3-D axisymmetric transonic shock solutions of the full Euler system in divergent nozzles
aa r X i v : . [ m a t h . A P ] A ug YONG PARK
Abstract.
We establish the stability of 3-D axisymmetric transonic shock solutions of thesteady full Euler system in divergent nozzles under small perturbations of an incoming radialsupersonic flow and a constant pressure at the exit of the nozzles. To study 3-D axisymmetrictransonic shock solutions of the full Euler system, we use a stream function formulation of thefull Euler system for a 3-D axisymmetric flow. We resolve the singularity issue arising in streamfunction formulations of the full Euler system for a 3-D axisymmetric flow. We develop a newscheme to determine a shock location of a transonic shock solution of the steady full Eulersystem based on the stream function formulation. Introduction
In [14, Chapter 147], authors, using an approximate model, describe a transonic shock phe-nomenon for a compressible invicid flow of an ideal polytropic gas in a convergent-divergenttype nozzle called de Laval nozzle: If a subsonic flow accelerating as it passes through the con-vergent part of the nozzle reaches the sonic speed at the throat of the nozzle, then it becomesa supersonic flow right after the throat of the nozzle. It further accelerates as it passes throughthe divergent part of the nozzle. If an appropriately large exit pressure p e is imposed at theexit of the nozzle, then at a certain place of the divergent part of the nozzle, a shock front in-tervenes, the flow is compressed and slowed down to subsonic speed. The position and strengthof the shock front are automatically adjusted so that the end pressure at the exit becomes p e . This phenomenon was rigorously studied using radial solutions of the full Euler systemin [31] (it was shown that in a divergent nozzle, for given a constant supersonic data on theentrance of the nozzle and an appropriately large constant pressure on the exit of the nozzle,there exists a unique radial transonic shock solution satisfying these conditions). Motivatedby this phenomenon, there were many studies on the stability of transonic shock solutions indivergent nozzles (structural stability of radial transonic shock solutions in divergent nozzlesunder multi-dimensional perturbations of an entrance supersonic data and exit pressure) andrelated problems.The stability of one-dimensional transonic shock solutions in flat nozzles was first studied.This subject was studied using the potential flow model in [7, 8, 9, 26, 27] and further studiedusing the full Euler system in [11, 29, 6, 5, 25, 30, 12, 13]. These results showed that one-dimensional transonic shock solutions in flat nozzles are not stable under a perturbation of aphysical boundary condition (supersonic data on the entrance or density, pressure or normalvelocity on the exit) and, even if one-dimensional transonic shock solutions in flat nozzles arestable, their shock locations are not uniquely determined unless there exists the assumption thata shock location passes through some point on the wall of the nozzle, as it can be expected fromthe behavior of one-dimensional transonic shock solutions in flat nozzles (as a shock locationchanges, the value of the subsonic part of an one-dimensional transonic shock solution in aflat nozzle does not change). After that, the stability of radial transonic shock solutions indivergent nozzles was studied. This subject was first studied using the full Euler system in[19, 28]. In these results, authors, by considering a perturbation of radial transonic shocksolutions in divergent nozzles, could show that a shock location is uniquely determined forgiven an exit pressure without the assumption that a shock location passes through some point Mathematics Subject Classification.
Key words and phrases. transonic shock, full Euler system, stream function, free boundary problem, inviscidcompressible flow, axisymmetric flow, divergent nozzle, elliptic system, exit pressure, nonzero vorticity. of the nozzle but they only had the result under the assumption that the tip angle of the nozzleis sufficiently small. After that, this subject without restriction on the tip angle of the nozzlewas studied. In [3], the authors studied this subject using the non-isentropic potential modelintroduced in [3]. And they obtained the stability result for radial transonic shock solutions indivergent nozzles. This subject was also studied using the full Euler system. This for the 2-Dcase was done in [24, 18, 21]. In these papers, the authors had the stability result for radialtransonic shock solutions in divergent nozzles. Especially, the authors in [21] had the resultfor flows having C ,α interior and C α up to boundary regularity, so that they could consider ageneral perturbation of a nozzle. This for the 3-D case for axisymmetric flows with zero angularmomentum components was done in [20]. The authors in this paper also had the same result.This for the general 3-D case was done in [10, 23]. The authors in [10, 23] also had the sameresult but under S-condition introduced in [10, 23]. Recently, this subject for the general 3-Dcase for flows having some friction term was studied in [32].In this paper, we study the stability of radial transonic shock solutions in divergent noz-zles under small perturbations of an incomming radial supersonic solution and a constant exitpressure using the full Euler system for the 3-D case for axisymmetric flows. We consideraxisymmetric flows with non-zero angular momentum components. (This is a difference from[20].) We consider a divergent nozzle having no restriction on the tip angle of the nozzle anddo not have any assumption on an incomming supersonic solution.The main new feature in this paper is to develop a new iteration scheme to determine a shocklocation for a transonic shock solution of the steady full Euler system in a divergent nozzle andresolve the singularity issue arising in the stream function formulations of the full Euler systemusing an elliptic system approach.To deal with the stability of 3-D axisymmetric transonic shock solutions of the full Eulersystem, we use a stream function formulation for the full Euler system for an axisymmetric flow.This formulation shows the fact that an initial shock position and a shape of a shock location(see the definitions below the proof of Theorem 2.17) are determined in different mechanismsclearly. Based on this formulation and using the fact that the entropy of the downstreamsubsonic solution of a radial transonic shock solution on a shock location monotonically increasesas a shock location moves toward the exit of the nozzle (see Lemma 2.10), we develop a newscheme to determine a shock location of a transonic shock solution of the full Euler system ina divergent nozzle: 1. Pseudo Free Boundary Problem 2. Determination of a shape of a shocklocation (see below the proof of Theorem 2.17).In technical part, we resolve the singularity issue arising in stream function formulationsof the full Euler system. A stream function formulation when it is formulated by using theStokes’ stream function (see (2.4.1)) has a singularity issue at the axis of symmetry. We resolvethis singularity issue by formulating a stream function formulation using the vector potentialform of the stream function (see § § m = 1 with a general domain (see Lemma 4.3)). Using the streamfunction formulation formulated by using the vector potential form of the stream function, weobtain the stability result for flows having C ,α interior and C α up to boundary regularity.This paper is organized as follows. In Section 2, we present definitions and a basic lemma usedthroughout this paper and introduce our problem and result. In this section, we introduce thestream function formulation used in this paper. In Section 3, we solve the Pseudo Free BoundaryProblem. In this section, we study a linear boundary value problem for a singular ellipticequation and an initial value problem of a transport equation appearing in the Pseudo FreeBoundary Problem, and prove the unique existence of solutions of the Pseudo Free BoundaryProblem. In Section 4, we show the existence and uniqueness of transonic shock solutions. InSection 5, we present some computations done by using the tensor notation given in § -D AXISYMMETRIC TRANSONIC SHOCK 3 Problem and Theorem
Preliminary.
In this paper, we consider a 3-D steady compressible invicid flow of an idealpolytropic gas. The motion of this flow is governed by the following full Euler system: div( ρ u ) = 0 , div( ρ u ⊗ u + p I ) = 0 , div( ρ u B ) = 0(2.1.1)where ρ , u and p are the density, velocity and pressure of the flow, I is the 3 × B is the Bernoulli invariant of the flow given by B = | u | γp ( γ − ρ (2.1.2)for a constant γ >
1. Types of the flow are classified by the quantity M := | u | c called Machnumber where c is the sound speed of the flow given by c := r γpρ for an ideal polytropic gas: if M >
1, then a flow is called supersonic, if M = 1, then it is calledsonic and if M <
1, then it is called subsonic. It is generally known that types of the systemvaries depending on the value of M . If M >
1, then the system is a hyperbolic system and if
M <
1, then the system is an elliptic-hyperbolic coupled system.When the flow passes through a domain having a certain geometric structure or satisfies acertain boundary condition, it may have a discontinuity across a surface in the domain in thedirection of the flow. Such a discontinuity is called a shock.A shock solution of (2.1.1) is defined as follows.
Definition 2.1 (Shock solution) . Let Ω be an open connected set in R . Assume that a C surface Γ in Ω divides Ω into two nonempty disjoint subset Ω ± such that Ω = Ω + ∪ Γ ∪ Ω − .Then a solution ( ρ, u , p ) of (2.1.1) is called a shock solution of (2.1.1) with a shock Γ if (i) ( ρ, u , p ) is in ( C (Ω ± ) ∩ C (Ω ± )) , (ii) ( ρ, u , p ) | Ω − ∩ Γ = ( ρ, u , p ) | Ω + ∩ Γ , (iii) ρ u | Ω − · ν = 0 on Γ , (iv) and ( ρ, u , p ) satisfies (2.1.1) pointwisely in Ω ± and the following Rankine-Hugoniot con-ditions: [ ρ u · ν ] Γ = [ u · τ ] Γ = [ u · τ ] Γ = [ ρ ( u · ν ) + p ] Γ = [ B ] Γ = 0(2.1.3) where ν is the unit normal vector field on Γ pointing toward Ω + and τ i for i = 1 , areunit tangent vector fields on Γ perpendicular to each other at each point on Γ and [ F ] Γ := F | Ω − ( x ) − F | Ω + ( x ) for x ∈ Γ . A shock solution is said to be physically admissible if it satisfies the following entropy condi-tion.
Definition 2.2 (Entropy condition) . Let ( ρ, u , p ) be a shock solution of (2.1.1) defined inDefinition 2.1. Without loss of generality, assume that u | Ω − · ν > on Γ . Then S | Ω + ∩ Γ >S | Ω − ∩ Γ where S := pρ γ is called the entropy condition. Definition 2.3.
In this paper, we call S the entropy of ( ρ, u , p ) . Using the definition of a shock solution, a transonic shock solution of (2.1.1) is defined asfollows.
Definition 2.4.
A shock solution of (2.1.1) in Definition 2.1 is called a transonic shock solutionif it satisfies
M > in Ω − and M < in Ω + or the otherway around. Remark 2.5.
If a transonic shock solution of (2.1.1) satisfies u | Ω − · ν > on Γ , M > in Ω − and M < in Ω + , then it satisfies the entropy condition. YONG PARK
In this paper, we deal with a 3-D axisymmetric transonic shock solution of (2.1.1). For precisestatement, we define an axisymmetric domain and axisymmetric functions used in this paper.For later use, we present a lemma used to deal with regularities of axisymmetric functions.In this paper, we use the spherical coordinate system ( r, θ, ϕ ) given by the relation( x, y, z ) = ( r sin θ cos ϕ, r sin θ sin ϕ, r cos θ )(2.1.4)where ( x, y, z ) is the cartesian coordinate system in R . The unit vectors in this coordinatesystem are given by e r = sin θ cos ϕ e + sin θ sin ϕ e + cos θ e , e θ = cos θ cos ϕ e + cos θ sin ϕ e − sin θ e , e ϕ = − sin ϕ e + cos ϕ e where e i for i = 1 , , x , y and z direction, respectively.Using this spherical coordinate system, an axisymmetric domain and axisymmetric functionsare defined as follows. Definition 2.6.
Let Ω ⊂ R . Ω is called axisymmetric if ( x, y, z ) ∈ Ω , then ( p x + y cos ϕ, p x + y sin ϕ, z ) ∈ Ω for ϕ ∈ [0 , π ) . A function f : Ω → R is called axisymmetric if f is independent of ϕ as afunction of the spherical coordinate system. A vector valued function u : Ω → R is calledaxisymmetric if u r = u · e r , u θ = u · e θ and u ϕ = u · e ϕ are axisymmetric. Definition 2.7.
In this paper, when a velocity field u is represented as u = u r e r + u θ e θ + u ϕ e ϕ , u ϕ e ϕ is called the angular momentum component of u . For later use, we present the following lemma that shows when an axisymmetric functionin C k as a function of the spherical coordinate system is in C k as a function of the cartesiancoordinate system. This lemma is obtained from [22, Corollary 1]. Lemma 2.8.
Let Ω be an axisymmetric connected open set in R that does not contain theorigin. Suppose that a function f: Ω → R is axisymmetric. Then (i) f and f e r are in C k (Ω) for k ∈ , , , . . . if and only if f is in C k as a function ofspherical coordinate system in Ω and ∂ m +1 θ f = 0 for all ≤ m ≤ ⌊ k − ⌋ . (ii) f e θ and f e ϕ are in C k (Ω) for k ∈ , , , . . . if and only if f is in C k as a function ofspherical coordinate system in Ω and ∂ mθ f = 0 for all ≤ m ≤ ⌊ k ⌋ . In this paper, we use the same function notation when we represent an axisymmetric functionas a function on the cartesian coordinate system or spherical coordinate system.2.2.
Radial transonic shock solution.
Let r , r and θ be constants such that 0 < r < r and 0 ≤ θ < π . Define a divergent nozzle by N := { ( x, y, z ) ∈ R | r < r < r , ≤ θ < θ } . (2.2.1)To introduce our problem and for our later analysis, we study a radial transonic shock solutionof (2.1.1) in N .Fix positive constants ( ρ in , u in , p in ) satisfying M in (:= u in / q γp in ρ in ) >
1. Let (¯ ρ, ¯ u e r , ¯ p ) bea radial shock solution of (2.1.1) in N with a shock Γ t := { r = t } ∩ N for some t ∈ [ r , r ]satisfying (¯ ρ, ¯ u e r , ¯ p ) = ( ρ in , u in e r , p in ) on Γ en := ∂ N ∩ { r = r , ≤ θ < θ } . (2.2.2)Then (¯ ρ, ¯ u, ¯ p ) is a solution of ( r ¯ ρ ¯ u ) ′ = 0 , ¯ ρ ¯ u ¯ u ′ + ¯ p ′ = 0 , ¯ ρ ¯ u ¯ B ′ = 0(2.2.3) -D AXISYMMETRIC TRANSONIC SHOCK 5 with (¯ ρ, ¯ u, ¯ p )( r ) = ( ρ in , u in , p in )(2.2.4)in D − t , where D − t := { r < r < t } , ′ is the derivative with respect to r and ¯ B := ¯ u + γ ¯ p ( γ − ρ ,and is a solution of (2.2.3) with ¯ ρ ¯ u ( t ) = ¯ ρ ¯ u | D − t ( t ) , (¯ ρ ¯ u + ¯ p )( t ) = (¯ ρ ¯ u + ¯ p ) | D − t ( t ) , ¯ B ( t ) = ¯ B | D − t ( t )(2.2.5)in D + t where D + t := { t < r < r } .By (2.2.3) and (2.2.4), a solution (¯ ρ, ¯ u, ¯ p ) of (2.2.3) with (2.2.4) satisfies r ¯ ρ ¯ u = m , ¯ S = S in , ¯ B = B (2.2.6)where m := r ρ in u in , ¯ S := ¯ p ¯ ρ γ , S in := p in ρ γin and B := u in + γp in ( γ − ρ in on the domain where (2.2.3)with (2.2.4) has a unique solution (¯ ρ, ¯ u, ¯ p ). By this fact, the local unique existence theoremfor ODE and the condition that M in >
1, (2.2.3) with (2.2.4) has a unique solution (¯ ρ, ¯ u, ¯ p )satisfying ¯ M > M ′ > r , r ] where ¯ M := ¯ u/ q γ ¯ p ¯ ρ . Thus, (2.2.5) is well-defined forany t ∈ [ r , r ]. From (2.2.5), we obtain¯ u ( t ) = ¯ K ¯ u (cid:12)(cid:12)(cid:12)(cid:12) D − t ( t )(2.2.7)and ¯ S ( t ) = ( g ( ¯ M ) ¯ S ) | D − t ( t )(2.2.8)where ¯ K := γ − γ +1 ¯ B and g ( x ) := 1 γ + 1 (2 γx − ( γ − (cid:18) γ − γ + 1 + 2 γ + 1 1 x (cid:19) γ . (2.2.9)Using (2.2.7), the third equations of (2.2.5) and (2.2.6) and ¯ M | D − t ( t ) >
1, one can check that(¯ ρ, ¯ u, ¯ p ) satisfying (2.2.5) satisfies ¯ M ( t ) <
1. By (2.2.3), the first and third equation of (2.2.5),(2.2.6) and (2.2.8), a solution (¯ ρ, ¯ u, ¯ p ) of (2.2.3) with (2.2.5) satisfies r ¯ ρ ¯ u = m , ¯ S = g ( ¯ M | D − t ( t )) S in , ¯ B = B (2.2.10)on the domain where (2.2.3) with (2.2.5) has a unique solution (¯ ρ, ¯ u, ¯ p ). By these two factsand the local unique existence theorem for ODE, (2.2.3) with (2.2.5) has a solution (¯ ρ, ¯ u, ¯ p )satisfying ¯ M < M ′ < D + t . Therefore, a radial shock solution (¯ ρ, ¯ u e r , ¯ p ) uniquelyexists in N for each t ∈ [ r , r ] and it is a radial transonic shock solution. Remark 2.9. By (2.2.3) and the fact that a solution (¯ ρ, ¯ u, ¯ p ) of (2.2.3) with (2.2.5) uniquelyexists in D + t satisfying (2.2.10) , we have that a solution (¯ ρ, ¯ u, ¯ p ) of (2.2.3) with (2.2.5) satisfies ¯ ρ ′ = 2¯ ρr ¯ M − ¯ M in D + t . From this fact and ¯ M | D + t < in D + t , we obtain ¯ ρ | ′ D + t > in D + t . YONG PARK
One can see that the values of (¯ ρ, ¯ u, ¯ p ) | D + t at a fixed location r in D + t are determined by thethree conserved quantities in the right-hand sides of the equations in (2.2.10). This combinedwith the fact that the conserved quantity for ¯ S in D + t given in (2.2.10) varies depending on t (obtained from (2.2.10) by using ( ¯ M | D − t ) ′ > D − t for any t ∈ [ r , r ]) implies that the valuesof (¯ ρ, ¯ u, ¯ p ) | D + t at a fixed location r in D + t vary depending on t . To represent this dependence,we write (¯ ρ, ¯ u, ¯ p ) | D + t ( r ) and ¯ S | D + t ( r ) as (¯ ρ, ¯ u, ¯ p ) | D + t ( r ; t ) and ¯ S | D + t ( r ; t ), respectively.The conserved quantity for ¯ S in D + t satisfies the following monotonicity. Lemma 2.10.
Let r , r , t be positive constants such that r ≤ t ≤ r . Suppose that (¯ ρ, ¯ u, ¯ p ) is as above. Then there holds d ¯ S | D + t ( t ; t ) dt > for any t ∈ [ r , r ] . Proof.
Differentiate ¯ S | D + t ( t ; t ) with respect to t . Then we have d ¯ S | D + t ( t ; t ) dt = dg ( x ) dx (cid:12)(cid:12)(cid:12)(cid:12) x =( ¯ M | D − t ) ( t ) d ( ¯ M | D − t ) ( t ) dt S in . (2.2.11)One can easily check that g (1) = 1 and g ′ ( x ) > x >
1. By this fact, ¯ M | D − t ( t ) > M | D − t ) ′ ( t ) > t ∈ [ r , r ], we obtain from (2.2.11) the desired result. (cid:3) From Lemma 2.10, we obtain the following result.
Proposition 2.11.
Suppose that r , r , t and (¯ ρ, ¯ u, ¯ p ) are as in Lemma 2.10. Then for any t ∈ [ r , r ] , d ¯ p | D + t dt ( r ; t ) < . Proof.
By the definitions of ¯ B and ¯ S and the first and third equation of (2.2.10), B = (cid:18)
12 ( m r ) ( ¯ S ¯ p ) γ + γγ − p − γ ¯ S γ (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) D + t ( r ; t ) . Differentiate this with respect to t . Then we get d ¯ p | D + t ( r ; t ) dt = − ( m r ) ( ¯ S ¯ p ) γ − + γγ − ¯ p − γ ¯ S γ − − ( m r ) ( ¯ S ¯ p ) γ + γ ¯ p − γ ¯ S γ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) D + t ( r ; t ) d ¯ S | D + t ( r ; t ) dt . (2.2.12)By ¯ M | D + t < D + t , Lemma 2.10 and the second equation of (2.2.10), we obtain from (2.2.12)the desired result. (cid:3) The above proposition implies that for any given p c ∈ [ p , p ] where p := p | D + r ( r ; r )and p := p | D + r ( r ; r ), there is a unique shock location Γ t in N such that ( ρ, u e r , p ) satisfies p | D + t ( r ; t ) = p c . Hereafter, we fix a constant p c ∈ ( p , p ) and denote t ∈ ( r , r ) such thata radial transonic shock solution of (2.1.1) satisfying (2.2.2) and having a shock location Γ t satisfies p ( r ) = p c by r s . Also, we denote a solution ( ρ, u, p ) of (2.2.3) with (2.2.4) and asolution ( ρ, u, p ) of (2.2.3) with (2.2.5) for t = r s by ( ρ − , u − , p − ) and ( ρ +0 , u +0 , p +0 ), respectively,and denote p +0 ρ +0 γ by S +0 . By the local unique existence theorem for ODE, there exists a positiveconstant δ such that ( ρ +0 , u +0 , p +0 ) uniquely exists in [ r s − δ , r s ] satisfying M +0 (= u +0 / r γp +0 ρ +0 ) < δ . -D AXISYMMETRIC TRANSONIC SHOCK 7 Problem.
Using the radial transonic shock solution given in the previous subsection, wepresent our problem.In this paper, we use the following weighted H¨older norm. For a bounded connected openset Ω ⊂ R n , let Γ be a closed portion of ∂ Ω. For x , y ∈ Ω, set δ x := dist(x , Γ) and δ x , y := min( δ x , δ y ) . For k ∈ R , α ∈ (0 ,
1) and m ∈ Z + , we define || u || ( k, Γ) m, , Ω := X ≤| β |≤ m sup x ∈ Ω δ max( | β | + k, | D β u (x) | [ u ] ( k, Γ) m,α, Ω := X | β | = m sup x , y ∈ Ω , x =y δ max( m + α + k, , y | D β u (x) − D β u (y) || x − y | α || u || ( k, Γ) m,α, Ω := || u || ( k, Γ) m, , Ω + [ u ] ( k, Γ) m,α, Ω where D β := ∂ β x . . . ∂ β n x n for a multi-index β = ( β , . . . , β n ) with β i ∈ Z + for i = 1 , . . . , n and | β | = P ni =1 β i . We denote the completion of a set of smooth functions under || · || ( k, Γ) m,α, Ω norm by C m,α ( k, Γ) (Ω).PSfrag replacements B = B M > M < ρ − , u − , p − ) ( ρ + , u + , p + )Γ f p = p ex N Our problem is given as follows.
Problem 1 (Transonic shock problem) . Given an axisymmetric supersonic solution ( ρ − , u − , p − ) of (2.1.1) in N satisfying the slip boundary condition u − · n w = 0 on Γ w := ∂ N ∩ { r < r < r , θ = θ } (2.3.1) where n w is the unit normal vector on Γ w , B = B on Γ en (2.3.2) and || ( ρ − , u − , p − ) − ( ρ − , u − e r , p − ) || ,α, N ≤ σ (2.3.3) and an axisymmetric exit pressure p ex on Γ ex := ∂ N ∩ { r = r , ≤ θ < θ } satisfying || p ex − p c || ( − α,∂ Γ ex )1 ,α, Γ ex ≤ σ (2.3.4) for a positive constant σ , find a shock location Γ f := N ∩ { r = f ( θ ) } and a correspondingsubsonic solution ( ρ + , u + , p + ) ∈ ( C ( N + f ) ∩ C ( N + f )) of (2.1.1) satisfying (i) the system (2.1.1) in N + f := N ∩ { r > f ( θ ) } , (ii) R-H conditions (2.1.3) on Γ f , (iii) the slip boundary condition u + · n w = 0 on Γ + w := Γ w ∩ { r > f ( θ ) } , (2.3.5)(iv) and the exit pressure condition p + = p ex on Γ ex . (2.3.6) Remark 2.12.
It is generally known that a supersonic solution of (2.1.1) is governed by ahyperbolic system. We assume that ( ρ − , u − , p − ) in Problem 1 exists. YONG PARK
Remark 2.13.
To simplify our argument, we assumed in Problem 1 that ( ρ − , u − , p − ) satis-fies (2.3.2) . This assumption will be used to reduce (2.1.1) and (2.1.3) (see § ( ρ − , u − e r , p − ) in Problem 1. We study Problem 1 using a stream function formulation of the full Euler system for anaxisymmetric flow. We introduce a stream function formulation used in this paper in the nextsubsection.2.4.
Stream function formulation.
Let Ω be an open simply connected axisymmetric set in R . Let ( ρ, u ) be axisymmetric C functions in Ω satisfying the first equation of (2.1.1) and | ρ u | >
0. For such ( ρ, u ), the Stokes’ stream function for an axisymmetric flow of the full Eulersystem is defined by V (x) = Z S x ρ u · ν dS for x ∈ Ω(2.4.1)where S x is a simply connected C surface in Ω whose boundary is a circle centered at z -axis,parallel to xy -plane and passing through x, and ν is the unit normal vector on S x pointingoutward direction with respect to the cone-like domain made by connecting ∂S x and the originby straight lines. By the first equation of (2.1.1), the value of this function at x is independentof the choice of S x . Since ∂S x is axisymmetric, V is axisymmetric in Ω.By the first equation of (2.1.1), V is a constant on each stream surface of ρ u in Ω. Here,the stream surfaces of a vector field ρ u in Ω by a set of surfaces made by collecting all thestreamlines of ρ u initiating from a point on a circle in Ω centered at z -axis and parallel to xy -plane. By | ρ u | > V is a constant on each stream surface of ρ u in Ω and V on eachdifferent stream surface of vector field ρ u in Ω is different from each other. From these facts,we have that if we apply ∇ ⊥ , where ∇ ⊥ = πr sin θ (cid:16) e r ∂ θ r − e θ ∂ r (cid:17) which satisfies ∇ ⊥ h · ∇ h = 0and ∇ ⊥ h · e ϕ = 0 for a scalar function h , to V , then we have a vector field in Ω tangent to thestream surfaces in Ω and having no e ϕ component. Apply ∇ ⊥ to V . Then we have ∇ ⊥ V = ρu r e r + ρu θ e θ (2.4.2)where u r = u · e r and u θ = u · e θ .Using (2.4.2), we can reformulate the full Euler system for an axisymmetric flow. But if wedo this, then there is a singularity issue that can be seen in the relation || ρu r e r + ρu θ e θ || α, N = ||∇ ⊥ V || α, N C || V || ,α, N for any constant C . To avoid this issue, we use the following form ofthe stream function.Let Φ e ϕ be an axisymmetric vector field in Ω satisfying I ∂S x Φ e ϕ · d r = Z S x ρ u · ν dS (2.4.3)where r is a parametrization of ∂S x in a counter clockwise direction. Then by the definitionsof Φ e ϕ and V , Φ = V πr sin θ (2.4.4)It is easily checked that ∇ × (Φ e ϕ ) = ∇ ⊥ V. (2.4.5)From this relation and (2.4.2) or (2.4.3) directly, we have ∇ × (Φ e ϕ ) = ρu r e r + ρu θ e θ . (2.4.6)We call Φ e ϕ the vector potential form of the stream function.We reformulate the full Euler system for an axisymmetric flow using (2.4.6). For our lateranalysis, when we reformulate the full Euler system using (2.4.6), we use the following form of -D AXISYMMETRIC TRANSONIC SHOCK 9 the full Euler system representing the relation between ∇ S and ∇ × u clearly div( ρ u ) = 0 , ( ∇ × u ) × u = ρ γ − γ − ∇ S − ∇ B,ρ u · ∇ B = 0(2.4.7)which is obtained under the assumption that ( ρ, u , p ) ∈ C and ρ > ρ, u , p ) in (2.4.7) is axisymmetric and reformulate (2.4.7) using (2.4.6).Rewrite (2.4.6) as u = ρ ∇ × (Φ e ϕ ) + u ϕ e ϕ . Substitute this into the second equation of (2.4.7).Then we obtain (cid:18) ∇ × (cid:18) ρ ∇ × (Φ e ϕ ) + u ϕ e ϕ (cid:19)(cid:19) × (cid:18) ρ ∇ × (Φ e ϕ ) + u ϕ e ϕ (cid:19) = ρ γ − γ − ∇ S − ∇ B. (2.4.8)From this equation, we can obtain three equations. From e ϕ -components of (2.4.8), we get( ∇ × ( u ϕ e ϕ )) × (cid:18) ρ ∇ × (Φ e ϕ ) (cid:19) = 0 . (2.4.9)Define L := 2 πr sin θu ϕ so that ∇ × ( u ϕ e ϕ ) = ∇ ⊥ L (see (2.4.4) and (2.4.5)). With this relation,write (2.4.9) as ∇ ⊥ L × (cid:18) ρ ∇ × (Φ e ϕ ) (cid:19) = 0 . From this equation, we obtain ∇ × (Φ e ϕ ) · ∇ L = 0 . (2.4.10)From e θ -components of (2.4.8), we have (cid:18) ∇ × (cid:18) ρ ∇ × (Φ e ϕ ) (cid:19)(cid:19) × (cid:18) ρ ∇ × (Φ e ϕ ) · e r (cid:19) e r + ( ∇ × ( u ϕ e ϕ ) · e r ) e r × u ϕ e ϕ = (cid:18) ρ γ − γ − ∂ θ Sr − ∂ θ Br (cid:19) e θ . With the definition of L , rewrite the above equation. Move the rewritten term into the right-hand side of the equation. And then multiply e r × to the resultant equation. Then we obtain(2.4.11) (cid:18) ρ ∇ × (Φ e ϕ ) · e r (cid:19) ∇ × (cid:18) ρ ∇ × (Φ e ϕ ) (cid:19) = (cid:18) L πr sin θ ∇ × (cid:18) L πr sin θ e ϕ (cid:19) · e r + ρ γ − γ − ∂ θ Sr − ∂ θ Br (cid:19) e ϕ . Using the third equation of (2.4.7), we obtain from u -components of (2.4.8) ρ u · ∇ S = 0 . (2.4.12)With the assumption that ( ρ, u , p ) are axisymmetric and (2.4.6), this equation can be writtenas ∇ × (Φ e ϕ ) · ∇ S = 0 . (2.4.13)Hence, we obtain from (2.4.8) three equations: (2.4.10), (2.4.11) and (2.4.13). Finally, in thesame way that we obtained (2.4.13) from (2.4.12), we obtain from the third equation of (2.4.7) ∇ × (Φ e ϕ ) · ∇ B = 0 . (2.4.14) Combining (2.4.10), (2.4.11), (2.4.13) and (2.4.14), we have the following stream function for-mulation of the full Euler system for an axisymmetric flow (cid:16) ρ ∇ × (Φ e ϕ ) · e r (cid:17) ∇ × (cid:16) ρ ∇ × (Φ e ϕ ) (cid:17) = (cid:16) L πr sin θ ∇ × (cid:0) L πr sin θ e ϕ (cid:1) · e r + ρ γ − γ − ∂ θ Sr − ∂ θ Br (cid:17) e ϕ , ∇ × (Φ e ϕ ) · ∇ L = 0 , ∇ × (Φ e ϕ ) · ∇ S = 0 , ∇ × (Φ e ϕ ) · ∇ B = 0 . (2.4.15)Note that the first equation of (2.1.1) is omitted in (2.4.15) because for a 3-D axisymmetric flow,the first equation of (2.1.1) is reduced to div( ρu r e r + ρu θ e θ ) = 0 and this equation is directlysatisfied by div( ∇ × (Φ e ϕ )) = 0 if Φ ∈ C (Ω). Also, note that when ( ρ, u , p ) in (2.1.1) is ina sufficiently small perturbation of ( ρ +0 , u +0 e r , p +0 ) or ( ρ − , u − e r , p − ), then unknowns of (2.4.15)can be (Φ e ϕ , L, S, B ). This fact (for the first case) is checked via the following lemma. Lemma 2.14.
