gui-qiang chen and ya-guang wang- existence and stability of compressible current-vortex sheets in...

8/3/2019 Gui-Qiang Chen and Ya-Guang Wang- Existence and Stability of Compressible Current-Vortex Sheets in Three-Dimensional Magnetohydrodynamics

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Digital Object Identifier (DOI) 10.1007/s00205-007-0070-8Arch. Rational Mech. Anal. 187 (2008) 369408

Existence and Stability of Compressible

Current-Vortex Sheets in Three-Dimensional

Magnetohydrodynamics

Gui-Qiang Chen & Ya-Guang Wang

Abstract

Compressible vortex sheets are fundamental waves, along with shocks and

rarefaction waves, in entropy solutions to multidimensional hyperbolic systems of

conservation laws. Understanding the behavior of compressible vortex sheets is an

important step towards our full understanding of fluid motions and the behavior

of entropy solutions. For the Euler equations in two-dimensional gas dynamics,the classical linearized stability analysis on compressible vortex sheets predicts

stability when the Mach number M >

2 and instability when M

2.

For the Euler equations in three dimensions, every compressible vortex sheet is

violently unstable and this instability is the analogue of the KelvinHelmholtz

instability for incompressible fluids. The purpose of this paper is to understand

whether compressible vortex sheets in three dimensions, which are unstable in the

regime of pure gas dynamics, become stable under the magnetic effect in three-

dimensional magnetohydrodynamics (MHD). One of the main features is that the

stability problem is equivalent to a free-boundary problem whose free boundary is a

characteristic surface, which is more delicate than noncharacteristic free-boundary

problems. Another feature is that the linearized problem for current-vortex sheets

in MHD does not meet the uniform KreissLopatinskii condition. These features

cause additional analytical difficulties and especially prevent a direct use of the

standard Picard iteration to the nonlinear problem. In this paper, we develop a

nonlinear approach to deal with these difficulties in three-dimensional MHD. We

first carefully formulate the linearized problem for the current-vortex sheets to

show rigorously that the magnetic effect makes the problem weakly stable and

establish energy estimates, especially high-order energy estimates, in terms of the

nonhomogeneous terms and variable coefficients. Then we exploit these results to

develop a suitable iteration scheme of the NashMoserHrmander type to deal with

the loss of the order of derivative in the nonlinear level and establish its convergence,


2/40

370 Gui-Qiang Chen & Ya-Guang Wang

which leads to the existence and stability of compressible current-vortex sheets,

locally in time, in three-dimensional MHD.

1. Introduction

We are concerned with the existence and stability of compressible current-

vortex sheets in three-dimensional magnetohydrodynamics (MHD). The motion of

inviscid MHD fluids is governed by the following system:

t + (v) = 0,t(v) + (v v H H) + (p + 12 |H|2) = 0,tH (v H) = 0,t

12 (|v|2 + |H|

2

) + e+ v 12 |v|2+e+ p + H (v H)=0,

(1.1)

and

H = 0, (1.2)where , v = (v1, v2, v3), H = (H1, H2, H3), and p are the density, velocity,magnetic field, and pressure, respectively; e = e(, S) is the internal energy; andS is the entropy. The Gibbs relation

dS = de p

2 d

implies the constitutive relations:

p = 2e (, S), = eS(, S),

where is the temperature. The quantity c = p (, S) is the sound speed of thefluid.

For smooth solutions, the equations in (1.1) are equivalent to

(t + v )p + c2 v = 0,(t + v )v + p ( H) H = 0,(t + v )H (H )v + H v = 0,(t + v )S = 0,

(1.3)

which can be written as an 88 symmetric hyperbolic system for U = (p, v, H, S)of the form:

B0(U)tU

+

3

j=1

Bj (U)xj U

=0. (1.4)

Compressible vortex sheets occur ubiquitously in nature (cf. [25, 11, 14, 24, 26,

27]) and are fundamental waves in entropy solutions to multidimensional hyper-

bolic systems of conservation laws. For example, the vortex sheets are one of the

core waves in the Mach reflection configurations when a planar shock hits a wedge


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Compressible Current-Vortex Sheets in Three-Dimensional MHD 371

and, more generally, in the two-dimensional Riemann solutions, which are buil-

ding blocks of general entropy solutions to the Euler equations in gas dynamics.

Therefore, understanding the existence and stability of compressible vortex sheets

is an important step towards our full understanding of fluid motions and the beha-

vior of entropy solutions to multidimensional hyperbolic systems of conservationlaws, along with the existence and stability of shocks and rarefaction waves (cf. [1,

14, 20, 21]; also see [12, 16]).

For the Euler equations in two-dimensional gas dynamics, the classical linea-

rized stability analysis on compressible vortex sheets predicts stability when the

Mach number M >

2 since there are no growing modes in this case, and insta-

bility when M

2 was recently established by Coulombel-Secchi

[9, 10] when the initial data is a small perturbation of a planar vortex sheet. In aseries of papers, Artola and Majda [24] studied the stability of vortex sheets by

using the argument of nonlinear geometric optics to analyze the interaction between

vortex sheets and highly oscillatory waves. They first observed the generation of

three distinct families of kink modes traveling along the slipstream bracketed by

shocks and rarefaction waves and provided a detailed explanation of the instability

of supersonic vortex sheets even when M >

2. Therefore, one cannot expect

nonlinear stability globally for two-dimensional compressible vortex sheets.

For the Euler equations in three-dimensional gas dynamics, the situation is even

more complicated. In fact, it is well known that every compressible vortex sheet

is violently unstable and this instability is the analogue of the KelvinHelmholtz

instability for incompressible fluids.

The purpose of this paper is to understand whether compressible vortex sheets

in three dimensions, which are unstable in the regime of pure gas dynamics, become

stable under the magnetic effect, that is, the nonlinear stability of current-vortex

sheets in three-dimensional MHD. One of the main features is that the stability

problem is equivalent to a free-boundary problem whose free boundary is a cha-

racteristic surface, which is more delicate than the noncharacteristic free-boundary

problems derived from the multidimensional shock problems that are endowed with

a strict jump of the normal velocity on the free boundary (cf. [12, 20, 21]; also [7,

11]). Another feature is that the linearized problem for current-vortex sheets in

MHD does not meet the uniform KreissLopatinskii condition (cf. [18, 20]); see

the analysis in Blokhin and Trakhinin [6] and Trakhinin [25]. These features

cause additional analytical difficulties and, in particular, they prevent a direct use

of the standard Picard iteration to prove the existence of solutions to the nonlinear

problem.

In this paper, we develop a nonlinear approach to deal with these difficulties

in three-dimensional MHD. We first carefully formulate the linearized problem for

the current-vortex sheet to show rigorously that, although any compressible vortexsheet may be violently unstable in the regime of pure gas dynamics and does not

meet the uniform KreissLopatinskii condition, the magnetic effect makes the li-

nearized problem weakly stable as observed in [6, 25]. In particular, we successfully

establish high-order energy estimates of the solutions for the linearized problem in


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terms of the nonhomogeneous terms and variable coefficients. Then we exploit these

results to develop a suitable iteration scheme of the NashMoserHrmander type

and establish its convergence, which leads to the existence and stability of com-

pressible current-vortex sheets, locally in time, in the three-dimensional MHD.

We observe that, even though the energy estimate given in Theorem 4.2 for thelinearized problem (3.1)(3.3) keeps the same order regularity for the solutions,

nonhomogeneous terms, and coefficients formally, in system (3.1), the nonhomo-

geneous term F depends on the first derivative ofU and the derivatives of up to order two, and the coefficient matrices depend on the first derivative of ,which leads to the loss of regularity in estimate (4.14). This observation motivates

us to develop the iteration scheme of the NashMoserHrmander type to deal with

the loss of regularity in the nonlinear problem.

We remark that, in order to establish the energy estimates, especially high-order

energy estimates, of solutions to the linearized problem derived from the current-vortex sheet, we have identified a well-structured decoupled formulation so that the

linear problem (3.1)(3.3) is decoupled into one standard initial-boundary value

problem (3.10) for a symmetric hyperbolic system and another problem (3.11)

for an ordinary differential equation for the front. This decoupled formulation is

much more convenient and simpler than that in (58)(60) given in [25] and is

essential for us to establish the desired high-order energy estimates of solutions,

which is one of the key ingredients for developing the suitable iteration scheme of

the NashMoserHrmander type that converges.

This paper is organized as follows. In Section 2, we first set up the current-vortex

sheet problem as a free-boundary problem, and then we reformulate this problem

into a fixed initial-boundary value problem and state the main theorem of this paper.

In Sections 34, we first carefully formulate the linearized problem for compressible

current-vortex sheets and identify the decoupled formulation; and then we establish

energy estimates, especially high-order energy estimates, of the solutions in terms

of the nonhomogeneous terms and variable coefficients. Then, in Section 5, we

analyze the compatibility conditions and construct the zeroth-order approximate

solutions for the nonlinear problem, which is a basis for our iteration scheme. In

Section 6, we develop a suitable iteration scheme of the NashMoserHrmander

type. In Section 7, we establish the convergence of the iteration scheme towards

a compressible current-vortex sheet. Finally, in Section 8, we complete several

necessary estimates of the iteration scheme used in Section 7.

2. Current-vortex sheets and main theorem

In this section, we first set up the current-vortex sheet problem as a free-

boundary problem and then reformulate this problem into a fixed initial-boundary

value problem, and finally we state the main theorem of this paper.Let a piecewise smooth function U(t,x) be an entropy solution to (1.1) with

the form:

U(t,x ) =

U+(t,x) for x1 > (t,x2,x3),U(t,x) for x1 < (t,x2,x3).

