Open Access

Diffraction problems for quasilinear parabolic systems with boundary intersecting interfaces

Boundary Value Problems20132013:99

DOI: 10.1186/1687-2770-2013-99

Received: 19 January 2013

Accepted: 9 April 2013

Published: 22 April 2013

Abstract

In this paper, we discuss the n-dimensional diffraction problem for weakly coupled quasilinear parabolic system on a bounded domain Ω, where the interfaces Γ k ( k = 1 , , K 1 ) are allowed to intersect with the outer boundary Ω and the coefficients of the equations are allowed to be discontinuous on the interfaces. The aim is to show the existence of solutions by approximation method. The approximation problem is a diffraction problem with interfaces, which do not intersect with Ω.

MSC:35R05, 35K57, 35K65.

Keywords

diffraction problem quasilinear parabolic system interface approximation method

1 Introduction

Let Ω be a bounded domain in R n with boundary Ω ( n 1 ), and let Ω be partitioned into a finite number of subdomains Ω k ( k = 1 , , K ) separated by Γ k , where Γ k , k = 1 , , K 1 , are interfaces, which do not intersect with each other. For any T > 0 , set
Q T : = Ω × ( 0 , T ] , S T : = Ω × [ 0 , T ] , Γ : = k = 1 K 1 Γ k , Γ T : = Γ × [ 0 , T ] .
In this paper, we consider the diffraction problem for quasilinear parabolic reaction-diffusion system in the form
{ u t l L l ( u l ) = g l ( x , t , u ) ( ( x , t ) Q T ) , [ u l ] Γ T = 0 , [ a i j l ( x , t , u l ) u x j l ν i ( x ) ] Γ T = 0 , u l = ψ l ( x , t ) ( ( x , t ) S T { Ω × { 0 } } ) , l = 1 , , N ,
(1.1)
where x = ( x 1 , , x n ) , u = ( u 1 , , u N ) , u t l : = u l / t , u x i l : = u l / x i , u x l : = ( u x 1 l , , u x n l ) ,
L l ( u l ) : = d d x i ( a i j l ( x , t , u l ) u x j l ) + b j l ( x , t , u l ) u x j l , l = 1 , , N ,
(1.2)

repeated indices i or j indicate summation from 1 to n, ν ( x ) : = ( ν 1 ( x ) , , ν n ( x ) ) is the unit normal vector to Γ (the positive direction of ν ( x ) is fixed in advance), the symbol [ ] Γ T denotes the jump of a quantity across Γ T , and the coefficients a i j l ( x , t , u l ) , b j l ( x , t , u l ) and g l ( x , t , u ) are allowed to be discontinuous on Γ T . In the following, we refer to the conditions on Γ T in (1.1) as diffraction conditions.

The diffraction problems often appear in different fields of physics, ecology, and technics. In some of them, the interfaces are allowed to intersect with the outer boundary Ω (see [15]). The linear diffraction problems have been treated by many researchers (see [110]). For the quasilinear parabolic and elliptic diffraction problems, when all of the interfaces Γ k do not intersect with Ω, the existence and uniqueness of the solutions have been investigated in [1114] by Leray-Schauder principle and the method of upper and lower solutions. In this paper, we investigate the existence of solutions of (1.1) when the interfaces are allowed to intersect with Ω. In this case, because of the existence of the intersection of Γ and Ω, the methods in [1114] can not be extended. We shall show the existence of solutions by approximation method. The approximation problem is a diffraction problem with interfaces which do not intersect with Ω.

The plan of the paper is as follows. In Sect. 2, we give the notations, hypotheses and an example, and state the existence theorem of the solutions. Section 3 is devoted to the proof of the existence theorem.

2 The hypotheses, main result and example

2.1 The notations, hypotheses and main result

First, let us introduce more notations and function spaces.

For any set S, S ¯ denotes its closure. The symbol Ω Ω means that Ω Ω and dist ( Ω , Ω ) > 0 .

Let
{ Γ 1 , , Γ K 1 } = { Γ 1 , , Γ K 0 1 } { Γ 1 , , Γ K K 0 } ,
where Γ k , k = 1 , , K 0 1 , intersect with the outer boundary Ω, and Γ k , k = 1 , , K K 0 do not intersect with Ω. Assume that the domain Ω is partitioned into subdomains Ω k , k = 1 , , K 0 , separated by interfaces Γ k , and partitioned into Ω k , k = 1 , , K K 0 + 1 , separated by Γ k . The interface of Ω k and Ω k + 1 is Γ k . Then Ω ¯ = k = 1 K 0 Ω ¯ k = k = 1 K K 0 + 1 Ω ¯ k . Set
Q k , T : = Ω k × ( 0 , T ] for  k = 1 , , K , Γ : = k = 1 K 0 1 Γ k , Γ : = k = 1 K K 0 Γ k , Γ T : = Γ × [ 0 , T ] , Γ T : = Γ × [ 0 , T ] , Q k , T : = Ω k × ( 0 , T ] for  k = 1 , , K 0 , Q k , T : = Ω k × ( 0 , T ] for  k = 1 , , K K 0 + 1 .

