Interaction of Rarefaction Waves of the Two-Dimensional Self-Similar Euler Equations

Arch. Rational Mech. Anal. 193 (2009) 623Ð657 Digital Object Identiﬁer (DOI) 10.1007/s00205-008-0140-6

Jiequan Li & Yuxi Zheng

Communicated by T.-P. Liu

Abstract

We construct classical self-similar solutions to the interaction of two arbitrary planar rarefaction waves for the polytropic Euler equations in two space dimensions. The binary interaction represents a major type of interaction in the two-dimensional Riemann problems, and includes in particular the classical problem of the expansion of a wedge of gas into vacuum. Based on the hodograph transformation, the method employed here involves the phase space analysis of a second-order equation and the 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, (ρ ) + (ρ 2 + ) + (ρ v) = , ⎪ u t u p x u y 0 (1) ⎩ 2 (ρv)t + (ρuv)x + (ρv + p)y = 0, where ρ is the density, (u,v) is the velocity and p is the pressure given by p(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, the solutions depend on the self-similar variables (ξ, η) = (x/t, y/t). The expansion of 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 two self-similar variables, resulting in a partial differential equation of second order for the 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 waves such as the constant states and some other types of waves resembling the simple waves 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 equation in the space (c, u,v), where he discovered a pair of variables resembling the well-known Riemann invariants together with their invariant regions and established the existence of a solution to the expansion of a wedge of gas into vacuum in the hodograph plane for wide ranges of the gas constant and the wedge angle. Recently, in 2006, paper [16], in an attempt to establish the inversion of the hodograph mapping, clarified the concept of simple waves for (1). We show in this paper that the hodograph transformation is non-degenerate precisely for non-simple waves, and all the solutions constructed in [14] in the hodograph plane can now be transformed back to the self-similar plane. Thus we complete the procedural circle of construction of solutions. We find that the circle of construction bears a very interesting similarity to the construction of centered rarefaction waves in the one-dimensional systems of conservation laws. The self-similar variable(s) in both cases decouple from the phase space(s), and the equations in the phase space(s) are solved first. The development of the solutions from the phase space onto the physical space(s) requires genuine non-linearity in the one-dimensional case and a non-degeneracy condition of the hodograph transformation in the two-dimensional case. The key ingredient of this paper is the simplification of the form of the equations in the phase space brought about by the employment of the inclination angles of characteristics, and the discovery of the precise forms of the equations for the second-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, for example, to establish the Lipschitz continuity and monotonicity behavior of the vacuum boundary in the problem of a wedge of gas into vacuum and establish the dependence of the location of the boundary on the wedge angle and gas constant. The approach is particularly suitable for studying two-dimensional Riemann problems [27], since the apparent nature of the solutions of a Riemann problem is piece-wise smooth. The expansion problem of gas into vacuum has been a favorite for a long time. 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 the interface of gas and vacuum by the method of unsteady PrandtlÐMeyer expansions and related it to the PSI approach in [22], from which Li [14] started with new motivation from the success on the pressure gradient system [7]. In the context of two-dimensional Riemann problems, the expansion problem of a wedge of gas into 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 hyperbolic region; this one expands without shocks but with a boundary degeneracy, while the 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 involve hyperbolic regions of non-constant continuous waves [1Ð3,8,9,11,15,23,28]show that 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 into vacuum (Theorem 7), and the detailed properties of the expansion (Theorems 8, 9). We provide some background information as well regarding the hodograph transformation and simple waves in Sections 2Ð5 for the convenience of non-expert readers. Section 6 is for the phase space analysis, while Section 7 handles the gas expansion 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 of the hodograph transformation with the associated trial-and-error process in finding the precise forms for the second-order equations suitable for bootstrapping [see formulae (82), (85)]. The difficulty of the inversion manifests itself in the fact that the characteristics in the phase space do not have a fixed convexity type although the corresponding characteristics in the physical space do (in most cases), see Sections 7.5 and 7.7. We mention additionally that hodograph transforms have been 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) the self-similar (or pseudo-steady) variables, ϕ pseudo-velocity potential, θ wedge half-angle, and U = u − ξ, V = v − η, κ = (γ − 1)/2, m = (3 − γ)/(γ + 1). 2m √ √ α1 = , tan n¯ = α1, tan m¯ = m. 3 + m + (3 + m)2 + 4m √ α − β UV ± c U 2 + V 2 − c2 Λ+ = tan α, Λ− = tan β, ω = ,Λ± = . 2 U 2 − c2 ± ∂ = ∂ξ + Λ±∂η,∂± = ∂u + λ±∂v,∂0 = ∂u,Λ±λ∓ =−1. ¯ ¯ ∂+ = (sin β,− cos β) · (∂u,∂v), ∂− = (sin α, − cos α) · (∂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, ( − ξ) + (v − η) + = , ⎩ u uξ uη iξ 0 (2) (u − ξ)vξ + (v − η)vη + iη = 0. 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 deduce the system, (2κ i − (u − ξ)2)uξ − (u − ξ)(v − η)(uη + vξ ) + (2κ i − (v − η)2)vη = 0, uη − vξ = 0, (4) 626 Jiequan Li & Yuxi Zheng supplemented by Bernoulli’s law

1 2 2 i + ((u − ξ) + (v − η) ) =−ϕ, ϕξ = u − ξ, ϕη = v − η. (5) 2 We remark that the difference between the pseudo-steady ﬂow (4) and the steady case (32) (see Section 7) is that the latter is self-contained, since the sound speed c can be expressed by a pointwise function of the velocity explicitly through Bernoulli’s law (33).

3. Concept of hodograph transformation

We introduce briefly the well-known hodograph transformation. The original form of a hodograph transformation is for a homogeneous quasi-linear system of two first-order equations for two known variables (u,v) in two independent variables (x, y). By regarding (x, y) as functions of (u,v) and assuming that the Jacobian does not vanish nor is infinity, one can re-write the system for the unknowns (x, y) in the variables (u,v), which is a linear system if the coefficients of the original system do not depend on (x, y). See the book of Courant and Friedrichs [5]. Specifically, consider the system of two equations of the form, u + ( ,v; , ) u = , v A u x y v 0 (6) x y where the coefficient matrix A(u,v; x, y) is a a A(u,v; x, y) = 11 12 . (7) a21 a22

The two eigenvalues, denoted by Λ±, satisﬁes,

2 Λ± − (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 −xv + A(u,v; x, y) = 0. (10) −yu xu

Its eigenvalues, denoted by λ±, satisfy

2 a12λ± − (a22 − a11)λ± − a21 = 0. (11) Obviously, if the coefﬁcient matrix A does not depend on (x, y),(10) becomes a linear system for the unknowns (x, y). The following proposition establishes the invariance of characteristics under the hodograph transformation. Interaction of Rarefaction Waves for Two-Dimensional Euler 627

