2. Main Results
Theorem 2.
Let A:D×D→E be L-ordering symmetric contraction operator, and there exists a α∈[0,1) such that
(1)A(x2,y2)-A(x1,y1)≥-α(x2-x1), u≤x1≤x2≤v, u≤y2≤y1≤v.
If condition (H_{1}) u≤A(u,v),A(v,u)≤v-α(v-u) or (H_{2}) u+α(v-u)≤A(u,v),A(v,u)≤v holds, then the following statements hold.

(C_{1}) A(x,x)=x has a unique solution x*∈D, and for any coupled solutions x,y∈D, x=y=x*.

(C_{2}) For any x0,y0∈D, we construct symmetric iterative sequences:
(2)xn=1α+1[A(xn-1,yn-1)+αxn-1],yn=1α+1[A(yn-1,xn-1)+αyn-1], hhhn=1,2,3,….
Then xn→x*, yn→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):
(3)∥xn(yn)-x*∥≤2N(α+βα+1)n∥u-v∥.

Proof.
Set B(x,y)=(1/(α+1))[A(x,y)+αx], and if condition (H_{1}) or (H_{2}) holds, then it is obvious that
(4)u≤B(u,v), B(v,u)≤v.
By (1), we easily prove that B:D×D→E is mixed monotone operator, and for any x,y∈D, u≤x≤y≤v,
(5)θ≤B(y,x)-B(x,y)≤H(y-x),
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
(6)θ≤Bn(y,x)-Bn(x,y)≤Hn(y-x), u≤x≤y≤v,
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
(7)∥Bn(y,x)-Bn(x,y)∥≤N∥Hn∥∥y-x∥, u≤x≤y≤v.

For any β∈(r(L),1), since limn→∞||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||Hm||<1. Considering mixed monotone operator Bm and constant N||Hm||, Bm(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 Bm(B(x*,x*),B(x*,x*))=B(Bm(x*,x*),Bm(x*,x*))=B(x*,x*), and the uniqueness of solution with Bm(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 Bm(x,x)=x; consequently, (C_{1}) has been proved.

Considering iterative sequence (2), we construct iterative sequences:
(8)un=B(un-1,vn-1), vn=B(vn-1,un-1),
where u0=u, v0=v, it is obvious that
(9)xn=B(xn-1,yn-1), yn=B(yn-1,xn-1),θ≤vn-un≤Hn(v-u),
by the mathematical induction and characterization of mixed monotone of B; then
(10)un≤x*≤vn, un≤xn≤vn, un≤yn≤vn.
Hence,
(11)∥xn(yn)-un∥≤N∥vn-un∥,∥x*-un∥≤N∥vn-un∥, n=1,2,3,….
Moreover, if n≥m, we get
(12)∥xn(yn)-x*∥≤2N∥vn-un∥ ≤2N∥Hn∥∥v-u∥≤2N(α+βα+1)n∥u-v∥.
Consequently, xn→x*,yn→x* (n→∞).

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

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.

(C_{3}) For any β∈(r(L),1) and α+β<1, we make iterative sequences:
(13)un=A(un-1,vn-1),vn=A(vn-1,un-1)+α(vn-1-un-1), hhhn=1,2,3,…,
or
(14)un=A(un-1,vn-1)-α(vn-1-un-1),vn=A(vn-1,un-1), hhhn=1,2,3,…,
where u0=u, v0=v.

Thus, un→x*,vn→x* (n→∞), and there exists a natural number m, and if n≥m, we have error estimates for iterative sequences (13) or (14):
(15)∥un(vn)-x*∥≤N(α+β)n∥u-v∥.

Proof.
By the character of mixed monotone of A, then (1) and (C_{1}), (C_{2}) [in (1), (C_{2}) where α=0] hold.

In the following, we will prove (C_{3}).

Consider iterative sequence (13); since u≤x*≤v, we get
(16)u1=A(u,v)≤A(x*,x*)=x*≤A(v,u)=v1-α(v-u)≤v1.
By the mathematical induction, we easily prove un≤x*≤vn, n≥1, hence
(17)θ≤x*-un≤vn-un, θ≤vn-x*≤vn-un.

