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Foundations of nabla fractional calculus on time scales and inequalities
Computers and Mathematics with Applications 59 (2010) 3750–3762
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Foundations of nabla fractional calculus on time scales and inequalities George A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA
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Article history: Received 31 January 2010 Accepted 30 March 2010
Here we develop the nabla fractional calculus on time scales. Then we produce related integral inequalities of types: Poincaré, Sobolev, Opial, Ostrowski and Hilbert–Pachpatte. Finally we give inequality applications on the time scales R, Z. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Fractional calculus on time scales Nabla Poincaré inequality Nabla Sobolev inequality Nabla Opial inequalities Nabla Ostrowski inequality Nabla Hilbert–Pachpatte inequality Fractional inequalities
1. Background and foundation results For the basics on time scales we follow [1–10]. Ref. [11] was an inspiration. By [12], p. 256, for µ, ν > 0 we have that x
Z t
(x − t )µ+ν−1 (x − s)µ−1 (s − t )ν−1 ds = , Γ (µ) Γ (ν) Γ (µ + ν)
(1)
where Γ is the gamma function. Here we consider time scales T such that Tk = T . Consider the coordinatewise ld-continuous functions b hα : T × T → R, α ≥ 0, such that b h0 (t , s) = 1,
b hα+1 (t , s) =
t
Z
b hα (τ , s) ∇τ ,
(2)
s
∀s, t ∈ T . Here ρ is the backward jump operator and ν (t ) = t − ρ (t ) . Furthermore for α, β > 1, we assume that Z t b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ = b hα+β−1 (t , ρ (u)) , ρ(u)
is valid for all u, t ∈ T : u ≤ t . In the case of T = R; then ρ (t ) = t, and b hk ( t , s ) = α
(t − s) b hα (t , s) = , Γ (α + 1)
(t −s)k , k!
α ≥ 0.
E-mail addresses: [email protected], [email protected]. 0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2010.03.072
k ∈ N0 = N ∪ {0}, and define
(3)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
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Notice that
(t − s)α+1 b (τ − s)α dτ = = hα+1 (t , s) , Γ (α + 1) Γ (α + 2)
t
Z s
fulfilling (2). Furthermore we observe that (α, β > 1) t
Z
(t − τ )α−1 (τ − u)β−1 dτ Γ (α) Γ (β) u (t − u)α+β−1 b = hα+β−1 (t , u) , Γ (α + β)
Z
b hα−1 (t , τ ) b hβ−1 (τ , u) dτ
=
u (by (1))
=
t
fulfilling (3). By Theorem 2.2 of [13], we have for k, m ∈ N0 that t
Z
b hk (t , ρ (τ )) b hm (τ , t0 ) ∇τ = b hk+m+1 (t , t0 ) .
(4)
t0
Let T = Z, then ρ (t ) = t −1, t ∈ Z. Define t 0 := 1, t k := t (t + 1) · · · (t + k − 1), k ∈ N, and by (2) we haveb hk ( t , s ) = s, t ∈ Z, Rk ∈ N0 . P t t Here t ∇τ = t0 + 1 . 0 Therefore by (4) we get
(t −s)k , k!
t X (t − τ + 1)k (τ − t0 )m (t − t0 )k+m+1 , = k! m! (k + m + 1)! τ =t +1 0
which results into t X (t − t0 + 1)k+m−1 (t − τ + 1)k−1 (τ − t0 + 1)m−1 = , (k − 1)! (m − 1)! (k + m − 1)! τ = t0
(5)
confirming (3). Next we follow [14]. Γ (t +α) Let a, α ∈ R, define t α = Γ (t ) , t ∈ R − {. . . , −2, −1, 0}, Na = {a, a ± 1, a ± 2, . . .}, notice N0 = Z, 0α = 0, t 0 = 1, and f : Na → R. Here ρ (s) = s − 1, σ (s) = s + 1, ν (t ) = 1. Also define
∇a−n f (t ) =
t X (t − ρ (s))n−1 s=a
(n − 1)!
f (s) ,
n ∈ N,
and in general
∇a−ν f (t ) =
t X (t − ρ (s))ν−1 s=a
Γ (ν)
f (s) ,
where ν ∈ R − {. . . , −2, −1, 0}. Here we set
(t − s)α b hα (t , s) = , Γ (α + 1)
α ≥ 0.
We need Lemma 1. Let α > −1, x > α + 1. Then
Γ (x) 1 = Γ (x − α) (α + 1) Proof. Obvious.
Γ (x + 1) Γ (x) − Γ (x − α) Γ (x − α − 1)
Proposition 2. Let α > −1. It holds t
Z s
(τ − s)α (t − s)α+1 ∇τ = , Γ (α + 1) Γ (α + 2)
That is b hα , α ≥ 0, on Na confirm (2).
t ≥ s.
.
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Proof. Let t > s. We have that t
Z s
t t X X 1 1 Γ (τ − s + α) (τ − s)α ∇τ = (τ − s)α = Γ (α + 1) Γ (α + 1) τ =s+1 Γ (α + 1) τ =s+1 Γ (τ − s)
=
1
t X
Γ (α + 1)
τ =s+1
t− s+α X Γ (τ − s + α) 1 Γ (x) = Γ (τ − s + α − α) Γ (α + 1) x=α+1 Γ (x − α)
(notice here τ − s ≥ 1 and x ≥ α + 1 > 0)
=
(
1
Γ (α + 1)
t −s+α
Γ (α + 1) +
X x=α+2
Γ (x) Γ (x − α)
) =1+
1
Γ (α + 1)
t −s+α
X x=α+2
Γ (x) Γ (x − α)
X Γ (x + 1) Γ ( x) 1 (by Lemma 1) − = 1+ Γ (α + 2) x=α+2 Γ (x − α) Γ (x − α − 1) 1 Γ (α + 4) =1+ − Γ (α + 3) (Γ (α + 3) − Γ (α + 2)) + Γ (α + 2) Γ (3) Γ (α + 5) Γ (α + 4) Γ (α + 6) Γ (α + 5) + − + − + ··· Γ (4) Γ (3) Γ (5) Γ (4) Γ (t − s + α) Γ (t − s + α − 1) Γ (t − s + α + 1) Γ (t − s + α) + − + − Γ (t − s − 1) Γ (t − s + α − 2) Γ (t − s) Γ (t − s − 1) t −s+α
(telescoping sum)
=1+ = That is proving the claim.
1
Γ (α + 2)
Γ (t − s + α + 1) − Γ (α + 2) Γ (t − s)
Γ (t − s + α + 1) (t − s)α+1 = . Γ (α + 2) Γ (t − s) Γ (α + 2)
Next for µ, ν > 1, τ < t, from the proof of Theorem 2.1 [14] we get that t X (t − ρ (s))ν−1 (s − ρ (τ ))µ−1 (t − ρ (τ ))ν+µ−1 = , Γ (ν) Γ (µ) Γ (µ + ν) s=τ
where τ ∈ {a, . . . , t }. So for t , t0 ∈ Na with t0 < t we obtain t X (t − τ + 1)ν−1 (τ − t0 + 1)µ−1 (t − t0 + 1)ν+µ−1 = , Γ (ν) Γ (µ) Γ (µ + ν) τ = t0
(6)
that is confirming (3) fractionally on the time scale T = Na . Notice also here that b X
b
Z
f (t ) ∇ t = a
f (t ) .
t =a+1
So fractional conditions (2) and (3) are very natural and common on time scales. For α ≥ 1 we define the time scale ∇ -Riemann–Liouville type fractional integral (a, b ∈ T ) Jaα f (t ) =
t
Z
b hα−1 (t , ρ (τ )) f (τ ) ∇τ ,
(7)
a
(by [15] the last integral is on (a, t] ∩ T ) Ja0 f (t ) = f (t ) , where f ∈ L1 ([a, b] R∩ T ) (Lebesgue ∇ -integrable functions on [a, b] ∩ T , see [8–10]), t ∈ [a, b] ∩ T . t Notice Ja1 f (t ) = a f (τ ) ∇τ is absolutely continuous in t ∈ [a, b] ∩ T , see [15]. Lemma 3. Let α > 1, f ∈ L1 ([a, b] ∩ T ). Assume that b hα−1 (s, ρ (t )) is Lebesgue ∇ -measurable on ([a, b] ∩ T )2 ; a, b ∈ T . Then α Ja f ∈ L1 ([a, b] ∩ T ) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
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Proof. Define K : Ω := ([a, b] ∩ T )2 → R, by K (s, t ) =
b hα−1 (s, ρ (t )) , if a ≤ t ≤ s ≤ b, 0, if a ≤ s < t ≤ b.
Clearly K (s, t ) is Lebesgue ∇ -measurable on ([a, b] ∩ T )2 . Then b
Z
K (s, t ) ∇ s =
Z [a,t )
a
K (s, t ) ∇ s =
b
Z
b hα−1 (s, ρ (t )) ∇ s t
t
Z
K ( s, t ) ∇ s t
b
Z =
b
Z
K (s, t ) ∇ s +
b
= ρ(t )
b hα−1 (s, ρ (t )) ∇ s −
Z
t
ρ(t )
b hα−1 (s, ρ (t )) ∇ s
hα−1 (t , ρ (t )) ∈ R. =b hα (b, ρ (t )) − ν (t ) b Next we consider the repeated double Lebesgue ∇ -integral b
Z a
b
Z
K (s, t ) |f (t )| ∇ s ∇ t =
b
Z
a
|f (t )|
b
Z
a
K (s, t ) ∇ s ∇ t a
b
Z
|f (t )| b hα (b, ρ (t )) − ν (t ) b hα−1 t , ρ (t ) ∇ t
= a b
Z
|f (t )| b hα (b, ρ (t )) ∇ t −
=
b
Z
a
|f (t )| ν (t ) b hα−1 (t , ρ (t )) ∇ t , a
which exists and is finite. Thus the function (s, t ) → K (s, t ) f (t ) is Lebesgue ∇ -integrable over Ω by Tonelli’s theorem. Let now the characteristic function
χ(a,s]∩T (t ) =
1, 0,
if t ∈ (a, s] ∩ T else,
where s ∈ [a, b] ∩ T . Then the function (s, t ) → χ(a,s]∩T (t ) K (s, t ) f (t ) is Lebesgue ∇ -integrable on Ω . Hence by Fubini’s theorem we get that b
Z
χ(a,s]∩T (t ) K (s, t ) f (t ) ∇ t = a
s
Z
b hα−1 (s, ρ (t )) f (t ) ∇ t = Jaα f (s)
a
is Lebesgue ∇ -integrable in s on [a, b] ∩ T , proving the claim.
