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Existence and uniqueness of solutions of Hahn difference equations
Advances in Difference Equations volume 2013, Article number: 316 (2013)
Abstract
Hahn introduced the difference operator {D}_{q,\omega}f(t)=(f(qt+\omega )f(t))/(t(q1)+\omega ) in 1949, where 0<q<1 and \omega >0 are fixed real numbers. This operator extends the classical difference operator {\mathrm{\Delta}}_{\omega}f(t)=(f(t+\omega )f(t))/\omega as the Jackson qdifference operator {D}_{q}f(t)=(f(qt)f(t))/(t(q1)).
In this paper, we present new results of the calculus based on the Hahn difference operator. Also, we establish an existence and uniqueness result of solutions of Hahn difference equations by using the method of successive approximations.
1 Introduction and preliminaries
Hahn introduced his difference operator which is defined by
at t\ne \omega /(1q) and the usual derivative at t=\omega /(1q), where 0<q<1 and \omega >0 are fixed real numbers [1, 2]. This operator unifies and generalizes two wellknown difference operators. The first is Jackson qdifference operator defined by
where q is fixed. Here f is supposed to be defined on a qgeometric set A\subset \mathbb{R}, for which qt\in A whenever t\in A, see [3–10]. The second operator is the forward difference operator
where \omega >0 is fixed, see [11–14].
In [15], Annaby et al. gave a rigorous analysis of the calculus associated with {D}_{q,\omega}. They stated and proved some basic properties of such a calculus. For instance, they defined a right inverse of {D}_{q,\omega} in terms of both the Jackson qintegral; see [9], which contains the right inverse of {D}_{q} and Nörlund sum; cf. [13], which involves the right inverse of {\mathrm{\Delta}}_{\omega}. Then they proved a fundamental theorem of Hahn’s calculus. An essential function which plays an important role in this calculus is h(t)=qt+\omega. This function is normally taken to be defined on an interval I, which contains the number \theta =\omega /(1q). One can see that the k th order iteration of h(t) is given by
The sequence {h}^{k}(t) is uniformly convergent to θ on I. Here {[k]}_{q} is defined by
Throughout this paper, I is any interval of ℝ containing θ and X is a Banach space.
Definition 1.1 Assume that f:I\to X is a function, and let a,b\in I. The q,\omegaintegral of f from a to b is defined by
where
provided that the series converges at x=a and x=b.
Definition 1.2 [15]
For certain z\in \mathbb{C}, the q,\omegaexponential functions {e}_{z}(t) and {E}_{z}(t) are defined by
and
To guarantee the convergence of the infinite product in (1.1) with t\in \mathbb{C}, we assume additionally that
see [16, 17]. For a fixed z\in \mathbb{C}, (1.2) converges for all t\in \mathbb{C}, defining an entire function of order zero. For the proofs of the equalities in (1.1) and (1.2), see [[18], Section 1.3] and [17]. Here the qshifted factorial {(b;q)}_{n} for a complex number b and n\in {\mathbb{N}}_{0} is defined to be
The following results were mentioned in [15], and we need them in this paper.
Lemma 1.3 Let f,g:I\to \mathbb{R} be q, ω differentiable at t\in I. Then the following statements are true:

(i)
{D}_{q,\omega}(f+g)(t)={D}_{q,\omega}f(t)+{D}_{q,\omega}g(t),

(ii)
{D}_{q,\omega}(fg)(t)={D}_{q,\omega}(f(t))g(t)+f(h(t)){D}_{q,\omega}g(t),

(iii)
for any constant c\in X, {D}_{q,\omega}(cf)(t)=c{D}_{q,\omega}(f(t)),

(iv)
{D}_{q,\omega}(f/g)(t)=({D}_{q,\omega}(f(t))g(t)f(t){D}_{q,\omega}g(t))/(g(t)g(h(t))) for g(t)g(h(t))\ne 0.
We notice that (ii) and (iv) are true even if f:I\to X. Also, (i) is true if f,g:I\to X.
Lemma 1.4 For a fixed z\in \mathbb{C}, the parametric q,\omegaexponential functions {e}_{z}(t) and {E}_{z}(t) are the unique solutions of the first order initial value problems
and
respectively.
Lemma 1.5 Let s\in I, s>\theta, f:I\to \mathbb{R} and g:I\to \mathbb{R} be q,\omegaintegrable on I. If f(t)\le g(t) for all t\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}, then for a,b\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}, a<b,
Theorem 1.6 Let f:I\to \mathbb{R} be continuous at θ. Define
Then F is continuous at θ. Furthermore, {D}_{q,\omega}F(x) exists for every x\in I and
Conversely,
Lemma 1.5 and Theorem 1.6 are also true if f is a function with values in X with replacing the norm \parallel \cdot \parallel instead of the modulus \cdot .
The aim of this paper is to establish an existence and uniqueness result of solutions of the first order abstract Hahn difference equations by using the method of successive approximations. This method is a very powerful tool that dates back to the works of Liouville [19] and Picard [20]. It is based on defining a sequence of functions {\{{\varphi}_{k}\}}_{k=0}^{\mathrm{\infty}} and showing that {\varphi}_{k} will successively approximate the solution ϕ in the sense that the ‘error’ between the two monotonically decreases as k increases. Also, it differs from fixed point methods and topological ideas that were used by some researchers to develop the existence and uniqueness of solutions.
We organize this paper as follows.
In Section 2, we prove Gronwall’s and Bernoulli’s inequalities with respect to the Hahn difference operator. In Section 3, we establish mean value theorems in the calculus based on this operator. In Sections 4 and 5, we apply the method of successive approximations to obtain the local and global existence and uniqueness theorem of first order Hahn difference equations in Banach spaces. Hence, we deduce this theorem for the n th order Hahn difference equations.
2 Gronwall’s and Bernoulli’s inequalities
In this section, I is a subinterval of [\theta \frac{1}{r(1q)},\theta +\frac{1}{r(1q)}], where r\in \mathbb{R}, and f is a real valued function defined on I.
Theorem 2.1 Assume that r\in \mathbb{R}, and y, f are continuous at θ. Then
for all t\in I implies that
Proof We have
Integrating both sides from θ to t in the inequality above, we obtain
This implies that
□
Theorem 2.2 (Gronwall’s inequality)
Let r\in {\mathbb{R}}^{\ge 0}, and y, f be continuous at θ. Then
for all t\in I implies that
Proof Set z(t)={\int}_{\theta}^{t}ry(\tau )\phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}\tau. Then z(\theta )=0, {D}_{q,\omega}z(t)=ry(t)\le r(f(t)+z(t)) and {D}_{q,\omega}z(t)\le rf(t)+rz(t). Consequently,
Hence,
□
Theorem 2.3 (Bernoulli’s inequality)
Let r\in \mathbb{R}. Then, {e}_{r}(t)\ge 1+r(t\theta ) for all t\in [\theta ,\theta +\frac{1}{r(1q)}].
Proof Let y(t)=r(t\theta ). Then,
Since y(\theta )=0, we have by Theorem 2.1
Therefore, {e}_{r}(t)\ge 1+r(t\theta ). □
3 Mean value theorems
Theorem 3.1 Let f:I\to X, g:I\to \mathbb{R} be q, ω differentiable on I and s\in I.
Assume that \parallel {D}_{q,\omega}f(t)\parallel \le {D}_{q,\omega}g(t) for all t\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}. Then \parallel f(b)f(a)\parallel \le g(b)g(a) for all a\le b, a,b\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}.
Proof The inequality \parallel {\int}_{a}^{b}{D}_{q,\omega}f(t)\phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}t\parallel \le {\int}_{a}^{b}{D}_{q,\omega}g(t)\phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}t implies that \parallel f(b)f(a)\parallel \le g(b)g(a) for all a\le b,a,b\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}. □
Corollary 3.2 Suppose that f,g:I\to X is q, ω differentiable. The following statements are true:

(i)
For every s\in I, the inequality \parallel f(b)f(a)\parallel \le {sup}_{t\in I}\parallel {D}_{q,\omega}f(t)\parallel (ba) holds for all a\le b, a,b\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}.

(ii)
If {D}_{q,\omega}f(t)=0 for all t\in I, then f is a constant function.

(iii)
If {D}_{q,\omega}f(t)={D}_{q,\omega}g(t) for all t\in I. Then g(t)=f(t)+c for all t\in I, where c is a constant in X.
Proof (i) Let g(t)={sup}_{t\in I}\parallel {D}_{q,\omega}f(t)\parallel (t\theta ). Then {D}_{q,\omega}g(t)={sup}_{t\in I}\parallel {D}_{q,\omega}f(t)\parallel \ge \parallel {D}_{q,\omega}f(t)\parallel for all t\in {\{s{q}^{k}+\omega {[k]}_{q}\}}_{k=0}^{\mathrm{\infty}}.
By Theorem 3.1,

(ii)
Statement (i) implies that f(s{q}^{k}+\omega {[k]}_{q})=f(s) for every s\in I and k=0,1,\dots . Letting k\to \mathrm{\infty}, we obtain f(s)=f(\theta ) for all s\in I.