Let Ω be an axisymmetric connected open subset of N + r s − δ . There exist positiveconstants δ , Ω and δ , Ω and a function ̺ : B (1) δ , Ω , Ω → B (2) δ , Ω , Ω , where B (1) δ, Ω := { ( ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ) | ∈ ( C (Ω)) | sup Ω {| ρu r e r + ρu θ e θ − ρ +0 u +0 e r | + | u ϕ e ϕ | + | S − S +0 | + | B − B |} ≤ δ } , and B (2) δ, Ω := { ρ ∈ C (Ω) | sup Ω | ρ − ρ +0 | ≤ δ } , such that ρ B − | ρu r e r + ρu θ e θ | − ρ | u ϕ e ϕ | − γSρ γ +1 γ − if and only if ρ = ̺ ( ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ) , (2.4.17) for all ( ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ) ∈ B (1) δ , Ω , Ω and ρ ∈ B (2) δ , Ω , Ω .Proof. Using the definition of B given in (2.1.2), we define(2.4.18) b ( ρ, ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ) = ρ B − | ρu r e r + ρu θ e θ | − ρ | u ϕ e ϕ | − γSρ γ +1 γ − . Then b ∈ C ∞ with respect to ( ρ, ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ), ∂ ρ b ( ρ +0 , ρ +0 u +0 e r , , S +0 , B ) = ρ +0 ( u +0 2 − γS +0 ρ +0 γ − ) < , and by the third equation of (2.2.10), b ( ρ +0 , ρ +0 u +0 e r , , S +0 , B ) = 0 . With these facts, we apply the implicit function theorem to b ( ρ, ρu r e r + ρu θ e θ , u ϕ e ϕ , S, B ).Then we obtain the desired result. (cid:3) Remark 2.15.
Hereafter, δ and δ denote some constants δ , Ω and δ , Ω in Lemma 2.14 for Ω = N + r s − δ . Hereafter, ̺ denotes ̺ in Lemma 2.14 for Ω = N + r s − δ . One can see that if Ω ⊂ N + r s − δ , then for all ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ) ∈ B (1) δ , Ω and ρ ∈ B (2) δ , Ω , (2.4.16) holds ifand only if (2.4.17) holds. -D AXISYMMETRIC TRANSONIC SHOCK 11 In order to study 3-D axisymmetric transonic shock solution of the full Euler system usingthe stream function formulation of the full Euler system for an axisymmetric flow above, wereformulate (2.1.3) with respect to the variables in (2.4.15).Assume that Γ in (2.1.3) is an axisymmetric C surface. Let τ and τ in (2.1.3) be e ϕ and the unit tangent vector field on Γ perpendicular to e ϕ and satisfying ν · ( τ × e ϕ ) > ν is the unit normal vector field on Γ pointing toward Ω + . By the definitionof V given in (2.4.1), the first equation of (2.1.3) is written as [ V ] Γ = 0. Rewrite this equationusing (2.4.4). Then we have [Φ e ϕ ] Γ = 0 . With (2.4.6), rewrite the second equation of (2.1.3). Then we get1 ρ ∇ × (Φ e ϕ ) · τ = u − · τ on Γ . By the definition of L , the third equation of (2.1.3) can be written as[ L ] Γ = 0 . From the second, third and fifth equation of (2.1.3), we can obtain[ B s ] Γ = 0(2.4.19)where B s := ( u · ν ) + γp ( γ − ρ . In the same way that (2.2.8) is obtained from (2.2.5), we obtainfrom the first and fourth equation of (2.1.3) and (2.4.19) S + = g (cid:18) u − · ν c − (cid:19) ! S − on Γwhere g ( x ) is a function defined in (2.2.9) and variables with lower indices ± denote variablesin Ω ± , respectively. Combining these reformulated equations of (2.1.3) and the fifth equationof (2.1.3), we have the following stream function formulation of (2.1.3): [Φ e ϕ ] Γ = 0 , ρ ∇ × (Φ e ϕ ) · τ = u − · τ on Γ , [ L ] Γ = 0 ,S + = g (cid:18)(cid:16) u − · ν c − (cid:17) (cid:19) S − on Γ , [ B ] Γ = 0 . (2.4.20)2.5. Restatement of Problem 1 using the stream function formulation and mainresult.
Using the stream function formulation in the previous subsection, we restate Problem1. We first reduce (2.1.1) and (2.1.3) in Problem 1.In Problem 1, we assumed that a supersonic solution ( ρ − , u − , p − ) of (2.1.1) satisfies (2.3.2).By this assumption, the third equation of (2.1.1) in N − f := N ∩ { r < f ( θ ) } and N + f , the fifthequation of (2.1.3), (2.3.1) and (2.3.5), ( ρ + , u + , p + ) we find in Problem 1 must satisfy B = B in N + f . To simplify our argument, we assume that ( ρ + , u + , p + ) in Problem 1 satisfies B = B in N + f . Under this assumption, (2.1.1) and (2.1.3) that ( ρ + , u + , p + ) in Problem 1 satisfiesare reduced to the first and second equation of (2.1.1) and the first, second, third and fourthequation of (2.1.3).Then we present the stream function formulations of (2.1.1) and (2.1.3) satisfied by ( ρ + , u + , p + )in Problem 1. By (2.4.15) and (2.4.20), the stream function formulations of (2.1.1) and (2.1.3)satisfied by ( ρ + , u + , p + ) in Problem 1 and reduced by using the assumption that ( ρ + , u + , p + ) satisfies B = B in N + f are given as (cid:18) ρ + ∇ × (Φ + e ϕ ) · e r (cid:19) ∇ × (cid:18) ρ + ∇ × (Φ + e ϕ ) (cid:19) (2.5.1) = L + πr sin θ ∇ × (cid:18) L + πr sin θ e ϕ (cid:19) · e r + ρ γ − γ − ∂ θ S + r ! e ϕ , ∇ × (Φ + e ϕ ) · ∇ L + = 0 , (2.5.2) ∇ × (Φ + e ϕ ) · ∇ S + = 0(2.5.3)in N + f and Φ + e ϕ = Φ − e ϕ on Γ f , (2.5.4) 1 ρ + ∇ × (Φ + e ϕ ) · τ f = u − · τ f on Γ f , (2.5.5) L + = L − on Γ f , (2.5.6) S + = g (cid:18) u − · ν f c − (cid:19) ! S − on Γ f (2.5.7)where ν f is the unit normal vector on Γ f pointing toward N + f and τ f is the unit tangentialvector on Γ f perpendicular to e ϕ and satisfying ν f · ( τ f × e ϕ ) >
0. Here, (Φ − e ϕ , L − , S − )and (Φ + e ϕ , L + , S + ) are (Φ e ϕ , L, S ) given by the definitions of Φ e ϕ , L and S for ( ρ, u , p ) =( ρ − , u − , p − ) and ( ρ + , u + , p + ), respectively.Determine ρ + in (2.5.1) and (2.5.5). In Problem 1, we consider the case that ( ρ − , u − , p − )and p ex are in sufficiently small perturbations of ( ρ − , u − e r , p − ) and p c so that ( ρ + , u + , p + ) inProblem 1 is in a small perturbation of ( ρ +0 , u +0 e r , p +0 ) such that ρ + is uniquely determined by( ρ + u + ,r e r + ρ + u + ,θ e θ , u + ,ϕ e ϕ , S + , B ) where u + ,r := u + · e r , u + ,θ := u + · e θ and u + ,ϕ = u + · e ϕ (see Lemma 2.14). To find such ( ρ + , u + , p + ) using (2.5.1)-(2.5.7), we set ρ + in (2.5.1) and(2.5.5) to be ρ + = ̺ ( ∇ × (Φ + e ϕ ) , L + πr sin θ e ϕ , S + , B )(2.5.8)where ̺ is a function given in Lemma 2.14. Hereafter, to simplify notation, we write ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ) as ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ).We will find a subsonic solution of (2.1.1) using (2.5.1)-(2.5.3). To do this, we define asubsonic solution of (2.5.1)-(2.5.3). Using the definition of subsonic solution of (2.1.1), we definea subsonic solution of (2.5.1)-(2.5.3) by a solution (Φ + e ϕ , L + , S + ) of (2.5.1)-(2.5.3) satisfying(2.5.9) | ̺ ( ∇ × (Φ + e ϕ ) , L + πr sin θ e ϕ , S + ) ∇ × (Φ + e ϕ ) + L + πr sin θ e ϕ | <γS + ( ̺ ( ∇ × (Φ + e ϕ ) , L + πr sin θ e ϕ , S + )) γ − . We rewrite the boundary conditions in Problem 1 with respect to the variables in (2.4.15).Denote V given by the definition of V for ( ρ, u , p ) = ( ρ − , u − , p − ) and ( ρ + , u + , p + ) by V − and V + , respectively. By (2.4.2), (2.3.1) and (2.3.5) can be written as V − = V − ( r , θ ) onΓ w and V + = V + ( f ( θ ) , θ ) on Γ + w . Rewrite this in the vector potential form. Then we haveΦ − e ϕ = r Φ − ( r ,θ ) r e ϕ on Γ w and Φ + e ϕ = f ( θ )Φ + ( f ( θ ) ,θ ) r e ϕ on Γ + w . Combine these relationswith (2.5.4) and the continuity condition for Φ e ϕ at Γ f ∩ Γ + w . Then we obtainΦ + e ϕ = r Φ − ( r , θ ) r e ϕ on Γ + w . (2.5.10)Using the definition of S and (2.5.8), rewrite (2.3.6). Then we have S + ( ̺ ( ∇ × (Φ + e ϕ ) , L + πr sin θ e ϕ , S + )) γ = p ex on Γ ex . (2.5.11) -D AXISYMMETRIC TRANSONIC SHOCK 13 Finally, we add some continuity condition for Φ + e ϕ in the restatement of Problem 1 usingthe stream function formulation. By (2.1.1) and (2.3.5), ( ρ + , u + ) we find in Prolem 1 mustsatisfy that the total outgoing flux for ρ + u + on Γ ex is equal to the total incomming flux for ρ + u + on any cross section of N + f whose all boundary points intersect with Γ + w . Using (2.4.1),this statement can be expressed as lim θ → θ V + ( r , θ ) = lim r → r V + ( r, θ ) . Rewrite this in thevector potential form. Then we havelim θ → θ (Φ + e ϕ )( r , θ ) = lim r → r (Φ + e ϕ )( r, θ ) . (2.5.12)Since this condition cannot be achieved from (2.5.11) (this will be seen in § Problem 2.
Given an axisymmetric supersonic solution ( ρ − , u − , p − ) of (2.1.1) and an ax-isymmetric exit pressure p ex as in Problem 1, find a shock location Γ f = N ∩ { r = f ( θ ) } and acorresponding subsonic solution (Φ + e ϕ , L + , S + ) of (2.5.1) - (2.5.3) satisfying (i) the system (2.5.1) - (2.5.3) in N + f , (ii) the R-H conditions (2.5.4) - (2.5.7) , (iii) the slip boundary condition (2.5.10) , (iv) the exit pressure condition (2.5.11) , (v) the compatibility condition (2.5.12) . Let S ,θ := { ( x, y, z ) ∈ R | r = 1 , ≤ θ < θ } . A function f representing a shock locationΓ f can be considered as a function on S ,θ . Using this fact and the stereographic projectionfrom (0 , , −
1) onto the plane z = 1 passing through S ,θ , we see that f can be regarded as afunction on Λ where Λ := { ( x, y ) ∈ R | p x + y < θ } . Thus, f can be regarded as afunction on Λ or (0 , θ ). In this paper, we regard f in both ways. To simplify our notation, weuse the same function notation when we represent f as a function on Λ or (0 , θ ).Hereafter, we denote Φ e ϕ , L and V given by the definitions of Φ e ϕ , L and V for ( ρ, u , p ) =( ρ ± , u ± e r , p ± ) by Φ ± e ϕ , L ± and V ± , respectively. To simplify our notation, hereafter, wedenote (Φ + e ϕ , L + , S + ) by (Φ e ϕ , L, S ).Our result of Problem 2, the main result in this paper, is given as follows. Theorem 2.16.
Let α ∈ ( , . There exists a positive constant σ depending on ( ρ in , u in , p in ) , p c , γ , r , r , θ and α such that if σ ∈ (0 , σ ] , then Problem 2 has a solution ( f, Φ e ϕ , L, S ) satisfying the estimate (2.5.13) || f − r s || ( − − α,∂ Λ)2 ,α, Λ + ||∇ × ((Φ − Φ +0 ) e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || L πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || S − S +0 || ( − α, Γ + w )1 ,α, N + f ≤ Cσ where C is a positive constant depending on ( ρ in , u in , p in ) , p c , γ , r , r , θ and α . Furthermore,this solution is unique in the class of functions satisfying (2.5.13) . Hereafter, we say that a constant depends on the data if a constant depends on ( ρ in , u in , p in ), p c , γ , r , r , θ and α .The following result of Problem 1 is obtained from Theorem 2.16. Theorem 2.17.
Let α ∈ ( , . There exists a positive constant σ depending on the data suchthat if σ ∈ (0 , σ ] , then Problem 1 has a solution ( f, ρ + , u + , p + ) satisfying the estimate (2.5.14) || f − r s || ( − − α,∂ Λ)2 ,α, Λ + || ρ + − ρ +0 || ( − α, Γ + w )1 ,α, N + f + || u + − u +0 e r || ( − α, Γ + w )1 ,α, N + f + || p + − p +0 || ( − α, Γ + w )1 ,α, N + f ≤ Cσ where C is a positive constant depending on the data. Furthermore, this solution is unique inthe class of functions satisfying (2.5.14) . Proof.
1. Let ( f, Φ e ϕ , L, S ) be a solution of Problem 2 given in Theorem 2.16 for σ ∈ (0 , σ ]where σ is a positive constant less than or equal to σ and to be determined later. Thensince ( f, Φ e ϕ , L, S ) is a solution of Problem 2, ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) is well-defined in N + f . Define ρ + := ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ), u + := ∇× (Φ e ϕ ) ̺ ( ∇× (Φ e ϕ ) , L πr sin θ e ϕ ,S ) and p + := S ( ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S )) γ . Then since (Φ e ϕ , L, S ) satisfies (2.5.1)-(2.5.3) in N + f , (2.5.4)-(2.5.7),(2.5.10), (2.5.11), and b ( ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) , ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ) = 0 in N + f where b is a funtion defined in (2.4.18), ( ρ + , u + , p + ) satisfies the second and third equation of(2.1.1) in N + f , (2.1.3) on Γ f , (2.3.5) and (2.3.6). Furthermore, since ∇ · ( ∇ × (Φ e ϕ )) = 0 and ∇ · ( ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) L πr sin θ e ϕ ) = 0 in N + f , ( ρ + , u + , p + ) satisfies the first equationof (2.1.1) in N + f . Since (Φ e ϕ , L, S ) is a subsonic solution of (2.5.1)-(2.5.3), ( ρ + , u + , p + ) is asubsonic solution of (2.1.1). Since ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) ∈ C ,α ( − α, Γ + w ) ( N + f ), ( ρ + , u + , p + ) ∈ C ,α ( − α, Γ + w ) ( N + f ). From these facts, we have that ( f, ρ + , u + , p + ) is a solution of Problem 1.Obtain (2.5.14). By Lemma 2.14, ρ +0 can be written as ρ +0 = ̺ ( ∇ × (Φ +0 e ϕ ) , L +0 πr sin θ e ϕ (=0) , S +0 ) . Using this expression, we write ρ + − ρ +0 as(2.5.15) Z ∇ ̺ ( t ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) + (1 − t )( ∇ × (Φ +0 e ϕ ) , , S +0 )) dt ( ∇ × ((Φ − Φ +0 ) e ϕ ) , L πr sin θ e ϕ , S − S +0 ) . Since b is a C ∞ function of ( ρ, ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ), ̺ is a C ∞ function of ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ). With this fact, the fact that ( ρ +0 , u +0 e r , p +0 ) ∈ ( C ∞ ( N + f )) and (2.5.13)satisfied by ( f, Φ e ϕ , L, S ) for σ ∈ (0 , σ ] for σ ≤ σ , we estimate (2.5.15) in C ,α ( − α, Γ + w ) ( N + f ).Then we obtain || ρ + − ρ +0 || ( − α, Γ + w )1 ,α, N + f ≤ Cσ (2.5.16)where C is a positive constant depending on the data. By this estimate, there exists a positiveconstant σ (1)2 depending on the data such that if σ ∈ (0 , σ (1)2 ], thensup N + f | ρ + | ≤ C (2.5.17)where C is a positive constant depending on the data. Take σ = min( σ , σ (1)2 ) so that (2.5.17)holds. With (2.5.13) satisfied by ( f, Φ e ϕ , L, S ), (2.5.16) and (2.5.17), we estimate u + − u +0 e r and p + − p +0 in C ,α ( − α, Γ + w ) ( N + f ). Then we obtain || u + − u +0 e r || ( − α, Γ + w )1 ,α, N + f + || p + − p +0 || ( − α, Γ + w )1 ,α, N + f ≤ Cσ where C is a positive constant depending on the data. Combining this estimate, (2.5.13) and(2.5.16), we obtain (2.5.14).2. Assume that for σ ∈ (0 , σ ] where σ is a positive constant to be determined later, thereexist two solutions ( f i , ρ i + , u i + , p i + ) for i = 1 , σ (1)2 depending on the data such that if σ ∈ (0 , σ (1)2 ], then 1)sup N + fi | ρ i + | ≤ C (2.5.18)for i = 1 , C is a positive constant depending on the data and 2) ρ i + for i = 1 , ρ i + ( u i + ,r e r + u i + ,θ e θ ) , u i + ,ϕ e ϕ , S i + , B ) where u i + ,r := u i + · e r , u i + ,θ := -D AXISYMMETRIC TRANSONIC SHOCK 15 u i + · e θ , u i + ,ϕ := u i + · e ϕ and S i + := p i + ( ρ i + ) γ for i = 1 , ρ i + , u i + , p i + ) are C ( N + f ) ∩ C ( N + f ) solutions of Problem 1, then ( ρ, u , p ) = ( ρ i + , u i + , p i + )satisfy B = B in N + f i ). Take σ = σ (1)2 so that 1) and 2) hold. Let (Φ i e ϕ , L i , S i ) for i = 1 , e ϕ , L, S ) given by the definitions of Φ, L and S for ( ρ, u , p ) = ( ρ i + , u i + , p i + ) for i = 1 , f i , ρ i + , u i + , p i + ) for i = 1 , f i , Φ i e ϕ , L i , S i ) for i = 1 , f i , ρ i + , u i + , p i + ) for i = 1 , f i , Φ i e ϕ , L i , S i ) for i = 1 , C replaced by C for some positive constant C depending on the data (here we used (2.5.18)). Take σ = min( σ (1)2 , Cσ C , σ ) where C is C in (2.5.13) so that ( ρ − , u − , p − ) and p ex satisfy (2.3.3) and (2.3.4), respectively, for σ ∈ (0 , σ ]and ( f i , Φ i e ϕ , L i , S i ) for i = 1 , σ ∈ (0 , σ ]. Then by Theorem 2.16,( f , Φ e ϕ , L , S ) = ( f , Φ e ϕ , L , S ). This implies ρ ( u ,r e r + u ,θ e θ ) = ρ ( u ,r e r + u ,θ e θ ) and u ,ϕ e ϕ = u ,ϕ e ϕ . By these relations, S = S and 2), we have that ρ = ρ . With this relation, we can concludethat ( ρ , u , p ) = ( ρ , u , p ). Let σ = min( σ , σ ). This finishes the proof. (cid:3) The rest of this paper is devoted to prove Theorem 2.16. For convenience, we describe ourmain process of proving Theorem 2.16 below.To describe our process of proving Theorem 2.16, we define some terminologies. Let f :[0 , θ ] → R be a function representing an axisymmetric shock location Γ f . Decompose f into f ( θ ) = f (0) + f s ( θ ). Then by f s (0) = 0, f s is uniquely determined by f ′ s . We call f (0) and f ′ s the initial shock position and the shape of a shock location, respectively.Using these terminologies, our process of proving Theorem 2.16 is described as follows.1. For given an incomming supersonic solution, an exit pressure and a shape of a shocklocation ( ρ − , u − , p − , p ex , f ′ s ) in a small perturbation of ( ρ − , u − e r , p − , p c , f (0) and a subsonic solution (Φ e ϕ , L, S )of (2.5.1)-(2.5.3) satisfying all the conditions in Problem 2 except (2.5.5), and that thissolution is unique in the class of functions in a small perturbation of ( r s , Φ +0 e ϕ , L +0 , S +0 ).2. For given an incomming supersonic solution and an exit pressure as in Step 1 or ina much small perturbation of ( ρ − , u − e r , p − , p c ) if necessary, show that there exists f ′ s in a small perturbation of 0 as in Step 1 such that ( f (0) , Φ e ϕ , L, S ) determined by( ρ − , u − , p − , p ex , f ′ s ) in Step 1 satisfies (2.5.5), and that for given ( ρ − , u − , p − , p ex ) in asmall perturbation of ( ρ − , u − e r , p − , p c ), a solution ( f, Φ e ϕ , L, S ) of Problem 2 is uniquein the class of functions in a small perturbation of ( r s , Φ +0 e ϕ , L +0 , S +0 ).Once Step 1 and Step 2 are done, then f = f (0) + f s and (Φ e ϕ , L, S ) obtained through Step 1and Step 2 satisfies all the conditions in Problem 2. Thus, Theorem 2.16 is proved if Step 1 andStep 2 are done. We will deal with Step 1 and Step 2 in Section 3 and Section 4, respectively.Note that the fact that for transonic shock solutions of the full Euler system, an initial shockposition and a shape of a shock location are determined in different mechanisms (an initialshock position is determined by the solvability condition related to the mass conservation lawand a shape of a shock location is determined by the R-H conditions) was pointed out in [24]and the authors in [24, 24, 18, 21, 20, 10, 23] prove the stability of transonic shock solutions ofthe full Euler system using iteration schemes based on this fact. In this paper, we also prove thestability of transonic shock solutions of the full Euler system using a scheme based on this fact.But we do this using a different scheme. In our scheme, a non-local elliptic equation appearingin [24, 18, 21, 20, 10, 23] does not appear.3. Pseudo Free Boundary Problem
As a first step to prove Theorem 2.16, we will solve the Pseudo Free Boundary Problembelow. This problem naturally arises from the requirement that a subsonic solution in Problem2 must satisfy (2.5.12). From the linearized equation of (2.5.11), it is seen that an iterationscheme for a fixed boundary problem does not give a subsonic solution satisfying (2.5.12) in general. Thus, an iteration scheme for a fixed boundary problem is not a proper scheme to finda subsonic solution in Problem 2. To find a subsonic solution satisfying (2.5.12), we let f (0)be an unknown to be determined simultaneously with a subsonic solution. Using this variable,we adjust the value of a subsonic solution so that this solution can satisfy (2.5.12). The mainingredient for this argument to hold is the monotonicity of the entropy of the downstreamsubsonic solution of a radial transonic shock solution on a shock location with respect to theshock location. This will be seen in the proof of Proposition 3.1. Problem 3 (Pseudo Free Boundary Problem) . Given an axisymmetric supersonic solution ( ρ − , u − , p − ) of (2.1.1) and exit pressure p ex as in Problem 1 and a shape of a shock location f ′ s ∈ C ,α ( − α, { θ = θ } ) ((0 , θ )) satisfying f ′ s ( θ ) = 0 at θ = 0 , θ and || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ σ (3.0.1) for a sufficiently small σ > , find an initial shock position f (0) and a corresponding subsonicsolution (Φ e ϕ , L, S ) of (2.5.1) - (2.5.3) satisfying (cid:18) ρ ∇ × (Φ e ϕ ) · e r (cid:19) ∇ × (cid:18) ρ ∇ × (Φ e ϕ ) (cid:19) = (cid:18) L πr sin θ ∇ × (cid:18) L πr sin θ e ϕ (cid:19) · e r + ρ γ − γ − ∂ θ Sr (cid:19) e ϕ in N + f (0)+ f s , Φ e ϕ = Φ − e ϕ on Γ f (0)+ f s , Φ e ϕ = r Φ − ( r , θ ) r e ϕ on Γ + w,f (0)+ f s ,Sρ γ = p ex on Γ ex , lim θ → θ (Φ e ϕ )( r , θ ) = lim r → r (Φ e ϕ )( r, θ ) , (A) where ρ = ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) , and ∇ × (Φ e ϕ ) · ∇ L = 0 in N + f (0)+ f s ,L = L − on Γ f (0)+ f s , ∇ × (Φ e ϕ ) · ∇ S = 0 in N + f (0)+ f s ,S = g (cid:18) u − · ν f c − (cid:19) ! S − on Γ f (0)+ f s (B) where f s denotes R θ f ′ s . Hereafter, we denote a set of functions in C ,α ( − α, { θ = θ } ) ((0 , θ )) having 0 value at θ = 0 , θ by C ,α ( − α, { θ = θ } ) , ((0 , θ )). Hereafter, f s denotes R θ f ′ s .Our result of Problem 3 is given as follows. Proposition 3.1.
Let α ∈ ( , . There exists a positive constant σ depending on the datasuch that if σ ∈ (0 , σ ] , then Problem 3 has a solution ( f (0) , Φ e ϕ , L, S ) satisfying (3.0.2) | f (0) − r s | + ||∇ × ((Φ − Φ +0 ) e ϕ ) || ( − α, Γ + w )1 ,α, N + f (0)+ fs + || L πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f (0)+ fs + || S − S +0 || ( − α, Γ + w )1 ,α, N + f (0)+ fs ≤ Cσ where C is a positive constant depending on the data. Furthermore, this solution is unique inthe class of functions satisfying (3.0.2) . We will prove Proposition 3.1 using a fixed point argument. To do this, we linearize (A)with respect to (A) satisfied by (Φ e ϕ , L, S ) = (Φ +0 e ϕ , L +0 , S +0 ) and reformulate (B) in terms of(Ψ e ϕ , A, T ) where (Ψ , A, T ) := (Φ − Φ +0 , L, S − S +0 ). -D AXISYMMETRIC TRANSONIC SHOCK 17 Linearization and reformulation of (A) and (B).
We linearize (A) with respect to(A) satisfied by (Φ +0 e ϕ , L +0 , S +0 ). Since ρ in the first and fourth equation of (A) is given usingan implicit relation, to obtain the linearized equations of (A), we first linearize ρ with respectto ρ +0 . Lemma 3.2.
Suppose that f : Λ → R is an axisymmetric function in C ,α ( − − α,∂ Λ) (Λ) satisfying || f − r s || ( − − α,∂ Λ)2 ,α, Λ ≤ δ . (3.1.1) Also, suppose that Ψ e ϕ : N + f → R , A : N + f → R and T : N + f → R are axisymmetric functionsin C ,α ( − − α, Γ + w ) ( N + f ) , C ,α ( − α, Γ + w ) ( N + f ) and C ,α ( − α, Γ + w ) ( N + f ) , respectively, and satisfy ||∇ × (Ψ e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || A πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || T || ( − α, Γ + w )1 ,α, N + f ≤ δ . (3.1.2) Let ρ = ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) . There holds ρ − ρ +0 = ∇ × (Φ +0 e ϕ ) ρ +0 ( u +0 2 − c +0 2 ) · ∇ × (Ψ e ϕ ) + γρ +0 γ ( γ − u +0 2 − c +0 2 ) T + g (Ψ e ϕ , A, T )(3.1.3) where g (Ψ e ϕ , A, T ) =(3.1.4) 1 ρ +0 ( u +0 2 − c +0 2 ) (cid:18)Z (cid:18) (2 ρ +0 B − γ ( γ + 1) γ − S +0 ρ +0 γ ) − (2( tρ + (1 − t ) ρ +0 ) B − γ ( γ + 1) γ − S +0 + T )( tρ + (1 − t ) ρ +0 ) γ ) (cid:19) dt ( ρ − ρ +0 )+ Z ( −∇ × (Φ +0 e ϕ ) + ( t ∇ × ((Φ +0 + Ψ) e ϕ ) + (1 − t ) ∇ × (Φ +0 e ϕ ))) dt · ∇ × (Ψ e ϕ )+ Z ρ +0 2 tA πr sin θ e ϕ dt · A πr sin θ e ϕ ! .g (Ψ e ϕ , A, T ) satisfies || g (Ψ e ϕ , A, T ) || ( − α, Γ + w )1 ,α, N + f ≤ C ( ||∇ × (Ψ e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || A πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || T || ( − α, Γ + w )1 ,α, N + f ) (3.1.5) where C is a positive constant depending on ( ρ +0 , u +0 , p +0 ) , γ , r s , r , α , δ and δ .Proof.
1. By the definition of ̺ , ρ = ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) satisfies b ( ρ, ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ) = 0where b is a function defined in (2.4.18). Subtract this equation from b ( ρ +0 , ∇× (Φ +0 e ϕ ) , , S +0 , B ) =0 obtained from the third equation of (2.2.10). And then linearize the resultant equation. Thenwe obtain (3.1.3).2. With the fact that ( ρ +0 , u +0 e r , p +0 ) ∈ ( C ∞ ( N + f )) and (3.1.2), estimate ρ − ρ +0 in C ,α ( − α, Γ + w ) ( N + f )in the way that we estimated ρ + − ρ +0 in the proof of Theorem 2.17. Then we have || ρ − ρ +0 || ( − α, Γ + w )1 ,α, N + f ≤ C ( ||∇ × (Ψ e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || A πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || T || ( − α, Γ + w )1 ,α, N + f )(3.1.6)where C is a positive constant depending on ( ρ +0 , u +0 , p +0 ), γ , r s , r , α , δ and δ . With this esti-mate, the fact that ( ρ +0 , u +0 e r , p +0 ) ∈ ( C ∞ ( N + f )) and (3.1.2), estimate (3.1.4) in C ,α ( − α, Γ + w ) ( N + f ).Then we obtain the desired result. (cid:3) Then we linearize (A).Subtract (2.5.1) satisfied by (Φ e ϕ , L, S ) = (Φ +0 e ϕ , L +0 , S +0 ) in N + f from (2.5.1) in N + f andthen linearize the resultant equation (in this process, we express ρ − ρ +0 using (3.1.3)). Thenwe obtain ∇× ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (Ψ e ϕ ) ! (3.1.7) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ Tr e ϕ + F (Ψ e ϕ , A, T ) in N + f (=: F (Ψ e ϕ , A, T ))where F (Ψ e ϕ , A, T ) =(3.1.8) ∇ × ρ +0 g ∇ × (Φ +0 e ϕ ) − ρρ +0 2 g ∇ × ((Φ +0 + Ψ) e ϕ ) + 1 ρ +0 2 g ∇ × (Ψ e ϕ ) ! − u +0 (cid:18) − ρρ +0 g ∇ × ((Φ +0 + Ψ) e ϕ ) · e r + 1 ρ +0 ∇ × (Ψ e ϕ ) · e r (cid:19) ∇ × (cid:18) − ρρ +0 g ∇ × ((Φ +0 + Ψ) e ϕ ) + 1 ρ +0 ∇ × (Ψ e ϕ ) (cid:19) + 1 u +0 (cid:18)Z ( tρ + (1 − t ) ρ +0 ) γ − dtg ∂ θ Tr (cid:19) e ϕ + Au +0 πr sin θ (cid:18) ∇ × ( A πr sin θ e ϕ ) · e r (cid:19) e ϕ where g and g are g (Ψ e ϕ , A, T ) given in (3.1.4) and the right-hand side of (3.1.3), respectively.By the definition of V ± and the first equations of (2.2.6) and (2.2.10), V +0 = V − on Γ f . Inthe vector potential form, this is written as Φ +0 e ϕ = Φ − e ϕ on Γ f . Subtract this equation from(2.5.4). Then we obtain Ψ e ϕ = (Φ − − Φ − ) e ϕ on Γ f . (3.1.9)By the definition of V ± and the first equations of (2.2.6) and (2.2.10), V +0 = V − ( r , θ ) onΓ + w . In the vector potential form, this is written as Φ +0 e ϕ = r Φ − ( r ,θ ) e ϕ r on Γ + w . By subtractingthis equation from (2.5.10), we obtainΨ e ϕ = r (Φ − − Φ − )( r , θ ) r e ϕ on Γ + w . (3.1.10)Rewrite (2.5.11) as ρ = ( p ex S ) γ on Γ ex . Subtract this equation from ρ +0 = ( p c S +0 ) γ on Γ ex . Andthen express ρ − ρ +0 using (3.1.3) and linearize ( p ex S ) γ − ( p c S +0 ) γ . Multiply u +0 2 − c +0 2 u +0 r sin θ andthen integrate the resultant equation from 0 to θ . After that, divide sin θ and multiply e ϕ onboth-hand sides of the equation. Then we obtain(3.1.11) Ψ e ϕ = (cid:18) r sin θ Z θ (cid:18) f ( T, p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ e ϕ , A, T ) (cid:19) r sin ξdξ (cid:19) e ϕ on Γ ex where f ( T, p ex ) = u +0 2 − c +0 2 u +0 (cid:18) S +0 + T (cid:19) γ ( p γ ex − p γ c )(3.1.12) -D AXISYMMETRIC TRANSONIC SHOCK 19 and(3.1.13) f (Ψ e ϕ , A, T ) = − p c γ ( u +0 2 − c +0 2 ) γu +0 Z (cid:18) tS + (1 − t ) S +0 (cid:19) γ +1 − (cid:18) S +0 (cid:19) γ +1 ! dtT − u +0 2 − c +0 2 u +0 g . By the definition of V ± and the first equation of (2.2.10), lim θ → θ V +0 ( r , θ ) = lim r → r V +0 ( r, θ ).In the vector potential form, this is written as lim θ → θ (Φ +0 e ϕ )( r , θ ) = lim r → r (Φ +0 e ϕ )( r, θ ).Subtract this equation from (2.5.12). Then we obtainlim θ → θ (Ψ e ϕ )( r , θ ) = lim r → r (Ψ e ϕ )( r, θ ) . Express this relation using (3.1.10) and (3.1.11). Then we have(3.1.14) 1 r sin θ Z θ (cid:18) f ( T, p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ e ϕ , A, T ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ = r (Φ − − Φ − )( r , θ ) r . Combining (3.1.7), (3.1.9), (3.1.10), (3.1.11) and (3.1.14), we have the following linearizedequations of (A): ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (Ψ e ϕ ) ! = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ Tr e ϕ + F (Ψ e ϕ , A, T ) in N + f (0)+ f s , Ψ e ϕ = (Φ − − Φ − ) e ϕ on Γ f (0)+ f s , Ψ e ϕ = r (Φ − − Φ − )( r , θ ) r e ϕ on Γ + w,f (0)+ f s , Ψ e ϕ = (cid:18) r sin θ Z θ (cid:18) f ( T, p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ e ϕ , A, T ) (cid:19) r sin ξdξ (cid:19) e ϕ on Γ ex , r sin θ Z θ (cid:18) f ( T, p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ e ϕ , A, T ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ = r (Φ − − Φ − )( r , θ ) r . (A ′ )For later use, we present the following estimates of F (Ψ e ϕ , A, T ) and f (Ψ e ϕ , A, T ). Lemma 3.3.
