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Continuous subsonic–sonic flows in a general nozzle
Available online at www.sciencedirect.com
ScienceDirect J. Differential Equations 259 (2015) 2546–2575 www.elsevier.com/locate/jde
Continuous subsonic–sonic flows in a general nozzle ✩ Chunpeng Wang School of Mathematics, Jilin University, Changchun 130012, China Received 17 August 2014 Available online 29 April 2015
Abstract This paper concerns continuous subsonic–sonic potential flows in a two dimensional finite nozzle with a general upper wall and a straight lower wall. We give a class of nozzles where continuous subsonic–sonic flows may exist. Consider a continuous subsonic–sonic flow in such a nozzle after rescaling the upper wall in a small scale. It is shown that for a given inlet and a fixed point at the upper wall, there exists uniquely a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the inlet and the sonic curve, which satisfies the slip conditions on the nozzle walls and whose sonic curve intersects the upper wall at the fixed point. Furthermore, the sonic curve of this flow is a free boundary, where the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. As the scale tends to zero, the precise convergent rate of the continuous subsonic–sonic flow converging to the sonic state is also determined. © 2015 Elsevier Inc. All rights reserved. MSC: 35R35; 35Q35; 76N10; 35J70 Keywords: Continuous subsonic–sonic flow; Free boundary; Degeneracy; Singularity
1. Introduction In this paper, we study the isentropic, irrotational, steady compressible Euler subsonic–sonic flows in general nozzles, which arise in the physical experiments and the engineering designs. ✩
Supported by the National Natural Science Foundation of China (Grant No. 1107222106). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.jde.2015.03.036 0022-0396/© 2015 Elsevier Inc. All rights reserved.
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Roughly speaking, there have been mainly two kinds of problems on subsonic–sonic flows. One is subsonic–sonic flows past a profile. In the significant work [1], Bers showed that for two dimensional potential flows past a profile, the whole flow field will be subsonic outside the profile if the Mach number of the freestream is small enough; furthermore, the maximum flow speed will tend to the sound speed as the freestream Mach number increases. Bers [1] did not consider the flow with the critical freestream Mach number. Based on the compensated compactness method, it was shown in [3] that the flows with sonic points past a profile may be realized as weak limits of sequences of strictly subsonic flows. The other kind is subsonic–sonic flows in an infinite nozzle. Bers [2] asserted that there is a global subsonic potential flow through an infinite nozzle for an appropriately given incoming mass flux, which was proved rigorously in [10]. It was shown that there exists a critical value for a general infinite nozzle such that a strictly subsonic flow exists uniquely as long as the incoming mass flux is less than the critical value. Furthermore, a class of subsonic–sonic flows can be obtained as the weak limits of strictly subsonic flows associated with the incoming mass fluxes increasing to the critical value. Moreover, Xie and Xin [11] considered subsonic–sonic flows in axially symmetric nozzles. It should be noted that the subsonic–sonic flows past a profile and the ones in an infinite nozzle obtained in [3,10] are both in the weak sense and have not been proved to be smooth yet, thus the locations and properties of the sonic states are unknown for such flows. A typical subsonic–sonic flow with precise regularity is a symmetric continuous subsonic–sonic flow in a finite straight nozzle, whose structural stability was studied in the recent works [7,8]. Precisely, for the given inlet, which is a perturbation of an arc centered at the vertex of the nozzle, and the given incoming mass flux, it was shown in [8] that there exists an open interval depending only on the adiabatic exponent and the length of the arc, such that a unique continuous subsonic–sonic flow, whose velocity vector is along the normal direction at the inlet and the outlet and which satisfies the slip conditions on the nozzle walls, exists as long as the incoming mass flux belongs to this interval and the perturbation of the inlet is sufficiently small; furthermore, the sonic curve of this continuous subsonic–sonic flow is a free boundary, where the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. Furthermore, Wang and Xin [9] studied smooth transonic flows of Meyer type in finite de Laval nozzles and showed that there exists uniquely a smooth transonic flow near the throat of the nozzle, whose sonic curve is located at the throat and whose acceleration is Lipschitz continuous, if the wall of the nozzle is sufficiently flat. It is shown from [8] that the existence of continuous subsonic–sonic flows depends on the geometry of the nozzle. A natural question is how to formulate the continuous subsonic–sonic flow problem in a general nozzle. In [8], the authors prescribed the incoming flow angle and the incoming mass flux, which are two physical quantities, to formulate the continuous subsonic–sonic flow problem in a straight nozzle; furthermore, the sonic curve of the continuous subsonic–sonic flow is a free boundary where the velocity vector is along the normal direction. Clearly, for a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the sonic curve, the second derivative of the nozzle wall at the intersecting point with the sonic curve must be zero. Therefore, it is not suitable to prescribe the incoming flow angle and the incoming mass flux to formulate the continuous subsonic–sonic flow problem in a general nozzle. In the present paper, we fix the intersecting point between the sonic curve and the upper wall to replace the incoming mass flux condition. That is to say, for a general nozzle with a given inlet, we seek a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the inlet and the sonic curve, which satisfies the slip conditions on the nozzle walls and whose sonic curve intersects the upper wall at a fixed point. Furthermore, the sonic curve of the flow, which is chosen to be the outlet of the flow, is a free boundary.
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Let us formulate the continuous subsonic–sonic flow problem in a general nozzle precisely. The flow is governed by the full potential equation div(ρ(|∇ϕ|2 )∇ϕ) = 0, where ϕ is the velocity potential, ρ is the density which is expressed in terms of the speed by γ − 1 2 1/(γ −1) ρ(q 2 ) = 1 − , q 2
0
2/(γ − 1)
with γ > 1 being the adiabatic exponent for a polytropic gas. The upper and lower walls of the nozzle are assumed to be given by up : y = f (x) ≥ 0,
−l0 ≤ x ≤ 0
and
low : y = −l1 ,
x ∈ R,
respectively, where l0 , l1 > 0 and f (0) = 0. The inlet is given by in : x = g(y),
−l1 ≤ y ≤ f (−l0 ).
The outlet is chosen to be the sonic curve, which is a free boundary from the origin at the upper wall to a point at the lower wall and is denoted by out : x = S(y),
−l1 ≤ y ≤ 0,
S(0) = 0.
The continuous subsonic–sonic problem is to seek (ϕ, S, Cin ) solving div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(x, y) ∈ ,
(1.1)
ϕ(g(y), y) = Cin , ∂ϕ (x, −l1 ) = 0, ∂y ∂ϕ ∂ϕ (x, f (x)) − f (x) (x, f (x)) = 0, ∂y ∂x |∇ϕ(S(y), y)| = 2/(γ + 1), ϕ(S(y), y) = 0, |∇ϕ(x, y)| < 2/(γ + 1),
−l1 < y < f (−l0 ),
(1.2)
g(−l1 ) < x < S(−l1 ),
(1.3)
−l0 < x < 0,
(1.4)
−l1 < y < 0,
(1.5)
(x, y) ∈ ,
(1.6)
where is the domain bounded by up , low , in and out . Since the sonic curve intersects the upper wall at the origin, f should satisfy f (0) = 0 according to the above discussion. Furthermore, it is assumed that f (−l0 ) < 0, which guarantees that the flow is accelerated at the inlet. The problem (1.1)–(1.6) is a free boundary problem of a quasilinear degenerate elliptic equation. It is well-known that both degeneracies and free boundaries are difficult topics for quasilinear elliptic equations. For the problem (1.1)–(1.6), the degeneracy occurs at the free boundary and the degeneracy is characteristic [6,12]. Particularly, if f takes the form of f (x) = −l2 x,
−l0 ≤ x ≤ 0,
(l2 > 0),
(1.7)
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then the nozzle is straight and the problem, where the incoming flow angle and the incoming mass flux are prescribed, is considered in [8]. For the problem (1.1)–(1.6) with (1.7) in a straight nozzle, the same discussion as Remark 2.12 in [8] shows that the special problem is solvable only when l2 is suitably small. Therefore, the problem (1.1)–(1.6) may be unsolvable. In order to avoid the unsolvability of the problem (1.1)–(1.6), we rescale the upper wall by fk (x) = kf (x),
−l0 ≤ x ≤ 0,
(0 < k ≤ 1)
and solve the problem (1.1)–(1.6) where f is replaced by fk . For a two-dimensional nozzle, the geometry of the nozzle determines whether there is a subsonic–sonic flow whose sonic curve extends from one wall of the nozzle to the other. A sufficient condition is shown to be √ |f (x)| ≤ τ −x,
−l0 ≤ x ≤ 0,
where τ > 0 is a suitable constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Then, it is proved in the present paper that there exists k0 ∈ (0, 1], depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that for each 0 < k ≤ k0 , the problem (1.1)–(1.6) with fk instead of f admits uniquely a solution. The regularity of the continuous subsonic–sonic flow is also studied and it is shown that the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. As k → 0+ , the nozzle will tend to a flat one and thus the continuous subsonic–sonic flow will tend to the sonic state in the whole nozzle. The precise convergent rate is determined in the paper. One may expect that there exists a critical exponent kc such that the problem (1.1)–(1.6) with fk instead of f is solvable when k < kc and unsolvable when k > kc . Unfortunately, this expectation is almost impossible since the criterion of the solvability of the problem (1.1)–(1.6) with fk instead of f is completely unknown. Furthermore, it is very difficult to compare the continuous subsonic–sonic flows as k changes since the domains are variable.
It was turn out in [8] that it is more convenient to investigate a subsonic–sonic flow in the potential plane than in the physical plane. The reason lies in that the shape of the sonic curve is unknown in the physical plane, while known in the potential plane. Moreover, it is more convenient to estimate the speed of the flow in the potential plane than in the physical plane since it is a solution in the potential plane, while the absolute value of the gradient of a solution in the physical plane. For the continuous subsonic–sonic flow problem in a general nozzle, we encounter the two main difficulties in [8]. One is that the problem is a free boundary problem of a quasilinear degenerate elliptic equation and the degeneracy is characteristic and occurs just at the free
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boundary. The other is that the boundary condition at the inlet is nonlinear and nonlocal. However, besides these difficulties, there is an additional disadvantage in this study. Different from the problem in [8], there are no background solutions, which play an important role to determine and control the rate of the flow from the subsonic region to the sonic state. Furthermore, the mass flux of the flow is free and the boundary condition at the upper wall is nonlinear and nonlocal. In the present paper, we use the Schauder fixed point theorem to prove the existence of continuous subsonic–sonic flows. For a given speed at the inlet and the upper wall, we solve a boundary problem of a quasilinear degenerate equation. If the solved speed at the inlet and the upper wall is just the given one, we get a solution. The key is precise estimates and optimal regularities of solutions to quasilinear degenerate elliptic equations. Particularly, we use the Harnack inequality to achieve compactness estimates. By a rescaling technique and a precise elliptic estimate, we get the optimal regularity for continuous subsonic–sonic flows. As for the uniqueness of the continuous subsonic–sonic flow, it is still the degeneracy, the free boundaries and the nonlocal boundary conditions that bring essential difficulties. By a suitable coordinates transformation, we fix the free boundaries into fixed boundaries and transform the nonlocal boundary conditions into common boundary conditions. Then, the uniqueness theorem is established by the method of energy estimate and a series of precise and complicated calculations. The paper is arranged as follows. In Section 2, we formulate the problem of continuous subsonic–sonic flows in a general nozzle and state the main results (existence, regularity and uniqueness) of the paper. Subsequently, we prove the existence, regularity and uniqueness results in the following three sections, respectively. 2. Formulation and main results Fix l0 , l1 > 0 and α ∈ (0, 1). Assume that f ∈ C 2,α ([−l0 , 0]) satisfying f (0) = 0 and f (x) < 0,
√ |f (x)| ≤ τ −x,
−l0 ≤ x ≤ 0,
(2.1)
where τ > 0 is a suitable constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . For 0 < k ≤ 1, set fk (x) = kf (x),
−l0 ≤ x ≤ 0
and gk (y) = −l0 −
fk (−l0 ) + l1 − fk (−l0 )
Rk2 − (y + l1 )2 ,
−l1 ≤ y ≤ fk (−l0 ),
where
Rk =
(fk (−l0 ) + l1 ) 1 + (fk (−l0 ))2 −fk (−l0 )
.
Consider a continuous subsonic–sonic flow in the nozzle whose upper and lower walls are up : y = fk (x),
−l0 ≤ x ≤ 0
and
low : y = −l1 ,
x ∈ R,
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respectively. The inlet is given by in : x = gk (y),
−l1 ≤ y ≤ fk (−l0 ).
The outlet is chosen to be the sonic curve, which is a free boundary from the origin at the upper wall to a point at the lower wall and is denoted by out : x = S(y),
−l1 ≤ y ≤ 0,
S(0) = 0.
We seek a continuous subsonic–sonic flow in the domain k bounded by up , low , in and out . Furthermore, the velocity vector of the flow is along the normal direction at the inlet and the sonic curve, which is the outlet of the flow and is a free boundary, the flow satisfies the slip conditions on the nozzle walls and the sonic curve intersects the upper wall at (0, 0).
2.1. Governing equations Consider the two dimensional compressible Euler system of steady flow ∂ ∂ (ρu) + (ρv) = 0, ∂x ∂y
(2.2)
∂ ∂ (P + ρu2 ) + (ρuv) = 0, ∂x ∂y
(2.3)
∂ ∂ (ρuv) + (P + ρv 2 ) = 0, ∂x ∂y
(2.4)
where (u, v), P and ρ represent the velocity, pressure and density of the flow, respectively. The flow is assumed to be isentropic so that P = P (ρ) is a smooth function. In particular, for a polytropic gas with adiabatic exponent γ > 1, P (ρ) =
1 γ ρ γ
is the normalized pressure. Assume further that the flow is irrotational, i.e.
(2.5)
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∂u ∂v = . ∂y ∂x
(2.6)
Summing up, we consider the flow governed by the system (2.2)–(2.6). The sound speed c is defined as c2 = P (ρ) = ρ γ −1 . Define a velocity potential ϕ(x, y) and a stream function ψ(x, y), respectively, by ∂ϕ = u, ∂x
∂ϕ = v, ∂y
∂ψ = −ρv, ∂x
∂ψ = ρu, ∂y
which are ∂ϕ = q cos θ, ∂x
∂ϕ = q sin θ, ∂y
∂ψ = −ρq sin θ, ∂x
∂ψ = ρq cos θ ∂y
in terms of polar coordinates in the velocity space, where q is the speed, while θ , which is called a flow angle, is the angle of the velocity inclination to the x-axis. Then, the system (2.2)–(2.6) is transformed into the full potential equation div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(2.7)
where ρ is expressed in terms of the speed q according to the Bernoulli law [4] γ − 1 2 1/(γ −1) ρ(q 2 ) = 1 − , 0 < q < 2/(γ − 1). q 2 √ At the sonic state, the sound speed is c∗ = 2/(γ + 1), which is also the critical speed in the sense that the flow is subsonic when q < c∗ , sonic when q = c∗ and supersonic when q > c∗ . By a standard process, (2.7) may be reduced to ∂θ ∂q = −A (q) , ∂ψ ∂ϕ
∂q ∂θ = B (q) ∂ϕ ∂ψ
and ∂ 2 A(q) ∂ 2 B(q) + =0 ∂ϕ 2 ∂ψ 2
(2.8)
in the potential-stream coordinates (ϕ, ψ), where q A(q) = c∗
ρ(s 2 ) + 2s 2 ρ (s 2 ) ds, sρ 2 (s 2 )
q B(q) =
ρ(s 2 ) ds, s
0
2/(γ − 1);
c∗
furthermore, A(·) and B(·) are both the strictly increasing functions in (0, c∗ ]. Note that, since the Jacobian
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∂(ϕ, ψ) ∂ϕ ∂ψ ∂ϕ ∂ψ = − = ρq 2 ∂(x, y) ∂x ∂y ∂y ∂x does not vanish if ρq = 0, the coordinates transformation between the Cartesian coordinates and the potential-stream coordinates is valid at least in the absence of stagnation points. It can be checked easily that both (2.7) and (2.8) are elliptic in the subsonic region (q < c∗ ) and degenerate in the sonic state (q = c∗ ). Therefore, (2.7) and (2.8) for subsonic–sonic flows are both second order quasilinear elliptic equations with degeneracies at sonic states. 2.2. Formation in the physical plane Similar to the problem (1.1)–(1.6), the continuous subsonic–sonic flow problem in k is formulated as follows div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(x, y) ∈ k ,
ϕ(gk (y), y) = Cin ,
−l1 < y < fk (−l0 ),
(2.10)
∂ϕ (x, −l1 ) = 0, ∂y
g(−l1 ) < x < S(−l1 ),
(2.11)
∂ϕ ∂ϕ (x, fk (x)) − fk (x) (x, fk (x)) = 0, ∂y ∂x
−l0 < x < 0,
(2.12)
|∇ϕ(S(y), y)| = c∗ , ϕ(S(y), y) = 0,
−l1 < y < 0,
(2.13)
|∇ϕ(x, y)| < c∗ ,
(x, y) ∈ k ,
(2.14)
(2.9)
where (ϕ, S, Cin ) is unknown. 2.3. Formulation in the potential plane Let us reformulate the problem (2.9)–(2.14) in the potential plane. Assume the speed of the flow at the inlet and at the upper wall to be denoted by q(gk (y), y) = Qin (y),
−l1 ≤ y ≤ fk (−l0 )
and q(x, fk (x)) = Qup (x),
−l0 ≤ x ≤ 0,
respectively. The incoming mass flux is given by fk(−l0 )
m= in
Qin (y)ρ((Qin (y))2 ) 1 + (gk (y))2 dy.
|∇ϕ(x, y)|ρ(|∇ϕ(x, y)|2 )dσ = −l1
Since the potential at the sonic curve is normalized to be zero, the potential at the inlet is given by
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−l0 ζ = Qup (x) 1 + (fk (x))2 dx. 0
At the inlet, the stream function is expressed as y in (y) =
2 Qin (s)ρ(Qin (s)) 1 + (gk (s))2 ds,
−l1 ≤ y ≤ fk (−l0 ).
−l1
At the upper wall, the potential function expressed as x up (x) =
Qup (s) 1 + (fk (s))2 ds,
−l0 ≤ x ≤ 0.