(2.1)


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Then, on the discontinuity front := {x1 = (t,x2,x3)}, U(t,x ) must satisfy theRankineHugoniot conditions:

[mN] = 0,[HN] = 0,mN[vN] + (1 + 2x2 + 2x3 )[q] = 0,mN[v] = HN[H],mN[H ] = HN[v],mN[e + 12 (|v|2 + |H|

2

)] + [qvN HN(H v)] = 0,

(2.2)

where [] denotes the jump of the function on the front , (vN, v) (resp.(HN, H)) are the normal and tangential components of v (resp. H) on , that

is,

vN := v1 x2 v2 x3 v3,v = (v1 , v2 ) := (x2 v1 + v2, x3 v1 + v3),HN := H1 x2H2 x3H3,H = (H1 , H2 ) := (x2H1 + H2, x3H1 + H3),

where mN = (vN t) is the mass transfer flux, and q = p + |H|2

2is the total

pressure.

IfmN = 0 on , then (U, ) is a contact discontinuity for (1.1). In this paper,we focus on the case:

H+N = HN = 0, (2.3)H+ H , (2.4)

for which U(t,x) in (2.1) is called a compressible current-vortex sheet. Then the

current-vortex sheet is determined by (2.3) and

t = v+N = vN, p + |H

|2

2 = 0, (2.5)

on which generically

([], [v], [S]) = 0. (2.6)When H 0, U(t,x) becomes a classical compressible vortex sheet in fluidmechanics without magnetic effect.

Since H+N = HN on = {x1 = (t,x2,x3)}, it is easy to see that condition(2.4) is equivalent to the condition:

H+2H+3

H2H3 on . (2.7)Indeed, when H+N = HN on , we have

H = ((1 + 2x2 )H2 + x2 x3H3 , x2 x3H2 + (1 + 2x3 )H3 ),


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which implies that H+ + H = 0 if and only if H+k + Hk = 0 hold fork = 2, 3.

Furthermore, system (1.1) has the following property: Let (U, ) be thecurrent-vortex sheet to (1.1)for t [0, T)for some T > 0. Then, if H(0,x ) =0, we have H(t,x) = 0 for all t [0, T). (2.8)This can be seen as follows: From (1.3), we find that, on both sides of ,

(t + v )( H) + ( v)( H) = 0. (2.9)On , we use t = vN to obtain

t + v = t + (vN + x2 v2 + x3 v3 )x1 + v2 x2 + v3 x3 = (t, ),

where the vectors = (1, vN + x2 v2 + x3 v3 , v2 , v3 ) are orthogonal tothe spacetime normal vector n = (t, 1, x2 , x3 ) on . Thus, t + v are tangential derivative operators to , and the equations in (2.9) are transport

equations for H on both sides of, which yields (2.8).Set

D(, U) := D(, U) 0

0 1

with D(, U) :=

1 c2

H O13H I3 I3031 I3 I3

.

As in [25], property (2.8) implies that system (1.1)(1.2) is equivalent to the follo-

wing system on both sides of:

D

, U

B0(U)tU+

3j=1

Bj (U)xj U

+G H = 0, (2.10)

provided H(0,x) = 0, whereG

= (1, 0, 0, 0, H, 0),

and = (v2 , v3 , H2 , H3 ) will be determined later so that

()2 0 such that

A0(U)tU

+

3

j=1

Aj (U)xj U

=0 for

(x1

(t,x2,x3)) > 0,

U|t=0 =

U+0 (x ) for x1 > 0(x2,x3),U0 (x ) for x1 < 0(x2,x3)

(2.13)

satisfying the jump conditions on :

t = v+N = vN, H+N = HN = 0, [p +|H|2

2] = 0, H+ H , (2.14)

where 0(x2,x3) = (0,x1,x2), H0 (x ) = 0, and condition (2.11) is satisfiedfor given by (3.9) later.

For a piecewise smooth solution U(t,x ) of problem (2.13)(2.14) with smooth

(U, ) for t [0, T), it can be similarly verified that H(t,x) = 0 for allt [0, T) since H0 (x) = 0. Hence the equations in (2.13) and (1.1)(1.2) areequivalent for such a solution in the domain := {(x1 (t,x2,x3)) > 0}.

Furthermore, unlike the shock case, the new difficulty here is that the free

boundary in problem (2.13)(2.14) is now a characteristic surface of the equations

in (2.13), rather than the noncharacteristic surface endowed with a strict jump of

the normal velocity as in the shock case (see [7, 11, 12, 20, 21]).To deal with such a free-boundary problem, it is convenient to employ the

standard partial hodograph transformation:t = t, x2 = x2, x3 = x3,x1 = (t, x1, x2, x3)

(2.15)

with satisfying

() x1 > 0,+| x1=0 = | x1=0 = (t, x2, x3)

(2.16)

for some constant > 0. Under (2.15), the domains are transformed into{ x1 > 0} and the free boundary into the fixed boundary { x1 = 0}.

Then we define U(t, x ) := U(t,x ). From the first jump condition in (2.14),the natural candidates for are those that satisfy the following eikonal equations:

t v1 + v2 x2 + v3 x3 = 0 in {x1 > 0}, (2.17)

where we drop the tildes in the formula here and from now on for simplicity of

notation.

It is easy to check that, under (2.15), U(t, x) = U(t,x ) satisfy

L

U,

U = 0 in {x1 > 0}, (2.18)


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the boundary conditions on {x1 = 0}:+

x1=0 =

x1=0 = ,

B U+, U, |x1=0 = 0,(2.19)

and the initial condition at {t = 0}:U,

|t=0 = U0 (x), 0(x2,x3) , (2.20)where the tildes have been also dropped for simplicity of notation,

L(U, )V = A0(U)tV + A1(U, ) x1 V +3

j=2Aj (U)xj V

with A1(U, ) = 1x1A1(U) tA0(U) 3j=2 xj Aj (U),

B(U+, U, ) =

t Uv,N, UH,N, q+ q

with Uv,N = U2 x2 U3 x3 U4 , UH,N = U5 x2 U6 x3 U7 , and

q = U1 +1

2|UH|2 for UH = (U5, U6, U7).

With these, the free-boundary problem has been reduced into the initial-boundaryvalue problem (2.17)(2.20) with a fixed boundary {x1 = 0}.

To solve (2.17)(2.20), as in [1, 8, 15], it is natural to introduce the tangential

vector M = (M1, M2, M3) of{x1 = 0}:M1 = (x1)x1 M2 = x2 , M3 = x3 ,

with

(x1)

:= x1 for 0 x1 1,

2 for x1 2,

and the weighted Sobolev spaces defined on T := {(t,x) [0, T]R3 : x1 > 0}:Bs(T) := {u L2(T) : etM kx1 u L2(T) for || + 2k s}

for all s N and > 0, imposed with the norms

us,,T := T

0

u(t, )2s,dt1/2

,

where

u(t, )2s, :=

||+2ks2(s||2k)etMkx1 u(t, )2L2 .


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We will also use similar notation as above for the spaces with = 0, Bs (T),whose norm is defined by

u

s,T

:= ||+2ks M k

x1u

2

L2

(T)

1/2

.

Also denote by bT := {(t,x2,x3) : t [0, T], (x2,x3) R2}.Then, when the initial data (U0 , 0) is a small perturbation of a planar current-

vortex sheet (U, ) for constant states U and = 0 satisfying (2.3)(2.5), wehave the following main theorem of this paper.

Theorem 2.1. (Main theorem). Assume that, for any fixed 14 and s [ +5, 2

9

], the initial data functions 0

H2s+3(R2) andU

0 U

B2(s+2)(R3

+)

satisfy the compatibility conditions of problem (2.17)(2.20) up to order s + 2 (cf.Section 5). Then there exists a solution (U, ) of the initial-boundary valueproblem (2.17)(2.20) such that

U Ua , a

B (T) and = |x1=0 H1 (bT)for some functions Ua and

a satisfying

Ua U, a

x1

Bs+3R+ R

3

+.We remark that, since () x1 > 0, the corresponding vector function of

the solution (U, ) in Theorem 2.1 under the inverse of the partial hodographtransform is a solution of the free-boundary problem that is the current-vortex sheet

problem, which implies the existence and stability of compressible current-vortex

sheets to (2.1)(2.2) under the initial perturbation. To achieve this, we start with the

linear stability in Sections 34 and then establish the nonlinear stability by develo-

ping nonlinear techniques in Sections 57. Some estimates used in Sections 57 for

the iteration scheme of the NashMoserHrmander type are given in Section 8.

3. Linear stability I: linearized problems

To study the linear stability of current-vortex sheets, we first derive a linearized

problem from the nonlinear problem (2.17)(2.20). By a direct calculation, we have

d

ds L(U + sV, + s)(U + sV)

|s=0

= L(U, )W + E(U, )W +

x1 (L(U, )U)x1 ,

where

W = V x1

Ux1


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is the good unknown as introduced in [1] (see also [12, 21]) and

E(U, )W = W U

A1(U, )

Ux1 +3

j=2W U Aj (U)Uxj .

