The Tricomi problem for a genuinely nonlinear Lavrentiev–Bitsadze equation of mixed type

J. Math. Anal. Appl. 398 (2013) 303–314

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The Tricomi problem for a genuinely nonlinear Lavrentiev–Bitsadze equation of mixed type Zhenguo Feng School of Mathematical Sciences, Fudan University, Shanghai, 200433, PR China

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Article history: Received 5 March 2012 Available online 4 September 2012 Submitted by Dehua Wang Keywords: Nonlinear mixed-type equation Tricomi problem Lavrentiev–Bitsadze equation

abstract In this paper, we consider the Tricomi problem for a genuinely nonlinear Lavrentiev– Bitsadze mixed-type equation

∂ 2u ∂ 2u + (sgn y)(1 + u2x ) 2 = 0, ∂ x2 ∂y whose coefficients depend on the first-order derivative of an unknown function. We prove the existence of a solution to this problem by a method which can be used to study more difficult problems for nonlinear mixed-type equations arising in gas dynamics. © 2012 Elsevier Inc. All rights reserved.

1. Introduction Equations of elliptic–hyperbolic mixed type arise naturally in gas dynamics [1,2]. One of the earliest equations of mixed type appeared in 1902, given by Chaplygin [3], in the following form: K (y)

∂ 2u ∂ 2u + 2 = 0, ∂ x2 ∂y

(1.1)

and it was later called the Chaplygin equation. Here K (y) is positive for y > 0 and negative for y < 0; therefore, Eq. (1.1) is elliptic for y > 0 and hyperbolic for y < 0. Boundary value problems for partial differential equations of mixed type were initiated in 1923 by F. G. Tricomi [4]. Consider the equation y

∂ 2u ∂ 2u + 2 =0 ∂ x2 ∂y

(1.2)

which corresponds to the case K (y) = y in (1.1), in a domain D bounded by a smooth arc g1 with endpoints A = (0, 0) and B = (1, 0) (this arc lies in the half-plane y > 0 outside these endpoints) and two characteristic curves g2 and g3 : g2 : x +

2 3 2

3

(−y) 2 = 1, 3

(−y) 2 = 0 3 for (1.2); F. G. Tricomi proved that there exists a unique solution satisfying Eq. (1.2) and the prescribed continuous boundary values on g1 and g3 . Eq. (1.2) was named the Tricomi equation by his successors, in view of his contributions to this area, while the related problem was called the Tricomi problem. g3 : x −

E-mail address: [email protected]. 0022-247X/$ – see front matter © 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2012.08.045

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Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

We emphasize the exceptional importance of the above-mentioned paper of Chaplygin, which was weakly understood for a long time. Only when the compressibility of air had been taken into account in aviation problems because of the high speeds achieved in the 1930s did this paper become the foundation of many works in gas dynamics. As is known, the two-dimensional, steady flow of inviscid fluid can be described by the system

 ∂(ρv1 ) ∂(ρv2 )   + = 0,  ∂x ∂ x2  1  (1.3) ∂ ∂ 1 ∂p    v1 + v2 vi + = 0, i = 1, 2, ∂ x1 ∂ x2 ρ ∂ xi − → where v = (v1 , v2 ) is the velocity vector, and ρ and p are the density and pressure respectively. If the flow is irrotational − → and isentropic, we can introduce the potential function φ by means of the formula ∇φ = v ; system (1.3) can be written in the form of the classical equation for gas dynamics:

 2

c −



∂φ ∂ x1

2 

    ∂ 2φ ∂φ 2 ∂ 2 φ ∂φ ∂φ ∂ 2 φ 2 + c − −2 = 0, ∂ x1 ∂ x2 ∂ x1 ∂ x2 ∂ x2 ∂ x21 ∂ x22 − →

(1.4)

− →

where c is the speed of sound. It is easy to see that Eq. (1.4) is elliptic if | v | < c and hyperbolic if | v | > c. As is mentioned in [1,2] and references therein, by introducing the stream function ψ satisfying ∇ψ = (−ρv2 , ρv1 )  and polar coordinates (q, θ ) through the formulas q =

v12 + v22 , θ = arctan( vv21 ), the system (1.3) can be reduced to the

Chaplygin equation

∂ 2ψ ∂ 2ψ + =0 ∂θ 2 ∂σ 2  ρ 2 q by defining σ = − q and M = c , where K (σ ) = 1−ρM and M is the Mach number. 2 When K (y) = sgn y in (1.1), the Chaplygin equation (1.1) becomes the Lavrentiev–Bitsadze equation K (σ )

(1.5)

uxx + (sgn y)uyy = 0.

