ON TRANSONIC SHOCKS IN A NOZZLE WITH VARIABLE END PRESSURES. Jun Li (Department of Mathematics & IMS, Nanjing University, Nanjing , P.R.

Transcription

1 ON TRANSONIC SHOCKS IN A NOZZLE WITH VARIABLE END PRESSURES Jun Li Department of Mathematics & IMS, Nanjing University, Nanjing , P.R.China Zhouping Xin Department of Mathematics IMS, CUHK, Shatin, N.T., Hong Kong Huicheng Yin Department of Mathematics & IMS, Nanjing University, Nanjing , P.R.China In the book [8], Courant Friedrichs described the following transonic shock phenomena in a de Laval nozzle: Given the appropriately large receiver pressure p r, if the upstream flow is still supersonic behind the throat of the nozzle, then at a certain place in the diverging part of the nozzle a shock front intervenes the gas is compressed slowed down to subsonic speed. The position the strength of the shock front are automatically adjusted so that the end pressure at the exit becomes p r. When the end pressure p r varies lies in an appropriate scope, in general, it is expected that a curved transonic shock is still formed in a nozzle. In this paper, we solve this problem for the two-dimensional steady Euler system with a variable exit pressure in a nozzle whose divergent part is an angular sector. Both existence uniqueness are established. Keywords: Steady Euler system, supersonic flow, subsonic flow, transonic shock, nozzle Mathematical Subject Classification: 35L70, 35L65, 35L67, 76N15 1. Introduction main results This paper concerns with the transonic shock problem in a nozzle when the given variable end pressure at the exit of the nozzle lies in an appropriate scope. In [22-25], the authors have studied the well-posedness or ill-posedness of a transonic shock for the supersonic flow through a general 2-D or 3-D slowly-varying nozzle with an appropriately large exit pressure. However, the end pressure or the position of the shock in [22-25] are either induced by the appropriate boundary conditions on the exit or determined by the ordinary differential equations which are resulted from the assumptions on symmetric properties of the supersonic incoming flow, the nozzle walls the end pressure. In this paper, under the more natural physical boundary condition i.e. the variable exit pressure in a suitable scope, we will study the transonic shock problem when a supersonic flow goes through a 2-D curved nozzle with a straight diverging part. In particular, we will verify the following transonic shock phenomena for the steady Euler flow as illustrated in [8]: Given the appropriately large receiver pressure p e x, if the upstream flow is still supersonic behind the throat of the nozzle, then at a certain place in the diverging part of the nozzle a shock front intervenes * Jun Li Huicheng Yin are supported by the National Natural Science Foundation of China No the National Basic Research Programm of China No.2006CB Zhouping Xin is supported in part by Zheng Ge Ru Foundation, Hong Kong RGC Earmarked Research Grants CUHK4028/04P, CUHK4040/06P RGC Central Allocation Grant CA05/06.SC01. 1 Typeset by AMS-TEX

2 the gas is compressed slowed down to subsonic speed, moreover, the position the strength of the shock front are automatically adjusted so that the end pressure at the exit becomes p e x. To simplify the presentation, we only consider the isentropic gases. In fact, by a slight modification, our discussions are also available to the non-isentropic case. The steady isentropic Euler system in two dimensional space is 1 ρu ρu 2 =0, 1 P + ρu ρu 1 u 2 =0, ρu 1 u P + ρu 2 2=0, where u =u 1,u 2,ρ P are the velocity, the density the pressure respectively. Moreover, the pressure function P = P ρ is smooth with P ρ > 0forρ>0, cρ = P ρ being the sound speed. For the ideal polytropic gas, the equation of state is given by P = Aρ γ, here A γ are positive constants, 1 <γ<3 especially γ 1.4 with respect to the air. Assume that the nozzle walls Γ 1 Γ 2 are C 4,α regular for 1 r = x x > 0 is a fixed constant the constant α 0, 1 Γ i consists of two curves Γ 1 i Γ2 i with Γ1 1 Γ 1 2 including the walls for the converging part of the nozzle, while Γ 2 1 Γ 2 2 being straight line segments so that the divergent part of the nozzle is part of a symmetric angular sector. Assume that Γ 2 i is represented by x 2 = 1 i tgθ 0 x 1 with x 1 > 0 <r< +1, where 0 <θ 0 < π 2 is sufficiently small. Furthermore, it is assumed that the C 4,α smooth supersonic incoming flow ρ 0 x,u 1,0 x,u 2,0 x is symmetric near r = so that ρ 0 x =ρ 0 r u i,0 x =U 0 rx i r i =1, 2 near r = this assumption can be easily realized by the hyperbolicity of the supersonic incoming flow the symmetric property of the nozzle walls for <r< + 1, one can see [13]. Suppose that the possible shock Σ the flow field state behind Σ are denoted by x 1 = ηx 2 ρ + x,u + 1 x,u+ 2 x respectively. Then the Rankine-Hugoniot conditions on Σ become [ρu 1 ] η x 2 [ρu 2 ]=0, [P + ρu 2 1] η x 2 [ρu 1 u 2 ]=0, 1.2 [ρu 1 u 2 ] η x 2 [P + ρu 2 2]=0. In addition, P + x satisfies the physical entropy condition see [8]: On the exit of the nozzle, the end pressure is prescribed by P + x >P x on x 1 = ηx P + x =P e + εp 0 θ on r = +1, 1.4 2

3 here ε>0 is sufficiently small, θ = arctan x 2 x 1, P 0 θ C 3,α [ θ 0,θ 0 ]with P 0±θ 0 =P 3 0 ±θ 0 =0, P 0 arctan x 2 x C C, 3,α {x 1 1,x 2: x 2 1 +x2 2 =X0+1, x2 x1tanθ0} the constant P e denotes the end pressure for which a symmetric shock lies at the position r = r 0 with r 0, + 1 the supersonic incoming flow is given by ρ 0 r,u 0 r, for more details, one can see Proposition 2.1 in 2. Since the flow is tangent to the nozzle walls x 2 = 1 i tgθ 0 x 1 i =1, 2, then u + 2 = 1i tgθ 0 u + 1 on x 2 = 1 i tgθ 0 x Finally, θ 0 are assumed to be suitably large small respectively so that η 0 θ 0 =1 2 <θ 0 <η with η 0 > 0 being a suitably small constant. It is noted here that the assumption 1.6 implies that the nozzle wall Γ 2 i : x 2 = 1 i tgθ 0 x 1 is close to the line segment x 2 = 1 i for r +1. As will be shown in 2, under the above assumptions on the nozzle the symmetric supersonic incoming flow near the throat of the nozzle, there exists a unique symmetric transonic shock solution for the given constant end pressure. Furthermore, the position of the shock location, r = r 0, depends monotonically on the given end pressure. This solution will be the background solution. Let P 0 + r,u+ 0 r be the subsonic part of the background solution for r 0 <r< + 1, which can be extended into the domain {r : r +1} the extension will be denoted by ˆP 0 + r, Û 0 + r. For more details, one can see Proposition 2.1 Remark 2.2 in 2. The first main result in this paper is Theorem 1.1. Uniqueness Let the assumptions above hold M0 U 0 X 0 cρ 0 > 2 γ+1 2 γ. Then there exists a constant ε 0 = 1 such that for all ε 0,ε X0 3 0 ], the problem has no more than one solution P + x,u + 1 x,u+ 2 x; ηx 2 with the following properties: i. ηx 2 C 4,α [x 1 2,x 2 2], ηx 2 r0 2 x2 2 L [x 1 2,x2 2 ] C 0 ε, ηx 2 r0 2 C 2 x2 3,α [x 1 2,x2 2 ] C 0 ε, where x i 1,x i 2i =1, 2 sts for the intersection points of x 1 = ηx 2 with x 2 = 1 i tgθ 0 x 1 for i =1, 2, C 0 is a positive constant depending on α the supersonic incoming flow. ii. P + x, u + 1 x,u+ 2 x C3,α Ω +, where Ω + is the subsonic region given by P +,u + 1,u+ 2 ˆP + 0, û+ 1,0, û+ 2 C 3,α Ω + C 0 ε, Ω + = {x 1,x 2 :ηx 2 <x 1 < +1 2 x 2 2, x 2 <tgθ 0 x 1 }, ˆP + 0, û+ 1,0, û+ 2 =ˆP + 0 r, Û + 0 r x r. Remark 1.1. Besides the uniqueness result described by Theorem 1.1, it will be also shown that the position of the shock depends on the given end pressure monotonically. This will be stated more precisely in Proposition 3.2 in 3. In addition, the order ε in the assumption of ηx 2 r 2 0 x2 2 L [x 1 2,x2 2 ] 3