For u ≤ t; u, tZ∈ T , we define u
ε (t , u) =
ρ(u)
b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ
= ν ( u) b hα−1 (t , ρ (u)) b hβ−1 (u, ρ (u)) ,
(8)
where α, β > 1. Next we notice for α, β > 1; a, b ∈ T , f ∈ L1 ([a, b] ∩ T ), and b hα−1 (s, ρ (t )) is continuous on ([a, b] ∩ T )2 for any α > 1, that Z τ Z t b b Jaα Jaβ f (t ) = hα−1 (t , ρ (τ )) ∇τ hβ−1 (τ , ρ (u)) f (u) ∇ u a
a
(by Fubini’s theorem) t
Z a
t
Z ·
ρ(u) (by (3))
Z t b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ = f (u) ∇ u u a Z u b b b b hα−1 (t , ρ (τ )) hβ−1 (τ , ρ (u)) ∇τ − hα−1 (t , ρ (τ )) hβ−1 (τ , ρ (u)) ∇τ
f (u) ∇ u
=
Z
Z
t
ρ(u)
t
= f ( u) ∇ u b hα+β−1 (t , ρ (u)) − ε (t , u) a Z t Z t b = hα+β−1 (t , ρ (u)) f (u) ∇ u − f (u) ε (t , u) ∇ u a a Z t = Jaα+β f (t ) − f (u) ε (t , u) ∇ u. a
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Hence Jaα Jaβ f (t ) +
t
Z
f (u) ε (t , u) ∇ u = Jaα+β f (t ) ,
∀t ∈ [a, b] ∩ T .
a
So we have proved the semigroup property Jaα Jaβ f (t ) +
t
Z
hβ−1 (u, ρ (u)) ∇ u = Jaα+β f (t ) , f (u) ν (u) b hα−1 (t , ρ (u)) b
(9)
a
∀t ∈ [a, b] ∩ T , with a, b ∈ T . We call the Lebesgue ∇ -integral Z t f (u) ν (u) b hα−1 (t , ρ (u)) b hβ−1 (u, ρ (u)) ∇ u, D (f , α, β, T , t ) =
(10)
a
t ∈ [a, b] ∩ T ; a, b ∈ T , the backward graininess deviation functional of f ∈ L1 ([a, b] ∩ T ) . If T = R, then D (f , α, β, R, t ) = 0. Putting things together we have Theorem 4. Let Tk = T , a, b ∈ T , f ∈ L1 ([a, b] ∩ T ); α, β > 1; b hα−1 (s, ρ (t )) is continuous on ([a, b] ∩ T )2 for any α > 1. Then Jaα Jaβ f (t ) + D (f , α, β, T , t ) = Jaα+β f (t ) ,
(11)
∀t ∈ [a, b] ∩ T . We make Remark 5. Let µ > 2 such that m − 1 < µ < m ∈ N, i.e. m = dµe (ceiling of the number), e ν = m − µ (0 < e ν < 1). m Let f ∈ Cldm ([a, b] ∩ T ). Clearly here [10] f ∇ is a Lebesgue ∇ -integrable function. We define the nabla fractional derivative on time scale T of order µ − 1 as follows:
Z t 1 e ν+1 ∇ m ∇m b ∇aµ− f t = J f t = he ( ) ( ) (τ ) ∇τ , ν (t , ρ (τ )) f ∗ a
(12)
a
∀t ∈ [a, b] ∩ T .
µ−1
Notice here that ∇a∗ f ∈ C ([a, b] ∩ T ) by a simple argument using dominated convergence theorem in Lebesgue ∇ -sense. If µ = m, then e ν = 0 and by (12) we get m
∇am∗−1 f (t ) = Ja1 f ∇ (t ) = f ∇
m−1
(t ) .
(13)
m−1
m
More generally, by [15], given that f ∇ is everywhere finite and absolutely continuous on [a, b] ∩ T , then f ∇ exists ∇ -a.e. and is Lebesgue ∇ -integrable on (a, t] ∩ T , ∀t ∈ [a, b] ∩ T , and one can plug it into (12). We observe that 1 Jaµ−1 ∇aµ− ∗ f (t )
ν+1 ∇ Jaµ−1 Jae f
m
(t ) Z t m ν ∇m Jaµ+e f f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he (t ) − ν (u, ρ (u)) ∇ u a Z t m m Jam f ∇ f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he (t ) − ν (u, ρ (u)) ∇ u.
= (by (11))
=
=
a
Hence µ−1
Ja
1 ∇aµ− ∗ f
(t ) +
Z a
=
m Jam f ∇
(t ) =
t
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u
Z
t
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ .
a
We have proved Theorem 6. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm ([a, b] ∩ T ), a, b ∈ T , Tk = T . Suppose b hµ−2 (s, ρ (t )), 2 b he ν (s, ρ (t )) to be continuous on ([a, b] ∩ T ) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
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Then t
Z
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ
a t
Z
m
he hµ−2 (t , ρ (u)) b f ∇ (u) ν (u) b ν (u, ρ (u)) ∇ u +
= a
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(14)
a
∀t ∈ [a, b] ∩ T . We need the nabla time scales Taylor formula Theorem 7 ([2]). Let f ∈ Cldm (T ), m ∈ N, Tk = T ; a, b ∈ T . Then m−1
f (t ) =
X
k b hk (t , a) f ∇ (a) +
t
Z
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ ,
(15)
a
k=0
∀t ∈ [a, b] ∩ T . Next we present the fractional time scales nabla Taylor formula Theorem 8. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ) , a, b ∈ T , Tk = T . Suppose b hµ−2 (s, ρ (t )), b he ν (s, ρ (t )) 2 to be continuous on ([a, b] ∩ T ) . Then m−1
f (t ) =
X
k b hk (t , a) f ∇ (a)
k=0 t
Z + a
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u +
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(16)
a
∀t ∈ [a, b] ∩ T . k
Corollary 9. All as in Theorem 8. Additionally suppose f ∇ (a) = 0, k = 0, 1, . . . , m − 1. Then
m
A (t ) := f (t ) − D f ∇ , µ − 1,e ν + 1, T , t
= f (t ) −
t
Z
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u
a t
Z =
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(17)
a
∀t ∈ [a, b] ∩ T . m
Notice here that D f ∇ , µ − 1,e ν + 1, T , t ∈ Cld ([a, b] ∩ T ). Also the R.H.S (17) is a continuous function in t ∈ [a, b] ∩ T .
2. Fractional nabla inequalities on time scales We present a Poincaré type related inequality. Theorem 10. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ), a, b ∈ T , a ≤ b, Tk = T . Suppose b hµ−2 k
2 ∇ he (s, ρ (mt )) , b ν (s, ρ (t )) to be continuous on ([a, b] ∩ T ) , and1 f 1(a) = 0, k = 0, 1, . . . , m − 1. Here A (t ) = f (t ) − ∇ D f , µ − 1,e ν + 1, T , t , t ∈ [a, b] ∩ T ; and let p, q > 1 : p + q = 1.
Then b
Z
|A (t )| ∇ t ≤ q
a
b
Z a
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
a
Proof. By Corollary 9 we get that A (t ) =
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ .
a
Hence
|A (t )| ≤
t
Z a
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ
pq
! Z
b
∇t a
µ−1 ∇ f (t ) q ∇ t . a∗
(18)
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
(by Hölder’s inequality) t
Z ≤
1p Z
p b hµ−2 (t , ρ (τ )) ∇τ
t
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a
a t
Z ≤
1p Z
p b hµ−2 (t , ρ (τ )) ∇τ
a
b
µ−1 ∇ f (τ ) q ∇τ a∗
1q
.
a
Therefore
|A (t )|q ≤
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
pq Z
b
µ−1 ∇ f (τ ) q ∇τ , a∗
(19)
a
a
∀t ∈ [a, b] ∩ T .
Next by integrating (19) we are proving the claim.
Next we give a related Sobolev inequality. Theorem 11. Here all as in Theorem 10. Let r ≥ 1 and denote b
Z
|f (t )|r ∇ t
kf kr =
1r
.
a
Then b
Z
t
Z
kAkr ≤ a
p b hµ−2 (t , ρ (τ )) ∇τ
! 1r
pr
µ−1
∇ f . a∗ q
∇t
a
(20)
Proof. As in the proof of Theorem 10 we have
|A (t )| ≤
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
b
µ−1 ∇ f (τ ) q ∇τ a∗
1q
.
a
a
Hence
|A (t )| ≤ r
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
pr Z
a
b
µ−1 ∇ f (t ) q ∇ t a∗
qr
,
a
and b
Z
|A (t )|r ∇ t ≤ a
b
Z
t
Z
a
p b hµ−2 (t , ρ (τ )) ∇τ
pr
b
Z ∇t
µ−1 ∇ f (t ) q ∇ t a∗
qr
.
(21)
a
a
Next raise (21) to power 1r . Hence proving the claim.
Next we give an Opial type related inequality.
µ−1
Theorem 12. Here all as in Theorem 10. Additionally assume that ∇a∗ f is increasing on [a, b] ∩ T . Then b
Z
1 1 q |A (t )| ∇aµ− ∗ f (t ) ∇ t ≤ (b − a)
b
Z
a
t
Z
a
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
Proof. As in the proof of Theorem 10 we get
|A (t )| ≤
Z
t
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a
a
Z
b
t
≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p
µ−1 1 ∇ f (t ) (t − a) q . a∗
a
Therefore
1 |A (t )| ∇aµ− ∗ f (t ) ≤
t
Z a
for all t ∈ [a, b] ∩ T .
p b hµ−2 (t , ρ (τ )) ∇τ
1p
2 1 1 q ∇aµ− ∗ f (t ) (t − a) ,
b
1 ∇aµ− ∗ f (t )
2q
1q ∇t
.
