Gui-Qiang Chen and Mikhail Feldman- Existence and Stability of Multidimensional Transonic Flows Through an Infinite Nozzle of Arbitrary Cross-Sections

8/3/2019 Gui-Qiang Chen and Mikhail Feldman- Existence and Stability of Multidimensional Transonic Flows Through an Infini

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EXISTENCE AND STABILITY OF MULTIDIMENSIONAL TRANSONIC

FLOWS THROUGH AN INFINITE NOZZLE OF ARBITRARY

CROSS-SECTIONS

GUI-QIANG CHEN AND MIKHAIL FELDMAN

Abstract. We establish the existence and stability of multidimensional steady transonic flowswith transonic shocks through an infinite nozzle of arbitrary cross-sections, including a slowlyvarying de Laval nozzle. The transonic flow is governed by the inviscid potential flow equationwith supersonic upstream flow at the entrance, uniform subsonic downstream flow at the exitat infinity, and the slip boundary condition on the nozzle boundary. Our results indicate that,if the supersonic upstream flow at the entrance is sufficiently close to a uniform flow, there

exists a solution that consists of a C1, subsonic flow in the unbounded downstream region,converging to a uniform velocity state at infinity, and a C1, multidimensional transonic shockdividing the subsonic flow from the supersonic upstream flow; the uniform velocity state atthe exit at infinity in the downstream direction is uniquely determined by the supersonicupstream flow; and the shock is orthogonal to the nozzle boundary at every point of theirintersection. In order to construct such a transonic flow, we reformulate the multidimensionaltransonic nozzle problem into a free boundary problem for the subsonic phase, in which theequation is elliptic and the free boundary is a transonic shock. The free boundary conditionsare determined by the Rankine-Hugoniot conditions along the shock. We further develop anonlinear iteration approach and employ its advantages to deal with such a free boundaryproblem in the unbounded domain. We also prove that the transonic flow with a transonicshock is unique and stable with respect to the nozzle boundary and the smooth supersonicupstream flow at the entrance.

1. Introduction

We are concerned with the existence and stability of multidimensional steady transonic flowswith transonic shocks through multidimensional infinite nozzles of arbitrary cross-sections. Suchproblems naturally arise in many physical situations, especially in the de Laval nozzles whichhave widely been used in the design of steam turbines and modern rocket engines (see Courant-Friedrichs [11], Whitham [47], and the references cited therein). Since the nozzles in applicationsare usually much longer with respect to their cross-sections, the problem is often formulatedmathematically as an infinite nozzle problem. Correspondingly, such a multidimensional infinitenozzle problem has extensively been studied experimentally, computationally, and asymptotically(see [11, 15, 18, 21, 22, 23, 47] and the references cited therein). The stability issue for transonic

gas flow through a nozzle for the one-dimensional model was analyzed in [40, 41]. The existenceand stability of multidimensional steady transonic flows through multidimensional infinite nozzleshas remained open; see [5, 11, 12, 43, 47].

In this paper, we focus on the infinity nozzle to establish the existence and stability of mul-tidimensional transonic flows with supersonic upstream flows at the entrance, uniform subsonicdownstream flows at the exit at infinity, and the slip boundary condition on the nozzle bound-ary. The potential flow equation for the velocity potential : Rn R is a second-order

Date: November 1, 2005.1991 Mathematics Subject Classification. 35M10,35J65,35R35,76H05,76L05,35B45.Key words and phrases. Elliptic-hyperbolic, nonlinear equations, second order, mixed type, multidimensional

transonic shocks, free-boundary problems, unbounded domains, infinite nozzles, cross-sections, existence, unique-ness, stability, Euler equations, inviscid potential flow.

1


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2 GUI-QIANG CHEN AND MIKHAIL FELDMAN

nonlinear mixed equation of elliptic-hyperbolic type:

div((

|D

|2)D) = 0, x

Rn, (1.1)

where the density function (q2) is

(q2) =

1 q2 12 (1.2)and = 1

2> 0 with adiabatic exponent > 1.

The nonlinear equation (1.1) is elliptic at D with |D| = q if(q2) + 2q2(q2) > 0 (1.3)

and hyperbolic if

(q2) + 2q2(q2) < 0. (1.4)

For this infinite nozzle problem, we first seek a multidimensional transonic flow containing amultidimensional transonic shock which divides a subsonic downstream flow from the supersonicupstream flow. In order to construct such a transonic flow, we first formulate this infinite nozzleproblem into a free boundary problem: The multidimensional transonic shock is a free boundarywhich divides two phases of C1, flow in the infinite nozzle, and the equation is hyperbolic in thesupersonic upstream phase and elliptic in the subsonic downstream phase. Since the existence ofthe supersonic upstream phase is a direct corollary of the standard local existence theory of theinitial-boundary value problem with the slip boundary condition for nonlinear wave equations (cf.[27, 28, 30]), we can further formulate the free boundary problem into a one-phase free boundaryproblem for a nonlinear elliptic equation by shiffmanization, a cut-off technique (see [4, 45]).

We further develop a nonlinear iteration approach and employ its advantages to deal with sucha free boundary problem in the unbounded domain and to solve the multidimensional transonicnozzle problem in a direct fashion. Our results indicate that, for the transonic nozzle problem,there exists a transonic flow such that the flow is divided into a C1, subsonic flow up to the nozzleboundary in the unbounded downstream region from the supersonic upstream flow by a C1,

multidimensional transonic shock that is orthogonal to the nozzle boundary at every intersectionpoint, and the uniform velocity state at the exit at infinity in the downstream direction is uniquelydetermined by the supersonic upstream flow at the entrance which is sufficiently close to a uniformflow. The uniform velocity state at the exit can not be apriori prescribed from the correspondingpressure for such a flow to exist. We further prove that the transonic flow with a transonic shockis unique and stable with respect to the nozzle boundary and the smooth supersonic upstreamflow at the entrance.

One of the advantages to employ our nonlinear iteration approach is that, as long as weknow how the corresponding fixed conormal problem for (1.1) can be solved and estimated, thesolution of the free boundary problem and its estimates directly follow, even for more complicatedgeometry of the domain under consideration. In this sense, the iteration approach is more efficientthan the partial hodograph approach developed in [6], since we do not need to change equation(1.1) with fine features. This indeed plays an important role to achieve the C1,regularity of boththe solution and the free boundary up to the nozzle boundary and to allow the nozzle boundary ofarbitrary cross-sections, by employing the features of equation (1.1). Also, this approach enablesus to deal with the multidimensional nozzle problem in a direct fashion, which especially appliesto the bounded nozzle problems with various different compatible boundary conditions at theexit by solving the elliptic problems with nonsmooth fixed boundaries, as discussed in Section 3and in [9, 10, 13, 24].

Another new ingredient of this paper is to develop a C1,estimate framework for solutionsof the conormal problem for elliptic equations of divergence form with Ccoefficients in thedomains with corners and wedges. In our nozzle problem, the corners and wedges are formed in


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MULTIDIMENSIONAL TRANSONIC FLOWS THROUGH AN INFINITE NOZZLE 3

the intersection of the free boundary with the fixed boundary, so it is necessary to develop suchan estimate framework to deal with the new difficulty of the intersection of a free boundary with

a fixed boundary, especially when n 3. In [5], we considered a transonic flow in an infinitecylinder with a cross-section of the form = (0, 1)n1, i.e., a flow in the domain (0, 1)n1 R,in which, by reflections, we can reduce the problem to the domain Tn1 R, where Tn1 isthe (n 1)-dimensional torus. Thus, in studying the free boundary problems in [4, 5], we wereable to avoid studying the intersection of the free boundary with the fixed boundary. In thecase of arbitrary smooth cross-section in dimension n 3, this can not be done even for theflow in the unperturbed cylinder: Indeed, if we locally flatten the boundary, the equation in thenew coordinates changes its structure and the reflection leads to an equation with discontinuouscoefficients. When we consider a flow in the nozzle (i.e. in a perturbed cylinder), we face a similarproblem. Thus, in the present paper, we have to consider the intersection of a free boundary witha fixed boundary, that is, a nonlinear conormal problem in the domain with nonsmooth boundaryfor any dimension (n 2). The problem in a fixed domain with wedges was considered in [36],where the C1,regularity of solutions near the wedges for a general oblique derivative problemfor linear elliptic equations was obtained. On the other hand, the results in [36] require certainconditions on the relative geometry of the wedge and the vector fields in the oblique derivativeconditions; and, moreover, the Holder exponent of the solution depends on the geometry ofwedges, in addition to the Holder exponent of the coefficients and the right-hand sides of theequation and boundary conditions. These features present the difficulties in using the results of[36] in constructing solutions for our nonlinear problem. Thus, in this paper, we first study theexistence and regularity of solutions, by restricting the class of equations and considering onlythe conormal boundary conditions and cylindrical domains, and then we obtain sharp resultsthat are used for constructing the solution of the free boundary problem for the nonlinear ellipticequation in the subsonic domain.

A further new ingredient is to develop a useful approach for obtaining uniform estimates ofsolutions in the (perturbed) cylinders of length R

and showing that the velocity (the

gradient of the solution) converges to an asymptotic constant state at infinity. In [5], we reducedthe problem to the domain Tn1 R so that the symmetry allows us to use the pointwiseestimates of solutions to study the asymptotics at infinity. In the present case, the domains areof more general geometry, so our approach is now to use natural integral estimates of solutions toachieve the asymptotics of solutions at infinity. This approach should be useful to other problemsin unbounded domains with similar features.

The uniqueness and stability of the nozzle problem is a question of great importance to knowunder what circumstances a steady transonic flow involving transonic shocks is uniquely deter-mined by the boundary condition and the conditions at the entrance and when further conditionsat the exit are appropriate (see Courant-Friedrichs [11]). In this paper, we identify that the steadytransonic flow involving a transonic shock is uniquely determined by the Cauchy data at the en-trance for the nonlinear wave equation and the uniform subsonic flow condition at the exit atinfinity, which are natural physical conditions. Note that the pressure can not be apriori pre-scribed at the exit at infinity even for the existence; otherwise, it is clear from our results thatthere is no weak solution for this problem (also see [5]).

Some efforts have been made for solving the nonlinear equation (1.1) of mixed type. In par-ticular, Shiffman [45], Bers [2], and Finn-Gilbarg [17] proved the existence and uniqueness ofsolutions for the problem of subsonic flows of (1.1) past an obstacle (also see [14]). Morawetzin [42] showed that the flows of (1.1) past the obstacle may contain transonic shocks in gen-eral. In [4, 5, 6], we developed two approaches to deal with multidimensional transonic shocks.Also see Canic-Keyfitz-Lieberman [3] for the transonic small disturbance equation and Zheng[51] for the pressure-gradient equations. For the bounded transonic nozzle problem, see [4] forthe multidimensional channel case and [48] for the 2-dimensional case. Non-transonic shocks


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(hyperbolic-hyperbolic shocks) were analyzed in [7, 8, 25, 32, 39, 44, 49] and the references citedtherein.

