In order to prove the solvability of problem (7a) and (7b), we first have to investigate the problem

depending on the parameter *λ*. Here, $\epsilon >0$ is from (H_{2}) and $n\in \mathbb{N}$.

The following result shows that the solvability of problem (17a) and (17b) is equivalent to the solvability of an integral equation in the set ${C}^{1}[0,T]$.

**Lemma 7** *Let* (H

_{1})

*hold*.

*Then* *u* *is a solution of problem* (17a)

*and* (17b)

*if and only if* *u* *is a solution of the integral equation* $u(t)=-t{\int}_{t}^{T}\frac{1}{{s}^{a+2}}\left({\int}_{0}^{s}{\xi}^{a+1}[\lambda {f}_{n}(\xi ,u(\xi ),{u}^{\prime}(\xi ))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$

(18)

*in the set* ${C}^{1}[0,T]$.

*Proof* Let

*u* be a solution of Eq. (

17a). Then

$u\in A{C}^{1}[0,T]$, and by Corollary 1 (with

${f}_{n}$ replaced by

$\lambda {f}_{n}-(1-\lambda )\epsilon $), there exists

$c\in \mathbb{R}$ such that the equation

$u(t)=t(c-{\int}_{t}^{T}\frac{1}{{s}^{a+2}}\left({\int}_{0}^{s}{\xi}^{a+1}[\lambda {f}_{n}(\xi ,u(\xi ),{u}^{\prime}(\xi ))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s)$

holds for $t\in [0,T]$. Hence, $u(0)=0$ and $u(T)=0$ if and only if $c=0$. Consequently, if *u* is a solution of problem (17a) and (17b), then *u* is a solution of Eq. (18) in ${C}^{1}[0,T]$.

Let *u* be a solution of Eq. (18) in ${C}^{1}[0,T]$. Then ${f}_{n}(t,u(t),{u}^{\prime}(t))\in {L}^{1}[0,T]$. Hence, Lemma 3(ii) (with *ρ* replaced by $-\lambda {f}_{n}+(1-\lambda )\epsilon $) guarantees that $u\in A{C}^{1}[0,T]$ and *u* is a solution of Eq. (17a). Moreover, $u(0)=u(T)=0$. Consequently, *u* is a solution of problem (17a) and (17b) which completes the proof. □

The following results provide bounds for solutions of problem (17a) and (17b).

**Lemma 8** *Let* (H

_{1})-(H

_{3})

*hold*.

*Then there exists a positive constant* *S* (

*independent of* $n\in \mathbb{N}$ *and* $\lambda \in [0,1]$)

*such that for all solutions* *u* *of problem* (17a)

*and* (17b),

*the estimates* *hold*.

*Moreover*,

*for any solution* *u* *of problem* (17a)

*and* (17b),

*there exists* $\xi \in (0,T)$ *such that* $|{u}^{\prime}(t)|\ge \frac{2\epsilon}{a+2}|t-\xi |\phantom{\rule{1em}{0ex}}\phantom{\rule{0.1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}t\in [0,T].$

(21)

*Proof* Let

*u* be a solution of problem (17a) and (17b). Then by Lemma 7, equality (18) holds for

$t\in [0,T]$. Since by (5),

$\lambda {f}_{n}(t,u(t),{u}^{\prime}(t))-(1-\lambda )\epsilon \le -\epsilon $ for a.e.

$t\in [0,T]$, the relation

$u(t)\ge \epsilon t{\int}_{t}^{T}\frac{1}{{s}^{a+2}}\left({\int}_{0}^{s}{\xi}^{a+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=\frac{\epsilon}{a+2}t(T-t)\phantom{\rule{1em}{0ex}}\text{for}t\in [0,T]$

follows from (18). Hence,

$g(u(t))\le g(\frac{\epsilon}{a+2}t(T-t))$ for

$t\in (0,T)$ because

*g* is nonincreasing on

${\mathbb{R}}_{+}$. Due to Remark 1,

$L={\parallel g(\frac{\epsilon}{a+2}t(T-t))\parallel}_{1}<\mathrm{\infty}$, which means that

${\int}_{0}^{T}g(u(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\le L.$

(22)

It is clear that *L* is independent of the choice of solution *u* to problem (17a) and (17b) and independent of $n\in \mathbb{N}$, $\lambda \in [0,1]$.