(22)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3757
Consequently we derive b
Z
1 |A (t )| ∇aµ− ∗ f (t ) ∇ t ≤
b
Z
"Z
p b hµ−2 (t , ρ (τ )) ∇τ
b
Z
t
Z
≤
1 ∇aµ− ∗ f
2
(t ) (t − a)
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
≤ ( b − a)
1 q
# ∇t
1 q
b
1 ∇aµ− ∗ f
(t )
2q
( t − a) ∇ t
1q
a b
Z
t
Z
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
proving the claim.
1p
a
a
a
t
b
1 ∇aµ− ∗ f
(t )
2q
1q ∇t
,
a
Next we give related Ostrowski type inequalities. Theorem 13. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ), a, b ∈ T , a ≤ b, Tk = T . Suppose b hµ−2 2 ∇k b [a (s, ρ (mt )) , heν (s, ρ (t )) to be continuous on ( , b] ∩ T ) , and f (a) = 0, k = 1, . . . , m − 1. Denote A (t ) = f (t ) − D f ∇ , µ − 1,e ν + 1, T , t , t ∈ [a, b] ∩ T . Then
Z 1 b − a
b a
1 A (t ) ∇ t − f (a) ≤ b−a
b
Z
t
Z
a
a
1 b hµ−2 (t , ρ (τ )) ∇τ ∇ t ∇aµ− ∗ f ∞,[a,b]∩T .
(23)
Proof. By (16) we get A (t ) − f (a) =
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
∀t ∈ [a, b] ∩ T .
a
Then
|A (t ) − f (a)| ≤
t
Z a
1 |b hµ−2 (t , ρ (τ )) |∇τ ∇aµ− ∗ f ∞,[a,b]∩T ,
∀t ∈ [a, b] ∩ T .
Therefore we obtain
Z 1 b − a
b a
≤ ≤ proving the claim.
b
Z
1 A (t ) ∇ t − f (a) = b−a 1 b−a 1 b−a
a b
Z
|A (t ) − f (a)| ∇ t Z b Z t
b hµ−2 (t , ρ (τ )) ∇τ ∇ t ∇ µ−1 f a
a∗
a
b a
a
∞,[a,b]∩T
,
Theorem 14. All as in Theorem 13. Let p, q > 1 :
Z 1 b − a
(A (t ) − f (a)) ∇ t
A (t ) ∇ t − f (a) ≤
1 b−a
1 p b
Z
+
1 q
Z
t
a
= 1. Then p b hµ−2 (t , ρ (τ )) ∇τ
1p
a
!
1 ∇ t ∇aµ− ∗ f q,[a,b]∩T .
Proof. By (16) we derive
|A (t ) − f (a)| ≤
Z
t
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ
a t
Z ≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
a
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a t
Z
t
≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p
µ−1
∇ f a∗
q,[a,b]∩T
.
µ−1
∇ f
q,[a,b]∩T
,
a
That is we have
|A (t ) − f (a)| ≤
t
Z a
p b hµ−2 (t , ρ (τ )) ∇τ
1p
a∗
∀t ∈ [a, b] ∩ T .
(24)
3758
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Therefore we derive
Z 1 b − a
b a
A (t ) ∇ t − f (a) ≤
1 b−a 1
≤ proving the claim.
b−a
b
Z
|A (t ) − f (a)| ∇ t a b
Z
Z
t
p b hµ−2 (t , ρ (τ )) ∇τ
1p
a
a
!
1 ∇ t ∇aµ− ∗ f q,[a,b]∩T ,
We finish general fractional nabla time scales inequalities with a related Hilbert–Pachpatte type inequality. Theorem 15. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; fi ∈ Cldm (Ti ), ai , bi ∈ Ti , ai ≤ bi , Tik = Ti time scale, i = 1, 2. (i)
(i)
k
2 ∇ Suppose b hµ−2 (si , ρi (ti )) , b he (ai ) = 0, k = 0, 1, . . . , m − 1; i = 1, 2. ν (si , ρi (ti )) to be continuous on ([ai , bi ] ∩ Ti ) , and fi
Here Ai (ti ) = fi (ti ) − Di (fi Call F (t1 ) =
t1
Z
∇m
, µ − 1,e ν + 1, Ti , ti ), ti ∈ [ai , bi ] ∩ Ti ; i = 1, 2, and p, q > 1 :
1 p
+
1 q
= 1.
p b(1) hµ−2 (t1 , ρ1 (τ1 )) ∇τ1 ,
a1
for all t1 ∈ [a1 , b1 ], and G (t2 ) =
t2
Z
a2
q b(2) hµ−2 (t2 , ρ2 (τ2 )) ∇τ2 , (i)
for all t2 ∈ [a2 , b2 ] (where b hµ−2 , ρi are the corresponding b hµ−2 , ρ to Ti , i = 1, 2). Then
Z
b1
Z
a1
|A1 (t1 )| |A2 (t2 )| ∇ t1 ∇ t2 ε + F (pt1 ) + G(qt2 )
b2 a2
≤ (b1 − a1 ) (b2 − a2 )
b1
Z
a1
µ−1 ∇ f1 (t1 ) q ∇ t1 a1 ∗
1q Z
b2 a2
µ−1 ∇ f2 (t2 ) p ∇ t2 a2 ∗
1p
.
(above double time scales Riemann nabla integration is considered in the natural iterative way). Proof. We notice that
λ (t1 ) =
|A2 (t2 )|
b2
Z
a2
ε+
F (t1 ) p
+
G(t2 ) q
∇ t2
is a Riemann ∇ -integrable function on [a1 , b1 ] ∩ T1 . k
Since fi∇ (ai ) = 0, k = 0, 1, . . . , m − 1; i = 1, 2, by Corollary 9 we get that Ai (ti ) =
ti
Z
ai
(i) 1 b hµ−2 (ti , ρi (τi )) ∇aµ− fi (τi ) ∇τi , i∗
∀ti ∈ [ai , bi ] ∩ Ti , where ai , bi ∈ Ti . Hence
|A1 (t1 )| ≤
Z
t1 a1
= F (t1 )
1p Z p b(1) hµ−2 (t1 , ρ1 (τ1 )) ∇τ1 1 p
t1
Z
a1
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q
t1 a1
µ−1 ∇ f1 (τ1 ) q ∇τ1
1q
a1 ∗
,
and
|A2 (t2 )| ≤
Z
t2 a2
1q Z q b(2) hµ−2 (t2 , ρ2 (τ2 )) ∇τ2 1
= G (t2 ) q
Z
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
.
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
(25)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3759
Young’s inequality for a, b ≥ 0 says that 1
1
ap bq ≤
a
b
+ .
p
q
Therefore we have 1
1
|A1 (t1 )| |A2 (t2 )| ≤ (F (t1 )) p (G (t2 )) q
t1
Z
a1
≤
F (t1 ) p
+
G (t2 )
t1
Z
q
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
a2
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
a1
t2
1q Z
t2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
a2
1p
.
The last gives (ε > 0)
|A1 (t1 )| |A2 (t2 )| ≤ ε + F (pt1 ) + G(qt2 )
Z
t1 a1
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
,
for all ti ∈ [ai , bi ] ∩ Ti , i = 1, 2. Next we observe that b1
Z
a1
|A1 (t1 )| |A2 (t2 )| ∇ t1 ∇ t2 ε + F (pt1 ) + G(qt2 )
b2
Z
a2
Z
b1
Z
t1
µ−1 ∇ f1 (τ1 ) q ∇τ1
≤ a1
1q
Z
b2
t2
Z
∇ t1 ·
a1 ∗
a1
!
a2
µ−1 ∇ f2 (τ2 ) p ∇τ2
1p ∇ t2
a2 ∗
a2
!
(by Hölder’s inequality)
Z
b1
Z
t1
≤ a1
Z
a1
b1
Z
b1
≤ a1
a1
Z 1q µ−1 1 ∇ f1 (τ1 ) q ∇τ1 ∇ t1 (b1 − a1 ) p · a1 ∗
µ−1 ∇ f1 (τ1 ) q ∇τ1 ∇ t1 a1 ∗
= (b1 − a1 ) (b2 − a2 )
Z
b1 a1
proving the claim.
1q
1 p
(b1 − a1 ) ·
b2 a2
Z
a2
b2
Z
a2
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
b2 a2
t2
Z
b2 a2
1p µ−1 1 ∇ f2 (τ2 ) p ∇τ2 ∇ t2 (b2 − a2 ) q a2 ∗ 1p µ−1 1 ∇ f2 (τ2 ) p ∇τ2 ∇ t2 (b2 − a2 ) q a2 ∗
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
,
3. Applications (I) Here T = R case. Let µ > 2 such that m − 1 < µ < m ∈ N, e ν = m − µ, f ∈ C m ([a, b]), a, b ∈ R. The nabla fractional derivative on R of order µ − 1 is defined as follows: 1 ∇aµ− ∗ f
∀t ∈ [a, b] .
e ν+1 (m)
( t ) = Ja
f
(t ) =
1
Γ (e ν + 1)
t
Z
(t − τ )eν f (m) (τ ) dτ ,
(26)
a
µ−1
Notice that ∇a∗ f ∈ C ([a, b]), and A (t ) = f (t ), ∀t ∈ [a, b] . We give a Poincaré type inequality.
Theorem 16. Let µ > 2, m − 1 < µ < m ∈ N, f ∈ C m (R), a, b ∈ R, a ≤ b. Suppose f (k) (a) = 0, k = 0, 1, . . . , m − 1. Let p, q > 1 : 1p + 1q = 1. Then b
Z
|f (t )|q dt ≤ a
Proof. By Theorem 10.
