Special Issues

Continuous subsonic-sonic flows in a two-dimensional semi-infinitely long nozzle

  • This paper focuses on two-dimensional continuous subsonic-sonic potential flows in a semi-infinitely long nozzle with a straight lower wall and an upper wall which is convergent at the outlet while straight at the far fields. It is proved that if the variation rate of the cross section of the nozzle is suitably small, there exists a unique continuous subsonic-sonic flows in the nozzle such that the sonic curve intersects the upper wall at a fixed point and the velocity of the flow is along the normal direction at the sonic curve. Furthermore, the sonic curve is free, where the flow is singular in the sense that the flow speed is only Hölder continuous and the flow acceleration blows up. Additionally, the asymptotic behaviors of the flow speed at the far fields is shown.

    Citation: Mingjun Zhou, Jingxue Yin. Continuous subsonic-sonic flows in a two-dimensional semi-infinitely long nozzle[J]. Electronic Research Archive, 2021, 29(3): 2417-2444. doi: 10.3934/era.2020122

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  • This paper focuses on two-dimensional continuous subsonic-sonic potential flows in a semi-infinitely long nozzle with a straight lower wall and an upper wall which is convergent at the outlet while straight at the far fields. It is proved that if the variation rate of the cross section of the nozzle is suitably small, there exists a unique continuous subsonic-sonic flows in the nozzle such that the sonic curve intersects the upper wall at a fixed point and the velocity of the flow is along the normal direction at the sonic curve. Furthermore, the sonic curve is free, where the flow is singular in the sense that the flow speed is only Hölder continuous and the flow acceleration blows up. Additionally, the asymptotic behaviors of the flow speed at the far fields is shown.



    The Euler system

    {x(ρu)+y(ρv)=0,x(p+ρu2)+y(ρuv)=0,x(ρuv)+y(p+ρv2)=0 (1)

    is usually used to describe the two-dimensional steady isentropic inviscid compressible flow, where (u,v), p and ρ represent the velocity, pressure and density of the flow, respectively, and p(ρ)=ργ/γ for a polytropic gas with the adiabatic exponent γ>1 after the nondimensionalization. Suppose that the flow is irrotational, i.e.,

    uy=vx. (2)

    Then the density ρ can be formulated as a function of the flow speed q=u2+v2 according to the Bernoulli law ([2]):

    ρ(q2)=(1γ12q2)1/(γ1),0<q<2/(γ1). (3)

    The sound speed c is defined as c2=p(ρ). At the sonic state, the flow speed is c=2/(γ+1), which is critical in the sense that the flow is subsonic when q<c, sonic when q=c, and supersonic when q>c. The system (1), (2) can be transformed into the full potential equation

    div(ρ(|φ|2)φ)=0, (4)

    where φ is the velocity potential with φ=(u,v), ρ is the function given by (3). It is noted that (4) is elliptic in the subsonic region, degenerate at the sonic state, while hyperbolic in the supersonic region.

    Subsonic-sonic flow is one of the most interesting aspects in the mathematical theory of compressible flows. The related problems are usually raised in physical experiments and engineering designs, and there are a lot of numerical simulations and rigorous theory involved in this field (see, e.g., [2,8,15]). Two kinds of subsonic-sonic flows have been intensively studied for decades: the flow past a profile and the flow in a nozzle. The outstanding work [1] by L. Bers proved that there exists a unique two-dimensional subsonic potential flow past a profile provided that the freestream Mach number is less than a critical value and the maximum flow speed tends to the sound speed as the freestream Mach number tends to the critical value. Later, the similar results for multi-dimensional cases were established in [13,9] by G. Dong, R. Finn and D. Gilbarg. These three works did not cover the flow with the critical freestream Mach number. It was shown in [3] based on a compensated compactness framework that the two-dimensional flow with sonic points past a profile may be realized as the weak limit of a sequence of strictly subsonic flows. However, all the subsonic-sonic flows above are obtained in the weak sense and their smoothness and uniqueness are unknown yet, so are the subsonic-sonic flows in an infinitely long nozzle. For a two-dimensional infinitely long nozzle, C. Xie et al. ([22]) proved that there exists a critical value such that a strictly subsonic flow exists uniquely as long as the incoming mass flux is less than the critical value, and a subsonic-sonic flow exists as the weak limit of a sequence of strictly subsonic flows. The multi-dimensional cases were investigated in [24,12,14]. A typical subsonic-sonic flow with precise regularity is a radially symmetric subsonic-sonic flow in a convergent straight nozzle. The structural stability was initially proved in [20] for the case of two-dimensional finitely long nozzle, and some new results can be found in [16,17,18,21,19]. In the recent decade, there are also some studies on rotational subsonic and subsonic-sonic flows, see [4,6,11,7,5,23] and the references therein.

    In the present paper, we would like to investigate the subsonic-sonic flow in a class of semi-infinitely long nozzles. Assume precisely that l0, l1>0 and α(0,1) are constants, and fC2,α((,0]) satisfies

    f(0)<f(0)=0,(x)1/2fL((l0,0]), (5)
    f(x)>0 for x(,0),f(x)=0 for x(,l0]. (6)

    The upper and lower wall of the nozzle are described as

    Γup:y=fk(x)(x(,0]),andΓlow:y=l1(xR),

    respectively, where k(0,1] and

    fk(x)=kf(x),x(,0].

    The sonic curve of the flow is a free boundary intersecting the upper wall at the origin, which is chosen as the outlet of the nozzle and is denoted by

    Γout:x=S(y),y[l1,0],S(0)=0.

    It is assumed further that the subsonic-sonic flow satisfies the slip conditions at Γup and Γlow, and its velocity is along the normal direction at Γout. See the following figure for an intuition.

    As in [18,21], the subsonic-sonic flow problem can be formulated in the physical plane as

    div(ρ(|φ|2)φ)=0,(x,y)Ωk, (7)
    φy(x,l1)=0,x(,S(l1)), (8)
    φy(x,fk(x))fk(x)φx(x,fk(x))=0,x(,0), (9)
    |φ(S(y),y)|=c,φ(S(y),y)=0,y(l1,0), (10)

