IJDE International Journal of Differential Equations 1687-9651 1687-9643 Hindawi Publishing Corporation 949860 10.1155/2014/949860 949860 Research Article On Inequality Applicable to Partial Dynamic Equations http://orcid.org/0000-0003-3763-4878 Pachpatte Deepak B. Wang Peiguang Department of Mathematics Dr. B.A.M. University Aurangabad Maharashtra 431004 India bamu.net 2014 1542014 2014 12 02 2014 26 03 2014 15 4 2014 2014 Copyright © 2014 Deepak B. Pachpatte. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

The main objective of the paper is to study new integral inequality on time scales which is used for the study of some partial dynamic equations. Some applications of our results are also given.

1. Introduction

During past few decades many authors have established various dynamic inequalities useful in the development of differential and integral equations. Mathematical inequalities on time scales play an important role in the theory of dynamic equations. The study of time scale was initiated by Hilger  in 1990 in his Ph.D. thesis which unifies continuous and discrete calculus. Since then, many authors have studied various properties of dynamic equations on time scales .

In what follows, let denotes the set of real numbers and let 𝕋 denote the arbitrary time scales. Let +=[0,), 𝕋1=[0,a], and 𝕋2=[0,b] be subsets of and Ω=𝕋1×𝕋2. Let Crd denote the set of rd-continuous function. The partial delta derivative of v(x,y) for (x,y)Ω with respect to x, y, and xy is denoted by vΔ1(x,y), vΔ2(x,y), and vΔ1Δ2(x,y)=vΔ2Δ1(x,y). We assume here understanding of time scales calculus and notations. Further information about time scales calculus can be found in [1, 5, 10].

We require the following lemmas given in [5, 6].

Lemma 1 (see [<xref ref-type="bibr" rid="B5">5</xref>], Theorem 2.6).

Let uCrd(𝕋,+), a+, and (1)uΔ(t)a(t)u(t), for all t𝕋k; then (2)u(t)u(t0)ea(t,t0), for all t𝕋k.

Lemma 2 (see [<xref ref-type="bibr" rid="B6">6</xref>], Lemma 2.1).

Let u,a,bCrd(Ω,+) and a(x,y) is nondecreasing in (x,y)Ω and (3)u(x,y)a(x,y)+x0xy0yb(s,t)u(s,t)ΔtΔs, for (x,y)Ω; then (4)u(x,y)a(x,y)eQ(x,y)(x,x0), where (5)Q(x,y)=y0yb(x,t)Δt, for (x,y)Ω.

2. Main Results

Now in this section we give our main results.

Theorem 3.

Let u(x,y), w(x,y), p(x,y), q(x,y), r(x,y)Crd(Ω,+) and suppose that (6)u(x,y)c+x0xw(s,y)u(s,y)Δs+s0sy0yp(s,t)×[u(s,t)+s0st0tq(ξ,τ)u(ξ,τ)ΔτΔξSDSS+a0ab0br(ξ,τ)u(ξ,τ)ΔτΔξ]ΔtΔs, for (x,y)Ω, where c0 is a constant. If (7)g=a0ab0br(ξ,τ)A(ξ,τ)eH(x,y)(ξ,τ0)ΔτΔξ, where (8)H(x,y)=τ0τA(x,t)[p(x,t)+q(x,t)]Δt,(9)A(x,y)=ew(x,y)(x,x0), for (x,y)Ω, then (10)u(x,y)c1-gA(x,y)eH(x,y)(x0,x), for (x,y)Ω.

Proof.

