To obtain the lower and upper solutions for problem (1.7)–(1.8), we consider the generalization case such as
$$ \textstyle\begin{cases} -u''(t)= f(t,u(t),u'(t)),\quad 0< t< 1, t\neq t_{1}, \\ \Delta u|_{t=t_{1}}=I(u(t_{1})), \\ u(0)=a\geq 0,\qquad u(1)=b\geq 0, \end{cases} $$
(2.1)
and
$$ \Delta u'|_{t=t_{1}}=N \bigl(u(t_{1}),u'(t_{1}-)\bigr). $$
(2.2)
The solution is defined as \(u\in PC^{1}([0,1],\mathbb{R})\cap PC^{2}((0,1),\mathbb{R})\) satisfying (2.1)–(2.2), where \(PC([0,1],\mathbb{R})=\{u:[0,1]\to \mathbb{R}:u(t)\text{ is continuous at } t\neq t_{1}, u(t_{1}-0)=\lim_{t\to t_{1}^{-}}u(t)\mbox{ and } u(t_{1}+0)=\lim_{t\to t_{1}^{+}}u(t)\mbox{ exist}\}\) (denoted by \(PC[0,1]\) simply), \(PC^{1}([0,1], \mathbb{R})=\{u\in PC[0,1]| u'(t)\mbox{ is continuous on } [0,t_{1})\cup (t_{1},1], \lim_{t\to t_{1}^{-}}u'(t)\mbox{ and } \lim_{t\to t_{1}^{+}}u'(t)\mbox{ exist}\}\) with the norm \(\|u\|_{1}=\max \{\sup_{t\in [0,1]}|u(t)|,\sup_{t\in [0,1]}|u'(t)| \}\) (denoted by \(PC^{1}[0,1]\) simply), and \(PC^{2}((0,1), \mathbb{R})=\{u\in PC([0,1],\mathbb{R})|u''(t)\mbox{ is continuous on }(0,t_{1})\cup (t_{1},1)\mbox{ and } \lim_{t\to t_{1}^{-}}u''(t), \lim_{t\to t_{1}^{+}}u''(t)\mbox{ exist}\}\) (denoted by \(PC^{2}(0,1)\) simply).
If \(\alpha \in PC^{1}[0,1]\cap PC^{2}(0,1)\) satisfies
$$ \textstyle\begin{cases} -\alpha ''(t)\leq f(t,\alpha (t),\alpha '(t)),\quad t \in (0,1), t\neq t_{1}, \\ \Delta \alpha |_{t=t_{1}}=I(\alpha (t_{1})), \\ \Delta \alpha '|_{t=t_{1}}\geq N(\alpha (t_{1}),\alpha '(t_{1}-)), \\ \alpha (0)\leq a,\qquad \alpha (1)\leq b, \end{cases} $$
it is a lower solution of problem (2.1)–(2.2), while the upper solution is defined as \(\beta \in PC^{1}[0,1]\cap PC^{2}(0,1)\) satisfying
$$ \textstyle\begin{cases} -\beta ''(t)\geq f(t,\beta (t),\beta '(t)),\quad t\in (0,1), t\neq t_{1}, \\ \Delta \beta |_{t=t_{1}}=I(\beta (t_{1})), \\ \Delta \beta '|_{t=t_{1}}\leq N(\beta (t_{1}),\beta '(t_{1}-)), \\ \beta (0)\geq a,\qquad \beta (1)\geq b. \end{cases} $$
Denote
$$ D_{\alpha }^{\beta }=\bigl\{ (t,u)\in (0,1)\times \mathbb{R}| \alpha (t)\leq u \leq \beta (t), t\in (0,1)\bigr\} , $$
where α, \(\beta \in PC^{1}([0,1],\mathbb{R})\) with \(\alpha (t)\leq \beta (t)\) for all \(t\in [0,1]\). Then we give the lower and upper solution theorem of problem (2.1)–(2.2).
Now we list a condition for convenience:
\((A_{0})\)\(I\in C(\mathbb{R},\mathbb{R})\) and \(u+I(u)\) is nondecreasing.
To establish a fundamental theorem of the upper and lower solution method for problem (2.1)–(2.2), we need to use a corresponding theorem for the following singular boundary value problem:
$$ \textstyle\begin{cases} -u''(t)=f(t,u(t),u'(t)),& t\in (0,1), \\ u(0)=a,\qquad u(1)=b. \end{cases} $$
(2.3)
The definitions of upper and lower solutions of problem (2.3) can be found in [11, 39]. Now we list the following lemma [45].
