In this section, we will establish the method of lower and upper solutions for (1.1), (1.2) in the case \(k_{1}<0<k_{2}\).
Denote
$$ g_{\alpha}(x)=L\bigl(\alpha(x)\bigr)f\bigl(x,\alpha(x)\bigr), \qquad g_{\beta}(x)=L\bigl(\beta(x)\bigr)f\bigl(x,\beta(x)\bigr), \quad x \in[0,1]. $$
(3.1)
Then
$$ g_{\alpha}(x)\leq0, \qquad g_{\beta}(x)\geq0, \quad x\in[0,1]. $$
(3.2)
Now let \(v_{\alpha}(x)\) be the solution of
$$ \begin{aligned} &L v_{\alpha}(x)=0, \quad x\in(0,1), \\ &v_{\alpha}(0)=\alpha(0), \qquad v_{\alpha}(1)=\alpha(1), \qquad v''_{\alpha}(0)=\alpha''(0), \qquad v''_{\alpha}(1)=\alpha''(1). \end{aligned} $$
(3.3)
Then \(v_{\alpha}(x)\) is uniquely determined as
$$ v_{\alpha}(x)= \alpha(0)w(x)+\alpha(1)w(1x)+\alpha''(0) \chi(x)+\alpha ''(1)\chi(1x), $$
(3.4)
where \(w(x)\) is the unique solution of the nonhomogeneous problem
$$ L(y)=0, \qquad y(0)=1, \qquad y''(0)=y(1)=y''(1)=0, $$
(3.5)
and it can be explicitly given by
$$ w(x)=\frac{m^{2}}{r^{2}+m^{2}}\frac{\sinh[r(1x)]}{\sinh r}+\frac {r^{2}}{r^{2}+m^{2}}\frac{\sin[m(1x)]}{\sin m}, $$
(3.6)
\(\chi(x)\) is the unique solution of the nonhomogeneous problem
$$ L(y)=0,\qquad y(0)=0, \qquad y''(0)=1, \qquad y(1)=y''(1)=0, $$
(3.7)
and it can be explicitly given by
$$\chi(x)=\frac{1}{(r^{2}+m^{2})}\frac{\sinh[r(1x)]}{\sinh r}\frac {1}{(r^{2}+m^{2})}\frac{\sin[m(1x)]}{\sin m}. $$
Let \(v_{\beta}(x)\) be the solution of
$$\begin{aligned} &L v_{\beta}(x)=0, \quad x\in(0,1), \\ &v_{\beta}(0)=\beta(0), \qquad v_{\beta}(1)=\beta(1), \qquad v''_{\beta}(0)=\beta''(0), \qquad v''_{\beta}(1)=\beta''(1). \end{aligned} $$
Then \(v_{\beta}(x)\) is uniquely determined as
$$v_{\beta}(x)= \beta(0)w(x)+\beta(1)w(1x)+\beta''(0) \chi(x)+\beta ''(1)\chi(1x). $$
Lemma 3.1

(1)
Let
\(0 < r < \infty\)
and
\(0< m<\pi\). Then
\(w(x)> 0\)
for
\(x\in(0,1)\).

(2)
Let
\(0 < r < \infty\)
and
\(0< m<\pi\). Then
\(\chi(x)< 0\)
for
\(x\in(0,1)\).
Proof
(1) Since \(r(1x)>0\) and \(\infty< m(1x)<\pi\) for \(x\in (0,1)\), it follows from (3.6) that \(w(x)> 0\) for \(x\in(0,1)\).
(2) Obviously, (3.7) is equivalent to the system
$$\begin{aligned}& L_{2}\chi=Z,\quad\chi(0)=0, \chi(1)=0, \end{aligned}$$
(3.8)
$$\begin{aligned}& L_{1}Z=0,\quad Z(0)=1,Z(1)=0. \end{aligned}$$
(3.9)
It is easy to see from (3.9) and the fact \(G_{1}(t,s)>0\) for \((t,s)\in (0,1)\times(0,1)\) that
$$Z(x)>0\quad\text{for } x\in[0,1). $$
Combining this with (3.8) and using the fact \(G_{2}(t,s)>0\) for \((t,s)\in (0,1)\times(0,1)\), we deduce that \(\chi(x)<0\) in \((0, 1)\). □
From Lemma 3.1 and the definitions of \(v_{\alpha}\) and \(v_{\beta}\), it follows that
$$ v_{\alpha}(x)\leq0, \qquad v_{\beta}(x)\geq0,\quad x\in[0,1]. $$
(3.10)
Now, for a lower solution α of (1.1), (1.2), we have the following implications:
$$\begin{aligned} &L(\alpha(x))=f\bigl(x, \alpha(x)\bigr)+g_{\alpha}(x)\\ &\quad \Rightarrow\quad \alpha(x)=v_{\alpha}(x)+ \int^{1}_{0} G(x, s)f\bigl(s, \alpha(s)\bigr)\,ds+ \int^{1}_{0} G(x, s)g_{\alpha}(s)\,ds \\ &\quad \Rightarrow\quad \alpha(x) \leq T\alpha(x)\quad\text{on } [0,1], \end{aligned} $$
and, by a similar way, we obtain \(\beta(x)\geq T\beta(x)\) on \([0, 1]\), where \(T: C[0, 1]\to C^{4}[0, 1]\) is the operator defined by
$$ T\phi(x)= \int^{1}_{0} G(x,s)f\bigl(s, \phi(s)\bigr)\,ds, \quad0 \leq x\leq1, $$
(3.11)
where the Green function G is as in (2.2). It is easy to check that (1.1), (1.2) is equivalent to the operator equation
As a direct consequence of the Schauder fixed point theorem [4, Theorem 5], we have the following lemma.