We see that Γ T = Γ T Γ T .

C α ( Q ¯ T ) is the spaces of Hölder continuous in Q ¯ T with exponent α ( 0 , 1 ) . W 2 1 ( Ω ) and W 2 1 , 1 ( Q T ) are the Hilbert spaces with scalar products ( v , w ) W 2 1 ( Ω ) = Ω ( v w + v x i w x i ) d x and ( v , w ) W 2 1 , 1 ( Q T ) = Q T ( v w + v t w t + v x i w x i ) d x d t , respectively. Let
W 2 1 ( Ω ) : = { v W 2 1 ( Ω ) , v | x Ω = 0 } , W 2 1 , 1 ( Q T ) : = { v W 2 1 , 1 ( Q T ) , v | ( x , t ) S T = 0 } .

For the vector functions with N-components we denote the above function spaces by C α ( Q ¯ T ) , W 2 1 ( Ω ) , W 2 1 , 1 ( Q T ) , W 2 1 ( Ω ) and W 2 1 , 1 ( Q T ) , respectively.

Moreover, we recall the following.

Definition 2.1 (see [13, 15])

Write u in the split form
u = ( u l , [ u ] a l , [ u ] b l ) .
The vector function g ( , u ) : = ( g 1 ( , u ) , , g N ( , u ) ) is said to be mixed quasimonotone in B R N with index vector ( a 1 , , a N ) if for each l = 1 , , N , there exist nonnegative integers a l , b l , satisfying
a l + b l = N 1 ,

such that g l ( , u l , [ u ] a l , [ u ] b l ) is nondecreasing in [ u ] a l , and is nonincreasing in [ u ] b l for all u B .

The following hypotheses will be used in this paper:
  1. (H)
    (i) Ω and Γ k , k = 1 , , K 1 , are of C 2 + α 0 for some exponent α 0 ( 0 , 1 ) and there exist θ 0 ( 0 , 1 ) and ρ 0 > 0 such that for every open ball K ρ centered at x 0 Ω and radius ρ ρ 0 ,
    mes ( K ρ Ω ) ( 1 θ 0 ) mes K ρ .
     
Assume that for each k = 1 , , K 0 1 ,
Γ k : φ k ( x ) = 0 ( x Ω ¯ ) ,
and
τ = 1 k Ω ¯ τ Γ k = { x : φ k ( x ) < 0 } Ω ¯ .
(2.1)
  1. (ii)
    Assume that
    { a i j l ( x , t , u l ) = a i j , k l ( x , t , u l ) , b j l ( x , t , u l ) = b j , k l ( x , t , u l ) , g l ( x , t , u ) = g k l ( x , t , u ) ( ( x , t ) Q k , T , u R N ) , k = 1 , , K 0 ,
    (2.2)
     
where a i j , k l ( x , t , u l ) and b j , k l ( x , t , u l ) are defined on Q ¯ T × R , g k l ( x , t , u ) are defined on Q ¯ T × R N , and all of them are allowed to be discontinuous on Γ T .
  1. (iii)
    There exist constant vectors M = ( M 1 , , M N ) and m = ( m 1 , , m N ) , m M , such that
    { g k l ( x , t , M l , [ M ] a l , [ m ] b l ) 0 ( ( x , t ) Q T ) , g k l ( x , t , m l , [ m ] a l , [ M ] b l ) 0 ( ( x , t ) Q T ) , m l ψ l ( x , t ) M l ( ( x , t ) S T { Ω × { 0 } } ) , k = 1 , , K 0 , l = 1 , , N ,
    (2.3)
     
where a l , b l are all independent of k . Let
S : = { u C ( Q ¯ T ) : m u M } .
The vector functions g k ( , u ) = ( g k 1 ( , u ) , , g k N ( , u ) ) , k = 1 , , K 0 , are mixed quasimonotone in S with the same index vector ( a 1 , , a N ) .
  1. (iv)
    For each k = 1 , , K 0 , k = 1 , , K K 0 + 1 , l = 1 , , N , a i j , k l ( x , t , u l ) , b j , k l ( x , t , u l ) C 1 + α 0 ( Q ¯ k , T × R ) ( i , j = 1 , , n ), g k l ( x , t , u ) C 1 + α 0 ( Q ¯ k , T × S ) . There exist a positive nonincreasing function ν ( θ ) and a positive nondecreasing function μ ( θ ) for θ [ 0 , + ) such that
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ6_HTML.gif
    (2.4)
     
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ7_HTML.gif
(2.5)
For each l = 1 , , N , ψ l ( x , t ) C α 0 ( Ξ ¯ × [ 0 , T ] ) W 2 1 , 1 ( Ξ × ( 0 , T ) ) for some domain Ξ with Ω Ξ , ψ l ( x , 0 ) C 2 + α 0 ( Ω ¯ k ) ( k = 1 , , K ), and the following compatibility condition on Γ holds:
[ a i j l ( x , 0 , ψ l ( x , 0 ) ) ψ x j l ( x , 0 ) ν i ( x ) ] Γ = 0 .
(2.6)

