Global existence of shock for the supersonic Euler flow past a curved 2-D wedge

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Global existence of shock for the supersonic Euler flow past a curved 2-D wedge Dian Hu Department of Mathematics, Shanghai University, Shanghai 200444, PR China

a r t i c l e

i n f o

Article history: Received 10 July 2012 Available online xxxx MSC: 35L65 35L67 35M10 35B35 76H05 76N10

a b s t r a c t In this paper, we study the global existence of the supersonic shock for the steady supersonic Euler flow past a curved 2-D wedge. By using the method of characteristic, we show that the shock exists globally and the flow between the shock and wedge is continuous provided the wedge is a small perturbation of a straight wedge under a weighted global Sobolev norm and the vertex angle is less than the extreme angle. © 2012 Elsevier Inc. All rights reserved.

Keywords: Full steady Euler system Compressible flow Free boundary Supersonic shock Global classical solution

1. Introduction This paper is concerned with the global existence of a shock wave arising in supersonic flow past an infinite curved wedge. Such a problem was studied by using the shock polar in [8] for the straight wedge. For the curved wedge, this problem has been widely studied by many authors. By using the methods of characteristic developed for the quasilinear hyperbolic system, the authors in [11,18,13] obtain the local existence of a shock attached at the edge for potential flow and Euler flow. The global existence of the shock has been proved by Chen [3–5] and Yin [19] for potential flow when the wedge is convex or a small perturbation of straight wedge with fast decay at the infinity. On the other hand, by using the Glimm scheme, the authors of [2,20,21] established the global existence and stability of

E-mail address: [email protected]. 0022-0396/$ – see front matter © 2012 Elsevier Inc. All rights reserved. http://dx.doi.org/10.1016/j.jde.2012.11.014

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a weak solution for supersonic flow past a wedge with small total curvature for potential flow [20,21] and Euler flow [2]. In our case, we use the 2-D steady full Euler system to describe the flow and construct the global solution by applying the methods of characteristic. We emphasize that, different from the results in [2], there are no other discontinuities in our solution besides the main curved shock. We first give a short description of the problem. For the 2-D steady full Euler flow, the system can be written as

⎧ ∂x (ρ u ) + ∂ y (ρ v ) = 0, ⎪ ⎪ ⎨ ∂  p + ρ u 2  + ∂ (ρ uv ) = 0, x  y  ⎪ ∂x (ρ uv ) + ∂ y p + ρ v 2 = 0, ⎪ ⎩ ∂x (ρ u E + pu ) + ∂ y (ρ v E + p v ) = 0,

(1.1)

where E = 12 (u 2 + v 2 ) + e and (u , v ), p , ρ and e represent the velocity, pressure, density and inner energy of the flow, respectively. For the polytropic gas, let γ be the adiabatic exponent; we have the state equation

p = A (S )ρ γ

(1.2)

with S being entropy. Suppose that there is a uniform supersonic flow with the state

U 1 = (u 1 , v 1 , p 1 , ρ1 ) coming from infinity and hitting a curved wedge with an angle. Because the flows on the upper part and on the lower part will not interact each other, we just consider the problem in the region above the upper side





W := (x, y ) y = ϕ (x), x  0



(1.3)

1 where ϕ ∈ C loc (R+ ) satisfies ϕ (0) = ϕ  (0) = 0. If the value | uv 11 | is less than the critical number, then there exists a supersonic shock front



S := (x, y ): y = φ(x), x  0



(1.4)

emanating from the wedge vertex (cf. Fig. 1). Across the shock front, the flow will be changed to U = (u , v , p , ρ ) which is supersonic as well. Along the shock front, the Rankine–Hugoniot condition

⎧ [ρ u ]φ  = [ρ v ], ⎪ ⎪ ⎨ p + ρ u 2 φ  = [ρ uv ],  ⎪ [ρ uv ]φ  = p + ρ v 2 , ⎪ ⎩  [ρ u E + pu ]φ = [ρ v E + p v ]

(1.5)

p > p1

(1.6)

and the entropy condition

are satisfied. Let Ω be the domain bounded by the shock front S and the wedge W

 Ω = (x, y ): ϕ (x) < y < φ(x), x > 0 .

(1.7)

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Fig. 1. Supersonic flow past a curved wedge.

Fig. 2. Characteristic.

If there exists no discontinuity for the flow behind the shock front, then the state U satisfies (1.1) in Ω with the wedge boundary condition

v − ϕ  u = 0 on W .

(1.8)

Therefore our main task is to solve the following free boundary problem

⎧ system (1.1) in Ω, ⎪ ⎨ (FB):

the Rankine–Hugoniot conditions (1.5) on S,

⎪ ⎩ the entropy condition (1.6) on S,

(1.9)

the wedge boundary condition (1.8) on W .

Next we give some comments about our proof. Different from the potential flow, for Euler system, there are three characteristics issuing from the shock to down stream, two genuinely nonlinear characteristics and one linearly degenerate characteristic (stream characteristic). The genuinely nonlinear characteristics finally hit the wedge ( P Q 1 , P Q 2 ), while the stream characteristic extends to infinity ( P Q ∞ ) (cf. Fig. 2). To obtain the results we need to establish some global estimates for the solution and its derivatives. For the quasilinear hyperbolic system without inhomogeneous term, there is one semi-global C 1 property, which is stated that, along the characteristic, the solution is proved to be C 1 in a given time period provided the first-order derivatives of the initial value are small. Thus, for the potential flow, if the perturbation of the wedge satisfies some decay in the order of the length of the genuinely nonlinear characteristic between the wedge and shock front, the solution can be proved C 1 between the shock and the wedge (see [4,19]). In our case, for the Euler flow, there exists another linearly degenerate characteristic which extends to infinity. For the full linearly degenerate hyperbolic system, it was proved that if we further assume that the initial value and its first-order derivatives satisfy some small L 1 property as well, then the solution can be proved C 1 global in the domain, even the linearly degenerate characteristic extends to infinity there (see [16,22,23], etc.). Indicated by these

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results, we suppose the perturbation along the wedge is given in a weighted global Sobolev norm, which guarantee not only the C 1 decay at infinity but also the small L 1 property, to obtain the global existence of the solution. Our paper is organized as follows. In Section 2, we use the shock polar to give some analysis for the simple case that the boundary of the wedge is straight, then we introduce the weighted global Sobolev space and state our main results in this paper. We begin our proof with Section 3, firstly Lagrange coordinate transformation is introduced and the main results is restated under the new coordinate (Section 3.1). Secondly we rewrite the system in characteristic form and introduce new variable to replace it by a decoupled form in which the genuinely nonlinear part and linear degenerate part are separated (Section 3.2). Then we use the boundary update scheme to reduce the free boundary problem to a fixed boundary problem (Section 3.3). Finally, we introduce two coordinate transformations to convert the fixed boundary value problem in a sector to a new boundary value problem in a standard strip (Section 3.4). In Section 4, to treat the nonlinear problem obtained in Section 3.4, the corresponding linearized problem is introduced, then we establish solvability of the linearized problem and obtain the a priori estimates. This section will play a key role in this paper. In Section 5, the nonlinear iteration and boundary update scheme are employed to prove our main result. 2. Some preliminaries and main theorem 2.1. The flat structure First, by (1.1) it can be proved that Bernoulli’s law holds along each stream line:

1 2 γp



u 2 + v 2 + i = const,

(2.1)

2

where i = (γ −1)ρ = γc−1 is the enthalpy with c being the sonic speed of the fluid. Moreover, by (1.5), we can also derive that (2.1) is invariant across shock front (see [8]). In our case, all the stream lines come from a region where the constant in (2.1) is the same even along different stream lines. Therefore, we may let 12 (u 2 + v 2 ) + i = κ0 in the whole domain. It turns out that we can eliminate the last equations in (1.1) and (1.5) to obtain the simple forms

⎧ ⎨ ∂x (ρ u ) + ∂ y(ρ v ) = 0, ∂x p + ρ u 2 + ∂ y (ρ uv ) = 0,   ⎩ ∂x (ρ uv ) + ∂ y p + ρ v 2 = 0

(2.2)

⎧  ⎨ [ρ u ]φ = [ρ v ], p + ρ u 2 φ  = [ρ uv ],  ⎩ [ρ uv ]φ  = p + ρ v 2

(2.3)

and

for the simplified state U = (u , v , p ). Denote

⎧ 1 ⎨ H (U , U 1 ) := [ρ v ]/[ρ u ], H 2 (U , U 1 ) := [ρ uv ]/ p + ρ u 2 ,  ⎩ 3 H (U , U 1 ) := p + ρ v 2 /[ρ uv ];

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the Rankine–Hugoniot condition (2.3) can be rewritten as

⎧ ⎨ φ  = H 1 (U , U 1 ), φ  = H 2 (U , U 1 ), ⎩  φ = H 3 (U , U 1 ).

(2.4)

If the wedge is straight, in other words ϕ = 0, let U 0 = (u 0 , v 0 , p 0 ) be the flow behind the straight shock S0 : y = φ0 x. By Rankine–Hugoniot equation (2.4) and the wedge boundary condition (1.8), U 0 and φ0 can be determined by the following algebraic equations:

⎧ 1 ⎪ ⎨ φ0 = H 2 (U 0 , U 1 ), φ0 = H (U 0 , U 1 ), 3 ⎪ ⎩ φ0 = H (U 0 , U 1 ), v0 = 0

(2.5)

u 20 + v 20 > c 02

(2.6)

p0 > p1 .

(2.7)

with the supersonic condition

and the entropy condition

As the analysis in [8], let (u , v , p ) be the state of the flow behind the shock front, the following relation can be extracted from (2.4)

tan arctan w − arctan

v1 u1

 =

γ

p −1 p1 p 2 M1 − p + 1 1

   (1 + μ2 )( M 12 − 1) − ( pp − 1) 1 · , p 2 + μ p1

where M 1 is the Mach number of the upstream flow, w =

( uv , p )

v u

and

μ2 =

γ −1 γ +1 . Let

(2.8)

γ0 be the curve

satisfying (2.8) must locate on it. For γ0 , there are generated by (2.8) on the ( w , p ) plane, two sonic points which are symmetrical with respect to the line w = uv 1 and below the points where 1

|θ − arctan uv 11 | takes local maximum θm . For the states P 0 = ( uv 00 , p 0 ), since v 0 = 0, it must be the

intersection of p-axis and γ0 . Moreover, by the supersonic condition and entropy condition, we can exclude the points which are above the sonic point and below P 1 ( uv 1 , p 1 ). Thus, it can be deduced 1

that if | uv 1 | is less than the critical angle θc , then (2.5) with (2.6) and (2.7) admits a unique solution U 0 1 (cf. Fig. 3). This flat wave structure U 1 , U 0 and S0 is called background solution. 2.2. Weighted norms

Before describing the main theorem, we define the weighted global Sobolev norms on sector. This kind of weighted norms is inductively defined by the norms on the strip via a coordinate transformation. Such weighted norms have been employed in [7,9] and they are analogous to the ones defined in [10,12,17]. Here we first introduce the global Sobolev norms on R and the strip [0, θ0 ] × R. Suppose l, m ∈ {0} ∪ N, for u ∈ C c∞ (R) and U ∈ C c∞ ([0, θ0 ] × R); we define the following global Sobolev norms

u  H l (R) = g

 k∈Z

u  H l ([k,k+1])

(2.9)

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Fig. 3. Shock polar.

and

U C m ([0,θ0 ]; H l (R)) = g

 k∈Z

sup

m   j  ∂ U (θ, ·)

θ ∈[0,θ0 ] j =0

θ

H l ([k,k+1])

(2.10)

.

Then we define the global Sobolev function spaces H lg (R) and C m ([0, θ0 ]; H lg (R)) to be the completion of C c∞ (R) and C c∞ ([0, θ0 ] × R) under the norms (2.9) and (2.10) respectively. We can check that

u  H l (R)  C u  H l (R) . g

On the other hand, by using Sobolev’s inequality, we have

u W l−1,1 (R)  C



u C l−1 ([k,k+1])  C

k∈Z



u  H l ([k,k+1])  C u  H l (R) . g

k∈Z

Moreover, we define the norm  · C l

g

U C l = U C 0 ([0,θ0 ]; H l (R)) + U C 1 ([0,θ g

l −1 0 ]; H g (R))

g

.

(2.11)

It can be verified that if l  2, then

U C 1 ([0,θ0 ]×R)  C U C l .

(2.12)

g

Proposition 2.1. Letting T > 0 and a ∈ R, for u ∈ H lg (R) and U ∈ C m ([0, θ0 ]; H lg (R)), we have the following properties

1 C 2 (T )

 k∈Z

u  H l ([a+kT ,a+(k+1)T ])  u  H l (R)  C 1 ( T ) g

 k∈Z

u  H l ([a+kT ,a+(k+1)T ])

(2.13)

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and



1 C 2 (T )

k∈Z

sup

m    j ∂ U (θ, ·) l θ H ([a+kT ,a+(k+1) T ])

θ ∈[0,θ0 ] j =0

 U C m ([0,θ0 ]; H l (R)) g

 C 1 (T )

 k∈Z

sup

m    j ∂ U (θ, ·)

θ ∈[0,θ0 ] j =0

θ

H l ([a+kT ,a+(k+1) T ])

,

(2.14)

where C 1 ( T ) = J T K + 2 and C 2 ( T ) = J T1 K + 2 with J T K denoting the integer part of T . Proof. Without loss of generality, assume that 0  a < T . For the interval [k, k + 1] where k ∈ Z, it ˜ , a + (k˜ + 1) T ]. In fact, we have [k, k + 1] ∈ can be covered by finite intervals of the form [a + kT

[ k+T1−a ]+2

a k˜ =[ k− ]−1 T

˜ , a + (k˜ + 1) T ]. It can be deduced that [a + kT

u  H l ([k,k+1]) 

 [ k+1−a ]+2 T 

 12

a k˜ =[ k− T ]−1

u 2 l ˜ ,a+(k˜ +1) T ]) H ([a+kT

[ k+T1−a ]+2





a k˜ =[ k− T ]−1

u  H l ([a+kT ˜ ,a+(k˜ +1) T ]) (by Cauchy’s inequality).

Summing these inequalities with respect to k ∈ Z, we have



u  H l ([k,k+1]) 

k∈Z



[ k+T1−a ]+2



k∈Z k˜ =[ k−a ]−1 T

u  H l ([a+kT ˜ ,a+(k˜ +1) T ]) .

Observing that for any k˜ ∈ Z the term u  H l ([a+kT ˜ ,a+(k˜ +1) T ]) is only summed for J T K + 2 times at most, we can obtain

 k∈Z

u  H l ([k,k+1]) 



[ k+T1−a ]+2



k∈Z k˜ =[ k−a ]−1 T

 C 1 (T )



u  H l ([a+kT ˜ ,a+(k˜ +1) T ])

u  H l ([a+kT ,a+(k+1)T ])

k∈Z

where C 1 ( T ) = J T K + 2. Hence, we have proved the right-hand side of (2.13). The left-hand side can be obtained in the same way. For (2.14) the previous discussion is also valid. 2 Proposition 2.2. Letting J ∈ N, T > 0 and a ∈ R, for u ∈ H lg (R) and U ∈ C m ([0, θ0 ]; H lg (R)), we have the following properties

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1 C( J)

u  H l ([a+kT ,a+(k+ J )T ]) 

k∈Z



u  H l ([a+kT ,a+(k+1)T ])

k∈Z

 C( J)



u  H l ([a+kT ,a+(k+ J )T ])

(2.15)

k∈Z

and



1 C( J)



k∈Z

m    j ∂ U (θ, ·) l θ H ([a+kT ,a+(k+ J ) T ])

sup

θ ∈[0,θ0 ] j =0



m   j  ∂ U (θ, ·)

sup

θ

θ ∈[0,θ0 ] j =0

k∈Z

 C( J)

 k∈Z

sup

H l ([a+kT ,a+(k+1) T ])

m   j  ∂ U (θ, ·) l , θ H ([a+kT ,a+(k+ J ) T ])

θ ∈[0,θ0 ] j =0

(2.16)

where C ( J ) > 0 is a constant just depending on J .

2

Proof. It is similar as Proposition 2.1. Suppose 0 < θ0 < π2 and let

 Ω = x ∈ R2 : 0 < r < ∞, 0 < θ < θ0 be an unbounded sector, where (r , θ) is the polar coordinate. The boundary of the domain Ω consists of two rays:



  Γ0 = x ∈ R2 : θ = 0, r ∈ R+ and Γ1 = x ∈ R2 : θ = θ0 , r ∈ R+ . We introduce the transformation P which blows up the vertex of the sector:

P:



t = ln r ,

θ = θ.

