Arch. Rational Mech. Anal. 187 (2008) 523–556 Digital Object Identifier (DOI) 10.1007/s00205-007-0079-z
Transonic Shocks in Compressible Flow Passing a Duct for Three-Dimensional Euler Systems Shuxing Chen∗ & Hairong Yuan Communicated by T.-P. Liu
Abstract In this paper we study the transonic shock in steady compressible flow passing a duct. The flow is a given supersonic one at the entrance of the duct and becomes subsonic across a shock front, which passes through a given point on the wall of the duct. The flow is governed by the three-dimensional steady full Euler system, which is purely hyperbolic ahead of the shock and is of elliptic–hyperbolic composed type behind the shock. The upstream flow is a uniform supersonic one with the addition of a three-dimensional perturbation, while the pressure of the downstream flow at the exit of the duct is assigned apart from a constant difference. The problem of determining the transonic shock and the flow behind the shock is reduced to a free-boundary value problem. In order to solve the free-boundary problem of the elliptic–hyperbolic system one crucial point is to decompose the whole system to a canonical form, in which the elliptic part and the hyperbolic part are separated at the level of the principal part. Due to the complexity of the characteristic varieties for the three-dimensional Euler system the calculus of symbols is employed to complete the decomposition. The new ingredient of our analysis also contains the process of determining the shock front governed by a pair of partial differential equations, which are coupled with the three-dimensional Euler system.
1. Introduction In this paper we study stationary three-dimensional compressible flow passing a duct. The flow is a given supersonic one at the entrance of the duct and becomes subsonic across a transonic shock front. Physical phenomena involving transonic ∗ The paper is partially supported by National Natural Science Foundation of China 10531020, the National Basic Research Program of China 2006CB805902, and the Doctorial Foundation of National Educational Ministry 20050246001.
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shocks often occur in gas dynamics [4–6, 8, 10, 12, 23]. Particularly, compressible flow passing a duct is one of the basic problems related to transonic flow and has been studied by many authors. For instance, Liu and Glaz [11, 16] studied the existence and stability of a transonic shock in the quasi-one-dimensional case. Recently, Chen and Feldman [2, 3] and Xin and Yin [24] studied the stability of a two-dimensional transonic shock by using a model of the potential flow equation. Another case for compressible flow passing a duct is that the upstream flow at the entrance is subsonic, while the downstream flow at the exit is supersonic. In this case the flow will be a smooth transonic flow generally, and the equation prescribing the flow becomes a mixed-type equation [1, 15, 18–21]. Since the full Euler system is a more precise description of the inviscid flow in gas dynamics, it is natural and expected to study the above problem by using the full Euler system. In [9] we studied the existence and stability of a transonic shock for the two-dimensional Euler system, while in this paper we develop our analysis for the three-dimensional Euler system. In our study the duct is assumed to be a cylinder with a square section and the upstream flow is a three-dimensional perturbation of a uniform supersonic flow. The initial unperturbed upstream flow is parallel to the axis of the duct. It passes a plane shock and becomes a uniform subsonic flow. The purpose of this paper is to prove the existence and the stability of the perturbed compressible flow including the transonic shock under some conditions. According to the character of the motion in our problem, the Euler system is purely hyperbolic ahead of the shock and is of elliptic–hyperbolic composed type behind the shock. The problem of determining the flow behind the shock will be reduced to a free-boundary value problem of the Euler system. The unknown shock front is the free boundary, on which the Rankine–Hugoniot conditions give the boundary conditions. On the wall the flow satisfies the impenetrability condition, that is, the normal component of the velocity vanishes along the wall. At the exit of the duct the pressure of the downstream flow is assigned apart from a constant difference. In other words, the pressure is given with freedom one. This means that the pressure is given in a space with codimension one, but not completely given; otherwise, there may not be any solution to such a problem. The way to give the boundary condition at the exit is similar to the case for the equivalued boundary problems of elliptic equations. As a complement to this freedom we let the unknown shock front pass a given point in the duct; such a restriction is necessary when dealing with a duct with constant section. We remark here that to give the pressure at the exit of a duct is more acceptable in physics. However, as we will see, even in the uniform case the value of the pressure at the exit is determined by the data of the supersonic flow at the entrance of the duct. Therefore, the pressure at the exit can at most be assigned up to a constant difference. The way to assign such a boundary condition at the exit essentially stems from the requirement for conservation of mass in the duct. This point will be explained in more detail in Section 4. In order to solve the free-boundary problem of the elliptic–hyperbolic system in the subsonic region one crucial point is to decompose the whole system into a canonical form in which the elliptic part and the hyperbolic part are separated at the principal part of the system. In [9] we introduced such a method. However, due to the
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complexity of the characteristic varieties for the three-dimensional Euler system we have to employ the calculus of symbols (see [14]) to complete the decomposition. Another new ingredient of our analysis is the determination of the function describing the shock front. From the Rankine–Hugoniot conditions we can obtain the slopes of the shock front with respect to two variables, which form an overdetermined system of partial differential equations. The two equations in the overdetermined system are coupled with the Euler system in the domain where the flow is defined. Meanwhile, the condition of solvability of the pair of differential equations leads to a new condition. To deal with the whole problem involving all parameters of the flow and the location of the shock we reduce the problem to its equivalent form, which is composed of several coupled subproblems, and is easier to linearize and solve. The next task is to establish all necessary estimates and use an iteration to obtain the solution to the nonlinear problems. Besides the assumption on the geometry of the duct, we also assume that the boundary conditions and the upstream flow have some symmetry. Due to the specific symmetry of the boundary conditions and the data our problem can be reduced to a periodic form, which is defined on the domain P := (−1, 1) × T2 , where T2 is a two-dimensional torus. Such assumptions allow us to avoid some technical difficulties caused by the intersection of the transonic shock and the wall of the duct. A study of a transonic shock for compressible flow passing a more general duct in the three-dimensional case will be given in near future. The remainder of this paper is arranged as follows. In Section 2 we give a description and mathematical formulation of the physical problem, which can also take a periodic form by symmetric extension. Moreover, by using the known result for hyperbolic systems the main part of the problem is reduced to a free-boundary problem in the subsonic region with the shock front as the free boundary. In Section 3 we temporarily fix the shock front and then concentrate on solving the fixedboundary value problem. The main work here is to reduce the Euler system to a canonical form in which the elliptic and hyperbolic parts are separated at the level of the principal part. In Section 4 the fixed-boundary value problem is again transformed to a new form, which is composed of four subproblems, and the equivalence is proved. Afterwards, we linearize these subproblems and solve them successively in Section 5, and then solve the nonlinear fixed-boundary problem in Section 6. Finally, the nonlinear free-boundary problem and the original physical problem in accordance, are solved in Section 7. To avoid interrupting our demonstration, in the proof we move some routine calculations to the Appendix.
2. Mathematical formulation and the free-boundary problem First, let us recall some basic facts on compressible flow. The full Euler system for steady compressible flow in three-dimensional space is 2 j=0
∂x j (ρu j ) = 0,
(1)
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∂x j (ρu k u j + pδk j ) = 0, k = 0, 1, 2,
(2)
∂x j (Eu j ) = 0,
(3)
j=0 2 j=0
where δi j =
1 0
i = j, i = j,
u i is the velocity component of the compressible flow along the xi axis (i = 0, 1, 2), and p, ρ, and S are the pressure, density, and entropy, respectively. For the polytropic gas S − S0 ρ γ , 1 < γ 3, = (γ − 1) exp Cν p ∂ p(ρ, S) = γ , a is the sound speed, = ∂ρ ρ 1 p = , e is the internal energy, γ −1ρ 1 1 γ = ρe + p + ρ |u|2 = p + ρ |u|2 , 2 γ −1 2
p a2 e E where |u|2 =
2
2 i=0 u i .
(4) (5) (6) (7)
Obviously, ρ can be written as a function of p and S ρ = ρ( p, S).
(8)
For convenience, we also write p = u 3 , S = u 4 , and let U = (u 0 , u 1 , u 2 , u 3 , u 4 )t denote the unknown functions of (1)–(3) in the flow field. For C 1 solutions, the system (1)–(3) can be written in the following symmetric form: 2 A j (U (x))∂x j U (x) = 0, (9) j=0
where
⎛
ρu j ⎜ 0 ⎜ A j (U ) = ⎜ ⎜ 0 ⎝ δ j0 0
0 ρu j 0 δ j1 0
0 δ j0 0 δ j1 δ j2 ρu j δ j2 ρ −1 a −2 u j 0 0
⎞ 0 0 ⎟ ⎟ 0 ⎟ ⎟ , j = 0, 1, 2. 0 ⎠ ρu j
(10)
If U has a discontinuity on a surface Σ:
x0 = f (x1 , x2 ),
(11)
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then the Rankine–Hugoniot conditions across Σ hold: [ρu 0 ] = ∂x1 f [ρu 1 ] + ∂x2 f [ρu 2 ],
(12)
[ρu 0 u 1 ] = ∂x1 f + p] + ∂x2 f [ρu 1 u 2 ], [ρu 0 u 2 ] = ∂x1 f [ρu 1 u 2 ] + ∂x2 f [ρu 22 + p],
(13) (14)
[ρu 20 + p] = ∂x1 f [ρu 0 u 1 ] + ∂x2 f [ρu 0 u 2 ], [Eu 0 ] = ∂x1 f [Eu 1 ] + ∂x2 f [Eu 2 ],
(15) (16)
[ρu 21
where [·] denotes the jump of the corresponding quantity across Σ. Consider a compressible flow in a duct D, which has uniform square section [0, d; 0, d] and locates at −1 x0 1. Throughout the paper we assume that the flow satisfies the impenetrability condition on the wall of the duct D. That is, denoting the wall of D as W0∗ = {(x0 , 0, x2 ) : x2 ∈ [0, d], x0 ∈ [−1, 1]}, Wd∗ = {(x0 , d, x2 ) : x2 ∈ [0, d], x0 ∈ [−1, 1]}, W∗0 = {(x0 , x1 , 0) : x1 ∈ [0, d], x0 ∈ [−1, 1]}, W∗d = {(x0 , x1 , d) : x1 ∈ [0, d], x0 ∈ [−1, 1]}, then the condition on the lateral boundary is
u1 = 0 u2 = 0
on W0∗ , Wd∗ , on W∗0 , W∗d .
