# Bounds of Double Integral Dynamic Inequalities in Two Independent Variables on Time Scales

Authors: S. H. Saker
Journal: Discrete Dynamics in Nature and Society
 (2011)

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## Abstract

Our aim in this paper is to establish some explicit bounds of the unknown function in a certain class of nonlinear dynamic inequalities in two independent variables on time scales which are unbounded above. These on the one hand generalize and on the other hand furnish a handy tool for the study of qualitative as well as quantitative properties of solutions of partial dynamic equations on time scales. Some examples are considered to demonstrate the applications of the results.


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## Body

## 1. Introduction

During the past decade, a number of dynamic inequalities have been established by some authors which are motivated by some applications, for example, when studying the behavior of solutions of certain classes of dynamic equations, the bounds provided by earlier inequalities are inadequate in applications and some new and specific type of dynamic inequalities on time scales are required. The general idea is to prove a result for a dynamic inequality where the domain of the unknown function is a so-called time scale𝕋, which may be an arbitrary closed subset of the real numbers ℝ. In [1, Theorem 6.1], it is proved that if u, f, and p∈Crd and p∈ℛ+, then(1.1)uΔ(t)≤f(t)+p(t)u(t),∀t∈[t0,∞)T,
implies(1.2)u(t)≤u(t0)ep(t,t0)+∫t0tep(t,σ(s))f(s)Δs,∀t∈[t0,∞)T,
where ℛ+:={f∈ℛ:1+μ(t)f(t)>0,t∈𝕋} and ℛ is the class of rd-continuous and regressive functions. A function f:𝕋→ℝ is said to be right-dense continuous (rd-continuous) provided f is continuous at right-dense points and at left-dense points in 𝕋, left-hand limits exist and are finite. The set of all such rd-continuous functions is denoted by Crd(𝕋). The graininess function μ for a time scale 𝕋 is defined by μ(t):=σ(t)-t, and, for any function f:𝕋→ℝ, the notation fσ(t) denotes f(σ(t)), where σ(t) is the forward jump operator defined by σ(t):=inf{s∈𝕋:s>t}. We say that a function f:𝕋→ℝ is regressive provided 1+μ(t)f(t)≠0,t∈𝕋. The set of all regressive functions on a time scale 𝕋 forms an Abelian group under the addition ⊕ defined by p⊕q:=p+q+μpq. Throughout this paper, we will assume that sup𝕋=∞ and define the time scale interval [t0,∞)𝕋 by [t0,∞)𝕋:=[t0,∞)∩𝕋. The exponential function ep(t,s) on time scales is defined by(1.3)ep(t,s)=exp⁡(∫stξμ(τ)(p(τ))Δτ),fort,s∈T,
where ξh(z) is the cylinder transformation, which is given by(1.4)ξh(z)={log⁡(1+hz)h,h≠0,z,h=0.
Alternatively, for p∈ℛ, one can define the exponential function ep(·,t0) to be the unique solution of the IVP xΔ=p(t)x, with x(t0)=1. If p∈ℛ, then ep(t,s) is real-valued and nonzero on 𝕋. If p∈ℛ+, then ep(t,t0) is always positive, ep(t,t)=1, and e0(t,s)=1. Note that(1.5)ep(t,t0)=exp⁡(∫t0tp(s)ds),ifT=R,ep(t,t0)=∏s=t0t-1(1+p(s)),ifT=N,ep(t,t0)=∏s=t0t-1(1+(q-1)sp(s)),ifT=qN0.
The book on the subject of time scales by Bohner and Peterson [1] summarizes and organizes much of time scale calculus. The three most popular examples of calculus on time scales are differential calculus, difference calculus, and quantum calculus (see [2]), that is, when 𝕋=ℝ,𝕋=ℕ, and 𝕋=qℕ0={qt:t∈ℕ0}, where q>1.In this paper, we will refer to the (delta) integral which we can define as follows: IfGΔ(t)=g(t), then the Cauchy (delta) integral of g is defined by ∫atg(s)Δs:=G(t)-G(a). It can be shown (see [1]) that if g∈Crd(𝕋), then the Cauchy integral G(t):=∫t0tg(s)Δs exists, t0∈𝕋, and satisfies GΔ(t)=g(t), t∈𝕋. There are applications of dynamic equations on time scales to quantum mechanics, electrical engineering, neural networks, heat transfer, and combinatorics. A recent cover story article in New Scientist [3] discusses several possible applications. Also, in [1, Theorem 6.4], it is proved that if u,a, and p∈Crd and p∈ℛ+, then(1.6)u(t)≤a(t)+∫t0tp(s)u(s)Δs,∀t∈[t0,∞)T,
implies that(1.7)u(t)≤a(t)+∫t0tep(t,σ(s))a(s)p(s)Δs,∀t∈[t0,∞)T.
