)

By Corollary 1, the signs of ∂+α and ∂−β are

∂+α < 0, ∂−β < 0, if θ > m;∂+α > 0, ∂−β > 0, if θ ∈ (n, m).

(168)

In view of Lemma 9, we have

∂+ X + sin β < 0, ∂− X + sin α > 0. (169)

Hence we conclude,(∂ξ +Λ+∂η

)Λ+ > 0,

(∂ξ +Λ−∂η

)Λ− < 0, for θ > m,(

∂ξ +Λ+∂η)Λ+ < 0,

(∂ξ +Λ−∂η

)Λ− > 0, for θ ∈ (n, m).

(170)

In sum, we have

Theorem 8. TheΛ±-characteristics in the wave interaction region D of the (ξ, η)plane have fixed convexity types, see Fig. 5:

(i) If θ > m, the Λ±-characteristics are convex and concave, respectively.(ii) If θ ∈ (n, m), the Λ±-characteristics are concave and convex, respectively.

(iii) If θ = m, the solution has the explicit form (128) with straight characteristics.


(a) (b)

(c)

, ,

, /Fig. 5. Convexity types of the characteristics and the vacuum boundaries. Dashed curvesare of minus family

7.7.2. Regularity of the vacuum boundary Recall that formulae (18) transformthe solution (α, β, c) in the (u, v) plane, back into the (ξ, η)-plane. Note that (α, β),and thus cu , cv , are uniformly bounded for 1 < γ < 3, and that c tends to zerowith a rate greater than cu , cv for γ 3. Thus, by using (18), we have

ξ = u + iu = u + c

κcu = u, η = v + iv = v + c

κcv = v. (171)

So we conclude that on the vacuum boundary, the (u, v) coordinates coincide withthe (ξ, η) coordinates.

Next, we prove that the vacuum boundary is Lipschitz continuous. Let us con-sider the curve (u, v) | i(u, v) = ε > 0 for all small positive ε. Differentiatingthe equation i(u(v), v) = ε with respect to v, we find

du

dv= − Y

X= − tan

α + β

2. (172)


Since |α + β| < π/2 uniformly with respect to ε > 0, the level curve i(u, v) = ε

has a bounded derivative and in the limit as ε → 0+ converges to a Lipschitzcontinuous vacuum boundary.

7.7.3. Relative location For the explicit solution with θ = m, the vacuum bound-ary is a vertical segment. Now we hold θ fixed and consider decreasing γ so thatθ < m (but θ > n). Then we find α and β lies on the left-hand side of the lineα−β = 2m in the (α, β) phase plane. By the formula iv = Y and the location of theboundary data, we have Y < 0 on the upper half of the wedge, thus i is monotonedecreasing in v on the upper half, hence the vacuum boundary is on the left of theSuchkov boundary and of a convex type (that is, bulging outward). Similarly, theother case θ > m has the opposite result.

Theorem 9. Let the vacuum boundary be represented as ξ = ξ(η). Then it isLipschitz continuous. It is less than the boundary of the Suchkov solution and isconvex if θ ∈ (n, m), but it is concave and greater than that of the Suchkov solutionfor θ > m.

7.7.4. Characteristics on the vacuum boundary We already know that thesound speed c attains zero in a finite range of u assuming γ > 1. Conversely,we deduce from (74) that the length of λ±-characteristics is finite. Further, in viewof (73) for γ ∈ (1, 3) it can be seen that on the vacuum boundary,

α − β = 2m. (173)

In fact, on the one hand, if (173) were not true, then ∂+α and ∂−β would becomeinfinite as (u, v) approaches the vacuum boundary, which would force (α, β) toreach the line α − β = 2m in the (α, β)-plane, see Fig. 3. On the other hand, theline α − β = 2m is the set of stationary points of (α, β). Thus once (173) holds,we have c = 0.

Hence it is clear how the characteristics behave at the vacuum boundary. Thatis, for 1 < γ < 3, aΛ+-characteristic line has a non-zero intersection angle with aΛ−-characteristic line; however, if γ 3, then there must be α = β on the vacuumboundary.

8. Summary remarks

We have considered the phase space equation (20)(2κ i − i2

u

)ivv + 2iuiviuv + (

2κ i − i2v

)iuu = i2

u + i2v − 4κ i (174)

known from 1958 for the enthalpy i with the inverse of the hodograph transforma-tion (18)

ξ = u + iu, η = v + iv (175)

for the two-dimensional self-similar isentropic irrotational Euler system. Uponintroducing the variables of inclination angles of characteristics and normalized


characteristic derivatives, we have changed the second-order phase space equationto a first order system (73)

∂+α = G(α, β, c), ∂−β = G(α, β, c), ∂0c = γ − 1

2· cos

α + β

2

/sin

α − β

2,

(176)

where

G(α, β, c) = 1 + γ

4c· sin(α − β) ·

[3 − γ

γ + 1− tan2 α − β

2

]

.

Derivatives of the variables α and β along directions not represented in (176) areprovided by the higher-order systems (78), (82), (85). We use these infrastructuresto construct solutions to binary interactions of planar waves in the phase space andshow that the Jacobian of the inverse of the hodograph transform does not vanish, sowe obtain in particular a global solution to the gas expansion problem with detailedshapes and positions of the vacuum boundaries and characteristics.

The invariant regions in the phase space revealed in the process have morepotential than what has been utilized here. For example, we will use them to handlebinary interactions of simple waves, which will lead to the eventual construction ofglobal solutions to some four-wave Riemann problems that will not have vacuumin their data, see a forthcoming paper [17]. However, being a well-known difficultproblem, the Euler system does not give in easily, which manifests in our inabilityto establish an invariant triangle (rather than the loose square) for the case θ < n,a technical blemish of the paper.

We have made a comparison of the pair (174), (175) to the pair of eigenvalueξ = λ(u) and wave curve system (λ − f ′(u))u′ = 0 for the one-dimensionalsystem ut + f (u)x = 0 from Lax [10]. The wave curves of the one-dimensionalcase correspond to surfaces in the phase space (i, u, v). It will be a very interestingnext step to find out the phase space structure that involves subsonic domains andshock waves as well as the hyperbolic surfaces.

Acknowledgments. Jiequan Li thanks the Department of Mathematics at Penn State Univer-sity, and Yuxi Zheng thanks the Mathematics Department at Capital Normal University forthe hospitality during their mutual visits when this work was done. Both authors appreciateShuxing Chen and Tong Zhang’s careful reading of the manuscript and valuable suggestions.Jiequan Li’s research is partially supported by 973 Key Program with no. 2006CB805902and the Key Program from the Beijing Educational Commission with no. KZ200510028018,Program for New Century Excellent Talents in University (NCET) and Funding Project forAcademic Human Resources Development in Institutions of Higher Learning Under theJurisdiction of Beijing Municipality (PHR-IHLB). Yuxi Zheng’s research is partially sup-ported by NSF-DMS-0305497, 0305114, 0603859.

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School of Mathematical Science,Capital Normal University,

100037 Beijing, China.e-mail: [email protected]: [email protected]

and

Department of Mathematics,The Pennsylvania State University,University Park, PA 16802, USA.

e-mail: [email protected]

(Received December 12, 2007 / Accepted March 4, 2008)Published online October 7, 2008 – © Springer-Verlag (2008)

Documents

Interaction of Rarefaction Waves of the Two-Dimensional …math0.bnu.edu.cn/~lijiequan/publications/ARMA...of two-dimensional Riemann problems, the expansion problem of a wedge of