0
The inverse functions of in and up are denoted by Yin and Xup , respectively. Similar to [8,9], the subsonic–sonic flow problem in the potential plane can be formulated as follows ∂ 2 A(q) ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(2.15)
∂A(q) 1 , (ζ, ψ) = 2 ∂ϕ Rk Qin (y)ρ(Qin (y)) y=Yin (ψ)
ψ ∈ (0, m),
(2.16)
∂q (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
(2.17)
fk (x) ∂B(q) , (ϕ, m) = ∂ψ (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ)
ϕ ∈ (ζ, 0),
(2.18)
q(0, ψ) = c∗ ,
ψ ∈ (0, m),
(2.19)
y ∈ [−l1 , fk (−l0 )],
(2.20)
x ∈ [−l0 , 0],
(2.21)
, Qin (y) = q(0, ψ) ψ=in (y) Qup (x) = q(ϕ, m) , ϕ=up (x)
where (q, m, ζ ) is unknown. Similar to Remark 2.16 in [8], one has Remark 2.1. A natural boundary condition at the inlet seems to be ∂A(q) 1 , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ)) y=Yin (ψ)
ψ ∈ (0, m),
(2.22)
d 1 instead of (2.16). However, since dq < 0 in q ∈ (0, c∗ ), it seems difficult to obtain the Rk qρ(q 2 ) uniqueness of the solution to the problem (2.15), (2.22), (2.17)–(2.19). Similarly, we use (2.18) instead of a Robin one.
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2.4. Main results One of the main results in this paper is the following existence theorem. Theorem 2.1. There exists k0 ∈ (0, 1], which depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that if 0 < k ≤ k0 then the problem (2.15)–(2.21) admits a solution (q, m, ζ ) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying max c∗ /2, c∗ − M1 −kϕ ≤ q(ϕ, ψ) ≤ c∗ − M2 −kϕ, (ϕ, ψ) ∈ (ζ, 0) × (0, m),
(2.23)
where 0 < M1 ≤ M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . The regularity of the solution obtained in Theorem 2.1 is as follows. Theorem 2.2. Let (q, m, ζ ) be a solution to the problem (2.15)–(2.21) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying (2.23). Then ∂q (ϕ, ψ) ≤ M3 k 1/4 (−ϕ)−1/2 , ∂ϕ ∂q √ (ϕ, ψ) ≤ M3 k −ϕ, (ϕ, ψ) ∈ (ζ, 0) × (0, m), ∂ψ
(2.24)
where M3 > 0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Furthermore, q ∈ C 1/2 ([ζ, 0] × [0, m]) and q ∈ / C β ([ζ, 0] × [0, m]) for each β ∈ (1/2, 1). Remark 2.2. The two powers of −ϕ and the second power of k in (2.24) are optimal due to (2.23) and (2.18). Moreover, the solution obtained in Theorem 2.1 is also unique. Theorem 2.3. There exists k˜0 ∈ (0, 1] such that if 0 < k ≤ k˜0 then the problem (2.15)–(2.21) admits at most one solution (q, m, ζ ) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying (2.23), where k˜0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . In terms of the physical variables, Theorems 2.1 and 2.3 can be transformed as Theorem 2.4. There exist k0 ∈ (0, 1] and 0 < M1 ≤ M2 such that if 0 < k ≤ k0 then the problem (2.9)–(2.14) admits uniquely a solution (ϕ, S, Cin ) with ϕ ∈ C 3 (k ) ∩ C 2 (k \ S) ∩ C 1 (k ) and S ∈ C 1 ([−l1 , 0]) satisfying max c∗ /2, c∗ − M1 kdistS (x, y) ≤ |∇ϕ(x, y)| ≤ c∗ − M2 kdistS (x, y),
(x, y) ∈ k ,
where distS (x, y) is the distance from (x, y) to S, and k0 , M1 and M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) .
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3. Proof of existence In order to solve the problem (2.15)–(2.21) by a fixed point argument, one needs to specify Qin and Qup in advance as follows: Qin ∈ C([−l1 , fk (−l0 )]) satisfies c∗ ≤ Qin (y) ≤ c∗ , 2
−l1 ≤ y ≤ fk (−l0 )
(3.1)
and Qup ∈ C([−l0 , 0]) satisfies c∗ ≤ Qup (x) ≤ c∗ , 2
−l0 ≤ x ≤ 0.
(3.2)
Then, m, ζ , in , up , Yin and Xup are determined, and one can solve the fixed boundary problem (2.15)–(2.19). Moreover, direct calculations show that −kf (−l0 ) 1 −kf (−l0 ) ≤ ≤ , l1 (f (−l0 ) + l1 ) 1 + (f (−l0 ))2 Rk 1 m1 = c∗ ρ(c∗2 /4)l1 ≤ m ≤ m2 = c∗ ρ(c∗2 ) 1 + (f (−l0 ))2 (f (−l0 ) + l1 ), 2 1 ζ2 = −c∗ 1 + f 2L∞ (−l0 ,0) l0 ≤ ζ ≤ ζ1 = − c∗ l0 , 2 1 (y) ≤ c∗ ρ(c∗2 ) 1 + (f (−l0 ))2 , −l1 ≤ y ≤ fk (−l0 ), c∗ ρ(c∗2 /4) ≤ in 2 1 c∗ ≤ up (x) ≤ c∗ 1 + f 2L∞ (−l0 ,0) , −l0 ≤ x ≤ 0. 2 3.1. Definition of weak solutions and comparison principle Consider the following boundary value problem ∂ 2 A(q) ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(3.3)
∂q (ϕ, 0) = h1 (ϕ), ∂ψ
ϕ ∈ (ζ, 0),
(3.4)
∂q (ϕ, m) = h2 (ϕ), ∂ψ
ϕ ∈ (ζ, 0),
(3.5)
∂A(q) (ζ, ψ) = h3 (ψ), ∂ϕ
ψ ∈ (0, m),
(3.6)
q(0, ψ) = h4 (ψ),
ψ ∈ (0, m),
(3.7)
where ζ < 0, m > 0, h1 , h2 ∈ L∞ (ζ, 0), h3 , h4 ∈ L∞ (0, m) and 0 < inf(0,m) h4 ≤ sup(0,m) h4 ≤ c∗ .
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Definition 3.1. A function q ∈ L∞ ((ζ, 0) × (0, m)) is said to be a weak supersolution (subsolution) to the problem (3.3)–(3.7), if 0 < inf(ζ,0)×(0,m) q ≤ sup(ζ,0)×(0,m) q ≤ c∗ and 0 m ∂ 2ξ ∂ 2ξ A(q(ϕ, ψ)) 2 (ϕ, ψ) + B(q(ϕ, ψ)) 2 (ϕ, ψ) dϕdψ ∂ϕ ∂ψ ζ
0
0 −
0 h1 (ϕ)ξ(ϕ, 0)dϕ +
ζ
h2 (ϕ)ξ(ϕ, m)dϕ ζ
m −
m h3 (ψ)ξ(ζ, ψ)dψ −
0
h4 (ψ)
∂ξ (0, ψ)dψ ≤ (≥)0 ∂ϕ
0
for any nonnegative function ξ ∈ C 2 ([ζ, 0] × [0, m]) with ∂ξ ∂ξ = =0 (·, 0) (·, m) (ζ,0) (ζ,0) ∂ψ ∂ψ
and
∂ξ = ξ(0, ·) = 0. (ζ, ·) (0,m) (0,m) ∂ϕ
Furthermore, q ∈ L∞ ((ζ, 0) × (0, m)) is said to be a weak solution to the problem (3.3)–(3.7) if q is both a weak supersolution and a weak subsolution. It follows from the proof of Proposition 3.1 in [8] that the following comparison principle holds. Lemma 3.1. Assume that q and q are a weak supersolution and a weak subsolution to the problem (3.3)–(3.7), respectively. Then q(ϕ, ψ) ≥ q(ϕ, ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
3.2. Well-posedness of the fixed boundary problem Proposition 3.1. There exist τ > 0 and k0 ∈ (0, 1], which depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that if 0 < k ≤ k0 then the problem (2.15)–(2.19) admits uniquely a weak solution. Furthermore, the solution satisfies max c∗ /2, c∗ − M1 −kϕ ≤ q(ϕ, ψ) ≤ c∗ − M2 −kϕ,
(ϕ, ψ) ∈ (ζ, 0) × (0, m), (3.8)
where 0 < M1 ≤ M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Proof. Denote −1 B (q) γ + 1 2 γ − 1 2 2/(γ −1)+1 − 1 , 0 < q < c∗ , = q q 1 − A (q) 2 2 −3 B (q) γ − 1 2 3/(γ −1)+1 4 γ +1 2 = (γ + 1)q − 1 , 0 < q < c∗ q q 1 − p(q) = A (q) 2 2 b(q) =
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and set N1 =
inf
(c∗ /2,c∗ )
−A(q)b(q),
N2 = sup
−A(q)b(q),
N3 = sup (−A(q))3/2 p(q).
(c∗ /2,c∗ )
(c∗ /2,c∗ )
First, let us construct a super and a sub solutions to the problem (2.15)–(2.19). Define √ μ1 2 q(ϕ, ψ) = A−1 kμ1 ϕ(1 − kμ2 ψ 2 ) + k ϕ , 2ζ2
(ϕ, ψ) ∈ [ζ, 0] × [0, m]
(3.9)
√ μ3 2 q(ϕ, ψ) = A−1 kμ3 ϕ(1 + kμ4 ψ 2 ) − k ϕ , 4ζ2
(ϕ, ψ) ∈ [ζ, 0] × [0, m],
(3.10)
and
where √ μ1 −f (−l0 ) μ1 = , , μ2 = 3/2 2 2 8N (−ζ 2 2) 2(f (−l0 ) + l1 ) 1 + (f (−l0 )) c∗ ρ(c∗ ) √ μ3 −4f (−l0 ) , μ = . μ3 = 4 2 l1 c∗ ρ(c∗ /4) 8N2 (−ζ2 )3/2 It is clear that μ1 < μ3 , μ2 < μ4 and 2kμ1 ≤
1 1 ≤ kμ3 , 2 y=Y (ψ) 2 Rk Qin (y)ρ(Qin (y)) in
ψ ∈ (0, m).
(3.11)
Let
kˆ0 = min 1,
1 2A(c∗ /2) , . 4μ24 m42 3μ3 ζ2
If 0 < k ≤ kˆ0 , one gets that 3 3 1 A(c∗ /2) ≤ kμ3 ϕ < kμ1 ϕ ≤ A(q(ϕ, ψ)) ≤ kμ1 ϕ, (ϕ, ψ) ∈ [ζ, 0] × [0, m], 2 2 2 3 3 A(c∗ /2) ≤ kμ3 ϕ ≤ A(q(ϕ, ψ)) ≤ kμ3 ϕ, (ϕ, ψ) ∈ [ζ, 0] × [0, m], 2 4 √ ∂A(q) μ1 ζ ≤ 2kμ1 , ψ ∈ (0, m), (ζ, ψ) = kμ1 (1 − kμ2 ψ 2 ) + k ∂ϕ ζ2 ∂A(q) ∂ϕ
(ζ, ψ) = kμ3 (1 +
√ μ3 ζ 1 kμ4 ψ 2 ) − k ≥ kμ3 , 2ζ2 2
For each 0 < k ≤ kˆ0 , direction calculations and (3.12) yield
ψ ∈ (0, m).
(3.12) (3.13) (3.14) (3.15)
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∂ 2 A(q) ∂ 2 B(q) (ϕ, ψ) + (ϕ, ψ) 2 ∂ϕ ∂ψ 2 =
∂A(q) 2 ∂ 2 A(q) ∂ 2 A(q) (ϕ, ψ) + b(q(ϕ, ψ)) (ϕ, ψ) + p(q(ϕ, ψ)) (ϕ, ψ) ∂ψ ∂ϕ 2 ∂ψ 2
2k 3/2 μ1 μ2 ϕ (−2k 3/2 μ1 μ2 ϕψ)2 μ1 + N3 − N2 √ , ζ2 (−kμ1 ϕ/2)3/2 −kμ1 ϕ/2 √ √ μ1 ≤k + 2 2kN2 μ2 −μ1 ζ2 + 8 2k 3/2 N3 μ22 m22 −μ1 ζ2 ζ2 (4 − √2)μ √ 1 =k + 8 2kN3 μ22 m22 −μ1 ζ2 , (ϕ, ψ) ∈ (ζ, 0) × (0, m) 4ζ2 ≤k
(3.16)
and ∂B(q) −2k 3/2 μ1 μ2 ϕm ∂A(q) (ϕ, m) = b(q(ϕ, m)) (ϕ, m) ≥ N1 √ ∂ψ ∂ψ −3kμ1 ϕ/2 2 √ ≥ √ kN1 μ1 c∗ μ2 m1 −Xup (ϕ), ϕ ∈ (ζ, 0). 3
(3.17)
Similarly, for each 0 < k ≤ kˆ0 , it follows from (3.13) that ∂ 2 A(q)
∂ 2 B(q)
2k 3/2 μ3 μ4 ϕ μ3 + N √ 2 2ζ2 ∂ϕ 2 ∂ψ 2 −3kμ3 ϕ/4 √ 4 3 − 3 μ3 μ3 − √ kN2 μ4 −μ3 ζ2 = − ≥ 0, (ϕ, ψ) ∈ (ζ, 0) × (0, m) k ≥ −k 2ζ2 6 ζ2 3 (ϕ, ψ) +
(ϕ, ψ) ≥ −k
(3.18)
and 2k 3/2 μ3 μ4 ϕm 2 √ ≤ − √ kN1 μ3 c∗ μ4 m1 −Xup (ϕ), (ϕ, m) ≤ N1 √ ∂ψ −3kμ3 ϕ/2 3
∂B(q)
ϕ ∈ (ζ, 0). (3.19)
Set 1 √ 3/2 τ = √ N1 μ1 μ2 c∗ m1 , 3
k0 = min kˆ0 ,
μ1 . 29 N32 μ42 m42 (−ζ2 )3
Then, for each 0 < k ≤ k0 , (3.11) and (3.14)–(3.19) show that q and q defined by (3.9) and (3.10) are a super and a sub solutions to the problem (2.15)–(2.19), respectively. Below we establish the existence of the weak solution to the problem (2.15)–(2.19) with 0 < k ≤ k0 by using a method of elliptic regularization. For each positive integer n, consider the following regularized problem
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∂ 2 A(qn ) ∂ 2 B(qn ) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(3.20)
∂A(qn ) 1 , (ζ, ψ) = 2 ∂ϕ Rk Qin (y)ρ(Qin (y)) y=Yin (ψ)
ψ ∈ (0, m),
(3.21)
∂qn (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
(3.22)
ϕ ∈ (ζ, 0),
(3.23)
ψ ∈ (0, m).
(3.24)
fk (x) ∂B(qn ) , (ϕ, m) = ∂ψ (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ) n c∗ , qn (0, ψ) = n+1
Note that (3.20) is a uniformly elliptic equation provided that 0 < inf(ζ,0)×(0,m) qn ≤ sup(ζ,0)×(0,m) qn < c∗ . Therefore, to get the existence of the weak solution to the problem (3.20)–(3.24), it suffices to construct an upper and a sub solutions q n and q n satisfying 0 < inf(ζ,0)×(0,m) q n ≤ sup(ζ,0)×(0,m) q n < c∗ . For εn > 0, set √ q n (ϕ, ψ) = A−1 kμ1 (ϕ − εn )(1 − kμ2 ψ 2 ) + k
μ1 (ϕ − εn )2 , (ϕ, ψ) ∈ [ζ, 0] × [0, m]. 2(ζ2 − εn )
For each 0 < k ≤ k0 and n ≥ 1, the above calculations show that q n is a supersolution to the problem (3.20)–(3.24) for sufficiently small εn > 0. For sufficiently large μ5 , μ6 > 0, it is not hard to show that q n (ϕ, ψ) = A−1 c∗ /2 + kμ5 ϕ(ϕ − 3ζ2 ) − kμ6 ψ 2 , (ϕ, ψ) ∈ [ζ, 0] × [0, m] is a subsolution to the problem (3.20)–(3.24) for each 0 < k ≤ k0 and n ≥ 1. Therefore, for each positive integer n, the problem (3.20)–(3.24) with 0 < k ≤ k0 admits a weak solution qn . Lemma 3.1 yields qn1 (ϕ, ψ) ≤ qn2 (ϕ, ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
n 1 ≤ n2 .
Set q(ϕ, ψ) = lim qn (ϕ, ψ), n→∞
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
Then, q is a weak solution to the problem (2.15)–(2.19) with 0 < k ≤ k0 . Since q and q defined by (3.9) and (3.10) are a super and a sub solutions to the problem (2.15)–(2.19), respectively, Lemma 3.1 yields (3.8). 2 Remark 3.1. According to the definition of τ , one gets from (2.1) that 1 f (x) ≤ f (−l0 ), 2
−l0 ≤ x ≤ 0.
Below, we establish the Hölder continuity of weak solutions to the problem (2.15)–(2.19).
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Proposition 3.2. Assume that q is a weak solution to the problem (2.15)–(2.19) satisfying (3.8). Then, there exist β ∈ (0, 1/2] and K > 0, which depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) and k, such that q ∈ C β ([ζ, 0] × [0, m]) and [q]β,(ζ,0)×(0,m) ≤ K.
(3.25)
Proof. In the proof we use Ki > 0 (1 ≤ i ≤ 6) and βi ∈ (0, 1) (i = 1, 2) to denote generic constants depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) and k. Thanks to (3.8) and the classical theory on uniformly elliptic equations, one can get that q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C([ζ, 0] ×[0, m]). Furthermore, the Harnack inequality [5] shows that q ∈ C β1 ([ζ, −δ0 ] ×[0, m]) and [q]β1 ,(ζ,−δ0 )×(0,m) ≤ K1 ,
(3.26)
where δ0 = min − ζ /2, m/2 . Fix 0 < δ < δ0 . Then q ∈ C 2 ([−2δ, −δ/4] × (0, m)) ∩ C([−2δ, −δ/4] × [0, m]) solves ∂ ∂q ∂ ∂q a(ϕ, ψ) + h(ϕ, ψ) = 0, ∂ϕ ∂ϕ ∂ψ ∂ψ
(ϕ, ψ) ∈ (−2δ, −δ/4) × (0, m),
∂q (ϕ, 0) = 0, ∂ψ
ϕ ∈ (−2δ, −δ/4),
∂q (ϕ, m) = e(ϕ), ∂ψ
ϕ ∈ (−2δ, −δ/4),
where a(ϕ, ψ) = A (q(ϕ, ψ)),
h(ϕ, ψ) = B (q(ϕ, ψ)),
(ϕ, ψ) ∈ [−2δ, −δ/4] × [0, m]
and e(ϕ) =
fk (x) 1 , · B (q(ϕ, m)) (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ)
ϕ ∈ [−2δ, −δ/4].