(1.6)

Like the well-known Tricomi equation, it is a model equation of mixed type. In 1950, Lavrentiev and Bitsadze [5] initiated the work on boundary value problems for Eq. (1.6). A complete investigation of the Tricomi problem and its generalizations for Eq. (1.6) was carried out by Bitsadze in [6–10]. Two applications of boundary value problems for the Lavrentiev–Bitsadze equation to gas dynamics are presented in [11]. Recently, in the study of the stability of wave structure in E–H-type Mach reflection, it was found that the E–H Mach configuration could not be described by (1.6), but could be reduced to a nonlinear equation of mixed type with discontinuous coefficients which has some similarity to the nonlinear Lavrentiev–Bitsadze equation; we refer the reader to [12,1] for details. As is said in [1], if the flow behind the reflected shock front is supersonic, then the model allows for nonlinear elliptic–hyperbolic equations with coefficients which are discontinuous along the slip line of the flow. Because of this, the author in [13] considered the following mixed-type equation: uxx + (sgn u)uyy = 0,

(1.7)

where the locus of transition of the type is u = 0. In the hyperbolic region, Eq. (1.7) becomes the d’Alembert equation; then we can reduce the problem in [13] to the familiar form by using the d’Alembert formula. In this paper, we will study the boundary value problem for the following mixed-type equation: 2 ∂ 2u 2 ∂ u + ( sgn y )( 1 + u ) = 0. x ∂ x2 ∂ y2

(1.8)

Eq. (1.8) is a nonlinear mixed-type equation and will change its type through y = 0. Here, since the equation in the hyperbolic region is the nonlinear hyperbolic equation of second order, we cannot directly use the method developed in [13]. Therefore, we call Eq. (1.7) pseudo-nonlinear, and call Eq. (1.8) genuinely nonlinear. As is said in [13], Eq. (1.7) is a prototype of mixed-type equations arising from the stability of the Mach configuration; therefore, Eq. (1.8) under review can be seen as another typical example in connection with Mach reflection. This is the motivation and background of our study in this paper. Here, we will consider the following Tricomi problem: u satisfies (1.8), in Ω , u = β(x), on Γ0 , u = ϕ(x), on Γ1



(1.9)

Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

305

in a neighborhood of a specific solution u(x, y) = y, where Ω is enclosed by a curve, Γ0 : y = γ0 (x), and two characteristic curves, Γ1 : y = γ1 (x) and Γ2 : y = γ2 (x), of Eq. (1.8) starting from (0, 0) and (σ , 0) respectively such that Γ1 and Γ2 meet at (x, y), where the rightward characteristic Γ1 starting from the origin is

  dy 

 = − 1 + u2x ,

(1.10)

dx y|x=0 = 0,

and the leftward characteristic Γ2 starting from the (σ , 0) is

  dy



1 + u2x , dx  y|x=σ = 0.

=

(1.11)

We emphasize that Γ1 and Γ2 depend on the solution u, so the domain Ω bounded by Γ0 , Γ1 , Γ2 also depends on the solution u. Throughout this paper, we assume that

 ϕ(x) ∈ C 1,α0 [0, σ ], β(x) ∈ C 1,α0 [0, σ ], γ (x) ∈ C 1,α0 [0, σ ],   0 γ0 (0) = γ0 (σ ) = β(0) = β(σ ) = ϕ(0), ′ γ0′ (σ ) < 0, γ0 (x) > 0 (0 < x < σ ),  γ0 (0) > 0, ∥γ0 − β∥C 1,α0 [0,σ ] < ε, ∥ϕ − γ1 ∥C 1,α0 [0,σ ] < ε,

(1.12)

where 0 < α0 < 1. The aim of this paper is to prove the existence of a solution to the problem (1.9). The main result is: Theorem 1.1. Assume that the functions γ0 (x), β(x), ϕ(x) satisfy the condition (1.12), and ε > 0 and σ > 0 are sufficiently small; then for the problem (1.9) there exists a unique solution in C 1,λ (Ω ), where 0 < λ 6 α0 depends on the given data. The main idea for proving Theorem 1.1 reads as follows. First, we take an appropriate function v , and replace the 1 + u2x in (1.8) with 1 + vx2 ; then the problem (1.9) becomes a linear problem which we are familiar with. Second, we use the Riemann method to give the formal solution in the hyperbolic region, and reduce the linear problem to a mixed boundary value problem in the elliptic region at the same time. Next, by the existence of the solution in the elliptic region, we can determine the solution in the whole domain to the linearized boundary value problem. Finally, by using the estimates of the linearized boundary value problem, we can prove that the problem (1.9) has a solution. We believe that the method developed in this paper could be used to study more difficult problems for nonlinear mixed-type equations arising in gas dynamics. 2. The linear boundary value problem First, we introduce a function v ∈ C 1,Λ satisfying

 v(0) = v(σ ) = 0, ∥v − y∥C 1,Λ ≤ δ <

1 2

(2.1)

,

where 0 < Λ < 1 will be determined later. In order to solve the problem (1.9), we consider the following linear problem:

 2 2 ∂ u  2 ∂ u   2 + (sgn y)(1 + vx ) 2 = 0, ∂x ∂y  u = β(x), on Γ0 ,   (v) u = ϕ (v) (x), on Γ1 ,

in Ω (v) , (2.2)