Lemma 2.1
(see [45])
Let \(u\in H^{2,1}[0,1]\)and \(M>0\)be a constant such that \(|u^{\prime \prime }(t)|\leq K|u'(t)|^{2}+f_{0}(t)\)with \(K>0\), \(f_{0}\in L^{1}[0,1]\)and \(\sup_{t\in [0,1]}|u(t)|\leq M\). Then there exists a constant \(k_{0}>0\)such that
$$ \vert u \vert _{H^{2,1}[0,1]}\leq k_{0}.$$
Lemma 2.2
Assume that the function \(f(t, u, v)\)is continuous in \((0,1)\times (0,+\infty )\times \mathbb{R}\). Furthermore, suppose that problem (2.3) has a pair of upper and lower solutions \(\beta (t)\)and \(\alpha (t) \)satisfying the conditions
-
(1)
α, \(\beta \in C^{1}[0,1]\cap C^{2}(0,1)\);
-
(2)
\(0<\alpha (t)\leq \beta (t)\), \(\forall t\in (0,1)\);
-
(3)
there exist a nonnegative function \(h\in L^{1}[0,1]\)and a positive constant \(K>0\)such that
$$ \bigl\vert f(t,u,v) \bigr\vert \leq h(t)+K \vert v \vert ^{2},\quad \forall (t,u)\in D^{\beta }_{\alpha }, v\in \mathbb{R}.$$
(2.4)
Then problem (2.3) has a solution \(u(t)\)belonging to \(C^{1}[0,1]\)and satisfying
$$ \alpha (t)\leq u(t)\leq \beta (t),\quad t\in [0,1]. $$
Proof
Let \(\overline{f}:(0,1)\times \mathbb{R}\times \mathbb{R}\) be defined by
$$\begin{aligned}& \overline{f}(t,u,v)\\& \quad = \textstyle\begin{cases} f(t,u,v),&\text{if } \alpha (t)\leq u\leq \beta (t); \\ f(t,\alpha (t),v)+\frac{1}{\alpha (t)}( \vert f(t,\alpha (t),0) \vert +1) \min \{\alpha (t),\alpha (t)-u\},&\text{if } u< \alpha (t); \\ f(t,\beta (t),v)-\frac{1}{\beta (t)}( \vert f(t,\beta (t),0) \vert +1) \min \{\beta (t),u-\beta (t)\},&\text{if } u>\beta (t). \end{cases}\displaystyle \end{aligned}$$
Now we consider the following problem:
$$ \textstyle\begin{cases} -u''(t)=\overline{f}(t,u(t),u'(t)),&t\in (0,1), \\ u(0)=a,\qquad u(1)=b. \end{cases} $$
(2.5)
First, we show that the solution of problem (2.5) is the solution of problem (2.3).
Now we prove that \(\alpha (t)\leq u(t)\leq \beta (t)\), \(t\in (0,1)\). In fact, if there is a \(t_{0}\in (0,1)\) with \(u(t_{0})<\alpha (t_{0})\), let \(t_{*}=\inf \{t< t_{0}|u(s)<\alpha (s), \forall s\in [t,t_{0}]\}\) and \(t^{*}=\sup \{t>t_{0}|u(s)<\alpha (s), \forall s\in [t_{0},t]\}\). Since \(\alpha (0)\leq a=u(0)\), \(\alpha (1)\leq b=u(1)\), we have \(u(t)<\alpha (t)\), \(\forall t\in (t_{*},t^{*})\) and \(u(t_{*})=\alpha (t_{*})\) and \(u(t^{*})=\alpha (t^{*})\). Suppose that \(t'\in (t_{*},t^{*})\) satisfies \(u(t')-\alpha (t')=\min_{t\in [t_{*},t^{*}]}(u(t)-\alpha (t))<0\). Then \(\alpha '(t')=u'(t')\). Hence,
$$\begin{aligned} 0 \geq& -\bigl(u\bigl(t'\bigr)-\alpha \bigl(t' \bigr)\bigr)'' \\ \geq& f\bigl(t',\alpha \bigl(t' \bigr),u'\bigl(t'\bigr)\bigr)+ \frac{1}{\alpha (t')} \bigl( \bigl\vert f\bigl(t', \alpha \bigl(t' \bigr),0\bigr) \bigr\vert +1\bigr)\min \bigl\{ \alpha \bigl(t' \bigr),\alpha \bigl(t'\bigr)-u\bigl(t'\bigr)\bigr\} \\ &{}-f\bigl(t',\alpha \bigl(t'\bigr), \alpha '\bigl(t'\bigr)\bigr) \\ =& f\bigl(t',\alpha \bigl(t'\bigr),\alpha '\bigl(t'\bigr)\bigr)+ \frac{1}{\alpha (t')}\bigl( \bigl\vert f\bigl(t',\alpha \bigl(t'\bigr),0\bigr) \bigr\vert +1\bigr)\min \bigl\{ \alpha \bigl(t'\bigr), \alpha \bigl(t'\bigr)-u\bigl(t'\bigr)\bigr\} \\ &{}-f \bigl(t',\alpha \bigl(t'\bigr),\alpha '\bigl(t'\bigr)\bigr) \\ >& 0. \end{aligned}$$
This is a contradiction. Thus, \(0<\alpha (t)\leq u(t)\) for all \(t\in [0,1]\).
