Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 82, pp. 1–21.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
SOLVABILITY OF FRACTIONAL ANALOGUES OF THE
NEUMANN PROBLEM FOR A NONHOMOGENEOUS
BIHARMONIC EQUATION
BATIRKHAN KH. TURMETOV
Abstract. In this article we study the solvability of some boundary value
problems for inhomogenous biharmobic equations. As a boundary operator
we consider the differentiation operator of fractional order in the Miller-Ross
sense. This problem is a generalization of the well known Neumann problems.
1. Introduction
Biharmonic equations appear in the study of mathematical models in several
real-life processes as, among others, radar imaging [3] or incompressible flows [11].
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 [2, 10]. By using numerical and iterative methods, Dirichlet and Neumann boundary problems for biharmonic equations
were studied in the papers [7, 8]. There are some works, for example [12], 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 [9], where regularity of solutions for nonlinear biharmonic equations was investigated. In [4] and the dissertation [13] various problems for complex biharmonic
and polyharmonic equations were investigated.
In this article 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 sense can be defined, in a sense to ([20], p.69), by the following
expression
Z r
1
α
J [u](x) =
(r − τ )α−1 u(τ θ)dτ .
Γ(α) 0
2000 Mathematics Subject Classification. 35J15, 35J25, 34B10, 26A33, 31A05, 31B05.
Key words and phrases. Biharmonic equation; fractional derivative; Miller-Ross operator;
Neumann problem.
c
2015
Texas State University - San Marcos.
Submitted January 29, 2015. Published April 3, 2015.
1
2
B. KH. TURMETOV
EJDE-2015/82
In what follows, we assume that J 0 [u](x) = u(x) for all x ∈ Ω. Let m − 1 < α ≤
m, m = 1, 2, . . . . The following expressions
dm u
dm m−α
J
[u](x), C Dα [u](x) = J m−α [ m ](x),
m
dr
dr
are called, respectively, derivatives of α order in Riemann-Liouville and Caputo
d
sense [20]. Here dr
is a differentiation operator of the form
RL D
α
[u](x) =
n
X xi ∂
d
=
,
dr
r ∂xi
i=1
d dk−1
dk
=
),
(
drk
dr drk−1
k = 2, 3, . . . .
Let the parameter j take one of the values, j = 0, 1, . . . , m and consider the set of
operators
dm−j
dj
Djα [u](x) = m−j J m−α j u(x).
dr
dr
d
If j ≥ 1 and D = dr
, then
Djα = |D · D {z
· · · · · D} ·C Dα−j .
m−j
This operator is called derivative of α order in Miller-Ross sense [24]. Denote
Bjα u(x) = rα Djα u(x),
Z 1
1
(1 − s)α−1 s−α u(sx)ds.
B −α u(x) =
Γ(α) 0
Let 0 < α ≤ 2. Consider the following problems in the domain Ω.
Problem 1.1. Let 0 < α < 2. Find a function u(x) ∈ C 4 (Ω) ∩ C(Ω̄) such that
B1α+k [u](x) ∈ C(Ω̄), k = 0, 1 satisfying the equation
∆2 u(x) = g(x),
x ∈ Ω,
(1.1)
x ∈ ∂Ω,
(1.2)
= f2 (x), x ∈ ∂Ω.
(1.3)
and the boundary value conditions:
D1α [u](x) = f1 (x),
D1α+1 [u](x)
Problem 1.2. Let 1 < α ≤ 2. Find a function u(x) ∈ C 4 (Ω) ∩ C(Ω̄) such that
B2α+k [u](x) ∈ C(Ω̄), k = 0, 1 satisfying equation (1.1) and the boundary value
condition:
D2α [u](x) = f1 (x),
D2α+1 [u](x)
= f2 (x),
x ∈ ∂Ω,
(1.4)
x ∈ ∂Ω.
(1.5)
Note that the boundary value problems with boundary operators of fractional
order for elliptic equations of the second order have been studied in [5, 17, 21, 22,
25, 26, 27, 28, 29, 32, 33]. Moreover, in [6] for the equation (1.1) the boundary-value
problem with the conditions
D0α [u](x) = f1 (x), D0α+1 [u](x) = f2 (x),
x ∈ ∂Ω,
0 < α ≤ 1 has been studied.
du
du
Note that for all x ∈ ∂Ω we have the equality r du
dr = dr = dν , where ν is
a vector of outward normal to ∂Ω. It is well known (see e.g. [15]) that for all
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SOLVABILITY OF FRACTIONAL ANALOGUES
3
k
d
d
d
d
x ∈ ∂Ω the operator r dr
(r dr
− 1) . . . (r dr
− k + 1) coincides with the operator dν
k,
k = 1, 2, . . . . Then in the case of α = 1 for all x ∈ ∂Ω we obtain
du(x)
d d
du 2 2
d2 u(x)
D11 u(x) =
=
r
r
=
, r Dj u(x) = r2
−
1
u(x).
dr
dν
dr2
dr dr
Consequently, for values α = 1 or α = 2, problems 1.1 and 1.2 are analogues of
the Neumann problem for the equation (1.1). The considered problems in the case
of α = 1 have been studied in [16], and in the case of α = 2 in [30]. It is proved
that in the case of α = 1 for solvability of the problem the following conditions are
necessary and sufficient:
Z
Z
1
2
(1 − |x| )g(x)dx =
[f2 (x) − f1 (x)]dsx ,
(1.6)
2 Ω
∂Ω
and in the case of α = 2,
Z
Z
1
(1 − |x|2 )Γ3 [g(x)]dx =
f2 (x)dSx ,
(1.7)
2 Ω
∂Ω
Z
Z
1
xk (1 − |x|2 )Γ4 [g](x)dx =
xk [f2 (x) − f1 (x)]dSx , k = 1, 2, . . . , n, (1.8)
2 Ω
∂Ω
∂
where Γc [u](x) = (r ∂r
+ c)u(x), c > 0.
Note that the Neumann problem in the case of polyharmonic equation was studied in [18, 19, 31].
2. Properties of the operators Bjα and B −α
We assume that the function u(x) is smooth enough in the domain Ω. The
following proposition can be proved by direct calculation.
d
d
Lemma 2.1. Let v1 (x) = r du(x)
dr , v2 (x) = r dr (r dr − 1)u(x). Then the following
equalities hold:
v1 (0) = v2 (0) = 0,
∂v2
(0) = 0,
∂xk
k = 1, 2, . . . , n.
(2.1)
(2.2)
Similar propositions hold for the function Bjα [u](x), j = 0, 1.
Lemma 2.2. Let 0 < α ≤ 2. Then the following equalities hold:
B1α [u](0) = 0,
B2α [u](0) = 0,
∂B2α [u](0)
∂xi
= 0,
(2.3)
i = 1, 2, . . . , n.