(v)

: y = γ1(v) (x) is the rightward characteristic starting from the origin defined by    dy = − 1 + vx2 , dx  y|x=0 = 0,

where Γ1

(v)

Ω (v) is enclosed by the curve Γ0 : y = γ0 (x) and two characteristic curves Γ1 equation in (2.2), and ϕ (v) ∈ C 1,α0 satisfies

(2.3)

: y = γ1(v) (x) and Γ2(v) : y = γ2(v) (x) of the

∥ϕ (v) − γ1(v) ∥C 1,α0 < ε. For convenience, in this section and Section 3, we write ϕ

(2.4) (v)

(v)

(v)

(x) as ϕ(x), γ1 (x) as γ1 (x), Γ1

as Γ1 , and Ω

(v)

as Ω .

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Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

Making the transformation of variables T : ξ = φ1 (x, y),

η = φ2 (x, y),

where φ1 , φ2 satisfy

(φix )2 − (1 + vx2 )(φiy )2 = 0,

i = 1, 2,

then we have that φ1 (x, y), φ2 (x, y) also satisfy

  φ1x  2   φ = − 1 + vx , 1y  φ2x    = 1 + vx2 . φ2y

(2.5)

∂(φ ,φ )

Writing J = ∂(1x,y)2 = φ1x φ2y − φ2x φ1y , by (2.5) we have J = φ1x φ2y − φ2x φ1y = 2φ1x φ2y = −2φ2x φ1y ̸= 0. By the implicit function theorem, the transformation inverse to T exists: T −1 : x = ψ1 (ξ , η),

y = ψ2 (ξ , η)

satisfying

 ∂ψ1 φ2y 1   = = ,  ∂ξ J 2φ1x ∂ψ2 φ2x 1   =− = ,  ∂ξ J 2φ1y

∂ψ1 φ1y 1 =− = , ∂η J 2φ2x ∂ψ2 φ1x 1 = = , ∂η J 2φ2y

(2.6)

and thus u(x, y) = u(ψ1 (φ1 (x, y), φ2 (x, y)), ψ2 (φ1 (x, y), φ2 (x, y))). Write U (ξ , η) = u(ψ1 (ξ , η), ψ2 (ξ , η)); then u(x, y) = U (φ1 (x, y), φ2 (x, y)). By such a transformation of variables, the equation in (2.2) y = 0 can be reduced to L(U ) = Uξ η − aUξ − bUη = 0, where U (ξ , η) is a new unknown function, a=−

φ1xx − (1 + vx2 )φ1yy φ1xx φ1yy =− − , 2(φ1x φ2x − (1 + vx )2 φ1y φ2y ) 4φ1x φ2x 4φ1y φ2y

b=−

φ2xx − (1 + vx2 )φ2yy φ2xx φ2yy =− − , 2(φ1x φ2x − (1 + vx )2 φ1y φ2y ) 4φ1x φ2x 4φ1y φ2y

and accordingly, in the (ξ , η) coordinates,

 (ψ1ξ ψ1η )ξ (ψ2ξ ψ2η )ξ   aˆ = 4(ψ )2 + 4(ψ )2 , 1ξ 2ξ (ψ ψ (ψ ψ )  1 ξ 1 η η 2 ξ 2η )η  bˆ = + . 4(ψ1η )2 4(ψ2η )2

(2.7)

Let Γ1 be given according to φ2 (x, y) = C2 ; therefore, the problem (2.2) for y = 0 becomes



ˆ η = 0, Uξ η − aˆ Uξ − bU U |η=C2 =  ϕ (ξ ),

(2.8)

where  ϕ (ξ ) = U (ξ , C2 ) = u(ψ1 (ξ , C2 ), ψ2 (ξ , C2 )) = ϕ(ψ1 (ξ , C2 )). Under the transformation T , y = 0 becomes the curve

 ξ = φ1 (t , 0) = α1 (t ), η = φ2 (t , 0) = α2 (t ), where 0 6 t 6 σ .

(2.9)

Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

307

Here, we take

 ξ = φ1 (x0 , 0), η0 = φ2 (0, 0) = C2 ,   0 R = (φ1 (0, 0), φ2 (0, 0)) = (φ1 (0, 0), C2 ),  Q = (φ1 (x0 , 0), φ2 (x0 , 0)), P = (ξ0 , η0 ) = (φ1 (x0 , 0), φ2 (0, 0)) = (φ1 (x0 , 0), C2 ).