4 comes essentially from the relation between the shock position the end pressure see Proposition 5.3 Remark 5.2 in 5. This implies that the shock position will move with the order Oε when the end pressure changes in an order Oε in 1.4. Remark 1.2. Due to the corners in the subsonic region, the requirement of C 3,α regularity for the uniqueness in Theorem 1.1 seems stringent. However, based on such a uniqueness, we can obtain the existence of a transonic solution in the same regularity class, see Theorem 1.2 below Appendix. Remark 1.3. The previous uniqueness results in [22, 24-25] require that either the transonic shock curve goes through a fixed point or one solution has special symmetry. Remark 1.4. The assumption that r 0, +1 is made to ensure a positive distance between the shock position the exit of the nozzle, this will be used in the analysis in 4. Based on Theorem 1.1, we can establish the following existence theorem. Theorem 1.2. Existence of a transonic shock solution For the nozzle the supersonic incoming flow defined as above, the problem 1.1 with has a unique transonic shock solution such that ρ +,u + 1,u+ 2 ; ηx 2 satisfies the estimates in Theorem 1.1. Remark 1.5. It should be noted that similar results as in Theorem 1.1 Theorem 1.2 hold for the 2-D Euler flows with the pressure depending on both the density the entropy. However, neither the uniqueness nor the existence as in Theorem 1.1 Theorem 1.2 holds for irrotational flows, see [22, 24]. Remark 1.6. Theorem 1.1 Theorem 1.2 can be extended into the 3-D nozzle case, which will be given in our forthcoming paper [15]. Remark 1.7. For nozzles with part of symmetric angular sector as the converging part, one can also establish a corresponding theory of uniqueness existence of a transonic shock lying in the converging part of the nozzle as Theorem However, it is shown in [25] that such a transonic shock is dynamically unstable. It is noted that there have been many works on the steady transonic problems see [3-9], [12], [14], [18-28] the references therein. In particular, in [25], it is shown that under the same assumptions as in this paper on the nozzle the supersonic incoming flows, there exist two constant pressures P 1 P 2 with P 1 <P 2, such that if the exit constant pressure P e P 1,P 2, then the symmetric transonic shock exists uniquely in the diverging part of the nozzle, the position the strength of the shock is completely determined by P e. More importantly, we establish the dynamically asymptotic stability of such a transonic shock for the unsteady Euler system in both 2-D 3-D cases. Various related results, such as uniqueness with additional assumptions, non-existence, compatibility conditions, many useful analytical tools on transonic shocks in slightly curved nozzles with given appropriate end pressure at the exit of the nozzle for either steady irrotational flows or steady Euler flows have been established in [22-25], see also [5-7, 26]. Next we would like to comment on the proofs of the main results in this paper. In should be noted that in almost all the previous works mentioned above except [15, 22], the uniqueness is obtained under the additional assumption that the shock curve goes through a fixed point in advance [7, 22-26]. This condition is crucial in the proofs. However, for non-flat nozzles, this additional condition may lead the transonic shock problem to be over-determined [22]. In the present paper, we have found a new way to determine the position of the transonic shock remove the undesired assumption that the shock curve goes through a fixed point so that the transonic shock problem as described by Courant-Friedrichs is well-posed. Besides the analytical tools developed in [22-25], the new key ingredients in this paper are to establish the monotonic property of the pressure along the nozzle walls to estimate the gradients of the solution instead of the solution itself so that one can avoid the difficulties induced by the unknown position of the shock, for more detailed explanations, see 5. It follows from this that the position of the shock can be uniquely determined in Theorem 1.1 when the end pressure is given, the continuous dependence monotonic property of the end pressure on the position of the shock are also derived. With these crucial results, we can complete the proofs on Theorem 1.1 Theorem 1.2. The rest of the paper is organized as follows. In 2, for reader s convenience, the description of the background solution is given although it has been done in [25]. In 3, we reformulate the 2-D problem 1.1 with the boundary conditions so that one can obtain a weakly coupled second order elliptic 4

5 equation for the density ρ with mixed boundary conditions, a 2 2 first order system for the angular velocity U 2 an algebraic equation on ρ, u 1,u 2 along a streamline. In 4, using the decomposition techniques in 3, we establish some a priori estimates on the derivatives of difference between two possible solutions. In 5, based on the estimates given in 4, through looking for an ordinary differential inequality along a nozzle wall, we can derive that the end pressure on the wall is monotonic with respect to the position of the shock, thus the position of the shock can be uniquely determined by the end pressure the proof of the uniqueness result in Theorem 1.1 is then completed. In 6, along the nozzle walls, by establishing the continuous dependence of the shock position on the end pressure, we can determine the position of the shock complete the proof on Theorem 1.2. Finally, in Appendix, we will give a proof on the existence of a transonic shock C 3,α solution when the transonic shock is assumed to go through a fixed point the end pressure is suitably adjusted by a constant. In what follows, we will use the following convention: Oε means that there exists a generic constant C 1 > 0 independent of ε such that Oε C 2,α C 1 ε. O 1 X m 0 O 1 X m 0 m >0 means that there exists a generic constant C 2 > 0 independent of ε such that C 1,α C2 X m The background solution. In this section, we will describe the transonic solution to the problem 1.1 with when the end pressure is a suitable constant P e under the assumptions on the nozzle walls the supersonic incoming flow in 1. Such a solution is called the background solution can be obtained by solving the related ordinary differential equations. In fact, the related analysis has been given in Section 147 of [8] the details can be seen in [22, 25]. But for the reader s convenience later computations in this paper, we still outline it here. Proposition 2.1. Existence of a transonic shock for the constant end pressure For the 2-D nozzle the supersonic incoming flow given in 1, there exist two constant pressures P 1 P 2 with P 1 <P 2, which are determined by the incoming flow the nozzle, such that if the end pressure P e P 1,P 2, then the system 1.1 has a symmetric transonic shock solution { P 0 r,u 1,0 P, u 1,u 2 = x,u 2,0 x, for r<r 0, P 0 + r,u+ 1,0 x,u+ 2,0 x, for r>r 0, here u ± i,0 x =U 0 ± xi r r i =1, 2, P 0 ± r,u± 0 r are C4,α smooth. Moreover, the position r = r 0 with <r 0 < +1 the strength of the shock are determined by P e. Proof. Let r = r 0 be the location of the shock which to be found. It follows from 1.1 that the supersonic incoming flow ρ 0 r,u 0 r <r<r 0 the subsonic flow ρ + 0 r,u+ 0 rr 0 <r< + 1 satisfy respectively { d dr rρ± 0 U 0 ± =0, U 0 ± 2 + hρ ± 0 = 1 2 U 0 ± r hρ ± 0 r 0, here hρ ± 0 is the enthalpy given by h ρ ± 0 = c2 ρ ± 0. ρ ± 0 The corresponding Rankine-Hugoniot conditions across the shock r = r 0 are { [ρ0 U 0 ]=0, [ρ 0 U0 2 + P 0 ]=0. As in the [22, 25], the proof of the Proposition can be divided into four steps. Step 1. For the supersonic state ρ 0 r 0,U 0 r 0, there exists a unique subsonic state ρ + 0 r 0, U + 0 r 0 satisfying

6 The proof can be found in Section 147 of [8], so it is omitted here. Step 2. For a given supersonic state ρ 0,U0, 2.1 has a unique supersonic solution ρ 0 r, U0 r for r [, +1]. In fact, it follows from 2.1 that { f1 ρ 0,U 0,r rρ 0 ru 0 r C 0 =0, f 2 ρ 0,U 0,r 1 2 U 0 r2 + hρ 0 r C 1 =0 with C 0 = ρ 0 U0 C1 = 1 2 U hρ 0. Since dρ 0 dr = ρ 0 U 0 2 ru0 2 c 2 ρ 0, du0 U = 0 c2 ρ 0 dr ru0 2 c 2 ρ 0 du0 2 c 2 ρ 0 2P ρ 0 = +ρ 0 P ρ 0 U0 2 dr ru0 2 c 2 ρ 0, then one has U0 r2 c 2 ρ 0 r 1 U0 2 2 c 2 ρ 0 > 0 for r In addition f 1,f 2 ρ 0,U 0 = r U0 r2 c 2 ρ 0 r f 1,f 2 ρ ρ 0,U 0 > 0. This, together with the implicit function theorem 2.3, yields that 2.1 has a unique supersonic solution 0 X0,U 0 X0,X0 ρ 0 r,u 0 r for r [, +1]. Step 3. For a given subsonic state ρ + 0,U 0 +, 2.1 has a unique subsonic solution ρ + 0 r,u+ 0 r for r [, +1]. The proof is similar to that in Step 2, so is omitted. Step 4. The shock position r 0 is a continuously decreasing function of P e when the end pressure P e lies in an appropriate scope. In fact, it follows from that for r [, +1] { rρ ± 0 ru 0 ± r C 0, 1 2 U 0 ± r2 + hρ ± 0 r C± 1, 2.4 here C 1 ± are the Bernoulli s constants. Note that C 1 C+ 1 are different in general, moreover, C+ 1 depends on the end pressure P =P e. Especially, { X0 ρ ± 0 U 0 ± C 0, 1 2 U 0 ± 2 + hρ ± 0 C 1 ±. 2.5 Next we derive the dependence of r 0 on the end pressure P =P e. It follows from the first equation in 2.4 the second equation in 2.5 that dρ ± 0 r 0U 0 ± r 0 dρ + 0 X = ρ ± 0 0 r 0U 0 ± r dr 0 0 r 0 dρ + 0, U 0 + r 0 du 0 + r 0 dρ c2 ρ + 0 r 0 dρ + ρ + 0 r 0 r 0 0 dρ + 0 = dc+ 1 dρ

