On the existence and regularity of solutions of semi-hyperbolic patches to 2-D Euler equations with van der Waals gas
OO N THE EXISTENCE AND REGULARITY OF SOLUTIONS OFSEMI - HYPERBOLIC PATCHES TO
ULER EQUATIONS WITHVAN DER W AALS GAS
Rahul Barthwal
Department of Mathematics,Indian Institute of Technology Kharagpur,Kharagpur, India [email protected]
T. Raja Sekhar
Department of Mathematics,Indian Institute of Technology Kharagpur,Kharagpur, India [email protected]
February 9, 2021 A BSTRACT
This article is concerned in establishing the existence and regularity of solution of semi-hyperbolicpatch problem for two-dimensional isentropic Euler equations with van der Waals gas. This type ofsolution appears in the transonic flow over an airfoil and Guderley reflection and is very commonin the numerical solution of Riemann problems. We use the idea of characteristic decompositionand bootstrap method to prove the existence of global smooth solution which is uniformly C , continuous up to the sonic curve. We also prove that the sonic curve is C , continuous. Further, weshow the formation of shock as an envelope for positive characteristics before reaching their sonicpoints. Cauchy problem in several space dimension for hyperbolic system of conservation laws is a very important butcomplicated open problem. A particular kind of Cauchy problem in two-dimensional case is the two-dimensionalRiemann problem which consists of initial data that are constant along any ray passing through origin. The study oftwo-dimensional Riemann problems are very interesting and challenging in the context of two-dimensional hyperbolicsystem of conservation laws. A significant research has been done for the two-dimensional compressible Euler systemand various other important models for a typical case of two-dimensional Riemann problems which is known as the fourwave Riemann problem. The four wave Riemann problem is an initial value problem where the initial data are constantin each of the four quadrants of the physical plane. For the two-dimensional compressible Euler system, a beautifulconjecture for the possible structures of solution for four wave Riemann problem was provided in the ground breakingpaper of Zhang and Zheng [1]. Several numerical results have been obtained in the field of two-dimensional Riemannproblems for gasdynamics equations and many small-scale structures have been observed in those numerical simulations[2, 3, 4]. One of such structures is the semi-hyperbolic patch which appears very often in many cases of two-dimensionalRiemann problems for compressible Euler system, pressure-gradient equations, magnetohydrodynamics and etc(Fordetails see [5, 6, 7, 8, 9]). The semi-hyperbolic patch is defined as a patch kind of solution in which one set ofcharacteristics starts on a sonic curve and ends on either a transonic shock wave or a sonic curve. These type of solutionsappear in many other situations too such as reflection of rarefaction wave along a compressive corner [10], transonicflow over an airfoil [11] and Guderley shock reflection of the von Neumann triple point paradox [12, 13]. These patchtype solutions are very meaningful and important for the construction of global solution of mixed-type equations infuture.The semi-hyperbolic patch, first time, has been identified among the small-scale structures in the work of Song andZheng [7] for pressure-gradient system. The same problem for isothermal Euler equations was studied by Hu et al.in [14] while for isentropic case by Li and Zheng in [8] and extended to magnetohydrodynamic system by Chen andLai in [9]. The regularity of the solution of semi-hyperbolic patch problems have also been widely discussed. The a r X i v : . [ m a t h . A P ] F e b PREPRINT - F
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9, 2021regularity of solution of the semi-hyperbolic patch problem for the pressure-gradient system was discussed in [15]and for isentropic Euler system in [16]. The regularity results for isothermal Euler equations have been discussedby Hu et al. [17]. The regularity results obtained for isentropic Euler system in [16] were improved by Hu et al. in[18]. For more details on ongoing research on two-dimensional Euler system and related models, we refer the readerto [19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34]. All the above works are based on the beautifulconcept of characteristic decomposition initiated in the work of Dai and Zhang [35]. The progress made in the field ofsemi-hyperbolic patch problems leads to a natural question of determining whether these results can be extended formore realistic gases, for instance, van der Waals gas. The main purpose of this article is to establish the existence andregularity results for the solution of the semi-hyperbolic patch problem for two-dimensional isentropic Euler equationswith van der Waals gas.Let us consider the two-dimensional isentropic compressible Euler equations [16] as follows: ρ t + ( ρu ) x + ( ρv ) y = 0 , ( ρu ) t + ( ρu + p ) x + ( ρuv ) y = 0 , ( ρv ) t + ( ρuv ) x + ( ρv + p ) y = 0 , (1.1)where ρ denotes the density, u and v denotes the flow velocity in the x and y direction, respectively and p denotes thepressure of the gas.We consider a polytropic van der Waals gas with the equation of state [36] as p ( τ ) = K ( τ − b ) γ +1 − aτ where γ is aconstant such that γ ∈ (0 , . The quantity τ = ρ is known as the specific volume of the gas, K is a positive constantwhich depends on the entropy of the system. The attraction between the gas molecules is represented by the positiveconstant a and the compressibility limit of these molecules in the gas is represented by the positive constant b . For thecase a = 0 , this corresponds to dusty gas while for a = b = 0 , this behaves as polytropic ideal gas.The expression for speed of sound is given by c ( τ ) = (cid:112) − τ p (cid:48) ( τ ) with p (cid:48) ( τ ) = − K ( γ + 1)( τ − b ) γ +2 + 2 aτ , p (cid:48)(cid:48) ( τ ) = K ( γ + 1)( γ + 2)( τ − b ) γ +3 − aτ . (1.2)We give the following list of notations, these are very important in the further discussion of the paper κ ( τ ) = − p (cid:48) ( τ )2 p (cid:48) ( τ ) + τ p (cid:48)(cid:48) ( τ ) = 2 − bτ − a ( τ − b ) γ +3 K ( γ +1) τ γ + bτ + a ( τ − b ) γ +3 K ( γ +1) τ ,m ( τ ) = κ ( τ ) − κ ( τ ) + 1 = 2 − γ − bτ − a ( τ − b ) γ +3 K ( γ +1) τ γ + 2 + a ( τ − b ) γ +3 K ( γ +1) τ ,µ ( τ ) = 11 + κ ( τ ) , Ω( τ, ω ) = m ( τ ) − tan ω, ¯ ∂ + = cos α∂ ξ + sin α∂ η , ¯ ∂ − = cos β∂ ξ + sin β∂ η ,σ = α + β , ω = α − β , tan α = λ + , tan β = λ − (1.3)where α and β are defined as characteristic angles and ¯ ∂ ± denotes the normalized directional derivatives along thecharacteristic directions in self-similar plane [25].From the expressions above we can observe that for sufficiently large τ > b , when τ > τ we have p (cid:48) ( τ ) < , p (cid:48)(cid:48) ( τ ) > , κ ( τ ) > and < m ( τ ) < . Further, for some technical reasons we adopt the hypothesis that κ (cid:48) ( τ ) > withoutloss of generality.This paper is organized as follows. In Section we give some preliminaries and mainly interested with characteristicdecompositions in terms of characteristic angles and the speed of sound. We define our problem precisely and establishthe boundary data estimates to prove the existence of local solution in Section . Section is devoted to constructthe uniform lower and upper bounds of the characteristic directional derivatives of speed of sound. We discuss theglobal existence of solution by extending the local solution up to the sonic boundary by solving many small Goursatproblems in each step of extension in Section . In Section we study the formation of shock as an envelope forpositive characteristics before reaching their sonic points. The regularity of solution in partial hodograph plane andself-similar plane is established in Section and , respectively. In Section we provide the concluding remarks.2 PREPRINT - F
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We denote U = ( u − ξ ) and V = ( v − η ) as pseudo-flow velocity. Then in the self-similar plane (cid:16) ξ = xt , η = yt (cid:17) , thereduced Euler equations can be written as ( ρU ) ξ + ( ρV ) η + 2 ρ = 0 ,U U ξ + V U η + τ p ξ + U = 0 ,U V ξ + V U η + τ p η + V = 0 . (2.1)Here we assume that the flow is irrotational which implies that v ξ = u η . Now we introduce a potential function φ ( ξ, η ) such that φ ξ = U and φ η = V . Then using the last two equations of system (2.1) we can easily obtain thepseudo-Bernoulli’s law U + V K ( τ − b ) γ (cid:18) γ + 1 γ + bτ − b (cid:19) − aτ + φ = a (2.2)where a is a constant which can be taken as throughout this article without loss of generality.The above system (2.1) can be reduced into matrix form as (cid:20) uv (cid:21) ξ + − U Vc − U c − V c − U − (cid:20) uv (cid:21) η = 0 , (2.3)which gives the eigenvalues λ ± = U V ± c √ U + V − c U − c with corresponding left eigenvectors l ± = (1 , λ ∓ ) .Multiplying l ± with the system (2.3) we obtain the system of characteristic equations as ¯ ∂ ± u + λ ∓ ¯ ∂ ± v = 0 . (2.4) Here we provide first order characteristic decompositions of characteristic angles without proof. The proofs of thesedecompositions can be found in [37].Using [37] we can obtain U = c cos σ sin ω and V = c sin σ sin ω with ¯ ∂ − c = µ ( τ )tan ω ( c ¯ ∂ − α − ω ) ,c ¯ ∂ − β = Ω( τ, ω ) cos ω ( c ¯ ∂ − α − ω ) = Ω( τ, ω )2 µ ( τ ) sin 2 ω ¯ ∂ − c, ¯ ∂ + c = − µ ( τ )tan ω ( c ¯ ∂ + β + 2 sin ω ) ,c ¯ ∂ + α = Ω( τ, ω ) cos ω ( c ¯ ∂ + β + 2 sin ω ) = − Ω( τ, ω )2 µ ( τ ) sin 2 ω ¯ ∂ + c,c ¯ ∂ ± ω = tan ω (1 + κ ( τ ) sin ω ) ¯ ∂ ± c + sin ω. (2.5) In this section we derive characteristic decomposition form for the variables α , β and c which is important and veryuseful for establishing a priori gradient estimates of solution. First we cite the following second order decompositionsof c from Lai [37]. Proposition 2.1.