(17)
Suppose that we have a uniform supersonic flow at the entrance of the duct as − − U = Ub− := (u − 0 , 0, 0, p , S ),
(18)
then such a uniform flow can be kept up to the exit of the duct. However, if there is a plane shock front x0 = 0 in the duct, then the flow will be changed to a subsonic + + one. The parameters of the subsonic flow Ub+ := (u + 0 , 0, 0, p , S ) behind the shock can be uniquely determined by the Rankine–Hugoniot conditions (12)–(16) and the entropy condition p + > p − . Throughout the paper we will call the solution Ub =
Ub− , if x0 < 0, Ub+ , if x0 > 0,
(19)
with the shock Σb : x0 = 0 as the background solution. Our goal is to study the stability of this solution under three-dimensional perturbation. Remark. In (19), the location of the shock front can be shifted to x0 = c with −1 < c < 1. Therefore, we have to assume that the shock is always passing through a fixed point (for example, the origin), even when considering a perturbed flow. The assumption is necessary for a duct with uniform section (also see [2, 3, 9]), and may be removed for some ducts with variable sections [24].
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Consider the perturbed case. Assume that the supersonic flow Ub− is perturbed at the entrance of the duct; how does the perturbation influence the flow in the whole duct? It is found that the flow also passes a transonic shock inside the duct and then becomes a subsonic flow. To determine the whole flow one should give some boundary condition at the exit of the duct, because for subsonic flow the downstream part is not completely determined by the upstream part. The main purpose of this paper is to prove that the background solution (and all solutions near to it) are stable under three-dimensional perturbation. It is found that, if the pressure of the subsonic flow at the exit is given apart from a constant difference, then in the whole duct a solution of the Euler system with a shock passing through a fixed point can be determined. This means that, if the pressure of the subsonic flow at the exit is given as a perturbation of a constant, and such a perturbation, as well as the perturbation of the flow at the entrance, are sufficiently small, then the flow with a transonic shock in the whole duct is also slightly perturbed. Meanwhile, the proof shows the stability of the background solution. Moreover, if all perturbations of the data at the entrance and the exit are sufficiently small, then the perturbed solution is sufficiently near to the background solution. From the proof we can also confirm that all solutions sufficiently near to the background solution are also stable. Since the flow satisfies the impenetrability condition (17) on the wall, we avoid the technical difficulties caused by the intersection of the shock front and the wall of the duct by restricting ourselves to the consideration of certain symmetric cases, so that the specific symmetric property can replace the effect of the lateral boundary of the duct. To this end we introduce the following terminology. Definition 2.1. A solution U = (u 0 , u 1 , u 2 , p, S) defined in {−1 x0 1, −∞ x1,2 ∞} is called properly symmetric with respect to {x1 = x10 , x2 = x20 }, if u 0 , u 2 , p, S are symmetric with respect to x1 = x10 , u 1 is antisymmetric with respect to x1 = x10 , u 0 , u 1 , p, S are symmetric with respect to x2 = x20 , u 2 is antisymmetric with respect to x2 = x20 . Obviously, if U is defined in {−1 x0 1, 0 x1,2 d}, and satisfies ∂x1 (u 0 , u 2 , p, S) = 0, u 1 = 0, on x1 = 0, d, ∂x2 (u 0 , u 1 , p, S) = 0, u 2 = 0, on x1 = 0, d,
(20) (21)
then one can define U for −∞ x1,2 ∞ by odd extension for u 1 and even extension for u 0 , u 2 , p, S with respect to x1 = kd, and also by odd extension for u 2 and even extension for u 0 , u 1 , p, S with respect to x2 = kd. Such an extension lets U be properly symmetric with respect to {x1 = 0, x2 = 0},{x1 = 0, x2 = d},{x1 = d, x2 = 0} and {x1 = d, x2 = d}. On the other hand, if U is properly symmetric with respect to {x1 = 0, x2 = 0},{x1 = 0, x2 = d},{x1 = d, x2 = 0} and {x1 = d, x2 = d}, then U is a periodic function of x1 , x2 with period (2d, 2d).
Transonic Shocks in Compressible Flow Passing a Duct
The perturbed transonic shock problem in the duct takes the form ⎧ (1), (2), (3) in D, ⎪ ⎪ ⎪ ⎪ ⎨ Rankine–Hugoniot conditions (12) − (16) on Σ, (D) : U = U (−1) on x0 = −1, ⎪ + ⎪ ⎪ ⎪ p = pb + g(x1 , x2 ) + const on x0 = 1, ⎩ (17).
529
(22)
Here U (−1) is a given datum prescribing a supersonic flow near to Ub− , g(x1 , x2 ) is a given function defined on [0, d] × [0, d] and const is an unknown constant, which will be determined together with the solution U . The main result of this paper can be written as follows: Theorem 1. Assume that U (−1) satisfies k ∂x1 u 1 (−1) = 0 on x0 = −1, x1 = 0, d, x2 ∈ [0, d], ∂xk2 u 2 (−1) = 0 on x0 = −1, x2 = 0, d, x1 ∈ [0, d],
for k = 0, 2, 4, 6.
(23) on x0 = −1, x1 = 0, d, x2 ∈ [0, d], on x0 = −1, x2 = 0, d, x1 ∈ [0, d], for k = 1, 3, 5,
∂xk1 (u 0 , u 2 , p, S)(−1) = 0 ∂xk2 (u 0 , u 1 , p, S)(−1) = 0
(24) and that g satisfies k ∂x1 g = 0 on x0 = 1, x1 = 0, d, x2 ∈ [0, d], ∂xk2 g = 0 on x0 = 1, x2 = 0, d, x1 ∈ [0, d],
for k = 1, 3,
then there exists a positive constant ε0 depending only on Ub , such that if U (−1) − U − 7 b H (D ,R5 ) ε ε0 , −1
gC 3,α ([0,d;0,d]) ε ε0 ,
for some α ∈ (0, 1),
(25)
(26) (27)
then the problem (D) has a unique entropy weak solution (U− , U+ ; Σ), where U− is supersonic, U+ is subsonic and Σ : x0 = f (x1 , x2 ), (x1 , x2 ) ∈ [0, d; 0, d],
f (0, 0) = 0
(28)
is the shock front separating U± . Furthermore, the number const in (D) is determined and the following estimates hold: U− − U − 3,α (29) b C (D− ,R5 ) C 0 ε, U+ − U + C0 ε, (30) b 3 f C 4,α ([0,d;0,d]) C0 ε, (31) |const| C0 ε,
(32)
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where the constant C0 depends only on background solution and α. The norm ·k (k = 2, 3) is defined by U k =
2 u j
2 u j
4 u j
j=0
j=0
j=3
+ C k,α (Σ)
+ C k−1,α (D+ )
C k,α (D+ )
,
(33)
with D+ (D− resp.) = {(x0 , x1 , x2 ) ∈ D : x0 f (x1 , x2 ) ( f (x1 , x2 ) resp.)}, and D−1 = {−1} × [0, d] × [0, d]. The problem (D) can be reduced to a new form, which has periodic property with respect to x1 , x2 , because the data can be extended to properly symmetric data in the sense of Definition 2.1. Let T2 be the torus (−d, d) × (−d, d), and P := (−1, 1) × T2 , Ps := {s} × T2 for − 1 s 1, then the data U (−1), g(x1 , x2 ) can be well defined on T2 because of their proper symmetry. Hence we can discuss the problem ⎧ (1), (2), (3) in P, ⎪ ⎪ ⎨ Rankine − Hugoniot conditions (12) − (16) on Σ, (34) (B) : U = U (−1) on x0 = −1, ⎪ ⎪ ⎩ p = pb+ + g(x1 , x2 ) + const on x0 = 1, in P instead of the problem (D) in D. Correspondingly, denote P+ (P− resp.) = {(x0 , x1 , x2 ) ∈ P : x0 f (x1 , x2 ) ( f (x1 , x2 ) resp.)}. Theorem 1 can be derived from the following theorem. Theorem 2. There exists a positive constant ε0 depending only on Ub such that if U (−1) − U − 7 (35) b H (P ,R5 ) ε ε0 , −1
gC 3,α (P1 ) ε ε0 , α ∈ (0, 1),
(36)
then the problem (B) has a unique entropy weak solution (U− , U+ ; Σ), where U− is supersonic, U+ is subsonic and Σ : x0 = f (x1 , x2 ), (x1 , x2 ) ∈ T2 ,
f (0, 0) = 0
(37)
is the shock front separating U± . Furthermore, the number const in (B) can be determined and the following estimates hold: U− − U − 3,α (38) b C (P− ,R5 ) C 0 ε, U+ − U + C0 ε, (39) b 3 f C 4,α (T2 ) C0 ε, (40) |const| C0 ε,
(41)
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where the constant C0 depends only on the background solution and α. The norm ·k is defined as 33 with D+ replaced by P+ Theorem 1 is a direct conclusion of Theorem 2, and we will explain it in more detail in Section 7. Next we shall endeavour to prove Theorem 2. To simplify computations we will take d = 1 without loss of generality. Notice that the system (9) [the equivalent form of (1)–(3) for C 1 flow] is a nonlinear symmetric hyperbolic system. The well-posedness of its Cauchy problem is known (see, for example, [7, 17]). The results there ensure the local existence of a C 1 supersonic flow in P, which is determined completely by the initial data U (−1) and independent of any condition on downstream. Moreover, if the initial data is a small perturbation of a constant state, then the existence can be semiglobal. It means that the lifespan of the smooth solution depends on the smallness of the perturbation of initial data. In other words, the lifespan can be larger than any given number provided the perturbation is small enough. The result can be precisely written as follows: Lemma 3. There exists a positive constant ε0 such that, for any given initial data U (−1) satisfying U (−1) − U − 7 (42) b H (P ,R5 ) ε ε0 , −1
the Cauchy problem
System (9) in P, U = U (−1) on x0 = −1
has a unique solution U− that satisfies U− − U − 3,α b C (P,R5 ) C 0 ε,
(43)
(44)
where the constants ε0 , C0 only depend on Ub− . In the sequel we use Nε and Oδ to denote the neighborhood of Ub− and Ub+ : Nε := {U− : U− − Ub− C 3,α (P,R5 ) ε}, (45) + Oδ := U : U − Ub 3 δ . (46) Then Lemma 3 means that, if the data U (−1) satisfies (42), then U− ∈ NC0 ε . According to Lemma 3 the solution U− ahead of the shock front Σ in Theorem 2 can be determined, and is independent of the location of the shock and the flow behind the shock. Therefore, to prove Theorem 2 we need only to solve a freeboundary problem defined in the domain Ω f : {(x0 , x1 , x2 ) ∈ P : f (x1 , x2 ) < x0 < 1}, which lies between the shock and the exit of the duct. Next we would like to write the free-boundary problem in a more explicit way, which will also make it easier to treat later. The Rankine–Hugoniot conditions (12)–(16) contain the derivatives of the unknown function f . This should be eliminated when we impose the boundary conditions on x = f (x1 , x2 ) in order to solve the unknown function U from the Euler
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system. Treating (13), (14) as a linear algebraic system of ∂x1 f and ∂x2 f we can solve these partial derivatives as: ∂xi f = −Φi /Φ0 , i = 1, 2,
(47)
Φ0 = Φ0 (U, U− ) := −[ρu 21 + p][ρu 22 + p] + [ρu 1 u 2 ]2 = 0, Φ1 = Φ1 (U, U− ) := [ρu 0 u 1 ][ρu 22 + p] − [ρu 0 u 2 ][ρu 1 u 2 ],
(48) (49)
Φ2 = Φ2 (U, U− ) := [ρu 0 u 2 ][ρu 21 + p] − [ρu 0 u 1 ][ρu 1 u 2 ],
(50)
where
provided that U− ∈ NC0 ε , U ∈ Oδ and ε, δ are sufficiently small. Substituting (47) into (12), (15) and (16) we obtain G 1 (U, U− ) :=
2 [ρu j ]Φ j = 0,
(51)
j=0
G 2 (U, U− ) :=
2 [ρu 0 u j + pδ j0 ]Φ j = 0,
(52)
j=0
G 3 (U, U− ) :=
2 [Eu j ]Φ j = 0.