In Section 2, we set up the infinite nozzle problem involving multidimensional transonic shocksand present the main theorems of this paper. In Section 3, we develop a C1, estimate frameworkup to the wedge boundary of the cylinder for the conormal problem for second-order linearelliptic equations of divergence form with C coefficients for subsequent development. In Section4, we reformulate the transonic nozzle problem into a free boundary problem by introducing amultidimensional transonic shock as a free boundary which divides the upstream and downstreamphases of C1, flow in the infinite nozzle. We introduce an iteration procedure to constructapproximate solutions of the free boundary problem in Section 4 and make uniform estimatesof the solutions on the bounded domains with wedge boundary in Section 6. In Section 7, weestablish the existence of multidimensional transonic flows with transonic shocks via solvingthe corresponding free boundary problem in the infinite nozzle. In Section 8, we establish theuniqueness and stability of multidimensional transonic flows with transonic shocks in the infinite

nozzle.

2. Transonic Shocks, Nozzle Problems, and Main Theorems

In this section, we first set up the infinite nozzle problem involving multidimensional transonicshocks and present the main theorems of this paper.

A function C0,1() is a weak solution of (1.1) in an unbounded domain if(i) |D(x)| 1/

a.e.

(ii) For any C0 (),

(|D|2)D D dx = 0. (2.1)

We are interested in weak solutions with shocks. Let + and be open subsets of suchthat

+ = , + = ,and S = + . Let C0,1() be a weak solution of (1.1) and be in C1( S) so thatD experiences a jump across S that is an (n 1)-dimensional smooth surface. Then satisfiesthe following Rankine-Hugoniot conditions on S:

[]S = 0,

(|D|2)D

S

= 0, (2.2)

where is the unit normal to S from to +, and the bracket denotes the difference betweenthe values of the function along S on the sides. Moreover, a function

C1(

S), which

satisfies |D| 1/, (2.2), and equation (1.1) in respectively, is a weak solution of (1.1),i.e., satisfies (2.1) in the whole domain .

Set = | . Then we can also write (2.2) as+ = on S (2.3)

and

(|D+|2)+ = (|D|2)D on S, (2.4)where + = D

+ is the normal derivative on the + side.Note that the function

(p) :=

1 p2

12 p (2.5)


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is continuous on

0,

1/

and satisfies

(p) > 0 for p 0,1/ , (0) = 1/ = 0, (2.6)0 < (p) < 1 on (0, c),

(p) < 0 on

c,

1/

, (2.7)

(p) < 0 o n ( 0, c], (2.8)

wherec =

1/( + 1) =

2/(+ 1) (2.9)

is the sonic speed, for which a flow is called supersonic if |D| > c and subsonic if |D| < c.Suppose that C0,1() C1( S) is a weak solution satisfying

|D| < c in +, |D| > c in , D |S > 0. (2.10)Then is a transonic shock solution with transonic shock S dividing into the subsonic region+ and the supersonic region and satisfying the physical entropy condition (see Courant-Friedrichs [11]):

(|D|2) < (|D+|2) along S.Note that equation (1.1) is elliptic in the subsonic region and hyperbolic in the supersonic region.

Let (x, xn) be the coordinates in Rn, where xn R and x = (x1, . . . , xn1) Rn1. Letq0

c, 1/

and 0 (x) := q0 xn. Then

0 is a supersonic solution in . According to

(2.6)(2.7), there exists a unique q+0 (0, c) such that

(q+0 )2

q+0 =

(q0 )2

q0 . (2.11)

Thus, the function

0(x) = q0 xn, x 0 := {x : xn < 0},q+0 xn, x

+0 :=

{x : xn > 0

}(2.12)

is a plane transonic shock solution in , +0 and 0 are its subsonic and supersonic regions

respectively, and S = {xn = 0} is a transonic shock.Defining +0 (x) := q

+0 xn in , we have

0(x) = min(+0 (x),

0 (x)) for x . (2.13)

In this paper, we focus on the following infinite nozzle with arbitrary smooth cross-sections:

= ( (, )) {xn 1}, (2.14)where Rn1 is an arbitrary open bounded, connected set with a smooth boundary, and : Rn Rn is a smooth map, which is close to the identity map. For simplicity, we assumethat

is in C[n2 ]+3, (2.15)

and

Id[n2 ]+3,,Rn (2.16)for some (0, 1) and small > 0, where [s] is the integer part of s, Id : Rn Rn is theidentity map, and um,,D is the norm in the Holder space Cm,(D) in the domain D. Suchnozzles especially include the slowly varying de Laval nozzles [11, 47]. For concreteness, we alsoassume that there exists L > L0 (say, L0 > 10 without loss of generality) such that

(x) = x for any x = (x, xn) with xn > L, (2.17)

that is, the nozzle slowly varies locally in a bounded domain as the de Laval nozzles.In the two-dimensional case, the domain defined above has the following simple form:

= {(x1, x2) : x1 1, f(x2) < x1 < f+(x2)},


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where f d4,,R and f d on [L, ) for some constants d satisfying d+ > d.For the multidimensional case, the geometry of the nozzles is much richer.

Note that our setup implies that = o l

with

l := [ (, )] {(x, xn) : xn > 1},o := ( (, )) {(x, xn) : xn = 1}.

Then our transonic nozzle problem can be formulated into the following form:

Problem (TN): Transonic Nozzle Problem. Given the supersonic upstreamflow at the entrance o:

= e , xn = e on o, (2.18)

the slip boundary condition on the nozzle boundary l: = 0 on l, (2.19)

and the uniform subsonic flow condition at the exit xn = :() xnC1({xn>R}) 0 as R , for some (0, c), (2.20)

find a solution of problem (1.1) and (2.18)(2.20) in .

As we will see below, the supersonic upstream part of the solution can be constructedin 1 := {xn < 1} from the data on the nozzle entrance (2.18) by using the standardresults on initial-boundary value problems for hyperbolic equations. Thus, assuming that theC1, supersonic solution is given in 1, we can formulate the transonic nozzle problem (TN)as the following one-phase free boundary problem.

Problem (FB): Free Boundary Problem. Given a supersonic upstream flow, a weak solution of (1.1) in 1, which is a C

1, perturbation of 0 :

0 1,,1 C0 (2.21)with > 0 small, for some constant C0, and satisfies (2.19), find a multidi-mensional subsonic flow + of (1.1) satisfying (2.19)(2.20) and identify a freeboundary xn = f(x

) dividing the subsonic flow + from the given supersonicflow so that

(x) =

+(x), xn > f(x

),(x), xn < f(x

)(2.22)

is a transonic shock solution.

Note that, for a solution of Problem (FB), the subsonic region + and supersonic region

in (2.10) are of the form

+() = {xn > f(x)}, () = {xn < f(x)}. (2.23)Our main theorem for the free boundary problem, Problem (FB), is the following.

Theorem 2.1 (Existence). There exist 0 > 0, C, and C, depending only on n, , , q+0 ,

C0, , and L, such that, for every (0, 0), any map satisfying (2.16)(2.17), and anysupersonic upstream flow of (1.1) satisfying (2.19) and (2.21), there exists a solution C0,1() C(+) of Problem (FB) such that

D q+0 en0,0,+ C. (2.24)Moreover, the solution satisfies the following properties:


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(i) The constant in (2.20) must be q+:

= q+, (2.25)

where q+ is the unique solution in the interval (0, c) of the equation

((q+)2)q+ = Q+ (2.26)

with

Q+ =1

||o

(|D|2)D dHn1. (2.27)Thus, satisfies

q+xnC1({xn>R}) 0 as R , (2.28)and q+ satisfies

|q+ q+0 | C ; (2.29)(ii) The function f(x) in (2.22) satisfies

f1,,Rn1 C, (2.30)and the surface S = {(x, f(x)) : x Rn1} is orthogonal to l at everyintersection point;

(iii) Furthermore, C1,(+) with q+xn1,,+ C. (2.31)

If the supersonic upstream flow has a higher regularity, then we have the following unique-ness theorem.

Theorem 2.2 (Uniqueness). There exists a constant 0 > 0 depending only on n,,,C0, ,L, and q+0 such that, if < 0 and the supersonic solution

in Problem (FB) additionallysatisfies

0 2,,1 C0, (2.32)then the solution of Problem (FB) satisfying (2.24) is unique.

The standard local existence theory of smooth solutions for the initial-boundary value problem(2.18)(2.19) for second-order quasilinear hyperbolic equations implies (see Appendix) that, as is sufficiently small in (2.16) and (2.21), there exists a supersonic solution of (1.1) in 1,which is a Ck+1 perturbation of 0 : For any (0, 1],

0 k,,1 C0, k = 1, 2, (2.33)for some constant C0 > 0, and satisfies

= 0 on l1, (2.34)

provided that (e , e ) on o satisfying

e q0 xnHs+k + e q0 Hs+k1 , k = 1, 2, (2.35)for some integer s > n/2 + 1 and the compatibility conditions up to the (s + k)thorder, wherethe norm Hs is the Sobolev norm with Hs = Ws,2. In particular, as a direct corollaryof Theorems 2.12.2 and Proposition A.1 in Appendix, we obtain the following existence anduniqueness result for the transonic nozzle problem, Problem (TN):

Theorem 2.3. Let q+0 (0, c) and q0

c, 1/

satisfy (2.11), and let 0 be the transonic

shock solution (2.12). Then there exist 0 > 0, C, and C, depending only on n, , , q+0 , ,

and L such that, for every (0, 0), any map satisfying (2.16)(2.17), and any supersonicupstream flow (e ,

e ) on o satisfying

e q0 xnHs+2 + e q0 Hs+1 (2.36)


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for s > n/2 + 1 and the compatibility conditions up to the (s + 2)thorder, there exists a uniquesolution C0,1() of Problem (TN) satisfying (2.23)(2.24) and

0 2,, C.

Moreover, this solution satisfies C0,1() C(+) and properties (i)(iii) of Theorem2.1, where (2.27) is replaced by

Q+ =1

||o

(|Dxe |2 + (e )2)e dHn1. (2.37)

Remark 2.1. The smoothness conditions (2.15)(2.16) are just for simplicity of presentationand can be relaxed: For example, in Theorems 2.12.2, we require only C2, in (2.15)(2.16).The localization condition (2.17) is not essential to achieve Theorem 2.1; In general, it can bereplaced by an appropriate decay condition.

Remark 2.2. When the initial data(e , e ) (e , e ) is constant and the nozzle {1 xn 1 + } = [1, 1 + ] for some > 0 as a de Laval nozzle, then the compatibilityconditions are automatically satisfied. In fact, in this case, (x) = e xn is a solution nearxn = 1 in the nozzle.Remark 2.3. When n = 2, condition (2.36) for the supersonic upstream flow (e ,

e ) on o

in Theorem 2.3 can be replaced by the C3-condition:

e q0 xnC3 + e q0 C2 . (2.38)This can be achieved by following the arguments in Li-Yu [33].

Remark 2.4. There exist 0 and C depending only on the data such that equation (2.26) with

Q+ defined by (2.27) has a unique solution q+ (0, c) satisfying (2.29).This can be seen as follows: From (2.16), it follows that | |o| || | C and | en| C

on o, and hence that | en 1| C on o. Then

|Q+ ((q+0 )2)q+0 | 1||

o

(|D|2)D ((q+0 )2)q+0 dx +1 |o|||

((q+0 )2)q+0=

1

||o

(|D|2)D ((q0 )2)q0 dx +1 |o|||

((q+0 )2)q+0 1

|

| o (|D|2)D (|D0 |2)D0 + ((q

0 )

2)q0 | en 1| dx + C

C,where we used (2.11) and (2.21). Thus, by (2.6)(2.7), we obtain that, if is small dependingonly on the data, then there exists a unique solution q+ (0, c) of equation (2.26) such that(2.29) holds.