The variable c satisfies the following characteristic decompositions c ¯ ∂ + ¯ ∂ − c = ¯ ∂ − c (cid:40) sin 2 ω + ¯ ∂ − c µ ( τ ) cos ω + (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) ¯ ∂ + c (cid:41) ,c ¯ ∂ − ¯ ∂ + c = ¯ ∂ + c (cid:40) sin 2 ω + ¯ ∂ + c µ ( τ ) cos ω + (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) ¯ ∂ − c (cid:41) . (2.6)3 PREPRINT - F
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9, 2021Using the above decompositions we are able to prove the following important decompositions for the variable c . Corollary 2.1. c satisfies the following second order decompositions ¯ ∂ + (cid:18) ¯ ∂ − cc (cid:19) = ¯ ∂ − cc (cid:40) (cid:18) ¯ ∂ + c + ¯ ∂ − c cµ ( τ ) cos ω (cid:19) + sin 2 ωc + (cid:0) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ ) sin ω ) (cid:1) ¯ ∂ + cc (cid:41) , (2.7) ¯ ∂ − (cid:18) ¯ ∂ + cc (cid:19) = ¯ ∂ + cc (cid:40) (cid:18) ¯ ∂ + c + ¯ ∂ − c cµ ( τ ) cos ω (cid:19) + sin 2 ωc + (cid:0) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ ) sin ω ) (cid:1) ¯ ∂ − cc (cid:41) . (2.8) Proof.
The proof of this corollary can be obtained by using direct calculations from the decompositions in proposition2.1. So we omit the details.
Corollary 2.2. c satisfies the following second order equations in homogeneous form c ¯ ∂ + (cid:18) − ¯ ∂ − c sin ω (cid:19) = (cid:18) − ¯ ∂ − c sin ω (cid:19) (cid:40) (cid:18) ¯ ∂ + c + ¯ ∂ − c µ ( τ ) cos ω (cid:19) − (cid:0) τ κ (cid:48) ( τ ) + 4 κ ( τ ) sin ω + 2 (cid:1) ¯ ∂ + c (cid:41) , (2.9) c ¯ ∂ − (cid:18) ¯ ∂ + c sin ω (cid:19) = (cid:18) ¯ ∂ + c sin ω (cid:19) (cid:40) (cid:18) ¯ ∂ + c + ¯ ∂ − c µ ( τ ) cos ω (cid:19) − (cid:0) τ κ (cid:48) ( τ ) + 4 κ ( τ ) sin ω + 2 (cid:1) ¯ ∂ − c (cid:41) . (2.10) Proof.
This corollary is a direct consequence of the proposition 2.1. Hence we omit its proof.
Proposition 2.2.
The variables α and β satisfy the following second order decompositions c ¯ ∂ + ¯ ∂ − α + Ψ ¯ ∂ − α = (cid:40) tan ω (1 − ω ) + 2 tan ωτ κ (cid:48) ( τ )Ω( τ, ω ) ( µ ( τ )) (cid:41) ¯ ∂ + α, (2.11) c ¯ ∂ − ¯ ∂ + β + Ψ ¯ ∂ + β = (cid:40) tan ω (1 − ω ) + 2 tan ωτ κ (cid:48) ( τ )Ω( τ, ω ) ( µ ( τ )) (cid:41) ¯ ∂ − β, (2.12) in which Ψ = sin ω (2 tan ω − Ω( τ, ω ) sin 2 ω ) + τ κ (cid:48) ( τ ) sin 2 ω ( µ ( τ )) − c sin 2 ω ¯ ∂ − α + (cid:40) ω − Ω( τ, ω ) sin 2 ω τ κ (cid:48) ( τ )tan ω ( µ ( τ )) (cid:41) c ¯ ∂ + β, and Ψ = sin ω (2 tan ω − Ω( τ, ω ) sin 2 ω ) + τ κ (cid:48) ( τ ) sin 2 ω ( µ ( τ )) + c sin 2 ω ¯ ∂ + β − (cid:40) ω − Ω( τ, ω ) sin 2 ω τ κ (cid:48) ( τ )tan ω ( µ ( τ )) (cid:41) c ¯ ∂ − α. Proof.
Using the decomposition of the variable c from (2.5) in (2.6) we obtain c ¯ ∂ + (cid:34) µ ( τ ) c ¯ ∂ − α tan ω − µ ( τ ) sin 2 ω (cid:35) = (cid:34) µ ( τ ) c ¯ ∂ − α tan ω − µ ( τ ) sin 2 ω (cid:35) × (cid:40) sin 2 ω + c ¯ ∂ − β Ω( τ, ω ) cos ω sin 2 ω − (cid:18) Ω( τ, ω ) cos 2 ω µ ( τ ) + 1 + τ κ (cid:48) ( τ ) (cid:19) cµ ( τ ) ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:41) (2.13)Now we compute ¯ ∂ + (cid:18) µ ( τ ) c tan ω (cid:19) = c tan ω (cid:0) µ ( τ ) (cid:1) (cid:48) ¯ ∂ + τ − µ ( τ ) c ω ( ¯ ∂ + α − ¯ ∂ + β ) + µ ( τ )tan ω ¯ ∂ + c = − τ ( µ ( τ )) ( κ ( τ )) (cid:48) c ¯ ∂ + α Ω( τ, ω ) tan ω sin 2 ω − µ ( τ ) − µ ( τ )) c ¯ ∂ + α Ω( τ, ω ) tan ω sin 2 ω + µ ( τ ) c ¯ ∂ + α Ω( τ, ω ) sin 2 ω sin ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) (2.14)4 PREPRINT - F
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9, 2021and c ¯ ∂ + (cid:0) µ ( τ ) sin 2 ω (cid:1) = c sin 2 ω ( µ ( τ )) (cid:48) ¯ ∂ + τ + µ ( τ ) c cos 2 ω [ ¯ ∂ + α − ¯ ∂ + β ]= − τ ( µ ( τ )) ( κ ( τ )) (cid:48) c ¯ ∂ + α Ω( τ, ω ) − µ ( τ ) sin ω + µ ( τ ) sin (2 ω ) − µ ( τ ) cos 2 ωc ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) . (2.15)so that the L.H.S. of (2.13) becomes c ¯ ∂ + (cid:34) µ ( τ ) c ¯ ∂ − α tan ω − µ ( τ ) sin 2 ω (cid:35) = µ ( τ ) c tan ω c ¯ ∂ + ¯ ∂ − α + c ¯ ∂ − α ¯ ∂ + (cid:18) µ ( τ ) c tan ω (cid:19) − c ¯ ∂ + (cid:0) µ ( τ ) sin 2 ω (cid:1) = µ ( τ ) c tan ω c ¯ ∂ + ¯ ∂ − α + τ ( µ ( τ )) ( κ ( τ )) (cid:48) c ¯ ∂ + α Ω( τ, ω ) + 2 µ ( τ ) sin ω − µ ( τ ) sin (2 ω )+ c ¯ ∂ − α (cid:40) µ ( τ ) c ¯ ∂ + α Ω( τ, ω ) sin 2 ω sin ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) − τ ( µ ( τ )) ( κ ( τ )) (cid:48) c ¯ ∂ + α Ω( τ, ω ) tan ω sin 2 ω − µ ( τ )) c ¯ ∂ + α Ω( τ, ω ) tan ω sin 2 ω − µ ( τ ) (cid:41) + 2 µ ( τ ) cos 2 ωc ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) . (2.16)While the R.H.S. of (2.13) is µ ( τ ) c tan ω ¯ ∂ − α (cid:40) sin 2 ω − ω + c ¯ ∂ − α sin 2 ω − (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) cµ ( τ ) ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:41) + (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) µ ( τ )) c ¯ ∂ + α Ω( τ, ω ) + 2 µ ( τ ) sin ω − µ ( τ ) sin (2 ω ) . (2.17)Comparing L.H.S. and R.H.S. of (2.13) we have c ¯ ∂ + ¯ ∂ − α + c ¯ ∂ − α (cid:40) tan ω ¯ ∂ + α Ω( τ, ω ) sin 2 ω sin ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) − µ ( τ ) ¯ ∂ + α Ω( τ, ω ) sin 2 ω − tan ωc − τ (( µ ( τ )) ( κ ( τ )) (cid:48) ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:41) + τ ( µ ( τ )) ( κ ( τ )) (cid:48) tan ω ¯ ∂ + α Ω( τ, ω ) + 2 tan ω cos 2 ω ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) = ¯ ∂ − α (cid:40) sin 2 ω − ω + c ¯ ∂ − α sin 2 ω − (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) cµ ( τ ) ¯ ∂ + α Ω( τ, ω ) sin 2 ω (cid:41) + (cid:18) τ, ω ) cos 2 ω µ ( τ ) + τ κ (cid:48) ( τ ) (cid:19) µ ( τ ) tan ω ¯ ∂ + α Ω( τ, ω ) , (2.18)which can be reduced as c ¯ ∂ + ¯ ∂ − α + Ψ ¯ ∂ − α = Υ ¯ ∂ + α (2.19)where the coefficient Ψ = (cid:18) sin 2 ω c ¯ ∂ + β ω (cid:19) (cid:34) tan ω sin ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) + Ω( τ, ω ) cos 2 ω + 2 τ κ (cid:48) ( τ )( µ ( τ )) (cid:35) + tan ω − sin 2 ω − c ¯ ∂ − α sin 2 ω . (2.20)By a direct calculation, we obtain the coefficient of c ¯ ∂ + β as ω − Ω( τ, ω ) sin 2 ω τ κ (cid:48) ( τ )tan ω ( µ ( τ )) . (2.21)5 PREPRINT - F
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9, 2021Also, we calculate directly sin 2 ω (cid:40) tan ω sin ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) + Ω( τ, ω ) cos 2 ω + 2 τ κ (cid:48) ( τ )( µ ( τ )) (cid:41) + tan ω − sin 2 ω − c ¯ ∂ − α sin 2 ω = sin ω (2 tan ω − Ω( τ, ω ) sin 2 ω ) + sin 2 ωτ κ (cid:48) ( τ )( µ ( τ )) − c ¯ ∂ − α sin 2 ω , (2.22)Using (2.21) and (2.22) in (2.20) we obtain the value of Ψ .Furthermore, the coefficient Υ = − ω cos 2 ω Ω( τ, ω ) sin 2 ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) + (cid:18) τ, ω ) cos 2 ω µ ( τ ) (cid:19) µ ( τ ) tan ω Ω( τ, ω )+ 2 tan ωτ κ (cid:48) ( τ )Ω( τ, ω ) ( µ ( τ )) (2.23)By straightforward computation we have − ω cos 2 ω Ω( τ, ω ) sin 2 ω (cid:18) tan ω − Ω( τ, ω ) sin 2 ω (cid:19) + (cid:18) τ, ω ) cos 2 ω µ ( τ ) (cid:19) µ ( τ ) tan ω Ω( τ, ω )= − ω Ω( τ, ω ) sin 2 ω (cid:104) µ ( τ ) sin 2 ω − Ω( τ, ω ) sin 2 ω cos 2 ω − cos 2 ω tan ω (cid:105) = 2 tan ω (cid:18) − cos 2 ω (cid:19) = tan ω (1 − ω ) . (2.24)Then the use of (2.24) in (2.23) yields the value of Υ .Using the values of Ψ and Υ in (2.19) we can obtain the proof of the first part of the Lemma. In a similar manner theother part of the Lemma can be proved. In this section we give a brief description of our problem for (1.1). Let us assume that R ( η ) be the planar rarefactionwave for (1.1) connecting two constant states ( ρ , , v ) and ( ρ , , ( c > c > , v > ) in the self-similar planewhich is defined as R ( η ) = η = c + v ( ρ ) , η ≤ η ≤ η v = (cid:90) ρρ cρ dρ, < ρ < ρ < ρ u = u = u = 0 , (3.1)in the region ξ > , where η i = v i + c i , i = 1 , .The value v is obtained from the solution and we denote A by (0 , v + c ) and B as the intersection point of positivecharacteristic passing through A and the bottom boundary in the rarefaction wave region. Now under these assumptionswe define our problem as follows. Let us consider a positive characteristic AB in a planar rarefaction wave region. Let BD be the tangential extensionof AB into a constant state such that both the points A and D are sonic. Let BC be a strictly convex negativecharacteristic such that the endpoint C is a sonic point. Then establish a solution in a maximal hyperbolic region withthe boundary data combined on the curve ABD and BC which means that the region starts from sonic points andends on either a sonic curve or an envelope of the positive family of characteristics. Further, establish the regularity ofsolution; see Figure . PREPRINT - F
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9, 2021Figure 1: The semi-hyperbolic patch.