Introducing ϕ˜ = δ −1 (ϕ + δ),
ϕ ∈ [−2δ, −δ/4],
ψ˜ = δ −3/4 ψ,
ψ ∈ [0, m]
and setting ˜ = q(−δ + δ ϕ, ˜ − c∗ , q( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ)
˜ ∈ [−1, 3/4] × [0, δ −3/4 m], (ϕ, ˜ ψ)
one can check that q˜ ∈ C 2 ([−1, 3/4] × (0, δ −3/4 m)) ∩ C([−1, 3/4] × [0, δ −3/4 m]) solves
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∂ q˜ ∂ −1/2 ∂ q˜ ∂ ˜ ˜ ˜ h(ϕ, ˜ ψ) = 0, a( ˜ ϕ, ˜ ψ) δ + ∂ ϕ˜ ∂ ϕ˜ ∂ ψ˜ ∂ ψ˜ ∂ q˜ (ϕ, ˜ 0) = 0, ∂ ψ˜ ∂ q˜ (ϕ, ˜ δ −3/4 m) = e( ˜ ϕ), ˜ ∂ ψ˜
(ϕ, ˜ ψ) ∈ (−1, 3/4) × (0, δ −3/4 m), ϕ˜ ∈ (−1, 3/4), ϕ˜ ∈ (−1, 3/4),
where ˜ = a(−δ + δ ϕ, ˜ a( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ), ˜ ϕ, ˜ ˜ = h(−δ + δ ϕ, h( ˜ ψ) ˜ δ 3/4 ψ),
˜ ∈ [−1, 3/4] × [0, δ −3/4 m] (ϕ, ˜ ψ)
and e( ˜ ϕ) ˜ = δ 3/4 e(−δ + δ ϕ), ˜
ϕ˜ ∈ [−1, 3/4].
Due to (3.8) and (2.1), one gets that √ √ ˜ ≤ K3 δ, K2 δ ≤ a( ˜ ϕ, ˜ ψ)
˜ ϕ, ˜ ≤ K3 , (ϕ, ˜ ∈ [−1, 3/4] × [0, δ −3/4 m], K2 ≤ h( ˜ ψ) ˜ ψ) √ q ˜ L∞ ((−1,3/4)×(0,δ −3/4 m)) ≤ K3 δ, e ˜ L∞ (−1,3/4) ≤ K3 δ 5/4 .
Therefore, the Harnack inequality [5] shows that q˜ ∈ C β2 ([−1/2, 1/2] × [0, δ −3/4 m]) and [q] ˜ β2 ,(−1/2,1/2)×(0,δ −3/4 m) ≤ K4 q ˜ L∞ (−1,3/4) ˜ L∞ ((−1,3/4)×(0,δ −3/4 m)) + e √ ≤ K5 δ. (3.27) Set β = min 1/2, β1 , β2 . Then, one gets from (3.8), (3.26) and (3.27) that |q(ϕ, ψ) − c∗ | ≤ K6 (−ϕ)β , [q]β,(ζ,−δ0 )×(0,m) ≤ K6 ,
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
[q]β,(−3δ/2,−δ/2)×(0,m) ≤ K6 ,
which yield q ∈ C β ([ζ, 0] × [0, m]) and (3.25).
0 < δ < δ0 ,
2
3.3. Proof of the existence theorem Proof of Theorem 2.1. Set
S = (Qin , Qup ) ∈ C([−l1 , fk (−l0 )]) × C([−l0 , 0]) : Qin satisfies (3.1) and Qup satisfies (3.2) with the norm
(Qin , Qup )S = max Qin L∞ (−l1 ,fk (−l0 )) , Qup L∞ (−l0 ,0) ,
(Qin , Qup ) ∈ S .
For given (Qin , Qup ) ∈ S , it follows from Propositions 3.1 and 3.2 that the problem (2.15)–(2.19) admits a unique weak solution q ∈ C β ([ζ, 0] × [0, m]) satisfying the estimates
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(3.8) and (3.25). Set ˆ in (y) = q(ζ, in (y)), Q
−l1 ≤ y ≤ fk (−l0 )
and ˆ up (x) = q(up (x), m), Q
−l0 ≤ x ≤ 0.
ˆ in ∈ C β ([−l1 , fk (−l0 )]) satisfies (3.1), It follows from q ∈ C β ([ζ, 0] × [0, m]) and (3.8) that Q β ˆ and Qup ∈ C ([−l0 , 0]) satisfies (3.2). Therefore, we can define a mapping J from S to itself as follows ˆ in , Q ˆ up ), J ((Qin , Qup )) = (Q
(Qin , Qup ) ∈ S .
Due to (3.25), J is compact. Similar to the discussion of Theorem 4.1 in [9], one can prove that J is also continuous from (3.25) and Lemma 3.1. Then, the Schauder fixed point theorem and (3.8) show that the problem (2.15)–(2.21) admits a weak solution (q, m, ζ ) with q ∈ C β ([ζ, 0] × [0, m]) satisfying (2.23). Furthermore, the classical theory on uniformly elliptic equations leads to q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]). 2 4. Proof of the regularity theorem In this section, let us prove Theorem 2.2. First we prove the following estimate by a rescaling technique. Proposition 4.1. Under the assumption of Theorem 2.2, ∂q (ϕ, ψ) ≤ Mk 1/4 (−ϕ)−1/2 , ∂ϕ ∂q √ (ϕ, ψ) ≤ M k(−ϕ)−1/4 , (ϕ, ψ) ∈ (ζ, 0) × (0, m), ∂ψ
(4.1)
where M > 0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Proof. In the proof we use Ci (1 ≤ i ≤ 14) to denote a generic positive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Note that q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) solves ∂ ∂q ∂ ∂q a(ϕ, ψ) + h(ϕ, ψ) = 0, ∂ϕ ∂ϕ ∂ψ ∂ψ ∂q (ζ, ψ) = r(ψ), ∂ϕ ∂q (ϕ, 0) = 0, ∂ψ ∂q (ϕ, m) = e(ϕ), ∂ψ
(ϕ, ψ) ∈ (ζ, 0) × (0, m), ψ ∈ (0, m), ϕ ∈ (ζ, 0), ϕ ∈ (ζ, 0),
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where a(ϕ, ψ) = A (q(ϕ, ψ)), r(ψ) = e(ϕ) =
h(ϕ, ψ) = B (q(ϕ, ψ)), 1
Rk
q(ζ, ψ)ρ(q 2 (ζ, ψ))A (q(ζ, ψ))
(ϕ, ψ) ∈ (ζ, 0) × (0, m), ,
ψ ∈ (0, m),
fk (Xup (ϕ)) , (1 + (fk (Xup (ϕ)))2 )3/2 q(ϕ, m)B (q(ϕ, m))
ϕ ∈ (ζ, 0).
Denote δ0 = min − ζ /2, m/2 . Introducing ϕˆ = k −1/4 ϕ,
ϕ ∈ [ζ, −δ0 /2]
and setting q( ˆ ϕ, ˆ ψ) = q(k 1/4 ϕ, ˆ ψ) − c∗ ,
(ϕ, ˆ ψ) ∈ [k −1/4 ζ, −k −1/4 δ0 /2] × [0, m],
one can check that qˆ ∈ C 2 ((ζ, −δ0 /2) × (0, m)) ∩ C 1 ([ζ, −δ0 /2] × [0, m]) solves ∂ −1/2 ∂ qˆ ∂ ˆ ∂ qˆ a( ˆ ϕ, ˆ ψ) k + h(ϕ, ˆ ψ) = 0, ∂ ϕˆ ∂ ϕˆ ∂ψ ∂ψ
(ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m),
∂ qˆ −1/4 ζ, ψ) = k 1/4 r(ψ), (k ∂ ϕˆ
ψ ∈ (0, m),
∂ qˆ (ϕ, ˆ 0) = 0, ∂ψ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2),
∂ qˆ (ϕ, ˆ m) = e( ˆ ϕ), ˆ ∂ψ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2),
where a( ˆ ϕ, ˆ ψ) = a(k 1/4 ϕ, ˆ ψ),
ˆ ϕ, h( ˆ ψ) = h(k 1/4 ϕ, ˆ ψ),
(ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m)
and e( ˆ ϕ) ˆ = e(k 1/4 ϕ), ˆ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2).
It follows from (2.23) that √ √ ˆ ϕ, C1 k ≤ a( ˆ ϕ, ˆ ψ) ≤ C2 k, C1 ≤ h( ˆ ψ) ≤ C2 , (ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m), √ √ ˆ L∞ (k −1/4 ζ,−k −1/4 δ0 /2) ≤ C2 k. q ˆ L∞ ((k −1/4 ζ,−k −1/4 δ0 /2)×(0,m)) ≤ C2 k, rL∞ (0,m) ≤ C2 k, e ˆ
Therefore, the Harnack inequality [5] shows that qˆ ∈ C β ([k −1/4 ζ, −3k −1/4 δ0 /4] × [0, m]) and
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[q] ˆ β;(k ˆ L∞ ((k −1/4 ζ,−k −1/4 δ0 /2)×(0,m)) + k 1/4 rL∞ (0,m) ˆ −1/4 ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C3 q √ + e ˆ L∞ (k −1/4 ζ,−k −1/4 δ0 /2) ≤ C4 k, (4.2) where βˆ ∈ (0, 1) depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Due to (4.2), one gets that √ √ ˆ ˆ −1/4 [a] ˆ β;(k ˆ −1/4 ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C5 k, [h]β;(k ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C5 k, √ [r]β;(0,m) ≤ C5 k, [e] ˆ α β;(k ˆ ˆ −1/4 ζ,−3k −1/4 δ0 /4) ≤ C5 k. ˆ
The classical Schauder theory shows that qˆ ∈ C 1,α β ([k −1/4 ζ, −k −1/4 δ0 ] × [0, m]) satisfies q ˆ 1,α β;(k ≤ C q ˆ L∞ ((k −1/4 ζ,−3k −1/4 δ0 /4)×(0,m)) + k 1/4 rα β;(0,m) −1/4 −1/4 6 ˆ ˆ ζ,−k δ0 )×(0,m) √ + e ˆ α β;(k (4.3) ˆ −1/4 ζ,−3k −1/4 δ0 /4) ≤ C7 k. Transforming (4.3) into the (ϕ, ψ) plane yields ∂q ≤ C7 k 1/4 , ∞ ∂ϕ L ((ζ,−δ0 )×(0,m))
∂q √ ≤ C7 k. ∞ ∂ψ L ((ζ,−δ0 )×(0,m))
(4.4)
Fix 0 < δ < δ0 . Introducing ϕ˜ = k −1/4 δ −1 (ϕ + δ),
ϕ ∈ [−2δ, −δ/4],
ψ˜ = δ −3/4 ψ,
ψ ∈ [0, m]
and setting ˜ = q(−δ + k 1/4 δ ϕ, ˜ − c∗ , q( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ)
˜ ∈ [−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m], (ϕ, ˜ ψ)
one can check that q˜ ∈ C 2 ([−k −1/4 , 3/4k −1/4 ] × (0, δ −3/4 m)) ∩ C 1 ([−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m]) solves ∂ q˜ ∂ −1/2 −1/2 ∂ q˜ ∂ ˜ ˜ ˜ h(ϕ, ˜ ψ) = 0, δ a( ˜ ϕ, ˜ ψ) k + ∂ ϕ˜ ∂ ϕ˜ ∂ ψ˜ ∂ ψ˜ ˜ ∈ (−k −1/4 , 3/4k −1/4 ) × (0, δ −3/4 m), (ϕ, ˜ ψ) ∂ q˜ (ϕ, ˜ 0) = 0, ∂ ψ˜
ϕ˜ ∈ (−k −1/4 , 3/4k −1/4 ),
∂ q˜ (ϕ, ˜ δ −3/4 m) = e( ˜ ϕ), ˜ ˜ ∂ψ
ϕ˜ ∈ (−k −1/4 , 3/4k −1/4 ),
˜ ∈ [−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m], where for (ϕ, ˜ ψ) ˜ = a(−δ + k 1/4 δ ϕ, ˜ a( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ),
˜ ϕ, ˜ = h(−δ + k 1/4 δ ϕ, ˜ ˜ δ 3/4 ψ), h( ˜ ψ)
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and e( ˜ ϕ) ˜ = δ 3/4 e(−δ + k 1/4 δ ϕ), ˜
ϕ˜ ∈ [−k −1/4 , 3/4k −1/4 ].
Due to (2.23) and (2.1), one gets that √ √ ˜ ϕ, ˜ ≤ C9 kδ, C8 ≤ h( ˜ ≤ C9 , (ϕ, ˜ ∈ (−k −1/4 , 3/4k −1/4 ) × (0, δ −3/4 m), C8 kδ ≤ a( ˜ ϕ, ˜ ψ) ˜ ψ) ˜ ψ) √ ˜ L∞ (−k −1/4 ,3/4k −1/4 ) ≤ C9 kδ 5/4 . q ˜ L∞ ((−k −1/4 ,3/4k −1/4 )×(0,δ −3/4 m)) ≤ C9 kδ, e ˜
Therefore, the Harnack inequality [5] shows that q˜ ∈ C β ([−1/2k −1/4 , 1/2k −1/4 ] × [0, δ −3/4 m]) and [q] ˜ β;(−1/2k ˜ L∞ ((−k −1/4 ,3/4k −1/4 )×(0,δ −3/4 m)) −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C10 q ˜ √ + e ˜ L∞ (−k −1/4 ,3/4k −1/4 ) ≤ C11 kδ, which leads to √ [a] ˜ β;(−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C12 kδ, ˜
√ ˜ ˜ [h] β;(−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C12 kδ, √ [e] ˜ α β;(−1/2k −1/4 ,1/2k −1/4 ) ≤ C12 kδ. ˜ ˜
The classical Schauder theory shows that q˜ ∈ C 1,α β ([−1/4k −1/4 , 1/4k −1/4 ] × [0, δ −3/4 m]) satisfies q ˜ 1,α β;(−1/4k ˜ L∞ ((−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m)) −1/4 ,1/4k −1/4 )×(0,δ −3/4 m) ≤ C13 q ˜ √ + e ˜ α β;(−1/2k (4.5) −1/4 ,1/2k −1/4 ) ≤ C14 kδ. ˜ Transforming (4.5) into the (ϕ, ψ) plane yields ∂q ∂q √ (ϕ, ψ) ≤ C14 k(−ϕ)−1/4 , (ϕ, ψ) ∈ (−δ0 , 0) × (0, m), (ϕ, ψ) ≤ C14 k 1/4 (−ϕ)−1/2 , ∂ϕ ∂ψ which and (4.4) lead to (4.1).
2
Remark 4.1. The powers of −ϕ and k in (4.1) are optimal when one uses a recalling technique. Owing to (2.16) and (2.18), the two powers of k in (4.1) are not greater than 1/2 and 1, respectively. As in the proof of Proposition 3.2 in [8], we introduce two functions to prove Theorem 2.2. The first one is E(s) = A(B −1 (s)), Clearly, E ∈ C ∞ ((−∞, 0)) satisfies
s < 0.
C. Wang / J. Differential Equations 259 (2015) 2546–2575
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γ + 1 2 γ − 1 2 −2/(γ −1)−1 E (s) = 1 − > 0, q q 1− q=B −1 (s) 2 2 γ − 1 2 −3/(γ −1)−2 E (s) = −(γ + 1)q 4 1 − < 0, q q=B −1 (s) 2 γ − 1 2 −4/(γ −1)−3 < 0, q E (s) = −(γ + 1)q 4 (4 + 3q 2 ) 1 − q=B −1 (s) 2
s < 0, s < 0, s < 0.
The other function is G(s) = E (E −1 (s)),
s < 0,
which satisfies G ∈ C ∞ ((−∞, 0)) and G (s) =
E (t) < 0, E (t) t=E −1 (s)
G (s) < −
(E (t))2 < 0, (E (t))3 t=E −1 (s)
s < 0.
Proof of Theorem 2.2. The former estimate in (2.24) follows from the first formula in (4.1). Below we prove the latter estimate in (2.24) and we use Ci (1 ≤ i ≤ 3) to denote a generic pos itive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . For each integer 2 2 n ≥ max − 2/ζ, 4M2 /c∗ , consider the following regularized problem ∂ 2 A(qn ) ∂ 2 B(qn ) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m), (4.6)
1 ∂A(qn ) , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, m),
(4.7)
∂qn (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, −1/n),
(4.8)
fk (Xup (ϕ + 1/n)) ∂B(qn ) , ϕ ∈ (ζ, −1/n), (ϕ, m) = ∂ψ (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m) √ qn (−1/n, ψ) = c∗ − M2 / n, ψ ∈ (0, m).
(4.9) (4.10)
Clearly, q is a supersolution to the problem (4.6)–(4.10). It is not hard to show that
q(ϕ, ψ) = A−1 c∗ /2 + μ1 ϕ(ϕ − 3ζ2 ) − μ2 ψ 2 ,
(ϕ, ψ) ∈ [ζ, 0] × [0, m]
is a subsolution to the problem (4.6)–(4.10), where μ1 and μ2 are sufficiently large positive constants independent of n. Therefore, the problem (4.6)–(4.10) admits a solution qn ∈ C 2 ((ζ, −1/n) × (0, m)) ∩ C 1 ([ζ, −1/n] × [0, m]) satisfying 0 < q(ϕ, ψ) ≤ qn (ϕ, ψ) ≤ q(ϕ, ψ) < c∗ , Set
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m].
(4.11)
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zn (ϕ, ψ) =
∂B(qn ) (ϕ, ψ), ∂ψ
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m].
Then zn solves the following problem ∂ 2 zn ∂ 2 zn ∂zn ∂zn + + jn,2 (ϕ, ψ) + jn,3 (ϕ, ψ) + jn,4 (ϕ, ψ)zn = 0, ∂ϕ ∂ψ ∂ϕ 2 ∂ψ 2 (ϕ, ψ) ∈ (ζ, −1/n) × (0, m), B (qn (ζ, ψ)) 1 ∂zn , (ζ, ψ) + jn,0 (ψ)zn (ζ, ψ) = ∂ϕ Rk A (qn (ζ, ψ)) q(ζ, ψ)ρ(q 2 (ζ, ψ)) ψ ∈ (0, m),
(4.13)
zn (ϕ, 0) = 0,
ϕ ∈ (ζ, −1/n),
(4.14)
fk (Xup (ϕ + 1/n)) , zn (ϕ, m) = (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m)
ϕ ∈ (ζ, −1/n),
(4.15)
zn (−1/n, ψ) = 0,
ψ ∈ (0, m),
(4.16)
jn,1 (ϕ, ψ)
(4.12)
where for (ϕ, ψ) ∈ (ζ, −1/n) × (0, m), jn,1 (ϕ, ψ) = G(A(qn (ϕ, ψ))),
jn,2 (ϕ, ψ) = 2G (A(qn (ϕ, ψ)))
∂A(qn (ϕ, ψ)) , ∂ϕ
∂B(qn (ϕ, ψ)) , ∂ψ ∂A(q (ϕ, ψ)) 2 n jn,4 (ϕ, ψ) = G (A(qn (ϕ, ψ))) , ∂ϕ jn,3 (ϕ, ψ) = −G (A(qn (ϕ, ψ)))
and jn,0 (ψ) =
E (A(qn (ζ, ψ))) 1 < 0, · 2 (E (A(qn (ζ, ψ)))) Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, min ).