(b − a)(µ−1)q (Γ (µ − 1))q (µ − 1) q ((µ − 2) p + 1)q−1
We give a Sobolev type inequality.
b
Z a
µ−1 ∇ f (t ) q dt . a∗
(27)
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Theorem 17. All as in Theorem 16. Let r ≥ 1. Then 1
1
(b − a)µ−2+ p + r 1 Γ (µ − 1) ((µ − 2) p + 1) p (µ − 2) r +
kf kr ≤
Proof. By Theorem 11.
r p
1r
µ−1
∇ f . a∗ q
(28)
+1
We continue with an Opial type inequality.
µ−1
Theorem 18. All as in Theorem 16. Assume ∇a∗ f is increasing on [a, b]. b
Z
1
1 |f (t )| ∇aµ− ∗ f (t ) dt ≤
a
Proof. By Theorem 12.
(b − a)µ− q
b
Z 1
Γ (µ − 1) [((µ − 2) p + 1) ((µ − 2) p + 2)] p
1 ∇aµ− ∗ f
2q (t ) dt
1q
.
(29)
a
Some Ostrowski type inequalities follow. Theorem 19. Let µ > 2, m − 1 < µ < m ∈ N, f ∈ C m (R), a, b ∈ R, a ≤ b. Suppose f (k) (a) = 0, k = 1, . . . , m − 1. Then
Z 1 b − a
b a
(b − a)µ−1 µ−1
∇ f f (t ) dt − f (a) ≤ . a∗ ∞,[a,b] Γ (µ + 1)
Proof. By Theorem 13.
Theorem 20. Here all as in Theorem 19. Let p, q > 1 :
Z 1 b − a
b a
f (t ) dt − f (a) ≤
Proof. By Theorem 14.
(30)
1 p
+
1 q
= 1. Then
1
µ−1 (b − a)µ− q −1
∇ f . a∗ 1 q,[a,b] 1 Γ (µ − 1) µ − q ((µ − 2) p + 1) p
(31)
We finish this subsection with a Hilbert–Pachpatte inequality on R. Theorem 21. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N, i = 1, 2; fi ∈ C m (R), ai , bi ∈ R, ai ≤ bi , fi k = 0, 1, . . . , m − 1; p, q > 1 : 1p + 1q = 1. Call F (t1 ) =
(k)
(ai ) = 0,
(t1 − a1 )(µ−2)p+1 , (Γ (µ − 1))p ((µ − 2) p + 1)
t1 ∈ [a1 , b1 ], and G (t2 ) =
(t2 − a2 )(µ−2)q+1 , (Γ (µ − 1))q ((µ − 2) q + 1)
t2 ∈ [a2 , b2 ]. Then
Z
b1
Z
a1
b2 a2
|f1 (t1 )| |f2 (t2 )| dt1 dt2 ε + F (pt1 ) + G(qt2 )
≤ (b1 − a1 ) (b2 − a2 )
Z
b1 a1
Proof. By Theorem 15.
µ−1 ∇ f1 (t1 ) q dt1 a1 ∗
1q Z
b2 a2
µ−1 ∇ f2 (t2 ) p dt2 a2 ∗
1p
.
(32)
(II) Here T = Z case. m Let µ > 2 such ν = m − µ, a, b ∈ Z, a ≤ b. Here f : Z → R, and f ∇ (t ) = ∇ m f (t ) = that m − 1 < µ < m ∈ N, e
Pm
k=0
(−1)k
m k
f (t − k) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3761
The nabla fractional derivative on Z of order µ − 1 is defined as follows: 1 e ν+1 ∇ mf ∇aµ− ∗ f ( t ) = Ja
(t ) =
1
t X
Γ (e ν + 1)
τ =a +1
(t − τ + 1)eν ∇ m f (τ ) ,
(33)
∀t ∈ [a, ∞) ∩ Z.
Notice here that ν (t ) = 1, ∀t ∈ Z, and A (t ) = f (t ) − D ∇ m f , µ − 1,e ν + 1, Z, t
= f (t ) −
t X
∇ m f (u)
(t − u + 1)µ−2 Γ (µ − 1)
u=a+1
,
(34)
∀t ∈ [a, ∞) ∩ Z. We give a discrete fractional Poincaré type inequality. Theorem 22. Let µ > 2, m − 1 < µ < m ∈ N, a, b ∈ Z, a ≤ b, f : Z → R. Assume ∇ k f (a) = 0, k = 0, 1, . . . , m − 1. Let p, q > 1 : 1p + 1q = 1. Then b X
|A (t )| ≤
1
b X
t X
(Γ (µ − 1))q
t =a+1
τ =a+1
q
t =a +1
Proof. By Theorem 10.
(t − τ + 1)(µ−2)p
!!
! b X µ−1 ∇ f (t ) q . a∗
(35)
t =a+1
We continue with a discrete fractional Sobolev type inequality. Theorem 23. Here all as in Theorem 22. Let r ≥ 1 and denote
kf kr =
b X
! 1r .
|f (t )|
r
t =a +1
Then
kAkr ≤
b X
t X
t =a+1
τ =a +1
1
Γ (µ − 1)
Proof. By Theorem 11.
! pr 1r
1 (t − τ + 1)(µ−2)p ∇aµ− ∗ f q.
(36)
Next we give a discrete fractional Opial type inequality.
µ−1
Theorem 24. Here all as in Theorem 22. Assume that ∇a∗ f is increasing on [a, b] ∩ Z. Then b X
1 |A (t )| ∇aµ− ∗ f (t )
t =a +1 1
( b − a) q ≤ Γ (µ − 1) Proof. By Theorem 12.
b X
t X
t =a+1
τ =a+1
(t − τ + 1)(µ−2)p
!! 1p
b X
! 1q 1 ∇aµ− ∗ f
(t )
2q
.
(37)
t =a +1
Next we give related discrete fractional Ostrowski type inequalities. Theorem 25. Let µ > 2, m − 1 < µ < m ∈ N, a, b ∈ Z, a ≤ b, f : Z → R. Assume ∇ k f (a) = 0, k = 1, . . . , m − 1. Then
b 1 X 1 A (t ) − f (a) ≤ b − a t =a +1 (b − a) Γ (µ − 1) Proof. By Theorem 13.
b X
t X
t =a +1
τ =a +1
!! (t − τ + 1)
µ−2
µ−1
∇ f . a∗ ∞,[a,b]∩Z
(38)
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Theorem 26. All as in Theorem 25. Let p, q > 1 :
1 p
+
1 q
= 1. Then
b b 1 X X 1 A (t ) − f (a) ≤ b − a t =a+1 (b − a) Γ (µ − 1) t =a+1 Proof. By Theorem 14.
t X
τ =a+1
! 1p
1 (t − τ + 1)(µ−2)p ∇aµ− ∗ f q,[a,b]∩Z .
(39)
We finish the article with a discrete fractional Hilbert–Pachpatte type inequality. Theorem 27. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N; i = 1, 2; fi : Z → R, ai , bi ∈ Z, ai ≤ bi . Suppose ∇ k fi (ai ) = 0, k = 0, 1, . . . , m − 1. Here Ai (ti ) = fi (ti ) − Call F (t1 ) = τ1
Pti
u i =a i +1
µ−2
i +1 ) , ∀ti ∈ [ai , ∞) ∩ Z; p, q > 1 : (∇ m f (ui )) (ti −Γu(µ− 1)
1 p
+
1 q
= 1.
t1 X (t1 − τ1 + 1)(µ−2)p , (Γ (µ − 1))p =a +1 1
∀t1 ∈ [a1 , ∞) ∩ Z, and G (t2 ) = τ2
t2 X (t2 − τ2 + 1)(µ−2)q , (Γ (µ − 1))q =a +1 2
∀t2 ∈ [a2 , ∞) ∩ Z. Then b1 X
b2 X
|A1 (t1 )| |A2 (t2 )| ≤ (b1 − a1 ) (b2 − a2 ) F (t1 ) G(t2 ) + t1 =a1 +1 t2 =a2 +1 ε + p q Proof. By Theorem 15.
b1 X µ−1 ∇ f1 (t1 ) q a1 ∗ t1 = a 1 + 1
! 1q
b2 X µ−1 ∇ f2 (t2 ) p a2 ∗
! 1p . (40)
t2 = a 2 + 1
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]
G. Anastassiou, Nabla time scales inequalities, 2009 (submitted for publication). D.R. Anderson, Taylor polynomials for nabla dynamic equations on times scales, Panamer. Math. J. 12 (4) (2002) 17–27. D. Anderson, J. Bullock, L. Erbe, A. Peterson, H. Tran, Nabla dynamic equations on time scales, Panamer. Math. J. 13 (1) (2003) 1–47. F. Atici, D. Biles, A. Lebedinsky, An application of time scales to economics, Math. Comput. Modelling 43 (2006) 718–726. M. Bohner, A. Peterson, Dynamic Equations on Time Scales: An Introduction with Applications, Birkhaüser, Boston, 2001. S. Hilger, Ein Maßketten kalkül mit Anwendung auf Zentrumsmannigfaltigkeiten, Ph.D. Thesis, Universität Würzburg, Germany (1988). N. Martins, D. Torres, Calculus of variations on time scales with nabla derivatives, Nonlinear Anal. 71 (12) (2009) 763–773. M. Bohner, G.S. Guseinov, Multiple Lebesgue integration on time scales, Adv. Difference Equ. (2006) 1–12 (Article ID 26391) doi:10.1155/ADE/2006/26391. M. Bohner, G. Guseinov, Double integral calculus of variations on time scales, Comput. Math. Appl. 54 (2007) 45–57. G. Guseinov, Integration on time scales, J. Math. Anal. Appl. 285 (2003) 107–127. Wenjun Liu, Quôc Anh Ngô, Wenbing Chen, Ostrowski type inequalities on time scales for double integrals, Acta Appl. Math. 110 (2010) 477–497. E.T. Whittaker, G.N. Watson, A Course in Modern Analysis, Cambridge University Press, 1927. M. Rafi Segi Rahmat, M. Salmi Md Noorani, Fractional integrals and derivatives on time scales with an application, Comput. Math. Appl., 2009, manuscript. F. Atici, P. Eloe, Discrete fractional Calculus with the nabla operator, Electron. J. Qual. Theory Differ. Equ. (1) (2009) 1–99. Spec. Ed. I, http://www.math.u-szeged.hu/ejqtde/. M. Bohner, H. Luo, Singular second-order multipoint dynamic boundary value problems with mixed derivatives, Adv. Difference Equ. (2006) 1–15 (Article ID 54989) doi:10.1155/ADE/2006/54989.
