Throughout this paper, \(L^{1}(I)\) will denote the Banach space consisting of all real functions defined and Lebesgue integrable on \(I:=[0,1]\), with the standard norm \(\Vert \cdot \Vert \); and \(L^{\infty}(I)\) will denote the Banach space consisting of all real functions defined and essentially bounded on I, with the standard norm \(\Vert \cdot \Vert _{\infty}\).
Lemma 3.1
A function
\(x=x(t)\)
is a solution of the boundary value problem (1)(2) if and only if
x
is a solution of the following integral equation:
$$\begin{aligned} x(t) = g\bigl(t,x(t)\bigr) + h\bigl(t,x(t)\bigr) \int_{0}^{1}G(t,s)\,f\bigl(s,x(s)\bigr) \,ds, \end{aligned}$$
(3.1)
where the Green’s function associated with (1.1)(1.2) is defined by
$$ G(t,s)= \textstyle\begin{cases} s(1t), & 0\leq s\leq t\leq1,\\ t(1s), & 0\leq t\leq s\leq1. \end{cases} $$
(3.2)
Proof
\(x:I\rightarrow\mathbb{R}\) is a solution of Eq. (1.1) if and only if it satisfies
$$ \frac{d}{dt} \biggl(\frac{x(t)g(t,x(t))}{h(t,x(t))}\biggr)= \int_{0}^{t} f\bigl(s,x(s)\bigr) \,ds + c_{1}, $$
it follows that
$$ \frac{x(t)g(t,x(t))}{h(t,x(t))} =  \int_{0}^{t} (ts)\, f\bigl(s,x(s)\bigr) \,ds + c_{1}t+c_{0}. $$
(3.3)
By choosing \(t=0\) and \(t=1\) in (3.3) respectively and applying the boundary condition (1.2), we get
$$ c_{0}=\frac{x(0)g(0,x(0))}{h(0,x(0))}=0,\qquad c_{1} = \int_{0}^{1}(1s)\, f\bigl(s,x(s)\bigr) \,ds. $$
(3.4)
From (3.3)(3.4) we deduce that
$$\begin{aligned} \frac{x(t)g(t,x(t))}{h(t,x(t))} =& \int_{0}^{t} s(1t)\, f\bigl(s,x(s)\bigr) \,ds + \int_{t}^{1} t(1s)\, f\bigl(s,x(s)\bigr) \,ds \\ =& \int_{0}^{1} G(t,s)\, f\bigl(s,x(s)\bigr) \,ds. \end{aligned}$$
Thus, problem (1.1)(1.2) has been transformed into the perturbed quadratic integral equations (3.1). □
Remark 3.2
Since the maximum value of the Green’s function G will be gotten as \(s=t\), we have \(\max_{(t,s)\in I^{2}}G(t,s) = \max_{t\in I} t(1t)= 1/4\). Thus, the linear operator \(\mathbb{G}\) defined by
$$ \mathbb{G} x(t):= \int_{0}^{1} G(t,s) x(s) \,ds, \quad \forall x\in L^{1}(I), $$
is bounded from \(L^{1}(I)\) into \(L^{\infty}(I)\). In fact, we have
$$ \Vert \mathbb{G} x \Vert _{\infty}\leqslant \frac{1}{4} \int_{0}^{1} \bigl\vert x(s) \bigr\vert \,ds \leqslant\frac{1}{4} \Vert x \Vert . $$
(3.5)
Definition 3.3
A function \(x\in L^{1}(I)\) is said to be a weak solution of problem (1.1)(1.2) if x satisfies Eq. (3.1) on the interval I.
We will consider (1.1)(1.2) under the following assumptions.
 (\(\mathcal{H}1\)):

The functions \(f:I\times\mathbb {R}\rightarrow\mathbb{R}\) and \(h:I\times\mathbb{R}\rightarrow \mathbb{R}\backslash\{0\}\) satisfy the Carathéodory conditions. Moreover, there exist functions \(f_{0}, h_{0} \in L^{1}_{+}(I)\) and positive numbers η and γ, respectively, such that
$$ \bigl\vert f(t,x) \bigr\vert \leqslant f_{0}(t)+ \eta \vert x \vert ,\qquad \bigl\vert h(t,x) \bigr\vert \leqslant h_{0}(t)+ \gamma \vert x \vert , \quad \forall x\in\mathbb{R} \text{ and a.e. } t\in I. $$
 (\(\mathcal{H}2\)):

The function \(g:I\times\mathbb {R}\rightarrow\mathbb{R}\) satisfies the Carathéodory conditions, and there exists a positive number β such that
$$ \bigl\vert g(t,x)g(t,y) \bigr\vert \leqslant\beta \vert xy \vert , \quad \forall x,y\in\mathbb{R} \text{ and a.e. } t\in I. $$
Moreover, \(g_{0}(t):= \vert g(t,0) \vert \) belongs to \(L^{1}_{+}(I)\).