Let δ be a positive constant ≤ δ such that for f as in Lemma 3.2, if (Ψ e ϕ , A, T ) satisfies ||∇ × (Ψ e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || A πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || T || ( − α, Γ + w )1 ,α, N + f ≤ δ , (3.1.15) then sup N + f | ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) | ≤ C and sup N + f | S | ≤ C where C s are positive constants depending on ( ρ +0 , u +0 , p +0 ) , γ , r s , r and δ . Suppose that f is as in Lemma 3.2. Also, suppose that (Ψ e ϕ , A, T ) are as in Lemma 3.2 and satisfy (3.1.15) . Then there hold || F (Ψ e ϕ , A, T ) || (1 − α, Γ + w ) α, N + f ≤ C ( || Ψ e ϕ || ( − − α, Γ + w )2 ,α, N + f + || A πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || T || ( − α, Γ + w )1 ,α, N + f ) (3.1.16) and || f (Ψ e ϕ , A, T ) || ( − α,∂ Γ ex )1 ,α, Γ ex ≤ C ( || Ψ e ϕ || ( − − α,∂ Γ ex )2 ,α, Γ ex + || A πr sin θ e ϕ ||| ( − α,∂ Γ ex )1 ,α, Γ ex + || T || ( − α,∂ Γ ex )1 ,α, Γ ex ) (3.1.17) where Cs are positive constants depending on ( ρ +0 , u +0 , p +0 ) , γ , r s , r , α , δ , δ and δ .Proof. With (3.1.5), (3.1.6), (3.1.15) and the fact that if an axisymmetric vector field a on anaxisymmetric connected open set Ω is in C k (Ω), then ∇ × a ∈ C k − (Ω) and a · e r ∈ C k (Ω) (thesecond one is obtained from Lemma 2.8), we estimate (3.1.8) and (3.1.13) in C α (1 − α, Γ + w ) ( N + f ),and C ,α ( − α,∂ Γ ex ) (Γ ex ), respectively. Then we obtain the desired result. (cid:3) Next, we reformulate (B) in terms of (Ψ e ϕ , A, T ). With the facts that L +0 = L − = 0 in N + f and S +0 = ( g ( M − ))( r s ) S in in N + f (see (2.2.10)), we reformulate (B) in terms of (Ψ e ϕ , A, T ).Then we obtain ∇ × ((Φ +0 + Ψ) e ϕ ) · ∇ A = 0 in N + f (0)+ f s ,A = A en,f (0)+ f s on Γ f (0)+ f s , ∇ × ((Φ +0 + Ψ) e ϕ ) · ∇ T = 0 in N + f (0)+ f s ,T = T en,f (0)+ f s on Γ f (0)+ f s (B ′ )where A en,f (0)+ f s := L − on Γ f (0)+ f s (3.1.18)and T en,f (0)+ f s := g (cid:18) u − · ν f (0)+ f s c − (cid:19) ! S − − (cid:16) g ( M − ) (cid:17) ( r s ) S in on Γ f (0)+ f s . (3.1.19)For later use, we present the following estimate of T en,f (0)+ f s . Lemma 3.4.
Let f (0) and f ′ s be a constant and a function in C ,α ( − α, { θ = θ } ) ((0 , θ )) , respectively.Let δ be a positive constant such that if ( ρ − , u − , p − ) satisfies (2.3.3) for σ ∈ (0 , δ ] , then sup N | c − | ≤ C where C is a positive constant depending on ( ρ − , u − , p − ) , γ , r , r and δ . Suppose that f = f (0)+ f s satisfies (3.1.1) . Also, suppose that ( ρ − , u − , p − ) is an axisymmetric supersonic solutionof (2.1.1) in N satisfying (2.3.3) for σ ∈ (0 , δ ] . Then there holds || T en,f || ( − α,∂ Γ f )1 ,α, Γ f ≤ C ( | f (0) − r s | + || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ) + Cσ (3.1.20) where C s are positive constants depending on ( ρ − , u − , p − ) , γ , r , r s , r , θ , α , δ and δ .Proof. In this proof, C s denote positive constants depending on the whole or a part of ( ρ − , u − , p − ), γ , r , r s , r , θ , α , δ and δ . Each C in different situations differs from each other. -D AXISYMMETRIC TRANSONIC SHOCK 21 By the fact that ( ρ − , u − , p − ) and f = f (0) + f s are axisymmetric, T en,f defined in (3.1.19)can be regarded as a function of θ . As a function of θ , T en,f can be written as T en,f = Z ( g ( M − )) ′ ( tf ( θ ) + (1 − t ) r s ) S in dt ( f ( θ ) − r s )+ g (cid:18) u − · ν f ( θ ) c − (cid:19) ! S − ! ( f ( θ ) , θ ) − (cid:16) g ( M − ) (cid:17) ( f ( θ ) , θ ) S in =: ( a ) + ( b ) . To estimate || T en,f || ( − α,∂ Γ f )1 ,α, Γ f , we estimate ( a ) and ( b ) in C ,α ( − α, { θ = θ } ) ((0 , θ )), respectively.Since an estimate of ( a ) in C ,α ( − α, { θ = θ } ) ((0 , θ )) is obtained directly: || ( a ) || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C ( | f (0) − r s | + || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) )(3.1.21)where we used || f s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) , we only estimate ( b ) in C ,α ( − α, { θ = θ } ) ((0 , θ )).Estimate of ( b ) in C ,α ( − α, { θ = θ } ) ((0 , θ )):Decompose ( b ) into two parts: g (cid:18) u − · e r c − (cid:19) ! S − ! ( f ( θ ) , θ ) − (cid:16) g ( M − ) (cid:17) ( f ( θ ) , θ ) S in =: ( b ) and g (cid:18) u − · ν f c − (cid:19) ! S − ! ( f ( θ ) , θ ) − g (cid:18) u − · e r c − (cid:19) ! S − ! ( f ( θ ) , θ ) =: ( b ) . With (2.3.3) for σ ∈ (0 , δ ] and (3.1.1), we estimate ( b ) in C ,α ( − α, { θ = θ } ) ((0 , θ )). Then we obtain || ( b ) || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ Cσ. (3.1.22)Write ( b ) as Z g ′ (cid:18) u − c − · ( t ν f + (1 − t ) e r ) (cid:19) ! (cid:18) u − c − · ( t ν f + (1 − t ) e r ) (cid:19) S − dt u − c − · ( ν f − e r ) . (3.1.23)By ν f = e r − f ′ f e θ q f ′ f ) , ν f − e r can be written as ν f − e r = Z − t (cid:18) f ′ f (cid:19) ! − dt (cid:18) f ′ f (cid:19) e r − f ′ f r (cid:16) f ′ f (cid:17) e θ . Substitute this expression of ν f − e r into ν f − e r in (3.1.23) and then estimate (3.1.23) in C ,α ( − α, { θ = θ } ) ((0 , θ )) with (2.3.3) for σ ∈ (0 , δ ] and (3.1.1). Then we obtain || ( b ) || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C ( || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ) + Cδ || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) . Using the fact that || f ′ s || ( − α, (0 ,θ ))1 ,α, (0 ,θ ) ≤ C || f − r s || ( − − α,∂ Λ)2 ,α, Λ ≤ Cδ , we get from this estimate || ( b ) || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) . (3.1.24)Combining (3.1.22) and (3.1.24), we obtain || ( b ) || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C || f ′ s || ( − α, (0 ,θ ))1 ,α, (0 ,θ ) + Cσ. (3.1.25)From the facts that ∂ θ T en,f (0) = 0, ( a ) ∈ C ,α ( − α, { θ = θ } ) ((0 , θ )) and ( b ) ∈ C ,α ( − α, { θ = θ } ) ((0 , θ ))(obtained from (3.1.21) and (3.1.25)), we see that T en,f ∈ C ,α ( − α,∂ Γ f ) (Γ f ) (see Lemma 2.8). Then || T en,f || ( − α,∂ Γ f )1 ,α, Γ f can be estimated by estimating || T en,f || ( − α, { θ = θ } )1 ,α, (0 ,θ ) . By this fact, (3.1.21) and(3.1.25), we have (3.1.20). This finishes the proof. (cid:3) From (A ′ ), (B ′ ), the Pseudo Free Boundary Problem is naturally derived. We explain thisbelow.For a given ( ρ − , u − , p − , p ex ), find (Ψ e ϕ , A, T ) satisfying (A ′ ), (B ′ ) using an iteration schemefor a fixed boundary problem (for example, in a fixed domain N + f (0)+ f s , for a given Ψ e ϕ , solve(B ′ ), substitute the resultant A and T and the previously given Ψ e ϕ into the right-hand sidesof (A ′ ), obtain a new Ψ e ϕ by solving this (A ′ ) and show that a new Ψ e ϕ is equal to the givenΨ e ϕ using a fixed point argument). Then since (Ψ e ϕ , A, T ) we find in this way does not satisfythe fifth equation of (A ′ ) in general, this iteration scheme does not give a subsonic solution of(2.5.1)-(2.5.3) satisfying (2.5.12) in general. From the facts that the entropy at a point on ashock location in the subsonic side is conserved along the streamline passing through that pointand the entropy of the downstream subsonic solution of a radial transonic shock solution in adivergent nozzle on a shock location monotonically increases as a shock location moves towardthe exit (see Lemma 2.10), we see that we can find (Ψ e ϕ , A, T ) satisfying the fifth equation of(A ′ ) by varying S on Γ ex by adjusting f (0). From this fact, Problem 3 is derived.(A ′ ) and (B ′ ) are of the form of one linear boundary value problem for a singular ellipticequation (this will be seen in the next subsection) and two initial value problems of a transportequation whose coefficient is an axisymmetric and divergence-free vector field, respectively. Wewill study these problems, seperately, in § § Linear boundary value problem for a singular elliptic equation.
Fix the right-hand sides of the first and fourth equation in (A ′ ) with the fifth equation in (A ′ ) satisfied. Thenwe obtain ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (Ψ e ϕ ) ! = F in N + f , (3.2.1) Ψ e ϕ = h e ϕ on Γ f , f ( θ ) h ( f ( θ ) ,θ ) r e ϕ on Γ + w ,h e ϕ on Γ ex (3.2.2)where f , F and h i e ϕ for i = 1 , (cid:18) − ρ +0 (cid:18) ∆Ψ − Ψ r sin θ (cid:19) + ∂ r ρ +0 ρ +0 2 r ∂ r ( r Ψ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) r (cid:18) θ ∂ θ (sin θ∂ θ Ψ) − Ψsin θ (cid:19) (cid:19) e ϕ = F , (3.2.1), (3.2.2) is a linear boundary value problem for a singular equation as a problem for Ψ.Thus, the standard elliptic theorems cannot be applied to this problem as a problem for Ψ. Weresolve the singularity issue in (3.2.1), (3.2.2) by dealing with (3.2.1), (3.2.2) as a boundaryvalue problem for an elliptic system.The following is the main result in this subsection. Lemma 3.5.
Let α ∈ ( , . Suppose that f is as in Lemma 3.2 and satisfy f ′ ( θ ) = 0 . Also,suppose that F : N + f → R is a function in C α (1 − α, Γ + w ) ( N + f ) having the form F = X i A i ∂ r B i e ϕ + X i C i ∂ θ D i e ϕ + E ∂ θ ( F sin θ )sin θ e ϕ (3.2.4) where A i , B i , C i , D i , E and F are axisymmetric functions satisfying A i ∈ C ,α ( − α, Γ + w ) ( N + f ) , B i e θ ∈ C ,α ( − α, Γ + w ) ( N + f ) , C i ∈ C ,α ( − α, Γ + w ) ( N + f ) , (3.2.5) D i ∈ C ,α ( − α, Γ + w ) ( N + f ) , E e ϕ ∈ C ,α ( N + f ) and F e ϕ ∈ C ,α ( N + f ) . -D AXISYMMETRIC TRANSONIC SHOCK 23 Finally, suppose that h e ϕ : Γ f → R and h e ϕ : Γ ex → R are axisymmetric functions in C ,α ( − − α,∂ Γ f ) (Γ f ) and C ,α ( − − α,∂ Γ ex ) (Γ ex ) , respectively, and satisfy f ( θ ) h ( f ( θ ) ,θ ) r = h ( r , θ ) .Then the boundary value problem (3.2.1) , (3.2.2) has a unique axisymmetric C ,α ( − − α, Γ + w ) ( N + f ) solution Ψ e ϕ . Furthermore, the solution Ψ e ϕ satisfies || Ψ e ϕ || ( − − α, Γ + w )2 ,α, N + f ≤ C || F || (1 − α, Γ + w ) α, N + f + X i =1 , , F i + || h e ϕ || ( − − α,∂ Γ f )2 ,α, Γ f + || h e ϕ || ( − − α,∂ Γ ex )2 ,α, Γ ex (3.2.6) where C is a positive constant depending only on ( ρ +0 , u +0 , p +0 ) , γ , r s , r , θ and α , and F = X i || A i || W , ( N + f ) ∩ L ∞ ( N + f ) || B i e θ || α, N + f F = X i || C i || W , ( N + f ) ∩ L ∞ ( N + f ) || D i || α, N + f ,F = || E ∂ θ ( F sin θ )sin θ e ϕ || L q ( N + f ) for q = − α with || · || W , ( N + f ) ∩ L ∞ ( N + f ) := || · || W , ( N + f ) + || · || L ∞ ( N + f ) . Remark 3.6.
The form of F given in (3.2.4) is obtained from F (Ψ e ϕ , A, T ) in (3.1.7) . Thisform will be used in the proof of Lemma 3.10. To avoid the singularity issue in (3.2.1), (3.2.2), we deal with (3.2.1), (3.2.2) as a boundaryvalue problem for a vector equation. From ∇ × ( ∇ × (Ψ e ϕ )) = − ∆(Ψ e ϕ ), we expected that(3.2.1) can be transformed into a form of an elliptic system. We, motivated by the work in[4], thought that if (3.2.1) can be transformed into a solvable elliptic system form, then theunique existence and regularity of solutions of (3.2.1), (3.2.2) can be obtained by obtainingthose of solutions of the elliptic system form of (3.2.1), (3.2.2) as a boundary value problem foran elliptic system.For this argument to hold, it is needed to find a solvable elliptic system form of (3.2.1). Forcomputational convenience to find such a form and for our later argument (reflection argumentin the proof of Lemma 3.9 and Lemma 3.10), we use the following tensor notation. Tensor notation
Let a ⊗ b = ab T for a , b ∈ R . Then a ⊗ b is a linear map from R to R and any linearmap from R to R can be represented using this operator. This notation can be extended sothat using the extension of this operator, we can represent any linear map from R × to R × .For any a , b , c , d ∈ R , let a ⊗ b ⊗ c ⊗ d be an operator satisfying( a ⊗ b ⊗ c ⊗ d )( e ⊗ f ) = ( d · e )( c · f ) a ⊗ b (3.2.7)where e , f ∈ R . Then a ⊗ b ⊗ c ⊗ d is a linear map from R × to R × and any linear mapfrom R × to R × can be represented using this operator.By direct computation done by using the above tensor notation, we found the following formof (3.2.1)(3.2.8) div c +0 2 ρ +0 ( c +0 2 − u +0 2 ) I − u +0 2 c +0 2 ( I ⊗ e r ⊗ e r ⊗ I ) ! D (Ψ e ϕ ) ! − ∂ r ρ +0 ρ +0 2 r Ψ e ϕ = − F in N + f where I is the identity map from R × to R × and I ⊗ e r ⊗ e r ⊗ I is a linear map from R × to R × satisfing ( I ⊗ e r ⊗ e r ⊗ I )( a ⊗ b ) = ( b · e r ) a ⊗ e r for any a , b ∈ R (see the definition of I ⊗ a ⊗ b ⊗ I for any a , b ∈ R in (5.0.4)). By M +0 < N + f and the boundedness of ( ρ +0 , u +0 , p +0 ) in N + f for N + f ⊂ N + r s − δ , there exist positive constants µ and M such that µ | ξ | ≤ X α,β,i,j =1 " c +0 2 ρ +0 ( c +0 2 − u +0 2 ) I − u +0 2 c +0 2 ( I ⊗ e r ⊗ e r ⊗ I ) ! αβij ξ iα ξ jβ ≤ M| ξ | in N + f (3.2.9)for any ξ ∈ R × where | ξ | = qP i,j =1 | ξ ij | with ξ = [ ξ ij ]. And by ∂ r ρ +0 > N + f for N + f ⊂ N + r s − δ (see Remark 2.9), ∂ r ρ +0 ρ +0 2 r > N + f . Hence, (3.2.8) is a form of a solvable elliptic system for a dirichlet boundary condition.We obtain the unique existence and regularity of solutions of (3.2.1), (3.2.2) by obtainingthose of solutions of (3.2.8), (3.2.2) as a boundary value problem for an elliptic system. Theresult of the unique existence and regularity of solutions of (3.2.8), (3.2.2) as a boundary valueproblem for an elliptic system is given in the following lemma.
Lemma 3.7.
Under the assumptions as in Proposition 3.5, the boundary value problem div c +0 2 ρ +0 ( c +0 2 − u +0 2 ) I − u +0 2 c +0 2 ( I ⊗ e r ⊗ e r ⊗ I ) ! D U ! − ∂ r ρ +0 ρ +0 2 r U = − F in N + f , (3.2.10) U = h e ϕ on Γ f , f ( θ ) h ( f ( θ ) ,θ ) r e ϕ on Γ + w ,h e ϕ on Γ ex , (3.2.11) has a unique C ,α ( − − α, Γ + w ) ( N + f ) solution U , and this solution U satisfies || U || ( − − α, Γ + w )2 ,α, N + f ≤ C || F || (1 − α, Γ + w ) α, N + f + X i =1 , , F i + || h e ϕ || ( − − α,∂ Γ f )2 ,α, Γ f + || h e ϕ || ( − − α,∂ Γ ex )2 ,α, Γ ex | {z } =: C ∗ (3.2.12) where C is a positive constant depending on ( ρ +0 , u +0 , p +0 ) , γ , r s , r , θ and α and F i for i = 1 , , are constants given in Lemma 3.10. This solution U is of the form Ψ( r, θ ) e ϕ . One can see that that Lemma 3.5 is obtained from Lemma 3.7. To prove Lemma 3.5, in theremainder of this subsection, we prove Lemma 3.7.Transform (3.2.10), (3.2.11) into the following boundary problem:div( A D U ♯ ) − d U ♯ = − F − div( A D h ) + d h (=: F ♯ ) in N + f , (3.2.13) U ♯ = 0 on ∂ N + f (3.2.14)where U ♯ := U − h , A = c +0 2 ρ +0 ( c +0 2 − u +0 2 ) (cid:18) I − u +0 2 c +0 2 ( I ⊗ e r ⊗ e r ⊗ I ) (cid:19) , d := ∂ r ρ +0 ρ +0 2 r and h := ( r − f ( θ )) r r h e ϕ +( r − r ) f ( θ ) r h e ϕ r − f ( θ ) . One can see that Lemma 3.7 can be proved by showing (3.2.13),(3.2.14) has a unique weak solution, the weak solution of (3.2.13), (3.2.14) is in C ,α ( − − α, Γ + w ) ( N + f )and the C ,α ( − − α, Γ + w ) ( N + f ) solution of (3.2.13), (3.2.14) is of the form Ψ( r, θ ) e ϕ . Hereafter, weprove these statements.We first prove the unique existence of weak solution of (3.2.13), (3.2.14). -D AXISYMMETRIC TRANSONIC SHOCK 25 Lemma 3.8.
Under the assumptions as in Proposition 3.5, the boundary value problem (3.2.13) , (3.2.14) has a unique weak solution U ♯ ∈ H ( N + f ) . Furthermore, U ♯ satisfies || U ♯ || H ( N + f ) ≤ C || F ♯ || L ( N + f ) where C is a positive constant depending on ( ρ +0 , u +0 , p +0 ) , γ and N + f .Proof. Write (3.2.13), (3.2.14) in the form( B [ U ♯ , ξ ] :=) Z N + f A D U ♯ D ξ + d U ♯ ξ = Z N + f F ♯ ξ (=: < F ♯ , ξ > )(3.2.15)for all ξ ∈ H ( N + f ). Then B is a bilinear map satisfying B [ U ♯ , ξ ] ≤ C || U ♯ || H ( N + f ) || ξ || H ( N + f ) for a constant C > µ || U ♯ || H ( N + f ) ≤ B [ U ♯ , U ♯ ]where µ = min( µ, min N + f d ) > µ given in (3.2.9). And by h e ϕ ∈ C ,α ( − − α,∂ Γ f ) (Γ f ), h e ϕ ∈ C ,α ( − − α,∂ Γ ex ) (Γ ex ) and F ∈ C α (1 − α, Γ + w ) ( N + f ) for α ∈ ( , F ♯ ∈ L ( N + f ) and thus < F ♯ , ξ > is a bounded linear functional on H ( N + f ). With these facts, we apply the Lax-Milgram Theorem to (3.2.15). Then we obtain that there exists a unique U ♯ ∈ H ( N + f ) suchthat (3.2.15) holds for all ξ ∈ H ( N + f ). This finishes the proof. (cid:3) We next prove that this weak solution is in C ,α ( − − α, Γ + w ) ( N + f ). For this, we prove that theweak solution of (3.2.13), (3.2.14) is in C β ( N + f ) for any β ∈ (0 ,
1) and C ,α ( N + f ). Lemma 3.9.
Under the assumptions as in Proposition 3.5, let U ♯ be a weak solution of theboundary value problem (3.2.13) , (3.2.14) . Then for any β ∈ (0 , , || U ♯ || β, N + f ≤ C (cid:16) || F ♯ || L p ( N + f ) + || U ♯ || H ( N + f ) (cid:17) for p = − β where C is a positive constant depending on µ , M , τ , || d || L ( N + f ) and N + f , and τ is the modulus of continuity of A in N + f given as τ ( t ) = sup x , y ∈N + f , | x − y |≤ t ( X | A αβij (x) − A αβij (y) | ) . (3.2.16) Lemma 3.10.
Under the assumptions as in Proposition 3.5, let U ♯ be a weak solution of (3.2.13) , (3.2.14) . Then || U ♯ || ,α, N + f ≤ C X i =1 , , F i + || h || ,α, N + f + || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) where C is a positive constant depending on µ , M , || A || α, N + f , || d || L q ( N + f ) with q = − α and N + f and F i for i = 1 , , are constants given in Lemma 3.10. We will prove Lemma 3.9 and Lemma 3.10 using the method of freezing the coefficients(Korn’s device of freezing the coefficients) (see [15, Chapter 3]). Since N + f is a Lipshitz domain, U ♯ ∈ C β ( N + f ) and U ♯ ∈ C ,α ( N + f ) can be proved by showing that (i) there are positiveconstants C and R such that Z D t (x ) | D U ♯ | ≤ Ct − β for any 0 < t < R (3.2.17) for all x ∈ N + f , and (ii) there are positive constants C and R such that Z D t (x ) | D U ♯ − ( D U ♯ ) x ,t | ≤ Ct α for any 0 < t < R (3.2.18)for all x ∈ N + f where D t (x ) := B t (x ) ∩ N + f with B t (x ) := { x ∈ R : | x − x | < t } and( D U ♯ ) x ,t := | D t (x ) | R D t (x ) D U ♯ . We prove (i) and (ii) by obtaining (3.2.17) and (3.2.18) ateach point x in N + f for C and R independent of x using the method of freezing the coefficients.When we do this, there exists some difficulty. For the case of x ∈ N + f or Γ f ∪ Γ + w ∪ Γ ex , we canobtain the integral estimates for the fixed coefficients equation using the Cacciopolli inequalityand the quotient difference method, and obtain (3.2.17) and (3.2.18) at x ∈ N + f or Γ f ∪ Γ + w ∪ Γ ex using these estimates and the method of freezing the coefficients (see [1, Chapter 6]). But forthe case of x ∈ Γ f ∩ Γ + w or Γ + w ∩ Γ ex , we cannot obtain the integral estimates for the fixedcoefficients equation using the Cacciopolli inequality and the quotient difference method. Thus,we cannot obtain (3.2.17) and (3.2.18) at x ∈ Γ f ∩ Γ + w or Γ + w ∩ Γ ex using the standard methodof freezing the coefficients. We resolve this difficulty by developing some reflection argumentthat holds for a linear boundary value problem on a Lipschitz domain whose all corners areperpendicular for an elliptic system whose the domain part of principal coefficients is diagonalwith respect to the coordinate systems representing the walls near the corners of the domain.This will be seen in the proof of Lemma 3.9 and Lemma 3.10.Hereafter, we use the following notation:div ( r,θ,ϕ ) := ( e ∂ r + e ∂ θ + e ∂ ϕ ) · , D ( r,θ,ϕ ) := e ∂ r + e ∂ θ + e ∂ ϕ T : a one dimensional torus with period 2 π, N + , ∗ a := { ( r, θ, ϕ ) ∈ R | a < r < r , < θ < θ , ϕ ∈ T } , Γ ∗ a := { ( r, θ, ϕ ) ∈ R | r = a, < θ < θ , ϕ ∈ T } , Γ + , ∗ w,a := { ( r, θ, ϕ ) ∈ R | a < r < r , θ = θ , ϕ ∈ T } for 0 < a < r Γ + , ∗ w := { ( r, θ, ϕ ) ∈ R | f ( θ ) < r < r , θ = θ , ϕ ∈ T } , Γ ∗ ex := { ( r, θ, ϕ ) ∈ R | r = r , < θ < θ , ϕ ∈ T } r , t : a radius of a ball in the spherical coordinate system ,B ∗ r (x ∗ ) := { ( r, θ, ϕ ) ∈ R + × [0 , π ] × T : | r − r ∗ | + | θ − θ ∗ | + | ϕ − ϕ ∗ | < r } ,D ∗ r (x ∗ ) := B ∗ r (x ∗ ) ∩ N + , ∗ f ( θ ) for x ∗ = ( r ∗ , θ ∗ , ϕ ∗ ) ∈ R + × (0 , π ) × T . To prove Lemma 3.9 and Lemma 3.10, we prove the following lemma.
Lemma 3.11.
Let x ∗ = ( f ( θ ) , θ , ϕ ) for some ϕ ∈ T . Let < r < min( θ , π, r − f ( θ )) .Suppose that W ∈ H ( D ∗ r (x ∗ )) is a weak solution of div ( r,θ,ϕ ) (cid:18) r sin θρ +0 (cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I (3.2.19) + sin θρ +0 ( c +0 2 c +0 2 − u +0 2 ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I + 1sin θρ +0 ( c +0 2 c +0 2 − u +0 2 ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I (cid:19) D ( r,θ,ϕ ) W = 0 in D ∗ r (x ∗ ) , W = on ∂D ∗ r (x ∗ ) ∩ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ) . (3.2.20) -D AXISYMMETRIC TRANSONIC SHOCK 27 Then for any t such that < t ≤ r , there hold Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) W | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) W | , (3.2.21) and Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗ x ∗ , t | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗ x ∗ , r | (3.2.22) where C is a positive constant depending on µ , M , f ( θ ) and θ and ( D ( r,θ,ϕ ) W ) ∗ x ∗ , r := | D ∗ r (x ∗ ) | R D ∗ r (x ∗ ) D ( r,θ,ϕ ) W .Proof. The result is obtained by using the reflection argument.Extend (3.2.19) in B ∗ r (x ∗ ):div ( r,θ,ϕ ) (cid:18) r sin θρ +0 (cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I (3.2.23) + sin θρ +0 ( c +0 2 c +0 2 − u +0 2 ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I + 1sin θρ +0 ( c +0 2 c +0 2 − u +0 2 ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = f ( θ ) ,θ = θ I ⊗ e ⊗ e ⊗ I (cid:19) D ( r,θ,ϕ ) W = 0 in B ∗ r (x ∗ ) . Extend a weak solution W of (3.2.19), (3.2.20) in B ∗ r (x ∗ ): W ext = W ( r, θ, ϕ ) in B ∗ r (x ∗ ) ∩ { r ≥ f ( θ ) , θ ≤ θ }− W (2 f ( θ ) − r, θ, ϕ ) in B ∗ r (x ∗ ) ∩ { r < f ( θ ) , θ ≤ θ }− W ( r, θ − θ, ϕ ) in B ∗ r (x ∗ ) ∩ { r ≥ f ( θ ) , θ > θ } W (2 f ( θ ) − r, θ − θ, ϕ ) in B ∗ r (x ∗ ) ∩ { r < f ( θ ) , θ > θ } Then W ext ∈ H ( B ∗ r (x ∗ )). There exists a unique weak solution of (3.2.23), W = W ext on ∂B ∗ r (x ∗ ) . (3.2.24)We denote the weak solution of (3.2.23), (3.2.24) by W .One can check that − W (2 f ( θ ) − r, θ, ϕ ) and − W ( r, θ − θ, ϕ ) are also weak solutions of(3.2.23), (3.2.24). By this fact and the uniqueness of weak solutions of (3.2.23), (3.2.24), W ( r, θ, ϕ ) = − W (2 f ( θ ) − r, θ, ϕ ) = − W ( r, θ − θ, ϕ ) in B ∗ r (x ∗ ) . (3.2.25)From (3.2.25), we have W = 0 on ∂D ∗ r (x ∗ ) ∩ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ). By this fact and the fact that W is a unique weak solution of (3.2.23), (3.2.24), W is a weak solution of (3.2.19), W = W on ∂D ∗ r (x ∗ ) \ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ) , ∂D ∗ r (x ∗ ) ∩ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ) . By the uniqueness of weak solutions of (3.2.19) satisfying W = W ext on ∂D ∗ r (x ∗ ) \ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ) and W = 0 on ∂D ∗ r (x ∗ ) ∩ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ), we have W = W in D ∗ r (x ∗ ) . (3.2.26)By [15, Theorem 2.1, Chapter 3], W satisfies for any 0 < t ≤ r , Z B ∗ t (x ∗ ) | D ( r,θ,ϕ ) W | ≤ C (cid:18) tr (cid:19) Z B ∗ r (x ∗ ) | D ( r,θ,ϕ ) W | (3.2.27) and Z B ∗ t (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗∗ ˜x ∗ , t | (3.2.28) ≤ C (cid:18) tr (cid:19) Z B ∗ r (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗∗ ˜x ∗ , r | where C is a positive constant depending on µ , M , f ( θ ) and θ and ( D ( r,θ,ϕ ) W ) ∗∗ x ∗ , r := | B ∗ r (x ∗ ) | R B ∗ r (x ∗ ) D ( r,θ,ϕ ) W . By (3.2.25) and (3.2.26), we obtain from (3.2.27) and (3.2.28) forany 0 < t ≤ r , 4 Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) W | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) W | and 4 Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗ x ∗ , t | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) W − ( D ( r,θ,ϕ ) W ) ∗ x ∗ , r | where we used( D ( r,θ,ϕ ) W ) ∗∗ x ∗ , r = 1 | B ∗ r (x ∗ ) | Z B ∗ r (x ∗ ) D ( r,θ,ϕ ) W = 1 | D ∗ r (x ∗ ) | Z D ∗ r (x ∗ ) D ( r,θ,ϕ ) W . This finishes the proof. (cid:3)
The following Corollary is obtained from Lemma 3.11 in the same way that Corollary 3.11 isobtained from Lemma 3.10 in [17]. We omit the proof.