Under such a transformation Ω becomes P (Ω) = {−∞ < t < ∞, 0 < θ < θ0 }. For any k ∈ R, m, l = ∞ (Ω) by 0, 1, . . . , we define weighted global H l norms of u ∈ C loc (k)

u 

m,l H g (Ω)

  = ekt u (t , θ)C m ([0,θ

(2.17)

l 0 ]; H g (R))

and the function spaces for our discussion by







,l (k) Hm = u: ekt u (t , θ) ∈ C m [0, θ0 ]; H lg (R) . g (Ω)

The weighted global H l norms of u on the boundaries Γ j ( j = 0, 1) are analogously defined by (k)

u 

H lg (Γ0 )

   = etk u et , 0  H l (R) g

(2.18)

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and (k)

u 

H lg (Γ1 )

   = etk u et cos θ0 , et sin θ0  H l (R) . g

(2.19)

It is trivial that (k)

u 

(k)

H lg (Γ j )

 u 

m,l

H g (Ω)

.

2.3. Main theorem With the defined weighted norms, we are now able to describe our main theorem. Theorem 2.3. Assume that U 1 , U 0 and S0 satisfying (2.5) with (2.6) and (2.7) form a background solution and the tangent line of γ0 on P 0 doesn’t parallel p-axis or w-axis (cf. Fig. 3); there exists a constant σ0 > 0 together with constants M > 0 and M S > 0 just depending on U 1 , U 0 and S0 , such that for any σ ∈ (0, σ0 ], if the boundary of the wedge satisfies the estimates: (−1) H 4g (R+ )

ϕ 

σ

(2.20)

then there exist U (x, y ) and S: y = φ(x) for x > 0 (i) (ii) (iii) (iv)

U satisfies system (2.2) in Ω ; U and U 1 satisfy the Rankine–Hugoniot conditions (2.4) and the entropy condition (1.6) on S; v − u ϕ  = 0 on the perturbed wedge W ; the following estimates hold:

  φ(·) − φ0 ·(−41) +  M S σ , H (R )

(2.21)

 (0) v  ( 0) ( 0)  − v0  +  p − p0  3 + S − S0  3 + S − S0 C 0 (Ω)  M σ . u u  3 C g (Ω) H g (S) 0 C g (Ω)

(2.22)

g

Remark 2.1. By the analysis in Section 2.1, the assumption for U 1 , U 0 and S0 in Theorem 2.3 is satisfied provided the angle of the incoming flow is less than the critical number θc . Remark 2.2. From (2.12) and the coordinate transformation P , it can be deduced that uv and p belong ¯ 0, 0)}) ∩ C 0 (Ω) ¯ , moreover we can further deduce S ∈ C 1 (Ω/ ¯ W ) ∩ C 0 (Ω) ¯ . For (u , v , p ), we to C 1 (Ω/{( ¯ with a crude estimate can deduce that u , v and p belong to C 1 (Ω) ∩ C 0 (Ω)

u − u 0 C 0 (Ω) +  v − v 0 C 0 (Ω) +  p − p 0 C 0 (Ω)  M σ .

(2.23)

3. Decoupling and linearization Starting from this section, we are going to prove Theorem 2.3. In this section, the problem (FB) will be converted to a fixed boundary value problem for hyperbolic system in a standard strip. In Section 3.1, the Lagrange transformation is introduced and Theorem 2.3 is restated under this coordinate. Then in Section 3.2, new variables are introduced and the system is rewritten in canonical form. In Section 3.3, by an application of the boundary update scheme, the free boundary problem is reduced to a fixed boundary problem in a sector. Finally in Section 3.4, by using the corner blowup coordinate transformation, we obtain a standard nonlinear boundary value problem in a strip which is equivalent to that obtained in Section 3.3.

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3.1. The reformulated problem under the Lagrange transformation We first introduce the Lagrange coordinate transformation to convert Euler system (2.2) to another form. This transformation, which has been used in many papers, straightens the stream lines. Especially, the curved wedge W will be transformed to be the axis of new coordinates. Consider the function η(x, y ) defined by (x, y )

η(x, y ) :=

ρ u (s, t ) dt − ρ v (s, t ) ds,

(3.1)

(0,0)

where the integration is proceeded along any smooth curve C connecting (0, 0) with (x, y ). Owing to the first equation of (2.2), it can be checked that the integration (3.1) is independent of the choice of C . So η(x, y ) is well defined. The Lagrangian transformation is defined as L : (x, y ) → (ξ, η):



ξ = x, η = η(x, y ).

(3.2)

Under this coordinate, the Euler system (2.2) becomes

⎧ ∂u ∂ p ∂p ⎪ ⎪ ρu + = 0, − ρv ⎪ ⎪ ∂ξ ∂ξ ∂η ⎪ ⎪ ⎨ ∂v ∂p = 0, + ⎪ ∂ξ ∂η ⎪ ⎪ ⎪ ⎪ 1 ∂u u ∂p ∂v ∂u ⎪ ⎩ −v = 0, + +u ρ ∂ξ a2 ρ 2 ∂ξ ∂η ∂η

(3.3)

which can also be written in the form of conservation laws

⎧ 

 ∂ pv p ∂ ⎪ ⎪ − = 0, u + ⎪ ⎪ ⎪ ∂ξ ρu ∂η u ⎪ ⎪ ⎨ ∂v ∂p = 0, + ⎪ ∂ξ ∂η ⎪ ⎪



 ⎪ ⎪ ∂ v 1 ∂ ⎪ ⎪ − + = 0. ⎩ ∂ξ ρu ∂η u Simultaneously, let

(3.4)

η = ψ(ξ ) be the shock front; the Rankine–Hugoniot conditions are reduced to ⎧     v 1 ⎪  ⎪ ⎪ / , = G ( U , U ) := − ψ 1 1 ⎪ ⎪ u ρu ⎨      pv 1   2 ⎪ ψ = G 2 (U , U 1 ) := − , / p + ρu ⎪ ⎪ u ρu ⎪ ⎪ ⎩  ψ = G 3 (U , U 1 ) := [ p ]/[ v ]

and we can deduce the relation between φ and ψ

φ  (x) =

dy dx

=

v u

dξ + ρ1u dη v 1  ψ (ξ ). = + dξ u ρu

(3.5)

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Furthermore, the condition on the wall takes the form

v u

(ξ, 0) = ϕ  (ξ ) for ξ > 0.

(3.6)

Now consider the (FB) in the Lagrange coordinate; we intend to solve the following (FB)L problem:

(FB)L :

⎧ system (3.3) in Ω L , ⎪ ⎪ ⎨

the Rankine–Hugoniot conditions (3.5) on SL ,

L ⎪ ⎪ ⎩ the entropy condition (1.6) on S ,

(3.7)

the wedge boundary condition (3.6) on W L : η = 0.

For (FB)L , we have the theorem: Theorem 3.1. Under the same assumption of Theorem 2.3, there exists a constant σ0 > 0 together with constants M > 0, M D > 0 and M S > 0, which are independent of σ0 , such that for any σ ∈ (0, σ0 ], if the boundary of the wedge is perturbed and satisfies the estimates:

ϕ (−41) +  σ H g (R )

(3.8)

then there exist SL : η = ψ(ξ ) for ξ > 0 (i) (ii) (iii) (iv)

U L (ξ, η) satisfies system (3.3) in Ω L ; U and U 1 satisfy the Rankine–Hugoniot conditions (3.5) and the entropy condition on SL ; the wedge boundary condition (3.6) is satisfied on the perturbed wedge W L ; the following estimates hold: (−1) H 4g (R+ )

ψ − ψ0 

 MSσ ,

  ( 0) v  ( 0) ( 0)  − v0   u u  3 L +  p − p 0 C 3g (Ω L ) + S − S0  H 3g (SL ) + S − S0 C 0 (Ω L )  M σ . 0 C g (Ω )

(3.9)

(3.10)

dξ + ρ1u dη is a total differential form due to the third  equation of (3.4). Hence y (ξ, η) is well defined by uv dξ + ρ1u dη and we can define the inverse L coordinate transformation Once Theorem 3.1 has been proved, then



v u

x = ξ, y = y (ξ, η).

(3.11)

Thus Theorem 2.3 is a direct conclusion of Theorem 3.1 and we will focus our attention on the latter following. 3.2. Decoupling In this subsection, (3.3) will be rewritten as characteristic form and then new variables will be introduced to replace (3.3) in decoupled form. For notational convenience, we omit the superscript L on Ω , W and S in the following discussion. Write (3.3) in the matrix form

A

∂U ∂U = 0, +B ∂ξ ∂η

(3.12)

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12

where

⎛ A=⎝

⎞

1

u

ρ

u



⎠,

1

u

ρ

c2 ρ 2

B=

−v u

 

 U=

,

−v u

u v p

;

then the characteristic polynomial of this system is

⎡

λ ρ +v

λu

D (λ) = det(λ A − B ) = det ⎣

⎤

λu −u ⎦ λu ρ + v −u c2 ρ 2

2 

λ . − λu u 2 + +v λ

= λ3 u 3

1 c2 ρ 2

ρ

(3.13)

2 If the flow is supersonic (u 2 + v 2 > c√ ), by a direct computation, it can be deduced that (3.13) has c 2 ρ v ±c ρ u u 2 + v 2 −c 2 with the corresponding left eigenvectors three roots λ0 = 0 and λ± = u 2 −c 2

i =

1

u

−

λi

ρ

  − v , 1, λ i ,

i = 0, ±.

Multiplying (3.12) by the left eigenvectors, it will be rewritten in the characteristic form

√ −v D ±u + u D ± v ± u

where D ± = D ξ + λ± D η . Let e =

√

u2 + v 2 − c2

ρc

D ± p = 0,

∂u ∂v 1 ∂p +v + = 0, ∂ξ ∂ξ ρ ∂ξ

u 2 + v 2 −c 2 cρ u2

and w =

v ; u

(3.14) (3.15)

(3.14) can be rewritten as

D ± w ± e D ± p = 0.

(3.16)

Moreover, by applying Bernoulli’s law (2.1), Eq. (3.15) can be replaced by the equation for the entropy S

∂S = 0. ∂ξ

(3.17)

Therefore, (3.14) and (3.15) have been converted to (3.16) and (3.17), which just contain the first derivatives for ( w , p , S ), respectively. On the other hand, in a neighborhood of U 0 , U , therefore e and λ± , can be expressed in terms of U¯ = ( w , p , S ). In fact, it can be deduced from Bernoulli’s law and the states equation that

   2κ − 2γ p 1− γ1 A − γ1 (S )  0 ( γ −1 ) u= 1 + w2

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13

and

   2κ − 2γ p 1− γ1 A − γ1 (S )  0 ( γ −1 ) v=w . 1 + w2 Thus if there is no discontinuity in Ω , then (FB)L is equivalent to the following free boundary value problem (FB)L for U¯



D± w ± e D± p = 0 ∂ξ S = 0

in Ω,

w = ϕ  on W , ⎧ ⎨ ψ  = G¯ 1 (U¯ , U 1 ) ψ  = G¯ (U¯ , U ) on S, ⎩  ¯2 ¯ 1 ψ = G 3 (U , U 1 )

(3.18) (3.19) (3.20)

where G¯ i (U¯ , U 1 ) := G i (U (U¯ ), U 1 ). 3.3. Fixing the free boundary To treat the free boundary condition (3.20), the boundary update scheme will be introduced. In fact we decompose the (FB)L into two problems: one is a fixed boundary value problem of (3.18) in a domain with temporary fixed boundary, the other is an initial value problem of an ordinary differential equation to update the approximate boundary. This scheme defines an update map of the approximate boundary. The existence of the solution to the free boundary value problem is a conclusion of the existence of a fixed point for this map. This scheme has been employed to treat the free boundary problem for many times (see [6,7,9], etc.). So we just give the main points here. First, in view of the reformed Rankine–Hugoniot conditions (3.20) on S we eliminate ψ to obtain the boundary value conditions on the temporarily fixed boundary:



G  (U¯ , U 1 ) = G¯ 1 (U¯ , U 1 ) − G¯ 2 (U¯ , U 1 ) = 0, G  (U¯ , U 1 ) = G¯ 1 (U¯ , U 1 ) − G¯ 3 (U¯ , U 1 ) = 0.

(3.21)

Let Γ be a perturbation curve with respect to the background solution shock S0 , which has the equation η = ψ(ξ ), then consider the temporary fixed boundary value problem

% (NL):

system (3.18) in Ωa , the reformed Rankine–Hugoniot conditions (3.21) on Γ, the wedge boundary condition (3.19) on W : η = 0,

(3.22)

where Ωa is the domain bounded by Γ and W . Second, once (NL) is solved, we solve the initial value problems of ordinary differential equation

%

to obtain the update boundary Ψ .

dΨ

= G¯ 1 (U¯ , U 1 ), dξ Ψ (0) = 0,

on Γ,

(3.23)

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Thus (3.22) and (3.23) define an update map from ψ to Ψ . The existence of the fixed point for this map leads to the existence of the solution for (FB)L . To give a precise description, we define a neighborhood of the unperturbed shock in some function space:





(−1)



K  = ψ(ξ ) ∈ L 1loc (R+ ); ψ(ξ ) − ψ0 ξ  H 4 (R+ )   . g

For given Γ = {(ξ, ψ(ξ )) | ψ ∈ K  and ξ  0}, ary value problem



 will be chosen properly later, (NL) is the fixed bound-

D± w ± e D± p = 0 ∂ξ S = 0



(3.24)

w = ϕ

in Ωa ,

on W ,

G  (U¯ , U

1) = 0 G  (U¯ , U 1 ) = 0

on Γ.

(3.25) (3.26) (3.27)

Since (3.23) can be solved just by integrating both sides, we will put main effort into solving the fixed boundary problem (3.25)–(3.27) following. 3.4. The standardized problem Consider (3.25)–(3.27), it’s a fixed boundary problem for hyperbolic system in a sector with a curved boundary. This kind problem has been widely studied (see [13–15], etc.). Here, in an absolutely different way, we will use the corner blowup coordinate transformation to convert (3.25)–(3.27) into a hyperbolic boundary value problem in a strip. This skill has been employed in the researches of the boundary value problem for elliptic equation in a sector (see [10,12,17], etc.). By the definitions in Section 2.2, the estimates of the solution in sector under the weighted global Sobolev norm will be obtained automatically from the corresponding estimates for the problem in the strip. We will not give any explanation for this following. Introduce coordinate transformation

P:

⎧  1  ⎪ ⎨ t = ln ξ 2 + η2 , 2

η ⎪ ⎩ θ = arctan , ξ

(3.28)

Ωa , Γ and W have been reduced to



Da = (t , θ) ∈ R2 t ∈ R, 0 < θ < θa (t ) , 

Σa = (t , θ) ∈ R2 t ∈ R, θ = θa (t ) ,

 Σ0 = (t , θ) ∈ R2 t ∈ R, θ = 0 respectively and (3.25)–(3.27) have been converted to



¯ ± w ± e D¯ ± p = 0 D in Da , ∂θ S − cot θ∂t S = 0   w (0, t ) = g W (t ) := ϕ  et on Σ0 ,   ¯ G (U , U 1 ) = 0 on Σa G  (U¯ , U 1 ) = 0 1+λ tan θ

± ¯ ± = ∂θ + λ¯ ± ∂t and λ¯ ± = where D λ± −tan θ .