 (\(\mathcal{H}3\)):

The following inequality holds:
$$ 4\beta+ \gamma \Vert f_{0} \Vert + \eta \Vert h_{0} \Vert + 2\sqrt{\gamma\eta\bigl( 4 \Vert g_{0} \Vert + \Vert f_{0} \Vert \Vert h_{0} \Vert \bigr)}< 4. $$
(3.6)
Remark 3.4
(1) Note that from (\(\mathcal{H}2\)) we deduce that \(\vert g(t,x) \vert \leqslant g_{0}(t)+\beta \vert x \vert \) for all \(x\in\mathbb{R}\) and a.e. \(t\in I\). Thus, by Theorem 2.5, (\(\mathcal{H}1\)) and (\(\mathcal{H}2\)) imply that the superposition operators \(\mathcal{N}_{f}\), \(\mathcal{N}_{g}\) and \(\mathcal {N}_{h}\), respectively, map \(L^{1}(I)\) into itself continuously. Further, according to [7, Lemma 3.2], \(\mathcal{N}_{f}\), \(\mathcal {N}_{g}\) and \(\mathcal{N}_{h}\) are wwcompact.
(2) It is easily deduced from inequality (3.6) of (\(\mathcal {H}3\)) that the following quadratic inequality about r
$$ \gamma\eta\mathbf{r}^{2}  \bigl(44\beta \gamma \Vert f_{0} \Vert  \eta \Vert h_{0} \Vert \bigr) \mathbf{r} +4 \Vert g_{0} \Vert + \Vert f_{0} \Vert \Vert h_{0} \Vert \leqslant0 $$
(3.7)
has positive solutions, since inequality (3.6) implies that
$$\begin{aligned}& \bigl(44\beta \gamma \Vert f_{0} \Vert  \eta \Vert h_{0} \Vert \bigr)^{2}4\gamma\eta\bigl(4 \Vert g_{0} \Vert + \Vert f_{0} \Vert \Vert h_{0} \Vert \bigr)>0, \\& \frac{44\beta \gamma \Vert f_{0} \Vert  \eta \Vert h_{0} \Vert }{2\gamma\eta}>0. \end{aligned}$$
Further, there is a certain solution \(\mathbf{r}_{0}\) of (3.7) such that
$$ 0< \mathbf{r}_{0} \leqslant\frac{44\beta \gamma \Vert f_{0} \Vert  \eta \Vert h_{0} \Vert }{2\gamma\eta}. $$
(3.8)
Theorem 3.5
Under assumptions (\(\mathcal{H}1\))(\(\mathcal{H}3\)), problem (1.1)(1.2) has at least one weak solution
\(x\in M\), where
\(M:=\{x: \Vert x \Vert \leqslant\mathbf{r}_{0}\}\)
is a closed ball of
\(L^{1}(I)\), and
\(\mathbf{r}_{0}\)
is a solution of (3.7) and satisfies (3.8) .
Proof
Let \(\mathcal{B}:=\mathcal{N}_{g}\). Define \(\mathcal{A}\) by \(\mathcal {A} x(t): =\mathcal{N}_{h} x(t)\cdot\mathbb{G}\mathcal{N}_{f} x(t)\) for \(x\in M\). For proving the operator equation \(x=\mathcal{A}x+\mathcal {B}x\) has a unique solution in \(L^{1}(I)\), our processes are divided into several steps.
(1). \(\mathcal{A}\) is wscompact.
For all \(x_{1},x_{2}\in M\), from (\(\mathcal{H}1\))(\(\mathcal{H}2\)), Remark 3.4 and Remark 3.2, we deduce that
$$\begin{aligned}& \Vert \mathcal{A} x_{1}  \mathcal{A} x_{2} \Vert \\& \quad\leqslant \bigl\Vert \mathbb{G}(\mathcal{N}_{f} x_{1}\mathcal {N}_{f} x_{2}) \bigr\Vert _{\infty}\cdot \Vert \mathcal{N}_{h} x_{1} \Vert + \Vert \mathbb{G}\mathcal{N}_{f} x_{2} \Vert _{\infty}\cdot \Vert \mathcal{N}_{h} x_{1}  \mathcal{N}_{h} x_{2} \Vert \\& \quad\leqslant \frac{1}{4} \bigl( \Vert h_{0} \Vert +\gamma \mathbf{ r}_{0} \bigr)\cdot \Vert \mathcal{N}_{f} x_{1}\mathcal{N}_{f} x_{2} \Vert + \frac{1}{4} \bigl( \Vert f_{0} \Vert +\eta \mathbf{ r}_{0}\bigr)\cdot \Vert \mathcal{N}_{h} x_{1} \mathcal{N}_{h} x_{2} \Vert . \end{aligned}$$
Thus, according to the continuity of the operators \(\mathcal{N}_{f}\) and \(\mathcal{N}_{h}\), we obtain that \(\mathcal{A}\) is continuous on M.