7 In addition, by C 1 + = C X0 2 hρ+ ρ+ 0 X , the second equation in , one has [ρ 0 U0 2 dr 0 ] r 0 dρ = ρ+ 0 r dc dρ = ρ+ 0 r 0c U ρ + 0 X Since [ρ 0 U0 2 ] < 0 due to [ρ 0 U0 2 + P 0 ]=0[P 0 ] > 0, then 2.7 implies that r 0 is a continuous strictly decreasing function of the end pressure P Next, we complete the proof on Proposition 2.1. For r 0 [, + 1, it follows from Step 2 that there exists a unique supersonic flow in [,r 0 ]. Moreover, due to Step 1 Step 3, there exists a unique shock at r 0 a unique subsonic flow in [r 0, + 1]. Thus one can define a function F r 0 =P for r 0, + 1. By Step 4, F r 0 is a strictly decreasing continuous differentiable function on P When r 0 = or r 0 = +1, one can obtain two different end pressures P 1 P 2 with P 1 <P 2. Therefore, by the monotonicity of F r 0, one can obtain a symmetric transonic shock for P P e P 1,P 2. Remark 2.1. By the assumption 1.6 the equation 2.4, one can easily conclude that there exists a constant C>0 independent of such that for r +1 d k U 0 + r dr k + d k P 0 + r dr k C X0 k, k =1, 2, 3, Remark 2.2. It follows from Step 2 Remark 2.1 that one can extend the subsonic flow ρ + 0 r,u+ 0 r into ˆρ + 0 r, Û + 0 r defined for r, +1 satisfies 2.1 on, +1. This extension will be often used later on. 3. The reformulation on problem 1.1 with In this section, the nonlinear problem 1.1 with will be reformulated so that one can obtain a second order elliptic equation for ρ + x two2 2 first order systems for the radial speed U 1 + the angular speed U 2 +. To this end, as in [22, 25], we firstly derive the relations between P +,U 1 + U 2 + on the shock Σ. It is more convenient to use the polar coordinates { x1 = rcosθ, 3.1 x 2 = rsinθ decompose u + 1,u+ 2 as { u + 1 = U 1 + cosθ U 2 + sinθ, u + 2 = U 1 + sinθ + U 2 + cosθ. 3.2 Then, become respectively r ρ + U r θρ + U ρ+ U + 1 r =0, r ρ + U P r θρ + U 1 + U ρ+ U U r =0, 3.3 r ρ + U 1 + U r θp + + ρ + U r ρ+ U 1 + U 2 + =0 [ρu 1 ] r θ rθ [ρu 2]=0, [ρu1 2 + P ] r θ rθ [ρu 1U 2 ]=0, [ρu 1 U 2 ] r θ rθ [P + ρu 2 2 ]=0, 7 3.4

8 where r = rθ sts for the shock Σ in the coordinate r, θ. In addition, for any C 1 solution, 3.3 is equivalent to r ρ + U r θρ + U ρ+ U + 1 r =0, U 1 + ru U + 2 r θ U rp + ρ U r =0, U + 1 ru U + 2 r θ U r θ P + ρ + + U + 1 U + 2 r = here It follows from 3.4 that on r = rθ { ρ + ˆρ + 0 r 0Û 0 + r 0+ˆρ + 0 r 0U 1 + Û 0 + r 0 = g 1, P + ˆP 0 + r 0 + ρ + ˆρ + 0 r 0U ˆρ + 0 r 0Û 0 + r 0U 1 + Û 0 + r 0 = g 2, g 1 = ρ+ 2 U + 1 U [P ]+ρ + U ρ 0 r 0U 0 r 0 ρ 0 U 0 ρ+ ˆρ + 0 r 0U + 1 Û + 0 r 0, g 2 = ρ+ U + 1 U [P ]+ρ + U P 0 r 0 P ρ 0 r 0U 0 r 0 2 ρ 0 U 0 2 ˆρ + 0 r 0U + 1 Û + 0 r 0 2. Thus, a direct computation yields that on r = rθ { U + 1 Û 0 + r 0= g 1 U 2 + 2,P0 P 0 r 0,U0 U 0 r 0, P + ˆP 0 + r 0= g 2 U 2 + 2,P0 P 0 r 0,U0 U 0 r with g i 0, 0, 0 = 0 for i =1, 2, here the quantities on the right h sides of the above will be suitably small. For convenience, the following transformation { y1 = r, y 2 = θ, will be used to change the fixed walls into y 2 = ±1 respectively. In addition, the superscripts + will be neglected for convenience in case no confusions. Then, can be rewritten respectively as y1 ρu 1 + y2 ρu 2 + ρu 1 =0, U 1 y1 U 1 + U 2 y2 U 1 + P ρ U 2 2 =0, U 1 y1 U 2 + U 2 y2 U 2 + y2 P ρ + U 1U 2 = [ρu 1 ] ξ y 2 [ρu ξy 2 2 ]=0, [ρu1 2 + P ] ξ y 2 [ρu ξy 2 1 U 2 ]=0, [ρu 1 U 2 ] ξ y 2 [P + ρu ξy ]=0, 3.9 where ξy 2 = r y2. 8

9 Note that the transformation between the coordinate systems x 1,x 2,y 2 keeps the equivalence of C 4,α norm independent of. So from now on, we will use the coordinate system,y 2 instead of x 1,x 2. Let y 2 = y 2,β be the characteristics starting from the point ξβ,β for the first order differential operator U 1 y1 + X0U2 y2.then { dy2,β d = U2 U 1 y1,y 2,β, y 2 ξβ,β=β, β [ 1, 1] It follows from the second the third equations in 3.8 that the following Bernoulli s law holds 1 2 U U hρ,y 2,β = G 0 β 3.11 with G 0 β = 1 2 U 1ξβ,β U 2ξβ,β 2 + hρξβ,β. Next, we derive the governing problems for U 1 U 2. It follows from 3.11 that U 1 y2 U 1 + U 2 y2 U 2 + y 2 P ρ = d dβ G 0β y2 β,y 2, 3.12 U 1 y1 U 1 + U 2 y1 U 2 + P ρ = X 0 U2 U,y 2,β d 1 dβ G 0β y2 β,y 2, 3.13 here β,y 2 denotes the inverse function of y 2 = y 2,β. Combining with the first equation the third equation in 3.8 yields { y1 U 1 = h 1 P, U 1,U 2, y1 P, y2 P, y2 U 1 = h 2 P, U 1,U 2, y1 P, y2 P 3.14 y1 U 2 = h 3 P, U 1,U 2, y1 P, y2 P, y2 U 2 = h 4 P, U 1,U 2, y1 P, y2 P, U 2, ±1 = 0, 3.15 here h i = Δi Δ 0 1 i 4 with 9

10 Δ 0 = U 1 2 +U 2 2, Δ 1 = U 2 X0 y2 P ρ + U 1U U1 U 1 y1 P ρ U1 y1 P + U 2 y2 P + U 2 d U 1 dβ G 0β,y 2 y2 β,y 2 U 2 2 ρc 2, ρ U1 Δ 2 = + U 2 2 d U 1 dβ G 0β,y 2 y2 β,y 2 y 2 P ρ U 2 d U 1 dβ G 0β,y y2 β,y 2 + U 2 2 X0 y2 P U 1 ρ Δ 3 = U 1U 2 ρc 2 ρ U 1 y1 P + U 2 y2 P U 1 y1 2ρ y 2 P U 2 y1 P ρ Δ 4 = U 1 2 ρc 2 ρ U 1 y1 P + U 2 y2 P y2 P U 2 ρ + U 1U 2 U1 2 P ρ U 1. + U 2 U1 + U 1U 2 Next, we derive the governing problem for the density ρ. It follows from U 2 =0ony 2 = ±1 the third equation in 3.8 that U1 ρc 2 y1 P + U 2 y2 P ρ + U 1 2 P U 2, ρ + U 2 U 1 d dβ G 0β,y 2 y2 β,y 2 + U 2 d dβ G 0β,y 2 y2 β,y 2 y2 ρ =0 on y 2 = ± Furthermore, applying the first order operator U 1 y1 + X0 U 2 y2 to the first equation in 3.8 subsequently subtracting y1 ρ {the second equation in 3.8} y2 ρ {the third equation in 3.8} yield the following boundary value problem for the density y1 U 2 1 c 2 ρ y1 ρ + X0U1U2 y2 ρ + X0 y2 U1 U 2 y1 ρ + X0 U 2 2 cρ 2 y2 ρ = y1 X0 ρu 1 y2 U 2 + ρu 2 1 X0ρU2 y2 U 1 + ρu 2 2 y2 ρu2 y1 U 1 ρu 1 y1 U 2 + y1 U 1 + X0 y2 U 2 U 1 y1 ρ + X0 U 2 y2 ρ + ρ y1 U 1 + X0 ρ y2 U 2 + ρu1, P ρ ˆP 0 + r 0= g 2 U 2 2,P P0 r 0,U U0 r 0 on = ξy 2, y2 ρ =0 on y 2 = ±1, P ρ =P e + εp 0 y2 on = +1. In addition, due to the third equation of 3.9, it holds that, 3.17 ξ y 2 = ξy 2[ρU 1 U 2 ] [ρu P ]. Therefore, in order to prove Theorem 1.1, it suffices to show Theorem 3.1. Let the assumptions of Theorem 1.1 hold. Then for ε< 1, the free boundary value X0 3 problem with 3.9 has no more than one solution P y,u 1 y,u 2 y; ξy 2 satisfying the following estimates with a uniform constant C>0 depending on α the supersonic incoming flow: i. ξy 2 C 4,α [ 1, 1], ξy 2 r 0 L [ 1,1] C X0 ε, 10 ξ y 2 C 3,α [ 1,1] Cε.