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 hodograph transform T . = ( ) dy = Λ v = v( ) Proof. Let y y x be a characteristic with dx ±. Its image is u under the hodograph transform (9). Then we have + · dv dy yu yv du = = Λ±, (12) dx + · dv xu xv du that is, dv Λ±x − y =− u u . (13) du Λ±xv − yv Using (10) we ﬁnd

(Λ±xu − yu) + λ±(Λ±xv − yv) = 0. (14) Therefore, we have dv = λ±( or λ∓), (15) du 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),sowe can still try to use (u,v)as the independent variables and regard (ξ, η) as functions of (u,v)and ultimately regard i as a function of (u,v). In this way we may obtain an equation for i = i(u,v) in the plane (u,v) which eliminates (ξ, η). This was done in 1958 in a paper [22]byPogodin et al. and has been referred to as the PSI approach. 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 + (−iξ uη + iηuξ )dv. (17) We insert this into the law of momentum conservation of (2) and use the irrotationality condition (3) to obtain ξ − u = i , u (18) η − v = iv. These interesting identities provide an explicit correspondence between the physical plane and the hodograph plane provided that the transformation T is not degenerate. Therefore, using (18), we convert (4)intoalinear (in fact, linearly degenerate, see Section 6)system 2κ i(u,v)− i2 ηv + i iv(ξv + η ) + 2κ i − i2 ξ = 0, u u u v u (19) ξv − ηu = 0. 628 Jiequan Li & Yuxi Zheng for the unknowns (ξ, η). The difﬁculty here is that i, as a function of u and v, cannot be determined explicitly and point-wise. We will remark more later in contrast with the steady case. We continue to differentiate (18) with respect to u and v:

ξ = 1 + i ,ξv = i v, u uu u (20) ηu = iuv,ηv = 1 + ivv, and inserting these into the ﬁrst equations of (19) to obtain,

κ − 2 + + κ − 2 = 2 + 2 − κ . 2 i iu ivv 2iuiviuv 2 i iv iuu iu iv 4 i (21) This is a very interesting second order partial differential equation for i alone. So the 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, instead of i = c2/(γ − 1),isused[13]. Then we can obtain a similar equation for i,

− 2 + + − 2 = 2 + 2 − . 1 iu ivv 2iuiviuv 1 iv iuu iu iv 2 (22) We will establish in the pseudo-steady case that the transform is not degenerate, that is, ∂(u,v) J (u,v; ξ,η) = = uξ vη − uηvξ = T ∂(ξ,η) 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 −1( ,v; ξ,η) = ξ η − ξ η = . JT u u v v u 0 (24) Noting that (4) and (19) are all two-by-two systems, we use Proposition 1 to assert that the characteristics of (4) are mapped into the characteristics of (19)by the hodograph transformation (16). Moreover, the eigenvalues of (4)are (u − ξ)(v − η) ± c (u − ξ)2 + (v − η)2 − c2 Λ± = , (25) (u − ξ)2 − c2 while the eigenvalues of (19)are

± 2 + 2 − 2 iuiv c iu iv c λ± = . 2 2 (26) c − iv By using (18), it is easy to see that 1 λ± =− . (27) Λ∓

Furthermore, there is a correspondence between Λ± and λ±. Indeed, let η = η(ξ) (ξ, η) dη = Λ be a characteristic curve in the plane with dξ + and be mapped onto a curve v = v(u). Then, using (20) and (27), we have dv dη ηu + ηv Λ = = du , (28) dξ ξ + ξ dv u v du Interaction of Rarefaction Waves for Two-Dimensional Euler 629 that is,

dv ξ Λ+ − η (1 + i )Λ+ − i v i + 1 + λ−i v =− u u =− uu u =− uu u . (29) du ξvΛ+ − ηv iuvΛ+ − (1 + ivv) iuv + λ−(ivv + 1)

We rewrite (21)as

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

Then we conclude dv = λ+. (31) du

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)u − uv(u + v ) + (c2 − v2)v = 0, x y x y (32) u y − vx = 0, where c is the sound speed, given by Bernoulli’s law

u2 + v2 c2 k + = 0 , (33) 2 γ − 1 2 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 + y ) + (2κ i − v2)x = 0, u u (34) xv − yu = 0.

The hodograph transform is valid in the region of non-simple waves. With the same set 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 the limit of unsteady flow (1)int →∞. Comparing (35) with (18), we see that it is much more difficult to convert the hodograph plane of the steady case back into the physical plane than the pseudo-steady case. However, system (34) has more advantage over (19) of the pseudo-steady case because i is expressed in an explicit form by Bernoulli’s law (33). 630 Jiequan Li & Yuxi Zheng

3.2. Similarity to one-dimensional problems

The current approach parallels the procedure that is used to find centered rarefaction waves to genuinely non-linear strictly hyperbolic systems of conservation laws in one space dimension. Recall for a one-dimensional system ut + f (u)x = 0ofn equations, a centered rarefaction wave takes the form ξ = λk(u) for a k ∈ (1, n) and the state variable u satisfies the system of ordinary differential equations ( f (u) − λk(u)I )uξ = 0, whose solutions are rarefaction wave curves in the phase space. The development (or inversion) of the phase space solutions onto the ξ-axis requires the monotonicity of λk(u) along the vector field of the k-th right eigenvector rk; that is, genuine non-linearity. For the self-similar two-dimensional Euler 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 differential system. For inversion to the physical space, we show that the Jacobian −1 JT 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 independent variables, ut + A(u)ux = 0, (36) where u = (u1,...,un) ,then × n matrix A(u) has real and distinct eigenvalues λ1 < ··· <λn for all u under consideration. They are deﬁned as a special family of 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 associated eigenvector. This concludes that in the (x, t) plane a simple wave is associated with a kind of characteristic ﬁeld, say, λk, and spans a domain in which characteristics of 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 following condition, rk · grad w = 0, (39) for all values of u, where rk is the k-th right eigenvector of A. Using the Riemann invariants, it can be shown that a state in a domain adjacent to a domain of constant state is always a simple wave. In general, system (36) cannot be diagonalized because it does not have a full coordinate system of Riemann invariants [6]. Note that in (36) the coefﬁcient matrix Interaction of Rarefaction Waves for Two-Dimensional Euler 631

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 techniques to 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 function l(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 − UVlη, −UVlξ + (2κ i − V 2)lη F = 0. (41) lη −lξ G

Here we use U := u −ξ,V := v −η for short. It turns out that (F , G ) = (0, 0) or there 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 2 2 2 2 (2κ i − U )lξ − 2UVlξ lη + (2κi − V )lη = 0; (42) that is, / lξ UV ±{2κ i(U 2 + V 2 − 2κ i)}1 2 − = =: Λ±, (43) lη U 2 − 2κ i 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. Then the pseudo-characteristics are deﬁned as √ dη UV ± c U 2 + V 2 − c2 = ≡ Λ±. (45) dξ U 2 − c2