    where (φ,S) is a solution and Ωk is the semi-infinitely long nozzle bounded by Γup, Γlow and Γout. The problem (7)–(10) is a free boundary problem of a quasilinear degenerate elliptic equation in an unbounded domain, whose degeneracy occurs at the free boundary and is characteristic. As mentioned in Remark 1.6 of [22], one can not require in advance that the flow tends to be uniformly subsonic at the far fields, otherwise, the elliptic problem may be overdetermined. In the paper, we prove that the subsonic-sonic flow in the nozzle is uniformly subsonic at the far fields, and the uniqueness of the flow results from this property. Similar to [20,18,16,17,21], we still solve the problem in the potential plane for the reason that the shape of the sonic curve is unknown in the physical plane while known in the potential plane, and the estimates of the flow speed can be made conveniently. In the potential plane, the subsonic-sonic flow problem (7)–(10) can be transformed into a quasilinear degenerate elliptic problem with free parameters and nonlocal boundary conditions in unbounded domain. The unboundedness of the domain makes the problem more difficulty than the ones in [20,18,16,17,21]. The Schauder fixed point theorem is employed to prove the existence of subsonic-sonic flows. For a given incoming mass flux and flow speed at the upper wall, we solve a fixed boundary problem of a quasilinear degenerate elliptic equation. If the solved incoming mass flux and flow speed at the upper wall are just the given ones, we get the solution. Note that the problem we concerns is in unbounded domain, we get the solution to the fixed boundary problem by taking limits of the sequences of the solutions to the truncated problems. Like that in [21], it seems very hard to construct appropriate super and sub solutions to prove the existence of solutions to truncated problems without sufficiently small (x)1/2fL((l0,0)). The method in [21] is used here: we first solve every regularized truncated problem when the flow speed at the outlet is suitable small and get priori estimates for the average and the derivatives of the solution, then we show the existence of the solution to the regularized truncated problem by use of the preliminaries obtained above, and finally we prove that their limit as the flow speed tends to be sonic at the outlet is a desired solution to the truncated problem. The difficulty here is that in order to get the solution to the fixed boundary problem by taking the limit of the solutions to the truncated problems, we must seek a suitable variation rate k0 such that the solutions to all the truncated problems exit provided that k(0,k0]. We overcome this difficulty by constructing complicated super and sub solutions to all the truncated problems. The Harnack's inequality is used to achieve the regularities and the asymptotic behaviors of the solution to the fixed boundary problem. As to the uniqueness of the subsonic-sonic flow, we first fix the free boundaries into fixed ones and transform the nonlocal boundary conditions into common ones by a proper coordinates transformation, and then we estiblish the uniqueness theorem by the energy estimates. Summing up, it is proved in this paper that if f satisfies (5) and (6), then there exists a unique subsonic-sonic flow to the problem (7)–(10) for suitably small k, and the flow speed is only C1/2 Hölder continuous and the flow acceleration blows up at the sonic curve. Furthermore, the flow is uniformly subsonic at the far fields.

    The paper is arranged as follows. In Section 2, we formulate the subsonic-sonic flow problem (7)–(10) in the potential plane. Then in Section 3, we solve the fixed boundary problem of a quasilinear degenerate elliptic equation in an unbounded domain. Finally in Section 4, we establish the well-posedness of the subsonic-sonic flow, and prove that the flow is uniformly subsonic at the far fields.

    Define a velocity potential φ and a stream function ψ, respectively, by

    φx=u=qcosθ,φy=v=qsinθ,ψx=ρv=ρqsinθ,ψy=ρu=ρqcosθ, (11)

    where θ is the flow angle. The system (1), (2) can be reduced to the Chaplygin equations ([2]):

    θψ+ρ(q2)+2q2ρ(q2)qρ2(q2)qφ=0,1qqψ1ρ(q2)θφ=0 (12)

    in the potential-stream coordinates (φ,ψ). The coordinates transformation (11) between the two coordinate systems are valid at least in the absence of stagnation points. Eliminating θ from (12) yields the following quasilinear equation of second order

    2A(q)φ2+2B(q)ψ2=0,

    where

    A(q)=qcρ(s2)+2s2ρ(s2)sρ2(s2)ds,B(q)=qcρ(s2)sds,0<q<2/(γ1).

    It is obvious that B() is strictly increasing in (0,2/(γ1)), while A() is strictly increasing in (0,c] and strictly decreasing in [c,2/(γ1)). It follows from [21] that there exist two constants 0<N1<N2 depending only on γ such that for c/6qc,

    N1(cq)A(q)N2(cq),N1B(q),A(q),B(q)N2, (13)
    N1(cq)E(B(q))N2(cq),N2E(B(q)),E(B(q))N1, (14)

    where E=AB1 and B1 is the inverse function of B. We use A1() to denote the inverse function of A()|(0,c] in this paper. Additionally, the flow angle at the upper and the lower wall are

    Θup(x)=arctanfk(x),x[l0,0]andΘlow(x)0,x(,0),

    respectively.

    As in [18,21], in order to describe the problem in the potential plane, we denote the flow speed at the upper wall by

    Qup(x)=q(x,fk(x)),x(,0],

    then the potential function at the upper wall is expressed by

    Φup(x)=x0Qup(s)(1+(fk(s))2)1/2ds={x0Qup(s)(1+(fk(s))2)1/2ds,if x[l0,0],ζ0+xl0Qup(s)ds,if x(,l0) (15)

    with

    ζ0=l00Qup(s)(1+(fk(s))2)1/2ds.

    The inverse function of Φup is denoted by Xup. The subsonic-sonic flow problem (7)–(10) can be formulated in the potential plane as follows:

    2A(q)φ2(φ,ψ)+2B(q)ψ2(φ,ψ)=0,(φ,ψ)(,0)×(0,m), (16)
    qψ(φ,0)=0,φ(,0), (17)
    B(q)ψ(φ,m)=fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ),φ(,0), (18)
    q(0,ψ)=c,ψ(0,m), (19)
    Qup(x)=q(φ,m)|φ=Φup(x),x(,0], (20)

    where (q,m) is a solution with m>0 being the incoming mass flux. Solutions to the problem (16)–(19) are defined as follows.

    Definition 2.1. For m>0, a function qL((,0)×(0,m)) is called a solution to the fixed boundary problem (16)–(19), if

    0<inf(,0)×(0,m)qsup(,0)×(0,m)qc

    such that the integral equation

    0m0(A(q(φ,ψ))2ξφ2(φ,ψ)+B(q(φ,ψ))2ξψ2(φ,ψ))dψdφ+0fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ)ξ(φ,m)dφ=0

    holds for any ξC2((,0)×[0,m]) which vanishes for large |φ| with

    ξψ(,0)|(,0)=ξψ(,m)|(,0)=ξ(0,)|(0,m)=0.

    The existence of solutions to the problem (16)–(20) will be proved by a fixed point argument. Give m and Qup in advance as follows:

    δ1mδ2 (21)

    with

    δ1=cρ(c2/4)l12,δ2=cρ(c2)(l1+f(l0)),

    while QupC1/4((,0]) satisfies

    max{c2,ck1/4}Qup(x)c for x(,0],[Qup]C1/4((,0])1. (22)

    For such Qup, it is clear that Φup and Xup are well determined. Direct calculations yield that

    δ4ζ0δ3,c2Φup(x)δ5,x(,0], (23)
    |fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ)|kδ6(φ)1/2χ[ζ0,0](φ),φ(,0], (24)

    where χ[ζ0,0](φ) is the characteristic function of the interval [ζ0,0], and

    δ3=cl02,δ4=cl0(1+f2L((l0,0)))1/2,δ5=c(1+f2L((l0,0)))1/2,δ6=(x)1/2fL((l0,0))(2c)3/2.

    For fC2,α((,0]) satisfying (5) and (6), it follows from [21] that there exists a constant ˜l0(0,l0) depending only on f(0) and (x)1/2fL(l0,0) such that 2f(0)f(x)f(0)/2 for x[˜l0,0], and hence there exist two constants 0<τ1τ2 depending only on ˜l0, l0, f(0), inf(l0,˜l0)f and sup(l0,˜l0)f such that

    τ1xf(x)τ2x,x[l0,0]. (25)

    In this section, we deal with the well-posedness of the fixed boundary problem. For the given m and QupC1/4((,0]) satisfy (21) and (22), respectively, we solve the degenerate elliptic problem (16)–(19). Since the problem is in an unbounded domain, we first deal with the truncated problem in [ζ0n,0]×[0,m] with any sufficient large positive integer n, and make some useful compact estimates. Then we solve the problem (16)–(19) by a limit process. The key of the proof is seeking the variation rate k, which ensures the solutions to the truncated problems exist, is independent of n.

