## I. INTRODUCTION $$ \left\{ \begin{array}{l} u^{(4)}(t) = \lambda f(t, u(t)), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\eta) = 0 \end{array} \right. $$ In this article, we aim to study the existence of a monotonic positive solution for fourth-order three-point Nonlinear BVP with changing sign Green's function $$ \left\{ \begin{array}{l} u^{(4)}(t) = \lambda f(t, u(t)), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\eta) = 0 \end{array} \right. \tag{1.1} $$ where α ∈ [0,2), f ∈ C(...) where $\alpha \in [0,2),f\in C([0,1]\times [0, + \infty),[0, + \infty))$ and $\eta \in [\frac{1}{2},1)$. By using iterative methods, We can still obtain the existence of a monotonic positive solution under certain suitable conditions of $f$. In recent decades, The differential equations come from various fields of a mathematical applied and physics, for example, in the deflection of curved beams with constant or varying cross-sections, triple-layer beams, electromagnetic waves or gravity-driven currents, etc. [1]. Recently, when the corresponding Green's function is undergoing sign conversion, there has been some work on the positive solution of the second and third-order BVP. For example, in [8] the existence of at least one positive solution of the following second-order periodic BVP with positive and negative transformation Green's function studied by Zhong and An $$ \left\{ \begin{array}{l} u ^ {\prime \prime} (t) + \rho^2 u = f (u), \quad 0 < t < T, \\u (0) = u (T), \\u ^ {\prime} (0) = u ^ {\prime} (T), \end{array} \right. $$ where $\eta \in (\frac{17}{24}, 1)$, Palamide and Smirlis [9] discussed the existence of at least one positive solution. Their technique is a combination of GuoKrasnosel'ski fixed point theory and the corresponding vector field characteristics. In 2012, Sun and Zhao [10], [11] obtained single or multiple positive solutions with three-point positive and negative BVP by applying the fixed point theory of Guo-Krasnosel'skii $$ \left\{ \begin{array}{l} u ^ {\prime \prime \prime} (t) = f (t, u (t)), \quad t \in [ 0, 1 ], \\ u (0) = u (1) = u ^ {\prime \prime} (\eta) = 0 \end{array} \right. $$ Motivated. Through the above work, this article will study BVP (1.1)Through an iterative method. Throughout this article, we always assume $\alpha \in [0,2)$ and $\eta \in [\frac{1}{2},1)$. Although the corresponding Green function is changing its sign, under certain suitable conditions, we can still obtain the existence of the monotonic positive solution of BVP (1.1) on $f$. Moreover, our iterative scheme starts with a zero function, This means that iterative scheme is feasible. $$ \left\{ \begin{array}{l} u^{(4)}(t) = \lambda f(t, u(t)), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\eta) = 