Here c is the speed of sound c2 = γ p/ρ. Regarding Λ± as simple straight functions of the three independent variables (U, V, c2),wehave ∂ Λ = Λ( Λ − )/Θ, ∂ Λ = ( − Λ)/Θ, ∂ Λ =−( + Λ2)/( Θ) U U V V V U c2 1 2 (46) where Θ := Λ(c2 − U 2) + UV. Then we further obtain ∂± + Λ ∂±v = , u ∓ 0 (47) ± ± ± ∂ c2 =−2κ U∂ u + V ∂ v , (48)

∂±Λ = ∂ Λ − Λ−1∂ Λ − κ( − /Λ )∂ Λ ∂± , ± U ± ∓ V ± 2 U V ∓ c2 ± u (49) 632 Jiequan Li & Yuxi Zheng

± 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 two and Λ−. The same is true for the plus family Λ+. Hence we have

Proposition 2. (Section 4,[16]) For the irrotational and isentropic pseudo-steady ﬂow (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 waves such that one family of characteristic curves are straight along which the physical quantities (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, isentropic and pseudo-steady fluid flow (2) be smooth at a point (ξ, η) = (ξ0,η0). Then the Jacobian JT (u,v; ξ,η) of the hodograph transformation (16) vanishes in a neighborhood 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 flow is a simple wave in the neighborhood.

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

JT (u,v; ξ,η) = uξ vη − uηvξ

1 2 2 2 =− · (c − U )uξ − 2UVuη uξ − uη (51) c2 − V 2

1 2 2 2 2 2 2 =− · (c − U )uξ − 2UVuξ uη + (c − V )uη . c2 − V 2 Therefore, the degeneracy of the transformation implies

2 2 2 2 2 2 (c − U )uξ − 2UVuξ uη + (c − V )uη = 0. (52)

It follows that uξ − = Λ±. (53) uη 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 whole Λ−-characteristic line through (ξ0,η0) in view of (50) in Proposition 2, and so do ∂+v and ∂+c. Therefore, we conclude that the wave is a simple wave associated with Λ+. Interaction of Rarefaction Waves for Two-Dimensional Euler 633

Conversely, if a point (ξ, η) = (ξ0,η0) is in the region of a simple wave, then Equation (54) holds either for the plus or minus families. From there we go up the derivation 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, the conclusion follows naturally.

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

1 + − + − JT (u,v; ξ,η) =− ∂ u · ∂ u =−∂ v · ∂ v. (55) Λ−Λ+

6. Phase space system of equations

In this section, we use the inclination angles of characteristics as useful variables to rewrite (21) in the hodograph plane. We proceed as follows. We ﬁrst transform the 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 X XY c2 − X 2 0 X ⎣ 010⎦ ⎣ Y ⎦ + ⎣ −100⎦ ⎣ Y ⎦ 001i 000i ⎡ ⎤ u v X 2 + Y 2 − 2c2 = ⎣ 0 ⎦ , (57) X where c2 = 2κ i. This system is equivalent to (21)forC1 solutions if the given datum for Y is compatible with the datum for iv. The characteristic equation is

(c2 − Y 2)λ2 − 2XYλ + c2 − X 2 = 0 (58) besides the trivial factor λ. This system has three eigenvalues

λ0 = 0, dv XY ± c2(X 2 + Y 2 − c2) c2 − X 2 (59) = λ± = = , du c2 − Y 2 XY ∓ c2(X 2 + Y 2 − c2) from which we deduce that (57) is hyperbolic if X 2 + Y 2 − c2 > 0 provided that i > 0 and c2 − Y 2 = 0(orc2 − X 2 = 0). If c2 − Y 2 = 0orc2 − X 2 = 0, we have planar rarefaction waves in the neighborhood. The three associated left eigenvectors with (59)are

l0 = (0, 0, 1), l∓ = (1,λ±, 0). (60) 634 Jiequan Li & Yuxi Zheng

We multiply (57) by the left eigen matrix M = (l+, l−, l0) (here and onward the superscript means transpose) from the left-hand side to obtain ⎧ ⎪ X 2 + Y 2 − 2c2 ⎪ X + λ−Y + λ+(Xv + λ−Yv) = , ⎨ u u c2 − Y 2 2 + 2 − 2 (61) ⎪ + λ + λ ( + λ ) = X Y 2c , ⎪ Xu +Yu − Xv +Yv ⎩⎪ c2 − Y 2 iu = X.

Introduce the inclination angles α, β (−π/2 <α, β<π/2) of Λ+ and Λ−-characteristics by tan α = Λ+, tan β = Λ−. (62) Note that, see (27), 1 1 Λ+ =− ,Λ− =− ; (63) λ− λ+ and denote A := tan(α/2), B := tan(β/2). (64) This explains the Riemann invariant introduced in [14]. Then we find that X, Y are related with A, B through the following identities, √ X − X 2 + Y 2 − c2 A = , − √c Y (65) X − X 2 + Y 2 − c2 B =− , c + Y or c(1 − AB) c(A + B) X = , Y = . (66) A − B A − B In terms of α, β,wehave α+β α+β cos sin X = c 2 , Y = c 2 , (67) sin ω sin ω for ω := (α − β)/2. (68) We observe that the variables α, β are Riemann invariants for (61). In fact, we can write (61)as + γ (α − β) 1 sin 2 ∂+α = · ·[m − tan ω], 4c sin β + γ (α − β) 1 sin 2 ∂−β = · ·[m − ω], α tan (69) 4c α+β sin cos ∂ c = κ 2 , 0 sin ω where we use the notations of directional derivatives, ∂ ∂ ∂ ∂ ∂ ∂+ = + λ+ ,∂− = + λ− ,∂= , (70) ∂u ∂v ∂u ∂v 0 ∂u Interaction of Rarefaction Waves for Two-Dimensional Euler 635 and keep the letter m for 1 − κ 3 − γ m = = . (71) 1 + κ 1 + γ We further introduce the normalized directional derivatives along characteristics, ¯ ¯ ∂+ = (sin β,− cos β) · (∂u,∂v), ∂− = (sin α, − cos α) · (∂u,∂v). (72) They are coordinate-free. Using them, we write (69)as, + γ ¯ 1 2 ∂+α = · sin(α − β) ·[m − tan ω]=:G(α, β, c), 4+cγ ¯ 1 2 ∂−β = · sin(α − β) ·[m − tan ω]≡G(α, β, c), (73) 4c α+β cos ∂ c = κ 2 . 0 sin ω In particular, we note that ¯ ¯ ∂+c =−κ, ∂−c = κ. (74) This highlights that the first two equations of (73) are entirely decoupled from the third c-equation. In addition, each of the first two equations of (73) is actually a decomposition of the second-order equation (21)forc. 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 that tan((α − β)/2) = m for 1 <γ <3, the first two equations become homogeneous equations α + λ+αv = 0, u (75) βu + λ−βv = 0, which always have a unique global continuous solution provided that the corresponding 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 of a 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 differential equations (69) or (73) provided that the transform (X, Y ) → (α, β) is invertible, that is, system (61) is hyperbolic. ¯ ¯ Regarding ∂−α and ∂+β, we have second-order equations although we are unable 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