    The truncated problem is written as

    2A(qn)φ2(φ,ψ)+2B(qn)ψ2(φ,ψ)=0,(φ,ψ)(ζ0n,0)×(0,m), (26)
    A(qn)φ(ζ0n,ψ)=0,ψ(0,m), (27)
    qnψ(φ,0)=0,φ(ζ0n,0), (28)
    B(qn)ψ(φ,m)=fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ),φ(ζ0n,0), (29)
    qn(0,ψ)=c,ψ(0,m). (30)

    Note that (26) is degenerate at qn=c, we replace (30) with the following boundary condition

    qn(0,ψ)=c,ψ(0,m), (31)

    where c[c/3,c) is a constant, and consider the regularized truncated problem (26)–(29), (31). Then we solve the problem (26)–(30) by a limit process.

    The proof can be divided into four steps.

    Step 1. Well-posedness of the problem (26)–(29), (31) for c[c/3,c/2].

    Lemma 3.1. Assume that n2δ4+1 and c[c/3,c/2]. There exists a constant k1(0,1] depending only on γ, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that if k(0,k1], then the problem (26)–(29), (31) admits a unique solution qn,cC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C([ζ0n,0]×[0,m]). Furthermore, qn,c satisfies

    c/6qn,c(φ,ψ)<c,(φ,ψ)[ζ0n,0]×[0,m], (32)
    qn,c(ζ0n,ψ)ck3/4,ψ[0,m]. (33)

    Proof. The uniqueness result follows from Proposition 3.2 in [20]. Set

    k1=min{(c6)4/3,(c48δ22δ34)2,(c96δ34)4,(A(c/4)A(c/6)8δ22δ34)2,(A(c/3)A(c/4)16δ34)4,(2δ1δ5/24B(5c/6)δ6)2,(2δ2δ5/24B(c/6)δ6A(c/6))2,(1δ22e2δ4)4,(3A(5c/6)4δ24e2δ4B(5c/6))2}.

    For k(0,k1], define

    ¯qn,c(φ,ψ)=23c+(k1/2ψ2+k1/4(φ2)eφ)Λ(φ),(φ,ψ)[ζ0n,0]×[0,m],q_n,c(φ,ψ)=A1(A(c/4)(k1/2ψ2+k1/4(φ2)eφ)Λ(φ)),(φ,ψ)[ζ0n,0]×[0,m],

    where

    Λ(φ)=max{0,(φ+2δ4)3},φ(,0].

    Thanks to (13), (14), (23) and (24), direct calculations show that

    c2¯qn,c(φ,ψ)5c6,c6q_n,c(φ,ψ)c3,(φ,ψ)[ζ0n,0]×[0,m],A(ˉqn,c)φ(ζ0n,ψ)=A(q_n,c)φ(ζ0n,ψ)=0,ψ(0,m),ˉqn,cψ(φ,0)=q_n,cψ(φ,0)=0,φ(ζ0n,0),
    B(ˉqn,c)ψ(φ,m)=2k1/2mB(¯qn,c(φ,m))Λ(φ)2k1/2δ2δ34B(5c/6)χ[ζ0,0](φ)kδ6(φ)1/2χ[ζ0,0](φ),φ(ζ0n,0),B(ˉqn,c)ψ(φ,m)=2k1/2mB(q_n,c(φ,m))A(q_n,c(φ,m))Λ(φ)2k1/2δ2δ34B(c/6)A(c/6)χ[ζ0,0](φ)kδ6(φ)1/2χ[ζ0,0](φ),φ(ζ0n,0),
    2A(ˉqn,c)φ2(φ,ψ)+2B(ˉqn,c)ψ2(φ,ψ)B(¯qn,c(φ,ψ))(A(¯qn,c(φ,ψ))B(¯qn,c(φ,ψ))2ˉqn,cφ2(φ,ψ)+2ˉqn,cψ2(φ,ψ))2k1/4B(¯qn,c(φ,ψ))(φ+2δ4)×(A(5c/6)B(5c/6)(3k1/4δ226e2δ4)+4k1/4δ24)χ[2δ4,0](φ)2k1/4B(¯qn,c(φ,ψ))(φ+2δ4)×(3e2δ4A(5c/6)B(5c/6)+4k1/4δ24)χ[2δ4,0](φ)0,(φ,ψ)(ζ0n,0)×(0,m),

    and

    2A(q_n,c)φ2(φ,ψ)+2B(q_n,c)ψ2(φ,ψ)2A(q_n,c)φ2(φ,ψ)+B(q_n,c(φ,ψ))A(q_n,c(φ,ψ))2A(q_n,c)ψ2(φ,ψ)2k1/4(φ+2δ4)(6e2δ43k1/4δ224k1/4δ24B(c/3)A(c/3))χ[2δ4,0](φ)2k1/4(φ+2δ4)(3e2δ44k1/4δ24B(c/3)A(c/3))χ[2δ4,0](φ)0,(φ,ψ)(ζ0n,0)×(0,m),

    where χ[2δ4,0](φ) is the characteristic function of the interval [2δ4,0]. Therefore, ¯qn,c and q_n,c are a supersolution and a subsolution to the problem (26)–(29), (31), respectively. Thanks to the comparison principle (Proposition 3.2 in [20]) and a standard argument in the classical theory for elliptic equations, one can complete the lemma.

    Step 2. A priori estimates of the average of solutions to the problem (26)–(29), (31).

    Lemma 3.2. Assume that n2δ4+1, c[c/3,c) and qn,cC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C([ζ0n,0)×[0,m]) is a solution to the problem (26)–(29), (31). Then

    1mm0A(qn,c(φ,ψ))dψ=1mm0A(qn,c(ζ0,ψ))dψ,φ[ζ0n,ζ0]. (34)

    Furthermore, there exist three constants k2(0,1] and 0<σ1σ2 depending only on γ, τ1, τ2 and fL((l0,0)) such that if k(0,k2], then

    A(c)kσ2min{φ,ζ0}1mm0A(qn,c(φ,ψ))dψA(c)kσ1min{φ,ζ0},φ[ζ0n,0]. (35)

    Proof. The proof is similar to the proof of Lemma 3.2 in [21]. Integrating (26) over (0,m) with respect to ψ and using (28) and (29) show that

    d2dφ2m0A(qn,c(φ,ψ))dψ=fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ),φ(ζ0n,0). (36)

    And (27) yields that

    ddφm0A(qn,c(ζ0n,ψ))dψ=0. (37)

    One gets from (6), (36) and (37) that

    ddφm0A(qn,c(φ,ψ))dψ=0,φ[ζ0n,ζ0], (38)

    and

    ddφm0A(qn,c(φ,ψ))dψ=φζ0fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(s)ds=Xup(φ)l0fk(x)Φup(x)(1+(fk(x))2)3/2Qup(x)dx=Xup(φ)l0(arctanfk(x))dx=arctanfk(Xup(φ)),φ[ζ0,0]. (39)

    Thus (34) follows from (38). As in the proof of Lemma 3.2 in [21], it follows from (15) and (39) that

    1mm0A(qn,c(φ,ψ))dψ=1mm0A(qn,c(0,ψ))dψ+1m0φarctanfk(Xup(˜φ))d˜φ=A(c)+1m0φarctanfk(Xup(˜φ))d˜φ=A(c)kcf(Xup(φ))+O(k5/4),φ[ζ0,0], (40)

    where O() depend only on fL((l0,0)). Using (25), (34) and (40), we can obtain (35).

