Acta Mathematica Scientia 2014,34B(6):1695–1706 http://actams.wipm.ac.cn

ON SOLVABILITY OF A BOUNDARY VALUE PROBLEM FOR A NONHOMOGENEOUS BIHARMONIC EQUATION WITH A BOUNDARY OPERATOR OF A FRACTIONAL ORDER∗ A.S. BERDYSHEV Department of Applied Mathematics and Informatics, Kazakh National Pedagogical University named after Abai, Almaty, Kazakhstan E-mail : [email protected]

A. CABADA Departmento de Analise Matematica, Facultade de Matematicas, University of Santiago de Compostela, Santiago de Compostela, Spain E-mail : [email protected]

B.Kh. TURMETOV Khoja Ahmet Yasawi International Kazakh-Turkish University, Kazakhstan E-mail : [email protected] Abstract This paper is concerned with the solvability of a boundary value problem for a nonhomogeneous biharmonic equation. The boundary data is determined by a differential operator of fractional order in the Riemann-Liouville sense. The considered problem is a generalization of the known Dirichlet and Neumann problems. Key words

biharmonic equation; boundary value problem; fractional derivative; the RiemannLiouville operator

2010 MR Subject Classification

1

31A30

Introduction

Biharmonic equations appear in the study of mathematical models in several real-life processes as, among others, radar imaging [1] or incompressible flows [2]. Omitting a huge amount of works devoted to the study of this kind of equations, we refer some of them regarding to their used methods. Difference schemes and variational methods were used in the works [3, 4]. By using numerical and iterative methods, Dirichlet and Neumann boundary problems for biharmonic equations were studied in the papers [5, 6]. ∗ Received

September 13, 2013; revised May 3, 2014. The research of A.Cabada was partially supported by Ministerio de Ciencia e Innovacion-SPAIN, and FEDER, project MTM2010-15314 and research of A.S. Berdyshev and B.Kh. Turmetov was supported by the Ministry of Science and Education of the Republic of Kazakhstan through the Project No. 0713 GF.

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There are some works, for example [7], where a computational method, based on the use of Haar wavelets was used for solving 2D and 3D Poisson and biharmonic equations. We also point out the work made in [8], where regularity of solutions for nonlinear biharmonic equations was investigated. In [9] and the dissertation [10] various problems for complex biharmonic and polyharmonic equations were investigated. Along the paper we refer to the domain Ω = {x ∈ Rn : |x| < 1}, as the unit ball. The dimension of the space is n ≥ 3 and it is denoted ∂Ω = {x ∈ Rn : |x| = 1} as the unit sphere. The usual Euclidean norm is written as |x|2 = x21 + x22 + · · · + x2n . Now, for any u : Ω → R smooth enough function and a given α > 0, denoting by r = |x| and θ = x/|x|, the appropriate integral operator of order α in the Riemann-Liouville can be defined, in a similar sense to [11, p.69], by the following expression Z r 1 α (r − τ )α−1 u(τ θ)dτ , x ∈ Ω. J [u](x) = Γ(1 − α) 0 In what follows, we suppose J 0 [u](x) = u(x) for all x ∈ Ω. Let m − 1 < α ≤ m for some m = 1, 2, · · · . The Riemann-Liouville derivative of order α is defined as, see [11, p.70], Dα [u](x) = where

∂ ∂r

∂ m m−α J [u](x), ∂rm

x ∈ Ω,

is the differential operator n

∂ 1X ∂ ∂ku ∂ = xj , = ∂r r j=1 ∂xj ∂rk ∂r



 ∂ k−1 u , k = 1, 2, · · · . ∂rk−1

Now, we introduce the following functionals: B α [u](x) = rα Dα [u](x), m − 1 < α ≤ m, m = 1, 2, · · · and B −α [u](x) =

1 Γ(α)

Z

1

(1 − s)α−1 s−α u(sx)ds,

0 < α ≤ 1.

0

It should be noted that some properties of the previous operators are studied for 0 < α < 1 in the work [12]. There it is proved that these operators are mutually inverse in the class of harmonic functions for 0 < α < 1.

