Open Access

On solvability of a boundary value problem for the Poisson equation with the boundary operator of a fractional order

Boundary Value Problems20132013:93

https://doi.org/10.1186/1687-2770-2013-93

Received: 4 December 2012

Accepted: 3 April 2013

Published: 17 April 2013

Abstract

In this work, we investigate the solvability of a boundary value problem for the Poisson equation. The considered problem is a generalization of the known Dirichlet and Neumann problems on operators of a fractional order. We obtain exact conditions for solvability of the studied problem.

MSC:35J05, 35J25, 26A33.

Keywords

the Poison equation boundary value problem fractional derivative the Riemann-Liouville operator the Caputo operator

1 Introduction

Let Ω = { x R n : | x | < 1 } be the unit ball, n 3 , Ω = { x R n : | x | = 1 } be the unit sphere, | x | 2 = x 1 2 + x 2 2 + + x n 2 . Further, let u ( x ) be a smooth function in the domain Ω, r = | x | , θ = x / | x | . For any 0 < α , the expression
J α [ u ] ( x ) = 1 Γ ( 1 α ) 0 r ( r τ ) α 1 u ( τ θ ) d τ
is called the operator of order α in the sense of Riemann-Liouville [1]. From here on, we denote J 0 [ u ] ( x ) = u ( x ) . Let m 1 < α m , m = 1 , 2 ,  ,
r = j = 1 n x j r x j , k u r k = r ( k 1 u r k 1 ) , k = 1 , 2 , .
The operators
D α [ u ] ( x ) = m r m J m α [ u ] ( x ) , D α [ u ] ( x ) = J m α [ m r m [ u ] ] ( x )
are called the derivative of order α in the sense of Riemann-Liouville and Caputo, respectively [1]. Further, let 0 β 1 , 0 < α 1 . Consider the operator
D α , β [ u ] ( x ) = J β ( 1 α ) d d r J ( 1 β ) ( 1 α ) u ( x ) .

D α , β is said to be the derivative of the order α in the Riemann-Liouville sense and of type β.

We note that the operator D α , β was introduced in [2]. Some questions concerning solvability for differential equations of a fractional order connected with D α , β were studied in [3, 4]. Introduce the notations:
B α , β [ u ] ( x ) = r α D α , β [ u ] ( x ) , B α [ u ] ( x ) = 1 Γ ( α ) 0 1 ( 1 s ) α 1 s α u ( s x ) d s .
In what follows, we denote
B α , 0 = B α , B α , 1 = B α .

2 Statement of the problem and formulation of the main result

Consider in the domain Ω the following problem:
Δ u ( x ) = g ( x ) , x Ω ,
(2.1)
D α , β [ u ] ( x ) = f ( x ) , x Ω .
(2.2)
We call a solution of the problem (2.1), (2.2) a function u ( x ) C 2 ( Ω ) C ( Ω ¯ ) such that B α , β [ u ] ( x ) C ( Ω ¯ ) , which satisfies the conditions (2.1) and (2.2) in the classical sense. If α = 1 , then the equality
B 1 , β = r r = ν

holds for all x Ω , here ν is the vector of the external normal to Ω. Therefore, the problem (2.1), (2.2) is represented the Neumann problem in the case of α = 1 , and the Dirichlet problem for the equation (2.1) in the case of α = 0 .

It is known, the Dirichlet problem is undoubtedly solvable, and the Neumann problem is solvable if and only if the following condition is valid [5]:
Ω f ( x ) d s x = Ω g ( x ) d x .
(2.3)
Problems with boundary operators of a fractional order for elliptic equations are studied in [6, 7]. The problem (2.1), (2.2) is studied for the Riemann-Liouville and Caputo operators in the case of the Laplace equation, i.e., when g ( x ) = 0 , in the same works [8, 9]. It is established that the problem (2.1), (2.2) is undoubtedly solvable for the case of the Riemann-Liouville operator
D α , 0 = D α , 0 < α < 1 ,
and in the case of the Caputo operator
D α , 1 = D α , 0 < α < 1 ,
the problem (2.1), (2.2) is solvable if and only if the condition
Ω f ( x ) d s x = 0

is valid, i.e., in this case, the condition for solvability of the problem (2.1), (2.2) coincides with the condition of the Neumann problem.