Corollary 3.12.
Suppose that W is as in Lemma 3.11. Let ˜ U ∗ be any function in H ( D ∗ r (x ∗ )) for < r < min( θ , π, r − f ( θ )) . Then for any t and r such that < t ≤ r < min( θ , π, r − f ( θ )) , there hold Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) ˜ U ∗ | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) ˜ U ∗ | + Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) ( ˜ U ∗ − W ) | ! and Z D ∗ t (x ∗ ) | D ( r,θ,ϕ ) ˜ U ∗ − ( D ( r,θ,ϕ ) ˜ U ∗ ) ∗ x ∗ , t | ≤ C (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) ˜ U ∗ − ( D ( r,θ,ϕ ) ˜ U ∗ ) ∗ x ∗ , r | + Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) ( ˜ U ∗ − W ) | ! where C is a positive constant depending on µ , M , f ( θ ) and θ . We first prove Lemma 3.9.
Proof of Lemma 3.9.
We prove Lemma 3.9 by proving that (3.2.17) holds for all x ∈ N + f forsome positive constants C and R .1. Transform (3.2.13), (3.2.14) into a problem in N + f ( θ ) .Define a map Π ∗ ab ( r, θ, ϕ ) := ( r − br − a ( r − a ) + b, θ, ϕ )(3.2.29)for 0 < a, b < r . Then Π ∗ ab ( r, θ, ϕ ) maps ( a, r ) × (0 , θ ) × T to ( b, r ) × (0 , θ ) × T . Π ∗ ab naturallyinduces a map from N + a to N + b as a map between two cartesian coordinate systems. Denote -D AXISYMMETRIC TRANSONIC SHOCK 29 this map by Π ab . Using Π := Π f ( θ ) f ( θ ) , we transform the boundary value problem (3.2.13),(3.2.14) into the following boundary value problemdiv y ( ˜ A D y ˜ U ♯ ) − ˜ d ˜ U ♯ = ˜ F ♯ in N + f ( θ ) , (3.2.30) ˜ U ♯ = on ∂ N + f ( θ ) (3.2.31)where ˜ A := ∂ Π ∂ x ) ( ∂ Π ∂ x ) T ( A ◦ Π − )( ∂ Π ∂ x ), ˜ d := d ◦ Π − det( ∂ Π ∂ x ) , ˜ U ♯ := U ♯ ◦ Π − , ˜ F ♯ := F ♯ ◦ Π − det( ∂ Π ∂ x ) and x andy are the cartesian coordinate systems for N + f and N + f ( θ ) , respectively.2. Transform the weak formulation of (3.2.30), (3.2.31) near x ∈ Γ f ( θ ) ∩ Γ + w into the weakformulation of the spherical coordinate representation of (3.2.30), (3.2.31).Write (3.2.30), (3.2.31) in the form Z N + f ( θ ˜ A D y ˜ U ♯ D y ξ + ˜ d ˜ U ♯ ξ = − Z N + f ( θ ˜ F ♯ ξ (3.2.32)for all ξ ∈ H ( N + f ( θ ) ). Let (˜ r, ˜ θ, ˜ ϕ ) be the spherical coordinate system for y and Ξ be the mapfrom y to (˜ r, ˜ θ, ˜ ϕ ). Choose any x ∗ := ( f ( θ ) , θ , ϕ ) for some ϕ ∈ T and set ξ = 0 outside ofΞ − ( D ∗ r (x ∗ )) for 0 < r < min( θ , π, r − f ( θ )) in (3.2.32). Then we obtain Z Ξ − ( D ∗ r (x ∗ )) ˜ A D y ˜ U ♯ D y ξ + ˜ d ˜ U ♯ ξ = − Z Ξ − ( D ∗ r (x ∗ )) ˜ F ♯ ξ for all ξ ∈ H (Ξ − ( D ∗ r (x ∗ ))). Using Ξ, transform this equation. Then we have Z D ∗ r (x ∗ ) M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ + ˜ d ∗ ˜ U ∗ ξ = − Z D ∗ r (x ∗ ) ˜ F ∗ ξ (3.2.33)for all ξ ∈ H ( D ∗ r (x ∗ )) where ˜ U ∗ = ˜ U ♯ ◦ Ξ − , ˜ d ∗ = ˜ d ◦ Ξ − det ˜ M , ˜ F ∗ = ˜ F ♯ ◦ Ξ − det ˜ M and ˜ M = ∂ Ξ ∂y .3. Obtain (3.2.17) at x ∈ Γ f ∩ Γ + w .Rewrite (3.2.33) as Z D ∗ r (x ∗ ) M ˜ M T A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ = Z D ∗ r (x ∗ ) ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M − M ˜ M T ˜ A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ − ˜ d ∗ ˜ U ∗ ξ − ˜ F ∗ ξ for all ξ ∈ H ( D ∗ r (x ∗ )). Fix the principal coefficients of the left-hand side of the resultantequation at x ∗ . Then we obtain(3.2.34) Z D ∗ r (x ∗ ) ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ = Z D ∗ r (x ∗ ) (( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) − M ˜ M T A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ + ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M − M ˜ M T ˜ A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ − ˜ d ∗ ˜ U ∗ ξ − ˜ F ∗ ξ for all ξ ∈ H ( D ∗ r (˜x ∗ )). Using the argument in Appendix, it can be checked that ( M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) is equal to the principal coefficients of the equation in (3.2.19).Let W be the weak solution ofdiv (˜ r, ˜ θ, ˜ ϕ ) (( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) D (˜ r, ˜ θ, ˜ ϕ ) W ) = 0 in D ∗ r (x ∗ ) , (3.2.35) W = ( ˜ U ∗ on ∂D ∗ r (x ∗ ) ∩ N + , ∗ f ( θ ) , on ∂D ∗ r (x ∗ ) ∩ (Γ ∗ f ( θ ) ∪ Γ + , ∗ w,f ( θ ) ) . (3.2.36) Subtract the weak formulation of (3.2.35), (3.2.36) from (3.2.34) and then take ξ = V to theresultant equation. Then we obtain Z D ∗ r (x ∗ ) ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) D (˜ r, ˜ θ, ˜ ϕ ) V D (˜ r, ˜ θ, ˜ ϕ ) V = Z D ∗ r (x ∗ ) (( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) − M ˜ M T A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) V + ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M − M ˜ M T ˜ A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) V − ˜ d ∗ ˜ U ∗ V − ˜ F ∗ V where V := ˜ U ∗ − W ∈ H ( D ∗ r (x ∗ )). Using the Sobolev and H¨older inequality, we obtain fromthis equation(3.2.37) Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) V | ≤ C ( τ ( r ) + τ ( r )) Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | +( Z D ∗ r (x ∗ ) | ˜ d ∗ | ) Z D ∗ r (x ∗ ) | ˜ U ∗ | + ( Z D ∗ r (x ∗ ) | ˜ F ∗ | ) ! where τ ( r ) = sup x ∗ , y ∗ ∈N + , ∗ f | x ∗ − y ∗ |≤ r | ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) − ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(y ∗ ) | and τ ( r ) := sup | θ − θ |≤ r (cid:8) | f ( θ ) − f ( θ ) | + | f ′ ( θ ) | (cid:9) (note that f ′ ( θ ) = 0). By Corollary 3.12 and (3.2.37), we have for any 0 < t ≤ r , Z D ∗ t (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | ≤ C (cid:18) tr (cid:19) + τ ( r ) + τ ( r ) ! Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | + ( Z D ∗ r (x ∗ ) | ˜ d ∗ | ) Z D ∗ r (x ∗ ) | ˜ U ∗ | + ( Z D ∗ r (x ∗ ) | ˜ F ∗ | ) ! . Using the H¨older and Poincar´e inequality, we get from this inequality(3.2.38) Z D ∗ t (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | ≤ C (cid:18) tr (cid:19) + τ ( r ) + τ ( r ) ! Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | +( Z D ∗ r (x ∗ ) | ˜ d ∗ | ) r Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | + ( Z D ∗ r (x ∗ ) | ˜ F ∗ | p ) p r − β ! where β = 2 − p ∈ (0 ,
1) if p ∈ ( , β , we consider two cases.Case 1: 3 − β ≤ R ∈ (0 , min( θ , π, r − f ( θ ))) such that for any 0 < r ≤ R , Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | ≤ C r − β Z D ∗ R (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | + ( Z D ∗ R (x ∗ ) | ˜ F ∗ | p ) p ! . (3.2.39)Case 2: 3 − β > -D AXISYMMETRIC TRANSONIC SHOCK 31 As in the Case 1, we apply Lemma 2.1 in [15, Chapter 3] to (3.2.38). Then we obtain thatthere exists R ∈ (0 , min( θ , π, r − f ( θ ))) such that for any 0 < r ≤ R , Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | ≤ C r Z D ∗ R (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | + ( Z D ∗ R (x ∗ ) | ˜ F ∗ | p ) p ! . Substitute this into (3.2.38). After then apply Lemma 2.1 in [15, Chapter 3] again. Then weobtain that there exists R ≤ R such that for any 0 < r ≤ R , (3.2.39) holds with R replacedby R .From this result, we obtain that for any β ∈ (0 , R > Z D t (x ) | D U ♯ | ≤ Ct − β (cid:18) || U ♯ || H ( N + f ) + || F ♯ || L p ( N + f ) (cid:19) for any 0 < t < R for x ∈ Γ f ∩ Γ + w . Since F ♯ ∈ C α (1 − α, Γ + w ) ( N + f ) for α ∈ ( , F ♯ ∈ L ( N + f ). This implies that F ♯ ∈ L p ( N + f ) for any β ∈ (0 ,
1) where p = − β . Hence, we obtain (3.2.17) with C replaced by C = C ( || U ♯ || H ( N + f ) + || F ♯ || L p ( N + f ) ) for x ∈ Γ f ∩ Γ + w .4. When x is in Γ + w ∩ Γ ex , we obtain (3.2.17) with C replaced by C using similar argumentswithout the process of transforming N + f into N + f ( θ ) . When x is in N + f or Γ f ∪ Γ + w ∪ Γ ex and faraway from the corners Γ f ∩ Γ + w and Γ + w ∩ Γ ex , we obtain (3.2.17) with C replaced by C usingthe standard method of freezing the coefficients. When x is in N + f or Γ f ∪ Γ + w ∪ Γ ex and nearΓ f ∩ Γ + w or Γ + w ∩ Γ ex , we obtain (3.2.17) with C replaced by C using the arguments in the proofof [16, Theorem 5.21] and arguments similar to the ones in Step 1-Step 3 above. Combiningthese results, we obtain that there exist a positive constant R such that U ♯ satisfies (3.2.17) forall x ∈ N + f for C = C . This finishes the proof. (cid:3) Next, we prove Lemma 3.10. We prove Lemma 3.10 using the method of freezing the co-efficients and the reflection arguments in the proof of Lemma 3.9. When we do this, thereexists some problem: since F ♯ is not in L p ( N + f ) for q = − α nor has the form div G with G ∈ C α ( N + f ), we cannot get the power of t required in (3.2.18) from the integral estimate of F ♯ directly. We obtain this power by delivering θ -derivatives imposed on some functions in F ♯ tothe functions multiplied to those functions in the integral form of F ♯ using integration by partsand estimating the resultant integral form of F ♯ . To make our argument clear, we present thedetailed proof. Proof of Lemma 3.10.
Using (3.2.4), write (3.2.13), (3.2.14) in the form Z N + f A D U ♯ D ξ + d U ♯ ξ = Z N + f X i A i ∂ r ( B i − B i (x )) ξ ϕ + X i C i ∂ θ ( D i − D i (x )) ξ ϕ + E ∂ θ ( F sin θ )sin θ ξ ϕ + div( A D h ) ξ − d hξ for all ξ ∈ H ( N + f ) where ξ ϕ = ξ · e ϕ . Using integration by parts, we change this equation into Z N + f A D U ♯ D ξ + d U ♯ ξ = Z N + f − X i (cid:18) ∂ r A i ( B i − B i (x )) ξ ϕ + A i ( B i − B i (x )) 1 r ∂ r ( r ξ ϕ ) (cid:19) − X i (cid:18) ∂ θ C i ( D i − D i (x )) ξ ϕ + C i ( D i − D i (x )) 1sin θ ∂ θ ( ξ ϕ sin θ ) (cid:19) + E ∂ θ ( F sin θ )sin θ ξ ϕ − A D h D ξ − d hξ for all ξ ∈ H ( N + f ). Using Π defined in Step 1 in the proof of Lemma 3.9, transform thisequation. Then we obtain Z N + f ( θ ˜ A D ˜ U ♯ D ξ + ˜ d ˜ U ♯ ξ = Z N + f ( θ ( a ) 1det( d y d x ) − ˜ A D ˜ h D ξ − ˜ d ˜ hξ for all ξ ∈ H ( N + f ( θ ) ) where( a ) = − X i (cid:18) ∂ ˜ r∂r ∂ ˜ r ˜ A i ( ˜ B i − B i (x )) ξ ϕ + ˜ A i ( ˜ B i − B i (x )) 1 r ∂ ˜ r∂r ∂ ˜ r ( r ξ ϕ ) (cid:19) − X i (cid:18) ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ ) ˜ C i ( ˜ D i − D i (x )) ξ ϕ + ˜ C i ( ˜ D i − D i (x )) 1sin ˜ θ ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ )( ξ ϕ sin ˜ θ ) (cid:19) + ˜ E θ ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ )( ˜ F sin ˜ θ ) ξ ϕ , ˜ A , ˜ d and ˜ U ♯ are functions given below (3.2.31),˜ A i := A i ◦ Π − , ˜ B i := B i ◦ Π − , ˜ C i := C i ◦ Π − , ˜ D i = D i ◦ Π − , ˜ E := E ◦ Π − , ˜ F := F ◦ Π − , ˜ h = h ◦ Π − , x is the cartesian coordinate representing N + f , y = Π(x) and ( r, θ, ϕ ) and (˜ r, ˜ θ, ˜ ϕ ) are thespherical coordinate systems for x and y, respectively. As we did in Step 2 in the proof of Lemma3.9, set ξ = 0 outside of Ξ − ( D ∗ r (x ∗ )) for 0 < r < min( θ , π, r − f ( θ )) where x ∗ = ( f ( θ ) , θ , ϕ )for some ϕ ∈ T and then transform this equation using Ξ. Then we obtain Z D ∗ r (x ∗ ) M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ + ˜ d ∗ ˜ U ∗ ξ = Z D ∗ r (x ∗ ) ( a ) ˜ r sin ˜ θ det( d y d x ) − (cid:18) M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ − ( 1det ˜ M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ )(x ∗ ) (cid:19) D (˜ r, ˜ θ, ˜ ϕ ) ξ − ˜ d ∗ ˜ h ∗ ξ for all ξ ∈ H ( D ∗ r (x ∗ )) where ˜ U ∗ , ˜ d ∗ , and ˜ M are functions defined below (3.2.33) and weused div (˜ r, ˜ θ, ˜ ϕ ) (cid:16) ( M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ )(x ∗ ) (cid:17) = 0. Fix the principal coefficients of theleft-hand side of the above equation. Then we get Z D ∗ r (x ∗ ) ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ = Z D ∗ r (x ∗ ) (( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) − M ˜ M T A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ + ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M − M ˜ M T ˜ A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) ξ − ˜ d ∗ ˜ U ∗ ξ + ( a ) ˜ r sin ˜ θ det( d y d x ) − (cid:18) M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ − ( 1det ˜ M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ )(x ∗ ) (cid:19) D (˜ r, ˜ θ, ˜ ϕ ) ξ − ˜ d ∗ ˜ h ∗ ξ for all ξ ∈ H ( D ∗ r (x ∗ )).Let W be the weak solution of (3.2.35), (3.2.36). As we did in Step 3 in the proof of Lemma3.9, subtracting the weak formulation of (3.2.35), (3.2.36) from the above equation. And then -D AXISYMMETRIC TRANSONIC SHOCK 33 take ξ = V where V = ˜ U ∗ − W to the resultant equation. Then we have Z D ∗ r (x ∗ ) ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) D (˜ r, ˜ θ, ˜ ϕ ) V D (˜ r, ˜ θ, ˜ ϕ ) V = Z D ∗ r (x ∗ ) (( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M )(x ∗ ) − M ˜ M T A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) V + ( 1det ˜ M ˜ M T A ◦ Ξ − ˜ M − M ˜ M T ˜ A ◦ Ξ − ˜ M ) D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ D (˜ r, ˜ θ, ˜ ϕ ) V − ˜ d ∗ ˜ U ∗ V − X i (cid:18) ∂ ˜ r∂r ∂ ˜ r ˜ A i ( ˜ B i − B i (x )) V ϕ + ˜ A i ( ˜ B i − B i (x )) 1 r ∂ ˜ r∂r ∂ ˜ r ( r V ϕ ) (cid:19) − X i (cid:18) ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ ) ˜ C i ( ˜ D i − D i (x )) V ϕ + ˜ C i ( ˜ D i − D i (x )) 1sin ˜ θ ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ )( V ϕ sin ˜ θ ) (cid:19) + ˜ E θ ( ∂ ˜ r∂θ ∂ ˜ r + ∂ ˜ θ )( ˜ F sin ˜ θ ) V ϕ ! ˜ r sin ˜ θ det( d y d x ) − (cid:18) M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ − ( 1det ˜ M ˜ M T ˜ A ◦ Ξ − ˜ M D (˜ r, ˜ θ, ˜ ϕ ) ˜ h ∗ )(x ∗ ) (cid:19) D (˜ r, ˜ θ, ˜ ϕ ) V − ˜ d ∗ ˜ h ∗ V where V ϕ = V · e ϕ . Using the Sobolev and H¨older inequality and the facts that A ∈ C α ( N + f ), f ∈ C ,α (Λ), h ∈ C ,α ( N + f ) and Z D ∗ r (x ∗ ) | r ∂ ˜ r ( r V ϕ ) | , Z D ∗ r (x ∗ ) | θ ∂ ˜ θ ( V ϕ sin ˜ θ ) | ≤ C Z D ∗ r (x ∗ ) | D ( V ϕ e ϕ ) | (See (5.0.7)) ≤ C Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) V | we obtain from the above equation(3.2.40) Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) V | ≤ C r α Z D ∗ r (x ∗ ) | D ( r,θ,ϕ ) ˜ U ∗ | +( Z D ∗ r (x ∗ ) | ˜ d ∗ | q ) q r α Z D ∗ r (x ∗ ) | ˜ U ∗ | + r α F ♭ ! for q = − α where F ♭ = X i || B i e θ || α, N + f || A i || W , ( N + f ) ∩ L ∞ ( N + f ) + X i || D i || α, N + f || C i || W , ( N + f ) ∩ L ∞ ( N + f ) + || E ∂ θ ( F sin θ )sin θ e ϕ || L q ( N + f ) + || h || ,α, N + f with || · || W , (Ω) ∩ L ∞ (Ω) := || · || W , (Ω) + || · || L ∞ (Ω) . By Corollory 3.12 and (3.2.40), we have forany 0 < t ≤ r ,(3.2.41) Z D ∗ t (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , t | ≤ C (cid:18) tr (cid:19) Z D ∗ r (˜x ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , r | + r α Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | + r α Z D ∗ r (x ∗ ) | ˜ U ∗ | + r α F ♭ ! . In the proof of Lemma 3.9, we showed that for any ε ∈ (0 , R > Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ | ≤ C r − ε ( || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) ) for any 0 < r ≤ R . Using this inequality and U ♯ ∈ C ( N + f ) obtained from Lemma 3.9, we apply Lemma 2.1 in [15,Chapter 3] to (3.2.41). Then we obtain(3.2.42) Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , r | ≤ C ( || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) + F ♭ ) r α − ε for any 0 < r < R for a constant R > ∗ is in Γ + . ∗ w ∩ Γ ∗ ex , we obtain (3.2.42) using similar argument without the process oftransforming N + f to N + f ( θ ) . When x ∗ is in N + , ∗ f ∩ { θ ≥ θ } or (Γ ∗ f ∪ Γ + , ∗ w ∪ Γ ∗ ex ) ∩ { θ ≥ θ } and far away from the corners Γ ∗ f ∩ Γ + , ∗ w and Γ + , ∗ w ∩ Γ ∗ ex , we obtain (3.2.42) using the standardmethod of freezing coefficients to the spherical coordinate representation of (3.2.13), (3.2.14)with integration by parts argument above. When x ∗ is in N + , ∗ f ∩{ θ ≥ θ } or Γ ∗ f ∪ Γ + , ∗ w ∪ Γ ∗ ex ∩{ θ ≥ θ } and near Γ ∗ f ∩ Γ + ∗ w or Γ + , ∗ w ∩ Γ ∗ ex , we obtain (3.2.42) using the arguments in [16, Theorem 5.21]and arguments similar to the ones above. Then we obtain D U ♯ ∈ C α − ε ([ f ( θ ) , r ] × [ θ , θ ] × T ).When x is in N + f ∩ { θ ≤ θ } or (Γ f ∪ Γ + w ∪ Γ ex ) ∩ { θ ≤ θ } , we obtain (3.2.18) with α and C replaced by α − ε for any ε ∈ (0 ,
1) and C ( || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) + F ♭ ) using the standardmethod of freezing coefficients to (3.2.13), (3.2.14) with integration by parts argument above.Note that if we estimate the integral form of F ♯ using integration by parts argument above,then there is no singularity issue. From this result, we obtain D U ♯ ∈ C α − ε ( N + f ∩ { θ ≤ θ } ).Combining these two regularity results for D U ♯ , we obtain D U ♯ ∈ C α − ε ( N + f ).Using the regularity result for D U ♯ and U ♯ ∈ C ( N + f ), we obtain from (3.2.41)(3.2.43) Z D ∗ t (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , t | ≤ C (cid:18) (cid:18) tr (cid:19) Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , r | + r α (cid:18) || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) + F ♭ (cid:19) (cid:19) (here x ∗ ∈ Γ ∗ f ∩ Γ + , ∗ w ). Apply Lemma 2.1 in [15, Chapter 3] to (3.2.43). Then we have(3.2.44) Z D ∗ r (x ∗ ) | D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ − ( D (˜ r, ˜ θ, ˜ ϕ ) ˜ U ∗ ) ∗ x ∗ , r | ≤ C ( || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) + F ♭ ) r α for any 0 < r < R for a constant R >
0. Using the arguments used when we obtained D U ∈ C α − ε ( N + f ), weobtain (3.2.44) at x ∗ ∈ N + , ∗ f ∩ { θ ≥ θ } and (3.2.18) with C replaced by C ( || U ♯ || H ( N + f ) + || F ♯ || L ( N + f ) + F ♭ ) at x ∈ N + f ∩ { θ ≤ θ } . From this result, we obtain the desired result. Thisfinishes the proof. (cid:3) Using the scailing argument given in the proof of Proposition 3.1 in [2] with the results inTheorem 5.21 in [16] and Lemma 3.10, we can obtain the following result. We omit the proof. -D AXISYMMETRIC TRANSONIC SHOCK 35
Lemma 3.13.
Under the assumption as in Lemma 3.5, let U ♯ be a weak solution of (3.2.13) , (3.2.14) . Then U ♯ ∈ C ,α ( − − α, Γ + w ) ( N + f ) . Furthermore, U ♯ satisfies || U ♯ || ( − − α, Γ + w )2 ,α, N + f ≤ C ( || F ♯ || (1 − α, Γ + w ) α, N + f + || U ♯ || ,α, N + f ) where C is a positive constant depending on ( ρ +0 , u +0 , p +0 ) , γ , N + f and α . Finally, we prove that the C ,α ( − − α, Γ + w ) ( N + f ) solution of (3.2.13), (3.2.14) is of the formΨ( r, θ ) e ϕ . This statement is proved using the argument as in the proof of Proposition 3.3in [4] (Method II). Although the arguments to prove this statement are almost same with thosein the proof of Proposition 3.3 in [4] (Method II), since (3.2.13), (3.2.14) is different from theproblem in Proposition 3.3 in [4] and similar arguments will be used later in the proof of Lemma4.3, we present the detailed proof. Lemma 3.14.
Under the assumption as in Lemma 3.5, the C ,α ( − − α, Γ + w ) ( N + f ) solution of (3.2.13) , (3.2.14) is of the form Ψ( r, θ ) e ϕ .Proof. Let U = U r e r + U θ e θ + U ϕ e ϕ be the C ,α ( − − α, Γ + w ) ( N + f ) solution of (3.2.13), (3.2.14). Then,( U r , U θ , U ϕ ) is in ( C ,α ( − − α, { θ = θ } ) ( N + , ∗ f )) satisfying || U r || ( − − α, { θ = θ } )2 ,α, N + , ∗ f , || U θ || ( − − α, { θ = θ } )2 ,α, N + , ∗ f , || U ϕ || ( − − α, { θ = θ } )2 ,α, N + , ∗ f ≤ C || U || ( − − α, Γ + w )2 ,α, N + f ≤ CC ∗ , (3.2.45)where C and C ∗ are positive constants depending on N + f and α and given in (3.2.12), respec-tively, and satisfies the following spherical coordinate representation of (3.2.13), (3.2.14) (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) (∆ U r − U r r − r sin θ ∂ θ ( U θ sin θ ) − r sin θ ∂ ϕ U ϕ ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U r ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU r ) r (cid:19) = 0 , (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) (∆ U θ − U θ r sin θ + r ∂ θ U r − θr sin θ ∂ ϕ U ϕ ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U θ ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU θ ) r (cid:19) = 0 , (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) (∆ U ϕ − U ϕ r sin θ + r sin θ ∂ ϕ U r + θr sin θ ∂ ϕ U θ ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U ϕ ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU ϕ ) (cid:19) = −F ( r, θ ) in N + , ∗ f , (3.2.46)( U r , U θ , U ϕ ) = (0 , ,
0) on Γ ∗ f , Γ + , ∗ w , Γ ∗ ex , (3.2.47)where F = F ♯ · e ϕ and ∆ U k = r ∂ r ( r ∂ r U k ) + r sin θ ∂ θ (sin θ∂ θ U k ) + r sin θ ∂ ϕ U k for k = r, θ, ϕ .Define U nk := 12 n n − X k =0 U k ( r, θ, ϕ + 2 πk n )for k = r, θ, ϕ . Then by the definition of U nk for k = r, θ, ϕ and (3.2.45), || U nr || ( − − α, { θ = θ } )2 ,α, N + , ∗ f , || U nθ || ( − − α, { θ = θ } )2 ,α, N + , ∗ f , || U nϕ || ( − − α, { θ = θ } )2 ,α, N + , ∗ f ≤ CC ∗ . (3.2.48)By (3.2.48) and C ,α ( − − α, { θ = θ } ) ( N + , ∗ f ) ⋐ C ,α ( − − α , { θ = θ } ) ( N + , ∗ f ), there exists a subsequence ( U n k r ,U n k θ , U n k ϕ ) of ( U nr , U nθ , U nϕ ) such that ( U n k r , U n k θ , U n k ϕ ) converges in C ,α ( − − α , { θ = θ } ) ( N + , ∗ f ) as n k →∞ . Denote its limit by ( U ∗ r , U ∗ θ , U ∗ ϕ ). Then ( U ∗ r , U ∗ θ , U ∗ ϕ ) is independent of ϕ and ( U ∗ r , U ∗ θ , U ∗ ϕ ) ∈ ( C ,α ( − − α, { θ = θ } ) ( N + , ∗ f )) . Since the coefficients of (3.2.46), F and the boundary conditions in (3.2.47) are independentof ϕ , ( U nr , U nθ , U nϕ ) satisfies (3.2.46), (3.2.47) for all n ∈ N ∪ { } . By this fact and the definitionof ( U ∗ r , U ∗ θ , U ∗ ϕ ), ( U ∗ r , U ∗ θ , U ∗ ϕ ) satisfies (3.2.46), (3.2.47). Since ( U ∗ r , U ∗ θ , U ∗ ϕ ) is independent of ϕ ,(3.2.46) satisfied by ( U ∗ r , U ∗ θ , U ∗ ϕ ) is given as (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) ( r ∂ r ( r ∂ r U ∗ r ) + r sin θ ∂ θ (sin θ∂ θ U ∗ r ) − U ∗ r r − r sin θ ∂ θ ( U ∗ θ sin θ )) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U ∗ r ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU ∗ r ) r (cid:19) = 0 , (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) ( r ∂ r ( r ∂ r U ∗ θ ) + r sin θ ∂ θ (sin θ∂ θ U ∗ θ ) − U ∗ θ r sin θ + r ∂ θ U ∗ r ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U ∗ θ ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU ∗ θ ) r (cid:19) = 0 , (cid:18) c +0 2 ρ +0 ( c +0 2 − u +0 2 ) ( r ∂ r ( r ∂ r U ∗ ϕ ) + r sin θ ∂ θ (sin θ∂ θ U ∗ ϕ ) − U ∗ ϕ r sin θ ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) 1 r ∂ r ( r ∂ r U ∗ ϕ ) − ∂ r ρ +0 ρ +0 2 ∂ r ( rU ∗ ϕ ) (cid:19) = −F ( r, θ ) in N + , ∗ f . (3.2.49)Note that the first and second equation of (3.2.49) is not coupled with the third equation of(3.2.49).Let Ψ = U ∗ ϕ . Using the third equation of (3.2.49) and the facts that U ∗ ϕ ∈ C ,α ( − − α, { θ = θ } ) ( N + , ∗ f )and F ♯ is an axisymmetric function in C α (1 − α, Γ + w ) ( N + f ), it can be checked that Ψ = ∂ θθ Ψ = 0 on θ = 0. By this fact and U ∗ ϕ ∈ C ,α ( − − α, { θ = θ } ) ( N + , ∗ f ), we have that Ψ e ϕ ∈ C ,α ( − − α, Γ + w ) ( N + f ). From(3.2.49) and (3.2.47), one can see that (0 , , U ∗ ϕ ) is a solution of (3.2.46), (3.2.47). Combiningthis fact with the fact that Ψ e ϕ ∈ C ,α ( − − α, Γ + w ) ( N + f ), we have that Ψ e ϕ is a C ,α ( − − α, Γ + w ) ( N + f )solution of (3.2.13), (3.2.14). By Lemma 3.8, a C ,α ( − − α, Γ + w ) ( N + f ) solution of (3.2.13), (3.2.14) isunique. Therefore, U = Ψ e ϕ . This finishes the proof. (cid:3) Initial value problem of a transport equation with an axisymmetric divergence-free coefficient.
The initial value problems of a transport equation in (B ′ ) are of the form ∇ × ((Φ +0 + Ψ) e ϕ ) · ∇ Q = 0 in N + f , (3.3.1) Q = Q en on Γ f (3.3.2)where Ψ e ϕ : N + f → R and Q en : Γ f → R are axisymmetric functions. Here, (3.3.1) is atransport equation whose coefficient is an axisymmetric and divergence-free vector field. Thus,the stream function of the coefficient vector field of (3.3.1) can be defined (see (2.4.1)). We find asolution of (3.3.1), (3.3.2) and obtain the regularity and uniqueness of solutions of (3.3.1), (3.3.2)using the stream function of the coefficient vector field of (3.3.1) and the solution expressiongiven by using the stream function in the following lemma. Lemma 3.15.
Suppose that f is as in Lemma 3.2. Let δ be a positive constant such that forsuch f , if ||∇ × (Ψ e ϕ ) || , , N + f ≤ δ , then (3.3.3) ∇ × ((Φ +0 + Ψ) e ϕ ) · e r > c ∗ in N + f and ∇ × ((Φ +0 + Ψ) e ϕ ) · ν f > c ∗ on Γ f where c ∗ is a positive constant depending on ρ +0 , u +0 , r s , r , δ and δ , and ν f is the unit normalvector on Γ f pointing toward N + f . Suppose that Ψ e ϕ : N + f → R is an axisymmetric functionin C ,α ( N + f ) satisfying Ψ = f ( θ )Ψ( f ( θ ) , θ ) r on Γ + w (3.3.4) -D AXISYMMETRIC TRANSONIC SHOCK 37 and ||∇ × (Ψ e ϕ ) || ,α, N + f ≤ δ . (3.3.5) Suppose finally that Q en : Γ f → R is an axisymmetric function in C ,α ( − α,∂ Γ f ) (Γ f ) . Then theproblem (3.3.1) , (3.3.2) has a unique axisymmetric C ( N + f ) ∩ C ( N + f ) solution Q = Q en ( L ) in N + f (3.3.6) where L = k − ◦ V with V = 2 πr sin θ (Φ +0 + Ψ) and k ( θ ) = V ( f ( θ ) , θ ) . Furthermore, thissolution Q satisfies || Q || ( − α, Γ + w )1 ,α, N + f ≤ C || Q en || ( − α,∂ Γ f )1 ,α, Γ f where C is a positive constant depending on ρ +0 , u +0 , r s , r , θ , α , δ and δ . Remark 3.16.
