Explicit Bounds to Some New Gronwall-Bellman-Type Delay Integral Inequalities in Two Independent Variables on Time Scales

Definition 1.1. The graininess μ ∈ (𝕋, ℝ₊) is defined by μ(t) = σ(t) − t.

Remark 1.2. Obviously, μ(t) = 0 if 𝕋 = ℝ while μ(t) = 1 if 𝕋 = ℤ.

Definition 1.3. A point t ∈ 𝕋 is said to be left-dense if ρ(t) = t and t ≠ inf 𝕋, right-dense if σ(t) = t and t ≠ sup 𝕋, left-scattered if ρ(t) < t, and right-scattered if σ(t) > t.

Definition 1.4. The set 𝕋^κ is defined to be 𝕋 if 𝕋 does not have a left-scattered maximum, otherwise it is 𝕋 without the left-scattered maximum.

Definition 1.5. A function f ∈ (𝕋, ℝ) is called rd-continuous if it is continuous at right-dense points and if the left-sided limits exist at left-dense points, while f is called regressive if 1 + μ(t)f(t) ≠ 0. C_rd denotes the set of rd-continuous functions, while ℜ denotes the set of all regressive and rd-continuous functions, and ℜ⁺ = {f∣f ∈ ℜ, 1 + μ(t)f(t) > 0, ∀t ∈ 𝕋}.

Definition 1.6. For some t ∈ 𝕋^κ and a function f ∈ (𝕋, ℝ), the delta derivative of f is denoted by f^Δ(t) and satisfies

where s ∈ 𝔘, and 𝔘 is a neighborhood of t which can depend on ɛ.

Similarly, for some y ∈ 𝕋^κ and a function f ∈ (𝕋 × 𝕋, ℝ), the partial delta derivative of f with respect to y is denoted by $$ or $$ and satisfies

where s ∈ 𝔘, and 𝔘 is a neighborhood of y which can depend on ɛ.

Remark 1.7. If 𝕋 = ℝ, then f^Δ(t) becomes the usual derivative f^′(t), while f^Δ(t) = f(t + 1) − f(t) if 𝕋 = ℤ, which represents the forward difference.

Definition 1.8. For a, b ∈ 𝕋 and a function f ∈ (𝕋, ℝ), the Cauchy integral of f is defined by

where F^Δ(t) = f(t),   t ∈ 𝕋^κ.

Similarly, for a, b ∈ 𝕋 and a function f ∈ (𝕋 × 𝕋, ℝ), the Cauchy partial integral of f with respect to y is defined by

where $$.

Definition 1.9. The cylinder transformation ξ_h is defined by

where Log is the principal logarithm function.

Definition 1.10. For p(x, y) ∈ ℜ with respect to y, the exponential function is defined by

Remark 1.11. If 𝕋 = ℝ, then for y ∈ ℝ the following formula holds:

If 𝕋 = ℤ, then, for y ∈ ℤ, $$ and s < y.

Theorem 1.12. If p(x, y) ∈ ℜ with respect to y, then the following conclusions hold:

• (i)

  e_p(y, y) ≡ 1 and e₀(s, y) ≡ 1,

• (ii)

  e_p(s, σ(y)) = (1 + μ(y)p(x, y))e_p(s, y),

• (iii)

  if p ∈ ℜ⁺ with respect to y, then e_p(s, y) > 0 for all s,   y ∈ 𝕋,

• (iv)

  if p ∈ ℜ⁺ with respect to y, then ⊖p ∈ ℜ⁺,

• (v)

  e_p(s, y) = 1/e_p(y, s) = e_⊖p(y, s), where (⊖p)(x, y) = −(p(x, y)/1 + μ(y)p(x, y)).

Theorem 1.13. If p(x, y) ∈ ℜ with respect to y,   y₀ ∈ 𝕋 is a fixed number, then the exponential function e_p(y, y₀) is the unique solution of the following initial value problem:

Lemma 2.1. Suppose that X ∈ 𝕋₀ is a fixed number and u(X, y),   b(X, y) ∈ C_rd,   m(X, y) ∈ ℜ₊ with respect to y, m(X, y) ≥ 0; then

implies where $$, and $$ is the unique solution of the following IVP

Lemma 2.2. Under the conditions of Lemma 2.1 and furthermore assuming that a(x, y) is nondecreasing in y for every fixed x,   b(x, y) ≡ 1, then one has

Proof. Since a(x, y) is nondecreasing in y for every fixed x, then from Lemma 2.1 we have

On the other hand, from [29, Theorems 2.39 and 2.36 (i)] we have $$.

Then collecting the above information we can obtain the desired inequality.