Definition 2.2 A function u is said to be a solution of (1.1) if u possesses the following properties: (i) For some α ( 0 , 1 ) , u C α ( Q ¯ T ) C 2 , 1 ( Q k , T ) , k = 1 , , K . For any given Ω Ω and t ( 0 , T ) , there exists α , 0 < α < 1 , such that u t C α ( Ω ¯ × [ t , T ] ) and u x j L 2 ( Q T ) C α ( ( Ω ¯ Ω ¯ k ) × [ t , T ] ) , k = 1 , , K , j = 1 , , n ; (ii) u satisfies the equations in (1.1) for ( x , t ) Q k , T , k = 1 , , K , the diffraction conditions for ( x , t ) Γ T Q T and the parabolic boundary conditions for ( x , t ) S T ( Ω × { 0 } ) .

The main result in this paper is the following existence theorem.

Theorem 2.1 Let Hypothesis (H) hold. Then problem (1.1) has a solution u in S .

2.2 An example

We next give an example satisfying the conditions in Hypothesis (H).

Example 2.1 In problem (1.1), let
n = 2 , φ = ( x 1 ) 2 + ( x 2 ) 2 100 , φ 1 = x 1 + ( x 2 ) 2 + 1 , φ 2 = x 1 ( x 2 ) 2 1 , φ 3 = ( x 1 4 ) 2 + ( x 2 ) 2 1 , Ω : φ = 0 , Γ 1 : φ 1 = 0 ( x 1 I 1 ) , Γ 2 : φ 2 = 0 ( x 1 I 2 ) , Γ 3 : φ 3 = 0 ,
where I 1 = [ ( 9 5 1 ) / 2 , 1 ] and I 2 = [ 1 , ( 9 5 1 ) / 2 ] , and let
Ω : φ < 0 , Ω 1 : φ < 0 , φ 1 < 0 , Ω 2 : φ < 0 , φ 1 > 0 , φ 2 < 0 , Ω 3 : φ < 0 , φ 2 > 0 , φ 3 > 0 , Ω 4 : φ 3 < 0 .
The outer boundary of domain is a circle of radius 10 with the center at the origin, whereas the interface curves are two parabolas and a smaller circle of radius 1 (see Figure 1). We see that Γ 1 and Γ 2 intersect with Ω, and Γ 3 does not.
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Fig1_HTML.jpg
Figure 1

The example of the domain and the interfaces for n = 2 .

For the coefficients of the equations and the boundary values in (1.1) we set
a i j l ( x , t , u l ) = { A k l E l ( u l ) , i = j , 0 , i j ( ( x , t ) Q k , T , u l R ) , k = 1 , 2 , 3 , 4 , i , j = 1 , 2 , b j l ( x , t , u l ) 0 ( ( x , t ) Q T , u l R ) , j = 1 , 2 , g l ( x , t , u ) = r k l u l f k l ( u ) ( ( x , t ) Q k , T , u R N ) , k = 1 , 2 , 3 , 4 , ψ l ( x , t ) o l , l = 1 , , N ,
where
f k l ( u ) = 1 l = 1 N δ l , k l u l for  l = 1 , , N 1 , f k N ( u ) = 1 + l = 1 N 1 δ l , k N u l δ N , k N u N ,

E l ( u l ) C 2 ( R ) with E l ( u l ) ν 0 , and ν 0 , A k l , r k l , δ l , k l and o l are all positive constants for k = 1 , 2 , 3 , 4 , l , l = 1 , , N .

Then
Γ 1 = Γ 1 , Γ 2 = Γ 2 , Ω 1 = Ω 1 , Ω 2 = Ω 2 , Ω 3 : φ < 0 , φ 2 > 0 , Γ 1 = Γ 3 , Ω 1 : φ < 0 , φ 3 > 0 , Ω 2 = Ω 4 .
For each l = 1 , , N , let
a i j , k l ( x , t , u l ) = 0 ( ( x , t ) Q T , u l R ) , i j , i , j = 1 , 2 , k = 1 , 2 , 3 , a i i , k l ( x , t , u l ) = A k l E l ( u l ) ( ( x , t ) Q T , u l R ) , i = 1 , 2 , k = 1 , 2 , a i i , 3 l ( x , t , u l ) = { A 3 l E l ( u l ) ( ( x , t ) Q 1 , T , u l R ) , A 4 l E l ( u l ) ( ( x , t ) Q 2 , T , u l R ) , i = 1 , 2 , g k l ( x , t , u ) = r k l u l f k l ( u ) ( ( x , t ) Q T , u R N ) , k = 1 , 2 , g 3 l ( x , t , u ) = { r 3 l u l f 3 l ( u ) ( ( x , t ) Q 1 , T , u R N ) , r 4 l u l f 4 l ( u ) ( ( x , t ) Q 2 , T , u R N ) .
We find that these functions satisfy (2.2) and the hypothesis (iv) of (H). Set m = ( 0 , , 0 ) . Then the requirements on M in (2.3) become
1 δ l , k l M l 0 , M l o l , l = 1 , , N 1 . 1 + l = 1 N 1 δ l , k N M l δ N , k N M N 0 , M N o N .

