In this paper, we prove the existence of a positive solution for some equations involving multiplication of concave (possibly nonlinear) operators. Also, we provide a successively sequence to approximate the solution for such equations. This kind of the solution is necessary for quadratic differential and integral equations.
Citation: Golnaz Pakgalb, Mohammad Jahangiri Rad, Ali Salimi Shamloo, Majid Derafshpour. Existence and uniqueness of a positive solutions for the product of operators[J]. AIMS Mathematics, 2022, 7(10): 18853-18869. doi: 10.3934/math.20221038
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In this paper, we prove the existence of a positive solution for some equations involving multiplication of concave (possibly nonlinear) operators. Also, we provide a successively sequence to approximate the solution for such equations. This kind of the solution is necessary for quadratic differential and integral equations.
In this paper we prove some fixed point theorems for the problem
x=A(x)+B(x)⋅C(x). | (1.1) |
However, these kind of theorems are related to some "quadratic" problems. Let us mention the quadratic integral equation
x(t)=g(t,x(t))+u(t,x(t))∫10K(t,s)f(s,x(s))ds. | (1.2) |
Special cases of Eq (1.2) were investigated in connection with the applications of some kind of problems in the theories of radiative transfer, neutron transport, and the kinetic theory of gases [4]. A more general problem (motivated by some practical interests in plasma physics) was investigated in [21]. See [11,24] for other applications.
So far, two methods have been proposed to solve Eq (1.1). In the former, the measure of non-compactness technique (see [2,8,9,10,14,17]) is used to prove the existence of a solution for Eq (1.1), and in the latter, Dhage [13] used the combining Schauder's fixed point theorem and Banach's contraction principle to prove the existence of a solution for Eq (1.1), also see [5]. Some authors have combined the two methods and have proved the existence of a solution for Eq (1.1).
In this paper, we prove the existence of a positive solution for the Eq (1.1) in which the operators A, B, and C are concave (or convex) or monotone. Also, we give a successively sequence to approximate it. But what is our motivation to prove the existence of a positive solution for the Eq (1.1) when the operators A, B, and C are concave (or convex) or monotone? The mentioned methods have not provided a way to approximate the solution for the Eq (1.1). Also, in the case that the operators A, B, and C are concave (or convex) or monotone, we do not have to suppose the continuity, compactness, and upper-lower assumptions for the operators A, B, and C. These assumptions play an important roles in order to prove the existence of positive solutions for nonlinear differential and integral equations and they are difficult to verify for real problems. Furthermore, there exist more extensively applied of the positive solution of nonlinear differential and integral equations in practical issues (see [3,12,28,29,30,31]).
The start of proving the existence of a positive solution for differential and integral equations can be found in the Picard investigation (see [25], p.129–138). Authors in [18,19,20] generalized theorems for abstract operator equations with special positive operators called u0-concave. After that, ordered concavity (convexity) and α-concavity (convexity) were introduced by Amann [1] in 1976 and Potter [26] in 1977. In [7,22,23,32,33,34], some others type of concave operators were investigated.
The paper is organized as follows: In Section 2, we introduce some of the preliminaries needed for the next sections. In Section 3, we prove some existential results for the Eq (1.1). Furthermore, we provide some examples that satisfy the main results. In Section 4, we prove the existence of positive solutions for nonlinear quadratic integral equations by theorems provided in the main results section. Section 5 is devoted to concluding and proposing new ideas.
Throughout this paper, we assume that E is a real Banach algebra. Which means, E is a real Banach space in which an operation of multiplication is defined, subject to the following properties (for all x,y,z∈E,λ∈R):
(1) (xy)z=x(yz);
(2) x(y+z)=xy+xz and (x+y)z=xz+yz;
(3) λ(xy)=(λx)y=x(λy);
(4) ||xy||≤||x||||y||.
Now let us recall the concepts of cone and partial order for a Banach algebra. A subset P of E is called a cone of E if
(1) P is a non-empty closed and θ∈P;
(2) λP+γP⊆P for all non-negative real numbers λ,γ;
(3) P2=P⋅P⊆P;
(4) P∩(−P)={θ},
where θ denotes the null of E. For a given cone P⊆E, we can define a partial ordering ≤ with respect to P by x≤y if and only if y−x∈P. x<y will stand for x≤y and x≠y. The cone P is called normal if there is a number M>m0 such that for all x,y∈E,
θ≤x≤y⇒||x||≤M||y||. |
The last positive number satisfying the above inequality is called the normal constant of P. In the following, we always assume that P is a cone in E and, ≤ is the partial ordering with respect to P. We call such space ordered Banach algebra and denote it by (OBA).