∂−λ+ − ∂+λ− ∂−∂+ I − ∂+∂− I = (∂− I − ∂+ I ). (76) λ− − λ+ 636 Jiequan Li & Yuxi Zheng

¯ Lemma 2. (commutator relation of ∂±) For any quantity I = I (u,v), there holds, ¯ ¯ ¯ ¯ ¯ ¯ ¯ ∂−∂+ I − ∂+∂− I = tan ω(∂− 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), ¯ ¯ ¯ (78) −∂−∂+β + W∂+β = Q(ω, c), where W(ω, c) and Q(ω, c) are

1 + γ W(ω, c) := [(m − tan2 ω)(3tan2 ω − 1) cos2 ω + 2tan2 ω], 4c (79) (1 + γ)2 Q(ω, c) := sin(2ω)(m − tan2 ω)(3tan2 ω − 1). 16c2 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 + tan ω(∂−α + ∂+α)∂−β. (81) ¯ ¯ Using the expressions of ∂+α and ∂−β in (73), we compute directly to yield the result in (78) and the proof of Theorem 3 is complete.

We can prove the next two theorems with straight computation.

Theorem 4. Assume that the solution of (73) (α, β) ∈ C2. Then we have ¯ ¯ ¯ ¯ ∂+∂−(α + β) + W∂−(α + β) = a(ω, c)∂+(α + β) ¯ ¯ ¯ ¯ (82) −∂−∂+(α + β) + W∂+(α + β) = a(ω, c)∂−(α + β), where γ + 1 a(ω, c) := [2tan2 ω − (m − tan2 ω) cos(2ω)] 4c γ + 1 (83) = cos2 ω(tan2 ω + α )(tan2 ω − α ), 4c 2 1 where 1 2 2m α2 := 3 + m + (3 + m) + 4m ,α1 := . 2 3 + m + (3 + m)2 + 4m (84) Interaction of Rarefaction Waves for Two-Dimensional Euler 637

Theorem 5. Assume that the solution of (73) (α, β) ∈ C2. Then we have γ + ¯ ¯ 1 2 ¯ (∂+ + W)(Z − ∂−α) = (tan ω + 1)(Z − ∂+β) 4c (85) γ + ¯ ¯ 1 2 ¯ (−∂− + W)(Z − ∂+β) = (tan ω + 1)(Z − ∂−α), 4c where γ + 1 Z := tan ω. (86) 2c We need Theorem 4 for lower bound and Theorem 5 for upper bound of the ¯ ¯ derivatives ∂−α and ∂+β.

7. The gas expansion problem

We now use the hodograph transformation and the decomposition of the previous section to study the expansion of a wedge of gas into vacuum. The problem was studied earlier in [12,14,20,25], and especially by Li in [13,14], but the solution of Li is in the hodograph plane for 1 γ<3, and the behavior of the vacuum boundary was left open. We continue the effort of Li and prove that the solution in the hodograph plane can be transformed back to the physical self-similar plane for all γ>1 and the vacuum boundary is a Lipschitz continuous curve which is monotone 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 the vertical position of the explicit solution of Suchkov [25]. Moreover, we can draw a clear picture of the distribution of characteristics. For notational simplicity in this section, we use m¯ , m0, and n¯ deﬁned by √ 2 2 tan m¯ = m, m0 = 1/ m, tan n¯ = α1 (87) for 1 <γ <3; and m¯ ≡ 0, n¯ ≡ 0forγ 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 , 0, 0), for n x + n y > 0, (ρ, u,v)(x, y, 0) = 1 1 2 (88) vacuum, for n1x + n2 y < 0, 2 + 2 = ρ where n1 n2 1, and 1 is a constant. The solution of (1) and (88) takes the form, see [15], ⎧ ⎨ (ρ1, 0, 0), ζ > 1, (ρ, ,v)( , , ) = (ρ, , v)(ζ ), − /κ ζ , u x y t ⎩ u 1 1 (89) vacuum,ζ<−1/κ, 638 Jiequan Li & Yuxi Zheng where ζ = n1ξ + n2η, (ξ, η) = (x/t, y/t), and the solution (c, u,v) has been normalized so that c1 = 1. The rarefaction wave solution (ρ, u, v)(ζ ) satisﬁes n n n n ζ = n u + n v + c, 1 c − u = 1 , 2 c − v = 2 . 1 2 κ κ κ 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 the x-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 a segment v ≡ 0, −1/κ u 0,onwhichwehave

1 2 i = (κu + 1) , iu = κu + 1, iuu = κ. (92) 2κ 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 also the derivatives with respect to v, on the segment v ≡ 0, −1/κ u 0. For this purpose, we insert (92)into(21) to obtain κ + 2 1 2 iv − iv =−(κ + 1)(κu + 1). (93) u κu + 1 2 Solving this equation in terms of iv yields, ⎧ ⎪ 1 1 1−κ ⎨ (κu + 1)2 + C2 − (κu + 1) κ , for γ = 3, 2 iv = m m (94) ⎩⎪ (1 + u)2[C2 − 2ln(1 + u)], for γ = 3, 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 sharp corner at the origin, as in Fig. 1a. This problem is then formulated mathematically as seeking the solution of (1) with the initial data, (i , u ,v ), −θ<δ<θ, (i, u,v)(t = 0, x, y) = 0 0 0 (95) (0, u¯, v),¯ otherwise, where i0 > 0, u0 and v0 are constant, (u¯, v)¯ is the velocity of the wave front, not being speciﬁed in the state of vacuum, δ = arctan y/x is the polar angle, and θ is the half-angle of the wedge restricted between 0 and π/2. This can be considered as 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 planar rarefaction waves, see Fig. 1b. We note that the solution we construct is valid for any portions of (95) as the solutions are hyperbolic. Interaction of Rarefaction Waves for Two-Dimensional Euler 639

, , , ,

(a) (b) Fig. 1. The expansion of a wedge of gas

The gas away from the sharp corner expands into the vacuum as planar rarefaction 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 assume that 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 computations [see also (2)]. We further assume that the initial sound speed is unit since the transformation (u,v,c,ξ,η) → c0(u,v,c,ξ,η)with c0 > 0 can make all variables dimensionless. Then the rarefaction waves R1, R2 emitting from the initial 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 at P = (1/ sin θ,0) in the (ξ, η) plane due to the presence of the sharp corner and a wave interaction region, called the wave interaction region D, is formed to separate from the planar rarefaction waves by k1, k2, : ( − κ2)ξ 2 − (κη + )2 = (1−κ)/κ (κη + )(κ+1)/κ k1 1 1 1 1 2 C 1 1 for ξ1 > 0, −1 η1 1/κ, : ( − κ2)ξ 2 − (κη + )2 = (1−κ)/κ (κη + )(κ+1)/κ (96) k2 1 2 2 1 2 C 2 1 for ξ2 > 0, −1/κ η2 1, where k1 and k2 are two characteristics from P, associated with the non-linear eigenvalues of system (2), see [15,27], and the constant C is (γ + )/(γ − ) 1 1 1 C = (γ + 1) √ γ(γ + 1) (97) × (3 − γ )(γ )−(γ +1)/(2(γ −1)) + (γ + 1)(γ )(γ −3)/(2(γ −1)) , and ξ = ξ cos θ + η sin θ, ξ = ξ cos θ − η sin θ, 1 2 (98) η1 =−ξ sin θ + η cos θ, η2 = ξ sin θ + η cos θ.