Then we obtain the following linearized problem of (2.17)(2.20):

L(U, )W + E(U, )W = F in {x1 > 0} (3.1)

with the boundary conditions on {x1 = 0}:

t

W2 x2 W3 x3 W4+ U3 x2 + U4 x3 = h1 ,

W5 x2 W6 x3 W7 U6 x2 U7 x3 = h2 ,

W+1 W1 +7

j=5(U+j W+j Uj Wj ) = h3,

(3.2)

and the initial condition at {t = 0}:

(W, )|t=0 = 0, (3.3)

for some functions F and h := (h1 , h2 , h3), where (,) = (,)|x1=0.To separate the characteristic and noncharacteristic components of the unknown

W, we introduce J = J(U, ) as an 8 8 regular matrix such that

X = (J)1W (3.4)

satisfies

X1 = W1 +7

j=5Uj W

j ,

X2 = W2 ()x2 W3 ()x3 W4 ,X5 = W5 ()x2 W6 ()x3 W7 ,(X

3

, X4

, X6

, X7

, X8

)=

(W3

, W4

, W6

, W7

, W8

).

Then, under transformation (3.4), problem (3.1)(3.3) for (W, ) is equivalent tothe following problem for (X, ):

L U, X + EU, X = F in {x1 > 0} (3.5)with the boundary conditions on {x1 = 0}:

t X2 + U3 x2 + U4 x3 = h1 ,

X5 U6 x2 U7 x3 = h2 ,X+1 X1 = h3,

(3.6)

and the initial condition at {t = 0}:X,

|t=0 = 0, (3.7)


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where F = (J) F,

L(U, ) = A0(U, )t +3

j=1Aj (U, )xj

with A1U,

= J A1U, J, AjU, = JAjUJ,j = 1, andE(U, )X = (J)E(U, )JX + (J)L(U, )JX.

For simplicity of notation, we will drop the tildes in (3.5). By a direct calculation,

we see that A1(U, ) can be decomposed into three parts:

A1(U, )

=A

,01

+A

,11

+A

,21

with

A,01 =

1

()x1

0 a

a O77

, A

,11 =

t Uv,N()x1

A,11 , A,21 =UH,N

()x1A,21 ,

where a = (1, 0, 0, , 0, 0, 0), Uv,N = U2 ()x2 U3 ()x3 U4 , andUH,N = U5 ()x2 U6 ()x3 U7 .

When the state (U, ) satisfies the boundary conditions given in (2.19), weknow from (3.5) that, on

{x1

=0

},

A1(U+, +) 00 A1(U

, )

X+X

,

X+X

= 2X1 [X2 X5], (3.8)

when [X1] = 0 on {x1 = 0}.Furthermore, from (3.6), we have

[X2 X5] = x2[U3 U6] + x3[U4 U7] [h1 + h2].Thus, from assumption (2.4) that is equivalent to (2.7), there exists a unique

= (v2 , v3 , H2 , H3 ) such thatv+2v+3

v2v3

= +

H+2H+3

H2H3

, (3.9)

that is, [U3 U6] = [U4 U7] = 0, which implies[X2 X5] = [h1 + h2].

Therefore, with the choice of in (3.9), problem (3.5)(3.7) is decomposedinto

L(U, )X + E(U, )X = F in {x1 > 0},[X2 X5] = [h1 + h2] on {x1 = 0},X+1 X1 = h3 on {x1 = 0},X|t0 = 0,

(3.10)


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which is maximally dissipative in the sense of Lax and Friedrichs [13], and on

{x1 = 0},

t = X2 (U3 , U4 )U+6 U

+7

U6 U7

1

X+5 h+2

X5 h2+ h1 ,

|t=0 = 0.(3.11)

Remark 3.1. To have the identities in (3.11) requires compatibility, that is, that the

right-hand side of (3.11) for the plus sign is equal to the term for the minus sign.

This is also guaranteed by the choice of in (3.9).

4. Linear stability II: energy estimates

In this section, we establish the energy estimates for the linear problem (3.1)

(3.3), which implies the existence and uniqueness of solutions. We know from

Section 3 that it suffices to study the linear problems (3.10)(3.11).

As noted as above, when the state (U, ) satisfies the eikonal equation(2.17) in {x1 > 0} and the boundary conditions on {x1 = 0} in (2.19), we knowthat the boundary {x1 = 0} is the uniform characteristic plane of multiplicities 12for the equations in (3.10).

First, we have the following elementary properties in the space Bs (T), whose

proof can be found in [1].

Lemma 4.1. (i) For any fixed s > 5/2, the identity mapping is a bounded embed-

ding from Bs (T) to L(T);

(ii) If s > 1 and u Bs (T) , then u|x1=0 Hs1(bT) , the classical Sobo-lev space. Conversely, if v H(bT) for a fixed > 0, then there existsu B+1(T) such that u|x1=0 = v.

Denote by coef(t,x ) all the coefficient functions appeared in the equations in

(3.10), and coef(t,x) = coef(t,x) coef(0). With the decoupled formulation(3.10)(3.11) for the linearized problem, we can now employ the LaxFriedrichs

theory [13] to establish the desired energy estimates, especially the high-order

energy estimates, of solutions for the linear problem.

Theorem 4.1. For any fixed s0 > 17/2, there exist constants C0 and0 depending

only on coefs0,T for the coefficient functions in (3.10) such that, for any s s0and 0, the estimate

max0tT X(t)2

s, + X2

s,,T

C0

F2s,,T + h2Hs+1 (bT)+ coef2s,,T

F2s0,T + h2Hs0+1(bT)

(4.1)


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holds, provided that the eikonal equations (2.17) and

UH,N := U5 ()x2 U6 ()x3 U7 = 0

are valid for (U, ) on {x1 = 0}, where h = (h1 , h2 , h3) and the norm inHs(bT) are defined as that of Bs(T) with functions independent of x1.

Proof. The proof is divided into two steps.

Step 1. We first study problem (3.10) with homogeneous boundary conditions:

h1 = h2 = h3 = 0 on {x1 = 0}. (4.2)

Define

P = I2 0 00 P1 0

0 0 I3

P2 00 I5

with

P1 = 0 1 00 0 1

1 0 0

, P2 =

1 0 00 1

0 0 1

.

Then

A1 = (P)A1 P =

A

,111 A

,112

A,121 A

,122

with

A,111 =

0 1

1 0

, A

,112 = A,121 = A,122 = 0 (4.3)

on {x1 = 0}.Thus, from (3.10) and (4.2), we find that the vector function

Y = (P)1X

satisfies the following problem:

L(U, )Y + E(U, )Y = F in {x1 > 0},[Y1] = [Y2] = 0 on {x1 = 0},Y|t0 = 0,

(4.4)

where F = ( P) F,

L(U, ) = A0 t +3

j=1A

j xj


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with Aj = (P)Aj P, 0 j 3, and

E = (P)E P + ( P)

3

j=0

Aj (P)xj .

L2-estimate: With setup (4.4), we can apply the LaxFriedrichs theory to obtain

the following L2-estimate:

max0tT

Y(t)20, + Y20,,T C0

F20,,T (4.5)

when 0 > 0, where

C0 = C0(coefL , coefL ).Estimates on the tangential derivatives.From(4.4),wefindthat M Y, || s,

satisfyLM Y + EMY = MF + [L+ E, M]Y in {x1 > 0},[M Y1] = [M Y2] = 0 on {x1 = 0},MY|t0 = 0,

(4.6)

which, by the LaxFriedrichs theory again, yields

max0tT

|Y(t)|2s, + |Y|2s,,T

C0

|F|2s,,T + (coefT)2|Y|2s,,T

+ Y2L | coef|2s,,T + |x1 YI |2s1,,T

,

(4.7)

where YI = (Y1 , Y2 ), the norms | |s, and | |s,,T are the special cases of thenorms s, and s,,T, respectively, without the normal derivatives

k

x1 , and

uT =

||2M uL(T) +

||1

Mx1 uL(T).

Estimates on the normal derivatives. Set Aj = diag[A+j ,Aj ], 0 j 3,with

Aj

= A

,j11 A

,j12

A,j

21 A,j

22 .

From (4.3)(4.4), we find that YI = (Y1 , Y2 ) and YII = (Y3 , . . . , Y8 ) satisfythe following equations:

A,111 x1 Y

I = coef MY + coef Y + FI A,112 x1 YII , (4.8)


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and3j=0 A

,j22 xj Y

II + C22YII = FII

3j=0 A

,j21 xj Y

I C21YI ,

YII |t0 = 0,(4.9)

where the linear operator

3j=0 A,j22 xj + C22 is symmetric hyperbolic and tan-

gential to {x1 = 0}.Thus, from (4.8), we obtain

|x1 YI (t)|s1, |FI (t)|s1, + coefLip(T)|Y(t)|s,+YLip(T)| coef(t)|s,,

(4.10)

with Lip(T) denoting the usual norm in the space of Lipschitz continuousfunctions in T and, from (4.9), we deduce the following estimate on x1 Y

II by

using the classical hyperbolic theory:max

0tT|x1 YII (t)|2s2, + |x1 YII |2s2,,T

C0

|x1FII |2s2,,T + coef2Lip(T)(|YI |2s,,T + |x1 YI |2s1,,T)

+ (coefT|YII |Ns,,T)2 + (YT| coef|Ns,,T)2

,

(4.11)

where

|u|N

s,,T = |u|s,,T + |ux1 |s2,,T.Combining estimates (4.7) with (4.10)(4.11), applying an inductive argument

on the order of normal derivatives kx1 , and noting that all the coefficient functions

ofL and E in problem (4.4) depending on (U, t,x ) and (U, t,x U,{1x1 2t,x2,x3 }11,|2|2,1+|2|2), respectively, we canconclude estimate (4.1)for the case of homogeneous boundary conditions by choosing s0 > 17/2.