Due to (2.23), (4.1) and (2.1), one gets that B (q (ζ, ψ)) 1 n ≤ C1 k 3/2 , ψ ∈ (0, m), 2 Rk A (qn (ζ, ψ)) q(ζ, ψ)ρ(q (ζ, ψ)) √ fk (Xup (ϕ + 1/n)) ≤ C1 k −ϕ, ψ ∈ (0, m). 2 3/2 (1 + (fk (Xup (ϕ + 1/n))) ) q(ϕ, m)
(4.17) (4.18)
Note that √ 1 jn,1 (ϕ, ψ)(−ϕ)−3/2 − jn,4 (ϕ, ψ) −ϕ ≥ −jn,1 (ϕ, ψ)j4 (ϕ, ψ)(−ϕ)−1/2 4 1 ≥ − jn,2 (ϕ, ψ)(−ϕ)−1/2 , (ϕ, ψ) ∈ (ζ, −1/n) × (0, m). 2 Owing to (4.17)–(4.19), one can verify that
(4.19)
C. Wang / J. Differential Equations 259 (2015) 2546–2575
√ zn (ϕ, ψ) = (2 −ζ + 1)C1 k −ϕ,
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(ϕ, ψ) ∈ [ζ, −1/n] × [0, m]
and √ zn (ϕ, ψ) = −(2 −ζ + 1)C1 k −ϕ,
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m]
are super and sub solutions to the problem (4.12)–(4.16). The classical comparison principle shows that √ |zn (ϕ, ψ)| ≤ (2 −ζ + 1)C1 k −ϕ,
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m),
which, together with (2.23), leads to ∂q √ n (ϕ, ψ) ≤ C2 k −ϕ, ∂ψ
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m).
(4.20)
Due to (4.6)–(4.10), one gets that m 0
m
fk (Xup (ϕ + 1/n)) ∂ 2 A(qn ) , (ϕ, ψ)dψ = − ∂ϕ 2 (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m) ∂A(qn ) 1 (ζ, ψ)dψ = ∂ϕ Rk
0
m
ϕ ∈ (ζ, −1/n),
1 dψ, q(ζ, ψ)ρ(q 2 (ζ, ψ))
0
m
√ A(qn )(−1/n, ψ)dψ = mA(c∗ − M2 / n),
0
which, together with (2.23) and (2.1), give m A(qn )(ϕ, ψ)dψ ≥ C3 ϕ,
ϕ ∈ (ζ, −1/n).
(4.21)
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m).
(4.22)
0
It follows from (4.20) and (4.21) that √ qn (ϕ, ψ) ≥ c∗ − C4 −ϕ,
Owing to (4.11), (4.22), (4.20) and the classical theory on uniformly elliptic equations, there exists a subsequence of {qn }, denoted by itself for convenience, and a function q˜ ∈ L∞ ((ζ, 0) × (0, m)) satisfying 0 < inf(ζ,0)×(0,m) q˜ ≤ sup(ζ,0)×(0,m) q˜ ≤ c∗ and √ q(ϕ, ˜ ψ) ≥ c∗ − C4 −ϕ, such that
∂ q˜ √ (ϕ, ψ) ≤ C2 k −ϕ, ∂ψ
(ϕ, ψ) ∈ (ζ, 0) × (0, m) (4.23)
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lim qn (ϕ, ψ) = q(ϕ, ˜ ψ),
n→∞
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
Then, it is not hard to show that q˜ solves ∂ 2 A(q) ˜ ˜ ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
1 ∂A(q) ˜ , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, m),
∂ q˜ (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
fk (Xup (ϕ)) ∂B(q) ˜ , (ϕ, m) = ∂ψ (1 + (fk (Xup (ϕ)))2 )3/2 q(ϕ, m)
ϕ ∈ (ζ, 0),
q(0, ˜ ψ) = c∗ ,
ψ ∈ (0, m).
Lemma 3.1 shows q(ϕ, ψ) = q(ϕ, ˜ ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
which, together with the second estimate in (4.23), leads to the latter estimate in (2.24). Due to / C β ([ζ, 0] × [0, m]) for each (2.23) and (2.24), it is clear that q ∈ C 1/2 ([ζ, 0] × [0, m]) and q ∈ β ∈ (1/2, 1). 2 5. Proof of the uniqueness In the last section, we prove Theorem 2.3. Similar to [8], not only the degeneracy and the free boundaries but also the nonlocal terms in (2.16) and (2.18) bring essential difficulties. Although these nonlocal terms arise from the coordinates transformation from the Cartesian coordinates to the potential-stream coordinates, it is not convenient to prove the uniqueness theorem in the Cartesian coordinates since the outlet of the problem (2.9)–(2.14) is a free boundary whose shape is unknown. In order to prove Theorem 2.3, we first introduce a suitable coordinates transformation to transform (2.16) and (2.18) into usual boundary conditions and to fix the two free boundaries into fixed ones. Let (qi , mi , ζi ) with qi ∈ C 2 ((ζi , 0) × (0, mi )) ∩ C 1 ([ζi , 0) × [0, mi ]) ∩ C([ζi , 0] × [0, mi ]) and (2.23) be a solution to the problem (2.15)–(2.21) for i = 1, 2. Denote in,i , up,i , Yin,i and Xup,i to be the associated functions for i = 1, 2. For i = 1, 2, introduce the coordinates transformation ϕ = up,i (x), −l0 ≤ x ≤ 0, x = Xup,i (ϕ), ζ ≤ ϕ ≤ 0, 0 ≤ ψ ≤ m, y = Yin,i (ψ), ψ = in,i (y), −l1 ≤ y ≤ fk (−l0 ). Define wi (x, y) = A(q(up,i (x), in,i (y))), Then, wi (i = 1, 2) solves
(x, y) ∈ [−l0 , 0] × [−l1 , fk (−l0 )],
i = 1, 2.
C. Wang / J. Differential Equations 259 (2015) 2546–2575
∂ Xi (x) ∂wi ∂ Yi (y) ∂B(A−1 (wi )) + = 0, ∂x Yi (y) ∂x ∂y Xi (x) ∂y
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(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )), (5.1)
1 Xi (−l0 ) ∂wi , (−l0 , y) = Yi (y) ∂x 2 Rk − (y + l1 )2
y ∈ (−l1 , fk (−l0 )),
(5.2)
∂wi (x, 0) = 0, ∂y
x ∈ (−l0 , 0),
(5.3)
fk (x) Yi (fk (−l0 )) ∂B(A−1 (wi )) , (x, fk (−l0 )) = Xi (x) ∂y 1 + (fk (x))2
x ∈ (−l0 , 0),
(5.4)
wi (0, y) = 0,
y ∈ (−l1 , fk (−l0 )),
(5.5)
where 1 Xi (x) = , (1 + (fk (x))2 A−1 (wi (x, fk (−l0 ))) Rk2 − (y + l1 )2 , Yi (y) = Rk A−1 (wi (−l0 , y))ρ((A−1 (wi (−l0 , y)))2 )
x ∈ [−l0 , 0],
(5.6)
y ∈ [−l1 , fk (−l0 )].
(5.7)
The problem (5.1)–(5.5) is a problem possessing common boundary conditions in a fixed domain. Proof of Theorem 2.3. In the proof we use Ci (1 ≤ i ≤ 5) to denote a generic positive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Set w(x, y) = w1 (x, y) − w2 (x, y),
(x, y) ∈ [−l0 , 0] × [−l1 , fk (−l0 )].
Due to (5.1)–(5.5), w solves ∂ X1 (x) ∂w ∂ Y1 (y) ∂w ∂ X(x) ∂w2 ∂ X2 (x)Y (y) ∂w2 + h + − ∂x Y1 (y) ∂x ∂y X1 (x) ∂y ∂x Y1 (y) ∂x ∂x Y1 (y)Y2 (y) ∂x −1 ∂ Y (y) ∂B(A (w2 )) ∂ X(x)Y2 (y) ∂B(A−1 (w2 )) ∂ Y1 (y) ∂h w + − = 0, + ∂y X1 (x) ∂y ∂y X1 (x) ∂y ∂y X1 (x)X2 (x) ∂y (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
(5.8)
where 1 h(x, y) =
B −1 (A (λw1 (x, y) + (1 − λ)w2 (x, y)))dλ, A
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
0
X(x) = X1 (x) − X2 (x),
x ∈ [−l0 , 0],
Y (y) = Y1 (y) − Y2 (y),
Due to (2.23) and (2.24), direct calculations show that
y ∈ [−l1 , fk (−l0 )].
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∂h C1 C2 C2 ≤ h(x, y) ≤ √ , (x, y) ≤ √ , (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )), √ ∂y −x −kx −kx (5.9) C2 w(x, fk (−l0 )), x ∈ [−l0 , 0], |Y (y)| ≤ C2 w(−l0 , y), y ∈ [−l1 , fk (−l0 )], |X(x)| ≤ √ −kx (5.10) ∂w ∂B(A−1 (w )) √ 2 2 (x, y) ≤ C2 k 3/4 , (x, y) ≤ C2 k −x, (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )). ∂x ∂y (5.11) Multiplying (5.8) by −w and then integrating over (−l0 , 0) × (−l1 , fk (−l0 )) by parts and using (5.2)–(5.5), one gets that 0
fk(−l0 )
−l0
−l1
X1 (x) ∂w 2 dxdy + Y1 (y) ∂x
−l0
0 fk(−l0 ) =− −l0
0
−l1
0
−l1
Y1 (y) ∂w 2 dxdy h X1 (x) ∂y
fk(−l0 ) −l1
0
Y1 (y) ∂h ∂w w dxdy − X1 (x) ∂y ∂y
−l0
fk(−l0 )
+ −l0
0
−l0
− 0
−l1
X(x) ∂w2 ∂w dxdy + Y1 (y) ∂x ∂x
fk(−l0 )
−l0
fk(−l0 )
−l1
X2 (x)Y (y) ∂w2 ∂w dxdy Y1 (y)Y2 (y) ∂x ∂x
fk(−l0 ) −l1
Y (y) ∂B(A−1 (w2 )) ∂w dxdy X1 (x) ∂y ∂y
X(x)Y2 (y) ∂B(A−1 (w2 )) ∂w dxdy, X1 (x)X2 (x) ∂y ∂y
which, together with (2.23), (2.24), (5.9)–(5.11), leads to 0
fk(−l0 )
∂w 2
−l0
−l1
≤ C3 k
1/4
∂x 0
−l0
1 dxdy + √ k
−l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y
fk(−l0 ) −l1
0 1 ∂w 3/4 dxdy + C3 k √ w(x, fk (−l0 )) ∂x −x −l0
0 fk(−l0 ) + C3 −l0
0
−l1
√ + C3 k
1 ∂w √ w dxdy + C3 k −x ∂y
−l0
0 fk(−l0 )
−l0
0 fk(−l0 )
−l1
∂w dxdy. w(x, fk (−l0 )) ∂y
−l1
fk(−l0 ) −l1
∂w dxdy w(−l0 , y) ∂x
√ ∂w −xw(−l0 , y) dxdy ∂y
C. Wang / J. Differential Equations 259 (2015) 2546–2575
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Then, the Hölder inequality gives 0
fk(−l0 )
∂w 2
−l0
∂x
−l1
√ ≤ C4 k
0
−l0
1 dxdy + √ k
0 −l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y fk(−l0 )
1 2 w (x, fk (−l0 ))dx + C4 k 3/2 −x
√ + C4 k
0 fk(−l0 )
−l0
−l1
w 2 (−l0 , y)dy −l1
1 √ w 2 dxdy + C4 k 5/2 −x
fk(−l0 )
w 2 (−l0 , y)dy −l1
0 + C4 k 3/2
w 2 (x, fk (−l0 ))dx.
(5.12)
−l0
Let us estimate the terms on the right side of (5.12). It follows from the Hölder inequality that 0 −l0
1 2 w (x, fk (−l0 ))dx −x
1 ≤ fk (−l0 ) + l1 1 ≤ fk (−l0 ) + l1 0
0 −l0
0
−l1 fk(−l0 )
−l0
fk(−l0 )
+ −l0
fk(−l0 )
−l1
−l1
1 2 w (x, y)dxdy + 2 −x
0
fk(−l0 )
−l0
1 2 w dxdy + −x
0
−l0
fk(−l0 ) −l1
−l1
1 ∂w (x, y)dxdy w(x, y) −x ∂y
1 w 2 dxdy (−x)3/2
1 ∂w 2 dxdy √ −x ∂y
(5.13)
and 0 w 2 (x, fk (−l0 ))dx −l0
≤
0
−l0
1 fk (−l0 ) + l1
0 fk(−l0 )
fk(−l0 )
|w(x, y)|dy + −l1
−l1
0
2 ∂w (x, y)dy dx ∂y fk(−l0 )
w dxdy + 2 l0 (fk (−l0 ) + l1 )
≤2 −l0
fk(−l0 )
2
−l1
−l0
−l1
1 ∂w 2 dxdy. √ −x ∂y
(5.14)
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C. Wang / J. Differential Equations 259 (2015) 2546–2575
Owing to (5.5) and the Hölder inequality, one gets that fk(−l0 )
fk(−l0 )
w (−l0 , y)dy ≤ 2
−l1
−l1
0 2 0 ∂w (x, y)dx dy ≤ l0 ∂x −l0
−l0
fk(−l0 )
∂w 2
−l1
dxdy
(5.15)
0 ≤ λ < 2.
(5.16)
∂x
and 0 −l0
fk(−l0 ) −l1
0
1 w 2 dxdy ≤ (−x)λ
fk(−l0 )
−l0
−l1
1 (−x)λ
x
0 ≤
0 2 ∂w (s, y)ds dxdy ∂x
0 1−λ
(−x)
∂w 2
dx
−l0
−l0
1 2−λ = l 2−λ 0
fk(−l0 )
0
−l0
∂x
−l1
dxdy
fk(−l0 )
∂w 2 ∂x
−l1
dxdy,
Substitute (5.13)–(5.16) into (5.12) to get 0 −l0
fk(−l0 )
∂w 2 ∂x
−l1
0 √ ≤ C5 k −l0
1 dxdy + √ k
fk(−l0 )
∂w 2
−l1
∂x
0 −l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y
√ dxdy + C5 k
0 fk(−l0 )
−l0
−l1
1 ∂w 2 dxdy. √ −x ∂y
(5.17)
Choose k˜0 = 1/(C52 + 1). For each 0 < k ≤ k˜0 , it follows from (5.17) that ∂w (x, y) = 0, ∂x
∂w (x, y) = 0, ∂y
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
which, together with (5.5), leads to w(x, y) = 0,
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )).
Therefore, (q1 , m1 , ζ1 ) and (q2 , m2 , ζ2 ) are the same solution to the problem (2.15)–(2.21). 2 References [1] L. Bers, Existence and uniqueness of a subsonic flow past a given profile, Comm. Pure Appl. Math. 7 (1954) 441–504. [2] L. Bers, Mathematical Aspects of Subsonic and Transonic Gas Dynamics, John Wiley & Sons, Inc./Chapman & Hall, Ltd., New York/London, 1958.
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[3] G.Q. Chen, C.M. Dafermos, M. Slemrod, D.H. Wang, On two-dimensional sonic-subsonic flow, Comm. Math. Phys. 271 (2007) 635–647. [4] R. Courant, K.O. Friedrichs, Supersonic Flow and Shock Waves, Interscience Publishers, Inc., New York, NY, 1948. [5] G.M. Lieberman, Local estimates for subsolutions and supersolutions of oblique derivative problems for general second order elliptic equations, Trans. Amer. Math. Soc. 304 (1987) 343–353. [6] O.A. Ole˘ınik, E.V. Radkevi˘c, Second Order Differential Equations with Nonnegative Characteristic Form, American Mathematical Society, Rhode Island and Plenum Press, New York, 1973. [7] C.P. Wang, Z.P. Xin, Optimal Hölder continuity for a class of degenerate elliptic problems with an application to subsonic–sonic flows, Comm. Partial Differential Equations 36 (5) (2011) 873–924. [8] C.P. Wang, Z.P. Xin, On a degenerate free boundary problem and continuous subsonic–sonic flows in a convergent nozzle, Arch. Ration. Mech. Anal. 208 (3) (2013) 911–975. [9] C.P. Wang, Z.P. Xin, Smooth transonic flows in de Laval nozzles, arXiv preprint, arXiv:1304.2473, 2013. [10] C.J. Xie, Z.P. Xin, Global subsonic and subsonic–sonic flows through infinitely long nozzles, Indiana Univ. Math. J. 56 (2007) 2991–3023. [11] C.J. Xie, Z.P. Xin, Global subsonic and subsonic–sonic flows through infinitely long axially symmetric nozzles, J. Differential Equations 248 (11) (2010) 2657–2683. [12] J.X. Yin, C.P. Wang, Evolutionary weighted p-Laplacian with boundary degeneracy, J. Differential Equations 237 (2) (2007) 421–445.
ScienceDirect J. Differential Equations 259 (2015) 2546–2575 www.elsevier.com/locate/jde
Continuous subsonic–sonic flows in a general nozzle ✩ Chunpeng Wang School of Mathematics, Jilin University, Changchun 130012, China Received 17 August 2014 Available online 29 April 2015
Abstract This paper concerns continuous subsonic–sonic potential flows in a two dimensional finite nozzle with a general upper wall and a straight lower wall. We give a class of nozzles where continuous subsonic–sonic flows may exist. Consider a continuous subsonic–sonic flow in such a nozzle after rescaling the upper wall in a small scale. It is shown that for a given inlet and a fixed point at the upper wall, there exists uniquely a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the inlet and the sonic curve, which satisfies the slip conditions on the nozzle walls and whose sonic curve intersects the upper wall at the fixed point. Furthermore, the sonic curve of this flow is a free boundary, where the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. As the scale tends to zero, the precise convergent rate of the continuous subsonic–sonic flow converging to the sonic state is also determined. © 2015 Elsevier Inc. All rights reserved. MSC: 35R35; 35Q35; 76N10; 35J70 Keywords: Continuous subsonic–sonic flow; Free boundary; Degeneracy; Singularity
1. Introduction In this paper, we study the isentropic, irrotational, steady compressible Euler subsonic–sonic flows in general nozzles, which arise in the physical experiments and the engineering designs. ✩
Supported by the National Natural Science Foundation of China (Grant No. 1107222106). E-mail address: [email protected].
http://dx.doi.org/10.1016/j.jde.2015.03.036 0022-0396/© 2015 Elsevier Inc. All rights reserved.