    Step 3. A priori derivative estimates of solutions to the problem (26)–(29), (31).

    Lemma 3.3. Assume that n2δ4+1, c[c/3,c), and qn,cC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C([ζ0n,0)×[0,m]) is a solution to the problem (26)–(29), (31) satisfying (32) and (33). Then for k(0,1],

    |qn,cψ(φ,ψ)|kσ3(min{φ,ζ0})1/2,(φ,ψ)(ζ0n,0)×(0,m), (41)
    |A(qn,c(φ1,ψ1))A(qn,c(φ2,ψ2))|kσ4(|φ1φ2|1/2+|ψ1ψ2|),(φ1,ψ1),(φ2,ψ2)[ζ0n,0]×[0,m], (42)

    where σ3 and σ4 are positive constants depending only on γ, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)).

    Proof. The proof is similar to Proposition 3.2 in [20]. Set

    z(φ,ψ)=B(qn,c)ψ(φ,ψ),(φ,ψ)[ζ0n,0]×[0,m].

    Then zC((ζ0n,0)×(0,m))C([ζ0n,0]×[0,m]) solves the problem

    j1(φ,ψ)2zφ2+2zψ2+j2(φ,ψ)zφ+j3(φ,ψ)zψ+j4(φ,ψ)z=0,(φ,ψ)(ζ0n,0)×(0,m), (43)
    zφ(ζ0n,ψ)=0,ψ(0,m), (44)
    z(φ,0)=0,φ(ζ0n,0), (45)
    z(φ,m)=fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ),φ(ζ0n,0), (46)
    z(0,ψ)=0,ψ(0,m), (47)

    where jiC((ζ0,0)×(0,m))(1i4) are defined by

    j1=E(B(qn,c))>0,j2=E(B(qn,c))E(B(qn,c))A(qn,c)φ,j3=E(B(qn,c))E(B(qn,c))B(qn,c)ψ,j4=(E(B(qn,c))(E(B(qn,c)))2(E(B(qn,c)))2(E(B(qn,c)))3)(A(qn,c)φ)2(E(B(qn,c)))2(E(B(qn,c)))3(A(qn,c)φ)20

    and E satisfies (14). It is clear that

    14j1(φ,ψ)(φ)3/2j4(φ,ψ)(φ)1/2j1(φ,ψ)j4(φ,ψ)(φ)1/212j2(φ,ψ)(φ)1/2,(φ,ψ)(ζ0n,0)×(0,m).

    Due to (24), one can show that

    z±(φ,ψ)=±kδ6(φ)1/2,(φ,ψ)[ζ0n,0]×[0,m]

    are a supersolution and a subsolution to the problem (43)–(47), respectively. The comparison principle (Proposition 3.2 in [20]) implies that

    |z(φ,ψ)|kδ6(φ)1/2,(φ,ψ)[ζ0n,0]×[0,m]. (48)

    Define

    ˜z±(φ,ψ)=±kδ6(ζ0)1/2,(φ,ψ)[ζ0n,ζ0]×[0,m].

    It is easy to verify that ˜z± are a supersolution and subsolution to the following problem

    j1(φ,ψ)2zφ2+2zψ2+j2(φ,ψ)zφ+j3(φ,ψ)zψ+j4(φ,ψ)z=0,(φ,ψ)(ζ0n,ζ0)×(0,m),zφ(ζ0n,ψ)=0,ψ(0,m),z(φ,0)=0,φ(ζ0n,ζ0),z(φ,m)=0,φ(ζ0n,ζ0),z(ζ0,ψ)=z(ζ0,ψ),ψ(0,m),

    respectively. The comparison principle shows that

    |z(φ,ψ)|kδ6(ζ0)1/2,(φ,ψ)[ζ0n,ζ0]×[0,m],

    which, together with (48), leads to (41). Finally, (42) can be proved in the same way as the proof of Proposition 3.2 in [20].

    Step 4. Well-posedness of the truncated problem (26)–(30).

    Lemma 3.4. Assume that n2δ4+1. There exists a constant 0<k3min{k1,k2} depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that if k(0,k3], then the problem (26)–(30) admits a unique solution qnC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C([ζ0n,0]×[0,m]) satisfies

    |qnψ(φ,ψ)|kσ3(min{φ,ζ0})1/2,(φ,ψ)(ζ0n,0)×(0,m), (49)
    |A(qn(φ1,ψ1))A(qn(φ2,ψ2))|kσ4(|φ1φ2|1/2+|ψ1ψ2|),(φ1,ψ1),(φ2,ψ2)[ζ0n,0]×[0,m], (50)
    cσ6k1/2(min{φ,ζ0})1/2qn(φ,ψ)cσ5k1/2(min{φ,ζ0})1/2,(φ,ψ)[ζ0n,0]×[0,m], (51)

    where 0<σ5σ6 are constants depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)).

    Proof. The uniqueness result follows from Proposition 3.2 in [20]. For 0<kmin{k1,k2}, set

    Ck={c[c/3,c):the problem (26)–(29), (31) admits a solutionqn,cC((ζ0n,0)×(0,m))C1([ζ0n,0]×[0,m])with (32) and (33)}.

    It follows from Lemma 3.1 and the comparison principle (Proposition 3.2 in [20]) that Ck is a nonempty interval. Assume that cCk. For φ[ζ0n,0], thanks to cCk, (13) and (35), there exists a number ψφ(0,m) such that

    qn,c(φ,ψφ)c(kσ1N2)1/2(min{φ,ζ0})1/2,

    which, together with (41), yields

    qn,c(φ,ψ)=qn,c(φ,ψφ)+ψψφqn,cψ(φ,˜ψ)d˜ψc((σ1N2)1/2k1/2σ3δ2)k1/2(min{φ,ζ0})1/2,(φ,ψ)[ζ0n,0]×[0,m]. (52)

    Choose

    σ5=(σ14N2)1/2,k3=min{k1,k2,σ14σ23δ22N2,σ45δ2416}.

    For 0<kk3, one gets from cCk, (23) and (52) that

    c/4qn,c(φ,ψ)cσ5k1/2(min{φ,ζ0})1/2,(φ,ψ)[ζ0n,0]×[0,m], (53)
    qn,c(ζ0n,ψ)c2k3/4,ψ[0,m]. (54)

    It follows from cCk, (31) and (42) that

    |A(qn,c(φ,ψ))A(c)|kσ4(φ)1/2,ψ[0,m]. (55)

    Thanks to (53)–(55), one can prove from the comparison principle (Proposition 3.2 in [20]) and the continuous dependence of solutions to the problem (26)–(29), (31) that Ck=[c/3,c) for 0<kk3.

    Let 0<kk3. For c/3c1<c2<c, the comparison principle (Proposition 3.2 in [20]) gives

    qn,c1(φ,ψ)qn,c2(φ,ψ),(φ,ψ)[ζ0n,0]×[0,m].

    Set

    qn(φ,ψ)=limccqn,c(φ,ψ),(φ,ψ)[ζ0n,0]×[0,m].

    Due to (41), (42) and (53), it is clear that qn is a solution to the problem (26)–(30), and qn satisfies (49), (50) and the second inequality in (51). For φ[ζ0n,0], it follows from (35) and (13) that there exists a number ˜ψφ(0,m) such that

    qn(φ,˜ψφ)c(kσ2N1)1/2(min{φ,ζ0})1/2.