11 ii. satisfies P y,u 1 y,u 2 y C 3,α ω + y k 1 P, U1 y ˆP 0 +, Û 0 + Cα ω + C X0 2, k =0, 1, 2, 3 y2 P, U 1 C 2,α ω + + U 2 C 3,α ω + Cε, where ω + = {,y 2 :ξy 2 < < +1, 1 <y 2 < 1}. Remark 3.1. It is noted that i ii in Theorem 3.1 are less restrictive than i ii in Theorem 1.1. To show Theorem 3.1, as in [25], one may reduce the free boundary problem with 3.9 into a fixed boundary value problem. Indeed, set Then the domain ω + becomes For convenience, one sets D 0 1 = 1 ξz 2 +z 1 +1 ξz 2, D 1 y1 = 1 +1 ξz 2 z 1, D 2 y2 = { z 1 = ξy 2 +1 ξy 2, z 2 = y E + = {z 1,z 2 :0<z 1 < 1, 1 <z 2 < 1} z 1 1ξ z 2 ξz 2 +z 1 +1 ξz 2 +1 ξz 2 z 1 + Then, can be rewritten respectively as [ρu 1 ]ξz 2,z 2 = ξ z 2 [ρu ξz 2 2 ]ξz 2,z 2, [ρu1 2 + P ]ξz 2,z 2 = ξ z 2 [ρu ξz 2 1 U 2 ]ξz 2,z 2, [ρu 1 U 2 ]ξz 2,z 2 = ξ z 2 [P + ρu ξz ]ξz 2,z 2, ξz 2 +z 1 +1 ξz 2 z { D1 U 1 = h 1 P, U 1,U 2,D 1 P, D 2 P, D 2 U 1 = h 2 P, U 1,U 2,D 1 P, D 2 P 3.23 D 1 U 2 = h 3 P, U 1,U 2,D 1 P, D 2 P, D 2 U 2 = h 4 P, U 1,U 2,D 1 P, D 2 P, U 2 z 1, ±1 = 0, 3.24 here h i i =1, 2, 3, 4 has the same property as h i. 11

12 In addition, 3.17 becomes D 1 U 2 1 c 2 ρd 1 ρ + U 1 U 2 D 2 ρ + D 2 U1 U 2 D 1 ρ +U2 2 c 2 ρd 2 ρ = D 1 D 0 ρu1 2 + D 0 ρu2 2 +D 0 ρu 1 D 2 U 2 D 1 ρu 1 D 2 U 2 + D 2 ρu 1 D 1 U 2 D 0 ρu 2 D 2 U 1 + D 1 ρu 2 D 2 U 1 D 2 ρu 2 D 1 U 1 +D 1 U 1 + D 2 U 2 U 1 D 1 ρ + U 2 D 2 ρ + ρd 1 U 1 + ρd 2 U 2 + D 0 ρu 1, P ρ ˆP 0 + r 0= g 2 U 2 2,P P0 r 0,U U0 r 0 on z 1 =0, z2 ρ =0 on z 2 = ±1, P ρ =P e + εp 0 z2 on z 1 = We conclude this section by noting that as a by-product of the analysis for Theorem 3.1 Theorem 1.1, one can further obtain more estimates on the location of the shock its monotonic dependence on the end pressure. Indeed, suppose that the problem 1.1 with has two C 3,α solutions ρ, U 1,U 2 ; ξ 1 q, V 1,V 2 ; ξ 2 which satisfy the exit pressure conditions P e + εp 01 θ P e + εp 02 θ at r = + 1 respectively admit the estimates in Theorem 1.1. In terms of the transformations 3.1, , the end conditions of P ρ P q can be written as P ρ =P e + εp 01 z2 P q =P e + εp 02 z2. Then we can arrive at Proposition 3.2. Under the assumptions of Theorem 1.1, if ρ, U 1,U 2 ; ξ 1 q, V 1,V 2 ; ξ 2 are defined as above, then the following estimates hold ξ 1z 2 ξ 2z 2 C 2,α [ 1,1] C 1 X0 2 ξ 1 1 ξ ε 1 P X 01 z 2 P 0 X 02 z 2 C 0 X 1,α [ 1,1] ρ q C 2,α E + + U 1,U 2 V 1,V 2 C 2,α E + C 1 ξ 1 1 ξ ε 1 P X 01 z 2 P 0 X 02 z 2 C 1,α [ 1,1], where E + is given in Furthermore, if P 01 z2 =P 02 z2 +C 1 with the constant C 1 > 0, then ξ 1 z 2 <ξ 2 z A priori estimates on the solutions of In this section, we establish some key a priori estimates on the gradients of the difference between two solutions ρ, U 1,U 2 ; ξ 1 z 2 q, V 1,V 2 ; ξ 2 z 2 to the problem with exit pressure conditions P e + εp 01 z 2 X P e + εp 02 z 2 0 X respectively. These estimates will play crucial roles in proving Theorem Theorem 1.2. For convenience, let Q = P q denote the pressure for the density q, D 0,D 1,D 2 D 0, D 1, D 2 will denote the expressions as in 3.21 corresponding to ρ, U 1,U 2 ; ξ 1 z 2 q, V 1,V 2 ; ξ 2 z 2 respectively. Set W i,w 3 z =U i,ρ ξ 1 z 2 +z 1 +1 ξ 1 z 2,z 2 V i,q ξ 2 z 2 +z 1 +1 ξ 2 z 2,z 2, i =1, 2, W 4 = ξ 1 z 2 ξ 2 z 2, M j = z1 W j, j =1, 2, 3, N k = z2 W k, k =1, 2, 3, 4. 12

13 A key technical point here is that we will focus on the estimates on M j j =1, 2, 3 N k k =1, 2, 3, 4 directly, which will be established in a series of Lemmas Lemma 4.1-Lemma 4.5. In the following lemmas, we always assume that the assumptions in Theorem 3.1 hold. Lemma 4.1. D 0 D 0 = O 1 W X0 2 4, D 1 D 1 = O1W 4 z1, 4.1 D 2 D 2 = OεW 4 z1 + O1N 4 z1 + O 1 W 4 z2. Proof. We only check the estimate on D 1 D 1 since the other terms can be treated similarly. Indeed, D 1 D ξ 1 z 2 ξ 2 z 2 1 = +1 ξ 1 z 2 +1 ξ 2 z 2 z 1 W 4 = +1 ξ 1 z 2 +1 ξ 2 z 2 z 1, so the second inequality in 4.1 holds since 1 +1 ξ 1 z 2 +1 ξ 2 z 2 C 1,α C ξ iz 2 isa small perturbation of r 0 due to the assumptions. Lemma 4.2. Estimate of N 4 It holds that N 4 C 2,α Cε W 1,W 2,W 3 C 1,α + W 4 L + N 1,ε 1 N 2,N 3 C 1,α. 4.2 Proof. It follows from 3.18 that ξ 1z 2 [ρu P ]ξ 1 z 2,z 2 =ξ 1 z 2 [ρu 1 U 2 ]ξ 1 z 2,z 2, ξ 2z 2 [qv2 2 + Q]ξ 2 z 2,z 2 =ξ 2 z 2 [qv 1 V 2 ]ξ 2 z 2,z 2. The difference of these two equations yields { N 4 z 2 =Oε W 1,W 2,W 3,W 4 +Oε N 1,ε 1 N 2,N 3,X0 1 N 4, N 4 ±1 = 0, here what follows, for notational convenience, Oε A denotes the inner product of vectors A Oε whose component is of order ε. 4.2 then follows from this initial value problem for an ODE. Lemma 4.3. Estimates of M i M 1,M 3 C 1,α C W 1,ε W 2,W 3,W 4 C 1,α + Cε N 1,ε 1 N 2,N 3,N 4 C 1,α, 4.3 M 2 C 1,α Cε W 1, ε 1 W 2,W 3,W 4 C 1,α + εn 1,N 2,ε 1 N 3, ε 1 N 4 C 1,α. 4.4 Proof. Rewrite 3.8 as D 1 ρu 1 + D 2 ρu 2 + D 0 ρu 1 =O 1 W 4 + OεN 4 D 1 P + ρu 1 D1 U 1 + ρu 2 D2 U 1 D 0 ρu 2 2 =O 1 W 4 + O ε N 4 ρu 1 D1 U 2 + ρu 2 D2 U 2 + D 2 P + D 0 ρu 1 U 2 =OεW 4 + O 1 N