(2.4)
Proof. Let 0 < α < 1. Then by the definition of the operator B1α for the function
B1α [u](x) we have
B1α [u](x)
Z r
Z r
rα
du
rα
d
du
(r − τ )−α (τ θ)dτ =
(r − τ )1−α (τ θ)dτ
Γ(1 − α) 0
dτ
Γ(2 − α) dr 0
dτ
Z r
h
i
α
1−α
τ
=r
r
d (r − τ )
=
u(τ θ)
+
(r − τ )−α u(τ θ)dτ
Γ(1 − α) dr
1−α
τ =0
0
Z 1
i
rα
d h r1−α
=
u(0) + r1−α
(1 − ξ)−α u(ξx)dξ
−
Γ(1 − α) dr
1−α
0
=
4
B. KH. TURMETOV
=−
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u(0)
du1 (x)
+ (1 − α)u1 (x) + r
,
Γ(1 − α)
dr
where
1
u1 (x) =
Γ(1 − α)
1
Z
(1 − ξ)−α u(ξx)dξ.
0
Therefore,
B1α [u](x) = −
u(0)
du1 (x)
+ (1 − α)u1 (x) + r
,
Γ(1 − α)
dr
x ∈ Ω.
(2.5)
Hence, by equality (2.1) we obtain
du1 (x)
u(0)
+ (1 − α) lim u1 (x) + lim r
x→0
x→0
Γ(1 − α)
dr
Z 1
u(0)
(1 − α)u(0)
=−
+
(1 − ξ)−α dξ
Γ(1 − α)
Γ(1 − α) 0
(1 − α)u(0)
u(0)
+
= 0.
=−
Γ(1 − α)
Γ(2 − α)
lim B1α [u](x) = −
x→0
Equality (2.3) is proved for the case 0 < α < 1.
No let 1 < α < 2 and j = 1. Then by definition of B1α we have
Z r
rα
d
du
α
B1 [u](x) =
(r − τ )1−α (τ θ)dτ
Γ(2 − α) dr 0
dτ
α
2 Z r
2−α
r
d
(r − τ )
du
=
(τ θ)dτ
Γ(1 − α) dr2 0
(2 − α) dτ
τ =r Z r
i
d2 h (r − τ )2−α
rα
1−α
u(τ
θ)
+
(r
−
τ
)
u(τ
θ)dτ
=
Γ(2 − α) dr2
2−α
τ =0
0
Z 1
i
α
2 h
2−α
r
d
r
2−α
1−α
=
−
u(0)
+
r
(1
−
ξ)
u(ξx)dξ
Γ(2 − α) dr2
2−α
0
du2 (x)
d2
(1 − α)u(0)
+ (1 − α)(2 − α)u2 (x) + 2(2 − α)r
+ r2 2 u2 (x),
=−
Γ(2 − α)
dr
dr
where
1
u2 (x) =
Γ(2 − α)
Z
1
(1 − ξ)1−α u(ξx)dξ.
0
Therefore,
(1 − α)
u(0) + (1 − α)(2 − α)u2 (x)
Γ(2 − α)
du2 (x)
d
d
+ 2(2 − α)r
+ r (r − 1)u2 (x), x ∈ Ω.
dr
dr dr
Then, taking into account the equalities (2.1) and (2.2), we obtain
B1α [u](x) = −
(1 − α)
u(0) + (1 − α)(2 − α) lim u2 (x)
x→0
Γ(2 − α)
Z
(1 − α)u(0) (1 − α)(2 − α)u(0) 1
+
(1 − ξ)1−α dξ
=−
Γ(2 − α)
Γ(2 − α)
0
(1 − α)u(0) (1 − α)(2 − α)u(0)
=−
+
= 0.
Γ(2 − α)
Γ(3 − α)
lim B1α [u](x) = −
x→0
(2.6)
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SOLVABILITY OF FRACTIONAL ANALOGUES
5
Equality (2.3) is proved for the case 1 < α < 2, j = 1.
Now we turn to the proof of the first equality of (2.4). By definition of B2α we
have
Z r
rα
d2 u
(r − τ )1−α 2 (τ θ)dτ
B2α [u](x) =
Γ(2 − α) 0
dτ
α
2 Z r
r
(r − τ )3−α d2 u
d
=
(τ θ)dτ
2
Γ(1 − α) dr 0 (2 − α)(3 − α) dτ 2
(1 − α)u(0)
r
du(0)
=−
−
+ (1 − α)(2 − α)u2 (x)
Γ(2 − α)
Γ(2 − α) dr
du2 (x)
d2 u2 (x)
+ 2(2 − α)r
.
+ r2
dr
dr2
Thus,
(1 − α)u(0)
r
du(0)
−
+ (1 − α)(2 − α)u2 (x)
Γ(2 − α)
Γ(2 − α) dr
du2 (x)
d d
+ 2(2 − α)r
+r
r − 1 u2 (x), x ∈ Ω.
dr
dr dr
Equalities (2.1) imply
d d
du2 (x) = 0, r
r − 1 u2 (x)
= 0.
r
dr x=0
dr dr
x=0
Then from representation (2.1) we obtain
B2α [u](x) = −
lim B2α [u](x) = −
x→0
(2.7)
(1 − α)u(0) (1 − α)(2 − α)u(0) Γ(2 − α)Γ(1)
+
= 0.
Γ(2 − α)
Γ(2 − α)
Γ(3 − α)
Further, we denote yi = τ θi , i = 1, 2, . . . , n. Then
n
n
X ∂u(τ θ)
du(τ θ) X ∂u(τ θ) dyi
=
=
θi
.
dτ
∂yi dτ
∂yi
i=1
i=1
Since θ = x/r, θi = xi /r, it follows that
n
n
X
X
r
du(0)
r
xi ∂u(τ θ) 1
∂u(0)
xi
=
=
.
Γ(2 − α) dτ
Γ(2 − α) i=1 r ∂yi τ =0
Γ(2 − α) i=1
∂yi
Thus, for any k = 1, 2, . . . , n,
r
du(0) i
1
∂u(0)
∂ h
−
=−
.
∂xk
Γ(2 − α) dτ
Γ(2 − α) ∂yk
It is obvious that
∂ 1−α
−
u(0) = 0.
∂xk
Γ(2 − α)
∂u ∂yk
∂u
Further, for any k = 1, 2, . . . , n, the equality ∂x∂ k u(ξx) = ∂y
= ξ ∂y
holds.
k ∂xk
k
Hence,
∂
∂u(0)
u(ξx)x=0 = ξ
.
∂xk
∂yk
Consequently,
∂
1
∂u(0)
u2 (x)x=0 =
.
∂xk
(3 − α)(2 − α)Γ(2 − α) ∂yk
6
B. KH. TURMETOV
d
2 (x)
Further, by the definition of r dr
we have r dudr
=
EJDE-2015/82
Pn
i=1
2 (x)
xi ∂u∂x
. Thus,
i
n
∂ du2 (x) X ∂ 2 u2 (x) ∂u2 (x)
r
=
+
.
xi
∂xk
dr
∂xk ∂xi
∂xk
i=1
Therefore,
n
X
du2 (x) ∂ 2 u2 (x) ∂u2 (x) ∂ =
2(2
−
α)[
2(2 − α)r
x
+
i
x=0
x=0
∂xk
dr
∂xk ∂xi
∂xk
i=1
=
2
∂u(0)
.
(3 − α)Γ(2 − α) ∂yk
Further, by (2.2), it follows that
d
∂ d
r (r − 1)u2 (x) x=0 = 0.