Contents lists available at ScienceDirect
Computers and Mathematics with Applications journal homepage: www.elsevier.com/locate/camwa
Foundations of nabla fractional calculus on time scales and inequalities George A. Anastassiou Department of Mathematical Sciences, University of Memphis, Memphis, TN 38152, USA
article
abstract
info
Article history: Received 31 January 2010 Accepted 30 March 2010
Here we develop the nabla fractional calculus on time scales. Then we produce related integral inequalities of types: Poincaré, Sobolev, Opial, Ostrowski and Hilbert–Pachpatte. Finally we give inequality applications on the time scales R, Z. © 2010 Elsevier Ltd. All rights reserved.
Keywords: Fractional calculus on time scales Nabla Poincaré inequality Nabla Sobolev inequality Nabla Opial inequalities Nabla Ostrowski inequality Nabla Hilbert–Pachpatte inequality Fractional inequalities
1. Background and foundation results For the basics on time scales we follow [1–10]. Ref. [11] was an inspiration. By [12], p. 256, for µ, ν > 0 we have that x
Z t
(x − t )µ+ν−1 (x − s)µ−1 (s − t )ν−1 ds = , Γ (µ) Γ (ν) Γ (µ + ν)
(1)
where Γ is the gamma function. Here we consider time scales T such that Tk = T . Consider the coordinatewise ld-continuous functions b hα : T × T → R, α ≥ 0, such that b h0 (t , s) = 1,
b hα+1 (t , s) =
t
Z
b hα (τ , s) ∇τ ,
(2)
s
∀s, t ∈ T . Here ρ is the backward jump operator and ν (t ) = t − ρ (t ) . Furthermore for α, β > 1, we assume that Z t b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ = b hα+β−1 (t , ρ (u)) , ρ(u)
is valid for all u, t ∈ T : u ≤ t . In the case of T = R; then ρ (t ) = t, and b hk ( t , s ) = α
(t − s) b hα (t , s) = , Γ (α + 1)
(t −s)k , k!
α ≥ 0.
E-mail addresses: [email protected], [email protected]. 0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2010.03.072
k ∈ N0 = N ∪ {0}, and define
(3)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3751
Notice that
(t − s)α+1 b (τ − s)α dτ = = hα+1 (t , s) , Γ (α + 1) Γ (α + 2)
t
Z s
fulfilling (2). Furthermore we observe that (α, β > 1) t
Z
(t − τ )α−1 (τ − u)β−1 dτ Γ (α) Γ (β) u (t − u)α+β−1 b = hα+β−1 (t , u) , Γ (α + β)
Z
b hα−1 (t , τ ) b hβ−1 (τ , u) dτ
=
u (by (1))
=
t
fulfilling (3). By Theorem 2.2 of [13], we have for k, m ∈ N0 that t
Z
b hk (t , ρ (τ )) b hm (τ , t0 ) ∇τ = b hk+m+1 (t , t0 ) .
(4)
t0
Let T = Z, then ρ (t ) = t −1, t ∈ Z. Define t 0 := 1, t k := t (t + 1) · · · (t + k − 1), k ∈ N, and by (2) we haveb hk ( t , s ) = s, t ∈ Z, Rk ∈ N0 . P t t Here t ∇τ = t0 + 1 . 0 Therefore by (4) we get
(t −s)k , k!
t X (t − τ + 1)k (τ − t0 )m (t − t0 )k+m+1 , = k! m! (k + m + 1)! τ =t +1 0
which results into t X (t − t0 + 1)k+m−1 (t − τ + 1)k−1 (τ − t0 + 1)m−1 = , (k − 1)! (m − 1)! (k + m − 1)! τ = t0
(5)
confirming (3). Next we follow [14]. Γ (t +α) Let a, α ∈ R, define t α = Γ (t ) , t ∈ R − {. . . , −2, −1, 0}, Na = {a, a ± 1, a ± 2, . . .}, notice N0 = Z, 0α = 0, t 0 = 1, and f : Na → R. Here ρ (s) = s − 1, σ (s) = s + 1, ν (t ) = 1. Also define
∇a−n f (t ) =
t X (t − ρ (s))n−1 s=a
(n − 1)!
f (s) ,
n ∈ N,
and in general
∇a−ν f (t ) =
t X (t − ρ (s))ν−1 s=a
Γ (ν)
f (s) ,
where ν ∈ R − {. . . , −2, −1, 0}. Here we set
(t − s)α b hα (t , s) = , Γ (α + 1)
α ≥ 0.
We need Lemma 1. Let α > −1, x > α + 1. Then
Γ (x) 1 = Γ (x − α) (α + 1) Proof. Obvious.
Γ (x + 1) Γ (x) − Γ (x − α) Γ (x − α − 1)
Proposition 2. Let α > −1. It holds t
Z s
(τ − s)α (t − s)α+1 ∇τ = , Γ (α + 1) Γ (α + 2)
That is b hα , α ≥ 0, on Na confirm (2).
t ≥ s.
.
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Proof. Let t > s. We have that t
Z s
t t X X 1 1 Γ (τ − s + α) (τ − s)α ∇τ = (τ − s)α = Γ (α + 1) Γ (α + 1) τ =s+1 Γ (α + 1) τ =s+1 Γ (τ − s)
=
1
t X
Γ (α + 1)
τ =s+1
t− s+α X Γ (τ − s + α) 1 Γ (x) = Γ (τ − s + α − α) Γ (α + 1) x=α+1 Γ (x − α)
(notice here τ − s ≥ 1 and x ≥ α + 1 > 0)
=
(
1
Γ (α + 1)
t −s+α
Γ (α + 1) +
X x=α+2
Γ (x) Γ (x − α)
) =1+
1
Γ (α + 1)
t −s+α
X x=α+2
Γ (x) Γ (x − α)
X Γ (x + 1) Γ ( x) 1 (by Lemma 1) − = 1+ Γ (α + 2) x=α+2 Γ (x − α) Γ (x − α − 1) 1 Γ (α + 4) =1+ − Γ (α + 3) (Γ (α + 3) − Γ (α + 2)) + Γ (α + 2) Γ (3) Γ (α + 5) Γ (α + 4) Γ (α + 6) Γ (α + 5) + − + − + ··· Γ (4) Γ (3) Γ (5) Γ (4) Γ (t − s + α) Γ (t − s + α − 1) Γ (t − s + α + 1) Γ (t − s + α) + − + − Γ (t − s − 1) Γ (t − s + α − 2) Γ (t − s) Γ (t − s − 1) t −s+α
(telescoping sum)
=1+ = That is proving the claim.
1
Γ (α + 2)
Γ (t − s + α + 1) − Γ (α + 2) Γ (t − s)
Γ (t − s + α + 1) (t − s)α+1 = . Γ (α + 2) Γ (t − s) Γ (α + 2)
Next for µ, ν > 1, τ < t, from the proof of Theorem 2.1 [14] we get that t X (t − ρ (s))ν−1 (s − ρ (τ ))µ−1 (t − ρ (τ ))ν+µ−1 = , Γ (ν) Γ (µ) Γ (µ + ν) s=τ
where τ ∈ {a, . . . , t }. So for t , t0 ∈ Na with t0 < t we obtain t X (t − τ + 1)ν−1 (τ − t0 + 1)µ−1 (t − t0 + 1)ν+µ−1 = , Γ (ν) Γ (µ) Γ (µ + ν) τ = t0
(6)
that is confirming (3) fractionally on the time scale T = Na . Notice also here that b X
b
Z
f (t ) ∇ t = a
f (t ) .
t =a+1
So fractional conditions (2) and (3) are very natural and common on time scales. For α ≥ 1 we define the time scale ∇ -Riemann–Liouville type fractional integral (a, b ∈ T ) Jaα f (t ) =
t
Z
b hα−1 (t , ρ (τ )) f (τ ) ∇τ ,
(7)
a
(by [15] the last integral is on (a, t] ∩ T ) Ja0 f (t ) = f (t ) , where f ∈ L1 ([a, b] R∩ T ) (Lebesgue ∇ -integrable functions on [a, b] ∩ T , see [8–10]), t ∈ [a, b] ∩ T . t Notice Ja1 f (t ) = a f (τ ) ∇τ is absolutely continuous in t ∈ [a, b] ∩ T , see [15]. Lemma 3. Let α > 1, f ∈ L1 ([a, b] ∩ T ). Assume that b hα−1 (s, ρ (t )) is Lebesgue ∇ -measurable on ([a, b] ∩ T )2 ; a, b ∈ T . Then α Ja f ∈ L1 ([a, b] ∩ T ) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
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Proof. Define K : Ω := ([a, b] ∩ T )2 → R, by K (s, t ) =
b hα−1 (s, ρ (t )) , if a ≤ t ≤ s ≤ b, 0, if a ≤ s < t ≤ b.