The existence of δ is obtained from the facts that ∇ × (Φ +0 e ϕ ) · e r = ρ +0 u +0 > in N + r s − δ and ν f · e r > on Γ f for f given in Lemma 3.2. Remark 3.17. If Ψ e ϕ satisfies (3.3.3) , then V = 2 πr sin θ (Φ +0 + Ψ) satisfies ∂ θ V > in N + f \ { θ = 0 } and ∂ θ ( V ( f ( θ ) , θ )) > for θ ∈ (0 , θ ) . This condition will be used to construct the stream surfaces of ∇ × ((Φ +0 + Ψ) e ϕ ) in N + f in theproof of Lemma 3.15.Proof of Lemma 3.15.
1. Construct the stream surfaces of ∇ × ((Φ +0 + Ψ) e ϕ ) in N + f .Let us define V := 2 πr sin θ (Φ +0 + Ψ). By the fact that Ψ e ϕ and Φ +0 e ϕ are axisymmetric func-tions in C ,α ( N + f ), V is an axisymmetric function in C ,α ( N + f ) (see Lemma 2.8). By (3.3.3), ∂ θ V > N + f \ { x = y = 0 } . Using these facts, we apply the implicit function theoremto V . Then we obtain that for any ( r ♯ , θ ♯ , ϕ ♯ ) ∈ N + , ∗ f , there exists a unique C surface θ = h ( r ♯ ,θ ♯ ,ϕ ♯ ) ( r, ϕ ) defined near ( r ♯ , ϕ ♯ ) such that θ ♯ = h ( r ♯ ,θ ♯ ,ϕ ♯ ) ( r ♯ , ϕ ♯ ) and V ( r, h ( r ♯ ,θ ♯ ,ϕ ♯ ) ( r, ϕ ) , ϕ ) = V ( r ♯ , θ ♯ , ϕ ♯ ). Since V is axisymmetric, this surface is axisymmetric. We denote θ = h ( r ♯ ,θ ♯ ,ϕ ♯ ) ( r, ϕ )by θ = h ( r ♯ ,θ ♯ ) ( r ).By V +0 = V +0 ( f ( θ ) , θ ) on Γ + w and (3.3.4), V = V ( f ( θ ) , θ ) on Γ + w . By this fact, V = 0on N + f ∩ { x = y = 0 } and ∂ θ V > N + f \ { x = y = 0 } , θ = h ( r ♯ ,θ ♯ ) ( r ) is defined untilit reaches Γ f or Γ ex not touching N + f ∩ { x = y = 0 } or Γ + w . Note that by the facts that ∂ θ ( V ( f ( θ ) , θ )) > θ ∈ (0 , θ ) and ∂ θ V ( r , θ ) > θ ∈ (0 , θ ) obtained from (3.3.3), thesurface θ = h ( r ♯ ,θ ♯ ) ( r ) intersects with Γ f and Γ ex once, respectively. Collect θ = h ( r ♯ ,θ ♯ ) ( r ) forall ( r ♯ , θ ♯ ) ∈ ( f ( θ ♯ ) , r ) × (0 , θ ), N + f ∩ { x = y = 0 } and Γ + w . Then we have the entire levelsurfaces of V in N + f . By ∂ θ V > N + f \ { x = y = 0 } , the values of V on distinct level surfacesof V in N + f are different from each other.2. Find a solution of (3.3.1), (3.3.2).By (2.4.5), (3.3.1) can be written as ∇ ⊥ V · ∇ Q = 0 . (3.3.7)Using this form of (3.3.1), it can be checked that 1) if Q is in C ( N + f ) and Q = constant onany curve on any level surface of V in N + f whose ϕ argument is fixed (in the case when a levelsurface of V is N + f ∩ { x = y = 0 } , Q = constant on N + f ∩ { x = y = 0 } ), then Q is a solutionof (3.3.1) and that 2) if Q is a C solution of (3.3.1), then Q = constant on any curve on anylevel surface of V in N + f whose ϕ argument is fixed. Denote the θ -argument of the intersectionpoints of Γ f and the level surface of V in N + f passing through x ∈ N + f by L (x). Since each level surface of V in N + f intersects with Γ f where the value of V on Γ f is equal to the value of V on the level surface, L (x) is given by L (x) := k − ◦ V (x)where k ( θ ) = V ( f ( θ ) , θ ).By 2), a C ( N + f ) ∩ C ( N + f ) solution of (3.3.1), (3.3.2) has the form Q ( r, θ, ϕ ) = ( Q en ( L ( r, θ ) , ϕ ) if ( r, θ ) ∈ [ f ( θ ) , r ] × (0 , θ ] Q en ( L ( r, ,
0) if ( r, θ ) ∈ [ f (0) , r ] × { } .Since Q en is axisymmetric, this can be written as Q = Q en ( L ) . Let Q = Q en ( L ). One can see that Q is a constant on any level surface of V in N + f and satisfies(3.3.2). Thus, by 1), if Q ∈ C ( N + f ) ∩ C ( N + f ), then Q is a C ( N + f ) ∩ C ( N + f ) solution of(3.3.1), (3.3.2).3. Estimate || Q || ( − α, Γ + w )1 ,α, N + f .Since L ∈ C ( N + f ) and Q en ∈ C (Γ f ), Q is in C ( N + f ). It directly follows from the definitionof Q || Q || , , N + f = || Q en || , , Γ f . Thus, to estimate || Q || ( − α, Γ + w )1 ,α, N + f , it is enough to estimate || DQ || (1 − α, Γ + w ) α, N + f . We estimate || DQ || (1 − α, Γ + w ) α, N + f .By direct computation, DQ = Q ′ en ( L ) D L = Q ′ en ( L ) DV ( ∂ r V ( f ( ϑ ) , ϑ ) f ′ ( ϑ ) + ∂ ϑ V ( f ( ϑ ) , ϑ )) | ϑ = L . To estimate || DQ || (1 − α, Γ + w ) α, N + f , we estimate || D L|| ,α, N + f . Write D L as DV π sin θ ( ∇ ⊥ V )( f ( ϑ ) , ϑ ) · ν f ( ϑ ) f ( ϑ ) r (cid:16) f ′ ( ϑ ) f ( ϑ ) (cid:17) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ϑ = L · sin θ sin L Here, we used the definition of ∇ ⊥ V and the spherical coordinate expression of ν f . Using (3.1.1)and (3.3.3), it is easily seen thatsup N + f (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( ∇ ⊥ V )( f ( ϑ ) , ϑ ) · ν f ( ϑ ) f ( ϑ ) s (cid:18) f ′ ( ϑ ) f ( ϑ ) (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ϑ = L (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) > c ∗∗ (3.3.8)where c ∗∗ is a positive constant depending on c ∗ , r s and δ . Using (3.1.1) and (3.3.5), it is alsoeasily seen that || DV π sin θ || ,α, N + f ≤ C (3.3.9)and || ( ∇ ⊥ V )( f ( ϑ ) , ϑ ) · ν f ( ϑ ) f ( ϑ ) s (cid:18) f ′ ( ϑ ) f ( ϑ ) (cid:19) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ϑ = L || ,α, N + f ≤ C || D L|| , , N + f (3.3.10)where Cs are positive constants depending on ρ +0 , u +0 , r s , α , δ and δ . By these three estimates,one can see that to estimate || D L|| ,α, N + f , it is enough to estimate || θ L || ,α, N + f . To estimate || D L|| ,α, N + f , we prove the following claim. -D AXISYMMETRIC TRANSONIC SHOCK 39 Claim.
There exists a positive constant C depending on ρ +0 , u +0 , r s , r , θ , α , δ and δ such that || θ L || ,α, N + f ≤ C. (3.3.11) Proof of Claim.
To simplify our argument, we assume that f = r s .First, we estimate || θ L || , , N + rs . By the definition of L , V ( r s , L ( r, θ )) = V ( r, θ )for all ( r, θ ) ∈ [ r s , r ] × [0 , θ ]. Write this as Z θ ( ρ +0 u +0 + ∇ × (Ψ e ϕ ) · e r )( r, ξ ) r sin ξdξ = Z L ( r,θ )0 ( ρ +0 u +0 + ∇ × (Ψ e ϕ ) · e r )( r s , ξ ) r s sin ξdξ. By (3.3.3), (3.3.5) and the first equation of (2.2.10), we have from this equation Z θ c ∗ r sin ξdξ ≤ Z L ( r,θ )0 ( m + δ r s ) sin ξdξ. Using sin θ θ ξ ≤ sin ξ ≤ ξ for ξ ∈ [0 , θ ], change sin ξ in the integrands in the left and right-handside of the above inequalities to sin θ θ ξ and ξ , respectively, and then integrate the resultantterms. Then we obtain sin θ θ c ∗ r θ ≤ ( m + δ r s ) L ( r, θ )2 . From this inequality, we have s sin θ θ c ∗ r s m + δ r s θ ≤ L ( r, θ ) . This holds for all ( r, θ ) ∈ [ r s , r ] × [0 , θ ]. Hence, || θ L || , , N + rs ≤ C (3.3.12)where C is a positive constant depending on ρ +0 , u +0 , r s , θ , δ and c ∗ .Next, we estimate [ θ L ] ,α, N + rs . By (3.3.12), we can obtain an estimate of [ θ L ] ,α, N + rs by obtaining (cid:12)(cid:12)(cid:12)(cid:12) θ L ( r, θ ) − θ ′ L ( r ′ , θ ′ ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Cε α (3.3.13)for all ( r, θ ), ( r ′ , θ ′ ) ∈ [ r s , r ] × [0 , θ ] satisfying ε ≤ ε for a positive constant ε and a positiveconstant C where ε := p | r ′ − r | + | θ ′ − θ | . We obtain this estimate. Hereafter, to simplifyour notation, we denote L ( r ′ , θ ′ ), L ( r, θ ) and Φ +0 + Ψ by L ′ , L and Φ, respectively.By the definition of L , V ( r s , L ′ ) − V ( r s , L ) = V ( r ′ , θ ′ ) − V ( r, θ ) . Write this as(3.3.14) Z L ′ L ( ∇ × (Φ e ϕ ) · e r )( r s , ξ ) r s sin ξdξ = Z θ ′ θ ( ∇ × (Φ e ϕ ) · e r )( r, ξ ) r sin ξdξ + Z r ′ r ( ∇ × (Φ e ϕ ))( t, θ ′ ) · e θ ( θ ′ ) tsinθ ′ dt. Using (3.3.14), we find an upper bound of L ′ . Using (3.3.5), the fact that Φ +0 e ϕ ∈ C ∞ ( N + r s )and sin θ θ ξ ≤ sin ξ ≤ ξ for ξ ∈ [0 , θ ], we have for L ′ ≥ L Z L ′ L (( ∇ × (Φ e ϕ ) · e r )( r s , L ) − C ( ξ − L ) α ) r s sin θ θ ξdξ ≤ ( ∇ × (Φ e ϕ ) · e r )( r, θ ) r θ ′ − θ Cr (cid:18) α + 1 | θ ′ − θ | α +1 θ ′ − α + 1)( α + 2) | θ ′ − θ | α +2 (cid:19) + ( ∇ × (Φ e ϕ ) · e θ )( r, θ ′ ) sin θ ′ r ′ − r C sin θ ′ (cid:18) α + 1 | r ′ − r | α +1 r ′ − α + 1)( α + 2) | r ′ − r | α +2 (cid:19) (=: ( u ))where the right-hand side is an upper bound of the right-hand side of (3.3.14) and C is a positiveconstant depending on ρ +0 , u +0 , N + r s , α and δ . From this inequality, we get( ∇ × (Φ e ϕ ) · e r )( r s , L ) sin θ θ r s L ′ − L − Cr s sin θ θ (cid:18) α + 1 ( L ′ − L ) α +1 L ′ − α + 1)( α + 2) ( L ′ − L ) α +2 (cid:19) ≤ ( u ) . Using this inequality, we have that for each ( r, θ ) ∈ [ r s , r ] × [0 , θ ], there exists positive constants ε ( r,θ ) < C ( r,θ ) such that for any ( r ′ , θ ′ ) satisfying ε ≤ ε ( r,θ ) , L ′ ≤ θ ′ (cid:18) L θ + C ( r,θ ) ε α (cid:19) . Note that when ( r, θ ) ∈ [ r s , r ] × { } , there exists a positive constant ε ( r,θ ) < r ′ , θ ′ ) satisfying ε ≤ ε ( r,θ ) , L ′ ≤ θ ′ s r ( ∇ × (Φ e ϕ ) · e r )( r, r s ( ∇ × (Φ e ϕ ) · e r )( r s ,
0) + C ( r,θ ) ε α ! . Let ε = inf ( r,θ ) ∈ [ r s ,r ] × [0 ,θ ] ε ( r,θ ) and C = sup ( r,θ ) ∈ [ r s ,r ] × [0 ,θ ] C ( r,θ ) . Then by the above state-ment, L ′ θ ′ ≤ L θ + Cε α for all ( r, θ ), ( r ′ , θ ′ ) satisfying ε ≤ ε . This gives θ L − θ ′ L ′ ≤ Cε α for all ( r, θ ), ( r ′ , θ ′ ) satisfying ε ≤ ε for a positive constant C .Similarly, we can obtain − Cε α ≤ θ L − θ ′ L ′ for all ( r, θ ), ( r ′ , θ ′ ) satisfying ε ≤ ε for a positive constant C and a positive constant ε . Let ε = min( ε , ε ). Choose ε ( r,θ ) so that it can depend on ρ +0 , u +0 , N + r s , α and δ . Then ε dependson ρ +0 , u +0 , N + r s , α and δ . In the same way, we have ε depends on ρ +0 , u +0 , N + r s , α and δ .Thus, ε depends on ρ +0 , u +0 , N + r s , α and δ . This finishes the proof of Claim. (cid:3) By (3.3.8)-(3.3.11), we have || D L|| ,α, N + f ≤ C for a positive constant C . It can be easily shown that there exists a positive constant C suchthat | θ − θ | ≤ C |L ( r, θ ) − θ | -D AXISYMMETRIC TRANSONIC SHOCK 41 for all ( r, θ ) ∈ [ f ( θ ) , r ] × [0 , θ ]. Using these two estimates, estimate || DQ || (1 − α, Γ + w ) α, N + f . Then weobtain || DQ || (1 − α, Γ + w ) α, N + f ≤ C where C is a positive constant depending on ρ +0 , u +0 , r s , r , α , δ and δ .4. By the result in Step 3, Q is a C ( N + f ) ∩ C ( N + f ) solution of (3.3.1), (3.3.2). By 2), a C ( N + f ) ∩ C ( N + f ) solution of (3.3.1), (3.3.2) is unique. This finishes the proof. (cid:3) Proof of Proposition 3.1.
Using the results in § § Proof of Proposition 3.1 (Existence).
Suppose that ( ρ − , u − , p − , p ex , f ′ s ) is as in Problem 3 for α ∈ ( ,
1) and σ ∈ (0 , σ ] where σ is a positive constant to be determined later. For the same α and σ and M > P ( M ) = { ( f (0) , Ψ e ϕ ) ∈ R × C ,α ( − − α, Γ + w,rs + fs ) ( N + r s + f s ) | Ψ e ϕ = f ( θ )Ψ( f ( θ ) , θ ) r e ϕ on Γ + w,r s + f s , | f (0) − r s | + || Ψ e ϕ || ( − − α, Γ + w,rs + fs )2 ,α, N + rs + fs ≤ M σ } . By the definition of P ( M ), P ( M ) is a compact convex subset of R × C , α ( − − α , Γ + w,rs + fs ) ( N + r s + f s ).We will prove the existence part of Proposition 3.1 by constructing a continuous map of P ( M )into itself as a map from R × C , α ( − − α , Γ + w,rs + fs ) ( N + r s + f s ) to R × C , α ( − − α , Γ + w,rs + fs ) ( N + r s + f s ) andapplying the Schauder fixed point theorem.In this proof, C s and C i for i = 1 , , . . . denote positive constants depending on the whole ora part of the data, δ , δ , δ , δ and δ unless otherwise specified. Each C in different situationsdiffers from each other.1. For a fixed ( f (0) , Ψ e ϕ ), solve (B ′ ).Take ( f (0) ∗ , ˜Ψ ∗ e ϕ ) ∈ P ( M ). By the definition of f s and the assumption that f ′ s satisfies(3.0.1), || f s || ( − − α,∂ Λ)2 ,α, Λ ≤ C σ. (3.4.1)Choose σ = min( δ M , δ C )(=: σ (1)3 ) so that | f (0) ∗ − r s | ≤ δ || f s || ( − − α,∂ Λ)2 ,α, Λ ≤ δ . (3.4.2)And then, using Π f (0) ∗ + f s r s + f s , extend ˜Ψ ∗ e ϕ to a function in N + f (0) ∗ + f s :Ψ ∗ e ϕ := W ∗ πr sin θ e ϕ where W ∗ := ˜ W ∗ (Π f (0) ∗ + f s r s + f s ) with ˜ W ∗ := 2 πr sin θ ˜Ψ ∗ and Π ab for 0 < a, b < r is a mapfrom N + a to N + b defined in Step 1 in the proof of Lemma 3.9. By (3.4.2) and the fact that || ˜Ψ ∗ e ϕ || ( − − α, Γ + w,rs + fs )2 ,α, N + rs + fs ≤ M σ , ||∇ × (Ψ ∗ e ϕ ) || ( − α, Γ + w,f (0) ∗ + fs )1 ,α, N + f (0) ∗ + fs ≤ C M σ. (3.4.3)Choose σ = min( δ C M , σ (1)3 )(=: σ (2)3 ) where δ is a positive constant given in Lemma 3.15 sothat Ψ e ϕ = Ψ ∗ e ϕ satisfies (3.3.5) for f = f (0) ∗ + f s . For ( f (0) , Ψ e ϕ ) = ( f (0) ∗ , Ψ ∗ e ϕ ), solve (B ′ ) : solve ( ∇ × ((Φ +0 + Ψ ∗ ) e ϕ ) · ∇ A = 0 in N + f (0) ∗ + f s ,A = A en,f (0) ∗ + f s on Γ f (0) ∗ + f s , (3.4.4) ( ∇ × ((Φ +0 + Ψ ∗ ) e ϕ ) · ∇ T = 0 in N + f (0) ∗ + f s ,T = T en,f (0) ∗ + f s on Γ f (0) ∗ + f s (3.4.5)where A en,f (0) ∗ + f s and T en,f (0) ∗ + f s are A en,f (0)+ f s and T en,f (0)+ f s given in (3.1.18) and (3.1.19),respectively, for f (0) = f (0) ∗ . By ( ρ − , u − , p − , f ′ s ) ∈ ( C ,α ( N )) × C ,α ( − α, { θ = θ } ) , ((0 , θ )), A en,f (0) ∗ + f s ∈ C ,α ( − − α,∂ Γ f (0) ∗ + fs ) (Γ f (0) ∗ + f s ) and T en,f (0) ∗ + f s ∈ C ,α ( − α,∂ Γ f (0) ∗ + fs ) (Γ f (0) ∗ + f s ) . Take σ = min( σ (2)3 , δ )(=: σ (3)3 ) where δ is a positive constant given in Lemma 3.4 so that ( ρ − , u − , p − )satisfies (2.3.3) for σ ≤ δ . Then by Lemma 3.4, (3.0.1) and the fact that f (0) ∗ satisfies | f (0) ∗ − r s | ≤ M σ , || T en,f ∗ || ( − α,∂ Γ f ∗ )1 ,α, Γ f ∗ ≤ CM σ + Cσ (3.4.6)where f ∗ := f ∗ (0) + f s . Apply Lemma 3.15 to (3.4.4), (3.4.5). Then we obtain that each (3.4.4)and (3.4.5) has a unique axisymmetric C ( N + f ) ∩ C ( N + f ) solution A ∗ = A en,f (0) ∗ + f s ( L ∗ ) and T ∗ = T en,f (0) ∗ + f s ( L ∗ ) , respectively, where L ∗ is L defined in Lemma 3.15 for V = 2 πr sin θ (Φ +0 +Ψ ∗ ) and f = f (0) ∗ + f s .Furthermore, we have that T ∗ satisfies || T ∗ || ( − α, Γ + w,f ∗ )1 ,α, N + f ∗ ≤ CM σ + Cσ (3.4.7)where we used (3.4.6). Using the solution expression of A ∗ , estimate A ∗ πr sin θ e ϕ in C ,α ( N + f ∗ ).Using (3.1.18), write A ∗ πr sin θ e ϕ as2 πf ∗ ( L ∗ ) sin( L ∗ ) u − ,ϕ ( L ∗ )2 πr sin θ e ϕ (3.4.8)where u − ,ϕ = u − · e ϕ . Using arguments similar to the ones in the proof of Claim in Lemma3.15, we can obtain || L ∗ θ || ,α, ( f ∗ ( θ ) ,r ) × (0 ,θ ) ≤ C. With this estimate, ||L ∗ || ,α, ( f ∗ ( θ ) ,r ) × (0 ,θ ) ≤ C , (2.3.3) and (3.4.2), we estimate (3.4.8) in C ,α ( N + f ∗ ). Then we have || A ∗ πr sin θ e ϕ || ,α, N + f ∗ ≤ Cσ. (3.4.9)2. By substituting an extension of (Ψ ∗ e ϕ , A ∗ , T ∗ ) into nonlinear parts of (A ′ ), obtain a linearproblem having unknowns ( f (0) , Ψ e ϕ ).Let f (0) ♯ be a point in [ r s − δ , r s + δ ]. By the choice of σ , || f ♯ − r s || ( − − α,∂ Λ)2 ,α, Λ ≤ δ where f ♯ := f (0) ♯ + f s . Extend Ψ ∗ e ϕ A ∗ and T ∗ to functions in N + f ♯ :Ψ ♯, ∗ e ϕ := W ♯, ∗ πr sin θ e ϕ , A ♯, ∗ := A ∗ (Π f ♯ f ∗ ) and T ♯, ∗ := T ∗ (Π f ♯ f ∗ )where W ♯, ∗ = W ∗ (Π f ♯ f ∗ ). By (3.4.3), (3.4.7) and (3.4.9), ||∇ × (Ψ ♯, ∗ e ϕ ) || ( − α, Γ + w,f♯ )1 ,α, N + f♯ + || A ♯, ∗ πr sin θ e ϕ || ,α, N + f♯ + || T ♯, ∗ || ( − α, Γ + w,f♯ )1 ,α, N + f♯ ≤ C M σ + C σ (3.4.10) -D AXISYMMETRIC TRANSONIC SHOCK 43 for all f ♯ (0) ∈ [ r s − δ , r s + δ ]. Take σ = min( σ (3)3 , δ C M + C )(=: σ (4)3 ) where δ is a positiveconstant given in Lemma 3.3 so that ̺ ( ∇ × ((Φ +0 + Ψ ♯, ∗ ) e ϕ ) , A ♯, ∗ πr sin θ e ϕ , S +0 + T ♯, ∗ ) is well-definedin N + f ♯ and ̺ ( ∇ × ((Φ +0 + Ψ ♯, ∗ ) e ϕ ) , A ♯, ∗ πr sin θ e ϕ , S +0 + T ♯, ∗ ) and S +0 + T ♯, ∗ are strictly positive in N + f ♯ for all f ♯ (0) ∈ [ r s − δ , r s + δ ]. By substituting f (0) ♯ and (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) into the placeof f (0) in (A ′ ) and (Ψ e ϕ , A, T ) in F (Ψ e ϕ , A, T ), f ( T, p ex ) and f (Ψ e ϕ , A, T ) in (A ′ ), we obtain(3.4.11) ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (Ψ e ϕ ) ! = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ Tr e ϕ + F (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) in N + f ♯ , Ψ e ϕ = (Φ − − Φ − ) e ϕ on Γ f ♯ , r (Φ − − Φ − )( r ,θ ) r e ϕ on Γ + w,f ♯ := Γ w ∩ { r > f ♯ } , (cid:16) r sin θ R θ (cid:0) f ( T ♯, ∗ , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) (cid:19) r sin ξdξ (cid:19) e ϕ on Γ ex , (3.4.12)(3.4.13) 1 r sin θ Z θ (cid:18) f ( T ♯, ∗ , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T + f (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ = r (Φ − − Φ − )( r , θ ) r . We let f (0) ♯ be an unknown in this problem and let T in (3.4.11)-(3.4.13) be a solution of ( ∇ × ((Φ +0 + Ψ ♯, ∗ ) e ϕ ) · ∇ T = 0 in N + f ♯ ,T = T en,f ♯ on Γ f ♯ . (3.4.14)Then since a solution T of (3.4.14) is uniquely determined by f (0) ♯ (see Step 3), unknownsof (3.4.11)-(3.4.13) become ( f ♯ (0) , Ψ e ϕ ). We denote an unknown Ψ e ϕ of (3.4.11)-(3.4.13) byΨ ♯ e ϕ .3. Find f (0) ♯ using (3.4.13).By the definition of Ψ ♯, ∗ e ϕ and (2.4.5), the transport equation in (3.4.14) can be written as ∇ ⊥ ( V +0 + W ♯, ∗ ) · ∇ T = 0 in N + f ♯ . From this form of the transport equation in (3.4.14), we see that the stream surface of the vectorfield ∇ × ((Φ +0 + Ψ ♯, ∗ ) e ϕ ) in N + f ♯ is obtained by stretching or contracting the stream surface ofthe vector field ∇ × ((Φ +0 + Ψ ∗ ) e ϕ ) in N + f ∗ in r -direction. Using this fact, we obtain that thesolution of (3.4.14) is given by T = T en,f ♯ ( L ♯ )where L ♯ = L ∗ (Π f ♯ f ∗ ). We denote this solution by T ♯ .By (3.1.19), T ♯ is expressed as T ♯ = g u − · ν f ♯ ( L ♯ ) c − ! S − ( f (0) ♯ + f s ( L ♯ ) , L ♯ ) − ( g ( M − ))( r s ) S in . Substituting this expression of T ♯ into the place of T in (3.4.13) using the fact that Π f ♯ f ∗ ( r , θ ) =( r , θ ), we obtain( L ) := 1 r sin θ Z θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = r ( a ) r sin ξdξ (3.4.15) = − r (Φ − − Φ − )( r , θ ) r + 1 r sin θ Z θ ( f ( T ∗ , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 (( a ) + ( a ) ) + f (Ψ ∗ e ϕ , A ∗ , T ∗ )) | r = r r sin ξdξ =: ( R )where ( a ) = ( g ( M − ))( f (0) ♯ + f s ( L ∗ ( r , θ ))) S in − ( g ( M − ))( r s + f s ( L ∗ ( r , θ ))) S in , ( a ) = g (cid:18) u − · ν f ♯ ( L ∗ ( r , θ )) c − (cid:19) ! S − ! ( f (0) ♯ + f s ( L ∗ ( r , θ )) , L ∗ ( r , θ )) − ( g ( M − ))( f (0) ♯ + f s ( L ∗ ( r , θ ))) S in , ( a ) = ( g ( M − ))( r s + f s ( L ∗ ( r , θ ))) S in − ( g ( M − ))( r s ) S in . We find f (0) ♯ satisfying (3.4.15). For this, we estimate | ( R ) | .Estimate | ( R ) | : With (2.3.3) and (3.0.1), we estimate ( a ) , ( a ) and r (Φ − − Φ − )( r ,θ ) r . Thenwe obtain sup θ ∈ (0 ,θ ) | ( a ) | ≤ Cσ | f (0) ♯ − r s | + Cσ ≤ C δ σ + Cσ (3.4.16)for all f ♯ (0) ∈ [ r s − δ , r s + δ ], sup θ ∈ (0 ,θ ) | ( a ) | ≤ Cσ (3.4.17)and | r (Φ − − Φ − )( r , θ ) r | ≤ Cσ. (3.4.18)With (2.3.4), we estimate f ( T ∗ , p ex )( r , θ ). Then we havesup θ ∈ (0 ,θ ) | f ( T ∗ , p ex )( r , θ ) | ≤ Cσ. (3.4.19)By Lemma 3.3 and using (3.4.3), (3.4.7) and (3.4.9), we estimate f (Ψ ∗ e ϕ , A ∗ , T ∗ )( r , θ ). Thenwe have sup θ ∈ (0 ,θ ) | f (Ψ ∗ e ϕ , A ∗ , T ∗ )( r , θ ) | ≤ C ( M + 1) σ . (3.4.20)With (3.4.16)-(3.4.20), we estimate | ( R ) | . Then we get | ( R ) | ≤ C ( M + 1) σ + C σ (3.4.21)for all f ♯ (0) ∈ [ r s − δ , r s + δ ].Then we find f (0) ♯ . By Lemma 2.10, there exists a positve constant λ such that( L ) ′ ( f (0) ♯ ) ≥ λ for all f ♯ (0) ∈ [ r s − δ , r s + δ ]. By this fact and ( L )( r s ) = 0,( L )( δ ≥ δ λ L )( − δ ≤ − δ λ . (3.4.22) -D AXISYMMETRIC TRANSONIC SHOCK 45 Choose σ = min( σ (4)3 , C C ( M +1) , δ λ C )(=: σ (5)3 ). Then since | ( R ) | ≤ δ λ f (0) ♯ ∈ [ r s − δ , r s + δ ] and (3.4.22) holds, by the intermediate value theorem, there exists f (0) ♯ satisfying (3.4.15) in [ r s − δ , r s + δ ]. Such f (0) ♯ is unique because ( L ) is a monotonefunction of f (0) ♯ . Since at f (0) ♯ where (3.4.15) holds there holds λ | f (0) ♯ − r s | ≤ | ( R )( f (0) ♯ ) | , there holds | f (0) ♯ − r s | ≤ C ( M + 1) σ + Cσ. (3.4.23)4. Find Ψ ♯ e ϕ .Fix f (0) ♯ obtained in Step 3. By (3.4.23), Lemma 3.4 and Lemma 3.15, T ♯ determined by f (0) ♯ satisfies || T ♯ || ( − α, Γ + w,f♯ )1 ,α, N + f♯ ≤ C ( M + 1) σ + Cσ. (3.4.24)Substitute this T ♯ into the place of T in (3.4.11) and (3.4.12). Then we obtain ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (Ψ e ϕ ) ! (3.4.25) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ T ♯ r e ϕ + F (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) in N + f ♯ , Ψ e ϕ = (Φ − − Φ − ) e ϕ on Γ f ♯ , r (Φ − − Φ − )( r ,θ ) r e ϕ on Γ + w,f ♯ , (cid:16) r sin θ R θ (cid:0) f ( T ♯, ∗ , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 T ♯ + f (Ψ ♯, ∗ e ϕ , A ♯, ∗ , T ♯, ∗ ) (cid:19) r sin ξdξ (cid:19) e ϕ on Γ ex . (3.4.26)Since f (0) ♯ is chosen for T = T ♯ to satisfy (3.4.13), (3.4.26) is a continuous boundary condition.We apply Lemma 3.5 to (3.4.25), (3.4.26) with (2.3.3), (2.3.4), (3.0.1), (3.4.10) and (3.4.24).Then we obtain that (3.4.25), (3.4.26) has a unique C ,α ( − − α, Γ f♯ ) ( N + f ♯ ) solution Ψ ♯ e ϕ and thissolution satisfies the estimate || Ψ ♯ e ϕ || ( − − α, Γ + w,f♯ )2 ,α, N + f♯ ≤ C ( M + 1) σ + Cσ. (3.4.27)5. Using Π r s + f s f ♯ , transform Ψ ♯ e ϕ into a function in N + r s + f s :˜Ψ ♯ e ϕ = ˜ W ♯ πr sin θ e ϕ where ˜ W ♯ = W ♯ (Π r s + f s f ♯ ) with W ♯ = 2 πr sin θ Ψ ♯ . By (3.4.23) and (3.4.27), || ˜Ψ ♯ e ϕ || ( − − α, Γ + w,rs + fs )2 ,α, N + rs + fs ≤ C ( M + 1) σ + Cσ.
