Some new generalized Volterra-Fredholm type discrete fractional sum inequalities and their applications

([16])

Let a be any real number, α be any positive real number, and \(\sigma(s)=s+1\). The α-th fractional sum (α-sum) of f is defined by Here f is defined for \(s=a\ (\operatorname{mod} 1)\), and \(\Delta_{a}^{-\alpha}f\) is defined for \(t=a+\alpha\ (\operatorname{mod} 1)\); in particular, \(\Delta_{a}^{-\alpha}\) maps functions defined on \(\mathbf{N}_{a}\) to functions defined on \(\mathbf{N}_{a+\alpha }\). We recall that the falling factorial is defined as \(t^{(\alpha )}=\frac{\Gamma(t+1)}{\Gamma(t-\alpha+1)}\).

([16])

The μ-th fractional difference is defined as where \(\mu>0\) and \(m-1<\mu<m\), where m denotes a positive integer, and \(-\nu=\mu -m\).

(Pachpatte [1], p.103)

Let \(u(t)\) and \(b(t)\) be nonnegative functions defined for \(t\in\mathbf{N}_{0}\), and c be a nonnegative constant. Let \(g(u)\) be a nondecreasing continuous function defined on \(\mathbf{R}_{+}\) with \(g(u)>0\) for \(u>0\). If for \(t\in\mathbf{N}_{0}\), then, for \(0\leq t\leq t_{1}\), \(t, t_{1}\in\mathbf{N}_{0}\), where \(G(r)=\int_{r_{0}}^{r}\frac{1}{g(s)}\, \mathrm{d}s\), \(r>0\), \(r_{0}>0\) is arbitrary, \(G^{-1}\) is the inverse of G, and \(t_{1}\in\mathbf {N}_{0}\) is chosen so that \(G(c)+\sum_{s=0}^{t-1}b(s)\in \operatorname{Dom} (G^{-1})\) for all \(t\in\mathbf{N}_{0}\) such that \(0\leq t \leq t_{1}\).

Assume that \(0<\alpha\leq1\) is a constant, \(u: \mathbf{N}_{\alpha-1}\rightarrow\mathbf{R}_{+}\), \(f, g:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) are functions, \(k\geq0\) is a constant, and \(p > q > 0\) are constants. Suppose that u satisfies If then where

Let \(k>0\). From (3.1) and (3.5) we have where Define Then \(z(n)\geq0\) is nondecreasing, and By the definitions of \(F(s,n)\) and \(t^{(\alpha)}\) we can easily get that \(F(s,n)\) is decreasing in n for each \(s\in\mathbf{N}_{0}\). So from a straightforward computation, for \(n\in I_{\alpha}\), we obtain that Using the monotonicity of z, we deduce So from (3.10) and (3.11) we have that is, On the other hand, by the mean value theorem we obtain So from (3.12) and (3.13) we obtain Setting \(n=s\) in inequality (3.14) and summing with respect to s from α to \(n-1\), we get that is, Then from inequality (3.15) we conclude that that is, By (3.8), (3.9), and (3.16) we get Therefore, using the inequality \((a+b)^{\mu}\leq2^{\mu-1}(a^{\mu}+b^{\mu})\), \(\mu\geq1\), we have Hence, in view of (3.2), we obtain where \(A(T)\) is defined as in (3.4). From (3.16) and (3.18) we get Using (3.9) and (3.19), we obtain If \(k=0\), then we carry out the above procedure with \(\varepsilon>0\) instead of k and subsequently let \(\varepsilon\rightarrow0\). This completes the proof. □