(2.10)

Then, by (A.8) of the Appendix, we get U (P ) =

1 2

UW (Q ) +



x0

B(α1 (s), α2 (s))α1′ (s) − A(α1 (s), α2 (s))α2′ (s) ds,





(2.11)

0

where α1 , α2 are defined by (2.9), and A, B by (A.3). By (A.3), we get

(B(ξ , η)α1′ (s) − A(ξ , η)α2′ (s))|(ξ ,η)=(φ1 (s,0),φ2 (s,0)) =

W

{Uξ φ1x (s, 0) − Uη φ2x (s, 0)}   ˆ Wξ + 2bW Wη + 2aˆ W ′ ′ − α1 (s) − α2 (s) u(s, 0) 2

2

=

W



Uξ φ1y

2

 − =

2

φ1x φ2x − Uη φ2y φ1y φ2y

 

ˆ Wξ + 2bW Wη + 2aˆ W ′ α1′ (s) − α2 (s) u(s, 0) 2 2

W

  {−Uξ φ1y 1 + vx2 − Uη φ2y 1 + vx2 } 2   ˆ Wξ + 2bW Wη + 2aˆ W ′ ′ − α1 (s) − α2 (s) u(s, 0) 2

=−

2

 W 2

 −

1 + vx2 (s, 0)uy (s, 0)



ˆ Wξ + 2bW Wη + 2aˆ W ′ α1′ (s) − α2 (s) u(s, 0), 2 2

(2.12)

where W = W (φ1 (s, 0), φ2 (s, 0), φ1 (x0 , 0), C2 ) depends on s and x0 . Writing

  W (φ1 (s, 0), φ2 (s, 0), φ1 (x0 , 0), C2 )   1 + vx2 (s, 0), g (s, x0 ) = −  2   ˆ Wξ + 2bW Wη + 2aˆ W ′  ′  α1 (s) − α2 (s) ,  e(s, x0 ) = − 2 2

(2.13)

(2.12) then becomes

(B(ξ , η)α1′ (x0 ) − A(ξ , η)α2′ (x0 ))|(ξ ,η)=(φ1 (x0 ,0),φ2 (x0 ,0)) = g (x0 , x0 )uy (x0 , 0) + e(x0 , x0 )u(x0 , 0).

(2.14)

By calculating directly, we obtain

(WU (Q ))x0 = (W (Q ))x0 u(x0 , 0) + W (Q )ux (x0 , 0), (U (P ))x0 = ϕ ′ (ψ1 (φ1 (x0 , 0), C2 ))

∂ψ1 (φ1 (x0 , 0), C2 ) φ1x . ∂ξ

(2.15) (2.16)

We take x0 = x, and set

 W (Q )  fˆ (x, x) = ,    2 gˆ (x, x) = g (x, x),    eˆ (x, x) = (W (Q ))x + e(x, x), 2

(2.17)

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Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

then differentiate (2.11) with respect to x and combine (2.14)–(2.16) with (2.17); we get fˆ (x, x)ux (x, 0) + gˆ (x, x)uy (x, 0) + eˆ (x, x)u(x, 0) +

x



gx (s, x)uy (s, 0) + ex (s, x)u(s, 0) ds





0

∂ψ1 (φ1 (x, 0), C2 ) = ϕ ′ (ψ1 (φ1 (x, 0), C2 )) φ1x , ∂ξ

(2.18)

and from (A.9), we get W0 (Q ) 2

ux (x, 0) −

 −

W0 (Q ) 2



1 + vx2 uy (x, 0)

ˆ 0 ′ W0ξ + 2bW W0η + 2aˆ W0 ′ (W0 (Q ))x α1 (x) − α2 (x) − 2 2 2

 u(x, 0)

x



gx (s, x)uy (s, 0) + ex (s, x)u(s, 0) ds



+



0

= ϕ ′ (ψ1 (φ1 (x, 0), C2 ))

∂ψ1 (φ1 (x, 0), C2 ) φ1x . ∂ξ

(2.19)

From (A.12), we know that −

W0 (Q ) = e

 φ2 (x,0) C2

aˆ (φ1 (x,0),t )dt

,

and, therefore,



(W0 (Q ))x = W0 (Q ) −ˆa(Q )α2 (x) − ′

−

 ˆaξ (φ1 (x, 0), t )dt α1′ (x)

C2

= W0ξ α1′ (x) + W0η α2′ (x), 

φ2 (x,0)



(2.20)

ˆ 0 ′ W0η + 2aˆ W0 ′ (W0 (Q ))x W0ξ + 2bW α1 (x) − α2 (x) − 2 2 2



  ˆ 0 α1′ (x) − (W0η + aˆ W0 )α2′ (x) = − bW ˆ 0 α1′ (x), = −bW

(2.21)

and then (2.19) becomes ux (x, 0) −

=



2 W0 (Q )

1 + vx2 uy (x, 0) +

2 W0 (Q )

ϕ ′ (ψ1 (φ1 (x, 0), C2 ))

x



{gx (s, x)uy (s, 0) + ex (s, x)u(s, 0)}ds 0

∂ψ1 (φ1 (x, 0), C2 ) φ1x + 2bα1′ (x)u(x, 0). ∂ξ

(2.22)

Then, the problem (2.2) above y = 0 can be reduced to

 2 ∂ 2u ∂ u    2 + (1 + vx2 ) 2 = 0, in S , ∂x ∂y  u = β( x ), on Γ 0,   u satisfies (2.22), on Σ ,