A same argument shows that \(u(t)\leq \beta (t)\) for all \(t\in [0,1]\).
Hence, \(\forall t\in (0,1)\), we have \(\alpha (t)\leq u(t)\leq \beta (t)\), which
$$ \overline{f}\bigl(t,u(t),u'(t)\bigr)=f\bigl(t,u(t),u'(t) \bigr),\quad t\in (0,1), $$
that is, \(u(t)\) is a solution of problem (2.3).
Next, we show that problem (2.5) has at least one solution.
It follows from \(h\in L[0,1]\), (2.4) and the definition of f̅ that
$$ \bigl\vert \overline{f}(t,u,v) \bigr\vert \leq 2h(t)+1+ \Vert \alpha \Vert _{1}+ \Vert \beta \Vert _{1}+K \vert v \vert ^{2}, \quad t\in [0,1],(u,v)\in \mathbb{R}^{2}. $$
Hence, for \(u\in C^{1}[0,1]\), the A can be defined well as
$$ (Au) (t)=\psi (t)+ \int _{0}^{1}G(t,s)\overline{f} \bigl(s,u(s),u'(s)\bigr)\,ds,\quad t \in [0,1], $$
where \(\psi (t)=a+t(b-a)\) and
$$ G(t,s)= \textstyle\begin{cases} s(1-t), &0\leq s\leq t\leq 1; \\ t(1-s), &0\leq t\leq s\leq 1. \end{cases} $$
It is easy to see that the fixed point of A is the solution of problem (2.5). Moreover, since
$$ \bigl\vert (Au)''(t) \bigr\vert \leq 2h(t)+1+ \Vert \alpha \Vert _{1}+ \Vert \beta \Vert _{1}+K \bigl\vert u'(t) \bigr\vert ^{2}, \quad t\in [0,1] $$
and f̅ is continuous, the Arzela–Ascoli theorem and the dominated convergence theorem guarantee that \(A:C^{1}[0,1]\to C^{1}[0,1] \) is compact.
For \(u\in C^{1}[0,1]\) and \(\lambda \in (0,1)\), if \(u=\lambda Au\), we claim that
$$ 0\leq u(t)\leq 2\max_{t\in [0,1]}\beta (t),\quad \forall t\in [0,1]. $$
(2.6)
(1) Suppose that there exists a \(t_{1}\in (0,1)\) such that \(u(t_{1})<0\). Then there exists a \(t'_{1}\in (0,1)\) such that \(u'(t'_{1})=\min_{t\in [0,1]}u(t)<0\) and \(u'(t'_{1})=0\), which together with \(u=\lambda Au\) implies that
$$\begin{aligned} 0 \geq& -u''\bigl(t'_{1} \bigr) \\ =&\lambda \overline{f}\bigl(t',u\bigl(t'_{1} \bigr),u'\bigl(t'_{1}\bigr)\bigr) \\ =&\lambda \biggl(f\bigl(t'_{1},\alpha \bigl(t'_{1}\bigr),0\bigr)+ \frac{1}{\alpha (t'_{1})}\bigl( \bigl\vert f\bigl(t'_{1},\alpha \bigl(t'_{1}\bigr),0\bigr) \bigr\vert +1\bigr)\min \bigl\{ \alpha \bigl(t'_{1}\bigr),\alpha \bigl(t'_{1}\bigr)-u\bigl(t'_{1} \bigr)\bigr\} \biggr) \\ =&\lambda \bigl(f\bigl(t'_{1},\alpha \bigl(t'_{1}\bigr),0\bigr)+ \bigl\vert f \bigl(t'_{1}, \alpha \bigl(t'_{1} \bigr),0\bigr) \bigr\vert +1\bigr) \\ >&0. \end{aligned}$$
This is a contradiction.