The boundary data on the characteristics AB and BC are [8] β | AB = 0 , π/ ≤ α | BC ≤ π + β C ,π/ ≤ α | AB ≤ π, β C ≤ β | BC ≤ , − π/ < β C < , π < α − β < π , ¯ ∂ + α | BA > , ¯ ∂ + c | BA > , − ¯ ∂ − β | BC < , − ¯ ∂ − c | BC > . (3.2)where β C is the inclination angle of the negative characteristic BC at point C .Hence our problem is a Goursat problem where the boundaries AB and BC are characteristic boundaries starting fromthe point B . To prove the existence of local solution, we parameterize the variables α , β and c on the positive and negativecharacteristics AB and BC as a function of parameter s . Further, we assume the point B as the origin ( s = 0 ). Thenusing (2.5) and (3.2) we obtain the boundary data in the form of parameter s as follows:7 PREPRINT - F
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9, 2021On AB we have π < α ( s ) < π, β ( s ) = 0 ,π < α ( s ) − β ( s )2 < π ,dαds > , dcds > ,c dαds = − sin 2 ω µ ( τ ) Ω( τ, ω ) dcds ,c dβds = − tan ωµ ( τ ) (cid:18) dcds (cid:19) − (cid:115)(cid:18) dξds (cid:19) + (cid:18) dηds (cid:19) sin ω (3.3)and on BC we have π < α ( s ) < π + β C , β C ≤ β ( s ) ≤ ,π < α ( s ) − β ( s )2 < π ,dβds < , dcds > ,c dβds = sin 2 ω µ ( τ ) Ω( τ, ω ) dcds ,c dαds = tan ωµ ( τ ) (cid:18) dcds (cid:19) + 2 (cid:115)(cid:18) dξds (cid:19) + (cid:18) dηds (cid:19) sin ω. (3.4)Using these boundary values we can prove the existence of local solution for our Goursat problem. We summarize thisin the following Lemma. Lemma 3.1. (Local solution) For sufficiently small (cid:15) > there exists a unique C solution for the Goursat problem (2.1) , (3.2) in a small domain D (cid:15) closed by the boundaries AB and BC and a level curve ω = (cid:15) . Further, this solutionsatisfies π < ω < π , ± ¯ ∂ ± c > , ± ¯ ∂ ± α > , ± ¯ ∂ ± β < in D (cid:15) . (3.5) Proof.
It is observed from (3.3) and (3.4) that the compatibility conditions hold at the point B. Therefore, using themethod of characteristics [38], we conclude that the Goursat problem (2.1), (3.2) admits a unique C local solution.From boundary estimate (3.2) we have ¯ ∂ + c | AB > and − ¯ ∂ − c | BC > .So using the characteristic decompositions from (2.6) we can conclude that ¯ ∂ + c > and − ¯ ∂ − c > in D (cid:15) .Now we prove ω > π in D (cid:15) using method of contradiction. Let us assume that there exists a point G such that ω G = π .Then using (2.5) we observe that ¯ ∂ + ω | G > which provides a contradiction, since ω | AB > π . Hence we have ω > π in D (cid:15) .Further, the fact ω > π and m ( τ ) < shows that Ω( τ, ω ) < in D (cid:15) . Hence, by characteristic decompositions (2.5)we have ¯ ∂ + α > , ¯ ∂ + β < , − ¯ ∂ − β < in D (cid:15) .From the boundary data (3.2) we have − ¯ ∂ − α | BC > . So using characteristic decomposition (2.11) we obtain − ¯ ∂ − α > in the domain D (cid:15) . 8 PREPRINT - F
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9, 2021Figure 2: Invariant triangleTherefore, using ¯ ∂ + α > and − ¯ ∂ − β < we obtain the invariant triangle, see Figure (For more details on invariantregions, see [39]) α > π/ , β < , π < ω < π . Hence the Lemma is proved.
Lemma 3.2.
If the Goursat problem (2.1) , (3.2) admits a unique C solution in the domain D (cid:15) , (cid:15) ∈ ( π/ , π/ . Thenthere exist a positive constant H such that || ( Dα, Dβ, Dc ) || D (cid:15) < H . (3.6) Proof.
Using the Lemma . , (2.5), relations sin β ¯ ∂ + − sin α ¯ ∂ − = − sin 2 ω∂ ξ and cos β ¯ ∂ + − cos α ¯ ∂ − = sin 2 ω∂ η ,we can easily prove this Lemma. ± ¯ ∂ ± c In this section, we give uniform lower and upper bounds of ¯ ∂ + c and − ¯ ∂ − c which are useful for further analysis in thesucceeding sections. ± ¯ ∂ ± c We provide the upper bounds of ¯ ∂ + c and − ¯ ∂ − c in the following Lemma: Lemma 4.1.
If the Goursat problem (2.1) , (3.2) admits a C solution in the domain D (cid:15) where (cid:15) ∈ (cid:0) π , π (cid:1) , then wehave (cid:18) ¯ ∂ + c sin ω , − ¯ ∂ − c sin ω (cid:19) ∈ (0 , M (cid:15) ] × (0 , M (cid:15) ] (4.1) where M (cid:15) = max (cid:40) sup BA (cid:15) (cid:18) ¯ ∂ + c sin ω (cid:19) , sup BC (cid:15) (cid:18) − ¯ ∂ − c sin ω (cid:19) (cid:41) such that A (cid:15) and C (cid:15) are points on AB and BC , respectivelywith ω ( A (cid:15) ) = ω ( C (cid:15) ) = (cid:15) .Proof. In order to prove this Lemma we need to prove that for any x > we have (cid:18) ¯ ∂ + c sin ω , − ¯ ∂ − c sin ω (cid:19) ∈ (0 , M (cid:15) + x ) × (0 , M (cid:15) + x ) in D (cid:15) . PREPRINT - F
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9, 2021Figure 3: (a) Invariant region of (cid:18) ¯ ∂ + c sin ω , − ¯ ∂ − c sin ω (cid:19) ; (b) Domain Σ E In the contrary let us assume that the Lemma is not valid. Then we must have at least one point E in D (cid:15) suchthat (cid:16) ¯ ∂ + c ( E )sin ω , − ¯ ∂ − c ( E )sin ω (cid:17) ∈ (cid:83) i =1 Γ i and (cid:16) ¯ ∂ + c sin ω , − ¯ ∂ − c sin ω (cid:17) ∈ (0 , M (cid:15) + x ) × (0 , M (cid:15) + x ) for any ( ξ, η ) ∈ Σ E / { E } where Γ = (0 , M (cid:15) + x ] × { M (cid:15) + x } and Γ = { M (cid:15) + x } × (0 , M (cid:15) + x ] with Σ E being the region bounded by BE , BE , EE and EE such that the point E is the point of intersection of the boundary AB and the negativecharacteristic passing through the point E and similarly the point E is the point of intersection of the boundary BC and the positive characteristic passing through the point E ; see Figure . Suppose that (cid:16) ¯ ∂ + c ( E )sin ω , − ¯ ∂ − c ( E )sin ω (cid:17) ∈ Γ , thenusing (2.9) we have ¯ ∂ + (cid:16) − ¯ ∂ − c ( E )sin ω (cid:17) < which leads to a contradiction since − ¯ ∂ − c ( E )sin ω = { M (cid:15) + x } . Similarly, if (cid:16) ¯ ∂ + c ( E )sin ω , − ¯ ∂ − c ( E )sin ω (cid:17) ∈ Γ , then again by exploiting (2.10), we have ¯ ∂ − (cid:16) ¯ ∂ + c ( E )sin ω (cid:17) < which leads to a contradictionsince ¯ ∂ + c ( E )sin ω = { M (cid:15) + x } . Therefore, our assumption is wrong which proves the Lemma. ± ¯ ∂ ± c In the following Lemma we find the lower bounds of ¯ ∂ + cc and − ¯ ∂ − cc which directly provides us lower bounds of ¯ ∂ + c and − ¯ ∂ − c . Lemma 4.2.
If the Goursat problem (2.1) , (3.2) admits a C solution in the domain D (cid:15) then the functions ¯ ∂ + cc and ¯ ∂ − cc satisfy < me − ˆ κd ≤ ¯ ∂ + cc , < ¯ me − ˆ κd ≤ − ¯ ∂ − cc where m = min { min BA ∗ ¯ ∂ + cc , min BC ∗ ( − ¯ ∂ − cc ) } > , d is the diameter of the domain D (cid:15) , ˆ κ ( τ ) = 3 + 4 κ ( τ )2 c , A ∗ is thepoint of intersection of negative characteristic and the boundary BA and C ∗ is the point of intersection of positivecharacteristic and the boundary BC such that both the characteristic curves start from a single point P in the domain D (cid:15) . PREPRINT - F
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9, 2021Figure 4: Characteristic curves passing through different points in D (cid:15) Proof.