We now show that inequality (21) holds for some

$\xi \in (0,T)$. Differentiation of (18) gives

$\begin{array}{rcl}{u}^{\u2033}(t)& =& -\frac{a}{{t}^{a+2}}{\int}_{0}^{t}{s}^{a+1}[\lambda {f}_{n}(s,u(s),{u}^{\prime}(s))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ +\lambda {f}_{n}(t,u(t),{u}^{\prime}(t))-(1-\lambda )\epsilon \phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [0,T].\end{array}$

(23)

Since

$a<0$, it follows from (5) and (23) that

${u}^{\u2033}(t)\le \frac{a\epsilon}{{t}^{a+2}}{\int}_{0}^{t}{s}^{a+1}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s-\epsilon =-\frac{2\epsilon}{a+2}\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [0,T].$

Hence,

${u}^{\prime}$ is decreasing on

$[0,T]$, and therefore

${u}^{\prime}$ vanishes at a unique point

$\xi \in (0,T)$ due to

$u(0)=u(T)=0$. The inequality (21) now follows from the relations

Hence,

$r(|{u}^{\prime}(t)|)\le r(\frac{2\epsilon}{a+2}|t-\xi |)$ on

$[0,T]\setminus \{\xi \}$, and

$\begin{array}{rcl}{\int}_{0}^{T}\left(r\left(|{u}^{\prime}(t)|\right)\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t& \le & {\int}_{0}^{\xi}\left(r(\frac{2\epsilon}{a+2}(\xi -t))\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t+{\int}_{\xi}^{T}\left(r(\frac{2\epsilon}{a+2}(t-\xi ))\right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ <& \frac{a+2}{\epsilon}{\int}_{0}^{(2\epsilon T)/(a+2)}r(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=V\phantom{\rule{1em}{0ex}}\text{for}t\in [0,T].\end{array}$

In particular,

${\parallel r\left(|{u}^{\prime}(t)|\right)\parallel}_{1}<V.$

(24)

Let

$W=\frac{a+3}{a+1}$. Taking into account (6), (9), (18), (22), (24), and Lemma 5, we obtain

$\begin{array}{rcl}|{u}^{\prime}(t)|& =& |-{\int}_{t}^{T}\frac{1}{{s}^{a+2}}\left({\int}_{0}^{s}{\xi}^{a+1}[\lambda {f}_{n}(\xi ,u(\xi ),{u}^{\prime}(\xi ))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ +\frac{1}{{t}^{a+1}}{\int}_{0}^{t}{s}^{a+1}[\lambda {f}_{n}(s,u(s),{u}^{\prime}(s))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s|\\ \le & W{\int}_{0}^{T}|\lambda {f}_{n}(t,u(t),{u}^{\prime}(t))-(1-\lambda )\epsilon |\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \le & W{\int}_{0}^{T}(\phi (t)h(1+{\parallel u\parallel}_{\mathrm{\infty}},1+{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}})+g(u(t))+r\left(|{u}^{\prime}(t)|\right))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t\\ \le & W(h(1+{\parallel u\parallel}_{\mathrm{\infty}},1+{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}}){\parallel \phi \parallel}_{1}+L+V),\phantom{\rule{1em}{0ex}}t\in [0,T].\end{array}$

It follows from

$u(t)={\int}_{0}^{t}{u}^{\prime}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$ for

$t\in [0,T]$,

${\parallel u\parallel}_{\mathrm{\infty}}\le T{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}},$

(25)

and therefore, we have

${\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}}\le Kh(1+T{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}},1+{\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}})+M,$

(26)

where

$K=W{\parallel \phi \parallel}_{1}$ and

$M=W(L+V)$. By (H

_{3}),

$\underset{z\to \mathrm{\infty}}{lim}(1/z)(Kh(1+Tz,1+z)+M)=0.$

Consequently, there exists $S>0$ such that $Kh(1+Tz,1+z)+M<z$ for $z\ge S$. Now, due to (26), ${\parallel {u}^{\prime}\parallel}_{\mathrm{\infty}}<S$, and therefore, by (25), ${\parallel u\parallel}_{\mathrm{\infty}}<ST$. □

We are now in the position to prove the existence result for problem (7a) and (7b).