∂xi dr dr
By using all these calculations, from the representation of the function B2α [u](x),
we obtain
∂u(0)
∂B2α [u](0)
∂u(0) 1
1 − α ∂u(0)
2
−
= 0.
=
+
+
∂xk
Γ(2 − α)
∂yk
(3 − α) ∂yk
(3 − α) ∂yk
d
d
2
If α = 1 or α = 2, then B11 u(x) = r du(x)
dr , B1 u(x) = r dr (r dr − 1)u(x), and for these
functions the statement of the lemma follows from the lemma 2.1.
The following proposition was proved in [27].
Lemma 2.3. Let 0 < α ≤ 1. Then for any x ∈ Ω the following equalities hold:
B −α [B1α [u]](x) = u(x) − u(0),
(2.8)
B1α [B −α [u]](x) = u(x).
(2.9)
and if u(0) = 0, then
A similar statement is true in the case of 1 < α < 2.
Lemma 2.4. Let 1 < α < 2, j = 1. Then equalities (2.8) and (2.9) hold.
Proof. Let us prove equality (2.8). Let x ∈ Ω and t ∈ (0, 1]. Consider the function
Z t
1
=t [u](x) =
(t − τ )α−1 τ −α B1α [u](τ x)dτ .
Γ(α) 0
By using the definition of B1α , we have
Z t
1
d
d
=t [u](x) =
(t − τ )α−1 J 2−α [ u](τ x)dτ .
Γ(α) 0
dτ
dτ
Integrating the above integral by parts, we obtain
Z
α−1 t
d
=t [u](x) =
(t − τ )α−2 J 2−α [ u](τ x)dτ
Γ(α) 0
dτ
Z t
1
d
=
(t − τ )α−2 J 2−α [ u](τ x)dτ =u(tx) − u(0).
Γ(α − 1) 0
dτ
If we put t = 1, then
1
u(x) = u(0) +
Γ(α)
Z
0
1
(1 − τ )α−1 τ −α B1α [u](τ x)dτ = u(0) + B −α [B1α [u]](x).
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SOLVABILITY OF FRACTIONAL ANALOGUES
7
Equality (2.8) is proved.
We turn to the proof of (2.9). Since u(0) = 0, then the operator B −α is determined for these functions, and, therefore, applying the operator B1α to the function
B −α [u](x), we have
d 2−α d −α
J
B [u](x)
dr
dr
α
2 Z r
(r − τ )2−α d −α
r
d
=
B [u](τ θ)dτ .
Γ(2 − α) dr2 0
2 − α dτ
B1α [B −α [u]](x) = rα
After the change of variables τ s = ξ, the function
Z 1
1
−α
(1 − s)α−1 s−α u(τ sθ)ds
B [u](τ θ) =
Γ(α) 0
will be represented as
B −α [u](τ θ) =
1
Γ(α)
Z
τ
(τ − ξ)α−1 ξ −α u(ξθ)dξ = J α [ξ −α u].
0
Then integrating by parts, we obtain
B1α [B −α [u]](x) = rα
2
d2 2−α α −α
α d
[J
[J
[ξ
u]]](x)
=
r
[J 2 [ξ −α u]](x) = u(x).
dr2
dr2
Lemma 2.5. Let 1 < α ≤ 2. Then for any x ∈ Ω the following equalities hold:
B −α [B2α [u]](x) = u(x) − u(0) −
n
X
i=1
and if u(0) = 0 and
∂u(0)
∂xi
xi
∂u(0)
,
∂xi
(2.10)
= 0 for i = 1, 2, . . . , n, then
B2α [B −α [u]](x) = u(x).
(2.11)
Proof. Let us prove equality (2.10). As in the proof of (2.8) we consider the function
Z t
1
=t [u](x) =
(t − τ )α−1 τ −α B2α [u](τ x)dτ, t ∈ (0, 1].
Γ(α) 0
By using the definition of B2α , we have
Z t
1
d2
(t − τ )α−1 J 2−α [ 2 u](τ x)dτ .
=t [u](x) =
Γ(α) 0
dτ
But this function by the definition of the fractional order integral has the form
Z
α−1 t
d
d2 (t − τ )α−2 J 2−α [ u](τ x)dτ = J α J 2−α [ 2 u] (x).
=t [u](x) =
Γ(α) 0
dτ
dτ
Since J α J 2−α = J α+2−α = J 2 ,
Z t
d2
d2
d
(t − τ ) 2 u(τ x)dτ = −t u(0) + u(tx) − u(0).
=t [u](x) = J 2 [ 2 u] =
dτ
dτ
dτ
0
Further, since
n
n
X ∂u(τ x) dyi X ∂u(τ x)
d
u(τ x) =
=
xi
,
dτ
∂yi dτ
∂yi
i=1
i=1
8
B. KH. TURMETOV
it follows that
n
EJDE-2015/82
n
X ∂u(0) X ∂u(0)
d
u(0) =
≡
.
xi
xi
dτ
∂yi
∂xi
i=1
i=1
If in the integral =t [u](x) we set t = 1, then
Z 1
n
X
∂u(0)
1
u(x)−u(0)−
(1 − τ )α−1 τ −α B2α [u](τ x)dτ ≡ B −α [B2α [u]](x).
xi
=
∂x
Γ(α)
i
0
i=1
Equality (2.10) is proved.
−α
is defined on
If u(0) = 0 and ∂u(0)
∂xi = 0, i = 1, 2, . . . , n, then the operator B
α
these functions. Applying B2 we obtain
Z r
rα
d2
α
−α
B2 [B [u]](x) =
(r − τ )1−α 2 B −α [u](τ θ)dτ.
Γ(2 − α) 0
dτ
We represent the function B −α [u](τ θ) as
Z 1
Z τ
1
1
B −α [u](τ θ) =
(1 − s)α−1 s−α u(sτ θ)ds =
(τ − ξ)α−1 ξ −α u(ξθ)dξ .
Γ(α) 0
Γ(α) 0
Since α − 1 > 0, the following equality holds
Z
α−1 τ
d −α
B [u](τ θ) =
(τ − ξ)α−2 ξ −α u(ξθ)dξ ≡ J α−1 [ξ −α u](τ θ).
dτ
Γ(α) 0
It is easy to check the following equalities:
Z r
d
(r − τ )2−α d α−1 −α
rα
α
−α
J
[ξ u](τ θ)dτ
B2 [B [u]](x) =
Γ(2 − α) dr 0
2 − α dτ
Z r
d
d
= rα J[ξ −α u](x) = rα
ξ −α u(ξθ)dξ
dr
dr 0
= rα r−α u(x) = u(x).
Let 0 < α ≤ 2, and consider the functions:
g1,α (x) = rα−5 J 1−α [r4 g](x)
Z r
rα−5
≡
(r − τ )−α τ 4 g(τ θ)dτ ,
Γ(1 − α) 0
0 < α ≤ 1.
g2,α (x) = rα−6 J 2−α [r4 g](x)
Z r
rα−6
(r − τ )1−α τ 4 g(τ θ)dτ , 1 < α ≤ 2.