(3.29) (3.30) (3.31)

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15

The boundary Σa , which is the image of the approximate boundary Γ , is variable because the approximate boundary Γ will be updated. To obtain the convergence of the solution sequence, we need to compare the norms between the solutions in different approximate domains. Thus we further use a coordinate transformation to convert Da to a standard strip. In fact, let

T:

⎧ ⎨ t˜ = t , θ0 θ , ⎩ θ˜ = θa (t )

where θ0 = arctan ψ0 . By this coordinate transformation, Da , Σa and Σ0 are standardized



 D = (t˜, θ˜ ) ∈ R2 t˜ ∈ R, 0 < θ˜ < θ0 ,

 Σ1 = (t˜, θ˜ ) ∈ R2 t˜ ∈ R, θ˜ = θ0 ,

 Σ0 = (t˜, θ˜ ) ∈ R2 t˜ ∈ R, θ˜ = 0 . Meanwhile (3.29)–(3.31) are reduced to

⎧ ⎪ ⎨ D˜ ± w ± e D˜ ± p = 0 

˜ a (t˜) ˜ a (t˜) θθ θθ  ˜ ˜ ⎪ ˜ ∂t˜ S = 0 ∂ S − θ ( t ) cot / θ + θθ ( t ) cot ⎩ θ˜ a 0 a θ0 θ0

in D ,

w (0, t˜) = g W (t˜) on Σ0 ,



G  (U¯ , U

1) = 0 G  (U¯ , U 1 ) = 0

(3.32)

(3.33) (3.34)

on Σ1 ,

˜ ± = ∂ ˜ + λ˜ ± ∂t˜ and where D θ

λ˜ ± =

˜ ˜ λ± θa (t˜) tan θ θθa0(t ) + θa (t˜) ˜

˜

˜

˜

˜ a (t˜) λ± θ0 − θ0 tan θ θθa0(t ) + λ± θ˜ θa (t˜) tan θθθa0(t ) + θθ

.

(3.35)

Therefore, we have obtained the ultimate form of the nonlinear problem in this section. 4. The solvability of the linearized problem In this section, to treat the nonlinear boundary value problem for the hyperbolic system (3.32)– (3.34) in a linear iteration scheme, we introduce the linearized problem and obtain the existence of solution with a priori estimates. The results deduced in this section will play a key role in our final derivation in Section 5. 4.1. The linearized problem First, by the equation for S in (3.32), we may assume that S has the expression

 ˜ a (t˜) θθ S (θ˜ , t˜) = S t˜ + ln sin θ0

(4.1)

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16

1 ˜ , p˜ , S˜ ) = ( w 0 , p 0 , S0 ) + where S ∈ C loc (R). To consider the linearized problem, assume that ( w

˜ , δ p˜ , δ S˜ ) is the approximate solution and δ S˜ has the expression (δ w

 θ˜ θ˜a (t˜) δ S˜(θ˜ , t˜) = δ S˜ t˜ + ln sin θ0

(4.2)

˜ δ θ˜a := θ˜a − θ0 ∈ C c∞ (R). We further assume that δθa := θa − θ0 ∈ ˜ , δ p˜ ∈ C c∞ ([0, θ0 ] × R); δ S, where δ w ∞ C c (R) in (3.32). The linearized problem of (3.32)–(3.34) is

⎧ ⎪ ⎨ D˜ ± δ w ± e D˜ ± δ p = 0 

˜ a (t˜) θ˜ θa (t˜) θθ  ˜ ˜ ⎪ ˜ ∂t˜ δ S = 0 ∂ δ S − θ ( t ) cot / θ + θθ ( t ) cot ⎩ θ˜ a 0 a θ0 θ0 

in D ,

δ w = g 0 on Σ0 ,

α δ w + β  δ p + γ δS = g  α δ w + β  δ p + γ δS = g 

(4.3)

(4.4) on Σ1

(4.5)

˜ , p˜ , S˜ ), λ± = λ± ( w ˜ , p˜ , S˜ ) and α  = ∂ w G  |U¯ 0 , β  = ∂ p G  |U¯ 0 , γ  = ∂S G  |U¯ 0 , α  = where e = e ( w ∂ w G  |U¯ 0 , β  = ∂ p G  |U¯ 0 , γ  = ∂S G  |U¯ 0 . Since γ  2 + γ  2 = 0, without loss of generality we assume γ  = 0 and rewrite (4.5) in another form



where

















α1 δ w + β1 δ p = g1 α2 δ w + β2 δ p + δ S = g2

α1 = αα  γγ  , β1 = ββ  γγ  , α2 =

on Σ1

(4.6)

α  , β = β  and g (i = 1, 2) are linear combinations of g  2 i γ γ

and g  . Replacing (4.5) by (4.6) it can be observed that the subsystem

˜ ± δ w ± e D˜ ± δ p = 0 in D , D

(4.7)

δ w = g 0 on Σ0 ,

(4.8)

α1 δ w + β1 δ p = g1 on Σ1

(4.9)

can be solved independently and then, with solved (δ w , δ p ), δ S can be obtained by solving the second equation of (4.3) with the second boundary condition of (4.6). Thus we will focus our attention on the above boundary value problem for 2 × 2 hyperbolic system. Our main task in this section is to prove the following theorem. Theorem 4.1. Let l = 2, 3 and consider the linearized problem (4.7)–(4.9); under the assumption of U 0 , U 1 and ˜ , δ p˜ ∈ C c∞ ([0, θ0 ] × R), δ S˜ , δ θ˜a , δθa ∈ C c∞ (R) S0 in Theorem 2.3, there exists a constant δ0 > 0 such that if δ w and satisfy

˜ C l + δ p˜ C l + δ S˜  H l (R) + δ θ˜a  H l+1 (R) + δθa  H l+1 (R)  δ0 , δ w g

g

g

g

g

(4.10)

then: (a) For any g 0 ∈ H lg (R), g 1 ∈ C c∞ (R) and that

 > 0, there exists an approximate solution for (4.7)–(4.9) such

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˜ ± δ w  ± e D˜ ± δ p  = 0 D

17

in D ,

(4.11)

δ w  = g 0 on Σ0 ,

α1 δ w  + β1 δ p  = g1

(4.12)

on Σ1 ,

(4.13)

where g 0 ∈ C c∞ (R) satisfies the approximate estimate

    g − g0  0

H lg (R)



(4.14)

and we have the estimate

  δ w 

C lg

      + δ p  C l  C  g 1  H l (R) +  g 0  H l (R) . g

g

(4.15)

g

(b) If g 0 ∈ C c∞ (R), g 1 ∈ C c∞ (R), μ± ∈ C c∞ ([0, θ0 ] × R) and δ w , δ p ∈ C c∞ ([0, θ0 ] × R) such that

˜ ± δ w ± e D˜ ± δ p = μ± D

in D ,

(4.16)

δ w = g 0 on Σ0 ,

(4.17)

α1 δ w + β1 δ p = g1 on Σ1 ,

(4.18)

then (δ w , δ p ) satisfies the estimate

δ w C l + δ p C l  C g



g

i =0,1

 g i  H l (R) + g

 i =±

 μi C 0 ([0,θ0 ]; H l (R)) g

(4.19)

where C is a constant just depending on the background solution U 0 and U 1 . Before the proof of Theorem 4.1, we derive a coefficient condition deduced from the assumption in Theorem 2.3. Lemma 4.2. If U 0 , U 1 and S0 form a background solution satisfying the assumption in Theorem 2.3, then α1 β1 = 0. Proof. Since the shock polar starting from ( w 1 , p 1 ) is determined by the relation



G  (U¯ , U 1 ) = 0, G  (U¯ , U 1 ) = 0,

the shock polar (2.8) can be expressed by the parameter S :









G  w (S ), p (S ), S , U 1 = 0,    G  w (S ), p (S ), S , U 1 = 0.

Differentiate the above equations with respect to S ; it yields

⎧               ⎪ ⎪ ⎨ G w w (S ), p (S ), S , U 1 w (S ) + G p w (S ), p (S ), S , U 1 p (S ) + G S w (S ), p (S ), S , U 1 = 0,             ⎪ G ⎪ w w (S ), p (S ), S , U 1 w  (S ) + G p w (S ), p (S ), S , U 1 p  (S ) + G S w (S ), p (S ), S , U 1 ⎩ = 0.

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18

Recalling the definitions of

α1 and β1 in (4.6), we have dw

= β1 . =  w (S ) S0 α1 p  (S )

dp w0

By the assumption of U 1 , U 0 and S0 in Theorem 2.3, the tangent line of the shock polar on point P 0 ( w 0 , p 0 ) does not parallel the coordinate axis, it yields that

dw dp

which implies

= 0 or ∞, w0

α1 β1 = 0. 2

Following, to avoid falling into technique details, we prove Theorem 4.1 in the special case that

δθa = δ θ˜a = 0 and l = 2. The general case can be proved in the same way by a more complicated computation. Under this assumption, the straighten coordinate transformation T in Section 3.4 is identical transformation, thus we may use the (θ, t ) coordinate to perform our discussion in this section following. As a result, in this simple case, (4.2) is reduced to

S˜ = S˜ (t + ln sin θ)

(4.20)

˜ ± in (4.3) has the expression and λ

λ˜ ± =

tan θ + λ± 1 − λ± tan θ

(4.21)

.

4.2. The proof of Theorem 4.1 The proof of Theorem 4.1 depends on a careful energy estimates of the solution (δ w , δ p ) for the following backward initial value problem for 2 × 2 hyperbolic system

˜ ± δ w ± e D˜ ± δ p = μ± D

in D ,

(4.22)

(δ w ± e δ p )(θ0 , t ) = ϕ± (t ) on Σ1 ,

(4.23)

where ϕ± ∈ C c∞ (R) and μ± ∈ C c∞ ([0, θ0 ] × R). ˜ , δ p˜ ∈ C c∞ ([0, θ0 ] × R), δ S˜ ∈ C c∞ (R) and satisfy For (4.22), under the assumption of Theorem 4.1 δ w the perturbation estimate

˜ C 2 + δ p˜ C 2 + δ S˜  H 2 (R)  δ0 . δ w g g g

(4.24)

Using the Sobolev inequalities, it yields that

˜ C 1 ([0,θ0 ]×R) + δ p˜ C 1 ([0,θ0 ]×R) + δ S˜ C 1 (R)  δ0 . δ w Thus, by the classical theory of the initial value problem for linear hyperbolic system (see [1,15], etc.), (4.22) with condition (4.23) admits a solution (δ w , δ p ) ∈ C c∞ ([0, θ0 ] × R) × C c∞ ([0, θ0 ] × R). To describe the energies of δ w and δ p, we define





0 ¯m t − (θ, t ) := λ −1 θ +t





¯ 0M + 1 θ + t and t + (θ, t ) := λ

(4.25)

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19

Fig. 4. Characteristic.

where 0 = λ¯ m

1 + λ− |U¯ 0 tan θ

inf

θ ∈[0,θ0 ]

λ− |U¯ 0 − tan θ

and

λ¯ 0M = sup

θ ∈[0,θ0 ]

1 + λ+ |U¯ 0 tan θ

λ+ |U¯ 0 − tan θ

.

By the entropy condition, we can deduce

λ− |U¯ 0 < 0 < tan θ0 < λ+ |U¯ 0 . Thus 0 −∞ < λ¯ m − 1 < λ˜ − < λ˜ + < λ¯ 0M + 1 < +∞,

(4.26)

provided δ0 in (4.24) is sufficiently small. Moreover, let f ± (τ ; θ, t ) be the characteristic curve with ˜ ± passing (θ, t ) (see Appendix A). By (4.26), for k ∈ Z, we have respect to λ







f ± (θ1 ; θ2 , t ): t ∈ t − (θ2 , k), t + (θ2 , k + 1)

 ⊆ t − (θ1 , k), t + (θ1 , k + 1)

(4.27)

where 0 < θ2  θ1  θ0 (cf. Fig. 4). Let k ∈ Z; the energies of δ w and δ p are defined in the following forms



t + (θ,  k +1 )



f i (θ, k) = eC δ0 θ



 12

2

∂ti δ w + e ∂ti δ p (θ, t ) dt

t + (θ,  k +1 )



 12 

2

∂ti δ w − e ∂ti δ p (θ, t ) dt

+

t − (θ,k)

,

t − (θ,k)

(4.28) furthermore we define

 g i (θ, k) = e

t + (θ,  k +1 )



C δ0 θ

∂ti

μ+ (θ, t ) dt

t − (θ,k)

F (θ, k) =



i =0,1,2



 12

2

t + (θ,  k +1 )



∂ti

+

2

μ− (θ, t ) dt

 12  ,

(4.29)

t − (θ,k)

f i (θ, k)

and

G (θ, k) =



i =0,1,2

g i (θ, k).

(4.30)

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Lemma 4.3 (Key step). Under the assumption of Theorem 4.1, consider the solution (δ w , δ p ) for (4.22)–(4.23); there exists an n0 ∈ N, such that Case 1: if e−n0  θ  τ  θ0 , then we have

τ f 0 (θ, k)  f 0 (τ , k) + C δ0

τ f 0 (s, k) ds +

θ

θ

τ

τ

f 1 (θ, k)  f 1 (τ , k) + C δ0

f 1 (s, k) ds + θ

τ f 2 (θ, k)  f 2 (τ , k) + C δ0



g 0 (s, k) ds,

(4.31)

g 1 (s, k) ds,

(4.32)

θ

τ



f 1 (s, k) + f 2 (s, k) ds +

g 2 (s, k) ds,

θ

(4.33)

θ

τ F (θ, k)  F (τ , k) + C δ0

τ F (s, k) ds +

θ

G (s, k) ds;

(4.34)

θ

Case 2: if e−n−1  θ  τ  e−n , then we have



n

f 0 (θ, k)  f 0 (τ , k) + C e ak−n + C δ0



n

f 1 (θ, k)  f 1 (τ , k) + C e ak−n + C δ0





τ

τ f 0 (s, k) ds +

θ

θ

τ

τ f 1 (s, k) ds +

θ



f 2 (θ, k)  f 2 (τ , k) + C en ak−n + C δ0



τ



g 0 (s, k) ds,

(4.35)

g 1 (s, k) ds,

(4.36)

θ



τ

f 1 (s, k) + f 2 (s, k) ds +

θ

g 2 (s, k) ds,

(4.37)

θ



F (θ, k)  F (τ , k) + C en ak−n + C δ0



τ

τ F (s, k) ds +

θ

G (s, k) ds

(4.38)

θ

where am = δ S˜  H 2 ([m−2,m+2]) and n0 is just depending on the background solution. Moreover for θ ∈ [0, θ0 ], we have

θ0 F (θ, k)  e

C δ0

F (θ0 , k) + e

C δ0

G (s, k) ds.

(4.39)

θ

Here and following, C denotes the constant just depending on the background solution. Proof. To show (4.31) and (4.35), we write (4.22) in another form

˜ ± (δ w ± e δ p ) = D

(δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e + μ± 2

and integrate this equation along the ±characteristic from θ to

τ , we have the expression

(4.40)

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21

  (δ w ± e δ p )(θ, t ) − (δ w ± e δ p ) τ , f ± (τ ; θ, t ) θ =

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ

(s, f ± (s;θ,t ))

τ

ds +

μ± |(s, f ± (s;θ,t )) ds.

(4.41)

τ

Let 0 < θ  τ  θ0 and consider the L 2 ([t − (θ, k), t + (θ, k + 1)]) norm of (δ w ± e δ p )(θ, ·); by integrating (4.41) with respect to t, we can deduce



 12

t + (θ,  k +1 ) 2

(δ w ± e δ p ) (θ, t ) dt t − (θ,k)



t + (θ,  k +1 )

  (δ w ± e δ p )2 τ , f ± (τ ; θ, t ) dt



 12

t − (θ,k)



t + (θ,  k+1) θ

+ τ

t − (θ,k)



ds

 12 dt

(s, f ± (s;θ,t ))

t + (θ,  k+1) θ

+

2

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2 2

(μ± )|(s, f ± (s;θ,t )) ds

 12 dt

(4.42)

.