Remark 2.5. In fact, the solution with a transonic shock of Problem (TN) in Theorem 2.3 isalso stable with respect to the nozzle boundary and the smooth supersonic upstream flow at theentrance; see Theorem 8.1.

Remark 2.6. For the isothermal gas = 1, the same results can be obtained by following thesame arguments in this paper.


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3. Conormal Problems for Linear Elliptic Equations in a Cylinder

In this section, we develop a C1, estimate framework up to the boundary of the cylinder fora conormal problem for linear elliptic equations of divergence form with C coefficients.

Let Rn1 be an open bounded, connected set with C2, boundary.We fix the set and a number (0, 1) through this section.For Q > 0, denote

CQ := (0, Q), 1Q := {xn = 0}, 2Q := (0, Q), 3Q := {xn = Q}.Consider the problem

ni,j=1

(aijuxj )xi = f in CQ, (3.1)

n

i,j=1

aijuxji = gk on kQ, k = 1, 2, (3.2)

u = h on 3Q, (3.3)

where is the inward unit normal on 1Q and 2Q to CQ.

The main issue in the argument below is to show the existence and uniqueness of solutionswhich are C1, up to the edge {0} of the cylinder. Lieberman [36] studied general obliquederivative problems for linear elliptic equations in the domains with wedges in which the C1, reg-ularity of solutions is obtained near the wedges. However, these results require certain conditionson the relative geometry of the wedge and the vector fields in the oblique derivative conditions;Moreover, in [36], the Holder exponent of the solution gradients depends on the geometry ofwedges and the coefficients of the equation and the right-hand sides of the boundary conditions,and < in general where is the Holder exponent of the coefficients and the right-hand sides.In this section, by restricting the class of equations and considering only conormal boundary con-

ditions and cylindrical domains, we obtain the sharp C1, regularity of solutions up to {0},i.e., = . This is important for our approach developed in Sections 48.

Theorem 3.1. Let 1, . . . , n, and are the constants satisfying 0 < i 1 for i =1, . . . , n. Then there exist > 0 and C > 0, depending only on n, , Q, and , such that, whenaij(x) satisfy

aij iij0,,CQ for i, j = 1, . . . , n , (3.4)and

f L(CQ), h H1(CQ), (3.5)g1 C0,(1Q), g2 C0,(2Q), (3.6)

there exists a unique weak solution u H1(CQ) of (3.1)(3.3) in the sense that (3.3) holds as thetrace and

CQ

(n

i,j=1

aijuxjwxi + f w)dx +2

k=1

kQ

gkw dHn1 = 0 (3.7)

for any functionw H1(CQ) with w = 0 on CQ {xn = Q}. Moreover, the solution u isC1,(CQ/2) and satisfies

u1,,CQ/2 C(uL2(CQ) + f0,0,CQ + g10,,1Q + g20,,2Q). (3.8)Theorem 3.1 is a direct corollary of Theorem 3.2 below, stated in more technical terms. In

order to state Theorem 3.2, we introduce some notations and weighted Holder norms.Denote

CQ = CQ \ {xn = Q}, (2Q) = 2Q \ {xn = Q}. (3.9)


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We will use the following weighted Holder semi-norms and norms in the cylinder CQ, in which theweight is the distance to the portion of the boundary 3Q =

CQ

\CQ. Denote x = dist(x,

3Q) =

Q xn for x = (x, xn) CQ and x,y = min(x, y) for x, y CQ. For k R, (0, 1), andm N (the set of nonnegative integers), define

|u|(k)m;0;CQ =

0jm

||=m

supxCQ

max(m+k,0)x |Du(x)|

, (3.10)

[[u]](k)m;;CQ

=||=m

supx,yCQ,x=y

max(m++k,0)x,y

|Du(x) Du(y)||x y|

,

|u|(k)m;;CQ = |u|(k)m;0;CQ

+ [[u]](k)m;;CQ

,

where D = 1x1 nxn , = (1, . . . , n) is a multi-index with j 0, j N, and || =1 + + n. The weighted Holder norms |u|(k)m;;(2Q) on the boundary part

2Q are defined

similarly.

Theorem 3.2. Let 1, . . . , n, and be the constants satisfying 0 < i 1 for i =1, . . . , n. Then there exist > 0 and C > 0, depending only on n, , Q, and , such that, whenaij(x) satisfy (3.4) and

f L2(CQ) Lloc(CQ), h H1(CQ), (3.11)g1 L2(1Q) C0,(1Q), g2 L2(2Q) C0,((2Q)) (3.12)

with

|f|(2+n/2)0,0,CQ + |g2|(1+n/2)

0,,(2Q) < ,

there exists a unique weak solution u H1(CQ) of (3.1)(3.3) in the sense that (3.3) holds asthe trace and (3.7) holds for any w

H1(

CQ) with w = 0 on

CQ

{xn = Q

}. Moreover, the

solution u is C1,(C) and satisfies|u|(n/2)1,,CQ C(uL2(CQ) + |f|

(2+n/2)0,0,CQ

+ g10,,1Q + |g2|(1+n/2)

0,,(2Q)). (3.13)

Now we prove Theorem 3.2. We first consider an auxiliary problem:

ni=1

iuxixi = f +ni=1

ixi in CQ, (3.14)ni=1

iuxii nuxn = g1 on 1Q, (3.15)

n

i=1

iuxii

n1

i=1

iuxii = g2 +n1

i=1

ii on 2Q, (3.16)

u = h on 3Q, (3.17)

where we used that = en on 1Q in (3.15) and en = 0 on 2Q in (3.16). Note that smoothsolutions of (3.14)(3.17) satisfy

CQ

(ni=1

iuxiwxi ni=1

iwxi + f w)dx

+

1Q

(g1 n) wdx

2Q

g2 w dHn1 = 0 (3.18)

for any function w H1(CQ) with w = 0 on CQ {xn = Q}.


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Lemma 3.1. Let1, . . . , n, and be the constants satisfying 0 < i 1 for i = 1, . . . , n .Let

f L2(CQ) Lloc(CQ), = (1, . . . , n) H1(CQ) C(CQ), h H1(CQ),

g1 L2(1Q) C0,(1Q), g2 C(CQ) L2(2Q) C0,(CQ)(3.19)

with

|f|(2+n/2)0,0,CQ + ||(1+n/2)0,,CQ

+ |g2|(1+n/2)0,,CQ < . (3.20)Then there exists a unique weak solution u H1(CQ) of (3.14)(3.17) in the sense that (3.17)holds as the trace and (3.18) holds for any w H1(CQ) with w = 0 on CQ {xn = Q}. Thesolution u is C1,(CQ) and satisfies

uH1(CQ) + |u|(n/2)1,,CQ

C(

fL2(C

Q) +

|f|(2+n/2)

0,0,C

Q

+

L2(C

Q) +

||(1+n/2)

0,,C

Q

+

g1

0,,1Q

+g2L2(2Q) + |g2|(1+n/2)

0,,(2Q) + hH1(CQ)) (3.21)

and

|u|(n/2)1,,CQ C(uL2(CQ) + |f|(2+n/2)0,0,CQ

+ ||(1+n/2)0,,CQ + g10,,1Q + |g2|(1+n/2)

0,,(2Q)), (3.22)

where C depends only on n, , Q, and .

Proof. We prove this lemma in six steps. The constant C in the argument below is a universalconstant depending only on n, , Q, and , and may be different at each occurrence.

Step 1. First we show the uniqueness of weak solutions. Suppose that u1 and u2 are twoweak solutions of (3.14)(3.17). Then u1 u2 = 0 on 3Q. Thus, writing the weak forms (3.18)for u1 and u2 respectively and subtracting the expressions with w = u1

u2 yield

0 =

CQ

ni,j=1

i|xi(u1 u2)|2dx.

Since u1 u2 = 0 on 3Q and i > 0, we conclude u1 = u2 a.e. in CQ.Step 2. We now focus on the existence of solutions and estimate (3.21). First we show that the

problem can be reduced to the case where the right-hand side of the boundary condition (3.15)vanishes. We can extend g1 from 1Q = {0} to the whole hyperplane {(x, 0) : x Rn1}such that the extension g1 C0 (Rn1) satisfies g10,,Rn1 Cg10,,. Now, as in [20, page124-125], we choose a nonnegative C20 (Rn1) with

Rn1

(y)dy = 1 and define

G(x, xn) = 1n xn Rn1

g1(x xny)(y)dy for xn > 0.

Then G C1,(Rn+) withG1,,CQ Cg10,,Rn1 Cg10,, (3.23)

and

nGxn = g1 = g1 on {xn = 0}.

Thus, u is a weak solution of (3.14)(3.17) if and only ifv = u G is a weak solution of the sameproblem with modified right-hand sides: f , , g1, and g2 are replaced by f = f,

i = i iGxi ,g1 = 0, and g2 = g2. Using (3.23), it is easy to check that (3.21)(3.22) for v in terms of f , , g1,and g2 implies (3.21)(3.22) for u in terms of f , , g1, and g2.

Thus, from now on, we assume

g1 = 0 on 1Q. (3.24)


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Step 3. We also note that we can assume without loss of generality that

n = 0 on 1

Q. (3.25)

Indeed, if u is a weak solution of (3.14)(3.17), then u is also a weak solution of (3.14)(3.17)

with n replaced by n: n(x, xn) = n(x, xn) n(x, 0) for (x, xn) CQ. Since

nL2(CQ) + ||(1+n/2)0,,CQ C(nL2(CQ) + ||(1+n/2)0,,CQ ),

it follows that (3.21)(3.22) in terms of the right-hand sides with n replaced by n implies(3.21)(3.22) in terms of the original functions.