If x1,x2∈E, the set [x1,x2]={x∈E|x1⩽x⩽x2} is called the order interval between x1 and x2. An operator A:E→E is called increasing (decreasing) if x⩽y implies Ax⩽Ay (Ax⩾Ay) where x,y∈E. For h>θ (i.e. h⩾θ and h≠θ), set
Ph={x|x∈E,∃λ(x)>0,μ(x)>0,s.t.λ(x)h⩽x⩽μ(x)h}. |
It is easy to notice that Ph⊆P.
Lemma 2.1. ([15]) The two following assumptions are equivalent:
(1) P is a normal cone,
(2) xn⩽zn⩽yn (n=1,2,3,...) and ||xn−x||→0, ||yn−x||→0, imply that ||zn−x||→0.
Definition 2.1. ([16]) Let α be a real number such that 0≤α<1. An operator A:P→P is called an α-concave ((-α)-convex) if it satisfies,
A(tx)⩾tαAx(A(tx)⩽t−αAx),∀t∈(0,1),x∈P. | (2.1) |
Theorem 2.2. ([6]) Assume that P is a normal cone and the operator T satisfies:
(D1) T:Ph→Ph is an increasing self-map in Ph;
(D2) For any x∈Ph and t∈(0,1), there exists β(t)∈(0,1) such that T(tx)⩾tβ(t)Tx;
(D3) For every x0∈P, there is a constant l≥0 suchthat x0∈[θ,lh].
Then, operator equation x=Tx+x0 has a unique solution in Ph.
Now the main results could be stated and proved.
Theorem 3.1. Let P be a normal cone, A:P→P is an increasing α-concave operator, B:P→P is an increasing γ1-concave operator, and C:P→P is anincreasing γ2-concave operator such that γ1+γ2=γ⩽1. Also, suppose that
(i) there exists h>θ such that h⋅h∈Ph, and Ah,Bh,Ch∈Ph;
(ii) there exists a constant δ0>0 such that for all x∈P, we have Ax⩾δ0Bx⋅Cx.
Then, the operator Eq (1.1) has a unique solution x∗ in Ph. Moreover, for the constructing successivelysequence yn=Ayn−1+Byn−1⋅Cyn−1(n=1,2,⋯) and for any initial value y0∈Ph, we have yn→x∗ as n→∞.
Proof. Since Ah,Bh,Ch∈Ph, there exist constants λ1,λ2,μ1,μ2,υ1,υ2>0 such that λ1h⩽Ah⩽λ2h,μ1h⩽Bh⩽μ2h,υ1h⩽Ch⩽υ2h. We have
λ1h+μ1υ1h⋅h⩽Ah+Bh⋅Ch⩽λ2h+μ2υ2h⋅h. |
By (i), there exist r,s>0 such that rh⩽h⋅h⩽sh. We get
(λ1+μ1υ1r)h⩽Ah+Bh⋅Ch⩽(λ2+μ2υ2s)h. |
Hence, we can write K1h⩽Ah+Bh⋅Ch⩽J1h, where K1=λ1+μ1υ1r and J1=λ2+μ2υ2s. Thus, Ah+Bh⋅Ch∈Ph. Define the operator T=A+B⋅C by Tx=Ax+Bx⋅Cx. Then T:P→P and Th∈Ph. Next, we show that T:Ph→Ph. By (2.1), for any t∈(0,1) and x∈P, we have
A(1tx)⩽1tαAx,B(1tx)⩽1tγ1Bx,C(1tx)⩽1tγ2Cx. |
For any x∈Ph, we can choose a sufficiently small number t0∈(0,1) such that
t0h⩽x⩽1t0h. | (3.1) |
Note that T:P→P is an increasing self-map and by (3.1),
Tx=Ax+Bx⋅Cx⩽A(1t0h)+B(1t0h)⋅C(1t0h)⩽1tα0Ah+1tγ0Bh⋅Ch⩽λ2tα0h+μ2υ2tγ0h⋅h=J2h, |
where J2=λ2tα0+μ2υ2tγ0s. Also,
Tx=Ax+Bx⋅Cx⩾A(t0h)+B(t0h)⋅C(t0h)⩾tα0Ah+tγ0Bh⋅Ch⩾λ1tα0h+μ1υ1tγ0h⋅h=K2h, |
where K2=λ1tα0+μ1υ1tγ0r. Thus Tx∈Ph. Henece, T:Ph→Ph. Moreover, A:Ph→Ph,B:Ph→Ph and C:Ph→Ph. In the following, we show that for any t∈(0,1), there exists β0(t)∈(α,1) with respect to t, such that for all x∈Ph,