≡
Γ(2 − α) 0
Since J 0 [r4 g](x) = r4 g(x), it follows that g1,1 (x) = g2,2 (x) = g(x).
(2.12)
(2.13)
Lemma 2.6. Let 0 < α ≤ 2, and ∆2 u(x) = g(x) for x ∈ Ω. Then for any x ∈ Ω
and j = 1, 2 the following statements hold:
(1) if 0 < α ≤ 1, then
∆2 B1α [u](x) = (1 − α)g1,α (x) + Γ4 [g1,α ](x);
(2.14)
(2) if 1 < α ≤ 2, j = 1, 2, then
2
∆ Bjα [u](x) = (1 − α)(2 − α)g2,α (x) + 2(2 − α)Γ4 [g2,α ](x) + Γ4 [Γ3 [g2,α ]](x). (2.15)
EJDE-2015/82
SOLVABILITY OF FRACTIONAL ANALOGUES
9
Proof. Note that for u(x) the following equality holds:
∆2 [r
d
d
d
u(x)] = r ∆2 u(x) + 4∆2 u(x) = (r + 4)∆2 u(x) ≡ Γ4 [∆2 u](x).
dr
dr
dr
Then when α = 1 we obtain
∆2 B11 [u](x) = Γ4 [g1,1 ](x) = Γ4 [g](x);
and when α = 2 we have
∆2 B22 [u](x) = Γ3 [Γ4 [g1,2 ]](x) = Γ4 [Γ3 [g]](x).
Consequently, in these two values of α the equalities (2.14) and (2.15) are proved.
Let 0 < α < 1. Using the representation of the function B1α [u](x) in (2.5), we
obtain
∆2 B1α [u](x) = (1 − α)∆2 u1 (x) + Γ4 [∆2 u1 ](x).
Since ∆2 u(x) = g(x),
∆2 u1 (x) =
1
Γ(1 − α)
Z
1
(1 − ξ)−α ξ 4 g(ξx)dξ =
0
rα−5
Γ(1 − α)
Z
r
(r − τ )−α τ 4 g(τ θ)dτ ;
0
i.e. ∆2 u1 (x) = g1,α (x). Thus, for the functions B1α [u](x) we obtain the equality
(2.14).
Let 1 < α < 2, j = 1. Then the representation (2.6) implies:
∆2 B1α [u](x) = (1 − α)(2 − α)∆2 u2 (x) + 2(2 − α)Γ4 [∆2 u2 ](x) + Γ4 [Γ3 [∆2 u2 ]](x)(x),
Further, taking into account ∆2 u(x) = g(x) for ∆2 u2 (x), we obtain
Z 1
1
∆2 u2 (x) =
(1 − ξ)1−α ξ 4 g(ξx)dξ
Γ(2 − α) 0
Z r
rα−6
=
(r − τ )1−α τ 4 g(τ θ)dτ = g2,α (x),
Γ(2 − α) 0
i.e. for the functions ∆2 B1α [u](x) the representation (2.15) holds.
Analogously, to the case 1 < α < 2 for j = 2, the representation (2.7), by the
equality
n
n
X
X
r
du(0)
r
xi ∂u(τ θ) 1
∂u(0)
=
=
xi
,
Γ(2 − α) dτ
Γ(2 − α) i=1 r ∂yi τ =0
Γ(2 − α) i=1
∂yi
yields the equality (2.15).
Lemma 2.7. Let 0 < α ≤ 2 and the functions g1,α (y), g2,α (y) be defined by the
equalities (2.12) and (2.13), respectively. Then for any x ∈ Ω and j = 1, 2 the
following equalities hold:
(1) if 0 < α ≤ 1, then
∆2 B1α [u](x) = |x|−4 B1α [|x|4 g](x);
(2.16)
(2) if 1 < α ≤ 2, then, for j = 1, 2,
∆2 Bjα [u](x) = |x|−4 Bjα [|x|4 g](x).
(2.17)
10
B. KH. TURMETOV
EJDE-2015/82
d
Proof. Since r dr
[|x|4 g] = |x|4 Γ4 [g](x), then we have the equality
d
[|x|4 g] = Γ4 [g](x) = Γ4 [∆2 u](x) = ∆2 Γ0 [u](x) ≡ ∆2 B11 [u](x).
dr
d
Further, if we denote r dr
[|x|4 g](x) = f (x), then
|x|−4 r
d2
d
d
[u](x)) = ∆2 (r (r − 1)[u](x))
dr2
dr dr
d
= Γ4 [Γ3 [∆2 u]](x) = Γ3 [Γ4 [g]](x) = (r + 3)(|x|−4 f )
dr
d
d
= r (|x|−4 f ) + 3(|x|−4 f ) = |x|−4 (r − 1)f.
dr
dr
∆2 B22 [u] = ∆2 (r2
Thus
d
d2
d
− 1)(r [|x|4 g])(x) = |x|−4 r2 2 [|x|4 g] = |x|−4 B22 [|x|4 g].
dr
dr
dr
Therefore, equalities (2.16) and (2.17) in the case of integer values of α is proved.
In the case of fractional values of α we use the equalities (2.14) and (2.15). To do it
we transform the functions gj,α (x), j = 1, 2. After changing the variable ξ = r−1 τ
the integral, representing the function g1,α (x), can be rewritten in the following
form
Z r
rα−5
g1,α (x) =
(r − τ )−α τ 4 g(τ θ)dτ .
Γ(1 − α) 0
Then
Z
1 − α α−5 r
(1 − α)g1,α (x) + Γ4 [g1,α ](x) =
r
(r − τ )−α τ 4 g(τ θ)dτ
Γ(1 − α)
0
Z r
d
rα
−4
=r
(r − τ )−α τ 4 g(τ θ)dτ .
Γ(1 − α) dr 0
∆2 B22 [u] = |x|−4 (r
We transform the above integral as follows:
Z r
rα
d
(r − τ )−α τ 4 g(τ θ)dτ
Γ(1 − α) dr 0
Z r
d
rα
d(r − τ )1−α
=
τ 4 g(τ θ)
Γ(1 − α) dr 0
−(1 − α)
Z r
n
o
α
1−α
r
d
(r − τ )
d 4
=
[τ g(τ θ)]dτ
Γ(1 − α) dr 0
(1 − α) dτ
Z r
rα
d
+
(r − τ )−α [τ 4 g(τ θ)]dτ ≡ B1α [|x|4 g](x).
Γ(1 − α) 0
dτ
Thus,
∆2 B1α [u](x) = |x|−4 B1α [|x|4 g](x), x ∈ Ω.
Let 1 < α < 2 and j = 1. Then after changing variables ξ = r−1 τ , for the function
g2,α (x) we obtain
Z r
rα−6
g2,α (x) =
(r − τ )1−α τ 4 g(τ θ)dτ .