Clearly, n satisfies (3.25). Thus we can assume (3.25) from now on.Step 4. With (3.24)(3.25), we extend problem (3.14)(3.17) to the cylinder C(Q,Q) =

(Q, Q). Set 2(Q,Q) := (Q, Q).Note that, with (3.24)(3.25), u is a weak solution of (3.14)(3.17) if (3.17) holds as the trace

and, for any w H1

(CQ) with w = 0 on CQ {xn = Q},CQ

(

ni=1

iuxiwxi ni=1

iwxi + f w)dx

2Q

g2w dHn1 = 0. (3.26)

We use the even reflection to extend u,f,h,g2, 1, . . . , n1 to C(Q,Q) and the odd reflection

to extend n to C(Q,Q). That is, for x and xn (0, Q), we use the same notations for theoriginal and extended functions to define

(x, xn) = (x, xn) for {u,f,h,g2, 1, . . . , n1},n(x, xn) = n(x, xn). (3.27)

Obviously, for u H1(CQ) and f,,h, and g2 as in Lemma 3.1 with (3.25) for n, the extendedfunctions satisfy

u H1(C(Q,Q)), f L2(C(Q,Q)) Lloc(C(Q,Q)), H1(C(Q,Q)) C(C(Q,Q)),

h H1(C(Q,Q)), g2 C(2(Q,Q)) L2(2(Q,Q)),where C(Q,Q) = C(Q,Q) 2(Q,Q). Moreover, defining the weighted Holder semi-norms andnorms | |(k)m;;C(Q,Q) and | |

(k)m;;2(Q,Q)

by (3.10) with the supremums taken over the domains

C(Q,Q) and 2(Q,Q) respectively and

x = dist(x, C(Q,Q) \ 2(Q,Q)) = min(Q xn, Q + xn) for x = (x, xn) C(Q,Q),we obtain

|f|(2+n/2)0,0,C(Q,Q) C|f|

(2+n/2)0,0,CQ , ||

(1+n/2)0,,C(Q,Q) C||

(1+n/2)0,,CQ ,

|g2|(1+n/2)0,,2(Q,Q) C|g2|(1+n/2)0,,CQ

,

(3.28)

where the norms on the right-hand sides are finite by (3.20).Then it follows from (3.26)(3.27) that the extended functions satisfy

C(Q,Q)

ni=1

iuxiwxi ni=1

iwxi + f w

dx

2(Q,Q)

g2w dHn1 = 0 (3.29)

for any w H1(C(Q,Q)) satisfying w = 0 on {xn = Q}. Indeed, (3.29) holds separately whenthe integration is over subdomains {xn > 0} and {xn < 0} (the last follows by the change of


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variable xn xn). Thus, the extended function u is a weak solution ofni=1

iuxixi = f +

ni=1

ixi in C(Q,Q), (3.30)ni=1

iuxii n1i=1

iuxii = g2 +

n1i=1

ii on 2(Q,Q), (3.31)

u = h on ( {Q}) ( {Q}). (3.32)Conversely, if u is a weak solution of (3.30)(3.32) (i.e. (3.29) holds for w as above), then usatisfies u(x, xn) = u(x, xn) for x and xn (0, Q). To see this, we first note that a weaksolution of (3.29) is unique: The proof similarly follows the proof of uniqueness for (3.1)(3.3)by using the fact that, if u1 and u2 are two weak solutions of (3.30)(3.32), then we can usew = u1 u2 in (3.29). Now, the properties in (3.27) for f,h,, and g2 imply that, ifu(x, xn) is asolution of (3.29), then u(x,

xn) is also a solution. Thus, by uniqueness, u(x

,

xn) = u(x

, xn).

Now, let w H1(CQ) with w = 0 on {xn = Q}. Extend w by the even reflection to C(Q,Q). Wecan use this extended w in (3.29). Then, using the properties in (3.27) for u,f,h,, and g2, wesee that, for all the integrals in (3.29), the integrals over subdomains {xn > 0} and {xn < 0}are equal. Thus, (3.29) implies (3.26). Then, in order to solve (3.14)(3.17), it suffices to solve(3.30)(3.32) for the extended right-hand sides.

Step 5. We now derive some estimates for the weak solution u H1(C(Q,Q)) of (3.30)(3.32).We first derive the energy estimate. We can use w = u h in (3.29) and then use the ellipticityand the Holder inequality to obtain

uH1(C(Q,Q)) C((f, )L2(C(Q,Q)) + hH1(C(Q,Q)) + g2L2(2(Q,Q))). (3.33)Note that, by the standard local regularity results for the conormal problem for elliptic equa-

tions in domains with C1, boundary (e.g., [35]), we have u

C1,(

C(Q,Q)).

Now we derive the estimates for the Holder norms of u. We first recall the standard estimatesfor the conormal problem for elliptic equations in a unit ball B1 = B1(0) and a half-ball B

+1 =

B1(0) {xn > 0}. If u H1(B1) is a weak solution of the equationn

i,j=1

(aijuxj )xi = f +ni=1

ixi in B1 (3.34)

with aij C(B1) satisfying the ellipticity condition: there exists some > 0 such that, for anyx B1,

||2 n

i,j=1

aij(x)ij 1||2 for any Rn, (3.35)

and, if f

L(B1) and

C(B1), then u

C1,(B1/2) with

u1,,B1/2 C(uL2(B1) + fL(B1) + 0,,B1), (3.36)where C depends only on n,, and aij0,,B1 . This follows from [20, Theorem 8.32], combinedwith [20, Theorem 8.17]. Similar estimates can be obtained for the conormal problem, in whichu H1(B+1 ) is a weak solution of (3.34) in B+1 with the conormal condition

nj=1

anjuxj = g + n on B+1 := B

+1 {xn = 0}, (3.37)

where g C(B+). Then u C1,(B+1/2) withu1,,B+1/2 C(uL2(B+1 ) + fL(B+1 ) + 0,,B+1 + g0,,B+1 ), (3.38)


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where C depends only on n,, and aij0,,B+1 . This can be inferred by simplifying the estimatesof [35] to the case of linear elliptic equations.

Let (0, 1/2). By scaling, we obtain the corresponding local estimates for weak solutions of(3.34) in B and for weak solutions of the conormal problem (3.34) and (3.37) in B+ . We write

only the estimate for the conormal problem in B+ , since the interior estimate in B differs onlyin that it does not have the terms with the norms of g on the right-hand side. Thus, if u is a

weak solution of (3.34) in B+/2 and satisfies (3.37) in B+ {xn = 0}, then u C1,(B+/2) with

u0,0,B+/2

+ Du0,0,B+/2

+ 1+[Du]0,,B+/2

C(n/2uL2(B+ ) + 2fL(B+ ) + 0,0,B++1+[]0,,B+ + g0,0,B+ + 1+[g]0,,B+ ).

(3.39)

This estimate is obtained by introducing the function v(x) = u(x) in B+1 , writing the equa-tion and the conormal condition satisfied by u in terms of v, using estimate (3.38) for v withcorresponding right-hand sides of the equation and conormal condition satisfied by v, and thenrewriting that estimate in terms of u.

Now we get similar estimates for weak solutions of (3.30)(3.32). Using that is C1,

and compact, there exits r > 0 and M > 0 such that, for any z 2(Q,Q) and any positive < min(r, z), there is a C

1, diffeomorphism F that flattens 2(Q,Q) in B(z) C(Q,Q) so thatB+/M F

B(z) C(Q,Q)

B+M and (F, F1)C1, depends only on . The transformedfunction v(y) := u(F1(y)) satisfies a conormal problem of the form (3.34) and (3.37) in B+/M,

where the equation is elliptic (with ellipticity constant /2 and the C norms of the coefficientsaij depending only on the C1

, norms of ) and the appropriate norms of right-hand sides ofthe original and transformed problems are estimated by one in terms of another with a constantdepending only on the C1, norms of. This can be seen by choosing the function w supported

in B(z) in (3.29) and changing the variables x y = F(x) in (3.29). Now we obtain estimate(3.39) for v, which implies the following estimate for u:u0,0,/2(z) + Du0,0,/2(z) + 1+[Du]0,,/2(z)

C(n/2uL2((z)) + 2fL((z)) + 0,0,(z) + 1+[]0,,(z)

+g20,0,(z) + 1+[g2]0,,(z))

(3.40)

for any z 2(Q,Q), 0 < < min(r, z), with C depending only on n, , L, and , where(z) := B(z) C(Q,Q) and (z) := B(z) 2(Q,Q). We also have the correspondinginterior estimates: For any z C(Q,Q) and 0 < < dist(z, C(Q,Q)), we obtain (3.40) for(z) = B(z) without the terms involving the norms of g2 on the right-hand side. Multiplying

the interior and boundary estimates (3.40) by

n/2

and using a standard argument (e.g., [20,Theorem 4.8]) yields

|u|(n/2)1,,C(Q,Q) C(uL2(C(Q,Q)) + |f|(2+n/2)0,0,C(Q,Q)

+ ||(1+n/2)0,,C(Q,Q) + |g2|(1+n/2)

0,,2(Q,Q)). (3.41)

Obviously, estimates (3.33) and (3.41) imply (3.21). Also, (3.41) implies (3.22).

Step 6. It remains to prove the existence of a weak solution u H1(C(Q,Q)) of (3.30)(3.32).We first assume that f, C(C(Q,Q)) and g2 C(2(Q,Q)). Then, since 2(Q,Q) is a C2,surface and g2 +

n1i=1

ii C1,(2(Q,Q)), we can find G C2,(C(Q,Q)) such thatn1i=1

iGxii = g2 +n1i=1

ii on 2(Q,Q). (3.42)


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To construct such a function G, we first extend the right-hand side of (3.42) to 2(2Q,2Q) =

(

2Q, 2Q) so that g

C1,(2(2Q,2Q)) with g = g2 + n1i=1 ii on 2(Q,Q). Then, for

every x (3Q/2, 3Q/2), we can locally flatten 2(2Q,2Q) by a C2, diffeomorphism : Rn Rn, i.e., for some r > 0, (Br(x) ( R)) = (Br(x)) {xn > 0} with(Br(x) ( R)) = (Br(x)) {xn = 0}. Then condition (3.42) is transformed into

ni=1

ci(x)Gxi = g on {xn = 0}, (3.43)

where ci C1,(Rn1) with cn min(1, . . . , n)/2 > 0 if r is chosen sufficiently small, and g C1,(Rn1) with compact support is obtained by transforming g by and extending to Rn1.Now, by the argument similar to [20, page 124-125], we choose a nonnegative C20 (Rn1) withRn1

(y)dy = 1, define the function

G(x, xn) = xn Rn1

g(x xny)cn(x xny)

(y)dy,

and show that G is C2,(Rn+) and satisfies (3.43). Transforming G back by 1, we obtain

G Gx C2,(( R) Br(x)) satisfying (3.42) and G 0 on 2(2Q,2Q) Br(x). Gluingthese local functions by using a partition of unity, and taking into account that the local functionsG vanish on 2(2Q,2Q), we obtain a C

2, function G in a neighborhood of the boundary part

2(Q,Q) in C, which satisfies (3.42) on 2(Q,Q). Finally, we extend G to the whole cylinderC(Q,Q) with G C2,(C(Q,Q)).

Now, denoting f = f+n

i=1(ixi iGxixi), we have f C(C(Q,Q)). Thus we can solve the

variational problem of minimizing

I[v] = C(Q,Q)(1

2

n

i=1

iv2xi + f v)dx

on the set {v H1(C(Q,Q)) : v = h G on {xn = Q}}. If v is a minimizer, thenu = v + G is a weak solution of (3.30)(3.32).

For general f, L2(C(Q,Q)) and g2 L2(2(Q,Q)), we approximate by fl, l C(C(Q,Q))and gl2 C(2(Q,Q)) such that (fl, l) (f, ) in L2(C(Q,Q)) and gl2 g2 in L2(2(Q,Q)).Then, for each l, we can find a weak solution ul H1(C(Q,Q)) of (3.30)(3.32) with functionsfl, l, and g

l2 on the right-hand side. Since uk ul satisfies (3.30)(3.32) with fk fl, k l,

gk2 gl2, and h = 0 on the right-hand side, we apply (3.33) to see that {ul} is a Cauchy sequencein H1(C(Q,Q)) and its limit u is a solution of (3.30)(3.32) with the original right-hand sides.Now Lemma 3.1 is proved.