T(tx)≥tβ0(t)Tx,∀t∈(0,1). | (3.2) |
By (ii), there exists δ0>0 such that Ax⩾δ0Bx⋅Cx. Consider the following function:
f(t)=tβ−ttα−tβ,∀t∈(0,1),whereβ∈(α,1). |
It is easy to prove that f is non-negative in (0,1). Especially, for any t∈(0,1) we have tβ>t and tα>tβ. Furthermore, for fixing t∈(0,1), we have limβ→1−f(t)=0. So, there exists β0(t)∈(α,1) with respect to t such that
tβ0(t)−ttα−tβ0(t)⩽δ0,t∈(0,1). |
Hence, we have
Ax⩾δ0Bx⋅Cx⩾tβ0(t)−ttα−tβ0(t)Bx⋅Cx,∀t∈(0,1),x∈Ph. |
Then, we can get
tαAx+tBx⋅Cx⩾tβ0(t)Ax+tβ0(t)Bx⋅Cx,∀t∈(0,1),x∈Ph. |
Consequently, for any t∈(0,1) and x∈Ph we have
T(tx)=A(tx)+B(tx)⋅C(tx)⩾tαAx+tγBx⋅Cx⩾tαAx+tBx⋅Cx⩾tβ0(t)Ax+tβ0(t)Bx⋅Cx,∀t∈(0,1),x∈Ph. |
Therefore,
T(tx)⩾tβ0(t)T(x),∀t∈(0,1),x∈Ph. |
Let x0=θ. Application of Theorem 2.2 implies that the equation Tx=x has a unique solution x∗ in Ph. It can be concluded that the operator Eq (1.1) has a unique solution x∗ in Ph. Now we can construct the successively sequence yn=Ayn−1+Byn−1⋅Cyn−1(n=1,2,⋯) for any initial point y0∈Ph. Since y0∈Ph and Ty0∈Ph, we can choose a sufficiently small number t0∈(0,1) such that
t0y0⩽Ty0⩽1t0y0. | (3.3) |
Note that 0<β0(t0)<1, and we can also take a positive integer k such that
k>11−β0(t0). | (3.4) |
Put u0=tk0y0,v0=1tk0y0. Evidently, u0,v0∈Ph and u0⩽y0⩽v0. By the monotonicity of T, we have Tu0⩽Tv0. Furthermore, by combining (3.2) and (3.3) we have
Tu0=T(tk0y0)=T(t0tk−10y0)⩾tβ0(t0)0T(t0tk−20y0)⩾tβ0(t0)0tβ0(t0)0T(tk−20y0)⩾⋯⩾(tβ0(t0)0)kTy0⩾(tβ0(t0)0)kt0y0=tkβ0(t0)+10y0. | (3.5) |
By (3.4), one can obtains that kβ0(t0)+1<k. Thus
tkβ0(t0)+10>tk0. | (3.6) |
Therefore, Tu0⩾tkβ0(t0)+10y0>tk0y0=u0. By (3.2),
T(1tx)⩽1tβ0(t)T(x),∀t∈(0,1),x∈Ph. |
Thus,
Tv0=T(1tk0y0)=T(1t01tk0y0)⩽1tβ0(t0)0T(1tk−10y0)=1tβ0(t0)0T(1t01tk−20y0)⩽1tβ0(t0)01tβ0(t0)0T(1tk−20y0)⩽⋯⩽1(tβ0(t0)0)kT(y0)⩽1tkβ0(t0)+10y0. | (3.7) |
The application of (3.6) implies Tv0⩽1tkβ0(t0)+10y0⩽1tk0y0=v0. Thus, u0⩽Tu0⩽Tv0⩽v0. For n=1,2,⋯, let un=Tun−1,vn=Tvn−1. Then, un⩽yn⩽vn (n=1,2,⋯). Similar to the proof of Theorem 1.3 of [35], there exists y∗∈Ph such that Ty∗=y∗ and limn→∞un=limvn=y∗. Then, by Lemma 2.1, yn→y∗(n→∞). Since the fixed point of the operator T in Ph is unique, we have x∗=y∗.
Example 3.1. Consider the Banach space E=C[0,1] with the supremum norm. E is a Banach algebra by the multiplication (x⋅y)(t)=x(t)⋅y(t). Assume that P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]}. Then E is an (OBA) by the cone P. Let us define operators A,B,C:P⟶P as the following
A(x)=√x+1,B(x)=1√x+1,C(x)=1. |
Assume that h(t)=1. We can prove that all of the assumptions of Theorem 3.1 are satisfied and the operator T=A+B⋅C has a unique positive solution.