(2) Suppose that there exists a \(t_{2}\in (0,1)\) such that \(u(t_{2})>2\max_{t\in [0,1]}\beta (t)\). Then there exists a \(t'_{2}\in (0,1)\) such that \(u'(t'_{2})=\max_{t\in [0,1]}u(t)>2\max_{t\in [0,1]}\beta (t)\) and \(u'(t'_{2})=0\), which together with \(u=\lambda Au\) implies that
$$\begin{aligned} 0 \leq &-u''\bigl(t'_{2} \bigr) \\ =&\lambda \overline{f}\bigl(t'_{2},u \bigl(t'_{2}\bigr),u' \bigl(t'_{2}\bigr)\bigr) \\ =&\lambda (f\bigl(t'_{2},\beta \bigl(t'_{2}\bigr),0\bigr)- \frac{1}{\beta (t'_{2})} \bigl( \bigl\vert f\bigl(t'_{2},\beta \bigl(t'_{2}\bigr),0\bigr) \bigr\vert +1\bigr)\min \bigl\{ \beta \bigl(t'_{2}\bigr),u\bigl(t'_{2} \bigr)-\beta \bigl(t'_{2}\bigr)\bigr\} \\ =&-\lambda \\ < &0. \end{aligned}$$
This is a contradiction also.
Then, if \(u=\lambda Au\), from (2.6) and
$$ \bigl\vert u''(t) \bigr\vert \leq 2h(t)+1+ \Vert \alpha \Vert _{1}+ \Vert \beta \Vert _{1}+K \bigl\vert u'(t) \bigr\vert ^{2}, $$
Lemma 2.1 means that there is a \(k_{0}>0\) such that \(|u|_{H^{2,1}[0,1]}\leq k_{0}\). The interpolation inequality lemma guarantees that there exists an \(R>0\) such that
$$ \Vert u \Vert _{1}< R. $$
The Leray–Schauder fixed point theorem guarantees that A has at least one fixed point u in \(C^{1}[0,1]\).
Consequently, problem (2.3) has at least one solution \(u\in C^{1}[0,1]\) such that \(\alpha (t)\leq u(t)\leq \beta (t)\) for all \(t\in [0,1]\). □
Remark 2.3
The idea of this lemma comes from [4, 6, 9, 10, 17, 52].
Remark 2.4
If \(f(t,u,v)=u^{-\frac{1}{3}}+u^{2}+v^{2}\), let \(\alpha (t)=\sin \pi t\), \(t\in [0,1]\) and \(\beta (t)\equiv 2\), \(t\in [0,1]\) and \(D_{\alpha }^{\beta }=\{(t,u)\in (0,1)\times \mathbb{R}| \alpha (t) \leq u\leq \beta (t)\}\). Then
$$ \bigl\vert f(t,u,v) \bigr\vert \leq \frac{1}{(\sin \pi t)^{\frac{1}{3}}}+4+v^{2}=h(t)+v^{2}. $$
Obviously, \(h\in L^{1}[0,1]\). Hence, (2.4) is true.
Lemma 2.5
Suppose that α, \(\beta \in PC^{1}[0, 1]\cap PC^{2}(0, 1)\)are, respectively, lower and upper solutions for problem (2.1)–(2.2) with \(\alpha (t)\leq \beta (t)\)and \(\alpha (t_{1})=\beta (t_{1})\). Moreover, u is a solution of problem (2.1) such that \(\alpha (t)\leq u(t)\leq \beta (t)\)for all \(t\in [0,1]\). Then u satisfies (2.2), i.e. \(u(t)\)is a solution of problem (2.1)–(2.2) such that \(\alpha (t)\leq u(t)\leq \beta (t)\)on \([0, 1]\).
Proof
Since \(\alpha (t_{1})=\beta (t_{1})\) and \(\alpha (t)\leq u(t)\leq \beta (t)\) for all \(t\in [0,1]\), we have \(u(t_{1})=\alpha (t_{1})=\beta (t_{1})\) and \(\alpha '(t_{1}-)\geq u'(t_{1}-)\), \(\alpha '(t_{1}+)\leq u'(t_{1}+)\) and \(\beta '(t_{1}-)\leq u'(t_{1}-)\), \(\beta '(t_{1}+)\geq u'(t_{1}+)\). Thus,
$$\begin{aligned} \Delta u'|_{t=t_{1}} =&u'(t_{1}+)-u'(t_{1}-) \\ \geq& \alpha '(t_{1}+)-\alpha '(t_{1}-) \\ \geq& N\bigl(\alpha (t_{1}),\alpha '(t_{1}-) \bigr) \\ \geq& N\bigl(u(t_{1}),u'(t_{1}-)\bigr) \end{aligned}$$
and
$$\begin{aligned} \Delta u'|_{t=t_{1}} =&u'(t_{1}+)-u'(t_{1}-) \\ \leq &\beta '(t_{1}+)-\beta '(t_{1}-) \\ \leq& N\bigl(\beta (t_{1}),\beta '(t_{1}-) \bigr) \\ \leq &N\bigl(u(t_{1}),u'(t_{1}-) \bigr), \end{aligned}$$
that is, \(u(t)\) satisfies (2.2). Consequently, \(u(t)\) is a solution of problem (2.1)–(2.2). □
Theorem 2.6
Assume
\((A_{0})\)
and also assume
-
(1)
\(f(t,u,v)\)is continuous in u, \(v\in \mathbb{R}\), \(t\neq t_{1}\). Furthermore, \(\lim_{t\to t_{1}-} f(t,u, v)=f(t_{1},u,v)\)and \(\lim_{t\to t_{1}+}f(t,u,v)\)exists.