    This estimate above and (49) yield

    qn(φ,ψ)=qn(φ,˜ψφ)+ψ˜ψφqnψ(φ,˜ψ)d˜ψc((σ2N1)1/2+k1/2σ3δ2)k1/2(min{φ,ζ0})1/2,(φ,ψ)[ζ0n,0]×[0,m].

    Hence the first inequality in (51) holds for σ6=(σ2/N1)1/2+σ3δ2. Finally, the Schauder theory for elliptic equations shows that qnC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C([ζ0n,0]×[0,m]).

    Let us establish the existence of the solution to the problem (16)–(19).

    Proposition 1. Assume that k(0,k3], then the problem (16)–(19) admits a solution qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) satisfies

    |qψ(φ,ψ)|kσ3(min{φ,ζ0})1/2,(φ,ψ)(,0)×(0,m), (56)
    |A(q(φ1,ψ1))A(q(φ2,ψ2))|kσ4(|φ1φ2|1/2+|ψ1ψ2|),(φ1,ψ1),(φ2,ψ2)(,0]×[0,m], (57)
    cσ6k1/2(min{φ,ζ0})1/2q(φ,ψ)cσ5k1/2(min{φ,ζ0})1/2,(φ,ψ)(,0]×[0,m], (58)

    where σ3, σ4, σ5 and σ6 are given in Lemmas 3.3 and 3.4. Furthermore,

    1mm0A(q(φ,ψ))dψ=A(q),φ(,ζ0], (59)

    where

    q=A1(1mm0A(q(ζ0,ψ))dψ)[cσ6k1/2(ζ0)1/2,cσ5k1/2(ζ0)1/2]. (60)

    Proof. For any n>2δ4+1, the truncated problem (26)–(30) admits a unique solution

    qnC((ζ0n,0)×(0,m))C1([ζ0n,0)×[0,m])C1/2([ζ0n,0]×[0,m])

    satisfying (49)–(51). Therefore, there exists a subsequence of {qn} weakly star convergenting to a function q in L((,0)×(0,m)), and q satisfies (58). It is not hard to check that q is a solution to the problem (16)–(19), and q satisfies (56)–(58). Finally, the Schauder theory for elliptic equations yields that

    qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]).

    Integrating (16) over (0,m) with respect to ψ and using (6), (17) and (18) lead to that

    d2dφ2m0A(q(φ,ψ))dψ=0,φ(,ζ0),

    and then there exists some constant C such that

    ddφm0A(q(φ,ψ))dψ=C,φ(,ζ0), (61)

    which implies that

    m0A(q(φ,ψ))dψ=m0A(q(ζ0,ψ))dψ+C(φζ0),φ(,ζ0). (62)

    It follows from (57) and (62) that

    |C||φζ0|m0|A(q(φ,ψ))A(q(ζ0,ψ))|dψkσ4δ2|φζ0|1/2,φ(,ζ0),

    that is,

    |C|kσ4δ2|φζ0|1/2,φ(,ζ0). (63)

    One can get C=0 by taking φ in (63), and then (61) implies that

    1mm0A(q(φ,ψ))dψ=1mm0A(q(ζ0,ψ))dψ,φ(,ζ0].

    Therefore, (59) holds.

    The solution to the problem (16)–(19) has the following regularity and asymptotic behavior.

    Proposition 2. Assume that q is a solution to the problem (16)–(19) satisfying Proposition 1. Then qC1/2([2ζ0,0]×[0,m]) and

    |qφ(φ,ψ)|σ7k1/4(φ)1/2,(φ,ψ)[2ζ0,0)×(0,m), (64)

    where σ7 is a positive constants depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)). Moreover, it holds that

    |qφ(φ,ψ)|σ8k1/2(φ)2,|qψ(φ,ψ)|σ8k(φ)2,(φ,ψ)(,2ζ0)×(0,m), (65)

    and hence

    q(φ,ψ)qL((,ζ)×(0,m))σ9k(ζ)2,ζ(,2ζ0), (66)

    where q is given in (60), and σ8,σ9>0 depend only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)).

    Proof. Similarly to the proof of Proposition 4.1 in [18], one can prove that qC1/2([2ζ0,0]×[0,m]) and satisfies (64).

    In the remaining of the proof, we use μi(1i11) to denote a generic positive constant depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)). It follows from (59) that for any φ(,ζ0], there exists a number ψφ(0,m) such that

    q(φ,ψφ)=q,

    which, together with (56), yields

    q(φ,ψ)qL((,ζ0)×(0,m))m0qψL((,ζ0)×(0,m))dψμ1k. (67)

    Note that qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) solves

    φ(a(φ,ψ)qφ)+ψ(b(φ,ψ)qψ)=0,(φ,ψ)(,ζ0)×(0,m),qψ(φ,0)=0,φ(,ζ0),qψ(φ,m)=0,φ(,ζ0),

    where

    a(φ,ψ)=A(q(φ,ψ)),b(φ,ψ)=B(q(φ,ψ)),(φ,ψ)(,ζ0)×(0,m).

    Fix integer n2. Introducing

    {ˆφ=k1/4(φnζ0)/n,φ[4nζ0,nζ0/2],ˆψ=ψ/n,ψ[0,m],

    and setting

    ˆq(ˆφ,ˆψ)=q(nζ0+k1/4nˆφ,nˆψ)q,(ˆφ,ˆψ)[3k1/4ζ0,k1/4ζ0/2]×[0,m/n].

    One can verify that

    ˆqC((3k1/4ζ0,k1/4ζ0/2)×(0,m/n))C1([3k1/4ζ0,k1/4ζ0/2]×[0,m/n])

    solves

    ˆφ(k1/2ˆa(ˆφ,ˆψ)ˆqˆφ)+ˆψ(ˆb(ˆφ,ˆψ)ˆqˆψ)=0,(ˆφ,ˆψ)(3k1/4ζ0,k1/4ζ0/2)×(0,m/n), (68)
    ˆqˆψ(ˆφ,0)=0,ˆφ(3k1/4ζ0,k1/4ζ0/2), (69)
    ˆqˆψ(ˆφ,m/n)=0,ˆφ(3k1/4ζ0,k1/4ζ0/2), (70)

    where

    ˆa(ˆφ,ˆψ)=a(nζ0+k1/4nˆφ,nˆψ),ˆb(ˆφ,ˆψ)=b(nζ0+k1/4nˆφ,nˆψ),(ˆφ,ˆψ)[3k1/4ζ0,k1/4ζ0/2]×[0,m/n].

    Extending the problem (68)–(70) into the domain [3k1/4ζ0,k1/4ζ0/2]×[0,2m] yields

    ˇφ(k1/2ˇa(ˇφ,ˇψ)ˇqˇφ)+ˇψ(ˇb(ˇφ,ˇψ)ˇqˇψ)=0,(ˇφ,ˇψ)(3k1/4ζ0,k1/4ζ0/2)×(0,2m),ˇqˇψ(ˇφ,0)=0,ˇφ(3k1/4ζ0,k1/4ζ0/2),ˇqˇψ(ˇφ,2m)=0,ˇφ(3k1/4ζ0,k1/4ζ0/2),

    where for (ˇφ,ˇψ)[3k1/4ζ0,k1/4ζ0/2]×[(i1)m/n,im/n](1i2n),

    ˇa(ˇφ,ˇψ)={ˆa(ˇφ,ˇψ(i1)m/n),if i is odd,ˆa(ˇφ,im/nˇψ),if i is even,ˇb(ˇφ,ˇψ)={ˆb(ˇφ,ˇψ(i1)m/n),if i is odd,ˆb(ˇφ,im/nˇψ),if i is even.