Combining this with (3.4.23), we have | f (0) ♯ − r s | + || ˜Ψ ♯ e ϕ || ( − − α, Γ + w,rs + fs )2 ,α, N + rs + fs ≤ C ( M + 1) σ + C σ. Take M = 2 C and σ = min( σ (5)3 , C C ( M +1) )(=: σ (6)3 ). And then define a map J from R × C , α ( − − α , Γ + w,rs + fs ) ( N + r s + f s ) to R × C , α ( − − α , Γ + w,rs + fs ) ( N + r s + f s ) by J ( f (0) ∗ , ˜Ψ ∗ e ϕ ) = ( f (0) ♯ , ˜Ψ ♯ e ϕ ) . By the choice of M and σ , J is a map of P ( M ) into itself. Using the standard argument, onecan easily check that J is continuous. Thus, the Schauder fixed point theorem can be appliedto J . We apply the Schauder fixed point theorem to J . Then we obtain that there exists afixed point ( f ♭ (0) , ˜Ψ ♭ e ϕ ) ∈ P ( M ) of J .One can see that if ( f (0) ∗ , ˜Ψ ∗ e ϕ ) = ( f (0) ♯ , ˜Ψ ♯ e ϕ ), then Ψ ∗ e ϕ = Ψ ♯, ∗ e ϕ = Ψ ♯ e ϕ , A ♯, ∗ = A ∗ and T ∗ = T ♯, ∗ = T ♯ . From this fact, we see that ( f (0) , Φ e ϕ , L, S ) = ( f (0) ♭ , (Φ +0 +Ψ ♭ ) e ϕ , A ♭ , S +0 + T ♭ ),where Ψ ♭ e ϕ := (2 πr sin θ ˜Ψ ♭ )(Π f (0) ♭ + fsrs + fs )2 πr sin θ e ϕ and A ♭ and T ♭ are solutions of (3.4.4) and (3.4.5) forgiven ( f (0) ∗ , Ψ ∗ ) = ( f (0) ♭ , Ψ ♭ ), respectively, is a solution of (A), (B). By ( f (0) ♭ , ˜Ψ ♭ e ϕ ) ∈ P ( M ),(3.4.7) and (3.4.9), | f (0) ♭ − r s | + ||∇ × (Ψ ♭ e ϕ ) || ( − α, Γ + w,f♭ )1 ,α, N + f♭ + || A ♭ πr sin θ || ,α, N + f♭ + || T ♭ || ( − α, Γ + w,f♭ )1 ,α, N + f♭ ≤ Cσ (3.4.28)where f ♭ := f (0) ♭ + f s . One can easily see that there exists a positive constant δ such thatfor any N + f ⊂ N + r s − δ , if ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S, B ) ∈ B (1) δ , N + f where B (1) δ, Ω for δ > ⊂ R is a neighborhood of ( ∇ × (Φ +0 e ϕ ) , , S +0 , B ) defined in Lemma 2.14, then (2.5.9) holdsin N + f . Let δ be such a constant. Take σ = min( σ (6)3 , δ C ) where C is C in (3.4.28). Then( f (0) ♭ , (Φ +0 + Ψ ♭ ) e ϕ , A ♭ , S +0 + T ♭ ) is a subsonic solution of (2.5.1)-(2.5.3) in N + f ♭ . Choose δ , δ , δ , δ , δ and δ so that they can depend on the data. Then M and σ depend on the data.This finishes the proof. (cid:3) Proof of Proposition 3.1 (Uniqueness).
Let α ∈ ( , σ be a positive constant ≤ σ andto be determined later. Suppose that there exist two solutions ( f i (0) , Φ i e ϕ , L i , S i ) for i = 1 , σ ≤ σ satisfying the estimate (3.0.2).Let (Ψ i , A i , T i ) := (Φ i − Φ +0 , L i , S i − S +0 ) for i = 1 ,
2. We will prove that there exists a positiveconstant σ such that if σ = σ , then( f (0) , Ψ e ϕ , A , T ) = ( f (0) , Ψ e ϕ , A , T )by constructing a contraction map in a low regularity space.In this proof, C s denote positive constants depending on the data unless otherwise specified.Each C in different situations differs from each other.1. By subtracting (A ′ ) satisfied by ( f (0) , Ψ e ϕ , A , T ) from (A ′ ) satisfied by ( f (0) , Ψ e ϕ ,A , T ), obtain the equations that will give a contraction map.Let f i := f i (0) + f s for i = 1 ,
2. From( ∇ ⊥ a )(Π f f ) = N ∇ ⊥ ( a (Π f f )) and ∇ × ( a e ϕ πr sin θ ) = ∇ ⊥ a for an axisymmetric scalar function a , we can obtain(3.4.29) ( ∇ × (Ψ e ϕ ))(Π f f ) = N ∇ × ( ˜Ψ e ϕ )and (cid:18) ∇ × (cid:18) A πr sin θ e ϕ (cid:19)(cid:19) (Π f f ) = N ∇ × ˜ A πr sin θ e ϕ ! , where ˜Ψ := ˜ W πr sin θ with ˜ W := W (Π f f ) and W := 2 πr sin θ Ψ , ˜ A := A (Π f f ) and N = r (Π rf f ) e r ⊗ e r − (cid:16) ∂ ˜ θ Π rf f (cid:17) (Π f f )(Π rf f ) r e r ⊗ e θ + (cid:16) ∂ ˜ r Π rf f (cid:17) (Π f f )Π rf f r e θ ⊗ e θ -D AXISYMMETRIC TRANSONIC SHOCK 47 with Π rf f and Π rf f , r -components of Π ∗ f f and Π ∗ f f , respectively (see the definition of Π ∗ ab in (3.2.29)), (˜ r, ˜ θ ), ( r, θ ) coordinates for the cartesian coordinate for N + f (0)+ f s , and ( r, θ ) =Π ∗ f f (˜ r, ˜ θ ). Using Π f f and (3.4.29), transform (A ′ ) satisfied by ( f (0) , Ψ e ϕ , A , T ). Andthen subtract the resultant equations from (A ′ ) satisfied by ( f (0) , Ψ e ϕ , A , T ). Then weobtain(3.4.30) ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ((Ψ − ˜Ψ ) e ϕ ) ! = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ( T − ˜ T ) r e ϕ + F in N + f (0)+ f s (Ψ − ˜Ψ ) e ϕ = (Φ − − Φ − ) e ϕ − Π rf f (Φ − − Φ − )(Π f f ) r e ϕ (=: h ) on Γ f (0)+ f s , on Γ + w,f (0)+ f s := Γ w ∩ { r > f (0) + f s } , (cid:18) r sin θ R θ (cid:18) f ( T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( T − ˜ T )+ f (Ψ e ϕ , A , T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:19) r sin ξdξ (cid:19) e ϕ (=: h ) on Γ ex , (3.4.31)(3.4.32) 0 = 1 r sin θ Z θ (cid:18) f ( T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( T − ˜ T )+ f (Ψ e ϕ , A , T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ where F = − ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ( ˜Ψ e ϕ ) ! + M ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ! (Π f f ) N ∇ × ( ˜Ψ e ϕ ) ! + ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) 1 r ∂ θ ˜ T e ϕ − ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) 1 r ! (Π f f ) (cid:18) ∂ Π rf f ∂ ˜ θ (Π f f ) ∂ r + ∂ θ (cid:19) ˜ T e ϕ + F (Ψ e ϕ , A , T ) − ˜ F ( ˜Ψ e ϕ , ˜ A , ˜ T ) , ˜ T = T (Π f f ), M = (cid:0) ∂ ˜ r Π rf f (cid:1) (Π f f ) e r ⊗ e r + (cid:16) ∂ ˜ θ Π rf f (cid:17) (Π f f )Π rf f e θ ⊗ e r + r Π rf f e θ ⊗ e θ + r Π rf f e ϕ ⊗ e ϕ and ˜ F is F changed by using the transformation Π f f . We will construct a contraction mapusing (3.4.30)-(3.4.32). For this, we estimate || T − ˜ T || ,β, N + f for β = 1 − q ∈ (0 , α ) where q ∈ (3 , − α ) and || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f .2. Estimate || T − ˜ T || ,β, N + f .Since T i for i = 1 , ∇ × ((Φ +0 + Ψ i ) e ϕ ) · ∇ T = 0 in N + f i (0)+ f s , T = T en,f i (0)+ f s on Γ f i (0)+ f s for i = 1 ,
2, respectively, where T en,f i (0)+ f s is T en,f (0)+ f s given in (3.1.19) for f (0) = f i (0), byLemma 3.15, T i = T en,f i (0)+ f s ( L i )(3.4.33)for i = 1 , L i are L given in Lemma 3.15 for V = V +0 + W i (=: V i ) and f = f i (0) + f s with W i := 2 πr sin θ Ψ i . By (3.1.19), (3.4.33) and the definition of ˜ T , T − ˜ T can be written as g (cid:18) u − · ν f ( L ) c − (cid:19) ! S − ! ( f ( L ) , L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L )where ν f i for i = 1 , f i pointing toward N + f i , respectively, and˜ L := L (Π f f ). This can be decomposed into g (cid:18) u − · ν f ( L ) c − (cid:19) ! S − ! ( f ( L ) , L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( L ) , L )+ g u − · ν f ( ˜ L ) c − ! S − ( f ( L ) , L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L )=: ( a ) + ( b ) . To estimate T − ˜ T in C β ( N + f ), we estimate ( a ) and ( b ) in C β ( N + f ), respectively. To obtainan estimate of ( a ) in C β ( N + f ) and later to obtain an estimate of || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f ,we prove the following claim.We take σ = min( σ , δ C )(=: σ (1)3 ) where C is C in (3.0.2) so that Ψ e ϕ = Ψ i e ϕ for i = 1 , f = f i (0) + f s . Claim.
Let h : [0 , θ ] → R be a function in C ,α ( − α, { θ = θ } ) ((0 , θ )) . There holds || Z h ′ ( t L + (1 − t ) ˜ L ) dt ( L − ˜ L ) || ,β, N + f ≤ C || h || ( − α, { θ = θ } )1 ,β, (0 ,θ ) || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . Proof of Claim.
By the definitions of L and ˜ L , L = k − ◦ V and ˜ L = ( k − ◦ V )(Π f f )where k i ( θ ) = V i ( f i ( θ ) , θ ) for i = 1 ,
2. By V +0 (Π f f ) = V +0 and ˜ W ( f ( θ ) , θ ) = W ( f ( θ ) , θ )where ˜ W is defined below (3.4.29), ˜ L can be written as˜ L = ˜ k − ◦ ˜ V where ˜ V = V +0 + ˜ W and ˜ k = ˜ V ( f ( θ ) , θ ). Since ˜ V ( r, θ ) ∈ [0 , V − ( r , θ )] for ( r, θ ) ∈ [ f ( θ ) , r ] × [0 , θ ] and V ( f ( θ ) , θ ) ∈ [0 , V − ( r , θ )] for θ ∈ [0 , θ ] where V − = 2 πr sin θ Φ − , k − ◦ ˜ V is well-defined in N + f . With this fact, we write L − ˜ L as L − ˜ L = ( k − ◦ V − k − ◦ ˜ V ) + ( k − ◦ ˜ V − ˜ k − ◦ ˜ V )= Z k ′ ( k − ◦ ( tV + (1 − t ) ˜ V )) dt ( V − ˜ V )+ Z k ′ (˜ k − ◦ ( t ˜ k ( ϑ ) + (1 − t ) k ( ϑ ))) dt (˜ k ( ϑ ) − k ( ϑ ))where ϑ = k − ◦ ˜ V and we used the fact that k − ◦ ˜ V − ˜ k − ◦ ˜ V = ˜ k − ◦ ˜ k ( ϑ ) − ˜ k − ◦ k ( ϑ ).Substitute the above expression of L − ˜ L into the place of L − ˜ L in R h ′ ( t L +(1 − t ) ˜ L ) dt ( L − -D AXISYMMETRIC TRANSONIC SHOCK 49 ˜ L ). Then we have Z h ′ ( t L + (1 − t ) ˜ L ) dt ( L − ˜ L )= Z h ′ ( t L + (1 − t ) ˜ L ) dt Z k ′ ( k − ◦ ( tV + (1 − t ) ˜ V )) dt ( V − ˜ V )+ Z h ′ ( t L + (1 − t ) ˜ L ) dt Z k ′ (˜ k − ◦ ( t ˜ k ( ϑ ) + (1 − t ) k ( ϑ ))) dt (˜ k ( ϑ ) − k ( ϑ ))=: ( c ) + ( d ) . To estimate || R h ′ ( t L + (1 − t ) ˜ L ) dt ( L − ˜ L ) || ,β, N + f , we estimate ( c ) and ( d ) in C β ( N + f ),respectively.Estimate ( c ) in C β ( N + f ): With the definitions of k , Ψ and ˜Ψ , write ( c ) as(3.4.34) ( c ) = Z ( θ − θ ) h ′ ( t L + (1 − t ) ˜ L ) dt Z r sin θJ sin( k − ◦ ( tV + (1 − t ) ˜ V )) dt Z ∂ θ (Ψ − ˜Ψ )( r, tθ + (1 − t ) θ ) dt where J := f s (cid:18) f ′ f (cid:19) ( · )( ∇ × ((Φ +0 + Ψ ) e ϕ ))( f ( · ) , · ) · ν f ( · ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · = k − ◦ ( tV +(1 − t ) ˜ V ) . By the choice of σ , Ψ e ϕ = Ψ i e ϕ for i = 1 , f = f i (0) + f s . By this fact and || f − r s || ( − − α,∂ Λ)2 ,α, Λ ≤ δ and the fact that k − ◦ ( tV + (1 − t ) ˜ V ) maps N + f to [0 , θ ], J > ˜ c ∗ in N + f (3.4.35)for all t ∈ [0 ,
1] for some positive constant ˜ c ∗ . Using arguments similar to the ones in the proofof Claim in Lemam 3.15, we can obtain || θ − θ ( t L + (1 − t ) ˜ L ) − θ || ,β, ( f ( θ ) ,r ) × (0 ,θ ) ≤ C (3.4.36)and || θk − ◦ ( tV + (1 − t ) ˜ V ) || ,β, ( f ( θ ) ,r ) × (0 ,θ ) ≤ C (3.4.37)for any t ∈ [0 , || ab || ,β, Ω ≤ || a || ,β, Ω || b || , , Ω + || a || , , Ω || b || ,β, Ω , (3.0.2)satisfied by ( f i (0) , Φ i e ϕ , L i , S i ) for i = 1 , h ′ ∈ C α (1 − α, { θ = θ } ) ((0 , θ )), ||L || , , ( f ( θ ) ,r ) × (0 ,θ ) ≤ C, || ˜ L || , , ( f ( θ ) ,r ) × (0 ,θ ) ≤ C (3.4.38)and || k − ◦ ( tV + (1 − t ) ˜ V ) || , , ( f ( θ ) ,r ) × (0 ,θ ) ≤ C , we estimate the right-hand side of (3.4.34)in C β ( N + f ). Then we obtain || ( c ) || ,β, N + f ≤ C || h || ( − α, { θ = θ } )1 ,β, (0 ,θ ) || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.39)( d ) can be estimated in C β ( N + f ) in a similar way. As we do this, ˜ k and k play the role of V and ˜ V in the estimate of ( e ) in C β ( N + f ) and ϑ is regarded as the argument of ˜ k and k .We have || ( d ) || ,β, N + f ≤ C || h || ( − α, { θ = θ } )1 ,β, (0 ,θ ) || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.40)Combining (3.4.39) and (3.4.40), we obtain the desired result. (cid:3) Note that if we change L − ˜ L in the way that we changed L − ˜ L in the estimate of R h ′ ( t L + (1 − t ) ˜ L ) dt ( L − ˜ L ) in the proof of Claim and estimate the resultant terms in C β ( N + f ) without changing V − ˜ V and ˜ k ( ϑ ) − k ( ϑ ) into 2 πr sin θ ( θ − θ ) R ∂ θ (Ψ − ˜Ψ )( r, tθ +(1 − t ) θ ) dt and 2 πf ( ϑ ) sin ϑ ( ϑ − θ ) R ∂ θ (( ˜Ψ − Ψ )( f ( θ ) , θ )) | θ = tϑ +(1 − t ) θ dt , then we obtain ||L − ˜ L || ,β, N + f ≤ C || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.41)With the above Claim, we estimate ( a ) in C β ( N + f ). Write ( a ) as(3.4.42) Z g ′ (cid:18) u − c − · ( t ν f ( L ) + (1 − t ) ν f ( ˜ L )) (cid:19) ! (cid:18) u − c − · ( t ν f ( L ) + (1 − t ) ν f ( ˜ L )) (cid:19) u − c − S − (cid:19) ( f ( L ) , L ) dt · ( ν f ( L ) − ν f ( ˜ L )) . By ν f i = e r − f ′ sfi (0)+ fs e θ r f ′ sfi (0)+ fs ) for i = 1 , ν f ( L ) − ν f ( ˜ L ) can written as t e r + t e θ (3.4.43)where t = Z ∇ ( f (0) ,f s ,f ′ s ) q f ′ s f (0)+ f s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( f (0) ,f s ,f ′ s )= χ dt · ( f (0) − f (0) , f s ( L ) − f s ( ˜ L ) , f ′ s ( L ) − f ′ s ( ˜ L ))and t = Z ∇ ( f (0) ,f s ,f ′ s ) − f ′ s f (0)+ f s q f ′ s f (0)+ f s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( f (0) ,f s ,f ′ s )= χ dt · ( f (0) − f (0) , f s ( L ) − f s ( ˜ L ) , f ′ s ( L ) − f ′ s ( ˜ L ))with χ = ( tf (0) + (1 − t ) f (0) , tf s ( L ) + (1 − t ) f s ( ˜ L ) , tf ′ s ( L ) + (1 − t ) f ′ s ( ˜ L )). To estimate(3.4.42) in C β ( N + f ), we estimate f s ( L ) − f s ( ˜ L ) and f ′ s ( L ) − f ′ s ( ˜ L ) in C β ( N + f ), respectively.Express f s ( L ) − f s ( ˜ L ) and f ′ s ( L ) − f ′ s ( ˜ L ) as Z f ′ s ( t L + (1 − t ) L ) dt ( L − ˜ L )(3.4.44)and Z f ′′ s ( t L + (1 − t ) ˜ L ) dt ( L − ˜ L ) , (3.4.45)respectively. Since f ′ s ∈ C α ([0 , θ ]), (3.4.44) can be estimated in C β ( N + f ) directly. With (3.0.1),(3.4.38) and (3.4.41), we estimate (3.4.44) in C β ( N + f ) directly. Then we obtain || f s ( L ) − f s ( ˜ L ) || ,β, N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.46)Since f ′′ s ∈ C α (1 − α, { θ = θ } ) ((0 , θ )), we cannot estimate (3.4.45) in C β ( N + f ) directly. With Claimand (3.0.1), we estimate (3.4.45) in C β ( N + f ). Then we obtain || f ′ s ( L ) − f ′ s ( ˜ L ) || ,β, N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.47) -D AXISYMMETRIC TRANSONIC SHOCK 51 With (2.3.3), (3.0.1), (3.4.38), the expression of ν f ( L ) − ν f ( ˜ L ) given in (3.4.43), (3.4.46)and (3.4.47), we estimate (3.4.42) in C β ( N + f ). Then we have || ( a ) || ,β, N + f ≤ Cσ | f (0) − f (0) | + Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.48)Next, we estimate ( b ) in C β ( N + f ). Divide ( b ) into two parts: g u − · ν f ( ˜ L ) c − ! S − ( f ( L ) , L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L )=: ( b ) and g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L )=: ( b ) . Write ( b ) and ( b ) as Z ∂ θ g u − · ν f ( ˜ L ) c − ! S − ( f ( θ ) , θ ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) θ = t L +(1 − t ) ˜ L dt ( L − ˜ L )and Z ∂ r g u − · ν f ( ˜ L ) c − ! S − ( r, ˜ L ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = tf ( ˜ L )+(1 − t ) f ( ˜ L ) dt ( f (0) − f (0)) . Integrands in both the expressions are in C β ( N + f ). Thus, ( b ) and ( b ) can be estimated in C β ( N + f ) directly. With (2.3.3), (3.0.1), (3.4.38) and (3.4.41), we estiamte ( b ) and ( b ) in C β ( N + f ), respectively. Then we obtain || ( b ) || ,β, N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f (3.4.49)and || ( b ) || ,β, N + f ≤ C | f (0) − f (0) | . (3.4.50)Combining (3.4.49) and (3.4.50), we have || ( b ) || ,β, N + f ≤ C | f (0) − f (0) | + Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.51)Finally, combine (3.4.48) and (3.4.51). Then we have || T − ˜ T || ,β, N + f ≤ C | f (0) − f (0) | + Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.52)3. Estimate || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f .Since A i for i = 1 , ∇ × ((Φ +0 + Ψ i ) e ϕ ) · ∇ A = 0 in N + f i (0)+ f s , A = A en,f i (0)+ f s on Γ f i (0)+ f s for i = 1 ,
2, respectively, where A en,f i is A en,f (0)+ f s given in (3.1.18) for f (0) = f i (0), by Lemma3.15, A i = A en,f i ( L i )for i = 1 ,
2. With these solution expressions, express ( A πr sin θ − ˜ A πr sin θ ) e ϕ as f ( L ) sin( L ) u − ,ϕ ( f ( L ) , L ) − f ( ˜ L ) sin( ˜ L ) u − ,ϕ ( f ( ˜ L ) , ˜ L ) r sin θ e ϕ (3.4.53) where u − ,ϕ = u − · e ϕ . With (2.3.3), (3.0.1), || L θ || ,β, N + f ≤ C and || ˜ L θ || ,β, N + f ≤ C obtained usingarguments similar to the ones in the proof of Claim in Lemma 3.15, ||L || ,β, ( f ( θ ) ,r ) × (0 ,θ ) ≤ C , || ˜ L || ,β, ( f ( θ ) ,r ) × (0 ,θ ) ≤ C and Claim, we estimate (3.4.53) in C β ( N + f ). Then we obtain(3.4.54) || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f ≤ Cσ | f (0) − f (0) | + Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f .
4. Estimate | f (0) − f (0) | .Substitute T − ˜ T = ( a ) + ( b ) + ( b ) into (3.4.32). Then we have1 r sin θ Z θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ (3.4.55) = 1 r sin θ Z θ (cid:16) f ( T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 (( a ) + ( b ) )+ f (Ψ e ϕ , A , T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:17)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ. Using (3.0.2) satisfied by ( f i (0) , Φ i e ϕ , L i , S i ) for i = 1 ,
2, we estimate f (Ψ e ϕ , A , T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T )in C β (Γ ex ). Then we obtain || f (Ψ e ϕ , A , T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) || ,β, Γ ex ≤ Cσ ( || (Ψ − ˜Ψ ) e ϕ || ,β, Γ ex + || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || ,β, Γ ex + || T − ˜ T || ,β, Γ ex ) . With this estimate, (2.3.3), (2.3.4), (3.0.1) and (3.0.2) satisfied by ( f i (0) , Φ i e ϕ , L i , S i ) for i =1 ,
2, (3.4.48), (3.4.49), (3.4.52), (3.4.54) and the fact that ( g ( M − )) ′ S in is strictly positive in[ r , r ] (see Lemma 2.10), we estimate f (0) − f (0) in (3.4.55). Then we get | f (0) − f (0) | ≤ C σ | f (0) − f (0) | + Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f where C is a positive constant depending on the data. Take σ = min( C , σ (1)3 )(=: σ (2)3 ). Thenwe have | f (0) − f (0) | ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . (3.4.56)5. With the fact that Ψ e ϕ and ˜Ψ e ϕ are in C ,α ( − − α, Γ + w,f ) ( N + f ), transform (3.4.30) into anelliptic system form (see (3.2.8)). And then transform the resultant equation with (3.4.31) intothe following boundary value problemdiv (cid:16) A D ((Ψ − ˜Ψ ) e ϕ − h ) (cid:17) − d ((Ψ − ˜Ψ ) e ϕ − h )(3.4.57)= − ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ( T − ˜ T ) r e ϕ − F | {z } F − div (cid:0) A D h (cid:1) + d h in N + f (0)+ f s , (Ψ − ˜Ψ ) e ϕ − h = on ∂ N + f (0)+ f s (3.4.58)where A and d are a (2 ,
2) tensor and scalar function defined below (3.2.14) and h := ( r − f ( θ )) r r h + ( r − r ) f ( θ ) r h r − f ( θ ) . -D AXISYMMETRIC TRANSONIC SHOCK 53 Here, F is of the form X i ( A i − A i ) ∂ r B i e ϕ + X i A i ∂ r ( B i − B i ) e ϕ + X i C i ∂ θ ( D i − D i ) e ϕ + ( E − E ) ∂ θ ( F sin θ )sin θ e ϕ + E ∂ θ (( F − F ) sin θ )sin θ e ϕ where A ij ∈ C ,α ( − α, Γ + w ) ( N + f ) , B ij e θ ∈ C ,α ( − α, Γ + w ) ( N + f ) , C ij ∈ C ,α ( − α, Γ + w ) ( N + f ) ,D ij ∈ C ,α ( − α, Γ + w ) ( N + f ) , E j e ϕ ∈ C ,α ( − − α, Γ + w ) ( N + f ) and F j e ϕ ∈ C ,α ( − − α, Γ + w ) ( N + f )for j = 1 ,
2. With arguments similar to the ones in the proof of Lemma 3.10 using (2.3.3),(2.3.4), (3.0.1), (3.0.2) satisfied by ( f i (0) , Φ i e ϕ , L i , S i ) for i = 1 ,
2, (3.4.56) and || T − ˜ T || ,β, N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f (3.4.59)and || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N f +2 (3.4.60)obtained from (3.4.52) and (3.4.54) using (3.4.56), we estimate (Ψ − ˜Ψ ) e ϕ in (3.4.57), (3.4.58)in C ,β ( N + f ) (in this argument, we only change R N + f P i A i ∂ r ( B i − B i ) e ϕ ξ , R N + f P i C i ∂ θ ( D i − D i ) e ϕ ξ and R N + f div( A D h ) ξ using integration by parts). Then we obtain || (Ψ − ˜Ψ ) e ϕ || ,β, N + f ≤ Cσ || (Ψ − ˜Ψ ) e ϕ || ,β, N + f . Take σ = min( C , σ (2)3 )(=: σ ). Then we have Ψ e ϕ = ˜Ψ e ϕ . Using this fact, we obtain from(3.4.56), (3.4.59) and (3.4.60) f (0) = f (0), T = ˜ T and A = ˜ A . One can see that σ dependson the data. This finishes the proof. (cid:3) Determination of a shape of a shock location
In the previous section, for given ( ρ − , u − , p − , p ex , f ′ s ) in a small perturbation of ( ρ − , u − e r , p − , p c , f (0) , Φ e ϕ , L, S ) satisfying all the conditions in Problem 2 except (2.5.5). Inthis section, to finish the proof of Theorem 2.16, for given ( ρ − , u − , p − , p ex ) in a small per-turbation of ( ρ − , u − e r , p − , p c ) as in the previous section or in a much small perturbation of( ρ − , u − e r , p − , p c ) if necessary, we find f ′ s in a small perturbation of 0 as in the previous sectionsuch that a solution of Problem 3 for given ( ρ − , u − , p − , p ex , f ′ s ) = ( ρ − , u − e r , p − , p c , f ′ s ) satisfies(2.5.5).4.1. Proof of Theorem 2.16 (Existence).
