Introduction

Fractional calculus has gained importance during the past three decades due to its applicability in diverse fields of science and engineering[ ¹ ]. The notions of fractional calculus may be traced back to the works of Euler, but the idea of fractional difference is very recent.

Diaz and Osler[ ² ] defined the fractional difference by the rather natural approach of allowing the index of differencing, in the standard expression for the n th difference, to be any real or complex number. Later, Hirota[ ³ ] defined the fractional order difference operator ∇ ^α , where α is any real number, using Taylor’s series. Nagai[ ⁴ ] adopted another definition for fractional difference by modifying Hirota’s[ ³ ] definition. Recently, Deekshitulu and Jagan Mohan[ ⁵ ] modified the definition of Nagai[ ⁴ ] and discussed some basic inequalities, comparison theorems, and qualitative properties of the solutions of fractional difference equations[ ⁵ – ¹⁰ ].

Discrete inequalities involving sequences of real numbers, which may be considered as discrete analogues of the Gronwall-Bellman inequality, have been extensively used in the analysis of finite difference equations. In the year 1973, Pachpatte[ ¹¹ ] established the following remarkable inequality:

Theorem 1

Let u ( n ), b ( n ), and c ( n ) be nonnegative real valued functions defined on N0+ and c ≥ 0 is a constant. If, for n∈N0+ ,

u(n)≤c+∑j=0n−1b(j)u(j)+∑k=0j−1c(k)u(k),

then

u(n)≤c1+∑j=0n−1b(j)∏k=0j−1[1+b(j)+c(j)].

Throughout the present paper, we use the following notations[ ¹² ]: N is the set of natural numbers including zero and Z is the set of integers. Na+={a,a+1,a+2,…} for a∈Z . Let u ( n ) be a real valued function defined on N0+ . Then, for all n1,n2∈N0+ , and n ₁ > n ₂ , ∑j=n1n2u(j)=0 and ∏j=n1n2u(j)=1 , i.e., empty sums and products are taken to be 0 and 1, respectively. If n and n + 1 are in N0+ , the backward difference operator ∇ is defined as ∇ u ( n ) = u ( n ) − u ( n − 1).

Now, we introduce some basic definitions and results concerning nabla discrete fractional calculus. The extended binomial coefficient an , ( a∈R , n∈Z ), is defined by[ ⁴ ]:

an=Γ(a+1)Γ(a−n+1)Γ(n+1)n>01n=00n<0.

In 2003, Nagai[ ⁴ ] gave the following definition for the fractional order difference operator:

Definition 1

Let

α∈R

and m be an integer such that m − 1 < α ≤ m . The difference operator ∇ of order α , with step length ϵ , is defined as

∇αu(n)=∇α−m[∇mu(n)]=ϵm−α∑j=0n−1α−mj(−1)j∇mu(n−j)α>0u(n)α=0ϵ−α∑j=0n−1αj(−1)ju(n−j)α<0.

The above definition of ∇ ^α u ( n ) given by Nagai[ ⁴ ] contains the ∇ operator and the term (−1) ^j inside the summation index, and hence, it becomes difficult to study the properties of the solution. Deekshitulu and Jagan Mohan[ ⁵ ] modified the above definition for α ∈ (0, 1) as follows:

Definition 2

The fractional sum operator of order α ( α∈R ) is defined as

∇−αu(n)=∑j=0n−1j+α−1ju(n−j)=∑j=1nn−j+α−1n−ju(j),

and the fractional difference operator of order α ( α∈R and 0 < α < 1) is defined as

∇αu(n)=∑j=0n−1j−αj∇u(n−j)=∑j=1nn−j−α−1n−ju(j)−n−α−1n−1u(0).

Remark 1

If we take α = 1in (2), using the definition of the generalized binomial coefficient, we have

∑j=0n−1j−1j∇u(n−j)=−10∇u(n)+∑j=1n−1j−1j∇u(n−j)=−10∇u(n)+∑j=1n−1j−1−1∇u(n−j)=∇u(n).

Gray and Zhang[ ¹³ ] gave the following definition:

Definition 3

For any complex numbers α and β ,

(α)β=Γ(α+β)Γ(α)whenαandα+βare neither zeronor negative integers1whenα=β=00whenα=0andβis neither zeronor a negative integerundefinedotherwise.

Remark 2

For any complex numbers α and β , when α , β , and α + β are neither zero nor negative integers,

(α+β)n=∑k=0nnk(α)n−k(β)k

for any positive integer n .