(2.23)

where S = Ω ∩ {y > 0}, Σ = {(x, y) : y = 0, 0 < x < σ }. 3. The solution to the linear boundary value problem The boundary value problem (2.23) is a mixed boundary value problem with a nonlocal boundary condition; by applying the results in [14,15], we can obtain the following lemma: Lemma 3.1. Suppose that the conditions of Theorem 1.1 hold, and v ∈ C 1,Λ (Ω ) satisfies (2.1); then there is a constant λ1 > 1 such that if 0 < λ < λ1 − 1, λ 6 α0 , the problem (2.23) has a unique solution u ∈ C 1,λ (S ) satisfying

∥u − y∥C 1,λ (S ) ≤ δ <

1 2

,

where λ1 depends on the given data.

(3.1)

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309

Before proving Lemma 3.1, we introduce a weighted Hölder space as in [16]. Let H0 (S ) denote the space C (S ) of continuous functions with the norm

|u|0 = sup |u(x)|. x∈S

If k < a 6 k + 1 for a nonnegative integer k, let Ha (S ) denote the set of functions in S that are k times continuously differentiable and such that



|u|a =

|Dα u|0 +

|α|6k



sup |Dα u(x) − Dα u(y)||x − y|k−a

|α|=k x,y∈S

is finite. Notice that the notation ∥u∥k,α , C k,α (S ) is usually used instead of |u|k+α , Hk+α (S ) for 0 < α 6 1, but Ha (S ) ̸= C a (S ) when a > 0 is an integer. Setting Sσ = {x ∈ S : dist(x, ∂ S ) > σ }, (b)

let Ha (S ) denote the set of functions in S belonging to Ha (Sσ ) for all σ > 0 and with finite norm

|u|(ab,S) = |u|(ab) = sup σ a+b |u|a,Sσ , σ >0

(b)

where a + b > 0. By [16], we know that the functions in Ha (S ) have the following property: (b′ ) a

(b)

Monotonicity |u|a 6 C |u|

(b) (b) if b > b′ , or |u|a′ 6 C |u|a if 0 6 a′ 6 a, a′ + b > 0 and b 6 0 is not an integer. (−b)

This monotonic property implies that Ha (−b)

⊂ Hb(−b) = Hb for a > b > 0, b a noninteger, so the upper and lower indices

in the norm |u|a describe, respectively, the global and interior regularity of u. Moreover, when we prove Lemma 3.1, we will use the main result in [14]. Here, we will state it as a lemma as follows. Lemma 3.2 (Lieberman, Theorem 1 in [14]). Define Lu = aij Dij u + bi Di u + cu, Mu = β Di u + γ u, i

in S ,

on Σ ,

where i, j = 1, 2. Suppose that aij , bi , c ∈ Hα ,

β i , γ ∈ H1+α ,

c 6 0,

γ 6 −γ0 ,

where γ0 is a positive constant. Then the problem Lu = f1

in S ,

Mu = f2

on Σ ,

u = f3

on ∂ S \ Σ

(3.2)

has a unique solution u ∈ C 2 (S ∪ Σ ) ∩ C (S ) for all bounded f1 ∈ Hα ,

f2 ∈ H1+α ,

f3 ∈ C (∂ S \ Σ ).

(3.3)

Moreover, when ∂ S \ Σ is nonempty, the result also holds even if the inequality γ ≤ −γ0 < 0 is relaxed to γ 6 0. Remark 3.3. If the conditions β i ∈ H1+α , f2 ∈ H1+α in Lemma 3.2 are relaxed to β i ∈ Hα , f2 ∈ Hα , we can get that the problem (3.2) has a unique solution u ∈ C 2 (S ) ∩ C 1 (S ∪ Σ ) ∩ C (S ). To confirm the weaker result under the relaxed assumptions given in the remark, we should check the details of the proof of Theorem 1 in [14]. Indeed, the existence theory established in [14] is based on the Perron process for oblique derivative problems in [17]. Besides the basic idea of the Perron process, the existence theorem obtained in [17] relies on an estimate for the solution to Lu = f1

in S ,

Mu = f2

on Σ .

(3.4)

The estimate is that under the assumptions of Lemma 3.2, b ≥ −2 − α , any solution u ∈ C (S ∪ Σ ) satisfies 2

) +b ) |u|(2b+α ≤ C1 (sup |db u| + |f1 |(α2+b) + |f2 |(11+α ).

(3.5)

The estimate follows the usual Schauder estimates for the elliptic derivative problem (see [18, Lemma 6.29]). By the intermediate Schauder estimates for oblique derivative problems given in [19], we can prove that even if β i only belongs to Hα , any solution u ∈ C 2 (S ) ∩ C 1 (S ∪ Σ ) ∩ C (S ) satisfies the estimate ) |u|(2b+α ≤ C1 (sup |db u| + |f1 |(α2+b) + |f2 |(α2+b) ).

