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

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.

1 Introduction

Let Ω={x R n :|x|<1} be the unit ball, n3, 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(τθ)dτ

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 m1<αm, 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 0β1, 0<α1. 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:

Δu(x)=g(x),xΩ,
(2.1)
D α , β [u](x)=f(x),xΩ.
(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 xΩ, 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)dx.
(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<α1, 0β1, 0<λ<1, f(x) C λ + 2 (Ω), g(x) C λ + 1 ( Ω ¯ ).

Then:

  1. (1)

    If 0<α<1, 0β<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<α1, 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)dx
    (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<α1, 0β1. 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 xΩ.

Proof

Denote

δ 1 = β ( 1 α ) , δ 2 = ( 1 β ) ( 1 α ) .

Let 0<α<1, 0β<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(sr), this integral can be represented in the form of

s r ( r τ ) δ 1 1 ( τ s ) δ 2 1 dτ= ( r s ) δ 1 + δ 2 1 0 1 ( 1 ξ ) δ 1 1 ξ δ 2 1 dξ.

Since

0 1 ( 1 ξ ) δ 1 1 ξ δ 2 1 dξ= Γ ( δ 1 ) Γ ( δ 2 ) Γ ( δ 1 + δ 2 ) , δ 1 + δ 2 =1α,

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 0β1, 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 x0, 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](τx)dτ
(3.3)

holds for any xΩ.

Proof Let xΩ and t(0,1]. Consider the function

t [u](x)= 1 Γ ( α ) 0 t ( t τ ) α 1 τ α B α [u](τx)dτ.

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 dτ=Γ(α)Γ(1α).

Then

t [u](x)= d d t 0 t u(ξx)dξ=u(tx).

If now we suppose t=1, then

u(x)= 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [u](τx)dτ.

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](τx)dτ
(3.4)

is valid for any xΩ.

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 xΩ.

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](τx)dτ.

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](sx)ds.

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 ] ds
(3.6)

holds for any xΩ.

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](sx)ds.
(3.7)

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

Corollary 3.8 Let 0<α1. Then for any xΩ, the representation

u(x)=u(0)+ 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ α B α [u](τx)dτ

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(τx)dτ.

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](tx)dt.

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 xΩ, 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<α1. Then for any xΩ, 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<α1 and u(0)=0, then

B α [ B α [ u ] ] (x)=u(x).

Lemma 3.11 Let

Δu(x)=g(x),xΩ,0<α1.

Then for any xΩ, 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(ξx)dξ+r d d r 0 1 ( 1 ξ ) α Γ ( 1 α ) u(ξx)dξ= I 1 (x)+ I 2 (x).

Since

Δu(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(ξx)dξ.

Further, if v(x) is a smooth function, then obviously,

Δ [ r r v ( x ) ] =r r Δv(x)+2Δv(x).

That is why

Δ I 2 (x)=r d d r 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g(ξx)dξ+2 0 1 ( 1 ξ ) α Γ ( 1 α ) ξ 2 g(ξx)dξ.

Consider the integral

0 1 ( 1 ξ ) α ξ 2 g(ξx)dξ.

After changing of variables ξr=τ, ξ= r 1 τ, the integral can be transformed to the following form:

0 1 ( 1 ξ ) α ξ 2 g(ξx)dξ= r α 3 0 r ( r τ ) α τ 2 g(τθ)dτ.

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 +2Δu(x)=r g ( x ) r +2g(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 functionsf(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)dy+ 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)dy;
    (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)dy+ 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)dy.

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)dy.
(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(ρξ)dρ,

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)dy= 0 1 ρ n 1 | ξ | = 1 g(ρξ)dξdρ= Ω g(y)dy.

The lemma is proved. □

5 The proof of the main proposition

Let 0<α<1, 0β<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

Δv(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(ξx)dξ,

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)dy.

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

Δu(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)dx.

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.

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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.

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Torebek, B.T., Turmetov, B.K. On solvability of a boundary value problem for the Poisson equation with the boundary operator of a fractional order. Bound Value Probl 2013, 93 (2013). https://doi.org/10.1186/1687-2770-2013-93

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Keywords

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