Assume that \(0<\alpha\leq1\) is a constant, \(u: \mathbf{N}_{\alpha-1}\rightarrow\mathbf{R}_{+}\), \(g, h:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) are functions, \(k\geq0\) is a constant, \(\varphi\in C^{1}(\mathbf{R}_{+},\mathbf{R}_{+})\) is an increasing function with \(\varphi(\infty)=\infty\) on \(\mathbf{R}_{+}\), and \(\psi_{i}:\mathbf{R}_{+}\rightarrow\mathbf{R}_{+}\) is a nondecreasing continuous function with \(\psi_{i}(u)>0\) for \(u>0\), \(i=1,2\). Suppose that there is a nondecreasing continuous function \(\psi:\mathbf {R}_{+}\rightarrow\mathbf{R}_{+}\) such that both \(\psi_{1}\) and \(\psi_{2}\) are less than or equal to ψ, \(\Psi(r)=\int_{r_{0}}^{r}\frac{1}{\psi(\varphi^{-1}(s))}\,\mathrm{d}s\), \(r\geq r_{0}>0\), with \(\lim_{r\rightarrow\infty}\Psi(r)=\infty\), and \(\Omega(t)=\Psi(2t-k)-\Psi(t)\) is increasing for \(t\geq k\). If u satisfies then where \(f:\mathbf{N}_{0}\rightarrow\mathbf{R}_{+}\) is a function such that both g and h are less than or equal to f, and \(\varphi^{-1}\), \(\Psi^{-1}\), \(\Omega^{-1}\) are the inverse functions of φ, Ψ, Ω, respectively.

Let \(k>0\). From the assumptions on g, h, \(\psi_{i}\) (\(i=1,2\)) and (3.21) we have where Define Then \(z(n)\geq0\) is nondecreasing, and By the definitions of \(F(s,n)\) and \(t^{(\alpha)}\), we can easily get that \(F(s,n)\) is decreasing in n for each \(s\in\mathbf{N}_{0}\). So from (3.24) and a straightforward computation, for \(n\in I_{\alpha}\), we obtain that Using the monotonicity of \(\varphi^{-1}\) and z, we deduce So from (3.27) and (3.28) we have that is, On the other hand, by the mean value theorem and the monotonicity of \(\varphi^{-1}\) and ψ we obtain So from (3.29) and (3.30) we obtain Setting \(n=s\) in inequality (3.31) and summing with respect to s from α to \(n-1\), we get that is, Then from inequality (3.32) we conclude that that is, By (3.24) we get that and then from (3.33) and (3.34) we have that is, Since \(\Omega(t)=\Psi(2t-k)-\Psi(t)\) is increasing for \(t\geq k\), and Ω has an inverse function \(\Omega^{-1}\), from (3.35) we get Substituting (3.36) into (3.33), we have Combining (3.37) with (3.26), we obtain the desired inequality (3.22). If \(k=0\), then we carry out the above procedure with \(\varepsilon>0\) instead of k and subsequently let \(\varepsilon\rightarrow0\). This completes the proof. □

Let α, u, and k be defined as in Theorem  3.2, and \(f:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) be a function. If u satisfies and then

From the definitions of Ψ and Ω, letting \(\psi(u)=u\), we obtain and hence \(\Psi^{-1}(r)=r_{0} \exp(r)\) and \(\Omega^{-1}(t)=\frac{k}{2- \exp(t)}\). From inequality (3.22) we obtain inequality (3.38). □

Assume that \(0<\alpha\leq1\) is a constant, \(u: \mathbf{N}_{\alpha-1}\rightarrow\mathbf{R}_{+}\), \(g, h:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) are functions, \(k\geq0\) is a constant, \(\varphi\in C^{1}(\mathbf{R}_{+},\mathbf{R}_{+})\) with \(\varphi(0)=0\), \(\varphi(\infty)=\infty\), and \(\varphi'(u)>0\) for \(u>0\), the derivative \(\varphi'\) is increasing on \(\mathbf{R}_{+}\), and \(\psi_{i}:\mathbf{R}_{+}\rightarrow\mathbf{R}_{+}\) are nondecreasing continuous functions with \(\psi_{i}(u)>0\) for \(u>0\), \(i=1,2\). Suppose that there is a nondecreasing continuous function \(\psi:\mathbf {R}_{+}\rightarrow\mathbf{R}_{+}\) such that both \(\psi_{1}\) and \(\psi_{2}\) are less than or equal to ψ, \(G(r)=\int_{r_{0}}^{r}\frac{1}{\psi(s)}\,\mathrm{d}s\), \(r\geq r_{0}>0\), with \(\lim_{r\rightarrow\infty}G(r)=\infty\), and \(\Omega(t)=G(\varphi^{-1}(2t-k))-G(\varphi^{-1}(t))\) is increasing for \(t\geq k\). If u satisfies then where \(f:\mathbf{N}_{0}\rightarrow\mathbf{R}_{+}\) is a function such that both g and h are less than or equal to f, and \(\varphi^{-1}\), \(G^{-1}\), and \(\Omega^{-1}\) are inverse functions of φ, G, and Ω, respectively.