For a constant σ >
0, we define B (1) σ := { ( ρ − , u − , p − ) ∈ ( C ,α ( N )) ||| ρ − − ρ − || ,α, N + || u − − u − e r || ,α, N + || p − − p − || ,α, N ≤ σ }B (2) σ := { p ex ∈ C ,α ( − α,∂ Γ ex ) (Γ ex ) | || p ex − p c || ( − α,∂ Γ ex )1 ,α, Γ ex ≤ σ } , B (3) σ := { h ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )) | || h || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ σ } , B (4) σ := B (1) σ × B (2) σ × B (3) σ . For given ( ρ − , u − , p − , p ex , f ′ s ) ∈ B (4) σ for σ ≤ σ , let ( f (0) , Φ e ϕ , L, S ) be a solution of Problem3 satisfying (3.0.2) given in Proposition 3.1. We define(4.1.1) A ( ρ − , u − , p − , p ex , f ′ s ):= ̺ ( ∇ × (Φ e ϕ ) , L πr sin θ e ϕ , S ) ∇ × (Φ e ϕ ) · τ f (0)+ f s − u − · τ f (0)+ f s !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ f (0)+ fs ◦ Π r s f (0)+ f s where τ f (0)+ f s is the unit tangent vector on Γ f (0)+ f s perpendicular to e ϕ and satisfying ( τ f (0)+ f s × e ϕ ) · ν f (0)+ f s > ν f (0)+ f s is the unit normal vector field on Γ f (0)+ f s . Then A satisfies A ( ρ − , u − e r , p − , p c ,
0) = 0 and A is a map from B (4) σ to B (3) Cσ where C is a positive constantdepending on the data. If for given ( ρ − , u − , p − , p ex ) ∈ B (1) σ × B (2) σ for σ ≤ σ , we find f ′ s ∈ B (3) Cσ for Cσ ≤ σ such that A ( ρ − , u − , p − , p ex , f ′ s ) = 0, then ( f (0) + f s , Φ e ϕ , L, S ) satisfies all theconditions in Problem 2 and thus the existence part of Theorem 2.16 is proved. We will findsuch f ′ s using the weak implicit function theorem introduced in [3]. To apply the weak im-plicit function theorem, we need to prove that A is continuous, A is Fr´echet differentiable at( ρ − , u − e r , p − , p c , ζ ) and the partial Fr´echet derivative of A with respect to f ′ s at ζ isinvertible. We will prove these in the following lemmas.We first prove that A is continuous. Lemma 4.1. A is continuous in B (4) σ for a positive constant σ ≤ σ in the sense that if ζ ( k ) :=( ρ ( k ) − , u ( k ) − , p ( k ) − , p ( k ) ex , ( f ′ s ) ( k ) ) ∈ B (4) σ converges to ζ ( ∞ ) := ( ρ ( ∞ ) − , u ( ∞ ) − , p ( ∞ ) − , p ( ∞ ) ex , ( f ′ s ) ( ∞ ) ) ∈ B (4) σ in ( C , α ( N )) × C , α ( − α ,∂ Γ ex ) (Γ ex ) × C , α ( − α , { θ = θ } ) , ((0 , θ )) , then A ( ζ ( k ) ) converges to A ( ζ ( ∞ ) ) in C , α ( − α , { θ = θ } ) , ((0 , θ )) .Proof. The result is obtained by using the standard argument.Let σ be a positive constant ≤ σ and to be determined later. Let ζ ( k ) for k = 1 , , . . . be a se-quence in B (4) σ that converges to ζ ( ∞ ) ∈ B (4) σ in ( C , α ( N )) × C , α ( − α ,∂ Γ ex ) (Γ ex ) × C , α ( − α , { θ = θ } ) , ((0 , θ )).By Proposition 3.1, for each given ( ρ − , u − , p − , p ex , f ′ s ) = ζ ( k ) for k = 1 , , . . . and ∞ , there existsa unique U k := ( f (0) ( k ) , Φ ( k ) e ϕ , L ( k ) , S ( k ) ) satisfying (A), (B) and the estimate (3.0.2) with σ replaced by σ . Let ( ˜Φ ( k ) , ˜ L ( k ) , ˜ S ( k ) ) := (Φ ( k ) , L ( k ) , S ( k ) )(Π r s f (0) ( k ) + f ( k ) s ) for k = 1 , , . . . and ∞ and ˜ U k := ( f (0) ( k ) , ˜Φ ( k ) e ϕ , ˜ L ( k ) , ˜ S ( k ) ) for k = 1 , , . . . and ∞ . To prove that A ( ζ ( k ) ) converges to A ( ζ ( ∞ ) ) in C , α ( − α , { θ = θ } ) , ((0 , θ )), we show that ˜ U k converges to ˜ U ∞ in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) .To do so, we show that if a subsequence of ˜ U k converges in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) , then it converges to ˜ U ∞ . Assume that a subsequence ˜ U k j of ˜ U k convergesto a function ˜ U ∗ = ( f (0) ∗ , ˜Φ ∗ e ϕ , ˜ L ∗ , ˜ S ∗ ) in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) . Let(A) ( k ) and (B) ( k ) be (A) and (B) satisfied by U k , respectively, and let ˜(A) ( k ) and ˜(B) ( k ) be(A) ( k ) and (B) ( k ) transformed by using Π r s f (0) ( k ) + f ( k ) s , respectively. Since U k j satisfies (A) ( k j ) ,(B) ( k j ) , ˜ U k j satisfies ˜(A) ( k j ) , ˜(B) ( k j ) . Take j → ∞ to ˜(A) ( k j ) , ˜(B) ( k j ) . Then since ζ ( k ) → ζ ( ∞ ) in( C , α ( N )) × C , α ( − α ,∂ Γ ex ) (Γ ex ) × C , α ( − α , { θ = θ } ) , ((0 , θ )) and ˜ U k j → ˜ U ∗ in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) , we obtain 1) ˜(A) ( ∞ ) , ˜(B) ( ∞ ) with ˜ U ∞ replaced by ˜ U ∗ . By the facts that U k -D AXISYMMETRIC TRANSONIC SHOCK 55 for k = 1 , , . . . satisfy (3.0.2) with σ replaced by σ and ( f ′ s ) ( k ) for k = 1 , , . . . are in B (3) σ , | f (0) ( k ) − r s | + ||∇ × (( ˜Φ ( k ) − ˜Φ + , ( k )0 ) e ϕ ) || ( − α, Γ + w,rs )1 ,α, N + rs + || ˜ L ( k ) π Π rr s f (0) ( k ) + f ( k ) s sin θ e ϕ || ( − α, Γ + w,rs )1 ,α, N + rs + || ˜ S ( k ) − ˜ S + , ( k )0 || ( − α, Γ + w,rs )1 ,α, N + rs ≤ Cσ where ( ˜Φ + , ( k )0 , ˜ S + , ( k )0 ) := (Φ +0 , S +0 )(Π r s f (0) ( k ) + f ( k ) s ) for k = 1 , , . . . and C is a positive con-stant depending on the data and independent of k . Using the facts that ( f ′ s ) ( k ) → ( f ′ s ) ( ∞ ) in C , α ( − α , { θ = θ } ) , ((0 , θ )) and ˜ U k j → ˜ U ∗ in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) , we obtainfrom this inequality | f (0) ∗ − r s | + ||∇ × (( ˜Φ ∗ − ˜Φ + , ( ∞ )0 ) e ϕ ) || ( − α, Γ + w,rs )1 ,α, N + rs + || ˜ L ∗ π Π rr s f (0) ( ∞ ) + f ( ∞ ) s sin θ e ϕ || ( − α, Γ + w,rs )1 ,α, N + rs + || ˜ S ∗ − ˜ S + , ( ∞ )0 || ( − α, Γ + w,rs )1 ,α, N + rs ≤ Cσ where ( ˜Φ + , ( ∞ )0 , ˜ S + , ( ∞ )0 ) := (Φ +0 , S +0 )(Π r s f (0) ( ∞ ) + f ( ∞ ) s ). From this inequality, we have | f (0) ∗ − r s | + ||∇ × ((Φ ∗ − Φ +0 ) e ϕ ) || ( − α, Γ + w,f (0) ∗ + f ( ∞ ) s )1 ,α, N + f (0) ∗ + f ( ∞ ) s + || L ∗ πr sin θ e ϕ || ( − α, Γ + w,f (0) ∗ + f ( ∞ ) s )1 ,α, N + f (0) ∗ + f ( ∞ ) s + || S ∗ − S +0 || ( − α, Γ + w,f (0) ∗ + f ( ∞ ) s )1 ,α, N + f (0) ∗ + f ( ∞ ) s ≤ C σ where (Φ ∗ , L ∗ , S ∗ ) := ( ˜Φ ∗ , ˜ L ∗ , ˜ S ∗ )(Π f (0) ∗ + f ( ∞ ) s r s ) and C is a positive constant depending onthe data. Take σ = min( Cσ C , σ ) where C is C in (3.0.2). Then 2) ( f (0) ∗ , Φ ∗ e ϕ , L ∗ , S ∗ )satisfies (3.0.2) with σ replaced by σ . By 1) and 2), ( f (0) ∗ , Φ ∗ e ϕ , L ∗ , S ∗ ) satisfies (A), (B) for( ρ − , u − , p − , f ′ s ) = ζ ( ∞ ) and (3.0.2) with σ replaced by σ . By Proposition 3.1, ( f (0) , Φ e ϕ , L, S )satisfying (A), (B) for ( ρ − , u − , p − , f ′ s ) = ζ ( ∞ ) and (3.0.2) with σ replaced by σ is unique.Therefore, ( f (0) ∗ , Φ ∗ e ϕ , L ∗ , S ∗ ) = ( f (0) ( ∞ ) , Φ ( ∞ ) e ϕ , L ( ∞ ) , S ( ∞ ) ). From this, we have that ˜ U ∗ =˜ U ∞ .Using the fact that we showed above, we prove that ˜ U k converges to ˜ U ∞ in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) . By R × C ,α ( − − α, Γ + w,rs ) ( N + r s ) × ( C ,α ( − α, Γ + w,rs ) ( N + r s )) ⋐ R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) , every subsequence of ˜ U k has a convergent subsequence in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) . By the fact that we showed above, this convergent subsequence must con-verge to ˜ U ∞ . Thus, we have ˜ U k → ˜ U ∞ in R × C , α ( − − α , Γ + w,rs ) ( N + r s ) × ( C , α ( − α , Γ + w,rs ) ( N + r s )) .Using this fact, we can conclude that A ( ζ ( k ) ) → A ( ζ ( ∞ ) ) in C , α ( − α , { θ = θ } ) , ((0 , θ )) as ζ ( k ) ∈B (4) σ converges to ζ ( ∞ ) ∈ B (4) σ in ( C , α ( N )) × C , α ( − α ,∂ Γ ex ) (Γ ex ) × C , α ( − α , { θ = θ } ) , ((0 , θ )). Thisfinishes the proof. (cid:3) Next, we prove that A is Fr´echet differentiable. Lemma 4.2. (i) The mapping A defined in (4.1.1) is Fr´echet differentiable at ζ := ( ρ − , u − , p − , p c , as a map from ( C ,α ( − α, Γ w ) ( N )) × C ,α ( − α,∂ Γ ex ) (Γ ex ) × C ,α ( − α, { θ = θ } ) , ((0 , θ )) to C ,α ( − α, { θ = θ } ) , ((0 , θ )) .(ii) The partial Fr´echet derivative of A with respect to f ′ s at ζ is given by D f ′ s A ( ζ ) ˜ f ′ s = 1 ρ +0 ∇ × ( ˜Ψ ( ˜ f ′ s ) e ϕ ) · e θ (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s (4.1.2) for ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )) where ˜Ψ ( ˜ f ′ s ) e ϕ is a solution of (4.1.17) , (4.1.18) for given ˜ f ′ s .Proof. In this proof, we prove that A is Fr´echet differentiable as a function of f ′ s at 0 with theother variables fixed at ( ρ − , u − e r , p − , p c ). The Fr´echet differentiability of A as a function of( ρ − , u − , p − , p ex , f ′ s ) at ζ can be proved in a similar way.In this proof, C s denote positive constants depending on the data. Each C in differentinequalities differs from each other.1. Let ( f (0) , Φ e ϕ , L, S ) be a solution of Problem 3 for σ ∈ (0 , σ ] satisfying (3.0.2). Let(Ψ , A, T ) = (Φ − Φ +0 , L, S − S +0 ) and ( ˜Ψ , ˜ A, ˜ T ) := (Ψ , A, T )(Π r s f (0)+ f s ). Find the partial Fr´echetderivatives of f (0), ˜Ψ e ϕ , ˜ A and ˜ T with respect to f ′ s at ζ .Let ˜ f ′ s be a function in C ,α ( − α, { θ = θ } ) , ((0 , θ )) satisfying || ˜ f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) = 1. By Proposition3.1, for each ( ρ − , u − , p − , p ex , f ′ s ) = ( ρ − , u − e r , p − , p c , ε ˜ f ′ s ) for ε ∈ [0 , σ ], there exists a unique( f ε (0) , Φ ε e ϕ , L ε , S ε ) satisfying (A), (B) and the estimate (3.0.2). Define (Ψ ε , A ε , T ε ) := (Φ ε − Φ +0 , L ε , S ε − S +0 ). Since each ( f ε (0) , Φ ε e ϕ , L ε , S ε ) satisfies (A), (B) for ( ρ − , u − , p − , p ex , f ′ s ) =( ρ − , u − e r , p − , p c , ε ˜ f ′ s ), each ( f ε (0) , Ψ ε e ϕ , A ε , T ε ) satisfies (A ′ ), (B ′ ) for ( ρ − , u − , p − , p ex , f ′ s ) =( ρ − , u − e r , p − , p c , ε ˜ f ′ s ). Let˜ f (0) ( ˜ f ′ s ) := lim ε → + f ε (0) − f (0) ε , ˜Ψ ( ˜ f ′ s ) e ϕ = lim ε → + ˜Ψ ε e ϕ − Ψ e ϕ ε , (4.1.3) ˜ A ( ˜ f ′ s ) := lim ε → + ˜ A ε − A ε and ˜ T ( ˜ f ′ s ) := lim ε → + ˜ T ε − T ε where ˜ f (0) ( ˜ f ′ s ) , ˜Ψ ( ˜ f ′ s ) e ϕ , ˜ A ( ˜ f ′ s ) and ˜ T ( ˜ f ′ s ) represent the Gˆateaux derivatives of f (0), ˜Ψ e ϕ , ˜ A and˜ T in the direction of ˜ f ′ s at ζ . To find the Fr´echet derivative of A with respect to f ′ s at ζ , wefind ˜ f (0) ( ˜ f ′ s ) , ˜Ψ ( ˜ f ′ s ) e ϕ , ˜ A ( ˜ f ′ s ) and ˜ T ( ˜ f ′ s ) .Transform (A ′ ), (B ′ ) satisfied by ( f ε (0) , Ψ ε e ϕ , A ε , T ε ) into the equations in N + r s or on a partof ∂ N + r s by using Π r s f ε where f ε := f ε (0) + ε ˜ f s with ˜ f s = R θ ˜ f ′ s . And then subtract the resultantequations from the same equations satisfied by ( f (0) , Ψ e ϕ , A , T ). Then we obtain ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (( ˜Ψ ε − Ψ ) e ϕ ) ! (4.1.4) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ( ˜ T ε − T ) r e ϕ + F + ˜ F ( ˜Ψ ε e ϕ , ˜ A ε , ˜ T ε ) in N + r s , ( ˜Ψ ε − Ψ ) e ϕ = on Γ r s , Γ + w,r s , r sin θ R θ (cid:18) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T ε − T )+ f ( ˜Ψ ε e ϕ , ˜ A ε , ˜ T ε ) (cid:19) r sin ξdξ e ϕ on Γ ex , (4.1.5)1 r sin θ Z θ − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T ε − T ) + f ( ˜Ψ ε e ϕ , ˜ A ε , ˜ T ε ) !(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ e ϕ = , (4.1.6) ( ( M Tε M ε ∇ × (( ˜Φ +0 + ˜Ψ ε ) e ϕ ) − ∇ × (Φ +0 e ϕ )) · ∇ ˜ A ε + ∇ × (Φ +0 e ϕ ) · ∇ ( ˜ A ε − A ) = 0 in N + r s , ˜ A ε − A = 0 on Γ r s -D AXISYMMETRIC TRANSONIC SHOCK 57 ( ( M Tε M ε ∇ × (( ˜Φ +0 + ˜Ψ ε ) e ϕ ) − ∇ × (Φ +0 e ϕ )) · ∇ ˜ T ε + ∇ × (Φ +0 e ϕ ) · ∇ ( ˜ T ε − T ) = 0 in N + r s , ˜ T ε − T = g (cid:0) ( M − e r · ν f ε ) (cid:1)(cid:12)(cid:12) Γ fε S in ◦ Π r s f ε − ( g ( M − ))( r s ) S in on Γ r s where F = ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ( ˜Ψ ε e ϕ ) ! − M ε ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ! (Π r s f ε ) M ε ∇ × ( ˜Ψ ε e ϕ ) ! − ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ˜ T ε r e ϕ + ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) 1 r ! (Π r s f ε ) (cid:16) ∂ ˜ θ Π rf ε r s (Π r s f ε ) ∂ r ˜ T ε + ∂ θ ˜ T ε (cid:17) e ϕ , ( ˜Ψ ε , ˜ A ε , ˜ T ε ) := (Ψ ε , A ε , T ε ) ◦ Π r s f ε , ˜Φ +0 = Φ +0 ◦ Π r s f ε , M ε = (cid:16) ∂ Π fεrs ∂ y (cid:17) (Π r s f ε ), y is the cartesiancoordinate system representing N + f ε , ˜ θ is θ coordinate for y, ( r, θ ) is ( r, θ ) coordinates for N + r s ,and Π rf ε r s is the r -component of Π ∗ f ε r s , ν f ε is the unit normal vector field on Γ f ε pointing toward N + f ε and ˜ F is F changed by using the transformation Π f ε r s . Divide the above equations by ε and formally take ε → + using( f ε (0) , ˜Ψ ε , ˜ A ε , ˜ T ε ) → ( r s , Ψ , A , T ) = ( r s , , ,
0) as ε → + , (4.1.7) ε ˜ f ′ s → ε → + and (4.1.3). Then we have ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ( ˜Ψ ( ˜ f ′ s ) e ϕ ) ! (4.1.8) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ˜ T ( ˜ f ′ s ) r e ϕ in N + r s , ˜Ψ ( ˜ f ′ s ) e ϕ = on Γ r s , Γ + w,r s , − r sin θ R θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ˜ T ( ˜ f ′ s ) r sin ξdξ e ϕ on Γ ex , (4.1.9) − r sin θ Z θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ˜ T ( ˜ f ′ s ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) r = r r sin ξdξ e ϕ = , (4.1.10) ( ∂ r ˜ A ( ˜ f ′ s ) = 0 in N + r s , ˜ A ( ˜ f ′ s ) = 0 on Γ r s , (4.1.11) ( ∂ r ˜ T ( ˜ f ′ s ) = 0 in N + r s , ˜ T ( ˜ f ′ s ) = ( g ( M − )) ′ ( r s ) S in ( ˜ f (0) ( ˜ f ′ s ) + ˜ f s ) on Γ r s . (4.1.12)We solve this system for given ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )).By (4.1.11) and (4.1.12), ˜ A ( ˜ f ′ s ) = 0 in N + r s (4.1.13)and ˜ T ( ˜ f ′ s ) = ( g ( M − )) ′ ( r s ) S in ( ˜ f (0) ( ˜ f ′ s ) + ˜ f s ) . (4.1.14) Substitute (4.1.14) into (4.1.10). And then find ˜ f (0) ( ˜ f ′ s ) in the resultant equation. Then weobtain ˜ f (0) ( ˜ f ′ s ) = − R θ ˜ f s sin ζdζ R θ sin ζdζ . (4.1.15)Substitute this into (4.1.14) again. Then we have˜ T ( ˜ f ′ s ) = ( g ( M − )) ′ ( r s ) S in ( − R θ ˜ f s sin ζdζ R θ sin ζdζ + ˜ f s ) in N + r s . (4.1.16)Substituting ˜ T ( ˜ f ′ s ) given in (4.1.16) into (4.1.8) and (4.1.9), we get ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ( ˜Ψ ( ˜ f ′ s ) e ϕ ) ! (4.1.17) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 )( g ( M − )) ′ ( r s ) S in ˜ f ′ s r e ϕ in N + r s , ˜Ψ ( ˜ f ′ s ) e ϕ = on Γ r s , Γ + w,r s , − r sin θ R θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( g ( M − )) ′ ( r s ) S in ( − R θ ˜ f s sin ζdζ R θ sin ζdζ + ˜ f s ) r sin ξdξ e ϕ on Γ ex . (4.1.18)Note that since (4.1.18) is a C boundary condition, Lemma 3.5 can be applied to (4.1.17),(4.1.18). Using Lemma 3.5 and the arguments used to prove Lemma 3.13 (here we can ob-tain more higher regularity of solutions of (4.1.17), (4.1.18) than that of solutions of (3.2.1),(3.2.2) in Lemma 3.5 because ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) ((0 , θ ))), we obtain that there exists a unique C ,α ( − − α, Γ + w,rs ) ( N + r s ) solution ˜Ψ ( ˜ f ′ s ) e ϕ of (4.1.17), (4.1.18) and this solution satisfies || ˜Ψ ( ˜ f ′ s ) e ϕ || ( − − α, Γ + w,rs )3 ,α, N + rs ≤ C (4.1.19)for a positive constant C independent of ˜ f ′ s .Using (3.0.2) satisfied by ( f ε (0) , Φ ε e ϕ , L ε , S ε ), (4.1.4)-(4.1.6), (4.1.8)-(4.1.10), (4.1.13), (4.1.15),(4.1.16), the solution expressions of ˜ A ε and ˜ T ε obtained by solving (B ′ ) for ( f (0) , Ψ) = ( f ε (0) , Ψ ε )and ( ρ − , u − , p − , p ex , f ′ s ) = ( ρ − , u − e r , p − , p c , ε ˜ f ′ s ) using Lemma 3.15, and Lemma 3.5, we can ob-tain | f ε (0) − f (0) − ε ˜ f (0) ( ˜ f ′ s ) | ≤ Cε , || ˜Ψ ε − Ψ − ε ˜Ψ ( ˜ f ′ s ) || ( − − α, Γ + w,rs )2 ,α, N + rs ≤ Cε , (4.1.20) || ˜ A ε − A − ε ˜ A ( ˜ f ′ s ) || ( − α, Γ + w,rs )1 ,α, N + rs = 0 , || ˜ T ε − T − ε ˜ T ( ˜ f ′ s ) || ( − α, Γ + w,rs )1 ,α, N + rs ≤ Cε for all ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )) satisfying || ˜ f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) = 1 and ε ∈ [0 , σ ]. By (4.1.13) and(4.1.15)-(4.1.18), ˜ f (0) ( ˜ f ′ s ) , ˜Ψ ( ˜ f ′ s ) e ϕ , ˜ A ( ˜ f ′ s ) and ˜ T ( ˜ f ′ s ) are bounded linear maps from C ,α ( − α, { θ = θ } ) , ((0 , θ ))to R , C ,α ( − − α, Γ + w,rs ) ( N + r s ), C ,α ( − α, Γ + w,rs ) ( N + r s ) and C ,α ( − α, Γ + w,rs ) ( N + r s ), respectively. From the aboveinequalities and this fact, we have that f (0), ˜Ψ e ϕ , ˜ A and ˜ T are Fr´echet differentiable as a func-tion of f ′ s at 0 with the other variables fixed at ( ρ − , u − e r , p − , p c ) and ˜ f (0) ( ˜ f ′ s ) , ˜Ψ ( ˜ f ′ s ) e ϕ , ˜ A ( ˜ f ′ s ) and ˜ T ( ˜ f ′ s ) are the partial Fr´echet derivatives of f (0), ˜Ψ e ϕ , ˜ A and ˜ T with respect to f ′ s at ζ ,respectively.2. Show that A is Fr´echet differentiable as a function of f ′ s at 0 with the other variables fixedat ( ρ − , u − e r , p − , p c ). -D AXISYMMETRIC TRANSONIC SHOCK 59 By definition, the Gˆateaux derivative of A in the direction (0 , , , , ˜ f ′ s ) at ζ is given bylim ε → + A ( ρ − , u − e r , p − , p c , ε ˜ f ′ s ) − A ( ρ − , u − e r , p − , p c , ε = lim ε → + ε ̺ ( ∇ × ((Φ +0 + Ψ ε ) e ϕ ) , A ε πr sin θ e ϕ , S +0 + T ε ) ∇ × ((Φ +0 + Ψ ε ) e ϕ ) · τ f ε (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Γ fε ◦ Π r s f ε − u − e r · τ f ε (cid:12)(cid:12) Γ fε ◦ Π r s f ε − ̺ ( ∇ × (Φ +0 e ϕ ) , , S +0 ) ∇ × (Φ +0 e ϕ ) · e θ (cid:12)(cid:12)(cid:12)(cid:12) Γ rs + u − e r · e θ (cid:12)(cid:12) Γ rs (cid:17) where τ f ε is the unit tangent vector on Γ f ε perpendicular to e ϕ and satisfying ( τ f ε × e ϕ ) · ν f ε > ν f ε is the unit normal vector field on Γ f ε . As we did in Step 1, formally take ε → + using(4.1.3) and (4.1.7) to the right-hand side of the above equation. Then we obtain1 ρ +0 ∇ × ( ˜Ψ ( ˜ f ′ s ) e ϕ ) · e θ (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s . Define a map L by L ˜ f ′ s := 1 ρ +0 ∇ × ( ˜Ψ ( ˜ f ′ s ) e ϕ ) · e θ (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s for ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )) satisfying || ˜ f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) = 1 where ˜Ψ ( ˜ f ′ s ) e ϕ is a solution of(4.1.17), (4.1.18) for given ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )). Then L is a bounded linear map from C ,α ( − α, { θ = θ } ) , ((0 , θ )) to C ,α ( − α, { θ = θ } ) , ((0 , θ )). Using (3.0.2) satisfied by ( f (0) , Φ ε e ϕ , L ε , S ε )and (4.1.20), one can check that ||A ( ρ − , u − e r , p − , p c , ε ˜ f ′ s ) − A ( ρ − , u − e r , p − , p c , − εL ˜ f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ Cε (4.1.21)for any ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )) satisfying || ˜ f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) = 1 and ε ∈ [0 , σ ]. Therefore, A isFr´echet differentiable as a function of f ′ s at 0 with the other variables fixed at ( ρ − , u − e r , p − , p c )and L is the partial Fr´echet derivative of A with respect to f ′ s at ζ . This finishes the proof. (cid:3) Finally, we prove that the partial Fr´echet derivative of A with respect to f ′ s at ζ is invertible.When we prove the invertibility of the partial Fr´echet derivative of A with respect to f ′ s at ζ ,we use eigenfunction expansions of ˜ f ′ s and ˜Ψ ( ˜ f ′ s ) . The eigenfunctions used to express ˜ f ′ s and˜Ψ ( ˜ f ′ s ) are eigenfunctions of the following eigenvalue problem ( θ ∂ θ (sin θ∂ θ q ) − q sin θ = − λq in θ ∈ (0 , θ ) ,q = 0 on θ = 0 , θ (4.1.22)that arises from θ -part of the spherical coordinate representation of (4.1.17), (4.1.18) (note thatthis is the associated Legendre equation of type m = 1 with a general domain that is a singularSturm-Liouville problem). To express ˜ f ′ s and ˜Ψ ( ˜ f ′ s ) using eigenfunctions of (4.1.22), we need toprove the orthogonal completeness of the set of eigenfunctions of (4.1.22). We prove this in thefollowing lemma. Lemma 4.3.
The eigenvalue problem (4.1.22) has infinitely countable eigenvalues λ j for j =1 , , . . . satisfying λ j → ∞ as j → ∞ and λ j > . A set of eigenfunctions of (4.1.22) forms acomplete orthorgonal set in L ((0 , θ ) , sin θdθ ) .Proof.
1. For given f ∈ L ((0 , θ ) , sin θdθ ), we consider ( θ ∂ θ (sin θ∂ θ q ) − q sin θ = − f in θ ∈ (0 , θ ) ,q = 0 on θ = 0 , θ . (4.1.23) Write this equation in the form Z θ ( ∂ θ q∂ θ ξ + qξ sin θ ) sin θdθ = Z θ f ξ sin θdθ (4.1.24)for all ξ ∈ H ((0 , θ ) , sin θdθ ). Assume for a moment that there exists a unique q ∈ H ((0 , θ ) , sin θdθ )satisfying (4.1.24) for all ξ ∈ H ((0 , θ ) , sin θdθ ) and this q satisfies || q || H ((0 ,θ ) , sin θdθ ) ≤ C || f || L ((0 ,θ ) , sin θdθ ) (4.1.25)for some positive constant C . Using this q , we define a map S : L ((0 , θ ) , sin θdθ ) → L ((0 , θ ) , sin θdθ )by S f = q. Then S is a self-adjoint and compact linear operator. Hence, by the spectral theorem forcompact self-adjoint operators, S has contable infinite eigenvalues µ j satisfying µ j → j → ∞ and the set of eigenfunctions q j of S corresponding to µ j forms a complete orthorgonal set in L ((0 , θ ) , sin θdθ ). From this fact, we obtain that (4.1.22) has infinitely countable eigenvalues λ j → ∞ as j → ∞ and the set of eigenfunctions of (4.1.22) forms a complete orthorgonal setin L ((0 , θ ) , sin θdθ ).2. Show that there exists a unique q ∈ H ((0 , θ ) , sin θdθ ) satisfying (4.1.24) for all ξ ∈ H ((0 , θ ) , sin θdθ ).For given f ∈ L ((0 , θ ) , sin θdθ ), we consider Z D δ U δ ξ = Z D f ( θ ) e ϕ ξ (4.1.26)for all ξ ∈ H ( D ) where U : D → R , ξ : D → R , D := { ( x, y, z ) ∈ S : z √ x + y + z ≥ cos θ } and δ is the covariant derivative on S . By the Lax-Milgram theorem, there exists a unique U ∈ H ( D ) satisfying (4.1.26) for all ξ ∈ H ( D ) and this U satisfies || U || H ( D ) ≤ C || f e ϕ || L ( D ) (4.1.27)for some positive constant C depending on D . Let U be a function in H ( D ) satisfying (4.1.26)for all ξ ∈ H ( D ). Using the standard argument, it can be shown that U satisfies || U || H ( D ) ≤ C || f e ϕ || L ( D ) . From this fact, we see that U satisfies Z D ∆ S U ξ = − Z D f ( θ ) e ϕ ξ for all ξ ∈ H ( D ). Using this fact and the fact that the coefficients of ∆ S U in the sphericalcoordinate system are independent of ϕ , we apply arguments similar to the ones in the proof ofLemma 3.14 to U (here, we use the facts that a bounded sequence in H ( D ) contains a weaklyconvergent subsequence and a H ( D ) function satisfying (4.1.26) for all ξ ∈ H ( D ) is unique).Then we have that U only has the form u ϕ ( θ ) e ϕ .One can see that if U = u ϕ ( θ ) e ϕ ∈ H ( D ) satisfies (4.1.26) for all ξ ∈ H ( D ), then u ϕ ∈ H ((0 , θ ) , sin θdθ ) satisfies (4.1.24) for all ξ ∈ H ((0 , θ ) , sin θdθ ) and that if u ϕ ∈ H ((0 , θ ) , sin θdθ ) satisfies (4.1.24) for all ξ ∈ H ((0 , θ ) , sin θdθ ), then U = u ϕ ( θ ) e ϕ ∈ H ( D )satisfies (4.1.26) for all ξ ∈ H ( D ) having the form ξ ( θ ) e ϕ . Using this fact, we can deduce thatthere exists a unique q ∈ H ((0 , θ ) , sin θdθ ) satisfying (4.1.24) for all ξ ∈ H ((0 , θ ) , sin θdθ ).By (4.1.27), this solution satisfies (4.1.25).3. Show that eigenvalues λ j of (4.1.22) are positive.If q is an eigenfunction of S corresponding to an eigenvalue µ = λ , then there holds Z D δ ( q e ϕ ) δ ξ = Z D λq e ϕ ξ (4.1.28)for all ξ ∈ H ( D ) having the form ξ ( θ ) e ϕ . Using the weak maximum principle, we can havethat for q e ϕ ∈ H ( D ) to be a nonzero function satisfying (4.1.28) for all ξ ∈ H ( D ) having the -D AXISYMMETRIC TRANSONIC SHOCK 61 form ξ ( θ ) e ϕ , λ must be positive. Hence, eigenvalues of (4.1.22) are positive. This finishes theproof. (cid:3) Then we prove the invertibility of the partial Fr´echet derivative of A with respect to f ′ s at ζ . Lemma 4.4.