So, the wave interaction region D is bounded by k1, k2 and the interface of gas with vacuum, connecting D and E, see Fig. 1b. The solution outside D consists of the constant state (i0, u0,v0), the vacuum and the planar rarefaction waves R1 and R2. Problem A. Findasolutionof(2) inside the wave interaction region D, subject to the boundary values on k1 and k2, which are determined continuously from the rarefaction waves R1 and R2. 640 Jiequan Li & Yuxi Zheng

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 associated problem in the hodograph plane and show that the hodograph transformation is invertible. Note that initial data (95) is irrotational, we conclude that the ﬂow is always irrotational provided that it is continuous. So the irrotationality condition (3) holds and 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 planar rarefaction waves R1 and R2 into (u,v)plane are exactly two segments H : u cos θ + v sin θ = 0,(− sin θ/κ u 0) and 1 (99) H2 : u cos θ − v sin θ = 0,(− sin θ/κ u 0).

The boundary values of c on H1, H2,are | = + κv =: 1, | = + κv =: 2, c H1 1 c0 c H2 1 c0 (100) where v = u sin θ − v cos θ and v = u sin θ + v cos θ. Obviously, 1, 2 . 0 c0 c0 1 (101)

Thus the wave interaction region Ω is bounded by H1, H2 and the interface of vacuum connecting D and E in the hodograph (u,v)-plane, see Fig. 2. We deﬁne Ω more precisely to contain the boundaries H1 and H2, but not the vacuum boundary c = 0. Boundary conditions. We need to derive the necessary boundary conditions on H1 and H2, respectively. This can be done simply by using coordinate transfor- mations for (92) and (94). Indeed, denote temporarily [compare (94)] ⎧ 1 ⎨ 1−κ 2 1 + C2 − 1 (κu + 1) κ , forγ = 3, Γ(u, C) := m m (102) ⎩ 1 [C2 − 2ln(1 + u)] 2 , forγ = 3.

Fig. 2. Wave interaction region in the hodograph plane Interaction of Rarefaction Waves for Two-Dimensional Euler 641

Then we have

iu = (1 − κv ) Γ(v, C1) cos θ + sin θ , , on H1 (103) iv = (1 − κv ) (Γ (v , C1) sin θ − cos θ , and

iu = (1 + κv ) Γ(v , C2) cos θ + sin θ , , on H2 (104) iv = (1 + κv ) −Γ(v , C2) sin θ + cos θ , 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 sin θ A|H = = tan(θ/2), 1 1 + cos θ −Γ(v , cot θ)sin θ + (1 + cos θ) B|H =− =: B , 1 Γ(v , cot θ)(1 + cos θ)+ sin θ 1 (106) −Γ(v , cot θ)sin θ + (1 + cos θ) A|H = =: A , 2 Γ(v , cot θ)(1 + cos θ)+ sin θ 2 1 + cos θ B|H =− = cot(θ/2). 2 sin θ

Thus the boundary values for α, β on H1 and H2 are α| = θ, β| = (− ), H1 H1 2arctan B1 α| = ( ), β| =−θ. (107) H2 2arctan A2 H2

The boundary values of c on H1 and H2 are given in (100). Now Problem A becomes: 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 boundaries Hi ,i = 1, 2, we have the following estimates:

(i) If θm,¯ there holds ¯ (α − β)| θ. 2m Hi 2 (109) 642 Jiequan Li & Yuxi Zheng

Proof. For the ﬁrst case, that is, θ

Therefore, θ (α − β)| ¯ . 2 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

− (θ/ ) | , | (θ/ ). tan 2 A H2 B H1 tan 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 compatibility condition to this problem, that is, α + β α + β 1 0 −1 1 0 −1 l · ∂+ K − κ cos sin ω = l · ∂− K − κ cos sin ω λ+ 2 λ− 2 (114) at (u,v) = (0, 0), where K = (α, β, c) and l0 = (0, 0, 1). That is, we need to check if there holds α + β α + β 1 −1 1 −1 ∂+c − κ cos sin ω = ∂−c − κ cos sin ω . (115) λ+ 2 λ− 2

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

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 of i comes from the norms of α and β, see the third equation of (69). Therefore, we need only the estimate on α and β. Recall that the derivation of (69) is based on the strict hyperbolicity of the ﬂow, i > 0. These will be achieved when we estimate the C0 norms of α and β, see Section 7.3. The main existence theorem is stated as follows. Let l be the interface of the gas with the vacuum. Interaction of Rarefaction Waves for Two-Dimensional Euler 643

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) in Ω. The vacuum interface l exists and is Lipschitz continuous.

We prove this theorem by two steps. We estimate the solution itself in Section 7.3 and then proceed with estimates on the gradients in Section 7.4. The proof of Theorem 6 is also given in Section 7.4. After we solve Problem B, we show the inversion of hodograph transformation 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 wave interaction region D in the self-similar (ξ, η)-plane for all γ 1 and all wedge half-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.We adopt the method of invariant regions [24]. The notations m¯ and n¯ are given in the beginning of this section.

Lemma 5. Suppose that there exists a C1 solution (α(u,v),β(u,v),c(u,v)) to problem (69), (100) and (107) in Ω. Then the C0-norms of α and β have uniform bounds: (i) If θ n,¯ then θ α 2m¯ − θ and −2m¯ + θ β −θ; (ii) If n¯ <θm,¯ then 2m¯ α − β 2θ,α θ,β −θ.