14 Then tedious but elementary computations using the assumptions in Theorem 3.1 show that U 1 D1 W 3 + ρ D 1 W 1 = OX0 1 W 1,ε W 2,W 3,W 4 +Oε M 2,εM 3 +Oε ε 1 N 2,N 3,N 4, c 2 ρ D 1 W 3 + ρu 1 D1 W 1 = OX0 1 W 1,ε W 2,W 3,W 4 +Oε 2 M 1 + OεN 1 + O ε N 4, ρu 1 D1 W 2 = Oε W 1, ε 1 W 2,W 3,W 4 +Oε εm 2,M 3 +OεN 2 + O1N 3 + O 1 N 4, 4.6 With respect to the variables D 1 W 1, D 1 W 2, D 1 W 3, the determinant of the coefficient matrix in 4.6 is ρ 2 U 1 c 2 ρ U1 2 > 0 for subsonic states, then a direct computation yields Lemma 4.3. Lemma 4.4. Estimates of N 1 N 2 For i =1, 2, N i C 1,α Cε W 1,W 2,W 3,W 4 C 1,α + C N 4 C 2,α + C N 3 C 1,α. 4.7 Proof. First, we deal with the term d dβ G 0βz 1,z 2 D 2 βz 1,z 2 contained in Let z2s; 1 z z2s; 2 z be the characteristics of 3.10 going through z 1,z 2 withz20; 1 z =β z20; 2 z = β corresponding to U 1,U 2 V 1,V 2 respectively, i.e., dz2s; 1 z = +1 ξ 1 z2 1 ds A U 2 ξ1 z s +1 ξ 1 z2,z 1 2 1, z2z 1 1 ; z =z 2, z20; 1 z =β, where A 1 =ξ 1 z 1 2+s +1 ξ 1 z 1 2U 1 + U 2 s 1ξ 1z 1 2. Similarly, one can define z 2 2s; z corresponding to V 1,V 2, ξ 2 β. Set ls; z =z 1 2s; z z 2 2s; z. Then one has { dl ds = Oεl + OεW 1s, z 1 2+O1W 2 s, z 1 2+OεW 4 z 1 2+Oε 2 N 4 z 1 2 l0; z =β β, lz 1 ; z = Since the solution has the C 3,α regularities, then all the coefficients in 4.8 are in C 2,α, this will lead to the C 2,α estimate of β β. By 4.8, one has β β L C εw 1,W 2,εW 4,ε 2 N 4 L { z1 β β = OεW1 t, z O1W 2 t, z2+oεw 1 4 z2+oε 1 2 N 4 z2+oεlt; 1 z dt, ls; z = s OεW1 z 1 t, z2+o1w 1 2 t, z2+oεw 1 4 z2+oε 1 2 N 4 z2+oεlt; 1 z dt, 4.9 This yields In addition, set z1 β, β C 2,α Cε, z2 β, β C 2,α C. β β C 2,α C εw 1,W 2,εW 4,ε 2 N 4 C 2,α { B1 = d dβ G 0βz 1,z 2 D 2 βz 1,z 2, B 2 = d G dβ 0 βz 1,z 2 D 2 βz1,z 2. 14

15 Then, direct computations using show that B 1 B 2 = O ε W 4 + Oε 2 N 4 + Oε 2 z1 β X β+oε z2 β β 0 + Oε W 1,W 2,W 3 0,β+O1 N 1,εN 2,N 3 0,β+Oεβ β. Thus, by use of 4.10, the following estimate holds B 1 B 2 C 1,α Cε W 1,W 2,W 3,X 1 0 W 4,εN 4 C 1,α + β β C 2,α+C N 1,εN 2,N 3 0,z 2 C 1,α [ 1,1] Cε W 1,W 2,W 3,X 1 0 W 4,εN 4 C 2,α + C N 1,εN 2,N 3 0,z 2 C 1,α [ 1,1] Note that the third equation in R-H conditions 3.9 implies that ξ 1 z 2 ξ 1z ξ 1 z 2 ξ 1 z 2 2 = [ρu2 2 + P ] τ ρu 1 U 2 ξ 1z 2 τ [ρu2 2 + P ], ξ 1 z 2 here τ denotes the tangent derivative of z 1 = ξ 1 z 2 similar expression holds for ξ 2 z 2. This, together with the first two equations in 3.9, yields for i =1, 3, N i 0,z 2 =Oε W 1,W 2,W 3, 1 W 4 +OεN 2 + O 1 N It follows from 4.11, that B 1 B 2 C 1,α Cε W 1,W 2,W 3,X 1 0 W 4 C 2,α + C N 4 C 1,α + Cε N 2 C 1,α Cε W 1,W 2,W 3,X0 1 W C 4 C 1,α + N 1,N 2,N 3 C 1,α + N 4 C 2,α By , one has { D1 W 1 = O 1 W 1,ε W 2,W 3,W 4 + Oε N3,X0 1 N 4,ε 1 M 3,B 1 B 2, D 2 W 1 = O 1 εw 1,W 2,ε W 3,W 4 + O1 N3,X0 1 N ,εM 3,B 1 B 2 { D1 W 2 = D 2 W 2 = Oε W 1, ε 1 W 2,W 3,W 4 + O1 N3,X0 1 4,εM 3,ε 2 B 1 B 2, O 1 W 1,ε W 2,W 3,W 4 + Oε N3,N 4,ε 1 M 3,B 1 B We can now estimate N 2. Since W 2 z 1, ±1 = 0, then there exists z 2 = z 2 z 1 such that N 2 z 1,z 2 z 1 = 0 holds. So, 4.15 implies that z1 N 2 = Oε W 1,W 2,W 3,W 4 + Oε N1, ε 1 N 2, ε 1 N 3,N 4,M 3 +O1 ε z1 N 3, z2 N 3,X0 1 z 2 N 4 + Oε 2 B 1 B 2, z2 B 1 B 2, z2 N 2 = Oε W 1,W 2,W 3,W 4 +OεM 3 + O 1 N 1,N 2,N 3,N 4 +Oε ε 1 z1 N 3, z2 N 3, z2 N 4, B 1 B 2, z2 B 1 B 2, N 2 z 1,z 2 z 1 = 0. 15

16 It should be emphasized here that instead of estimating W 2 in 4.15 directly, one differentiates 4.15 with respect to z 2 uses the structure of the background solution to derive the desired system for N 2 with order Oε coefficientsforw i i =1, 2, 3, 4 M 3. This will make it possible to obtain a control on N 2 C 1,α in terms of ε W 1,W 2,W 3,W 4 C 1,α, X0 1 N 1 C 1,α + N 4 C 2,α, N 3 C 1,α. Indeed, it follows from this system for N 2, , 4.13, a direct computation that N 2 C 1,α C z N 2 C α Cε W 1,W 2,W 3,W 4 C 1,α + C N 1 C 1,α + N 4 C 2,α+C N 3 C 1,α Next, note that 4.14 shows that z1 N 1 = Oε W 1,W 2,W 3,W 4 + O 1 εx 0 M 3,N 1,ε N 2,N 3,N 4 +O1 z1 N 3 + Oε z2 N 3,X 1 0 z2 N 4,B 1 B 2, z2 B 1 B 2, z2 N 1 = Oε W 1,W 2,W 3,W 4 + O 1 εx 0 M 3,N 1,N 2,ε N 3,N 4 +Oε z1 N 3 + O1 z2 N 3 + O 1 z2 N 4 + OεB 1 B 2 +O1 z2 B 1 B 2, N 1 z 1, ±1 = 0, where N 1 0, ±1 = 0 follows from the compatibility condition derived in [22]. Then one can estimate N 1 as above to obtain N 1 C 1,α Cε W 1,W 2,W 3,W 4 C 1,α + C N 2 C 1,α + N 4 C 2,α+C N 3 C 1,α. Combining this with 4.16 shows Lemma 4.4. Finally, we estimate N 3. Lemma 4.5. Estimate of N 3 N 3 satisfies N 3 C 1,α Cε W 1,W 2,W 3,W 4 C 1,α + C N1,N 2 C 1,α + N 4 C 2,α + Cε 1 P 01 z 2 P 02 z 2 C 1,α [ 1,1] Proof. Due to 3.25, one has by a direct computation that D 1 U 2 1 c 2 ρ D 1 W 3 + U 1 U 2 D2 W 3 + D2 U1 U 2 D1 W 3 +U2 2 c 2 ρ D 2 W 3 = O 1 W X0 2 1,ε W 2,W 3,W 4 + O 1 M 1,ε M 2,M 3 + OεN1 + O 1 N 2 +OεN 3 + O ε N 4 + O 1 z2 N 4, W 3 1,z 2 =P 1 P e + εp 01 z2 P 1 P e + εp 02 z2, N 3 z 1, ±1 = 0. It follows from this 4.12 that D 1 U 2 1 c 2 ρ D 1 N 3 + U 1 U 2 D2 N 3 + D2 U1 U 2 D1 N 3 +u 2 2 c 2 ρ D 2 N 3 = z1 Oε M3,N 3 + z2 Oε N1, ε 1 N 2,N 3,X0 1 N 4+O 1 z2 N 4 +Oε W 1,W 2,W 3,W 4 + Oε M1,M 2,εM 3 + OX 2 0 N 1,ε N 2,N 3,N 4 +O 1 z1 N 1 + Oε z1 N 2 + O 1 z1 N 3, N 3 0,z 2 =Oε W 1,W 2,W 3,X0 1 W 4 + OεN2 + O 1 N 4, N 3 1,z 2 =OεW 3 1,z 2 +O ε P 01 z2 P 02 z2, N 3 z 1, ±1 =

