In this section we consider the auxiliary singular differential equation,

${u}^{\u2033}(t)+\frac{a}{t}{u}^{\prime}(t)-\frac{a}{{t}^{2}}u(t)=f(t,u(t)),$

(3)

where $a<-1$. For $k:=-a$ and $v(t):={t}^{-k}u(t)$, (3) becomes a special case of (1) and therefore, results obtained for (3) apply for (1). For the further analysis, we assume that *f* satisfies the following conditions:

(H_{1}) $f\in Car([0,T]\times [0,\mathrm{\infty}))$,

(H_{2}) $0<f(t,x)$ for a.e. $t\in [0,T]$ and all $x\in [0,\mathrm{\infty})$,

(H

_{3})

$f(t,x)$ is increasing in

*x* for a.e.

$t\in [0,T]$ and

$\underset{x\to \mathrm{\infty}}{lim}\frac{1}{x}{\int}_{0}^{T}f(t,x)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=0.$

We now study (3) subject to the boundary conditions

$u(0)=0,\phantom{\rule{2em}{0ex}}u(T)=0,$

(4)

and require that

${u}^{\prime}(T)=-c,\phantom{\rule{1em}{0ex}}c\ge 0,$

(5)

holds. We call a function $u:[0,T]\to \mathbb{R}$ a *positive solution of the Dirichlet problem* (3), (4) if $u\in A{C}^{1}[0,T]$, $u>0$ on $(0,T)$, *u* satisfies the boundary conditions (4), and (3) holds for a.e. $t\in [0,T]$. Clearly, for each positive solution *u* of (3), (4), there exists $c\ge 0$ such that (5) is satisfied.

We now denote by $\mathcal{S}$ the set of all positive solutions of problem (3), (4), and ${\mathcal{S}}_{c}=\{u\in \mathcal{S}:{u}^{\prime}(T)=-c\}$, where $c\ge 0$.

In the following lemma we cite those results from [20] which will be used in the analysis of problem (1), (2).

**Lemma 1** *Let* (H

_{1})-(H

_{3})

*hold*.

*Then the following statements hold*:

- (a)
*For each* $c\ge 0$ *the set* ${\mathcal{S}}_{c}$ *is nonempty and there exist functions* ${u}_{c,min},{u}_{c,max}\in {\mathcal{S}}_{c}$ *such that* ${u}_{c,min}(t)\le u(t)\le {u}_{c,max}(t)$ *for* $t\in [0,T]$ *and* $u\in {\mathcal{S}}_{c}$.

- (b)
*If* ${c}_{1}>{c}_{2}\ge 0$, ${u}_{i}\in {\mathcal{S}}_{{c}_{i}}$, $i=1,2$, *then* ${u}_{1}(t)>{u}_{2}(t)$ *for* $t\in (0,T)$.

- (c)
*If* $c\ge 0$, ${u}_{i}\in {\mathcal{S}}_{c}$, $i=1,2$, *and* ${u}_{1}({t}_{0})>{u}_{2}({t}_{0})$ *for some* ${t}_{0}\in (0,T)$, *then either* ${u}_{1}(t)>{u}_{2}(t)$ *for* $t\in (0,T)$ *or there exists* ${t}_{\ast}\in ({t}_{0},T)$ *such that* ${u}_{1}(t)>{u}_{2}(t)$ *for* $t\in (0,{t}_{\ast})$ *and* ${u}_{1}(t)={u}_{2}(t)$ *for* $t\in [{t}_{\ast},T]$.

- (d)
*For each* ${t}_{0}\in (0,T)$ *and each* $A>{u}_{0,max}({t}_{0})$ *there exists* $u\in \mathcal{S}$ *satisfying* $u({t}_{0})=A$.

- (e)
${\mathcal{S}}_{c}$ *is a one*-*point set for each* $c\in [0,\mathrm{\infty})\setminus \mathrm{\Gamma}$, *where* $\mathrm{\Gamma}\subset [0,\mathrm{\infty})$ *is at most countable*.

- (f)
*For each* $0\le K\le Q<\mathrm{\infty}$, *the set* ${\bigcup}_{K\le c\le Q}{\mathcal{S}}_{c}$ *is compact in* ${C}^{1}[0,T]$.