Define a function z(x,y) by (11)z(x,y)=c+x0xy0yp(s,t)×[u(s,t)×s0st0tq(ξ,τ)u(ξ,τ)ΔτΔξfdfdf+s0st0tr(ξ,τ)u(ξ,τ)ΔτΔξ]ΔtΔs. Then (6) is (12)u(x,y)z(x,y)+x0xw(s,y)u(s,y)Δs. It is easy to see that z(x,y) is nonnegative, rd-continuous, and nondecreasing function for (x,y)Ω. Treating y fixed and using Lemma 1 we get (13)u(x,y)A(x,y)z(x,y), for (x,y)Ω, where A(x,y) is defined by (9). From (11), (12), and the fact that A(x,y)1, we have (14)z(x,y)c+x0xy0yp(s,t)gfdgfgfdgfd×[z(ξ,τ)ΔτΔξs0st0tq(ξ,τ)A(ξ,τ)A(s,t)z(s,t)gfgfgfffffffffffff+s0st0tq(ξ,τ)A(ξ,τ)gfgfdgfffffffffffff×z(ξ,τ)ΔτΔξgfgfgfffffffffffff+a0ab0br(ξ,τ)A(ξ,τ)fdsfdsfdfdfdfdddddd×z(ξ,τ)ΔτΔξs0st0tq(ξ,τ)A(ξ,τ)]ΔtΔsc+x0xy0yp(s,t)A(s,t)gfdgfdgfdgdf×[z(s,t)+s0st0tq(ξ,τ)A(ξ,τ)fdfdfdfdfdfdfdfdfdfdfdfddfdfdf×z(ξ,τ)ΔτΔξdfsdfdfdfdddfdd+a0ab0bh(ξ,τ)A(ξ,τ)fdfdfdfdfdfdfdfdfdf×z(ξ,τ)ΔτΔξz(s,t)+s0st0tq(ξ,τ)A(ξ,τ)]ΔtΔs. Define a function v(x,y) by right hand side of (14). Then v(0,y)=v(x,0)=c, z(x,y)v(x,y). One has (15)vΔ2Δ1=p(x,y)A(x,y)×[z(x,y)+x0xy0yq(ξ,τ)A(ξ,τ)z(ξ,τ)ΔτΔξfdfds+a0ab0br(ξ,τ)A(ξ,τ)z(ξ,τ)ΔτΔξ]p(x,y)A(x,y)×[v(x,y)+x0xy0yq(ξ,τ)A(ξ,τ)v(ξ,τ)ΔτΔξfdsfd+a0ab0br(ξ,τ)A(ξ,τ)v(ξ,τ)ΔτΔξ]. Define a function f(x,y) by (16)f(x,y)=v(x,y)+x0xy0yq(ξ,τ)A(ξ,τ)v(ξ,τ)ΔτΔξ+a0ab0br(ξ,τ)A(ξ,τ)v(ξ,τ)ΔτΔξ; then v(x,y)f(x,y), vΔ2Δ1(x,y)p(x,y)A(x,y)f(x,y), (17)f(x0,y)=f(x,y0)=c+a0ab0br(ξ,τ)A(ξ,τ)v(ξ,τ)ΔτΔξ=M(say),(18)fΔ2Δ1(x,y)=vΔ2Δ1(x,y)+q(x,y)A(x,y)v(x,y)p(x,y)A(x,y)f(x,y)+q(x,y)A(x,y)f(x,y)=A(x,y)[p(x,y)+q(x,y)]f(x,y). By keeping x fixed in (18), taking y=t and delta integrating with second variable from y0 to y. Using the fact that fΔ1(x,y0)=0 and f(x,y) is nondecreasing in (x,y)Ω, we have (19)fΔ1(x,y)y0yA(x,t)[p(x,t)+q(x,t)]f(x,t)Δtf(x,y)y0yA(x,t)[p(x,t)+q(x,t)]Δt. Let (20)Q¯(x,y)=y0yA(x,t)[p(x,t)+q(x,t)]Δt; then (20) gives (21)fΔ1(x,y)f(x,y)Q¯(x,y). Now treating y fixed in (21) and applying Lemma 1, we have (22)f(x,y)MeQ¯(x,y)(x,x0). From (18), (22), and (7), it is easy to see that (23)Mc1-g. Using (23) in (22) and the fact that z(x,y)v(x,y) and z(x,y)A(x,y)v(x,y) we get the inequality in (10).

This completes the proof.

3. Applications

Now we give some application of theorem to study properties of solutions of initial value problem: (24)uΔ2Δ1(x,y)=(w(x,y)u(x,y))Δ2+G(a0ab0bh(x,y,ξ,τ,u(ξ,τ))ΔτΔξx,y,u(x,y),FDSFDDa0ab0bh(x,y,ξ,τ,u(ξ,τ))ΔτΔξ),u(x,y0)=α(x),u(x0,y)=β(y),α(x0)=β(y0)=0, where αCrd(𝕋1,), βCrd(𝕋2,) for 0ξx, 0τy, hCrd(Ω2×,), GCrd(Ω×2,), pCrd(Ω,) is delta differentiable with respect to y.