τ

t − (θ,k)

For the first term on the right-hand side of (4.42), recalling (4.27) and using the estimates of f ± (τ ; θ, t ), (A.10) in Appendix A, it can be obtained that



t + (θ,  k +1 )



 (δ w ± e δ p ) τ , f ± (τ ; θ, t ) dt 2

 12

 e

C δ0 (τ −θ )

 12

t + ( τ ,k+1) 2

(δ w ± e δ p ) (τ , t ) dt

.

t − (τ ,k)

t − (θ,k)

(4.43) Similarly, for the third term, we have



t + (θ,  k+1) θ

2 (μ± )|(s, f ± (s;θ,t )) ds

t − (θ,k)



 12 dt

t + (θ,  k+1) θ

=

τ

t − (θ,k)

τ 

  (μ± ) s, f ± (s; θ, t ) ds

τ 



 (μ± ) s, f ± (s; θ, t ) dt 2

θ

 12 dt

τ

t + (θ,  k +1 )



2

 12 ds

t − (θ,k)

 eC δ0 (s−θ )

θ

where Minkowski’s inequality is used at the first inequality.

t + ( s,k+1)

(μ± )2 (s, t ) dt t − (s,k)

 12 ds

(4.44)

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22

For the second term on the right-hand side of (4.42), by the expression

˜ ± ln e = (ln e ) ˜ ∂θ δ S˜ + (ln e ) ˜ λ¯ ± ∂t δ S˜ + (ln e ) w˜ D˜ ± δ w ˜ + (ln e ) p˜ D˜ ± δ p˜ , D S S

(4.45)

we deduce



t + (θ,  k+1) θ

ds

 12 

dt

(s, f ± (s;θ,t ))

τ

t − (θ,k)

2

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

4 

Ii,

(4.46)

i =1

where



t + (θ,  k+1) θ

(δ w + e δ p ) − (δ w − e δ p )

I1 = τ

t − (θ,k)



2

t + (θ,  k+1) θ

I2 =



2

t + (θ,  k+1) θ

I3 = t − (θ,k)

 I4 =

 12

ds

dt

2

 ˜ (ln e )δ w˜ D˜ ± δ w

,

ds

 12 dt

,

(s, f ± (s;θ,t ))

(δ w + e δ p ) − (δ w − e δ p ) 2

τ

t − (θ,k)

,

(s, f ± (s;θ,t ))

2

t + (θ,  k+1) θ

dt

2

 (ln e )δ S˜ λ¯ ± ∂t δ S˜

(δ w + e δ p ) − (δ w − e δ p )

τ

ds

 12

(s, f ± (s;θ,t ))

(δ w + e δ p ) − (δ w − e δ p )

τ

t − (θ,k)

2

 (ln e )δ S˜ ∂θ δ S˜

2

 ˜ (ln e )δ p˜ D ± δ p˜

ds

 12 dt

.

(s, f ± (s;θ,t ))

First, for I 2 , it can be computed that

 I 2  C ∂t δ S˜  L ∞ (R)

t + (θ,  k+1) θ

τ

t − (θ,k)

   C δ S˜  L ∞ (R)

 C δ S˜  H 2g (R)

τ 

t + (θ,  k+1)

θ

t − (θ,k)

 e

C δ0 (s−θ )

 C δ0 θ

C δ0 (s−θ )

dt

t + ( s,k+1)

 12

2

dt

ds

(s, f ± (s;θ,t ))

(δ w + e δ p ) − (δ w − e δ p )

 12

2 (s, t ) dt

ds

t − (s,k)

 e

(δ w + e δ p ) − (δ w − e δ p )

2

θ

τ

ds

 12

(s, f ± (s;θ,t ))

2

τ

2

 (δ w + e δ p ) − (δ w − e δ p ) 2

t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p ) 2

t − (s,k)

 12

2 (s, t ) dt

ds.

(4.47)

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23

Second, by the perturbation estimate (4.24), the similar discussions are also valid for I 3 and I 4 , in fact, we can prove that

τ I 3 , I 4  C δ0

 eC δ0 (s−θ )

t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p ) 2

θ

 12

2 (s, t ) dt

ds.

(4.48)

t − (s,k)

Finally, for I 1 , we have



t + (θ,  k+1) θ

I1  C τ

t − (θ,k)

τ 

t + (θ,  k+1)

C

(δ w + e δ p ) − (δ w − e δ p ) 2 (δ w + e δ p ) − (δ w − e δ p ) 2

θ

2

t − (θ,k)



 · ln sin s + f ± (s; θ, t ) dt 

τ C

(s, f ± (s;θ,t ))

e

C δ0 (s−θ )

2

 12 dt

cot2 sδ S˜  2 (s, f ± (s;θ,t ))

 12 ds

t + ( s,k+1)

cot s

(δ w + e δ p ) − (δ w − e δ p )

2

2

θ



 cot sδ S ln sin s + f ± (s; θ, t ) ds ˜

(s, t )δ S˜  2 (ln sin s + t ) dt

 12 ds.

t − (s,k)

(4.49) For (4.49), we can’t obtain a similar estimate as (4.47) and (4.48) while θ, τ ∈ (0, θ0 ], because lims→0+ cot s = +∞. The estimate will be obtained in different forms depending on the interval which θ and τ belong to. In fact, if e−n0  θ  τ , then we have:



I 1  C cot e

−n0

   δ S˜ 



τ e

L ∞ (R)

C δ0 (s−θ )

 C cot e

−n0

δ S˜ 

 e

H 2g (R)

 12

2 (s, t ) dt

ds

t − (s,k)

τ



(δ w + e δ p ) − (δ w − e δ p ) 2

θ



t + ( s,k+1)

C δ0 (s−θ )

t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p )

(s, t ) dt

2

θ

 12

2

ds

t − (s,k)

(4.50) where n0 ∈ N will be determined later. If n  n0 and e−n−1  θ  τ  e−n , then we can deduce that

  I 1  C cot e−n−1

·

e θ

t ∈[t − (e−n−1 ,k)+ln sin(e−n−1 ),t + (e−n ,k+1)+ln sin(e−n )]



τ

C δ0 (s−θ )

 δ S˜ (t )

sup t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p ) 2

t − (s,k)

 12

2 (s, t ) dt

ds.

(4.51)

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24

To give a careful estimate, we focus our attention on the interval [t − (e−n−1 , k) + ln sin(e−n−1 ), t + (e−n , k + 1) + ln sin(e−n )]. Recall the definition of t ± (θ, t ) (4.25); it can be proved that

lim

 

   t − e−n−1 , k + ln sin e−n−1 − (k − n − 1)

n→+∞



= lim

n→+∞

   0 − 1 e−n−1 + k + ln sin e−n−1 − (k − n − 1) λ¯ m

=0

(4.52)

and

 

lim

n→+∞

   t + e−n , k + 1 + ln sin e−n − (k − n + 1)

= lim



n→+∞

   λ¯ 0M + 1 e−n + k + 1 + ln sin e−n − (k − n + 1)

= 0.

(4.53)

From the above observation, we can choose n0 ∈ N such that for any n  n0

 

       t − e−n−1 , k + ln sin e−n−1 , t + e−n , k + 1 + ln sin e−n ⊂ [k − n − 2, k − n + 2].

Note that n0 is just depending on the background solution. Thus recalling the definition of am in Lemma 4.3, we can give the estimate

  I 1  C cot e−n−1 τ ·

t ∈[k−n−2,k−n+2]

 eC δ0 (s−θ )

θ

 δ S˜ (t )

sup

t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p ) 2

 12

2 (s, t ) dt

ds

t − (s,k)

τ  C en ak−n θ

 eC δ0 (s−θ )

t + ( s,k+1)

(δ w + e δ p ) − (δ w − e δ p ) 2

 12

2 (s, t ) dt

ds.

(4.54)

t − (s,k)

Therefore from (4.42)–(4.44), (4.46)–(4.48), (4.50) and (4.54), we conclude that if e−n0  θ  τ  θ0 , then

τ f 0 (θ, k)  f 0 (τ , k) + C δ0

τ f 0 (s, k) ds +

θ

g 0 (s, k) ds; θ

if n0  n and e−n−1  θ  τ  e−n , then



n

f 0 (θ, k)  f 0 (τ , k) + C e ak−n + C δ0



τ

τ f 0 (s, k) ds +

θ

g 0 (s, k) ds. θ

Hence (4.31) and (4.35) have been proved. Next, for f 1 (θ, k) and f 2 (θ, k), we differentiate (4.40) with respect to t; it follows that

˜ ± δ w t ± e D˜ ± δ pt = −∂t λ˜ ± δ w t ∓ ∂t e (δ p θ + λ˜ ± δ pt ) ∓ e ∂t λ˜ ± δ pt + ∂t μ± D

(4.55)

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25

and

˜ ± δ w tt ± e D˜ ± δ ptt = −∂t λ˜ ± δ w tt ∓ ∂t e (δ p θ t + λ˜ ± δ ptt ) ∓ e ∂t λ˜ ± δ ptt − ∂tt λ˜ ± δ w t D − ∂t λ˜ ± δ w tt ∓ ∂tt e (δ p θ + λ˜ ± δ pt ) ∓ ∂t e (δ p θ t + λ˜ ± δ ptt + ∂t λ˜ ± δ pt ) ∓ ∂t e ∂t λ˜ ± δ pt ∓ e ∂tt λ˜ ± δ pt ∓ e ∂t λ˜ ± δ ptt + ∂tt μ± .

(4.56)

We observe that by Eq. (4.22), δ p θ can be expressed in terms of δ w t ± e δ pt . Moreover, by Eqs. (4.22) and (4.55) δ p θ t can be expressed in terms of δ w t ± e δ pt and δ w tt ± e δ ptt . Therefore (4.55) and (4.56) can be rewritten in the similar form as (4.40). Thus, as the proof of (4.31) and (4.35), we can deduce (4.32), (4.33), (4.36) and (4.37). By the definition of F (θ, k), (4.34) and (4.38) can be obtained by adding up the estimates for f i (θ, k). To prove (4.39), applying the Gronwall inequality for e−n−1  θ < τ  e−n to (4.38), we obtain

τ

τ

θ

n

e C (e

F (s, k) ds  F (τ , k)

ak−n +δ0 )(s−θ )

τ ds +

θ

τ

θ

 C e−n eC (ak−n +δ0 e

−n )

n

e C (e

G (s, k) ds

ak−n +δ0 )(s−θ )

ds

θ

−n F (τ , k) + C e−n eC (ak−n +δ0 e )

τ G (s, k) ds. θ

Therefore, substituting this estimate into (4.38), we deduce

  −n F (θ, k)  F (τ , k) + C ak−n + δ0 e−n eC (ak−n +δ0 e ) F (τ , k) 

+ C ak−n + δ0 e

−n



e

C (ak−n +δ0 e−n )

τ

τ G (s, k) ds +

θ

   −n  1 + C ak−n + δ0 e−n eC (ak−n +δ0 e ) F (τ , k) 



+ 1 + C ak−n + δ0 e

−n



e

C (ak−n +δ0 e−n )

G (s, k) ds θ

τ G (s, k) ds θ

e

C (ak−n +δ0 e−n )

F (τ , k) + e

C (ak−n +δ0 e−n )

τ G (s, k) ds. θ

Iterating this inequality, we have that for e−n−1  θ  e−n

F (θ, k)  e

C

&n

m=n0

(ak−m +δ0 e−m )



F e

−n0



,k + e

C (ak−n +δ0 e−n )

e−n G (s, k) ds θ

+e

C

1−n

&n

−m ) m=n−1 (ak−m +δ0 e

e

G (s, k) ds + · · · + e e−n

e

C(

&+∞

m=−∞ am +δ0

&+∞

m=n0

e−m )

C

&n

m=n0

(ak−m +δ0 e−m )

e−n0

e−n0 −1



F e−n0 , k



G (s, k) ds

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26

+e

C(

&+∞

m=−∞ am +δ0

&+∞

m=n0

e−m )

e−n G (s, k) ds + e

C(

&+∞

m=−∞ am +δ0

&+∞

m=n0

e−m )

+ ··· + e

&+∞

m=−∞ am +δ0

G (s, k) ds e−n

θ

C(

e1−n

e−n0

&+∞

m=n0

e−m )



G (s, k) ds

e−n0 −1

e

C (δ S˜ 

H2 g (R)

+δ0 e−n0 )



F e

−n0



,k + e

C (δ S˜ 

H2 g (R)

+δ0 e−n0 )

e−n0

G (s, k) ds θ

   eC δ0 F e−n0 , k + eC δ0

e−n0



G (s, k) ds

(4.57)

θ

where we use the inequality +∞  m=−∞

am  C δ S˜  H 2 (R) , g

which can be proved as Proposition 2.1. On the other hand, for (4.34), by a direct application of the Gronwall inequality, it can be deduced that



F e

−n0



θ0

,k  e

C δ0

F (θ0 , k) + e

C δ0

G (s, k) ds.

(4.58)

e−n0

Combining (4.57) with (4.58), we have proved (4.39) for θ ∈ (0, θ0 ]. In our case, since δ w and δ p belong to C c∞ ([0, θ0 ] × R), we can extend the estimates (4.39) to θ = 0. We have thus completed our proof for this lemma. 2 Theorem 4.4. Under the assumption of Theorem 4.1, consider the solution for (4.22) with (4.23); we have the following estimate

δ w C 2g + δ p C 2g  C

 i =±

ϕi  H 2g (R) +

 i =±

 μi C 0 ([0,θ0 ]; H 2g (R)) .

(4.59)

Proof. Recall the definition of the norm  · C l ; we first prove g

δ w C 0 ([0,θ0 ]; H 2g (R)) + δ p C 0 ([0,θ0 ]; H 2g (R))  C

 i =±

ϕi  H 2g (R) +

 i =±

 μi C 0 ([0,θ0 ]; H 2g (R)) . (4.60)

In fact, we can deduce

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27

δ w C 0 ([0,θ0 ]; H 2g (R)) + δ p C 0 ([0,θ0 ]; H 2g (R))  C sup F (θ, k) k∈Z

C

θ ∈[0,θ0 ]

'

(

F (θ0 , k) + sup G (θ, k)

C

(by Lemma 4.3)

θ ∈[0,θ0 ]

k∈Z

 i =±

ϕi  H 2g (R) +

 i =±

 μi C 0 ([0,θ0 ]; H 2g (R)) ,

the first and third inequalities can be obtained as Proposition 2.1. By Eq. (4.22), ∂θ δ w and ∂θ δ p can be written in terms of ∂t δ w and ∂t δ p; it implies that

δ w C 1 ([0,θ0 ]; H 1g (R)) + δ p C 1 ([0,θ0 ]; H 1g (R))

   C δ w C 0 ([0,θ0 ]; H 2g (R)) + δ p C 0 ([0,θ0 ]; H 2g (R)) + μi C 0 ([0,θ0 ]; H 2g (R)) .

(4.61)

i =±

Combining (4.60) with (4.61), (4.59) has been proved.

2

Before starting our proof of Theorem 4.1, we deduce some useful corollaries of Lemma 4.3 and Theorem 4.4. Assume μ± = 0 in (4.41) and consider the expression for (δ w ± e δ p )(0, t ); we have

  (δ w ± e δ p )(0, t ) = (δ w ± e δ p ) θ0 , f ± (θ0 ; 0, t ) *+ , ) A

0 +

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ0

)

*+

ds .

(s, f ± (s;0,t ))

(4.62)

,

B

From this formula, we can treat

 (δ w +eδ p )(0,t )  (δ w −e δ p )(0,t )

as a linear map of the initial data

 (δ w +eδ p )(θ0 ,t ) 

fact, in (4.62), define the operators

(δ w −e δ p )(θ0 ,t )

. In



  (δ w + e δ p )(θ0 , ·) (δ w + e δ p )(0, t ) (t ) := , (δ w − e δ p )(θ0 , ·) (δ w − e δ p )(0, t )

  (δ w + e δ p )(θ0 , ·) (δ w + e δ p )(θ0 , f + (θ0 ; 0, t )) A (t ) := (δ w − e δ p )(θ0 , ·) (δ w − e δ p )(θ0 , f − (θ0 ; 0, t )) T

(4.63)

(4.64)

and

B

(δ w + e δ p )(θ0 , ·) (δ w − e δ p )(θ0 , ·)

 0

 (t ) :=

θ0 (

0

θ0 (

(δ w +e δ p )−(δ w −e δ p ) 2

(δ w +e δ p )−(δ w −e δ p ) 2

˜ + ln e )|(s, f + (s;0,t )) ds D ˜ − ln e )|(s, f − (s;0,t )) ds D

 .