Theorem 2

Let u ( n ) and v ( n ) be real valued functions defined on

N0+

and α ,

β∈R

such that 0 < α , β , α + β < 1 and c , d are constants . Then ,

∇ ^α [ cu ( n ) + dv ( n )] = c ∇ ^α u ( n ) + d ∇ ^α v ( n ).
∇ ^−
α ∇ ^α u ( n ) = u ( n ) − u (0).
∇ ^α u (0) = 0 and ∇ ^α u (1) = u (1) − u (0) = ∇ u (1).

Proof

Consider
∇−α∇αu(n)=∇−α∇αu(n)=∑j=1nn−j+α−1n−j∇αu(j)=∑j=1nn−j+α−1n−j∑k=0j−1k−αk∇u(j−k)=∑j=1nn−j+α−1n−j∑i=1jj−i−αj−i∇u(i)

=∑j=1n∑i=1jn−j+α−1n−jj−i−αj−i∇u(i)=∑i=1n∑j=inΓ(n−j+α)Γ(n−j+1)Γ(α)Γ(j−i−α+1)Γ(j−i+1)Γ(−α+1)∇u(i)=∑i=1n∑j=0n−iΓ(n−j−i+α)Γ(n−j−i+1)Γ(α)Γ(j−α+1)Γ(j+1)Γ(−α+1)∇u(i)=∑i=1n∇u(i)Γ(n−i+1)∑j=0n−iΓ(n−i+1)Γ(n−i−j+1)Γ(j+1)×Γ(n−i−j+α)Γ(α)Γ(j−α+1)Γ(−α+1)=∑i=1n∇u(i)Γ(n−i+1)∑j=0n−in−ij(α)n−i−j(−α+1)j=∑i=1n∇u(i)Γ(n−i+1)(1)n−i=∑i=1nΓ(n−i+1)Γ(n−i+1)Γ(1)∇u(i)=u(n)−u(0).

The proofs of 1 and 3 are clear from Definition 2. □

Definition 4

Let f(n,r):N0+×R→R . Then, a nonlinear difference equation of order α , 0 < α < 1, together with an initial condition is of the form

∇αu(n+1)=f(n,u(n)),u(0)=u0.

Now, we consider (1) and replace u ( n )by ∇ ^α u ( n ), and we have

∇−α[∇αu(n)]=∑j=1nn−j+α−1n−j[∇αu(j)]oru(n)−u(0)=∑j=1nn−j+α−1n−j[∇αu(j)]oru(n)=u0+∑j=0n−1n−j+α−2n−j−1[∇αu(j+1)]oru(n)=u0+∑j=0n−1B(n−1,α;j)f(j,u(j)),

where

B(n,α;j)=n−j+α−1n−j

for 0 ≤ j ≤ n . The above recurrence relation shows the existence of the solution of (3).

Recently, the authors have established the following fractional discrete Gronwall-Bellman-type inequality[ ¹⁰ ]:

Theorem 3

Let u ( n ), a ( n ), and b ( n ) be nonnegative real valued functions defined on

N0+

. If , for n∈N0+

∇αu(n+1)≤a(n)u(n)+b(n),

then

u(n)≤u(0)∏j=0n−11+B(n−1,α;j)a(j)×∑j=0n−1B(n−1,α;j)b(j)∏k=j+1n−11+B(n−1,α;k)a(k).

Main Results

In this section, we shall establish some new fractional order difference inequalities of Gronwall-Bellman type analogous to the inequality (Theorem 1) given by Pachpatte[ ¹¹ ]. Let u ( n ), b ( n ), c ( n ), p ( n ), and q ( n ) be nonnegative real valued functions defined on

N0+

and u (0) ≥ 0 be a constant.

Theorem 4

If a ( n ) is a positive, monotonic, and nondecreasing real valued function defined on N0+ and

u(n)≤a(n)+∑j=0n−1B(n−1,α;j)b(j)×u(j)+∑k=0j−1B(j−1,α;k)c(k)u(k)

for n∈N0+ , then

u(n)≤a(n)1+∑j=0n−1B(n−1,α;j)b(j)×∏k=0j−11+B(j−1,α;k)[b(k)+c(k)]

for n∈N0+ .

Proof

Since a ( n ) is a positive, monotonic, and nondecreasing real valued function, from (6), we observe that

u(n)a(n)≤1+∑j=0n−1B(n−1,α;j)b(j)×u(j)a(n)+∑k=0j−1B(j−1,α;k)c(k)u(k)a(n)≤1+∑j=0n−1B(n−1,α;j)b(j)×u(j)a(j)+∑k=0j−1B(j−1,α;k)c(k)u(k)a(k).