2

Statement of the Problem and Formulation of the Main Result Let 0 < α ≤ 1 be fixed and consider the following boundary value problem  2  x ∈ Ω,   ∆ u(x) = f (x), B α [u] (x) = ϕ1 (x),

  

B

α+1

[u] (x) = ϕ2 (x),

(1)

x ∈ ∂Ω,

(2)

x ∈ ∂Ω,

(3)

where f , ϕ1 and ϕ2 are given functions. A solution of the problem (1)–(3) will be a function ¯ such that B α+k [u] ∈ C(Ω), k = 0, 1, and satisfies the equation (1) coupled u ∈ C 4 (Ω) ∩ C(Ω) with the boundary conditions (2) and (3).

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Denoting as ν outer normal vector to ∂Ω, we have that for all x ∈ ∂Ω it is satisfied the ∂u ∂u ∂u 0 1 equality ∂u ∂r = ∂ν . In particular, if α = 0 we have that B u(x) = u(x) and B u(x) = r ∂r = ∂ν for all x ∈ ∂Ω. Moreover,   2 ∂ ∂u 2 2∂ u B u(x) = r =r r −u . ∂r2 ∂r ∂r 2

∂ ∂ ∂ It is known [13], that for all x ∈ ∂Ω, the operator r ∂r (r ∂r − 1) coincides with the operator ∂ν 2. Hence, the problem (1)–(3) represents the Dirichlet problem in the case of α = 0, and the Neumann problem when α = 1. For the case of α = 1, the problem (1)–(3) was studied in [14], where it is proved that the Neumann problem is solvable if and only if the following condition Z Z |x|2 − 1 [ϕ1 (x) − ϕ2 (x)] dsx = f (x)dx (4) 2 ∂Ω Ω

holds. Properties of a volume potential were used for studying this problem. In the present work, a complete investigation of the problem (1)–(3) is carried out, allowing in this case dependence on the order α from 0 to 1 of the boundary operators. In particular, it is developed a different approach in the study of the Neumann problem when α = 1. Let v be a solution of the Dirichlet problem    ∆2 v(x) = g(x), x ∈ Ω,    v(x) = ψ1 (x), x ∈ ∂Ω, (5)    ∂v(x)   = ψ2 (x), x ∈ ∂Ω. ∂ν The main result of the present work is the following ¯ ϕ1 ∈ C λ+3 (∂Ω) and ϕ2 ∈ C λ+2 (∂Ω). Theorem Let 0 < α ≤ 1, 0 < λ < 1, f ∈ C λ+1 (Ω), Then 1) if 0 < α < 1 then Problem (1)–(3) admits a unique solution and it is represented in the form of u(x) = B −α [v] (x),

x ∈ Ω,

where v is a solution of the problem (5) with   g(x) = |x|−4 B α |x|4 f (x), ψ1 (x) = ϕ1 (x) and ψ2 (x) = ϕ2 (x) + αϕ1 (x).

(6)

2) if α = 1 then Problem (1)–(3) is solvable if and only if condition (4) is fulfilled. In such a case, it is unique up to a constant summand and is represented by (6), where v is such that v(0) = 0 and it is a solution of the problem (5) with   g(x) = |x|−4 B α |x|4 f (x), ψ1 (x) = ϕ1 (x) and ψ2 (x) = ϕ2 (x) + ϕ1 (x). ¯ 3) if u is a solution of the Problem (1)–(3) then u ∈ C λ+4 (Ω).

3

Properties of the Operators B α and B −α In what follows, we assume that u is a smooth function in the domain Ω.

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Lemma 1 Let 0 < α < 1 and u ∈ C 1 (Ω). Then for any x ∈ Ω, the following equality holds Z 1 1 α−1 −α α u(x) = (1 − τ ) τ B [u] (τ x)dτ . (7) Γ(α) 0 Let x ∈ Ω and t ∈ (0, 1]. Consider the function Z t 1 ℑt [u] (x) = (t − τ )α−1 τ −α B α [u] (τ x)dτ . Γ(α) 0

Proof

It is obvious that it can be rewritten as Z t  (t − τ )α −α α 1 d τ B [u] (τ x)dτ . ℑt [u] (x) = Γ(α) dt α 0