Proof of Theorem 3.2. If is a small constant depending only on , then (3.4) implies the

ellipticity (3.35) with34 instead of . We choose such below.First we show the uniqueness of weak solutions. Suppose that u1 and u2 are two weak solutions

of (3.1)(3.3). Then u1 u2 = 0 on 3Q. Writing the weak form (3.7) for u1 and u2 respectively,subtracting the expressions with w = u1 u2, and then using the ellipticity yields

0 =

CQ

ni,j=1

aijxi(u1 u2)xj (u1 u2)dx 3

4

CQ

|D(u1 u2)|2dx.

Since u1 u2 = 0 on 3Q, we conclude u1 = u2 a.e. in CQ.We now apply the Banach contraction fixed point theorem to prove the existence and regularity

of the weak solution u H1(CQ) of (3.1)(3.3) for sufficiently small > 0 in the space:K = {v H1(CQ) C1,(CQ) : vK := vH1(CQ) + |v|(n/2)1,,CQ < }.


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We now define a mapping J : K K. For v K, consider problem (3.14)(3.17) with functionsf, h, g1, g2, and on the right-hand sides, where

f = f, h = h, g2 = g2, (3.44)

g1(x) = g1(x) +

ni,j=1

(iji aij(x))vxj (x)i(x) for x 1Q = {0}, (3.45)

i(x) =n

j=1

(iji aij(x))vxj (x) for x CQ, i = 1, . . . , n , (3.46)

with f, g1, g2, and h from (3.1)(3.3). Since = en on 1Q, we have

g1 = g1 +

nj=1

(njn anj)vxj g1 + n on x 1Q. (3.47)

By (3.4) and v K, the functions g1 and satisfy the conditions of Lemma 3.1. Thus, by Lemma3.1, there exists a unique weak solution u of (3.14)(3.17) with the right-hand sides describedabove, which satisfies (3.21) and thus implies u K. We define the mapping J : K K bysetting Jv = u.

Now we show that J is a contraction mapping in the norm K if > 0 is small. Letv1, v2 K and uk = Jvk for k = 1, 2. Then u1 u2 is a weak solution of (3.14)(3.17) with thefollowing functions f, h, g1, g2, and on the right-hand sides:

f = 0, g2 = 0, h = 0,

i(x) =n

j=1

(iji aij(x))(v1xj (x) v2xj (x)) for x CQ, i = 1, . . . , n ,

g1(x) =n

j=1

(nj

n anj(x))(v

1

xj(x)

v2

xj(x)) for x

1

Q.

By (3.4), we have

L2(CQ) CDv1 Dv2L2(CQ) Cv1 v2K,||(1+n/2)0,,CQ C|Dv

1 Dv2|(1+n/2)0,,CQ C|v1 v2|(n/2)1,,CQ Cv

1 v2K,g10,,1Q CDv1 Dv20,,1Q C|Dv1 Dv2|

(1+n/2)0,,CQ

C|v1 v2|(n/2)1,,CQ Cv1 v2K.

Thus, from (3.21),

u1 u2K Cv1 v2K.Therefore, the mapping J : K K is a contraction mapping in the norm K if < 1/C, i.e., is small depending only on n, , Q, and .

For such , there exists a fixed point u K satisfying Ju = u. Then u is a weak solution of(3.1)(3.3). Indeed, since Ju = u, then u satisfies (3.18) with right-hand sides given by (3.44)(3.46) computed with v = u. Rearranging and taking into account the last expression in (3.47)yields (3.7). Also, since u satisfies (3.18) with right-hand sides given by (3.44)(3.46), u satisfies(3.22) with these right-hand sides, which implies

|u|(n/2)1,,CQ C(uL2(CQ) + |f|(2+n/2)0,0,CQ

+ g10,,1Q + |g2|(1+n/2)

0,,(2Q)) + C|u|(n/2)1,,CQ ,

by estimating and g1 similar to the estimates of and g1 above and with v = u. If is small,this implies (3.13). Then Theorem 3.2 is proved.


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Proposition 3.2. Let 1, . . . , n, and be as in Theorem 3.2. Let L > 0. Then there exist > 0, depending only on n, , , and L, such that for any Q L, if aij(x) satisfy (3.4) andf, g1, g2, h satisfy (3.5), (3.6), there exists a unique weak solution u H

1

(CQ) of (3.1)(3.3).Proof. By Theorem 3.2, if > 0 is small depending only on n, , , and L, there exists u1 H1(CL) C1,(CL) which is a weak solution of (3.1) in CL, (3.2) on kL, and u1 = 0 on 3L, inthe sense of (3.7) with Q replaced by L. Then we can extend u1 from CL/2 to CQ so that theextended function u1 satisfies u1 C1,(CQ) and u1 = u1 on CL/2.

Denote

g2 := g2 n

i,j=1

aij(u1)xji on 2Q.

Then g2 C0,(2Q), and g2 0 on 2L/2. Now, repeating the construction of Step 6 of proof ofLemma 3.1, we show existence of G C1,(CQ) satisfying

ni,j=1

aijGxji = g2 and G = 0 on 2Q. (3.48)

Also, choosing by C(R1) satisfying 0 on (, L/4) and 1 on (L/2, ), and usingthat g2 0 on 2L/2 and G 0 on 2Q, we see that the function G(x) := (xn)G(x) also satisfies(3.48). Thus, replacing G by G, we get a function G C1,(CQ) satisfying (3.48) and

ni,j=1

aijGxji = 0 on 1Q.

Now, for f, i C0,(CQ) for i = 1, . . . , n, and h H1(CQ), we construct a weak solutionv H1(CQ) of the problem

ni,j=1

(aijvxj )xi = f +

ni=1

ixi in CQ,n

i,j=1

aijvxji = 0 on 1Q,

2Q, v = h on

3Q

in the sense that v = h on 3Q holds as the trace andCQ

(n

i,j=1

aijvxjwxi ni=1

iwxi + f w)dx +2

k=1

kQ

ni=1

ii w dHn1 = 0 (3.49)

for any function w H1(CQ) with w = 0 on 3Q. First we assume i C1,(CQ) for i = 1, . . . , n.Then, denoting f = f+ni=1 ixi , we have f

C(

CQ). Then we find a weak solution v of (3.49)

as a minimizer of

I[v] =

CQ

(1

2

ni,i=1

aijvxivxj + f v)dx

on the set {v H1(CQ) : v = h on 3Q}. In general case, assuming that i C0,(CQ) andusing w = v h in (3.49), and using ellipticity, we obtain

vH1(C(Q,Q)) C(fL2(C(Q,Q)) + C0(C(Q,Q)) + hH1(C(Q,Q))), (3.50)where C depends only on n, ,Q,,. Then approximating = (1, . . . , n) by l C1,(CQ)such that l in C0,(CQ), denoting by vl a weak solution of (3.49) with l instead of ,and using (3.50) and the linearity of the problem, we find that {vl}l=1 is a Cauchy sequence inH1(CQ), and its limit is a weak solution of (3.49) with C0,(CQ).


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Now let v H1(CQ) be a weak solution of (3.49) with f as in Proposition, and

i

= n

j=1aij(u1 + G)xj , h = h u1 G,

where h is as in Proposition. Note that i C0,(CQ) and h H1(C(Q,Q)), thus such v exists,Then function u = u1 + G + v is a weak solution of (3.1)(3.3)

Next we prove some regularity properties of weak solutions near 3Q, more specifically near

the wedge W := {xn = Q}. For a < b, denote C(a,b) = (a, b) and 2(a,b) = (a, b).Proposition 3.3. Let > 0 and b > 0 be constants. Let the coefficients aij(x) satisfy theellipticity condition (3.35) and let aij0,,(Qb,Q) 1. Letu H1(C(Qb,Q)) be a weaksolution of the conormal problem

n

i,j=1

(aijuxj )xi = 0 in

C(Qb,Q),

n

i,j=1

aijuxji = 0 on 2(Qb,Q), (3.51)

u = 0 on {xn = Q} (3.52)in the sense that (3.52) holds as the trace and

C(Qb,Q)

ni,j=1

aijuxjwxidx = 0 (3.53)

for any function w H1(C(Qb,Q)) with w = 0 on C(Qb,Q) \ 2(Qb,Q). Then there exists (0, 1) and C > 0, depending only on the data (i.e., on n, , b, , , ) such that u C1,(C(Qb/2,Q) \ W) and

|Du(x)| CuL2((Qb,Q))(dist(x, W))1 for any x C(Qb/2,Q), (3.54)

where W := {xn = Q}.Proof. The regularity u C1,(C(Qb/2,Q) \ W) follows from the standard interior estimates forlinear elliptic equations and the estimates for the Dirichlet and conormal problems. Thus itsuffices to prove (3.54).

DenoteD(Qb,Q) := C(Qb,Q) \ 2(Qb,Q).

We first note that any weak solution u of (3.51) (without requiring (3.52)) satisfies the followinglocal estimates: For any B2R = B2R(y) Rn with R (0, b/10), denote M = supB2RD(Qb,Q) u

+

and m = infB2RD(Qb,Q)u, where in the case B2R D(Qb,Q) = we set m = and M = .

Define the functions u+M and um in C(,) by

u+

M(x) := sup(u(x), M), x C(Qb,Q),M, x C(,) \ C(Qb,Q), um(x) := inf(u(x), m), x C(Qb,Q),m, x C(,) \ C(Qb,Q).

Then, for any p > 1,

supBR(y)C(,)

u+M CRn/pu+MLp(B2R(y)C(,)).

If u is nonnegative in B4R(y) C(Qb,Q), then, for any 1 p n/(n 2), the weak Harnackinequality holds:

infBR(y)C(,)

um CRn/pumLp(B2R(y)C(,)),where the constant C depends only on , b , , and p. The proof follows the proofs of [20, Thm8.17, 8.18, 8.25, 8.26] since, in the weak formulation (3.53), we can use the same test functionsas in [20, Thm 8.17, 8.18, 8.25, 8.26] and obtain the same expressions in the result, with the only


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difference that we integrate over C(,) instead of Rn. Also, in our case, the test functionsmay be nonzero on the boundary part 2(Qb,Q), and thus we use the Sobolev inequalities for the

functions that vanish only on a part of the boundary (cf. [34, Thm. 6.37]).Once the weak Harnack inequality is obtained, we follow the proofs of [20, Thm 8.22, 8.27]

to obtain the existence of (0, ) and C > 0 depending only on the data such that, if u is aweak solution of (3.51)(3.52), then

|u(x)| CuL(C(Q3b/4,Q))|Q xn| for any x C(Qb/2,Q).We also obtain uL(C(Q3b/4,Q)) CuL2((Qb,Q)) by applying the local estimates to uwith p = 2 and using (3.52). Then, using 0 Q xn dist(x, W) for x C(Qb/2,Q), we have

|u(x)| CuL2((Qb,Q))(dist(x, W)) for all x C(Qb/2,Q). (3.55)Now we follow the argument of Step 5 of the proof of Lemma 3.1. For x Rn and > 0,

denote (x) :=

C(Qb,Q)

B(x). Let x

{xn = Q

}and = min(b, dist(x, W))/10. Then

B(x) C(Qb,Q) {xn = Q}. Thus, applying the standard estimates for the Dirichletproblem [20, Thm 8.32, 8.33], properly scaled, we obtain

u0,0,/2(x) + Du0,0,/2(x) + 1+[Du]0,,/2(x) Cn/2uL2((x)), (3.56)where C depends only on the data. Now let x 2(Qb/2,Q) and = min(b, dist(x, W))/10. ThenB(x)C(Qb,Q) 2(Qb,Q). Thus the proof of (3.40) yields (3.56) in the present case. Finally,if x C(Qb/2,Q), = min(b, dist(x, W))/40, and B(x) C(Qb,Q), then the interior estimates[20, Thm 8.32] also yield (3.56). Note that, ifz C(Qb/2,Q) \ W, 0 < r min(b, dist(x, W))/10,and y r(z), then 12 dist(z, W) dist(y, W) 2dist(z, W). Then (3.55) and (3.56) imply

Du0,0,/2(x) CuL2((Qb,Q))(dist(x, W))1 (3.57)

for all choices of (x, ) discussed above. Moreover, the sets /2(x), with (x, ) specified above,cover C(Qb/2,Q). Then we use again that 12 dist(x, W) dist(y, W) 2dist(x, W) for all y /2(x) to have (3.54).