    Duo to (13), (51) and (67), one gets that

    μ2k1/2ˇa(ˇφ,ˇψ)μ3k1/2,μ2ˇb(ˇφ,ˇψ)μ3,(ˇφ,ˇψ)[4k1/4,3k1/4ε/(4n)]×[0,2m],

    and

    ˇqL((3k1/4ζ0,k1/4ζ0/2)×(0,2m))μ1k.

    It follows from the Hölder continuity estimates for uniformly elliptic equations that there exists a number β(0,1) such that

    [ˇq]β;(5k1/4ζ0/2,k1/4ζ0/4)×(0,2m)μ4ˇqL((3k1/4ζ0,k1/4ζ0/2)×(0,2m))μ5k,

    which implies

    [ˇa]β;(5k1/4ζ0/2,k1/4ζ0/4)×(0,2m)μ6k,[ˇb]β;(5k1/4ζ0/2,k1/4ζ0/4)×(0,2m)μ6k.

    The Schauder estimates on uniformly elliptic equations imply that

    ˇqC1,β((2k1/4ζ0,k1/4ζ0/8)×(0,2m))μ7ˇqL((5k1/4ζ0/2,k1/4ζ0/4)×(0,2m))μ8k. (71)

    Transforming (71) into the (φ,ψ) plane, one can get that

    qφL((3nζ0,3nζ0/4)×(0,m))μ9k3/4n1,qψL((3nζ0,3nζ0/4)×(0,m))μ9kn1. (72)

    Similar to (67), we have from (72) that

    q(φ,ψ)qL((3nζ0,3nζ0/4)×(0,m))m0qnψL((3nζ0,3nζ0/4)×(0,m))dψμ10kn1. (73)

    Using (73) and the same operation on q leads to that

    qφL((2nζ0,nζ0)×(0,m))μ11k1/2n2,qψL((2nζ0,nζ0)×(0,m))μ11kn2,

    Then the arbitrariness of n2 leads to (65), and hence (66) holds.

    Remark 1. Through the similar process of the proof of Proposition 2, one can show that for any positive integer λ, it holds that

    |qφ(φ,ψ)|σ8k1λ/4(φ)λ,|qψ(φ,ψ)|σ8k(φ)λ,(φ,ψ)(,2ζ0)×(0,m)

    and

    q(φ,ψ)qL((,ζ)×(0,m))σ9k(ζ)λ,ζ(,2ζ0),

    where σ8,σ9>0 depend only on λ, γ, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)).

    The solution to the problem (16)–(19) is also unique for small k.

    Proposition 3. There exists a constant k4(0,1] depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that if k(0,k4], then the problem (16)–(19) admits at most one solution qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) satisfying (58).

    Proof. In the proof, we use νi(1i5) to denote a generic positive constant depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)). Let q(1),q(2)C((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) be two solution to the problem (16)–(19) satisfying (58). Define

    wi(φ,ψ)=A(q(i)(φ,ψ)),(φ,ψ)(,0]×[0,m],i=1,2.

    Then wi(i=1,2) solves

    2wiφ2+2B(A1(wi))ψ2=0,(φ,ψ)(,0)×(0,m),wiψ(φ,0)=0,φ(,0),B(A1(wi))ψ(φ,m)=fk(x)(1+(fk(x))2)3/2Qup(x)|x=Xup(φ),φ(,0),wi(0,ψ)=0,ψ(0,m).

    Set

    w(φ,ψ)=w1(φ,ψ)w2(φ,ψ),(φ,ψ)(,0]×[0,m].

    It is easy to show that w solves

    2wφ2+2ψ2(h(φ,ψ)w)=0,(φ,ψ)(,0)×(0,m), (74)
    wψ(φ,0)=0,φ(,0), (75)
    (hw)ψ(φ,m)=0,φ(,0), (76)
    w(0,ψ)=0,ψ(0,m), (77)

    where

    h(φ,ψ)=10B(A1(ηw1(φ,ψ)+(1η)w2((φ,ψ))))A(A1(ηw1(φ,ψ)+(1η)w2((φ,ψ))))dη,(φ,ψ)(,0)×(0,m).

    Thanks to (56), (58), (64) and (65), direct calculations yield

    ν1k1/2φ1/2h(φ,ψ)ν1k1/2φ1/2,(φ,ψ)(,0)×(0,m), (78)
    |hψ(φ,ψ)|{ν2(φ)1/2,(φ,ψ)[2ζ0,0)×(0,m),ν2(φ)2,(φ,ψ)(,2ζ0)×(0,m), (79)
    |wφ(φ,ψ)|ν2k(φ)2,(φ,ψ)(,2ζ0)×(0,m), (80)

    where φ=min{φ,2ζ0}. Fix ζ<2ζ01. Multiplying (74) by w, then integrating over (ζ,0)×(0,m) by parts and using (75)–(77), we have

    0ζm0(wφ)2dψdφ+0ζm0h(φ,ψ)(wψ)2dψdφ=0ζm0hψ(φ,ψ)wwψdψdφm0w(ζ,ψ)wφ(ζ,ψ)dψ,

    which, together with (78)–(80), yields

    0ζm0(wφ)2dψdφ+k1/20ζm0φ1/2(wψ)2dψdφν302ζ0m0(φ)1/2|wwψ|dψdφ+ν32ζ0ζm0(φ)2|wwψ|dψdφ+ν3k(ζ)2m0|w(ζ,ψ)|dψ.

    Then the Hölder's inequality gives

    0ζm0(wφ)2dψdφ+k1/20ζm0φ1/2(wψ)2dψdφν4k1/202ζ0m0(φ)1/2w2dψdφ+ν4k1/22ζ0ζm0(φ)4w2dψdφ+ν4k(ζ)2m0|w(ζ,ψ)|dψ. (81)

    It follows from the Hölder's inequality and Cauchy inequality that

    02ζ0m0(φ)1/2w2dψdφ02ζ0m0(φ)1/2(0φwφ(s,ψ)ds)2dψdφ02ζ0(φ)1/2dφ0ζ0m0(wφ)2dφdψ(2ζ0)3/20ζm0(wφ)2dφdψ, (82)
    2ζ0ζm0(φ)4w2dψdφ2ζ0ζm0(φ)4(0φwφ(s,ψ)ds)2dψdφ2ζ0ζ(φ)3dφ0ζm0(wφ)2dψdφ(2ζ0)20ζm0(wφ)2dψdφ, (83)

    and

    m0|w(ζ,ψ)|dψm2+12m0w2(ζ,ψ)dψδ22+m0(0ζ|wφ|dφ)2dψδ22+(ζ)0ζm0(wφ)2dφdψ. (84)

    Substituting (82)–(84) into (81) to get

    0ζm0(wφ)2dψdφ+k1/20ζm0φ1/2(wψ)2dψdφν5k1/20ζm0(wφ)2dψdφ+ν5k(ζ)2+ν5k(ζ)10ζm0(wφ)2dφdψ2ν5k1/20ζm0(wφ)2dψdφ+ν5k(ζ)2. (85)

    Choose k4=1/(16ν25+1). For any k(0,k4], (85) implies

    0ζm0(wφ)2dψdφ+k1/20ζm0φ1/2(wψ)2dψdφ2ν5k1/2(ζ)2. (86)

    Taking ζ in (86) to get

    0m0(wφ)2dψdφ+k1/20m0φ1/2(wψ)2dψdφ0,

    which implies

    wφ(φ,ψ)=wψ(φ,ψ)=0,(φ,ψ)(,0)×(0,m). (87)

    It follows (77) and (87) that

    w(φ,ψ)=0,(φ,ψ)(,0]×[0,m].