**Lemma 9** *Let* (H_{1})-(H_{3}) *hold*. *Then for each* $n\in \mathbb{N}$, *problem* (7a) *and* (7b) *has a solution* *u* *satisfying inequalities* (19)-(21), *where* *S* *is a positive constant independent of* *n*.

*Proof* Let

*S* be a positive constant in Lemma 8 and let us define

$\mathrm{\Omega}:=\{x\in {C}^{1}[0,T]:\parallel x\parallel <ST,\parallel {x}^{\prime}\parallel <S\}.$

Then Ω is an open and bounded subset of the Banach space

${C}^{1}[0,T]$. Keeping in mind Lemma 3, define an operator

$\mathcal{K}:[0,T]\times \overline{\mathrm{\Omega}}\to {C}^{1}[0,T]$ by the formula

$\mathcal{K}(\lambda ,x)(t)=-t{\int}_{t}^{T}\frac{1}{{s}^{a+2}}\left({\int}_{0}^{s}{\xi}^{a+1}[\lambda {f}_{n}(\xi ,x(\xi ),{x}^{\prime}(\xi ))-(1-\lambda )\epsilon ]\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

(27)

By Lemma 7, any fixed point of the operator

$\mathcal{K}(1,\cdot )$ is a solution of problem (7a) and (7b). In order to show the existence of a fixed point of

$\mathcal{K}(1,\cdot )$, we apply Lemma 1 with

$E={C}^{1}[0,T]$,

$U=\mathrm{\Omega}$,

$\mathcal{F}=\mathcal{K}(1,\cdot )$ and

$\ell =\frac{\epsilon}{a+2}t(T-t)$. Especially, we show that

- (i)
$\mathcal{K}(1,\cdot ):\overline{\mathrm{\Omega}}\to {C}^{1}[0,T]$ is a compact operator, and

- (ii)
$\mathcal{K}(\lambda ,x)\ne x$ for each $\lambda \in (0,1]$ and $x\in \partial \mathrm{\Omega}$.

We first verify that

$\mathcal{K}(1,\cdot )$ is a continuous operator. To this end, let

$\{{x}_{m}\}\subset \overline{\mathrm{\Omega}}$ be a convergent sequence, and let

${lim}_{m\to \mathrm{\infty}}{x}_{m}=x$ in

${C}^{1}[0,T]$. Let

${r}_{m}(t):={f}_{n}(t,{x}_{m}(t),{x}_{m}^{\prime}(t))-{f}_{n}(t,x(t),{x}^{\prime}(t))\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [0,T].$

It follows from Lemma 5 and (9) that

for

$t\in [0,T]$. Here

$\mathcal{K}{(1,x)}^{\prime}=\frac{\mathrm{d}}{\mathrm{d}t}\mathcal{K}(1,x)$. In particular, for

$m\in \mathbb{N}$,

$\begin{array}{r}{\parallel \mathcal{K}(1,{x}_{m})-\mathcal{K}(1,x)\parallel}_{\mathrm{\infty}}\le \frac{2T{\parallel {r}_{m}\parallel}_{1}}{a+1},\\ {\parallel \mathcal{K}{(1,{x}_{m})}^{\prime}-\mathcal{K}{(1,x)}^{\prime}\parallel}_{\mathrm{\infty}}\le \frac{(a+3){\parallel {r}_{m}\parallel}_{1}}{a+1}.\end{array}$

(28)

Since

${lim}_{m\to \mathrm{\infty}}{f}_{n}(t,{x}_{m}(t),{x}_{m}^{\prime}(t))={f}_{n}(t,x(t),{x}^{\prime}(t))$ for a.e.

$t\in [0,T]$ and there exists

$\rho \in {L}^{1}[0,T]$ such that

$\left|{f}_{n}(t,{x}_{m}(t),{x}_{m}^{\prime}(t))\right|\le \rho (t)\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [0,T]\text{and all}m\in \mathbb{N},$

we have

${lim}_{m\to \mathrm{\infty}}{\parallel {r}_{m}\parallel}_{1}=0$ by the Lebesgue dominated convergence theorem. Hence, by (28),

$\mathcal{K}(1,\cdot )$ is a continuous operator. We now show that the set

$\mathcal{K}(1,\overline{\mathrm{\Omega}})$ is relatively compact in

${C}^{1}[0,T]$. It follows from

${f}_{n}\in Car([0,T]\times {\mathbb{R}}^{2})$ and

$\overline{\mathrm{\Omega}}$ bounded in

${C}^{1}[0,T]$ that there exists

$\mu \in {L}^{1}[0,T]$ such that

$\left|{f}_{n}(t,x(t),{x}^{\prime}(t))\right|\le \mu (t)\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [0,T]\text{and all}x\in \overline{\mathrm{\Omega}}.$