Γ(2 − α) 0
Further, if f (x) is a smooth function then
r
d α−6
d
[r
f ] = rα−6 r + α − 6 f (x).
dr
dr
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SOLVABILITY OF FRACTIONAL ANALOGUES
11
Thus,
(1 − α)(2 − α)g2,α (x) + 2(2 − α)Γ4 [g2,α ](x) + Γ4 [Γ3 [g2,α ]](x)
d
d
= (r + 4) rα−6 (r + 3 + 2(2 − α) + α − 6)J 2−α [τ 4 g] (x)
dr
dr
Z r
i
d2 h
1
+ rα−4 2
(r − τ )1−α τ 4 g(τ θ)dτ .
dr Γ(2 − α) 0
We transform the above integral as follows:
Z r
1
(r − τ )1−α τ 4 g(τ θ)dτ
Γ(2 − α) 0
Z r
d(r − τ )2−α
1
τ 4 g(τ θ)
=
Γ(2 − α) 0
−(2 − α)
Z r
τ =r
2−α
1
d
(r − τ )
+
(r − τ )1−α [τ 4 g(τ θ)]dτ
= −τ 4 g(τ θ)
(2 − α)Γ(2 − α) τ =0 Γ(2 − α) 0
dτ
≡ D1α [τ 4 g(τ θ)].
Hence,
∆2 B1α [u](x) = |x|−4 B1α [|x|4 g].
Similarly, we consider the case 1 < α < 2, j = 2.
3. Some properties of the solution of the Dirichlet problem
Consider the Dirichlet problem
∆2 v(x) = g1 (x),
x∈Ω
(3.1)
dv(x)
= ϕ2 (x), x ∈ ∂Ω.
dν
It is known that (see e.g. [1]), if g1 (x), ϕ1 (x) and ϕ2 (x) are smooth functions, then
the solution of (3.1) exists and is unique. The solution of (3.1) is represented as:
Z
v(x) =
G2,n (x, y)g1 (y)dy + w(x),
(3.2)
v(x) = ϕ1 (x),
Ω
where G2,n (x, y) is the Green function of (3.1), and w(x) is a solution of (3.1) when
g1 (x) = 0; i.e.,
∆2 w(x) = 0,
dw(x)
= ϕ2 (x),
dν
w(x) = ϕ1 (x),
Denote
x ∈ Ω,
x ∈ ∂Ω.
Z
v1 (x) =
G2,n (x, y)g1 (y)dy.
Ω
The explicit form of the Green’s function for the Dirichlet problem is obtained
for the cases n ≥ 2 in [14]. For example, in the case when n is odd or n is even and
n > 4, the Green’s function of the problem (3.1) follows from the expression
h
y 4−n
G2,n (x, y) = d2,n |x − y|4−n − x|y| −
|y|
i
n y 2−n
(2 − ) x|y| −
(1 − |x|2 )(1 − |y|2 ) ,
2
|y|
12
B. KH. TURMETOV
EJDE-2015/82
n/2
1
2π
where d2,n = ω1n 2(n−4)(n−2)
and ωn = Γ(n/2)
− is area of the unit sphere.
Furthermore, for convenience, we consider only the case when n-odd or n-even
and n > 4. The following proposition was proved in [25].
Lemma 3.1. Let ϕ1 (x), ϕ2 (x) be smooth functions. Then the following equalities
hold:
Z
1
[2ϕ1 (y) − ϕ2 (y)]dSy ,
(3.3)
w(0) =
2ωn ∂Ω
Z
n
∂w(0)
=
yk [3ϕ1 (y) − ϕ2 (y)]dSy , k = 1, 2, . . . , n.
(3.4)
∂xk
2ωn ∂Ω
Lemma 3.2. Let g2 (x) be a smooth function. Then
(1) if g1 (x) = Γ4 [g2 ](x), then
Z
1
v1 (0) =
(1 − |y|2 )g2 (y)dy;
4ωn Ω
(2) if g1 (x) = Γ3 [Γ4 [g2 ]](x) then
Z
∂v1 (0)
n
=
yk (1 − |y|2 )Γ4 [g](y)dy,
∂xk
4ωn Ω
Now we study the values of v1 (0) and
∂v1 (0)
∂xk ,
k = 1, 2, . . . , n.
(3.5)
(3.6)
k = 1, 2, . . . , n, when
g1 (x) = (1 − α)g1,α (x) + Γ4 [g1,α ](x), and
g1 (x) = (1 − α)(2 − α)g2,α (x) + 2(2 − α)Γ4 [g2,α ](x) + Γ4 [Γ3 [g2,α ]](x).
(3.7)
(3.8)
Lemma 3.3. Let 0 < α ≤ 2, j = 1, 2, g(x) be a smooth function, and gj,α (x) be
defined by (2.12) or (2.13). Then
(1) if 0 < α ≤ 1 and g1 (x) is defined by (3.7), then
Z
Z h
1
1 − |y|2
1−α
v1 (0) =
g1,α (y)dy +
|y|4−n − 1
2ωn Ω
2
ωn 2(n − 2)(n − 4) Ω
(3.9)
i
n
2
+ (2 − )(1 − |y| ) g1,α (y)dy;
2
(2) if 1 < α < 2, j = 1 and the function g1 (x) is defined by (3.8), then
Z
Z
1
1 − |y|2
2(2 − α)
1 − |y|2
v1 (0) =
Γ3 [g2,α ](y)dy +
g2,α (y)dy
2ωn Ω
2
2ωn
2
Ω
(3.10)
Z
(1 − α)(2 − α)
n
4−n
2
+
[|y|
− 1 + (2 − )(1 − |y| )]g2,α (y)dy;
ωn 2(n − 2)(n − 4) Ω
2
(3) if 1 < α ≤ 2, j = 2 and g1 (x) is defined by (3.8), then for v1 (0) we have the
equality (3.10), moreover
Z
∂v1 (0)
n
=
yk (1 − |y|2 )Γ4 [g2,α ](y)dy
∂xk
4ωn Ω
Z
1 2(2 − α)
2−n
+
yk |y|2−n − 1 +
(1 − |y|2 ) Γ4 [g2,α ](y)dy (3.11)
2ωn (n − 2) Ω
2
Z
(1 − α)(2 − α)
2
−n
+
yk [|y|2−n − 1 +
(1 − |y|2 )]g2,α (y)dy,
2ωn (n − 2)
2
Ω
k = 1, . . . , n.
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SOLVABILITY OF FRACTIONAL ANALOGUES
13
Proof. When α is an integer, equalities (3.9) and (3.11) follows from Lemma 3.3.
Let 0 < α < 1, j = 1 and g1 (x) be represented in the form (3.7). Then
Z
v1 (x) =
G2,n (x, y)g1 (y)dy
Ω
Z
Z
= (1 − α)
G2,n (x, y)g1,α (y)dy
+
G2,n (x, y)Γ4 [g1,α ](y)dy.
Ω
Ω
From the first statement of Lemma 3.2, for the second integral of the above equality
we obtain
Z
Z
1
1 − |y|2
G2,n (0, y)Γ4 [g1,α ](y)dy =
g1,α (y)dy.
2ωn Ω
2
Ω
For the first integral, using the representation of the functions G2,n (x, y), we have
Z
Z
n
G2,n (0, y)g1,α (y)dy = d2,n
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g1,α (y)dy.