Further, using the definition of the operator B α , we obtain Z t  Z τ 1 d (t − τ )α d −α ℑt [u] (x) = (τ − ξ) u(ξx)dξdτ Γ(α)Γ(1 − α) dt α dτ 0 0 1 = , Γ(α)Γ(1 − α) τ =t Z t  Z τ + (t − τ )α−1 (τ − ξ)−α u(ξx)dξdτ (τ − ξ) u(ξx)dξ 0 0 0 τ =0 Z t  Z τ 1 d = (t − τ )α−1 (τ − ξ)−α u(ξx)dξdτ Γ(α)Γ(1 − α) dt 0 0 Z t  Z t 1 d α−1 −α = u(ξx) (t − τ ) (τ − ξ) dτ dξ . Γ(α)Γ(1 − α) dt 0 ξ d dt



(t − τ )α α

Z

τ

−α

It is easy to show that Z

t

(t − τ )α−1 (τ − ξ)−α dτ = Γ(α)Γ(1 − α).

ξ

Then d ℑt [u] (x) = dt

Z

1 u(x) = ℑ1 [u] (x) = Γ(α)

Z

t

u(ξx)dξ = u(tx). 0

In particular, 1

(1 − τ )α−1 τ −α B α [u] (τ x)dτ

0

and this proves the Lemma 1. Lemma 2

are valid. Proof



Let 0 < α < 1. Then for any x ∈ Ω, the equalities   B −α [B α [u]] (x) = B α B −α [u] (x) = u(x)

Let’s prove the first equality. By definition of B −α , we have Z 1 1 B −α [B α [u]] (x) = (1 − τ )α−1 τ −α B α [u] (τ x)dτ . Γ(α) 0

But by virtue of the equality (7), the last integral is equal to u(x), i.e., B −α [B α [u]] (x) = u(x),

x ∈ Ω.

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Now let’s prove the second equality. We get Z r   rα d B α B −α [u] (x) = (r − τ )−α B −α [u] (τ θ)dτ Γ(1 − α) dr 0 Z r  Z 1 rα d = (r − τ )−α (1 − s)α−1 s−α u(sτ θ)dsdτ . Γ(α)Γ(1 − α) dr 0 0

sx x Further, by taking into account that θ = |x| = |sx| , it is not difficult to verify the following equalities: Z r Z rs rα d rα d ξ dξ −α (r − τ ) u(sτ θ)dτ = (r − )−α u(ξθ) Γ(1 − α) dr 0 s s {sτ =ξ} Γ(1 − α) dr 0 Z rs α α−1 r s d = (sr − ξ)−α u(ξθ)dξ Γ(1 − α) dr 0 Z rs (sr)α d = (sr − ξ)−α u(ξθ)dξ Γ(1 − α) d(sr) 0 = B α [u] (sx).

Therefore   B α B −α [u] (x) =

1 Γ(α)

Z

1

(1 − s)α−1 s−α B α [u] (sx)ds.

0

Hence, using the equality (7), we obtain B α [B −α [u]] (x) = u(x), and the conclusions of the Lemma hold.  In [15], it is proved the following Lemma 3 Let u ∈ C 1 (Ω). Then for any x ∈ Ω, the following equality is fulfilled # Z 1 "X n ∂u(sx) u(x) = u(0) + xk ds. ∂xk 0

(9)

k=1

Further, since

n X

k=1

xk

∂u(x) ∂u(x) =r = B 1 [u](x), ∂xk ∂r

the equality (9) can be represented in the form of Z 1 u(x) = u(0) + s−1 B 1 [u](sx)ds.

(10)

0

Note that if u(0) 6= 0 then the operator B −1 is not defined in such functions. Obviously, B 1 [u](0) = 0, so one can consider actions of B −1 to the function B 1 [u](x). By definition of B −1 , we have Z 1   B −1 B 1 [u] (x) = s−1 B 1 [u](sx)ds. 0

By virtue of (10), 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)

is correct. Conversely, let u(0) = 0. Then the operator B −1 is defined for such functions, and Z 1  Z r    ∂ ∂ B 1 B −1 [u] (x) = r s−1 u(sx)ds = r ξ −1 u(ξθ)dξ = u(x). ∂r 0 ∂r 0

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Thus, we have proved the following result Lemma 4 For any x ∈ Ω, the following equalities hold: 1)   B −1 B 1 [u] (x) = u(x) − u(0);

2) If u(0) = 0 then

Lemma 5

  B 1 B −1 [u] (x) = u(x).