The partial Fr´echet derivative of A with respect to f ′ s at ζ given by (4.1.2) is aninvertible map from C ,α ( − α, { θ = θ } ) ((0 , θ )) to C ,α ( − α, { θ = θ } ) ((0 , θ )) .Proof. By (4.1.19) and C ,α ( − − α, { θ = θ } ) ((0 , θ )) ⋐ C ,α ( − α, { θ = θ } ) ((0 , θ )), D f ′ s A ( ζ ) given by (4.1.2)is of the form cI − K where c is a constant and K is a compact linear map from C ,α ( − α, { θ = θ } ) , ((0 , θ ))to C ,α ( − α, { θ = θ } ) , ((0 , θ )). By the Fredholm alternative, this implies that if ker D f ′ s A ( ζ ) = { } ,then D f ′ s A ( ζ ) is invertible. In this proof, we show that ker D f ′ s A ( ζ ) = { } . Since it is obviousthat D f ′ s A ( ζ )0 = 0, we show that D f ′ s A ( ζ ) ˜ f ′ s = 0 only if ˜ f ′ s = 0.1. Assume that for a nonzero ˜ f ′ s ∈ C ,α ( − α, { θ = θ } ) , ((0 , θ )), D f ′ s A ( ζ ) ˜ f ′ s = 0. Then there holds1 ρ +0 ∇ × (Ψ ( ˜ f ′ s ) e ϕ ) · e θ (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s = 0(4.1.29)where ˜Ψ ( ˜ f ′ s ) e ϕ is the C ,α ( − − α, Γ + w,rs ) ( N + r s ) solution of − ρ +0 r ∂ r ( r ∂ r ˜Ψ ( ˜ f ′ s ) ) + ∂ r ρ +0 ρ +0 2 r ∂ r ( r ˜Ψ ( ˜ f ′ s ) ) − ρ +0 r c +0 2 c +0 2 − u +0 2 ! θ ∂ θ (sin θ∂ θ ˜Ψ ( ˜ f ′ s ) ) − ˜Ψ ( ˜ f ′ s ) sin θ ! (4.1.30)= ρ +0 γ − ( γ − u +0 γu +0 2 c +0 2 − u +0 2 ! (cid:16) g ( M − ) (cid:17) ′ ( r s ) S in r ˜ f ′ s in N + , ∗ r s , ˜Ψ ( ˜ f ′ s ) = ∗ r s , Γ + , ∗ w,r s , − r sin θ R θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 (cid:16) g ( M − ) (cid:17) ′ ( r s ) S in (cid:18) − R θ ˜ f s sin ζdζ R θ sin ζdζ + ˜ f s (cid:19) r sin ξdξ on Γ ∗ ex , (4.1.31)which is the spherical coordinate representation of (4.1.17), (4.1.18), for given ˜ f ′ s . Using C ,α ( − α, { θ = θ } ) ((0 , θ )) ⊂ L ((0 , θ ) , sin θdθ ) and Lemma 4.3, we express ˜ f ′ s as˜ f ′ s = ∞ X j =1 c j q j (4.1.32)where c j are constants and q j are eigenfunctions of (4.1.22) corresponding to eigenvalues λ j of(4.1.22). Define ˜ f ′ s,m := m X j =1 c j q j (4.1.33)and ˜ f s,m := R θ ˜ f ′ s,m . We consider (4.1.30), (4.1.31) for given ˜ f ′ s = ˜ f ′ s,m and ˜ f s = ˜ f s,m .Using the fact that q j ∈ C ∞ ([0 , θ ]) and θ ∂ θ (sin θ∂ θ q ) − q sin θ = ∂ θ (cid:16) ∂ θ ( q sin θ )sin θ (cid:17) , we can have˜ f s,m = m X j =1 − c j λ j (cid:18) ∂ θ ( q j sin θ )sin θ − ∂ θ ( q j sin θ )sin θ (cid:12)(cid:12)(cid:12)(cid:12) θ =0 (cid:19) . (4.1.34) Here, ∂ θ ( q j sin θ )sin θ (cid:12)(cid:12)(cid:12) θ =0 is bounded because q j = 0 on θ = 0 and q j ∈ C ∞ ([0 , θ ]). Substitute˜Ψ m = P mj =1 p j ( r ) q j ( θ ), ˜ f ′ s,m given in (4.1.33) and ˜ f s,m given in (4.1.34) into the places of ˜Ψ ( ˜ f ′ s ) ,˜ f ′ s and ˜ f s in (4.1.30), (4.1.31), respectively. Then we obtain( L :=) − r ( 1 ρ +0 r p ′ j ) ′ + ( 1 ρ +0 r (1 + u +0 2 c +0 2 − u +0 2 ) λ j + ∂ r ρ +0 ρ +0 2 r ) p j (4.1.35) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) (cid:0) g ( M − ) (cid:1) ′ ( r s ) S in r c j in ( r s , r ) ,p j = r = r s , ρ +0 (( γ − u +0 2 + c +0 2 )( g ( M − )) ′ ( r s ) S in r γ ( γ − u +0 S +0 λ j c j on r = r (4.1.36)for j = 1 , . . . , m . One can see that L is of the form L − c I where L (= r ( ρ +0 r p ′ j ) ′ ) isan invertible operator and c is a positive constant. Using the Fredholm alternative and themaximum principle, we can obtain that (4.1.35), (4.1.36) for each j has a unique C ∞ ([ r s , r ])solution. Let p j be the solution of (4.1.35), (4.1.36). Then by the fact that q j e ϕ ∈ C ∞ ( D )where D is a domain defined in the proof of Lemma 4.3, we have that ˜Ψ m e ϕ = P mj =1 p j q j e ϕ isa C ∞ ( N + r s ) solution of (4.1.17), (4.1.18) for given ˜ f ′ s = ˜ f ′ s,m and ˜ f s = ˜ f s,m .2. Show that there exists a subsequence ˜Ψ m l e ϕ of ˜Ψ m e ϕ such that ˜Ψ m l e ϕ and D ( ˜Ψ m l e ϕ )weakly converge to ˜Ψ ( ˜ f ′ s ) e ϕ and D ( ˜Ψ ( ˜ f ′ s ) e ϕ ) in L ( N + r s ), respectively.Since ˜Ψ m e ϕ is a C ∞ ( N + r s ) solution of (4.1.17), (4.1.18) for given ˜ f ′ s = ˜ f ′ s,m and ˜ f s = ˜ f s,m ,(4.1.17), (4.1.18) satisfied by ˜Ψ m e ϕ can be transformed into the following boundary value prob-lem for an elliptic system(4.1.37) div (cid:16) A D ( ˜Ψ m e ϕ ) (cid:17) − d ( ˜Ψ m e ϕ )= − ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 )( g ( M − )) ′ ( r s ) S in ˜ f ′ s,m r e ϕ in N + r s , where A and d are a (2 , m e ϕ = on Γ r s , Γ + w,r s , − r sin θ R θ ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( g ( M − )) ′ ( r s ) S in ( − R θ ˜ f s,m sin ζdζ R θ sin ζdζ + ˜ f s,m ) r sin ξdξ e ϕ (=: h m e ϕ ) on Γ ex . (4.1.38)Set h m := ( r − r s ) r ( r − r s ) r h m e ϕ . Transform (4.1.37), (4.1.38) into a boundary value problem bysubstituting ˜Ψ ∗ m e ϕ + h m into the place of ˜Ψ m e ϕ in (4.1.37), (4.1.38). Write the resultantproblem in the following form(4.1.39) Z N + rs A D ( ˜Ψ ∗ m e ϕ ) D ξ + d ˜Ψ ∗ m e ϕ ξ = Z N + rs ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 )( g ( M − )) ′ ( r s ) S in ˜ f ′ s,m r e ϕ ξ − A D h m D ξ − d h m ξ for all ξ ∈ H ( N + r s ) where ˜Ψ ∗ m e ϕ := ˜Ψ m e ϕ − h m . Using Lemma 3.8, we obtain from (4.1.39) || ˜Ψ ∗ m e ϕ || H ( N + rs ) ≤ C || ˜ f ′ s,m e ϕ || L ( N + rs ) . -D AXISYMMETRIC TRANSONIC SHOCK 63 Since || ˜ f ′ s,m e ϕ || L ( N + rs ) ≤ || ˜ f ′ s e ϕ || L ( N + rs ) , ˜Ψ ∗ m e ϕ is a bounded sequence in H ( N + r s ). Hence, thereexists a subsequence ˜Ψ ∗ m l e ϕ of ˜Ψ ∗ m e ϕ and some function ˜Ψ ∗ e ϕ ∈ H ( N + r s ) such that ˜Ψ ∗ m l e ϕ and D ( ˜Ψ ∗ m l e ϕ ) weakly converge to ˜Ψ ∗ e ϕ and D ( ˜Ψ ∗ e ϕ ) in L ( N + r s ), respectively. Take l → ∞ to(4.1.39) for m = m l . Then by ˜Ψ ∗ m l e ϕ ⇀ ˜Ψ ∗ e ϕ in L ((0 , θ ) , sin θdθ ), D ( ˜Ψ ∗ m l e ϕ ) ⇀ D ( ˜Ψ ∗ e ϕ )in L ((0 , θ ) , sin θdθ ) and ˜ f ′ s,m → ˜ f ′ s in L ((0 , θ ) , sin θdθ ), one has (4.1.39) with ˜ f ′ s,m , ˜ f s,m and˜Ψ ∗ m replaced by ˜ f ′ s , ˜ f s and ˜Ψ ∗ e ϕ , respectively. Thus, ˜Ψ ∗ e ϕ is a H ( N + r s ) function satisfying(4.1.39) with ˜ f ′ s,m and ˜ f s,m replaced by ˜ f ′ s and ˜ f s , respectively, for all ξ ∈ H ( N + r s ). Let h be h m with ˜ f ′ s,m and ˜ f s,m replaced by ˜ f ′ s and ˜ f s , respectively. One can see that ˜Ψ ( ˜ f ′ s ) e ϕ − h is a H ( N + r s ) function satisfying (4.1.39) with ˜ f ′ s,m and ˜ f s,m replaced by ˜ f ′ s and ˜ f s , respectively, for all ξ ∈ H ( N + r s ). By Lemma 3.8, a H ( N + r s ) function satisfying (4.1.39) with ˜ f ′ s,m and ˜ f s,m replacedby ˜ f ′ s and ˜ f s , respectively, for all ξ ∈ H ( N + r s ) is unique. Hence, ˜Ψ ∗ e ϕ = ˜Ψ ( ˜ f ′ s ) e ϕ − h . Fromthis, we can conclude that ˜Ψ m l e ϕ and D ( ˜Ψ m l e ϕ ) weakly converge to ˜Ψ ( ˜ f ′ s ) e ϕ and D ( ˜Ψ ( ˜ f ′ s ) e ϕ )in L ( N + r s ), respectively.3. Show that D f ′ s A ( ζ ) ˜ f ′ s = 0 only if ˜ f ′ s = 0.Since ˜ f ′ s = 0 by the assumption, there exists k ∈ N such that c k > c k < f ′ s in (4.1.32). Without loss of generality, assume that c k > k ∈ N .Then since λ k > p k , that is, the solution of (4.1.35), (4.1.36) for j = k , satisfies p k ( r ) >
0. Using this fact, p k ( r s ) = 0 and the form of (4.1.35), we can deduce that p k ≥ r s , r ]. Thus, p ′ k ( r s ) ≥ Z θ − ρ +0 r ∂ r ( r ˜Ψ ( ˜ f ′ s ) ) (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s ! ξ sin θdθ = 0for all ξ ∈ L ((0 , θ ) , sin θdθ ). Rewrite this as(4.1.40) Z θ − ρ +0 m l X j =1 ∂ r ( rp j ) r q j − ( u − − u +0 )( r s ) m l X j =1 c j q j r s − ρ +0 r ∂ r ( r ˜Ψ ( ˜ f ′ s ) ) (cid:12)(cid:12)(cid:12)(cid:12) r = r s − ( u − − u +0 )( r s ) ˜ f ′ s r s − − ρ +0 m l X j =1 ∂ r ( rp j ) r q j − ( u − − u +0 )( r s ) m l X j =1 c j q j r s ξ sin θdθ = 0for all ξ ∈ L ((0 , θ ) , sin θdθ ). Since p ′ k ( r s ) ≥ p k ( r s ) = 0, ( u − − u +0 )( r s ) >
0, ˜Ψ m l e ϕ ⇀ ˜Ψ ( ˜ f ′ s ) e ϕ in L ( N + r s ) and D ( ˜Ψ m l e ϕ ) ⇀ D ( ˜Ψ ( ˜ f ′ s ) e ϕ ) in L ( N + r s ), for a sufficiently large l such that m l ≥ k ,if we take ξ = q k , then the left-hand side of (4.1.40) becomes a negative number (here we usedthe trace theorem). This contradicts to the assumption that D f ′ s A ( ζ ) ˜ f ′ s = 0. This finishes ourproof. (cid:3) Applying the weak implicit function theorem introduced in [3] with the results in Lemma 4.1,Lemma 4.2 and Lemma 4.4, we obtain the result of the existence part of Theorem 2.16.4.2.
Proof of Theorem 2.16 (Uniqueness).
Finally, we prove the uniqueness part of The-orem 2.16.
Proof of Theorem 2.16 (Uniqueness).
Let σ be a positive constant ≤ σ obtained in the pre-vious subsection such that if σ ≤ σ , then Problem 2 has a solution satisfying(4.2.1) | f (0) − r s | + || f ′ s || ( − α, { θ = θ } )1 ,α, (0 ,θ ) + ||∇ × ((Φ − Φ +0 ) e ϕ ) || ( − α, Γ + w )1 ,α, N + f + || L πr sin θ e ϕ || ( − α, Γ + w )1 ,α, N + f + || S − S +0 || ( − α, Γ + w )1 ,α, N + f ≤ Cσ where C is a positive constant depending on the data. Let σ be a positive constant ≤ σ andto be determined later. Suppose that there exist two solutions ( f i , Φ i e ϕ , L i , S i ) for i = 1 , σ ≤ σ satisfing (4.2.1).We will prove that there exists a positive constant σ ≤ σ such that if σ = σ , then( f , Φ e ϕ , L , S ) = ( f , Φ e ϕ , L , S ) . (4.2.2)In this proof, C s and C i for i = 1 , , . . . denote positive constants depending on the data. Each C in different inequalities differs from each other. In this proof, when we estimate quantities,we will use all or a part of the conditions ( ρ − , u − , p − ) ∈ B (1) σ , p ex ∈ B (3) σ , (4.2.1) satisfied by( f i , Φ i e ϕ , L i , S i ) for i = 1 , σ ≤ σ without mentioning that we use these conditions.Case 1. f s, = f s, where f s,i := f i − f i (0) for i = 1 , f i , Φ i e ϕ , L i , S i ), || f ′ s,i || ( − α, { θ = θ } )1 ,α, (0 ,θ ) ≤ C σ for i = 1 ,
2. Choose σ = min( σ , C σ C , σ C )(=: σ (1)1 ) where C and C are C in (4.2.1) and(3.0.2), respectively. Then ( ρ − , u − , p − , p ex , f ′ s ) ∈ B (4) σ for σ ∈ (0 , σ ] and ( f i (0) , Φ i e ϕ , L i , S i )satisfies (3.0.2) for f s = f s,i , so the hypothesis in Proposition 3.1 is satisfied. If f s, = f s, , thenby Proposition 3.1, ( f (0) , Φ e ϕ , L , S ) = ( f (0) , Φ e ϕ , L , S ). From this, we have (4.2.2).Case 2. General case.1. Let (Ψ i , A i , T i ) := (Φ i − Φ +0 , L i , S i − S +0 ) and ( ˜Ψ i , ˜ A i , ˜ T i ) := ( ˜ W i πr sin θ , A i (Π r s f i ) , T i (Π r s f i ))for i = 1 , W i := W i (Π r s f i ) and W i := 2 πr sin θ Ψ i . Show that(4.2.3) | f (0) − f (0) | + || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs + || ˜ A πr sin θ − ˜ A πr sin θ || ,β, N + rs + || ˜ T − ˜ T || ,β, N + rs ≤ C || f ′ s, − f ′ s, || ,β, (0 ,θ ) where β is a positive constant given in the proof of the uniqueness part of Proposition 3.1.By the assumption, ( f (0) , f s , Ψ e ϕ , L, T ) = ( f i (0) , f s,i , Ψ i e ϕ , A i , T i ) for i = 1 , ′ ).Transform (A ′ ) satisfied by ( f i (0) , f s,i , Ψ i e ϕ , A i , T i ) for i = 1 , N + r s or on a partof ∂ N + r s by using Π r s f i in the way that we transformed (A ′ ) satisfied by ( f (0) , Ψ e ϕ , A , T ) inthe proof of the uniqueness part of Proposition 3.1. Then we obtain M i ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 )(Π r s f i ) N i ∇ × ( ˜Ψ i e ϕ ) ! (4.2.4) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ! (Π r s f i ) ( ∂ ˜ θ i Π ∗ ,rf i r s )(Π r s f i ) ∂ r ˜ T i + ∂ θ ˜ T i Π ∗ ,rr s f i e ϕ + ˜ F i ( ˜Ψ i e ϕ , ˜ A i , ˜ T i ) in N + r s , ˜Ψ i e ϕ = Π ∗ ,rrsfi (Φ − − Φ − )(Π rsfi ) r e ϕ on Γ r s , r (Φ − − Φ − )( r ,θ ) r e ϕ on Γ + w,r s := Γ w ∩ { r > r s } , (cid:16) r sin θ R θ (cid:16) f ( ˜ T i , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ˜ T i + f ( ˜Ψ i e ϕ , ˜ A i , ˜ T i ) (cid:19) r sin ξdξ (cid:19) e ϕ on Γ ex , (4.2.5) -D AXISYMMETRIC TRANSONIC SHOCK 65 (4.2.6) 1 r sin θ Z θ f ( ˜ T i , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ˜ T i + f ( ˜Ψ i e ϕ , ˜ A i , ˜ T i ) ! r sin ξdξ = r (Φ − − Φ − )( r , θ ) r for i = 1 , M i = (cid:16) ∂ Π firs ∂ y i (cid:17) (Π r s f i ) for i = 1 , N i = r (Π ∗ ,rrsfi ) e r ⊗ e r − (cid:16) ∂ ˜ θi Π ∗ ,rfirs (cid:17) (Π rsfi )(Π ∗ ,rrsfi ) r e r ⊗ e θ + (cid:16) ∂ ˜ ri Π ∗ ,rfirs (cid:17) (Π rsfi )Π ∗ ,rrsfi r e θ ⊗ e θ , y i for i = 1 , N + f i , respectively, (˜ r i , ˜ θ i ) are ( r, θ ) coordinates for y i , respectively, ( r, θ ) = Π ∗ f i r s (˜ r i , ˜ θ i ), Π ∗ ,rr s f i andΠ ∗ ,rf i r s are the r -components of Π ∗ r s f i and Π ∗ f i r s , respectively, and ˜ F i for i = 1 , F changed byusing the transformation Π r s f i for i = 1 , ∇× (Ψ i e ϕ ))(Π r s f i ) = N i ∇× ( ˜Ψ i e ϕ )and (cid:0) ∇ × (cid:0) A πr sin θ e ϕ (cid:1)(cid:1) (Π r s f i ) = N i ∇ × (cid:16) ˜ A i πr sin θ e ϕ (cid:17) . Subtract (4.2.4)-(4.2.6) for i = 1 from thesame equations for i = 2. Then we have ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (( ˜Ψ − ˜Ψ ) e ϕ ) ! (4.2.7) = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ( ˜ T − ˜ T ) r e ϕ + A − A + B − B + ˜ F ( ˜Ψ e ϕ , ˜ A , ˜ T ) − ˜ F ( ˜Ψ e ϕ , ˜ A , ˜ T )(=: ˜ F ) in N + r s , ( ˜Ψ − ˜Ψ ) e ϕ = (cid:18) Π ∗ ,rrsf (Φ − − Φ − )(Π rsf ) − Π ∗ ,rrsf (Φ − − Φ − )(Π rsf ) r (cid:19) e ϕ (=: ˜ h ) on Γ r s , on Γ + w,r s , (cid:16) r sin θ R θ (cid:16) f ( ˜ T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T − ˜ T )+ f ( ˜Ψ e ϕ , ˜ A , ˜ T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:17) r sin ξdξ (cid:17) e ϕ (=: ˜ h ) on Γ ex , (4.2.8)(4.2.9) 1 r sin θ Z θ (cid:16) f ( ˜ T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T − ˜ T )+ f ( ˜Ψ e ϕ , ˜ A , ˜ T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:17) r sin ξdξ = 0where A i = ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × ( ˜Ψ i e ϕ ) ! − M i ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 )(Π r s f i ) N i ∇ × ( ˜Ψ i e ϕ ) ! and B i = − ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ˜ T i r e ϕ + ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ! (Π r s f i ) ( ∂ ˜ θ i Π ∗ ,rf i r s )(Π r s f i ) ∂ r ˜ T i + ∂ θ ˜ T i Π rr s f i e ϕ for i = 1 ,
2. We will estimate || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs using (4.2.7)-(4.2.9). For this, we estimate || ˜ A πr sin θ − ˜ A πr sin θ || , , N + rs and || ˜ T − ˜ T || ,β, N + rs .Estimate || ˜ A πr sin θ − ˜ A πr sin θ || ,β, N + rs and || ˜ T − ˜ T || ,β, N + rs :Since A i and T i for i = 1 , ′ ) for (Ψ , f (0) , f s ) = (Ψ i , f i (0) , f s,i ), by Lemma3.15, A i and T i are represented as A i = 2 πf i ( L i ) sin L i u − ,ϕ ( f i ( L i ) , L i )and T i = g (cid:18) u − · ν f i ( L i ) c − (cid:19) ! S − ! ( f i ( L i ) , L i ) − ( g ( M − ))( r s ) S in where u − ,ϕ = u − · e ϕ , L i for i = 1 , L given in Lemma 3.15 for V = 2 πr sin θ (Φ +0 + Ψ i )and f = f i , respectively, and ν f i for i = 1 , f i pointing toward N + f i , respectively. Using these solution expressions, express ( ˜ A πr sin θ − ˜ A πr sin θ ) e ϕ and ˜ T − ˜ T as πf ( ˜ L ) sin ˜ L u ϕ, − ( f ( ˜ L ) , ˜ L )2 πr sin θ − πf ( ˜ L ) sin ˜ L u ϕ, − ( f ( ˜ L ) , ˜ L )2 πr sin θ ! e ϕ (4.2.10)and g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L ) − g u − · ν f ( ˜ L ) c − ! S − ( f ( ˜ L ) , ˜ L ) , (4.2.11)respectively, where ˜ L i := L i (Π r s f i ) for i = 1 ,
2. Using arguments similar to the ones used toestimate || ( A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + f and || T − ˜ T || ,β, N + f in the proof of the uniqueness partof Proposition 3.1, we estimate (4.2.10) and (4.2.11) in C β ( N + r s ). Then we obtain || ( ˜ A πr sin θ − ˜ A πr sin θ ) e ϕ || , , N + rs ≤ Cσ (cid:16) || f − f || ,β, Λ + || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs (cid:17) (4.2.12)and || ˜ T − ˜ T || ,β, N + rs ≤ C || f − f || ,β, Λ + Cσ || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs . (4.2.13)Using these estimates, we estimate || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs . Substitute (4.2.11) into (4.2.9).And then using (4.2.12) and (4.2.13), estimate f (0) − f (0) in the resultant equation (see Step4 in the proof of the uniqueness part of Proposition 3.1). Then we obtain | f (0) − f (0) | ≤ C σ | f (0) − f (0) | + C || f ′ s, − f ′ s, || ,β, (0 ,θ ) + Cσ || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs . Take σ = min( σ (1)1 , C )(=: σ (2)1 ). Then we have | f (0) − f (0) | ≤ C || f ′ s, − f ′ s, || ,β, (0 ,θ ) + Cσ || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs . (4.2.14)Using (4.2.14), we obtain from (4.2.12) and (4.2.13) || ˜ A πr sin θ − ˜ A πr sin θ || ,β, N + rs ≤ Cσ (cid:16) || f ′ s, − f ′ s, || ,β, (0 ,θ ) + || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs (cid:17) (4.2.15)and || ˜ T − ˜ T || ,β, N + rs ≤ C || f ′ s, − f ′ s, || ,β, (0 ,θ ) + Cσ || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs . (4.2.16)Using these two estimates, (4.2.14) and arguments similar to the ones in Step 5 in the proof ofthe uniqueness part of Proposition 3.1, we estimate ( ˜Ψ − ˜Ψ ) e ϕ in (4.2.7), (4.2.8) in C ,β ( N + r s ).Then we obtain || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs ≤ C σ || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs + C || f ′ s, − f ′ s, || ,β, (0 ,θ ) . -D AXISYMMETRIC TRANSONIC SHOCK 67 Take σ = min( σ (2)1 , C )(=: σ (3)1 ). Then we have || ( ˜Ψ − ˜Ψ ) e ϕ || ,β, N + rs ≤ C || f ′ s, − f ′ s, || ,β, (0 ,θ ) (4.2.17)Combining (4.2.14), (4.2.15), (4.2.16) and (4.2.17), we obtain (4.2.3).2. Using the arguments in Step 1 in the proof of Lemma 4.2, we can see that the sys-tem (4.1.8)-(4.1.12) has a unique solution for given ˜ f ′ s = f ′ s, − f ′ s, ∈ C β ([0 , θ ]) := { f ∈ C β ([0 , θ ]) | f ′ (0) = f ′ ( θ ) = 0 } : f (0) ( f ′ s, − f ′ s, ) = − R θ ( f s, − f s, ) sin ζdζ R θ sin ζdζ , (4.2.18) ˜ A ( f ′ s, − f ′ s, ) = 0 , (4.2.19) ˜ T ( f ′ s, − f ′ s, ) = ( g ( M − )) ′ ( r s ) S in − R θ ( f s, − f s, ) sin ζdζ R θ sin ζdζ + f s, − f s, ! (4.2.20)and ˜Ψ ( f ′ s, − f ′ s, ) e ϕ , the unique C ,β ( − − α, Γ + w,rs ) ( N + r s ) solution of (4.1.17), (4.1.18) for given ˜ f ′ s = f ′ s, − f ′ s, ∈ C β ([0 , θ ]) (here we had ˜Ψ ( f ′ s, − f ′ s, ) e ϕ ∈ C ,β ( − − α, Γ + w,rs ) ( N + r s ) because f ′ s, − f ′ s, ∈ C β ([0 , θ ])). Subtract (4.1.8)-(4.1.10) for given ˜ f ′ s = f ′ s, − f ′ s, ∈ C β ([0 , θ ]) from (4.2.7)-(4.2.9).Then we obtain(4.2.21) ∇ × ρ +0 (1 + u +0 e r ⊗ u +0 e r c +0 2 − u +0 2 ) ∇ × (( ˜Ψ − ˜Ψ − ˜Ψ ( f ′ s, − f ′ s, ) ) e ϕ ) ! = ρ +0 γ − ( γ − u +0 (1 + γu +0 2 c +0 2 − u +0 2 ) ∂ θ ( ˜ T − ˜ T − ˜ T ( f ′ s, − f ′ s, ) ) r e ϕ + A − A + B − B + ˜ F ( ˜Ψ e ϕ , ˜ A , ˜ T ) − ˜ F ( ˜Ψ e ϕ , ˜ A , ˜ T ) in N + r s , (4.2.22)( ˜Ψ − ˜Ψ − ˜Ψ ( f ′ s, − f ′ s, ) ) e ϕ = (cid:18) Π ∗ ,rrsf (Φ − − Φ − )(Π rsf ) − Π ∗ ,rrsf (Φ − − Φ − )(Π rsf ) r (cid:19) e ϕ on Γ r s , on Γ + w,r s , r sin θ R θ (cid:18) f ( ˜ T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T − ˜ T − ˜ T ( f ′ s, − f ′ s, ) )+ f ( ˜Ψ e ϕ , ˜ A , ˜ T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:19) r sin ξdξ ! e ϕ on Γ ex , (4.2.23) 1 r sin θ Z θ (cid:18) f ( ˜ T , p ex ) − f ( ˜ T , p ex ) − ρ +0 (( γ − u +0 2 + c +0 2 ) γ ( γ − u +0 S +0 ( ˜ T − ˜ T − ˜ T ( f ′ s, − f ′ s, ) )+ f ( ˜Ψ e ϕ , ˜ A , ˜ T ) − f ( ˜Ψ e ϕ , ˜ A , ˜ T ) (cid:19) r sin ξdξ = 0 . Estimate || ˜Ψ − ˜Ψ − ˜Ψ ( f ′ s, − f ′ s, ) || ,β, N + rs using (4.2.21)-(4.2.23).Write ˜ T ( f ′ s, − f ′ s, ) as ( g ( M − )) ′ ( r s ) S in (cid:16) f (0) ( f ′ s, − f ′ s, ) + f s, − f s, (cid:17) . (4.2.24) Substitute this ˜ T ( f ′ s, − f ′ s, ) and (4.2.11) into the places of ˜ T ( f ′ s, − f ′ s, ) and ˜ T − ˜ T in (4.2.23),respectively. And then using (4.2.3), estimate f (0) − f (0) − f (0) ( f ′ s, − f ′ s, ) in the resultantequation. Then we obtain | f (0) − f (0) − f (0) ( f ′ s, − f ′ s, ) | ≤ Cσ || f ′ s, − f ′ s, || ,β, (0 ,θ ) . Using this estimate and (4.2.3), we estimate ˜ T − ˜ T − ˜ T ( f ′ s, − f ′ s, ) given by (4.2.11) and (4.2.24)in C β ( N + r s ). Then we obtain || ˜ T − ˜ T − ˜ T ( f ′ s, − f ′ s, ) || ,β, N + rs ≤ Cσ || f ′ s, − f ′ s, || ,β, (0 ,θ ) . (4.2.25)Using arguments similar to the ones in Step 5 in the proof of the uniqueness part of Proposition3.1 with (4.2.3) and (4.2.25), we estimate ˜Ψ − ˜Ψ − ˜Ψ ( f ′ s, − f ′ s, ) in (4.2.21), (4.2.22) in C ,β ( N + r s ).Then we have || ˜Ψ − ˜Ψ − ˜Ψ ( f ′ s, − f ′ s, ) || ,β, N + rs ≤ Cσ || f ′ s, − f ′ s, || ,β, (0 ,θ ) . (4.2.26)3. By the assumption, ( f i , Φ i e ϕ , L i , S i ) for i = 1 , A ( ρ − , u − , p − , p ex , f ′ s,i ) = 0 for i = 1 , A is a map defined in (4.1.1). Subtract thesetwo equations. Then we have A ( ρ − , u − , p − , p ex , f ′ s, ) − A ( ρ − , u − , p − , p ex , f ′ s, ) = 0 . Write this as 0 = D f ′ s A ( ζ )( f ′ s, − f ′ s, ) + R (4.2.27)where R = A ( ρ − , u − , p − , p ex , f ′ s, ) − A ( ρ − , u − , p − , p ex , f ′ s, ) − D f ′ s A ( ζ )( f ′ s, − f ′ s, )and D f ′ s A ( ζ ) is a map given in (4.1.2) as a map from C β ([0 , θ ]) to C β ([0 , θ ]) (here D f ′ s A ( ζ )takes C β ([0 , θ ]) functions and ˜Ψ ( f ′ s, − f ′ s, ) e ϕ in D f ′ s A ( ζ )( f ′ s, − f ′ s, ) is the C ,β ( − − α, Γ + w,rs ) ( N + r s )solution of (4.1.17), (4.1.18) for given ˜ f ′ s = f ′ s, − f ′ s, ∈ C β ([0 , θ ])).Using (4.2.3) and (4.2.26), we can obtain || R || ,β, (0 ,θ ) ≤ Cσ || f ′ s, − f ′ s, || ,β, (0 ,θ ) . (4.2.28)In the way that we proved the invertiblity of D f ′ s A ( ζ ) as a map from C ,α ( − α, { θ = θ } ) ((0 , θ )) to C ,α ( − α, { θ = θ } ) ((0 , θ )) in the the proof of Lemma 4.4, we can prove that D f ′ s A ( ζ ) is an invertiblemap as a map from C β ([0 , θ ]) to C β ([0 , θ ]). Using this fact and (4.2.28), we obtain from(4.2.27) || f ′ s, − f ′ s, || ,β, (0 ,θ ) ≤ C σ || f ′ s, − f ′ s, || ,β, (0 ,θ ) . Take σ = min( σ (3)1 , C )(=: σ ). Then we have f ′ s, = f ′ s, . One can see that σ depends onthe data. This finishes the proof. (cid:3) Appendix
In this section, we present some computations done by using the tensor notation given in § q , q , q ) be an orthogonal coordinate system in R . The unit vectors in this coordinatesystem in the direction of q i for i = 1 , , h i ∂ x ∂q i (=: e q i ) for i = 1 , , x = x e + y e + z e and h i := | ∂ x ∂q i | . By ∇ = P i =1 e qi h i ∂ q i , ∇ U where U : R → R can bewritten as ∇ U = X i =1 ∂ q i U h i ⊗ e q i . (5.0.1) -D AXISYMMETRIC TRANSONIC SHOCK 69 Here, a ⊗ b for a , b ∈ R denotes ab T . By (3.2.7), the multiplication of the tensor notationdefined in § a ⊗ b ⊗ c ⊗ d )( e ⊗ f ⊗ g ⊗ h ) = ( d · e )( c · f )( a ⊗ b ⊗ g ⊗ h )(5.0.2)where a , b , c , d , e , f , g , h ∈ R . By (5.0.1) and (5.0.2), M in ∇ U = M ∇ ( q ,q ,q ) U , where ∇ ( q ,q ,q ) U = P i =1 ∂ q i U ⊗ e i and M = ∂ ( q ,q ,q ) ∂ ( x,y,z ) , can be expressed as M = X i =1 h i I ⊗ e q i ⊗ e i ⊗ I . (5.0.3)Here, I ⊗ a ⊗ b ⊗ I for a , b ∈ R is a linear map from R × to R × defined by I ⊗ a ⊗ b ⊗ I := X i =1 e i ⊗ a ⊗ b ⊗ e i . (5.0.4)(Note that I ⊗ a ⊗ b ⊗ I satisfies( I ⊗ a ⊗ b ⊗ I )( c ⊗ d ) = ( b · d ) c ⊗ a (5.0.5)for c , d ∈ R .) In the standard way, we transform div( B ∇ U ) = 0 in the cartesian coordinatesystem where B is a linear map from R × to R × into an equation in ( q , q , q )-coordinatesystem. Then we obtaindiv ( q ,q ,q ) (cid:18) M M T B ◦ χ − M ∇ ( q ,q ,q ) ( U ◦ χ − ) (cid:19) = 0(5.0.6)where div ( q ,q ,q ) := ∇ ( q ,q ,q ) · , ∇ ( q ,q ,q ) and M are as above, χ is the map from ( x, y, z ) to( q , q , q ) and det M = h h h . Using (5.0.2) and (5.0.3), we compute M T B ◦ χ − M . Thenwe have the explicit form of (5.0.6). In this way, we obtain the explicit form of the sphericalcoordinate representation of (3.2.30).Next, we show that (3.2.8) is equivalent to (3.2.1).By (5.0.1), ∇ (Ψ e ϕ ) = ∂ r Ψ e ϕ ⊗ e r + ∂ θ Ψ r e ϕ ⊗ e θ − Ψ r e r ⊗ e ϕ − cos θr sin θ Ψ e θ ⊗ e ϕ . (5.0.7)Using (5.0.5) and (5.0.7), we have c +0 2 ρ +0 ( c +0 2 − u +0 2 ) I − u +0 2 c +0 2 ( I ⊗ e r ⊗ e r ⊗ I ) ! ∇ (Ψ e ϕ )= c +0 2 ρ +0 ( c +0 2 − u +0 2 ) ∇ (Ψ e ϕ ) − u +0 2 ρ +0 ( c +0 2 − u +0 2 ) ∂ r Ψ e ϕ ⊗ e r . Using this relation and the relation div( a ⊗ b ) = ∇ ab + a div b , we compute the left-hand sideof (3.2.8). Then we get a ( r )∆(Ψ e ϕ ) + ∇ (Ψ e ϕ ) ∇ a ( r ) − ∇ e ϕ b ( r ) ∂ r Ψ e r − e ϕ div( b ( r ) ∂ r Ψ e r ) − ∂ r ρ +0 ρ +0 2 r Ψ e ϕ (5.0.8)where a ( r ) := c +0 2 ρ +0 ( c +0 2 − u +0 2 ) and b ( r ) := u +0 2 ρ +0 ( c +0 2 − u +0 2 ) . By direct computation, one can check that ∇ e ϕ b ( r ) ∂ r Ψ e r = 0. Using this fact and (5.0.7), we compute (5.0.8). Then we have (cid:18) a ( r )(∆Ψ − Ψ r sin θ ) + a ′ ( r ) ∂ r Ψ − b ( r )( 1 r ∂ r ( r ∂ r Ψ)) − b ′ ( r ) ∂ r Ψ (cid:19) e ϕ − ∂ r ρ +0 ρ +0 2 r Ψ e ϕ . (5.0.9) By a − b = ρ +0 , ( a − b ) ′ = − ∂ r ρ +0 ρ +0 2 . Substitute this relation into (5.0.9) and then express theresultant equation in the spherical coordinate system. Then we obtain ρ +0 r ∂ r ( r ∂ r Ψ) − ∂ r ρ +0 ρ +0 2 r ∂ r ( r Ψ) + c +0 2 ρ +0 ( c +0 2 − u +0 2 ) r (cid:18) θ ∂ θ (sin θ∂ θ Ψ) − Ψsin θ (cid:19)! e ϕ . This after multiplied by ( −
1) is equal to the left-hand side of (3.2.3). By this fact, (3.2.8) isequivalent to (3.2.1).
Acknowledgements:
The author would like to thank Myoungjean Bae for her valuable advicein doing this work. The research of Yong Park was supported in part by Samsung Science andTechnology Foundation under Project Number SSTF-BA1502-02.
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Yong Park, Department of Mathematics, POSTECH, San 31, Hyojadong, Namgu, Pohang, Gyung-buk, Republic of Korea 37673
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