(53)
j=0
Then the free-boundary value problem (FB) is formulated as follows. ⎧ (9), in P+ , ⎪ ⎪ ⎨ p = pb+ + g(x1 , x2 ) + const on x0 = 1, (FB) : G (U, U− ) = 0, on Σ, i = 1, 2, 3, ⎪ ⎪ ⎩ i f satisfies (47), f (0, 0) = 0.
(54)
By using Lemma 3 it is obvious that Theorem 2 is the conclusion of the following theorem. Theorem 4. Under the assumptions of Theorem 1 the problem (FB) has a unique solution (U+ ; Σ), where the shock Σ : x0 = f (x1 , x2 ),
f (0, 0) = 0
(55)
is a part of the boundary of the domain for the unknown functions U+ . Furthermore, the number const satisfies (56) |const| C0 ε, and the following estimates hold: U+ − U + C0 ε, b 3 f C 4,α (T2 ) C0 ε, with C0 depending only on Ub± and α.
(57) (58)
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3. Fixed-boundary value problem 3.1. Reduction to a fixed-boundary value problem The problem (FB) is a free-boundary value problem. To solve it we will reduce the free-boundary value problem to a fixed-boundary value problem determining the downstream flow field and a problem modifying the shape of the shock front, which is a part of the boundary of the domain. Such an outline has been employed in the papers [5, 6, 9], where the authors also studied free-boundary value problems involving shock fronts. Set Sη = { f (x1 , x2 ) ∈ C 3,α (T2 ) : f (0, 0) = 0, f C 3,α (T2 ) η},
(59)
η < η0 < 1/2,
(60)
with where η0 is a constant depending only on the background solution and will be chosen later. For any f ∈ Sη , denoting Ω f = {(x0 , x1 , x2 ) ∈ P : f (x1 , x2 ) < x0 < 1}, Σ f = {(x0 , x1 , x2 ) ∈ P : f (x1 , x2 ) = x0 }, we can define a fixed-boundary problem (C f ) in Ω f as a part of the problem (FB). The unknown function U has to satisfy all conditions listed in (FB). Moreover, in view of (37), Φi (i = 0, 1, 2) should also satisfy the following additional condition G 4 (U, U− ) := −D1 (Φ2 /Φ0 ) + D2 (Φ1 /Φ0 ) = 0,
(61)
Dk := ∂xk + (∂xk f )∂x0 , k = 1, 2,
(62)
where Since f is well defined on T2 (equivalently, is periodic with period 2d), then 1 1 Φ1 ( f (s, x2 ), s, x2 ) Φ2 ( f (x1 , s), x1 , s) ds = 0, ds = 0. (63) Φ ( f (s, x ), s, x ) 0 2 2 −1 −1 Φ0 ( f (x 1 , s), x 1 , s) Due to the periodicity of Φi (i = 0, 1, 2) and Green’s theorem, it is enough to have 1 1 Φ1 ( f (s, 0), s, 0) Φ2 ( f (0, s), 0, s) ds = 0, ds = 0, (64) Φ ( f (s, 0), s, 0) −1 0 −1 Φ0 ( f (0, s), 0, s) which play the role to ensure the solvability of the function describing the shock front later. Based on the above analysis we can set up a fixed-boundary value problem (C f ) in Ω f for any fixed f (x1 , x2 ) in Sη . That is ⎧ ⎪ ⎪ System+(9), in Ω f , ⎨ p = pb + g(x1 , x2 ) + const, on x0 = 1, (65) (C f ) : (U, U− ) = 0, on Σ f , i = 1, 2, 3, 4, G ⎪ ⎪ ⎩ i (64),
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where const is a constant to be determined. On x0 = 1 the pressure is given apart from a constant difference. Later we will see that the freedom of one dimension originates from the requirement of conservation of mass. Once the above problem (C f ) is solved, we solve the following problem for f¯(x1 , x2 ): Φ ( f (x1 , x2 ), x1 , x2 ) , on T2 , i = 1, 2, ∂xi f¯(x1 , x2 ) = − i Φ0 ( f (x1 , x2 ), x1 , x2 ) (66) f¯(0, 0) = 0. The system (66) is linear and overdetermined. The condition G 4 = 0 ensures its solvability on the simply connected domain [−d, d] × [−d, d], while (65) [or (64)] indicates that the solution f¯(x1 , x2 ) can be well defined on T2 . Later we will show in Section 7 that f → f¯ determines a contract mapping S: Sη → Sη , provided ε0 in Theorem 2 is sufficiently small. Then by the contraction mapping principle we will obtain a unique fixed point f of S which is the shock front required in problem (B). Correspondingly, the solution U f of the problem (C f ) is also obtained in the meantime, which is exactly the desired subsonic state behind the shock front. 3.2. Transformation of the domains In the process of solving the free-boundary value problem (FB), we have to solve a series of fixed-boundary value problems (Cf ), for which the domain of definition Ω f for U depends on the function f (x1 , x2 ). The change of boundary causes some trouble when establishing necessary estimates of solutions in the process of iteration. To overcome the difficulty we introduce a C 3,α homeomorphism to fix the boundary. The homeomorphism is ⎧ x0 − f (x1 , x2 ) ⎪ ⎪ , ⎨ y0 = 1 − f (x1 , x2 ) Ψf : (67) y1 = x1 , ⎪ ⎪ ⎩y = x , 2
2
which transforms x ∈ Ω f to y ∈ Ω0 , and transforms the boundaries Σ f and x0 = 1 to y0 = 0 and y0 = 1, respectively. In the sequel we use the notation Σs = {(y0 , y1 , y2 ) : (y1 , y2 ) ∈ T2 , y0 = s} for − 1 s 1
(68)
and often simply denote Ω0 by Ω. Denoting ∂i = we have
∂ , i = 0, 1, 2. ∂ yi
⎧ ⎪ ⎨ ∂x0 =
1 ∂0 , 1 − f (y1 , y2 ) y0 − 1 ⎪ ⎩ ∂xi = (∂i f )∂0 + ∂i , i = 1, 2. 1 − f (y1 , y2 )
(69)
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Hence the operator Dk defined on y0 = 0 by (62), is nothing but ∂k . Moreover, from the following proposition, we know that the norm of u defined in Ω and the norm of the function u ◦ Ψ f defined on Ω f are equivalent and their ratio depends only on sup f ∈Sη f C 3,α (T2 ) , but not on the specific form of f . Proposition 5. Let Ω1 , Ω2 ∈ Rn be open bounded sets. Let u ∈ C k,α (Ω¯ 2 ) and Φ : Ω1 → Ω2 satisfy Φ ∈ C k,α (Ω¯ 1 ; Rn ). Then u ◦ Φ ∈ C k,α (Ω1 ) and u ◦ Φ C k,α (Ω1 ) C u C k,α (Ω2 ) ,
(70)
where k = 2, 3 and C = C(n, Φ C k,α (Ω1 ;Rn ) ). The proposition can be verified by direct calculation. By the transformation (67) the problem (C f ) can be written in (y)-coordinates. First, the system (9) is transformed to 2
B j (U (y))∂ j U (y) = 0 in Ω,
(71)
j=0
where
B0 (U ) = A0 (U ) + (y0 − 1)A j (U )∂ j f (y), k = 1, 2. Bk (U ) = (1 − f (y))Ak (U ),
(72)
In (72) the repeated indices indicate the summation from 1 to 2. In the sequel we will always use this convention without further explanation. The boundary conditions on Σ0 and Σ1 take the same form as in subsection 3.1. Therefore, we obtain a problem in the (y)-coordinate system, which is still called (C f ). Notice now that (65) is simply 1 1 Φ1 (0, s, 0) Φ2 (0, 0, s) ds = 0, ds = 0. (73) Φ (0, s, 0) −1 0 −1 Φ0 (0, 0, s) Then the problem C f in (y)-coordinates is ⎧ System (71), in Ω0 , ⎪ ⎪ ⎨ p = pb+ + g(x1 , x2 ) + const, on Σ1 , G (U, U− ) = 0, on Σ0 , i = 1, 2, 3, 4, ⎪ ⎪ ⎩ i (73),
(74)
where const is a constant to be determined. In order to solve the problem (C f ) for U− satisfying (33) and f ∈ Sη with ε ε0 , η η0 , we will use linear approximation and iteration methods. To this end we often write nonlinear equations in a form, called an L-like form, where the linear principal part of a nonlinear equation is kept in the left-hand side, and all higher-order terms are removed to right-hand side of the equation. Moreover, to present our proof more clearly we often use the following terminology. Definition 3.2. A term involving U , f , and their derivatives is called a good small term, if it satisfies the following conditions:
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1. As a function of the derivatives of U and f , the term only involves ∇U, ∇ f, ∇ 2 f , ∇ 2 S, and ∇ 2 p. 2. The term is a sum of quantities with factor (U− − Ub− ) and quantities with quadratic factors as f, U − Ub+ u 1 , u 2 , ∇ f, ∇U, ∇ 2 f , ∇ 2 S, and ∇ 2 p, where f and its derivatives appear at most once. Obviously, for any U− satisfying (33) and f ∈ Sη , U ∈ Oδ all good small terms are dominated by C(ε + δ 2 + δη). 3.3. Reformulation of boundary conditions In this subsection we write the boundary conditions on Σ0 in an L-like form in preparation for the linearization in Section 5. Because of G i (Ub+ , Ub− ) = 0 for 1 i 3, the condition G i (U, U− ) = 0 can be written as ∇+ G i (Ub+ , Ub− ) · (U − Ub+ )
= ∇+ G i (Ub+ , Ub− ) · (U − Ub+ ) − (G i (U, Ub− ) − G i (Ub+ , Ub− )) +(G i (U, Ub− ) − G i (U, U− ))
:= gi (U, U− ),
(75)
where ∇+ G i (U, U− ) is the gradient of G i (U, U− ) with respect to the variable U . Direct calculation shows that ∂(G 1 , G 2 , G 3 )(U, U− ) + − ∂U (U ,U ) b
b
⎛
⎞ [ p]2 pu 0 ⎜ Cν a 2 ⎟ ⎜ ⎟ ⎜ ⎟ 2 pu 2 ⎟ ⎜ [ p] 0 2 2 2 ⎜ ⎟. =⎜ −2ρu 0 [ p] 0 0 −[ p] (1 + (u 0 /a) ) Cν a 2 ⎟ ⎜ ⎟ ⎜ ⎟ 3 3 2 2 2 ⎝ [ p] u 0 [ p] pu 0 ⎠ γ [ p] u 0 3 2 γp 2 )00− − −[ p] ( ρu 0 + 2 γ −1 γ −1 2a 2 2Cν a 2 −ρ[ p]2
00
−u 0 [ p]2 /a 2
(76)
The determinant of the submatrix composed of the first, fourth, and fifth columns is pρu 0 [ p]6 (u 20 − a 2 ) , (77) (γ − 1)Cν a 2 + − (Ub ,Ub )
which is not zero. Therefore, for U ∈ Oδ the corresponding submatrix of (76) is also nonsingular, provided that ε0 and δ0 are small. It turns out that one can derive from (75) that (78) u i − (u i )+ b = h i (U, U− ), i = 0, 3, 4, where h i (i = 0, 3, 4) are linear combinations of g j ( j = 1, 2, 3) with constant coefficients depending only on Ub± .