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Roughly speaking, there have been mainly two kinds of problems on subsonic–sonic flows. One is subsonic–sonic flows past a profile. In the significant work [1], Bers showed that for two dimensional potential flows past a profile, the whole flow field will be subsonic outside the profile if the Mach number of the freestream is small enough; furthermore, the maximum flow speed will tend to the sound speed as the freestream Mach number increases. Bers [1] did not consider the flow with the critical freestream Mach number. Based on the compensated compactness method, it was shown in [3] that the flows with sonic points past a profile may be realized as weak limits of sequences of strictly subsonic flows. The other kind is subsonic–sonic flows in an infinite nozzle. Bers [2] asserted that there is a global subsonic potential flow through an infinite nozzle for an appropriately given incoming mass flux, which was proved rigorously in [10]. It was shown that there exists a critical value for a general infinite nozzle such that a strictly subsonic flow exists uniquely as long as the incoming mass flux is less than the critical value. Furthermore, a class of subsonic–sonic flows can be obtained as the weak limits of strictly subsonic flows associated with the incoming mass fluxes increasing to the critical value. Moreover, Xie and Xin [11] considered subsonic–sonic flows in axially symmetric nozzles. It should be noted that the subsonic–sonic flows past a profile and the ones in an infinite nozzle obtained in [3,10] are both in the weak sense and have not been proved to be smooth yet, thus the locations and properties of the sonic states are unknown for such flows. A typical subsonic–sonic flow with precise regularity is a symmetric continuous subsonic–sonic flow in a finite straight nozzle, whose structural stability was studied in the recent works [7,8]. Precisely, for the given inlet, which is a perturbation of an arc centered at the vertex of the nozzle, and the given incoming mass flux, it was shown in [8] that there exists an open interval depending only on the adiabatic exponent and the length of the arc, such that a unique continuous subsonic–sonic flow, whose velocity vector is along the normal direction at the inlet and the outlet and which satisfies the slip conditions on the nozzle walls, exists as long as the incoming mass flux belongs to this interval and the perturbation of the inlet is sufficiently small; furthermore, the sonic curve of this continuous subsonic–sonic flow is a free boundary, where the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. Furthermore, Wang and Xin [9] studied smooth transonic flows of Meyer type in finite de Laval nozzles and showed that there exists uniquely a smooth transonic flow near the throat of the nozzle, whose sonic curve is located at the throat and whose acceleration is Lipschitz continuous, if the wall of the nozzle is sufficiently flat. It is shown from [8] that the existence of continuous subsonic–sonic flows depends on the geometry of the nozzle. A natural question is how to formulate the continuous subsonic–sonic flow problem in a general nozzle. In [8], the authors prescribed the incoming flow angle and the incoming mass flux, which are two physical quantities, to formulate the continuous subsonic–sonic flow problem in a straight nozzle; furthermore, the sonic curve of the continuous subsonic–sonic flow is a free boundary where the velocity vector is along the normal direction. Clearly, for a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the sonic curve, the second derivative of the nozzle wall at the intersecting point with the sonic curve must be zero. Therefore, it is not suitable to prescribe the incoming flow angle and the incoming mass flux to formulate the continuous subsonic–sonic flow problem in a general nozzle. In the present paper, we fix the intersecting point between the sonic curve and the upper wall to replace the incoming mass flux condition. That is to say, for a general nozzle with a given inlet, we seek a continuous subsonic–sonic flow whose velocity vector is along the normal direction at the inlet and the sonic curve, which satisfies the slip conditions on the nozzle walls and whose sonic curve intersects the upper wall at a fixed point. Furthermore, the sonic curve of the flow, which is chosen to be the outlet of the flow, is a free boundary.
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Let us formulate the continuous subsonic–sonic flow problem in a general nozzle precisely. The flow is governed by the full potential equation div(ρ(|∇ϕ|2 )∇ϕ) = 0, where ϕ is the velocity potential, ρ is the density which is expressed in terms of the speed by γ − 1 2 1/(γ −1) ρ(q 2 ) = 1 − , q 2
0
2/(γ − 1)
with γ > 1 being the adiabatic exponent for a polytropic gas. The upper and lower walls of the nozzle are assumed to be given by up : y = f (x) ≥ 0,
−l0 ≤ x ≤ 0
and
low : y = −l1 ,
x ∈ R,
respectively, where l0 , l1 > 0 and f (0) = 0. The inlet is given by in : x = g(y),
−l1 ≤ y ≤ f (−l0 ).
The outlet is chosen to be the sonic curve, which is a free boundary from the origin at the upper wall to a point at the lower wall and is denoted by out : x = S(y),
−l1 ≤ y ≤ 0,
S(0) = 0.
The continuous subsonic–sonic problem is to seek (ϕ, S, Cin ) solving div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(x, y) ∈ ,
(1.1)
ϕ(g(y), y) = Cin , ∂ϕ (x, −l1 ) = 0, ∂y ∂ϕ ∂ϕ (x, f (x)) − f (x) (x, f (x)) = 0, ∂y ∂x |∇ϕ(S(y), y)| = 2/(γ + 1), ϕ(S(y), y) = 0, |∇ϕ(x, y)| < 2/(γ + 1),
−l1 < y < f (−l0 ),
(1.2)
g(−l1 ) < x < S(−l1 ),
(1.3)
−l0 < x < 0,
(1.4)
−l1 < y < 0,
(1.5)
(x, y) ∈ ,
(1.6)
where is the domain bounded by up , low , in and out . Since the sonic curve intersects the upper wall at the origin, f should satisfy f (0) = 0 according to the above discussion. Furthermore, it is assumed that f (−l0 ) < 0, which guarantees that the flow is accelerated at the inlet. The problem (1.1)–(1.6) is a free boundary problem of a quasilinear degenerate elliptic equation. It is well-known that both degeneracies and free boundaries are difficult topics for quasilinear elliptic equations. For the problem (1.1)–(1.6), the degeneracy occurs at the free boundary and the degeneracy is characteristic [6,12]. Particularly, if f takes the form of f (x) = −l2 x,
−l0 ≤ x ≤ 0,
(l2 > 0),
(1.7)
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then the nozzle is straight and the problem, where the incoming flow angle and the incoming mass flux are prescribed, is considered in [8]. For the problem (1.1)–(1.6) with (1.7) in a straight nozzle, the same discussion as Remark 2.12 in [8] shows that the special problem is solvable only when l2 is suitably small. Therefore, the problem (1.1)–(1.6) may be unsolvable. In order to avoid the unsolvability of the problem (1.1)–(1.6), we rescale the upper wall by fk (x) = kf (x),
−l0 ≤ x ≤ 0,
(0 < k ≤ 1)
and solve the problem (1.1)–(1.6) where f is replaced by fk . For a two-dimensional nozzle, the geometry of the nozzle determines whether there is a subsonic–sonic flow whose sonic curve extends from one wall of the nozzle to the other. A sufficient condition is shown to be √ |f (x)| ≤ τ −x,
−l0 ≤ x ≤ 0,
where τ > 0 is a suitable constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Then, it is proved in the present paper that there exists k0 ∈ (0, 1], depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that for each 0 < k ≤ k0 , the problem (1.1)–(1.6) with fk instead of f admits uniquely a solution. The regularity of the continuous subsonic–sonic flow is also studied and it is shown that the flow is singular in the sense that the speed is only C 1/2 Hölder continuous and the acceleration blows up at the sonic state. As k → 0+ , the nozzle will tend to a flat one and thus the continuous subsonic–sonic flow will tend to the sonic state in the whole nozzle. The precise convergent rate is determined in the paper. One may expect that there exists a critical exponent kc such that the problem (1.1)–(1.6) with fk instead of f is solvable when k < kc and unsolvable when k > kc . Unfortunately, this expectation is almost impossible since the criterion of the solvability of the problem (1.1)–(1.6) with fk instead of f is completely unknown. Furthermore, it is very difficult to compare the continuous subsonic–sonic flows as k changes since the domains are variable.
It was turn out in [8] that it is more convenient to investigate a subsonic–sonic flow in the potential plane than in the physical plane. The reason lies in that the shape of the sonic curve is unknown in the physical plane, while known in the potential plane. Moreover, it is more convenient to estimate the speed of the flow in the potential plane than in the physical plane since it is a solution in the potential plane, while the absolute value of the gradient of a solution in the physical plane. For the continuous subsonic–sonic flow problem in a general nozzle, we encounter the two main difficulties in [8]. One is that the problem is a free boundary problem of a quasilinear degenerate elliptic equation and the degeneracy is characteristic and occurs just at the free
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boundary. The other is that the boundary condition at the inlet is nonlinear and nonlocal. However, besides these difficulties, there is an additional disadvantage in this study. Different from the problem in [8], there are no background solutions, which play an important role to determine and control the rate of the flow from the subsonic region to the sonic state. Furthermore, the mass flux of the flow is free and the boundary condition at the upper wall is nonlinear and nonlocal. In the present paper, we use the Schauder fixed point theorem to prove the existence of continuous subsonic–sonic flows. For a given speed at the inlet and the upper wall, we solve a boundary problem of a quasilinear degenerate equation. If the solved speed at the inlet and the upper wall is just the given one, we get a solution. The key is precise estimates and optimal regularities of solutions to quasilinear degenerate elliptic equations. Particularly, we use the Harnack inequality to achieve compactness estimates. By a rescaling technique and a precise elliptic estimate, we get the optimal regularity for continuous subsonic–sonic flows. As for the uniqueness of the continuous subsonic–sonic flow, it is still the degeneracy, the free boundaries and the nonlocal boundary conditions that bring essential difficulties. By a suitable coordinates transformation, we fix the free boundaries into fixed boundaries and transform the nonlocal boundary conditions into common boundary conditions. Then, the uniqueness theorem is established by the method of energy estimate and a series of precise and complicated calculations. The paper is arranged as follows. In Section 2, we formulate the problem of continuous subsonic–sonic flows in a general nozzle and state the main results (existence, regularity and uniqueness) of the paper. Subsequently, we prove the existence, regularity and uniqueness results in the following three sections, respectively. 2. Formulation and main results Fix l0 , l1 > 0 and α ∈ (0, 1). Assume that f ∈ C 2,α ([−l0 , 0]) satisfying f (0) = 0 and f (x) < 0,
√ |f (x)| ≤ τ −x,
−l0 ≤ x ≤ 0,
(2.1)
where τ > 0 is a suitable constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . For 0 < k ≤ 1, set fk (x) = kf (x),
−l0 ≤ x ≤ 0
and gk (y) = −l0 −
fk (−l0 ) + l1 − fk (−l0 )
Rk2 − (y + l1 )2 ,
−l1 ≤ y ≤ fk (−l0 ),
where
Rk =
(fk (−l0 ) + l1 ) 1 + (fk (−l0 ))2 −fk (−l0 )
.
Consider a continuous subsonic–sonic flow in the nozzle whose upper and lower walls are up : y = fk (x),
−l0 ≤ x ≤ 0
and
low : y = −l1 ,
x ∈ R,
C. Wang / J. Differential Equations 259 (2015) 2546–2575
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respectively. The inlet is given by in : x = gk (y),
−l1 ≤ y ≤ fk (−l0 ).
The outlet is chosen to be the sonic curve, which is a free boundary from the origin at the upper wall to a point at the lower wall and is denoted by out : x = S(y),
−l1 ≤ y ≤ 0,
S(0) = 0.
We seek a continuous subsonic–sonic flow in the domain k bounded by up , low , in and out . Furthermore, the velocity vector of the flow is along the normal direction at the inlet and the sonic curve, which is the outlet of the flow and is a free boundary, the flow satisfies the slip conditions on the nozzle walls and the sonic curve intersects the upper wall at (0, 0).
2.1. Governing equations Consider the two dimensional compressible Euler system of steady flow ∂ ∂ (ρu) + (ρv) = 0, ∂x ∂y
(2.2)
∂ ∂ (P + ρu2 ) + (ρuv) = 0, ∂x ∂y
(2.3)
∂ ∂ (ρuv) + (P + ρv 2 ) = 0, ∂x ∂y
(2.4)
where (u, v), P and ρ represent the velocity, pressure and density of the flow, respectively. The flow is assumed to be isentropic so that P = P (ρ) is a smooth function. In particular, for a polytropic gas with adiabatic exponent γ > 1, P (ρ) =
1 γ ρ γ
is the normalized pressure. Assume further that the flow is irrotational, i.e.
(2.5)
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∂u ∂v = . ∂y ∂x
(2.6)
Summing up, we consider the flow governed by the system (2.2)–(2.6). The sound speed c is defined as c2 = P (ρ) = ρ γ −1 . Define a velocity potential ϕ(x, y) and a stream function ψ(x, y), respectively, by ∂ϕ = u, ∂x
∂ϕ = v, ∂y
∂ψ = −ρv, ∂x
∂ψ = ρu, ∂y
which are ∂ϕ = q cos θ, ∂x
∂ϕ = q sin θ, ∂y
∂ψ = −ρq sin θ, ∂x
∂ψ = ρq cos θ ∂y
in terms of polar coordinates in the velocity space, where q is the speed, while θ , which is called a flow angle, is the angle of the velocity inclination to the x-axis. Then, the system (2.2)–(2.6) is transformed into the full potential equation div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(2.7)
where ρ is expressed in terms of the speed q according to the Bernoulli law [4] γ − 1 2 1/(γ −1) ρ(q 2 ) = 1 − , 0 < q < 2/(γ − 1). q 2 √ At the sonic state, the sound speed is c∗ = 2/(γ + 1), which is also the critical speed in the sense that the flow is subsonic when q < c∗ , sonic when q = c∗ and supersonic when q > c∗ . By a standard process, (2.7) may be reduced to ∂θ ∂q = −A (q) , ∂ψ ∂ϕ
∂q ∂θ = B (q) ∂ϕ ∂ψ
and ∂ 2 A(q) ∂ 2 B(q) + =0 ∂ϕ 2 ∂ψ 2
(2.8)
in the potential-stream coordinates (ϕ, ψ), where q A(q) = c∗
ρ(s 2 ) + 2s 2 ρ (s 2 ) ds, sρ 2 (s 2 )
q B(q) =
ρ(s 2 ) ds, s
0
2/(γ − 1);
c∗
furthermore, A(·) and B(·) are both the strictly increasing functions in (0, c∗ ]. Note that, since the Jacobian
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∂(ϕ, ψ) ∂ϕ ∂ψ ∂ϕ ∂ψ = − = ρq 2 ∂(x, y) ∂x ∂y ∂y ∂x does not vanish if ρq = 0, the coordinates transformation between the Cartesian coordinates and the potential-stream coordinates is valid at least in the absence of stagnation points. It can be checked easily that both (2.7) and (2.8) are elliptic in the subsonic region (q < c∗ ) and degenerate in the sonic state (q = c∗ ). Therefore, (2.7) and (2.8) for subsonic–sonic flows are both second order quasilinear elliptic equations with degeneracies at sonic states. 2.2. Formation in the physical plane Similar to the problem (1.1)–(1.6), the continuous subsonic–sonic flow problem in k is formulated as follows div(ρ(|∇ϕ|2 )∇ϕ) = 0,
(x, y) ∈ k ,
ϕ(gk (y), y) = Cin ,
−l1 < y < fk (−l0 ),
(2.10)
∂ϕ (x, −l1 ) = 0, ∂y
g(−l1 ) < x < S(−l1 ),
(2.11)
∂ϕ ∂ϕ (x, fk (x)) − fk (x) (x, fk (x)) = 0, ∂y ∂x
−l0 < x < 0,
(2.12)
|∇ϕ(S(y), y)| = c∗ , ϕ(S(y), y) = 0,
−l1 < y < 0,
(2.13)
|∇ϕ(x, y)| < c∗ ,
(x, y) ∈ k ,
(2.14)
(2.9)
where (ϕ, S, Cin ) is unknown. 2.3. Formulation in the potential plane Let us reformulate the problem (2.9)–(2.14) in the potential plane. Assume the speed of the flow at the inlet and at the upper wall to be denoted by q(gk (y), y) = Qin (y),
−l1 ≤ y ≤ fk (−l0 )
and q(x, fk (x)) = Qup (x),
−l0 ≤ x ≤ 0,
respectively. The incoming mass flux is given by fk(−l0 )
m= in
Qin (y)ρ((Qin (y))2 ) 1 + (gk (y))2 dy.
|∇ϕ(x, y)|ρ(|∇ϕ(x, y)|2 )dσ = −l1
Since the potential at the sonic curve is normalized to be zero, the potential at the inlet is given by
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−l0 ζ = Qup (x) 1 + (fk (x))2 dx. 0
At the inlet, the stream function is expressed as y in (y) =
2 Qin (s)ρ(Qin (s)) 1 + (gk (s))2 ds,
−l1 ≤ y ≤ fk (−l0 ).
−l1
At the upper wall, the potential function expressed as x up (x) =
Qup (s) 1 + (fk (s))2 ds,
−l0 ≤ x ≤ 0.
0
The inverse functions of in and up are denoted by Yin and Xup , respectively. Similar to [8,9], the subsonic–sonic flow problem in the potential plane can be formulated as follows ∂ 2 A(q) ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(2.15)
∂A(q) 1 , (ζ, ψ) = 2 ∂ϕ Rk Qin (y)ρ(Qin (y)) y=Yin (ψ)
ψ ∈ (0, m),
(2.16)
∂q (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
(2.17)
fk (x) ∂B(q) , (ϕ, m) = ∂ψ (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ)
ϕ ∈ (ζ, 0),
(2.18)
q(0, ψ) = c∗ ,
ψ ∈ (0, m),
(2.19)
y ∈ [−l1 , fk (−l0 )],
(2.20)
x ∈ [−l0 , 0],
(2.21)
, Qin (y) = q(0, ψ) ψ=in (y) Qup (x) = q(ϕ, m) , ϕ=up (x)
where (q, m, ζ ) is unknown. Similar to Remark 2.16 in [8], one has Remark 2.1. A natural boundary condition at the inlet seems to be ∂A(q) 1 , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ)) y=Yin (ψ)
ψ ∈ (0, m),
(2.22)
d 1 instead of (2.16). However, since dq < 0 in q ∈ (0, c∗ ), it seems difficult to obtain the Rk qρ(q 2 ) uniqueness of the solution to the problem (2.15), (2.22), (2.17)–(2.19). Similarly, we use (2.18) instead of a Robin one.
