Electronic Journal of Differential Equations, Vol. 2010(2010), No. 135, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
A BOUNDARY VALUE PROBLEM OF FRACTIONAL ORDER
AT RESONANCE
NICKOLAI KOSMATOV
Abstract. We establish solvability of a boundary value problem for a nonlinear differential equation of fractional order by means of the coincidence degree
theory.
1. Introduction
This article is a study of the boundary value problem of fractional order with
non-local conditions
Dα u(t) = f (t, u(t), u0 (t)),
α−2
D0+
u(0)
= 0,
a. e. t ∈ (0, 1),
ηu(ξ) = u(1),
where 1 < α < 2, 0 < ξ < 1 and ηξ α−1 = 1. It will be shown that, with the present
choice of boundary conditions, the boundary value problem is at resonance. We
apply a well-known degree theory theorem for coincidences due to Mawhin [16].
The monographs [10, 20, 21, 22] are commonly cited for the theory of fractional
derivatives and integrals and applications to differential equations of fractional order. Contributions to the theory of initial and boundary value problems for nonlinear differential equations of fractional order have been made by several authors
including a recent monograph [13] and the papers [1, 2, 9, 15, 24]. Although an application of the coincidence degree theory to a fractional order problem is not known
to the author, we can account for several results that have been devoted to both theoretical developments [5, 17, 19] and applications [23] to various types of boundary
and initial value problems. A broad range of scenarios of resonant problems were
studied in the framework of ordinary differential and difference equations [17] (more
generally, dynamic equations on time scales [3, 11]) on bounded and unbounded [12]
domains with periodic [18], non-local boundary conditions [4, 6, 7, 8, 23] as well as
boundary value problems with impulses [14].
2. Technical preliminaries
We start out by introducing the reader to the fundamental tools of fractional
calculus and the coincidence degree theory.
2000 Mathematics Subject Classification. 34A34, 34B10, 34B15.
Key words and phrases. Carathéodory conditions; resonance; Riemann-Liouville derivative;
Riemann-Liouville integral.
c 2010 Texas State University - San Marcos.
Submitted June 8, 2010. Published September 20, 2010.
1
2
N. KOSMATOV
EJDE-2010/135
The Riemann-Liouville fractional integral of order α > 0 of a function u ∈
Lp [0, 1], 1 ≤ p < ∞, is the integral
Z t
1
α
I u(t) =
(t − s)α−1 u(s) ds.
(2.1)
Γ(α) 0
The Riemann-Liouville fractional derivative of order α > 0, n = [α] + 1, is defined
by
Z t
1
dn
Dα u(t) =
(t − s)n−α−1 u(s) ds.
(2.2)
Γ(n − α) dtn 0
Let AC[0, 1] denote the space of absolutely continuous functions on the interval
[0, 1] and AC n [0, 1] = {u ∈ AC[0, 1] : u(n) ∈ AC[0, 1]}, n = 0, 1, 2, . . . . We make
use of several relationships between (2.1) and (2.2) that are stated in the next two
theorems (see [10, 20, 22]).
Theorem 2.1.
(a) The equality Dα I α g = g holds for every g ∈ L1 [0, 1];
1
(b) For u ∈ L [0, 1], n = [α] + 1, β > 0, if I n−α u ∈ AC n−1 [0, 1], then
I β Dα u(t) = Dα−β u(t) −
n−1
X
k=0
tβ−k−1 dn−k−1 n−α I
u (0).
Γ(β − k) dtn−k−1
For α < 0, we introduce the notation I α = D−α .
Theorem 2.2. If β, α + β > 0 and g ∈ L1 [0, 1], then the equality
I α I β g . = I α+β g
Definition 2.3. Let X and Z be real normed spaces. A linear mapping L :
dom L ⊂ X → Z is called a Fredholm mapping if the following two conditions hold:
(i) ker L has a finite dimension, and
(ii) Im L is closed and has a finite codimension.
If L is a Fredholm mapping, its (Fredholm) index is the integer Ind L = dim ker L −
codim Im L.