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Let e4 be a constant, and consider G 4 ((U, U− ) = e4 instead of G 4 = 0, then by straightforward calculation we have ∂2 Φ1 − ∂1 Φ2 = e4 Φ0 + (Φ1 ∂2 Φ0 − Φ2 ∂1 Φ0 )/Φ0 .
(79)
Direct calculation yields ∂2 Φ1 = [ρu 22 + p]ρu 0 ∂2 u 1 + [ρu 22 + p]∂2 (ρu 0 )u 1 − [ρu 22 + p] − 2 ×∂2 (ρ − u − 0 u 1 ) + [ρu 0 u 1 ]∂2 [ρu 2 + p] − ∂2 ([ρu 0 u 2 ][ρu 1 u 2 ]), ∂1 Φ2 = [ρu 21 + p]ρu 0 ∂1 u 2 + [ρu 21 + p]∂1 (ρu 0 )u 2 − [ρu 21 + p] − 2 ×∂1 (ρ − u − 0 u 2 ) + [ρu 0 u 2 ]∂1 [ρu 1 + p] − ∂1 ([ρu 0 u 1 ][ρu 1 u 2 ]). Then by denoting
λ4 = [ p]ρu 0 (U + ,U − ) ,
(80)
∂2 u 1 − ∂1 u 2 = e4 Φ0 (U, U− )/λ4 + f 4 (U, U− ) on Σ0 ,
(81)
b
b
(79) can be rewritten as
where f 4 (U, U− ) = λ−1 4
Φ1 ∂2 Φ0 /Φ0 − Φ2 ∂1 Φ0 /Φ0 + [ρu 21 + p]∂1 (ρu 0 )u 2
− 2 − [ρu 21 + p]∂1 (ρ − u − 0 u 2 )+[ρu 0 u 2 ]∂1 [ρu 1 + p]−∂1 ([ρu 0 u 1 ][ρu 1 u 2 ])
− + ∂2 ([ρu 0 u 2 ][ρu 1 u 2 ])−[ρu 0 u 1 ]∂2 [ρu 22 + p]+[ρu 22 + p]∂2 (ρ − u − 0 u1 )
− [ρu 22 + p]∂2 (ρu 0 )u 1 + (λ4 − [ρu 22 + p]ρu 0 )∂2 u 1 − (λ4 − [ρu 21 + p]ρu 0 )∂1 u 2 ,
(82)
which is a good small term in the sense of Definition 3.2. The corresponding verification is left to the Appendix. We remark here that the constant e4 will vanish for the nonlinear problem, because T2 is a compact manifold without boundary and G 4 has divergence form. However, it generally may not vanish in the process of iteration in order to let the corresponding linearized problem be solvable. 3.4. Decomposition of the system The Euler system (71) in the subsonic region is an elliptic–hyperbolic system, whose characteristic polynomial has a real eigenvalue with multiplicity three, and a pair of conjugate complex eigenvalues. To solve its boundary value problem we will reduce the system to a canonical form in which the elliptic part and the hyperbolic part are separated at their principal level. Because the system involves three variables, we will employ the calculus of symbols to proceed with such a separation. Write the symbol of the differential operator in (71) as b(y, ξ ) =
2 j=0
Bjξj.
(83)
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Shuxing Chen & Hairong Yuan
In the sequel for notational simplicity we take ξk as the symbol of ∂k rather than 1 √ ∂k . The generalized eigenvalues of b(y, ξ ) is the root of the determinant −1 ˆ 0 B0 + B j ξ j ) = 0, det(−λξ which has one real root λˆ h with multiplicity three and two conjugate complex roots √ λˆ ± = λˆ R ± −1λˆ I . Direct calculation shows λh (ξ ) = λˆ h ξ0 = ρ(1 − f )ξi u i /β,
(84)
where β = ρu 0 + (y0 − 1)ρu i ∂i f (it does not vanish in the whole region), and λ± (ξ ) = λˆ ± ξ0 = λ R (ξ ) ±
√
with λ R (ξ ) = −B(ξ1 , ξ2 )/A, λ I (ξ ) =
−1λ I (ξ )
(85)
AC − B 2 /A,
where A = A(U, f ) = −1 + β 2 /(ρa)2 − (y0 − 1)2 ((∂1 f )2 + (∂2 f )2 ), B = B(ξ1 , ξ2 ) = (1 − f ) (y0 − 1)(∂i f )ξi + βu i ξi /(ρa 2 ) , C = C(ξ1 , ξ2 ) = (1 − f )2 (u i ξi )2 /a 2 − (ξ12 + ξ22 ) . + 2 2 We notice that, due to ((u 0 )+ b /ab ) − 1 < 0, the discriminant AC − B is positive, so that λ I is real-valued provided δ0 , η0 are small enough. Corresponding to these eigenvalues the (generalized) left eigenvectors are √ (i) (86) lh (ξ ) (i = 1, 2, 3) and l± (ξ ) = l R (ξ ) ± −1l I (ξ ),
where (1) lh = −λh (ξ )(y0 − 1)∂1 f + (1 − f )ξ1 λh 0 0 0 , (2) lh = −λh (ξ )(y0 − 1)∂2 f + (1 − f )ξ2 0 λh 0 0 , (3) lh = 0 0 0 0 1 , l R (ξ ) = (( f − 1)ρu 0 )−1 (λ R (ξ ) ( f − 1)ξ1 + (y0 − 1)∂1 f λ R (ξ ) ×( f − 1)ξ2 + (y0 − 1)∂2 f λ R (ξ ) (1 − f )ρu i ξi − βλ R (ξ ) 0), l I (ξ ) = (( f −1)ρu 0 )−1 λ I (ξ ) (y0 −1)∂1 f λ I (ξ ) (y0 −1)∂2 f λ I (ξ ) −βλ I (ξ ) 0 .
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Notice that all elements of l I (ξ ) have a common factor λ I (ξ ). By dividing out λ I (ξ ), we obtain l˜I = (( f − 1)ρu 0 )−1 1 (y0 − 1)∂1 f (y0 − 1)∂2 f −β 0 . Since l R (ξ ) is a polynomial of (ξ1 , ξ2 ), then the operator l R (∂) is a corresponding differential operator. By the definition of eigenvectors we have (l R B0 )(ξ ) l R (ξ ) ξ0 + λ R (ξ ) −λ2I (ξ ) b(y, ξ ) = , (87) 1 ξ0 + λ R (ξ ) l˜I (l˜I B0 )(ξ ) where 1 (ρu i ξi (y0 − 1)∂1 fρu i ξi − βξ1 (y0 − 1)∂2 fρu i ξi ρu 0 −βξ2 0 0) , 1 000 A0 , (l˜I B0 )(ξ ) = (1 − f )ρu 0
(l R B0 )(ξ ) = −
(88) (89)
and λ2I (ξ ) = (AC − B 2 )/A2 is also a polynomial of ξ1 , ξ2 , so that the operator L with symbol λ2I (ξ ) is a differential operator of second order. Direct computations give us (90) (l R B0 )(∂)U = ∂1 u 1 + ∂2 u 2 − f 3 (U, f ), where ui y0 − 1 ∂i u 0 + (∂1 f u 2 ∂2 u 1 −∂2 f u 2 ∂1 u 1 +∂2 f u 1 ∂1 u 2 −∂1 f u 1 ∂2 u 2 ). u0 u0 (91) Similarly, from (89)
f 3 (U, f ) =
(l˜I B0 )(∂)U =
A p, (1 − f )ρu 0
(92)
By acting l˜I and l R (∂) from the left on (71) we derive the elliptic part of (71). Denoting (93) v = (l R B0 )(∂)U = ∂1 u 1 + ∂2 u 2 − f 3 (U, f ), we obtain v+ Let
A (∂0 + λ R (∂)) p = 0, (1 − f )ρu 0
λ2 = −A/(ρu 0 ) f =0,U =U + > 0, b
(94)
(95)
then (93) can be written as
where
v − λ2 ∂0 p = f 2 (U, f ),
(96)
A A f 2 (U, f ) = − ∂0 p λ R (∂) p − λ2 + (1 − f )ρu 0 (1 − f )ρu 0
(97)
is a good small term.
