### Theorem 1

*Suppose we have the existence of ordered functions* \(\eta _{1},\eta _{2}\in C[0,1]\), *such that*, \(\eta _{1}(0)<\eta _{2}(0)\)*and* \(\eta _{1}(1)<\eta _{2}(1)\),

$$\begin{aligned}& \max \bigl\{ \bigl\vert \eta _{2}(1)-\eta _{1}(0) \bigr\vert ; \bigl\vert \eta _{1}(1)-\eta _{2}(0) \bigr\vert \bigr\} < a, \\& \Upsilon _{1}\bigl(\eta _{1}(0),t,w\bigr)\geq 0, \quad \forall (t,w)\in \bigl[\eta _{1}(0), \eta _{2}(0)\bigr] \times [-a,a], \\& \Upsilon _{1}\bigl(\eta _{2}(0),t,w\bigr)\leq 0, \quad \forall (t,w)\in \bigl[\eta _{1}(0), \eta _{2}(0)\bigr] \times [-a,a], \\& \eta _{1}(1)-\Upsilon _{2}\bigl(x,\eta _{1}(1),z\bigr)\leq 0,\quad \forall x,z\in \bigl[ \eta _{1}(0),\eta _{2}(0)\bigr], \\& \eta _{2}(1)-\Upsilon _{2}\bigl(x,\eta _{2}(1),z\bigr)\geq 0,\quad \forall x,z\in \bigl[ \eta _{1}(0),\eta _{2}(0)\bigr]. \end{aligned}$$

(10)

*Then the problem* (1)*–*(2) *has at least one solution*, *such that* \(\eta _{1}(0)-at\leq x(t)\leq \eta _{2}(0)+at\), *for all* \(t \in [0,1]\).

### Proof

Suppose we have the following modified BVP:

$$\begin{aligned} (\phi \bigl(x'(t)\bigr)'=G(x) (t), \quad \mbox{a.e. }t\in [0,1], \end{aligned}$$

(11)

with modified BCs

$$\begin{aligned}& x(0)=\mathbf{A}(x), \\& x(1)=\mathbf{B}(x), \end{aligned}$$

(12)

□

where

$$\begin{aligned} G(x) (t)=g_{1} \bigl(t,\rho \bigl(t,x(t) \bigr),\varrho \bigl(x'(t)\bigr) \bigr), \end{aligned}$$

with

$$\begin{aligned}& \rho \bigl(t,x(t)\bigr) = \max \bigl\{ \eta _{1}(0)-at,\min \bigl\{ x,\eta _{2}(0)+at\bigr\} \bigr\} , \\& \varrho \bigl(x'(t)\bigr) = \max \bigl\{ -a,\min \{x,a\}\bigr\} , \end{aligned}$$

and

$$\begin{aligned}& \mathbf{A}(x) = \sigma \bigl(x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr) \bigr), \\& \mathbf{B}(x) = \chi \biggl(\frac{1}{2}x(1)+\frac{1}{2}\Upsilon _{2} \bigl(x(0),x(1),x'(1) \bigr) \biggr), \end{aligned}$$

with

$$\begin{aligned}& \sigma (x) = \max \bigl\{ \eta _{1}(0),\min \bigl\{ x,\eta _{2}(0)\bigr\} \bigr\} , \\& \chi (x) = \max \bigl\{ \eta _{1}(1),\min \bigl\{ x,\eta _{2}(1)\bigr\} \bigr\} . \end{aligned}$$

For simplicity, we divide the proof in three steps.

*Step 1:* The solution of the problem (11)–(12) is equivalent to find a fixed points of the operator, \(\Omega :C^{1}[0,1]\longrightarrow C^{1}[0,1]\) defined as

$$\begin{aligned} \Omega (x) (t)=\mathbf{A}(x)+ \int _{0}^{t}\phi ^{-1} \biggl[\lambda _{x}+ \int _{0}^{s}G(x) (x)\,dx \biggr]\,ds,\quad \forall t\in [0,1]. \end{aligned}$$

(13)