-
(2)
\(N(u,v)\in C(\mathbb{R}\times \mathbb{R},\mathbb{R})\)is nondecreasing in v.
-
(3)
α and β are, respectively, lower and upper solutions of problem (2.1)–(2.2) such that \(\alpha (t)\leq \beta (t)\)for all \(t\in [0, 1]\).
-
(4)
There exists a function \(h\in L^{1}[0,1]\)such that \(|f(t,u,v)| \leq h(t)+K|v|^{2}\)for all \((t, u)\in D_{\alpha }^{\beta }\)and \(v\in \mathbb{R}\).
Then problem (2.1)–(2.2) has at least one solution u such that \(\alpha (t)\leq u(t)\leq \beta (t)\)for all \(t\in [0,1]\).
Proof
Let \(C\in [\alpha (t_{1}),\beta (t_{1})]\). Condition (3) implies that α and β are, respectively, lower and upper solutions of the following problem:
$$ \textstyle\begin{cases} -u^{\prime \prime }(t)=f(t,u,u'),\quad t\in (0,t_{1}), \\ u(0)=a,\qquad u(t_{1})=C. \end{cases} $$
(2.7)
Thus by Lemma 2.2, problem (2.7) has a solution \(v\in C^{1}[0,t_{1}]\cap C^{2}(0,t_{1})\) satisfying
$$ \alpha (t)\leq v(t)\leq \beta (t), \quad \forall t\in [0,t_{1}]. $$
(2.8)
By \((A_{0})\), we get
$$ \alpha (t_{1}+)=\alpha (t_{1})+I\bigl(\alpha (t_{1})\bigr)\leq \beta (t_{1})+I\bigl( \beta (t_{1})\bigr)=\beta (t_{1}+). $$
Define
$$ \tilde{\alpha }(t)= \textstyle\begin{cases} \alpha (t_{1}+),& t=t_{1}, \\ \alpha (t),& t\in (t_{1},1] \end{cases} $$
and
$$ \tilde{\beta }(t)= \textstyle\begin{cases} \beta (t_{1}+),& t=t_{1}, \\ \beta (t),& t\in (t_{1},1]. \end{cases} $$
It is easy to see that α̃ and β̃ are lower and upper solutions of the following problem:
$$ \textstyle\begin{cases} -u''(t)=f(t, u, u'),\quad t\in (t_{1}, 1), \\ u(t_{1})=C+I(C),\qquad u(1) =b. \end{cases} $$
(2.9)
Thus again by Lemma 2.2, problem (2.9) has a solution \(w\in C^{1}[t_{1}, 1]\cap C^{2}(t_{1}, 1)\) satisfying
$$ \tilde{\alpha }(t)\leq w(t)\leq \tilde{\beta }(t), \quad \forall t\in [t_{1}, 1]. $$
(2.10)
Define
$$ u(t)= \textstyle\begin{cases} v(t),& t\in [0,t_{1}], \\ w(t),& t\in (t_{1},1], \end{cases} $$
which guarantees that
$$ \Delta u|_{t=t_{1}}=u(t_{1}+)-u(t_{1})= w(t_{1})-v(t_{1})=I(C)=I\bigl(v(t_{1}) \bigr)=I\bigl(u(t_{1})\bigr).$$
From (2.8)–(2.10), u is a solution of (2.1) satisfying \(u(t_{1})=C\) and \(\alpha (t)\leq u(t)\leq \beta (t)\) for all \(t\in [0,1]\).
Consequently, for each \(C\in [\alpha (t_{1}),\beta (t_{1})]\), there exists a solution \(u_{C}\) of problem (2.1) satisfying \(u_{C}(t_{1})=C\).