In this note we are concerned with a Fredholm mapping of index zero. From
Definition 2.3 it follows that there exist continuous projectors P : X → X and
Q : Z → Z such that
Im P = ker L,
ker Q = Im L,
X = ker L ⊕ ker P,
Z = Im L ⊕ Im Q
and that the mapping
L|dom L∩ker P : dom L ∩ ker P → Im L
is one-to-one and onto. The inverse of L|dom L∩ker P we denote by KP : Im L →
dom L ∩ ker P . The generalized inverse of L denoted by KP,Q : Z → dom L ∩ ker P
is defined by KP,Q = KP (I − Q).
If L is a Fredholm mapping of index zero, then, for every isomorphism J :
Im Q → ker L, the mapping JQ + KP,Q : Z → dom L is an isomorphism and, for
every u ∈ dom L,
(JQ + KP,Q )−1 u = (L + J −1 P )u.
EJDE-2010/135
A BOUNDARY VALUE PROBLEM
3
Definition 2.4. Let L : dom L ⊂ X → Z be a Fredholm mapping, E be a metric
space, and N : E → Z be a mapping. We say that N is L-compact on E if
QN : E → Z and KP,Q N : E → X are continuous and compact on E. In addition,
we say, that N is L-completely continuous if it is L-compact on every bounded
E ⊂ X.
The existence of a solution of the equation Lu = N u will be shown using [16,
Theorem IV.13].
Theorem 2.5. Let Ω ⊂ X be open and bounded, L be a Fredholm mapping of index
zero and N be L-compact on Ω. Assume that the following conditions are satisfied:
(i) Lu 6= λN u for every (u, λ) ∈ ((dom L\ ker L) ∩ ∂Ω) × (0, 1);
(ii) N u ∈
/ Im L for every u ∈ ker L ∩ ∂Ω;
(iii) deg(JQN |ker L∩∂Ω , Ω∩ker L, 0) 6= 0, with Q : Z → Z a continuous projector
such that ker Q = Im L and J : Im Q → ker L is an isomorphism.
Then the equation Lu = N u has at least one solution in dom L ∩ Ω.
Suppose now that the function f satisfies the Carathéodory conditions with
respect to Lp [0, 1], p ≥ 1; that is, the following conditions hold:
(C1) for each z ∈ Rn , the mapping t 7→ f (t, z) is Lebesgue measurable;
(C2) for a. e. t ∈ [0, 1], the mapping z 7→ f (t, z) is continuous on Rn ;
(C3) for each r > 0, there exists a nonnegative φr ∈ Lp [0, 1] such that, for a. e.
t ∈ [0, 1] and every z such that |z| ≤ r, we have |f (t, z)| ≤ φr (t).
3. Main results
Consider the differential equation
Dα u(t) = f (t, u(t), u0 (t)),
a. e. t ∈ (0, 1),
(3.1)
of fractional order 1 < α < 2, subject to the boundary conditions
Dα−2 u(0) = 0,
(3.2)
ηu(ξ) = u(1),
(3.3)
where 0 < ξ < 1 and
ηξ α−1 = 1.
We let the following assumption stand throughout this article:
p
1
(P) p > α−1
and q = p−1
.
(3.4)
Let ACloc (0, 1] be the space consisting of functions that are absolutely continuous
on every interval [a, 1] ⊂ (0, 1]. We introduce the space
X0 = {u : u ∈ AC[0, 1], u0 ∈ ACloc (0, 1], Dα u ∈ Lp [0, 1]}.
Let
X = {u ∈ C[0, 1] ∩ C 1 (0, 1] : lim t2−α u0 (t) exists}
t→0+
with the weighted norm kuk = max{kuk0 , kt2−α u0 k0 }, where k · k0 is the max-norm
and kt2−α vk0 = supt∈(0,1] |t2−α v(t)|. Let Z = Lp [0, 1] with the usual norm k · kp ,
where p satisfies (P ). Define the mapping L : dom L ⊂ X → Z with
dom L = {u ∈ X0 : u satisfies (3.2) and (3.3)}
α
and Lu(t) = D u(t).
4
N. KOSMATOV
EJDE-2010/135
Define the mapping N : X → Z by
N u(t) = f (t, u(t), u0 (t)).
Lemma 3.1. The mapping L : dom L ⊂ X → Z is a Fredholm mapping of index
zero.