The uniform lower bounds of ¯ ∂ + cc and − ¯ ∂ − cc are established using the new variables R = A ( κ ) sin ω (cid:113) κ ( τ ) sin ω ¯ ∂ + cc > , S = − A ( κ ) sin ω (cid:113) κ ( τ ) sin ω ¯ ∂ − cc > (4.2)where A ( κ ( τ )) = κ ( τ ) exp (cid:18)(cid:90) − κ (cid:48) ( τ ) sin ω κ ( τ ) sin ω ) dτ (cid:19) > in the region D (cid:15) .Then using (2.7) and (2.8) we obtain ¯ ∂ − R = RΠ, ¯ ∂ + S = SΠ (4.3)where Π = ( R − S )(1 + κ ( τ )) A ( κ ( τ )) sin 2 ω cos ω (cid:113) κ ( τ ) sin ω + sin 2 ω c (cid:18) κ ( τ ) sin ω κ ( τ ) sin ω (cid:19) . (4.4)Now we consider the following two cases. Case A:
Let us assume that S ≥ R holds entirely in the region P A ∗ BC ∗ P for each point P .Then using (4.3) we get ¯ ∂ − ln R ≤ κ ( τ )2 c ≤ κ ( τ )2 c = ˆ κ ( τ ) . (4.5)Then integrating (4.5) along the negative characteristic from A ∗ to P yields S P > R P ≥ R A ∗ e − ˆ κ ( η P − η A ∗ ) ≥ me − ˆ κd . Case B:
Let us assume that there exists a point Q in P A ∗ BC ∗ P such that S < R at Q . Now we draw a positivecharacteristic curve Γ +1 starting from Q to Q ∗ which lies on the boundary BC ∗ .If Π ≥ for all the points on Γ +1 then using (4.3) we have ¯ ∂ + S > on Γ +1 PREPRINT - F
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9, 2021which implies that R Q > S Q ≥ S Q ∗ ≥ m ≥ me − ˆ κd . Otherwise there exists a point on Γ +1 such that Π < at that point. But Π > at Q so we use continuity of Π to seethat Π > holds for all points in N ∩ Γ +1 for some neighbourhood N of Q and Π = 0 at R = ∂N ∩ Γ +1 .So we see that ¯ ∂ + S > on N ∩ Γ +1 ,which gives R Q > S Q > S R > R R . (4.6)The last inequality holds because of the fact Π = 0 at R = ∂N ∩ Γ +1 .Now we draw a negative characteristic Γ − from R to R ∗ on the boundary BA ∗ . We further investigate the analysisof this case by considering the following two subcases. Subcase i :
Assume that S ≥ R holds on every point of Γ − then using ¯ ∂ − R = RΠ leads to − ¯ ∂ − ln R ≥ − ˆ κ. Integrating the above from R ∗ to R we obtain R R ≥ R R ∗ e − ˆ κd ≥ me − ˆ κd . (4.7)By using (4.6) and (4.7) we have R Q > S Q ≥ me − ˆ κd . (4.8) Subcase ii:
Assume that
S < R holds for some point on Γ − . Then using the fact S > R at R we observe that S > R holds for all the points in N ∩ Γ − for some neighbourhood N of R and S = R at Q = ∂N ∩ Γ − .Then integrating (4.3) along negative characteristic from Q to R and using (4.6) yields R Q > S Q > S R > R R ≥ R Q e − ˆ κd = S Q e − ˆ κd . (4.9)Now from Q we draw a positive characteristic Γ +2 up to Q ∗ on the boundary BC ∗ . At Q we have Π > . Again, if Π ≥ holds on Γ +2 completely then S Q > S Q ∗ ≥ m, which in the view of (4.9) gives R Q > S Q ≥ me − ˆ κd . Otherwise we have a point on Γ +2 such that Π < at that point and ¯ ∂ + S > or S Q > S R in the neighbourhood of Q where R is the point on Γ +2 such that Π > on Q R ∩ Γ +2 and Π = 0 at R . Again using (4.3) and (4.9) yields R Q > S Q ≥ S R e − ˆ κd . Since at R , Π = 0 so we have R R < S R . Then we draw a negative characteristic from R . The repetition of the above process completes the proof of theLemma. In this section, we try to extend our ideas to obtain the global solution from the local solution by solving many localGoursat problems in each step of extension. Let us assume that the Goursat problem (2.1), (3.2) admits a unique C solution in D (cid:15) . Let XZ and XY be positive( C + ) and negative( C − ) characteristics in D (cid:15) , respectively. Then, weprescribe ( α, β, c ) = (cid:26) ( α | XZ , β | XZ , c | XZ ) , on XZ ( α | XY , β | XY , c | XY ) , on XY (5.1)where ( α | XY , β | XY , c | XY ) and ( α | XZ , β | XZ , c | XZ ) are the values of the solution ( α, β, c ) on XY and XZ , respec-tively.We then have the following Lemma. 12 PREPRINT - F
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Lemma 5.1. (Curved quadrilateral building block) If the arc lengths of XY and XZ are less than ν where ν = c sin ( π − (cid:15) ) M (cid:15) tan ( π + (cid:15) ) (cid:0) π − (cid:15) (cid:1) µ ( τ ) , then the Goursat problem (2.1) , (5.1) admits a global C solution on a curvedquadrilateral domain bounded by XY , XZ, T Y and
T Z where
T Y is the positive characteristic passing through Y , T Z is the negative characteristic passing through Z . Further, this solution satisfies (cid:15) < ω < π + (cid:15) .Proof. We know by Lemma . that sup XZ (cid:16) ¯ ∂ + c sin ω (cid:17) ≤ M (cid:15) and sup XY (cid:16) − ¯ ∂ − c sin ω (cid:17) ≤ M (cid:15) . Then using a similar argument asin the proof of Lemma . , one can prove that the solution of Goursat problem (2.1), (5.1) in the quadrilateral regionbounded by XY , XZ, T Y and
T Z satisfies Lemma . and ( ¯ ∂ + c, − ¯ ∂ − c ) ∈ (0 , M (cid:15) ] × (0 , M (cid:15) ] . (5.2)Thus using (2.5) we can easily obtain < ¯ ∂ + α ≤ M (cid:15) tan ω µ ( τ ) c , − M (cid:15) tan ω µ ( τ ) c ≤ − ¯ ∂ − β < . (5.3)We now prove that, when the arc lengths of XZ and XY are less than ν the solution of the Goursat problem satisfies ω < π (cid:15) , < ¯ ∂ + α ≤ M (cid:15) tan ( π + (cid:15) )2 µ ( τ ) c , − M (cid:15) tan ( π + (cid:15) )2 µ ( τ ) c ≤ − ¯ ∂ − β < . (5.4)We prove it using the method of contradiction. Let us consider an arbitrary point H in the domain of the solutionof Goursat problem such that ω ( H ) = (cid:0) π + (cid:15) (cid:1) and let the positive characteristic passing through H intersects theboundary XY at a point H and negative characteristic passing through H intersects the boundary XZ at a point H .We draw a straight line passing through H having a slope tan α ( H ) and another straight line with slope tan β ( H ) passing through H . Let us consider that these two lines intersect at a point J ; see Figure . By the estimate (5.3) wesee that the arcs HH and HH lie inside the triangle JH H . Also, their arc lengths are less than the sum of the arclengths of JH , JH and H H . Therefore the arc lengths of HH and HH are less than ν sin( π − (cid:15) ) .Then we integrate (5.3) from H to H along positive characteristic and from H to H along negative characteristic toobtain α ( H ) < α ( H ) + (cid:32) M (cid:15) tan (cid:0) π + (cid:15) (cid:1) µ ( τ ) c (cid:33) (cid:18) ν sin( π − (cid:15) ) (cid:19) < α ( X ) + π − (cid:15) (5.5) β ( H ) > β ( H ) − (cid:32) M (cid:15) tan (cid:0) π + (cid:15) (cid:1) µ ( τ ) c (cid:33) (cid:18) ν sin( π − (cid:15) ) (cid:19) > β ( X ) − π (cid:15) (5.6)Thus, we have ω ( H ) < π + (cid:15) which leads to contradiction.In order to obtain a priori uniform C norm estimates of the solution to the Goursat problem we follow the ideas usedin the proof of the Lemma . and . . Hence using the theory of global classical solutions for quasilinear hyperbolicequations we can obtain the proof of the Lemma . [40]. Theorem 5.1. (Global solution) The semi-hyperbolic patch problem with the boundary data (3.2) admits a uniqueglobal C solution in the region ABC where the curve AC is sonic.Proof. Let Y = C (cid:15) , Y , Y , Y , ........, Y n = A (cid:15) be n + 1 different points on the level curve ω = (cid:15) . From the point Y i we draw a C + characteristic curve which intersects the C − characteristic curve passing through Y i +1 at a point X i ,where i = 0 , , , ....., n − . Due to the fact ¯ ∂ ± ω (cid:54) = 0 , the level curve ω = (cid:15) is a non-characteristic curve. Hence, X i (cid:54) = Y i and X i (cid:54) = Y i +1 for any i = 0 , , , ......, n − . For sufficiently close Y i and Y i +1 , the arc lengths of X i Y i and X i Y i +1 are less than ν . Therefore, using Lemma . , we know that the Goursat problem with the characteristicboundaries X i Y i and X i Y i +1 admits a C solution in the quadrilateral domain bounded by X i Y i , X i Y i +1 , Y i Z i +1 and Y i +1 Z i +1 where Y i Z i +1 is the C − characteristic curve passing through Y i and Y i +1 Z i +1 is the C + characteristic curvepassing through Y i +1 .Let Z = A and Z n +1 = C , then for every i = 0 , , , ....., n , there exists a ω i , (cid:15) < ω i < π , suchthat the Goursat problem for system (2.1) admits a unique C solution in the domain closed by Y i Z i +1 , Y i Z i and the level curve ω ( ξ, η ) = ω i with Y i Z i and Y i Z i +1 as the characteristic boundaries. Let ω e =min { ω , ω , ω , ......., ω n , ω ( Z ) , ω ( Z ) , ......, ω ( Z n ) } . Using the fact ¯ ∂ + ω > we see that ω e > (cid:15) . Then weconstruct the solution of Goursat problem in the domain D ω e . Repeating the same process, we can construct the globalsolution in the whole domain ABC which proves the theorem.13
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9, 2021Figure 5: Left: A curved quadrilateral building block; Right: Global solutionFigure 6: Envelope formation
In this section, we discuss the formation of the envelope for positive characteristics passing through strictly convexcurve BC . Further, we prove that the envelope forms before the sonic points of positive characteristics. Theorem 6.1. (Envelope formation) For a given strictly monotonically convex curve BC , we draw the positivecharacteristics passing through the curve BC which are moving towards downward; see Figure . Then positivecharacteristics form an envelope before their sonic points. PREPRINT - F
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Proof.