We now show that one of the solutions \(u_{C}\) satisfies condition (2.2). Let \(X(C) =\{u| u\mbox{ is a solution of problem (2.1) such that }u(t_{1})=C\mbox{ and }\alpha (t)\leq u(t)\leq \beta (t)\mbox{ for all } t\in [0, 1]\}\). Then, for each \(C\in [\alpha (t_{1}),\beta (t_{1})]\), \(X(C)\neq\emptyset \). There are two cases: (A) \(\alpha (t_{1})=\beta (t_{1})=C\); (B) \(\alpha (t_{1})<\beta (t_{1})\).
First we consider the case (A). Since \(\alpha (t_{1})=\beta (t_{1})=C\), by Lemma 2.3, \(u_{C}(t)\) is a solution of problem (2.1)–(2.2).
Now we consider the case (B). Since \(\alpha (t_{1})<\beta (t_{1})\), we define \(S=\{\overline{C}\in [\alpha (t_{1}), \beta (t_{1})): C\in ( \overline{C},\beta (t_{1}))\mbox{ implies } \Delta u'_{C}|_{t=t_{1}}< N(u_{C}(t_{1}), u'_{C}(t_{1}-)),\mbox{ for all }u_{C}\in X(C)\}\). There are two cases also: (\(B_{1}\)) \(S\neq\emptyset \); (\(B_{2}\)) \(S=\emptyset \).
Now we consider the case (\(B_{1}\)). Let \(C^{*}=\inf S\), then \(\alpha (t_{1})\leq C^{*}<\beta (t_{1})\).
If \(C^{*}>\alpha (t_{1})\), by the definition of \(C^{*}\), we can choose sequences \(C_{n}\in (\alpha (t_{1}), C^{*})\) and \(u_{C_{n}}\in X(C_{n})\) such that \(C_{n}\to C^{*}\) and
$$ \Delta u'_{C_{n}}\geq N \bigl(u_{C_{n}}(t_{1}),u'_{C_{n}}(t_{1}-) \bigr).$$
(2.11)
Obviously \(u_{C_{n}}(t)\) satisfies
$$ u_{C_{n}}(t)= \int _{0}^{t_{1}}G_{1}(t,s)f \bigl(s,u_{C_{n}}(s),u'_{C_{n}}(s)\bigr)\,ds+a+ \frac{C_{n}-a}{t_{1}}t,\quad t\in [0,t_{1}],$$
(2.12)
and
$$\begin{aligned} u_{C_{n}}(t) =& \int _{t_{1}}^{1}G_{2}(t,s)f \bigl(s,u_{C_{n}}(s),u'_{C_{n}}(s)\bigr)\,ds+b \\ &{}+ \frac{b-I(C_{n})-C_{n}}{1-t_{1}}(1-t),\quad t\in (t_{1},1], \end{aligned}$$
(2.13)
where \(G_{1}(t, s)\) and \(G_{2}(t, s)\) are the Green’s functions of linear homogeneous problem with Dirichlet boundary condition corresponding to (2.7) and (2.9). From \(\alpha (t)\leq u_{C_{n}}(t)\leq \beta (t)\), \(t\in [0,1]\), Lemma 2.1 together with condition (4) of Theorem 2.6 guarantees that there is a \(k_{0}>0\) such that
$$ \vert u_{C_{n}} \vert _{H^{2,p}(0,1)}\leq k_{0}.$$
Now the Embedding theorem implies that there is an \(R>0\) such that \(|u'_{C_{n}}(t)|\leq R\), \(\forall t\in [0,1]\). Condition (4) implies that
$$ \bigl\vert f\bigl(x,u_{C_{n}}(t),u'_{C_{n}}(t) \bigr) \bigr\vert \leq h(t)+KR^{2}$$
also. Since \(h\in L^{1}[0,1]\), the functions belonging to \(\{u_{C_{n}}(t)\}\) and the functions belonging to \(\{u'_{C_{n}}(t)\}\) are equicontinuous on \([0,t_{1}]\) and \((t_{1},1]\), respectively.