Proof. It is easy to see that ker L = {ctα−1 : c ∈ R}. We claim that
Im L = {g ∈ Z : ηI α g(ξ) = I α g(1)}.
Let g ∈ Z and
u(t) = I α g(t) + ctα−1 ,
c ∈ R.
α
Then D u(t) = g(t), a. e. in (0, 1). By Theorem 2.2,
Dα−2 u(t) = I 2−α u(t)
= I 2−α I α g(t) + cI 2−α (tα−1 )
= I 2 g(t) + cΓ(α)t,
so that Dα−2 u(0) = 0. One can readily verify that, in view of (3.4), u satisfies (3.3)
provided ηI α g(ξ) = I α g(1). It is obvious that u ∈ AC[0, 1]. Then u0 exists, for a.
e. t ∈ (0, 1], and, by Theorem 2.2,
u0 (t) = I α−1 g(t) + c(α − 1)tα−2 .
Moreover,
lim t2−α u0 (t) = c(α − 1)
t→0+
since
lim t2−α |I α−1 g(t)| ≤ lim
t→0+
t→0+
t1/q kgkp
= 0.
Γ(α − 1)((α − 2)q + 1)1/q
Let t1 , t2 ∈ (0, 1) and t1 < t2 . Then
|I α−1 g(t2 ) − I α−1 g(t1 )|
Z t2
Z t1
1
(t2 − s)α−2 g(s) ds −
(t1 − s)α−2 g(s) ds
=
Γ(α) 0
0
Z t2
Z t1
1
=
(t2 − s)α−2 g(s) ds +
(t2 − s)α−2 − (t1 − s)α−2 g(s) ds
Γ(α) t1
0
Z t2
Z t1
1
1
≤
(t2 − s)α−2 |g(s)| ds +
(t1 − s)α−2 − (t2 − s)α−2 |g(s)| ds
Γ(α) t1
Γ(α) 0
h Z t1
q i1/q
1
α−2+ q
≤ C1 (t2 − t1 )
kgkp + C1
(t1 − s)α−2 − (t2 − s)α−2 ds
kgkp
0
h Z t1 i1/q
α−2+ q1
≤ C1 (t2 − t1 )
kgkp + C1
(t1 − s)(α−2)q − (t2 − s)(α−2)q ds
kgkp
0
α−2+ 1
q
≤ C1 (t2 − t1 )
kgkp
1/q
(α−2)q+1
(α−2)q+1
− t2
+ (t2 − t1 )(α−2)q+1
kgkp ,
+ C1 t1
where C1 is a generic constant that depends only on α and p. Thus, u0 ∈ ACloc (0, 1].
Combining the preceding observations, we obtain that u ∈ dom L. So, {g ∈ Z :
ηI α g(ξ) = I α g(1)} ⊆ Im L.
EJDE-2010/135
A BOUNDARY VALUE PROBLEM
5
Let u ∈ dom L. Then, for Dα u ∈ Im L, we have, by Theorem 2.1(b) and (3.2),
I α Dα u(t) = u(t) −
Dα−1 u(0) α−1 Dα−2 u(0) α−2
Dα−1 u(0) α−1
t
t
t
−
= u(t) −
,
Γ(α)
Γ(α − 1)
Γ(α)
which, due to the boundary conditions (3.2), (3.3) together with (3.4), implies
that Dα u satisfies ηI α Dα u(ξ) = I α Dα u(1). Hence, Im L ⊆ {g ∈ Z : ηI α g(ξ) =
I α g(1)}. Therefore, Im L = {g ∈ Z : ηI α g(ξ) = I α g(1)}.
Define Q : Z → Z by
Qg(t) = κ (ηI α g(ξ) − I α g(1)) tα−1 ,
where
κ=
Γ(2α)
.
Γ(α)(ξ α − 1)
Then
Q2 g(t) = κ (ηI α Qg(ξ) − I α Qg(1)) tα−1
Z 1
η Z ξ
1
α−1
(ξ − s)
Qg(s) ds −
(1 − s)α−1 Qg(s) ds tα−1
=κ
Γ(α) 0
Γ(α) 0
Z 1
η Z ξ
1
(ξ − s)α−1 sα−1 ds −
(1 − s)α−1 sα−1 ds Qg(t)
=κ
Γ(α) 0
Γ(α) 0
Γ(α)
(ηξ 2α−1 − 1)Qg(t)
=κ
Γ(2α)
= Qg(t)
in view of (3.4). Therefore, Q : Z → Z is a continuous linear projector with
Ker Q = Im L.