We exploit (2.6) to prove this theorem. Since ¯ ∂ + c = 0 in the region of simple waves with the positivecharacteristics, so from (2.6) we obtain ¯ ∂ + ¯ ∂ − c = sin 2 ωc ¯ ∂ − c + ( ¯ ∂ − c ) µ ( τ ) c cos ω . So that − ¯ ∂ + (cid:18) − ¯ ∂ − c (cid:19) = sin 2 ωc (cid:18) − ¯ ∂ − c (cid:19) − µ ( τ ) c cos ω , (6.1)which can be written as − ¯ ∂ + (cid:18) − ¯ ∂ − c exp (cid:18)(cid:90) BD sin 2 ωc ds (cid:19)(cid:19) = − µ ( τ ) c cos ω exp (cid:18)(cid:90) BD sin 2 ωc ds (cid:19) . (6.2)Using the boundedness of sin 2 ω and c , we see that exp (cid:18)(cid:90) BD sin 2 ωc ds (cid:19) remains bounded on the characteristicextension BD which means that the right hand side of (6.2) remains negative while on the boundary BC we have − ¯ ∂ − c > . Which clearly shows that the function − ¯ ∂ − c is decreasing along the positive characteristics in the directionfrom B to D . Further, the R.H.S of (6.2) blows up quadratically at least as shown in [8]. So / ( − ¯ ∂ − c ) approachesto zero before the characteristic reaches to its sonic point which means that positive characteristics form an envelopebefore their sonic points. We use the partial hodograph mapping ( ξ, η ) −→ ( z, t ) as in [18]. We define t = cos ω ( ξ, η ) , z = φ ( ξ, η ) , where φ is the potential function used in pseudo-Bernoulli’s law (2.2).From the definition of transformation we have the Jacobian J as J = ∂ ( z, t ) ∂ ( ξ, η ) = − (1 + κ ( τ ) sin ω )( ¯ ∂ + c − ¯ ∂ − c )2 cos ω (cid:54) = 0 in the entire domain ABC .Let us assume that A (cid:48) B (cid:48) C (cid:48) is the image of the domain ABC in the z − t plane. Then, we transform the normalizedderivatives in the new coordinate system ( z, t ) . Using the expression of ¯ ∂ ± we obtain ¯ ∂ + = − (cid:40) (1 + κ ( τ )(1 − t ))(1 − t ) ¯ ∂ + cct + (cid:112) (1 − t ) c (cid:41) ∂∂t + ct √ − t ∂∂z , ¯ ∂ − = − (cid:40) (1 + κ ( τ )(1 − t ))(1 − t ) ¯ ∂ − cct + (cid:112) (1 − t ) c (cid:41) ∂∂t + ct √ − t ∂∂z . (7.1)Using (7.1) we obtain c t = − ct (1 + κ ( τ )(1 − t ))(1 − t ) , c z = t − c (1 + κ ( τ )(1 − t )) . (7.2)Using the uniform boundedness of c, κ ( τ ) and t = cos ω , we can verify that c t and c z are also uniformly bounded(Recall c < c < c ).For convenience in the further calculations, we use R ( z, t ) = ¯ ∂ + cc and S ( z, t ) = ¯ ∂ − cc .By exploiting (7.1) in (2.7) and (2.8) we have R t − cf t S − tg R z = − f R √ − t S − tg (cid:40) t √ − t c + R + S µ t + (cid:110) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ )(1 − t )) (cid:111) St (cid:41) ,S t − cf t R − tg S z = − f S √ − t R − tg (cid:40) t √ − t c + R + S µ t + (cid:110) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ )(1 − t )) (cid:111) Rt (cid:41) , (7.3)15 PREPRINT - F
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9, 2021where f ( t, τ ) = 1(1 + κ ( τ )(1 − t )) (cid:112) (1 − t ) and g ( t, τ ) = − (1 − t ) fc .We directly compute f t = (cid:16) κ ( τ ) t (cid:112) (1 − t ) + 3 t (cid:112) − t − κ (cid:48) ( τ )(1 − t ) τ t (cid:17) f , f z = − (1 − t ) f κ (cid:48) ( τ ) τ z ,g t = 4 t (1 − t ) f − (1 − t ) f t c + (1 − t ) fc c t , g z = g (cid:18) f z f − c z c . (cid:19) where τ t = − τ κ ( τ ) c c t and τ z = − τ κ ( τ ) c c z .The expressions of f ( t, τ ) , g ( t, τ ) , c t , c z , τ t , τ z , f t , f z , g t and g z clearly show us that these functions are uniformlybounded in the region A (cid:48) B (cid:48) C (cid:48) near sonic boundary, i.e., t = 0 .Further, we set (cid:98) R ( z, t ) = 1 R ( z, t ) and (cid:98) S ( z, t ) = − S ( z, t ) .Using Lemma . we see that the functions (cid:98) R and (cid:98) S are uniformly bounded up to sonic curve.Now (7.3) can be transformed as follows: (cid:98) R t + cf t (cid:98) S tg (cid:98) S (cid:98) R z = f √ − t tg (cid:98) S (cid:40) (cid:98) R − (cid:98) S µ t − t √ − t (cid:98) R (cid:98) Sc + (cid:110) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ )(1 − t )) (cid:111) (cid:98) Rt (cid:41) , (cid:98) S t − cf t (cid:98) R − tg (cid:98) R (cid:98) S z = f √ − t − tg (cid:98) R (cid:40) (cid:98) S − (cid:98) R µ t + 2 t √ − t (cid:98) R (cid:98) Sc + (cid:110) τ κ (cid:48) ( τ ) − (1 + 2 κ ( τ )(1 − t )) (cid:111) (cid:98) St (cid:41) . (7.4)We denote Λ + = cf t (cid:98) S tg (cid:98) S , Λ − = − cf t (cid:98) R − tg (cid:98) R .
Further, we denote ∂ ± = ∂ t + Λ ± ∂ z , Ξ = ∂ + (cid:98) R − ∂ − (cid:98) R, Θ = ∂ + (cid:98) S − ∂ − (cid:98) S. Then we have Λ + − Λ − = cf ( (cid:98) R + (cid:98) S ) t (1 − tg (cid:98) R )(1 + tg (cid:98) S ) , (cid:98) R z = ΞΛ + − Λ − , (cid:98) S z = ΘΛ + − Λ − . (7.5)Now we use the commutator relation [24] ∂ − ∂ + − ∂ + ∂ − = ∂ − Λ + − ∂ + Λ − Λ + − Λ − ( ∂ + − ∂ − ) , and arrives at ∂ + Ξ = ∂ − Λ + − ∂ + Λ − Λ + − Λ − Ξ + ( ∂ + ∂ + (cid:98) R − ∂ − ∂ + (cid:98) R ) ,∂ − Θ = ∂ − Λ + − ∂ + Λ − Λ + − Λ − Θ + ( ∂ + ∂ − (cid:98) S − ∂ − ∂ − (cid:98) S ) , (7.6)A straightforward calculation leads to ∂ − Λ + − ∂ + Λ − Λ + − Λ − = 2 t + h ( z, t ) , (7.7)where h ( z, t ) = f √ − t (1 − tg (cid:98) R )(1 + tg (cid:98) S ) (cid:40) ( (cid:98) R − (cid:98) S ) g µ − t √ − t g (cid:98) R (cid:98) Sc + t (cid:104) τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t ) (cid:105)(cid:41) + ( g + tg t )( (cid:98) R − (cid:98) S + 2 tg (cid:98) R (cid:98) S )(1 − tg (cid:98) R )(1 + tg (cid:98) S ) + f t f + tg + cf t g z (cid:98) R (cid:98) S (1 − tg (cid:98) R )(1 + tg (cid:98) S ) . PREPRINT - F
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9, 2021Also, we calculate ( ∂ + (cid:98) R ) z = f (cid:98) R z + f (cid:98) S z + f , (7.8)where f ( z, t ) = 12 t − ( κ ( τ ) + 1) g (cid:98) S − t (1 + κ ( τ )(2 − t ))(1 + tg (cid:98) S )2(1 − t )(1 + tg (cid:98) S )(1 + κ ( τ )(1 − t )) , + f √ − t tg (cid:98) S (cid:34) t ( τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t )) − t √ − t (cid:98) Sc (cid:35) f ( z, t ) = − t + ( κ ( τ ) + 1) g (cid:98) S − t (1 + κ ( τ )(2 − t ))(1 + tg (cid:98) S )2(1 + κ ( τ )(1 − t ))(1 − t )(1 + tg (cid:98) S ) − t f (1 − t ) (cid:98) Rc (1 + tg (cid:98) S ) − f g √ − t (1 + tg (cid:98) S ) (cid:34) (cid:98) R − (cid:98) S µ − t √ − t (cid:98) R (cid:98) Sc + (cid:0) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1)(cid:1) (cid:98) Rt (cid:35) ,f ( z, t ) = ( (cid:98) R − (cid:98) S )2 t (cid:32) κ (cid:48) ( τ ) τ z κ ( τ ) + f z f (1 + tg (cid:98) S ) (cid:33) + ( (cid:98) S − (cid:98) R ) (cid:32) κ (cid:48) ( τ ) τ z κ ( τ ) + f z f (1 + tg (cid:98) S ) (cid:33) (cid:34) ( κ ( τ ) + 1) g (cid:98) S − t (1 + κ ( τ )(2 − t ))(1 + tg (cid:98) S )2(1 + κ ( τ )(1 − t ))(1 − t )(1 + tg (cid:98) S ) (cid:35) + f √ − t tg (cid:98) S (cid:34) − t √ − t (cid:98) R (cid:98) Sτ κ ( τ ) c + (cid:98) Rt (cid:104) κ (cid:48) ( τ )(2 t −
1) + τ κ (cid:48)(cid:48) ( τ ) (cid:105)(cid:35) τ z + f z √ − t (1 + tg (cid:98) S ) (cid:34) − t √ − t (cid:98) R (cid:98) Sc + (cid:104) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1) (cid:105) (cid:98) Rt (cid:35) + f g (cid:98) S √ − t c (1 + tg (cid:98) S ) (cid:34) (cid:98) R − (cid:98) S µ − t √ − t (cid:98) R (cid:98) Sc + (cid:104) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1) (cid:105) (cid:98) Rt (cid:35) c z . Similarly ( ∂ − (cid:98) S ) z = g (cid:98) S z + g (cid:98) R z + g , (7.9)where g ( z, t ) = 12 t + ( κ ( τ ) + 1) g (cid:98) R + t (cid:0) κ ( τ )(2 − t ) (cid:1) (1 − tg (cid:98) R )2(1 − t )(1 − tg (cid:98) R ) (1 + κ ( τ )(1 − t ))+ f √ − t − tg (cid:98) R (cid:34) t (cid:0) τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t ) (cid:1) + 2 t √ − t (cid:98) Rc (cid:35) ,g ( z, t ) = − t − ( κ ( τ ) + 1) g (cid:98) R + t (cid:0) κ ( τ )(2 − t ) (cid:1) (1 − tg (cid:98) R )2(1 − t )(1 − tg (cid:98) R ) (1 + κ ( τ )(1 − t )) + 2 t f (1 − t ) (cid:98) Sc (1 − tg (cid:98) R ) − f g √ − t (1 − tg (cid:98) R ) (cid:34) (cid:98) R − (cid:98) S µ − t √ − t (cid:98) R (cid:98) Sc − (cid:0) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1)(cid:1) (cid:98) St (cid:35) ,g ( z, t ) = (cid:98) S − (cid:98) R t (cid:32) κ (cid:48) ( τ ) τ z κ ( τ ) + f z f (1 − tg (cid:98) R ) (cid:33) + ( (cid:98) S − (cid:98) R ) (cid:32) κ (cid:48) ( τ ) τ z κ ( τ ) + f z f (1 − tg (cid:98) R ) (cid:33) (cid:34) ( κ ( τ ) + 1) g (cid:98) R + t (cid:0) κ ( τ )(2 − t ) (cid:1) (1 − tg (cid:98) R )2(1 − t )(1 − tg (cid:98) R ) (1 + 2 κ ( τ )(1 − t )) (cid:35) PREPRINT - F