(4.65)

The estimate (4.59) implies that T can be extended to a linear map from H 2g (R) × H 2g (R) to H 2g (R) × H 2g (R). For A and B, on one hand, by the expression of (4.64) it can be known that the operator A consists of coordinate transformations with respect to the components of the vector

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28

 (δ w +eδ p )(θ0 ,t )  (δ w −e δ p )(θ0 ,t )

. In view of this, by the characteristic estimate (A.10), we can directly deduce that

A is an invertible map from H 2g (R) × H 2g (R) to H 2g (R) × H 2g (R). Moreover it can be verified that

A −1



(δ w + e δ p )(0, ·) (δ w − e δ p )(0, ·)

 (t ) =

 (δ w + e δ p )(0, f + (0; θ0 , t )) . (δ w − e δ p )(0, f − (0; θ0 , t ))

(4.66)

On the other hand, for the operator B, we claim that it satisfies the estimate



  (δ w + e δ p )(θ0 , ·)  B   (δ w − e δ p )(θ0 , ·) 

   (δ w + e δ p )(θ0 , ·)     C δ0  . (4.67) (δ w − e δ p )(θ0 , ·)  H 2 (R)× H 2 (R) H 2 (R)× H 2 (R) g

g

g

g

To see this, consider the L 2 ([k, k + 1]) norm of

0

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ0

ds. (s, f ± (s;0,t ))

We have

 k+1 0

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ0

k

θ0 k+1 

2

k

θ0 k+1  sup

θ ∈(0,θ0 )

θ

C δ0

sup

θ ∈(0,θ0 )

 C eC δ0

θ ∈(0,θ0 )

 Ce

 12 dt

ds

(s, f ± (s;0,t ))

 12

2 (s, t ) dt

ds

t − (s,k)

θ

sup

θ ∈(0,θ0 )

C δ0

2

(δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e

θ0   (δ w + e δ p )(s, ·) 2 sup L ([t

θ0  Ce

ds

2

θ

  + (δ w − e δ p )(s, ·) L 2 ([t C δ0

dt (s, f ± (s;0,t ))

k t + ( s,k+1)

dt

 12

(δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ0 e

ds

 12

(s, f ± (s;0,t ))

(δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

0

2

− (s,k),t + (s,k+1)])



− (s,k),t + (s,k+1)])

  ( D˜ ± ln e )(s, ·) ∞ L ([t

− (s,k),t + (s,k+1)])

ds



˜ ± ln e )(s, ·) ∞ F (s, k)( D ds L ([t − (s,k),t + (s,k+1)])

θ

θ0   ( D˜ ± ln e )(s, ·) ∞ F (θ0 , k) sup L ([t θ ∈(0,θ0 )

ds

(by Lemma 4.3)

θ

θ0    C eC δ0 F (θ0 , k) ( D˜ ± ln e )(s, ·) L ∞ ([t 0

− (s,k),t + (s,k+1)])

− (s,k),t + (s,k+1)])

ds.

(4.68)

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To estimate

 θ0 0

29

( D˜ ± ln e )(s, ·) L ∞ ([t− (s,k),t+ (s,k+1)]) ds, by the expression (4.45), we have

θ0   ( D˜ ± ln e )(s, ·)

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0      (ln e ) ˜ ∂θ δ S˜ (s, ·) S

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0    +  (ln e ) ˜ λ˜ ± ∂t δ S˜ (s, ·) S

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0    ˜ (s, ·) +  (ln e ) w˜ D˜ ± δ w

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0    +  (ln e ) p˜ D˜ ± δ p˜ (s, ·)

L ∞ ([t − (s,k),t + (s,k+1)])

ds.

(4.69)

0

On one hand, for the first term on the right-hand side of (4.69), it can be treated as

θ0     (ln e ) ˜ ∂θ δ S˜ (s, ·) S

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0      =  (ln e ) ˜ (s, ·) cot(s) δ S˜  (· + ln sin s) S

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

θ0      C cot(s) δ S˜  (· + ln sin s)

L ∞ ([t − (s,k),t + (s,k+1)])

ds

0

 θ0 =C

    cot(s) δ S˜  (· + ln sin s)

L ∞ ([t − (s,k),t + (s,k+1)])

ds

e−n0

e−n ∞      cot(s) δ S˜  (· + ln sin s) ∞ + L ([t n=n0 −n−1 e

 − (s,k),t + (s,k+1)])

ds



 ∞     C δ S˜ L ∞ (R) + ak−n  C δ S˜  H 2 (R)  C δ0 . g n=n0

On the other hand, for the last three terms in (4.69), it can be directly obtained

(4.70)

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30

θ0     (ln e ) ˜ λ˜ ± ∂t δ S˜ (s, ·) S

L ∞ ([t − (s,k),t + (s,k+1)])

ds  C δ0 ,

0

θ0     (ln e ) w˜ D˜ ± δ w ˜ (s, ·)

L ∞ ([t − (s,k),t + (s,k+1)])

ds  C δ0 ,

0

θ0     (ln e ) p˜ D˜ ± δ p˜ (s, ·)

L ∞ ([t − (s,k),t + (s,k+1)])

ds  C δ0 .

(4.71)

0

Thus, from (4.68)–(4.71), we have that

 k+1 0 k

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

θ0

2 ds

 12  C δ0 eC δ0 F (θ0 , k).

dt

(s, f ± (s;0,t ))

Adding up the above inequalities with respect to k ∈ Z, it can be deduced that

 0     (δ w + e δ p ) − (δ w − e δ p ) ˜   D ± ln e ds   2 (s, f ± (s;0,·))  θ0

   (δ w + e δ p )(θ0 , ·)     C δ0  , (δ w − e δ p )(θ0 , ·)  H 2 (R)× H 2 (R) g

H 0g (R)

g

in other words



  (δ w + e δ p )(θ0 , ·)  B   (δ w − e δ p )(θ0 , ·) 

   (δ w + e δ p )(θ0 , ·)    C δ0  .  (δ w − e δ p )(θ0 , ·)  2 H 0 (R)× H 0 (R) H (R)× H 2 (R)

0

g

To consider the higher norms of θ ( 0 spect to t, we have

d

0

g

g

(δ w +e δ p )−(δ w −e δ p ) 2

θ0

0 =

˜ ± ln e )|(s, f ± (s;0,t )) ds, differentiating it with reD

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

dt

f ± t ( s ; 0, t )

ds (s, f ± (s;0,t ))

 ∂t (δ w + e δ p ) − ∂t (δ w − e δ p ) ˜ D ± ln e 2

+

 (δ w + e δ p ) − (δ w − e δ p ) ˜ f ± t ( s ; 0, t ) D ± ∂t (ln e ) 2

θ0

0 +

f ± t ( s ; 0, t )

θ0

ds

(s, f ± (s;0,t ))

θ0

0

(4.72)

g

(δ w + e δ p ) − (δ w − e δ p ) 2

ds

(s, f ± (s;0,t ))

 ∂t λ˜ ± ∂t (ln e )

ds (s, f ± (s;0,t ))

(4.73)

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31

and

d2

0

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ln e 2

dt 2 θ0

0 =

ds

(s, f ± (s;0,t ))



∂t (δ w + e δ p ) − ∂t (δ w − e δ p ) ˜ f ± tt (s; 0, t ) D ± ln e 2

θ0

0 +

f ± tt (s; 0, t )

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ∂t (ln e ) 2

0

f ± tt (s; 0, t )

(δ w + e δ p ) − (δ w − e δ p ) 2

 ∂t λ˜ ± ∂t (ln e )

ds

(s, f ± (s;0,t ))

θ0

0

ds

(s, f ± (s;0,t ))

θ0

+

ds (s, f ± (s;0,t ))



∂tt (δ w + e δ p ) − ∂tt (δ w − e δ p ) ˜ f ± t ( s ; 0, t ) D ± ln e 2

+

2

θ0

0

2 f± ( s ; 0, t ) t

+

 (δ w + e δ p ) − (δ w − e δ p ) ˜ D ± ∂tt (ln e ) 2

0

2 f± ( s ; 0, t ) t

(δ w + e δ p ) − (δ w − e δ p ) 2

 ∂tt λ˜ ± ∂t (ln e )

ds

(s, f ± (s;0,t ))

θ0

0

ds

(s, f ± (s;0,t ))

θ0

+

ds (s, f ± (s;0,t ))

 ∂t (δ w + e δ p ) − ∂t (δ w − e δ p ) ˜ f ± t ( s ; 0, t ) D ± ∂t (ln e ) 2

+2

2

0

2 f± ( s ; 0, t ) t

+2

∂t (δ w + e δ p ) − ∂t (δ w − e δ p ) 2

 ∂t λ˜ ± ∂t (ln e )

θ0

0

2 f± ( s ; 0, t ) t

+2 θ0

(δ w + e δ p ) − (δ w − e δ p ) 2

ds

(s, f ± (s;0,t ))

θ0

 ∂t λ˜ ± ∂tt (ln e )

ds (s, f ± (s;0,t ))

ds.

(4.74)

(s, f ± (s;0,t ))

Since we only suppose l = 2 in this section, the third and fifth terms in the right-hand side of (4.74) can’t be treated directly to obtain the required estimates. But we can integrate by parts and rewrite this term in another form as Lemma 3.2 in [1] (we omit here). Then, by a similar derivation as (4.72), we can deduce the estimate (4.67). From the above arguments, we can conclude that the operator T can be decomposed into two operators: One is invertible and has an inverse expression ( A) while the other is a small norm operator (B).

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32

Proof of Theorem 4.1(a). Now let’s come back to our original problem (4.7)–(4.9), to solve it, in other words, is to determine the data ϕ± ∈ H 2g (R) such that the solution of (4.22) and (4.23) with μ± = 0 satisfies

(δ w + e δ p ) + (δ w − e δ p ) 2

(0, t ) = δ w (0, t ) = g 0 (t )

(4.75)

and

(δ w + e δ p ) + (δ w − e δ p )

α1

2

(θ0 , t ) + β1

(δ w + e δ p ) − (δ w − e δ p ) 2e

(θ0 , t )

= α1 δ w (θ0 , t ) + β1 δ p (θ0 , t ) = g 1 (t ).

(4.76)

From this observation, we will use (4.75) and (4.76) to relate the initial data (δ w ± e δ p )(θ0 , t ) with g 0 and g 1 in a functional equation, then the solvability of this functional equation can be obtained by the invertibility of the corresponding operator. On one hand, by using the condition (4.76), we can derive

(δ w + e δ p )(θ0 , t )  

α1 e − β1 2e (θ0 , t )(δ w − e δ p )(θ0 , t ) + (θ0 , t ) g 1 (t ). =− α1 e + β1 α1 e + β1  (δ w +eδ p )(θ

,t ) 

Applying this formula to eliminate (δ w + e δ p )(θ0 , t ), we rewrite (δ w −eδ p )(θ0 ,t ) 0

(4.77)

as

 (δ w + e δ p )(θ0 , t ) (δ w − e δ p )(θ0 , t ) 

α1 e−β1 −( α1 e+β1 )(θ0 , t )(δ w − e δ p )(θ0 , t ) + ( α1 e2e+β1 )(θ0 , t ) g 1 (t ) = (δ w − e δ p )(θ0 , t )  

α1 e−β1 −( α1 e+β1 )(θ0 , t )(δ w − e δ p )(θ0 , t ) ( α1 e2e+β1 )(θ0 , t ) g 1 (t ) + . = 0 (δ w − e δ p )(θ0 , t )

(4.78)

Then, recalling the definitions of A and B in (4.64) and (4.65) respectively, we have the expression



(δ w + e δ p )(0, t ) (δ w − e δ p )(0, t )

= I + II

(4.79)

where

I= A

1 −( αα11 ee−β +β1 )(θ0 , t )(δ w − e δ p )(θ0 , t )





(δ w − e δ p )(θ0 , t )

+B

1 −( αα11 ee−β +β1 )(θ0 , t )(δ w − e δ p )(θ0 , t )



(δ w − e δ p )(θ0 , t )

(4.80) and

II = A

( α1 e2e+β1 )(θ0 , t ) g 1 (t ) 0



+B

( α1 e2e+β1 )(θ0 , t ) g 1 (t ) 0

 .

(4.81)

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33

On the other hand, by use of the expression (4.79), (4.75) is reduced to the functional equation

1 1

,

2 2

 · (I + II) = g 0 (t )

(4.82)

for the unknown function (δ w − e δ p )(θ0 , t ). Write this equation in another form

1 1



· I = g 0 (t ) −

,

2 2

1 1



,

2 2

· II;

(4.83)

˜ g 0 (t ) − ( 1 , 1 ) · II just depends on g 0 and g 1 while ( 1 , 1 ) · I ˜ , δ p˜ and δ S, we can see that for given δ w 2 2 2 2 just depends on the unknown function (δ w − e δ p )(θ0 , t ). Since g 0 ∈ H 2g (R) and g 1 ∈ C c∞ (R), by the discussion for the operators A and B before this proof, we know that g 0 (t ) − ( 12 , 12 ) · II belongs to H 2g (R). For the left-hand side of (4.83), let ϕ (t ) = (δ w − e δ p )(θ0 , t ) and define the operator

T : H 2g (R) → H 2g (R) by

T (ϕ )(t ) =

 · I(t )

,

=

1 1 2 2 1 1

,

2 2



·A

1 −( αα11 ee−β +β1 )(θ0 , ·)ϕ (·)

 (t ) +

ϕ (·)

1 1



,

2 2

·B

1 −( αα11 ee−β +β1 )(θ0 , ·)ϕ (·)

ϕ (·)

 (t ).

The solvability of the functional equation (4.83) for any g 0 , g 1 ∈ H 2g (R) is equivalent to the invertibility of T as an operator on H 2g (R). Indeed, we claim that T is invertible and satisfies the estimates

T  H 2g (R)→ H 2g (R)  C

and

 −1  T  2  C, H (R)→ H 2 (R) g

g

(4.84)

where C is a constant depending on the background solution. It then follows that for given g 0 ∈ H 2g (R) and g 1 ∈ C c∞ (R) there exists ϕ ∈ H 2g (R) such that

T (ϕ ) = g 0 − Thus we choose

1 1



,

2 2

· II.

(4.85)

.

(4.86)

ϕ  ∈ C c∞ (R) to make sure    ϕ − ϕ 

H 2g (R)



 C

Let

g 0 := T

   1 1 · II ϕ + ,

(4.87)

2 2

and combine (4.85) with (4.87); we have

  T ϕ  − ϕ = g 0 − g 0 . By (4.86) and the operator estimate (4.84), we deduce

    g − g0  0

H 2g (R)

     = T ϕ  − ϕ  H 2 (R)  C ϕ  − ϕ  H 2 (R)   . g

g

(4.88)

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34

Therefore, let

(δ w  + e δ p  )(θ0 , t ) (δ w  − e δ p  )(θ0 , t )



=

2e  1 −( αα11 ee−β +β1 )(θ0 , t )ϕ (t ) + ( α1 e +β1 )(θ0 , t ) g 1 (t )



ϕ  (t )

=

 1 −( αα11 ee−β +β1 )(θ0 , t )ϕ (t )



ϕ  (t )

∞

+

( α1 e2e+β1 )(θ0 , t ) g 1 (t )



0

∞

∈ C c (R) × C c (R)

(4.89)

˜ ± δ w  ± e D˜ ± δ p  = 0 in D . D

(4.90)

and solve

It can be deduced that

  δ w 

C 2g

  + δ p  C 2 g

    C  δ w  + e δ p  (θ0 , ·)    C ϕ  

H 2g (R)

+  g 1  H 2g (R)

 



 

 −1   1 1  · II  g0 − ,  C T  2 2

  

   1 1 C  · II g − ,   0 2 2

    +  δ w  − e δ p  (θ0 , ·) H 2 (R) (by Theorem 4.4)

H 2g (R)

H 2g (R)

by (4.89)

 H 2g (R)

     1 1   + ,  2 2 · II H 2g (R)

   C g 

 

0 H 2g (R)

+  g 1  H 2g (R)

+  g 1  H 2g (R) 

+  g 1  H 2g (R)

   C g  0

g









by (4.87)



by (4.84)

 H 2g (R)

+  g 1  H 2g (R) 

by (4.81) .

(4.91)

(4.12) and (4.13) can be deduced by (4.87) and (4.89) respectively. Meanwhile, by (4.88), approximate condition (4.15) is obtained as well. We complete our proof of Theorem 4.1(a). It remains to show that T is invertible and verify (4.84). For this purpose, we need to give a careful analysis of the operator T . Define the operators

A(ϕ )(t ) :=

1 1



,

2 2

·A

1 −( αα11 ee−β +β1 )(θ0 , ·)ϕ (·)



ϕ (·)

(t )

(4.92)

(t ).

(4.93)

and

B(ϕ )(t ) := Hence T = A + B .