Let v ( x ) be a solution of the Dirichlet problem
{ Δ v ( x ) = g 1 ( x ) , x Ω , v ( x ) = f ( x ) , x Ω .
(2.4)

The main result of the present work is the following.

Theorem 2.1 Let 0 < α 1 , 0 β 1 , 0 < λ < 1 , f ( x ) C λ + 2 ( Ω ) , g ( x ) C λ + 1 ( Ω ¯ ) .

Then:
  1. (1)
    If 0 < α < 1 , 0 β < 1 , then a solution of the problem (2.1), (2.2) exists, is unique and is represented in the form of
    u ( x ) = B α [ v ] ( x ) ,
    (2.5)
     
where v ( x ) is the solution of the problem (2.4) with the function
g 1 ( x ) = | x | 2 B α , β [ | x | 2 g ] ( x ) .
  1. (2)
    If 0 < α 1 , β = 1 , then the problem (2.1), (2.2) is solvable if and only if the condition
    Ω f ( x ) d s x = Ω | x | 2 n 1 n 2 | x | 2 B α , 1 [ | x | 2 g ] ( x ) d x
    (2.6)
     

is satisfied.

If a solution of the problem exists, then it is unique up to a constant summand and is represented in the form of (2.5), where v ( x ) is the solution of the problem (2.4) with the function
g 1 ( x ) = | x | 2 B α , 1 [ | x | 2 g ] ( x ) ,
satisfying to the condition v ( 0 ) = 0 .
  1. (3)

    If the solution of the problem exists, then it belongs to the class C λ + 2 ( Ω ¯ ) .

     

3 Properties of the operators B α , β and B α

It should be noted that properties and applications of the operators B α , B α and B α in the class of harmonic functions in the ball Ω are studied in [10]. Later on, we assume that u ( x ) is a smooth function in the domain Ω. The following proposition establishes a connection between operators B α , β and B α .

Lemma 3.1 Let 0 < α 1 , 0 β 1 . Then the equalities
B α , β [ u ] ( x ) = { B α [ u ] ( x ) , 0 β < 1 , 0 < α < 1 , B α [ u ] ( x ) u ( 0 ) Γ ( 1 α ) , β = 1 , 0 < α 1 ,
(3.1)

hold for any x Ω .

Proof

Denote
δ 1 = β ( 1 α ) , δ 2 = ( 1 β ) ( 1 α ) .
Let 0 < α < 1 , 0 β < 1 . Using definition of the operator B α , β , we obtain
B α , β [ u ] ( x ) = r α J β ( 1 α ) d d r J ( 1 β ) ( 1 α ) u ( x ) = r α { 1 Γ ( δ 1 ) 0 r ( r τ ) δ 1 1 1 Γ ( δ 2 ) d d τ 0 τ ( τ s ) δ 2 1 u ( s θ ) d s d τ } = r α { 1 Γ ( δ 1 ) d d r 0 r ( r τ ) δ 1 δ 1 1 Γ ( δ 2 ) d d τ 0 τ ( τ s ) δ 2 1 u ( s θ ) d s d τ } = r α 1 Γ ( δ 1 ) 1 Γ ( δ 2 ) d d r { 0 r ( r τ ) δ 1 1 0 τ ( τ s ) δ 2 1 u ( s θ ) d s d τ } = r α 1 Γ ( δ 1 ) 1 Γ ( δ 2 ) d d r { 0 r u ( s θ ) s r ( r τ ) δ 1 1 ( τ s ) δ 2 1 d τ d s } .
Consider the inner integral. If we change variables τ = r + ξ ( s r ) , this integral can be represented in the form of
s r ( r τ ) δ 1 1 ( τ s ) δ 2 1 d τ = ( r s ) δ 1 + δ 2 1 0 1 ( 1 ξ ) δ 1 1 ξ δ 2 1 d ξ .
Since
0 1 ( 1 ξ ) δ 1 1 ξ δ 2 1 d ξ = Γ ( δ 1 ) Γ ( δ 2 ) Γ ( δ 1 + δ 2 ) , δ 1 + δ 2 = 1 α ,
we have
B α , β [ u ] ( x ) = r α Γ ( δ 1 + δ 2 ) d d r { 0 r ( r s ) δ 1 + δ 2 1 u ( s θ ) d s } = r α Γ ( 1 α ) d d r { 0 r ( r s ) α u ( s θ ) d s } = B α [ u ] ( x ) .