4. Free boundary problems in the infinite nozzle

In order to solve Problem (FB), we first reformulate it into a free boundary problem for thesubsonic part of the solution, since Problem (FB) is originally hyperbolic-elliptic mixed.

We first modify equation (1.1) to make it uniformly elliptic so that it coincides with the originalequation in the range of D in the subsonic region + for satisfying (2.31) with sufficientlysmall . The details of the truncation procedure are in [4, Section 4.2].

Let = (c q+0 )/2. Then there exist C1,1([0, )) and cj > 0, j = 1, 2, 3, depending onlyon q+0 and , such that

(q2) = (q2) if 0 q < c , (4.1)(q2) = c0 +

c1q

if q > c , (4.2)

0 < c0

(q2)q c2 for q (0, ). (4.3)

Then the equation

L := div((|D|2)D) = 0 (4.4)is uniformly elliptic, whose ellipticity constants depend only on q+0 and .

To formulate the problem into a free boundary problem, the main point is to replace thepointwise gradient condition {|D(x)| < c} defining +() by a condition stated in terms of so that our problem is formulated into the framework of free boundary problems.


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To do that, we first need to construct a local C1, supersonic solution in the domain1 := {1 < xn < 1} of the initial-boundary value problem (2.18)(2.19) for the nonlinearhyperbolic equation (1.1) such that

0 1,,1 C1 (4.5)when is sufficiently small. This is ensured by the local existence theorem, Proposition A.1 inAppendix. By the standard extension argument (see [5] and Section 7.2), we can extend tothe whole infinite nozzle such that

0 1,, C0, |l = 0. (4.6)Then the following heuristic observation motivates our formulation: By (2.13) and q0 > q

+0 ,

we have +(0) = {x : 0(x) < 0 (x)}. Since is a small C1, perturbation of 0and q0 > q

+0 , then we expect that

+ is close to +0 in C1,(+()) so that we can expect that

+() = {x : (x) < (x)}.We also perform the corresponding truncation of the free boundary condition (2.4):

(|D|2) = (|D|2)D on S. (4.7)On the right-hand side of (4.7), we use the original function since = on the range of|D|2. Note that (4.7), with the right-hand side considered as a known function, is the conormalboundary condition for the uniformly elliptic equation (4.4).

We first solve the following free boundary problem (TFB), which is a truncated version ofProblem (FB), in which the gradient condition that determines + in Problem (FB) is alsoreplaced by the condition in with the definition + := { < } .

Problem (TFB): Truncated Free Boundary Problem. Given a supersonicupstream flow C1,(1) of (1.1) satisfying (4.5) and (2.18)(2.19) for small > 0 and some constant C0, find C() such that

(i) satisfies

in (4.8)and conditions (2.18)(2.19) on the boundary and (2.20) at infinity;

(ii) C1,(+) C2(+) is a solution of (4.4) in + := { < } , thenon-coincidence set;

(iii) the free boundary S = + is given by the equation xn = f(x) so that+ = {xn > f(x)} with f C1,;

(iv) the free boundary condition (4.7) holds on S.

In the next sections we first solve Problem (TFB) by an iteration procedure using uniformestimates at infinity and the regularity estimates near the nonsmooth boundary. Finally, we usean estimate for |D| to conclude that the solution of the truncated problem, Problem (TFB), isactually a solution of Problem (FB) and thus Problem (TN).

5. Iteration Procedure and Uniform EstimatesWe now introduce an iteration procedure to construct approximate solutions of Problem (TFB)

in the domain and make uniform estimates of the solutions on the unbounded domain withnonsmooth boundary.

5.1. Iteration set. Let M 1. We setKM :=

C1,() : q+xn1,, M

, (5.1)

where q+ is defined by equation (2.26) (see Remark 2.4) and (0, 1). By definition, KM isconvex. Now we will show that KM is compact in the weighted Holder space C1,/2(1) () as definedbelow.


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For (0, 1), m N, and k R, we define the following spaces:Cm,

(k)

() :={

u

Cm,() :

u

(k)

m,,

> 0, thenKM is a compact subset of C1,(1)().Proof. Let j KM for j = 1, 2, . . . . By a standard argument, we can extract a subsequence (stilldenoted by) j, which converges in C1, on every compact subset of to the limit C1,().Then q+xn1,, M and thus KM. It remains to show that j (1)1,, 0 asj .

Fix 0 < < 1. Then j(1)1,,{xn>1/} (M + q+), and the same estimate holdsfor . Also, since j in C1,( {xn 2/}), there exists j0 such that, for j > j0,j (1)1,,{xn2/} . Then, for j > j0, we have j

(1)1,, C, and the assertion is

proved.

5.2. Construction of the iteration scheme. Let KM. Since q0 > q+0 , it follows that, if

q0 q+0

C(M + 1)(5.3)

with large C depending only on n, then (4.5) implies

( )xn(x) q0 q+0

2> 0 in (5.4)

and > on {xn > 1}. Then the set +() := { < } has the form:+() = {xn > f(x)} with f1,,Rn1 CM (5.5)

for C depending only on q0 q+0 . The inward unit normal on S := {xn = f(x)} to + is

(x) =D(x) D(x)|D(x) D(x)| for x S. (5.6)

By the definition of KM and (5.3), formula (5.6) also defines (x) on 1 and en0,,1 CM with C = C(q+0 , q0 ). (5.7)

Motivated by (4.7), we define the function

G(x) := (|D(x)|2)D(x) (x) on 1. (5.8)Then we consider the following problem in the domain +():

div((|D|2)D) = 0 in +(), (5.9)(|D|2) = G(x) on S := {xn = f(x)}, (5.10) = 0 on l

+ := +() l, (5.11)limR

q+xnL(+(){xn>R}) = 0, (5.12)show that there exists a unique solution (x), and extend the solution to the whole domain sothat KM.


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We rewrite problem (5.9)(5.12) in terms of the function u(x) = (x) q+xn. We first notethat the boundary condition (5.11) is equivalent to

(|D|2) = 0 on l+.Now u(x) is a solution of the following problem:

div A(Du) = 0 in +(), (5.13)

A(Du) = g(x) on S, (5.14)A(Du) = ((q+)2)q+ en on l+, (5.15)

limR

uL(+(){xn>R}) = 0, (5.16)where

A(P) = (|P + q+en|2)(P + q+en) ((q+)2)q+en for P Rn, (5.17)g(x) = G(x) ((q+)2)q+ en. (5.18)

Thus, u(x) satisfies the uniformly elliptic equation with the same ellipticity constants as in (4.4):

||2 n

i,j=1

AiPj (P)ij 1||2 for any P, Rn. (5.19)

Moreover, from the definition of A(P) and (4.1)(4.2), A(P) satisfies

A(0) = 0, (1 + |P|)|DPAiPj (P)| C. (5.20)Finally, we state a linear problem corresponding to (5.13)(5.16). Namely, we use (5.20) to

find that, for i = 1, . . . , n,

Ai(Du(x)) =

n

j=1aij(x)uxj (x), aij(x) =

10

Aipj (tDu(x))dt.

We replace u q+xn in the definition of the coefficients aij by q+xn for KM todefine

aij(x) = a()ij (x) =

10

Aipj

t(D(x) q+en)

dt for x , i, j = 1, . . . , n . (5.21)

The ellipticity (5.19) of A(P) implies that the coefficients a()ij (x) satisfy the ellipticity condition

(3.35) for any x .Also, from (5.17) and (5.21),

a()ij (x) =

10

(|tD(x) + (1 t)q+en|2)ji (5.22)

+2(|tD(x) + (1

t)q+en

|2)(txi(x) + (1

t)q+n

i

)(txj (x) + (1

t)q+n

j

)dtfor x , where ji = 1 if i = j and ji = 0 if i = j. In particular, we have aij = aji .

We note that, for 0 = q+xn, the corresponding coefficients aij defined by (5.22) are constants

and satisfy

aij = iji for i, j = 1, . . . , n , (5.23)

where i are i =

(q+)2

for i = 1, . . . , n 1, and n = (q+) for (s) = (s2)s. By(4.1)(4.3), we have

i 1 for i = 1, . . . , n .Now, for KM, we use (5.22) and (4.1)(4.3) to obtain

a()ij aij0,, CM. (5.24)


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Thus, we formulate the following conormal fixed boundary elliptic problem:

ni,j=1

(a()ij uxj )xi = 0 in +(), (5.25)

ni,j=1

a()ij uxji = g(x) on S, (5.26)

ni,j=1

a()ij uxji = ((q+)2)q+ en on l+(), (5.27)

limR

uC(+(){xn>R}) = 0. (5.28)

Since the coefficients are only C, we can expect to find only a weak solution u C1,(+())of (5.25)(5.28) in the sense that u(x) satisfies (5.28) and, for any w C1c (Rn),

+()

ni,j=1

a()ij uxjwxi dx +

S

g w dHn1 (5.29)

l+()

((q+)2)q+ en w dHn1 = 0.

We will determine the iteration map J() = by solving (5.25)(5.28) for u, extending ufrom +() to so that u + q+xn KM, and defining = u + q+xn. A fixed point of this mapis obviously a solution of Problem (TFB) in . In Sections 67, we prove the existence of sucha map J, as well as its fixed point.

6. Fixed Boundary Problems in a bounded domain +R()

In order to find a solution of (5.25)(5.28) in the unbounded domain +(), we first solve thecorresponding problem in the bounded domain

+R() := +() {xn < R}, R > L,

and then pass to the limit as R . Thus we consider the following problem:n

i,j=1

(a()ij uxj )xi = 0 in

+R(), (6.1)

ni,j=1

a()ij uxji = g(x) on S, (6.2)

ni,j=1

a()

ij uxji = ((q+

)2

)q+

en on l+R(), (6.3)

u = 0 on +R() {xn = R}, (6.4)where a

()ij are defined by (5.21) and l

+R() = l

+ +R().We first note several properties of g, which will be used later.

Lemma 6.1. There exists C > 0, depending only on the data and independent of M and R,such that

g0,,1 C. (6.5)This is proved in [5, (4.48), page 334]. Another property of g is


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Lemma 6.2.

Sf

gd

Hn1 = ((q+)2)q+

l+R

end

Hn1.