Since (1.7) provides an explicit bound to the unknown function u(t) and a tool to the study of many qualitative as well as quantitative properties of solutions of dynamic equations, it has become one of the very few classic and most influential results in the theory and applications of dynamic inequalities. Because of its fundamental importance, over the years, many generalizations and analogous results of (1.7) have been established. Since the discovery of the inequalities (1.1)–(1.7), much work has been done, and many papers which deal with various generalizations and extensions have appeared in the literature, we refer the reader to [4–9] and the references cited therein. On the other hand, a few authors have focused on the theory of partial dynamic equations on time scales [10–15]. However, to the best of author’s knowledge, only [16–19] have studied integral inequalities useful in the theory of partial dynamic equations on time scales. Before, we give a brief summary of some of the results of dynamic inequalities in two independent variables, we present some basic definitions about calculus in two variables on time scales (for more details, we refer to [12]).Let𝕋1 and 𝕋2 be two time scales with at least two points, and consider the time scale intervals Ω1=[t0,∞)∩𝕋1 and Ω2=[s0,∞)∩𝕋2 for t0∈𝕋1 and s0∈𝕋2. Let σ1,ρ1,Δ1 and σ2,ρ2,Δ2 denote the forward jump operators, backward jump operators, and the delta differentiation operator, respectively, on 𝕋1 and 𝕋2. We say that a real valued function f on 𝕋1×𝕋2 at (t,s)∈Ω≡Ω1×Ω2 has a Δ1 partial derivative fΔ1(t,s) with respect to t if for each ϵ>0 there exists a neighborhood Ut of t such that(1.8)|f(σ1(t),s)-f(η,s)-fΔ1(t,s)[σ1(t)-η]|≤ϵ|σ(t)-η|,∀η∈Ut.
In this case, we say fΔ1(t,s) is the (partial delta) derivative of f(t,s) at t. We say that a real valued function f on 𝕋1×𝕋2 at (t,s)∈Ω1×Ω2 has a Δ2 partial derivative fΔ2(t,s) with respect to s if for each ϵ>0 there exists a neighborhood Us of s such that(1.9)|f(t,σ2(s))-f(t,ξ)-fΔ2(t,s)[σ2(t)-ξ]|≤ϵ|σ(t)-ξ|,∀ξ∈Us.
In this case, we say fΔ2(t,s) is the (partial delta) derivative of f(t,s) at s. The function f is called rd-continuous in t if for every α2∈𝕋2 the function f(t,α2) is rd-continuous on 𝕋1. The function f is called rd-continuous in s if for every α1∈𝕋1 the function f(α1,s) is rd-continuous on 𝕋2. Now, we are ready to present some results for dynamic inequalities in two independent variables on times scales which are related to the main results in our paper. In [18], the authors proved that if a,f, and u are positive rd-continuous functions and a is nonnegative and nondecreasing in each of its variables, then(1.10)u(x,y)≤a(x,y)+∫x0x∫y0yf(s,t)u(s,t)ΔtΔs,
for all (x,y)∈[x0,∞)𝕋×[y0,∞)𝕋, implies(1.11)u(x,y)≤a(x,y)eF(x,x0),whereF=∫y0yf(x,t)Δt.
In [19], the author proved that if a,b,g,h, and u are positive continuous real functions defined on 𝕋×𝕋 and γ>1 is a real constant, then(1.12)uγ(x,y)≤a(x,y)+b(x,y)∫x0x∫y0y[g(s,t)uγ(s,t)+h(s,t)u(s,t)]ΔtΔs,
implies(1.13)u(x,y)≤[a(x,y)+b(x,y)m(x,y)eG(t,t0)]1/γ,
for all (x,y)∈[x0,∞)𝕋×[y0,∞)𝕋, where(1.14)m(x,y)=∫x0x∫y0y[a(s,t)g(s,t)+(γ-1γ+a(s,t)γ)h(s,t)]ΔtΔs,G(s,t)=∫y0y[g(x,t)+h(x,t)γ]b(x,t)Δt.
In this paper, we are concerned with bounds of the double integral nonlinear dynamic inequality in two independent variables(1.15)uγ(t,s)≤a(t,s)+b(t,s)∫t0t∫s0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ,
for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋. For (1.15), we will assume the following hypotheses:(1.16)(H){u,a,b,f,andgarerd-continuouspositivefunctionson⁡Ω1×Ω2,α,δ,λ,andγarepositiveconstants.The main aim in this paper is to establish some explicit bounds of the unknown functionu(t,s) of the inequality (1.15). Our results not only complement the results in [18, 19] but also improve the results in [19], in the sense that the results can be applied in the cases when γ≤1. The main results will be proved by employing the Bernoulli inequality [20, Bernoulli’s inequality]:(1.17)(1+x)γ≤1+γx,for0<γ≤1,x>-1,
the Young inequality [20]:(1.18)ab≤app+bqq,wherea,b≥0,p>1,1p+1q=1,
and the algebraic inequalities [20]:(1.19)(a+b)λ≤2λ-1(aλ+bλ),fora,b≥0,λ≥1,(1.20)(a+b)λ≤aλ+bλ,fora,b≥0,0≤λ≤1.
Some examples are considered to illustrate the main results.


## 2. Main Results

Before, we stated and proved the main results and we proved some Lemmas which play important roles in the proofs of the main results. We will assume that the equations or the inequalities possess such nontrivial solutions.Lemma 2.1.