Step 2. With Step 1, for the case of nonhomogeneous boundary conditions in

(3.10) with (h1 , h2 , h3) Hs+1([0, T] R2), it suffices to study (3.10) when

F = 0.By Lemma 4.1, there exists X0 Bs+2(T) satisfying

[X0,2 X0,5] = [h1 + h2], [X0,1] = h3 on {x1 = 0}. (4.12)Thus, from (3.10), we know that Y := X X0 satisfies the following

problem:

L(U, )Y + E(U, )Y = f in {x1 > 0},[Y1] = [Y2 Y5] = 0 on {x1 = 0},Y|t0 = 0,

(4.13)

with f = L(U, )X0 E(U, )X0 Bs (T).Employing estimate (4.1) in the case h = 0 for problem (4.13) established in

Step 1 yields an estimate on Y, which implies an estimate of X = Y + X0 .Combining this estimate with that for the case of homogeneous boundary conditions

yields (4.1) for the general case.


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By choosing first 1 and then T so that T 1 in (4.1), we can directlyconclude:

Theorem 4.2. For any fixed s0 > 17/2, there exists a constant C0 > 0 depending

only on coefs0,T such that, for any s s0, the following estimate holds:X2s,T C0

F2s,T + h2Hs+1(bT)+ coef2s,T(F2s0,T + h2Hs0+1(bT))

.

(4.14)

5. Nonlinear stability I: construction of the zeroth-order approximate

solutions

With the linear estimates in Section 4, we now establish the local existence of

current-vortex sheets of problem (2.17)(2.20) under the compatibility conditions

on the initial data. From now on, we focus on the initial data (U0 , 0) that is asmall perturbation of a planar current-vortex sheet (U, ) with constant statesU and = 0 satisfying (2.3)(2.5).

We start with the compatibility conditions on {t = x1 = 0}. For fixed k N,it is easy to formulate the j th-order compatibility conditions, 0 j k, for the

initial data (U0 , 0) of problem (2.17)(2.20), from which we determine the data:

{ jt U|t=x1=0}0jk and { jt |t=0}0jk+1. (5.1)

For any fixed integer s > 9/2 and given data 0 Hs1(R2) and U0 with

U0 = U0 U Bs (R3+),

by using Lemma 4.1 for T = 0, we can extend 0 to be 0 Bs (R3+) satisfying

U0,5 (0 )x2 U0,6 (0 )x3 U0,7 = 0 in {x1 > 0}, (5.2)

which is possible by using assumption (2.4). Let 0 = 0 x1 be the initial datafor solving from the eikonal equations (2.17).

Let Uj (x) = jt U

|t=0 and j = jt

|t=0 for any j 0.For all k [s/2], from problem (2.17)(2.20), we determine

{Uj }1jk and {j }0jk+1 (5.3)

in terms of(U0 ,

0 ) andUj Bs2j (R3+), 1 j k,1 Bs1(R3+), j Bs2(j1)(R3+), 2 j k+ 1.

(5.4)


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18/40


Then it follows from Lemma 5.1 and (5.11) that problem (2.17)(2.20) is equivalent

to the following fixed initial-boundary value problem for (V, ):

L(V, )V = fa in {t > 0, x1 > 0},E(V, ) = 0 in {t > 0, x1 > 0},+|x1=0 = |x1=0 = ,B(V+, V, ) = 0 on {x1 = 0},V|t0 = 0, |t0 = 0,

(5.13)

where fa = L(Ua , a )Ua ,L(V, )V = L(Ua + V, a + )(Ua + V) L(Ua , a )Ua ,

E(V, )=

t

V2

+(V3

+Ua,3)x2

+(V4

+Ua,4)x3

+(a )x2 V3 + (a )x3 V4 ,and

B(V+, V, ) = B(U+a + V+, Ua + V, a + ).

6. Nonlinear stability II: iteration scheme

From Theorem 4.1, we observe that the high-order energy estimates of solutionsfor the linearized problem (3.10) in terms of the nonhomogeneous terms F and thevariable coefficients coef keep the same order in the linear level. Based on theseestimates and the structure of the nonlinear system, we now develop a suitable

iteration scheme of the NashMoserHrmander type (cf. [17]) to deal with the

loss of regularity for the nonlinear problem (5.13).

To do this, we first recall a standard family of smoothing operators (cf. [1, 10]):

{S } >0 : B0(T) s0Bs(T) (6.1)

satisfying

S us,T C (s)+u,T for all s, 0,S u us,T C su,T for all s [0, ], d

dS us,T C s1u,T for all s, 0,

(6.2)

and

(S u+ S u)|x1=0Hs (bT) C (s+1)+(u+ u)|x1=0,T (6.3)

for all s, 0.

Similarly, one has a family of smoothing operators (still denoted by) {S } >0acting on Hs (bT), satisfying (6.2) as well for the norms ofH

s (bT) (cf. [1, 10]).

Now we construct the iteration scheme for solving the nonlinear problem (5.13)

in R+ R3.


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Iteration scheme: Let V,0 = ,0 = 0. Assume that (V,k, ,k) havebeen known for k = 0, . . . , n, and satisfy

(V,k, ,k) = 0 in {t 0},

+,k

= ,k

= k

on {x1 = 0}.(6.4)

Denote the (n + 1)th approximate solutions to (5.13) in R+ R3 byV,n+1 = V,n + V,n , ,n+1 = ,n + ,n , n+1 = n + n . (6.5)

Let 0 1 and n =

20 + n for any n 1. Let Sn be the associated smoo-thing operator defined as above. We now determine the problem of the increments

(V,n , ,n ) as follows.Construction ofV,n: First, it is easy to see

L(V,n+1, ,n+1)V,n+1 L(V,n , ,n )V,n

= L e,(Ua +V,n+

12 ,a +Sn ,n )

V,n + e,n ,(6.6)

where V,n+12 is the modified state ofSn V

,n defined in (6.24)(6.25) later, whichguarantees that the boundary {x1 = 0} is the uniform characteristic plane of constantmultiplicity at each iteration step,

L e,(Ua +V,n+

1

2 ,a +Sn ,n )

V,n

= L(Ua + V,n+12 , a + Sn ,n)V,n

+ E(Ua + V,n+12 , a + Sn ,n )V,n

(6.7)

is the effective linear operator,

V,n = V,n ,n (Ua + V,n+

12 )x1

(a + Sn ,n )x1(6.8)

is the good unknown, and

e,n =4

j=1e

(j ),n (6.9)

is the total error with the first error resulting from the Newton iteration scheme:

e(1),n =L(Ua + V,n+1, a + ,n+1)(Ua + V,n+1)

L(Ua + V,n, a + ,n )(Ua + V,n )

L

(Ua +

V

,n ,a +

,n )(V,n , ,n ),

(6.10)

the second error resulting from the substitution:

e(2),n =L (Ua +V,n ,a +,n )(V

,n, ,n )

L (Ua +Sn V,n ,a +Sn ,n )

(V,n , ,n ),(6.11)


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the third error resulting from the second substitution:

e(3),n =L (Ua +Sn V,n ,a +Sn ,n )(V

,n, ,n )

L

(Ua +V,n

+1

2 ,a +Sn ,n )(V,n , ,n ),

(6.12)

and the remaining error:

e(4),n =

,na +Sn ,n

x1

LUa + V,n+

12 , a + Sn ,n

Ua +V,n+

12

x1.

(6.13)

Similarly, we have

BV+,n+1, V,n+1, n+1BV+,n, V,n , n= B

(Ua +V,n+

12 ,a+Sn n )

(V+,n , V,n, n ) + en, (6.14)

where

en =4

j=1e

(j )n (6.15)

with the first error resulting from the Newton iteration scheme:

e(1)n

=B(U+a

+V+,n+1, Ua

+V,n+1, a

+n+1)

B(U+a + V+,n , Ua + V,n , a + n) B

(Ua +V,n ,a+n )(V

+,n , V,n , n),(6.16)


e(2)n =B(Ua +V,n ,a+n )(V+,n , V,n, n )

B(U

a +Sn V,n ,a+Sn n )

(V+,n , V,n, n ),(6.17)


e(3)n =BU

a +Sn V,n ,a+Sn n

V+,n , V,n , n B

Ua +V,n+12 ,a+Sn n

V+,n, V,n , n , (6.18)and the remaining error:

e(4)n =BUa +V,n+12 ,a+Sn n

V+,n , V,n , n

BUa +V,n+

12 ,a+Sn n

V+,n, V,n , n . (6.19)Using (6.6) and (6.14) and noting that

L

V,0, ,0

V,0 = 0, BV+,0, V,0, 0 = 0,


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we obtain

L

V,n+1, ,n+1

V,n+1

=n

j=0

L e,(U

a +V

,j+ 12 ,

a +S

j

,j )

V,j + e,j ,B(V+,n+1, V,n+1, n+1)

= nj=0 B(U

a +V,j+

12 ,a+Sj j )

(V+,j , V,j , j ) + ej

.