0 \end{array} \right. \tag{1.1} $$ In this article, by applying an iterative approach, we always assume that $\alpha \in [0,2)$ and $\eta \in [\frac{1}{2},1)$. We first recall the following fixed point of Krasnoselskii's type. Theorem 1.1. Let $E$ be a Banach space and $K$ be a cone in $E$. Assume that $\Omega_1$ and $\Omega_1$ are bounded open subsets of $E$ such that $0 \in \Omega_1$, $\overline{\Omega}_1 \subset \Omega_2$, and let $A: K \cap (\Omega_2 \setminus \Omega_1) \to K$ be a completely continuous operator such that either (1) $\| Au\| \leq \| Au\|$ for $u\in K\cap \partial \Omega_1$ and $\| Au\| \geq \| u\|$ for $u\in K\cap \partial \Omega_2$ or (2) $\| Au\| \geq \| Au\| for u\in K\cap \partial \Omega_1$ and $\| Au\| \leq \| u\|$ for $u\in K\cap \partial \Omega_{2}$ * A. has a fixed point in $K \cap (\overline{\Omega}_2 \setminus \Omega_1)$. ## II. PRELIMINARIES For the BVP $$ \left\{ \begin{array}{l} u^{(4)}(t) = \lambda f(t, u(t)), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\eta) = 0 \end{array} \right. \tag{2.1} $$ we have the following lemma Lemma 2.1. The BVP (2.1) has only trivial solution. Proof. It is simple to check. for any $y \in C[0,1]$, we consider the boundary value problems $$ \left\{ \begin{array}{l} u^{(4)}(t) = \lambda f(t, u(t)), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\eta) = 0 \end{array} \right. \tag{2.2} $$ After a direct computation, one may obtain the expression of Greens function $G(t,s)$ of the BVP (2.2) as following: Proof. Integrating four times the linear problem gives us that $$ u ^ {\prime \prime \prime} (t) = u ^ {\prime \prime \prime} (0) + \int_ {0} ^ {t} y (s) d s, $$ $$ u ^ {\prime \prime} (t) = u ^ {\prime \prime} (0) + t u ^ {\prime \prime \prime} (0) + \int_ {0} ^ {t} (t - s) y (s) d s, $$ $$ u ^ {\prime} (t) = u ^ {\prime} (0) + t u ^ {\prime \prime} (0) + \frac {t ^ {2}}{2} u ^ {\prime \prime \prime} (0) + \frac {1}{2} \int_ {0} ^ {t} (t - s) ^ {2} y (s) d s, $$ $$ u (t) = u (0) + t u ^ {\prime} (0) + \frac {t ^ {2}}{2} u ^ {\prime \prime} (0) + \frac {t ^ {3}}{6} u ^ {\prime \prime \prime} (0) + \frac {1}{6} \int_ {0} ^ {t} (t - s) ^ {3} y (s) d s. $$ The conditions $u'(0) = u'''(0) = 0$ implies that $$ u (t) = u (0) + \frac {t ^ {2}}{2} u ^ {\prime \prime} (0) + \frac {1}{6} \int_ {0} ^ {t} (t - s) ^ {3} y (s) d s, $$ and the conditions $u(1) = 0$ this means $$ u (1) = u (0) + \frac {1}{2} u ^ {\prime \prime} (0) + \frac {1}{6} \int_ {0} ^ {1} (1 - s) ^ {3} y (s) d s = 0, $$ Next, $\alpha u(0) + u''(\eta) = 0$ is rewritten Such as $$ u ^ {\prime \prime} (0) + \int_ {0} ^ {\eta} (\eta - s) y (s) d s - \frac {\alpha}{2} u ^ {\prime \prime} (0) - \frac {\alpha}{6} \int_ {0} ^ {1} (1 - s) ^ {3} y (s) d s = 0, $$ whence $$ u ^ {\prime \prime} (0) = \frac {\alpha}{3 (2 - \alpha)} \int_ {0} ^ {1} (1 - s) ^ {3} y (s) d s - \frac {2}{2 - \alpha} \int_ {0} ^ {\eta} (\eta - s) y (s) d s. \tag {2.3} $$ form The conditions $u(1) = 0$ we have $$ u (0) = - \frac {1}{2} u ^ {\prime \prime} (0) - \frac {1}{6} \int_ {0} ^ {1} (1 - s) ^ {3} y (s) d s $$ If we substitute(1.3) with the expression from above and simplify, we get that $$ u (0) = \frac {1}{2 - \alpha} \int_ {0} ^ {\eta} (\eta - s) y (s) d s - \frac {1}{3 (2 - \alpha)} \int_ {0} ^ {1} (1 - s) ^ {3} y (s) d s. \tag {2.4} $$ Finally, we obtain that $$ \begin{array}{l} { u ( t ) } { = } { \frac { 1 } { 2 - \alpha } \int _ { 0 } ^ { \eta } ( \eta - s ) y ( s ) d s - \frac { 1 } { 3 ( 2 - \alpha ) } \int _ { 0 } ^ { 1 } ( 1 - s ) ^ { 3 } y ( s ) d s } \\{ + } { \frac { \alpha t ^ { 2 } } { 6 ( 2 - \alpha ) } \int _ { 0 } ^ { 1 } ( 1 - s ) ^ { 3 } y ( s ) d s - \frac { t ^ { 2 } } { 2 - \alpha } \int _ { 0 } ^ { \eta } ( \eta - s ) y ( s ) d s } \\+ \frac {1}{6} \int_ {0} ^ {t} (t - s) ^ {3} y (s) d s. \\\end{array} $$ As a result, we have that For $s \geq \eta$ $$ G (t, s) = \left\{ \begin{array}{l l} \frac {- (2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} & 0 \leq t \leq s, \\ \frac {(t - s) ^ {3}}{6} - \frac {(2 - \alpha t ^ {3}) (1 - s) ^ {3}}{6 (2 - \alpha)} & s \leq t \leq 1 \end{array} \right. $$ and $s < \eta$ $$ G (t, s) = \left\{ \begin{array}{l l} \frac {6 (1 - t ^ {2}) (\eta - s) - (2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} & 0 \leq t \leq s, \\ \frac {(t - s) ^ {3}}{6} + \frac {6 (1 - t ^ {2}) (\eta - s) - (2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} & s \leq t \leq 1 \end{array} \right. $$ Remark 2.1. It is not difficult to verify that $G(t, s)$ has the following characteristics: $$ G (t, s) \geq 0 \text{for} 0 \leq s \leq \eta \text{and} G (t, s) \leq 0 \text{for} \eta \leq s \leq 1. $$ Moreover, if $s \geq \eta$, then $$ \max G (t, s): t \in [ 0, 1 ] = G (1, s) = 0, $$ $$ \min G (t, s): t \in [ 0, 1 ] = G (0, s) = \frac {- (1 - s) ^ {3}}{3 (2 - \alpha)} \geq \frac {- (1 - \eta) ^ {3}}{3 (2 - \alpha)} $$ if $s < \eta$, then $$ \max G (t, s): t \in [ 0, 1 ] = G (0, s) = \frac {(s ^ {3} - 3 s ^ {2}) + (3 \eta - 1)}{3 (2 - \alpha)} \leq \frac {(\eta^ {3} - 3 \eta^ {2}) + (3 \eta - 1)}{3 (2 - \alpha)}, $$ $$ \min G (t, s): t \in [ 0, 1 ] \quad = G (1, s) = 0 $$ therefore, if we let $\delta = \max |G(t,s)|:t,s\in [0,1]$ then $$ \delta = \max \left\{\frac {- (1 - \eta) ^ {3}}{3 (2 - \alpha)}, \frac {\left(\eta^ {3} - 3 \eta^ {2}\right) + (3 \eta - 1)}{3 (2 - \alpha)} \right\} < \frac {\eta - s}{(2 - \alpha)} $$ Now, let Banach space $E = C[0,1]$ is equipped with the $\| u\| = \max_{t\in [0,1]}|u(t)|$. $K = \{y\in C[0,1]:y(t)\}$ is