Let𝕋 be an unbounded time scale with (t0,s0) and (t,s)∈𝕋×𝕋.  Let gi:ℝ→ℝ for i=1,2,…,n be functions with gi(x1(t,s))≤gi(x2(t,s)) for i=1,2,…,n, where xi(t,s):𝕋×𝕋→ℝ for i=1,2, whenever x1≤x2. Let v,w:𝕋×𝕋→ℝ be differentiable with
(2.1)vΔtΔs(t,s)≤∑i=1nai(t,s)gi(v(t,s)),wΔtΔs(t,s)≥∑i=1nai(t,s)gi(w(t,s)),
for all (t,s)∈𝕋×𝕋. Then, v(t0,s0)<w(t0,s0) implies v(t,s)≤w(t,s) for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋.Proof.
The proof is by induction and similar to the proof of Theorem 6.9 in [1] and hence is omitted.Lemma 2.2.
Let𝕋 be an unbounded time scale with (t0,s0) and (t,s)∈𝕋×𝕋. Suppose that gi:ℝ→ℝ is nondecreasing for i=1,2,…,n and y:𝕋×𝕋→ℝ+ is such that gi(y) is rd-continuous. Let pi be rd-continuous and positive for i=1,2,…,n and f:𝕋×𝕋→ℝ+ differentiable. Then,
(2.2)y(t,s)≤f(t,s)+∑i=1n∫t0t∫s0spi(η,τ)gi(y(η,s))ΔηΔτ,∀(t,s)∈T×T,
implies y(t,s)≤x(t,s) for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋,  where x(t,s) solves the initial value problem
(2.3)xΔtΔs(t,s)=fΔtΔs(t,s)+∑i=1npi(t,s)gi(x(t,s)),x(t0,s0)>f(t0,s0)>0.Proof.
Let(2.4)v(t,s):=f(t,s)+∑i=1n∫t0t∫t0spi(τ,η)gi(y(τ,η))ΔηΔτ,
for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋. Then,
(2.5)vΔtΔs(t,s):=fΔtΔs(t,s)+∑i=1npi(t,s)gi(y(t,s)),
for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋 and y(t,s)≤v(t,s) so that
(2.6)vΔtΔs(t,s)≤fΔtΔs(t,s)+∑i=1npi(t,s)gi(v(t,s)),
for all (t,s)∈[t0,∞)𝕋×[s0,∞)𝕋. Since v(t0,s0)=f(t0,s0)<x0=x(t0,s0), the comparison Lemma 2.1 yields v(t,s)≤x(t,s) for all (t,s)≥(t0,s0), where x(t,s) solves the initial value problem (2.3). Hence, since y(t,s)≤v(t,s), we obtain y(t,s)≤x(t,s). The proof is complete.Now, we are ready to state and prove the main results in this paper. First, we consider the case whenλ≥1 and α,δ≤γ. For simplicity, we introduce the following notations:(2.7)F(t,s):=22(λ-1)∫t0t∫s0s[fλ(τ,η)[aδ/γ(τ,η)]λ]ΔηΔτ+22(λ-1)∫t0t∫s0s[gλ(τ,η)[aα/γ(τ,η)]λ]ΔηΔτ,G(t,s):=22(λ-1)(fλ(t,s)[δγa(δ/γ)-1(t,s)]λ+gλ(t,s)[αγa(α/γ)-1(t,s)]λ).Theorem 2.3.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  λ,γ≥1  and α,δ≤γ. Then,
(2.8)uγ(t,s)≤a(t,s)+b(t,s)∫t0t∫s0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ,
for (t,s)∈Ω,  implies that
(2.9)u(t,s)≤a1/γ(t,s)+1γa(1/γ)-1(t,s)b(t,s)w(t,s),∀(t,s)∈Ω,
where  w(t) solves the initial value problem
(2.10)wΔtΔs(t,s)=FΔtΔs(t,s)+bλ(t,s)G(t,s)wλ(t,s),w(t0,s0)>0.Proof.
Define a functiony(t,s) by
(2.11)y(t,s):=∫t0t∫t0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ.
This reduces (2.8) to
(2.12)uγ(t,s)≤a(t,s)+b(t,s)y(t,s),for(t,s)∈Ω.
This implies that
(2.13)u(t,s)≤(a(t,s)+b(t,s)y(t,s))1/γ,for(t,s)∈Ω.
Applying the inequality (1.17) (noting that 1/γ≤1), we see that
(2.14)u(t,s)≤a1/γ(t,s)+1γa(1/γ)-1(t,s)b(t,s)y(t,s),for(t,s)∈Ω.
From (2.13), we obtain
(2.15)uα(t,s)≤aα/γ(t,s)[1+b(t,s)y(t,s)a(t,s)]α/γ,for(t,s)∈Ω.
Applying inequality (1.17) on (2.15) (where α≤γ), we obtain for (t,s)∈Ω that
(2.16)uα(t,s)≤aα/γ(t,s)[1+αγb(t,s)a(t,s)y(t,s)]=aα/γ(t,s)+αγa(α/γ)-1(t,s)b(t,s)y(t,s).
Also, from (2.13), we obtain
(2.17)uδ(t,s)≤aδ/γ(t,s)[1+b(t,s)y(t,s)a(t,s)]δ/γ,for(t,s)∈Ω.