Let g ∈ Z be written as g = (g − Qg) + Qg with g − Qg ∈ Ker Q = Im L and
Qg ∈ Im Q. Hence, Z = Im L + Im Q. Let g ∈ Im L ∩ Im Q and set g(t) = ctα−1 to
obtain that
c
cΓ(α)
0 = γI α g(ξ) − I α g(1) =
(ηξ 2α−1 − 1) = ,
Γ(2α)
κ
which implies that c = 0. Hence {0} = Im L ∩ Im Q and so Z = Im L ⊕ Im Q.
Note that Ind L = dim ker L − codim Im L = 0; that is, L is a Fredholm mapping
of index zero.
Define P : X → X by
P u(t) =
1
Dα−1 u(0)tα−1 .
Γ(α)
Since 0 < α − 1 < 1,
Dα−1 u(t) =
1
d
Γ(2 − α) dt
Z
t
(t − s)1−α u(s) ds.
0
Then
1
Dα−1 (P u)(0)tα−1
Γ(α)
Z
1
1 d t
=
(t − s)1−α sα−1 ds
Γ(2 − α) Γ(α) dt 0
= P u(t).
P 2 u(t) =
P u(t)
t=0
6
N. KOSMATOV
EJDE-2010/135
We have that P : X → X is a continuous linear projector. Note that ker P = {u ∈
X : Dα−1 u(0) = 0}. For u ∈ X,
1
kP uk0 =
|Dα−1 u(0)|
Γ(α)
and
kt2−α (P u)0 k0 =
1
|Dα−1 u(0)|.
Γ(α − 1)
Hence,
1
|Dα−1 u(0)|.
Γ(α)
Define KP : Im L → dom L ∩ ker P by
kP uk =
KP g(t) = I α g(t),
(3.5)
t ∈ (0, 1).
For g ∈ Im L,
LKP g(t) = Dα I α g(t) = g(t)
by Theorem 2.1(a). For u ∈ dom L ∩ ker P , we have Dα−2 u(0) = 0 and Dα−1 u(0) =
0. Hence, by Theorem 2.1(b),
KP Lu(t) = I α Dα u(t)
= u(t) −
Dα−1 u(0) α−1 Dα−2 u(0) α−2
t
−
t
Γ(α)
Γ(α − 1)
= u(t).
Thus,
−1
KP = L|dom L∩ker P
.
Furthermore, using (P), we have
kt2−α (KP g)0 k0 = max t2−α (KP g)0 (t)
t∈(0,1]
t2−α
t∈(0,1] Γ(α − 1)
Z
≤ max
t
(t − s)α−2 |g(s)| ds
0
Z
1/q
t2−α t
≤ max
(t − s)(α−2)q ds
kgkp
t∈(0,1] Γ(α − 1)
0
α−1
1
=
kgkp .
Γ(α) ((α − 2)q + 1)1/q
Similarly,
kKP gk0 = max |KP g(t)|
t∈[0,1]
1
t∈[0,1] Γ(α)
Z
≤ max
t
(t − s)α−1 |g(s)| ds
0
Z
1/q
1 t
≤ max
(t − s)(α−1)q ds
kgkp
t∈[0,1] Γ(α)
0
1
1
=
kgkp .
Γ(α) ((α − 1)q + 1)1/q
Hence
kKP gk ≤ Λkgkp ,
(3.6)
EJDE-2010/135
A BOUNDARY VALUE PROBLEM
where
Λ=
n
o
1
1
α−1
max
,
.
Γ(α)
((α − 1)q + 1)1/q ((α − 2)q + 1)1/q
7
(3.7)
We introduce
QN u(t) = κ(ηI α N u(ξ) − I α N u(1))tα−1
Z ξ
κ η
(ξ − s)α−1 f (s, u(s), u0 (s)) ds
=
Γ(α)
0
Z 1
−
(1 − s)α−1 f (s, u(s), u0 (s)) ds tα−1
0
and
KP,Q N u(t) = KP (I − Q)N u(t) =
κ
Γ(α)
Z
t
(t − s)α−1 (N u(s) − QN u(s)) ds.