Firstly, we ensure the existence of \(\lambda _{x}\in \mathbb{R}\). For this, we claim that for each \(x\in C^{1}[0,1]\), there exists a unique \(\lambda _{x}\in \mathbb{R}\), such that

$$\begin{aligned} \int _{0}^{1} \biggl(\phi ^{-1} \biggl[ \lambda _{x}+ \int _{0}^{t}G(x) (s)\,ds \biggr] \biggr)\,dt= \mathbf{B}(x)-\mathbf{A}(x). \end{aligned}$$

(14)

Let

$$\begin{aligned} k(t)= \int _{0}^{t}G(x) (s)\,ds\in C[0,1] \end{aligned}$$

(15)

and

$$\begin{aligned} g=\mathbf{B}(x)-\mathbf{A}(x). \end{aligned}$$

(16)

Obviously,

$$\begin{aligned} \vert g \vert \leq \bigl\vert \mathbf{B}(u)-\mathbf{A}(u) \bigr\vert \leq \max \bigl\vert \beta (1)-\alpha (0); \alpha (1)-\beta (0) \bigr\vert < a. \end{aligned}$$

(17)

Consequently, \(\lambda _{x}\in \mathbb{R}\) exists by Lemma 2.1.

Since \(G(x)\) is bounded and continuous on \([0,1]\), and the integral is a continuous function on \([0,s]\). Furthermore, \(\lambda _{x}\) exists, *ϕ* is a homomorphism and its inverse exists. Also, \(\phi ^{-1} [\lambda _{x}+\int _{0}^{s}G(x)(x)\,dx ]\) is continuous, and its integral exists. Therefore \(\Omega (x)\) is continuous on \([0,1]\). Further, the class \(\{\Omega (x):x\in C^{1}[0,1]\}\) is uniformly bounded and equicontinuous. Therefore in view of the Arzelà–Ascoli theorem \(\{\Omega (x):x\in C^{1}[0,1]\}\) is relatively compact. Consequently Ω is a compact map. Now the Schauder fixed point theorem guarantees the existence of at least a fixed point since Ω is continuous and compact.

*Step 2:* If \(x(t)\), \(t\in [0,1]\) be a solution of the problem (11)–(12), then it must satisfy

$$\begin{aligned} \eta _{1}(0)-at< x(t)< \eta _{2}(0)+at,\quad \forall t\in [0,1]. \end{aligned}$$

(18)

Since \(x(0)=\mathbf{A}(x)\),

$$\begin{aligned} x(0) =&\sigma \bigl(x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr) \bigr) \\ =&\max \bigl\{ \eta _{1}(0),\min \bigl\{ x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr),\eta _{2}(0)\bigr\} \bigr\} \\ =&\eta _{2}(0). \end{aligned}$$

(19)

Now, we show that

$$\begin{aligned} \eta _{1}(0)\leq x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr) \leq \eta _{2}(0). \end{aligned}$$

(20)

Suppose on the contrary that

$$\begin{aligned}& \eta _{2}(0)< x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr), \\& \text{then} \\& \eta _{2}(0)-x(0)< \Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr), \\& 0< \Upsilon _{1} \bigl(\eta _{2}(0),x(1),x'(0) \bigr), \end{aligned}$$

which implies a contradiction, because \(\Upsilon _{1} (\eta _{2}(0),x(1),x'(0) )\ngtr 0\). Hence

$$\begin{aligned} x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr)\leq \eta _{2}(0). \end{aligned}$$

(21)

Along the same lines, we can show that

$$\begin{aligned} \eta _{1}(0)\leq x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr). \end{aligned}$$

(22)

Using (21) and (22), we have

$$\begin{aligned} \eta _{1}(0)\leq x(0)\leq \eta _{2}(0). \end{aligned}$$

(23)

Now using the second boundary condition, \(x(1)=\mathbf{B}\), we may show that

$$\begin{aligned} \eta _{1}(1)\leq x(1)\leq \eta _{2}(1). \end{aligned}$$

(24)

Since \(\| x' \| _{\infty }< a\),

$$\begin{aligned} x(0)-at< x(t)< x(0)+at,\quad \forall t\in [0,1]. \end{aligned}$$

(25)