(3.6)

Combining the proof of Theorem 1 in [14] with the above additional explanation for the Perron process for the oblique derivative problem, we are led to the weaker result given in this remark.

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Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

Proof of Lemma 3.1. A usual characteristic of (2.22) is the nonlocalness of the boundary condition; we consider the linear condition defined as follows: u(xn) (x, 0) −



∂ψ1 (φ1 (x, 0), C2 ) φ1x W0 (Q ) ∂ξ  x   gx (s, x)u(yn−1) (s, 0) + ex (s, x)u(n−1) (s, 0) ds.

1 + vx2 u(yn) (x, 0) = 2bα1′ (x)u(n−1) (x, 0) +

−

2 W0 (Q )

2

ϕ ′ (ψ1 (φ1 (x, 0), C2 ))

(3.7)

0

We can start from u(0) = y to establish a sequence of solutions u(n) to the problem

 2 (n) ∂ u ∂ 2 u(n)   + (1 + vx2 ) = 0,  2 ∂x ∂ y2 (n)   u(n) = β(x) on Γ0 , u satisfies (3.7), on Σ .

in S , (3.8)

Set w (n) = u(n) − y; then the condition (3.7) becomes



wx(n) (x, 0) −

1 + vx2 wy(n) (x, 0) = Ψ (x),

(3.9)

where

 ∂ψ1 (P ) φ1x + 1 + vx2 W0 (Q ) ∂ξ  x   gx (s, x)u(yn−1) (s, 0) + ex (s, x)u(n−1) (s, 0) ds.

Ψ (x) = 2bα1′ (x)u(n−1) (x, 0) +

−

2 W0 (Q )

2

ϕ ′ (ψ1 (P ))

(3.10)

0

Then the problem (3.8) is reduced to

 2 (n) ∂ w ∂ 2 w (n)   + (1 + vx2 ) = 0,  2 ∂x ∂ y2 (n)   w(n) = β(x) − γ0 (x), on Γ0 , w satisfies (3.9), on Σ .

in S , (3.11)

The boundary value problem (3.11) is a mixed boundary value problem; from Remark 3.3, we know that the problem (3.11) has a unique solution u(1) ∈ C 2 (S ) ∩ C 1 (S ∪ Σ ) ∩ C (S ). To establish the estimate (3.1), we use some results on the optimal Hölder regularity for mixed boundary value problems given in [15]. Here, Σ ∈ C 1 , ∂ S \ Σ ∈ C 1,α0 . Notice that for the domain S, there exists a number η ∈ (θi , π ) such that



1 6 − 1 + vx2 cot η. (−λ)

Then by Theorem 4 and the remark on Theorem 4 in [15], we can give an estimate of |u|a type. Here, we can take a = 2 + Λ and find a constant λ1 > 1 depending on θi (i = 1, 2) and η, such that the solution w (n) ∈ C 2 (S )∩ C 1 (S ∪ Σ )∩ C (S ), satisfying 1) 6 C (∥Ψ ∥C α0 + ∥β − γ0 ∥C 1,α0 ), |w (n) |(−λ− 2 +Λ

(3.12)

where θ1 > 0 and θ2 > 0 are the angles between Γ0 and y = 0 at (0, 0) and (σ , 0). Write

    F ( x) = −

2

x



{gx (s, x)u(yn−1) (s, 0) + ex (s, x)u(n−1) (s, 0)}ds,  ∂ψ1 (P )   G(x) = ϕ (ψ1 (P )) φ1x + 1 + vx2 ; W0 (Q ) ∂ξ

(3.13)

Ψ (x) = 2bα1′ (x)u(n−1) (x, 0) + F (x) + G(x).

(3.14)

W0 (Q ) 2 ′

0

then

In the following, we will prove that

∥Ψ ∥C α0 6 δ.

Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

First, for the term F (x), we take the term

 x    (n−1)   g ( s , x ) u ( s , 0 ) ds x y   0

C α0

x

(n−1)

(s, 0)ds as an example for proving the result:  x    = sup  gx (s, x)uy(n−1) (s, 0)ds 06x6σ 0   x  x1  (n−1) (s, 0)ds − 0 2 gx (s, x2 )u(yn−1) (s, 0)ds  0 gx (s, x1 )uy + sup |x1 − x2 |α0 06x1 ,x2 6σ   x   6 sup  gx (s, x)uy(n−1) (s, 0)ds 06x6σ 0    x1  (n−1) (s, 0) − gx (s, x2 )uy(n−1) (s, 0) ds  0 gx (s, x1 )uy + sup |x1 − x2 |α0 06x1 ,x2 6σ    x2  (n−1) (s, 0) ds  x1 gx (s, x2 )uy + sup , |x1 − x2 |α0 06x1 ,x2 6σ 0