0
Now we are in position to prove the existence results. We impose the condiitions
(H1) there exists a positive constant K such that u ∈ dom L \ Ker L with
mint∈[0,1] |Dα−1 (t)| > K implies QN u(t) 6= 0 on (0, 1];
(H2) there exist δ, β, tα−2 γ, ρ ∈ Lp [0, 1] and a continuous nondecreasing function
φ : [0, ∞) → [0, ∞) and x0 > 0 with the properties:
(a)
Γ(α)
;
kβkp + ktα−2 γkp <
1 + Γ(α)Λ
(b) for all x ≥ x0
x≥
K + (1 + Γ(α)Λ)kδkp
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
(1 + Γ(α)Λ)kρkp
φ(x);
+
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
(3.8)
(c) f : [0, 1] × R2 → R satisfies
|f (t, x, y)| ≤ δ(t) + β(t)|x| + γ(t)|y| + ρ(t)φ(|x|);
(H3) there exists a constant B > 0 such that, for every c ∈ R satisfying |c| > B
we have
sgn [c(ηIuc (ξ) − Iuc (1))] 6= 0,
α−1
where uc (t) = ct
.
Theorem 3.2. If the hypotheses (P), (H1)-(H3) are satisfied, then the boundary
value problem (3.1)-(3.4) has a solution.
Proof. Let Ω1 = {u ∈ dom L \ Ker L : Lu = λN ufor some λ ∈ (0, 1)}. Applying
(H1), QN u(t) = 0 for all t ∈ [0, 1]. Hence there exists t0 ∈ (0, 1] such that
|Dα−1 (t0 )| ≤ K. By Theorem 2.1 with β = 1,
IDα u(t0 ) = Dα−1 u(t0 ) − Dα−1 u(0) − Dα−2 u(0)t−1
0
= Dα−1 u(t0 ) − Dα−1 u(0)
since u ∈ dom L. That is,
D
α−1
u(0) = D
α−1
Z
u(t0 ) −
0
t0
Dα u(s) ds,
8
N. KOSMATOV
EJDE-2010/135
which implies
|D
α−1
u(0)| ≤ |D
α−1
Z
u(t0 )| +
t0
|Dα u(s)| ds
0
≤ K + kLuk
< K + kN ukp .
By (3.5),
1
1
|Dα−1 u(0)| <
(K + kN ukp ).
Γ(α)
Γ(α)
Since (I − P )u ∈ dom L ∩ Ker P = Im KP , for u ∈ Ω1 , k(I − P )uk < ΛkN ukp by
(3.6) and (3.7). Also P u ∈ Im P = Ker L ⊂ dom L and, therefore,
1
K
+
+ Λ kN ukp .
kuk ≤ kP uk + k(I − P )uk <
Γ(α)
Γ(α)
kP uk =
From (H2) and the previous inequality, it follows that
1
K
kt2−α u0 k0 <
+
+ Λ kδkp + kβkp kkuk0
Γ(α)
Γ(α)
+ ktα−2 γkp kt2−α u0 k0 + kρkp φ(kuk0 )
or
(1 + Γ(α)Λ)kβkp
K + (1 + Γ(α)Λ)kδkp
+
kuk0
α−2
Γ(α) − (1 + Γ(α)Λ)kt
γkp
Γ(α) − (1 + Γ(α)Λ)ktα−2 γkp
(1 + Γ(α)Λ)kρkp
+
φ(kuk0 ).
Γ(α) − (1 + Γ(α)Λ)ktα−2 γkp
(3.9)
Combining the above inequality with
1
K
kuk0 <
+
+ Λ kδkp + kβkp kkuk0
Γ(α)
Γ(α)
α−2
+ kt
γkp kt2−α u0 k0 + kρkp φ(kuk0 )
kt2−α u0 k0 <
we obtain
kuk0 <
K + (1 + Γ(α)Λ)kδkp
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
(1 + Γ(α)Λ)kρkp
φ(kuk0 ),
+
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
for all u ∈ Ω1 . Suppose that Ω1 is unbounded. If {kt2−α u0 k0 : u ∈ Ω1 } is unbounded, then, by (3.9), so is {kuk0 : u ∈ Ω1 }. So, it suffices to consider the
case that {kuk0 : u ∈ Ω1 } is unbounded. Then, in view of (3.8), we arrive at a
contradiction. Therefore, Ω1 is bounded.