17 As in [22], it can be verified that the compatible condition holds at the corner points 0, ±1 1, ±1. Furthermore, these compatible conditions guarantee the C 1,α regularity of solution. So by the regularity estimates of second order elliptic equations with divergence forms in [1-2], one can arrive at from that N 3 C 1,α Cε W 1,W 2,W 3,W 4 C 1,α + C N1,N 2 C 1,α + N 4 C 2,α + Cε 1 P 01 z 2 P 02 z 2 C 1,α [ 1,1], which shows Lemma 4.5. Finally, we point out that all the estimates above can be improved. Remark 4.1. Let ρ, U 1,U 2 ; ξ 1 z 2 = ˆρ + 0 r 0 + z 1 +1 r 0, Û + 0 r 0 + z 1 +1 r 0,O; r 0 be the background solution q, V 1,V 2 ; ξ 2 z 2 be any solution to the problem as before. Then the corresponding estimates in Lemma 4.2 Lemma 4.3 Lemma 4.5 can be improved C 3,α C 2,α respectively. This fact is used to get the high regularity estimates in Theorem 1.2. Indeed, by Proposition 2.1, in this case, ρ, U 1,U 2 ; ξ 1 z 2 is C 4,α -smooth. It follows that the coefficients of ls; z in 4.8 are C 3,α -smooth. Hence, one may obtain the desired higher regularity estimates just following the proofs of Lemma 4.2 Lemma 4.5. Based on Lemma 4.1 Lemma 4.5, the uniqueness result of Theorem 3.1 thus Theorem 1.1 will be proved in next section. 5. Proof of the Theorem 1.1. Due to the equivalence between Theorem 1.1 Theorem 3.1, it suffices to prove Theorem 3.1 only. In 4, we have established a priori estimates for the gradients of the solution instead of the solution itself. If trying to derive a priori estimates on the solution itself, one then can obtain from that M 3 C 1,α C 1 N 2 C 1,α + some positive terms with small coefficients N 2 C 1,α C 2 M 3 C 1,α + some positive terms with small coefficients, with C 1 C 2 being some order one positive constants. Thus, it seems difficult to get any useful information on M 3 N 2. To overcome this difficulty, we derive the gradient estimates on the solution instead of the solution itself. Furthermore, we also estimate N 3 instead of M 3 from the corresponding second order elliptic equation 4.18 to avoid the difficulties caused by the constant P e in the variable end pressure. Combining these estimates with properties of the background solution, we can derive the monotonic continuous dependence between the shock position the exit pressure along the nozzle wall, which will be crucial in proving Theorem 1.1 Theorem 1.2. Assume that there exist two solutions ρ, U 1,U 2 ; ξ 1 q, V 1,V 2 ; ξ 2 to the problem First, we intend to show ξ 1 1 = ξ 2 1 holds by contradiction. Otherwise, without loss of generality, one may assume ξ 1 1 <ξ Under this assumption, it will be shown that the corresponding end pressures are different, which is contradictory with 1.4. Indeed, we have first Lemma 5.1. Under the assumption 5.1 M0 U 0 cρ 0 > 2 γ+1 2, it holds that γ ρξ 1 1, 1 >qξ 2 1,

18 Proof. Note that the background supersonic solution ρ 0,U0 satisfies dρ 0 = ρ 0 M 0 2 d 1 M0 2, du0 U = 0 d 1 M0 2, dm0 = M γ 1 M0 2 2 d 1 M0, here M0 = U 0 cρ 0 y denotes the Mach number of the supersonic incoming flow. 1 This yields that for large y 1, + 1, ρ 0,U 0,M 0 =ρ 0,U 0,M 0 +O 1. In addition, dρ 0 U P0 = ρ 0 U 0 2 d < 0, dρ 0 U 0 = ρ 0 U 0 d < 0. Next, we analyze the relation between the density ρ, 1 the shock position, 1. Since U 2 =0fory 2 = ±1, then the Rankine-Hugoniot conditions 3.9 imply that 5.4 This yields for the polytropic gas, [ρu 1 ], 1 = 0, [ρu P ], 1 = 0. Aρ, 1 γ+1 B ρ, 1 + C =0, 5.5 with B =P0 +ρ 0 U0 2 C =ρ 0 U0 2. It follows from 5.5 that dρ, 1 = ρ, 1B C d ρ, 1c 2 ρ, 1 U1 2. In addition, 5.4 shows that ρ, 1B C = ρ 0 U 0 2 2ρ 0 ρ, 1. Next, elementary calculations show that 2ρ 0 <ρ, 1 for, + 1. Indeed, set fx =Ax γ+1 B x + C, 5.6 then by the expressions of B C, it holds that fρ 0 = 0, f ρ 0 < 0, f x > 0forx>0, f+ =+, so there exists a unique point ρ, 1 >ρ 0 such that fρ, 1 = 0, 18

19 namely, 5.5 holds. On the other h, noting that M 0 >M 0 > 1 due to 5.3, one has for large f2ρ 0 = ρ 0 P 0 2γ+1 2 γ where one has used the assumption M0 > 2 γ+1 2 γ. Thus, one concludes that ρ, 1 > 2ρ 0. This implies that dρy1,1 d < 0, consequently ρξ 1 1, 1 >qξ 2 1, 1. M 0 2 < 0, Remark 5.1. It follows from the assumption 5.1 Lemma 5.1 that W 3 0, 1 > 0 holds in 4. This property will play an important role in proving Theorem 3.1. Next, we establish some estimates which will be used to derive the monotonic property of the shock position on the end pressure. Lemma 5.2. For ε 0 1 in Proposition 3.2, the following estimates hold X0 3 W 1,W 3,X 1 0 W 4,M 1,M 3 C 1,α C W 3 0, 1 + Cε 1 P 01 z2 P 02 z2 C 1,α [ 1,1], W 2,M 2 C 1,α 4 i=1 C W 3 0, 1 + Cε 1 P 01 z2 P 02 z2 C 1,α [ 1,1], N i C 1,α C W 3 0, 1 + Cε 1 P 01 z 2 P 02 z 2 C 1,α [ 1,1]. 5.8 Proof. As in the derivation of 5.5, one may obtain that Same expression holds for q, V 1,ξ 2. Then, Aρξ 1 1, 1 γ+1 Bξ 1 1ρξ 1 1, 1 + Cξ 1 1 = Aρξ 1 1, 1 γ+1 ρξ 1 1, 1Aqξ 2 1, 1 γ qξ 2 1, 1V 2 1 ξ 2 1, 1ρξ 1 1, 1 qξ 2 1, 1 = ρξ 1 1, 1Bξ 1 1 Bξ 2 1 Cξ 1 1 Cξ 2 1. This, together with 5.4 the definitions of B C, yields for some ξ between ξ 1 1 ξ 2 1, a 0 W 3 0, 1 = ρ 0 ξu 0 2 ξ 2ρ ξ 0 ξ ρξ 1 1, 1W 4 1, 5.10 a 0 = ρξ 11, 1P ρξ 1 1, 1 P qξ 2 1, 1 ρξ 1 1, 1 qξ 2 1, 1 qξ 21, 1P ρξ 1 1, 1 P qξ 2 1, 1 ρξ 1 1, 1 qξ 2 1, 1 > qξ 2 1, 1V 2 1 ξ 2 1, 1 qξ 2 1, 1V 2 1 ξ 2 1, 1