Let ∆2 u(x) = g(x), x ∈ Ω, 0 < α ≤ 1. Then for any x ∈ Ω, it is satisfied that h i −4 4 ∆2 B α [u] (x) = |x| B α |x| g (x). (11)

Proof Let 0 < α < 1. After a change of variables, the function B α [u] (x) can be represented in the form of Z 1 Z 1 1−α d (1 − ξ)−α B α [u] (x) = (1 − ξ)−α u(ξx)dξ + r u(ξx)dξ Γ(1 − α) 0 dr 0 Γ(1 − α) = I1 (x) + I2 (x). Since ∆2 u(x) = g(x), it is easy to show that ∆2 I1 (x) =

1−α Γ(1 − α)

Z

1

(1 − ξ)−α ξ 4 g(ξx)dξ.

0

Further, if v is a smooth function, then it is evident that   ∂ ∂ 2 ∆ r v(x) = r ∆2 v(x) + 4∆2 v(x). ∂r ∂r That’s why ∆2 I2 (x) = r

d dr R1

Z

1 0

(1 − ξ)−α 4 ξ g(ξx)dξ + 4 Γ(1 − α)

Z

0

1

(1 − ξ)−α 4 ξ g(ξx)dξ. Γ(1 − α)

To solve the integral 0 (1 − ξ)−α ξ 4 g(ξx)dξ, by making use of the change of variables, it can be transformed in to the following form: Z 1 Z r −α 4 α−5 (1 − ξ) ξ g(ξx)dξ = r (r − τ )−α τ 4 g(τ θ)dτ . 0

0

Then Z 1 (1 − ξ)−α 4 d ξ g(ξx)dξ dr 0 Γ(1 − α)   Z r d =r rα−5 (r − τ )−α τ 4 g(τ θ)dτ dr 0 Z r Z r d = (α − 5)rα−5 (r − τ )−α τ 4 g(τ θ)dτ + rα−4 (r − τ )−α τ 4 g(τ θ)dτ . dr 0 0 r

Hence, ∆2 I1 (x) + ∆2 I2 (x) Z Z 1 − α α−5 r (α − 5) α−5 r = r (r − τ )−α τ 4 g(τ θ)dτ + r (r − τ )−α τ 4 g(τ θ)dτ Γ(1 − α) Γ(1 − α) 0 0 Z r Z r rα−4 d 4 + (r − τ )−α τ 4 g(τ θ)dτ + rα−5 (r − τ )−α τ 4 g(τ θ)dτ Γ(1 − α) dr 0 Γ(1 − α) 0

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Z r rα−4 d = (r − τ )−α τ 4 g(τ θ)dτ Γ(1 − α) dr 0 Z r rα d = r−4 (r − τ )−α τ 4 g(τ θ)dτ Γ(1 − α) dr 0   = r−4 B α |x|4 g (x).

Let now α = 1. In this case B 1 [u](x) = r ∂u ∂r (x), and therefore   ∂u ∂∆2 u ∂g ∆2 B 1 [u](x) = ∆2 r (x) = r (x) + 4∆2 u(x) = r (x) + 4g(x). ∂r ∂r ∂r On the other hand,      ∂ ∂  4 + 4 g(x) = |x|−4 r r g(x) = |x|−4 B 1 |x|4 g (x), r ∂r ∂r

and Lemma 5 is proved.

4



Some Properties of a Solution of the Dirichlet Problem

To study the problem (1)–(3), we need some knowledge on the Green’s function of the Dirichlet problem for the equation (5). First, we enunciate a known result given in [16, 17]. Lemma 6 There exists a unique function G4,n (x, y) such that 1) It satisfies to the conditions ∂G ∆G4,n (x, y) = δ(x − y), G4,n (x, y)|∂Ω = 0, (x, y) = 0, ∂ν ∂Ω where δ(x − y) is the Dirac delta-function; 2) For any g ∈ L2 (Ω), the unique solution of the problem 2

∆ u(x) = g(x), x ∈ Ω,

u(x)|∂Ω = 0,

is represented in the form of u(x) =

Z

∂u (x) = 0 ∂ν ∂Ω

(12)

G4,n (x, y)g(y)dy.