The first equality from (3.1) is proved.

If β = 1 , then
B α , 1 [ u ] ( x ) = r α Γ ( 1 α ) { 0 r ( r τ ) α d d τ u ( τ θ ) d τ } = r α Γ ( 1 α ) d d r { 0 r ( r τ ) 1 α 1 α d d τ u ( τ θ ) d τ } = r α Γ ( 1 α ) d d r { r 1 α 1 α u ( 0 ) + 0 r ( r τ ) α u ( τ θ ) d τ } = r α Γ ( 1 α ) { r α u ( 0 ) + d d r 0 r ( r τ ) α u ( τ θ ) d τ } = B α [ u ] ( x ) u ( 0 ) Γ ( 1 α ) .

The lemma is proved. □

This lemma implies that for any 0 β 1 , the problem (2.1), (2.2) can be always reduced to the problems with the boundary Riemann-Liouville or Caputo operators.

Corollary 3.2 If 0 < α < 1 , β = 1 , then the equality
B α [ u ] ( x ) = B α [ u ] ( x ) u ( 0 ) Γ ( 1 α )
(3.2)

is correct.

Lemma 3.3 If 0 < α < 1 , β = 1 , then the equality
B α [ u ] ( 0 ) = 0

holds.

Proof

Since
B α [ u ] ( x ) = r α Γ ( 1 α ) d d r 0 r ( r s ) α u ( s θ ) d s = 1 α Γ ( 1 α ) 0 1 ( 1 ξ ) α u ( ξ x ) d ξ + r Γ ( 1 α ) d d r 0 1 ( 1 ξ ) α u ( ξ x ) d ξ ,

by virtue of smoothness of the function u ( x ) at x 0 , the second integral converges to zero.

Then the equality (3.2) implies
lim x 0 B α [ u ] ( x ) = lim x 0 [ B α [ u ] ( x ) u ( 0 ) Γ ( 1 α ) ] = 1 α Γ ( 1 α ) lim x 0 0 1 ( 1 ξ ) α u ( ξ x ) d ξ u ( 0 ) Γ ( 1 α ) = 1 α Γ ( 1 α ) u ( 0 ) 0 1 ( 1 ξ ) α d ξ u ( 0 ) Γ ( 1 α ) = 0 .

The lemma is proved. □

Lemma 3.4 Let 0 < α < 1 . Then the equality
u ( x ) = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ
(3.3)

holds for any x Ω .

Proof Let x Ω and t ( 0 , 1 ] . Consider the function
t [ u ] ( x ) = 1 Γ ( α ) 0 t ( t τ ) α 1 τ α B α [ u ] ( τ x ) d τ .
Represent t [ u ] ( x ) in the form of
t [ u ] ( x ) = 1 Γ ( α ) d d t { 0 t ( t τ ) α α τ α B α [ u ] ( τ x ) d τ } .
Further, using definition of the operator B α , we have
t [ u ] ( x ) = 1 Γ ( 1 α ) d d t { 0 t ( t τ ) α α Γ ( α ) τ α τ α d d τ 0 τ ( τ ξ ) α u ( ξ x ) d ξ d τ } = 1 Γ ( α ) Γ ( 1 α ) d d t { ( t τ ) α α 0 τ ( τ ξ ) α u ( ξ x ) d ξ | τ = 0 τ = t + 0 t ( t τ ) α 1 0 τ ( τ ξ ) α u ( ξ x ) d ξ d τ } = 1 Γ ( α ) Γ ( 1 α ) d d t { 0 t ( t τ ) α 1 0 τ ( τ ξ ) α u ( ξ x ) d ξ d τ } = 1 Γ ( α ) Γ ( 1 α ) d d t { 0 t u ( ξ x ) ξ t ( t τ ) α 1 ( τ ξ ) α d τ d ξ } .
It is easy to show that
ξ t ( t τ ) α 1 ( τ ξ ) α d τ = Γ ( α ) Γ ( 1 α ) .
Then
t [ u ] ( x ) = d d t 0 t u ( ξ x ) d ξ = u ( t x ) .
If now we suppose t = 1 , then
u ( x ) = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ .

The lemma is proved. □

Using connection between operators B α and B α , one can prove the following.

Lemma 3.5 Let 0 < α < 1 . Then the representation
u ( x ) = u ( 0 ) + 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ
(3.4)

is valid for any x Ω .

Proof Using the equality (3.3), taking into account (3.2), we obtain
u ( x ) = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α [ u ( 0 ) Γ ( 1 α ) + B α [ u ] ( τ x ) ] d τ = u ( 0 ) + 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ .

The lemma is proved. □

Lemma 3.6 Let 0 < α < 1 . Then the equalities
B α [ B α [ u ] ] ( x ) = B α [ B α [ u ] ] ( x ) = u ( x )
(3.5)

hold for any x Ω .

Proof Let us prove the first equality. Apply to the function B α [ u ] , the operator B α . By definition of B α , we have
B α [ B α [ u ] ] ( x ) = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ .
But by virtue of the equality (3.3), the last integral is equal to u ( x ) , i.e.,
B α [ B α [ u ] ] ( x ) = u ( x ) .
Now let us prove the second equality. Applying the operator B α to the function B α [ u ] ( x ) , we obtain
B α [ B α [ u ] ] ( x ) = r α Γ ( 1 α ) d d r 0 r ( r τ ) α B α [ u ] ( τ x ) d τ = r α Γ ( α ) Γ ( 1 α ) d d r { 0 r ( r τ ) α 0 1 ( 1 s ) α 1 s α u ( s τ θ ) d s d τ } .
Further, it is not difficult to verify correctness of the following equalities:
r α Γ ( 1 α ) d d r 0 r ( r τ ) α u ( s τ θ ) d τ = s τ = ξ r α Γ ( 1 α ) d d r 0 r s ( r ξ s ) α u ( ξ θ ) d ξ s = r α s α 1 Γ ( 1 α ) d d r 0 r s ( s r ξ ) α u ( ξ θ ) d ξ = ( s r ) α Γ ( 1 α ) d d ( s r ) 0 r s ( s r ξ ) α u ( ξ θ ) d ξ = B α [ u ] ( s x ) .
Here, it is taken into account θ = x | x | = s x | s x | . Therefore,
B α [ B α [ u ] ] ( x ) = 1 Γ ( α ) 0 1 ( 1 s ) α 1 s α B α [ u ] ( s x ) d s .
Hence, using the equality (3.3), we obtain
B α [ B α [ u ] ] ( x ) = u ( x ) .

The lemma is proved. □

In [11], the following is proved.

Lemma 3.7 Let u ( x ) C 1 ( Ω ) . Then the equality
u ( x ) = u ( 0 ) + 0 1 [ k = 1 n x k u ( s x ) x k ] d s
(3.6)

holds for any x Ω .

Since
k = 1 n x k u ( x ) x k = r u ( x ) r = B 1 [ u ] ( x ) = B 1 [ u ] ( x ) ,
the equality (3.5) can be represented in the form of
u ( x ) = u ( 0 ) + 0 1 s 1 B 1 [ u ] ( s x ) d s .
(3.7)

Then Lemma 3.5 and the equality (3.6) imply the following.

Corollary 3.8 Let 0 < α 1 . Then for any x Ω , the representation
u ( x ) = u ( 0 ) + 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ u ] ( τ x ) d τ
is valid. Hence, as the inverse operator to B 1 , we can consider the following operator:
B 1 [ u ] ( x ) = 0 1 τ 1 u ( τ x ) d τ .
Note that if
u ( 0 ) 0 ,
then the operator B 1 is not defined in such functions. Let u ( x ) be a smooth function. Obviously,
B 1 [ u ] ( 0 ) = 0 .
Consider action of the operator B 1 to the function B 1 [ u ] ( x ) . By definition of the operator B 1 , we have
B 1 [ B 1 [ u ] ] ( x ) = 0 1 s 1 B 1 [ u ] ( t x ) d t .
By virtue of (3.7), the value of the last integral is equal to u ( x ) u ( 0 ) . Thus, the equality
B 1 [ B 1 [ u ] ] ( x ) = u ( x ) u ( 0 )

holds.