20 Then under the assumptions of lemma 5.1, it follows from 5.10 the R-H conditions for ρ, U 1 q, V 1 respectively that W 1 0, 1 C W 3 0, 1, W 4 1 C W 3 0, 1, W 3 0, 1 C W Since W 1 C 1,α W 1 0, 1 + M 1 C 1,α + N 1 C 1,α W 2 C 1,α M 2 C 1,α + N 2 C 1,α, W 3 C 1,α W 3 0, 1 + M 3 C 1,α + N 3 C 1,α, W 4 C 1,α W N 4 C 1,α, then one has by 5.11 that W 1 C 1,α C W 3 0, 1 + M 1 C 1,α + N 1 C 1,α, W 2 C 1,α M 2 C 1,α + N 2 C 1,α, W 3 C 1,α W 3 0, 1 + M 3 C 1,α + N 3 C 1,α, W 4 C 1,α C W 3 0, 1 + N 4 C 1,α On the other h, it follows from lemma 4.2-lemma 4.5 that N 1,N 2,N 3 C 1,α + N 4 C 2,α Cε W 1,W 2,W 3,W 4 C 1,α + M 1,M 2,M 3 C 1,α + Cε P X 01 z 2 P 0 X 02 z 2 C 0 X 1,α [ 1,1], 0 M 1,M 3 C 1,α C W 1,ε W 2,W 3,W 4 C 1,α + Cε M 2 C 1,α + C N1,ε N 2,N 3 C 1,α + N 4 C 2,α, M 2 C 1,α Cε W 1,W 3,W 4 C 1,α + C W 2 C 1,α + Cε M 1,M 3,N 1,N 2 C 1,α + N 4 C 2,α+ C N 2 C 1,α. This implies that N 1,N 2,N 3 C 1,α + N 4 C 2,α Cε W 1,W 2,W 3,W 4 C 1,α + Cε 1 P 1 z 2 P 0 X 02 z 2 C 1,α [ 1,1], 0 M 1,M 3 C 1,α C W 1,W 3,W 4 C 1,α + Cε W 2 C 1,α + Cε 1 P 1 z 2 P 0 X 02 z 2 C 1,α [ 1,1], 0 M 2 C 1,α Cε W 1,W 3,W 4 C 1,α + C W 2 C 1,α + Cε 1 P 1 z 2 P 0 X 02 z 2 C 0 X 1,α [ 1,1]. 0 Consequently, combining this with 5.12 yields 5.8, which completes the proof of Lemma 5.2. Now we are ready to prove Theorem 3.1. Proof of Theorem 3.1. Under the assumption of P 01 z2 =P 02 z2, it suffices to prove W 1 = W 2 = W 3 = W 4 =0. 20

21 It can be verified directly that 4.6 may be rewritten as U 1 D1 W 3 + ρ D 1 W 1 = a 1 W 4 + O 1 W 1,ε W 2,W 3 +Oε M 2,εM 3 +O1N 2 + Oε N 3,N 4, c 2 ρ D 1 W 3 + ρu 1 D1 W 1 = a 2 W 4 + O 1 W 1,ε W 2,W 3 +Oε 2 M 1 +OεN 1 + O ε N 4, where a 1 = 1 +1 ξ 1 z 2 +1 ξ 2 z 2 z 1 ρu z 1 ρu1 2 + ξ 1 z 2 +z 1 +1 ξ 1 z 2 ξ 2 z 2 +z 1 +1 ξ 2 z 2 + O ε, 1 a 2 = +1 ξ 1 z 2 +1 ξ 2 z 2 c2 ρ z1 ρ + ρu 1 z1 U 1 +O ε. Then, it follows from that for every z 2 [ 1, 1], where z1 W 3 =azw 4 + O 1 W 1,ε W 2,W 3 +Oε εm 1,M 2,εM 3 +O1N 2 + Oε N 1,N 3,N 4, 5.14 az = +1 ξ 2 1 c 2 ρ U1 2 a 2 U 1 a 1 z1 ρ = +1 ξ 1 z 2 +1 ξ 2 11 z 1 ρu 3 1 c 2 ρ U 2 1 ξ 1z 2 +z 1 +1 ξ 1 z 2 ξ 2 z 2 +z 1 +1 ξ 2 z 2 + O ε. Under the assumptions of Theorem 3.1, we have z1 ρ>0, z1 ρ = O 1, U 1 > 0, U 1 = O1, c 2 ρ U 2 1 > 0, c 2 ρ U 2 1 = O1. Hence, az is a negative function in the subsonic domain. Then it follows from Remark 5.1, Lemma that for every z 2 [ 1, 1] z1 W 3 = azw 4 z 2 +bzw 3 0, with bz L O 1. In addition, W 4 1 < 0 due to assumption 5.1. This means that the term azw 4 1 is always nonnegative. Therefore, along the line z 2 = 1, 5.15 yields { z1 W 3 bz 1, 1W 3 0, 1, W 3 0, 1 > 0. Thus, for suitably large W 3 z 1, 1 >C 1 W 3 0, 1 > 0 21

22 for some constant C 1 > 0. However, this contradicts to W 3 1, 1 = 0 due to the end pressure condition 1.4. This implies that the assumption 5.1 is not right. Thus, ξ 1 1 = ξ 2 1 holds, namely, W 4 0, 1 = 0. As a consequence of this 5.11, W 3 0, 1 = 0. This, together with Lemma 5.2, yields W 1 = W 2 = W 3 = W 4 0, which completes the proof of Theorem 3.1. Proof of Proposition in Proposition 3.2 follow immediately from Lemma 5.2. So it suffices to show By P 01 z2 =P 02 z2 +C 1,weget P 01 z2 P 02 z2 C 1,α [ 1,1] = 0 in 5.8. Thus, it follows from the third inequality in that there exits a generic constant C>1such that 1 C W 30, 1 W 3 0,z 2 C W 3 0, 1. First, we claim that ξ 1 1 < ξ 2 1 holds. Otherwise, it follows from the proof of Lemma 5.1 that W 4 1 0W 3 0, 1 0. This, together with 5.15, shows that { z1 W 3 bz 1, 1W 3 0, 1 on z 2 =1, 5.16 W 3 0, 1 0 with bz 1, 1 L = O 1. Hence W 3 1, 1 0 for suitably large, which contradicts to P 01 z2 >P 02 z2. Similarly, one can obtain ξ 1 1 <ξ 2 1. Next, we show that ξ 1 z 2 <ξ 2 z 2 forz 2 [ 1, 1]. Note that W 4 z 2 =W N 4 z 2 z 2 1 for some z 2 [z 2, 1]. By , N 4 L CX0 1 W 30, 1 CX0 2 W 41. Hence W 4 z 2 < 0 holds for all z 2 [ 1, 1] for suitably large since W 4 1 < 0. So we complete the proof of Proposition 3.2. In the end of this section, based on Lemma 5.1 Remark 2.1-Remark 2.2 in 2, one can estimate the differences of two shock positions the related subsonic flows in the domain {r, θ : r +1, θ 0 θ θ 0 } corresponding to two different background transonic shock solutions. Proposition 5.3. Let ˆρ + 0,i, Û + 0,i r, i =1, 2, r [, +1] be two subsonic flows with corresponding shock location r 0,i constant exit pressures P i,e i =1, 2 respectively as described in Remark 2.2. Then for large, it holds that { + ˆP 0,2 r, Û 0,2 + r ˆP 0,1 + r, Û 0,1 + r C 4,α [,+1] C P 2,e P 1,e, 5.17 r 0,2 r 0,1 C P 2,e P 1,e. Remark 5.2. It follows from Proposition 5.3 that if the difference of two end pressures is of order Oε, then the differences of related shock positions extended subsonic flows will be of order Oε Oε respectively. In addition, it also implies that the assumptions in Theorem 1.1 are plausible although the actual shock position further the related background transonic flow are not known in advance for such an end pressure condition P e + Oε. Proof. Without loss of generality, we assume that <r 0,2 <r 0,1 < +1. Then r 0,1 < r 0,2 < r 0,1 +1 r 0,2 r 0,2 Denoted by L the interval [, 1]. As in 4, we set +1 r 0,2 W 1 z 1 =Û 0,2 + r 0,2 + z 1 +1 r 0,2 Û 0,1 + r,1 + z 1 +1 r 0,1, z 1 L, W 3 z 1 =ˆρ + 0,2 r 0,2 + z 1 +1 r 0,2 ˆρ + 0,1 r 0,1 + z 1 +1 r 0,1, z 1 L, W 4 = r 0,2 r 0,1. 22

