Fixed Point Theorems on Nonlinear Binary Operator Equations with Applications

Definition 1. Let A : D × D → E be a binary operator. A is said to be L-ordering symmetric contraction operator if there exists a bounded linear and positive operator L : E → E, where spectral radius r(L) < 1 such that A(y, x) − A(x, y) ≤ L(y − x) for any x, y ∈ D,  x ≤ y, where L is called a contraction operator of A.

Theorem 2. Let A : D × D → E be L-ordering symmetric contraction operator, and there exists a α ∈ [0,1) such that

If condition (H₁) u ≤ A(u, v), A(v, u) ≤ v − α(v − u) or (H₂) u + α(v − u) ≤ A(u, v), A(v, u) ≤ v holds, then the following statements hold.

(C₁) A(x, x) = x has a unique solution x^* ∈ D, and for any coupled solutions x, y ∈ D,  x = y = x^*.

(C₂) For any x₀, y₀ ∈ D, we construct symmetric iterative sequences:

Then x_n → x^*,  y_n → x^*  (n → ∞), and for any β ∈ (r(L), 1), there exists a natural number m; and if n ≥ m, we get error estimates for iterative sequences (2):

Proof. Set B(x, y) = (1/(α + 1))[A(x, y) + αx], and if condition (H₁) or (H₂) holds, then it is obvious that

By (1), we easily prove that B : D × D → E is mixed monotone operator, and for any x, y ∈ D,  u ≤ x ≤ y ≤ v, where H = (1/(α + 1))(L + αI) is a bounded linear and positive operator and  I, is identical operator.

By the mathematical induction, we easily prove that

where Bⁿ(x, y) = B(Bⁿ⁻¹(x, y), Bⁿ⁻¹(y, x)),  x, y ∈ D,  n ≥ 2.

By the character of normal cone P, it is shown that

For any β ∈ (r(L), 1), since lim⁡_n→∞| | Hⁿ | |^1/n = r(H)≤(α + r(L))/(α + 1)<(α + β)/(α + 1) < 1, there exists a natural number m, and if n ≥ m, we have | | Hⁿ|| < ((α+β)/(α+1))ⁿ, and N| | H^m|| < 1. Considering mixed monotone operator B^m and constant N| | H^m||, B^m(x, x) = x has a unique solution x^* and for any coupled solution x, y ∈ D, such that x = y = x^* by Theorem 3 in [3].

From B^m(B(x^*, x^*), B(x^*, x^*)) = B(B^m(x^*, x^*), B^m(x^*, x^*)) = B(x^*, x^*), and the uniqueness of solution with B^m(x, x) = x, then we have B(x^*, x^*) = x^* and A(x^*, x^*) = x^*.

We take note of that A(x, x) = x and B(x, x) = x have the same coupled solution; therefore, a coupled solution for B(x, x) = x must be a coupled solution for B^m(x, x) = x; consequently, (C₁) has been proved.

Considering iterative sequence (2), we construct iterative sequences:

where u₀ = u, v₀ = v, it is obvious that by the mathematical induction and characterization of mixed monotone of B; then Hence, Moreover, if n ≥ m, we get Consequently, x_n → x^*, y_n → x^*  (n → ∞).

Remark 3. When α = 0, Theorem 1 in [4] is a special case of this paper Theorem 2 under condition (H₁) or (H₂).

Corollary 4. Let A : D × D → E be L-ordering symmetric contraction operator; if there exists a α ∈ [0,1) such that A satisfies condition of Theorem 2, the following statement holds.

(C₃) For any β ∈ (r(L), 1) and α + β < 1, we make iterative sequences:

or where u₀ = u,  v₀ = v.

Proof. By the character of mixed monotone of A, then (1) and (C₁), (C₂) [in (1), (C₂) where α = 0] hold.

In the following, we will prove (C₃).

Consider iterative sequence (13); since u ≤ x^* ≤ v, we get

By the mathematical induction, we easily prove u_n ≤ x^* ≤ v_n, n ≥ 1, hence

It is clear that

For any β ∈ (r(L), 1), α + β < 1, since

there exists a natural number m, if n ≥ m, such that

Moreover,

Consequently, u_n → x^*, v_n → x^*, (n → ∞).