Conversely, let u ( 0 ) = 0 . Then the operator B 1 is defined for such functions, and
B 1 [ B 1 [ u ] ] ( x ) = r r [ 0 1 s 1 u ( s x ) d s ] = r r [ 0 r ξ 1 u ( ξ θ ) d ξ ] = u ( x ) .
It means that
B 1 [ B 1 [ u ] ] ( x ) = u ( x ) .

Thus, we prove the following.

Lemma 3.9 For any x Ω , the following equalities are valid:
  1. (1)

    B 1 [ B 1 [ u ] ] ( x ) = u ( x ) u ( 0 ) ;

     
  2. (2)
    if u ( 0 ) = 0 , then
    B 1 [ B 1 [ u ] ] ( x ) = u ( x ) .
     

Using Lemma 3.9 and connection between operators B α and B α , we get the following.

Corollary 3.10 Let 0 < α 1 . Then for any x Ω , the following equalities hold: if 0 < α < 1 , then
B α [ B α [ u ] ] ( x ) = u ( x ) u ( 0 ) Γ ( 1 α ) ; B 1 [ B 1 [ u ] ] ( x ) = u ( x ) u ( 0 ) ;
if 0 < α 1 and u ( 0 ) = 0 , then
B α [ B α [ u ] ] ( x ) = u ( x ) .
Lemma 3.11 Let
Δ u ( x ) = g ( x ) , x Ω , 0 < α 1 .
Then for any x Ω , the equality
Δ B α [ u ] ( x ) = | x | 2 B α [ | x | 2 g ] ( x )
(3.8)

holds.

Proof Let 0 < α < 1 . After changing of variables, the function B α [ u ] ( x ) can be represented in the form of
B α [ u ] ( x ) = 1 α Γ ( 1 α ) 0 1 ( 1 ξ ) α u ( ξ x ) d ξ + r d d r 0 1 ( 1 ξ ) α Γ ( 1 α ) u ( ξ x ) d ξ = I 1 ( x ) + I 2 ( x ) .
Since
Δ u ( x ) = g ( x ) ,
it is easy to show that
Δ I 1 ( x ) = 1 α Γ ( 1 α ) 0 1 ( 1 ξ ) α ξ 2 g ( ξ x ) d ξ .
Further, if v ( x ) is a smooth function, then obviously,
Δ [ r r v ( x ) ] = r r Δ v ( x ) + 2 Δ v ( x ) .
That is why
Δ I 2 ( x ) = r d d r 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g ( ξ x ) d ξ + 2 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g ( ξ x ) d ξ .
Consider the integral
0 1 ( 1 ξ ) α ξ 2 g ( ξ x ) d ξ .
After changing of variables ξ r = τ , ξ = r 1 τ , the integral can be transformed to the following form:
0 1 ( 1 ξ ) α ξ 2 g ( ξ x ) d ξ = r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ .
Then
r d d r 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g ( ξ x ) d ξ = r d d r [ r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ ] = ( α 3 ) r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ + r α 2 d d r 0 r ( r τ ) α τ 2 g ( τ θ ) d τ .
Hence,
Δ I 1 ( x ) + Δ I 2 ( x ) = 1 α Γ ( 1 α ) r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ + ( α 3 ) Γ ( 1 α ) r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ + r α 2 Γ ( 1 α ) d d r 0 r ( r τ ) α τ 2 g ( τ θ ) d τ + 2 Γ ( 1 α ) r α 3 0 r ( r τ ) α τ 2 g ( τ θ ) d τ = r α 2 Γ ( 1 α ) d d r 0 r ( r τ ) α τ 2 g ( τ θ ) d τ = r 2 r α Γ ( 1 α ) d d r 0 r ( r τ ) α τ 2 g ( τ θ ) d τ = r 2 B α [ | x | 2 g ] ( x ) .
Let now α = 1 . In this case,
B 1 [ u ] ( x ) = r u ( x ) r ,
and therefore.
Δ B 1 [ u ] ( x ) = Δ [ r u ( x ) r ] = r Δ u ( x ) r + 2 Δ u ( x ) = r g ( x ) r + 2 g ( x ) .
On the other hand,
( r r + 2 ) g ( x ) = | x | 2 r r [ r 2 g ( x ) ] = | x | 2 B 1 [ | x | 2 g ] ( x ) .