Clearly K (s, t ) is Lebesgue ∇ -measurable on ([a, b] ∩ T )2 . Then b
Z
K (s, t ) ∇ s =
Z [a,t )
a
K (s, t ) ∇ s =
b
Z
b hα−1 (s, ρ (t )) ∇ s t
t
Z
K ( s, t ) ∇ s t
b
Z =
b
Z
K (s, t ) ∇ s +
b
= ρ(t )
b hα−1 (s, ρ (t )) ∇ s −
Z
t
ρ(t )
b hα−1 (s, ρ (t )) ∇ s
hα−1 (t , ρ (t )) ∈ R. =b hα (b, ρ (t )) − ν (t ) b Next we consider the repeated double Lebesgue ∇ -integral b
Z a
b
Z
K (s, t ) |f (t )| ∇ s ∇ t =
b
Z
a
|f (t )|
b
Z
a
K (s, t ) ∇ s ∇ t a
b
Z
|f (t )| b hα (b, ρ (t )) − ν (t ) b hα−1 t , ρ (t ) ∇ t
= a b
Z
|f (t )| b hα (b, ρ (t )) ∇ t −
=
b
Z
a
|f (t )| ν (t ) b hα−1 (t , ρ (t )) ∇ t , a
which exists and is finite. Thus the function (s, t ) → K (s, t ) f (t ) is Lebesgue ∇ -integrable over Ω by Tonelli’s theorem. Let now the characteristic function
χ(a,s]∩T (t ) =
1, 0,
if t ∈ (a, s] ∩ T else,
where s ∈ [a, b] ∩ T . Then the function (s, t ) → χ(a,s]∩T (t ) K (s, t ) f (t ) is Lebesgue ∇ -integrable on Ω . Hence by Fubini’s theorem we get that b
Z
χ(a,s]∩T (t ) K (s, t ) f (t ) ∇ t = a
s
Z
b hα−1 (s, ρ (t )) f (t ) ∇ t = Jaα f (s)
a
is Lebesgue ∇ -integrable in s on [a, b] ∩ T , proving the claim.
For u ≤ t; u, tZ∈ T , we define u
ε (t , u) =
ρ(u)
b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ
= ν ( u) b hα−1 (t , ρ (u)) b hβ−1 (u, ρ (u)) ,
(8)
where α, β > 1. Next we notice for α, β > 1; a, b ∈ T , f ∈ L1 ([a, b] ∩ T ), and b hα−1 (s, ρ (t )) is continuous on ([a, b] ∩ T )2 for any α > 1, that Z τ Z t b b Jaα Jaβ f (t ) = hα−1 (t , ρ (τ )) ∇τ hβ−1 (τ , ρ (u)) f (u) ∇ u a
a
(by Fubini’s theorem) t
Z a
t
Z ·
ρ(u) (by (3))
Z t b hα−1 (t , ρ (τ )) b hβ−1 (τ , ρ (u)) ∇τ = f (u) ∇ u u a Z u b b b b hα−1 (t , ρ (τ )) hβ−1 (τ , ρ (u)) ∇τ − hα−1 (t , ρ (τ )) hβ−1 (τ , ρ (u)) ∇τ
f (u) ∇ u
=
Z
Z
t
ρ(u)
t
= f ( u) ∇ u b hα+β−1 (t , ρ (u)) − ε (t , u) a Z t Z t b = hα+β−1 (t , ρ (u)) f (u) ∇ u − f (u) ε (t , u) ∇ u a a Z t = Jaα+β f (t ) − f (u) ε (t , u) ∇ u. a
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Hence Jaα Jaβ f (t ) +
t
Z
f (u) ε (t , u) ∇ u = Jaα+β f (t ) ,
∀t ∈ [a, b] ∩ T .
a
So we have proved the semigroup property Jaα Jaβ f (t ) +
t
Z
hβ−1 (u, ρ (u)) ∇ u = Jaα+β f (t ) , f (u) ν (u) b hα−1 (t , ρ (u)) b
(9)
a
∀t ∈ [a, b] ∩ T , with a, b ∈ T . We call the Lebesgue ∇ -integral Z t f (u) ν (u) b hα−1 (t , ρ (u)) b hβ−1 (u, ρ (u)) ∇ u, D (f , α, β, T , t ) =
(10)
a
t ∈ [a, b] ∩ T ; a, b ∈ T , the backward graininess deviation functional of f ∈ L1 ([a, b] ∩ T ) . If T = R, then D (f , α, β, R, t ) = 0. Putting things together we have Theorem 4. Let Tk = T , a, b ∈ T , f ∈ L1 ([a, b] ∩ T ); α, β > 1; b hα−1 (s, ρ (t )) is continuous on ([a, b] ∩ T )2 for any α > 1. Then Jaα Jaβ f (t ) + D (f , α, β, T , t ) = Jaα+β f (t ) ,
(11)
∀t ∈ [a, b] ∩ T . We make Remark 5. Let µ > 2 such that m − 1 < µ < m ∈ N, i.e. m = dµe (ceiling of the number), e ν = m − µ (0 < e ν < 1). m Let f ∈ Cldm ([a, b] ∩ T ). Clearly here [10] f ∇ is a Lebesgue ∇ -integrable function. We define the nabla fractional derivative on time scale T of order µ − 1 as follows:
Z t 1 e ν+1 ∇ m ∇m b ∇aµ− f t = J f t = he ( ) ( ) (τ ) ∇τ , ν (t , ρ (τ )) f ∗ a
(12)
a
∀t ∈ [a, b] ∩ T .
µ−1
Notice here that ∇a∗ f ∈ C ([a, b] ∩ T ) by a simple argument using dominated convergence theorem in Lebesgue ∇ -sense. If µ = m, then e ν = 0 and by (12) we get m
∇am∗−1 f (t ) = Ja1 f ∇ (t ) = f ∇
m−1
(t ) .
(13)
m−1
m
More generally, by [15], given that f ∇ is everywhere finite and absolutely continuous on [a, b] ∩ T , then f ∇ exists ∇ -a.e. and is Lebesgue ∇ -integrable on (a, t] ∩ T , ∀t ∈ [a, b] ∩ T , and one can plug it into (12). We observe that 1 Jaµ−1 ∇aµ− ∗ f (t )
ν+1 ∇ Jaµ−1 Jae f
m
(t ) Z t m ν ∇m Jaµ+e f f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he (t ) − ν (u, ρ (u)) ∇ u a Z t m m Jam f ∇ f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he (t ) − ν (u, ρ (u)) ∇ u.
= (by (11))
=
=
a
Hence µ−1
Ja
1 ∇aµ− ∗ f
(t ) +
Z a
=
m Jam f ∇
(t ) =
t
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u
Z
t
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ .
a
We have proved Theorem 6. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm ([a, b] ∩ T ), a, b ∈ T , Tk = T . Suppose b hµ−2 (s, ρ (t )), 2 b he ν (s, ρ (t )) to be continuous on ([a, b] ∩ T ) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
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Then t
Z
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ
a t
Z
m
he hµ−2 (t , ρ (u)) b f ∇ (u) ν (u) b ν (u, ρ (u)) ∇ u +
= a
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(14)
a
∀t ∈ [a, b] ∩ T . We need the nabla time scales Taylor formula Theorem 7 ([2]). Let f ∈ Cldm (T ), m ∈ N, Tk = T ; a, b ∈ T . Then m−1
f (t ) =
X
k b hk (t , a) f ∇ (a) +
t
Z
m b hm−1 (t , ρ (τ )) f ∇ (τ ) ∇τ ,
(15)
a
k=0
∀t ∈ [a, b] ∩ T . Next we present the fractional time scales nabla Taylor formula Theorem 8. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ) , a, b ∈ T , Tk = T . Suppose b hµ−2 (s, ρ (t )), b he ν (s, ρ (t )) 2 to be continuous on ([a, b] ∩ T ) . Then m−1
f (t ) =
X
k b hk (t , a) f ∇ (a)
k=0 t
Z + a
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u +
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(16)
a
∀t ∈ [a, b] ∩ T . k
Corollary 9. All as in Theorem 8. Additionally suppose f ∇ (a) = 0, k = 0, 1, . . . , m − 1. Then
m
A (t ) := f (t ) − D f ∇ , µ − 1,e ν + 1, T , t
= f (t ) −
t
Z
m
f ∇ (u) ν (u) b hµ−2 (t , ρ (u)) b he ν (u, ρ (u)) ∇ u
a t
Z =
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
(17)
a
∀t ∈ [a, b] ∩ T . m
Notice here that D f ∇ , µ − 1,e ν + 1, T , t ∈ Cld ([a, b] ∩ T ). Also the R.H.S (17) is a continuous function in t ∈ [a, b] ∩ T .
2. Fractional nabla inequalities on time scales We present a Poincaré type related inequality. Theorem 10. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ), a, b ∈ T , a ≤ b, Tk = T . Suppose b hµ−2 k
2 ∇ he (s, ρ (mt )) , b ν (s, ρ (t )) to be continuous on ([a, b] ∩ T ) , and1 f 1(a) = 0, k = 0, 1, . . . , m − 1. Here A (t ) = f (t ) − ∇ D f , µ − 1,e ν + 1, T , t , t ∈ [a, b] ∩ T ; and let p, q > 1 : p + q = 1.
Then b
Z
|A (t )| ∇ t ≤ q
a
b
Z a
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
a
Proof. By Corollary 9 we get that A (t ) =
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ .
a
Hence
|A (t )| ≤
t
Z a
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ
pq
! Z
b
∇t a
µ−1 ∇ f (t ) q ∇ t . a∗
(18)
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
(by Hölder’s inequality) t
Z ≤
1p Z
p b hµ−2 (t , ρ (τ )) ∇τ
t
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a
a t
Z ≤
1p Z
p b hµ−2 (t , ρ (τ )) ∇τ
a
b
µ−1 ∇ f (τ ) q ∇τ a∗
1q
.
a
Therefore
|A (t )|q ≤
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
pq Z
b
µ−1 ∇ f (τ ) q ∇τ , a∗
(19)
a
a
∀t ∈ [a, b] ∩ T .
Next by integrating (19) we are proving the claim.
Next we give a related Sobolev inequality. Theorem 11. Here all as in Theorem 10. Let r ≥ 1 and denote b
Z
|f (t )|r ∇ t
kf kr =
1r
.
a
Then b
Z
t
Z
kAkr ≤ a
p b hµ−2 (t , ρ (τ )) ∇τ
! 1r
pr
µ−1
∇ f . a∗ q
∇t
a
(20)
Proof. As in the proof of Theorem 10 we have
|A (t )| ≤
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
b
µ−1 ∇ f (τ ) q ∇τ a∗
1q
.
a
a
Hence
|A (t )| ≤ r
t
Z
p b hµ−2 (t , ρ (τ )) ∇τ
pr Z
a
b
µ−1 ∇ f (t ) q ∇ t a∗
qr
,
a
and b
Z
|A (t )|r ∇ t ≤ a
b
Z
t
Z
a
p b hµ−2 (t , ρ (τ )) ∇τ
pr
b
Z ∇t
µ−1 ∇ f (t ) q ∇ t a∗
qr
.