(6.20)

Observe that, if the limit of(V,n, ,n) exists which is expected to be a solutionto problem (5.13), then the left-hand side in the first equation of (6.20) should

converge to fa , and the left-hand side in the second one of (6.20) should tend tozero when n . Thus, from (6.20), it suffices to study the following problem:

L e,(Ua +V,n+

12 ,a +Sn ,n )V

,n = fn in T,B

(Ua +V,n+12 ,a+Sn n )

(V+,n, V,n, n ) = gn on bT,V,n|t0 = 0,

(6.21)

where fn and gn are defined by

n

j=0fj + Sn (

n1

j=0e,j ) = Sn fa ,

n

j=0gj + Sn (

n1

j=0ej ) = 0, (6.22)

by induction on n, with f0 = S0 fa and g0 = 0.We now define the modified state V,n+

12 to guarantee that the boundary {x1 =

0} is the uniform characteristic plane of constant multiplicity at each iteration step(6.21). To achieve this, we require

B(V+,n+12 , V,n+

12 , Sn

n)

i|x1=0 = 0 for i = 1, 2, all n N, (6.23)

which leads one to define

V+,n+ 12j = Sn V,nj for j {3, 4, 6, 7}, (6.24)and

V,n+ 12

2 = (a + Sn ,n )x2 V,n+ 12

3 + (a + Sn ,n )x3 V,n+ 12

4

+(Sn ,n )t + Ua,3(Sn ,n )x2 + Ua,4(Sn ,n )x3 ,

V,n+ 12

5 = (a + Sn ,n )x2 V,n+ 12

6 + (a + Sn ,n )x3 V,n+ 12

7

+Ua,6(Sn

,n )x2+

Ua,7(Sn ,n )x3 .

(6.25)

Construction of ,n satisfying ,n|x1=0 = n . Clearly, we haveE(V,n+1, ,n+1)E(V,n, ,n ) = E

(V,n+ 1

2 ,Sn ,n )

(V,n , ,n )+ e,n,(6.26)


22/40


where

E(V,)(W

, ) =t W2 + (Ua,3 + V3 )x2 + (Ua,4 + V4 )x3

+ (a + )x2 W3 + (a + )x3 W4(6.27)is the linearized operator ofE and

e,n =4

j=1e

(j ),n (6.28)

with the first error resulting from the Newton iteration scheme:

e(1),n = E(V,n+1, ,n+1) E(V,n, ,n) E(V,n ,,n )(V,n, ,n )

= (,n

)x2 V,n

3 + ( ,n

)x3 V,n

4 ,(6.29)


e(2),n =E(V,n ,,n )(V,n, ,n) E(Sn V,n ,Sn ,n )(V

,n , ,n )

=((I Sn ),n )x2 V,n3 + ((I Sn ),n)x3 V,n4+ (I Sn )V,n3 (,n )x2 + (I Sn )V,n4 (,n)x3 ,

(6.30)


e(3),n = E(Sn V,n ,Sn ,n )(V

,n , ,n ) E(V

,n+ 12 ,Sn

,n )(V,n , ,n ),

(6.31)

which vanishes due to (6.24) and that all the coefficients of the linearized operator

E(V,) are independent of(V

2 , V

5 ), and the remaining error:

e(4),n = E

(V,n+ 1

2 ,Sn ,n )

(V,n, ,n) E(V

,n+ 12 ,Sn

,n )(V,n , ,n ).

(6.32)

Thus, from (6.26), we have

E(V,n+1, ,n+1) =n

j=0

E

(V

,j+ 12 ,Sj

,j )(V,j , ,j ) + e,j

, (6.33)

which leads to define ,n to be governed by the following problem:E

(V,n+ 1

2 ,Sn ,n )

(V,n , ,n ) = hn on T, ,n

|t0

=0,

(6.34)

where hn is defined by

nk=0

hk + Sn n1

k=0e,k

= 0 (6.35)


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by induction on n starting with h0 = 0.By comparing (6.34) with (6.21), it is easy to verify

,n|x1=0 = n. (6.36)

From (5.10)(5.11) and (6.34), we know that the compatibility conditions hold for

the initial-boundary value problem (6.21) for all n 0.

The steps for determining (V,n , ,n ) are to solve first V,n from (6.21)and then ,n from (6.34), and to obtain finally V,n from (6.8).

7. Nonlinear stability III: convergence of the iteration scheme and existence

of the current-vortex sheet

Fix any s0 9, s0 + 5, and s1 [ + 5, 2 s0]. Let the zeroth-orderapproximate solutions for the initial data (U0 , 0) constructed in Section 5 satisfy

Ua s1+3,T +a s1+3,T +fa s14,T , fa +1,T/ is small, (7.1)

for some small constant > 0.

Before we prove the convergence of the iteration scheme (6.21) and (6.34), we

first introduce the following lemmas, under the inductive assumption that (7.10)

hold for all 0 k n 1.

Lemma 7.1. For any k = 0, 1, . . . , n, we have(V,k, ,k),T + kH1(bT) C log k,(V,k, ,k)s,T + kHs1(bT) C

(s)+k ,

(7.2)

for s [s0, s1] with s = ,(SkV,k, Sk,k),T + SkkH1(bT) C log k,

(SkV

,k, Sk,k)

s,T

+ Sk

k

Hs

1(b

T) C

(s)+k

,(7.3)

for s s0 with s = , and

((I Sk)V,k, (I Sk),k)s,T +(I Sk)kHs1(bT) C sk (7.4)

for s [s0, s1].

These results can be easily obtained by using the triangle inequality, the classi-

cal comparison between series and integrals, and the properties in (6.2) of the

smoothing operators S .

Lemma 7.2. If 4 s0 and 5 , then, for the modified state V,n+ 12 , we

have

V,n+ 12 Sn V,ns,T C s+1n for any s [4, s1 + 2]. (7.5)


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Lemma 7.3. Let s0 5 and s0 + 3. For all k n 1, we have

e,ks,T C2L1(s)

k k for s [s0, s1 4],

e

,k

s,T C

2L2(s)k k for s

[s0, s1

3

],

ekHs (bT) C2L3(s)k k for s [s0 1, s1 4],(7.6)

where k = k+1 k and

L1(s)=

max((s + 2 )++s01, s+s0+42) if = s + 2, s + 4,max(s0 , 2(s0 ) + 4) if = s + 4,s0 + 2 if = s + 2,

L2(s)=max((s+2)++2(s0), s+s0 + 2 2) if = s + 2, s + 3,s0 1 if =s+3,s0 if = s + 2,

and

L3(s)=

max((s+3)++2(s0), s + s0 + 3 2) if = s + 3, s + 4,s0 1 if = s + 4,s0 if = s + 3.

Denote the accumulated errors by

E,n =n1k=0

e,k, En =n1k=0

ek, E,n =n1k=0

e,k. (7.7)

Then, as a corollary of Lemma 7.3, we have

Lemma 7.4. Let s0 5, s0 + 3, and s1 2 s0 1. Then

E

,n

s,T C

2n for s [

s0, s1

4],

E,ns,T 2 for s [s0, s1 3], EnHs (bT) C2n for s [s0 1, s1 4].

(7.8)

Lemma 7.5. For any s0 5, s0 + 5, and s1 [ + 5, 2 s0 1], we have

fn s,T Cn s1n (fa ,T + 2),gnHs+1(bT) C2n s1n ,hn s,T C2n s1n

(7.9)

for all s s0.

The proofs of Lemmas 7.2 7.3 and 7.5 will be given in Section 8. With these

lemmas, we can now prove the following key result for the convergence of the

iteration scheme.


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Proposition 7.1. For the solution sequence (Vk, ,k, k) given by (6.21)and(6.34), we have

(V,k, ,k)s,T+ kHs1(bT) s1k k for s [s0, s1],L(V

,k

, ,k

)V,k

fa s,T 2s

1

k for s [s0, s1 4],B(V+,k, V,k, k)Hs1(bT) s1k for s [s0, s1 2]

(7.10)

for any k 0, where k = k+1 k.

Proof. Estimate (7.10) is proved by induction on k 0.

Step 1. Verification of (7.10) for k = 0. We first notice from (5.8)(5.9) that(Ua ,

a ) satisfies the RankineHugoniot conditions, and V

,0 = ,0 = 0imply that V,

12 = 0. Thus, V,0 satisfies the following problem:

L e,(U

a ,

a )

V,0 = S0 fa in T,B

(Ua ,a )

(V+,0, V,0, 0) = 0 on bT,V,0|t0 = 0.

(7.11)

Applying Theorem 4.2 to problem (7.11), we have

V,0s,T C (s1)+0 fa +1,T for all s [s0, s1]. (7.12)

Similarly, from (6.34), we obtain

,0s,T Ch0 s,T + Ua s,Th0 s0,T

, (7.13)

where h0 = V2,0 (a )x2 V3

,0 (a )x3 V4,0

satisfies

h0 s,T CV,0s,T C0(s1)+0 fa +1,T. (7.14)

Thus, we have

,0

s,T C

1

(s1)+0

fa +1,T

. (7.15)

Since

V,0 = V,0 ,0 (Ua )x1

(a )x1,

we use (7.12) and (7.15) to find

V,0s,T C2 (s1)+0 fa +1,T. (7.16)

From (7.15)(7.16), we deduce

(V,0, ,0)s,T s10 0 for any s [s0, s1], (7.17)

provided that fa +1,T/ is small.The other two inequalities in (7.10) hold obviously when k = 0.


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Step 2. Suppose that (7.10) holds for all k n 1, we now verify (7.10) fork = n. First, we note that, when s0 + 3,

(Ua + V,n+12 , a + Sn ,n )s0+2,T

(Ua , a )s0+2,T + V,n+ 12 Sn V,ns0+2,T+ (Sn V,n, Sn ,n )s0+2,T

C

1 + s0+3n + (s0+2)+n C,

(7.18)

by using assumption (7.1) and Lemmas 7.1 7.2. Applying Theorem 4.2 to problem

(6.21), we have

V,ns,T C

fn s,T + gnHs+1(bT) + coefs,T(fn s0,T + gnHs0+1(bT)) C

fn s,T + gnHs+1(bT)+ (Ua + V,n+

12 , a + Sn ,n)s+2,T(fn s0,T + gnHs0+1(bT))

.

