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On solvability of a boundary value problem for the Poisson equation with the boundary operator of a fractional order
Boundary Value Problems volume 2013, Article number: 93 (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.
1 Introduction
Let be the unit ball, , be the unit sphere, . Further, let be a smooth function in the domain Ω, , . For any , the expression
is called the operator of order α in the sense of Riemann-Liouville [1]. From here on, we denote . Let , ,
The operators
are called the derivative of order α in the sense of Riemann-Liouville and Caputo, respectively [1]. Further, let , . Consider the operator
is said to be the derivative of the order α in the Riemann-Liouville sense and of type β.
We note that the operator was introduced in [2]. Some questions concerning solvability for differential equations of a fractional order connected with were studied in [3, 4]. Introduce the notations:
In what follows, we denote
2 Statement of the problem and formulation of the main result
Consider in the domain Ω the following problem:
We call a solution of the problem (2.1), (2.2) a function such that , which satisfies the conditions (2.1) and (2.2) in the classical sense. If , then the equality
holds for all , 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 , and the Dirichlet problem for the equation (2.1) in the case of .
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]:
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 , 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
and in the case of the Caputo operator
the problem (2.1), (2.2) is solvable if and only if the condition
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 be a solution of the Dirichlet problem
The main result of the present work is the following.
Theorem 2.1 Let , , , , .
Then:
-
(1)
If , , then a solution of the problem (2.1), (2.2) exists, is unique and is represented in the form of
(2.5)
where is the solution of the problem (2.4) with the function
-
(2)
If , , then the problem (2.1), (2.2) is solvable if and only if the condition
(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 is the solution of the problem (2.4) with the function
satisfying to the condition .
-
(3)
If the solution of the problem exists, then it belongs to the class .
3 Properties of the operators and
It should be noted that properties and applications of the operators , and in the class of harmonic functions in the ball Ω are studied in [10]. Later on, we assume that is a smooth function in the domain Ω. The following proposition establishes a connection between operators and .
Lemma 3.1 Let , . Then the equalities
hold for any .
Proof
Denote
Let , . Using definition of the operator , we obtain
Consider the inner integral. If we change variables , this integral can be represented in the form of
Since
we have
The first equality from (3.1) is proved.
If , then
The lemma is proved. □
This lemma implies that for any , 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 , , then the equality
is correct.
Lemma 3.3 If , , then the equality
holds.
Proof
Since
by virtue of smoothness of the function at , the second integral converges to zero.
Then the equality (3.2) implies
The lemma is proved. □
Lemma 3.4 Let . Then the equality
holds for any .
Proof Let and . Consider the function
Represent in the form of
Further, using definition of the operator , we have
It is easy to show that
Then
If now we suppose , then
The lemma is proved. □
Using connection between operators and , one can prove the following.
Lemma 3.5 Let . Then the representation
is valid for any .
Proof Using the equality (3.3), taking into account (3.2), we obtain
The lemma is proved. □
Lemma 3.6 Let . Then the equalities
hold for any .
Proof Let us prove the first equality. Apply to the function , the operator . By definition of , we have
But by virtue of the equality (3.3), the last integral is equal to , i.e.,
Now let us prove the second equality. Applying the operator to the function , we obtain
Further, it is not difficult to verify correctness of the following equalities:
Here, it is taken into account . Therefore,
Hence, using the equality (3.3), we obtain
The lemma is proved. □
In [11], the following is proved.
Lemma 3.7 Let . Then the equality
holds for any .
Since
the equality (3.5) can be represented in the form of
Then Lemma 3.5 and the equality (3.6) imply the following.
Corollary 3.8 Let . Then for any , the representation
is valid. Hence, as the inverse operator to , we can consider the following operator:
Note that if
then the operator is not defined in such functions. Let be a smooth function. Obviously,
Consider action of the operator to the function . By definition of the operator , we have
By virtue of (3.7), the value of the last integral is equal to . Thus, the equality
holds.
Conversely, let . Then the operator is defined for such functions, and
It means that
Thus, we prove the following.