Then by Lemma 5 and (9), the inequalities

$|\mathcal{K}(1,x)(t)|\le \frac{2T{\parallel \mu \parallel}_{1}}{a+1},\phantom{\rule{2em}{0ex}}|\mathcal{K}{(1,x)}^{\prime}(t)|\le \frac{(a+3){\parallel \mu \parallel}_{1}}{a+1}$

are satisfied for

$t\in [0,T]$ and

$x\in \overline{\mathrm{\Omega}}$, and therefore, the set

$\mathcal{K}(1,\overline{\mathrm{\Omega}})$ is bounded in

${C}^{1}[0,T]$. Moreover, the relation

$\begin{array}{rcl}|\mathcal{K}{(1,x)}^{\u2033}(t)|& =& |-\frac{a}{{t}^{a+2}}{\int}_{0}^{t}{s}^{a+1}{f}_{n}(s,x(s),{x}^{\prime}(s))\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{f}_{n}(t,x(t),{x}^{\prime}(t))|\\ \le & \frac{|a|}{{t}^{a+2}}{\int}_{0}^{t}{s}^{a+1}\mu (s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+\mu (t)\in {L}^{1}[0,T]\end{array}$

holds for a.e. $t\in [0,T]$ and all $x\in \overline{\mathrm{\Omega}}$ (*cf.* (9)). Consequently, the set $\{\mathcal{K}{(1,x)}^{\prime}:x\in \overline{\mathrm{\Omega}}\}$ is equicontinuous on $[0,T]$. Hence, the set $\mathcal{K}(1,\overline{\mathrm{\Omega}})$ is relatively compact in ${C}^{1}[0,T]$ by the Arzelà-Ascoli theorem. As a result, $\mathcal{K}(1,\cdot )$ is a compact operator and the condition (i) follows.

Due to the fact that by Lemma 7 any fixed point *u* of the operator $\mathcal{K}(\lambda ,\cdot )$ is a solution of problem (17a) and (17b), Lemma 8 guarantees that *u* satisfies inequality (20). Therefore, $\mathcal{K}$ has property (ii). Consequently, by Lemmas 1 and 8, for each $n\in \mathbb{N}$, problem (7a) and (7b) has a solution *u* satisfying estimates (19)-(21). □

Let ${u}_{n}$ be a solution of problem (7a) and (7b) for $n\in \mathbb{N}$. The following property of the sequence $\{|{f}_{n}(t,{u}_{n}(t),{u}_{n}^{\prime}(t))|\}$ is an important prerequisite for solving problem (1a) and (1b).

**Lemma 10** *Let* (H_{1})-(H_{3}) *hold*. *Let* ${u}_{n}$ *be a solution of problem* (7a) *and* (7b) *for* $n\in \mathbb{N}$. *Then the sequence* $\{|{f}_{n}(t,{u}_{n}(t),{u}_{n}^{\prime}(t))|\}$ *is uniformly integrable on* $[0,T]$.

*Proof*

By Lemma 9, the inequalities

hold, where

*S* is a positive constant and

${\xi}_{n}\in (0,T)$. Hence, by (3) and (4),

$0<-{f}_{n}(t,{u}_{n}(t),{u}_{n}^{\prime}(t))\le \phi (t)h(1+ST,1+S)+{g}_{\ast}(t)+r(\frac{2\epsilon}{a+2}|t-{\xi}_{n}|)$

(32)

for a.e. $t\in [0,T]$, where ${g}_{\ast}(t)=g(\frac{\epsilon}{a+2}t(T-t))\in {L}^{1}[0,T]$, see Remark 1. Since the sequence $\{r(\frac{2\epsilon}{a+2}|t-{\xi}_{n}|)\}$ is uniformly integrable on $[0,T]$ (*cf.* [[20], criterion A.4], [21, 22]), it follows from (32) that $\{|{f}_{n}(t,{u}_{n}(t),{u}_{n}^{\prime}(t))|\}$ is uniformly integrable on $[0,T]$ and the result follows. □