We observe that (24) is equivalent to (25)u(x,y)=F(x,y)+x0xw(s,y)u(s,y)Δs+x0xy0yG(a0ab0bh(s,t,ξ,τ,u(ξ,τ))s,t,u(s,t),fdfdfdddffffffa0ab0bh(s,t,ξ,τ,u(ξ,τ))ΔτΔξa0ab0bh(s,t,ξ,τ,u(ξ,τ)))ΔtΔs, where (26)F(x,y)=α(x)+β(y)-x0xp(s,y0)α(s)Δs. The following theorem deals with estimate on solution (24).

Theorem 4.

Suppose (27)|F(x,y)|c,|h(x,y,s,t,u)|k(x,y)r(s,t)|u|,|G(x,y,u,u¯)|p(x,y)(|u|+|u¯|), where p, r, c which are as in Theorem 3 and k(x,y) is rd-continuous function defined on Ω such that k(x,y)1. Let (28)g0=a0ab0br(ξ,τ)A(ξ,τ)eH¯(x,y)(ξ,τ0)ΔτΔξ, where (29)H¯(x,y)=t0τA(x,t)p(x,t)k(x,t)Δt,(30)A¯(x,y)=ew(r,y)(x,x0), for (x,y)Ω. If u(x,y) is any solution of (24), then (31)u(x,y)c1-g0A¯(x,y)eH¯(x,y)(x,x0), where (x,y)Ω.

Proof.

The solution u(x,y) of (24) satisfies (25). Using (27) in (25) we have (32)|u(x,y)|c+x0x|w(s,y)||u(s,y)|Δs+x0xy0yp(s,t)fdfdfdfdfd×[a0ab0bk(s,t)r(ξ,τ)|u(ξ,τ)|ΔτΔξ|u(s,t)|fddddddddddff+a0ab0bk(s,t)r(ξ,τ)fddddddddddddddddd×|u(ξ,τ)|ΔτΔξa0ab0bk(s,t)r(ξ,τ)]ΔtΔsc+x0x|w(s,y)||u(s,y)|Δs+x0xy0yp(s,t)k(s,t)fdfddddddd×[+a0ab0br(ξ,τ)|u(ξ,τ)|ΔτΔξ|u(s,t)|fdfdddddddddd+a0ab0br(ξ,τ)fdfdddddfdfdfdddffddfdf×|u(ξ,τ)|ΔτΔξa0ab0br(ξ,τ)]ΔtΔs. Now an application of Theorem 3 (with g=0) to (32) yields (30).

This completes the proof.

Now we establish the uniqueness of solutions of (24).

Theorem 5.

Suppose that (33)|h(x,y,s,t,u)-h(x,y,s,t,u¯)|k(x,y)r(s,t)|u-u¯|,|G(x,y,u,u¯)-G(x,y,v,v¯)|p(x,y)(|u-v|+|u¯-v¯|), where k, p, and r are as in Theorem 4. Let g0 and A¯(x,y) be as in (28) and (30). Then (24) has at most one solution on G.

Proof.

Let u(x,y) and v(x,y) be two solutions of (24) on Ω; then we have (34)u(x,y)-v(x,y)=x0xw(x,y){u(s,y)-v(s,y)}Δs+x0xy0y{G(a0ab0bh(s,t,ξ,τ,u(ξ,τ))ΔτΔξs,t,u(s,t),fdfdfdfdfdfdfdDa0ab0bh(s,t,ξ,τ,u(ξ,τ))ΔτΔξ)-G(a0ab0bh(s,t,ξ,τ,v(ξ,τ))ΔτΔξs,t,v(s,t),fsdfdddda0ab0bh(s,t,ξ,τ,v(ξ,τ))ΔτΔξ)}ΔtΔs. From (34) and (33) we obtain (35)|u(x,y)-v(x,y)|x0x|w(s,y)||u(s,y)-v(s,y)|Δs+x0xy0yp(s,t)fdfdfdfddd×(b0br(ξ,τ)|u(ξ,τ)-v(ξ,τ)|ΔτΔξ|u(s,t)-v(s,t)|+k(s,t)fgdfgdfgfgffff×a0ab0br(ξ,τ)gfgfgfgfgfgffffffffff×|u(ξ,τ)-v(ξ,τ)|ΔτΔξa0ab0br(ξ,τ)|u(ξ,τ)-v(ξ,τ)|)ΔtΔs. Applying Theorem 3 (with c=0, g=0) yields (36)|u(x,y)-v(x,y)|0. Therefore u(x,y)v(x,y); there is at most one solution of (24) in Ω.

Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

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