Lemma 3.9 For any , the following equalities are valid:
-
(1)
;
-
(2)
if , then
Using Lemma 3.9 and connection between operators and , we get the following.
Corollary 3.10 Let . Then for any , the following equalities hold: if , then
if and , then
Lemma 3.11 Let
Then for any , the equality
holds.
Proof Let . After changing of variables, the function can be represented in the form of
Since
it is easy to show that
Further, if is a smooth function, then obviously,
That is why
Consider the integral
After changing of variables , , the integral can be transformed to the following form:
Then
Hence,
Let now . In this case,
and therefore.
On the other hand,
The lemma is proved. □
4 Some properties of a solution of the Dirichlet problem
Let be a solution of the problem (2.4). It is known (see [12]), if functions and are sufficiently smooth, then a solution of the problem (2.4) exists and is represented in the form of
here is the area of the unit sphere, is the Green function of the Dirichlet problem for the Laplace equation, and is the Poisson kernel.
In addition, the representations
take place.
Lemma 4.1 Let be a solution of the problem (2.4).
Then
-
(1)
if , then
(4.2) -
(2)
if the equality (4.2) is valid, then the condition 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
and
Then
Hence,
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 be a solution of the problem (2.4), and the function be represented in the form of
Then the condition (4.2) can be represented in the form of
Proof Using representation of the function , we have
Then
Consider . After integrating by parts, we get
what follows
Hence,
The lemma is proved. □
5 The proof of the main proposition
Let , , and be a solution of the problem (2.1), (2.2). In this case, by Lemma 3.1, . Apply to the function the operator , and denote
Then, using the equality (3.8), we obtain
Since , it is obviously,
Thus, if is a solution of the problem (2.1), (2.2), then we obtain for the function
the problem (2.4) with
Further, since
for , we have .
Then for , , a solution of the problem (2.4) exists and belongs to the class (see, for example, [13]).
Further, applying to the equality
the operator , by virtue of the first equality of the formula (3.5), we obtain
The last function satisfies to all the conditions of the problem (2.1), (2.2).
Really,
Now, using the second equality from (3.5), we obtain
So, the function satisfies equation (2.1) and the boundary condition (2.2). Let now , , and be a solution of the problem (2.1), (2.2). Apply to the function the operator , and denote . In this case, we obtain for the function the problem (2.4) with the function
Since , the function must satisfy in addition to the condition .
Arbitrary solution of the problem (2.4) at smooth and is represented in the form of (4.1). And in order that this solution satisfies to the condition , 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
In this case, 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 , and the function
satisfies to all conditions of the problem (2.1), (2.2). Let us check these conditions. Fulfillment of the condition
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 and , we get
Hence,
If , then
and
In this case,
Then by virtue of Lemma 4.1, the solvability condition of the problem (2.1), (2.2) can be rewritten in the form of
It is the solvability condition for the Neumann problem. Further, since , the function also belongs to the class . The theorem is proved.
6 Example
Example Let , and
Then
Since
we have
Then the solvability condition for the problem (2.1), (2.2) has in this case the form
For example, if , this condition is not fulfilled. If
then the solvability condition of the problem is carried out. In this case, solving the Dirichlet problem (2.4) with the functions
we obtain (see [14])
Using the formula (2.5), we obtain the solution of the problem (2.1), (2.2)
Thus, the solution of the problem (2.1), (2.2) has the form
References
Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. In Mathematics Studies. Elsevier, Amsterdam; 2006:204.
Hilfer R: Fractional calculus and regular variation in thermodynamics. In Applications of Fractional Calculus in Physics. Edited by: Hilfer R. World Scientific, Singapore; 2000.
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.
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.
Sobolev SL: Equations of Mathematical Physics. Nauka, Moscow; 1977. [in Russian]
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.
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)
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)
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)
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)
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)
Bitsadze AV: Equations of Mathematical Physics. Nauka, Moscow; 1981. [in Russian]
Gilbarg D, Trudinger N: Elliptical Partial Differential Equations of the Second Order. Nauka, Moscow; 1989. [in Russian]
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)
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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DOI: https://doi.org/10.1186/1687-2770-2013-93