Applying inequality (1.17) on (2.17) (where δ≤γ), we have for (t,s)∈Ω that
(2.18)uδ(t,s)≤aδ/γ(t,s)[1+δγb(t,s)a(t,s)y(t,s)]=aδ/γ(t,s)+δγa(δ/γ)-1(t,s)b(t,s)y(t,s).
Combining (2.11), (2.16), and (2.18) and applying the inequality (1.19) (noting that λ≥1), we have
(2.19)y(t,s)=∫t0t∫t0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ≤2λ-1∫t0t∫t0s[f(τ,η)uδ(τ,η)]λΔηΔτ+2λ-1∫t0t[g(τ,η)uα(τ,η)]λΔηΔτ≤2λ-1∫t0t∫t0sfλ(τ,η)[aδ/γ(τ,η)+δγa(δ/γ)-1(τ,η)b(τ,η)y(τ,η)]λΔηΔτ+2λ-1∫t0t∫t0sgλ(τ,η)[aα/γ(τ,η)+αγa(α/γ)-1(τ,η)b(τ,η)y(τ,η)]λΔηΔτ.
This implies that
(2.20)y(t,s)≤22(λ-1)∫t0t∫t0sfλ(τ,η)[a(δ/γ)(τ,η)]λΔηΔτ+22(λ-1)∫t0t∫t0sfλ(τ,η)[δγa(δ/γ)-1(τ,η)b(τ,η)]λyλ(τ,η)ΔηΔτ+22(λ-1)∫t0t∫t0sgλ(τ,η)[aα/γ(τ,η)]λΔηΔτ+22(λ-1)∫t0t∫t0sgλ(τ,η)[αγa(α/γ)-1(τ,η)b(τ,η)]λyλ(τ,η)ΔηΔτ=F(t,s)+∫t0t∫t0sG(τ,η)yλ(τ,η)ΔηΔτ,for(t,s)∈Ω.
Now an application of Lemma 2.2 (with n=1 and g(y)=yλ) gives that
(2.21)y(t,s)<w(t,s),for(t,s)∈Ω,
where w(t,s) solves the initial value problem (2.10). Substituting (2.21) into (2.14), we obtain the desired inequality (2.9). The proof is complete.Theorem 2.4.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  λ,γ≥1 and α,δ≤γ. Then, (2.8) implies
(2.22)u(t,s)≤a1/γ(t,s)+b1/γ(t,s)w1/γ(t,s),∀(t,s)∈Ω,
where  w(t) solves the initial value problem
(2.23)wΔtΔs(t,s)=FΔtΔs(t,s)+G1(t,s)wλ(δ/γ)(t,s)+G2(t,s)wλ(α/γ)(t,s),w(t0,s0)>0,
where  F(t) is defined as in (2.7) and
(2.24)G1(t,s):=22(λ-1)fλ(t,s)bλδ/γ(t,s),G2:=22(λ-1)gλ(t,s)bλα/γ(t,s).Proof.
Define a functiony(t,s) by (2.11) and proceed as in the proof of Theorem 2.3 to obtain
(2.25)u(t,s)≤(a(t,s)+b(t,s)y(t,s))1/γ,for(t,s)∈Ω.
Applying the inequality (1.20), we see that
(2.26)u(t,s)≤a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s),for(t,s)∈Ω.
From (2.25), we obtain
(2.27)uα(t,s)≤(a(t,s)+b(t,s)y(t,s))α/γ,for(t,s)∈Ω.
Applying inequality (1.20) on (2.27) (where α≤γ), we obtain for (t,s)∈Ω that
(2.28)uα(t,s)≤aα/γ(t,s)+bα/γ(t,s)yα/γ(t,s).
Also, from (2.25), we have by (1.20) that
(2.29)uδ(t,s)≤aδ/γ(t,s)+bδ/γ(t,s)yδ/γ(t,s),for(t,s)∈Ω.
Combining (2.11), (2.28), and (2.29) and applying the inequality (1.19) (noting that λ≥1), we have
(2.30)y(t,s)=∫t0t∫t0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ≤2λ-1∫t0t∫t0s[f(τ,η)uδ(τ,η)]λΔηΔτ+2λ-1∫t0t[g(τ,η)uα(τ,η)]λΔηΔτ≤2λ-1∫t0t∫t0sfλ(τ,η)[aδ/γ(τ,η)+bδ/γ(τ,η)yδ/γ(τ,η)]λΔηΔτ+2λ-1∫t0t∫t0sgλ(τ,η)[aα/γ(τ,η)+bα/γ(τ,η)yα/γ(τ,η)]λΔηΔτ.