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2.4. Main results One of the main results in this paper is the following existence theorem. Theorem 2.1. There exists k0 ∈ (0, 1], which depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that if 0 < k ≤ k0 then the problem (2.15)–(2.21) admits a solution (q, m, ζ ) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying max c∗ /2, c∗ − M1 −kϕ ≤ q(ϕ, ψ) ≤ c∗ − M2 −kϕ, (ϕ, ψ) ∈ (ζ, 0) × (0, m),
(2.23)
where 0 < M1 ≤ M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . The regularity of the solution obtained in Theorem 2.1 is as follows. Theorem 2.2. Let (q, m, ζ ) be a solution to the problem (2.15)–(2.21) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying (2.23). Then ∂q (ϕ, ψ) ≤ M3 k 1/4 (−ϕ)−1/2 , ∂ϕ ∂q √ (ϕ, ψ) ≤ M3 k −ϕ, (ϕ, ψ) ∈ (ζ, 0) × (0, m), ∂ψ
(2.24)
where M3 > 0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Furthermore, q ∈ C 1/2 ([ζ, 0] × [0, m]) and q ∈ / C β ([ζ, 0] × [0, m]) for each β ∈ (1/2, 1). Remark 2.2. The two powers of −ϕ and the second power of k in (2.24) are optimal due to (2.23) and (2.18). Moreover, the solution obtained in Theorem 2.1 is also unique. Theorem 2.3. There exists k˜0 ∈ (0, 1] such that if 0 < k ≤ k˜0 then the problem (2.15)–(2.21) admits at most one solution (q, m, ζ ) with q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) satisfying (2.23), where k˜0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . In terms of the physical variables, Theorems 2.1 and 2.3 can be transformed as Theorem 2.4. There exist k0 ∈ (0, 1] and 0 < M1 ≤ M2 such that if 0 < k ≤ k0 then the problem (2.9)–(2.14) admits uniquely a solution (ϕ, S, Cin ) with ϕ ∈ C 3 (k ) ∩ C 2 (k \ S) ∩ C 1 (k ) and S ∈ C 1 ([−l1 , 0]) satisfying max c∗ /2, c∗ − M1 kdistS (x, y) ≤ |∇ϕ(x, y)| ≤ c∗ − M2 kdistS (x, y),
(x, y) ∈ k ,
where distS (x, y) is the distance from (x, y) to S, and k0 , M1 and M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) .
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3. Proof of existence In order to solve the problem (2.15)–(2.21) by a fixed point argument, one needs to specify Qin and Qup in advance as follows: Qin ∈ C([−l1 , fk (−l0 )]) satisfies c∗ ≤ Qin (y) ≤ c∗ , 2
−l1 ≤ y ≤ fk (−l0 )
(3.1)
and Qup ∈ C([−l0 , 0]) satisfies c∗ ≤ Qup (x) ≤ c∗ , 2
−l0 ≤ x ≤ 0.
(3.2)
Then, m, ζ , in , up , Yin and Xup are determined, and one can solve the fixed boundary problem (2.15)–(2.19). Moreover, direct calculations show that −kf (−l0 ) 1 −kf (−l0 ) ≤ ≤ , l1 (f (−l0 ) + l1 ) 1 + (f (−l0 ))2 Rk 1 m1 = c∗ ρ(c∗2 /4)l1 ≤ m ≤ m2 = c∗ ρ(c∗2 ) 1 + (f (−l0 ))2 (f (−l0 ) + l1 ), 2 1 ζ2 = −c∗ 1 + f 2L∞ (−l0 ,0) l0 ≤ ζ ≤ ζ1 = − c∗ l0 , 2 1 (y) ≤ c∗ ρ(c∗2 ) 1 + (f (−l0 ))2 , −l1 ≤ y ≤ fk (−l0 ), c∗ ρ(c∗2 /4) ≤ in 2 1 c∗ ≤ up (x) ≤ c∗ 1 + f 2L∞ (−l0 ,0) , −l0 ≤ x ≤ 0. 2 3.1. Definition of weak solutions and comparison principle Consider the following boundary value problem ∂ 2 A(q) ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(3.3)
∂q (ϕ, 0) = h1 (ϕ), ∂ψ
ϕ ∈ (ζ, 0),
(3.4)
∂q (ϕ, m) = h2 (ϕ), ∂ψ
ϕ ∈ (ζ, 0),
(3.5)
∂A(q) (ζ, ψ) = h3 (ψ), ∂ϕ
ψ ∈ (0, m),
(3.6)
q(0, ψ) = h4 (ψ),
ψ ∈ (0, m),
(3.7)
where ζ < 0, m > 0, h1 , h2 ∈ L∞ (ζ, 0), h3 , h4 ∈ L∞ (0, m) and 0 < inf(0,m) h4 ≤ sup(0,m) h4 ≤ c∗ .
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Definition 3.1. A function q ∈ L∞ ((ζ, 0) × (0, m)) is said to be a weak supersolution (subsolution) to the problem (3.3)–(3.7), if 0 < inf(ζ,0)×(0,m) q ≤ sup(ζ,0)×(0,m) q ≤ c∗ and 0 m ∂ 2ξ ∂ 2ξ A(q(ϕ, ψ)) 2 (ϕ, ψ) + B(q(ϕ, ψ)) 2 (ϕ, ψ) dϕdψ ∂ϕ ∂ψ ζ
0
0 −
0 h1 (ϕ)ξ(ϕ, 0)dϕ +
ζ
h2 (ϕ)ξ(ϕ, m)dϕ ζ
m −
m h3 (ψ)ξ(ζ, ψ)dψ −
0
h4 (ψ)
∂ξ (0, ψ)dψ ≤ (≥)0 ∂ϕ
0
for any nonnegative function ξ ∈ C 2 ([ζ, 0] × [0, m]) with ∂ξ ∂ξ = =0 (·, 0) (·, m) (ζ,0) (ζ,0) ∂ψ ∂ψ
and
∂ξ = ξ(0, ·) = 0. (ζ, ·) (0,m) (0,m) ∂ϕ
Furthermore, q ∈ L∞ ((ζ, 0) × (0, m)) is said to be a weak solution to the problem (3.3)–(3.7) if q is both a weak supersolution and a weak subsolution. It follows from the proof of Proposition 3.1 in [8] that the following comparison principle holds. Lemma 3.1. Assume that q and q are a weak supersolution and a weak subsolution to the problem (3.3)–(3.7), respectively. Then q(ϕ, ψ) ≥ q(ϕ, ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
3.2. Well-posedness of the fixed boundary problem Proposition 3.1. There exist τ > 0 and k0 ∈ (0, 1], which depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) , such that if 0 < k ≤ k0 then the problem (2.15)–(2.19) admits uniquely a weak solution. Furthermore, the solution satisfies max c∗ /2, c∗ − M1 −kϕ ≤ q(ϕ, ψ) ≤ c∗ − M2 −kϕ,
(ϕ, ψ) ∈ (ζ, 0) × (0, m), (3.8)
where 0 < M1 ≤ M2 depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Proof. Denote −1 B (q) γ + 1 2 γ − 1 2 2/(γ −1)+1 − 1 , 0 < q < c∗ , = q q 1 − A (q) 2 2 −3 B (q) γ − 1 2 3/(γ −1)+1 4 γ +1 2 = (γ + 1)q − 1 , 0 < q < c∗ q q 1 − p(q) = A (q) 2 2 b(q) =
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and set N1 =
inf
(c∗ /2,c∗ )
−A(q)b(q),
N2 = sup
−A(q)b(q),
N3 = sup (−A(q))3/2 p(q).
(c∗ /2,c∗ )
(c∗ /2,c∗ )
First, let us construct a super and a sub solutions to the problem (2.15)–(2.19). Define √ μ1 2 q(ϕ, ψ) = A−1 kμ1 ϕ(1 − kμ2 ψ 2 ) + k ϕ , 2ζ2
(ϕ, ψ) ∈ [ζ, 0] × [0, m]
(3.9)
√ μ3 2 q(ϕ, ψ) = A−1 kμ3 ϕ(1 + kμ4 ψ 2 ) − k ϕ , 4ζ2
(ϕ, ψ) ∈ [ζ, 0] × [0, m],
(3.10)
and
where √ μ1 −f (−l0 ) μ1 = , , μ2 = 3/2 2 2 8N (−ζ 2 2) 2(f (−l0 ) + l1 ) 1 + (f (−l0 )) c∗ ρ(c∗ ) √ μ3 −4f (−l0 ) , μ = . μ3 = 4 2 l1 c∗ ρ(c∗ /4) 8N2 (−ζ2 )3/2 It is clear that μ1 < μ3 , μ2 < μ4 and 2kμ1 ≤
1 1 ≤ kμ3 , 2 y=Y (ψ) 2 Rk Qin (y)ρ(Qin (y)) in
ψ ∈ (0, m).
(3.11)
Let
kˆ0 = min 1,
1 2A(c∗ /2) , . 4μ24 m42 3μ3 ζ2
If 0 < k ≤ kˆ0 , one gets that 3 3 1 A(c∗ /2) ≤ kμ3 ϕ < kμ1 ϕ ≤ A(q(ϕ, ψ)) ≤ kμ1 ϕ, (ϕ, ψ) ∈ [ζ, 0] × [0, m], 2 2 2 3 3 A(c∗ /2) ≤ kμ3 ϕ ≤ A(q(ϕ, ψ)) ≤ kμ3 ϕ, (ϕ, ψ) ∈ [ζ, 0] × [0, m], 2 4 √ ∂A(q) μ1 ζ ≤ 2kμ1 , ψ ∈ (0, m), (ζ, ψ) = kμ1 (1 − kμ2 ψ 2 ) + k ∂ϕ ζ2 ∂A(q) ∂ϕ
(ζ, ψ) = kμ3 (1 +
√ μ3 ζ 1 kμ4 ψ 2 ) − k ≥ kμ3 , 2ζ2 2
For each 0 < k ≤ kˆ0 , direction calculations and (3.12) yield
ψ ∈ (0, m).
(3.12) (3.13) (3.14) (3.15)
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∂ 2 A(q) ∂ 2 B(q) (ϕ, ψ) + (ϕ, ψ) 2 ∂ϕ ∂ψ 2 =
∂A(q) 2 ∂ 2 A(q) ∂ 2 A(q) (ϕ, ψ) + b(q(ϕ, ψ)) (ϕ, ψ) + p(q(ϕ, ψ)) (ϕ, ψ) ∂ψ ∂ϕ 2 ∂ψ 2
2k 3/2 μ1 μ2 ϕ (−2k 3/2 μ1 μ2 ϕψ)2 μ1 + N3 − N2 √ , ζ2 (−kμ1 ϕ/2)3/2 −kμ1 ϕ/2 √ √ μ1 ≤k + 2 2kN2 μ2 −μ1 ζ2 + 8 2k 3/2 N3 μ22 m22 −μ1 ζ2 ζ2 (4 − √2)μ √ 1 =k + 8 2kN3 μ22 m22 −μ1 ζ2 , (ϕ, ψ) ∈ (ζ, 0) × (0, m) 4ζ2 ≤k
(3.16)
and ∂B(q) −2k 3/2 μ1 μ2 ϕm ∂A(q) (ϕ, m) = b(q(ϕ, m)) (ϕ, m) ≥ N1 √ ∂ψ ∂ψ −3kμ1 ϕ/2 2 √ ≥ √ kN1 μ1 c∗ μ2 m1 −Xup (ϕ), ϕ ∈ (ζ, 0). 3
(3.17)
Similarly, for each 0 < k ≤ kˆ0 , it follows from (3.13) that ∂ 2 A(q)
∂ 2 B(q)
2k 3/2 μ3 μ4 ϕ μ3 + N √ 2 2ζ2 ∂ϕ 2 ∂ψ 2 −3kμ3 ϕ/4 √ 4 3 − 3 μ3 μ3 − √ kN2 μ4 −μ3 ζ2 = − ≥ 0, (ϕ, ψ) ∈ (ζ, 0) × (0, m) k ≥ −k 2ζ2 6 ζ2 3 (ϕ, ψ) +
(ϕ, ψ) ≥ −k
(3.18)
and 2k 3/2 μ3 μ4 ϕm 2 √ ≤ − √ kN1 μ3 c∗ μ4 m1 −Xup (ϕ), (ϕ, m) ≤ N1 √ ∂ψ −3kμ3 ϕ/2 3
∂B(q)
ϕ ∈ (ζ, 0). (3.19)
Set 1 √ 3/2 τ = √ N1 μ1 μ2 c∗ m1 , 3
k0 = min kˆ0 ,
μ1 . 29 N32 μ42 m42 (−ζ2 )3
Then, for each 0 < k ≤ k0 , (3.11) and (3.14)–(3.19) show that q and q defined by (3.9) and (3.10) are a super and a sub solutions to the problem (2.15)–(2.19), respectively. Below we establish the existence of the weak solution to the problem (2.15)–(2.19) with 0 < k ≤ k0 by using a method of elliptic regularization. For each positive integer n, consider the following regularized problem
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∂ 2 A(qn ) ∂ 2 B(qn ) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
(3.20)
∂A(qn ) 1 , (ζ, ψ) = 2 ∂ϕ Rk Qin (y)ρ(Qin (y)) y=Yin (ψ)
ψ ∈ (0, m),
(3.21)
∂qn (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
(3.22)
ϕ ∈ (ζ, 0),
(3.23)
ψ ∈ (0, m).
(3.24)
fk (x) ∂B(qn ) , (ϕ, m) = ∂ψ (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ) n c∗ , qn (0, ψ) = n+1
Note that (3.20) is a uniformly elliptic equation provided that 0 < inf(ζ,0)×(0,m) qn ≤ sup(ζ,0)×(0,m) qn < c∗ . Therefore, to get the existence of the weak solution to the problem (3.20)–(3.24), it suffices to construct an upper and a sub solutions q n and q n satisfying 0 < inf(ζ,0)×(0,m) q n ≤ sup(ζ,0)×(0,m) q n < c∗ . For εn > 0, set √ q n (ϕ, ψ) = A−1 kμ1 (ϕ − εn )(1 − kμ2 ψ 2 ) + k
μ1 (ϕ − εn )2 , (ϕ, ψ) ∈ [ζ, 0] × [0, m]. 2(ζ2 − εn )
For each 0 < k ≤ k0 and n ≥ 1, the above calculations show that q n is a supersolution to the problem (3.20)–(3.24) for sufficiently small εn > 0. For sufficiently large μ5 , μ6 > 0, it is not hard to show that q n (ϕ, ψ) = A−1 c∗ /2 + kμ5 ϕ(ϕ − 3ζ2 ) − kμ6 ψ 2 , (ϕ, ψ) ∈ [ζ, 0] × [0, m] is a subsolution to the problem (3.20)–(3.24) for each 0 < k ≤ k0 and n ≥ 1. Therefore, for each positive integer n, the problem (3.20)–(3.24) with 0 < k ≤ k0 admits a weak solution qn . Lemma 3.1 yields qn1 (ϕ, ψ) ≤ qn2 (ϕ, ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
n 1 ≤ n2 .
Set q(ϕ, ψ) = lim qn (ϕ, ψ), n→∞
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
Then, q is a weak solution to the problem (2.15)–(2.19) with 0 < k ≤ k0 . Since q and q defined by (3.9) and (3.10) are a super and a sub solutions to the problem (2.15)–(2.19), respectively, Lemma 3.1 yields (3.8). 2 Remark 3.1. According to the definition of τ , one gets from (2.1) that 1 f (x) ≤ f (−l0 ), 2
−l0 ≤ x ≤ 0.
Below, we establish the Hölder continuity of weak solutions to the problem (2.15)–(2.19).
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Proposition 3.2. Assume that q is a weak solution to the problem (2.15)–(2.19) satisfying (3.8). Then, there exist β ∈ (0, 1/2] and K > 0, which depend only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) and k, such that q ∈ C β ([ζ, 0] × [0, m]) and [q]β,(ζ,0)×(0,m) ≤ K.
(3.25)
Proof. In the proof we use Ki > 0 (1 ≤ i ≤ 6) and βi ∈ (0, 1) (i = 1, 2) to denote generic constants depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) and k. Thanks to (3.8) and the classical theory on uniformly elliptic equations, one can get that q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C([ζ, 0] ×[0, m]). Furthermore, the Harnack inequality [5] shows that q ∈ C β1 ([ζ, −δ0 ] ×[0, m]) and [q]β1 ,(ζ,−δ0 )×(0,m) ≤ K1 ,
(3.26)
where δ0 = min − ζ /2, m/2 . Fix 0 < δ < δ0 . Then q ∈ C 2 ([−2δ, −δ/4] × (0, m)) ∩ C([−2δ, −δ/4] × [0, m]) solves ∂ ∂q ∂ ∂q a(ϕ, ψ) + h(ϕ, ψ) = 0, ∂ϕ ∂ϕ ∂ψ ∂ψ
(ϕ, ψ) ∈ (−2δ, −δ/4) × (0, m),
∂q (ϕ, 0) = 0, ∂ψ
ϕ ∈ (−2δ, −δ/4),
∂q (ϕ, m) = e(ϕ), ∂ψ
ϕ ∈ (−2δ, −δ/4),
where a(ϕ, ψ) = A (q(ϕ, ψ)),
h(ϕ, ψ) = B (q(ϕ, ψ)),
(ϕ, ψ) ∈ [−2δ, −δ/4] × [0, m]
and e(ϕ) =
fk (x) 1 , · B (q(ϕ, m)) (1 + (fk (x))2 )3/2 Qup (x) x=Xup (ϕ)
ϕ ∈ [−2δ, −δ/4].
