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(Tt)\phantom{\rule{1em}{0ex}}\text{for}t\in [0,T]
follows from (18). Hence, g(u(t))\le g(\frac{\epsilon}{a+2}t(Tt)) for t\in (0,T) because g is nonincreasing on {\mathbb{R}}_{+}. Due to Remark 1, L={\parallel g(\frac{\epsilon}{a+2}t(Tt))\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(Tt). 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 ArzelàAscoli 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(Tt))\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. □