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Shuxing Chen & Hairong Yuan
On the other hand, acting l R (∂) from the left on (71) gives l R (∂)
2
B j ∂ j U = 0.
(98)
j=0
The second-order terms on the left-hand side are (∂0 + λ R (∂))v −
A Lp (1 − f )ρu 0
(99)
apart from certain first-order terms. Then we have − where T2 =
1− f ρu 0 A
1− f 2 A Lp = (∂ + ∂22 ) p + T2 , (1 − f )ρu 0 ρu 0 1
(100)
βu j βu i (y0 − 1)∂ j f + ∂i j p −u i u j + (y0 − 1)∂i f + 2 ρa ρa 2
apart from first-order terms. In the expression of l R (∂)
2
B j ∂ j U , all first-order terms are
j=0
T1 := (∂ξi l R )
2
(∂i B j (U ))∂ j U,
(101)
j=0
where (∂ξi l R ) is independent of ξ , so that it is simply a multiplier. Therefore, acting l R (∂) from the left on (74) yields (∂0 + λ R (∂))v +
(1 − f ) (∂11 + ∂22 ) p + f˜1 = 0, (ρu 0 )
(102)
where f˜1 is a good small term. In view of (94) A λ R (∂)v = λ R (∂) (∂0 + λ R (∂)) p (1 − f )ρu 0 is also a good small term, then by setting λ1 =
1 , ρb+ (u 0 )+ b
(103)
we are led to ∂0 v + λ1 (∂11 + ∂22 ) p = f 1 (U, f ),
(104)
where f 1 (U, f ) is a good small term in the sense of Definition 3.2. Summing up the above analysis we have separated the elliptic part (94) and (104) from (71). The hyperbolic part separated from (71) will be given in the next section.
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4. Equivalent form of problem (C f ) In this section we decompose problem (C f ) into four coupled problems (I– IV) which are easier to handle via standard theory and the methods of differential equations. Problem I.
⎧ ∂0 v + λ1 (∂11 + ∂22 ) p = f 1 (U, f ) in Ω, [see (104)] ⎪ ⎪ ⎪ ⎪ v − λ2 ∂0 p = f 2 (U, f ) in Ω, [see(97)] ⎨ p − pb+ = h 3 (U, U− ) on Σ0 , [see (78)] ⎪ ⎪ p − pb+ = g + const on Σ1 , [see(74)] ⎪ ⎪ ⎩ Σ0 v(0, y1 , y2 ) dy1 dy2 = − Σ0 f 3 (U, f ) dy1 dy2 ,
(105)
where const ∈ R is a constant to be determined. The last equality in (105) comes from the integration of equation (94) on Σ0 . The latter has been employed in deduction of the first equation in (105). In view of (94) the last equation is an identity if U is the solution of (C f ). However, if all the right-hand sides are given except the unknown constant const, then the problem (105) becomes a linear problem. In this case the last equation plays a role of compensating for the freedom of the boundary condition on Σ1 . Such a setting of boundary conditions is also adopted in the equivalued boundary problems for elliptic problems (see, for example, [22]). Later on we can see that, if (105) is solved, then the unknown number const must take the value 1 − f 3 dy1 dy2 − f 2 dy0 dy1 dy2 − λ2 g dy1 dy2 4λ2 Σ0 Ω Σ1 1 y0 + λ2 h 3 dy1 dy2 + f 1 (U (s, y1 , y2 ), f (y1 , y2 )) dy1 dy2 ds dy0 . Σ0
0 0
Σs
(106) Problem II. Set u˜ 1 (y1 , y2 ) = u 1 (0, y1 , y2 ) − u 1 (0, 0, 0), u˜ 2 (y1 , y2 ) = u 2 (0, y1 , y2 ) − u 2 (0, 0, 0), then u˜ 1 , u˜ 2 on Σ0 should satisfy ⎧ ⎨ ∂1 u˜ 1 + ∂2 u˜ 2 = v + f 3 (U, f ), [ see (94) ] ∂2 u˜ 1 − ∂1 u˜ 2 = e4 Φ0 (U, U− )/λ4 + f 4 (U, U− ), [see (81)] ⎩ u˜ 1 (0, 0) = u˜ 2 (0, 0) = 0, where we take
(107)
(108)
e4 = −λ4
Σ0
Σ0
f 4 (U, U− ) dy1 dy2 Φ0 (U, U− ) dy1 dy2
.
(109)
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Later we denote c1 = u 1 (0, 0, 0), c2 = u 2 (0, 0, 0), and use the notation U˜ as U˜ = U − 0 c1 c2 0 0 + + = (u 0 )+ b + h 0 (U, U− ) u˜ 1 u˜ 2 pb + h 3 (U, U− ) (Sb ) + h 4 (U, U− ) . Problem III. The constants c1 = c1 (U˜ , U− ), c2 = c2 (U˜ , U− ) are determined by (73). That is, ⎧ 1 Φ1 (U, U− ) ⎪ ⎪ ⎪ ds = 0, ⎨ F1 := Φ0 (U, U− ) y0 =y2 =0 −1 1 Φ (U, U ) ⎪ 2 − ⎪ ⎪ ds = 0, ⎩ F2 := Φ (U, U − ) y0 =y1 =0 −1 0
(110)
where U = U˜ + (0, c1 , c2 , 0, 0). To ensure the solvability of (110) for c1 , c2 , by implicit function theorem [25], we have to verify ∂(F1 , F2 ) = 0, det ∂(c1 , c2 ) Ub
(111)
where ∗ Ub means the value of ∗ at U− = Ub− , U = Ub+ . According to the expressions of Φ0 , Φ1 , Φ2 we have Φ0 (U, U− ) Ub = −[ p]2 = 0, Φ1 (U, U− ) Ub = Φ2 (U, U− ) Ub = 0, ∂Φ1 ∂Φ1 Ub = U = ρu 0 [ p], ∂c1 ∂u 1 b ∂Φ1 ∂Φ2 Ub = U = 0, ∂c2 ∂c1 b ∂Φ2 ∂Φ2 Ub = U = ρu 0 [ p], ∂c2 ∂u 2 b then
Φ0
∂Φ1 ∂Φ0 − Φ1 2ρu 0 ∂c1 ∂c1 , ds = − [ p] Φ02
∂ F1 = ∂c1 −1 ∂ F1 ∂ F2 = = 0, ∂c2 ∂c1 ∂Φ ∂Φ 1 Φ0 2 − Φ2 0 ∂ F2 2ρu 0 ∂c2 ∂c2 = ds = − , ∂c2 [ p] Φ02 −1 which yields (111).
1
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Problem IV. Denote by D = D(U, f ) the differential operator (ρu 0 + (y0 − 1)ρu i ∂i f )∂0 + (1 − f )ρu i ∂i ,
(112)
then the equation for conservation of energy (or invariance of entropy after shocks) and conservation of momentum in (71) can be written in a simpler form by using the transport operator D. Combining with the data on Σ0 we obtain the following four Cauchy problems. DS = 0 in Ω, (113) S − (S)+ b = h 4 (U, U− ) on Σ0 , Du 0 + ∂0 p = 0 in Ω, (114) u 0 − (u 0 )+ b = h 0 (U, U− ) on Σ0 , Du 1 + (1 − f )∂1 p + (y0 − 1)∂1 f ∂0 p = 0 in Ω, (115) u 1 = u˜ 1 + c1 on Σ0 , Du 2 + (1 − f )∂2 p + (y0 − 1)∂2 f ∂0 p = 0 in Ω, (116) u 2 = u˜ 2 + c2 on Σ0 . Lemma 6. The above Problems I–IV are equivalent to problem (C f ) if ε0 , δ0 , η0 are small, which depends only on the background solution. Proof. Due to the deduction of the Problems I–IV it is obvious that any solution of problem (C f ) satisfies the four problems. Therefore, we only need to show that U obtained from the Problems I–IV satisfies (C f ). Indeed, (113)–(116) show the validity of equations of conservation laws of momentum and entropy, as well as the boundary condition (78), for i = 0, 4. The third equation in (105) is (78) for i = 3, and the second in (108) is exactly G 4 (U, U− ) = e4 , while e4 must vanish due to the divergence form of G 4 . Thus G i (U, U− ) = 0 for 1 i 4 hold. The fourth equation in (105) is the same as the condition imposed at the exit of the duct for Problem (C f ). Furthermore, (73) coincides with (110) in Problem III. What remains is to show that the solution of Problems I–IV satisfies the conservation of mass, or equivalently to show that the function v obtained in (105) satisfies in Ω. (117) v = ∂1 u 1 + ∂2 u 2 − f 3 (U, f ) Denote by E the left-hand side of the fourth equation of the system (71), and ψ = v − (∂1 u 1 + ∂2 u 2 − f 3 (U, f )). We are going to prove ψ = 0. Act l R (∂) on the system (71). Since only the fourth equation in (71) (that is, the equation of conservation of mass) remains to be verified, we only write the result related to the fourth row. That is 1 ((1 − f )ρu i ∂i − βλ R (∂))E ( f − 1)ρu 0 = (∂0 + λ R (∂))(∂1 u 1 + ∂2 u 2 − f 3 ) − λ I (∂)
A λ I (∂) p . (1 − f )ρu 0
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Shuxing Chen & Hairong Yuan
Using the fact that (v, p) is the solution of (105) we have (∂0 + λ R (∂))ψ +
1 ((1 − f )ρu i ∂i − βλ R (∂))E = 0. ( f − 1)ρu 0
(118)
Similarly, acting the vector (93) on (71) yields −
β E + ψ = 0. ( f − 1)ρu 0
(119)
Since β is bounded away from zero, then by substituting (119) into (118) we get a first-order linear equation of E: β 1 (∂0 + λ R (∂)) E + ((1 − f )ρu i ∂i − βλ R (∂))E = 0. ( f − 1)ρu 0 ( f − 1)ρu 0 (120) Recall that the first equation in (108) reads ψ =0
on Σ0 ,
thus E = 0 on Σ0 , so that by the uniqueness of solutions of linear equations we obtain E = 0 and ψ = 0 in Ω as desired. Remark. Let us explain here in more detail why we have to introduce an undetermined constant const in Problems (D) and (B), or equivalently, in the fourth equation in (105). This is to compensate for the restriction added by the last equation in (105), which is required, in turn, to make the first equation of (108) be solvable. Moreover, the first equation of (108) directly shows the validity of (117) on Σ0 , which can be regarded as the initial condition of (117) in Ω. When the equations for conservation of momentum and energy have been applied as in the equations of Problem IV, equation (117) is then equivalent to the equation of conservation of mass. Hence we see that the way to give the pressure at the exit as in Problem (B) essentially stems from the conservation law of mass. In Section 2 we have also seen that even for uniform flow the pressure of the flow at the exit cannot be completely given. We also indicate here that in most cases the constant const introduced above is not zero. 5. Solution of the linearized problem To solve the nonlinear coupled Problems I–IV we use the nonlinear iteration method. To this end we first study the linearized problems corresponding to the nonlinear Problems I, II, and IV. ¯ v¯ satisfying Problem I . The Problem I is to find p, ⎧ ∂0 v¯ + λ1 (∂11 + ∂22 ) p¯ = f 1 , in Ω, ⎪ ⎪ ⎪ ⎪ v¯ − λ2 ∂0 p¯ = f 2 , in Ω, ⎪ ⎪ ⎨ p¯ − pb+ = h 3 , on Σ0 , + ⎪ ⎪ p¯ − pb = g + const, on Σ1, ⎪ ⎪ ⎪ ⎪ v(0, ¯ y1 , y2 ) dy1 dy2 = − f 3 dy1 dy2 , ⎩ Σ0
Σ0
(121)
Transonic Shocks in Compressible Flow Passing a Duct
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where the nonhomogeneous terms are assumed to satisfy f 1 ∈ C k−2,α (Ω), f 2 ∈ C k−1,α (Ω), f 3 ∈ C k−1,α (Σ0 ), h 3 , g ∈ C k,α (T2 ) (122) with k = 2 or 3. The problem (121) is solvable. Indeed, by substituting the second equation of (121) into its first equation we obtain a second-order elliptic equation of p¯ λ1 (∂11 + ∂22 ) p¯ + λ2 ∂00 p¯ = f 1 − ∂0 f 2 in Ω.
