Skip to main content

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.

References

  1. Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. In Mathematics Studies. Elsevier, Amsterdam; 2006:204.

    Google Scholar 

  2. Hilfer R: Fractional calculus and regular variation in thermodynamics. In Applications of Fractional Calculus in Physics. Edited by: Hilfer R. World Scientific, Singapore; 2000.

    Google Scholar 

  3. Hilfer R, Luchko Y, Tomovski Z: Operational method for the solution of fractional differential equations with generalized Riemann-Liouville fractional derivatives. Fract. Calc. Appl. Anal. 2009, 12(3):299-318.

    MathSciNet  Google Scholar 

  4. Shinaliyev KM, Turmetov BK, Umarov SR: A fractional operator algorithm method for construction of solutions of fractional order differential equations. Fract. Calc. Appl. Anal. 2012, 15(2):267-281.

    Article  MathSciNet  Google Scholar 

  5. Sobolev SL: Equations of Mathematical Physics. Nauka, Moscow; 1977. [in Russian]

    Google Scholar 

  6. Umarov SR, Luchko YF, Gorenflo R: On boundary value problems for elliptic equations with boundary operators of fractional order. Fract. Calc. Appl. Anal. 2000, 3(4):454-468.

    MathSciNet  Google Scholar 

  7. Kirane M, Tatar N-E: Nonexistence for the Laplace equation with a dynamical boundary condition of fractional type. Sib. Mat. Zh. 2007, 48(5):1056-1064. [in Russian]; English transl.: Siberian Mathematical Journal. 48(5), 849-856 (2007)

    Article  MathSciNet  Google Scholar 

  8. Turmetov BK: On a boundary value problem for a harmonic equation. Differ. Uravn. 1996, 32(8):1089-1092. [in Russian]; English transl.: Differential equations. 32(8), 1093-1096 (1996)

    MathSciNet  Google Scholar 

  9. Turmetov BK: On smoothness of a solution to a boundary value problem with fractional order boundary operator. Mat. Tr. 2004, 7(1):189-199. [in Russian]; English transl.: Siberian Advances in Mathematics. 15 (2), 115-125 (2005)

    MathSciNet  Google Scholar 

  10. Karachik VV, Turmetov BK, Torebek BT: On some integro-differential operators in the class of harmonic functions and their applications. Mat. Tr. 2011, 14(1):99-125. [in Russian]; English transl.: Siberian Advances in Mathematics. 22(2), 115-134 (2012)

    MathSciNet  Google Scholar 

  11. Bavrin II: Operators for harmonic functions and their applications. Differ. Uravn. 1985, 21(1):9-15. [in Russian]; English transl.: Differential Equations. 21(1), 6-10 (1985)

    MathSciNet  Google Scholar 

  12. Bitsadze AV: Equations of Mathematical Physics. Nauka, Moscow; 1981. [in Russian]

    Google Scholar 

  13. Gilbarg D, Trudinger N: Elliptical Partial Differential Equations of the Second Order. Nauka, Moscow; 1989. [in Russian]

    Google Scholar 

  14. Karachik VV: Construction of polynomial solutions to some boundary value problems for Poisson’s equation. Ž. Vyčisl. Mat. Mat. Fiz. 2011, 51(9):1674-1694. [in Russian]; English transl.: Computational Mathematics and Mathematical Physics. 51(9), 1567-1587 (2011)

    MathSciNet  Google Scholar 

Download references

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.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Berikbol T Torebek.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors completed the paper together. All authors read and approved the final manuscript.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and permissions

About this article

Cite this article

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

Download citation

  • Received:

  • Accepted:

  • Published:

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

Keywords