The lemma is proved. □

4 Some properties of a solution of the Dirichlet problem

Let v ( x ) be a solution of the problem (2.4). It is known (see [12]), if functions f ( x ) and g 1 ( x ) are sufficiently smooth, then a solution of the problem (2.4) exists and is represented in the form of
v ( x ) = 1 ω n Ω G ( x , y ) g 1 ( y ) d y + 1 ω n Ω P ( x , y ) f ( y ) d s y ,
(4.1)

here ω n is the area of the unit sphere, G ( x , y ) is the Green function of the Dirichlet problem for the Laplace equation, and P ( x , y ) is the Poisson kernel.

In addition, the representations
G ( x , y ) = 1 n 2 [ | x y | 2 n | | y | x y | y | | 2 n ] , P ( x , y ) = 1 | x | 2 | x y | n

take place.

Lemma 4.1 Let v ( x ) be a solution of the problem (2.4).

Then
  1. (1)
    if v ( 0 ) = 0 , then
    Ω f ( y ) d s y = Ω | y | 2 n 1 n 2 g 1 ( y ) d y ;
    (4.2)
     
  2. (2)

    if the equality (4.2) is valid, then the condition v ( 0 ) = 0 is fulfilled for a solution of the problem (2.4).

     
Proof Let a solution of the problem (2.4) exist. Represent it in the form of (4.1). We have from the representation of the function G ( x , y )
G ( 0 , y ) = 1 n 2 [ | y | 2 n 1 ]
and
P ( 0 , y ) = 1 .
Then
0 = v ( 0 ) = 1 ω n Ω G ( 0 , y ) g 1 ( y ) d y + 1 ω n Ω P ( 0 , y ) f ( y ) d s y .
Hence,
Ω f ( y ) d s y = Ω | y | 2 n 1 n 2 g 1 ( y ) d y .

The equality (4.2) is proved. The second assertion of the lemma is proved in the inverse order. The lemma is proved. □

Lemma 4.2 Let v ( x ) be a solution of the problem (2.4), and the function g 1 ( x ) be represented in the form of
g 1 ( y ) = ( ρ ρ + 2 ) g ( y ) , ρ = | y | .
Then the condition (4.2) can be represented in the form of
Ω f ( y ) d s y = Ω g ( y ) d y .
(4.3)
Proof Using representation of the function g 1 ( y ) , we have
Ω | y | 2 n 1 n 2 g 1 ( y ) d y = 0 1 ρ n 1 | ξ | = 1 ρ 2 n 1 n 2 ( ρ ρ + 2 ) g ( ρ ξ ) d ξ d ρ .
Then
0 1 ρ n 1 ρ 2 n 1 n 2 ( ρ ρ + 2 ) g ( ρ ξ ) d ρ = 1 n 2 0 1 [ ρ 2 ρ n ] ρ [ g ] ( ρ ξ ) d ρ + 2 n 2 0 1 [ ρ ρ n 1 ] g ( ρ ξ ) d ρ = I 1 + I 2 .
Consider I 1 . After integrating by parts, we get
I 1 = 1 n 2 0 1 [ n ρ n 1 2 ρ ] g ( ρ ξ ) d ρ ,
what follows
I 1 + I 2 = 1 n 2 { 0 1 ( n 2 ) ρ n 1 g ( ρ ξ ) d ρ 2 0 1 ρ g ( ρ ξ ) d ρ + 2 0 1 ρ g ( ρ ξ ) d ρ } = 0 1 ρ n 1 g ( ρ ξ ) d ρ .
Hence,
Ω | y | 2 n 1 n 2 g 1 ( y ) d y = 0 1 ρ n 1 | ξ | = 1 g ( ρ ξ ) d ξ d ρ = Ω g ( y ) d y .