Define a function

z(n)=1+∑j=0n−1B(n−1,α;j)b(j)×u(j)a(j)+∑k=0j−1B(j−1,α;k)c(k)u(k)a(k).

Then, z (0) = 1, u(n)a(n)≤z(n) , and using (4), we get

∇αz(n+1)=b(n)u(n)a(n)+∑k=0n−1B(n−1,α;k)c(k)u(k)a(k)≤b(n)z(n)+∑k=0n−1B(n−1,α;k)c(k)z(k).

Let

v(n)=z(n)+∑k=0n−1B(n−1,α;k)c(k)z(k).

Then, v (0) = z (0) = 1, z ( n ) ≤ v ( n ), and ∇ ^α v ( n + 1) = ∇ ^α z ( n + 1) + c ( n ) z ( n ) ≤ [ b ( n ) + c ( n )] v ( n ). Now, an application of Theorem 3 yields

v(n)≤∏j=0n−11+B(n−1,α;j)[b(j)+c(j)].

Then, from (9) and (11), we have

∇αz(n+1)≤b(n)∏j=0n−11+B(n−1,α;j)[b(j)+c(j)].

Now, again by application of Theorem 3, we get

z(n)≤z(0)+∑j=0n−1B(n−1,α;j)×b(j)∏k=0j−11+B(j−1,α;k)[b(k)+c(k)].

Using (13) in u(n)a(n)≤z(n) , we get the required inequality (7). □

Theorem 5

If a ( n ) is a nonnegative function defined on N0+ and for n∈N0+ ,

u(n)≤u(0)+∑j=0n−1B(n−1,α;j)a(j)×u(j)+b(j)+∑k=0j−1B(j−1,α;k)c(k)u(k),

then

u(n)≤u(0)+∑j=0n−1B(n−1,α;j)a(j)[b(j)+A(j)],

where

A(n)=u(0)∏j=0n−11+B(n−1,α;j)[a(j)+c(j)]×∑j=0n−1B(n−1,α;j)a(j)b(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)+c(k)].

Proof

Define a function

z(n)=u(0)+∑j=0n−1B(n−1,α;j)a(j)×u(j)+b(j)+∑k=0j−1B(j−1,α;k)c(k)u(k).

Then, z (0) = u (0), u ( n ) ≤ z ( n ), and

∇αz(n+1)≤a(n)b(n)+a(n)z(n)+∑k=0n−1B(n−1,α;k)c(k)z(k).

Let

v(n)=z(n)+∑k=0n−1B(n−1,α;k)c(k)z(k).

Then, v (0) = z (0), z ( n ) ≤ v ( n ), and ∇ ^α v ( n + 1) = ∇ ^α z ( n + 1) + c ( n ) z ( n ) ≤ a ( n ) b ( n ) + [ a ( n ) + c ( n )] v ( n ). Now, an application of Theorem 3 yields

v(n)≤v(0)∏j=0n−11+B(n−1,α;j)[a(j)+c(j)]+∑j=0n−1B(n−1,α;j)a(j)b(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)+c(k)]=A(n)

Then, from (17) and (19), we have

∇αz(n+1)≤a(n)[b(n)+A(n)].

Now, again by application of Theorem 3, we get

z(n)≤z(0)+∑j=0n−1B(n−1,α;j)a(j)[b(j)+A(j)].

Using (21) in u ( n ) ≤ z ( n ), we get the required inequality (15). □

Theorem 6

Let a ( n ) be a nonnegative real valued function defined on N0+ . Suppose the following inequality holds for all n∈N0+ :

u(n)≤u(0)+∑j=0n−1B(n−1,α;j)a(j)∑k=0j−1B(j−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)c(l)u(l).

If [ 1 − B ( n − 1, α ; j ) a ( j )] ≥ 0 and [1 + B ( n − 1, α ; j )[ a ( j ) − b ( j )]] ≥ 0 for all 0 ≤ j ≤ ( n − 1), then, for n∈N0+ ,

u(n)≤u(0)∏j=0n−11−B(n−1,α;j)a(j)×∑j=0n−1B(n−1,α;j)a(j)C(j)×∏k=j+1n−11−B(n−1,α;k)a(k),

where

B(n)=∏j=0n−11+B(n−1,α;j)[a(j)+b(j)+c(j)]

and

C(n)=∏j=0n−11+B(n−1,α;j)[a(j)−b(j)]×∑j=0n−1B(n−1,α;j)b(j)B(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)−b(k)].