Example 3.2. Consider the Banach space E=L∞[0,1] with the L∞ norm. E is a Banach algebra by the multiplication (x⋅y)(t)=x(t)⋅y(t). Assume that P={x(t)∈L∞[0,1]|x(t)⩾0,∀t∈[0,1]}. Then E is an (OBA) by the cone P. For any x∈P and t∈[0,1], let us define
A(x)=1√x+1,B(x)=√x+1,C(x)=1,h(t)=1. |
It is easy to prove that for any x∈P, A(x),B(x),C(x)∈P. Also we have h>0. Therefore, A,B,C,h:P⟶P. It is easily noticed that, A is an α-concave operator (for some 0<α<1), B is a γ1 concave operator (for some 0⩽γ1<1), and C is a γ2 concave operator (for some 0⩽γ2<1) such that γ1+γ1⩽1. Now let us consider that
x(t)={nt=1n,0t∈[0,1]∖{1n}n. |
Then, x(t)∈P and for t=1n we have
A(x(1n))=1√x(1n)+1=1√n+1,B(x(1n))=√x(1n)+1=√n+1. |
Therefore, there is no δ0>0 such that for all t∈[0,1] we have Ax(t)⩾δ0Bx(t)⋅Cx(t). Hence, the assumption (ii) of Theorem 3.1 is not satisfied. It is easy to prove that the rest of the assumptions of Theorem 3.1 are satisfied and T=A+B⋅C has a unique positive solution.
Example 3.3. Consider E, that is defined in Example 3.1. Let us define operators A,B,C:P⟶P as the following
A(x)=11+sin(x),B(x)=2+sin(x),C(x)=2+cos(x). |
Suppose that h(t)=π4. We have A,B,C:P⟶P and h⋅h>0. Also, for all x∈P, Ax⩾Bx⋅Cx. None of the functions A,B, and C are concave. But the operator T=A+B⋅C has a unique positive solution.
Example 3.4. Consider the Banach space E=C[0,1] with the supremum norm. Let us consider the multiplication
(f⋅g)(t)=∫x0f(t)g(x−t)dt |
for any x,y∈E. E is a Banach algebra([27]). Assume that, P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]}. Then, E is an (OBA) by the cone P. Now suppose that, h(t)=1>0. We have h⋅h(x)=x. Then, h⋅h∉Ph.
Example 3.5. Consider E that is defined in Example 3.4. Let h(t)=1>0. Let us define the operators A,B,C:P⟶P as the following
A(x)=√x+1,B(x)=1√x+1+1,C(x)=1. |
We can prove that all of the assumptions of Theorem 3.1, except h⋅h>0, are satisfied. Operator T=A+B⋅C has no positive solution.
Let the operators A,B,C:P⟶P be defined as the following
A(x)=√x+1,B(x)=1,C(x)=12. |
It is easy to prove that all of the assumptions of the Theorem 3.1, except the assumption h⋅h>0, are satisfied. The equation T=A+B⋅C has a unique positive solution.
Lemma 3.2. Assume that, P is a normal cone and the operator A satisfies the following conditions:
(D1) A:Ph→Ph is decreasing in Ph;
(D2) For any x∈Ph and t∈(0,1), there exists α(t)∈(0,1) such that A(tx)⩽t−α(t)Ax.
Then, there exist u0,v0∈Ph such that u0<v0, u0⩽Av0⩽Au0≤v0.
Proof. Since Ah∈Ph, we can select a sufficiently small number t0 such that
t0h⩽Ah⩽1t0h. | (3.8) |
Note that 0<α(t0)<1, and we can choose a positive integer k such that
k>11−α(t0). | (3.9) |
Set u0=tk0h and v0=1tk0h. Evidently, u0,v0∈Ph and u0<v0. By the monotonicity of A, Au0⩾Av0. Furthermore, by (D2) and (3.8) we have,
Au0=A(tk0h)=A(t0tk−10h)⩽t−α(t0)0A(tk−10h)⩽⋯⩽t−kα(t0)0A(t0h)⩽t−kα(t0)01t0h=t−kα(t0)−10h. | (3.10) |
By (3.10), we get that kα(t0)+1<k. Thus,
t−kα(t0)−10<t−k0. | (3.11) |
Hence,
Au0⩽t−kα(t0)−10h<t−k0h<v0. |
By (D2),
tα(t)A(x)⩽A(xt)∀t∈(0,t),x∈Ph. | (3.12) |
Thus,
Av0=A(1tk0h)=A(1t01tk−10h)⩾tα(t0)0A(1tk−10h)=tα(t0)0A(1t01tk−20h)⩾tα(t0)0tα(t0)0A(1tk−20h)⩾⋯⩾tkα(t0)0A(1t0h)=tkα(t0)+10h. |
Application of (3.9) implies that Av0⩾tkα(t0)+10h⩾tk0h=u0. So we have
u0⩽Av0⩽Au0⩽v0. | (3.13) |
Theorem 3.3. Assume that P is normal cone, the operator T satisfies (D1) and(D2) of Lemma 3.2, and there is a constant l⩾0 such that x0∈[θ,lh].
Then, the operator equation Tx+x0=x has a unique solutionin Ph.