    Therefore, q(1)=q(2).

    First we prove the existence of the solution to the problem (16)–(20) by a fixed point argument.

    Theorem 4.1. Assume that fC2,α([l0,0]) satisfies (5) and (6). There exists a constant k0(0,1] depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that if k(0,k0], then the problem (16)–(20) admits a solution (q,m) satisfying

    qC((,0)×(0,m))C1((,0)×[0,m])C1/2((,0]×[0,m])|qψ(φ,ψ)|kσ3(min{φ,ζ0})1/2,(φ,ψ)(,0)×(0,m), (88)
    |A(q(φ1,ψ1))A(q(φ2,ψ2))|kσ4(|φ1φ2|1/2+|ψ1ψ2|),(φ1,ψ1),(φ2,ψ2)(,0]×[0,m], (89)
    cσ6k1/2(min{φ,ζ0})1/2q(φ,ψ)cσ5k1/2(min{φ,ζ0})1/2,(φ,ψ)(,0]×[0,m], (90)

    where

    m=qρ(q2)(fk(l0)+l1),cσ6k1/2(ζ0)1/2qcσ5k1/2(ζ0)1/2, (91)

    and σ3, σ4, σ5, σ6 are given in Lemmas 3.3 and 3.4. Furthermore,

    |qφ(φ,ψ)|σ7k1/4(φ)1/2,(φ,ψ)[2ζ0,0)×(0,m), (92)

    and for any positive integer λ, it holds that

    |qφ(φ,ψ)|σ8k1λ/4(φ)λ,|qψ(φ,ψ)|σ8k(φ)λ,(φ,ψ)(,2ζ0)×(0,m), (93)

    and

    q(φ,ψ)qL((,ζ)×(0,m))σ9k(ζ)λ,ζ(,2ζ0), (94)

    where σ7, σ8 and σ9 are given in Proposition 2 and Remark 1. Therefore, the flow is uniformly subsonic at the far fields.

    Proof. Choose

    k0=min{k3,k4,c24σ26δ4,1σ46δ24,N12σ4δ1/25}.

    For k(0,k0], set

    Q={(m,Qup)[δ1,δ2]×C1/4((,0]):Qup satisfies (22)}

    with the norm

    (m,Qup)Q=max{m,QupL(,0)}.

    For a given (m,Qup)Q, it is clear that Φup, Xup and q are well determined, and it follows from Propositions 1–3 that the problem (16)–(19) admits a unique solution qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) satisfying (56)–(58) and (66). Set

    ˆm=qρ(q2)(fk(l0)+l1),ˆQup(x)=q(Φup(x),m),x(,0].

    From (56)–(58), (66) and the choice of k0, it is easy to verify that (ˆm,ˆQup)Q and

    K:QQ,(m,Qup)(ˆm,ˆQup).

    is a self-mapping. Furthermore, one can prove the compactness of K by using (56)–(58), and the continuity of K by using its compactness and the uniqueness result for the problem (16)–(19). Therefore, the Schauder fixed point theorem shows that the problem (16)–(20) admits a solution (q,m) such that qC((,0)×(0,m))C1((,0)×[0,m])C1/2((,0]×[0,m]) satisfies (88)–(94).

    From Theorem 4.1, for k(0,k0], the problem (16)–(20) admits a solution (q,m) satisfying qC((,0)×(0,m))C1((,0)×[0,m])C1/2((,0]×[0,m]),

    max{c2,cM1k1/2(min{φ,ζ0})1/2}q(φ,ψ)cM2k1/2(min{φ,ζ0})1/2,(φ,ψ)(,0)×(0,m) (95)

    and

    q(φ,ψ)qL((,ζ)×(0,m)M3k(ζ)2,ζ<2ζ0,

    where

    m=qρ(q2)(fk(l0)+l1),max{c2,cM1k1/2(ζ0)1/2}qcM2k1/2(ζ0)1/2,

    and M1, M2, M3 are positive constants. Indeed, this solution is also unique if k is suitably small.

    Theorem 4.2. Assume that fC2,α([l0,0]) satisfies (5) and (6). There exists a constant k0(0,1] depending only on γ, l0, l1, f(l0), fL((l0,0)), (x)1/2fL((l0,0)), M1 and M2, such that if k(0,k0], then there is at most one solution (q,m) to the problem (16)–(20) such that qC((,0)×(0,m))C1((,0)×[0,m])C((,0]×[0,m]) and q satisfies (95).

    Proof. In the proof, we use Ci(1i5) to denote a generic positive constant depending only on γ, l0, l1, f(l0), fL((l0,0)), (x)1/2fL((l0,0)), M1 and M2. Let (q(1),m(1)) and (q(2),m(2)) be two solutions to the problem (16)–(20) such that q(i)C((,0)×(0,m(i)))C1((,0)×[0,m(i)])C((,0]×[0,m(i)]) and satisfies (95) for i=1,2. Denote Φup,i and Xup,i to be the associated functions defined in Section 2 corresponding to q(i) for i=1,2. For i=1,2, introduce the new coordinates transformations

    {x=Xup,i(φ),φ(,0],y=ψm(i),ψ[0,m(i)],{φ=Φup,i(x),x(,0],ψ=m(i)y,y[0,1].

    Define

    Wi(x,y)=A(q(i)(Φup,i(x),m(i)y)),(x,y)(,0]×[0,1],i=1,2.

    Then Wi satisfies

    x(m(i)Xi(x)Wix)+y(1m(i)Xi(x)B(A1(Wi))y)=0,(x,y)(,0)×(0,1), (96)
    Wiy(x,0)=0,x(,0), (97)
    1m(i)Xi(x)B(A1(Wi))y(x,1)=fk(x)1+(fk(x))2,x(,0), (98)
    Wi(0,y)=0,y(0,1), (99)

    where

    Xi(x)=1(1+(fk(x))2)1/2A1(Wi(x,fk(L0))),x(,0].

    Set

    W(x,y)=W1(x,y)W2(x,y),(x,y)(,0]×[0,1].

    One can verify from that W satisfies

    x(m(1)X1(x)Wx)+y(1m(1)X1(x)H(x,y)Wy)+x(m(1)X(x)W2x)+x(mX2(x)W2x)+y(1m(1)X1(x)Zy(x,y)W)y(mm(1)m(2)X1(x)B(A1(W2))y)y(X(x)m(2)X1(x)X2(x)B(A1(W2))y)=0,(x,y)(,0)×(0,1), (100)

    where

    m=m(1)m(2),X(x)=X1(x)X2(x),x(,0],H(x,y)=10B(A1(ηW1(x,y)+(1η)W2(x,y)))A(A1(ηW1(x,y)+(1η)W2(x,y)))dη,(x,y)(,0)×(0,1).