Similarly, we can prove (14).

Theorem 5. Let A : D × D → E be a L-ordering symmetric contraction operator; if there exists a α ∈ [0,1) such that (1 − α)u ≤ A(u, v), A(v, u)≤(1 − α)v, then the following statements hold.

(C₄) Operator equation A(x, x) = (1 − α)x has a unique solution x^* ∈ D, and for any coupled solutions x, y ∈ D, x = y = x^*.

(C₅) For any x₀, y₀, w₀, z₀ ∈ D, we make symmetric iterative sequences

Then x_n → x^*, y_n → x^*, w_n → x^*, z_n → x^*  (n → ∞), and for any β ∈ (r(L), 1),   α + β < 1, there exists a natural number m, and if n ≥ m, then we have error estimates for iterative sequences (22) and (23), respectively,

Proof. Set B(x, y) = (1/(1 − α))A(x, y) or C(x, y) = A(x, y) + αx; we can prove that this theorem imitates proof of Theorem 2.

Theorem 6. Let A : D × D → E be L-ordering symmetric contraction operator; if there exists a α ∈ [0,1) such that u + αv ≤ A(u, v),  A(v, u) ≤ v + αu, then the following statements hold.

(C₆) Equation A(x, x) = (1 + α)x has a unique solution x^* ∈ D, and for any coupled solutions x, y ∈ Dx = y = x^*.

(C₇) For any x₀, y₀ ∈ D, we make symmetric iterative sequence:

Then x_n → x^*, y_n → x^*  (n → ∞); moreover, β ∈ (r(L), 1), and there exists natural number m, and if n ≥ m, then we have error estimates for iterative sequence (25):

(C₈) For any β ∈ (r(L), 1)(α + β < 1), w₀, z₀ ∈ D, we make symmetry iterative sequence w_n = A(w_n−1, z_n−1) − αz_n−1, z_n = A(z_n−1, w_n−1) − αw_n−1,  n ≥ 1; then w_n → x^*,  z_n → x^*  (n → ∞), and there exists a natural number m, and if n ≥ m, we have error estimates for iterative sequence (24).

Remark 7. When α = 0, Corollary 2 in [4] is a special case of this paper Theorems 2–6.

Remark 8. The contraction constant of operator in [5] is expand into the contraction operator of this paper.

Remark 9. Operator A of this paper does not need character of mixed monotone as operator in [6].

Theorem 10. Let a(t), b(t) be nonnegative continuous function in [0,1],  p = max⁡_t∈[0,1]a(t), q = max⁡_t∈[0,1]b(t). If p < 1, mp + q < 2, then boundary value problem (23) has a unique solution x^*(t) such that 0 ≤ x^*(t) ≤ 1  (t ∈ [0,1]). Moreover, for any initial function x₀(t), y₀(t), such that

we make iterative sequence: Then x_n(t) and y_n(t) are all uniformly converge to x^*(t) on [0,1], and we have error estimates:

Proof. Let E = C[0,1],  P = {x ∈ E∣x(t) ≥ 0, t ∈ [0,1]},  ∥x∥ = max⁡_t∈[0,1]|x(t)| denote norm of; then E has become Banach space, P is normal cone of E, and its normal constant N = 1. It is obvious that integral Equation (24) transforms to operator equation A(x, x) = x, where

Set u = u(t) ≡ 0,  v = v(t) ≡ 1; then D = [0,1] denote ordering interval of E, A : D × D → E is mixed monotone operator, and 0 ≤ A(0,1), A(1,0)≤(1 + p)/2 < 1.

Set

Then L : E → E is bounded linear operator, its spectral radius r(L)≤(mp + q)/2 < 1, and for any x, y ∈ E,   0 ≤ x(t) ≤ y(t) ≤ 1 such that 0 ≤ A(y, x)(t) − A(x, y)(t) ≤ L(y − x)(t), A is L-ordering symmetric contraction operator, by Theorem 2 (where α = 0); then Theorem 10 has been proved.