Introducing ϕ˜ = δ −1 (ϕ + δ),
ϕ ∈ [−2δ, −δ/4],
ψ˜ = δ −3/4 ψ,
ψ ∈ [0, m]
and setting ˜ = q(−δ + δ ϕ, ˜ − c∗ , q( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ)
˜ ∈ [−1, 3/4] × [0, δ −3/4 m], (ϕ, ˜ ψ)
one can check that q˜ ∈ C 2 ([−1, 3/4] × (0, δ −3/4 m)) ∩ C([−1, 3/4] × [0, δ −3/4 m]) solves
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∂ q˜ ∂ −1/2 ∂ q˜ ∂ ˜ ˜ ˜ h(ϕ, ˜ ψ) = 0, a( ˜ ϕ, ˜ ψ) δ + ∂ ϕ˜ ∂ ϕ˜ ∂ ψ˜ ∂ ψ˜ ∂ q˜ (ϕ, ˜ 0) = 0, ∂ ψ˜ ∂ q˜ (ϕ, ˜ δ −3/4 m) = e( ˜ ϕ), ˜ ∂ ψ˜
(ϕ, ˜ ψ) ∈ (−1, 3/4) × (0, δ −3/4 m), ϕ˜ ∈ (−1, 3/4), ϕ˜ ∈ (−1, 3/4),
where ˜ = a(−δ + δ ϕ, ˜ a( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ), ˜ ϕ, ˜ ˜ = h(−δ + δ ϕ, h( ˜ ψ) ˜ δ 3/4 ψ),
˜ ∈ [−1, 3/4] × [0, δ −3/4 m] (ϕ, ˜ ψ)
and e( ˜ ϕ) ˜ = δ 3/4 e(−δ + δ ϕ), ˜
ϕ˜ ∈ [−1, 3/4].
Due to (3.8) and (2.1), one gets that √ √ ˜ ≤ K3 δ, K2 δ ≤ a( ˜ ϕ, ˜ ψ)
˜ ϕ, ˜ ≤ K3 , (ϕ, ˜ ∈ [−1, 3/4] × [0, δ −3/4 m], K2 ≤ h( ˜ ψ) ˜ ψ) √ q ˜ L∞ ((−1,3/4)×(0,δ −3/4 m)) ≤ K3 δ, e ˜ L∞ (−1,3/4) ≤ K3 δ 5/4 .
Therefore, the Harnack inequality [5] shows that q˜ ∈ C β2 ([−1/2, 1/2] × [0, δ −3/4 m]) and [q] ˜ β2 ,(−1/2,1/2)×(0,δ −3/4 m) ≤ K4 q ˜ L∞ (−1,3/4) ˜ L∞ ((−1,3/4)×(0,δ −3/4 m)) + e √ ≤ K5 δ. (3.27) Set β = min 1/2, β1 , β2 . Then, one gets from (3.8), (3.26) and (3.27) that |q(ϕ, ψ) − c∗ | ≤ K6 (−ϕ)β , [q]β,(ζ,−δ0 )×(0,m) ≤ K6 ,
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
[q]β,(−3δ/2,−δ/2)×(0,m) ≤ K6 ,
which yield q ∈ C β ([ζ, 0] × [0, m]) and (3.25).
0 < δ < δ0 ,
2
3.3. Proof of the existence theorem Proof of Theorem 2.1. Set
S = (Qin , Qup ) ∈ C([−l1 , fk (−l0 )]) × C([−l0 , 0]) : Qin satisfies (3.1) and Qup satisfies (3.2) with the norm
(Qin , Qup )S = max Qin L∞ (−l1 ,fk (−l0 )) , Qup L∞ (−l0 ,0) ,
(Qin , Qup ) ∈ S .
For given (Qin , Qup ) ∈ S , it follows from Propositions 3.1 and 3.2 that the problem (2.15)–(2.19) admits a unique weak solution q ∈ C β ([ζ, 0] × [0, m]) satisfying the estimates
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(3.8) and (3.25). Set ˆ in (y) = q(ζ, in (y)), Q
−l1 ≤ y ≤ fk (−l0 )
and ˆ up (x) = q(up (x), m), Q
−l0 ≤ x ≤ 0.
ˆ in ∈ C β ([−l1 , fk (−l0 )]) satisfies (3.1), It follows from q ∈ C β ([ζ, 0] × [0, m]) and (3.8) that Q β ˆ and Qup ∈ C ([−l0 , 0]) satisfies (3.2). Therefore, we can define a mapping J from S to itself as follows ˆ in , Q ˆ up ), J ((Qin , Qup )) = (Q
(Qin , Qup ) ∈ S .
Due to (3.25), J is compact. Similar to the discussion of Theorem 4.1 in [9], one can prove that J is also continuous from (3.25) and Lemma 3.1. Then, the Schauder fixed point theorem and (3.8) show that the problem (2.15)–(2.21) admits a weak solution (q, m, ζ ) with q ∈ C β ([ζ, 0] × [0, m]) satisfying (2.23). Furthermore, the classical theory on uniformly elliptic equations leads to q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]). 2 4. Proof of the regularity theorem In this section, let us prove Theorem 2.2. First we prove the following estimate by a rescaling technique. Proposition 4.1. Under the assumption of Theorem 2.2, ∂q (ϕ, ψ) ≤ Mk 1/4 (−ϕ)−1/2 , ∂ϕ ∂q √ (ϕ, ψ) ≤ M k(−ϕ)−1/4 , (ϕ, ψ) ∈ (ζ, 0) × (0, m), ∂ψ
(4.1)
where M > 0 depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Proof. In the proof we use Ci (1 ≤ i ≤ 14) to denote a generic positive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Note that q ∈ C 2 ((ζ, 0) × (0, m)) ∩ C 1 ([ζ, 0) × [0, m]) ∩ C([ζ, 0] × [0, m]) solves ∂ ∂q ∂ ∂q a(ϕ, ψ) + h(ϕ, ψ) = 0, ∂ϕ ∂ϕ ∂ψ ∂ψ ∂q (ζ, ψ) = r(ψ), ∂ϕ ∂q (ϕ, 0) = 0, ∂ψ ∂q (ϕ, m) = e(ϕ), ∂ψ
(ϕ, ψ) ∈ (ζ, 0) × (0, m), ψ ∈ (0, m), ϕ ∈ (ζ, 0), ϕ ∈ (ζ, 0),
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where a(ϕ, ψ) = A (q(ϕ, ψ)), r(ψ) = e(ϕ) =
h(ϕ, ψ) = B (q(ϕ, ψ)), 1
Rk
q(ζ, ψ)ρ(q 2 (ζ, ψ))A (q(ζ, ψ))
(ϕ, ψ) ∈ (ζ, 0) × (0, m), ,
ψ ∈ (0, m),
fk (Xup (ϕ)) , (1 + (fk (Xup (ϕ)))2 )3/2 q(ϕ, m)B (q(ϕ, m))
ϕ ∈ (ζ, 0).
Denote δ0 = min − ζ /2, m/2 . Introducing ϕˆ = k −1/4 ϕ,
ϕ ∈ [ζ, −δ0 /2]
and setting q( ˆ ϕ, ˆ ψ) = q(k 1/4 ϕ, ˆ ψ) − c∗ ,
(ϕ, ˆ ψ) ∈ [k −1/4 ζ, −k −1/4 δ0 /2] × [0, m],
one can check that qˆ ∈ C 2 ((ζ, −δ0 /2) × (0, m)) ∩ C 1 ([ζ, −δ0 /2] × [0, m]) solves ∂ −1/2 ∂ qˆ ∂ ˆ ∂ qˆ a( ˆ ϕ, ˆ ψ) k + h(ϕ, ˆ ψ) = 0, ∂ ϕˆ ∂ ϕˆ ∂ψ ∂ψ
(ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m),
∂ qˆ −1/4 ζ, ψ) = k 1/4 r(ψ), (k ∂ ϕˆ
ψ ∈ (0, m),
∂ qˆ (ϕ, ˆ 0) = 0, ∂ψ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2),
∂ qˆ (ϕ, ˆ m) = e( ˆ ϕ), ˆ ∂ψ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2),
where a( ˆ ϕ, ˆ ψ) = a(k 1/4 ϕ, ˆ ψ),
ˆ ϕ, h( ˆ ψ) = h(k 1/4 ϕ, ˆ ψ),
(ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m)
and e( ˆ ϕ) ˆ = e(k 1/4 ϕ), ˆ
ϕˆ ∈ (k −1/4 ζ, −k −1/4 δ0 /2).
It follows from (2.23) that √ √ ˆ ϕ, C1 k ≤ a( ˆ ϕ, ˆ ψ) ≤ C2 k, C1 ≤ h( ˆ ψ) ≤ C2 , (ϕ, ˆ ψ) ∈ (k −1/4 ζ, −k −1/4 δ0 /2) × (0, m), √ √ ˆ L∞ (k −1/4 ζ,−k −1/4 δ0 /2) ≤ C2 k. q ˆ L∞ ((k −1/4 ζ,−k −1/4 δ0 /2)×(0,m)) ≤ C2 k, rL∞ (0,m) ≤ C2 k, e ˆ
Therefore, the Harnack inequality [5] shows that qˆ ∈ C β ([k −1/4 ζ, −3k −1/4 δ0 /4] × [0, m]) and
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[q] ˆ β;(k ˆ L∞ ((k −1/4 ζ,−k −1/4 δ0 /2)×(0,m)) + k 1/4 rL∞ (0,m) ˆ −1/4 ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C3 q √ + e ˆ L∞ (k −1/4 ζ,−k −1/4 δ0 /2) ≤ C4 k, (4.2) where βˆ ∈ (0, 1) depends only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . Due to (4.2), one gets that √ √ ˆ ˆ −1/4 [a] ˆ β;(k ˆ −1/4 ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C5 k, [h]β;(k ζ,−3k −1/4 δ0 /4)×(0,m) ≤ C5 k, √ [r]β;(0,m) ≤ C5 k, [e] ˆ α β;(k ˆ ˆ −1/4 ζ,−3k −1/4 δ0 /4) ≤ C5 k. ˆ
The classical Schauder theory shows that qˆ ∈ C 1,α β ([k −1/4 ζ, −k −1/4 δ0 ] × [0, m]) satisfies q ˆ 1,α β;(k ≤ C q ˆ L∞ ((k −1/4 ζ,−3k −1/4 δ0 /4)×(0,m)) + k 1/4 rα β;(0,m) −1/4 −1/4 6 ˆ ˆ ζ,−k δ0 )×(0,m) √ + e ˆ α β;(k (4.3) ˆ −1/4 ζ,−3k −1/4 δ0 /4) ≤ C7 k. Transforming (4.3) into the (ϕ, ψ) plane yields ∂q ≤ C7 k 1/4 , ∞ ∂ϕ L ((ζ,−δ0 )×(0,m))
∂q √ ≤ C7 k. ∞ ∂ψ L ((ζ,−δ0 )×(0,m))
(4.4)
Fix 0 < δ < δ0 . Introducing ϕ˜ = k −1/4 δ −1 (ϕ + δ),
ϕ ∈ [−2δ, −δ/4],
ψ˜ = δ −3/4 ψ,
ψ ∈ [0, m]
and setting ˜ = q(−δ + k 1/4 δ ϕ, ˜ − c∗ , q( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ)
˜ ∈ [−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m], (ϕ, ˜ ψ)
one can check that q˜ ∈ C 2 ([−k −1/4 , 3/4k −1/4 ] × (0, δ −3/4 m)) ∩ C 1 ([−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m]) solves ∂ q˜ ∂ −1/2 −1/2 ∂ q˜ ∂ ˜ ˜ ˜ h(ϕ, ˜ ψ) = 0, δ a( ˜ ϕ, ˜ ψ) k + ∂ ϕ˜ ∂ ϕ˜ ∂ ψ˜ ∂ ψ˜ ˜ ∈ (−k −1/4 , 3/4k −1/4 ) × (0, δ −3/4 m), (ϕ, ˜ ψ) ∂ q˜ (ϕ, ˜ 0) = 0, ∂ ψ˜
ϕ˜ ∈ (−k −1/4 , 3/4k −1/4 ),
∂ q˜ (ϕ, ˜ δ −3/4 m) = e( ˜ ϕ), ˜ ˜ ∂ψ
ϕ˜ ∈ (−k −1/4 , 3/4k −1/4 ),
˜ ∈ [−k −1/4 , 3/4k −1/4 ] × [0, δ −3/4 m], where for (ϕ, ˜ ψ) ˜ = a(−δ + k 1/4 δ ϕ, ˜ a( ˜ ϕ, ˜ ψ) ˜ δ 3/4 ψ),
˜ ϕ, ˜ = h(−δ + k 1/4 δ ϕ, ˜ ˜ δ 3/4 ψ), h( ˜ ψ)
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and e( ˜ ϕ) ˜ = δ 3/4 e(−δ + k 1/4 δ ϕ), ˜
ϕ˜ ∈ [−k −1/4 , 3/4k −1/4 ].
Due to (2.23) and (2.1), one gets that √ √ ˜ ϕ, ˜ ≤ C9 kδ, C8 ≤ h( ˜ ≤ C9 , (ϕ, ˜ ∈ (−k −1/4 , 3/4k −1/4 ) × (0, δ −3/4 m), C8 kδ ≤ a( ˜ ϕ, ˜ ψ) ˜ ψ) ˜ ψ) √ ˜ L∞ (−k −1/4 ,3/4k −1/4 ) ≤ C9 kδ 5/4 . q ˜ L∞ ((−k −1/4 ,3/4k −1/4 )×(0,δ −3/4 m)) ≤ C9 kδ, e ˜
Therefore, the Harnack inequality [5] shows that q˜ ∈ C β ([−1/2k −1/4 , 1/2k −1/4 ] × [0, δ −3/4 m]) and [q] ˜ β;(−1/2k ˜ L∞ ((−k −1/4 ,3/4k −1/4 )×(0,δ −3/4 m)) −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C10 q ˜ √ + e ˜ L∞ (−k −1/4 ,3/4k −1/4 ) ≤ C11 kδ, which leads to √ [a] ˜ β;(−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C12 kδ, ˜
√ ˜ ˜ [h] β;(−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m) ≤ C12 kδ, √ [e] ˜ α β;(−1/2k −1/4 ,1/2k −1/4 ) ≤ C12 kδ. ˜ ˜
The classical Schauder theory shows that q˜ ∈ C 1,α β ([−1/4k −1/4 , 1/4k −1/4 ] × [0, δ −3/4 m]) satisfies q ˜ 1,α β;(−1/4k ˜ L∞ ((−1/2k −1/4 ,1/2k −1/4 )×(0,δ −3/4 m)) −1/4 ,1/4k −1/4 )×(0,δ −3/4 m) ≤ C13 q ˜ √ + e ˜ α β;(−1/2k (4.5) −1/4 ,1/2k −1/4 ) ≤ C14 kδ. ˜ Transforming (4.5) into the (ϕ, ψ) plane yields ∂q ∂q √ (ϕ, ψ) ≤ C14 k(−ϕ)−1/4 , (ϕ, ψ) ∈ (−δ0 , 0) × (0, m), (ϕ, ψ) ≤ C14 k 1/4 (−ϕ)−1/2 , ∂ϕ ∂ψ which and (4.4) lead to (4.1).
2
Remark 4.1. The powers of −ϕ and k in (4.1) are optimal when one uses a recalling technique. Owing to (2.16) and (2.18), the two powers of k in (4.1) are not greater than 1/2 and 1, respectively. As in the proof of Proposition 3.2 in [8], we introduce two functions to prove Theorem 2.2. The first one is E(s) = A(B −1 (s)), Clearly, E ∈ C ∞ ((−∞, 0)) satisfies
s < 0.
C. Wang / J. Differential Equations 259 (2015) 2546–2575
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γ + 1 2 γ − 1 2 −2/(γ −1)−1 E (s) = 1 − > 0, q q 1− q=B −1 (s) 2 2 γ − 1 2 −3/(γ −1)−2 E (s) = −(γ + 1)q 4 1 − < 0, q q=B −1 (s) 2 γ − 1 2 −4/(γ −1)−3 < 0, q E (s) = −(γ + 1)q 4 (4 + 3q 2 ) 1 − q=B −1 (s) 2
s < 0, s < 0, s < 0.
The other function is G(s) = E (E −1 (s)),
s < 0,
which satisfies G ∈ C ∞ ((−∞, 0)) and G (s) =
E (t) < 0, E (t) t=E −1 (s)
G (s) < −
(E (t))2 < 0, (E (t))3 t=E −1 (s)
s < 0.
Proof of Theorem 2.2. The former estimate in (2.24) follows from the first formula in (4.1). Below we prove the latter estimate in (2.24) and we use Ci (1 ≤ i ≤ 3) to denote a generic pos itive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ) and f L∞ (−l0 ,0) . For each integer 2 2 n ≥ max − 2/ζ, 4M2 /c∗ , consider the following regularized problem ∂ 2 A(qn ) ∂ 2 B(qn ) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m), (4.6)
1 ∂A(qn ) , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, m),
(4.7)
∂qn (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, −1/n),
(4.8)
fk (Xup (ϕ + 1/n)) ∂B(qn ) , ϕ ∈ (ζ, −1/n), (ϕ, m) = ∂ψ (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m) √ qn (−1/n, ψ) = c∗ − M2 / n, ψ ∈ (0, m).
(4.9) (4.10)
Clearly, q is a supersolution to the problem (4.6)–(4.10). It is not hard to show that
q(ϕ, ψ) = A−1 c∗ /2 + μ1 ϕ(ϕ − 3ζ2 ) − μ2 ψ 2 ,
(ϕ, ψ) ∈ [ζ, 0] × [0, m]
is a subsolution to the problem (4.6)–(4.10), where μ1 and μ2 are sufficiently large positive constants independent of n. Therefore, the problem (4.6)–(4.10) admits a solution qn ∈ C 2 ((ζ, −1/n) × (0, m)) ∩ C 1 ([ζ, −1/n] × [0, m]) satisfying 0 < q(ϕ, ψ) ≤ qn (ϕ, ψ) ≤ q(ϕ, ψ) < c∗ , Set
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m].
(4.11)
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zn (ϕ, ψ) =
∂B(qn ) (ϕ, ψ), ∂ψ
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m].
Then zn solves the following problem ∂ 2 zn ∂ 2 zn ∂zn ∂zn + + jn,2 (ϕ, ψ) + jn,3 (ϕ, ψ) + jn,4 (ϕ, ψ)zn = 0, ∂ϕ ∂ψ ∂ϕ 2 ∂ψ 2 (ϕ, ψ) ∈ (ζ, −1/n) × (0, m), B (qn (ζ, ψ)) 1 ∂zn , (ζ, ψ) + jn,0 (ψ)zn (ζ, ψ) = ∂ϕ Rk A (qn (ζ, ψ)) q(ζ, ψ)ρ(q 2 (ζ, ψ)) ψ ∈ (0, m),
(4.13)
zn (ϕ, 0) = 0,
ϕ ∈ (ζ, −1/n),
(4.14)
fk (Xup (ϕ + 1/n)) , zn (ϕ, m) = (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m)
ϕ ∈ (ζ, −1/n),
(4.15)
zn (−1/n, ψ) = 0,
ψ ∈ (0, m),
(4.16)
jn,1 (ϕ, ψ)
(4.12)
where for (ϕ, ψ) ∈ (ζ, −1/n) × (0, m), jn,1 (ϕ, ψ) = G(A(qn (ϕ, ψ))),
jn,2 (ϕ, ψ) = 2G (A(qn (ϕ, ψ)))
∂A(qn (ϕ, ψ)) , ∂ϕ
∂B(qn (ϕ, ψ)) , ∂ψ ∂A(q (ϕ, ψ)) 2 n jn,4 (ϕ, ψ) = G (A(qn (ϕ, ψ))) , ∂ϕ jn,3 (ϕ, ψ) = −G (A(qn (ϕ, ψ)))
and jn,0 (ψ) =
E (A(qn (ζ, ψ))) 1 < 0, · 2 (E (A(qn (ζ, ψ)))) Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, min ).