This implies that
(2.31)y(t,s)≤22(λ-1)∫t0t∫t0sfλ(τ,η)[aδ/γ(τ,η)]λΔηΔτ+22(λ-1)∫t0t∫t0sgλ(τ,η)[aα/γ(τ,η)]λΔηΔτ+22(λ-1)∫t0t∫t0sfλ(τ,η)[bδ/γ(τ,η)]λyλ(δ/γ)(τ,η)ΔηΔτ+22(λ-1)∫t0t∫t0sgλ(τ,η)[bα/γ(τ,η)]λyλ(α/γ)(τ,η)ΔηΔτ=F(t,s)+∫t0t∫t0s[G1(τ,η)yλ(δ/γ)(τ,η)+G2(τ,η)yλ(α/γ)(τ,η)]ΔηΔτ,
for (t,s)∈Ω. Now, an application of Lemma 2.2 (with n=2,g1(y)=yλ(δ/γ), and g2(y)=yλ(α/γ)) gives that
(2.32)y(t,s)<w(t,s),for(t,s)∈Ω,
where w(t,s) solves the initial value problem (2.23). Substituting (2.32) into (2.26), we obtain the desired inequality (2.22). The proof is complete.As in the proof of Theorem2.3 by employing the inequality (1.20) instead of the inequality (1.19), we can obtain an explicit bound for u(t) when 0≤λ≤1. This will be presented below in Theorem 2.5 without proof since the proof is similar to the proof of Theorem 2.3. For simplicity, we introduce the following notations:(2.33)F1(t,s):=∫t0t∫s0s[fλ(τ,η)aλ(δ/γ)(τ,η)]ΔηΔτ+∫t0t∫s0s[gλ(τ,η)aλ(α/γ)(τ,η)]ΔηΔτ,G3(t,s):=(fλ(t,s)[δγa(δ/γ)-1(t,s)]λ+gλ(t,s)[αγa(α/γ)-1(t,s)]λ).Theorem 2.5.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  γ≥1,0<λ≤1,δ≤γ, and α≤γ. Then, (2.8) implies that
(2.34)u(t,s)≤a1/γ(t,s)+1γa(1/γ)-1(t,s)b(t,s)z(t,s),for(t,s)∈Ω,
where  z(t) solves the initial value problem
(2.35)zΔtΔt(t)=F1ΔtΔs(t)+G3(t,s)bλ(t,s)zλ(t,s),z(t0,s0)>0.In the following, we apply the Young inequality (1.18) to find a new explicit upper bound for u(t) of (2.8) when λ≥1 and 0≤λ≤1. First, we consider the case when λ≥1 and assume that λ(α/γ)<1 and λ(δ/γ)<1.Theorem 2.6.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  γ,λ≥1and  α,δ≤γ such that  (λα/γ)<1 and  (λδ/γ)<1. Then, (2.8) implies that
(2.36)u(t,s)≤a1/γ(t,s)+b1/γ(t,s)F31/γ(t,s),∀(t,s)∈Ω,
where
(2.37)F3(t,s):=F0(t,s)+eβ(s-s0)(t,t0),β=λ[αγ+δγ],F0(t,s):=F(t,s)+(γ-λδ)γ∫t0t(G1(τ,η))γ/(γ-λδ)ΔηΔτ+(γ-λα)γ∫t0t(G2(τ,η))γ/(γ-λα)ΔηΔτ,
and  F,G1, and  G2 are defined as in (2.7) and (2.24).Proof.
Define a functiony(t,s) by (2.11) and proceed as in the proof of Theorem 2.4 to obtain
(2.38)u(t,s)≤a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s),for(t,s)∈Ω,(2.39)y(t,s)≤F(t,s)+∫t0t∫t0s[G1(τ,η)yλ(δ/γ)(τ,η)+G2(τ,η)yλ(α/γ)(τ,η)]ΔηΔτ,
where F,G1, and G2 are defined as in (2.7) and (2.24). Applying the Young inequality (1.18) on the term G1yλ(δ/γ) with q=γ/λδ>1 and p=γ/(γ-λδ)>1, we see that
(2.40)G1yλ(δ/γ)≤(γ-λδ)γ(G1)γ/(γ-λδ)+(λδγ)y.
Again applying the Young inequality (1.18) on the term G2yλ(α/γ) with q=γ/λα>1 and p=γ/(γ-λα)>1, we see that
(2.41)G2yλ(α/γ)≤(γ-λα)γ(G2)γ/(γ-λα)+(λαγ)y.
Substituting (2.40) and (2.41) into (2.39), we have
(2.42)y(t,s)≤F(t,s)+(γ-λδ)γ∫t0t∫t0s(G1(τ,η))γ/(γ-λδ)ΔηΔτ+(γ-λα)γ∫t0t∫t0s(G2(τ,η))γ/(γ-λα)ΔηΔτ+[λαγ+λδγ]∫t0t∫s0sy(τ,η)ΔηΔτ,∀(t,s)∈Ω.
From the definitions of F0(t) and β, we get that
(2.43)y(t,s)≤F0(t,s)+β∫t0t∫t0sy(τ,η)ΔηΔτ,for(t,s)∈Ω.
Applying the inequality (1.10) with f(t,s)=β, we have
(2.44)y(t,s)<F0(t,s)+eβ(s-s0)(t,t0),∀(t,s)∈Ω.