(7.19)

On the other hand, similar to (7.18), from assumption (7.1) and Lemmas 7.1

7.2, we have

(Ua +V,n

+1

2 , a +Sn ,n

)s+2,T C1+ s+3n + (s+2)+n , (7.20)which implies

(Ua + V,n+12 , a + Sn ,n )s+2,T

fn s0,T + gnHs0+1(bT) C s1n n

fa ,T + 2(7.21)

for all s [s0, s1] by using s0 + 3 and (7.9).Thus, from (7.19), we conclude

V,ns,T C s1n nfa ,T + 2 for s [s0, s1]. (7.22)

For problem (6.34), we can easily obtain the following estimate:

,ns,T Chn s,T + Ua + Sn V,ns,Thn s0,T

for any s s0,

(7.23)

where C > 0 depends only on Ua + Sn V,ns0,T, and

h

n =h

n +

V

,n2

(

a +

Sn

,n)x2

V,n

3 (

a +S

n

,n)x3

V,n

4. (7.24)

Obviously, we have

hn s,T hn s,T +

1 + a + Sn ,ns0,TV,ns,T

+ a + Sn ,ns+1,TV,ns0,T,


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which implies

hn s,T Cn s1nfa ,T + 2 (7.25)

by using (7.22) and Lemmas 7.1 and 7.5.

Substituting (7.25) into (7.23), we find

,ns,T Cn s1nfa ,T + 2 for all s [s0, s1]. (7.26)

Together (7.22) with (7.26), we obtain (7.10) for (V,n , ,n ) by using

V,n = V,n + (Ua + V,n+

12 )x1

(a + Sn ,n )x1,n

and choosing

fa

,T/ and > 0 to be small.

Finally, we verify the other inequalities in (7.10). First, from (6.21)(6.22), wehave

L(V,n , ,n )V,n fa = (Sn1 I) fa +(ISn1 )E,n1+e,n1. (7.27)

From Lemma 7.4, we have

(I Sn1 )E,n1s,T C ssn E,n1s,T C ss+1n 2 C s1n 2(7.28)

by choosing

s

=

+2.

From Lemma 7.3, we obtain

e,n1s,T CL1(s)

n 2 C s1n

2 for all s [s0, s1 4], (7.29)

by using s0 + 5.When s + 1, we have

(Sn1 I) fa s,T C s1n fa +1,T; (7.30)

while s

(+

1, s1

4], we have

(Sn1 I) fa s,T Sn1 fa s,T + fa s,T C s1n fa +1,T + (7.31)

by using (6.2) and (7.1).

Substitution of (7.28)(7.31) into (7.27) yields

L(V,n, ,n )V,n fa s,T 2 s1n for s [s0, s1 4],

provided that fa s14,T and fa +1,T/ is small.Similarly, we have

B(V+,n, V,n , n )Hs1(bT) s1n . (7.32)

Thus, we obtain (7.10) for k = n.


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Convergence of the Iteration Scheme: From Proposition 7.1, we have

n0

(V,n, ,n),T + nH1(bT) < , (7.33)

which implies that there exists (V, ) B (T) with H1(bT) suchthat

(V,n, ,n) (V, ) in B (T) B (T),n in H1(bT).

(7.34)

Thus, we conclude

Theorem 7.1. Let 14 and s1 [ + 5, 2 9]. Let 0 H2s1+3(R2) andU0 U B2(s1+2)(R3+) satisfy the compatibility conditions of problem (2.17)(2.20) up to order s1 + 2, and let conditions (2.3)(2.4) and(7.1) be satisfied. Thenthere exists a solution (V, ) B (T) with H1(bT) to problem(5.13).

Then Theorem 2.1 (main theorem) in Section 2 directly follows fromTheorem 7.1.

8. Error estimates: proofs of Lemmas 7.27.3 and 7.5

In this section, we study the error estimates for the iteration scheme (6.21)

(6.22) and (6.34)(6.35) to provide the proofs for Lemmas 7.27.3 and 7.5 under

the assumption that (7.10) holds for all 0 k n 1. We start with the proof ofLemma 7.2.

Proof of Lemma 7.2. Denote by

E,n1 := E(V,n, ,n), (8.1)

and

E,n2 :=V,n5 (a + ,n )x2 V,n6 (a + ,n )x3 V,n7

Ua,6(,n )x2 Ua,7(,n )x3(8.2)

as the extension of(B(V,n , n))2 in T.


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By the definition in (6.25) for V,n+12 , we have

V,n+ 12

2 Sn V,n2= Sn

E

,n

1 + [t, Sn ],n

+(a + Sn ,n)x2 Sn V,n3 Sn ((a + ,n )x2 V,n3 )+(a + Sn ,n)x3 Sn V,n4 Sn ((a + ,n)x3 V,n4 )+(Sn ,n )x2 Ua,3 Sn ((,n )x2 Ua,3)+(Sn ,n )x3 Ua,4 Sn ((,n )x3 Ua,4),

V,n+ 12

5 Sn V,n5= (a + Sn ,n )x2 Sn V,n6 Sn ((a + ,n)x2 V,n6 )

+(

a

+Sn

,n)x3 Sn V,n

7

Sn ((

a

+,n)x3 V

,n7 )+(Sn ,n )x2 Ua,6 Sn ((,n )x2 Ua,6))

+((Sn ,n )x3 Ua,7 Sn ((,n )x3 Ua,7) SnE,n2 .

(8.3)

On the other hand, we have

E,n1 = E,n11 + ( ,n1)t + (a + ,n )x2 V,n13

+(a + ,n)x3 V,n14 V,n12

+( ,n1)x2 (U

a,3

+V

,n13 )

+(,n1)x3 (U

a,4

+V

,n14 ),

E,n2 = E,n12 (a + ,n )x2 V,n16(a + ,n)x3 V,n17 + V,n15( ,n1)x2 (Ua,6 + V,n16 ) + (,n1)x3 (Ua,7 + V,n17 ),

(8.4)

which implies

(E,n1 , E,n2 )s0,T C s01n (8.5)

by using n = O(1n ), the inductive assumption for (7.10), and Lemma 7.1. Thus

we deduce Sn (E,n1 , E,n2 )s,T C s1n for any s s0. (8.6)

The discussion for the commutators in (8.3) follows an argument from

Coulombel and Secchi [10]. We now analyze the third term ofV,n+ 12

2 Sn V,n2given by (8.3) in detail.

When s [ + 1, s1 + 2], we have

(a

+Sn

,n)x2 Sn V,n

3

s,T

C(a + Sn ,n)x2LSn V,n3 s,T

+a + Sn ,ns+1,TSn V,n3 L

C2 s+1n ,


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and

Sn

(a + ,n )x2 V,n3s,T

C sn

(a + ,n)x2LV,n

3 ,T

+

a +,n

+

1,T

V

,n3

L

C2 s+1n

by using (7.1), the induction assumption for (7.10), and Lemma 7.1, which implies

(a + Sn ,n )x2 Sn V,n3 Sn

(a + ,n)x2 V,n3s,T C2 s+1n .

(8.7)

When s [s0, ], we have

(a

+Sn

,n)x2 Sn V,n

3

Sn (a +

,n )x2 V,n

3 s,T (I Sn )

(a + ,n )x2 V,n3

s,T+(a + ,n)x2 (Sn I)V,n3 s,T+(Sn I),nx2 Sn V,n3 s,T C sn

(a +,n )x2 V,n3 ,T+(a + ,n )x2LV,n3 ,T

+C 3n

V

,n3

,T

a

+,n

s+1,T

+((Sn I),n )x2 Sn V,n3 s,T C2 s+1n .

(8.8)

Together (8.7) with (8.8), it follows that

(a +Sn ,n )x2 Sn V,n3 Sn

(a +,n )x2 V,n3s,T C2 s+1n (8.9)

for all s [s0, s1 + 2].The other commutators appearing in (8.3) satisfy estimates similar to the above.

Thus, we complete the proof.

To show the error estimates given in Lemma 7.3, we first show several lemmas

dealing with different types of errors.

Lemma 8.1. Let s0 + 1 6. For the quadratic errors, we have

e(1),ks,T C2L1(s)k k for s [s0 2, s1 2],e(1),ks,T C2 s2+s02k k for s [s0 1, s1 1],

e(1)

k Hs

(bT)

C

2

s

2

+s0

1

k k for s [s0 2, s1 2]

(8.10)

for all k n 1, where

L1(s) = max((s+2)++2(s0)3, s+s012, s+2s033, s03).(8.11)


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Proof. The proof is divided into three steps.

Step 1. From the definition ofe(1),k, we first have

e(1)

,k

= 1

0 (1

)L (Ua

+V

,k

+ V

,k

;a

+

,k

+

,k)

(V,k, ,k),(V,k, ,k) d. (8.12)From (7.1) and Lemma 7.1, we find

sup01

Ua + V,k + V,kW1,(T) + a + ,k + ,kW1,(T)

C.

(8.13)


L (U,)

(V,1, ,1), (V,2, ,2)

s,T C

(U, )s+2,T(V,1, ,1)W1,(V,2, ,2)W1,+ (V,1, ,1)s+2,T(V,2, ,2)W1,+ (V,1, ,1)W1,(V,2, ,2)s+2,T

.

(8.14)

Therefore, we obtain

e(1),ks,T C

2

2(s01)k

2k

+ (V,k, ,k)s+2,T + (V,k, ,k)s+2,T

+ s01k k(V,k, ,k)s+2,T

,

which implies

e(1),ks,T C2

L(s)k k when s + 2 = , s s1 2,

e(1),ks,T C2max(s03,2s022)k k when s + 2 = ,

(8.15)

by Lemma 7.1, the inductive assumption for (7.10), and k = O(k1), where

L(s)=max((s + 2 )+ + 2(s0 ) 3, s + s0 1 2, s + 2s0 3 3).