2
Ω
Ω
Thus, for v1 (0) we obtain the equality (3.9). Let 1 < α < 2, j = 1. Then, using the
equality (3.8), we obtain
Z
v1 (x) =
G2,n (x, y)g1 (y)dy
Ω
Z
= (1 − α)(2 − α)
G2,n (x, y)g2,α (y)dy
Ω
Z
Z
+ 2(2 − α)
G2,n (x, y)Γ4 [g2,α ](y) +
G2,n (x, y)Γ4 [Γ3 [g2,α ]](y)dy
Ω
Ω
By (3.5), for the second and third integrals of the last equality we obtain
Z
Z
1
1 − |y|2
G2,n (0, y)Γ4 [g2,α ](y)dy =
g2,α (y)dy,
2ωn Ω
2
Ω
Z
Z
1
1 − |y|2
Γ3 [g2,α ](y)dy.
G2,n (0, y)Γ4 [Γ3 [g2,α ]](y)dy =
2ωn Ω
2
Ω
For the first integral we have
Z
Z
n
G2,n (0, y)g1,α (y)dy = d2,n [|y|4−n − 1 + (2 − )(1 − |y|2 )]g2,α (y)dy.
2
Ω
Ω
Therefore, for v1 (0) we obtain the equality (3.10).
No let 1 < α < 2, j = 2. Since in this case g1 (x) has the form (3.8), for v1 (0)
again we obtain (3.10). Further, we obtain
Z
v1,1 (x) = (1 − α)(2 − α)
G2,n (x, y)g2,α (y)dy,
Ω
Z
v1,2 (x) = 2(2 − α)
G2,n (x, y)Γ4 [g2,α ](y)dy,
Ω
Z
v1,3 (x) =
G2,n (x, y)Γ4 [Γ3 [g2,α ]](y)dy.
Ω
Using (3.6), for the function v1,3 (x) we obtain
Z
∂v1,3 (0)
n
=
yk (1 − |y|2 )Γ4 [g2,α ](y)dy,
∂xk
4ωn Ω
k = 1, 2, . . . , n.
14
B. KH. TURMETOV
EJDE-2015/82
Since
∂
4−n
|x − y|4−n =
|x − y|2−n 2(xk − yk )x=0 = −(4 − n)|y|2−n yk ,
∂xk
2
y 4−n
4−n
∂ y 2−n
yk
x|y| −
=
|x|y| −
|
2(xk |y| −
)|y|
∂xk
|y|
2
|y|
|y|
x=0
= −(4 − n)yk ,
y 2−n
∂ |x|y| −
|
(1 − |x|2 )
∂xk
|y|
2 − n y −n
yk
y 2−n
=
x|y| −
2(xk |y| −
(−2xk )x=0
)|y|(1 − |x|2 ) + x|y| −
2
|y|
|y|
|y|
= −(2 − n)yk ,
it follows that for
∂G2,n (x, y) ,
∂xk
x=0
we obtain
1
2−n
1 ∂G2,n (x, y) =
yk |y|2−n − yk +
yk (1 − |y|2 )
x=0
∂xk
2ωn (n − 2)
2
Then
Z
∂v1,1 (0)
1 (1 − α)(2 − α)
2−n
=
yk [|y|2−n − 1 +
(1 − |y|2 )]g2,α (y)dy,
∂xk
2ωn
(n − 2)
2
Ω
Z
∂v1,2 (0)
2−n
1 2(2 − α)
yk [|y|2−n − 1 +
(1 − |y|2 )]Γ4 [g2,α ](y)dy.
=
∂xk
2ωn (n − 2) Ω
2
Hence, for
∂v1 (0)
∂xk
we obtain (3.11).
4. Main results
Let g1,α (x) and g2,α (x), x ∈ Rn be defined by (2.12) and (2.13), and let n be
odd, or n be even with n > 4.
Theorem 4.1. Let 0 < α < 2, g(x), f1 (x) and f2 (x) be smooth functions.
(1) If 0 < α ≤ 1 and j = 1, then problem 1.1 is solvable if and only if
Z
[f2 (y) + (α − 2)f1 (y)]dSy
∂Ω
Z
1 − |y|2
=
g1,α (y)dy
2
Ω
Z
1−α
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g1,α (y)dy
(n − 2)(n − 4) Ω
2
(2) If 1 < α < 2 and j = 1, then problem 1.1 is solvable if and only if
Z
[f2 (y) + (α − 2)f1 (y)]dSy
∂Ω
Z
Z
1 − |y|2
1 − |y|2
=
Γ3 [g2,α ](y)dy + 2(2 − α)
g2,α (y)dy
2
2
Ω
Ω
Z
(1 − α)(2 − α)
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g2,α (y)dy .
(n − 2)(n − 4) Ω
2
(4.1)
(4.2)
EJDE-2015/82
SOLVABILITY OF FRACTIONAL ANALOGUES
15
(3) If the solution of the problem 1.1 exists then it is unique up to a constant
term and can be represented as
u(x) = C + B −α [v](x),
(4.3)
where v(x) is a solution of (3.1), satisfying the condition v(0) = 0, with the functions
ϕ1 (x) = f1 (x),
ϕ2 (x) = f2 (x) + αf1 (x)
g1 (x) = |x|
−4
(4.4)
Bjα [|x|4 g](x).
(4.5)
Proof. Let u(x) be a solution of problem 1.1. Apply the operator B1α to the function
u(x), and denote v(x) = B1α [u](x). Then in the case 0 < α ≤ 1, using (2.16) from
lemma 2.7, we obtain
∆2 v(x) = ∆2 B1α [u](x) = |x|−4 B1α [|x|4 g](x) ≡ g1 (x),
0 < α ≤ 1.
and if 1 < α < 2, then by (2.17), we have
∆2 v(x) = ∆2 B1α [u](x) = |x|−4 B1α [|x|4 g](x) ≡ g1 (x),
Then by assumption,
1 < α < 2.
B1α [u](x)
∈ C(Ω̄). Therefore, v(x) ∈ C(Ω̄) and
v(x) = f1 (x) ≡ ϕ1 (x).
∂Ω
Further, if 0 < α ≤ 1, then by the definition of B1α+1 ,
d
d
d
d
B1α+1 [u](x) = rα+1 J 2−(α+1) [ u](x) = rα+1 J 1−α [ u](x)
dr
dr
dr
dr
d α
α+1 d
−α
α
=r
[r · B1 [u]](x) = r B1 [u](x) − αB1α [u](x).
dr
dr
Therefore, the boundary condition (2.3) of the problem 1.1 implies the condition
∂v(x) = f2 (x) + αf1 (x) ≡ ϕ2 (x).
∂ν ∂Ω
Similarly, in the case 1 < α < 2, j = 1 from definition of B1α+1 , we have
d
d2 3−(α+1) d
d d
J
[ u](x) = rα+1 [ J 2−α [ u]](x)
dr2
dr
dr dr
dr
d
d
= rα+1 [r−α B1α [u]](x) = r B1α [u](x) − αB1α [u](x).
dr
dr
Consequently, in this case,
∂v(x) = f2 (x) + αf1 (x) ≡ ϕ2 (x).
∂ν ∂Ω
Thus, if u(x) is a solution of problem 1.1, then for the function v(x) = B1α [u](x) we
obtain the problem (3.1) with the functions (4.4) and (4.5).