(21)
a
a
Next raise (21) to power 1r . Hence proving the claim.
Next we give an Opial type related inequality.
µ−1
Theorem 12. Here all as in Theorem 10. Additionally assume that ∇a∗ f is increasing on [a, b] ∩ T . Then b
Z
1 1 q |A (t )| ∇aµ− ∗ f (t ) ∇ t ≤ (b − a)
b
Z
a
t
Z
a
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
Proof. As in the proof of Theorem 10 we get
|A (t )| ≤
Z
t
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a
a
Z
b
t
≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p
µ−1 1 ∇ f (t ) (t − a) q . a∗
a
Therefore
1 |A (t )| ∇aµ− ∗ f (t ) ≤
t
Z a
for all t ∈ [a, b] ∩ T .
p b hµ−2 (t , ρ (τ )) ∇τ
1p
2 1 1 q ∇aµ− ∗ f (t ) (t − a) ,
b
1 ∇aµ− ∗ f (t )
2q
1q ∇t
.
(22)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3757
Consequently we derive b
Z
1 |A (t )| ∇aµ− ∗ f (t ) ∇ t ≤
b
Z
"Z
p b hµ−2 (t , ρ (τ )) ∇τ
b
Z
t
Z
≤
1 ∇aµ− ∗ f
2
(t ) (t − a)
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
≤ ( b − a)
1 q
# ∇t
1 q
b
1 ∇aµ− ∗ f
(t )
2q
( t − a) ∇ t
1q
a b
Z
t
Z
1p Z p b hµ−2 (t , ρ (τ )) ∇τ ∇ t
a
a
proving the claim.
1p
a
a
a
t
b
1 ∇aµ− ∗ f
(t )
2q
1q ∇t
,
a
Next we give related Ostrowski type inequalities. Theorem 13. Let µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; f ∈ Cldm (T ), a, b ∈ T , a ≤ b, Tk = T . Suppose b hµ−2 2 ∇k b [a (s, ρ (mt )) , heν (s, ρ (t )) to be continuous on ( , b] ∩ T ) , and f (a) = 0, k = 1, . . . , m − 1. Denote A (t ) = f (t ) − D f ∇ , µ − 1,e ν + 1, T , t , t ∈ [a, b] ∩ T . Then
Z 1 b − a
b a
1 A (t ) ∇ t − f (a) ≤ b−a
b
Z
t
Z
a
a
1 b hµ−2 (t , ρ (τ )) ∇τ ∇ t ∇aµ− ∗ f ∞,[a,b]∩T .
(23)
Proof. By (16) we get A (t ) − f (a) =
t
Z
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ ,
∀t ∈ [a, b] ∩ T .
a
Then
|A (t ) − f (a)| ≤
t
Z a
1 |b hµ−2 (t , ρ (τ )) |∇τ ∇aµ− ∗ f ∞,[a,b]∩T ,
∀t ∈ [a, b] ∩ T .
Therefore we obtain
Z 1 b − a
b a
≤ ≤ proving the claim.
b
Z
1 A (t ) ∇ t − f (a) = b−a 1 b−a 1 b−a
a b
Z
|A (t ) − f (a)| ∇ t Z b Z t
b hµ−2 (t , ρ (τ )) ∇τ ∇ t ∇ µ−1 f a
a∗
a
b a
a
∞,[a,b]∩T
,
Theorem 14. All as in Theorem 13. Let p, q > 1 :
Z 1 b − a
(A (t ) − f (a)) ∇ t
A (t ) ∇ t − f (a) ≤
1 b−a
1 p b
Z
+
1 q
Z
t
a
= 1. Then p b hµ−2 (t , ρ (τ )) ∇τ
1p
a
!
1 ∇ t ∇aµ− ∗ f q,[a,b]∩T .
Proof. By (16) we derive
|A (t ) − f (a)| ≤
Z
t
1 b hµ−2 (t , ρ (τ )) ∇aµ− ∗ f (τ ) ∇τ
a t
Z ≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p Z
a
µ−1 ∇ f (τ ) q ∇τ a∗
1q
a t
Z
t
≤
p b hµ−2 (t , ρ (τ )) ∇τ
1p
µ−1
∇ f a∗
q,[a,b]∩T
.
µ−1
∇ f
q,[a,b]∩T
,
a
That is we have
|A (t ) − f (a)| ≤
t
Z a
p b hµ−2 (t , ρ (τ )) ∇τ
1p
a∗
∀t ∈ [a, b] ∩ T .
(24)
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G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Therefore we derive
Z 1 b − a
b a
A (t ) ∇ t − f (a) ≤
1 b−a 1
≤ proving the claim.
b−a
b
Z
|A (t ) − f (a)| ∇ t a b
Z
Z
t
p b hµ−2 (t , ρ (τ )) ∇τ
1p
a
a
!
1 ∇ t ∇aµ− ∗ f q,[a,b]∩T ,
We finish general fractional nabla time scales inequalities with a related Hilbert–Pachpatte type inequality. Theorem 15. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N, e ν = m − µ; fi ∈ Cldm (Ti ), ai , bi ∈ Ti , ai ≤ bi , Tik = Ti time scale, i = 1, 2. (i)
(i)
k
2 ∇ Suppose b hµ−2 (si , ρi (ti )) , b he (ai ) = 0, k = 0, 1, . . . , m − 1; i = 1, 2. ν (si , ρi (ti )) to be continuous on ([ai , bi ] ∩ Ti ) , and fi
Here Ai (ti ) = fi (ti ) − Di (fi Call F (t1 ) =
t1
Z
∇m
, µ − 1,e ν + 1, Ti , ti ), ti ∈ [ai , bi ] ∩ Ti ; i = 1, 2, and p, q > 1 :
1 p
+
1 q
= 1.
p b(1) hµ−2 (t1 , ρ1 (τ1 )) ∇τ1 ,
a1
for all t1 ∈ [a1 , b1 ], and G (t2 ) =
t2
Z
a2
q b(2) hµ−2 (t2 , ρ2 (τ2 )) ∇τ2 , (i)
for all t2 ∈ [a2 , b2 ] (where b hµ−2 , ρi are the corresponding b hµ−2 , ρ to Ti , i = 1, 2). Then
Z
b1
Z
a1
|A1 (t1 )| |A2 (t2 )| ∇ t1 ∇ t2 ε + F (pt1 ) + G(qt2 )
b2 a2
≤ (b1 − a1 ) (b2 − a2 )
b1
Z
a1
µ−1 ∇ f1 (t1 ) q ∇ t1 a1 ∗
1q Z
b2 a2
µ−1 ∇ f2 (t2 ) p ∇ t2 a2 ∗
1p
.
(above double time scales Riemann nabla integration is considered in the natural iterative way). Proof. We notice that
λ (t1 ) =
|A2 (t2 )|
b2
Z
a2
ε+
F (t1 ) p
+
G(t2 ) q
∇ t2
is a Riemann ∇ -integrable function on [a1 , b1 ] ∩ T1 . k
Since fi∇ (ai ) = 0, k = 0, 1, . . . , m − 1; i = 1, 2, by Corollary 9 we get that Ai (ti ) =
ti
Z
ai
(i) 1 b hµ−2 (ti , ρi (τi )) ∇aµ− fi (τi ) ∇τi , i∗
∀ti ∈ [ai , bi ] ∩ Ti , where ai , bi ∈ Ti . Hence
|A1 (t1 )| ≤
Z
t1 a1
= F (t1 )
1p Z p b(1) hµ−2 (t1 , ρ1 (τ1 )) ∇τ1 1 p
t1
Z
a1
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q
t1 a1
µ−1 ∇ f1 (τ1 ) q ∇τ1
1q
a1 ∗
,
and
|A2 (t2 )| ≤
Z
t2 a2
1q Z q b(2) hµ−2 (t2 , ρ2 (τ2 )) ∇τ2 1
= G (t2 ) q
Z
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
.
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
(25)
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3759
Young’s inequality for a, b ≥ 0 says that 1
1
ap bq ≤
a
b
+ .
p
q
Therefore we have 1
1
|A1 (t1 )| |A2 (t2 )| ≤ (F (t1 )) p (G (t2 )) q
t1
Z
a1
≤
F (t1 ) p
+
G (t2 )
t1
Z
q
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
a2
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
a1
t2
1q Z
t2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
a2
1p
.
The last gives (ε > 0)
|A1 (t1 )| |A2 (t2 )| ≤ ε + F (pt1 ) + G(qt2 )
Z
t1 a1
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
t2 a2
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
,
for all ti ∈ [ai , bi ] ∩ Ti , i = 1, 2. Next we observe that b1
Z
a1
|A1 (t1 )| |A2 (t2 )| ∇ t1 ∇ t2 ε + F (pt1 ) + G(qt2 )
b2
Z
a2
Z
b1
Z
t1
µ−1 ∇ f1 (τ1 ) q ∇τ1
≤ a1
1q
Z
b2
t2
Z
∇ t1 ·
a1 ∗
a1
!
a2
µ−1 ∇ f2 (τ2 ) p ∇τ2
1p ∇ t2
a2 ∗
a2
!
(by Hölder’s inequality)
Z
b1
Z
t1
≤ a1
Z
a1
b1
Z
b1
≤ a1
a1
Z 1q µ−1 1 ∇ f1 (τ1 ) q ∇τ1 ∇ t1 (b1 − a1 ) p · a1 ∗
µ−1 ∇ f1 (τ1 ) q ∇τ1 ∇ t1 a1 ∗
= (b1 − a1 ) (b2 − a2 )
Z
b1 a1
proving the claim.