It follows from these inequalities that there exist positive constant vector M, such that m and M satisfy (2.3). Furthermore, the vector functions g k ( , u ) = ( g k 1 ( , u ) , , g k N ( , u ) ) , k = 1 , 2 , 3 , are mixed quasimonotone in S with the same index vector ( 0 , , 0 , N 1 ) . The above arguments show that the conditions in Hypothesis (H) can be satisfied.

3 The proof of the existence theorem

3.1 Preliminaries

Lemma 3.1 The following statements hold true:
  1. (i)
    For any given x Ω ¯ , if φ k 0 ( x ) 0 for some k 0 { 1 , , K 0 2 } , then
    φ θ ( x ) < 0 for all θ { k 0 + 1 , , K 0 1 } .
     
  2. (ii)
    There exists a positive number ε 0 such that for any given k { 2 , , K 0 1 } , if 1 θ k 1 , then
    φ θ ( x ) ε 0 for all x { y : φ k ( y ) 0 } Ω ¯ .
     

Proof By (2.1), if x Ω ¯ and φ k 0 ( x ) 0 , then x τ = 1 k 0 Ω ¯ τ . Thus for each θ = k 0 + 1 , , K 0 1 , x τ = 1 θ Ω ¯ τ Γ θ . Again by (2.1) we get φ θ ( x ) < 0 . This proves the result in (i).

For any given k { 2 , , K 0 1 } , if x Ω ¯ and φ k ( x ) 0 , then it follows from (i) that φ θ ( x ) > 0 for all θ { 1 , , k 1 } . Since φ θ C 2 + α 0 , there exist positive constants ε k , θ such that
φ θ ( x ) ε k , θ for all  x { y : φ k ( y ) 0 } Ω ¯ .

Hence, the conclusion in (ii) follows from the above relation by taking ε 0 : = min k , θ ε k , θ . □

For an arbitrary ε, 0 < ε < ε 0 , let s ε = s ε ( θ ) be smooth function with values between 0 and 1 such that | d d θ s ε ( θ ) | C / ε for all θ R , s ε ( θ ) = 1 for θ 0 and s ε ( θ ) = 0 for θ ε . Define
z ε , k ( x ) : = { τ = 1 K 0 1 s ε ( φ τ ( x ) ) ( x Ω ¯ ) , k = 1 , ϑ = 1 K 0 1 [ 1 s ε ( φ ϑ ( x ) ) ] ( x Ω ¯ ) , k = K 0 , τ = k K 0 1 s ε ( φ τ ( x ) ) ϑ = 1 k 1 [ 1 s ε ( φ ϑ ( x ) ) ] ( x Ω ¯ ) , k = 2 , , K 0 1 .
(3.1)
Lemma 3.2 z ε , k ( x ) , k = 1 , , K 0 , are smooth functions with values between 0 and 1, and possess the property
k = 1 K 0 z ε , k ( x ) = 1 ( x Ω ¯ ) .
(3.2)
Let functions η k ( x ) , k = 1 , , K 0 , be defined on Ω ¯ , and let
η ε ( x ) = k = 1 K 0 η k ( x ) z ε , k ( x ) ( x Ω ¯ ) .
(3.3)
Then for any x Ω ¯ ,
η ε ( x ) = { η 1 ( x ) 0 if φ 1 ( x ) 0 , η K 0 ( x ) if φ K 0 1 ( x ) ε , η k ( x ) , if φ k 1 ( x ) ε and φ k ( x ) 0 for some k { 2 , , K 0 1 } , η k 1 ( x ) s ε ( φ k 1 ( x ) ) + η k ( x ) [ 1 s ε ( φ k 1 ( x ) ) ] if 0 < φ k 1 ( x ) < ε for some k { 2 , , K 0 1 } .
(3.4)

Proof Since (3.2) is a special case of (3.4) with η k ( x ) 1 for all k { 1 , , K 0 } , we only prove (3.4).

Case 1. If φ 1 ( x ) 0 , then the conclusion of (i) in Lemma 3.1 implies that φ k ( x ) 0 and s ε ( φ k ( x ) ) = 1 for all k { 1 , , K 0 1 } . (3.1) yields that z ε , 1 ( x ) = 1 and z ε , k ( x ) = 0 for k 2 . These, together with (3.3), imply that η ε ( x ) = η 1 ( x ) .

Case 2. If φ K 0 1 ( x ) ε , then the conclusion of (ii) in Lemma 3.1 shows that φ k ( x ) ε and s ε ( φ k ( x ) ) = 0 for all k { 1 , , K 0 1 } . Hence, z ε , K 0 ( x ) = 1 and z ε , k ( x ) = 0 for all k { 1 , , K 0 1 } . Again by (3.3) we get η ε ( x ) = η K 0 ( x ) .