By (2.3), the additional condition v(0) = 0 holds. For smooth enough functions
g1 (x), ϕ1 (x) and ϕ2 (x) the solution of (3.1) exists, is unique and can be represented
as in (3.2).
Let 0 < α ≤ 1. Then, using the representation of the function g1 (x) as (2.14),
and from (3.3) and (3.9), we obtain
Z
Z
1
1 − |y|2
1
[2ϕ1 (y) − ϕ2 (y)]dSy +
g1,α (y)dy
v(0) =
2ωn ∂Ω
2ωn Ω
2
Z
1−α
n
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g1,α (y)dy.
+
ωn 2(n − 2)(n − 4) Ω
2
B1α+1 [u](x) = rα+1
16
B. KH. TURMETOV
EJDE-2015/82
Hence, the condition v(0) = 0 holds if
Z
−
[2ϕ1 (y) − ϕ2 (y)]dSy
∂Ω
Z
1 − |y|2
=
g1,α (y)dy
2
Ω
Z
1−α
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g1,α (y)dy.
(n − 2)(n − 4) Ω
2
Since
2ϕ1 (y) − ϕ2 (y) = 2f1 (y) − f2 (y) − αf1 (y) = −[f2 (y) + (α − 2)f1 (y)],
this condition can be rewritten as (4.1). Therefore, necessity of condition (4.1) is
proved.
Applying the equality v(x) = B1α [u](x), the operator B −α , by (2.8), yields
B −α [v](x) = B −α [B1α [u]](x) = u(x) − u(0),
i.e. if the solution of problem 1.1 exists, and can be represented as in (4.2). Now we
show that condition (4.1) is also sufficient for the existence of solutions of problem
1.1. Indeed, if condition (4.1) holds, then for solutions of problem (3.1) with functions (4.4) and (4.5), condition v(0) = 0 holds. Then for such functions the operator
B −α is defined and we can consider the function u(x) = C + B −α [v](x). This function satisfies all conditions of the problem 1.1. Indeed, since ∆2 v(x) = g1 (x) and
g1 (x) = (1 − α)g1,α (x) + Γ4 [g1,α ](x), then, using (2.16) we can write the equalities
∆2 u(x) = ∆2 [C + B −α [v](x)]
Z 1
1
=
(t − τ )α−1 τ −α ∆2 v(τ x)dτ
Γ(α) 0
Z 1
1
=
(t − τ )α−1 τ 4−α [|τ x|−4 B1α [τ 4 g]](τ x)dτ
Γ(α) 0
Z
|x|−4 1
=
(t − τ )α−1 τ −α B1α [τ 4 g](τ x)dτ
Γ(α) 0
= |x|−4 B −α [B1α [|x|4 g]](x) = |x|−4 |x|4 g(x) = g(x).
Using (2.9), we obtain
D1α [u](x)∂Ω = B1α [u](x)∂Ω = B1α [C + B −α [v]](x)∂Ω
= v(x) = ϕ1 (x) = f1 (x),
∂Ω
∂
D1α+1 [u](x)∂Ω = B1α+1 [u](x)∂Ω = r B1α [u](x) − αB1α [u](x)∂Ω
∂r
∂
= r v(x) − αv(x)∂Ω = ϕ2 (x) − αϕ1 (x)
∂r
= f2 (x) + αf1 (x) − αf1 (x) = f2 (x).
Consequently, the function u(x) = C + B −α [v](x) satisfies all conditions of the
problem 1.1.
Let 1 < α < 2, j = 1. In this case v(x) = B1α [u](x) will be a solution of problem
(3.1) with functions ϕ1 (x) = f1 (x), ϕ2 (x) = f2 (x) + αf1 (x) and
g1 (x) = |x|−4 B1α [|x|4 g](x)
EJDE-2015/82
SOLVABILITY OF FRACTIONAL ANALOGUES
17
≡ (1 − α)(2 − α)g2,α (x) + 2(2 − α)Γ4 [g2,α ](x) + Γ4 [Γ3 [g2,α ]](x).
By (2.6), the condition v(0) = 0, holds additionally. Then, using (3.3) and (3.10),
we have
Z
Z
1
2(2 − α)
1 − |y|2
v(0) =
[2ϕ1 (y) − ϕ2 (y)]dSy +
g1,α (y)dy
2ωn ∂Ω
2ωn
2
Ω
Z
1
1 − |y|2
+
Γ3 [g2,α ](y)dy
2ωn Ω
2
Z
(1 − α)(2 − α)
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g2,α (y)dy.
ωn 2(n − 2)(n − 4) Ω
2
Thus, for the condition v(0) = 0, the following equality is necessary
Z
[2ϕ1 (y) − ϕ2 (y)]dSy
−
∂Ω
Z
Z
1 − |y|2
1 − |y|2
g2,α (y)dy +
Γ3 [g2,α ](y)dy
= 2(2 − α)
2
2
Ω
Ω
Z
(1 − α)(2 − α)
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g2,α (y)dy.
(n − 2)(n − 4) Ω
2
Since 2ϕ1 (y) − ϕ2 (y) = −[f2 (y) + (α − 2)f1 (y)], this condition can be rewritten as
(4.3). Therefore, necessity of the condition (4.3) is proved. Further, by repetition
of the argument in the case 0 < α < 1, one can show the rest of the theorem. Theorem 4.2. Let 1 < α ≤ 2, j = 2, g(x), f1 (x) and f2 (x) be smooth functions.
Then problem 1.2 is solvable if and only if:
Z
[f2 (y) + (α − 2)f1 (y)]dSy
∂Ω
Z
Z
1 − |y|2
1 − |y|2
= 2(2 − α)
g2,α (y)dy +
Γ3 [g2,α ](y)dy
(4.6)
2
2
Ω
Ω
Z
(1 − α)(2 − α)
n
+
[|y|4−n − 1 + (2 − )(1 − |y|2 )]g2,α (y)dy,
(n − 2)(n − 4) Ω
2
and
Z
yk [f2 (y) + (α − 3)f1 (y)]dSy
Z
1
=
yk (1 − |y|2 )Γ4 [g2,α ](y)dy
2 Ω
Z
2(2 − α)
2−n
+
yk [|y|2−n − 1 +
(1 − |y|2 )]Γ4 [g2,α ](y)dy
n(n − 2) Ω
2
Z
(1 − α)(2 − α)
2−n
yk [|y|2−n − 1 +
(1 − |y|2 )]g2,α (y)dy,
+
n(n − 2)
2
Ω
∂Ω
(4.7)
for k = 1, . . . , n.
If a solution of the problem 1.2 exists, then it is unique up to a first order
polynomial and can be represented as
u(x) = c0 +
n
X
i=1
ci xi +B −α [v](x),
(4.8)
18
B. KH. TURMETOV
EJDE-2015/82
where ci , i = 0, 1, . . . , n are arbitrary constants, and v(x) is a solution of the problem
(3.1) with functions g1 (x) = |x|−4 B2α [|x|4 g](x), ϕ1 (x) = f1 (x) and ϕ2 (x) = f2 (x) +
αf1 (x), and which satisfies conditions v(0) = 0, ∂v(0)
∂xi = 0, i = 1, 2, . . . , n.