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9, 2021 + f √ − t − tg (cid:98) R (cid:34) t √ − t (cid:98) R (cid:98) Sτ κ ( τ ) c + (cid:98) St (cid:104) κ (cid:48) ( τ )(2 t −
1) + τ κ (cid:48)(cid:48) ( τ ) (cid:105)(cid:35) τ z + f z √ − t (1 − tg (cid:98) R ) (cid:34) t √ − t (cid:98) R (cid:98) Sc + (cid:0) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1)(cid:1) (cid:98) St (cid:35) + f g (cid:98) R √ − t c (1 − tg (cid:98) R ) (cid:34) (cid:98) R − (cid:98) S µ − t √ − t (cid:98) R (cid:98) Sc − (cid:0) τ κ (cid:48) ( τ ) − (cid:0) κ ( τ )(1 − t ) (cid:1)(cid:1) (cid:98) St (cid:35) c z . Usage of (7.5) in (7.8) and (7.9) leads to (cid:40) ∂ + ∂ + (cid:98) R − ∂ − ∂ + (cid:98) R = (Λ + − Λ − )( ∂ + (cid:98) R ) z = f Ξ + f Θ + (Λ + − Λ − ) f ,∂ + ∂ − (cid:98) S − ∂ − ∂ − (cid:98) S = (Λ + − Λ − )( ∂ − (cid:98) S ) z = g Θ + g Ξ + (Λ + − Λ − ) g . (7.10)Thus, exploiting (7.6), (7.7) and (7.10), we obtain ∂ + Ξ = (cid:18) t + ˆ f ( z, t ) (cid:19) Ξ + ˆ f ( z, t ) Θ2 t + ˆ f ( z, t ) t ,∂ − Θ = (cid:18) t + ˆ g ( z, t ) (cid:19) Θ + ˆ g ( z, t ) Ξ2 t + ˆ g ( z, t ) t , (7.11)where ˆ f = h + (cid:18) f − t (cid:19) , ˆ f = 2 tf , ˆ f = 2 t (Λ + − Λ − ) f , ˆ g = h + (cid:18) g − t (cid:19) , ˆ g = 2 tg , ˆ g = 2 t (Λ + − Λ − ) g . Noting the expressions of h, f i , g i ( i = 1 , , and using the Lemma . and . , we observe that the functions ˆ f , ˆ f , ˆ f , ˆ g , ˆ g and ˆ g are uniformly bounded near sonic boundary, i.e., t = 0 . Also, we see that ˆ f → − , ˆ g → − as t → .Further, we introduce new variables (cid:98) Ξ = Ξ t , (cid:98)
Θ = Θ t to transform the system (7.11) into ∂ + (cid:98) Ξ = (cid:40) t + ˆ f ( z, t ) (cid:41)(cid:98) Ξ + ˆ f ( z, t ) (cid:98) Θ2 t + ˆ f ( z, t )2 ,∂ − (cid:98) Θ = (cid:40) t + ˆ g ( z, t ) (cid:41) (cid:98) Θ + ˆ g ( z, t ) (cid:98) Ξ2 t + ˆ g ( z, t )2 , or ∂ + (cid:16) t − (cid:98) Ξ (cid:17) = t − (cid:40) t ˆ f ( z, t ) (cid:98) Ξ + 12 ˆ f (cid:98) Θ + t ˆ f ( z, t )2 (cid:41) ,∂ − (cid:16) t − (cid:98) Θ (cid:17) = t − (cid:40) t ˆ g ( z, t ) (cid:98) Θ + 12 ˆ g (cid:98) Ξ + t ˆ g ( z, t )2 (cid:41) . (7.12) R , S and W in partial hodograph plane In this subsection, we are interested to establish the regularity of solution near sonic boundary AC , i.e., near t = 0 . Weuse (7.12) to derive the regularity results in partial hodograph plane. Let F = ( φ , be any point on the line segment A (cid:48) C (cid:48) where A (cid:48) C (cid:48) is the image of sonic boundary AC in z − t plane. We take a new point I = ( φ , t m ) where t m isvery small positive number such that I = ( φ , t m ) remains in the domain A (cid:48) B (cid:48) C (cid:48) . Then from the point I we can drawpositive and negative characteristic curves φ + ( I ) and φ − ( I ) up to the line segment A (cid:48) C (cid:48) at I and I , respectively.Since R, − S are uniformly bounded and positive in the domain ABC , we see that the functions ˆ f , ˆ f , ˆ f , ˆ g , ˆ g and ˆ g are uniformly bounded near sonic boundary, i.e., t = 0 in a small subdomain with ˆ f → − , ˆ g → − as t → .18 PREPRINT - F
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9, 2021Figure 7: Domain of ∆( φ , .Then for any constant (cid:37) ∈ (0 , we can choose sufficiently small t m < such that t | ˆ f | ≤ (cid:37) , t | ˆ f | ≤ (cid:37) , | ˆ f | ≤ (cid:37),t | ˆ g | ≤ (cid:37) , t | ˆ g | ≤ (cid:37) , | ˆ g | ≤ (cid:37) (7.13)hold in the domain II I . Let ∆( φ , be the domain bounded by II , II and the positive and negative characteristicsstarting from F = ( φ , . Further, we draw a negative characteristic up to the point p ( z p , t p ) on the boundary II anda positive characteristic up to the point q ( z q , t q ) on the boundary II starting from an arbitrary point ( z, t ) .Let us denote M = max (cid:26) max ∆( φ , (cid:12)(cid:12)(cid:12)(cid:98) Ξ( z q , t q ) (cid:12)(cid:12)(cid:12) + 1 , max ∆( φ , (cid:12)(cid:12)(cid:12) (cid:98) Θ( z p , t p ) (cid:12)(cid:12)(cid:12) + 1 (cid:27) . Since t p and t q are strictly positive, so using the definitions of (cid:98) Ξ and (cid:98) Θ we observe that M is well-defined and uniformlybounded in the domain ∆( φ , .Now for any fixed δ ∈ (0 , t m ) , define ¯∆ δ = (cid:8) ( φ, t ) | δ ≤ t ≤ t m , φ − ( I ) ≤ z ≤ φ + ( I ) (cid:9) ∩ ∆( φ , and M δ = max ¯∆ δ {| t (cid:37) (cid:98) Ξ | , | t (cid:37) (cid:98) Θ |} . Then we provide the bounds of | t (cid:37) (cid:98) Ξ | and | t (cid:37) (cid:98) Θ | in the following Lemma. Lemma 7.1. If ( φ , is an arbitrary fixed point on the line segment A (cid:48) C (cid:48) and (cid:37) ∈ (0 , be any constant. Then thereexists a uniform positive constant (cid:102) M such that the following inequalities hold (cid:12)(cid:12)(cid:12) t (cid:37) (cid:98) Ξ (cid:12)(cid:12)(cid:12) ≤ (cid:102) M , | t (cid:37) (cid:98) Θ | ≤ (cid:102) M ∀ ( z, t ) ∈ ∆( φ , . Proof.
Suppose that for every δ ∈ (0 , t m ) , M δ ≤ M then the Lemma holds true. Otherwise, there exists a δ ∈ (0 , t m ) such that M δ > M .For any point ( z δ , δ ) ∈ ∆( φ , , we integrate (7.12) along the positive characteristic from t ( ≥ δ ) to t q and use(7.13) to obtain 19 PREPRINT - F
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9, 2021 (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:98) Ξ( z, t ) t (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:98) Ξ( z q , t q ) t q + (cid:90) t q t t ˆ f (cid:98) Ξ + ˆ f (cid:98) Θ + t ˆ f t dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:98) Ξ( z q , t q ) t q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:90) t q t (cid:0) (cid:37) (cid:1) (cid:98) Ξ + (cid:0) (cid:37) (cid:1) (cid:98) Θ + (cid:37) t dt (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:98) Ξ( z q , t q ) t q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 + 3 (cid:37) M δ (cid:90) t q t t + (cid:37) dt = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) (cid:98) Ξ( z q , t q ) t q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) + 2 + 3 (cid:37) M δ (cid:0) + (cid:37) (cid:1) (cid:32) t + (cid:37) − t + (cid:37)q (cid:33) < (cid:37) (cid:37) M δ t − − (cid:37) < M δ t − − (cid:37) , which proves the inequality | (cid:98) Ξ( z, t ) | t = δ < M δ δ − (cid:37) , (7.14)holds on the line segment { t = δ } ∩ ¯∆ δ . Similarly, we can prove that | (cid:98) Θ( z, t ) | t = δ < M δ δ − (cid:37) . (7.15)According to (7.14) and (7.15) we notice that | t (cid:37) (cid:98) Ξ | and | t (cid:37) (cid:98) Θ | do not attain the maximum values on the line segment t = δ , which means that the maximum value is attained in the interior, i.e., for δ < t ≤ t m in the domain ¯∆ δ . Theabove assertion holds in a larger domain ¯∆ δ (cid:48) , where δ (cid:48) < δ . We can extend the domain larger and larger in each stepto the whole domain ∆( φ , to complete the proof of the Lemma.Next, for any point ( φ , ∈ A (cid:48) C (cid:48) let r be a positive constant such that ( φ − r, φ + r ) ⊂ A (cid:48) C (cid:48) . Further, assumethat the intersection points of negative and positive characteristics passing through points F = ( φ − r, and F = ( φ + r, are F ∗ and F ∗ , respectively. Then we can extend the inequality in a larger domain IF ∗ F F F ∗ usingthe same arguments as in Lemma . . Then we have the following Lemma. Lemma 7.2. If ( φ , is an arbitrary fixed point on the line segment A (cid:48) C (cid:48) and (cid:37) ∈ (1 , be any constant. Then thereexists a uniform positive constant (cid:102) M depending only on (cid:37) and r such that the following inequalities hold | t (cid:37) (cid:98) Ξ | ≤ (cid:102) M , | t (cid:37) (cid:98) Θ | ≤ (cid:102) M ∀ ( z, t ) ∈ IF ∗ F F F ∗ . We now prove the uniform boundedness of the function (cid:99) W = (cid:98) R − (cid:98) St . Lemma 7.3.