Proof

Define a function

z(n)=u(0)+∑j=0n−1B(n−1,α;j)a(j)∑k=0j−1B(j−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)c(l)u(l).

Then, z (0) = u (0), u ( n ) ≤ z ( n ), and

∇αz(n+1)≤a(n)∑k=0n−1B(n−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)c(l)z(l).

Adding a ( n ) z ( n ) to both sides of the above inequality, we have

∇αz(n+1)+a(n)z(n)≤a(n)z(n)+∑k=0n−1B(n−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)c(l)z(l).

Let

v(n)=z(n)+∑k=0n−1B(n−1,α;k)b(k)∑l=0k−1B(k−1,α;l)c(l)z(l).

Then, v (0) = z (0), z ( n ) ≤ v ( n ), and

∇αv(n+1)=∇αz(n+1)+b(n)∑l=0n−1B(n−1,α;l)c(l)z(l).

Using the facts that ∇ ^α z ( n + 1) ≤ a ( n ) v ( n ) and z ( n ) ≤ v ( n ), we get

∇αv(n+1)≤a(n)v(n)+b(n)∑l=0n−1B(n−1,α;l)c(l)z(l).

Adding b ( n ) v ( n ) to both sides of the above inequality, we have

∇αv(n+1)+b(n)v(n)≤a(n)v(n)+b(n)v(n)+∑l=0n−1B(n−1,α;l)c(l)z(l).

Let

w(n)=v(n)+∑l=0n−1B(n−1,α;l)c(l)z(l).

Then, w (0) = v (0), v ( n ) ≤ w ( n ), and

∇αw(n+1)≤∇αv(n+1)+c(n)w(n).

Now, from (30) and (31), we have

∇αv(n+1)≤∇αv(n+1)+b(n)v(n)≤[a(n)+b(n)]w(n).

Using (34) in (33), we get

∇αw(n+1)≤[a(n)+b(n)+c(n)]w(n).

Now, an application of Theorem 3 yields

w(n)≤w(0)∏j=0n−11+B(n−1,α;j)[a(j)+b(j)+c(j)]=w(0)B(n).

Then, from (31) and (35), we have

∇αv(n+1)≤[a(n)−b(n)]v(n)+w(0)b(n)B(n).

Now, again by application of Theorem 3, we get

v(n)≤v(0)∏j=0n−11+B(n−1,α;j)[a(j)−b(j)]+w(0)∑j=0n−1B(n−1,α;j)b(j)B(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)−b(k)]=v(0)C(n).

Then, from (27) and (37), we get

∇αz(n+1)≤−a(n)z(n)+v(0)a(n)C(n).

Now, again by application of Theorem 3, we get

z(n)≤z(0)∏j=0n−11−B(n−1,α;j)a(j)+v(0)∑j=0n−1B(n−1,α;j)a(j)C(j)×∏k=j+1n−11−B(n−1,α;k)a(k).

Using (38) in u ( n ) ≤ z ( n ), we get the required inequality (23). □

Theorem 7

Let a ( n ) be a nonnegative real valued function defined on N0+ . Suppose the following inequality holds for all n∈N0+ :

u(n)≤u(0)+∑j=0n−1B(n−1,α;j)a(j)∑k=0j−1B(j−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)[c(l)u(l)+p(l)].

If [ 1 − B ( n − 1, α ; j ) a ( j ) ] ≥ 0 and [1 + B ( n − 1, α ; j )[ a ( j ) − b ( j )] ] ≥ 0 for all 0 ≤ j ≤ ( n − 1), then , for n∈N0+ ,

u(n)≤u(0)∏j=0n−11−B(n−1,α;j)a(j)+∑j=0n−1B(n−1,α;j)a(j)E(j)×∏k=j+1n−11−B(n−1,α;k)a(k),

where

D(n)=u(0)∏j=0n−11+B(n−1,α;j)[a(j)+b(j)+c(j)]+∑j=0n−1p(j)∏k=j+1n−11+B(n−1,α;k)[a(k)+b(k)+c(k)]

and

E(n)=u(0)∏j=0n−11+B(n−1,α;j)[a(j)−b(j)]+∑j=0n−1B(n−1,α;j)b(j)D(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)−b(k)].

Theorem 8

Let a ( n ) be a nonnegative real valued function defined on N0+ . Suppose the following inequality holds for all n∈N0+ :

u(n)≤p(n)+q(n)∑j=0n−1B(n−1,α;j)a(j)×∑k=0j−1B(j−1,α;k)b(k)∑l=0k−1B(k−1,α;l)c(l)u(l).