    It follows from (13), (59), (88) and (90)–(93) that

    C1k1/2x1/2H(x,y)C2k1/2x1/2,(x,y)(,0)×(0,1), (101)
    |Hy(x,y)|{C2(x)1/2,(x,y)[L0,0)×(0,1),C2(x)2,(x,y)(,L0)×(0,1), (102)
    |Wix(x,y)|{C2k3/4,(x,y)[L0,0)×(0,1),C2k(x)2,(x,y)(,L0)×(0,1),i=1,2, (103)
    |B(A1(W2))y(x,y)|{C2k(x)1/2,(x,y)[L0,0)×(0,1),C2k(x)2,(x,y)(,L0)×(0,1), (104)
    |X(x)|{C2k1/2(x)1/2|W(x,1)|,(x,y)[L0,0)×(0,1),C2k1/2|W(x,1)|,(x,y)(,L0)×(0,1), (105)
    |m|C2(0L010(Wx)2dydx)1/2, (106)

    where

    x=min{x,L0},L0=3l0(1+f2L((l0,0)))1/2.

    Fix L>L0. Multiplying (100) by W and then integrating by parts over (L,0)×(0,1), one gets from (96)–(99) that

    0L10m(1)X1(x)(Wx)2dydx+0L101m(1)X1(x)H(x,y)(Wy)2dydx=0L10m(1)X(x)WxW2xdydx0L10mX2(x)WxW2xdydx0L101m(1)X1(x)Hy(x,y)WWydydx+0L10mm(1)m(2)X1(x)B(A1(W2))yWydydx+0L10X(x)m(2)X1(x)X2(x)B(A1(W2))yWydydx+10W(L,y)(m(1)X1(L)W1x(L,y)m(2)X2(L)W2x(L,y))dy,

    which, together with (23), (90), (101) and (103), yields

    0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydxC30L10|X(x)WxW2x|dydxJ1+C30L10|mWxW2x|dydxJ2+C30L10|Hy(x,y)WWy|dydxJ3+C30L10|mB(A1(W2))yWy|dydxJ4+C30L10|X(x)B(A1(W2))yWy|dydxJ5+C3k(L)210|W(L,y)|dyIL. (107)

    Below, let us make estimates on Ji(1i5) and IL in (107). The following five inequalities are necessary. From the Hölder's inequality and (99), it follows

    0L010(x)ϑ1W2dydx0L010(x)ϑ1(0x|Wx(s,y)|ds)2dydx0L0(x)1ϑ1dx0L010(wx)2dydxL2ϑ102ϑ10L10(Wx)2dydx,ϑ1[0,2), (108)

    and

    L0L10(x)ϑ2W2dydxL0L10(x)ϑ2(0x|Wx(s,y)|ds)2dydxL0L(x)1ϑ2dx0L10(Wx)2dydxL2ϑ20ϑ220L10(Wx)2dydx,ϑ2(2,+). (109)

    Then from the Cauchy's inequality, (108) and (109), we have

    0L0W2(x,1)dx0L010W2dydx+20L010|WWy|dydxL200L10(Wx)2dydx+k1/2L1/200L010W2dydx+k1/20L010(x)1/2(Wy)2dydx(L20+L5/20)0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydx, (110)
    0L0(x)1W2(x,1)dx0L010(x)1W2dydx+20L010(x)1|WWy|dydxL00L010(Wx)2dydx+k1/20L010(x)3/2W2dydx+k1/20L010(x)1/2(Wy)2dydx(L0+2L1/20)0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydx, (111)

    and

    L0L(x)4W2(x,1)dxL0L10(x)4W2dydx+2L0L10(x)4|WWy|dydxL200L10(Wx)2dydx+k1/2L1/20L0L10(x)8W2dydx+k1/2L1/20L0L10(Wy)2dydx(L20+L5/20)0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydx. (112)

    It follows from Cauchy's inequality with ε, (102)–(106) and (108)–(112) that

    J1ε0L10(Wx)2dydx+1ε0L10|X(x)|2|W2x|2dydxε0L10(Wx)2dydx+C22k1/2ε0L0(x)1W2(x,1)dydx+C22kεL0L(x)4W2(x,1)dydxC4(ε+k1/2ε)0L10(Wx)2dydx+C4k1/2εk1/20L10x1/2(Wy)2dydx, (113)
    J2ε0L10(Wx)2dydx+1ε0L10m2(W2x)2dydxε0L10(Wx)2dydx+C22εm2(k3/2+L0k2L0L(x)4dx)C4(ε+k1/2ε)0L10(Wx)2dydx, (114)
    J3C20L010(x)1/2|WWy|dydx+C2L0L10(x)2|WWy|dydxC2k1/2ε0L010(x)1/2W2dydx+C2L1/20k1/2εL0L10(x)4W2dydx+C2εk1/20L10x1/2(Wy)2dydxC4k1/2ε0L10(Wx)2dydx+C4εk1/20L10x1/2(Wy)2dydx, (115)
    J4C2k0L010(x)1/2|mWy|dydx+C2kL0L10(x)2|mWy|dydxC2k5/2εm2(0L0(x)3/2dx+L0L(x)4dx)+C2εk1/20L10x1/2(Wy)2dydxC4k1/2ε0L10(Wx)2dydx+C4εk1/20L10x1/2(Wy)2dydx, (116)

    and

    J5C22k1/20L010|W(x,1)Wy|dydx+C22k1/2L0L10(x)2|W(x,1)Wy|dydxC22L1/20k3/2ε0L0W2(x,1)dx+C22L1/20k3/2εL0L(x)4W2(x,1)dx+C22εk1/20L10x1/2(Wy)2dydxC4k1/2ε0L10(Wx)2dydx+C4(ε+k1/2ε)k1/20L10x1/2(Wy)2dydx, (117)

    where ε>0 is to be determined. Additionally,

    IL1+10W2(L,y)dy1+10(0L|Wx|dx)2dy1+(L)0L10(Wx)2dydx. (118)

    Substituting (113)–(118) into (107) to get

    0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydxC5(ε+k1/2ε)(0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydx)+C5(L)1+C5k0L10(Wx)2dydx. (119)

    Choose ε=(4C5)1 and k0=min{(16C25+1)1,(4C5+1)1}. For any k(0,k0], (119) implies

    0L10(Wx)2dydx+k1/20L10x1/2(Wy)2dydx2C5(L)1. (120)

    Taking L+ in (120), we obtain that

    010(Wx)2dydx+k1/2010x1/2(Wy)2dydx0,

    which shows that

    Wx(x,y)=Wy(x,y)=0,(x,y)(,0)×(0,1).

    Then W(x,y)=0 follows from (99), and hence (q(1),m(1))=(q(2),m(2)).

    In terms of the physical variables, Theorems 4.1 and 4.2 can be transformed as

    Theorem 4.3. Assume that fC2,α([l0,0]) satisfies (5) and (6). There exist four constants ˜k0(0,1] and ˜M1,˜M2>0 depending only on γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that if k(0,˜k0] then the problem (7)–(10) admits a unique solution (φ,S,m) satisfying φC3(Ωk)C2(¯ΩkS)C1(¯Ωk), SC1([l1,0]),

    max{c2,c˜M2(kdistS(x,y))1/2}|φ(x,y)|c˜M1(kdistS(x,y))1/2,(x,y)Ωk,

    where distS(x,y) is the distance from (x,y) to S and x=max{x,l0}. Moreover, for any positive integer λ, there exists a constant ˜M3>0 depending only on λ, γ, τ1, τ2, l0, l1, f(l0), fL((l0,0)) and (x)1/2fL((l0,0)), such that

    φ(x,y)qxC1(Ωk{x<R})˜M3kRλ,R>l0,

    where

    max{c2,c˜M2(kl0)1/2}qc˜M1(kl0)1/2.

    Therefore, the flow is uniformly subsonic at the far fields.



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