The function (cid:99) W is uniformly bounded up to the sonic boundary A (cid:48) C (cid:48) .Proof. We exploit the values of (cid:98) R t , (cid:98) S t , (cid:98) Ξ and (cid:98) Θ to prove the uniform boundedness of (cid:99) W near sonic boundary, i.e., t = 0 . Therefore, we compute (cid:99) W t = θ (cid:99) W + θ , (7.16)where θ = (1 + κ ( τ )) g ( (cid:98) S − (cid:98) R ) + 2 t (1 + κ ( τ )(2 − t )) + 2 g (1 + κ ( τ )(1 − t ))(1 − t )( (cid:98) R − (cid:98) S + tg (cid:98) R (cid:98) S )2(1 − tg (cid:98) R )(1 + tg (cid:98) S )(1 + κ ( τ )(1 − t ))(1 − t ) ,θ = f √ − t (1 − tg (cid:98) R )(1 + tg (cid:98) S ) (cid:34) ( tg ( ˜ R − ˜ S ) − (cid:40) t √ − t (cid:98) R (cid:98) Sc (cid:41)(cid:35) + f √ − t (1 − tg (cid:98) R )(1 + tg (cid:98) S ) (cid:34) ( τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t ))( (cid:98) R − (cid:98) S − tg ( (cid:98) R + (cid:98) S )) (cid:35) − (cid:34) (cid:98) Ξ (cid:98) S (1 − tg (cid:98) R ) + (cid:98) Θ (cid:98) R (1 + tg (cid:98) S ) (cid:98) R + (cid:98) S (cid:35) . For sufficiently small t , we see that θ and θ are uniformly bounded. Thus, integrating (7.16) yields that the function (cid:99) W is uniformly bounded near t = 0 . 20 PREPRINT - F
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9, 2021Since (cid:98) R ( z, t ) = R ( z,t ) and (cid:98) S ( z, t ) = − S ( z,t ) then using the fact that the function (cid:99) W ( z, t ) = (cid:98) R − (cid:98) St ( z, t ) is uniformlybounded, we can easily obtain the bound of W ( z, t ) = R + St ( z, t ) which follows that there exists a uniform positiveconstant M such that (cid:12)(cid:12)(cid:12)(cid:12) R + St ( z, t ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ M ∀ ( z, t ) ∈ A (cid:48) B (cid:48) C (cid:48) . (7.17)So using (7.3) we can prove the uniform boundedness of | ¯ ∂ + R | and | ¯ ∂ − S | in the entire domain A (cid:48) B (cid:48) C (cid:48) including thesonic boundary A (cid:48) C (cid:48) , i.e., there exists a uniform positive constant M such that | ¯ ∂ + R | < M , | ¯ ∂ − S | < M ∀ ( z, t ) ∈ A (cid:48) B (cid:48) C (cid:48) . (7.18)We now develop the uniform regularity of functions R, S and W in partial hodograph plane up to the degenerate linesegment A (cid:48) C (cid:48) . Lemma 7.4.
The functions R ( z, t ) , S ( z, t ) and W ( z, t ) are uniformly Lipschitz continuous in the region A (cid:48) B (cid:48) C (cid:48) upto the degenerate line segment A (cid:48) C (cid:48) .Proof. Using (7.5) we obtain R z = (cid:98) Ξ( R − tg )( S − tg ) R cf ( R − S ) (cid:18) R + St (cid:19) , S z = (cid:98) Θ( R − tg )( S − tg ) S cf ( S − R ) (cid:18) R + St (cid:19) (7.19)Using (7.17), Lemma . and Lemma . we observe that the functions R z and S z are uniformly bounded in the domain A (cid:48) B (cid:48) C (cid:48) up to the degenerate line segment A (cid:48) C (cid:48) . Therefore, by (7.18) the functions R t and S t are also uniformlybounded in the entire region A (cid:48) B (cid:48) C (cid:48) .These observations implies the uniform Lipschitz continuity of the function R ( z, t ) in the entire domain A (cid:48) B (cid:48) C (cid:48) .Similarly, we can prove the uniform Lipschitz continuity of the function S ( z, t ) in the whole domain A (cid:48) B (cid:48) C (cid:48) .Now we compute ¯ ∂ + W and ¯ ∂ − W as follows ¯ ∂ + W = − f (cid:112) − t (cid:40) t √ − t ( R + S − ( R + S ) tg ) c ( S − tg )( R − tg ) + RS (cid:0) τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t ) (cid:1) ( R + S − tg )( R − tg )( S − tg ) (cid:41) + cf ( S − R )( R − tg )( S − tg ) ( tS z ) + (1 + κ ( τ )(2 − t )) tW (1 − t )(1 + κ ( τ )(1 − t )) − (1 + κ ( τ ))( R + S − tg )( W − g ) W − t )(1 + κ ( τ )(1 − t ))( S − tg )( R − tg ) , and ¯ ∂ − W = − f (cid:112) − t (cid:40) t √ − t ( R + S − ( R + S ) tg ) c ( S − tg )( R − tg ) + RS (cid:0) τ κ (cid:48) ( τ ) − − κ ( τ )(1 − t ) (cid:1) ( R + S − tg )( R − tg )( S − tg ) (cid:41) − cf ( S − R )( R − tg )( S − tg ) ( tR z ) + (1 + κ ( τ )(2 − t )) tW (1 − t )(1 + κ ( τ )(1 − t )) − (1 + κ ( τ ))( R + S − tg )( W − g ) W − t )(1 + κ ( τ )(1 − t ))( S − tg )( R − tg ) , which clearly indicates that the functions ¯ ∂ + W and ¯ ∂ − W are uniformly bounded in the domain A (cid:48) B (cid:48) C (cid:48) up to thedegenerate line segment A (cid:48) C (cid:48) . Therefore, a similar proof as of uniform Lipschitz continuity of the functions R and S provide us the uniform Lipschitz continuity of W in z − t plane. Hence the Lemma is proved. Using the results obtained in the preceding section we now prove that the physical variables ( ρ, u, v )( ξ, η ) are uniformly C , continuous in the region ABC up to the sonic boundary AC and also the sonic boundary is C , continuous.To check the regularity of solution in the entire region ABC , we first consider the level curves l ϑ ( ξ, η ) = 1 − sin ω = ϑ, where ϑ is a positive constant. In particular for ω = π , this level curve represents the sonic boundary AC .21 PREPRINT - F
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9, 2021Now exploiting ¯ ∂ + , ¯ ∂ − and (2.5) we obtain l ϑξ = sin σ (1 + κ ( τ ) sin ω )( ¯ ∂ + c − ¯ ∂ − c )2 c − sin ω cos σ (1 + κ ( τ ) sin ω )( ¯ ∂ + c + ¯ ∂ − c )2 c cos ω − sin ω cos σcl ϑη = − cos σ (1 + κ ( τ ) sin ω )( ¯ ∂ + c − ¯ ∂ − c )2 c − sin σ sin ω (1 + κ ( τ ) sin ω )( ¯ ∂ + c + ¯ ∂ − c )2 c cos ω − sin ω sin σc . (8.1)From (8.1) we obtain ( l ϑξ ) + ( l ϑη ) = (cid:18) (1 + κ ( τ ) sin ω )( ¯ ∂ + c − ¯ ∂ − c )2 c (cid:19) + (cid:18) sin ω (1 + κ ( τ ) sin ω )( ¯ ∂ + c − ¯ ∂ − c )2 c cos ω + sin ωc (cid:19) . Hence, by Lemma . and Lemma . we have < m e − κd < ( l ϑξ ) + ( l ϑη ) ≤ (1 + κ ( τ ) M (cid:48) ) + (cid:34) c + 1 + κ ( τ )2 M (cid:48) (cid:35) , (8.2)where M (cid:48) is a uniform positive constant which is the upper bound of ¯ ∂ + cc and − ¯ ∂ − cc .Now we prove the regularity result in the following four steps. ( ξ, η ) → ( z, t ) is injective We prove this by the method of contradiction. Let us assume that there exist two distinct points ( ξ , η ) and ( ξ , η ) in theregion ABC such that t = t and z = z . Which implies that cos ω ( ξ , η ) = cos ω ( ξ , η ) and φ ( ξ , η ) = φ ( ξ , η ) such that both the points ( ξ , η ) and ( ξ , η ) lie on the same level curve l ϑ = 1 − sin ω ( ξ, η ) = ϑ ≥ . Now from(8.1) we obtain ∇ φ. ( l ϑη , − l ϑξ ) = c sin ω (cid:40) − (1 + κ ( τ ) sin ω ( ¯ ∂ + c − ¯ ∂ − c )2 c (cid:41) < , so φ ( ξ, η ) is strictly monotonically decreasing along each level curve l (cid:36) ( ξ, η ) = (cid:36) ≥ which contradicts theassumption φ ( ξ , η ) = φ ( ξ , η ) . Hence the mapping is injective. C continuity of the function ω ( ξ, η ) Using (8.1) we have cos ωω ξ = − l ϑξ , cos ωω η = − l ϑη (8.3)which gives us | cos ωω ξ | + | cos ωω η | ≤ κ ( τ ) M (cid:48) ) + (cid:34) c + (1 + κ ( τ )) M (cid:48) (cid:35) ≤ C (8.4)where C is a positive constant. From (8.4) we obtain (cid:12)(cid:12)(cid:12)(cid:16) π − ω (cid:17) ω ξ (cid:12)(cid:12)(cid:12) + (cid:12)(cid:12)(cid:12)(cid:16) π − ω (cid:17) ω η (cid:12)(cid:12)(cid:12) ≤ π − ω sin ( π − ω ) C ≤ C which follows that the function ( π − ω ) is uniformly Lipschitz continuous in ( ξ, η ) plane, which means that for anytwo points T = ( ξ , η ) and T = ( ξ , η ) in the domain ABC we have (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) − (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) ≤ C | ( ξ , η ) − ( ξ , η ) | (8.5)Since π < ω < π , we observe that (cid:16) π − ω ( ξ , η ) (cid:17) and (cid:16) π − ω ( ξ , η ) (cid:17) are positive so that (cid:12)(cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) − (cid:16) π − ω ( ξ , η ) (cid:17) (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) − (cid:16) π − ω ( ξ , η ) (cid:17)(cid:12)(cid:12)(cid:12) . (cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) + (cid:16) π − ω ( ξ , η ) (cid:17)(cid:12)(cid:12)(cid:12) ≥ (cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) − (cid:16) π − ω ( ξ , η ) (cid:17)(cid:12)(cid:12)(cid:12) . (cid:12)(cid:12)(cid:12)(cid:16) π − ω ( ξ , η ) (cid:17) − (cid:16) π − ω ( ξ , η ) (cid:17)(cid:12)(cid:12)(cid:12) = | ω ( ξ , η ) − ω ( ξ , η ) | PREPRINT - F