Proof. Note that

0 =

R

(xn) =

{xn=R}

en dx Sf

en dHn1 l

+R

en dHn1,

where is the inward unit normal. Since = en on {xn = R}, then, from the last equality,we have

Sf

en dHn1 +l

+R

en dHn1 = ||. (6.6)

Also, since G is defined by (5.8) and C1,( \ +R()) is a weak solution of (1.1) in

\ +R() and satisfies the boundary condition (2.19), we findSf

GdHn1 = o

(|D|2)D dHn1. (6.7)Then we have

Sf

gdHn1 ((q+)2)q+l

+R

endHn1

=

Sf

GdHn1 ((q+)2)q+

Sf

endHn1 +l

+R

endHn1

=

Sf

GdHn1 ((q+)2)q+|| = 0,

where we used (6.7) and the fact that q+ satisfies (2.26)(2.27).

We now fix KM and write +

R for

+

R() and aij for a

()

ij to simplify the presentation.Also, C denotes a universal constant, which may be different at each occurrence, and dependsonly on the data and is independent of M and R. Furthermore, there exists 0 > 0 dependingonly on M and the data such that, when 0, S {1/10 < xn < 1/10} by (5.5); we alwaysassume 0 below.Lemma 6.3. Let (0, 0) be sufficiently small, depending only on the data and M. Letu H1(+R), R > L, be a weak solution of (6.1)(6.4) in the sense that

(i) u = 0 on +R {xn = R} as the trace;(ii) For any w H1(+R) satisfying w = 0 on +R {xn = R} in the sense of traces,

+R

n

i,j=1aijuxjwxi dx +

Sf

g w dHn1 l

+R

((q+)2)q+ en w dHn1 = 0. (6.8)

ThenDuL2(+R) C. (6.9)

Proof. We use the approach in the proof of [5, Lemma 4.2]. We first choose w = u in (6.8) toobtain

+R

ni,j=1

aijuxjuxi dx = Sf

gu dHn1 +l

+R

((q+)2)q+ en u dHn1.

Now, if Q is a constant which will be chosen below, we use Lemma 6.2 to obtain+R

ni,j=1

aijuxjuxi dx = Sf

g(u Q) dHn1 l

+L

((q+)2)q+ en (u Q) dHn1,


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where we used that en = 0 on l+R \ l+L by (2.14) and (2.17).Choosing Q = (u)L :=

1

|+L ()| +L() u(x)dx and using the L

2 estimate of the boundary

traces of functions in the Sobolev space H1(+L()), (6.5), and | en| C on l+L by (2.16),we find

+R

ni,j=1

aijuxjuxi dx

Sf

g2dHn11/2

Sf

(u (u)L)2dHn11/2

+((q+)2)q+

l+L

| en|2dHn11/2

l+L

(u (u)L)2dHn11/2

C +L()

(u (u)L)2 + |Du|2

dx1/2

C

+L() |Du

|2dx1/2,

where, in the last estimate, we used the Poincare inequality in the domain +L(), and thus theconstant C depends only on n, L, the norms in (2.15)(2.16), and f1,,. On the other hand,it follows from (5.5) that if < 1/M, then f1,, C with C > 0 independent of M, R, and KM. Now, using (3.35), we have

+R

|Du|2dx 1

+R

ni,j=1

aijuxjuxi dx C

+R

|Du|2dx1/2.This completes the proof.

By [35], the weak solution u of (6.1)(6.4) is in C1, +R \ (S {xn = R}), since l

+R

is C1,.

Lemma 6.4. Let and u be as in Lemma 6.3. Then, for any xn (L, R), there exists x such that u(x, xn) = 0.

Proof. We follow the scheme of the proof [5, Lemma 4.3, Step 2]. Fix z (L, R). Suppose thatthere is no x such that u(x, z) = 0. Then we can assume u(x, z) > 0 for all x , sincethe case u(x, z) < 0 for all x can be handled similarly. From the compactness of andcontinuity of u, there exists a constant > 0 such that

u(x, z) > 0 for all x .Consider the domain D := (z, R). Then, using Proposition 3.3, we get u C1,(D \ W),with W = l

{xn = R

}. Moreover, u is a weak solution of

ni,j=1

(aijuxj )xi = 0 in D,

ni,j=1

aijuxji = 0 on lD = D l {xn < R},

u > 0 on D {xn = z},u = 0 on D {xn = R}.

Thus, by the maximum principle, u 0 in D. In particular,u = uxn 0 on (D {xn = R}) \ W. (6.10)


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By the strong maximum principle [20, Theorem 8.19], u > 0 in the interior of D. Also, denotingD = {x D : dist(x, l) > }, where > 0 is fixed to be small so that D = , we have

aij C0,(D/2), u C1,(D/2).Thus, by [5, Lemma A.1],

u = uxn > 0 on D {xn = R}. (6.11)Since u = 0 on {xn = R} and ann > 0 from the ellipticity, we conclude from (6.10)(6.11)that

ni,j=1

aijuxji = annuxn

0 on (D {xn = R}) \ W,

> 0 on D {xn = R}.Since ann and uxn are continuous on D {xn = R}, we conclude

{xn=R}

n

i,j=1

aij

uxj

i

dH

n1 > 0. (6.12)

On the other hand, we show that{xn=R}

ni,j=1

aijuxji dHn1 = 0, (6.13)

thus contradicting (6.12). To prove (6.13), we choose (L, R), and use in (6.8) the test functionw(x) = (( xn)/), where > 0, C(R) with 0 on R, 1 on (1, ) and 0 on(, 1/2). Note that w C(R) and w 0 on {xn = R}. Thus we use this w in (6.8) andu C1,(D \ W) and send 0+ to obtain

{

xn=

}

n

i,j=1

anjuxj dHn1 + Sf

g dHn1 l

+

((q+)2)q+ en dHn1 = 0. (6.14)

Since > L and en = 0 on l {xn > L}, we find from (6.14) and Lemma 6.2 that{xn=}

ni,j=1

anjuxj dHn1 = 0. (6.15)

From u C1,(D\W) and aij C(), we haven

i,j=1(anjuxj )(x, ) ni,j=1(anjuxj )(x, R)

as R for every x . Moreover, from (3.54) of Proposition 3.3, there exist C > 0 and (0, 1) such that, for each ((R + L)/2, R) and x , we have

|Du(x, )| C(dist((x, ), W))1 C(dist(x, ))1,where the last inequality holds since dist(x, W) dist(x, ) for any x = (x, xn) R. Since(dist(

, ))1

L1(), we use the dominated convergence theorem to send to the limit as

R in (6.15) and use that = en on D {xn = R} to obtain (6.13).Thus, we conclude a contradiction with (6.12), which completes the proof.

From (2.14)(2.17), there exist r > 0 and C > 0 with the following properties: For anyx0 l{xn > 1/2}, there exist an orthonormal coordinate system (y1, . . . , yn) with yn = xnand a function : Rn1 R such that

B10r(x0) = {y1 > (y2, . . . , yn)} B10r(x0), 1,,Rn1 C, D (x0) = 0. (6.16)Proposition 6.5. Letu(x) be as in Lemma 6.3. If is sufficiently small, depending only on the

data and M, then, for any x0 = (x01, . . . , x0n) +R10r with L + 1 x0n < R 10r,

Du0,,Br(x0)+R CDuL2(B4r(x0)+R). (6.17)


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Proof. We use the fact that u K also satisfies equation (6.1) and the conormal boundaryconditions (6.2)(6.3) for any constant K R.

When x0 satisfies B2r(x0) +R, we apply [26, Theorem 3.13] to u K and obtain

Du0,,Br(x0)+R Cu KL2(B4r(x0)+R). (6.18)When x0 l+ with L + 1 x0n < R 10r, then, since the right-hand side of (6.3) vanishes

on l+ {xn > L}, we use (5.24) and (a linear version of) the estimates for the conormal

derivative problem [35], applied to u K, to obtainDu0,,Br(x0)+R Cu KC0(B2r(x0)+R). (6.19)

Now we use the standard estimates [20, Theorem 8.17] for the equations of divergence form(extended to the case of local estimates near the boundary for the conormal problem, see e.g. [37,Chapter 4, Section 10] where these modifications are indicated in the parabolic case; note thatthe Lipschitz regularity of the boundary in our case allows to use the weak form of the conormalproblem) to obtain

u KC0(B2r(x0)+R) Cu KL2(B4r(x0)+R).That is, (6.18) is satisfied for the case x0 l+R.

Now choosing

K =1

|B4r(x0) +R|B4r(x0)

+R

udx,

using the Poincare inequality, and noting that the constant in the Poincare inequality can bechosen independent of x0 +R by (6.16), we complete the proof. Proposition 6.6. Let and u(x) be as in Lemma 6.3. Then

uC0(+R)

C.

Proof. Lemma 6.3 and Proposition 6.5 imply that|Du| C on +R {L + 1 < xn < R 10r}.

Combining this with Lemma 6.4, we have

|u| C on +R {xn (L + 1, R 10r)}.To extend this bound to the domains +R {R 10r < xn < R} = (R 10r, R) and +L+1,we note that these domains are of the fixed size and structure (the cylinder (a, b) and itssmall perturbation, respectively). Note that, on the boundary parts {xn = L + 1} +L+1 and{xn = R 10r} ( (R 10r, R)), we have |u| C, as we proved above. Moreover, theright-hand sides of (6.2)(6.3) are estimated in L by C from (2.16) and (6.5), respectively.Thus we can use the standard estimates [20, Theorem 8.15] for the equations of divergence form(extended to the case when we have the conormal boundary conditions on a part of the boundary,

see e.g. [37, Chapter 4, Section 10]). Proposition 6.7. If is sufficiently small depending only on the data and M, there exists a

unique weak solution u H1(+R) of (6.1)(6.4). Furthermore, u C1,(+R10r) satisfies (6.9)and

u1,,+R10r C. (6.20)Proof. Since the map in (2.14) satisfies (2.16), then, if is sufficiently small, there exists theinverse map : Rn Rn of . The map satisfies

Id2,,Rn C. (6.21)Then, from (2.14),

(l) R. (6.22)


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Note that, if is small, then, by (2.16), (4.5), and (5.4), we conclude that the set ( +()) ={ < } ( R) has the form

(+()) = {xn > f(x)} ( R) with f1,,Rn1 CM. (6.23)Now we consider a map

= 1 : Rn Rn for 1(x, xn) = (x, xn (xn)f(x)), (6.24)where C(R) is nonincreasing and satisfies 1 on (, 1/8) and 0 on (1/4, ).From this definition, we have

Id1,,Rn CM.Thus, for small , the map is invertible and

1 Id1,,Rn CM.In particular, for small ,

Id

1,,Rn +

1

Id

1,,Rn

1/10. (6.25)

Moreover, if is small, then

(0, ) = +R {xn < } for > L/4. (6.26)Define the function v(y) := u(x) = u(1(y)) for y CR := (0, R). Then u is a weak

solution of the conormal problem (6.1)(6.4) if and only if v H1(CR) is a weak solution of thecorresponding problem of the form (3.1)(3.3) in CR. To obtain the expressions of the coefficientsaij and the right-hand sides f, g1, and g2 of problem (3.1)(3.2) for v, in terms of the data of

problem (6.1)(6.3) and the map , we make the change of variables y = (x) in (6.8). Thenwe have

aij iji 0,,CR CM for i, j = 1, . . . , n , (6.27)with i defined by (5.23), where we used (5.24) and (6.25). In this case, the right-hand side in(3.1) is f = 0; to estimate g1 and g2 in (3.2), we use (2.16), (6.5), and (6.25) to obtain

gk0,,k C, k = 1, 2. (6.28)Also, from (6.4) and (6.26), we see that h = 0 in (3.3) for v.