Proof. Case (i). Consider using system (73). We construct a square, shown in Fig. 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 and L2 to H2.Wehave G(α, β, c)>0, on L1, L2. (116) We have the opposite on the other two sides. Note that the vector (sin β,− cos β) on H1 points toward the interior of Ω, and the vector (sin α, − cos α) on H2 point towards outside of Ω, see Fig. 2. Thus, such a square bounded by L1 and L2 is invariant. [The square exists and remains in the fourth quadrant and is invariant for all the cases θ 2m¯ , but we have smaller invariant regions (that is, invariant triangles) for all θ>n¯.] Case (ii). We show in this case that the upper triangle in Fig. 3a is invariant. Let L denote the diagonal line α − β = 2m¯ . We show that the solution remains above L. ¯ ¯ Consider using Equation (82) to establish ∂±(α+β) 0. We note that ∂+α>0 ¯ and ∂−β>0 before the solution hits L,soα θ and β −θ, thus ω>n¯ and 644 Jiequan Li & Yuxi Zheng

(a) (b) (c) Fig. 3. Invariant regions a(ω, c)>0 for the solution, recalling that n¯ is such that a = 0atθ =¯n.In ¯ ¯ addition, there holds ∂+(α + β) > 0onH2 and ∂−(α + β) > 0onH1.Asimple ¯ bootstrapping argument on Equation (82) implies that both ∂±(α + β) > 0inside the domain. Thus, we ﬁnd γ + ¯ ¯ ¯ ¯ 1 2 ∂+(α − β) =−∂+(α + β) + 2∂+α 2∂+α = sin(2ω)(m − tan ω) 2c (117) or ¯ 2 c∂+ψ −(γ + 1)(tan ω)ψ −(3 − γ)ψ (118) for ψ := m − tan2 ω. (119) Inequality (118) implies that ψ>0 in the domain c > 0. Thus the upper triangle is invariant. Case (iii). First let γ ∈[1, 3) so that θ>m¯ > 0. We show that the lower triangle is invariant, see Fig. 3b. The proof is similar. Again L denotes the line α − 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 not go above the line L. ¯ ¯ Consider using Equation (82) to establish ∂±(α+β) 0. We note that ∂+α<0 ¯ and ∂−β<0 before the solution hits L,soα θ and β −θ, thus ω<θand ω>m¯ > n¯ and a(ω, c)>0 for the solution before the solution hits L. In addition, ¯ ¯ there hold ∂+(α +β) < 0onH2 and ∂−(α +β) < 0onH1. A simple bootstrapping ¯ argument on Equation (82) implies that both ∂±(α + β) < 0 inside the domain. Thus, we ﬁnd γ + ¯ ¯ ¯ ¯ 1 2 ∂+(α − β) =−∂+(α + β) + 2∂+α 2∂+α = sin(2ω)(m − tan ω) 2c (121) or ¯ 2 2 c∂+ψ −(γ + 1)(tan ω)ψ −(γ + 1)(tan θ)ψ (122) Interaction of Rarefaction Waves for Two-Dimensional Euler 645 for ψ := m − tan2 ω. (123) Inequality (122) implies that ψ<0 in the domain c > 0. Thus the upper triangle is invariant. Lastly, we consider γ 3 and m¯ = 0. We show that the lower triangle is invariant and so α − β>0, see Fig. 3c. Let L denote the line α − β = 0. Similar to the previous case, we need to show that the solution does not cross L.Wehave ¯ similarly obtained ∂±(α + β) < 0 inside the domain. Thus, we ﬁnd γ + ¯ ¯ ¯ ¯ 1 2 ∂+(α − β) =−∂+(α + β)+ 2∂+α 2∂+α = sin(2ω)(m − tan ω) (124) 2c or ¯ 2 c∂+(α − β) −(γ + 1) sin ω cos ω(tan ω − m), (125) or ¯ 2 c∂+ sin ω −(γ + 1) sin ω(tan θ +|m|). (126) Inequality (126) implies that α − β>0 in the domain c > 0. Thus the upper triangle is invariant.

Corollary 1. For solutions (α, β, c) of (73), (107) and (100), we have: ¯ ¯ (i) If n¯ <θ0 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 3 − γ tan2 θ = , (127) γ + 1 for 1 <γ <3, that is, θ =¯m, then boundary value (107) becomes constant (α, β)| = (θ, −θ) = , H j , 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 the boundaries H1, H2. We can use (75) to get an explicit solution, κ c = 1 + u, (128) sin θ where − sin θ/κ u 0. We further use (18) to get an explicit solution for the original gas expansion problem, κ(ξ sin θ − 1) c = 1 + , κ + sin2 θ sin θ(ξ sin θ − 1) (129) u = , κ + sin2 θ v = η.

This solution was ﬁrst observed in [25]. 646 Jiequan Li & Yuxi Zheng

Remark 2. In the proof of Lemma 5, we observe that cos((α + β)/2) >δ (130) sin ω 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, there exists a curve u = u(v) such that c(u(v), v) = 0 where u = u(v) is well-deﬁned in 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), 1 J(X, Y ; α, β) =− · cot ω. (132) sin2 ω In view of Lemma 5, we obtain the conclusion.

Corollary 2 shows that we can convert system (57) into system (69) and, therefore, 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 interaction 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 the interface l and the level set of c = ε, ε>0. The remaining sub-domain is denoted by Ωε, in which c >ε. We ﬁrst show that there is a unique solution on Ωε. Then we 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) in Ωε, then the C1 norm of α and β has a uniform bound, which only depends on the C0 and C1 norms of boundary values (107). That is, there is a constant C > 0, depending only on the boundary data (107) and (100), but not on ε, such that (α, β) /ε2, C1(Ωε) C (133) · 1 where C1(Ωε) represents the C -norm. ¯ ¯ Proof. We use (78)tointegrate∂−α and ∂+β along λ+ and λ−-characteristics, respectively. Noting (74), we know that the integral path has a limited length. Also ¯ we note that Q has a uniform bound C/ε2 in Ωε. Then we deduce that ∂−α and ¯ ∂+β are uniformly bounded in Ωε, ¯ 2 ¯ 2 |∂−α| < C/ε , |∂+β| < C/ε . (134) Interaction of Rarefaction Waves for Two-Dimensional Euler 647

¯ ¯ On the other hand, since G has a bound C/ε in Ωε [see (73)], so are ∂+α and ∂−β, ¯ ¯ |∂+α| < C/ε, |∂−β| < C/ε. (135) Hence using the identities, −1 ¯ ¯ −1 ¯ ¯ ∂u =−sin (2ω)(cos α ∂+ −cos β ∂−), ∂v =−sin (2ω)(sin α ∂+ −sin β ∂−), (136) and using the hyperbolicity α = β in Ωε, we conclude that ∂uα, ∂vα, ∂uβ and ∂vβ are uniformly bounded in Ωε, as expressed in (133).