Substituting (2.44) into (2.38), we get the desired inequality (2.36). The proof is complete.Theorem 2.7.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  γ≥1,0<λ≤1, and  α,δ≤γ. Then, (2.8) implies that
(2.45)u(t,s)≤a1/γ(t,s)+1γa(1/γ)-1(t,s)b(t,s)F4(t,s),for(t,s)∈Ω,
where
(2.46)F4(t,s):=F2(t,s)+eλ(s-s0)(t,t0),F2(t,s):=F1(t,s)+(1-λ)∫t0t∫s0s(G3(τ,η))1/(1-λ)ΔηΔτ,
and  F1 and  G3 are defined as in (2.33).Proof.
Define a functiony(t,s) by (2.11) and proceed as in the proof of Theorem 2.3 to obtain
(2.47)u(t,s)≤a1/γ(t,s)+1γa(1/γ)-1(t,s)b(t,s)y(t,s),for(t,s)∈Ω,(2.48)y(t,s)≤F1(t,s)+∫t0t∫s0sG3(s)yλ(τ,η)ΔηΔτ,for(t,s)∈Ω,
where F1 and G3 are defined in (2.33). Applying the Young inequality (1.18) on the term G3yλ with q=1/λ>1 and p=1/(1-λ)>1, we see that
(2.49)G3yλ≤(1-λ)(G3)1/1-λ+λy.
This and (2.48) imply that
(2.50)y(t,s)≤F1(t,s)+(1-λ)∫t0t∫s0s(G3(τ,η))1/(1-λ)ΔηΔτ+∫t0t∫s0sλy(τ,η)ΔηΔτ.
Using the definition of F2(t,s), we get that
(2.51)y(t,s)≤F2(t,s)+λ∫t0t∫s0sy(τ,η)ΔηΔτ,for(t,s)∈Ω.
Applying the inequality (1.10) with f(t,s)=λ, we have
(2.52)y(t,s)<F2(t,s)+eλ(s-s0)(t,t0),∀(t,s)∈Ω.
Substituting (2.52) into (2.47), we get the desired inequality (2.45). The proof is complete.Next in the following, we consider the case whenγ≤1 and establish some new explicit bounds of the unknown function u(t,s) of (2.8).Theorem 2.8.
Let𝕋 be an unbounded time scale with  (t0,s0)∈𝕋×𝕋. Assume that  (H) holds,  λ≤1,γ≤1, and αλ,δλ≤γ. Then, (2.8) implies that
(2.53)uγ(t,s)≤a(t,s)+b(t,s)[H(t,s)+eβ1(s-s0)(t,t0)],for(t,s)∈Ω,
where  β1=(λα/γ)+(λδ/γ) and
(2.54)H(t,s)=2λδ((1/γ)-1)∫t0t∫t0sfλ(τ,η)aλδ/γ(τ,η)ΔηΔτ+2λα((1/γ)-1)∫t0t∫t0sgλ(τ,η)aλα/γ(τ,η)ΔηΔτ+2λδ((1/γ)-1)(γ-λδ)γ∫t0t∫t0s(fλ(τ,η)bλδ/γ(τ,η))γ/(γ-λδ)ΔηΔτ+2λα((1/γ)-1)(γ-λα)γ∫t0t∫t0s(gλ(τ,η)bλα/γ(τ,η))γ/(γ-λα)ΔηΔτ.Proof.
Define a functiony(t,s) by
(2.55)y(t,s):=∫t0t∫t0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ.
This reduces (2.8) to
(2.56)uγ(t,s)≤a(t,s)+b(t,s)y(t,s),for(t,s)∈Ω.
This implies that
(2.57)u(t,s)≤(a(t,s)+b(t,s)y(t,s))1/γ,for(t,s)∈Ω.
Applying the inequality (1.19) (noting that γ≤1), we see that
(2.58)u(t,s)≤2(1/γ)-1[a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s)],for(t,s)∈Ω.
From (2.58), we obtain
(2.59)uα(t,s)≤2α((1/γ)-1)[a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s)]α,for(t,s)∈Ω.
Also, from (2.58), we obtain
(2.60)uδ(t,s)≤2δ((1/γ)-1)[a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s)]δ,for(t,s)∈Ω.
Combining (2.55), (2.59), and (2.60) and applying the inequality (1.19) (noting that λ≤1), we have
(2.61)y(t,s)=∫t0t∫t0s[f(τ,η)uδ(τ,η)+g(τ,η)uα(τ,η)]λΔηΔτ≤∫t0t∫t0s[f(τ,η)uδ(τ,η)]λΔηΔτ+∫t0t[g(τ,η)uα(τ,η)]λΔηΔτ≤2λδ((1/γ)-1)∫t0t∫t0sfλ(τ,η)[a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s)]λδΔηΔτ+2λα((1/γ)-1)∫t0t∫t0sgλ(τ,η)[a1/γ(t,s)+b1/γ(t,s)y1/γ(t,s)]αλΔηΔτ.
This implies (noting that αλ≤1 and δλ≤1) that
(2.62)y(t,s)≤2λδ((1/γ)-1)∫t0t∫t0sfλ(τ,η)aλδ/γ(τ,η)ΔηΔτ+2λδ((1/γ)-1)∫t0t∫t0sfλ(τ,η)bλδ/γ(τ,η)yλδ/γ(τ,η)ΔηΔτ+2λα((1/γ)-1)∫t0t∫t0sgλ(τ,η)aλα/γ(τ,η)ΔηΔτ+2λα((1/γ)-1)∫t0t∫t0sgλ(τ,η)bλα/γ(τ,η)yλα/γ(τ,η)ΔηΔτ.