1q
1 p
(b1 − a1 ) ·
b2 a2
Z
a2
b2
Z
a2
µ−1 ∇ f1 (τ1 ) q ∇τ1 a1 ∗
1q Z
b2 a2
t2
Z
b2 a2
1p µ−1 1 ∇ f2 (τ2 ) p ∇τ2 ∇ t2 (b2 − a2 ) q a2 ∗ 1p µ−1 1 ∇ f2 (τ2 ) p ∇τ2 ∇ t2 (b2 − a2 ) q a2 ∗
µ−1 ∇ f2 (τ2 ) p ∇τ2 a2 ∗
1p
,
3. Applications (I) Here T = R case. Let µ > 2 such that m − 1 < µ < m ∈ N, e ν = m − µ, f ∈ C m ([a, b]), a, b ∈ R. The nabla fractional derivative on R of order µ − 1 is defined as follows: 1 ∇aµ− ∗ f
∀t ∈ [a, b] .
e ν+1 (m)
( t ) = Ja
f
(t ) =
1
Γ (e ν + 1)
t
Z
(t − τ )eν f (m) (τ ) dτ ,
(26)
a
µ−1
Notice that ∇a∗ f ∈ C ([a, b]), and A (t ) = f (t ), ∀t ∈ [a, b] . We give a Poincaré type inequality.
Theorem 16. Let µ > 2, m − 1 < µ < m ∈ N, f ∈ C m (R), a, b ∈ R, a ≤ b. Suppose f (k) (a) = 0, k = 0, 1, . . . , m − 1. Let p, q > 1 : 1p + 1q = 1. Then b
Z
|f (t )|q dt ≤ a
Proof. By Theorem 10.
(b − a)(µ−1)q (Γ (µ − 1))q (µ − 1) q ((µ − 2) p + 1)q−1
We give a Sobolev type inequality.
b
Z a
µ−1 ∇ f (t ) q dt . a∗
(27)
3760
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Theorem 17. All as in Theorem 16. Let r ≥ 1. Then 1
1
(b − a)µ−2+ p + r 1 Γ (µ − 1) ((µ − 2) p + 1) p (µ − 2) r +
kf kr ≤
Proof. By Theorem 11.
r p
1r
µ−1
∇ f . a∗ q
(28)
+1
We continue with an Opial type inequality.
µ−1
Theorem 18. All as in Theorem 16. Assume ∇a∗ f is increasing on [a, b]. b
Z
1
1 |f (t )| ∇aµ− ∗ f (t ) dt ≤
a
Proof. By Theorem 12.
(b − a)µ− q
b
Z 1
Γ (µ − 1) [((µ − 2) p + 1) ((µ − 2) p + 2)] p
1 ∇aµ− ∗ f
2q (t ) dt
1q
.
(29)
a
Some Ostrowski type inequalities follow. Theorem 19. Let µ > 2, m − 1 < µ < m ∈ N, f ∈ C m (R), a, b ∈ R, a ≤ b. Suppose f (k) (a) = 0, k = 1, . . . , m − 1. Then
Z 1 b − a
b a
(b − a)µ−1 µ−1
∇ f f (t ) dt − f (a) ≤ . a∗ ∞,[a,b] Γ (µ + 1)
Proof. By Theorem 13.
Theorem 20. Here all as in Theorem 19. Let p, q > 1 :
Z 1 b − a
b a
f (t ) dt − f (a) ≤
Proof. By Theorem 14.
(30)
1 p
+
1 q
= 1. Then
1
µ−1 (b − a)µ− q −1
∇ f . a∗ 1 q,[a,b] 1 Γ (µ − 1) µ − q ((µ − 2) p + 1) p
(31)
We finish this subsection with a Hilbert–Pachpatte inequality on R. Theorem 21. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N, i = 1, 2; fi ∈ C m (R), ai , bi ∈ R, ai ≤ bi , fi k = 0, 1, . . . , m − 1; p, q > 1 : 1p + 1q = 1. Call F (t1 ) =
(k)
(ai ) = 0,
(t1 − a1 )(µ−2)p+1 , (Γ (µ − 1))p ((µ − 2) p + 1)
t1 ∈ [a1 , b1 ], and G (t2 ) =
(t2 − a2 )(µ−2)q+1 , (Γ (µ − 1))q ((µ − 2) q + 1)
t2 ∈ [a2 , b2 ]. Then
Z
b1
Z
a1
b2 a2
|f1 (t1 )| |f2 (t2 )| dt1 dt2 ε + F (pt1 ) + G(qt2 )
≤ (b1 − a1 ) (b2 − a2 )
Z
b1 a1
Proof. By Theorem 15.
µ−1 ∇ f1 (t1 ) q dt1 a1 ∗
1q Z
b2 a2
µ−1 ∇ f2 (t2 ) p dt2 a2 ∗
1p
.
(32)
(II) Here T = Z case. m Let µ > 2 such ν = m − µ, a, b ∈ Z, a ≤ b. Here f : Z → R, and f ∇ (t ) = ∇ m f (t ) = that m − 1 < µ < m ∈ N, e
Pm
k=0
(−1)k
m k
f (t − k) .
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
3761
The nabla fractional derivative on Z of order µ − 1 is defined as follows: 1 e ν+1 ∇ mf ∇aµ− ∗ f ( t ) = Ja
(t ) =
1
t X
Γ (e ν + 1)
τ =a +1
(t − τ + 1)eν ∇ m f (τ ) ,
(33)
∀t ∈ [a, ∞) ∩ Z.
Notice here that ν (t ) = 1, ∀t ∈ Z, and A (t ) = f (t ) − D ∇ m f , µ − 1,e ν + 1, Z, t
= f (t ) −
t X
∇ m f (u)
(t − u + 1)µ−2 Γ (µ − 1)
u=a+1
,
(34)
∀t ∈ [a, ∞) ∩ Z. We give a discrete fractional Poincaré type inequality. Theorem 22. Let µ > 2, m − 1 < µ < m ∈ N, a, b ∈ Z, a ≤ b, f : Z → R. Assume ∇ k f (a) = 0, k = 0, 1, . . . , m − 1. Let p, q > 1 : 1p + 1q = 1. Then b X
|A (t )| ≤
1
b X
t X
(Γ (µ − 1))q
t =a+1
τ =a+1
q
t =a +1
Proof. By Theorem 10.
(t − τ + 1)(µ−2)p
!!
! b X µ−1 ∇ f (t ) q . a∗
(35)
t =a+1
We continue with a discrete fractional Sobolev type inequality. Theorem 23. Here all as in Theorem 22. Let r ≥ 1 and denote
kf kr =
b X
! 1r .
|f (t )|
r
t =a +1
Then
kAkr ≤
b X
t X
t =a+1
τ =a +1
1
Γ (µ − 1)
Proof. By Theorem 11.
! pr 1r
1 (t − τ + 1)(µ−2)p ∇aµ− ∗ f q.
(36)
Next we give a discrete fractional Opial type inequality.
µ−1
Theorem 24. Here all as in Theorem 22. Assume that ∇a∗ f is increasing on [a, b] ∩ Z. Then b X
1 |A (t )| ∇aµ− ∗ f (t )
t =a +1 1
( b − a) q ≤ Γ (µ − 1) Proof. By Theorem 12.
b X
t X
t =a+1
τ =a+1
(t − τ + 1)(µ−2)p
!! 1p
b X
! 1q 1 ∇aµ− ∗ f
(t )
2q
.
(37)
t =a +1
Next we give related discrete fractional Ostrowski type inequalities. Theorem 25. Let µ > 2, m − 1 < µ < m ∈ N, a, b ∈ Z, a ≤ b, f : Z → R. Assume ∇ k f (a) = 0, k = 1, . . . , m − 1. Then
b 1 X 1 A (t ) − f (a) ≤ b − a t =a +1 (b − a) Γ (µ − 1) Proof. By Theorem 13.
b X
t X
t =a +1
τ =a +1
!! (t − τ + 1)
µ−2
µ−1
∇ f . a∗ ∞,[a,b]∩Z
(38)
3762
G.A. Anastassiou / Computers and Mathematics with Applications 59 (2010) 3750–3762
Theorem 26. All as in Theorem 25. Let p, q > 1 :
1 p
+
1 q
= 1. Then
b b 1 X X 1 A (t ) − f (a) ≤ b − a t =a+1 (b − a) Γ (µ − 1) t =a+1 Proof. By Theorem 14.
t X
τ =a+1
! 1p
1 (t − τ + 1)(µ−2)p ∇aµ− ∗ f q,[a,b]∩Z .
(39)
We finish the article with a discrete fractional Hilbert–Pachpatte type inequality. Theorem 27. Let ε > 0, µ > 2, m − 1 < µ < m ∈ N; i = 1, 2; fi : Z → R, ai , bi ∈ Z, ai ≤ bi . Suppose ∇ k fi (ai ) = 0, k = 0, 1, . . . , m − 1. Here Ai (ti ) = fi (ti ) − Call F (t1 ) = τ1
Pti
u i =a i +1
µ−2
i +1 ) , ∀ti ∈ [ai , ∞) ∩ Z; p, q > 1 : (∇ m f (ui )) (ti −Γu(µ− 1)
1 p
+
1 q
= 1.
t1 X (t1 − τ1 + 1)(µ−2)p , (Γ (µ − 1))p =a +1 1
∀t1 ∈ [a1 , ∞) ∩ Z, and G (t2 ) = τ2
t2 X (t2 − τ2 + 1)(µ−2)q , (Γ (µ − 1))q =a +1 2
∀t2 ∈ [a2 , ∞) ∩ Z. Then b1 X
b2 X
|A1 (t1 )| |A2 (t2 )| ≤ (b1 − a1 ) (b2 − a2 ) F (t1 ) G(t2 ) + t1 =a1 +1 t2 =a2 +1 ε + p q Proof. By Theorem 15.
b1 X µ−1 ∇ f1 (t1 ) q a1 ∗ t1 = a 1 + 1
! 1q
b2 X µ−1 ∇ f2 (t2 ) p a2 ∗
! 1p . (40)
t2 = a 2 + 1
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