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29) on parameters. 6. 35) 0 x 0 y 0 in which M1 , M2 are nonnegative constants. 29) respectively. 5). 36) 20 Proof. Multidimensional Integral Equations and Inequalities Let w(x, y) = |u1 (x, y) − u2 (x, y)|, (x, y) ∈ E. 29) and hypotheses, we have x w(x, y) 0 x + 0 x y + 0 0 x y + 0 0 |g(x, y, ξ , u2 (ξ , y), μ ) − g(x, y, ξ , u2 (ξ , y), μ0 )| d ξ |h(x, y, σ , τ , u1 (σ , τ ), μ ) − h(x, y, σ , τ , u2 (σ , τ ), μ )| d τ d σ |h(x, y, σ , τ , u2 (σ , τ ), μ ) − h(x, y, σ , τ , u2 (σ , τ ), μ0 )| d τ d σ x 0 x y + 0 0 |g(x, y, ξ , u1 (ξ , y), μ ) − g(x, y, ξ , u2 (ξ , y), μ )|d ξ q(x, y, ξ )w(ξ , y) d ξ + r(x, y, σ , τ )w(σ , τ ) d τ d σ + x 0 p1 (x, y, ξ )|μ − μ0 |d ξ x 0 y 0 p2 (x, y, σ , τ )|μ − μ0 |d τ d σ (M1 + M2 )|μ − μ0 | x + 0 x q(x, y, ξ )w(ξ , y) d ξ + 0 y 0 r(x, y, σ , τ )w(σ , τ ) d τ d σ .

4) and fk ∈ C(E0 × R3 , R), αk ∈ C(Ia , R), βk ∈ C(Ib , R) for k = 1, 2, . .. 2. 3. 7) and there exist constants εk 0, δk 0 (k = 1, 2, . 22). 1. If uk (x, y) (k = 1, 2, . 2) on E0 , then uk (x, y) → u(x, y) as k → ∞. Proof. For k = 1, 2, . 2 hold. 25) 0 for (x, y) ∈ E0 . 25). 3. 2). 28) f ∈ C(E × Rn × Rn × R , R ), g ∈ C(E2 × Rn × Rn , Rn ), σ , τ ∈ C(R+ , Rn ). Obviously, M0 = x y 0 0 g(x, y, m, n, 0, 0) dn dm. 27). For z, D2 D1 z ∈ C(E, Rn ), we denote by |z(x, y)|0 = |z(x, y)| + |D2 D1 z(x, y)|.

1 with suitable modifications. Here, we omit the details. 1. 1) in S. 1). 2. 2. 9) where b, f , c, g ∈ C(E, R+ ). 15) respectively. 1) has at most one solution on E in Rn . Proof. 1). Then, we have u1 (x, y) − u2 (x, y) = + x y {F(x, y, s,t, u1 (s,t)) − F(x, y, s,t, u2 (s,t))}dt ds 0 0 ∞ ∞ 0 0 {G(x, y, s,t, u1 (s,t)) − G(x, y, s,t, u2 (s,t))}dt ds. 10) and using hypotheses, we have |u1 (x, y) − u2 (x, y)| x y b(x, y) + c(x, y) 0 0 ∞ ∞ 0 0 f (s,t)|u1 (s,t) − u2 (s,t)|dt ds g(s,t)|u1 (s,t) − u2 (s,t)|dt ds.

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