23 Since ˆρ + 0,i r, Û + 0,i ri =1, 2 satisfy dˆρ + 0,i r ˆρ + 0,i = rû + 0,i r2 dr rû + 0,i r2 c 2 ˆρ + 0,i r, r [, +1] dû + 0,i r Û + 0,i = rc2 ˆρ + 0,i r dr rû + 0,i r2 c 2 ˆρ + 0,i r, r [, +1], then it follows from Remark 2.2 a direct computation that dw 1 dz 1 dw 3 dz 1 = OX 1 0 W 1,W 3 +b 1 z 1 W 4, z 1 L, = OX 1 0 W 1,W 3 +b 3 z 1 W 4, z 1 L, 5.18 where +1 Û 0,1 b + 1 z 1 = c2 ˆρ + 0,1 r 0,2 + z 1 +1 r 0,2 r 0,1 + z 1 +1 r 0,1 Û 0,1 + 2 c 2 ˆρ + 0,1 +1 ˆρ + 0,1 b 3 z 1 = Û 0,1 + 2 r 0,2 + z 1 +1 r 0,2 r 0,1 + z 1 +1 r 0,1 Û 0,1 + 2 c 2 ˆρ + 0,1. Obviously, b 3 z 1 < 0b i z 1 =OX0 1 i =1, 3 for large. As a consequence of W 4 < 0, Lemma , one has W 3 0 > 0, W 1 0 C W 3 0, W 4 C W 3 0, W 3 0 C W Thus, it follows from the positivity of b 3 z 1 W 4 that W 3 z 1 CW 3 0 X 1 0 W 1 L [0,1], z 1 [0, 1], W 1 L [0,1] CX 1 0 W 3 L [0,1] + W 3 0, W 3 L [0,1] CX 1 0 W 1 L [0,1] + W 3 0. This yields that { W3 z 1 CW 3 0 X0 2 W 3 L, z 1 [0, 1], W 3 L [0,1] CW 3 0. Therefore, W 3 z 1 CW 3 0 for z 1 0 further W 3 0 CW 3 1 hold true. On the other h, by 5.18, one has { W1 C 3,α L C W X0 1 W 4 + X0 1 W 3 C 2,α L, W 3 C 3,α L C W X0 1 W 4 + X0 1 W 1 C 2,α L Combining 5.20 with <W 3 0 CW 3 1 yields This, together with W 4 C W 3 0 in 5.19, shows W 1,W 3 C 3,α L C W W 4 C W 3 1, 23

24 which is the second inequality in Next, we prove the first estimate in In fact, Remark 2.1 in 2 implies that Thus, ˆP + 0,1, Û + 0,1 C 3,α [,+1] CX 1 0. ˆP 0,2 + r ˆP 0,1 + r C 3,α [,+1] ˆP 0,2 + r r 0,2 r0, r 0,2 +1 r ˆP + r r 0,2 0,1 r0, r 0,1 C 0,2 +1 r 3,α [,+1] 0,2 + ˆP 0,1 + r r r 0,2 0, r 0,1 +1 r ˆP 0,1 + r r r 0,1 0, r 0,1 C 3,α [X 0,2 +1 r 0,+1] 0,1 C W 3 C 3,α L + ˆP 0,1 + C 3,α [,+1] W 4 C W 3 1. Analogously, Û 0,2 + Û 0,1 + can be estimated. Thus the proof of Proposition 5.3 is completed. 6. Proof of Theorem In this section, based on Theorem 1.1 the related estimates given in 4-5, we will show the existence result in Theorem 1.2. First, note that if the transonic shock is required to go through some fixed point on the wall, then as in [26], one can prove that the problem 1.1 with has a unique transonic shock solution when the end pressure P e + εp 0 θ in 1.4 is adjusted by an appropriate constant. It follows from this that if one can show that there exists a point at the wall such that the shock goes through this point the corresponding adjustment constant on the end pressure is zero, then Theorem 1.2 will be proved. Next we state an existence result when the shock is assumed to go through a fixed point on the wall, whose proof will be given in Appendix. Consider the 2-D nozzle the supersonic incoming flow as given in 1. Let x 0 1,x 0 1tanθ 0 be a given point on the wall of the nozzle with r 0 = x1 0 1+tan2 θ 0, + 1. Denote by P e P 1,P 2 the constant exit pressure when the shock position is given by r = r 0 with P 1 P 2 being given in Proposition 2.1 of 2. Then one has Theorem 6.1. Under the assumptions as in Theorem 1.1, there exists a constant C 0 such that the transonic shock problem 1.1 with 1.2, has a solution with the following properties ρ +,u + 1,u+ 2 ; ηx 2 C 3,α, 6.1 ηx 0 1tanθ 0 =x 0 1, 6.2 P + = P e + εp 0 θ+c 0 on r = Moreover, the solution satisfies the analogous estimates in Theorem 1.1. In particular, ηx 2 r 0 Cε holds true. In terms of the coordinates,y 2 in 3.7, Theorem 6.1 can be restated equivalently as follows Theorem 6.1. Under the assumptions in Theorem 6.1, there exists an appropriate constant C 0 such that the free boundary value problem , 3.9, 3.14 has a C 3,α solution ρ, U 1,U 2 ; ξ satisfying ξ1 = r 0, 6.4 P ρ =P e + εp 0 y 2 +C 0 on =

25 Moreover, the solution admits the same estimates as in Theorem 3.1. In particular, ηy 2 r 0 Cε. It follows from Proposition 3.2 that the adjustment constant C 0 in Theorem 6.1 Theorem 6.1 depends continuously on the position where the shock intersects with the wall of the nozzle. More precisely, one has Lemma 6.1. Continuity uniqueness i Assume that two variable exit pressures P 1 P 2 have the form 1.4 satisfy P 1 = P 2 + C 0 with a constant C 0. Let ρ, U 1,U 2 ; ξ 1 q, V 1,V 2 ; ξ 2 be solutions to the free boundary value problems , corresponding to the exit pressure P 1 P 2 respectively satisfy the corresponding estimates in Theorem 3.1. Then C 0 C ξ 1 1 ξ with a uniform constant C. ii. If the transonic shock goes through a fixed point on the wall of the nozzle, then the corresponding exit pressure is uniquely determined. Namely, if there exist two constants C 1 C 2 such that the end pressures of two solutions are P 1 = P e + εp 0 θ+c 1 P 2 = P e + εp 0 θ+c 2, then C 1 C 2. Remark 6.1. By Theorem 3.1, the solutions corresponding to the variable exit pressure P 1 P 2 are unique respectively. Based on Lemma 6.1, we now prove Theorem 1.2. Proof of Theorem 1.2. Denote by P 1 = P e ε P 2 = P e + ε the exit pressures of the symmetric transonic shock solutions with corresponding shocks at = r 1 = r 2 respectively. Then it follows from Remark 5.1 the uniqueness result in Theorem 1.1 that r 1 >r 2. For each fixed point, 1 with [r 2,r 1 ], it follows from Theorem 6.1 that there exists a constant C 0 such that problem 1.1 with , 1.5 the end pressure P = P e + εp 0 θ+c 0 has a unique solution ρ, U 1,U 2 ; ξy 2 which satisfies ξ1 = the estimates in Theorem 3.1. If = r 2, then this corresponding adjustment constant, C 0, must be positive. Indeed, if not, then C 0 0. Applying the estimate 3.27 to ρ, U 1,U 2 ; ξy 2 the background solution q, V 1,V 2 ; r 2 which corresponds to the constant end pressure P 2, noting that W 4 1 = 0, one has W 3 L Cε 1 P 0 z 2 C 1,α <Cε. 6.7 On the other h, W 3 1, 1 = P 2 P e + εp 0 1 C ε > Cε, which contradicts to 6.7 for large. Hence, C 0 > 0. Similarly, for = r 2, the corresponding adjustment constant, C 0,mustbe negative. It follows from Theorem 6.1 Lemma 6.1 that C 0 is a Lipschitz continuous function of, i.e. C 0 = C 0. We have shown that C 0 r 2 > 0C 0 r 1 < 0, thus there exists a y 0 1 r 2,r 1 such that C 0 y 0 1 = 0. Consequently, it follows from Theorem 6.1 that the problem has a transonic shock solution P y,u 1 y,u 2 y; ξy the transonic shock passes through y 0 1, 1. By Theorem 1.1 such a solution is unique. Thus Theorem 1.2 is proved. Appendix In this section, we will focus on the proof on Theorem 6.1. In [26], for almost parallel nozzle walls a special exit pressure boundary condition, when the shock is required to go through a fixed point, it is proved that the problem 1.1 with the related boundary conditions has a solution in some weighted Hölder space if the exit pressure is adjusted by an appropriate constant. It should be noted that the exit boundary in [26] is straight, this makes it possible to straighten out both the solid walls the exit of the nozzle simultaneously by a Langrange transformation. This ingredient plays an important role in the proof of the main result in [26]. However, in our case, the exit of the nozzle is curved, so it is related to the solution itself under a Langrange transformation. Thus, in order to overcome this difficulty also obtain higher 25