Let \(k>0\). From the assumptions on g, h, \(\psi_{i}\) (\(i=1,2\)) and (3.39) we have where Define Then \(z(n)\geq0\) is nondecreasing, and By the definition of \(F(s,n)\) and \(t^{(\alpha)}\), we can easily get that \(F(s,n)\) is decreasing in n for each \(s\in\mathbf{N}_{0}\). So from (3.42) and a straightforward computation, for \(n\in I_{\alpha}\), we obtain that Using the monotonicity of \(\varphi'\), \(\varphi^{-1}\), and z, we deduce So from (3.45) and (3.46) we have that is, On the other hand, by the mean value theorem and the monotonicity of \(\varphi'\) and \(\varphi^{-1}\) we have, for \(n\in I_{\alpha}\), So from (3.47) and (3.48) we obtain Setting \(n=s\) in inequality (3.49) and summing with respect to s from α to \(n-1\), we get Now by applying Lemma 2.1 to the function \(\varphi^{-1}(z(n-1))\) we have that is, where \(G(r)=\int_{r_{0}}^{r}\frac{1}{\psi(s)}\,\mathrm{d}s\). By (3.42) we get and then from (3.50) and (3.51) we have that is, Since \(\Omega(t)=G(\varphi^{-1}(2t-k))-G(\varphi^{-1}(t))\) is increasing for \(t\geq k\) and Ω has an inverse function \(\Omega^{-1}\), from (3.52) we get Substituting (3.53) into (3.50) we have Combining (3.44) with (3.54), we obtain the desired inequality (3.40). If \(k=0\), then we carry out the above procedure with \(\varepsilon>0\) instead of k and subsequently let \(\varepsilon\rightarrow0\). This completes the proof. □

Let α, u, k, g, h, f, \(\psi_{1}\), \(\psi_{2}\), ψ, and G be defined as in Theorem  3.3, and \(p\geq1\) be a constant. If u satisfies and \(\Omega(t)=G((2t-k)^{\frac{1}{p}})-G(t^{\frac{1}{p}})\) is increasing for \(t\geq k\), then where \(G^{-1}\) and \(\Omega^{-1}\) are inverse functions of G and Ω, respectively.

For Eq. (4.1), assume that there exist functions \(f, g:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) and a constant q satisfying \(p>q>0\) such that and If u is any solution of Eq. (4.1), then where \(A(T)\) is as in Theorem  3.1.

From (4.1) and (4.2) we get Now, a suitable application of the inequality given in Theorem 3.1 to (4.4) yields the desired result. This completes the proof. □

For Eq. (4.5), assume that there exists a function \(f:\mathbf {N}_{0}\rightarrow\mathbf{R}_{+}\) satisfying \(\exp (\sum_{s=\alpha }^{T}f(s-\alpha) )<2\) and Then Eq. (4.5) has at most one solution.

Suppose that Eq. (4.5) has two solutions \(u_{1}(n)\) and \(u_{2}(n)\). Then we have Furthermore, From (4.6) we have With respect to the function \(|u_{1}(n)-u_{2}(n)|\), by a suitable application of Corollary 3.1 to (4.7) we can deduce that \(|u_{1}(n)-u_{2}(n)|\leq0\), which implies \(u_{1}(n)\equiv u_{2}(n)\). The proof is complete. □