Proof. For all x∈Ph, we have Tx∈Ph. Then, there exist real numbers λ,μ>0 such that λh⩽Tx⩽μh. Thus,
λh⩽Tx+x0⩽(μ+l)h. | (3.14) |
Hence,
Tx+x0∈Ph,∀x∈Ph. | (3.15) |
Define the operator F by,
Fx=Tx+x0,∀x∈Ph. | (3.16) |
By (3.15), and considering the monotonicity of the operator T, the operator F:Ph→Ph is decreasing. Furthermore, for all x∈Ph and t∈(0,1), we have
F(tx)=T(tx)+x0⩽t−α(t)T(x)+t−α(t)x0⩽t−α(t)F(x). | (3.17) |
Lemma 3.2 implies that, there exist u0,v0∈Ph such that
u0<v0,u0⩽Fv0⩽Fu0⩽v0. | (3.18) |
Construct the successively sequences
un=Fun−1,vn=Fvn−1,n=1,2,⋯. |
By the monotonicity of F, we have v1=Fv0⩽Fu0=u1. Similarly, we have
u0⩽v1⩽u2⩽v2⩽u1⩽v0. | (3.19) |
By continuing this process, for n=1,2,⋯, we get
u0⩽v1⩽u2⩽⋯⩽u2n⩽v2n+1⩽u2n+1⩽v2n⩽⋯⩽v2⩽u1⩽v0. | (3.20) |
Therefore, {u2n}, {v2n+1} are the increasing and {u2n+1}, {v2n} are decreasing sequences. By (3.20), for n=1,2,⋯, we have
u2n⩽v2n,v2n+1⩽u2n+1. | (3.21) |
Assume that,
t2n=sup{t|tv2n⩽u2n},t2n+1=sup{t|tu2n+1⩽v2n+1}. |
Thus, for n=1,2,⋯, we have u2n⩾t2nv2n and v2n+1⩾t2n+1v2n+1. Then,
u2n+1⩾u2n⩾t2nv2n⩾t2nv2n+1n=1,2,⋯, |
v2m⩾v2m+1⩾t2m+1u2m+1⩾t2m+1u2mm=1,2,⋯. |
Therefore, tn+1⩾tn, i.e. tn is an increasing sequence such that tn⊆(0,1]. If tn→t∗ as n→∞, then t∗=1. Otherwise, 0<t∗<1. We distinguish two cases:
Case (i): There exists an integer N such that tN=t∗. In this case, we know tn=t∗, for all n⩾N. So for n⩾N, we have
u2n+1=Fu2n⩽F(t∗v2n)⩽(t∗)−α(t∗)F(v2n)=(t∗)−α(t∗)v2n+1, |
i.e. (t∗)α(t∗)u2n+1⩽v2n+1. By the definition of t2n+1, we have t2n+1=t∗⩾(t∗)α(t∗)>t∗, which is a contradiction.
Case (ii): For all integer n, tn<t∗. Then,
v2n+1=F(v2n)⩾F(t−12nu2n)=F(t∗t2nu2nt∗)=F(t∗t2nu2nt∗)⩾(t∗)α(t∗)F(t∗t2nu2n)=(t∗)α(t∗)F(u2nt2nt∗)⩾(t∗)α(t∗)(t2nt∗)(α(t2nt∗))Fu2n⩾(t∗)α(t∗)(t2nt∗)u2n+1=t2n(t∗)α(t∗)−1u2n+1. | (3.22) |
By the definition of tn, we have t2n+1⩾t2n(t∗)α(t∗)−1. If n→∞, we get t∗⩾(t∗)α(t∗)>t∗, which is a contradiction. Thus, limn→∞tn=1. For any natural number p, we have
θ⩽u2(n+p)−u2n⩽v2(n+p)−t2nv2n⩽v2n−t2nv2n=(1−t2n)v2n⩽(1−t2n)v0, |
θ⩽v2n−v2(n+p)⩽v2n−u2n⩽v0−u2n⩽v0−t2nv0⩽(1−t2n)v0. |
Since P is normal, we have
||u2(n+p)−u2n||⩽N(1−t2n)||v0||→0(n→∞), |
||v2n−v2(n+p)||≤N(1−t2n)||v0||→0(n→∞), |
where N is the normal constant. Hence, we can claim that u2n and v2n are Cauchy sequences. Since E is a complete space, there exist u∗ and v∗ such that u2n→u∗, v2n→v∗ as n→∞. By (3.20), we know that u2n⩽u∗⩽v∗⩽v2n where u∗,v∗∈Ph. Then
θ⩽v∗−u∗⩽v2n−u2n⩽(1−t2n)v0. |
Furthermore,
||v∗−u∗||⩽N(1−t2n)||v0||→0(n→∞). |
Thus, u∗=v∗. Let x∗=u∗=v∗. Also, by (3.20), we have
θ⩽v2n+1−u2n⩽v2n−u2n, |
θ⩽u2n+1−v2n+1⩽v2n−v2n+1. |
Then v2n+1→x∗ and u2n+1→x∗ as n→∞. By the inequality u2n⩽x∗⩽v2n for n=1,2,⋯, we have
v2n+1=Fv2n⩾Fx∗⩾Fu2n=u2n+1. |
If n→∞, we get x∗=Fx∗. That is, x∗ is a fixed point of F in Ph. In the following, we prove that x∗ is the unique fixed point of F in Ph. Let ˉx be any fixed point of F in Ph. Set r1=sup{r>0|rx∗⩽ˉx⩽1rx∗}. Evidently, 0<r1<∞ and r1x∗⩽ˉx⩽1r1x∗. Next, we prove that r1⩾1. If 0<r1<1,
rα(r1)1x∗⩽F(x∗r1)⩽ˉx=Fˉx⩽F(r1x∗)⩽r−α(r1)1x∗. |
However, by rα(r1)1x∗⩽ˉx, rα(r1)1⩽r1. Since rα(r1)1>r1, we get a contradiction. Hence, r1⩾1 and we get ˉx⩾r1x∗⩾x∗. Similarly, we can prove that x∗⩾ˉx, and x∗=ˉx. Therefore, F has a unique fixed point x∗ in Ph. That is to say, Tx+x0=x has a unique solution in Ph.