Lemma 2.3 (see [30].)Assume that a ≥ 0,   p ≥ q ≥ 0, and p ≠ 0, then for any K > 0

Theorem 2.4. Suppose that $$ and a, b are nondecreasing. p, q, r, m are constants, and p ≥ q ≥ 0,   p ≥ r ≥ 0,   p ≥ m ≥ 0,     p ≠ 0.  τ₁ ∈ (𝕋₀, 𝕋),   τ₁(x) ≤ x,    − ∞ < α = inf {τ₁(x), x ∈ 𝕋₀} ≤ x₀. $$. ϕ ∈ C_rd(([α, x₀]×[β, y₀])⋂𝕋², ℝ₊). If for $$, u(x, y) satisfies the following inequality:

with the initial condition then where

Proof. Let the right side of (2.7) be v(x, y). Then

If τ₁(x) ≥ x₀ and τ₂(y) ≥ y₀, then $$, and since a, b are nondecreasing we have If τ₁(x) ≤ x₀ or τ₂(y) ≤ y₀, then from (2.8) we have From (2.14) and (2.15) we have Fix X ∈ 𝕋₀, and let $$; then From Lemma 2.3, we have So combining (2.17) and (2.18), it follows that where B₁(x, y), B₂(x, y) are defined in (2.10) and (2.11), respectively. Considering $$, by application of Lemma 2.1, we have Since X ∈ 𝕋₀ is arbitrary, then in fact (2.20) holds for all x ∈ 𝕋₀, that is, Combining (2.13) and (2.21), we obtain which is the desired inequality.

Theorem 2.5. Suppose that u, f, g, h, a, p, q,   r, m, τ₁, τ₂, α, β, ϕ are defined as in Theorem 2.4. If that for $$ satisfies the following inequality:

with the initial condition (2.8), then where

Corollary 2.6. Under the conditions of Theorem 2.4, if, for $$, u(x, y) satisfies the following inequality:

with the initial condition (2.8) (p=1), then where

Corollary 2.7. Under the conditions of Theorem 2.5, if, for $$, u(x, y) satisfies the following inequality:

with the initial condition (2.8) (p = 1), then where

Theorem 2.8. Suppose that $$, f, g, h, τ₁, τ₂ are defined as in Theorem 2.4, and τ₁(x) ≥ x₀, τ₂(y) ≥ y₀. If, for $$ satisfies the following inequality:

then u(x, y) ≡ 0.

Theorem 2.9. Suppose that $$ are the same as in Theorem 2.4, and $$ are two fixed numbers. If, for (x, y)∈([x₀, M]⋂𝕋)×([y₀, N]⋂𝕋),   u(x, y) satisfies the following inequality:

with the initial condition (2.8), then one has provided that $$, where

Proof. Let the right side of (2.33) be v(x, y) and

Then Similar to the process of (2.14)–(2.16) one has Fix X ∈ [x₀, M]⋂𝕋, and let x ∈ [x₀, X]⋂𝕋,   y ∈ [y₀, N]⋂𝕋. Then Considering the structure of (2.45) is similar to (2.17), then following in a same manner as the process of (2.17)–(2.20) we can deduce where $$ are defined in (2.36) and (2.37), respectively.

Since X is selected from [x₀, M]⋂𝕋 arbitrarily, then in fact (2.46) holds for all x ∈ 𝕋₀, that is,

where $$ are defined in (2.38) and (2.39), respectively.

On the other hand, from (2.18), (2.42), and (2.44) we obtain

where λ is defined in (2.35). Then using (2.47) in (2.48) yields which implies Combining (2.43), (2.47), and (2.50) we can obtain the desired inequality (2.34).

Theorem 2.10. Under the conditions of Theorem 2.9, if, for (x, y)∈([x₀, M]⋂𝕋)×([y₀, N]⋂𝕋), u(x, y) satisfies (2.33) with the initial condition (2.8), then the following inequality holds:

provided that $$, where

Theorem 2.11. If, for (x, y)∈([x₀, M]⋂𝕋)×([y₀, N]⋂𝕋),   u(x, y) satisfies (2.53), then the following inequality holds:

provided that $$, where

Proof. Let the right side of (2.53) be v(x, y) and

Then Similar to the process of (2.14)–(2.16) we have Fix X ∈ [x₀, M]⋂𝕋, and let x ∈ [x₀, X]⋂𝕋,   y ∈ [y₀, N]⋂𝕋. Then From Lemma 2.3, we have Combining (2.65) and (2.66), it follows that where $$ are defined in (2.56) and (2.57), respectively.