(iii)
Can be deduced immediately from (ii). □
As a direct consequence of Theorem 1.6, we get the following result.
Theorem 3.3 Let f:I\to X be continuous at θ. Then
4 Successive approximations and local results
Throughout the remainder of the paper,
and R is the rectangle
where a and b are fixed positive numbers.
Theorem 4.1 Assume that f:R\to X satisfies the following conditions:

(i)
f(t,x) is continuous at t=\theta for every x\in S({x}_{0},b).

(ii)
There is a positive constant A such that the following Lipschitz condition \parallel f(t,x)f(t,y)\parallel \le A\parallel xy\parallel for all x,y\in X is satisfied.
Then there is h>0 such that the following Cauchy problem
has a unique solution x(t) on [\theta ,\theta +h].
Proof Since f is continuous at t=\theta, there exists \gamma >0 such that \parallel f(t,x)f(\theta ,x)\parallel <1 for all t with t\theta <\gamma. Hence,
for all t\in I, x\in X such that t\theta <\gamma and \parallel x{x}_{0}\parallel <b. Define the following constants K, h to be
We establish the existence of the solution \varphi (t) of (4.1) by the method of successive approximations.
We consider the sequence defined as follows:
The proof of the existence consists of four steps.

(i)
{\varphi}_{k}(t)\in S({x}_{0},b), k\ge 0, t\in I. Indeed, we have \parallel {\varphi}_{0}(t){x}_{0}\parallel =0\le b. Assume that the inequality \parallel {\varphi}_{k}(t){x}_{0}\parallel \le b holds. This implies that
\begin{array}{rl}\parallel {\varphi}_{k+1}(t){x}_{0}\parallel & \le {\int}_{\theta}^{t}\parallel f(s,{\varphi}_{k}(s))\parallel \phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}s\\ \le K{\int}_{\theta}^{t}{d}_{q,\omega}s\\ \le b,\end{array}
i.e., {\varphi}_{k}(t)\in S({x}_{0},b), k\in {\mathbb{Z}}^{\ge 0}. Also, each {\varphi}_{k}(t) is continuous at t=\theta.

(ii)
Now, we show by induction that
\parallel {\varphi}_{m+1}(t){\varphi}_{m}(t)\parallel \le K{A}^{m}{h}^{m+1},\phantom{\rule{1em}{0ex}}t\in [\theta ,\theta +h].(4.3)
First, \parallel {\varphi}_{1}(t){\varphi}_{0}(t)\parallel \le K{\int}_{\theta}^{t}{d}_{q,\omega}s\le Kh.
Assume that (4.3) is true. Hence,
Thus, inequality (4.3) is true for every m\in {\mathbb{Z}}^{\ge 0}.

(iii)
We show that {\varphi}_{m}(t) converges uniformly to a function \varphi (t) on [\theta ,\theta +h].
We write {\varphi}_{m}(t) as the following sum
So,
provided that (4.4) converges.
Thus, the convergence of the sequence of functions {\varphi}_{m}(t) is equivalent to the convergence of the righthand side of (4.4).
To prove the latter, we use estimate (4.3), and then apply the Weierstrass Mtest. The convergence of the series {\sum}_{m=0}^{\mathrm{\infty}}{A}^{m}{h}^{m+1} implies the uniform convergence of the series {\sum}_{i=1}^{\mathrm{\infty}}\parallel {\varphi}_{i}(t){\varphi}_{i1}(t)\parallel on [\theta ,\theta +h]. This implies that the righthand side of (4.4) is uniform convergent to some function \varphi (t). That is, there exists a function \varphi (t) such that {lim}_{m\to \mathrm{\infty}}{\varphi}_{m}(t)=\varphi (t). Obviously, \varphi (t)\in S({x}_{0},b) since \parallel \varphi (t){x}_{0}\parallel ={lim}_{m\to \mathrm{\infty}}\parallel {\varphi}_{m}(t){x}_{0}\parallel \le b.