1 1

,

2 2



·B

1 −( αα11 ee−β +β1 )(θ0 , ·)ϕ (·)

ϕ (·)



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35

For A, we have

A(ϕ )(t ) =

1 1

,

2 2



·A

1 −( αα11 ee−β +β1 )(θ0 , ·)ϕ (·)

 (t )

ϕ (·)

  α1 e−β1 −( α1 e+β1 )(θ0 , f + (θ0 ; 0, t ))ϕ ( f + (θ0 ; 0, t )) · 2 2 ϕ ( f − (θ0 ; 0, t ))

     1   1 α1 e − β1 θ0 , f + (θ0 ; 0, t ) ϕ f + (θ0 ; 0, t ) + ϕ f − (θ0 ; 0, t ) . =− 2 α1 e + β1 2

=

1 1

,

Here we further decompose A into two operators

  A1 (ϕ )(t ) = K˜ (t )ϕ f + (θ0 ; 0, t )

(4.94)

and

A2 (ϕ )(t ) =

1  2



ϕ f − (θ0 ; 0, t )

(4.95)

where

K˜ (t ) = −



 α1 e − β1  θ0 , f + (θ0 ; 0, t ) . 2 α1 e + β1 1

(4.96)

By the assumption (4.24) and the characteristic estimates (A.10), it can be directly verified that A1 and A2 are both invertible operators on H 2g (R) and have the expressions 1 A− 1 (ϕ )(t ) =



1 K˜ ( f + (0; θ0 , t ))



ϕ f + (0; θ0 , t )

(4.97)

and

  1 A− 2 (ϕ )(t ) = 2ϕ f − (0; θ0 , t ) .

(4.98)

Let

e 0 = e |U¯ 0



α1 e0 − β1 ; and K 0 = − 2 α1 e 0 + β1 1

it can be deduced that | K 0 | = 12 , because α1 β1 = 0 by Lemma 4.2. By choosing δ0 sufficiently small in (4.24), we can obtain the invertibility of A. In fact we can check that

% A− 1 =

&∞

−1

&∞

−1

 ±1  A 

H 2g (R)→ H 2g (R)

(Id + (Id +

n=1 (−A1 n=1 (−A2

1 A2 )n )A− if | K 0 | > 12 , 1 1 A1 )n )A− if | K 0 | < 2

1 2

(4.99)

and

 C,

where C is a constant just depending on the background solution.

(4.100)

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36

For B , by the estimate (4.67), we have

  B(ϕ )

 

α1 e−β1   1 1 −( α1 e+β1 )(θ0 , ·)ϕ (·)    · = , B  2 2  2 H 2g (R) ϕ (·) H g (R)  C δ0 ϕ  H 2g (R) .

(4.101)

Summarize the previous arguments for A and B ; T can be decomposed into an invertible operator (A) and a small norm operator (B ). Then similar as (4.99) it can be deduced that T is invertible and

 T

−1

= Id +

∞  

−A

−1

B

n

 A− 1 .

(4.102)

n =1

Then (4.84) can be obtained by substituting the estimates (4.100) and (4.101) into (4.102).

2

Proof of Theorem 4.1(b). To prove Theorem 4.1(b), we first consider the following backward initial value problem for (δ w 1 , δ p 1 )

˜ ± δ w 1 ± e D˜ ± δ p 1 = μ± D

in D ,

(4.103)

(δ w 1 ± e δ p 1 )(θ0 , t ) = 0 on Σ1 .

(4.104)

By Theorem 4.4, there exists a solution (δ w 1 , δ p 1 ) ∈ C c∞ ([0, θ0 ] × R) × C c∞ ([0, θ0 ] × R) satisfying (4.103) and (4.104) with

δ w 1 C 2g + δ p 1 C 2g  C

 i =±

μi C 0 ([0,θ0 ]; H 2g (R)) .

(4.105)

Second, let δ w 2 := δ w − δ w 1 and δ p 2 := δ p − δ p 1 ; by (4.16)–(4.18), (4.103) and (4.104), we know that (δ w 2 , δ p 2 ) belongs to C c∞ ([0, θ0 ] × R) × C c∞ ([0, θ0 ] × R) and satisfies

˜ ± δ w 2 ± e D˜ ± δ p 2 = 0 in D , D

(4.106)

δ w 2 = g 0 − δ w 1 on Σ0 ,

(4.107)

α1 δ w 2 + β1 δ p 2 = g1 − α1 δ w 1 − β1 δ p 1 on Σ1 .

(4.108)

Following the idea in the proof of Theorem 4.1(a), (δ w 2 − e δ p 2 )(θ0 , t ) satisfies the functional equation (4.82) where g 0 and g 1 have been replaced by g 0 − δ w 1 and g 1 − α1 δ w 1 − β1 δ p 1 respectively, thus we have

  (δ w 2 ± e δ p 2 )(θ0 , ·)

H 2g (R)

   C  g 0  H 2g (R) +  g 1  H 2g (R) + δ w 1 C 2g + δ p 1 C 2g . (4.109)

By an application of Theorem 4.4, it yields that

  δ w 2 C 2g + δ p 2 C 2g  C  g 0  H 2g (R) +  g 1  H 2g (R) + δ w 1 C 2g + δ p 1 C 2g . Combining (4.105) with (4.110), we finally obtain the estimate

(4.110)

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δ w C 2g + δ p C 2g 

 i =1,2

δ w i C 2g + δ p i C 2g

37



   C  g 0  H 2g (R) +  g 1  H 2g (R) + δ w 1 C 2g + δ p 1 C 2g

   C  g 0  H 2g (R) +  g 1  H 2g (R) + μi C 0 ([0,θ0 ]; H 2g (R)) . i =±

Hence we have completed our proof.

2

Remark 4.1. We point out that the idea in our proof is not similar as that given in [13–15]. We didn’t use the boundary value on Σ0 to calculate explicit expression of the solution in D by characteristic method. The boundary value on Σ0 is attained indirectly here. 5. Proof of the main theorem In this section, we prove that the nonlinear boundary value problem (NL) admits a solution (Lemma 5.1) and the update map has a fixed point (Lemma 5.2). The proof is based on the theorems established for the linearized problem in Section 4. For given ψ ∈ K M S σ , to solve the corresponding fixed boundary problem under the straighten boundary coordinate (θ˜ , t˜) (3.32)–(3.34), we need to determine θa which is depending on the approximate boundary ψ . We claim that, for l = 2, 3

θa − θ0  H l+1 (R)  C ψ − ψ0 (−l+11)

(5.1)

H g (R)

g

provided

ψ − ψ0 (−l+11)

H g (R)

 δ0 ,

(5.2)

where C and δ0 are constants just depending on ψ0 , in other words depending on the background solution. In fact, θa (t ) is determined by the equation

⎧  1  ⎪ ⎪ ⎨ t = ln ξ 2 + ψ 2 (ξ ) , 2

(5.3)

ψ(ξ ) ⎪ ⎪ ⎩ θa (t ) = arctan ξ where ψ satisfies

 −·  ·   e ψ e − ψ0 

H lg+1 (R)

= ψ − ψ0 (−l+11)

H g (R+ )

.

(5.4)

To obtain (5.1), let ξ = es in (5.3); it yields

⎧   1  ⎪ ⎨ t (s) = ln e2s + ψ 2 es , 2

s ⎪ ⎩ θ t (s) = arctan ψ(e ) . a s

e

Combining (5.4) with (5.5), we can deduce

(5.5)

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38

     t (·) − · − 1 ln 1 + ψ 2  0   2

H lg+1 (R)

  1    1   −2 · 2 · 2  − = 1 + e ψ e 1 + ψ ln ln 0   2

2

     C e−· ψ e· − ψ0 

H lg+1 (R)

(5.6)

H lg+1 (R)

and

    θa t (·) − θ0 

    ψ(e· )   = arctan − arctan ψ 0  H l+1 (R) · e

g

     C e−· ψ e· − ψ0 

H lg+1 (R)

H lg+1 (R)

(5.7)

,

provided

 −·  ·   e ψ e − ψ0 

H lg+1 (R)

 δ0 .

(5.8)

By (5.6), using the implicit function theorem, we solve the first equation of (5.5) to express s as a function of t. For s(t ) we have the estimate

     s(·) − · − 1 ln 1 + ψ 2  0   2

H lg+1 (R)

     C e−· ψ e· − ψ0  H l+1 (R) .

(5.9)

g

Since θa (t ) = θa (t (s(t ))), combine (5.7) with (5.9); it yields (5.1). (n) To solve (3.32)–(3.34), we introduce the approximate iteration scheme as follows. Let {θa } be a (n) (n) ∞ sequence of functions converging to θa such that δθa := θa − θ0 ∈ C c (R) with the approximate estimates

 (n)  θa − θa 

H 4g (R)

   MSσ =  θa(n) − θ0 − (θa − θ0 ) H 4 (R)  n , g

(5.10)

2

and denote

U¯ (0) = U¯ 0 ,

in Ω,

U¯ (n) = U¯ 0 + δ U¯ (n) ,

(5.11)

in Ω.

(5.12)

δ U¯ (n) is inductively defined by the following scheme:

⎧ (n) (n) (n+1) ⎪ ± e (n) D˜ ± δ p (n+1) = 0 ⎪ D˜ ± δ w ⎪ ⎨ (n+1) ˜ ˜ (n+1) ˜ ˜  θa (t )θ θ (t )θ (n+1)  cos a θ ∂δ S (n+1) cos θa (t˜) ∂δ S (n+1) θ0 θ0 0 ˜ ⎪ θ / (n+1) + =0 − ⎪ (n+1) ˜ ˜ (n+1) ⎪ θ (t )θ ⎩ ∂ t˜ ∂ θ˜ θa (t˜) sin θa (t˜)θ˜ θa(n+1) (t˜) sin a θ0

in D ,

(5.13)

θ0

δ w (n+1) = g (Wn+1) on Σ0 ,

⎧  (n+1) ⎪ α δw + β  δ p (n+1) + γ  δ S (n+1)  ⎪  ⎪ ⎨ = α  δ w (n) + β  δ p (n) + γ  δ S (n) − G  U (n) , U 1 ⎪ α  δ w (n+1) + β  δ p (n+1) + γ  δ S (n+1) ⎪ ⎪   ⎩ = α  δ w (n) + β  δ p (n) + γ  δ S (n) − G  U (n) , U 1

(5.14)

on Σ1

(5.15)

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39

where

˜ ± = ∂ ˜ + λ˜ ± ∂t˜ , D θ (n)

(n)

(n+1)

˜

(n) = λ˜ ±

(n+1)

and g W

θa(n+1) (t˜) tan θ θa θ0 (n)

˜

(n+1)

(t˜)

(t˜)

(5.16)

(n) + θa(n+1) (t˜)λ±

(n+1) 

(n+1)

˜

+ θa (t˜) tan θθa θ0   (n) λ± = λ± w (n) , p (n) , S (n)

θ0 − λ± θ0 tan θ θa θ0

(t˜)

(n+1) 

+ θa

(t˜)λ±

(n)

,

(5.17)

(5.18)

∈ H 3g (R) is undetermined and satisfies

 (n+1)  g − gW  W

H 3g (R)



σ 2n+1

(5.19)

.

This scheme operates by using Theorem 4.1(a) for the linearized problem in Section 4. In fact, we replace (5.15) by an equivalent form

%





α1 δ w (n+1) + β1 δ p (n+1) = α1 δ w (n) + β1 δ p (n) − G1 U (n) , U 1 ,   α2 δ w (n+1) + β2 δ p (n+1) + δ S (n+1) = α2 δ w (n) + β2 δ p (n) + δ S (n) − G2 U (n) , U 1

where G1 (U , U 1 ) = γ  G  (U , U 1 ) − γ  G  (U , U 1 ) and G2 (U , U 1 ) =

(5.20)

G  (U ,U 1 )

. Obviously, α1 = ∂ w G1 |U¯ 0 , γ β1 = ∂ p G1 |U¯ 0 , 0 = ∂S G1 |U¯ 0 , α2 = ∂ w G2 |U¯ 0 , β2 = ∂ p G2 |U¯ 0 , 1 = ∂S G2 |U¯ 0 . Consider the first part of (5.13)

with the first boundary condition of (5.20) and the exactly wedge boundary value condition

δ w (n+1) = g W , we can apply Theorem 4.1(a) in the case

δ p (n+1)

(n+1)

C ∞ ([0, θ

=

σ

2n+1

to obtain the approximate solution δ w (n+1) ,

C c∞ (R) satisfying the estimate (5.19). Next since the second

∈ c ∈ 0 ] × R) with g W part of (5.13) is an ordinary differential equation, which can be solved independently, we may assume that δ S (n) has the expression

 ˜ (n) (t˜) θθ ˜ t˜) = δ S (n) t˜ + ln sin a δ S (n) (θ, θ0

(5.21)

where δ S (n) would be determined by the second condition of (5.20). We can also check that δ S (n) ∈ C c∞ (R). By (5.21), the iteration scheme (5.13)–(5.15) can be viewed for the functions δ w (n) , δ p (n) and δ S (n) . Let







L  = (δ w , δ p , δ S ) δ w C 3 + δ p C 3 + δ S  H 3 (R)   ; g

g

g

we have Lemma 5.1. For any given M S > 0, there exist σ0 > 0 and M > 0, independent of M S , such that for any 0 < σ  σ0 the iteration scheme is well defined in L M σ and admits a limit under the lower norm. Proof. Following, C represents the constants only depending on the background solution. By inductive (n) (n+1) condition, δ w (n) , δ p (n) ∈ C c∞ ([0, θ0 ] × R) and δ S (n) , δθa , δθa ∈ C c∞ (R) satisfy

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40

 (n)          δ w  3 + δ p (n)  3 + δ S (n)  3 + δθa(n)  4 + δθa(n+1)  4  C ( M + M S )σ . C (R) C (R) H (R) H (R) H (R) g

g

g

g

g

Since

   α1 δ w (n) + β1 δ p (n) − G1 U¯ (n) , U 1 

H 3g (Σ1 )

  1      ¯ (n)  2 ( n ) ( n ) = δ U ∇U¯ G1 U¯ 0 + t δ U¯ , U 1 · δ U¯ dt    0

 2  C δ U¯ (n)  3

H g (Σ1 )

H 3g (Σ1 )

 C M2σ 2

and



 g W  H 3g (Σ0 )  C σ



by (3.30) and (2.20) ,

(5.22) (n+1)

we can use Theorem 4.1(a) to solve δ w (n+1) , δ p (n+1) ∈ C c∞ ([0, θ0 ] × R) with g W and obtain the estimate

 (n+1)  δ w 

C 3g

satisfying (5.19)

    + δ p (n+1) C 3  C M 2 σ + 1 σ . g

(5.23)

To derive the estimate for δ S (n+1) , we consider the second condition of (5.20), since

   α2 δ w (n) + β2 δ p (n) + δ S (n) − G2 U¯ (n) , U 1 

H 3g (Σ1 )

 C M2σ 2,

(5.24)

thus we can deduce that

  (n+1)   δ S (θ0 , ·) H 3 (R)  C M 2 σ + 1 σ . g

(5.25)

Recall the expression of δ S (n+1)

(n+1) ˜  (t ) θ˜ θa δ S (n+1) (θ˜ , t˜) = δ S (n+1) t˜ + ln sin θ0 (n+1)

and the estimate for δθa

(5.26)

in (5.10); we can obtain that

 (n+1)      δ S  3  C δ S (n+1) (θ0 , ·) 3  C M 2 σ + 1 σ . H (R) H (R)

(5.27)

 (n+1)        δ w  3 + δ p (n+1)  3 + δ S (n+1)  3  C M 2 σ + 1 σ . C C H (R)

(5.28)

g

g

Therefore

g

g

g

For any given M > 0, σ can be chosen sufficiently small to make sure M 2 σ < 1. Hence the left-hand side of the above inequality can be further estimated by 2C σ where C is a constant on the right-hand side. As a result, we can choose M = 2C , which is a constant depending on the background solution, such that the iteration scheme (5.13)–(5.15) is well defined for sufficiently small σ > 0. To prove {(δ w (n) , δ p (n) , δ S (n) )} is convergent under the lower norm, we will apply the part (b) of Theorem 4.1. Consider the system for δ U¯ (n)

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41

⎧ (n−1) (n−1) (n) ˜ ± δ w (n) ± e (n−1) D˜ ± ⎪ D δp = 0 ⎪ ⎪ ⎨ (n) ˜ ˜ (n) 

θa (t )θ θ (t˜)θ˜ (n)  cos a θ ∂δ S (n) cos θ0 θa (t˜) ∂δ S (n) θ0 0 ˜ ⎪ θ / (n) + =0 ⎪ (n) (n) ⎪ ∂ θ˜ − (n) θ (t˜)θ˜ ⎩ ∂ t˜ θa (t˜) sin θa (t˜)θ˜ θa (t˜) sin a θ0

(5.29)

θ0

(n)

%

in D ,

δ w (n) = g W

on Σ0 ,

(5.30)





α1 δ w (n) + β1 δ p (n) = α1 δ w (n−1) + β1 δ p (n−1) − G1 U (n−1) , U 1   α2 δ w (n) + β2 δ p (n) + δ S (n) = α2 δ w (n−1) + β2 δ p (n−1) + δ S (n−1) − G2 U (n−1) , U 1

on Σ1 (5.31)

and combine with (5.13)–(5.15); it can be deduced that (n) 



(n) 

˜ ± δ w (n+1) − δ w (n) ± e (n) D˜ ± δ p (n+1) − δ p (n) D



 (n−1) (n)  = λ˜ ± − λ˜ ± ∂t˜ δ w (n)     (n−1) (n)  ± e (n−1) − e (n) ∂θ˜ δ p (n) + e (n−1) λ˜ ± − e (n) λ˜ ± ∂t˜ δ p (n) in D , (n+1)

δ w (n+1) − δ w (n) = g W 

α1 δ w (n+1) − δ w

 (n)

(n)

− gW

on Σ0 ,



(5.32) (5.33)

 (n)

+ β1 δ p (n+1) − δ p     = α1 δ w (n) − δ w (n−1) + β1 δ p (n) − δ p (n−1)      on Σ1 . − G1 U (n) , U 1 − G1 U (n−1) , U 1

(5.34)

To use Theorem 4.1(b), we need to estimate the C 0 ([0, θ0 ]; H 2g (R)) norm of the inhomogeneous terms in (5.32). In fact

 (n−1)  (n)   λ˜ ∂t˜ δ w (n)  − λ˜ ± ±

C 0 ([0,θ0 ]; H 2g (R))

 (n−1)    (n)   C sup  λ˜ ± − λ˜ ± (θ˜ , ·) H 2 (R) · ∂t˜ δ w (n) C 2 . g

˜ 0,θ0 ] θ∈[

(5.35)

By the inductive assumption (δ w (n) , δ p (n) , δ S (n) ) ∈ L M σ , we have

  ∂˜ δ w (n)  t

C 2g

   C δ w (n) C 3  C M σ .