311

gx (s, x)uy

(3.15)

and by (2.13) and the properties of the function W introduced in the Appendix, and as ∥g ∥C 1,α0 is bounded, we can get

 x    (n−1)   g ( s , x ) u ( s , 0 ) ds x y   0

C α0

6 δ,

(3.16)

when σ is sufficiently small. For the term 2bα1′ (x)u(n−1) (x, 0), we can get the result obviously when σ is sufficiently small. Last, because



1 + vx2 = −

φ1x , φ1y

from (3.13), we only consider the term 2 W0 (Q )

ϕ ′ (ψ1 (P ))

∂ψ1 (P ) 1 − . ∂ξ φ1y

Under the condition (2.4), we only prove

   2 1  ∂ψ1 (P ) ′   6 δ, γ − (ψ ( P ))  W (Q ) 1 1 ∂ξ φ1y C α0 0 and using (2.3) and (2.5), 2 W0 (Q )

γ1′ (ψ1 (P ))

1 2 φ1x (ψ1 (φ1 (x, 0), C2 )) ∂ψ1 (φ1 (x, 0), C2 ) 1 ∂ψ1 (P ) − = − , ∂ξ φ1y W0 (Q ) φ1y (ψ1 (φ1 (x, 0), C2 )) ∂ξ φ1y (x, 0)

(3.17)

and from (2.6), when σ is sufficiently small, we can get the result. By the monotonic property of weighted Hölder space and (1.12), (3.12), when ε > 0 is sufficiently small, problem (3.11) has a unique solution w (n) ∈ C 1,λ (S ) and satisfies

∥w (n) ∥C 1,λ (S ) ≤ δ <

1 2

;

(3.18)

then the problem (3.8) has a unique solution u(n) ∈ C 1,λ (S ) and satisfies

∥u(n) − y∥C 1,λ (S ) ≤ δ <

1 2

.

(3.19)

Therefore, there is a unique function u ∈ C 1,λ (S ) such that for any 0 < α < λ, 1) |u(n) − u|(−α− → 0, 2+α

as n → +∞.

Obviously, u satisfies the equation and the Dirichlet boundary condition and the nonlocal boundary condition in (2.23); from (3.19), u still satisfies (3.1). Hence the lemma is proved.  Remark 3.4. From the proof of Lemma 3.1, we can choose Λ = λ in (2.1).

312

Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

Obviously, from Lemma 3.1 and (2.11), we have the following lemma: Lemma 3.5. Suppose that the conditions of Theorem 1.1 hold, and v ∈ C 1,Λ satisfies (2.1); then there is a constant λ1 > 1 such that if 0 < λ < λ1 − 1, λ 6 α0 , the problem (2.2) has a unique solution u ∈ C 1,λ (Ω ) satisfying

∥u − y∥C 1,λ (Ω ) ≤ δ <

1 2

,

(3.20)

where λ1 is defined in Lemma 3.1. 4. Proof of the main theorem Now we prove the main result— Theorem 1.1. According to Section 2, we can start from u(0) = y to establish a sequence of solutions u(n) to the problem

 2 (n) 2 (n) ∂ u (n−1) 2 ∂ u   + ( sgn y )( 1 + ( u ) ) = 0,  x ∂ x2 ∂ y2 (n)   u(n) = β((nx−),1) on Γ0 , (n−1) u =ϕ (x), on Γ1 ,

in Ω (n−1) , (4.1)

(n−1)

: y = γ1(n−1) (x) is the rightward characteristic starting from the origin defined by    dy = − 1 + (u(xn−1) )2 , dx  y|x=0 = 0,

where Γ1

(4.2)

ϕ (n−1) ∈ C 1,α0 satisfies ∥ϕ (n−1) − γ1(n−1) ∥C 1,α0 < ε,

(4.3) (n−1)

and Ω (n−1) depending on u(n−1) is enclosed by the curves Γ0 , Γ1 From Lemma 3.5, for any n, there exists a unique solution u

∥u(n) − y∥C 1,λ (Ω (n−1) ) ≤ δ <

1 2

.

(n)

, Γ2(n−1) . ∈ C 1,λ (Ω (n−1) ) satisfying (4.4)

(n−1)

, Γ2(n−1) , for ∀n, the domain Ω (n−1) contains the domain Ω + bounded by Γ0 and y = 0; therefore, there exists a function u ∈ C 1,λ (Ω + ) such that for any 0 < α < λ,

By the properties of Γ1

∥u(n) − u∥C 1,α (Ω + ) → 0,

as n → +∞,

(4.5)

and u also satisfies

∥u − y∥C 1,λ (Ω + ) ≤ δ <

1 2

.

(4.6)

For u above, we know the values of u(x, 0) and uy (x, 0); then we can use the classical theory of hyperbolic equations to

prove that u ∈ C 1,λ (Ω − ) and

∥u − y∥C 1,λ (Ω − ) ≤ δ <

1

(4.7)

2

where Ω − = Ω ∩ {y < 0}. Hence Theorem 1.1 is proved.