(123)
Temporarily fixing const as e, we can obtain a unique solution p¯ to the Dirichlet problem ⎧ ⎨ Equation (123), in Ω, p¯ − pb+ = h 3 , on Σ0 , (124) ⎩ p¯ − pb+ = g + e, on Σ1 , according to the standard theory of elliptic equation of second-order [13]. Now taking the value of e as 1 e= − f 3 dy1 dy2 − f 2 dy0 dy1 dy2 − λ2 g(y1 , y2 ) dy1 dy2 λ2 Σ0 Ω Σ1 1 y0 (125) + λ2 h 3 dy1 dy2 + f 1 dy1 dy2 ds dy0 , Σ0
0 0
Σs
and letting v¯ = f 2 + λ2 ∂0 p, ¯
(126)
then (v, ¯ p) ¯ satisfies (121). In fact, (123), (126) imply the first two equations in (121). To verify the last condition in (121) we integrate the first equation on Σs and then obtain d v¯ dy1 dy2 = f 1 dy1 dy2 . ds Σs Σs Therefore,
Ω
Σ y0
v¯ dy1 dy2 =
v¯ dy0 dy1 dy2 =
Σ0
Σ0
v¯ dy1 dy2 +
y0 0
v¯ dy1 dy2 +
f 1 dy1 dy2 ds,
1 y0 0 0
On the other hand, (126) implies v¯ dy0 dy1 dy2 = f 2 dy0 dy1 dy2 + λ2 Ω
Σs
Ω
Σs
f 1 dy1 dy2 ds dy0 .
Σ1
p¯ dy1 dy2 −
Hence (125) and the boundary conditions satisfied by p¯ imply v¯ dy1 dy2 = − f 3 dy1 dy2 . Σ0
Σ0
Σ0
p¯ dy1 dy2 .
(127)
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As for the estimate of the solution (v, ¯ p), ¯ the standard Schauder estimates give p¯ − p + k,α b C (Ω) C f 1 C k−2,α (Ω) + f 2 C k−1,α (Ω) + h 3 C k,α (T2 ) + gC k,α (T2 ) + |e| . (128) Then from the expression of v¯ we have
v ¯ C k−1,α (Ω) f 2 C k−1,α (Ω) + C p¯ − pb+ C k,α (Ω) .
(129)
Furthermore, (125) implies |e| C h 3 L ∞ (Σ0 ) + g L ∞ (Σ1 )
+ f 1 L ∞ (Ω) + f 2 L ∞ (Ω) + f 3 L ∞ (Σ0 ) ,
(130)
and we obtain p¯ − p + k,α ¯ C k−1,α (Ω) C f 1 C k−2,α (Ω) b C (Ω) + v
+ f 2 C k−1,α (Ω) + f 3 C k−1,α (Σ0 ) + h 3 C k,α (T2 ) + gC k,α (T2 ) .
(131)
Problem II . This linearized problem of Problem II is to find u˜ 1 , u˜ 2 satisfying ⎧ ⎪ ⎨ ∂1 u˜ 1 + ∂2 u˜ 2 = v¯e4+ f 3 , ∂2 u˜ 1 − ∂1 u˜ 2 = Φ0 + f 4 , (132) λ4 ⎪ ⎩ u˜ 1 (0, 0) = u˜ 1 (0, 0) = 0,
where
Σ0
f 4 dy1 dy2
Σ0
Φ0 dy1 dy2
e4 = −λ4
,
(133)
and the nonhomogeneous terms satisfy f 4 , Φ0 ∈ C k−1,α (Σ0 ), with |Φ0 | C0 > 0. Since the right-hand sides of (132) satisfy e4 (v¯ + f 3 ) dy1 dy2 = 0, Φ0 + f 4 dy1 dy2 = 0 Σ0 Σ0 λ4
(134)
(135)
due to the last equation of (121) and (133), then (132) has a unique solution u˜ 1 , u˜ 2 on T2 , and u˜ 1 C k,α (Σ0 ) + u˜ 2 C k,α (Σ0 ) C f 3 C k−1,α (Σ0 ) + f 4 C k−1,α (Σ0 ) + v ¯ C k−1,α (Σ0 ) + e4 Φ0 C k−1,α (Σ0 ) . (136) By using (131), (134), and (135) |e4 | C f 4 L ∞ (Σ0 ) ,
(137)
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then we have u˜ 1 C k,α (Σ0 ) + u˜ 2 C k,α (Σ0 ) C gC k,α (T2 ) + f 1 C k−2,α (Ω) + f 2 C k−1,α (Ω) + f 3 C k−1,α (Σ0 ) (138) + f 4 C k−1,α (Σ0 ) (1 + Φ0 C k−1,α (Σ0 ) ) + h 3 C k,α (T2 ) . Problem III . Same as Problem III the Problem III is to find c1 , c2 from F1 (U˜ , U− , c1 , c2 ) = 0, F2 (U˜ , U− , c1 , c2 ) = 0
(139)
for given U− ∈ NC0 ε , and + + U˜ = (u 0 )+ b + h 0 u˜ 1 u˜ 2 pb + h 3 (Sb ) + h 4
(140)
where u˜ 1 , u˜ 2 are the solution of Problem II’. In view of (111) we can solve c1 , c2 from (139) by using implicit function theorem [25], provided ε0 , η0 , δ0 are sufficiently small. Moreover, ˜ + − (141) + U− − Ub L ∞ (Σ ) . |c1 | + |c2 | C U − Ub ∞ L (Σ0 )
0
Problem IV . For f ∈ Sη , U ∈ Oδ and D(U, f ) defined in (112), the linearized form of Problem IV takes the form D S¯ = f 5 in Ω, (142) S¯ − (S)+ b = h 4 on Σ0 , D u¯ 0 + ∂0 p¯ = f 6 in Ω, (143) u¯ 0 − (u 0 )+ b = h 0 on Σ0 , D u¯ 1 + (1 − f )∂1 p¯ + (y0 − 1)∂1 f ∂0 p¯ = f 7 in Ω, (144) u¯ 1 = u˜ 1 + c1 on Σ0 , D u¯ 2 + (1 − f )∂2 p¯ + (y0 − 1)∂2 f ∂0 p¯ = f 8 in Ω, (145) u¯ 2 = u˜ 2 + c2 on Σ0 , where f 5 , f 6 , f 7 , f 8 are possible terms arising in the iterative approximation. Assume f 5 ∈ C k,α (Ω), f j ∈ C k−1,α (Ω) ( j = 6, 7, 8), h 0 , h 4 ∈ C k,α (Σ0 ),
(146)
then by the characteristic method for first-order linear equations the Cauchy Prob¯ u¯ 0 , u¯ 1 , u¯ 2 and satisfy the estimates lems (142)–(146) have unique solutions S, S¯ − S + k,α (147) b C (Ω) C h 4 C k,α (Σ0 ) + f 5 C k,α (Ω) , + u¯ 0 − (u 0 )+ k,α b C (Σ0 ) + u¯ 0 − (u 0 )b C k−1,α (Ω) (148) C h 0 C k,α (Σ0 ) + f 6 C k−1,α (Ω) + p¯ − pb+ C k,α (Ω) , u¯ 1 C k,α (Σ0 ) + u¯ 1 C k−1,α (Ω) C u˜ 1 + c1 C k,α (Σ0 ) + f 7 C k−1,α (Ω) + p¯ − pb+ C k,α (Ω) , (149) u¯ 2 C k,α (Σ0 ) + u¯ 2 C k−1,α (Ω) C u˜ 2 + c2 C k,α (Σ0 ) + f 8 C k−1,α (Ω) + p¯ − pb+ C k,α (Ω) . (150)
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Thus we finally obtain for k = 2, 3 ⎛ U¯ −U + C ⎝ O(ε) + f 1 k−2,α C (Ω) + f 2 C k−1,α (Ω) b k + f 3 C k−1,α (Σ0 ) + f 4 C k−1,α (Σ0 ) (1+Φ0 C k−1,α (Σ0 ) )+gC k,α (T2 ) ⎞ 8 f j k−1,α h j k,α ⎠ . (151) + f 5 C k,α (Ω) + + C (Ω) C (Σ ) 0
j=6
j=0,3,4
Here the term O(ε) on the right-hand side comes from the estimate U− −Ub− L ∞ (Σ ) 0 in (141).