The lemma is proved. □

5 The proof of the main proposition

Let 0 < α < 1 , 0 β < 1 , and u ( x ) be a solution of the problem (2.1), (2.2). In this case, by Lemma 3.1, B α , β = B α . Apply to the function u ( x ) the operator B α , and denote
v ( x ) = B α [ u ] ( x ) .
Then, using the equality (3.8), we obtain
Δ v ( x ) = Δ B α [ u ] ( x ) = | x | 2 B α [ | x | 2 g ] ( x ) g 1 ( x ) .
Since B α , β = B α , it is obviously,
Thus, if u ( x ) is a solution of the problem (2.1), (2.2), then we obtain for the function
v ( x ) = B α [ u ] ( x )
the problem (2.4) with
g 1 ( x ) = | x | 2 B α [ | x | 2 g ] ( x ) .
Further, since
g 1 ( x ) = | x | 2 B α [ | x | 2 g ] ( x ) = [ r d d r + 3 α ] 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g ( ξ x ) d ξ ,

for g ( x ) C λ + 1 ( Ω ¯ ) , we have g 1 ( x ) C λ ( Ω ¯ ) .

Then for g 1 ( x ) C λ ( Ω ¯ ) , f ( x ) C λ + 2 ( Ω ) , a solution of the problem (2.4) exists and belongs to the class C λ + 2 ( Ω ¯ ) (see, for example, [13]).

Further, applying to the equality
v ( x ) = B α [ u ] ( x )
the operator B α , by virtue of the first equality of the formula (3.5), we obtain
u ( x ) = B α [ v ] ( x ) .

The last function satisfies to all the conditions of the problem (2.1), (2.2).

Really,
Δ u ( x ) = Δ B α [ v ] ( x ) = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ 2 α Δ v ( τ x ) d τ = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ 2 α g 1 ( τ x ) d τ = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ 2 α τ 2 | x | 2 B α [ | x | 2 g ] ( τ x ) d τ = 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α | x | 2 B α [ | x | 2 g ] ( τ x ) d τ = | x | 2 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [ | x | 2 g ] ( τ x ) d τ = | x | 2 | x | 2 g ( x ) = g ( x ) .
Now, using the second equality from (3.5), we obtain
So, the function u ( x ) = B α [ v ] ( x ) satisfies equation (2.1) and the boundary condition (2.2). Let now 0 < α < 1 , β = 1 , and u ( x ) be a solution of the problem (2.1), (2.2). Apply to the function u ( x ) the operator B α , 1 = B α , and denote v ( x ) = B α [ u ] ( x ) . In this case, we obtain for the function v ( x ) the problem (2.4) with the function
g 1 ( x ) = | x | 2 B α [ | x | 2 g ] ( x ) .

Since B α [ u ] ( 0 ) = 0 , the function v ( x ) must satisfy in addition to the condition v ( 0 ) = 0 .

Arbitrary solution of the problem (2.4) at smooth f ( x ) and g ( x ) is represented in the form of (4.1). And in order that this solution satisfies to the condition v ( 0 ) = 0 , according to Lemma 3.11, it is necessary and sufficient fulfillment of the condition (4.2).

In our case, the condition (4.2) has the form
Ω f ( y ) d s y = Ω | y | 2 n 1 n 2 | y | 2 B α [ | y | 2 g ] ( y ) d y .

In this case, B α , 1 = B α and, therefore, the condition (4.2) coincides with the condition (2.6).

Thus, necessity of (2.6) is proved. This condition is also sufficient condition for existence of a solution for the problem (2.1), (2.2).

In fact, if the condition (2.6) holds, then v ( 0 ) = 0 , and the function
u ( x ) = B α [ v ] ( x ) + C
satisfies to all conditions of the problem (2.1), (2.2). Let us check these conditions. Fulfillment of the condition
Δ u ( x ) = g ( x )
can be checked similarly as in the case of the proof of the first part of the theorem. Further, using the equality (3.5) and connection between operators B α and B α , we get
B α [ u ] ( x ) = B α [ B α [ v ] + C ] ( x ) = B α [ B α [ v ] ] ( x ) + B α [ C ] = B α [ B α [ v ] ] ( x ) + B α [ v ] ( 0 ) Γ ( 1 α ) = v ( x ) .
Hence,
If α = 1 , then
D α , 1 = D α = r
and
B α , 1 = r r .
In this case,
| x | 2 B α , 1 [ | x | 2 g ] ( x ) = ( r r + 2 ) g ( x ) .
Then by virtue of Lemma 4.1, the solvability condition of the problem (2.1), (2.2) can be rewritten in the form of
Ω f ( x ) d s x = Ω g ( x ) d x .