Proof. Let u(x) be a solution of problem 1.2. Apply to the function u(x) the
operator B2α , and denote it by v(x) = B2α [u](x). Then (2.12) and
d 3−(α+1) d2
d
d2
J
[ 2 u](x) = rα+1 J 2−α [ 2 u](x)
dr
dr
dr
dr
d α
α+1 d
−α
α
=r
[r · B2 [u]](x) = r B2 [u](x) − αB2α [u](x).
dr
dr
imply that the function v(x) is a solution of the problem (3.1) with functions
g1 (x) = |x|−4 B2α [|x|4 g](x), ϕ1 (x) = f1 (x), ϕ2 (x) = f2 (x) + αf1 (x).
Moreover, by lemma 2.2, the function v(x) = B2α [u](x) should satisfy conditions
v(0) = 0, ∂v(0)
∂xk = 0, k = 1, 2, . . . , n.
For enough smooth functions g1 (x), ϕ1 (x) and ϕ2 (x) the solution of problem
(3.1) exists, is unique and can be represented as (3.2).
Further, using the representation of the function g1 (x) in the form (2.15), by
similar arguments, as in the case 1 < α < 2, j = 1, one can show that the equality
v(0) = 0 holds if the condition (4.4) holds.
Now we check that the equalities ∂v(0)
∂xk = 0, k = 1, 2, . . . , n hold if condition (4.5)
holds. To do it we use the representation of the function v(x) in the form (3.2) and
the lemma 3.3. Since the function g1 (x) = |x|−4 B2α [|x|4 g](x) can be represented as
(3.8), then by (3.4) and (3.11), we obtain
Z
Z
n
n
∂v(0)
=
yk [3ϕ1 (y) − ϕ2 (y)]dSy +
yk (1 − |y|2 )Γ4 [g2,α ](y)dy
∂xk
2ωn ∂Ω
4ωn Ω
Z
2−n
1 2(2 − α)
yk [|y|2−n − 1 +
(1 − |y|2 )]Γ4 [g2,α ](y)dy
+
2ωn (n − 2) Ω
2
Z
1 (1 − α)(2 − α)
2−n
+
yk [|y|2−n − 1 +
(1 − |y|2 )]g2,α (y)dy,
2ωn
(n − 2)
2
Ω
B2α+1 [u](x) = rα+1
for k = 1, . . . , n. Consequently, equalities ∂v(0)
∂xk = 0, k = 1, 2, . . . , n hold if
Z
yk [ϕ2 (y) − 3ϕ1 (y)]dSy
∂Ω
Z
1
=
yk (1 − |y|2 )Γ4 [g2,α ](y)dy
2 Ω
Z
2(2 − α)
2−n
+
yk [|y|2−n − 1 +
(1 − |y|2 )]Γ4 [g2,α ](y)dy
n(n − 2) Ω
2
Z
(1 − α)(2 − α)
2−n
+
yk [|y|2−n − 1 +
(1 − |y|2 )]g2,α (y)dy,
n(n − 2)
2
Ω
for k = 1, . . . , n.
Since ϕ2 (y) − 3ϕ1 (y) = f2 (x) + αf1 (x) − 3f1 (x) = f2 (x) + (α − 3)f1 (x), the above
condition can be rewritten as (4.7).
Applying the operator B −α to the equality v(x) = B1α [u](x), by (2.10), we obtain
B −α [v](x) = B −α [B1α [u]](x) = u(x) − u(0) −
n
X
i=1
xi
∂u(0)
.
∂xi
EJDE-2015/82
SOLVABILITY OF FRACTIONAL ANALOGUES
19
Denoting
∂u(0)
, i = 1, 2, . . . , n,
∂xi
we obtain the representation (4.8). Therefore, if solution of the problem 1.2 exists,
then it can be represented as (4.6).
Now we show that conditions (4.6) and (4.7) are also sufficient for existence of
a solution of the problem 1.2. Indeed, if conditions (4.6) and (4.7) hold, then for a
solution of the problem (3.1) with functions
c0 = u(0),
g1 (x) = |x|−4 B1α [|x|4 g](x),
ci =
ϕ1 (x) = f1 (x),
ϕ2 (x) = f2 (x) + αf1 (x)
the conditions
∂v(0)
= 0, i = 1, 2, . . . , n,
∂xi
hold. Then in the class of such functions the operator B −α is defined, and we can
consider the function
n
X
u(x) = c0 +
ci xi +B −α [v](x).
v(0) = 0,
i=1
We show that this function satisfies all conditions of the problem 1.1. Indeed, since
∆2 v(x) = g1 (x) ≡ |x|−4 B1α [|x|4 g](x),
it follows that
n
i
h
X
∆2 u(x) = ∆2 c0 +
ci xi +B −α [v](x) =
−4
= |x|
i=1
−α
B [B2α [|x|4 g]](x).
1
Γ(α)
Z
1
(t − τ )α−1 τ −α ∆2 v(τ x)dτ
0
The above expression, by (2.11), equals to g(x). Further, using (2.11), we obtain
n
X
D2α [u](x)∂Ω = B2α [u](x)∂Ω = B2α [c0 +
ci xi +B −α [v]](x)∂Ω
i=1
= v(x)
∂Ω
= ϕ1 (x) = f1 (x),
D2α+1 [u](x)∂Ω
d
d2
= B2α+1 [u](x)∂Ω = rα+1 J 3−(α+1) 2 u(x)
dr
dr
2
d α
α+1 d 2−α d
=r
J
u(x) = r B1 [u](x) − αB1α [u](x)∂Ω
2
dr
dr
dr
n
n
X
X
d
= r B1α [c0 +
ci xi +B −α [v]](x) − αB1α [c0 +
ci xi +B −α [v]](x)∂Ω
dr
i=1
i=1
d α
dv(x)
B1 [u](x) − αB1α [u](x)∂Ω = r
− αv(x)|∂Ω
dr
dr
= ϕ2 (x) − αϕ1 (x) = f2 (x) + αf1 (x) − αf1 (x) = f2 (x).
Pn
Consequently, the function c0 + i=1 ci xi +B −α [v] satisfies all conditions of the
problem 1.2.
=r
20
B. KH. TURMETOV
EJDE-2015/82
Remark 4.3. If in (4.1) α = 1, then condition on solvability of the problem
1.1 coincides with the condition (1.6). Similarly, in the case α = 2 condition on
solvability of the problem 1.2 coincides with the conditions (1.7) and (1.8).
Acknowledgements. The author would like to thank the editor and referees for
their valuable comments and remarks, which led to a great improvement of the
article. This research is financially supported by a grant from the Ministry of
Science and Education of the Republic of Kazakhstan (Grant No. 0819/GF4).
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Batirkhan Kh. Turmetov
Institute of Mathematics and Mathematical Modeling, Ministry of Education and Science Republic of Kazakhstan, 050010,Almaty, Kazakhistan.
Department of Mathematics, Akhmet Yasawi International Kazakh-Turkish University,
161200, Turkistan, Kazakhistan
E-mail address: batirkhan.turmetov@iktu.kz