Lemma 3.2
Let there exist a constant
M
such that
$$\bigl\vert f(x,y) \bigr\vert \leq M $$
for
\((x, y)\in[0, 1]\times\mathbb{R}\). Then (1.1), (1.2) has a solution.
Theorem 3.1
Let
\(k_{1}<0<k_{2}<\pi^{2}\). Suppose that for problem (1.1), (1.2) there exist a lower solution
α
and an upper solution
β
such that
$$\alpha(x) \leq\beta(x) \quad \textit{for } x \in[0, 1]. $$
If
\(f: [0, 1]\times\mathbb{R}\to\mathbb{R}\)
is continuous and satisfies
$$ f(x,u_{1}) \leq f(x,u_{2} )\quad\textit{for } \alpha(x) \leq u_{1} \leq u_{2} \leq\beta(x), \textit{ and } x \in[0,1], $$
(3.13)
then there exists a solution
\(y(x)\)
for (1.1), (1.2) satisfying
$$ \alpha(x)\leq y(x)\leq\beta(x)\quad\textit{for } 0\leq x\leq1. $$
(3.14)
Proof
Define the function F on \([0, 1] \times\mathbb{R}\) by setting
$$F(x,y)= \textstyle\begin{cases} f(x, \beta(x)), &y>\beta(x),\\ f(x,y), &\alpha(x)\leq y\leq\beta(x),\\ f(x, \alpha(x)), &y< \alpha(x). \end{cases} $$
Since F is continuous and bounded on \([0, 1]\times\mathbb{R}\), by Lemma 3.2, there exists a solution y of the problem
$$\begin{aligned} &L(y) = F(x, y), \\ &y(0)=y(1)=y''(0)=y''(1)=0. \end{aligned} $$
We now show that inequality (3.14) is true. We have
$$L \bigl(y(x)\beta(x) \bigr)=L\bigl(y(x)\bigr)L\bigl(\beta(x)\bigr)\leq F \bigl(x, y(x)\bigr)f\bigl(x, \beta (x)\bigr)\leq0. $$
Thus \(L (y(x)\beta(x) )=h_{1}(x)\leq0\) for \(x\in[0,1]\), that is, from Theorem 2.1 and (3.10)
$$y(x)\beta(x)=v_{\beta}(x)+ \int^{1}_{0} G(t,s)h_{1}(s)\,ds\leq0 \quad \text{for } x\in[0,1]. $$
By a similar way,
$$L \bigl(y(x)\alpha(x) \bigr)=L\bigl(y(x)\bigr)L\bigl(\alpha(x)\bigr)\geq F \bigl(x, y(x)\bigr)f\bigl(x, \alpha(x)\bigr)\geq0. $$
Thus \(L (y(x)\alpha(x) )=h_{2}(x)\geq0\) for \(x\in[0,1]\), that is, from Theorem 2.1 and (3.10)
$$y(x)\alpha(x)=v_{\alpha}(x)+ \int^{1}_{0} G(t,s)h_{2}(s)\,ds\geq0 \quad \text{for } x\in[0,1]. $$
Therefore, \(\alpha(x)\leq y(x)\leq\beta(x)\) for \(x\in[0,1]\), and accordingly, y is a solution of (1.1), (1.2). □
Remark 3.1
It is worth remarking that if (3.13) is not valid, then the existence of a lower solution α and an upper solution β with \(\alpha (x)\leq\beta(x)\) in \([0,1]\) cannot guarantee the existence of solutions in the order interval \([\alpha(x), \beta(x)]\). Let us see the counterexample in Cabada et al. [5, Remark 3.1].
Remark 3.2
In the case \(\vert k_{1} \vert > k_{2}\), the assertions of Theorem 3.1 can be deduced from Habets and Sanchez [24, Theorem 4.1].
Remark 3.3
Let us consider the problem
$$ \begin{aligned} &u^{(4)}(x)4u''(x)+3u(x)=u^{3}+ \sin x, \quad x\in(0,1), \\ &u(0)=u(1)=u''(0)=u''(1)=0. \end{aligned} $$
(3.15)
It is easy to verify that \(f(x,u)=u^{3}+\sin x\), \(k_{1}=1\) and \(k_{2}=4\), and
$$\alpha(x)\equiv1, \qquad\beta(x)\equiv1 $$
satisfy all of the conditions in Theorem 3.1. Therefore, (3.15) has a solution u satisfying
$$1\leq u(x)\leq1,\quad x\in[0,1]. $$