Using Eqs. (23)–(25), we have

$$\begin{aligned} \eta _{1}(0)-at< x(t)< \eta _{2}(0)+at,\quad \forall t\in [0,1]. \end{aligned}$$

(26)

*Step3:* If \(x(t)\) is a solution of the problem (11)–(12), then it must satisfy the BCs

$$\begin{aligned} \textstyle\begin{cases} \Upsilon _{1}(x(0),x(1),x'(0))=0, \\ \Upsilon _{2}(x(0),x(1),x'(1))=0. \end{cases}\displaystyle \end{aligned}$$

(27)

To satisfy the BCs (27), it is sufficient to show that

$$\begin{aligned} \textstyle\begin{cases} \eta _{1}(0)\leq x(0)+\Upsilon _{1}(x(0),x(1),x'(0))\leq \eta _{2}(0), \\ \eta _{1}(1)\leq \frac{1}{2}x(1)+\frac{1}{2}\Upsilon _{2}(x(0),x(1),x'(1)) \leq \eta _{2}(1). \end{cases}\displaystyle \end{aligned}$$

(28)

Suppose on the contrary that

$$\begin{aligned} \eta _{1}(0)\geq x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0)\bigr). \end{aligned}$$

(29)

Then

$$\begin{aligned} x(0) =&\sigma \bigl(x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr) \bigr) \\ =&\max \bigl\{ \eta _{1}(0),\min \bigl\{ x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0) \bigr),\eta _{2}(0)\bigr\} \bigr\} \\ =&\alpha (0). \end{aligned}$$

(30)

Using (29) and (30), we have

$$\begin{aligned} \Upsilon _{1} \bigl(\eta _{1}(0),x(1),x'(0) \bigr)< 0. \end{aligned}$$

(31)

This is a contradiction, because \(\Upsilon _{1} (\eta _{1}(0),x(1),x'(0) )\nless 0\). Hence

$$\begin{aligned} \eta _{1}(0)\leq x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0)\bigr). \end{aligned}$$

(32)

Similarly, we can show that

$$\begin{aligned} \eta _{2}(0)\geq x(0)+\Upsilon _{1} \bigl(x(0),x(1),x'(0)\bigr). \end{aligned}$$

(33)

Now, assume on contrary that

$$\begin{aligned} \eta _{2}(1)< \frac{1}{2}x(1)+\frac{1}{2} \Upsilon _{2}\bigl(x(0),x(1),x'(1)\bigr). \end{aligned}$$

(34)

Then

$$\begin{aligned} x(1) =&\mathbf{B}(x) \\ =&\chi \biggl(\frac{1}{2}x(1)+\frac{1}{2}\Upsilon _{2}\bigl(x(0),x(1),x'(1)\bigr) \biggr) \\ =&\max \biggl\{ x(1),\min \biggl\{ \frac{1}{2}x(1)+ \frac{1}{2}\Upsilon _{2}\bigl(x(0),x(1),x'(1)\bigr), \eta _{2}(1)\biggr\} \biggr\} \\ =&\eta _{2}(1). \end{aligned}$$

(35)

Using Eqs. (34) and (35), we have

$$\begin{aligned} \eta _{2}(1)-\Upsilon _{2}\bigl(x(0),\eta _{2}(1),x'(1)\bigr)< 0, \end{aligned}$$

(36)

which implies a contradiction, because \(\eta _{2}(1)-\Upsilon _{2}(x(0),\eta _{2}(1),x'(1))\nless 0\). Hence

$$\begin{aligned} \eta _{2}(1)\geq \frac{1}{2}x(1)+ \frac{1}{2}\Upsilon _{2}\bigl(x(0),x(1),x'(1)\bigr). \end{aligned}$$

(37)

Along the same line, we can show that

$$\begin{aligned} \eta _{1}(1)\leq \frac{1}{2}x(1)+ \frac{1}{2}\Upsilon _{2}\bigl(x(0),x(1),x'(1)\bigr). \end{aligned}$$

(38)

Using Eqs. (32), (33), (37), and (38), the BCs (27) are satisfied.

Hence, \(x(t)\), \(t\in [0,1]\) is a solution of the modified BVP (11)–(12) which leads to the solution of the BVP (1)–(2).