Further, let us take a weakly convergent sequence \((x_{n})_{n\in\mathbb {N}}\) from M. Then, for any measurable subset D of the interval I, from (3.5) we obtain the following estimate:
$$\begin{aligned} \int_{D} \bigl\vert \mathcal{A}x_{n}(t) \bigr\vert \,dt \leqslant& \Vert \mathbb{G}\mathcal{N}_{f} x_{n} \Vert _{\infty}\int _{D} \bigl\vert \mathcal{N}_{h} x_{n}(t) \bigr\vert \,dt \\ \leqslant& \frac{1}{4} \bigl( \Vert f_{0} \Vert +\eta \mathbf {r}_{0}\bigr) \int_{D} \bigl\vert \mathcal{N}_{h} x_{n}(t) \bigr\vert \,dt \quad (n=1,2,\ldots). \end{aligned}$$
(3.9)
Since the sequence \((\mathcal{N}_{h} x_{n})_{n\in\mathbb{N}}\) is relatively weakly compact (cf. Remark 3.4(1)), then applying formula (2.1) we deduce from (3.9) that \((\mathcal{A} x_{n})_{n\in\mathbb{N}}\) is relatively weakly compact as well.
In what follows, let us fix a number \(\varepsilon>0\). According to Theorem 2.2, we can choose a number \(\delta>0\) such that, for any measurable subset \(D_{\delta}\) of the interval I with \(\operatorname{meas}(D_{\delta})\leqslant\delta\), we have
$$ \int_{D_{\delta}} \bigl\vert \mathcal{A} x_{n}(s) \bigr\vert \,ds \leqslant\frac{\varepsilon}{3}, \quad m=1,2,\ldots. $$
(3.10)
According to Remark 3.4(1), \((\mathcal{N}_{f} x_{n})_{n\in\mathbb {N}}\) has a weakly convergent subsequence, say \((\mathcal{N}_{f} x_{n_{k}})_{k\in\mathbb{N}}\). Since the continuity of the linear operator \(\mathbb{G}\) implies its weak continuity on \(L^{1}(I)\) for a.e. \(t\in I\), then \((\mathbb{G}\mathcal{N}_{f} x_{n_{k}})_{k\in\mathbb{N}}\) converges pointwise for a.e. \(t\in I\). Now, applying Egoroff’s theorem, there exists a measurable subset \(I_{0}\subseteq I\) with \(\operatorname{meas}(I\backslash I_{0})\leqslant\delta\) such that \((\mathbb {G}\mathcal{N}_{f} x_{n_{k}})_{k\in\mathbb{N}}\) is uniformly convergent on \(I_{0}\).
On the other hand, \((\mathcal{N}_{h} x_{n_{k}})_{k\in\mathbb{N}}\) has a weakly convergent subsequence as well according to Remark 3.4(1). Without loss of generality, we can assume that it is still \((\mathcal {N}_{h} x_{n_{k}})_{k\in\mathbb{N}}\). Thus, the sequence \((\mathcal{N}_{h} x_{n_{k}}\cdot\mathbb{G}\mathcal{N}_{f} x_{n_{k}})_{k\in\mathbb{N}}\), that is, \((\mathcal{A} x_{n_{k}})_{k\in\mathbb{N}}\) is strongly convergent in \(L^{1}(I_{0})\) (cf. [14, Proposition 3.5, p. 58]). Therefore, \((\mathcal{A} x_{n_{k}})_{k\in\mathbb{N}}\) satisfies the Cauchy criterion on \(I_{0}\), i.e., there exists \(k_{0}\in\mathbb{N}\) such that for arbitrary natural numbers \(j,k\geqslant k_{0}\) the following inequality holds:
$$ \int_{I_{0}} \bigl\vert \mathcal{A} x_{n_{j}}(t) \mathcal{A} x_{n_{k}}(t) \bigr\vert \,dt \leqslant\frac{\varepsilon}{3}. $$
(3.11)
Thus, we deduce from (3.10) and (3.11) that
$$\begin{aligned} \Vert \mathcal{A} x_{n_{j}}\mathcal{A} x_{n_{k}} \Vert &\leqslant \int_{I_{0}} \bigl\vert \mathcal{A} x_{n_{j}}(t) \mathcal{A} x_{n_{k}}(t) \bigr\vert \,dt \\ &\quad{} + \int_{I \backslash I_{0}} \bigl\vert \mathcal{A} x_{n_{j}}(t) \bigr\vert \,dt + \int_{I \backslash I_{0}} \bigl\vert \mathcal{A} x_{n_{k}}(t) \bigr\vert \,dt \\ &\leqslant\varepsilon, \end{aligned} $$
which implies that the sequence \((\mathcal{A} x_{n_{k}})_{k\in\mathbb {N}}\) is convergent in \(L^{1}(I)\), and then \(\mathcal{A}\) maps relatively weakly compact subsets of M into relatively strongly compact ones.