Comment 3.4. In [26], Theorem 3.3 is proved by Hilbert'sprojective metric method where, α is constant function. But, the successively sequence that converges to the fixed pointhas not been obtained.
Theorem 3.5. Consider that P is a normal cone, A,B:P→P aredecreasing operators, and C:P→P is a decreasing(-α)-convex operator. Assume that
(i) there exists h>θ such that Ah∈Ph,Bh∈Ph and Ch∈Ph;
(ii) h⋅h∈Ph.
Then, the operator Eq (1.1) has a unique solution x∗ in Ph. Moreover, for the constructing successivelysequence yn=Ayn−1+Byn−1⋅Cyn−1,n=1,2,... and for any initial value y0∈Ph, we have yn→x∗ as n→∞.
Proof. Since Ah,Bh,Ch∈Ph, there exist constants λ1,λ2,μ1,μ2,υ1,υ2>0 such that λ1h⩽Ah⩽λ2h,μ1h⩽Bh⩽μ2h,υ1h⩽Ch⩽υ2h. Similar to the proof of the Theorem 3.1, we can prove that, if the operator T=A+B⋅C is defined by Tx=Ax+Bx⋅Cx. Then, T:P→P and Th∈Ph. Next, we show that T:Ph→Ph. By (2.1) and the monotonicity of A,B, we have
A(1tx)⩾Ax,B(1tx)⩾Bx,C(1tx)≥tαAx,t∈(0,1),x∈P. |
For any x∈Ph, we can select a sufficiently small number t0∈(0,1) such that
t0h⩽x⩽1t0h. | (3.23) |
T:P→P is decreasing and by (3.23) we have,
Tx=Ax+Bx⋅Cx⩾A(1t0h)+B(1t0h)⋅C(1t0h)⩾Ah+tα0Bh⋅Ch⩾λ2h+μ2υ2tα0h⋅h. |
By (ii), there exist s>0 such that 1sh⩽h⋅h⩽sh. So, we have Tx⩾J2h where J2=λ2+μ2υ21stα0. Also,
Tx=Ax+Bx⋅Cx⩽A(t0h)+B(t0h)⋅C(t0h)⩽Ah+t−α0Bh⋅Ch⩽λ1h+μ1υ1t−α0h⋅h. |
Hence, we have Tx⩽K2h, where K2=λ1+μ1υ1t−α0s. Thus, Tx∈Ph. So, T:Ph→Ph. Moreover, A:Ph→Ph,B:Ph→Ph, and C:Ph→Ph. On the other hand, for any t∈(0,1) and x∈Ph,
Ax+t−αBx⋅Cx⩽t−α(Ax+Bx⋅Cx). |
Then,
T(tx)=A(tx)+B(tx)⋅C(tx)⩽t−αT(x)∀t∈(0,1),x∈Ph. |
Therefore, T is the (-α)-convex operator. Let, x0=θ. Application of Theorem 3.3 implies that the equation Tx=x has a unique solution x∗ in Ph. That is, the operator Eq (1.1) has a unique solution x∗ in Ph. Now, we construct successively the sequence yn=Ayn−1+Byn−1⋅Cyn−1(n=1,2,⋯) for any initial point y0∈Ph. Since y0∈Ph and Ty0∈Ph, we can choose a sufficiently small number t0∈(0,1) such that
t0y0⩽Ty0⩽1t0y0. | (3.24) |
Since 0<α(t0)<1, we can also take a positive integer k such that
k>11−α(t0). | (3.25) |
Set u0=tk0y0,v0=1tk0y0. Let un+1=Tun, vn+1=Tvn(n=1,2,⋯). By Theorem 3.3, un→x∗ and vn→x∗(as n→∞). By (3.25), we have u0⩽y1⩽v0. Let us define yn+1=Tyn. Since T is monotone decreasing and by (3.20), we get
v2n−1⩽y2n⩽u2n−1,v2n⩽y2n+1⩽u2n,n=1,2,⋯. | (3.26) |
Then by Lemma 2.1, y2n→x∗, y2n+1→x∗ as n→∞. Thus, for ε>0, there exists an integer N such that for n⩾N,
||y2n−x∗||<ε,||y2n+1−x∗||<ε. | (3.27) |
Therefore, (3.27) show that yn→x∗ as n→∞.