It is clear that
(18)θ≤vn-un≤(L+αI)(vn-1-un-1)=(L+αI)n(v-u), n≥1.

For any β∈(r(L),1), α+β<1, since
(19)limn→∞∥(L+αI)n∥1/n=r(L+αI)≤r(L)+α<α+β<1,
there exists a natural number m, if n≥m, such that
(20)∥(L+αI)n∥<(α+β)n.

Moreover,
(21)∥un(vn)-x*∥≤N∥(L+αI)n∥∥u-v∥ ≤N(α+β)n∥u-v∥, (n≥m).
Consequently, un→x*, vn→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.

(C_{4}) 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*.

(C_{5}) For any x0,y0,w0,z0∈D, we make symmetric iterative sequences
(22)xn=11-αA(xn-1,yn-1),yn=11-αA(yn-1,xn-1), hhn=1,2,3,…,(23)wn=A(wn-1,zn-1)+αwn-1, zn=A(zn-1,wn-1)+αzn-1, hhln=1,2,3,….
Then xn→x*, yn→x*, wn→x*, zn→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,
(24)∥xn(yn)-x*∥≤2N(β1-α)n∥u-v∥,∥wn(zn)-x*∥≤2N(α+β)n∥u-v∥.

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.

Similarly, we can prove the following theorems.

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.

(C_{6}) Equation A(x,x)=(1+α)x has a unique solution x*∈D, and for any coupled solutions x,y∈Dx=y=x*.

(C_{7}) For any x0,y0∈D, we make symmetric iterative sequence:
(25)xn=11+αA(xn-1,yn-1),yn=11+αA(yn-1,xn-1), hhn=1,2,3,….
Then xn→x*, yn→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):
(26)∥xn(yn)-x*∥≤2N(βα+1)n∥u-v∥,

(C_{8}) For any β∈(r(L),1)(α+β<1),w0,z0∈D, we make symmetry iterative sequence wn=A(wn-1,zn-1)-αzn-1,zn=A(zn-1,wn-1)-αwn-1, n≥1; then wn→x*, zn→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].

3. Application
We consider that two-point boundary value problem for two-degree super linear ordinary differential equations:
(27)x′′+a(t)xm+11+b(t)x=0, t∈[0,1], (m≥2)x(0)=x′(1)=0.

Let k(t,s) be Green function with boundary value problem (23); that is,
(28)k(t,s)=min{t,s}={t,t≤ss,s<t.

Then the solution with boundary value problem (23) and solution for nonlinear integral equation with type of Hammerstein(29)x(t)=∫01k(t,s){a(s)[x(s)]m+11+b(s)x(s)}ds
are equivalent, where maxt∈[0,1]∫01k(t,s)ds=1/2.

Theorem 10.
Let a(t), b(t) be nonnegative continuous function in [0,1], p=maxt∈[0,1]a(t), q=maxt∈[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
(30)0≤x0(t)≤1, 0≤y0(t)≤1 (t∈[0,1]),
we make iterative sequence:
(31)xn(t)=∫01k(t,s){a(s)[xn-1(s)]m+11+b(s)yn-1(s)}ds,yn(t)=∫01k(t,s){a(s)[yn-1(s)]m+11+b(s)xn-1(s)}ds, hhhhhhhhhlln=1,2,3,….
Then xn(t) and yn(t) are all uniformly converge to x*(t) on [0,1], and we have error estimates:
(32)|xn(t)(yn(t))-x*(t)|≤2(mp+q2)n, hlt∈[0,1], n=1,2,3,….

Proof.
Let E=C[0,1], P={x∈E∣x(t)≥0,t∈[0,1]}, ∥x∥=maxt∈[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
(33)A(x,y)(t) =∫01k(t,s){a(s)[x(s)]m+11+b(s)y(s)}ds, t∈[0,1].

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
(34)Lx(t)=∫01k(s,t)[ma(s)+b(s)]x(s)ds, t∈[0,1].

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.