Therefore, without loss of generality, we assume that
$$ u_{C_{n}}(t)\rightrightarrows u_{0}(t), \qquad u'_{C_{n}}(t) \rightrightarrows u'_{0}(t),\quad \text{as }n\to +\infty .$$
Letting \(n\to +\infty \) in (2.12) and (2.13), we have
$$ u_{0}(t)= \int _{0}^{t_{1}}G_{1}(t,s)f \bigl(s,u_{0}(s),u'_{0}(s)\bigr)\,ds+a+ \frac{C^{*}-a}{t_{1}}t,\quad t\in [0,t_{1}], $$
and
$$ u_{0}(t)= \int _{t_{1}}^{1}G_{2}(t,s)f \bigl(s,u_{0}(s),u'_{0}(s)\bigr)\,ds+b+ \frac{b-I(C^{*})-C^{*}}{1-t_{1}}(1-t),\quad t\in (t_{1},1].$$
Obviously, \(u_{0}\) is a solution of problem (2.1) satisfying \(u_{0}(t_{1})=C^{*}\) and \(\alpha (t)\leq u_{1}(t)\leq \beta (t)\) on \([0,1]\). And from (2.12) and (2.13), we get
$$\begin{aligned} \Delta u'_{C_{n}}|_{t=t_{1}} =&u'_{C_{n}}(t_{1}+)-u'_{C_{n}}(t_{1}-) \\ =&\frac{b-I(C_{n})-C_{n}}{1-t_{1}}+\frac{1}{1-t_{1}} \int _{t_{1}}^{1}(1-s)f\bigl(s,u_{C_{n}}(s),u'_{C_{n}}(s) \bigr)\,ds \\ &{} -\biggl(\frac{C_{n}-t_{1}}{t_{1}}- \int _{0}^{t_{1}}sf\bigl(s,u_{C_{n}}(s),u'_{C_{n}}(s) \bigr)\,ds\biggr). \end{aligned}$$
Letting \(n\to +\infty \), we have
$$ \begin{aligned}[b] \lim_{n\to +\infty } \Delta u'_{C_{n}}|_{t=t_{1}} ={}& \frac{b-I(C^{*})-C^{*}}{1-t_{1}}+\frac{1}{1-t_{1}} \int _{t_{1}}^{1}(1-s)f\bigl(s,u_{0}(s),u'_{0}(s) \bigr)\,ds \\ & {}-\biggl(\frac{C^{*}-t_{1}}{t_{1}}- \int _{0}^{t_{1}}sf\bigl(s,u_{0}(s),u'_{0}(s) \bigr)\,ds\biggr) \\ =&\Delta u'_{0}|_{t=t_{1}}. \end{aligned} $$
(2.14)
Letting \(n\to +\infty \) in (2.11), we obtain
$$ \lim_{n\to +\infty }\Delta u'_{C_{n}}|_{t=t_{1}} \geq \lim_{n\to + \infty }N\bigl(u_{C_{n}}(t_{1}), u'_{C_{n}}(t_{1}-)\bigr)=N \bigl(u_{0}(t_{1}),u'_{0}(t_{1}-) \bigr). $$
(2.15)
Equations (2.14) and (2.15) indicate
$$ \Delta u'_{0}|_{t=t_{1}}\geq N \bigl(u_{0}(t_{1}), u'_{0}(t_{1}-) \bigr),$$
(2.16)
which implies that \(u_{0}\) is a lower solution of problem (2.1)–(2.2) satisfying \(u_{0}(t_{1})=C^{*}\).
By the definition of \(C^{*}\), for \(D_{n}\in [C^{*},\beta (t_{1})]\) with \(D_{n}\to C^{*}\) as \(n\to +\infty \), we have
$$\begin{aligned}& \textstyle\begin{cases} -u^{\prime \prime }_{D_{n}}(t)=f(t,u_{D_{n}},u'_{D_{n}}),\quad t\in (0,t_{1}), \\ u_{D_{n}}(0)=a,\qquad u_{D_{n}}(t_{1})=D_{n}, \end{cases}\displaystyle \\& \textstyle\begin{cases} -u^{\prime \prime }_{D_{n}}(t)=f(t,u_{D_{n}},u'_{D_{n}}),\quad t\in (t_{1},1), \\ u_{D_{n}}(t_{1})=D_{n}+I(D_{n}),\qquad u_{D_{n}}(1)=b, \end{cases}\displaystyle \end{aligned}$$
and
$$ \Delta u_{D_{n}}|_{t=t_{1}}< N\bigl(u_{D_{n}}(t_{1}), u'_{D_{n}}(t_{1}-)\bigr).$$
By a similar argument to the construction of \(u_{0}\), we obtain \(u_{1}\), a solution of problem (2.1) satisfying \(u_{1}(t_{1})=C^{*}\), \(u_{0}(t)\leq u_{1}(t)\leq \beta (t)\) on \([0,1]\), \(u_{D_{n}}(t)\rightrightarrows u_{1}(t)\), \(u'_{D_{n}}(t)\rightrightarrows u'_{1}(t)\) as a subsequence if necessary and
$$ \Delta u'_{1}|_{t=t_{1}} \leq N \bigl(u_{1}(t_{1}), u'_{1}(t_{1}-) \bigr). $$
Thus \(u_{1}\) is an upper solution of problem (2.1)–(2.2).