Now the existence a solution v(y) follows from Proposition 3.2 if > 0 is sufficiently smalldepending only on the data and L, so that the right-hand side of (6.27) is smaller than inProposition 3.2.

Thus u(x) := v((x)) is a weak solution of (6.1)(6.4).Since aij satisfy ellipticity condition, uniqueness of weak solution u of problem (6.1)(6.4)

follows from the uniqueness argument in the proof of Theorem 3.2.Estimate (6.9) for u(x) follows from Lemma 6.3.

It remains to prove (6.20). For any x +R {L/2 xn R 10r}, the standard interiorestimates [20, Theorem 8.32] and the estimates for oblique derivative problems [35] imply

u1,,Br(x0)+R C(uC0(B4r(x0)+R) + ((q+)2)q+ en0,,Br(x0)+R) C, (6.29)where we used (2.16) and Proposition 6.6. Thus it suffices to obtain an estimate in +R{xn L}.By (6.25)(6.26), it suffices to show that

v1,,(0,L) C, (6.30)since the similar estimate then holds for u1,,+R{xnL}.

To show (6.30), we note that using (6.25)(6.26) and Proposition 6.6 yields

vL2((0,2L)) 2uL2(+R{xn


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7. Existence of Solutions of the Free Boundary Problem

We now show the existence of solutions of the free boundary problem. In this section, Cdenotes a universal constant, which may be different at each occurrence and depends only on thedata.

7.1. Solutions of the fixed boundary problem in the infinite nozzle. We first have

Proposition 7.1. If is sufficiently small depending only on the data and M, then there existsa unique weak solution u C1,(+()) of (5.25)(5.28). This solution satisfies

DuL2(+()) C, (7.1)u1,,+() C, (7.2)lim

u1,,{xn>} = 0. (7.3)Moreover, we have

(i) u = limRuR, where uR is the unique solution of (6.1)(6.4) for each R > 0 and the

convergence is in C1, on any compact subset of +() for any 0 < ;(ii) For every R L, there exists x such that u(x, R) = 0.

Proof. We assume that is sufficiently small to satisfy the conditions stated in Sections 56.We first show that any weak solution u C1,(+()) of (5.25)(5.28) satisfies (7.3). From

(5.28), for any > 0, there exists > L + 2r such that

uC({xn>}) .Note that (2.17) implies that the right-hand side of (5.27) vanishes on l

+() {xn > }.Then the standard estimates [20, Theorem 6.2, Lemma 6.29] imply that, for x = (x, xn) withxn > L + 2r,

u1,,Br(x) Cu0,0,B2r(x) C.Thus, (7.3) follows.

Now we prove the uniqueness of weak solutions of (5.25)(5.28) in C1,(+()). Let u, v C1,(+()) be two solutions of (5.25)(5.28). Then, for any R > L + diam(), define

(x) := R(x)(u(x) v(x)),where R(x) = (|x|/R) and Cc (R+) is a nonnegative function such that 1 on [0, 1].Then C1c (+). Using as a test function in (5.29) for u and v respectively and subtractingthem, we have

0 =

+

ni,j=1

a()ij

(uxi vxi)(uxj vxj )R + (u v)(uxj vxj )xiR

dx.

Thus,+

ni,j=1

a()ij (uxi vxi)(uxj vxj )Rdx

C

R

+(B2R(0)\BR(0))

(|Du| + |Dv|)(|u| + |v|)dx.

Note that |+ (B2R(0) \ BR(0))| 2R||. Thus, using the ellipticity of a()ij , we find

+BR(0)

|Du Dv|2dx C(u1,0,\BR(0) + v1,0,\BR(0))

for large R. Now we send R and use (7.3) to obtain

+R

|Du Dv|2dx = 0.


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Thus, u v = const., which yields u = v by using (7.3).It remains to prove the existence of a weak solution of (5.25)(5.28) satisfying (7.1)(7.2) and

properties (i)(ii) of Proposition 7.1.The existence of a unique solution uR C1,(+R10r) H1(+R) of (6.1)(6.4) for any R > L

follows from Proposition 6.7.Now, let Rj . Using estimates (6.9) and (6.20) for uR, which hold by Proposition

6.7, we can extract from {uRj} a subsequence converging in C1,/2 on compact subsets of to a function u C1,() that satisfies (7.1). Indeed, this can be achieved by extracting asubsequence converging in C1,/2(+L+1), a further subsequence converging in C

1,/2(+L+2),etc., and by using the diagonal procedure. Obviously, u is a weak solution of (5.25)(5.27), since

we can pass to the limit in (5.29) for any fixed w C1c (+).Moreover, this solution u satisfies property (ii): Indeed, for fixed R0 L, Rj R0 for

sufficiently large j. For such j, by Lemma 6.4, there exists xj such that uRj (xj , R0) = 0.

Then there exists a subsequence of {x

j} converging to some point x

. Since uRj (, R0) u(, R0) uniformly in , we have u(x, R0) = 0. Thus, the solution u satisfies properties (i)(ii).Also, since each uR satisfies (6.9) and (6.20) with C independent of R, then (7.1)(7.2) hold.Now we prove that u satisfies (5.28). From (7.1), for any > 0, there exists > L such that

DuL2(+{xn>}) .Then, by Proposition 6.5 (which holds for u as uR),

Du0,,+{xn>+1} C.From this estimate and property (ii) proved above, we have

uC0(+{xn>+1}) C.Now (5.28) is proved.

7.2. Iteration map. We first define an extension operator P, similar to [4, Proposition 4.5]and [5, Proposition 5.1], to obtain

Proposition 7.2. Let > 0 be sufficiently small depending only on the data and M, and

satisfy (4.5). Then, for any KM, there exists an extension operator P : C1,(+()) C1,() satisfying the following two properties:

(i) There exists C depending only on n,,q+0 , and (but independent of M,R,, and )

such that, for the solution u C1,(+()) of problem (5.25)(5.28) and = u + q+xn,P q+xn1,, C; (7.4)

(ii) Let (0, ). If a sequence k KM converges to C1,() in the norm (1)1,,defined in(5.2), then KM; Furthermore, if uk C1,(+(k)) andu C1,(+())are the solutions of problems (5.25)(5.28) for

kand , respectively, and if

k= u

k+

q+xn and = u + q+xn, thenPkk P in the norm (1)1,,.Proof. Let KM. We first define an appropriate extension operator P : C1,(+()) C1,().

Let C1,(+(j)) and u = q+xn. Then v := u 1 satisfies v C1,(C(0,)) withC(a,b) = (a, b), where is defined by (6.24).

We first define the extension operator E1 : C1,(C(0,)) C1,(C(2,)) for any (0, 1).Let v C1,(C(0,)). Define E1v = v in C(0,). For (x, xn) C(1,0), define

E1v(x, xn) =2

i=1

civ(x, xn

i),


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where c1 = 3 and c2 = 4, which are determined by2

i=1 ci1i m = 1 for m = 0, 1. Note that,

for sufficiently small , we have

(

C(2,)) by (6.25). Thus, (

E1v)

C1,().

Finally, we defineP()(x) = (E1v) (x) + q+xn.

Obviously, P() C1,(). Also, since (+()) = C(0,), we haveP()(x) = v (x) + q+xn = u(x) + q+xn = (x) for x +(),

i.e., P : C1,(+()) C1,() is an extension operator.Now, using (6.25), we obtain the following estimate as in the proof of [4, Proposition 4.5]:

P q+xn1,, C q+xn1,,+(). (7.5)Then, ifu and are those defined in assertion (i), we have (7.2) for u = q+xn by Proposition7.1, and thus (7.5) implies (7.4). Assertion (i) is proved.

Now we prove assertion (ii). First, by Lemma 5.1, KM.By (7.4) applied to j and by Lemma 5.1, there exists a subsequence (still denoted by) j(x)

such that

Pjj in C1,(1)() (7.6)for some C1,().

Denote k := k and := . Then, by (2.16), k in C1,( [1, 1]). Denote byfk(x

) the function from (6.23) for k(x). Then, by (2.16), (4.5), and (5.4), fk f in C1,().Now let k and be the maps from (6.24) for fk and f, respectively. Then k and and their

inverse maps satisfy (6.25), and k and 1k 1 in C1, on compact subsets of Rn.Note vk := uk 1k C1,( [0, )). Recall that uk is a weak solution of (5.25)(5.28) in thedomain +(k), with coefficients a

(k)ij defined by (5.22) for k and corresponding right-hand

sides (5.18) and (5.27) for k. Then, by Proposition 7.1, each uk satisfies (7.1)(7.2). Therefore,

DvkL2((0,)) C, (7.7)vk1,,(0,) C for k = 1, . . . . (7.8)

Moreover, vk is a weak solution of a conormal problem

ni,j=1

(b(k)ij xjvk)xi = 0 in (0, ), (7.9)

ni,j=1

b(k)ij xjvki = gk(x) on {0}, (7.10)

ni,j=1

b(k)ij xjvki = hk(x) on (0, ), (7.11)

where b(k)ij , gk , and hk are defined by the coefficients and right-hand sides of (5.25)(5.27) for

k and by 1k . In particular, since a

(k)ij a()ij in C1, on compact subsets of as k ,

which follows from (5.22), and since 1k 1 in C1, on compact subsets of Rn), it is easyto show that, for any compact K [0, ),

b(k)ij b()ij in C1,(K) and (gk , hk) (g, h) in C(K),

where b()ij , g, and h are the coefficients and right-hand sides of (7.9)(7.11) for . By (7.8),

there exists a subsequence vkm which converges in C1, on the compact subsets of [0, ) to


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some v C1,( [0, )). Then, choosing w C1c (Rn) in the weak formulation (5.29) for theconormal problem (7.9)(7.11):

(0,)

ni,j=1

b(k)ij vxkwxi dx +

{0}

gkw dx

(0,)

hkw dHn1 = 0,

we pass to the limit for the subsequence {km} and conclude that v is a weak solution of problem(7.9)(7.11) with coefficients b

()ij and right-hand sides g and h. Note also that v satisfies (7.7)

since each vkm satisfies (7.7). Transforming back, we conclude that u := v is a weak solutionof problem (5.25)(5.27) for and satisfies (7.1). Now the proof of Proposition 7.1 implies that usatisfies (5.28). Then the uniqueness assertion of Proposition 7.1 implies u = u on +(). Thus,

P() = (E1v) + q+xn. (7.12)Finally,

Pkm (km) = (E1vkm) km + q+xn (E1v) + q+xn in C1,(K)for any compact K , since (k, vkm) (, v) in C1,(K) and E1 : C1,(C(0,)) C1,(C(2,))is continuous. From this, using (7.6) and (7.12), we obtain that Pkm (km) P() in C1,(1)().Moreover, by the same argument, from any subsequence of Pk(k), we can extract a furthersubsequence converging in C1,(1)() to the same limit P(). Thus the whole sequence Pk(k)converges to this limit. Proposition 7.2 is proved.

Now we can define the iteration map J : KM C1,() byJ := P (7.13)

with

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