Due to (2.23), (4.1) and (2.1), one gets that B (q (ζ, ψ)) 1 n ≤ C1 k 3/2 , ψ ∈ (0, m), 2 Rk A (qn (ζ, ψ)) q(ζ, ψ)ρ(q (ζ, ψ)) √ fk (Xup (ϕ + 1/n)) ≤ C1 k −ϕ, ψ ∈ (0, m). 2 3/2 (1 + (fk (Xup (ϕ + 1/n))) ) q(ϕ, m)
(4.17) (4.18)
Note that √ 1 jn,1 (ϕ, ψ)(−ϕ)−3/2 − jn,4 (ϕ, ψ) −ϕ ≥ −jn,1 (ϕ, ψ)j4 (ϕ, ψ)(−ϕ)−1/2 4 1 ≥ − jn,2 (ϕ, ψ)(−ϕ)−1/2 , (ϕ, ψ) ∈ (ζ, −1/n) × (0, m). 2 Owing to (4.17)–(4.19), one can verify that
(4.19)
C. Wang / J. Differential Equations 259 (2015) 2546–2575
√ zn (ϕ, ψ) = (2 −ζ + 1)C1 k −ϕ,
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(ϕ, ψ) ∈ [ζ, −1/n] × [0, m]
and √ zn (ϕ, ψ) = −(2 −ζ + 1)C1 k −ϕ,
(ϕ, ψ) ∈ [ζ, −1/n] × [0, m]
are super and sub solutions to the problem (4.12)–(4.16). The classical comparison principle shows that √ |zn (ϕ, ψ)| ≤ (2 −ζ + 1)C1 k −ϕ,
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m),
which, together with (2.23), leads to ∂q √ n (ϕ, ψ) ≤ C2 k −ϕ, ∂ψ
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m).
(4.20)
Due to (4.6)–(4.10), one gets that m 0
m
fk (Xup (ϕ + 1/n)) ∂ 2 A(qn ) , (ϕ, ψ)dψ = − ∂ϕ 2 (1 + (fk (Xup (ϕ + 1/n)))2 )3/2 q(ϕ, m) ∂A(qn ) 1 (ζ, ψ)dψ = ∂ϕ Rk
0
m
ϕ ∈ (ζ, −1/n),
1 dψ, q(ζ, ψ)ρ(q 2 (ζ, ψ))
0
m
√ A(qn )(−1/n, ψ)dψ = mA(c∗ − M2 / n),
0
which, together with (2.23) and (2.1), give m A(qn )(ϕ, ψ)dψ ≥ C3 ϕ,
ϕ ∈ (ζ, −1/n).
(4.21)
(ϕ, ψ) ∈ (ζ, −1/n) × (0, m).
(4.22)
0
It follows from (4.20) and (4.21) that √ qn (ϕ, ψ) ≥ c∗ − C4 −ϕ,
Owing to (4.11), (4.22), (4.20) and the classical theory on uniformly elliptic equations, there exists a subsequence of {qn }, denoted by itself for convenience, and a function q˜ ∈ L∞ ((ζ, 0) × (0, m)) satisfying 0 < inf(ζ,0)×(0,m) q˜ ≤ sup(ζ,0)×(0,m) q˜ ≤ c∗ and √ q(ϕ, ˜ ψ) ≥ c∗ − C4 −ϕ, such that
∂ q˜ √ (ϕ, ψ) ≤ C2 k −ϕ, ∂ψ
(ϕ, ψ) ∈ (ζ, 0) × (0, m) (4.23)
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lim qn (ϕ, ψ) = q(ϕ, ˜ ψ),
n→∞
(ϕ, ψ) ∈ (ζ, 0) × (0, m).
Then, it is not hard to show that q˜ solves ∂ 2 A(q) ˜ ˜ ∂ 2 B(q) + = 0, ∂ϕ 2 ∂ψ 2
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
1 ∂A(q) ˜ , (ζ, ψ) = ∂ϕ Rk q(ζ, ψ)ρ(q 2 (ζ, ψ))
ψ ∈ (0, m),
∂ q˜ (ϕ, 0) = 0, ∂ψ
ϕ ∈ (ζ, 0),
fk (Xup (ϕ)) ∂B(q) ˜ , (ϕ, m) = ∂ψ (1 + (fk (Xup (ϕ)))2 )3/2 q(ϕ, m)
ϕ ∈ (ζ, 0),
q(0, ˜ ψ) = c∗ ,
ψ ∈ (0, m).
Lemma 3.1 shows q(ϕ, ψ) = q(ϕ, ˜ ψ),
(ϕ, ψ) ∈ (ζ, 0) × (0, m),
which, together with the second estimate in (4.23), leads to the latter estimate in (2.24). Due to / C β ([ζ, 0] × [0, m]) for each (2.23) and (2.24), it is clear that q ∈ C 1/2 ([ζ, 0] × [0, m]) and q ∈ β ∈ (1/2, 1). 2 5. Proof of the uniqueness In the last section, we prove Theorem 2.3. Similar to [8], not only the degeneracy and the free boundaries but also the nonlocal terms in (2.16) and (2.18) bring essential difficulties. Although these nonlocal terms arise from the coordinates transformation from the Cartesian coordinates to the potential-stream coordinates, it is not convenient to prove the uniqueness theorem in the Cartesian coordinates since the outlet of the problem (2.9)–(2.14) is a free boundary whose shape is unknown. In order to prove Theorem 2.3, we first introduce a suitable coordinates transformation to transform (2.16) and (2.18) into usual boundary conditions and to fix the two free boundaries into fixed ones. Let (qi , mi , ζi ) with qi ∈ C 2 ((ζi , 0) × (0, mi )) ∩ C 1 ([ζi , 0) × [0, mi ]) ∩ C([ζi , 0] × [0, mi ]) and (2.23) be a solution to the problem (2.15)–(2.21) for i = 1, 2. Denote in,i , up,i , Yin,i and Xup,i to be the associated functions for i = 1, 2. For i = 1, 2, introduce the coordinates transformation ϕ = up,i (x), −l0 ≤ x ≤ 0, x = Xup,i (ϕ), ζ ≤ ϕ ≤ 0, 0 ≤ ψ ≤ m, y = Yin,i (ψ), ψ = in,i (y), −l1 ≤ y ≤ fk (−l0 ). Define wi (x, y) = A(q(up,i (x), in,i (y))), Then, wi (i = 1, 2) solves
(x, y) ∈ [−l0 , 0] × [−l1 , fk (−l0 )],
i = 1, 2.
C. Wang / J. Differential Equations 259 (2015) 2546–2575
∂ Xi (x) ∂wi ∂ Yi (y) ∂B(A−1 (wi )) + = 0, ∂x Yi (y) ∂x ∂y Xi (x) ∂y
2571
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )), (5.1)
1 Xi (−l0 ) ∂wi , (−l0 , y) = Yi (y) ∂x 2 Rk − (y + l1 )2
y ∈ (−l1 , fk (−l0 )),
(5.2)
∂wi (x, 0) = 0, ∂y
x ∈ (−l0 , 0),
(5.3)
fk (x) Yi (fk (−l0 )) ∂B(A−1 (wi )) , (x, fk (−l0 )) = Xi (x) ∂y 1 + (fk (x))2
x ∈ (−l0 , 0),
(5.4)
wi (0, y) = 0,
y ∈ (−l1 , fk (−l0 )),
(5.5)
where 1 Xi (x) = , (1 + (fk (x))2 A−1 (wi (x, fk (−l0 ))) Rk2 − (y + l1 )2 , Yi (y) = Rk A−1 (wi (−l0 , y))ρ((A−1 (wi (−l0 , y)))2 )
x ∈ [−l0 , 0],
(5.6)
y ∈ [−l1 , fk (−l0 )].
(5.7)
The problem (5.1)–(5.5) is a problem possessing common boundary conditions in a fixed domain. Proof of Theorem 2.3. In the proof we use Ci (1 ≤ i ≤ 5) to denote a generic positive constant depending only on γ , l0 , l1 , f (−l0 ), f (−l0 ), f L∞ (−l0 ,0) , M1 and M2 . Set w(x, y) = w1 (x, y) − w2 (x, y),
(x, y) ∈ [−l0 , 0] × [−l1 , fk (−l0 )].
Due to (5.1)–(5.5), w solves ∂ X1 (x) ∂w ∂ Y1 (y) ∂w ∂ X(x) ∂w2 ∂ X2 (x)Y (y) ∂w2 + h + − ∂x Y1 (y) ∂x ∂y X1 (x) ∂y ∂x Y1 (y) ∂x ∂x Y1 (y)Y2 (y) ∂x −1 ∂ Y (y) ∂B(A (w2 )) ∂ X(x)Y2 (y) ∂B(A−1 (w2 )) ∂ Y1 (y) ∂h w + − = 0, + ∂y X1 (x) ∂y ∂y X1 (x) ∂y ∂y X1 (x)X2 (x) ∂y (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
(5.8)
where 1 h(x, y) =
B −1 (A (λw1 (x, y) + (1 − λ)w2 (x, y)))dλ, A
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
0
X(x) = X1 (x) − X2 (x),
x ∈ [−l0 , 0],
Y (y) = Y1 (y) − Y2 (y),
Due to (2.23) and (2.24), direct calculations show that
y ∈ [−l1 , fk (−l0 )].
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∂h C1 C2 C2 ≤ h(x, y) ≤ √ , (x, y) ≤ √ , (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )), √ ∂y −x −kx −kx (5.9) C2 w(x, fk (−l0 )), x ∈ [−l0 , 0], |Y (y)| ≤ C2 w(−l0 , y), y ∈ [−l1 , fk (−l0 )], |X(x)| ≤ √ −kx (5.10) ∂w ∂B(A−1 (w )) √ 2 2 (x, y) ≤ C2 k 3/4 , (x, y) ≤ C2 k −x, (x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )). ∂x ∂y (5.11) Multiplying (5.8) by −w and then integrating over (−l0 , 0) × (−l1 , fk (−l0 )) by parts and using (5.2)–(5.5), one gets that 0
fk(−l0 )
−l0
−l1
X1 (x) ∂w 2 dxdy + Y1 (y) ∂x
−l0
0 fk(−l0 ) =− −l0
0
−l1
0
−l1
Y1 (y) ∂w 2 dxdy h X1 (x) ∂y
fk(−l0 ) −l1
0
Y1 (y) ∂h ∂w w dxdy − X1 (x) ∂y ∂y
−l0
fk(−l0 )
+ −l0
0
−l0
− 0
−l1
X(x) ∂w2 ∂w dxdy + Y1 (y) ∂x ∂x
fk(−l0 )
−l0
fk(−l0 )
−l1
X2 (x)Y (y) ∂w2 ∂w dxdy Y1 (y)Y2 (y) ∂x ∂x
fk(−l0 ) −l1
Y (y) ∂B(A−1 (w2 )) ∂w dxdy X1 (x) ∂y ∂y
X(x)Y2 (y) ∂B(A−1 (w2 )) ∂w dxdy, X1 (x)X2 (x) ∂y ∂y
which, together with (2.23), (2.24), (5.9)–(5.11), leads to 0
fk(−l0 )
∂w 2
−l0
−l1
≤ C3 k
1/4
∂x 0
−l0
1 dxdy + √ k
−l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y
fk(−l0 ) −l1
0 1 ∂w 3/4 dxdy + C3 k √ w(x, fk (−l0 )) ∂x −x −l0
0 fk(−l0 ) + C3 −l0
0
−l1
√ + C3 k
1 ∂w √ w dxdy + C3 k −x ∂y
−l0
0 fk(−l0 )
−l0
0 fk(−l0 )
−l1
∂w dxdy. w(x, fk (−l0 )) ∂y
−l1
fk(−l0 ) −l1
∂w dxdy w(−l0 , y) ∂x
√ ∂w −xw(−l0 , y) dxdy ∂y
C. Wang / J. Differential Equations 259 (2015) 2546–2575
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Then, the Hölder inequality gives 0
fk(−l0 )
∂w 2
−l0
∂x
−l1
√ ≤ C4 k
0
−l0
1 dxdy + √ k
0 −l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y fk(−l0 )
1 2 w (x, fk (−l0 ))dx + C4 k 3/2 −x
√ + C4 k
0 fk(−l0 )
−l0
−l1
w 2 (−l0 , y)dy −l1
1 √ w 2 dxdy + C4 k 5/2 −x
fk(−l0 )
w 2 (−l0 , y)dy −l1
0 + C4 k 3/2
w 2 (x, fk (−l0 ))dx.
(5.12)
−l0
Let us estimate the terms on the right side of (5.12). It follows from the Hölder inequality that 0 −l0
1 2 w (x, fk (−l0 ))dx −x
1 ≤ fk (−l0 ) + l1 1 ≤ fk (−l0 ) + l1 0
0 −l0
0
−l1 fk(−l0 )
−l0
fk(−l0 )
+ −l0
fk(−l0 )
−l1
−l1
1 2 w (x, y)dxdy + 2 −x
0
fk(−l0 )
−l0
1 2 w dxdy + −x
0
−l0
fk(−l0 ) −l1
−l1
1 ∂w (x, y)dxdy w(x, y) −x ∂y
1 w 2 dxdy (−x)3/2
1 ∂w 2 dxdy √ −x ∂y
(5.13)
and 0 w 2 (x, fk (−l0 ))dx −l0
≤
0
−l0
1 fk (−l0 ) + l1
0 fk(−l0 )
fk(−l0 )
|w(x, y)|dy + −l1
−l1
0
2 ∂w (x, y)dy dx ∂y fk(−l0 )
w dxdy + 2 l0 (fk (−l0 ) + l1 )
≤2 −l0
fk(−l0 )
2
−l1
−l0
−l1
1 ∂w 2 dxdy. √ −x ∂y
(5.14)
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C. Wang / J. Differential Equations 259 (2015) 2546–2575
Owing to (5.5) and the Hölder inequality, one gets that fk(−l0 )
fk(−l0 )
w (−l0 , y)dy ≤ 2
−l1
−l1
0 2 0 ∂w (x, y)dx dy ≤ l0 ∂x −l0
−l0
fk(−l0 )
∂w 2
−l1
dxdy
(5.15)
0 ≤ λ < 2.
(5.16)
∂x
and 0 −l0
fk(−l0 ) −l1
0
1 w 2 dxdy ≤ (−x)λ
fk(−l0 )
−l0
−l1
1 (−x)λ
x
0 ≤
0 2 ∂w (s, y)ds dxdy ∂x
0 1−λ
(−x)
∂w 2
dx
−l0
−l0
1 2−λ = l 2−λ 0
fk(−l0 )
0
−l0
∂x
−l1
dxdy
fk(−l0 )
∂w 2 ∂x
−l1
dxdy,
Substitute (5.13)–(5.16) into (5.12) to get 0 −l0
fk(−l0 )
∂w 2 ∂x
−l1
0 √ ≤ C5 k −l0
1 dxdy + √ k
fk(−l0 )
∂w 2
−l1
∂x
0 −l0
fk(−l0 ) −l1
1 ∂w 2 dxdy √ −x ∂y
√ dxdy + C5 k
0 fk(−l0 )
−l0
−l1
1 ∂w 2 dxdy. √ −x ∂y
(5.17)
Choose k˜0 = 1/(C52 + 1). For each 0 < k ≤ k˜0 , it follows from (5.17) that ∂w (x, y) = 0, ∂x
∂w (x, y) = 0, ∂y
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )),
which, together with (5.5), leads to w(x, y) = 0,
(x, y) ∈ (−l0 , 0) × (−l1 , fk (−l0 )).
Therefore, (q1 , m1 , ζ1 ) and (q2 , m2 , ζ2 ) are the same solution to the problem (2.15)–(2.21). 2 References [1] L. Bers, Existence and uniqueness of a subsonic flow past a given profile, Comm. Pure Appl. Math. 7 (1954) 441–504. [2] L. Bers, Mathematical Aspects of Subsonic and Transonic Gas Dynamics, John Wiley & Sons, Inc./Chapman & Hall, Ltd., New York/London, 1958.
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[3] G.Q. Chen, C.M. Dafermos, M. Slemrod, D.H. Wang, On two-dimensional sonic-subsonic flow, Comm. Math. Phys. 271 (2007) 635–647. [4] R. Courant, K.O. Friedrichs, Supersonic Flow and Shock Waves, Interscience Publishers, Inc., New York, NY, 1948. [5] G.M. Lieberman, Local estimates for subsolutions and supersolutions of oblique derivative problems for general second order elliptic equations, Trans. Amer. Math. Soc. 304 (1987) 343–353. [6] O.A. Ole˘ınik, E.V. Radkevi˘c, Second Order Differential Equations with Nonnegative Characteristic Form, American Mathematical Society, Rhode Island and Plenum Press, New York, 1973. [7] C.P. Wang, Z.P. Xin, Optimal Hölder continuity for a class of degenerate elliptic problems with an application to subsonic–sonic flows, Comm. Partial Differential Equations 36 (5) (2011) 873–924. [8] C.P. Wang, Z.P. Xin, On a degenerate free boundary problem and continuous subsonic–sonic flows in a convergent nozzle, Arch. Ration. Mech. Anal. 208 (3) (2013) 911–975. [9] C.P. Wang, Z.P. Xin, Smooth transonic flows in de Laval nozzles, arXiv preprint, arXiv:1304.2473, 2013. [10] C.J. Xie, Z.P. Xin, Global subsonic and subsonic–sonic flows through infinitely long nozzles, Indiana Univ. Math. J. 56 (2007) 2991–3023. [11] C.J. Xie, Z.P. Xin, Global subsonic and subsonic–sonic flows through infinitely long axially symmetric nozzles, J. Differential Equations 248 (11) (2010) 2657–2683. [12] J.X. Yin, C.P. Wang, Evolutionary weighted p-Laplacian with boundary degeneracy, J. Differential Equations 237 (2) (2007) 421–445.