Lemma 7. (modulus estimate) Assume that the solution (α, β) ∈ C2(Ωε). Then we have the following modulus estimate,

(α, β) , < /ε2, C1 1(Ωε) C (137) , , · , 1 1 1 1(Ω ) where C1 1(Ωε) represents the C -norm, and C ε 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)by using (78). The veriﬁcation process is speciﬁed as follows. For any point (u¯, v)¯ inside Ωε, we denote v = v(u;¯u, v)¯ the λ+-characteristic curve from (u¯, v)¯ . Then we 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 (u¯1, v¯1) and (u¯2, v¯2) (compare Fig. 2). We want to use (78) to show the Lipschitz continuity of ¯ ¯ ∂−α. The same is true for ∂+β. Denote u¯ Θ(u,v(u;¯u, v))¯ = exp W(q,v(q;¯u, v))¯ dq Q(u,v(u;¯u, v)).¯ (138) u ¯ Recall that on H1, ∂−α(u,v) ≡ 0. Then we obtain by integrating (78) along the λ+-characteristics, u ¯ 1 ∂−α(u1,v1) = Θ(u,v(u; u1,v1)) du, u¯1 (139) u ¯ 2 ∂−α(u2,v2) = Θ(u,v(u; u2,v2)) du. u¯2 Therefore, we proceed to obtain ¯ ¯ |∂−α(u1,v1) − ∂−α(u2,v2)| u u 1 2 = Θ(u,v(u; u1,v1)) du − Θ(u,v(u; u2,v2)) du u¯1 u¯2 u 1 [Θ(u,v(u; u1,v1)) − Θ(u,v(u; u2,v2))] du (140) u¯2 u¯ u 1 1 + Θ(u,v(u; u1,v1)) du + Θ(u,v(u; u2,v2)) du u¯2 u2

=: T1 + T2 + T3. 648 Jiequan Li & Yuxi Zheng

2 Obviously, since |Θ(u,v(u; u2,v2))| C/ε for some constant C,wehave

T3 C|u1 − u2|. (141)

v = v( ;¯, v)¯ dv(u;¯u,v)¯ = λ ( ,v( ;¯, v))¯ Toestimate T1, we use the deﬁnition of u u : du + u u u and obtain d ∂v(u;¯u, v)¯ ∂λ+ ∂v(u;¯u, v)¯ = (u,v(u;¯u, v))¯ · . (142) du ∂v¯ ∂v ∂v¯ Integration along v = v(u;¯u, v)¯ yields, ∂v(u;¯u, v)¯ u ∂λ+ = exp (q,v(q;¯u, v))¯ dq . (143) ∂v¯ u¯ ∂v

Recall again that λ+ =−cot β and apply Lemma 6. Then we deduce: ∂v(u;¯u, v)¯ C/ε2. ∂v¯ (144)

Noting that ∂v(u;¯u, v)¯ ∂v(u;¯u, v)¯ =−λ+(u,v(u;¯u, v))¯ · , (145) ∂v¯ ∂u¯ we obtain ∂v(u;¯u, v)¯ C/ε2. (146) ∂u¯ Therefore, we have u1 ∂Θ ∂Θ T (u − u ) + (v − v ) u C/ε2(|u − u |+|v − v |), 1 ∂ ¯ 1 2 ∂v¯ 1 2 d 1 2 1 2 u¯2 u (147) whereweuse(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, by the Gronwall inequality, that

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

So we have 2 T2 C/ε (|u1 − u2|+|v1 − v2|). (149) Hence we conclude that

¯ ¯ 2 |∂−α(u1,v1) − ∂−α(u2,v2)| C/ε (|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. Interaction of Rarefaction Waves for Two-Dimensional Euler 649

Proof of Theorem 6. With the classical technique in [18]or[7], we obtain the global 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

dv cu α + β s0 := =− =−cot . (151) du cv 2 Then we have ω ω 1 − 1 = sin > , 1 − 1 =− sin < . α+β 0 α+β 0 (152) s λ− α s λ+ β 0 cos 2 cos 0 cos 2 cos

This shows that the level set Υc is not a characteristic and λ±-characteristics always points 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 solutions in Ωε. Owing to the arbitrariness of width ε>0, we use the contradiction argument to 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 Jacobian −1( ,v; ξ,η) JT u in (24) does not vanish: −1( ,v; ξ,η) = ξ η − ξ η = ( + )( + ) − 2 = . JT u u v v u 1 iuu 1 ivv iuv 0 (153)

We calculate, on the one hand, multiplying (21) with (1 + iuu),

κ − 2 2 + ( + ) + κ − 2 ( + )2 2 i iu iuv 2iuiviuv 1 iuu 2 i iv 1 iuu = ( κ − 2) 2 − ( + )( + ) . (154) 2 i iu iuv 1 iuu 1 ivv On the other hand, from (26) and (58)wehave

2 2 2 2 2κi − i i + 2i ivi v(1 + i ) + 2κi − i (1 + i ) u uv u u uu v uu 2 = 2κi − iv (∂+iu + 1)(∂−iu + 1). (155) Then we obtain ¯ ¯ − (∂+ X + 1)(∂− X + 1) (∂+ X + sin β)(∂− X + sin α) J 1(u,v; ξ,η) =− =− , T λ−λ+ cos α cos β (156) ¯ by using the deﬁnition of ∂±,see(72). This is parallel to (55). Therefore, in order −1( ,v; ξ,η) to show that JT u does not vanish, it is equivalent to prove that: 650 Jiequan Li & Yuxi Zheng

−1( ,v; ξ,η) Lemma 8. The non-degeneracy of the Jacobian JT u 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

α+β + κ ¯ cos 2 1 2 ∂+ X + sin β =−κ − cot ω cos β[m − tan ω] sin ω 2 c cos α ¯ + sin β + ∂+β, 2 sin2 ω α+β + κ (158) ¯ cos 2 1 2 ∂− X + sin α = κ + cot ω cos α[m − tan ω] sin ω 2 c cos β ¯ + sin α − ∂−α. 2 sin2 ω

They are easily simpliﬁed to be

¯ 1 + κ c cos α ¯ c cos α ¯ ∂+ X + sin β =− cos α + ∂+β = [∂+β − Z], sin(α − β) 2 sin2 ω 2 sin2 ω ¯ 1 + κ c cos β ¯ c cos β ¯ ∂− X + sin α = cos β − ∂−α =− [∂−α − Z], sin(α − β) 2 sin2 ω 2 sin2 ω (159) where Z = (1 + γ)tan ω/(2c), as denoted in (86) before. Note that on the bound- ¯ ¯ aries H1 and H2, the values ∂+β and ∂−α are, respectively, ∂¯ β| ≡ , ∂¯ α| ≡ . + H2 0 − H1 0 (160) Therefore, (157) follows from the following Lemma.