Case 3. If φ k ( x ) 0 and φ k 1 ( x ) ε for some k { 2 , , K 0 1 } , then Lemma 3.1 yields that φ τ ( x ) 0 , s ε ( φ τ ( x ) ) = 1 for all τ { k , , K 0 1 } , and that φ τ ( x ) ε , s ε ( φ τ ( x ) ) = 0 for all τ { 1 , , k 1 } . Hence, z ε , k ( x ) = 1 and z ε , τ ( x ) = 0 for τ k . Therefore, η ε ( x ) = η k ( x ) .

Case 4. If 0 < φ k 1 ( x ) < ε for some k { 2 , , K 0 1 } , then it follows from Lemma 3.1 that φ τ ( x ) > ε and s ε ( φ τ ( x ) ) = 0 for all τ { 1 , , k 2 } , and that φ k ( x ) < 0 . Again by the conclusion of (i) in Lemma 3.1 we have φ τ ( x ) < 0 and s ε ( φ τ ( x ) ) = 1 for all τ { k , , K 0 1 } . Hence, z ε , k ( x ) = 1 s ε ( φ k 1 ( x ) ) , z ε , k 1 ( x ) = s ε ( φ k 1 ( x ) ) and z ε , τ ( x ) = 0 for τ k , k 1 . Thus, η ε ( x ) = η k 1 ( x ) s ε ( φ k 1 ( x ) ) + η k ( x ) [ 1 s ε ( φ k 1 ( x ) ) ] . □

3.2 The approximation problem of (1.1)

In this subsection, we construct a problem to approximate (1.1).

For each l = 1 , , N , let
{ a i j ε l = a i j ε l ( x , t , u l ) : = k = 1 K 0 a i j , k l ( x , t , u l ) z ε , k ( x ) , b j ε l = b j ε l ( x , t , u l ) : = k = 1 K 0 b j , k l ( x , t , u l ) z ε , k ( x ) , g ε l = g ε l ( x , t , u ) : = k = 1 K 0 g k l ( x , t , u ) z ε , k ( x ) ( ( x , t ) Q T ) .
(3.5)
It follows from hypothesis (iv) of (H), (3.2) and (3.5) that a i j ε l ( x , t , u l ) , b j ε l ( x , t , u l ) are in C 1 + α 0 ( Q ¯ k , T × R ) ( i , j = 1 , , n ), g ε l ( x , t , u ) is in C 1 + α 0 ( Q ¯ k , T × S ) ( k = 1 , , K K 0 + 1 ), the vector function g ε ( , u ) = ( g ε 1 ( , u ) , , g ε N ( , u ) ) is mixed quasimonotone in S with index vector ( a 1 , , a N ) , and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ14_HTML.gif
(3.6)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ15_HTML.gif
(3.7)

We note that the functions a i j ε l ( x , t , u l ) , b j ε l ( x , t , u l ) and g ε l ( x , t , u ) are continuous on Γ T , and are allowed to be discontinuous on Γ T .

For each k = 1 , , K K 0 , there exists Ω τ k such that Γ k Ω τ k . Take two subdomains B k , 1 , B k , 2 satisfying Γ k B k , 1 B k , 2 Ω τ k . Let λ k = λ k ( x ) be an arbitrary smooth function taking values in [ 0 , 1 ] such that λ k = 0 for x Ω τ k and λ k = 1 for x B k , 2 . Set
ψ ε l = ψ ε l ( x , t ) : = | x y | ε ω ( | x y | ) ( 1 k = 1 K K 0 λ k ( y ) ) ψ l ( y , t ) d y + k = 1 K K 0 λ k ( x ) ψ l ( x , t )
(3.8)
with a sufficiently smooth nonnegative averaging kernel ω ( | ξ | ) that is equal to zero for | ξ | 1 and is such that | ξ 1 ω ( ξ ) d ξ = 1 . Then from the hypothesis (iv) of (H) and [[1], Chapter II] we know that for each l = 1 , , N , ψ ε l ( x , t ) is in C α 0 ( Q ¯ T ) W 2 1 , 1 ( Q T ) , ψ ε l ( x , 0 ) is in C 2 + α 0 ( Ω ¯ k ) ( k = 1 , , K K 0 + 1 ), ψ ε l ψ l in C α 0 ( Q ¯ T ) and ψ ε l ψ l in W 2 1 , 1 ( Q T ) . Thus,
ψ ε l ( x , t ) C α 0 ( Q ¯ T ) + ψ ε l W 2 1 , 1 ( Q T ) μ 1 ,
(3.9)
where μ 1 is a positive constant, independent of ε. Furthermore, (3.4), (3.5) and (3.8) show that for small enough ε,
a i j ε l ( x , t , u l ) = a i j , τ k l ( x , t , u l ) , ψ ε l ( x , t ) = ψ l ( x , t ) ( ( x , t ) B k , 1 × [ 0 , T ] ) , k = 1 , , K K 0 .
These, together with (2.6), imply that
[ a i j ε ( x , 0 , ψ ε l ( x , 0 ) ) ψ ε x j l ( x , 0 ) ν i ] Γ = 0 .
(3.10)
For any given ε, 0 < ε < ε 0 , consider the approximation diffraction problem of (1.1)
{ u t l L ε l ( u l ) = g ε l ( x , t , u ) ( ( x , t ) Q T ) , [ u l ] Γ T = 0 , [ a i j ε l ( x , t , u l ) u x j l ν i ( x ) ] Γ T = 0 , u l = ψ ε l ( x , t ) ( ( x , t ) S T { Ω × { 0 } } ) , l = 1 , , N ,
(3.11)
where
L ε l ( u l ) : = d d x i ( a i j ε l ( x , t , u l ) u x j l ) + b j ε l ( x , t , u l ) u x j l .