Consequently, \(u_{0}\) and \(u_{1}\) are lower and upper solutions of problem (2.1)–(2.2), respectively, satisfying \(u_{0}(t_{1})= C^{*}=u_{1}(t_{1})\) and \(u_{0}(t)\leq u_{1}(t)\) on \([0,1]\).
Therefore, by the argument of (A), problem (2.1)–(2.2) has at least one solution between \(u_{0}\) and \(u_{1}\).
If \(C^{*}=\alpha (t_{1})\), by the definition of \(C^{*}\), we can choose sequences \(C_{n}\in (\alpha (t_{1}), \beta (t_{1}))\) and \(u_{C_{n}}\in X(C_{n})\) such that \(C_{n}\to \alpha (t_{1})+\) as \(n\to +\infty \) and
$$ \Delta u'_{C_{n}}< N \bigl(u_{C_{n}}(t_{1}),u'_{C_{n}}(t_{1}-) \bigr).$$
(2.17)
By the same argument as for \(u_{1}\) and \(\{u_{D_{n}}\}\), there is a \(u_{2}\), a solution of problem (2.1) satisfying \(u_{2}(t_{1})=\alpha (t_{1})\), \(\alpha (t)\leq u_{2}(t)\leq \beta (t)\) on \([0,1]\), and \(u_{C_{n}}(t)\rightrightarrows u_{2}(t)\), \(u'_{C_{n}}(t)\rightrightarrows u'_{2}(t)\) as a subsequence if necessary and
$$ \Delta u'_{2}|_{t=t_{1}} \leq N \bigl(u_{2}(t_{1}), u'_{2}(t_{1}-) \bigr). $$
Consequently, α and \(u_{2}\) are lower and upper solutions of problem (2.1)–(2.2), respectively, satisfying \(\alpha (t_{1})=C^{*}=u_{2}(t_{1})\) and \(\alpha (t)\leq u_{2}(t)\) on \([0,1]\).
Therefore, by the argument of (A), problem (2.1)–(2.2) has at least one solution between α and \(u_{2}\).
The above proof shows that problem (2.1)–(2.2) has at least one solution if \(S\neq\emptyset \).
Next, we consider (\(B_{2}\)). Since \(S=\emptyset \), we may choose sequences \(C_{n}\in (\alpha (t_{1}),\beta (t_{1}))\) and \(u_{C_{n}}\in X(C_{n})\) satisfying \(C_{n}\to \beta (t_{1})\) as \(n\to +\infty \) and \(\Delta u'_{C_{n}}|_{t=t_{1}}\geq N(u_{C_{n}}(t_{1}), u'_{C_{n}}(t_{1}-))\).
By a similar limit argument to the one before, we obtain \(u_{3}\), a solution of problem (2.1) satisfying \(u_{3}(t_{1})=\beta (t_{1})\), \(\alpha (t)\leq u_{3}(t)\leq \beta (t)\), \(t\in [0, 1]\), \(u_{C_{n}}(t)\rightrightarrows u_{3}(t)\) and \(u'_{C_{n}}(t)\rightrightarrows u'_{3}(t)\) as a subsequence if necessary and
$$ \Delta u'_{3}|_{t=t_{1}}\geq N \bigl(u_{3}(t_{1}), u'_{3}(t_{1}-) \bigr). $$
Therefore, \(u_{3}\) and β are lower and upper solutions of problem (2.1)–(2.2). By the proof of (A) also, problem (2.1)–(2.2) has at least one solution between \(u_{3}\) and β.
Consequently, problem (2.1)–(2.2) has at least one solution if \(S=\emptyset \).
The proof is completed. □
Remark 2.7
The idea of our theorem comes from [23].
Remark 2.8
In the problem (1.1), let \(f(t,u,v)=\lambda u^{p}-K(t)u^{-q}-|v|^{\mu }\). Since \(\mu \in [0,2]\), there is a \(C'>0\) such that
$$ \bigl\vert f(t,u,v) \bigr\vert \leq \lambda \bigl\vert u^{p} \bigr\vert + \bigl\vert K(t)u^{-q} \bigr\vert +C'+ \vert v \vert ^{2}.$$
(2.18)
In Sects. 3 and 4, we will obtain the positive solutions of problem (1.1) by constructing a pair of lower and upper solutions which satisfy Theorem 2.6.
Throughout the paper, we always assume that \(\mu \in [0,2]\), \(p\in (0,1)\), \(q\in (0,1)\).