Theorem 4.1. Assume that E=C[0,1], P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]},
(H1) g(t,x):[0,1]×[0,∞)→[0,∞),u(t,x):[0,1]×[0,∞)→[0,∞) and f(t,x):[0,1]×[0,∞)→[0,∞) are increasing operators with respect to x;
(H2) there exists h>θ in P such that g(t,h),u(t,h)∈Ph and h⋅h∈Ph;
(H3) g(t,x) is an α-concave operator, f(t,x) is a γ1-concave operator, and u(t,x) is a γ2-concave operator with respect to x such that γ1+γ2=γ⩽1;
(H4) G(t,s) is non-negative for any t,s∈[0,1]. Also, forany fixed t∈[0,1], the function G(t,s) is boundedin [0,1] and for any fixed s∈[0,1], we have G(t,s)∈Ph;
(H5) there exists δ0>0 such that for any t∈[0,1] and any y∈[0,∞) we have
g(t,y)⩾δ0u(t,y)⋅∫10G(t,s)f(s,y)ds. |
Then the problem
x(t)=g(t,x(t))+u(t,x(t))∫10G(t,s)f(s,x(s))ds |
has a unique positive solution x∗ in Ph. Moreover, forany x0∈Ph and for the constructing successivelysequence
xn+1(t)=g(t,xn(t))+u(t,xn(t))∫10G(t,s)f(s,xn(s))ds,n=0,1,2,⋯, |
we have ||xn−x∗||→0 as n→∞.
Proof. Let us define the operators A:P→E, B:P→E, and C:P→E as the following:
(Ax)(t)=g(t,x(t)),(Bx)(t)=u(t,x(t)),(Cx)(t)=∫10G(t,s)f(s,x(s))ds. |
It can easily be noticed that x∗ is a solution of the problem (4.1) if x∗=Ax∗+Bx∗⋅Cx∗. By (H1) and (H4), we notice that A:P→P, B:P→P and C:P→P. By (H3), for any λ∈(0,1) and x∈P, we have
C(λx)(t)=∫10G(t,s)f(s,λx(s))ds⩾λγ1∫10G(t,s)f(s,x(s))ds=λγ1C(x)(t). |
Then, C is a γ1-concave operator with respect to x. By (H2), Ah∈Ph and Bh∈Ph. By (H4), for any s∈[0,1] there exist λ(s),μ(s)>0 such that
λ(s)h(t)⩽G(t,s)⩽μ(s)h(t). |
Since G(t,s) is bounded, λ(s),μ(s) are bounded positive real numbers(for any s∈[0,1]). Therefore,
Ch(t)=∫10G(t,s)f(s,h(s))ds⩽∫10μ(s)h(t)f(s,1)ds=h(t)∫10μ(s)f(s,1)ds |
and
Ch(t)=∫10G(t,s)f(s,h(s))ds⩾∫10λ(s)h(t)f(s,0)ds=h(t)∫10λ(s)f(s,0)ds. |
Thus, Ch∈Ph. Hence, the condition (i) of Theorem 3.1 is satisfied. By (H2), the condition (iii) of Theorem 3.1 is satisfied and by (H5) the condition (ii) of Theorem 3.1 is satisfied. Then, Theorem 4.1 follows from Theorem 3.1.
Example 4.1. Assume that E=C[0,1], P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]}. Let us define
g(t,x(t))=√x(t)1+√x(t),u(t,x(t))=x(t)√1+x2(t),f(t,x(t))=√1+x(t) |
for t∈[0,1] and x∈P. It is easy to prove that g is α−concave (for α=12). Also, f is a γ1-concave operator (for γ1=0) and u is a γ2-concave operator (for γ2=1) with respect to x. Suppose that G(t,s)=e−ts1+ts (for t,s∈[0,1]). It is easy to see that g, u, f, and G are satisfied in all assumptions of Theorem 4.1 for h(t)=1. Hence, the problem
x(t)=√x(t)1+√x(t)+x(t)√1+x2(t)∫10e−ts√1+x(s)1+tsds |
has a unique positive solution.