It is the solvability condition for the Neumann problem. Further, since v ( x ) C λ + 2 ( Ω ¯ ) , the function u ( x ) = B α [ v ] ( x ) also belongs to the class C λ + 2 ( Ω ¯ ) . The theorem is proved.

6 Example

Example Let 0 < α < 1 , β = 1 and
g ( x ) = | x | 2 k , k = 0 , 1 , .
Then
| x | 2 B α , 1 [ | x | 2 g ] ( x ) = | x | α 2 Γ ( 1 α ) 0 r ( r τ ) α τ τ 2 k + 2 d τ = ( 2 k + 2 ) | x | α 2 Γ ( 1 α ) 0 r ( r τ ) α τ 2 k + 2 1 d τ = ( 2 k + 2 ) | x | α 2 Γ ( 1 α ) | x | 2 k + 2 α 0 1 ( 1 ξ ) α ξ 2 k + 1 d ξ = ( 2 k + 2 ) | x | 2 k Γ ( 1 α ) Γ ( 1 α ) Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) = Γ ( 2 k + 3 ) Γ ( 2 k + 3 α ) | x | 2 k .
Since
0 1 r 2 k + n 1 ( r 2 n 1 ) d r = 0 1 ( r 2 k + 1 r 2 k + n 1 ) d r = 1 2 k + 2 1 2 k + n = n 2 ( 2 k + 2 ) ( 2 k + n ) ,
we have
Ω | y | 2 n 1 n 2 | y | 2 B α [ | y | 2 g ] ( y ) d y = | ξ | = 1 0 1 r 2 n 1 n 2 r 2 B α [ | y | 2 g ] ( r ξ ) d r d ξ = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) ω n ( 2 k + n ) .
Then the solvability condition for the problem (2.1), (2.2) has in this case the form
Ω f ( y ) d s y = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) ω n ( 2 k + n ) .
For example, if f ( x ) = 1 , this condition is not fulfilled. If
f ( x ) = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) 1 ( 2 k + n ) ,
then the solvability condition of the problem is carried out. In this case, solving the Dirichlet problem (2.4) with the functions
g 1 ( x ) | x | 2 B α , 1 [ | x | 2 g ] ( x ) = Γ ( 2 k + 3 ) Γ ( 2 k + 3 α ) | x | 2 k , f ( x ) = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) 1 ( 2 k + n ) ,
we obtain (see [14])
v ( x ) = Γ ( 2 k + 3 ) Γ ( 2 k + 3 α ) | x | 2 k + 2 ( 2 k + 2 ) ( 2 k + n ) = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) | x | 2 k + 2 2 k + n .
Using the formula (2.5), we obtain the solution of the problem (2.1), (2.2)
u ( x ) = B α [ v ] ( x ) = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) | x | 2 k + 2 2 k + n 1 Γ ( α ) 0 1 ( 1 s ) α 1 s 2 k + 2 α d s = Γ ( 2 k + 2 ) Γ ( 2 k + 3 α ) | x | 2 k + 2 2 k + n 1 Γ ( α ) Γ ( α ) Γ ( 2 k + 3 α ) Γ ( 2 k + 3 ) = | x | 2 k + 2 ( 2 k + 2 ) ( 2 k + n ) .
Thus, the solution of the problem (2.1), (2.2) has the form
u ( x ) = | x | 2 k + 2 ( 2 k + 2 ) ( 2 k + n ) + C .

Declarations

Acknowledgements

This paper is financially supported by the grant of the Ministry of Science and Education of the Republic of Kazakhstan (Grant No. 0830/GF2). The authors would like to thank the editor and referees for their valuable comments and remarks, which led to a great improvement of the article.

Authors’ Affiliations

(1)
Department of Mathematics, Akhmet Yasawi International Kazakh-Turkish University

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