6. Solving the fixed-boundary problem 6.1. Boundedness in X 3 In this section we will establish the unique solution U f to the nonlinear fixedboundary problem (C f ) for any f ∈ Sη . To this end we introduce the Banach spaces (152) X k := U ∈ C k−1,α (Ω) : u 3,4 ∈ C k,α (Ω), u 0,1,2 |Σ0 ∈ C k,α (Σ0 ) with the norm U k ( = U X k ) :=
2 u j
C k,α (Σ0 )
j=0
+
2 u j
C k−1,α (Ω)
j=0
+
4 u j
C k,α (Ω)
j=3
(153) for k = 2, 3. By Proposition 5 this is equivalent to (33). Our next demonstration is based on the following theorem. Theorem 7. Let B be a bounded closed subset of X 3 . Suppose T : B → B and that there exists a constant θ < 1 such that, for any u, v ∈ B, T u − T v2 θ u − v2 ,
(154)
then T has a unique fixed point in B. Proof. By the Banach contraction mapping principle we need only show that B is complete in X 2 . Due to the convexity properties of Hölder norms uλa+(1−λ)b C uaλ u1−λ , 0 a b < ∞, λ ∈ (0, 1), b any bounded sequence in C b that converges in C a will also converge in C d , for any d ∈ [a, b). Thus by the definition of Hölder norms it is straightforward to check that the limit still lies in C b . This fact leads to the completeness of B.
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549
By virtue of Theorem 7 to prove the existence of a solution to (C f ) in X 3 , we only need to show the boundedness of T in X 3 and its contraction in X 2 . For given f ∈ Sη , we take B = Oδ [defined in (46)]. Then, for any U ∈ B, we define the terms on the right-hand sides of linearized problems in Section 5 as follows: ⎧ ⎪ ⎪ f 1 = f 1 (U, f ) [see (104)], ⎪ ⎪ f 2 = f 2 (U, f ) [see (97)] ⎪ ⎪ ⎪ ⎪ ⎨ f 3 = f 3 (U, f ) [see (91)], f 4 = f 4 (U, U− ) [see (82)], (155) ⎪ ⎪ ⎪ f k = 0, k = 5, 6, 7, 8, ⎪ ⎪ ⎪ ⎪ h j = h j (U, U− ), j = 0, 3, 4 [see (78)], ⎪ ⎩ Φ0 = Φ0 (U, U− ) [see (48)]. Then by using (151) and the estimates for the nonlinear terms established in Appendix we know that Problems I –IV admit a solution U¯ = T U , satisfying U¯ − U + C(δ 2 + δη + ε), (156) b 3 where C is independent of ε0 , δ0 , η0 . Therefore, if we choose δ = 3Cε, δ0 one gets
1 1 , η0 , 3C 3C
U¯ − U + δ. b 3
(157) (158)
This means that T is an inner mapping from Oδ to Oδ , provided that ε0 , η0 , δ0 are small. 6.2. Contraction in X 2 To show the contraction of the mapping T , we consider the corresponding Problems (I )–(IV ) for Uˆ = U¯ (1) − U¯ (2) = T U (1) − T U (2) for any U (i) ∈ Oδ , f (i) ∈ Sη , i = 1, 2, and then estimate the norm of Uˆ in X 2 . Remark. We remark that for the proof of the solvability of the problem (C f ) the boundary of the domain Ω f is fixed, so that f (1) = f (2) . However, here we would rather let f (1) and f (2) be different in Sη , because the proof with such a change can also be applied to deal with the free-boundary problems in the next section. Problem I for Uˆ is ⎧ ∂0 vˆ + λ1 (∂11 + ∂22 ) pˆ = f 1 (U (1) , f (1) ) − f 1 (U (2) , f (2) ) in Ω, ⎪ ⎪ ⎪ ⎪ vˆ − λ2 ∂0 pˆ = f 2 (U (1) , f (1) ) − f 2 (U (2) , f (2) ) in Ω, ⎪ ⎪ ⎪ (1) (2) ⎪ ⎪ pˆ = h 3 (U (1) , U− ) − h 3 (U (2) , U− ) on Σ0 , ⎪ ⎨ pˆ = const on Σ1 , ⎪ ⎪ ⎪ v(0, ˆ y1 , y2 ) dy1 dy2 = − f 3 (U (1) , f (1) ) dy1 dy2 ⎪ ⎪ ⎪ Σ Σ 0 0 ⎪ ⎪ ⎪ ⎪ + f 3 (U (2) , f (2) ) dy1 dy2 , ⎩ Σ0
where
(i) U−
= U− ( f (i) (x1 , x2 ), x1 , x2 ), and const is an unknown constant.
(159)
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Shuxing Chen & Hairong Yuan
Similar to the discussions in Section 5, we can establish the following estimate (1) (1) (2) (2) pˆ 2,α +vˆ 1,α C(ε+δ +η) f . U −U + − f C (Ω) C (Ω) 3,α 2 2
C
(T )
(160) Problem II is ⎧ ⎪ ∂1 u˜ˆ 1 + ∂2 u˜ˆ 2 = vˆ + f 3 (U (1) , f (1) ) − f 3 (U (2) , f (2) ), ⎪ ⎪ ⎨ ˜ (1) (1) ∂2 uˆ 1 − ∂1 u˜ˆ 2 = e4 (U (1) , U− )Φ0 (U (1) , U− )/λ4 (2) (2) (1) (2) ⎪ −e4 (U (2) , U− )Φ0 (U (2) , U− )/λ4 + f 4 (U (1) , U− ) − f 4 (U (2) , U− ), ⎪ ⎪ ⎩˜ uˆ 1 (0, 0) = u˜ˆ 1 (0, 0) = 0, (161) (1) (2) with e4 defined by (109) and u˜ˆ i = u˜ i − u˜ i , i = 1, 2. Similar to (136), using the estimates (194)–(197) in Section 8 we have ˜ + u˜ˆ 2 2,α uˆ 1 2,α C (Σ0 ) C (Σ0 ) (1) (1) (2) (2) C(ε + δ + η) U − U + f − f 3,α 2 . (162) 2
C
(T )
Problem III is ( j)
( j)
( j)
F(U¯ ( j) , U− , c1 , c2 ) = 0,
j = 1, 2.
(163)
The implicit function theorem shows that the functions c1 (U˜ , U− ), c2 (U˜ , U− ) are at least C 1 with respect to their arguments, if ε0 , δ0 , η0 are small. Thus, 2 ˜ (1) (2) (1) (1) (2) (2) ¯ − U˜¯ U + − U . (164) U ci − ci C − − i=1
∞,Σ0
∞,Σ0
Then by the estimate (160), (162), and (190) we obtain 2 (1) (2) ci − ci C(ε + δ + η) U (1) − U (2) + f (1) − f (2) i=1
2
C 3,α (T2 )
.
(165) For Problem IV, the transport operator D depends on U and f . As said in the Remark in this section we allow f (1) and f (2) to be different. Denoting D (i) = D(U (i) , f (i) ) we have (1) D Sˆ = f 5 := −(D (1) − D (2) ) S¯ (2)) in Ω, (166) (1) (2) on Σ0 , Sˆ = h 4 (U (1) , U− ) − h 4 (U (2) , U− D (1) uˆ 0 + ∂0 pˆ = f 6 := −(D (1) − D (2) )u¯ (2) in Ω, 0 (167) (1) (2) (1) (2) uˆ 0 = h 0 (U , U− ) − h 0 (U , U− ) on Σ0 , ⎧ (1) D uˆ 1 + (1 − f (1) )∂1 pˆ + (y0 − 1)∂1 f (1) ∂0 pˆ = f 7 ⎪ ⎪ ⎪ (2) ⎨ := −(D (1) − D (2) )u¯ 1 − ∂1 p¯ (2) ( f (2) − f (1) ) (168) ⎪ − (y0 − 1)∂0 p¯ (2) (∂1 f (1) − ∂1 f (2) ) in Ω, ⎪ ⎪ ⎩ (1) (2) uˆ 1 = u˜ˆ 1 + c1 − c1 on Σ0 ,
Transonic Shocks in Compressible Flow Passing a Duct
551
⎧ (1) (1) (1) ⎪ D uˆ 2 + (1 − f )∂2 pˆ + (y0 − 1)∂2 f ∂0 pˆ = f 8 ⎪ ⎪ (2) ⎨ := −(D (1) − D (2) )u¯ − ∂ p¯ (2) ( f (2) − f (1) ) 2 2 (2) (∂ f (1) − ∂ f (2) ) ⎪ − (y − 1)∂ p ¯ in Ω, ⎪ 0 0 2 2 ⎪ ⎩ (1) (2) ˜ on Σ0 . uˆ 2 = uˆ 2 + c − c
(169)
2
2
Thus by using (160), (162), (165) and the fact that U (i) ∈ Oδ , we obtain ˆ (1) (1) (2) (2) . U C(ε + δ + η) U − U + f − f 3,α 2
2
C
(Σ0 )
(170)
Particularly, when f (1) = f (2) , by choosing δ, δ0 , η0 as in (157) and ε0 small we obtain the contraction of T : 1
ˆ (171) U C ε0 U (1) − U (2) U (1) − U (2) . 2 2 2 2 By applying Theorem 7, we are led to Theorem 8. For any f ∈ Sη , there exists a unique solution U f ∈ Oδ to problem (C f ), provided that δ = C0 ε and η0 , ε0 are small. Proof. We need only to show e4 = 0 now. Recall that (81) is equivalent to (79) or e4 = −∂1 (Φ2 /Φ0 ) + ∂2 (Φ1 /Φ0 ),
(172)
then by integrating the equality on T2 we obtain e4 = 0. 7. Solution to the free-boundary problem For f ∈ Sη , we have constructed solution U f to problem (C f ) satisfying U f − U + δ = C0 ε. (173) b 3 Now we can solve ⎧ ⎨
Φ (U , U ) ∂ yi f¯ = − i f − on Σ0 , i = 1, 2, Φ0 (U f , U− ) ⎩ f¯(0, 0) = 0.