Applying the Young inequality (1.18) on the term H1yλ(δ/γ) with q=γ/λδ>1 and p=γ/(γ-λδ)>1, we see that
(2.63)H1yλ(δ/γ)≤(γ-λδ)γ(H1)γ/(γ-λδ)+(λδγ)y,
where H1=fλbλδ/γ. Again applying the Young inequality (1.18) on the term H2yλ(α/γ) with q=γ/λα>1 and p=γ/(γ-λα)>1, we see that
(2.64)H2yλ(α/γ)≤(γ-λα)γ(H2)γ/(γ-λα)+(λαγ)y,
where H2=gλbλα/γ. Combining (2.62)–(2.64), we have
(2.65)y(t,s)≤H(t,s)+β1∫t0t∫t0sy(τ,η)ΔηΔτ.
Applying the inequality (1.10) on (2.65) with f(t,s)=β1, we have
(2.66)y(t,s)<H(t,s)+eβ1(s-s0)(t,t0),∀(t,s)∈Ω.
Substituting (2.66) into (2.56), we get the desired inequality (2.45). The proof is complete.Remark 2.9.
The above results can be extended to the general inequality(2.67)uγ(t,s)≤a(t,s)+b(t,s)∫t0t∫s0sL(τ,η,u(τ,η))ΔηΔτ,
when
(2.68)L(τ,η,u)-L(τ,η,v)≤[L1(τ,η,v)(u-v)δ+L2(τ,η,v)(u-v)α]λ.
In fact, by using the new substitution
(2.69)y(t,s)=∫t0t∫s0sL(τ,η,u(τ,η))ΔηΔτ,
the inequality (2.67) can be written as
(2.70)uγ(t,s)≤a(t,s)+b(t,s)y(t,s),
and then, since γ≥1, we have
(2.71)u(t,s)≤A(t,s)+B(t,s)y(t,s),
where A(t,s)=a1/γ(t,s) and B(t,s)=a(1/γ)-1(t,s)b(t,s). This implies that
(2.72)y(t,s)=∫t0t∫s0sL(τ,η,u(τ,η))ΔηΔτ≤∫t0t∫s0s[L(τ,η,A(τ,η)+B(τ,η)y(τ,η))-L(τ,η,A(τ,η))]ΔηΔτ+∫t0t∫s0sL(τ,η,A(τ,η))ΔηΔτ.
Using (2.68) in the last inequality, we get an inequality similar to the inequality (2.8) and then follow the proof of the above results to find some explicit bounds of (2.8). The details are left to the reader.


## 3. Applications

In this section, we present some applications of our main results.Example 3.1.
Consider the partial dynamic equation on time scales(3.1)(uγ(t,s))ΔtΔs=H(t,s,u(t,s))+h(t,s),(t,s)∈Ω≡[t0,∞)T×[s0,∞)T,
with initial boundary conditions
(3.2)u(t,s0)=a(t)>,u(t0,s)=b(s)>,a(t0)=b(s0)=0,
where γ≥1 is a constant and H and h are rd-continuous functions on Ω,a:[t0,∞)𝕋→ℝ+ and b:[t0,∞)𝕋→ℝ+ are rd-continuous functions. Assume that
(3.3)|H(t,s,u)|≤f(t,s)|u(t,s)|δ+g(t,s)|u(t,s)|α,
where f(t,s) and g(t,s) are nonnegative rd-continuous functions for (t,s)∈Ω and α,δ<γ. If u(t,s) is a solution of (3.1)-(3.2), then u(t,s) satisfies
(3.4)|u(t,s)|≤a1/γ(t,s)+A1/γ(t,s),∀(t,s)∈Ω,
where
(3.5)a(t,s)=aγ(t)+bγ(s)+∫t0t∫s0s|h(τ,η)|ΔηΔτ,A(t,s):=H0(t,s)+eβ(s-s0)(t,t0),β=[αγ+δγ],H0(t,s)=∫t0t∫s0s[f(τ,η)aδ/γ(τ,η)]ΔηΔτ+∫t0t∫s0s[g(τ,η)aα/γ(τ,η)]ΔηΔτ+(γ-δ)γ∫t0t∫s0s(f(τ,s))γ/(γ-δ)ΔηΔτ+(γ-α)γ∫t0t∫s0s(g(τ,η))γ/(γ-α)ΔηΔτ.
In fact, the solution of (3.1)-(3.2) satisfies
(3.6)|u(t,s)|γ=aγ(t)+bγ(t)+∫t0t∫s0sh(τ,η)ΔηΔτ+∫t0t∫s0sH(τ,η,u(τ,η))ΔηΔτ,
for (t,s)∈Ω.Therefore,
(3.7)|u(t,s)|γ≤a(t,s)+∫t0t∫s0s|H(τ,η,u(τ,η))|ΔηΔτ,for(t,s)∈Ω.