Lemma 9. There holds ¯ ¯ ∂+β

Proof. By identities (85), the boundary condition on Z, and (160) we obtain the inequalities (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 ﬁrst level 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 integrate identities (85) in the domain ε

While the non-vanishing Jacobian guarantees local one-to-one, we need global one-to-one, which is guaranteed by the monotonicity of ξ and η along characteristics ¯ for the cases θ ∈ (0, 2m¯ ). In fact, ∂±ξ = 1+∂± X = 0, or ±∂±ξ<0 more precisely, following the above lemma. From (20), (30)wehave∂+ξ =−λ−∂+η, ∂−ξ = −λ+∂−η. Thus ξ and η have the same monotonicity along a plus characteristic Interaction of Rarefaction Waves for Two-Dimensional Euler 651 curve since −λ− > 0, but opposite monotonicity along a minus characteristics since −λ+ < 0 for the cases θ ∈ (0, 2m¯ ) for which α>0 and β<0. For any two points in the interaction zone in the (u,v)plane, there exist two characteristic curves connecting the two points. Either ξ or η is monotone along the connecting path. 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 ξ remains monotone decreasing along both characteristics regardless of the signs of ¯ α or β because of ±∂±ξ<0. See Fig. 4b where we have indicated three points a, b1, and b2.Ifb1 is located above the plus characteristic curve passing a, the variable ξ is monotone decreasing from a to b1 because it is so along both characteristic curves 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 the fact from next subsections [that the two characteristic curves in the (ξ, η) plane are convex and concave, respectively] so that points a and b2 will not give the same value of η. (We have drawn two tangent-lines at the intersection point to show that a 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 convert Problem A into Problem B. Since the region D in Fig. 1b is a wave interaction region, the Jacobian JT (u,v; ξ,η) does not vanish in view of Theorem 1,sothe hodograph 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 in the entire domain of interaction, using properties to be established in the next Section 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) 652 Jiequan Li & Yuxi Zheng

7.7. Properties of the solutions 7.7.1. Convexity of characteristics in the physical plane Now we discuss the convexity of Λ±-characteristics in the mixed wave region D,inthe(ξ, η) plane. It is 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 chain rule, that, ∂ξ ∂ξ ∂η ∂η ∂ + λ+∂v = + λ+ ∂ξ + + λ+ ∂η. (162) u ∂u ∂v ∂u ∂v We rewrite (19)as ∂ξ ∂ξ ∂η ∂η + λ+ =−λ− + λ+ . (163) ∂u ∂v ∂u ∂v Using (18), we have ∂ξ ∂ξ + λ+ = ∂+ X + 1. (164) ∂u ∂v 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)onΛ+ and (166)onΛ− as well as using the deﬁnition of α, β (that is, Λ+ = tan α, Λ− = tan β), we obtain ¯ − ¯ (∂ξ + Λ+∂η)Λ+ = (1 + tan2 α) · (∂+ X + sin β) 1 · ∂+α, ¯ − ¯ (167) (∂ξ + Λ−∂η)Λ− = (1 + tan2 β) · (∂− X + sin α) 1 · ∂−β. ¯ ¯ By Corollary 1, the signs of ∂+α and ∂−β are ¯ ¯ ∂+α<0, ∂−β<0, if θ>m¯ ; ¯ ¯ (168) ∂+α>0, ∂−β>0, if θ ∈ (n¯, m¯ ). In view of Lemma 9,wehave ¯ ¯ ∂+ 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¯ , (170) ∂ξ + Λ+∂η Λ+ < 0, ∂ξ + Λ−∂η Λ− > 0, for θ ∈ (n¯, m¯ ). In sum, we have

Theorem 8. The Λ±-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¯ ),theΛ±-characteristics are concave and convex, respectively. (iii) If θ =¯m, the solution has the explicit form (128) with straight characteristics. Interaction of Rarefaction Waves for Two-Dimensional Euler 653

(a) , (b) ,

Fig. 5. Convexity types of the characteristics and the vacuum boundaries. Dashed curves are of minus family

7.7.2. Regularity of the vacuum boundary Recall that formulae (18) transform the 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 zero with a rate greater than cu, cv for γ 3. Thus, by using (18), we have

c c ξ = u + i = u + c = u,η= v + iv = v + cv = v. u κ u κ (171)

So we conclude that on the vacuum boundary, the (u,v)coordinates coincide with the (ξ, η) coordinates. Next, we prove that the vacuum boundary is Lipschitz continuous. Let us consider the curve {(u,v)| i(u,v) = ε>0} for all small positive ε. Differentiating the equation i(u(v), v) = ε with respect to v, we ﬁnd

du Y α + β =− =−tan . (172) dv X 2 654 Jiequan Li & Yuxi Zheng

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 Lipschitz continuous vacuum boundary.

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

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

7.7.4. Characteristics on the vacuum boundary We already know that the sound 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 view of (73)forγ ∈ (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 become inﬁnite as (u,v) approaches the vacuum boundary, which would force (α, β) to reach the line α − β = 2m¯ in the (α, β)-plane, see Fig. 3. On the other hand, the line 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. That is, for 1 <γ <3, a Λ+-characteristic line has a non-zero intersection angle with a Λ−-characteristic line; however, if γ 3, then there must be α = β on the vacuum boundary.

8. Summary remarks

We have considered the phase space equation (20)

κ − 2 + + κ − 2 = 2 + 2 − κ 2 i iu ivv 2iuiviuv 2 i iv iuu iu iv 4 i (174) known from 1958 for the enthalpy i with the inverse of the hodograph transformation (18) ξ = u + iu,η= v + iv (175) for the two-dimensional self-similar isentropic irrotational Euler system. Upon introducing the variables of inclination angles of characteristics and normalized Interaction of Rarefaction Waves for Two-Dimensional Euler 655 characteristic derivatives, we have changed the second-order phase space equation to a ﬁrst order system (73) ¯ ¯ γ − 1 α + β α − β ∂+α = G(α, β, c), ∂−β = G(α, β, c), ∂ c = · cos sin , 0 2 2 2 (176) where 1 + γ 3 − γ α − β G(α, β, c) = · sin(α − β) · − tan2 . 4c γ + 1 2 Derivatives of the variables α and β along directions not represented in (176)are provided by the higher-order systems (78), (82), (85). We use these infrastructures to construct solutions to binary interactions of planar waves in the phase space and show that the Jacobian of the inverse of the hodograph transform does not vanish, so we obtain in particular a global solution to the gas expansion problem with detailed shapes and positions of the vacuum boundaries and characteristics. The invariant regions in the phase space revealed in the process have more potential than what has been utilized here. For example, we will use them to handle binary interactions of simple waves, which will lead to the eventual construction of global solutions to some four-wave Riemann problems that will not have vacuum in their data, see a forthcoming paper [17]. However, being a well-known difﬁcult problem, the Euler system does not give in easily, which manifests in our inability to establish an invariant triangle (rather than the loose square) for the case θ

Acknowledgments. Jiequan Li thanks the Department of Mathematics at Penn State Univer- sity, and Yuxi Zheng thanks the Mathematics Department at Capital Normal University for the hospitality during their mutual visits when this work was done. Both authors appreciate Shuxing 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. 2006CB805902 and the Key Program from the Beijing Educational Commission with no. KZ200510028018, Program for New Century Excellent Talents in University (NCET) and Funding Project for Academic Human Resources Development in Institutions of Higher Learning Under the Jurisdiction of Beijing Municipality (PHR-IHLB). Yuxi Zheng’s research is partially supported by NSF-DMS-0305497, 0305114, 0603859.

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School of Mathematical Science, Capital Normal University, 100037 Beijing, China. e-mail: [email protected] e-mail: [email protected] and Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, USA. e-mail: [email protected]