We note that the interfaces in (3.11) are Γ k ( k = 1 , , K K 0 ) which do not intersect with Ω. In view of (3.10), the compatibility condition on Γ holds.

Proposition 3.1 Problem (3.11) has a unique piecewise classical solution u ε = u ε ( x , t ) in S possessing the following properties:
u ε C α ( Q ¯ T ) , u ε t C α , α / 2 ( Q ¯ T ) , u ε x j C α , α / 2 ( Q ¯ k , T ) ( α ( 0 , 1 ) ) , u ε x j t L 2 ( Q T ) , u ε x i x j C ( Q k , T ) , k = 1 , , K K 0 + 1 .
(3.12)

Proof Problem (3.11) is a special case of [[13], problem (1.1)] without time delays. Formulas (2.3) and (3.5) show that u ˜ = M , u ˆ = m are a pair of bounded and coupled weak upper and lower solutions of (3.11) in the sense of [[13], Definition 2.2]. We find that the conditions of [[13], Theorem 4.1] are all fulfilled. Then from [[13], Theorem 4.1], we obtain that problem (3.11) has a unique piecewise classical solution u ε = u ε ( x , t ) in S possessing the properties in (3.12). □

3.3 The uniform estimates of u ε

In the following discussion, let K ρ be an arbitrary open ball of radius ρ with center at x 0 , and let Q ρ be an arbitrary cylinder of the form K ρ × [ t 0 ρ 2 , t 0 ] .

For each l = 1 , , N , consider the equality t 0 t Ω [ u ε t l L ε l ( u ε l ) ] η l d x d t = t 0 t Ω g ε l ( x , t , u ε ) η l d x d t for any function η l = η l ( x , t ) from W 2 1 , 1 ( Q T ) with ess sup Q T | η l | < and for any t 0 , t from [ 0 , T ] . In view of u ε S , it follows from (3.6), (3.7), (3.9) and the formula of integration by parts that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ21_HTML.gif
(3.13)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ22_HTML.gif
(3.14)
Similarly, for any ϕ l W 2 1 ( Ω ) and for every t [ 0 , T ] we get
Ω a i j ε l ( x , t , u ε l ) u ε x j l ϕ x i l d x = Ω [ u ε t l b j ε l ( x , t , u ε l ) u ε x j l + g ε l ( x , t , u ε ) ] ϕ l d x d t .
(3.15)
Lemma 3.3 There exist constants α 1 ( 0 < α 1 < 1 ) and C depending only on M 0 ( : = max ( | M | , | m | ) ), ρ 0 , θ 0 , α 0 , ν ( M 0 ) , μ ( M 0 ) and μ 1 , independent of ε, such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ24_HTML.gif
(3.16)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ25_HTML.gif
(3.17)

Proof (3.16) follows from (3.14), (3.6), (3.7), (3.9) and [[1], Chapter V, Theorem 1.1 and Remark 1.2]. Setting η l = u ε l ψ ε l in (3.14) and using Cauchy’s inequality, we can obtain (3.17). □

Lemma 3.4 For any given k 1 { 1 , , K 0 } , let D 1 Ω k 1 and t ( 0 , T ) . Then there exist positive constants α 2 ( 0 < α 2 < 1 ) and C ( d 1 , t ) depending only on d 1 ( : = dist ( D 1 , Ω k 1 ) ), t and the parameters M 0 , ρ 0 , θ 0 , α 0 , ν ( M 0 ) , μ ( M 0 ) and μ 1 , independent of ε, such that for any Ω k satisfying D 1 Ω k ,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ26_HTML.gif
(3.18)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ27_HTML.gif
(3.19)
For any given k { 1 , , K } , let Ω Ω k and t ( 0 , T ) . Then there exist positive constants α 3 ( 0 < α 3 < 1 ) and C ( d , t ) depending only on d ( : = dist ( Ω , Ω k ) ), t and the parameters M 0 , ρ 0 , θ 0 , α 0 , ν ( ( M 0 ) ) , μ ( M 0 ) and μ 1 , such that
u ε l C 2 + α 3 , 1 + α 3 / 2 ( Ω ¯ × [ t , T ] ) C ( d , t ) , l = 1 , , N .
(3.20)
Proof Choose a subdomain B satisfying D 1 B Ω k 1 . (3.4) and (3.5) show that for small enough ε,
{ a i j ε l ( x , t , u l ) = a i j , k 1 l ( x , t , u l ) , b j ε l ( x , t , u l ) = b j , k 1 l ( x , t , u l ) , g ε l ( x , t , u ) = g k 1 l ( x , t , u ) ( ( x , t ) B × ( 0 , T ] ) , l = 1 , , N .
(3.21)