(5.36)

g

˜ (±n−1) − λ˜ (±n) , recalling the expression for λ˜ (±n) in (5.17), we can deduce To estimate λ

 (n−1)  (n)  ˜  λ˜ (θ , ·) − λ˜ ±

±

H 2 (R)

  (n+1)   (n−1)  (n+1)  (n)  (n)   (n)   C θa − θa  H 2 (R) + θa − θa  H 2 (R) +  λ± − λ± (θ˜ , ·) H 2 (R)   (n−1)    (n)  ˜ (θ, ·) H 2 (R) .  C θa(n+1) − θa(n)  H 3 (R) +  λ± − λ± (5.37) By (5.10), it can be directly obtained that

 (n+1) (n)  θa − θa 

H 3 (R)

 (n+1) C MSσ (n)   θa − θa  H 3 (R)  . n g

2

(5.38)

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42

(n−1)

For λ±

(n)

− λ± , by (5.18) we can deduce

  (n−1)   λ ˜ ·) − λ(n) (θ, ±

±

H 2 (R)

       C  δ w (n) − δ w (n−1) (θ˜ , ·) H 2 (R) +  δ p (n) − δ p (n−1) (θ˜ , ·) H 2 (R)     +  δ S (n) − δ S (n−1) (θ˜ , ·) H 2 (R) . (5.39)

Recall (5.21); we have

˜ t˜) − δ S (n−1) (θ, ˜ t˜) δ S (n) (θ,



 ˜ a(n) (t˜) ˜ a(n−1) (t˜) θθ θθ (n) ˜ (n−1) ˜ = δS − δS t + ln sin t + ln sin θ0 θ0





(n) ˜ a(n) (t˜) θθ θ˜ θa (t˜) − δ S (n−1) t˜ + ln sin = δ S (n) t˜ + ln sin θ0 θ0





(n) ˜ a(n−1) (t˜) θ˜ θa (t˜) θθ − δ S (n−1) t˜ + ln sin . + δ S (n−1) t˜ + ln sin θ0 θ0

(5.40)

On one hand





 ˜ a(n) (·) ˜ a(n) (·)   (n) θθ θθ (n−1) δ S  · + − δ · + ln sin S ln sin   θ θ 0

0

   C δ S (n) − δ S (n−1)  H 2 (R)    C δ S (n) − δ S (n−1)  H 2 (R) ,

H 2 (R)

(5.41)

g

on the other hand





 ˜ a(n) (·) ˜ a(n−1) (·)   (n−1) θθ θθ (n−1) δ S  · + − δ · + ln sin S ln sin   2 θ0 θ0 H (R)  (n−1)   (n−1)  ( n )  3 θa  C δ S − θa  H 2 (R) H (R)    (n−1) (n)   C δ S (n−1)  H 3 (R) θa − θa  H 4 (R) . g

(5.42)

g

Thus by (5.40) 2   (n)   δ S (θ˜ , ·) − δ S (n−1) (θ˜ , ·) 2  C δ S (n) − δ S (n−1)  2 + C M S M σ . H (R) H (R) n g

2

Substitute this inequality into (5.39); it yields that

 (n−1)  (n)   λ − λ (θ˜ , ·) ±

±

H 2 (R)

    C  δ w (n) − δ w (n−1) (θ˜ , ·)

H 2 (R)

   +  δ p (n) − δ p (n−1) (θ˜ , ·) H 2 (R)

   M S Mσ 2 + δ S (n) − δ S (n−1)  H 2 (R) + C n 2

(5.43)

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43

     C δ w (n) − δ w (n−1) C 2 + δ p (n) − δ p (n−1) C 2 g

  + δ S (n) − δ S (n−1) 

g



+C

H 2g (R)

M S Mσ

2

2n

(5.44)

.

Combining the above estimate with (5.37), we have

 (n−1)  (n)   λ˜ − λ˜ ± ∂t˜ δ w (n)  ±

C 0 ([0,θ0 ]; H 2g (R))

   C M σ δ w (n) − δ w (n−1)  +C

M S ( M + 1)σ

C 2g

     + δ p (n) − δ p (n−1) C 2 + δ S (n) − δ S (n−1)  H 2 (R) g

g

2

(5.45)

.

2n

Similarly, for the rest inhomogeneous terms in (5.32) and the right-hand side terms in (5.33) and (5.34), we have

 (n−1)   e − e (n) ∂ ˜ δ p (n)  θ

C 0 ([0,θ0 ]; H 2g (R))

   C M σ δ w (n) − δ w (n−1)  +C

M S ( M + 1)σ

C 2g

     + δ p (n) − δ p (n−1) C 2 + δ S (n) − δ S (n−1)  H 2 (R) g

g

2

(5.46)

,

2n

 (n−1) (n−1)   e − e (n) λ˜ (n) ∂˜ δ p (n)  λ˜ ±

±

M S ( M + 1)σ

2

t

C 0 ([0,θ0 ]; H 2g (R))

        C M σ δ w (n) − δ w (n−1) C 2 + δ p (n) − δ p (n−1) C 2 + δ S (n) − δ S (n−1)  H 2 (R) g

+C

2n

g

g

(5.47)

,

 (n+1)  g − g (n)  W

W

H 3g (R)

C

σ 2n



by (5.19)



(5.48)

and

  (n)           α1 δ w − δ w (n−1) + β1 δ p (n) − δ p (n−1) − G1 U¯ (n) , U 1 − G1 U¯ (n−1) , U 1 (θ0 , ·)     C M σ  U¯ (n) − U¯ (n−1) (θ0 , ·)

H 2g (R)

    C M σ  δ w (n) − δ w (n−1) (θ0 , ·)   + δ S (n) − δ S (n−1) 

 H 2g (R)

   C M σ δ w (n) − δ w (n−1) 

+C

C 2g

H 2g (R)

H 2g (R)

   +  δ p (n) − δ p (n−1) (θ0 , ·) H 2 (R) g

M S ( M + 1)σ 2 2n

  + δ p (n) − δ p (n−1) 

   M S ( M + 1)σ 2 + δ S (n) − δ S (n−1)  H 2 (R) + C . n g

By Theorem 4.1(b), we can deduce that

2

C 2g

(5.49)

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44

 (n+1)    δ w − δ w (n) C 2 + δ p (n+1) − δ p (n) C 2 g g    (n)  (n) (n−1)     C Mσ δ w − δ w + δ p − δ p (n−1) C 2 C2 g

  + δ S (n) − δ S (n−1) 

 H 2g (R)

g

+C

M S ( M + 1)σ + σ 2

2n

(5.50)

.

Moreover for δ S (n+1) − δ S (n) , we have

 (n+1)  δ S − δ S (n)  H 2 (R) g     (n)  C M σ δ w − δ w (n−1) C 2 + δ p (n) − δ p (n−1) C 2 g

  + δ S (n) − δ S (n−1) 

 H 2g (R)

g

+C

M S ( M + 1)σ 2 + σ 2n

(5.51)

.

Combining (5.50) with (5.51), we have proved

 (n+1)  δ w − δ w (n) 

    + δ p (n+1) − δ p (n) C 2 + δ S (n+1) − δ S (n)  H 2 (R) g g     (n)  (n)  (n)  (n−1)  (n−1)      C Mσ δ w − δ w + δp − δp + δ S − δ S (n−1)  H 2 (R) C2 C2 C 2g

g

+C

M S ( M + 1)σ + σ

g

g

2

2n

(5.52)

.

Define













n := δ w (n) − δ w (n−1) C 2g + δ p (n) − δ p (n−1) C 2g + δ S (n) − δ S (n−1)  H 2g (R) ;

(5.53)

(5.52) can be rewritten as

n+1  C M σ n + C By choosing

M S ( M + 1)σ 2 + σ 2n

.

(5.54)

σ sufficiently small we have that for some constant ν ∈ (0, 1)

n+1  νn + C

M S ( M + 1)σ 2 + σ 2n

.

(5.55)

Therefore ∞ 

n < +∞;

(5.56)

n =1

it implies that {(δ w (n) , δ p (n) , δ S (n) )} is a Cauchy sequence and converges under the lower norm. Meanwhile the boundary value condition on Σ0 is attained by the approximate condition (5.19). Summarizing the arguments above, we have proved that (3.32)–(3.34) admits a solution ( w , p , S ). 2 Thus, for given ψ ∈ K M S σ we have constructed the solution for the nonlinear fixed boundary value problem (3.32)–(3.34). Next we will use (3.23) to update the approximate boundary. If the update scheme has a fixed point, Theorem 3.1 has been proved. Thus Theorem 3.1 is a conclusion of the following lemma.

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Lemma 5.2. There exists an M S > 0 such that for sufficiently small defined on K M S σ and it is contractive with respect to the lower norm.

45

σ the update boundary scheme is well

Proof. To prove that the update scheme of the boundary is well defined for some M S > 0, with U¯ had been solved at hands by Lemma 5.1, we should come back to the Lagrange coordinate (ξ, η) and obtain the updated boundary Γ ∗ : η = ψ ∗ (ξ ), in fact

ξ

∗

ψ (ξ ) =









G¯ 1 U¯ (ξ,η) s, ψ(s) , U 1 ds.

(5.57)

0

Here we use notation U¯ (·,·) to represent U¯ in (·,·) coordinate. First, by Lemma 5.1, in (θ˜ , t˜) coordinate, we have the estimate

  (θ,   G¯ 1 U¯ ˜ t˜) (θ0 , ·), U 1 − ψ0 

H 3g (R)

 ˜˜   C U¯ (θ,t ) (θ0 , ·) − U¯ 0  H 3 (R)  C M σ . g

(5.58)

We define the function





˜˜ g 1 (t˜) = G¯ 1 U¯ (θ,t ) (θ0 , t˜), U 1 .

(5.59)

Then, in the (θ, t ) coordinate, we have









G¯ 1 U¯ (θ,t ) θa (t ), t , U 1 = g 1 (t ),

(5.60)

on the other hand, coming back to the (ξ, η) coordinate, we have

















G¯ 1 U¯ (ξ,η) ξ, ψ(ξ ) , U 1 = G¯ 1 U¯ (θ,t ) θa (t ), t , U 1 = g 1 (t )

(5.61)

with the relation

t=

1 2





ln ξ 2 + ψ 2 (ξ ) .

(5.62)

Therefore









G¯ 1 U¯ (ξ,η) ξ, ψ(ξ ) , U 1 = g 1

1 2







ln ξ 2 + ψ 2 (ξ )

(5.63)

and

∗

ξ

ψ (ξ ) =

g1

1 2

  ln s + ψ (s) ds. 

2

2

(5.64)

0

To consider the weighted norm of ψ ∗ (ξ ) − ψ0 ξ , by the definition, it is equivalent to estimate the norm of

H 4g (R)

e

−t

∗

 t

ψ e − ψ0 = e

−t

et g1 0

1 2





ln s2 + ψ 2 (s)

 − ψ0 ds.

(5.65)

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46

Differentiate (5.65); it yields

 d  −t ∗  t  e ψ e − ψ0 = −e−t dt

et g1



1



2

0

+ g1

1 2



ln s2 + ψ 2 (s)

  



ln e2t + ψ 2 et

− ψ0 ds

− ψ0 .

(5.66)

 et For g 1 ( 12 ln(e2t + ψ 2 (et ))) and e−t 0 g 1 ( 12 ln(s2 + ψ 2 (s))) − ψ0 ds, we have the expression

g1

1 2



ln e

2t

2

  t 

+ψ e

= g1 t +

1 2



ln 1 + e

−2t

2

  t 

ψ e

(5.67)

and

e

−t

et g1

1 2

  ln s + ψ (s) − ψ0 ds 

2

2

0

1 =

g1

1 2



2 2t

2



t

+ ψ re

ln r e

 

− ψ0 dr

0

1 =

g 1 t + ln r +

1 2





ln 1 + e−2(t +ln r ) ψ 2 et +ln r

 

− ψ0 dr .

(5.68)

0

Since ψ ∈ K M S σ , we have

 −·  ·   e ψ e − ψ0  Thus for any M S > 0 if

H 4 (R)

 MSσ .