Ω

G4,n (x, y) is called the Green’s function related to the problem (5). In works [16, 17] the explicit form of the Green’s function of the Dirichlet problem is obtained for the cases n ≥ 2. For example, in the case when n is odd or n is even and n > 4, the Green’s function of the problem (5) follows the expression 4−n # 2−n  4 − n y y 4−n G4,n (x, y) = c4,n |x − y| − x|y| − + x|y| − (1 − |x|2 )(1 − |y|2 ), (13) |y| 2 |y| 1 where c4,n = 2(4−n)(2−n) · ω1n and ωn = Γ( n2 )/(2 · π 2 ) is the area of the unit sphere. Let v be a solution of the Dirichlet problem (5). It is easy to show that if v1 and v2 are two solutions of the problems   ∆v (x) = 0, x ∈ Ω, 1  v1 (x) = ψ1 (x), x ∈ ∂Ω, n

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and

    ∆v2 (x) = 0,     v2 (x) = 1 ψ2 (x) − ∂v1 (x) ,  2 ∂ν

Vol.34 Ser.B

x ∈ Ω, x ∈ ∂Ω,

then any solution of the problem (5) is represented in the form of Z 2 v(x) = G4,n (x, y)g1 (y)dy + v1 (x) + (|x| − 1)v2 (x),

x ∈ Ω.

(14)

Ω

The following proposition establishes some properties of the function defined in (14). Lemma 7 Let v a be solution of problem (5). Then v(0) = 0 if and only if the following equality holds  Z  Z 1 1 ψ1 (y) − ψ2 (y) dSy = − G4,n (0, y)g(y)dy. (15) ωn ∂Ω 2 Ω Proof

Let v be solution of the problem (5). From (14) we have that Z v(0) = G4,n (0, y)g(y)dy + v1 (0) − v2 (0). Ω

Taking into account the Poisson integral expression for functions v1 and v2 ,  Z Z 2 2  1 1 − |x| 1 1 − |x| ∂v1 v1 (x) = ψ1 (y)dSy , v2 (x) = ψ2 (y) − (y) dSy , ωn ∂Ω |x − y|n 2ωn ∂Ω |x − y|n ∂ν we deduce that  Z 1 1 ∂v1 ψ1 (y) − ψ2 (y) dSy + (y)dSy . 2 2ωn ∂Ω ∂ν ∂Ω R On the other hand, since function v1 is harmonic in the ball Ω we know that ∂Ω = 0, in consequence  Z  1 1 v1 (0) − v2 (0) = ψ1 (y) − ψ2 (y) dSy , ωn ∂Ω 2 v1 (0) − v2 (0) =

1 ωn

Z



and the result holds.

∂v1 ∂r (y)dSy



Using the representation of the Green’s function given in (13), it is easy to find the values of G4,n (0, y). In what follows, for convenience, we will consider only the case of n odd or n even and n > 4. Then   4−n 4−n 2 G4,n (0, y) = c4,n |y| −1+ (1 − |y| ) . 2 ∂ Lemma 8 If we consider problem (5) with g(x) = (r ∂r + 4)f (x), then the condition (15) can be rewritten as Z Z |y|2 − 1 [2ψ1 (y) − ψ2 (y)] dSy = f (y)dy. (16) 2 ∂Ω Ω

Proof Let v be such that v(0) = 0, and it is given by expression (11) with g(x) = ∂ (r ∂r + 4)f (x) , with f ∈ C λ+1 (Ω). We remark that such a function exists because of Lemmas 5 and 7. From (10) we have that Z 1 v(x) = s−1 B 1 [v](sx)ds. 0

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Consider the integral I= and denote ρ = |y|. Then 1 2(2 − n)(4 − n) 1 = 2(2 − n)(4 − n)

I=

Z

G4,n (0, y)g(y)dy, Ω

  Z  1 4−n ∂f |y|4−n − 1 + (1 − |y|2 ) |y| (y) + 4f (y) dy ωn Ω 2 ∂|y| 1 n · [I + In2 ] , ωn 1 ·

where In1

Z 

 4−n ∂f 2 |y| −1+ (1 − |y| ) |y| (y)dy = 2 ∂ |y| Ω  Z Z 1  4−n ∂f (ρξ) (1 − ρ2 ) dρdξ, = ρn ρ4−n − 1 + 2 ∂ρ |ξ|=1 0  Z  4−n 4−n 2 In2 = 4 |y| −1+ (1 − |y| ) f (y)dy. 2 Ω 4−n