Theorem 4.2. Assume that E=C[0,1], P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]},
(H1) g(t,x):[0,1]×[0,∞)→[0,∞),u(t,x):[0,1]×[0,∞)→[0,∞) and f(t,x):[0,1]×[0,∞)→[0,∞) are decreasing with respect to x;
(H2) there exists h>θ in P such that g(t,h),u(t,h)∈Ph, and h⋅h∈Ph;
(H3) f(t,x) is (-α)-convex with respect to x;
(H4) G(t,s) is non-negative for any t,s∈[0,1], for anyfixed t∈[0,1], the function G(t,s) is boundedin [0,1], and for any fixed s∈[0,1], we have G(t,s)∈Ph.
Then, the problem
x(t)=g(t,x(t))+u(t,x(t))∫10G(t,s)f(s,x(s))ds, |
has a unique positive solution x∗ in Ph. Moreover, forany x0∈Ph and for the constructing successivelysequence
xn+1(t)=g(t,xn(t))+u(t,xn(t))∫10G(t,s)f(s,xn(s))ds,n=0,1,2,⋯ |
we have ||xn−x∗||→0 as n→∞.
Proof. Let us define the operators A:P→E, B:P→E and C:P→E as the following:
(Ax)(t)=g(t,x(t)),(Bx)(t)=u(t,x(t)),(Cx)(t)=∫10G(t,s)f(s,x(s))ds. |
It is easily noticed that if x∗=Ax∗+Bx∗⋅Cx∗, x∗ is a solution of the problem (4.1). By (H1) and (H4), we have A:P→P, B:P→P and C:P→P. By (H3), for any λ∈(0,1) and x∈P we have,
C(λx)(t)=∫10G(t,s)f(s,λx(s))ds⩾λ−α∫10G(t,s)f(s,x(s))ds=λ−αC(x)(t). |
Hence, C is an (-α)-convex operator. By (H2), we have Ah∈Ph and Bh∈Ph. By (H4), for any s∈[0,1], there exist λ(s) and μ(s)>0 such that
λ(s)h(t)⩽G(t,s)⩽μ(s)h(t). |
Since G(t,s) is bounded, λ(s) and μ(s) are bounded positive real numbers (for any s∈[0,1]). Therefore, we have
Ch(t)=∫10G(t,s)f(s,h(s))ds⩽∫10μ(s)h(t)f(s,0)ds |
=h(t)∫10μ(s)f(s,0)ds |
and
Ch(t)=∫10G(t,s)f(s,h(s))ds⩾∫10λ(s)h(t)f(s,1)ds |
=h(t)∫10λ(s)f(s,1)ds. |
Thus, Ch∈Ph. Hence, the assumption (i) of Theorem 3.5 is satisfied. By (H2), the assumption (iii) of Theorem 3.5 is satisfied. Then, Theorem 4.2 follows from Theorem 3.5.
Example 4.2. Assume that E=C[0,1], P={x(t)∈C[0,1]|x(t)⩾0,∀t∈[0,1]}. Let us define,
g(t,x(t))=11+x2(t),u(t,x(t))=arccot(x(t)−4),f(t,x(t))=1√1+x(t), |
for t∈[0,1] and x∈P. It can be proved that f is (-α)-convex (for α=12). Also u,g are convex sub-homogeneous in x. Suppose that G(t,s)=e−ts1+ts(for t,s∈[0,1]). We can see that g, u, f, and G are satisfied in all assumptions of Theorem 4.2 for h(t)=1. Then the problem
x(t)=11+x2(t)+arccot(x(t)−4)∫10e−ts1+ts1√1+x(t)ds |
has a unique positive solution.
In this paper, we firstly proved the existence of a positive solution for the Eq (1.1) and approximated it by the constructing successively sequence, where A is an α-concave operator, B:P→P is an increasing γ1-concave operator and C:P→P is an increasing γ2-concave operator such that γ1+γ2=γ⩽1.
Secondly, we proved the existence of a positive solution for the Eq (1.1) and approximated it by the constructing successively sequence, where A,C are decreasing operators and C is a (-α)-convex operator.
Thirdly, we proved the existence a positive solution for some nonlinear integral equations and approximated it by the constructing successively sequence(especially in the case of quadratic integral equation).
Remark 5.1. It is suggested that the Theorems 3.1 and 3.5 be proved without the assumption h⋅h>θ, and also the Theorem 3.1 be proved without assumption (ii). Another interesting topic can be the comparison of the results of Theorems 3.1 and 3.5 with the results of theorems that are proven by the measure of non-compactness [2] and Dhage's techniques [13].
The authors would like to thank the anonymous referee for his/her comments that helped us improve this article.
Authors state no conflicts of interest.
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