(174)
Since Φi are bounded and Φ0 is away from 0 [see (182), (184)], then the solution uniquely exists and satisfies f¯
C C 4,α (T2 )
2
Φi C 3,α (T2 ) C(δ + ε) C1 ε.
(175)
i=1
So let η = C1 ε and choose ε0 small, so that C1 ε0 η0 . We can construct a mapping S: Sη → Sη by f → f¯. Now suppose f (1) , f (2) ∈ Sη , denote S( f (i) ) = f¯(i) and (i)
U− = U− ( f (i) (x1 , x2 ), x1 , x2 ), U (i) = U f (i) , i = 1, 2.
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Shuxing Chen & Hairong Yuan
Then by using the estimates derived in Appendix we have ¯(1) f − f¯(2)
C 3,α (T2 )
C
2 (1) (2) Φi (U (1) , U− ) − Φi (U (2) , U− ) i=1
C U (1) − U (2) + ε f (1) − f (2)
C 3,α (T2 )
2
C 2,α (Σ0 )
.
(176)
Moreover, (170) yields (notice that Uˆ = U (1) − U (2) now) (1) U − U (2) C(δ + ε + η) f (1) − f (2)
C 3,α (T2 )
2
.
(177)
So let δ, δ0 , η0 satisfy (157) and η = C1 ε; by choosing ε0 small we get ¯(1) f − f¯(2)
C 3,α (T2 )
1 (1) f − f (2) 3,α 2 . C (T ) 2
(178)
Thus by the Banach contraction mapping principle we obtain the following theorem: Theorem 9. The free-boundary problem (FB) has a unique solution f ∈ Sη , U f ∈ Oδ with η = C1 ε, δ = C0 ε if ε0 is sufficiently small. This is actually the Theorem 4, which also leads us to the conclusion of Theorem 2. We thus obtain the solvability of Problem (B). Proof of Theorem 1. The conclusion of Theorem 1 can be derived from Theorem 2. To this end we only need to show that the lateral boundary conditions
u1 = 0 u2 = 0
x0 ∈ [−1, 1], x1 = 0, d, x2 ∈ [0, d], x0 ∈ [−1, 1], x1 ∈ [0, d] x2 = 0, d
(179)
are satisfied. This can be derived from the uniqueness of the solution to Problem (B). Indeed, by extension of U (−1), g(x1 , x2 ) to properly symmetric data in the sense of Definition 2.1, and then regard them as periodic data with period (2d, 2d), Problem (D) is reduced to Problem (B). By using Theorem 2.2 we obtain a solution U defined in P. Define U # by ⎧ # u 0 (x0 , x1 , x2 ) = u 0 (x0 , −x1 , x2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u #1 (x0 , x1 , x2 ) = −u 1 (x0 , −x1 , x2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ u # (x0 , x1 , x2 ) = u 2 (x0 , −x1 , x2 ), 2 ⎪ p # (x0 , x1 , x2 ) = p(x0 , −x1 , x2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ S # (x0 , x1 , x2 ) = S(x0 , −x1 , x2 ), ⎪ ⎪ ⎪ ⎪ # ⎩ f (x1 , x2 ) = f (−x1 , x2 ),
(180)
Transonic Shocks in Compressible Flow Passing a Duct
and define U by
⎧ u 0 (x0 , x1 , x2 ) = u 0 (x0 , x1 , −x2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ u 1 (x0 , x1 , x2 ) = u 1 (x0 , x1 , −x2 ), ⎪ ⎪ ⎪ ⎪ ⎨ u (x , x , x ) = −u (x , x , −x ), 2
0
1
2
2
0
1
2
⎪ ⎪ p (x0 , x1 , x2 ) = p(x0 , x1 , −x2 ), ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ S (x0 , x1 , x2 ) = S(x0 , x1 , −x2 ), ⎪ ⎪ ⎪ ⎩ f (x1 , x2 ) = f (x1 , −x2 ).
553
(181)
Then one may directly check that (U # , f # ) and (U , f ) also solve the Euler system and satisfy all boundary conditions in P. Thus all (U, f ), (U # , f # ) and (U , f ) are solutions to Problem (B). By uniqueness, they must be the same. This fact indicates (179). Theorem 1 is thus proved.
8. Appendix: Estimates of nonlinear terms In this appendix we provide the estimates needed in Sections 6 and 7. 1. It is straightforward to verify that f i (U, U− ) with 1 i 4 are good small terms in the sense of Definition 3.2. For instance, we estimate f 4 (U, U− ) with expression (82). For small ε0 , η0 , δ0 , 0 < C Φ0 (U, U− )C 3,α (Σ0 ) C, ∇ y Φ0 (U (0, y1 , y2 ), U− (0, y1 , y2 )) 2,α C(δ + ε), C (Σ ) 0
2
Φi (U, U− )C 3,α (Σ0 ) C(δ + ε),
(182) (183) (184)
i=1
thus Φ1 ∂2 Φ0 /Φ0 − Φ2 ∂1 Φ0 /Φ0 C 2,α (Σ0 ) C(δ + ε)2 C(δ 2 + ε), 2 [ρu 1 + p]∂1 (ρu 0 )u 2 2,α C (Σ0 ) 2 ∂1 (ρu 0 )C 2,α (Σ0 ) u 2 C 2,α (Σ0 ) Cδ 2 , (185) [ρu 1 + p] 2,α C
(Σ0 )
Other terms can be treated in the same way. Then we have f 4 (U, U− )C 2,α (Σ0 ) C(δ 2 + ε). Similarly, we can establish the estimates f 1 (U, f )C 1,α (Ω) C(δ + η)δ, f 2 (U, f )C 2,α (Ω) C(δ + η)δ, f 3 (U, f )C 2,α (Σ0 ) Cδ 2 .
(186)
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Shuxing Chen & Hairong Yuan
2. To estimate h j (U, U− ) with j = 0, 3, 4 we use (75). For i = 1, 2, 3, gi (U, U− ) = (∇+ G i (Ub+ , Ub− ) − ∇+ G i (θUb+ + (1 − θ )U, Ub− ))
· (U − Ub+ ) + ∇− G i (U, ϑUb− + (1 − ϑ)U− ) · (Ub− − U− ), (187)
where θ, ϑ ∈ (0, 1), and ∇− (G i (U, U− )) is the gradient of G i (U, U− ) with respect to U− . Thus for j = 0, 3, 4 h j (U, U− )
C 3,α (Σ0 )
C
3
gi (U, U− )C 3,α (Σ0 )
i=1
2 C(U − Ub+ C 3,α (Σ ) + U− − Ub− C 3,α (Σ ) ) 0
0
C(δ 2 + ε).
(188)
3. For f (i) ∈ Sη (i = 1, 2), we denote U−(i) (0, y1 , y2 ) = U− ( f (i) (x1 , x2 ), x1 , x2 )
(189) (i)
and estimate the difference of corresponding nonlinear functions. In view of U− ∈ Nε we have for some θ ∈ (0, 1) (1) (2) C ∂x0 U− (θ f (1) + (1 − θ ) f (2) , x1 , x2 ) 2,α U− − U− 2,α C (Σ0 ) C (P) (1) (2) · f − f 2,α 2 C (T ) (1) (2) Cε f − f 2,α 2 . (190) C
(T )
4. For the operator D defined by (112), if wC 2,α (Ω) ¯ Cδ, then (D(U (1) , f (1) ) − D(U (2) , f (2) ))w 1,α C (Ω) (1) Cδ U − U (2) + f (1) − f (2) 3,α 2
C
(T2 )
.
(191)
Besides, for j = 1, 2, 3 (1)
(2)
g j (U (1) , U− ) − g j (U (2) , U− )
(1) (1) = ∇+ G j (Ub+ , Ub− ) · (U (1) − U (2) ) − G j (U (1) , U− ) − G j (U (2) , U− ) − G j (U (2) , U−(1) ) − G j (U (2) , U−(2) ) (1) = ∇+ G j (Ub+ , Ub− ) − ∇+ G j (θU (1) + (1 − θ )U (2) , U− ) · (U (1) − U (2) ) (1)
(2)
(1)
(2)
−∇− G j (U (2) , ϑU− + (1 − ϑ)U− ) · (U− − U− ),
(192)
Transonic Shocks in Compressible Flow Passing a Duct
with θ, ϑ ∈ (0, 1), hence (75) and (78) imply that (1) (2) h k (U (1) , U− ) − h k (U (2) , U− ) 2,α C (Σ0 ) C(ε + δ) U (1) − U (2) + f (1) − f (2)
555
(193)
C 3,α (T2 )
2
for k = 0, 3, 4. 5. The following estimates can also be verified by straightforward calculations: (1) (2) , f i (U (1) , U− ) − f i (U (2) , U− ) α C (Ω) C(η + δ) U (1) − U (2) + f (1) − f (2) 3,α 2 , i = 1, 2 (194) 2 C (T ) (1) (2) (1) (2) f 3 (U , U− ) − f 3 (U , U− ) 1,α C (Σ0 ) (1) (1) (2) (2) (195) Cδ U − U + f − f 3,α 2 , 2 C (T ) (1) (2) Φi (U (1) , U− ) − Φi (U (2) , U− ) 2,α C (Σ0 ) (1) (1) (2) (2) C U − U + U− − U2 2,α 2 , 2 C (T ) (1) (2) (1) (2) (196) C U − U + ε f − f 3,α 2 , i = 0, 1, 2. 2
C
(T )
Combining with the estimates (182)–(185) we have (1) (2) f 4 (U (1) , U− ) − f 4 (U (2) , U− ) 1,α C (Σ0 ) (1) C(ε + δ) U − U (2) + f (1) − f (2) 2
C 3,α (T2 )
.
(197)
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[email protected] (Received November 19, 2006 / Accepted December 5, 2006) Published online November 3, 2007 – © Springer-Verlag (2007)
