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,ℝ+) and suppose that
(6)u(x,y)≤c+∫x0xw(s,y)u(s,y)Δs+∫s0s∫y0yp(s,t) ×[u(s,t)+∫s0s∫t0tq(ξ,τ)u(ξ,τ)Δτ ΔξSDSS+∫a0a∫b0br(ξ,τ)u(ξ,τ)Δτ Δξ]Δt Δs,
for (x,y)∈Ω, where c≥0 is a constant. If
(7)g=∫a0a∫b0br(ξ,τ)A(ξ,τ)eH(x,y)(ξ,τ0)Δτ Δξ,
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+∫x0x∫y0yp(s,t) ×[u(s,t)×∫s0s∫t0tq(ξ,τ)u(ξ,τ)Δτ Δξfdfdf+∫s0s∫t0tr(ξ,τ)u(ξ,τ)Δτ Δξ]Δt Δs.
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+∫x0x∫y0yp(s,t)gfdgfgfdgfd×[z(ξ,τ)ΔτΔξ∫s0s∫t0tq(ξ,τ)A(ξ,τ)A(s,t)z(s,t)gfgfgfffffffffffff+∫s0s∫t0tq(ξ,τ)A(ξ,τ)gfgfdgfffffffffffff×z(ξ,τ)Δτ Δξgfgfgfffffffffffff+∫a0a∫b0br(ξ,τ)A(ξ,τ)fdsfdsfdfdfdfdddddd×z(ξ,τ)Δτ Δξ∫s0s∫t0tq(ξ,τ)A(ξ,τ)]Δt Δs≤c+∫x0x∫y0yp(s,t)A(s,t)gfdgfdgfdgdf×[z(s,t)+∫s0s∫t0tq(ξ,τ)A(ξ,τ)fdfdfdfdfdfdfdfdfdfdfdfddfdfdf×z(ξ,τ)Δτ Δξdfsdfdfdfdddfdd+∫a0a∫b0bh(ξ,τ)A(ξ,τ)fdfdfdfdfdfdfdfdfdf×z(ξ,τ)Δτ Δξz(s,t)+∫s0s∫t0tq(ξ,τ)A(ξ,τ)]Δt Δs.
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)+∫x0x∫y0yq(ξ,τ)A(ξ,τ)z(ξ,τ)Δτ Δξfdfds+∫a0a∫b0br(ξ,τ)A(ξ,τ)z(ξ,τ)Δτ Δξ]≤p(x,y)A(x,y) ×[v(x,y)+∫x0x∫y0yq(ξ,τ)A(ξ,τ)v(ξ,τ)Δτ Δξfdsfd+∫a0a∫b0b r(ξ,τ)A(ξ,τ)v(ξ,τ)Δτ Δξ].
Define a function f(x,y) by
(16)f(x,y)=v(x,y)+∫x0x∫y0yq(ξ,τ)A(ξ,τ)v(ξ,τ)Δτ Δξ +∫a0a∫b0br(ξ,τ)A(ξ,τ)v(ξ,τ)Δτ Δξ;
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+∫a0a∫b0br(ξ,τ)A(ξ,τ)v(ξ,τ)Δτ Δξ=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)Δt≤f(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)M≤c1-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.