Thus, we conclude the first result in (8.10).

Step 2. Obviously, from the definition of e(1)

,k in (6.29), we have

e(1),ks,T CV,ks,T,kW1, + V,kL,ks+1,T

,

(8.16)

which implies the second result in (8.10).


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Step 3. Since e(1),k|x1=0 = (ek)1 , we obtain that (ek)1 satisfies the estimate

given in (8.10). Moreover, from the definition of e(1)k , we have

( e(1)k )2 = ( k)x2 V,k6 ( k)x3 V,k7 ,(e

(1)k )3 = 12 (|V

+,kH |2 |V,kH |2), (8.17)

which implies

(e(1)k )2 Hs (bT) C kHs+1(bT)V,kL

+ kW1,(bT)V,kHs (bT)

,

(e(1)k )3Hs (bT) CV,kLV,kHs (bT),(8.18)

yielding the last estimate of (8.10) by the inductive assumption for (7.10).

For the first substitution errors, we have

Lemma 8.2. Let s0 + 1 6. Then, for all k n 1, we have

e(2),ks,T C2L 2(s)

k k for s [s0 2, s1 2],e(2),ks,T C2 s2+s0k k for s [s0 1, s1 1],e

(2)k Hs (bT) C

2

s

2+

s0+

1k k for s [s0 2, s1 2],

(8.19)

where

L2(s) =

(s + 2 )+ + s0 1 for = s + 2,max(5 , s0 1 ) for = s + 2.

(8.20)


Step 1. From the definition ofe(2),k, we have

e(2),k =

T0

L (Ua +SkV,k+ (1Sk)V,k;a +Sk,k+ (1Sk),k)

((V,k, ,k),((1 Sk)V,k, (1 Sk),k)) d.(8.21)

As in (8.13), from the inductive assumption for (7.10), we find

sup01

Ua + SkV,k + (1 Sk)V,kW1,(T) C, (8.22)

and

sup01

a + Sk,k + (1 Sk),kW1,(T) C. (8.23)


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Then we use (8.14) in (8.21) to obtain

e(2),ks,T C(V,k, ,k)W1,(1 Sk)(V,k, ,k)W1,

(

(

Ua ,

a )

s

+2,T

+ Sk(V

,k, ,k)

s

+2,T

+ (1 Sk)(V,k, ,k)s+2,T)+ (V,k, ,k)s+2,T(1 Sk)(V,k, ,k)W1,+ (V,k, ,k)W1,(1 Sk)(V,k, ,k)s+2,T

.

(8.24)

Using the properties of the smoothing operators, the inductive assumption for

(7.10), and Lemma 7.1 in (8.24), we conclude the first estimate in (8.19) when

s s1 2.Step 2. From the definition of e

(2)

,k in (6.30), we have

e(2),ks,T CV,ks,T(x2,x3)(1 Sk),kL+ (1 Sk),ks+1,TV,kL

+ (1 Sk)V,ks,T(x2,x3),kL+ ,ks+1,T(1 Sk)V,kL

,

(8.25)

which implies the second result in (8.19) when s s1

1 by the inductive assump-

tion for (7.10) and Lemma 7.1 in (8.25).

Step 3. Noting that e

(2)k

1

= e(2),k|x1=0

from the above discussion, we conclude that (e(2)k )

1 satisfy the estimate given in

(8.19) when s s1 2.From the definition of e

(2)k , we have

(e(2)k )

2 =

(Sk I)k

x2

V,k6 +

(Sk I)k

x3V,k7

+(Sk I)V,k6 ( k)x2 + (Sk I)V,k7 ( k)x3 ,(e

(2)k )3 = (I Sk)V+,kH V+,kH (I Sk)V,kH V,kH ,

(8.26)

which implies

(e(2)k )2 Hs (bT) CV,kHs (bT)(1 Sk)kW1,+(1 Sk)kHs+1(bT)V,kL+(1 Sk)V,kHs (bT) kW1,+ kHs+1(bT)(1 Sk)V,kL

,

(e(2)k )3Hs (bT) CV,kHs (bT)(1 Sk)V,kL

+(1 Sk)V,kHs (bT)V,kL

.

(8.27)


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Using the inductive assumption for (7.10) and Lemma 7.1 in (8.27), we conclude

the last result in (8.19) when s s1 2.

The representations of the second substitution errors (e(3),k, e

(3)k ) are similar to

those of (e(2),k, e(2)k ). Thus, using Lemma 7.2 and the same argument as the proofof Lemma 8.2, we conclude

Lemma 8.3. Let s0 + 1 6. For the second substitution errors, we havee(3),ks,T C2L 3

(s)k k for s [s0 2, s1 2],

e(3)k Hs (bT) C2 s+s0+22k k for s [s0 2, s1 2](8.28)

for all k n 1, where

L3(s)=max

s+s0+2 2, s + 2s0 + 3 3, (s + 2 )+ + 2(s0 ) + 1

.

Lemma 8.4. Let s0 5 and s0 + 3. For the last errors, we have

e(4),ks,T C2L 4(s)

k k for s [s0, s1 4],e(4),ks,T C2

L 5(s1)k k for s [s0, s1 3],

e(4)k Hs (bT) C2L5(s)k k for s [s0 1, s1 4]

(8.29)

for all k n

1, where

L4(s) =

max((s + 2 )+ + 2 + 2(s0 ), s + 4 + s0 2)if = s + 2, s + 4,

max(s0 , 2(s0 ) + 4) if = s + 4,s0 + 2 if = s + 2,

and

L5(s) =

max((s

+3

)+

+2(s0

), s

+3

+s0

2)

if = s + 3, s + 4,s0 if = s + 3,s0 1 if = s + 4.


Step 1. Set

Rk =L(Ua + V,k+

12 , a + Sk,k)(Ua + V,k+

12 )

x1

. (8.30)

Then we have

Rk s,T L(Ua + V,k+12 , a + Sk,k)(Ua + V,k+

12 )

L(Ua + V,k, a + ,k)(Ua + V,k)s+2,T+ L(V,k, ,k)V,k fa s+2,T,


35/40


which implies that, for all s [s0 2, s1 4],

Rk s,T

CV,k+ 12 V,kW1,Ua + V,k+ 12 s+2,T + a + Sk,ks+4,T+ Ua + V,ks+4,T

V,k+ 12 V,kL + (I Sk),kW1,+ Ua + V,kW1,

V,k+ 12 V,ks+2,T + (I Sk),ks+4,T+ V,k+ 12 V,ks+4,T + L(V,k, ,k)V,k fa s+2,T

C2

(s+4)++s0+1k + C s+5k for s + 4 = ,

C2s0+2k + Ck for s + 4 = .

(8.31)Thus we find that

e(4),k =

Rk ,k

(a + Sk,k)x1

satisfy

e(4),ks,T CRk s02,T s1k k + s01k k( + (s+2)+k )+ s01k kRk s,T

,

(8.32)

when s + 2 = , and

e(4),ks,T CRk s02,T

s1k k + s01k k( + log k)

+ s0

1

k kRk s,T,(8.33)

when s + 2 = , which yields the first estimate in (8.29) for any s [s0, s1 4],provided s0 + 3 by using (8.31).

Step 2. Set

Rbk = B(Ua + V,k+12 , a + Skk). (8.34)

Then we have

RbkHs (bT) B(Ua + V,k+

12 , a + Skk) B(Ua + V,k, a + k)Hs (bT)

+ B(V,k, k)Hs (bT),


36/40


which implies

(Rbk)1 Hs (bT) C(Sk 1)kHs+1(bT) + V

,k+ 122

V

,k

2 Hs

(bT)

+ (Sk I)V,kLa + SkkHs+1(bT)+ (Sk I)V,kHs (bT)a + SkkW1,+ (Sk 1)kHs+1(bT)Ua + V,kL+ (Sk 1)kW1,(bT)Ua + V,kHs (bT)+ B(V,k, k)Hs (bT)

.

(8.35)

Therefore, we have the estimate

(Rbk)1 Hs (bT)

Cmax((s+2)++s0,s+2)k if = s + 1, s + 2,

Ck if = s + 1,C if = s + 2

(8.36)

for all s [s0 1, s1 2]. Thus we find that

(e(4)

k )1 = ((Rbk)

1 )x1

(a + Sk,k)x1 |x1=0 k

satisfy

(e(4)k )1 Hs (bT) C(Rbk)1 H4(bT)

sk k + 2 s01k k

(s+3)+k

+2 s01k k(Rbk)1 Hs+2(bT)

,

(8.37)

when s

+3

=, and

(e(4)k )1 Hs (bT) C(Rbk)1 H4(bT)

sk k + 2 s01k k log k

+2 s01k k(Rbk)1 Hs+2(bT)

,

(8.38)

when s + 3 = , which yields the estimate of (e(4)k )1 given in (8.29) for anys [s0 1, s1 4] by using (8.36).

Estimate (8.29) for the other components of e(4)k can be proved by the same

argument as above.

Step 3. Noting that the restriction of e(4),k on {x1 = 0} is the same as (e

(4)k )1 ,

we conclude the estimate of e(4),k in (8.29) for any s [s0, s1 3].

Proof of Lemma 7.3 . Summarizing all the results in Lemmas 8.18.4, we obtain

the estimates given in Lemma 7.3.