(iv)
The last part of the proof of the existence is to show that \varphi (t) is a solution of (4.1).
From the Lipschitz condition,
for all t\in [\theta ,\theta +h] and m\in \mathbb{N}.
Thus, for every \u03f5>0, \mathrm{\exists}{n}_{0}\in \mathbb{N} such that
So,
This implies that
uniformly on [\theta ,\theta +h]. Taking k\to \mathrm{\infty} in (4.2), it follows that
Consequently, we obtain {D}_{q,\omega}\varphi (t)=f(t,\varphi (t)). Clearly, \varphi (\theta )={x}_{0}.
Uniqueness
To prove the uniqueness, let us assume that \psi (t) is another solution which is valid in t\in [\theta ,\theta +h]. We have
and
From which we deduce
Taking \sigma (t)=\parallel \varphi (t)\psi (t)\parallel, we get
i.e.,
Hence,
By taking the limit as k\to \mathrm{\infty}, we get
Since \sigma (\theta )=0, then \sigma (t)=0, which completes the proof. □
Corollary 4.2 Let I be an interval containing \theta ,{f}_{i}(t,{x}_{1},{x}_{2},\dots ,{x}_{n}):I\times {\prod}_{i=1}^{n}{S}_{i}({y}_{i},{b}_{i})\to X such that the following conditions are satisfied:

(i)
For {x}_{i}\in {S}_{i}({y}_{i},{b}_{i}), 1\le i\le n, {f}_{i}(t,{x}_{1},{x}_{2},\dots ,{x}_{n}) are continuous at t=\theta.

(ii)
There is a positive constant A such that, for t\in I, {x}_{i},{x}_{i}^{\prime}\in {S}_{i}({y}_{i},{b}_{i}), 1\le i\le n, the following Lipschitz condition is satisfied:
\parallel {f}_{i}(t,{x}_{1},{x}_{2},\dots ,{x}_{n}){f}_{i}(t,{x}_{1}^{\prime},{x}_{2}^{\prime},\dots ,{x}_{n}^{\prime})\parallel \le A\sum _{i=1}^{n}\parallel {x}_{i}{x}_{i}^{\prime}\parallel .
Then there exists a unique solution of the initial value problem
Proof Let {x}_{0}={({y}_{1},{y}_{2},\dots ,{y}_{n})}^{T} and b={({b}_{1},{b}_{2},\dots ,{b}_{n})}^{T}.
Define f:I\times {\prod}_{i=1}^{n}{S}_{i}({y}_{i},{b}_{i})\to {X}^{n} by
System (4.6) yields
First f is continuous at t=\theta, since each {f}_{i} is continuous at t=\theta.
Second f satisfies a Lipschitz condition because for x,{x}^{\prime}\in {\prod}_{i=1}^{n}{S}_{i}({y}_{i},{b}_{i})
Applying Theorem 4.1, then there exist h>0 such that the initial value problem (4.7) has a unique solution on [\theta ,\theta +h]. It is easy to show that (4.7) is equivalent to the initial value problem (4.6). □
The following corollary shows that Corollary 4.2 can be applied to indicate the required conditions for the existence and uniqueness of solutions of the Cauchy problem
The proof of this corollary depends on converting the n th order Hahn difference equation (4.8) to a first order system. Clearly, the Cauchy problem (4.8) is equivalent to the first order system
in the sense that {\{{\varphi}_{i}(t)\}}_{i=1}^{n} is a solution of (4.9) if and only if {\varphi}_{1}(t) is a solution of (4.8). Here,
This leads us to state the following theorem.
Theorem 4.3 Let f(t,{x}_{1},\dots ,{x}_{n}) be a function defined on I\times {\prod}_{i=1}^{n}{S}_{i}({y}_{i},{b}_{i}) such that the following conditions are satisfied:

(i)
For any values of {x}_{r}\in {S}_{r}({y}_{r},{b}_{r}), f is continuous at t=\theta.

(ii)
f satisfies a Lipschitz condition
\parallel f(t,{x}_{1},\dots ,{x}_{n})f(t,{x}_{1}^{\prime},\dots ,{x}_{n}^{\prime})\parallel \le A\sum _{i=1}^{n}\parallel {x}_{i}{x}_{i}^{\prime}\parallel ,
where A>0, {x}_{i},{x}_{i}^{\prime}\in {S}_{i}({y}_{i},{b}_{i}), i=1,\dots ,n and t\in I.
Then the Cauchy problem (4.8) has a unique solution which is valid on [\theta ,\theta +h].
The following corollary gives us the required conditions for the existence of solutions of the Cauchy problem
Corollary 4.4 Assume that the functions {a}_{j}(t):I\to \mathbb{C} (0\le j\le n) and b(t):I\to X satisfy the following conditions:

(i)
{a}_{j}(t) (j=1,\dots ,n) and b(t) are continuous at θ with {a}_{0}(t)\ne 0 \mathrm{\forall}t\in I.