Acknowledgments The paper was partially supported by the National Natural Science Foundation of China 11031001 and the Doctoral Foundation of the National Educational Ministry 20090071110002. Appendix. Riemann’s method The Riemann method [20] is used to solve the problem of Cauchy for a linear equation of hyperbolic type; it depends on finding a particular solution, known as the Riemann–Green function, of the characteristic boundary value problem.

Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

313

Let

ˆ η = 0; LU = Uξ η − aˆ Uξ − bU

(A.1)

the adjoint equation of Eq. (A.1) is

ˆ )η = 0, M (W ) = Wξ η + (ˆaW )ξ + (bW

(A.2)

where aˆ , bˆ are defined by (2.7). Set A=

WUη − UWη 2

− aˆ UW ,

B=

WUξ − UWξ 2

ˆ − bUW ;

(A.3)

then Aξ + Bη = WL(U ) − UM (W ).

(A.4)

This identity (A.4) implies that the line integral



{Bdξ − Adη}

I =

(A.5)

vanishes around closed paths lying in the interior of a domain within which the functions U and W are regular solutions of L(U ) = 0 and M (W ) = 0, respectively. P



B(ξ , η0 )dξ −

I = R

Q



A(ξ0 , η)dη + P

R



{Bdξ − Adη} = 0,

(A.6)

Q

where W (ξ , η, ξ0 , η0 ) satisfies

 ˆ )η = 0,  W + (ˆaW )ξ + (bW   ξη ˆ = 0, on η = η0 , Wξ + bW  Wη + aˆ W = 0, on ξ = ξ0 ,  W (ξ0 , η0 , ξ0 , η0 ) = 1,

(A.7)

is called the Riemann–Green function. Then U (P ) =

1 2

(UW (Q ) + UW (R)) +

Q



{Bdξ − Adη}.

(A.8)

R

We try to find a solution of the following form: W =

∞  Γj Wj , j!j! j =0

Γ = (ξ − ξ0 )(η − η0 ).

(A.9)

Differentiating (A.9) term by term, we get M (W ) =

∞ ∞   Γj Γ j −1 M (Wj ) + ((ξ − ξ0 )Wjξ + (η − η0 )Wjη ) j!j! j!(j − 1)! j =0 j =1

+

∞  j =1

∞  Γ j −1 Γ j −1 ((ξ − ξ0 )bˆ + (η − η0 )ˆa)Wj + Wj . j!(j − 1)! (j − 1)!(j − 1)! j =1

Since M (W ) = 0, equating to zero the coefficients of powers of Γ , we obtain

ˆ j + (η − η0 )ˆaWj = 0 jM (Wj−1 ) + jWj + (ξ − ξ0 )Wjξ + (η − η0 )Wjη + (ξ − ξ0 )bW

(A.10)

for j = 1, 2, . . .. The coefficient W0 is at our disposal; once it is fixed, Eq. (A.10) gives successively the coefficients W1 , W2 , . . .. For η = η0 , we have W = W0 and Wξ = W0ξ . Hence

∂ ˆ 0 (ξ , η0 ) = 0. W0 (ξ , η0 ) + bW ∂ξ If we take W0 = 1 at (ξ0 , η0 ), we get W0 (ξ , η0 ) = e

ξ − ξ bˆ (s,η0 )ds 0 .

(A.11)

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Z. Feng / J. Math. Anal. Appl. 398 (2013) 303–314

Similarly, W0 (ξ0 , η) = e

 − ηη aˆ (ξ0 ,t )dt 0

.

(A.12)

Thus we know W0 on the characteristics, and have to assign W0 elsewhere. There are many ways of doing this. We follow Hadamard and define W0 = e

 (ξ ,η) − (ξ ,η ) bˆ (s,t )ds+ˆa(s,t )dt 0 0 ,

(A.13)

where the integration is along the straight line from (ξ0 , η0 ) to (ξ , η). If ξ = ξ0 + r cos θ , η = η0 + r sin θ , we have s = ξ0 + τ cos θ ,

t = η0 + τ sin θ ,

where θ is constant and τ varies from 0 to r. (A.13) becomes W0 = e−

ˆ a(s,t )} dττ 0 {(s−ξ0 )b(s,t )+(t −η0 )ˆ

r

.

(A.14)

If we denote the value of W0 given by (A.14) by Ω0 , we get

Ω0 ∂ Ω0 = (−(ξ − ξ0 )bˆ − (η − η0 )ˆa). ∂r r

(A.15)

Using polar coordinates, (A.10) becomes r

∂ Wj r ∂ Ω0 − Wj + jWj = −jM (Wj−1 ), ∂r Ω0 ∂ r

or

∂ ∂r

 r

j

Wj



Ω0

=−

jr j−1

Ω0

M (Wj−1 ).

Hence since Wj is finite when r = 0, Wj = −

jΩ0 rj

 0

r

τ j −1 M (Wj−1 )dτ , Ω0

(A.16)

where, in M (Wj−1 ), we have s = ξ0 + τ cos θ , t = η0 + τ sin θ . (A.16) determines successively all the coefficients Wj . References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20]

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