Consider the inner integral in In1 .  Z 1  ∂f (ρξ) 4−n (1 − ρ2 ) dρ ρn ρ4−n − 1 + 2 ∂ρ 0 ρ=1   4−n n 4 n 2 = ρ −ρ + ρ (1 − ρ ) f (ρξ) 2 ρ=0  Z 1 4 − n n−1 4−n − 4ρ3 − nρn−1 + nρ (1 − ρ2 ) − · 2ρn+1 f (ρξ)dρ 2 2 0 Z 1 Z 1 = −4 ρ4−n f (ρξ)ρn−1 dρ + n f (ρξ)ρn−1 dρ 0

0

4−n − ·n 2

Z

1

2

(1 − ρ )f (ρξ)ρ

0

n−1

(4 − n) · 2 dρ + 2

Z

1

ρ2 f (ρξ)ρn−1 dρ.

0

Hence, we get for In1 Z Z Z (4 − n)n In1 = −4 |y|4−n f (y)dy + n f (y)dy − (1 − |y|2 )f (y)dy 2 Ω Ω Ω Z (4 − n) · 2 2 + |y| f (y)dy. 2 Ω

It means that In1

Hence

+

In2

Z (4 − n)4 − (4 − n)n = (n − 4) f (y)dy + (1 − |y|2 )f (y)dy 2 Ω Ω Z (4 − n) · 2 + |y|2 f (y)dy 2 Ω Z Z (4 − n)(4 − n) = (n − 4) (1 − |y|2 )f (y)dy + + (1 − |y|2 )f (y)dy 2 Ω Ω Z (4 − n)(2 − n) 2 = (1 − |y| )f (y)dy. 2 Ω Z

1 ωn

Z

∂Ω



 1 ψ1 (y) − ψ2 (y) dSy 2

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=−

1 1 (4 − n)(2 − n) · · 2(2 − n)(4 − n) ωn 2

or, which is the same, Z

[2ψ1 (y) − ψ2 (y)] dSy = −

∂Ω

1 2

Z

Z

Vol.34 Ser.B

(1 − |y|2 )f (y)dy,

Ω

2

(1 − |y| )f (y)dy,

Ω

i.e., we obtain the equality (16).

5



The Proof of the Main Assertion

Let 0 < α < 1 and u be a solution of the problem (1)–(3). We denote v(x) = B α [u] (x). Then, using the equality (11), we obtain from Lemma 5   ∆2 v(x) = ∆2 B α [u] (x) = |x|−4 B α |x|4 f (x) ≡ g(x). Further, since

 ∂ α ∂  −α α α D [u](x) = rα+1 r · r D [u](x) ∂r ∂r  −α α α  ∂ α α α+1 ∂ =r r · r D [u](x) = r [r D [u]] (x) − αrα Dα [u](x) ∂r ∂r ∂ = r B α [u](x) − αB α [u](x), ∂r

B α+1 [u](x) = rα+1

we have v(x)|∂Ω = B α [u] (x)|∂Ω = ϕ1 (x) and

∂v(x) = B α+1 [u] (x) ∂Ω = ϕ2 (x) + αϕ1 (x). ∂ν ∂Ω

Thus, if u is a solution of the problem (1)–(3), then we obtain that v is a solution of problem (5) with   g(x) = |x|−4 B α |x|4 f (x), ψ1 (x) = ϕ1 (x), ψ2 (x) = ϕ2 (x) + αϕ1 (x). Further, since