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Proof of Lemma 7.5 . From the definitions of ( fn , gn , hn ) given in (6.22) and

(6.35), respectively, we have

fn = (Sn Sn1 ) fa (Sn Sn1 )E,n1 Sn e,n1,gn = (Sn Sn1 ) En1 Sn en1,hn = (Sn1 Sn ) E,n1 Sn e,n1.

(8.39)

Using the properties of the smoothing operators, we have

(Sn Sn1 ) fa s,T C ss1n nfa s,T for s 0,(Sn Sn1 )E,n1s,T C ss1n nE,n1s,T

C2 ssn n for s [s0, s1 4],

Sn e

,n

1

s,T C

(ss)+n

e

,n

1

s,T

C2 (ss)++L1(s)n n for s [s0, s1 4].(8.40)

Estimate (8.40)3 can be represented as the following three cases.

Case 1. s = s0. For this case, if we choose s [s0, 5], then

L1(s) = max(s0 1, s + s0 + 4 2) = s0 1,

which implies from (8.40)3 that

Sn e,n1s0,T C2 s01n n . (8.41)

Case 2. s = s0 + 1 or s = s0 + 2. For these two cases, if we choose s = 4 s0 + 1, then

L1(s) = max(s0 , 2(s0 ) + 4) = s0 ,

which implies

Sn e,n1s,T C2 s1n n . (8.42)

Case 3. s s0 + 3. For this case, if we choose s = 2 s0, then(s s)+ + L1(s) = (s + 2 )+ + s0 + 2 s 1

by using s0 + 5, which implies from (8.40)3 that

Sn e,n1s,T C2 s1n n . (8.43)


(Sn Sn1 )E,n1s,T C2

s

1n n (8.44)

by setting s = + 1 in (8.40)2 ifs1 + 5.From (8.41)(8.44) with (8.39), it follows that

fn s,T C s1n n (2 + fa ,T) for all s s0. (8.45)


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Similarly, we have

(Sn Sn1 ) En1Hs+1(bT) C ssn n En1Hs (bT) C2 s+1sn n ,

Sn en1Hs+1(bT) C(s

+1

s)+n en1Hs (bT)

C(s+1s)++L3(s)n

2n

(8.46)

for s [s0 1, s1 4].Estimate (8.46)2 can be represented as the following three cases.

Case 1. s = s0. For this case, if we choose s = 4, then L3( 4) = s0 1,it follows from (8.46)3 that

Sn

en1

Hs0+1(bT) C

2 s01n n;

Case 2. s0 +1 s 3. For this case, if we choose s = 3, then L3( 3) =s0 , which implies

Sn en1Hs+1(bT) C2 s0n n C2 s1n n;

Case 3. s 2. For this case, if we choose s = 5 s0, then L3(s) =s0 2 , which implies

Sn en1Hs+1(bT) C2(s+1s)++L3(s)n n C

2 s1n n .

In summary, we obtain

Sn en1Hs+1(bT) C2 s1n n for all s s0. (8.47)From the assumption s1 + 6, it is possible to let s = + 2 in (8.46)1,

which yields

gnHs+1(bT) C2 s1n n for all s s0, (8.48)by using (8.46)1 and (8.47) in (8.39)2.

Furthermore, we have(Sn Sn1 ) E,n1s,T C ss1n n E,n1s,T C2 ss1n n ,Sn e,n1s,T C (ss)+n e,n1s,T C (ss)++L

2(s)n

2n(8.49)

for s [s0, s1 3].Estimate (8.49)2 can be represented as the following three cases.

Case 1. s = s0. For this case, if we choose s = 3, then L2(s) = s0 1,which implies from (8.49)2 that

Sn e,n1s0,T C2

s

0

1n n;

Case 2. s = s0 + 1. For this case, if we choose s = 2, then L2(s) = s0 ,which implies

Sn e,n1s0+1,T C2 s0n n;


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Case 3. s s0 + 2. For this case, if we choose s = 4 s0, then L2(s) =s0 2 , which implies

Sn e,n1s,T C2 s1n n.

In summary, we obtain

Sn e,n1s,T C2 s1n n for all s s0. (8.50)Letting s = in (8.49)1, we find

hn s,T C2 s1n n for all s s0, (8.51)by using (8.50) in (8.39)3.

Acknowledgment. The research of Gui-Qiang Chen was supported in part by the NationalScience Foundation under grants DMS-0720925, DMS-0505473, DMS-0244473, and anAlexandre von Humboldt Foundation fellowship. The research of Ya-Guang Wang wassupported in part by a key project from the NSFC under grant 10531020, a joint project fromthe NSAF under grant 10676020 and a Post-Qimingxing Fund from the Shanghai Scienceand Technology Committee under grant 03QMH1407. The second author would like toexpress his gratitude to the Department of Mathematics of Northwestern University (USA)for their hospitality, where this work was initiated when he visited during the Spring Quarter2005.

References

1. Alinhac, S.: Existence dondes de rarfaction pour des systmes quasi-linaireshyperboliques multidimensionnels. Commun. Partial Differ. Eqs. 14, 173230 (1989)

2. Artola, M., Majda, A.: Nonlinear development of instability in supersonic vortexsheets, I: The basic kink modes. Phys. D. 28, 253281 (1987)

3. Artola, M., Majda, A.: Nonlinear development of instability in supersonic vortexsheets. II SIAM J. Appl. Math. 49, 13101349 (1989)

4. Artola, M., Majda, A.: Nonlinear kind modes for supersonic vortex sheets. Phys.Fluids 1A, 583596 (1989)

5. Begelman, M.C., Blandford, R.D., Rees, M.J.: Theory of exagalactic radio sources.Rev. Mod. Phys. 56, 255351 (1984)

6. Blokhin, A., Trakhinin, Y.: Stability of strong discontinuities in fluids and MHD,In: Handbook of Mathematical Fluid Dynamics. Vol. I, pp. 545652, North-Holland,Amsterdam, 2002

7. Chen, G.-Q., Feldman, M.: Multidimensional transonic shocks and free boundaryproblems for nonlinear equations of mixed type. J. Am. Math. Soc. 3, 461494 (2003)

8. Chen, S.: On the initial-boundary value problems for quasilinear symmetric hyperbolicsystems with characteristic boundary (in Chinese). Chinese Anal. Math. 3, 222232(1982)

9. Coulombel, J.F., Secchi, P.: Stability of compressible vortex sheet in two spacedimensions. Indiana Univ. Math. J. 53, 9411012 (2004)

10. Coulombel, J.F., Secchi, P.: Nonlinear compressible vortex sheets in two spacedimensions. Preprint, 2006

11. Courant, R., Friedrichs, K.: Supersonic Flow and Shock Waves. Springer, New York1948

12. Francheteau, J., Mtivier, G.: Existence de chocs faibles pour des systmesquasi-linaires hyperboliques multidimensionnels. Astrisque 268, 1198 (2000)


40/40


13. Friedrichs, K.O., Lax, P.D.: Boundary value problems for first order operators.Comm. Pure Appl. Math. 18, 355388 (1965)

14. Glimm, J., Majda, A.:Multidimensional Hyperbolic Problems and Computations, TheIMA Volumes in Mathematics and its Applications, Vol. 29. Springer, New York, 1991

15. Gus, O.: Problme mixte hyperbolique quasi-linaire caractristique. Commun. Partial

Differ. Eqs. 15, 595645 (1990)16. Gus, O., Mtivier, G., Williams, M., Zumbrun, K.: Existence and stability of

multidimensional shock fronts in the vanishing viscosity limit. Arch. Ration. Mech.Anal. 175, 151244 (2005)

17. Hrmander, L.: The boundary problems of physical geodesy, Arch. Ration. Mech.Anal. 62, 152 (1982)

18. Kreiss, H.-O.: Initial boundary value problems of hyperbolic systems. Comm. PureAppl. Math. 23, 277296 (1970)

19. Lions, J.L., Magenes, E.: Non-Homogeneous Boundary Value Problems and Applica-tions, Vol. 13, Springer, New York, 1972

20. Majda, A.: Compressible Fluid Flow and Systems of Conservation Laws in Several

Space Variables. Springer, New York, 198421. Mtivier, G.: Stability of multimensioanl shocks. In:Advances in The Theory of Shock

Waves, Vol. 47, pp. 25103. PNDEA, Birkhuser, Boston, 200122. Miles, J.W.: On the reflection of sound at an interface of relative motion. J. Acoust.

Soc. Am. 29, 226228 (1957)23. Miles, J.W.: On the disturbed motion of a plane vortex sheet.J. Fluid Mech. 4, 538552

(1958)24. Smart, L.L., Norman, M.L., Winkler, K.A.: Shocks, interfaces, and patterns in

supersonic jets. Physica 12D, 83106 (1984)25. Trakhinin, Y.: Existence of compressible current-vortex sheets: variable coefficients

linear analysis. Arch. Ration. Mech. Anal. 177, 331366 (2005)

26. Van Dyke, M.: An Album of Fluid Motion. Parabolic, Stanford, 198227. von Neumann, J.: Collected Works, Vol. 5. Pergamon, New York, 1963

School of Mathematical Sciences, Fudan University,Shanghai 200433, Peoples Republic of China

and

Department of Mathematics,Northwestern University,

2033 Sheridan Road, Evanston, IL 60208, USA.

e-mail: [email protected]

and

Department of Mathematics,Shanghai Jiao Tong University,

Shanghai 200240, Peoples Republic of China.e-mail: [email protected]

(Received June 29, 2007 / Accepted July 2, 2007)Published online October 23, 2007 Springer-Verlag (2007)

gui-qiang chen and ya-guang wang- existence and stability of compressible current-vortex sheets in...

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