nonnegative and decreasing on [0, 1]. Then $K$ is a cone in $C[0,1]$. Note that this order relationship is inducesan in $E$ by defining $uv$ if and only if $u - v\in K$ In the remainder of this paper, we always assume that $f: C[0,1] \times [0, +\infty) \to [0, +\infty)$ is continuous and satisfies the following conditions: (F1) For each $u \in [0, +\infty)$, the mapping $t \mapsto f(t,u)$ is decreasing; (F2) For each $t \in [0,1]$, the mapping $u \mapsto f(t,u)$ is increasing. $$ (A u) (t) = \int_ {0} ^ {1} G (t, s) f (s, u (s)) d s, t \in [ 0, 1 ] \tag {2.5} $$ Obviously, if $u$ is a fixed point of $A$ in $K$, then $u$ is a nonnegative and decreasing solution of the BVP (1.1). Lemma 2.2. Let $A:K\to K$ is completely continuous. Proof. let $u \in K$. Then, for $0 \leq t \leq \eta$, we have $$ (Au)(t) = \int_0^t \left[ \frac{(t-s)^3}{6} + \frac{6(1-t^2)(\eta-s) - (2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] y(s) ds + \int_t^\eta \left[ \frac{6(1-t^2)(\eta-s) - (2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] y(s) ds + \int_\eta^1 \frac{-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} y(s) ds $$ which together with $(F1)$ and $(F2)$ implies that $$ \begin{array}{l} (A u) ^ {\prime} (t) = \int_ {0} ^ {\eta} \frac {3 t ^ {2} (2 - \alpha) + 2 \alpha t (1 - s) ^ {3} - 1 2 t (\eta - s)}{6 (2 - \alpha)} y (s) d s \\+ \frac {1}{2} \int_ {0} ^ {t} (s ^ {2} - 2 t s) y (s) d s + \frac {t ^ {2}}{2} \int_ {t} ^ {\eta} y (s) d s \\+ \int_ {\eta} ^ {1} \frac {\alpha t (1 - s) ^ {3}}{3 (2 - \alpha)} y (s) d s \\\leq y (\eta) \left[ \int_ {0} ^ {\eta} \frac {3 t ^ {2} (2 - \alpha) + 2 \alpha t (1 - s) ^ {3} - 1 2 t (\eta - s)}{6 (2 - \alpha)} \right. \\+ \frac {1}{2} \int_ {0} ^ {t} (s ^ {2} - 2 t s) + \frac {t ^ {2}}{2} \int_ {t} ^ {\eta} + \int_ {\eta} ^ {1} \frac {\alpha t (1 - s) ^ {3}}{3 (2 - \alpha)} \bigg ] d s \\\leq t y (\eta) \left[ \frac {4 \eta t (2 - \alpha) + (\alpha - 8 \eta)}{1 2 (2 - \alpha)} - \frac {5 t ^ {2}}{6} + \frac {t \eta}{2} \right] \\\leq b t y (\eta) \left[ \frac {4 \eta t (2 - \alpha) + (\alpha - 8 \eta)}{1 2 (2 - \alpha)} - \frac {\eta^ {8}}{6} \right] \\\leq 0 \\\end{array} $$ At the same time, $\eta >\frac{1}{2}$ shows that $$ \begin{array}{l} (A u) ^ {\prime \prime} (t) = \int_ {0} ^ {\eta} \frac {6 t (2 - \alpha) + 2 \alpha (1 - s) ^ {3} - 1 2 (\eta - s)}{6 (2 - \alpha)} y (s) d s \\- \int_ {0} ^ {t} s y (s) d s + t \int_ {t} ^ {\eta} y (s) d s \\+ \int_ {\eta} ^ {1} \frac {\alpha (1 - s) ^ {3}}{3 (2 - \alpha)} y (s) d s \\\leq y (\eta) \left[ \int_ {0} ^ {\eta} \frac {6 t (2 - \alpha) + 2 \alpha (1 - s) ^ {3} - 1 2 (\eta - s)}{6 (2 - \alpha)} - \int_ {0} ^ {t} s d s \right. \\\left. + t \int_ {t} ^ {\eta} + \int_ {\eta} ^ {1} \frac {\alpha (1 - s) ^ {3}}{3 (2 - \alpha)} \right] d s \\\leq y (\eta) \left[ \frac {\alpha (3 t - 2 \eta) + 6 (\eta - t)}{2 (2 - \alpha)} - 2 \eta^ {2} + \alpha \right] \\\leq y (\eta) \left[ \frac {\eta (\alpha - 2 \eta)}{2 (2 - \alpha)} + \alpha \right] \\\leq 0 \quad t \in (0, \eta) \\\end{array} $$ For $t\in [\eta,1]$, we have $$ \begin{array}{l} (A u) (t) = \int_ {0} ^ {\eta} \left[ \frac {(t - s) ^ {3}}{6} - \frac {6 (1 - t ^ {2}) (\eta - s) - (2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} \right] y (s) d s \\+ \int_ {\eta} ^ {t} \left[ \frac {(t - s) ^ {3}}{6} - \frac {(2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} \right] y (s) d s \\+ \int_ {t} ^ {1} \left[ \frac {- (2 - \alpha t ^ {2}) (1 - s) ^ {3}}{6 (2 - \alpha)} \right] y (s) d s \\\end{array} $$ which together with $(F1)$ and $(F2)$ implies that $$ (Au)'(t) = \int_0^\eta \frac{3t^2(2-\alpha) + 2\alpha t(1-s)^3 - 12t(\eta-s)}{6(2-\alpha)} ds + \frac{1}{2} \int_0^\eta (s^2 - 2ts)y(s)ds + \int_\eta^t \frac{(t-s)^2}{2} + \int_\eta^1 \frac{2\alpha t(1-s)^3}{6(2-\alpha)} y(s)ds \leq y(\eta) \left[ \int_0^\eta \frac{3t^2(2-\alpha) + 2\alpha t(1-s)^3 - 12t(\eta-s)}{6(2-\alpha)} + \frac{1}{2} \int_0^\eta (s^2 - 2ts) + \int_\eta^t \frac{(t-s)^2}{2} + \int_\eta^1 \frac{\alpha t(1-s)^3}{6(2-\alpha)} \right] ds = ty(\eta) \left[ \frac{(\alpha - 12\eta)}{12(2-\alpha)} + t\eta - \frac{\eta^2}{2} + 1 + \frac{\eta^3}{6} - \eta \right] \leq ty(\eta) \left[ \frac{(\alpha - 12\eta)}{12(2-\alpha)} + \frac{\eta^3}{6} + \frac{\eta^2}{2} - \eta + 1 \right] \leq 0 \quad t \in (\eta, 1) $$ So, $(Au)(t)$ is decreasing on $[0,1]$. At the same time, since $(Au)(1) = 0$, we know that $(Au)(t)$ is nonnegative on $[0,1]$. This indicates that $(Au)(t) \in K$. Furthermore, although $G(t,s)$ is not continuous, it follows from known text book results, for example, see [12], that $A:K \to K$ is completely continuous Theorem 2.3. Assume that $f(t,0) \neq 0$ for $t \in [0,1]$ and there exist two positive constants $a$ and $b$ such that the following conditions are satisfied: $$ \begin{array}{l} (H 1) f (0, a) \leq 6 (2 - \alpha) a, \\(H 2) b \left(u _ {2} - u _ {1}\right) \leq f (t, u _ {2}) - f (t, u _ {1}) \geq 2 b \left(u _ {2} - u _ {1}\right), \quad t \in [ 0, 1 ], \\0 \leq u _ {1} \leq u _ {2} \leq a. \\\end{array} $$ If we construct a iterative sequence $v_{n+1} = Av_n$ and $n = 0,1,2, \dots$, where $v_0(t) \equiv 0$ for $t \in [0,1]$, then $v_{n,n=1}^{\infty}$ converges to $v^{\dagger}$ in $E$ and $v^{\dagger}$ is a decreasing positive solution of the BVP (1.1) Proof. Let $K_{a} = u \in K: \|u\| \leq a$. Then we may assert that $A: K_{a} \to K_{a}$. In fact, if $u \in K_{a}$, then it follows from (H1) that $$ \begin{array}{l} 0 \leq (A u) (t) = \int_ {0} ^ {1} G (t, s) f (s, u (s)) d s \\\leq \int_ {0} ^ {1} | G (t, s) | f (0, a) d s \\\leq 6 (2 - \alpha) a \delta \\< a, \quad t \in [ 0, 1 ], \\\end{array} $$ which indicate that$\| Au\| \leq a$ $$ so A: K_a \to K_a. $$ Now, we prove that $v_{n n=1}^{\infty}$ converges to $v^{\dagger}$ in $E$ and $v^{\dagger}$ is a decreasing positive solution of (1.1). Indeed, in view of $v_0 \in K_a$ and $A: K_a \to K_a$, we have $v_n \in K_a$, $n = 1,2, \dots$. Since the set $v_{n n=0}^{\infty}$ is bounded and $A$ is completely continuous, we know that the set $v_{n n=1}^{\infty}$ is relatively compact. In what follows, we prove that $v_{n n=0}^{\infty}$ is monotone by induction. First, it is explicit that $v_1 - v_0 = v_1 \in K$, Which indicates this $v_1 v_0$. Subsequently, we suppose that $v_{k-1} v_k$. Then $v_k - v_{k-1}$ is decreasing and $0 \leq v_{k-1}(t) \leq v_k(t) \leq a, 0 \leq t \leq 1$. So, it follows from (H2) that for $0 \leq t \leq \eta$ $$ v_{k+1}^\prime(t) - v_k^\prime(t) = \frac{1}{6(2-\alpha)} \int_0^\eta 3t^2(2-\alpha) + 2\alpha t(1-s)^3 - 12t(\eta-s) [f(s,v_k(s)) - f(s,v_{k-1}(s))] ds + \frac{1}{2} \int_0^\eta (s^2 - 2ts) [f(s,v_k(s)) - f(s,v_{k-1}(s))] ds + \frac{1}{2} \int_\eta^t (t-s)^2 [f(s,v_k(s)) - f(s,v_{k-1}(s))] ds + \frac{\alpha t}{3(2-\alpha)} \int_\eta^1 (1-s)^3 [f(s,v_k(s)) - f(s,v_{k-1}(s))] ds \leq \frac{b}{6(2-\alpha)} \int_0^\eta 3t^2(2-\alpha) + 2\alpha t(1-s)^3 - 12t(\eta-s) [v_k(s) - v_{k-1}(s)] ds + \frac{b}{2} \int_0^\eta (s^2 - 2ts) [v_k(s) - v_{k-1}(s)] ds + \frac{b}{2} \int_\eta^t (t-s)^2 [v_k(s) - v_{k-1}(s)] ds + \frac{\alpha t}{6(2-\alpha)} \int_\eta^1 (1-s)^3 [v_k(s) - v_{k-1}(s)] ds $$ $$ \begin{array}{l} \leq b \left[ v _ {k} (\eta) - v _ {k - 1} (\eta) \right] \\\times \left[ \frac{1}{6 (2 - \alpha)} \int_ {0} ^ {\eta} 3 t ^ {2} (2 - \alpha) + 2 \alpha t (1 - s) ^ {3} - 1 2 t (\eta - s) \\+ \frac{1}{2} \int_ {0} ^ {\eta} (s ^ {2} - 2 t s) + \frac{1}{2} \int_ {\eta} ^ {t} (t - s) ^ {2} + \frac{\alpha t}{3 (2 - \alpha)} \int_ {\eta} ^ {1} (1 - s) ^ {3} \right] d s \\= b \left[ v _ {k} (\eta) - v _ {k - 1} (\eta) \right] t \left[ \frac{ (\alpha - 1 2 \eta) }{1 2 (2 - \alpha)} + t \eta - \frac{\eta^ {2}}{2} + 1 + \frac{\eta^ {3}}{6} - \eta \right] \\\leq b \left[ v _ {k} (\eta) - v _ {k - 1} (\eta) \right] t \left[ \frac{ (\alpha - 1 2 \eta) }{1 2 (2 - \alpha)} + \frac{\eta^ {3}}{6} + \frac{\eta^ {2}}{2} - \eta + 1 \right] \\\leq 0 \quad t \in (\eta , 1) \\\end{array} $$ And therefore, $$ v_{k+1}^\prime(t) - v_k^\prime(t) \leq 0, \quad v_{k+1}^\prime\prime(t) - v_k^\prime\prime(t) \leq 0 \quad t \in [0,1] $$ This together with $$ v_{k+1}^\prime(t) - v_k^\prime(t) = \int_0^1 G(1,s)[f(s,v_k(s)) - f(s,v_{k-1}(s))]\,ds,\,t\in[0,1]. $$ $v_{k + 1}(t) - v_k(t)\geq 0,\quad t\in [0,1]$ Subsequently, given the above (1.7) and (1.8) that $v_{k + 1} - v_k\in K$, Which shows $v_{k + 1}v_{k}\in K$ Thus, we have shown that $v_{k+1}v_k \in K$, $n = 0,1,2\ldots$. Since $v_{n_{n=1}}^{\infty}$ Relatively compact and monotonous, there exists a $v^{\dagger} \in K_a$ such that $\lim_{n \to \infty} v_n = v^{\dagger}$, which together with the continuity of $A$ and the fact that $v_{n+1} = Av_n$ It means that $v^{\dagger} = Av^{\dagger}$. This indicates that $v^{\dagger}$ is a decreasing non-negative solution of (1.1). Moreover, in view of $f(t,0) \neq 0$, $t \in [0,1]$, we know that zero function is not a solution of (1.1), which indicates that $v^{\dagger}$ is a positive solution of (1.1). ## III. AN EXAMPLE Consider the boundary value problem: $$ \left\{ \begin{array}{l} u^{(4)}(t) = \frac{1}{8} u^2(t) + u(t) + (1 - t), \quad t \in [0, 1], \\ u'(0) = u'''(0) = u(1) = 0, \\ \alpha u(0) + u''(\frac{1}{2}) = 0 \end{array} \right. \tag{3.1} $$ If we let $\eta = \frac{1}{2},\alpha = 1$ and $f(t,u) = \frac{1}{8} u^{2}(t) + u(t) + (1 - t),(t,u)\in [0,1]\times$ $[0, + \infty)$, Then all the assumptions of Theorem 2.2 $a = 2$ and $b = 1$. It follows from Theorem 2.2 that (3.1) has a decreasing positive solution $v^{\dagger}$. Moreover, the iterative scheme is $v_{0}(t)\equiv 0$ for $t\in [0,1]$ $$ v_{n+1}(t) = \left\{ \begin{array}{l l} \int_0^t \left[ \frac{(t-s)^3}{6} + \frac{6(1-t^2)(\eta-s)-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \qquad + \int_t^\eta \left[ \frac{6(1-t^2)(\eta-s)-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \qquad + \int_\eta^1 \frac{-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \text{if } t \in [0, \frac{1}{2}] \quad n = 0, 1, 2, 3, 4, \dots \\ \int_0^\eta \left[ \frac{(t-s)^3}{6} - \frac{6(1-t^2)(\eta-s)-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \qquad + \int_\eta^t \left[ \frac{(t-s)^3}{6} - \frac{(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \qquad + \int_t^1 \left[ \frac{-(2-\alpha t^2)(1-s)^3}{6(2-\alpha)} \right] \left[ \frac{1}{8} u_n^2(s) + u_n(s) + (1-t) \right] ds \\ \text{if } t \in [\frac{1}{2}, 1] \quad n = 0, 1, 2, 3, \dots \end{array} \right. $$ ## IV. CONCLUSION in this paper, when $\alpha \in [0,2)$ and $\eta \in [\frac{1}{2},1)$, we have successfully constructed an animation sequence whose limit is just the ideal monotonic positive solution of boundary value problem (1.1). In addition, a zero function started with the iterative scheme, which shows that the iterative scheme is feasible. ### ACKNOWLEDGMENT This paper is supported by the National Natural Science Foundation of China(no.11961060), The Key Project of Natural Sciences Foundation of Gansu Province(no.18JR3RA084).

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