 Z 1   d (1 − ξ)−α 4 g(x) = |x|−4 B α |x|4 f (x) = r +5−α ξ f (ξx)dξ, dr 0 Γ(1 − α)

we have that if f ∈ C λ+1 (Ω) then g ∈ C λ (Ω). But in this case, it is well known (see [18]) that if g ∈ C λ (Ω), ψ1 ∈ C λ+3 (∂Ω) and ψ2 ∈ λ+2 C (∂Ω), then the problem (5) has a solution that belongs to the class C λ+4 (Ω). To finish the proof of the main result we must verify that u(x) = B −α [v] (x) is a solution of problem (1)–(3). Indeed, for all x ∈ Ω it is satisfied that Z 1 1 2 2 −α ∆ u(x) = ∆ B [v] (x) = (1 − τ )α−1 τ 4−α ∆2 v(τ x)dτ Γ(α) 0 Z 1 1 = (1 − τ )α−1 τ 4−α g(τ x)dτ Γ(α) 0 Z 1   1 = (1 − τ )α−1 τ 4−α τ −4 |x|−4 B α |x|4 f (τ x)dτ Γ(α) 0

No.6

A.S. Berdyshev et al: ON SOLVABILITY OF A BOUNDARY VALUE PROBLEM

1 = Γ(α)

Z

1

0

|x|−4 = Γ(α)

Z

0

1

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  (1 − τ )α−1 τ −α |x|−4 B α |x|4 f (τ x)dτ   (1 − τ )α−1 τ −α B α |x|4 f (τ x)dτ

= |x|−4 |x|4 f (x) = f (x).

Further, using the second equality in (8), for all x ∈ ∂Ω we have   B α [u] (x)|∂Ω = B α B −α [v] (x) ∂Ω = v(x)|∂Ω = ϕ1 (x),

∂ B α+1 [u](x) ∂Ω = r B α [u](x) − αB α [u](x)|∂Ω ∂r     ∂ = r B α B −α [v] (x) − αB α B −α [v] (x) ∂Ω ∂r ∂ = r v(x) − αv(x)|∂Ω = ϕ2 (x) + αϕ1 (x) − αϕ1 (x) = ϕ2 (x). ∂r So, the function u(x) = B −α [v] (x) satisfies both the equation (1) and the boundary conditions (2) and (3) with 0 < α < 1. Let now α = 1 and let u be a solution of the problem (1)-(3). Define function v(x) = B 1 [u] (x). As in the previous case, one can verify that v is a solution of problem (5) with   ∂ + 4)f (x). g(x) = |x|−4 B 1 |x|4 f (x) = (r ∂r ∂u Since B 1 [u] (0) = r ∂r (x) x=0 = 0, the function v must satisfy, in addition, that v(0) = 0. As we have remarked above, any solution of problem (5) at smooth ψ1 , ψ2 and g, is given by expression (14). From Lemma 8 we know that v(0) = 0 if and only if condition (15) is fulfilled. ∂ In our case g(x) = (r ∂r + 4)f (x), and therefore condition (15) is rewritten as (16). Since 2ψ1 (x) − ψ2 (x) = 2ϕ1 (x) − ϕ2 (x) − ϕ1 (x) = ϕ1 (x) − ϕ2 (x), we obtain condition (4), which, from [12], is necessary and sufficient to ensure the existence of a solution for the problem (1)–(3). Let’s see know what happens with the solutions of problem (5). Notice that if condition (4) holds then v(0) = 0 and the function B −1 [v] (x) is well defined. In the sequel we verify that for any real constant C, the function u(x) = B −1 [v] (x) + C is a solution of problem (1)–(3). As in the previous case, one can verify that ∆2 u(x) = f (x) in Ω. Further, using the second equality of Lemma 4, we have   B 1 [u] (x) = B 1 B −1 [v] + C (x) = v(x). Hence,

B 1 [u] (x) ∂Ω = v(x)|∂Ω = ϕ1 (x),

and

    2 ∂ ∂ ∂ 2∂ u B [u](x) ∂Ω = r (x) = r −1 u(x) = r − 1 v(x) 2 ∂r ∂r ∂r ∂r ∂Ω ∂Ω ∂Ω ∂v(x) = − v(x) = ϕ2 (x) + ϕ1 (x) − ϕ1 (x) = ϕ2 (x) ∂ν ∂Ω 2

and the result is proved. Remark The main Theorem implies that for all α ∈ [0, 1) the solution of problem (1)– (3) satisfies the properties of a solution of the Dirichlet problem, and it has the properties of a solution of the Neumann problem in the case of α = 1.

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ACTA MATHEMATICA SCIENTIA

Vol.34 Ser.B

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