Thus, we complete the proof that \(\mathcal{A}\) is wscompact.
(2). \(\mathcal{A}\) is ωcontractive.
Let D be a measurable subset of the interval I, and let S be a nonempty subset of M. From Remark (3.4) and Remark (3.2), for all \(x\in S\), we deduce that
$$ \int_{D} \bigl\vert \mathcal{A}x(t) \bigr\vert \,dt \leqslant \Vert \mathbb{G}\mathcal{N}_{f} x \Vert _{\infty}\int_{D} \bigl\vert \mathcal{N}_{h} x(t) \bigr\vert \,dt \leqslant\frac{1}{4}\bigl( \Vert f_{0} \Vert +\eta \Vert x \Vert \bigr) \int _{D}\bigl(h_{0}(t)+\gamma \bigl\vert x(t) \bigr\vert \bigr) \,dt. $$
Taking into account the facts that the set of a single element is weakly compact and \(\Vert x \Vert \leqslant\mathbf{r}_{0}\), the use of formula (2.1) leads to
$$ \omega\bigl(\mathcal{A}(S)\bigr)\leqslant \frac{( \Vert f_{0} \Vert +\eta\mathbf{r_{0}})\gamma}{4} \omega(S) = \alpha \omega(S), $$
(3.12)
where \(\alpha:=( \Vert f_{0} \Vert +\eta\mathbf {r_{0}})\gamma/4\). Applying inequality (3.8), we have
$$ \alpha< \frac{\gamma}{4} \biggl( \Vert f_{0} \Vert +\frac {44\beta \gamma \Vert f_{0} \Vert  \eta \Vert h_{0} \Vert }{\gamma}\biggr)< 1\beta. $$
(3.13)
Thus, (3.12) implies that \(\mathcal{A}\) is ωcontraction with α.
(3). If \(y=\mathcal{B}y + \mathcal{A}x\) for \(x\in M\), then \(y\in M\).
If \(y\in L^{1}(I)\) satisfies \(y=\mathcal{B}y + \mathcal{A}x\) for \(x\in M\), then we have
$$\begin{aligned} \bigl\vert y(t) \bigr\vert \leqslant& \bigl\vert \mathcal{N}_{g} y(t) \bigr\vert + \bigl\vert \mathcal{N}_{h} x(t) \bigr\vert \cdot \bigl\vert \mathbb{G}\mathcal{N}_{f} x(t) \bigr\vert \\ \leqslant& g_{0}(t)+\beta \bigl\vert y(t) \bigr\vert + \bigl(h_{0}(t)+\gamma \bigl\vert x(t) \bigr\vert \bigr)\cdot \frac{1}{4} \bigl( \Vert f_{0} \Vert +\eta \Vert x \Vert \bigr), \end{aligned}$$
for a.e. \(t\in I\), it follows that
$$ 4 \Vert y \Vert \leqslant4 \Vert g_{0} \Vert + 4\beta \Vert y \Vert + \bigl( \Vert f_{0} \Vert +\eta \Vert x \Vert \bigr) \bigl( \Vert h_{0} \Vert + \gamma \Vert x \Vert \bigr). $$
Noting that \(\Vert x \Vert \leqslant\mathbf{r_{0}}\) and applying inequality (3.7), we deduce from the above that
$$ \Vert y \Vert \leqslant\frac{4 \Vert g_{0} \Vert + \Vert f_{0} \Vert \Vert h_{0} \Vert + (\gamma \Vert f_{0} \Vert + \eta \Vert h_{0} \Vert ) \mathbf{r}_{0} + \gamma\eta\mathbf{r}^{2}_{0}}{4(1\beta)}\leqslant \mathbf{r}_{0}, $$
which implies \(y\in M\).
(4). Conclusion.
The condition (i) of Theorem 2.7 is verified in (1)(2), and the condition (iii) of Theorem 2.7 is verified in (3). Moreover, \(\mathcal{B}=\mathcal{N}_{g}\) is contractive with β by (\(\mathcal{H}2\)), and \(\mathcal{B}\) is wwcompact by Remark 3.4(1). Then the condition (ii) of Theorem 2.7 is satisfied. The estimate \(\alpha+\beta<1\) is from (3.13).
Now, according to Theorem 2.7, we obtain that Eq. (3.1) has at least one solution in M, and then the existence of weak solutions in \(L^{1}(I)\) for problem (1.1)(1.2) is proved. □