(5.69)

σ is sufficiently small

         g 1 · + 1 ln 1 + e−2· ψ 2 e·  − ψ 0   2

H 3g (R)

        C 1 + e−· ψ e· − ψ0  H 4 (R)  g 1 (·) − ψ0  H 3 (R) . g

(5.70) To estimate the H 3g (R) norm of (5.65), we compute the H 3 ([k, k + 1]) norm of (5.68):

 1       1    −2(·+ln r ) 2 ·+ln r − ψ0 dr  ψ e  g 1 · + ln r + ln 1 + e   2 0

H 3 ([k,k+1])

 e−n  

∞       1    −2(·+ln r ) 2 ·+ln r − ψ0 dr   g 1 · + ln r + ln 1 + e ψ e    2 n =0

e−(n+1)

H 3 ([k,k+1])

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  e−n  ∞        g 1 · + ln r + 1 ln 1 + e−2(·+ln r ) ψ 2 e·+ln r  − ψ  0  2 n=0 −(n+1) e

dr H 3 ([k,k+1])

       1 + e−· ψ e· − ψ0  H 3 (R) e−n ∞     g 1 (·) − ψ0  · n=0 −(n+1) e

H 3 ([k−n−2+ 12 ln(1+ψ02 ),k−n+2+ 12 ln(1+ψ02 )])

dr

       C 1 + e−· ψ e· − ψ0  H 4 (R) g

·

∞  n =0





e−n  g 1 (·) − ψ0  H 3 ([k−n−2+ 1 ln(1+ψ 2 ),k−n+2+ 1 ln(1+ψ 2 )]) . 2

0

(5.71)

0

2

By the definition of H 3g (R), we know that

 1       1    −2(·+ln r ) 2 ·+ln r − ψ0 dr  ψ e  g 1 · + ln r + ln 1 + e   2 0

      C 1 + e−· ψ e· − ψ0  H 3 (R) 

·

∞  k=Z n=0



H 3g (R)



e−n  g 1 (·) − ψ0  H l ([k−n−2+ 1 ln(1+ψ 2 ),k−n+2+ 1 ln(1+ψ 2 )]) 2

0

2

0

       C 1 + e−· ψ e· − ψ0  H 4 (R) g

·

∞  n =0

e−n

   g 1 (·) − ψ0  k=Z

H 3 ([k−n−2+ 12 ln(1+ψ02 ),k−n+2+ 12 ln(1+ψ02 )])

∞          C 1 + e−· ψ e· − ψ0  H 4 (R) e−n  g 1 (·) − ψ0  H 3 (R) g

g

n =0

        C 1 + e−· ψ e· − ψ0  H 4 (R)  g 1 (·) − ψ0  H 3 (R) . g

g

(5.72)

Therefore, by the expressions (5.65), (5.66) and the estimates (5.70), (5.72)

 −· ∗  ·   e ψ e − ψ0 

H 4g (R)

   C (1 + M S σ ) g 1 (·) − ψ0  H 4 (R)  C (1 + M S σ ) M σ . g

(5.73)

For any given M S > 0, choose σ small enough such that M S σ < 1, the left-hand side of (5.73) is less than 2C M σ . Recall Lemma 5.1, M is just depending on the background solution. So we can let M S = 2C M in (5.73), where C is a constant just depending on the background solution, to guarantee the update scheme is well defined. To prove the update map is contractive with respect to the lower norm, we solve the new fixed boundary problem with respect to the updated boundary ψ ∗ and obtain the new solution U¯ ∗ and the updated boundary ψ ∗∗ . To compare ψ ∗ − ψ ∗∗ with ψ − ψ ∗ , we use the approximate scheme again. Let θa , θaa be the image functions with respect to the boundary ψ , ψ ∗ under the (θ, t ) coordinate respectively; we have the estimates

θa − θ0  H 4g (R) , θaa − θ0  H 4g (R)  C M S σ ,

(5.74)

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U¯ satisfies

⎧ ⎪ ⎨ D˜ ± w ± e D˜ ± p = 0 

˜ (t˜) θ˜ θa (t˜) θθ ˜ a (t˜) cot a ⎪ ∂t˜ S = 0 / θ0 + θθ ⎩ ∂θ˜ S − θa (t˜) cot θ0 θ0 %

w (0, t˜) = g W (t˜)

 







in D ,

on Σ0 ,

G U¯ (θ0 , t˜), U 1 = 0 G  U¯ (θ0 , t˜), U 1 = 0

(5.75)

(5.76)

on Σ1 ;

(5.77)

U¯ ∗ = ( w a , pa , Sa ) satisfies

⎧ ⎪ ⎨ D˜ a ± δ w a ± ea D˜ a ± δ pa = 0 

˜ aa (t˜) ˜ aa (t˜) θθ θθ  ˜ ˜ aa ⎪ ∂t˜ Sa = 0 / θ0 + θθ (t ) cot ⎩ ∂θ˜ Sa − θaa (t˜) cot θ0 θ0 %

in D ,

δ w a (0, t˜) = g W (t˜) on Σ0 ,   G  U¯ ∗ (θ0 , t˜), U 1 = 0 on Σ1 .   G  U¯ ∗ (θ0 , t˜), U 1 = 0

(5.78)

(5.79) (5.80)

For any  > 0, by the proof of Lemma 5.1, we can choose the approximate boundaries  , θˆ  with θ  − θ , θˆ  − θ , θ  − θ , θˆ  − θ ∈ C ∞ (R) and the approximate solutions θa , θˆa , θaa 0 a 0 aa 0 aa 0 aa a c ˆ  , δ pˆ  , δ w a , δ pa , δ w ˆ a , δ pˆ a ∈ C c∞ ([0, θ0 ] × R), δ S  , δ Sˆ  , δ S a , δ Sˆ a ∈ C c∞ (R) with δ w  , δ p , δ w

 ˜  (t˜) θθ ˜ t˜) = δ S  t˜ + ln sin a , δ S  (θ, θ0

 θ˜ θˆ  (t˜) ˜ t˜) = δ Sˆ  t˜ + ln sin a δ Sˆ (θ, θ0

 ˜  (t˜) θθ , δ Sa (θ˜ , t˜) = δ S a t˜ + ln sin aa θ0

 ˜ ˆ ˜ ˜ t˜) = δ Sˆ a t˜ + ln sin θ θaa (t ) δ Sˆa (θ, θ0

and

satisfy the uniform estimates

            δ w  3 , δ w ˆ C 3 , δ p C 3 , δ pˆ C 3 , δ S  H 3 (R) , δ Sˆ   H 3 (R)  M σ , Cg g g g g g             δ w  3 , δ w ˆ a C 3 , δ pa C 3 , δ pˆ a C 3 , δ S a  H 3 (R) , δ Sˆ a  H 3 (R)  M σ , a C g

g

g

g

g

g

(5.81) (5.82)

the approximate estimates

   θ − θa 

       ˆ − δ w C 2   , , θˆa − θa  H 4 (R) , δ w  − δ w C 2 , δ w g g g             δ p − δ p  2 , δ pˆ − δ p  2 , δ S − δ S  2 , δ Sˆ − δ S  2   , Cg Cg H g (R) H g (R)             θ − θaa  4 , θˆ − θaa  4 , δ w − δ w a  2 , δ w ˆ a − δ w a C 2   , aa aa a H g (R) H g (R) Cg g          δ p − δ pa  2 , δ pˆ  − δ pa  2 , δ S  − δ S a  2 , δ Sˆ  − δ S a  2   , a a a a C C H (R) H (R) a

H 4g (R)

g

g

g

g

(5.83)

(5.84)

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49

     g − gW  3 , g − gW  3   W Wa H (R) H (R) g

(5.85)

g

and the equations

⎧  ˆ  ± e D ˜ ± δ pˆ  = 0 ⎪ ⎨ D˜ ± δ w 

θ˜ θˆ  (t˜) θ˜ θˆ  (t˜) ⎪ ∂t˜ δ Sˆ = 0 ⎩ ∂θ˜ δ Sˆ  − θˆa (t˜) cot a / θ0 + θ˜ θˆa  (t˜) cot a θ0 θ0 %

in D ,

ˆ  (0, t˜) = g W (t˜) on Σ0 , δw

(5.86)

(5.87)

  α  δ wˆ  + β  δ pˆ  + γ  δ Sˆ = α  δ w  + β  δ p  + γ  δ S  − G  U¯  , U 1   α  δ wˆ  + β  δ pˆ  + γ  δ Sˆ = α  δ w  + β  δ p  + γ  δ S  − G  U¯  , U 1

on Σ1 ,

(5.88)

where

˜ ± = ∂ ˜ + λ˜ ± ∂t˜ , D θ

(5.89)

θ˜ θˆ (t˜) θˆa (t˜) tan θa0 + θˆa (t˜)λ± 

λ˜ ± =

˜ ˆ ˜

˜ ˆ ˜

θ0 − λ± θ0 tan θ θθa0(t ) + θˆa  (t˜) tan θ θθa0(t ) + θˆa  (t˜)λ±   λ± = λ± w  , p  , S  ;

(5.90)

,

(5.91)

⎧  ⎪ ˆ a ± ea D˜ a ± δ pˆ a = 0 ⎨ D˜ a ± δ w 

˜ ˆ ˜ ˜ ˆ ˜  (t˜) cot θ θaa (t ) / θ + θ˜ θˆ   (t˜) cot θ θaa (t ) ∂ δ Sˆ  = 0 ⎪ ⎩ ∂θ˜ δ Sˆa − θˆaa 0 aa a t˜ θ0 θ0 %

in D ,

ˆ a (0, t˜) = g W a (t˜) on Σ0 , δw

(5.92)

(5.93)





α  δ wˆ a + β  δ pˆ a + γ  δ Sˆa = α  δ w a + β  δ pa + γ  δ Sa − G  U¯ ∗ , U 1   α  δ wˆ a + β  δ pˆ a + γ  δ Sˆa = α  δ w a + β  δ pa + γ  δ Sa − G  U¯ ∗ , U 1

on Σ1 ,

(5.94)

where

˜ a ± = ∂ ˜ + λ˜ a ± ∂t˜ , D θ λ˜ a± =

(5.95)

  (t˜) tan θ˜ θˆaa (t˜) + θˆ  (t˜)λ θˆaa aa a± θ0  (t˜)  (t˜) θ˜ θˆaa θ˜ θˆaa    ˜   (t˜)λ ˆ θ0 − λa± θ0 tan θ0 + θaa (t ) tan θ0 + θˆaa a±

,

  λa± = λ± w a , pa , Sa .

(5.96) (5.97)

Keeping in mind that the higher norm is bounded, by a similar discussion as Lemma 5.1, we can obtain the estimates

   δ w ˆ  ˆ − δw

    + δ pˆ a − δ pˆ  C 2 + δ Sˆ a − δ Sˆ   H 2 (R) g g         C M σ δ w a − δ w  C 2 + δ pa − δ p  C 2 + δ S a − δ S   H 2 (R) g g g       − δθa  H 3 (R) + C M σ δ θˆaa − δ θˆa  H 3 (R) . + C M σ δθaa a

C 2g

g

g

(5.98)

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50

Using the approximate estimates (5.83)–(5.85), we have

δ w a − δ w C 2g + δ pa − δ p C 2g + δ S a − δ S  H 2g (R)    C M σ δ w a − δ w C 2g + δ pa − δ p C 2g + δ S a − δ S  H 2g (R) + C M σ δθaa − δθa  H 3g (R) + C  . By choosing

(5.99)

σ small enough, we derive δ w a − δ w C 2g + δ pa − δ p C 2g + δ S a − δ S  H 2g (R)  C M σ δθaa − δθa  H 3g (R) + C  .

(5.100)

For θaa − θa , as the derivation of (5.1), it can be deduced that

  (−1) (−1) δθaa − δθa  H 3g (R)  C ψ ∗ − ψ  H 3 (R+ ) + C M S σ ψ ∗ − ψ  H 3 (R+ ) . g

g

(5.101)

On the other hand, as the derivation of (5.73), for ψ ∗∗ − ψ ∗ , we have

 ∗∗    ψ − ψ ∗ (−31) +  C δ w a − δ w  2 + δ pa − δ p  2 + δ S a − δ S  2 Cg Cg H g (R) H (R ) g

(−1)  + C M σ ψ ∗ − ψ  H 3 (R+ ) .

(5.102)

g

From the above three inequalities, we can conclude

 ∗∗    ψ − ψ ∗ (−31) +  C M σ ψ ∗ − ψ (−31) + + C  . H (R ) H (R )

(5.103)

 ∗∗    ψ − ψ ∗ (−31) +  C M σ ψ ∗ − ψ (−31) + . H (R ) H (R )

(5.104)

g

Letting

g

 → 0, we have g

g

So if σ is small enough, the update scheme is contractive with respect to the lower norm, we have completed our proof. 2 Acknowledgments The author is grateful to Peng Qu for many stimulating and fruitful discussions and Hui-Cheng Yin for helpful comments. The author also thanks referees for patient reading and encouraging comments. The work was partly supported by NSFC 19701130. Appendix A. Some estimates for the characteristic Assume that λ(θ, t ) ∈ C ∞ ([0, θ0 ]; R) is a perturbation of λ0 (θ) ∈ C ∞ ([0, θ0 ]) and satisfies

  λ(θ, ·) − λ0 (θ)

H 2 (R)

 δ  1.

(A.1)

Let

ξ = f (τ ; θ, t )

(A.2)

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be the characteristic curve corresponding to λ, passing through the point (θ, t ); we have

d f (τ ; θ, t ) dτ

  = λ τ , f (τ ; θ, t )

(A.3)

and

f (θ; θ, t ) = t .

(A.4)

Differentiating (A.3) and (A.4) with respect to t, we deduce

∂t f (τ ; θ, t ) = e

τ

∂t λ(s, f (s;θ,t )) ds

θ

(A.5)

.

From this formula, since

    λt (θ, ·) ∞  C λ(θ, ·) − λ0 (θ) 2  δ, L (R) H (R)

(A.6)

it can be easily verified that

∂t f (τ ; θ, t )  eC δ|τ −θ | .

(A.7)

Differentiating (A.5) with respect to t, we obtain the expression

∂tt f (τ ; θ, t ) = e

τ θ

∂ξ λ(s, f (s;θ,t )) ds

τ

  ∂tt λ s, f (s; θ, t ) ∂t f (s; θ, t ) ds.

(A.8)

θ

Considering the L 2 norm of ∂tt f (τ ; θ, t ), we have

  ∂tt f (τ ; θ, ·)

 eC δ|τ −θ | L 2 (R)

τ

  λ(s, ·) − λ0 (s)

H 2 (R)

ds.

(A.9)

θ

Moreover, for any l ∈ N ∪ {0}, we have the characteristic estimates

  ∂tt f (τ ; θ, ·)

H l (R)

 Ce

C δ|τ −θ |

τ

  λ(s, ·) − λ0 (s)

H l+2 (R)

ds

(A.10)

θ

where the constant C just depends on l. References [1] A. Bressan, Hyperbolic Systems of Conservation Laws. The One-Dimensional Cauchy Problem, Oxford University Press, Oxford, 2000. [2] G.Q. Chen, Y.Q. Zhang, D.W. Zhu, Existence and stability of supersonic Euler flows past Lipschitz wedges, Arch. Ration. Mech. Anal. 181 (2006) 261–310. [3] S.X. Chen, Supersonic flow past a concave wedge, Sci. China Ser. A 10 (1997) 903–910. [4] S.X. Chen, Asymptotic behavior of supersonic flow past a convex combined wedge, Chin. Ann. Math. Ser. B 19 (1998) 255–264. [5] S.X. Chen, Global existence of supersonic flow past a curved convex wedge, J. Partial Differ. Equ. 11 (1998) 73–82. [6] S.X. Chen, Stability of Mach configuration, Comm. Pure Appl. Math. 59 (2006) 1–35. [7] S.X. Chen, B.X. Fang, Stability of reflection and refraction of shocks on interface, J. Differential Equations 244 (2008) 1946– 1984.

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[8] R. Courant, K.O. Friedrichs, Supersonic Flow and Shock Waves, Interscience Publ. Inc., New York, 1948. [9] B.X. Fang, Stability of transonic shocks for the full Euler system in supersonic flow past a wedge, Math. Methods Appl. Sci. 29 (2006) 1–26. [10] P. Grisvard, Elliptic Problems in Nonsmooth Domains, Monogr. Stud. Math., vol. 24, Pitman, London, 1985. [11] C.H. Gu, A method for solving the supersonic flow past a curved wedge, Fudan J. (Nature Sci.) 7 (1962) 11–14. [12] V.A. Kozlov, V.G. Maz’ya, J. Rossmann, Elliptic Boundary Value Problems in Domains With Point Singularities, Math. Surveys Monogr., vol. 52, AMS, 1997. [13] T.T. Li, On a free boundary problem, Chin. Ann. Math. 1 (1980) 351–358. [14] T.T. Li, Global Classical Solutions for Quasilinear Hyperbolic Systems, Res. Appl. Math., vol. 34, Wiley, Masson, New York, Paris, 1994. [15] T.T. Li, W.C. Yu, Boundary Value Problem for Quasilinear Hyperbolic System, Duke Univ. Math. Ser., vol. 5, 1985. [16] T.T. Li, Y. Zhou, D.X. Kong, Weak linear degeneracy and global classical solutions for general quasilinear hyperbolic systems, Comm. Partial Differential Equations 19 (1994) 1263–1317. [17] V.G. Maz’ya, B.A. Plamenevskiˇı, Estimates in L p and in Hölder classes and the Miranda–Agmon maximum principle for solutions of elliptic boundary value problems in domains with singular points on the boundary, in: Amer. Math. Soc. Transl., vol. 123, 1984, pp. 1–56. [18] D.G. Schaeffer, Supersonic flow past a nearly straight wedge, Duke Math. J. 43 (1976) 637–670. [19] H.C. Yin, Global existence of a shock for the supersonic flow past a curved wedge, Acta Math. Sin. (Engl. Ser.) 22 (2006) 1425–1432. [20] Y.Q. Zhang, Global existence of steady supersonic potential flow past a curved wedge with piecewise smooth boundary, SIAM J. Math. Anal. 31 (1999) 166–183. [21] Y.Q. Zhang, Steady supersonic flow past an almost straight wedge with large vertex angle, J. Differential Equations 192 (2003) 1–46. [22] Y. Zhou, Global classical solutions to quasilinear hyperbolic systems with weak linear degeneracy, Chin. Ann. Math. Ser. B 25 (2004) 37–56. [23] Y. Zhou, Point-wise decay estimate for the global classical solutions to quasilinear hyperbolic systems, Math. Methods Appl. Sci. 32 (2009) 1669–1680.