It follows from (3.3) and (3.7) that
(3.8)|u(t,s)|γ≤a(t,s)+∫t0t∫s0sf(τ,s)|u(τ,η)|δ+g(τ,η)|u(τ,η)|αΔηΔτ,
for (t,s)∈Ω. Applying Theorem 2.6 on (3.8) with λ=1 and b(t,s)=1, we obtain (3.4).Example 3.2.
Consider the equation(3.9)uγ(t,s)=H(t,s,u(t,s))+h(t,s),(t,s)∈Ω≡[t0,∞)T×[s0,∞)T,
where γ≥1 is a constant and H and hare rd-continuous on Ω,a:[t0,∞)𝕋→ℝ+ and b:[t0,∞)𝕋→ℝ+ are rd-continuous functions. Assume that
(3.10)|H(t,s,u)|≤f(t,s)|u(t,s)|δ+g(t,s)|u(t,s)|α,
where f(t,s) and g(t,s) are nonnegative rd-continuous functions for (t,s)∈Ω and α,δ<γ. If u(t,s) is a solution of (3.1)-(3.2), then u(t,s) satisfies
(3.11)|u(t,s)|≤|h(t,s)|1/γ+1γ|h(t,s)|(1/γ)-1B1/γ(t,s),∀(t,s)∈Ω,
where
(3.12)B(t,s):=F*(t,s)+(1-λ)∫t0t∫s0t(G*(τ,η))1/(1-λ)ΔηΔτ+eλ(s-s0)(t,t0),F*(t,s):=∫t0t∫s0s[fλ(τ,η)aλ(δ/γ)(τ,η)+gλ(τ,η)aλ(α/γ)(τ,η)]ΔηΔτ,G*(t,s):=(fλ(t,s)[δγa(δ/γ)-1(t,s)]λ+gλ(t,s)[αγa(α/γ)-1(t,s)]λ).
In fact, the solution of (3.9) satisfies
(3.13)|u(t,s)|γ≤|h(t,s)|+∫t0t∫s0s|H(τ,η,u(τ,η))|ΔηΔτ,for(t,s)∈Ω.
It follows from (3.10) and (3.13) that
(3.14)|u(t,s)|γ≤|h(t,s)|+∫t0t∫s0s[f(τ,s)|u(τ,η)|δ+g(τ,η)|u(τ,η)|α]λΔηΔτ,
for (t,s)∈Ω. Applying Theorem 2.7 on (3.14) with b(t,s)=1, we obtain (3.11).Example 3.3.
Assume that𝕋=ℝ and consider the partial differential equation
(3.15)∂∂s(uγ-1(t,s)∂∂tu(t,s))+H(t,s,u(t,s))=h(t,s),(t,s)∈Ω*,
where Ω*=[0,∞)×[0,∞), with initial boundary conditions
(3.16)u(t,0)=a(t)>0,u(0,s)=b(s)>0,a(0)=b(0)=0.
Assume that γ≥1 is a constant and H:[0,∞)×[0,∞)×ℝ→ℝ and h:[0,∞)ℝ×[0,∞)ℝ→ℝ, a:ℝ→ℝ+ and b:ℝ→ℝ+ are continuous functions. Also, we assume that
(3.17)|H(t,s,u)|≤f(t,s)|u(t,s)|δ+g(t,s)|u(t,s)|α,
where f(t,s) and g(t,s) are nonnegative continuous functions for (t,s)∈Ω* and α,δ<γ. If u(t,s) is a solution of (3.1)-(3.2), then u(t,s) satisfies
(3.18)|u(t,s)|≤a1/γ(t,s)+γ1/γB1/γ(t,s),∀(t,s)∈Ω*,
where
(3.19)a(t,s)=aγ(t)+bγ(s)+γ∫t0t∫s0s|h(τ,η)|ΔηΔτ,B(t,s):=H0(t,s)+eβ(s-s0)(t,t0),β=[αγ+δγ],H0(t,s)=∫t0t∫s0s[f(τ,η)aδ/γ(τ,η)]ΔηΔτ+∫t0t∫s0s[g(τ,η)aα/γ(τ,η)]ΔηΔτ+(γ-δ)γ∫t0t∫s0s(f(τ,s))γ/(γ-δ)ΔηΔτ+(γ-α)γ∫t0t∫s0s(g(τ,η))γ/(γ-α)ΔηΔτ.
In fact, the solution formula of (3.15)-(3.16), after integration twice, is given by
(3.20)|u(t,s)|γ-aγ(t)-bγ(t)+γ∫t0t∫s0sH(τ,η,u(τ,η))ΔηΔτ=γ∫t0t∫s0sh(τ,η)ΔηΔτ,for(t,s)∈Ω*.
Therefore,
(3.21)|u(t,s)|γ≤a(t,s)+γ∫t0t∫s0s|H(τ,η,u(τ,η))|ΔηΔτ,for(t,s)∈Ω*.
It follows from (3.17) and (3.21) that
(3.22)|u(t,s)|γ≤a(t,s)+γ∫t0t∫s0s[f(τ,s)|u(τ,η)|δ+g(τ,η)|u(τ,η)|α]ΔηΔτ,
for (t,s)∈Ω*. Applying Theorem 2.6 on (3.22) with λ=1 and b(t,s)=γ, we obtain (3.18).



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