We notice the structure of (2.67) is similar to (2.19), so following in a same manner as in (2.19)–(2.21) we obtain

where $$ are defined in (2.58) and (2.59), respectively.

On the other hand, from (2.62), (2.64), and (2.66) we have

where $$ is defined in (2.55). Then using (2.68) in (2.69) yields which implies Combining (2.63), (2.68), and (2.71), we obtain the desired inequality (2.54).

Remark 2.12. The established above inequalities generalize many known results including both integral inequalities for continuous functions and discrete inequalities. For example, if we take 𝕋 = ℝ,   p = q = 1,   g(x, y) = h(x, y) ≡ 0, then Theorem 2.4 reduces to [1, Theorem 2.2], which is one case of integral inequality for continuous function. If we take 𝕋 = ℝ,   h(x, y) ≡ 0,   a(x, y) ≡ C, then Corollary 2.7 reduces to [2, Theorem 3 (c1)], which is another case of inequality for continuous function. If we take 𝕋 = ℝ,   p = q = 1, g₁(x, y) = g₂(x, y) = f₂(x, y) = h₁(x, y) = h₂(x, y) ≡ 0, then Theorem 2.10 reduces to [1, Theorem 2.2] with slight difference. If we take 𝕋 = ℤ,   g₁(x, y) = h₁(x, y) = f₂(x, y) = h₂(x, y) ≡ 0,   τ₁(x) = x,   τ₂(y) = y, then Theorem 2.10 reduces to [3, Theorem 2.1], which is a discrete inequality.

Remark 2.13. Since 𝕋 is an arbitrary time scale, then if we take 𝕋 for some peculiar cases, such as 𝕋 = ℝ or 𝕋 = ℤ, we can deduce a series of corollaries according to Theorem 2.4–2.11. Due to the limited space, we omit them here.

Example 3.1. Consider the following delay dynamic differential equation:

with the initial condition where $$, a ∈ C_rd(𝕋₀, ℝ), $$, b is delta differential, and $$, p > 0 is a constant, $$. α, β, τ₁, τ₂ are the same as in Theorem 2.4.

Theorem 3.2. Suppose that u(x, y) is a solution of (3.1)-(3.2), |a(x) + b(y)| ≤ k(x, y), and |F(s, t, x, y)| ≤ f(s, t) | x|^q+|y|, |W(ξ, η, x)| ≤ h(ξ, η) | x|^m, where f, h, q, m are defined as in Theorem 2.4; then

where and B₂(x, y) is defined as in Theorem 2.4 (with g(x, y) ≡ 0).

Proof. The equivalent integral equation of (3.1) can be denoted by

Then and a suitable application of Theorem 2.4 to (3.6) yields the desired inequality (3.3).

Theorem 3.3. Under the conditions of Theorem 3.2, one has

where B₁, B₂ are defined as in Theorem 3.2.

Proof. The desired inequality can be obtained by an application of Theorem 2.5 to (3.6).

Example 3.4. Consider the following delay dynamic integral equation:

with the initial condition where $$, p > 0 is a constant, C = u^p(x₀, y₀), $$ are two fixed numbers. $$. α, β, τ₁, τ₂ are the same as in Theorem 2.4.

Theorem 3.5. Suppose u(x, y) is a solution of (3.8)-(3.9) and |F_i(s, t, x, y)| ≤ L(s, t, |x|)+|y | ,    | W_i(ξ, η, x)| ≤ h_i(ξ, η) | x|^q,   i = 1,2, where L, h_i,   i = 1,2,   q are defined the same as in Theorem 2.11; then the following inequality holds:

provided that $$, where $$ are defined the same as in Theorem 2.11, and

Proof. From (3.8) we have

So by use of Theorem 2.11 we obtain the desired inequality (3.10).

Example 3.6. Consider the following delay dynamic integral equation:

where $$, $$. τ₁, τ₂ are the same as in Theorem 2.4.

Theorem 3.7. Assume that |F(s, t, u₁, v₁) − F(s, t, u₂, v₂)| ≤ f(s, t)|u₁ − u₂| + |v₁ − v₂|,    | W(s, t, u₁) − W(s, t, u₂)| ≤ h(s, t) | u₁ − u₂|, where f,   h are defined as in Theorem 2.4, and; furthermore, assume that τ₁(x) ≥ x₀,   τ₂(y) ≥ y₀, then (3.13) has at most one solution.

Proof. Suppose that u₁(x, y),   u₂(x, y) are two solutions of (3.13); then we have

Then a suitable application of Theorem 2.8 yields |u₁(x, y) − u₂(x, y)| ≤ 0, that is, u₁(x, y) ≡ u₂(x, y), and the proof is complete.