Set Ω2 = {u ∈ ker L : N u ∈ Im L}. Hence uc ∈ ker L is given by uc (t) = ctα−1 ,
c ∈ R. Then (QN )(ctα−1 ) = 0, since N u ∈ Im L = ker Q. It follows from (H3)
that kuc k = max{kuc k0 , kt2−α u0c k0 } = max{|c|, (α − 1)|c|} = |c| ≤ B; that is, Ω2 is
bounded.
Define the isomorphism J : Im Q → ker L by Juc = uc , uc (t) = ctα−1 for c ∈ R.
Let Ω3 = {u ∈ ker L : −λJ −1 u + (1 − λ)QN u = 0, λ ∈ [0, 1]}, if sgn[c(ηIuc (ξ) −
Iuc (1))] = −1. Then u ∈ Ω3 implies λc = (1−λ) (ηIuc (ξ) − Iuc (1)). If λ = 1, then
EJDE-2010/135
A BOUNDARY VALUE PROBLEM
9
c = 0 and, if λ ∈ [0, 1) and |c| > B, then 0 < λc2 = (1 − λ)c (ηIuc (ξ) − Iuc (1)) < 0,
which is a contradiction. Let Ω3 = {u ∈ ker L : λJ −1 u+(1−λ)QN u = 0, λ ∈ [0, 1]}
if sgn [c (ηIuc (ξ) − Iuc (1))] = 1, and we arrive at a contradiction, again. Thus,
kuc k ≤ B, for all uc ∈ Ω3 .
Let Ω be open and bounded such that ∪3i=1 Ωi ⊂ Ω. Then the assumptions (i)
and (ii) of Theorem 2.5 are fulfilled. It is a straightforward exercise to show that
the mapping N is L-compact on Ω. Lemma 3.1 establishes that L is a Fredholm
mapping of index zero.
Define
H(u, λ) = ±λ Id u + (1 − λ)JQN u.
By the degree property of invariance under a homotopy, if u ∈ ker L ∩ ∂Ω, then
deg(JQN |ker L∩∂Ω , Ω ∩ ker L, 0) = deg(H(·, 0), Ω ∩ ker L, 0)
= deg(H(·, 1), Ω ∩ ker L, 0)
= deg(± Id, Ω ∩ ker L, 0) 6= 0.
Therefore, the assumption (iii) of Theorem 2.5 is fulfilled and the proof is completed.
Suppose that the hypothesis (H2) is replaced by
(H2’) there exist δ, β, tα−2 γ, tα−2 ρ ∈ Lp [0, 1] and a continuous nondecreasing
function φ : [0, ∞) → [0, ∞) and y0 > 0 with the properties:
(a)
Γ(α)
kβkp + ktα−2 γkp <
;
1 + Γ(α)Λ
(b) for all y ∈ [0, ∞) and t ∈ [0, 1],
t2−α φ(y) ≤ φ(t2−α y);
(c) for all y ≥ y0 ,
y≥
K + (1 + Γ(α)Λ)kδkp
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
+
(1 + Γ(α)Λ)ktα−2 ρkp
φ(y);
Γ(α) − (1 + Γ(α)Λ)(kβkp + ktα−2 γkp )
(d) f : [0, 1] × R2 → R satisfies
|f (t, x, y)| ≤ δ(t) + β(t)|x| + γ(t)|y| + ρ(t)φ(|y|).
Then we have the following existence criterion whose proof is analogous to that
of Theorem 3.2.
Theorem 3.3. If the hypotheses (P), (H1), (H2’), (H3) are satisfied, then the
boundary value problem (3.1)-(3.4) has a solution.
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Nickolai Kosmatov
Department of Mathematics and Statistics, University of Arkansas at Little Rock,
Little Rock, AR 72204-1099, USA
E-mail address: nxkosmatov@ualr.edu