If [ 1 − B ( n − 1, α ; j ) a ( j ) ] ≥ 0 and [1 + B ( n − 1, α ; j )[ a ( j ) − b ( j )] ] ≥ 0 for all 0 ≤ j ≤ ( n − 1), then , for n∈N0+ ,

u(n)≤p(n)+q(n)∑j=0n−1B(n−1,α;j)a(j)G(j)×∏k=j+1n−11−B(n−1,α;k)a(k),

where

F(n)=∑j=0n−1B(n−1,α;j)[c(j)p(j)]×∏k=j+1n−11+B(n−1,α;k)[a(k)+b(k)+c(k)q(k)]

and

G(n)=∑j=0n−1B(n−1,α;j)b(j)F(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)−b(k)].

Proof

Define a function z ( n )by

z(n)=∑j=0n−1B(n−1,α;j)a(j)×∑k=0j−1B(j−1,α;k)b(k)∑l=0k−1B(k−1,α;l)c(l)u(l).

Then, z (0) = 0, u ( n ) ≤ p ( n ) + q ( n ) z ( n ), and using the same argument as in the proof of Theorem 6, we obtain

∇αz(n+1)≤a(n)∑k=0n−1B(n−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Adding a ( n ) z ( n ) to both sides of the above inequality, we have

∇αz(n+1)+a(n)z(n)≤a(n)z(n)+∑k=0n−1B(n−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Let

v(n)=z(n)+∑k=0n−1B(n−1,α;k)b(k)×∑l=0k−1B(k−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Then, v (0) = z (0), z ( n ) ≤ v ( n ), and

∇αv(n+1)=∇αz(n+1)+b(n)∑l=0n−1B(n−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Using the facts that ∇ ^α z ( n + 1) ≤ a ( n ) v ( n ) and z ( n ) ≤ v ( n ), we get

∇αv(n+1)≤a(n)v(n)+b(n)∑l=0n−1B(n−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Adding b ( n ) v ( n ) to both sides of the above inequality, we have

∇αv(n+1)+b(n)v(n)≤a(n)v(n)+b(n)v(n)+∑l=0n−1B(n−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Let

w(n)=v(n)+∑l=0n−1B(n−1,α;l)[c(l)p(l)+c(l)q(l)z(l)].

Then, w (0) = v (0), v ( n ) ≤ w ( n ), and

∇αw(n+1)≤∇αv(n+1)+[c(n)p(n)+c(n)q(n)w(n)].

Now, from (53) and (54), we have

∇αv(n+1)≤∇αv(n+1)+b(n)v(n)≤[a(n)+b(n)]w(n).

Using (56) in (55), we get

∇αw(n+1)≤[a(n)+b(n)+c(n)q(n)]w(n)+c(n)p(n).

Now, an application of Theorem 3 yields

w(n)≤∑j=0n−1B(n−1,α;j)[c(j)p(j)]×∏k=j+1n−11+B(n−1,α;k)[a(k)+b(k)+c(k)q(k)]=F(n).

Then, from (53) and (58), we have

∇αv(n+1)≤[a(n)−b(n)]v(n)+b(n)F(n).

Now, again by application of Theorem 3, we get

v(n)≤∑j=0n−1B(n−1,α;j)b(j)F(j)×∏k=j+1n−11+B(n−1,α;k)[a(k)−b(k)]=G(n).

Then, from (49) and (59), we get

∇αz(n+1)≤−a(n)z(n)+a(n)G(n).

Now, again by application of Theorem 3, we get

z(n)≤∑j=0n−1B(n−1,α;j)a(j)G(j)×∏k=j+1n−11−B(n−1,α;k)a(k).

Using (60) in u ( n )≤ p ( n ) + q ( n ) z ( n ), we get the required inequality (44). □

Acknowledgements

The authors are grateful to the referees for their suggestions and comments which considerably helped improve the content of this paper.

Competing interests

Authors’ contributions

Both authors gave excellent contributions to the final manuscript. Both authors read and approved the final manuscript.

Some new fractional difference inequalities of Gronwall-Bellman type

Abstract

Abstract

MSC

Introduction

Theorem 1

Definition 1

Definition 2

Remark 1

Definition 3

Remark 2

Theorem 2

Proof

Definition 4

Theorem 3

Main Results

Theorem 4

Proof

Theorem 5

Proof

Theorem 6

Proof

Theorem 7

Theorem 8

Proof

Conclusions

Acknowledgements

Competing interests

Authors’ contributions

References