Then the same proofs as those of [[13], formulas (3.30) and (3.31)] give (3.18) and (3.19). If Ω Ω k , then Ω Ω k 1 Ω k for some k 1 { 1 , , K 0 } , k { 1 , , K K 0 + 1 } . Hence, the conclusion in (3.20) follows from (3.18), (3.19), (3.21) and the same argument as that for [[13], formula (3.37)]. □

In the rest of this subsection, let k 2 be an arbitrary fixed number in { 1 , , K 0 1 } , and let D 2 Ω be an arbitrary fixed subdomain satisfying D 2 Γ k 2 , D ¯ 2 ( Γ k 2 1 Γ k 2 + 1 ) = and D ¯ 2 Γ = . We next investigate the uniform estimates in the neighborhood of Γ k 2 D ¯ 2 . Let x 0 be any point of Γ k 2 D ¯ 2 . [[2], Chapter 3, Section 16] and [13] show that there exists a ball K ρ with center at x 0 such that we can straighten Γ k 2 K ρ out by introducing a local coordinate system y = y ( x ) . Our assumptions concerning Γ imply that we can divide Γ k 2 D ¯ 2 into a finite number of pieces and to introduce for each of them coordinates y. Since the investigations in the rest of this subsection are local properties, we can assume without loss of generality that the interface Γ k 2 lies in the plane x n = 0 . Then by (3.4), when ( x , t ) D 2 × [ 0 , T ] the coefficients of problem (3.11) can be represented in the form
{ a i j ε l ( x , t , u l ) = a i j , k 2 l ( x , t , u l ) s ε ( x n ) + a i j , k 2 + 1 l ( x , t , u l ) [ 1 s ε ( x n ) ] , b j ε l ( x , t , u l ) = b j , k 2 l ( x , t , u l ) s ε ( x n ) + b j , k 2 + 1 l ( x , t , u l ) [ 1 s ε ( x n ) ] , g ε l ( x , t , u ) = g k 2 l ( x , t , u ) s ε ( x n ) + g k 2 + 1 l ( x , t , u ) [ 1 s ε ( x n ) ] , l = 1 , , N ,
(3.22)
and the diffraction conditions on Γ T in problem (1.1) can be represented in the form
[ u l ] Γ T = 0 , [ a n j l ( x , t , u l ) u x j l ] Γ T = 0 , l = 1 , , N .
(3.23)
Lemma 3.5 Let t ( 0 , T ) . Then there exist positive constants α 4 ( 0 < α 4 < 1 ) and C ( d 2 , t ) depending only on d 2 ( : = min { dist ( D 2 , Ω ) , dist ( D 2 , Γ k 2 1 Γ k 2 + 1 ) , dist ( D 2 , Γ ) } ), t , and the parameters M 0 , ρ 0 , θ 0 , α 0 , ν ( M 0 ) , μ ( M 0 ) and μ 1 , independent of ε, such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ32_HTML.gif
(3.24)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ33_HTML.gif
(3.25)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ34_HTML.gif
(3.26)
Proof It follows from (3.22) and Hypothesis (H) that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-99/MediaObjects/13661_2013_Article_353_Equ35_HTML.gif
(3.27)
and from the equations in (3.11) that
| d d x n ( a n j ε l ( x , t , u ε l ) u ε x j l ) | C ( | u ε t l | + s = 1 n 1 j = 1 n | u ε x j x s l | + | u ε x l | 2 + 1 ) ( ( x , t ) D ¯ 2 × [ 0 , T ] ) , l = 1 , , N .
(3.28)

Then using (3.13), (3.15), (3.22), (3.27) and (3.28), we can prove (3.24)-(3.26) by a slight modification of the proofs of [[13], formulas (3.30) and (3.31)]. The detailed proofs are omitted. □

3.4 The proof of Theorem 2.1

From estimates (3.16), (3.17) and the Arzela-Ascoli theorem it follows that we can find a subsequence (we retain the same notation for it) { u ε } such that { u ε } converges in C ( Q ¯ T ) to u and { u ε x j } converges weakly in L 2 ( Q T ) to u x j for each j = 1 , , n . Then u C α 1 ( Q ¯ T ) and u x j L 2 ( Q T ) . Furthermore, the parabolic boundary conditions for u ε in (3.11) imply that u satisfies the parabolic boundary conditions in (1.1).

For any given k { 1 , , K } , and for any Ω Ω k , t ( 0 , T ) , (3.20) yields that there exists a subsequence { u ε } (denoted by { u ε } still) such that { u ε } converges in C 2 , 1 ( Ω ¯ × [ t , T ] ) to u. By letting ε 0 , from (3.21) and the equations u ε t l L ε l ( u ε l ) = g ε l ( x , t , u ε ) in (3.11) we get that
u t l L l ( u l ) = g l ( x , t , u ) ( ( x , t ) Ω × [ t , T ] ) , l = 1 , , N .