(ii)
{a}_{j}(t)/{a}_{0}(t) is bounded on I, j\in \{1,\dots ,n\}.
Then, for any elements {y}_{r}\in X, equation (4.10) has a unique solution on a subinterval J\subset I containing θ.
Proof Dividing by {a}_{0}(t), we get the equation
where {A}_{j}(t)={a}_{j}(t)/{a}_{0}(t), and B(t)=b(t)/{a}_{0}(t). Since {A}_{j}(t) and B(t) are continuous at t=\theta, then the function f(t,{x}_{1},{x}_{2},\dots ,{x}_{n}) defined by
is continuous at t=\theta. Furthermore, {A}_{j}(t) is bounded on I. Consequently, there is A>0 such that {A}_{j}(t)\le A, t\in I. We can easily see that f satisfies a Lipschitz condition with Lipschitz constant A. Thus, the function f(t,{x}_{1},\dots ,{x}_{n}) satisfies the conditions of Theorem 4.3. Hence, there exists a unique solution of (4.11) on a subinterval J of I containing θ. □
5 Nonlocal results
Theorem 4.1 is called a local existence theorem, because it guarantees the existence of a solution x(t) defined for t\in R, which is close to the initial point θ. However, in many situations, a solution will actually exist on the entire interval I=[\theta ,\theta +a]. We now show that if f satisfies a Lipschitz condition on a strip
rather than on the rectangle R, which is given in Section 4, then solutions will exist on the entire interval I=[\theta ,\theta +a].
Theorem 5.1 Let f be continuous on the strip S, and suppose that there exists a constant A>0 such that \parallel f(t,x)f(t,y)\parallel \le A\parallel xy\parallel for all (t,x),(t,y)\in S, where A<\frac{1}{a(1q)}. Then the successive approximations that are given in (4.2) exist on the entire interval [\theta ,\theta +a] and converge there uniformly to the unique solution of (4.1).
Proof There is a constant M such that \parallel f(t,{x}_{0})\parallel \le M for all t\in I. We will show that the inequality
holds for every k\in N. Inequality (5.1) is true at k=1. Indeed, we have
and
Assume that inequality (5.1) is true. Now, we have
We see that
which converges to (M/A)({e}_{A}(t)1), t\in I. This convergence enables us to apply the Weierstrass Mtest to conclude that a series
converges uniformly on I. Also, we can write {\varphi}_{k}(t) as the following sum
Consequently, the absolute uniform convergence of the series in (5.3) on I implies that {\varphi}_{k}(t) converges uniformly to some function \varphi (t) on I.
Our objective now is to show that \varphi (t) is a solution of (4.1) for all t\in I.
From the Lipschitz condition
for all t\in [\theta ,\theta +a]. This implies that {lim}_{m\to \mathrm{\infty}}f(t,{\varphi}_{m}(t))=f(t,\varphi (t)) uniformly on I and {lim}_{m\to \mathrm{\infty}}{\int}_{\theta}^{t}f(s,{\varphi}_{m}(s))\phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}s={\int}_{\theta}^{t}f(s,\varphi (s))\phantom{\rule{0.2em}{0ex}}{d}_{q,\omega}s uniformly on I.
Taking m\to \mathrm{\infty} in (4.2), it follows that
Consequently, {D}_{q,\omega}\varphi (t)=f(t,\varphi (t)). Clearly, \varphi (\theta )={x}_{0}.
Uniqueness
Let x(t) and y(t) be two solutions of (4.1) for all t\in I. We show that x(t)\equiv y(t) on I.
For t\in I, consider
Thus, from Gronwall’s inequality, we have z(t)=0. Consequently, x(t)=y(t) for all t\in I. □
Corollary 5.2 Let f be continuous on the halfplane
Assume that f satisfies a Lipschitz condition
on each strip
where {L}_{\theta ,a} is a constant that may depend on θ and a. Then, the initial value problem (4.1) has a unique solution that exists on the whole halfline [\theta ,\mathrm{\infty}).
Proof The proof involves showing that the conditions of Theorem 5.1 hold on every strip of {S}_{\theta ,a} and is omitted for brevity. □
6 Conclusion and future directions
This article was devoted to establish the method of successive approximations in proving the existence and uniqueness of solutions of the initial value problems associated with Hahn difference operators. Also, some new results of the calculus based on this operator like a mean value theorem were obtained. In one direction, one should ask about the qω Taylor’s theorem. In this respect, we point out that qTaylor’s theorem has been established in [5]. Another direction, is to study in more details the theory of Hahn difference equations, based on Hahn difference operator, and the stability of its solutions.
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Hamza, A.E., Ahmed, S.M. Existence and uniqueness of solutions of Hahn difference equations. Adv Differ Equ 2013, 316 (2013). https://doi.org/10.1186/168718472013316
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DOI: https://doi.org/10.1186/168718472013316