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9, 2021Hence by (8.5), we obtain | ω ( ξ , η ) − ω ( ξ , η ) | ≤ (cid:112) C | ( ξ , η ) − ( ξ , η ) | (8.6)which proves the uniform C continuity of the function ω ( ξ, η ) in the entire domain ABC . ¯ ∂ + cc , ¯ ∂ − cc and (cid:16) ¯ ∂ + c +¯ ∂ − cc cos ω (cid:17) in the entire region ABC of ( ξ, η ) plane From subsection . we know that the mapping ( ξ, η ) → ( z, t ) is injective, so for any two distinct points ( ξ , η ) and ( ξ , η ) we have two different images, say T (cid:48) = ( z , t ) and T (cid:48) = ( z , t ) in the region A (cid:48) B (cid:48) C (cid:48) . Then in this subsectionwe prove the uniform C continuity of the functions (cid:101) R ( ξ, η ) := R ( z, t ) , (cid:101) S ( ξ, η ) := S ( z, t ) and (cid:102) W ( ξ, η ) := W ( z, t ) where (cid:101) R ( ξ, η ) = ¯ ∂ + c/c , (cid:101) S ( ξ, η ) = ¯ ∂ − c/c and (cid:102) W ( ξ, η ) = ( ¯ ∂ + c + ¯ ∂ − c ) /c cos ω in ξ − η plane.We use the uniform Lipschitz continuity of R ( z, t ) from Lemma . to obtain | (cid:101) R ( ξ , η ) − (cid:101) R ( ξ , η ) | = | R ( z , t ) − R ( z , t ) | ≤ C | ( z , t ) − ( z , t ) | = C (cid:110) (cos ω ( ξ , η ) − cos ω ( ξ , η )) + ( φ ( ξ , η ) − φ ( ξ , η )) (cid:111) (8.7)for some uniform positive constant C .Now using (8.6) we have | cos ω ( ξ , η ) − cos ω ( ξ , η ) | ≤ (cid:12)(cid:12)(cid:12)(cid:12) (cid:18) ω ( ξ , η ) − ω ( ξ , η )2 (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) ≤ | ω ( ξ , η ) − ω ( ξ , η ) | ≤ (cid:112) C | ( ξ , η ) − ( ξ , η ) | . So, the uniform Lipschitz continuity of potential function φ ( ξ, η ) in the entire region ABC and (8.7) yields | (cid:101) R ( ξ , η ) − (cid:101) R ( ξ , η ) | ≤ C (cid:110) C | ( ξ , η ) − ( ξ , η ) | + C | ( ξ , η ) − ( ξ , η ) | (cid:111) ≤ C | ( ξ , η ) − ( ξ , η ) | where C and C are positive constants such that C = max { C √ C , C C } .The above result concludes the uniform C continuity of the function (cid:101) R ( ξ, η ) in the entire region ABC . In the samemanner, we can prove that the functions (cid:101) S ( ξ, η ) and (cid:102) W ( ξ, η ) are also uniformly C continuous. ( ρ, u, v ) in the entire region ABC of ( ξ, η ) plane Now we derive the uniform regularity of solution ( ρ, u, v ) and sonic boundary AC using the uniform regularity offunctions (cid:101) R, (cid:101) S and (cid:102) W in the entire region ABC in ( ξ, η ) plane. To prove this we first prove the uniform regularity of σ ( ξ, η ) .Using (2.5) we obtain σ ξ = sin σ sin ωc + (cid:16) sin σ cos ω ( (cid:101) R + (cid:101) S ) − cos σ sin ω ( (cid:101) R − (cid:101) S ) (cid:17) κ ( τ )2 ,σ η = − cos σ sin ωc − (cid:16) cos σ cos ω ( (cid:101) R + (cid:101) S ) + sin σ sin ω ( (cid:101) R − (cid:101) S ) (cid:17) κ ( τ )2 (8.8)which clearly shows that the functions σ ( ξ, η ) , cos σ ( ξ, η ) and sin σ ( ξ, η ) are uniformly Lipschitz continuous in thewhole region ABC . Also, using the fact that (cid:101) R, (cid:101) S, ω ∈ C and (8.8) we see that σ ( ξ, η ) ∈ C , and eventually cos σ ( ξ, η ) , sin σ ( ξ, η ) ∈ C , . Using this and (8.1) one can prove that the function sin ω ( ξ, η ) is uniformly C , continuous in the whole domain ABC .Further, using the pseudo-Bernoulli’s law we notice that ρ ( ξ, η ) is uniformly C , continuous in the whole region ABC which means that the function c ( ξ, η ) is uniformly C , continuous. The expressions for pseudo-velocities leadto uniform C , continuity of the functions u ( ξ, η ) and v ( ξ, η ) in the entire region ABC .23
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9, 2021Also from (8.1) and (8.2) we notice that l ϑξ and l ϑη are C continuous and ( l ϑξ ) + ( l ϑη ) is bounded. Therefore, thesonic boundary AC is C , continuous.We summarize the regularity results in the following theorem. Theorem 8.1.
If the angle β C ∈ ( − π , , then the Goursat problem (2.1) and (3.2) admits a global smooth solutionin the region ABC where the curve AC is the sonic boundary. Further, this solution is uniformly C , continuous upto the sonic boundary AC and the sonic boundary AC is C , continuous. In this paper, we considered a special case of convex pressure and proved that the global solution to the semi-hyperbolicpatch problem for two-dimensional compressible Euler equations with van der Waals gas exists and these solutionsare uniformly C , continuous and also the sonic boundary is C , continuous. The study of semi-hyperbolic patchproblem opens door to extend the solution into the subsonic domain, which we will try to tackle in future. References [1] T. Zhang, Y. Zheng, Conjecture on structure of solutions of Riemann problem for 2-D gasdynamic system, SIAMJournal on Mathematical Analysis 21 (3) (1990) 593–630.[2] J. Glimm, X. Ji, J. Li, X. Li, P. Zhang, T. Zhang, Y. Zheng, Transonic shock formation in a rarefaction Riemannproblem for the 2D compressible Euler equations, SIAM Journal on Applied Mathematics 69 (3) (2008) 720–742.[3] C. W. Schulz-Rinne, J. P. Collins, H. M. Glaz, Numerical solution of the Riemann problem for two-dimensionalgas dynamics, SIAM Journal on Scientific Computing 14 (6) (1993) 1394–1414.[4] A. Kurganov, E. Tadmor, Solution of two-dimensional Riemann problems for gas dynamics without Riemannproblem solvers, Numerical Methods for Partial Differential Equations 18 (5) (2002) 584–608.[5] Y. Zheng, Systems of conservation laws: two-dimensional Riemann problems, Vol. 38, Springer Science &Business Media, 2012.[6] J. Li, T. Zhang, S. Yang, The two-dimensional Riemann problem in gas dynamics, Vol. 98, CRC Press, 1998.[7] K. Song, Y. Zheng, Semi-hyperbolic patches of solutions of the pressure gradient system, Discrete & ContinuousDynamical Systems-A 24 (4) (2009) 1365–1380.[8] M. Li, Y. Zheng, Semi-hyperbolic patches of solutions to the two-dimensional Euler equations, Archive forRational Mechanics and Analysis 201 (3) (2011) 1069–1096.[9] J. Chen, G. Lai, Semi-hyperbolic patches of solutions to the two-dimensional compressible magnetohydrodynamicequations, Communications on Pure & Applied Analysis 18 (2) (2019) 943–958.[10] W. Sheng, G. Wang, T. Zhang, Critical transonic shock and supersonic bubble in oblique rarefaction wavereflection along a compressive corner, SIAM Journal on Applied Mathematics 70 (8) (2010) 3140–3155.[11] R. Courant, K. O. Friedrichs, Supersonic flow and shock waves, Vol. 21, Springer Science & Business Media,1999.[12] A. M. Tesdall, R. Sanders, B. L. Keyfitz, Self-similar solutions for the triple point paradox in gasdynamics, SIAMJournal on Applied Mathematics 68 (5) (2008) 1360–1377.[13] A. M. Tesdall, R. Sanders, B. L. Keyfitz, The triple point paradox for the nonlinear wave system, SIAM Journalon Applied Mathematics 67 (2) (2007) 321–336.[14] Y. Hu, J. Li, W. Sheng, Degenerate Goursat-type boundary value problems arising from the study of two-dimensional isothermal Euler equations, Zeitschrift für angewandte Mathematik und Physik 63 (6) (2012)1021–1046.[15] Q. Wang, Y. Zheng, The regularity of semi-hyperbolic patches at sonic lines for the pressure gradient equation ingas dynamics, Indiana University Mathematics Journal 63 (2) (2014) 385–402.[16] K. Song, Q. Wang, Y. Zheng, The regularity of semi-hyperbolic patches near sonic lines for the 2-D Euler systemin gas dynamics, SIAM Journal on Mathematical Analysis 47 (3) (2015) 2200–2219.24
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