Since Ω and t are arbitrary, then u satisfies the equations in (3.11) for ( x , t ) Q k , T .

For any given k 1 { 1 , , K 0 } and for any D 1 Ω k 1 , t ( 0 , T ) , we see from (3.18), (3.19) that there exists a subsequence { u ε } (denoted by { u ε } still) such that for each j = 1 , , n and for any Ω k satisfying D 1 Ω k , { u ε x j } converges in C ( ( D 1 Ω k ¯ ) × [ t , T ] ) to u x j , and { u ε t } converges in C ( D ¯ 1 × [ t , T ] ) to u t . Hence
u x j C α 2 ( ( D 1 Ω k ¯ ) × [ t , T ] ) , u t C α 2 ( D ¯ 1 × [ t , T ] ) .
(3.29)
By letting ε 0 we conclude from (3.21) and the diffraction conditions on Γ T for u ε in (3.11) that
[ u l ] Γ T Q T = 0 , [ a i j l ( x , t , u l ) u x j l ν i ( x ) ] Γ T Q T = 0 , l = 1 , , N .
(3.30)
For any given k 2 { 1 , , K 0 1 } and D 2 Ω satisfying D 2 Γ k 2 , D ¯ 2 ( Γ k 2 1 Γ k 2 + 1 ) = and D ¯ 2 Γ = , the estimates (3.24)-(3.26) imply that for any given t ( 0 , T ) there exists a subsequence { u ε } (denoted by { u ε } still) such that for each s = 1 , , n 1 , l = 1 , , N ,
u ε x s l u x s l , u ε t l u t l , ϖ ε , n l = a n j ε l ( x , t , u ε l ) u ε x j l ϖ l in  C ( D ¯ 2 × [ t , T ] ) .
(3.31)
Then
u x s l , u t l , ϖ l C α 4 ( D ¯ 2 × [ t , T ] ) .
(3.32)
We next show that ϖ l = a n j l ( x , t , u l ) u x j l . For any η = η ( x , t ) L 2 ( D 2 × ( t , T ) ) ,
t T D 2 [ a n j ε l ( x , t , u ε l ) u ε x j l a n j l ( x , t , u l ) u x j l ] η d x d t = t T D 2 ( a n j ε l ( x , t , u ε l ) a n j ε l ( x , t , u l ) ) u ε x j l η d x d t + t T D 2 ( a n j ε l ( x , t , u l ) a n j l ( x , t , u l ) ) u ε x j l η d x d t + t T D 2 a n j l ( x , t , u l ) ( u ε x j l u x j l ) η d x d t : = I ε , 1 l + I ε , 2 l + I ε , 3 l .
By (3.27), (3.17), we get
| I ε , 1 l | C ( u ε l u l ) η L 2 ( D 2 × ( t , T ) ) u ε x j l L 2 ( D 2 × ( t , T ) ) C ( u ε l u l ) η L 2 ( D 2 × ( t , T ) ) 0 as  ε 0 ,
and by (2.2), (3.22),
| I ε , 2 l | = | t T D 2 { x | 0 x n ε } ( a n j , k 2 l ( x , t , u l ) a n j , k 2 + 1 l ( x , t , u l ) ) s ε ( x n ) u ε x j l η d x d t | C u ε x j l L 2 ( D 2 × ( 0 , T ) ) { t T D 2 { x | 0 x n ε } η 2 d x d t } 1 / 2 C { t T D 2 { x | 0 x n ε } η 2 d x d t } 1 / 2 0 as  ε 0 .
Since { u ε x j } converges weakly in L 2 ( Q T ) to u x j for each j = 1 , , n , then I ε , 3 l 0 as ε 0 . Hence, ϖ ε , n l = a n j ε l ( x , t , u ε l ) u ε x j l converges weakly in L 2 ( D 2 × ( t , T ) ) to a n j l ( x , t , u l ) u x j l for each j = 1 , , n . This, together with (3.31), implies that
ϖ l = a n j l ( x , t , u l ) u x j l C α 4 ( D ¯ 2 × [ t , T ] ) ,
(3.33)

and u satisfies the diffraction conditions on Γ T Q T in (3.23).

In view of (3.30) u satisfies the diffraction conditions on Γ T Q T in (1.1). Furthermore, (3.29), (3.32) and (3.33) imply that for any k { 1 , , K } , Ω Ω ,
u x j C α ( ( Ω Ω k ¯ ) × [ t , T ] ) , u t C α ( Ω ¯ × [ t , T ] ) , j = 1 , , n

for some α ( 0 , 1 ) . Therefore, u is a solution of (1.1). This completes the proof of Theorem 2.1.

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

The authors would like to thank the reviewers and the editors for their valuable suggestions and comments. The work was supported by the research fund of Department of Education of Sichuan Province (10ZC127) and the research fund of Chengdu Normal University (CSYXM12-06).

Authors’ Affiliations

(1)
Department of Mathematics, Chengdu Normal University

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© Tan and Pan; licensee Springer. 2013

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