- (g)
*If* $c\ge 0$,

*then* $u\in {\mathcal{S}}_{c}$ *if and only if it is a solution of the equation* $u(t)=t\frac{c{T}^{a+1}}{|a+1|}({T}^{-a-1}-{t}^{-a-1})+t{\int}_{t}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$

(6)

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

We now formulate new results which complete those from [20]. We first establish a relation between ${\mathcal{S}}_{0}$ and the set $\{(t,x)\in {\mathbb{R}}^{2}:0\le t\le T,{u}_{0,min}(t)\le x\le {u}_{0,max}(t)\}$ if its interior is nonempty. This question was a short time ago an open problem [[20], Remark 4.4]. We note that the relation between ${\mathcal{S}}_{c}$ and the set $\{(t,x)\in {\mathbb{R}}^{2}:0\le t\le T,{u}_{c,min}(t)\le x\le {u}_{c,max}(t)\}$ with $c>0$ having nonempty interior is described in Lemma 1(d).

**Theorem 1** *Let* (H_{1})-(H_{3}) *hold*. *Let us assume that there exists* ${t}_{0}\in (0,T)$ *such that* ${u}_{0,min}({t}_{0})<{u}_{0,max}({t}_{0})$. *Then for any* $Q\in ({u}_{0,min}({t}_{0}),{u}_{0,max}({t}_{0}))$ *there exists* $u\in {\mathcal{S}}_{0}$ *satisfying* $u({t}_{0})=Q$.

*Proof* *Step* 1. *Auxiliary Dirichlet problem*.

Choose

$Q\in ({u}_{0,min}({t}_{0}),{u}_{0,max}({t}_{0}))$. Consider (3) subject to the Dirichlet conditions

$u({t}_{0})=Q,\phantom{\rule{2em}{0ex}}u(T)=0.$

(7)

We claim that there exists a solution

*v* to problem (3), (7) such that

${u}_{0,min}(t)\le v(t)\le {u}_{0,max}(t)\phantom{\rule{1em}{0ex}}\text{for}t\in [{t}_{0},T].$

(8)

We show this result utilizing the method of lower and upper functions. It follows from (H

_{1}) that there exists

$\phi \in {L}^{1}[{t}_{0},T]$ such that

$|\frac{a}{{t}^{2}}x+f(t,x)|\le \phi (t)\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [{t}_{0},T]\text{and}{u}_{0,min}(t)\le x\le {u}_{0,max}(t).$

(9)

Let

$W={\int}_{{t}_{0}}^{T}\phi (t)\phantom{\rule{0.2em}{0ex}}\mathrm{d}t,\phantom{\rule{2em}{0ex}}S=max\{{\parallel {u}_{0,min}^{\prime}\parallel}_{\mathrm{\infty}},{\parallel {u}_{0,max}^{\prime}\parallel}_{\mathrm{\infty}},{\left(\frac{T}{{t}_{0}}\right)}^{|a|}W\}+1,$

and let

$\chi :\mathbb{R}\to [0,1]$ be given as

$\chi (x)=\{\begin{array}{cc}1\hfill & \text{if}|x|\le S,\hfill \\ 2-\frac{|x|}{S}\hfill & \text{if}S|x|2S,\hfill \\ 0\hfill & \text{if}|x|\ge 2S.\hfill \end{array}$

We now consider the auxiliary differential equation

${v}^{\u2033}(t)=\chi ({v}^{\prime}(t))(-\frac{a}{t}{v}^{\prime}(t)+\frac{a}{{t}^{2}}v(t)+f(t,v(t))).$

(10)

It is not difficult to verify that

$\chi (y)|y|\le S$ for

$y\in \mathbb{R}$. Hence,

*cf.* (9),

$\chi (y)|-\frac{a}{t}y+\frac{a}{{t}^{2}}x+f(t,x)|\le \chi (y)\frac{|ay|}{t}+\chi (y)\phi (t)\le \frac{|a|S}{t}+\phi (t)$

(11)

for a.e.

$t\in [{t}_{0},T]$ and all

${u}_{0,min}(t)\le x\le {u}_{0,max}(t)$,

$y\in \mathbb{R}$. Since

$\chi ({u}_{0,min}^{\prime}(t))=1$ and

$\chi ({u}_{0,max}^{\prime}(t))=1$ for

$t\in [{t}_{0},T]$,

${u}_{0,min}(T)={u}_{0,max}(T)=0$,

${u}_{0,min}({t}_{0})<Q<{u}_{0,max}({t}_{0})$, and

${u}_{0,min}$,

${u}_{0,max}$ solve (3) on

$[0,T]$, we conclude that

${u}_{0,min}$ and

${u}_{0,max}$ are lower and upper functions of problem (3), (7) (see

*e.g.* [

21] or [

22]). This fact together with (11) implies the existence of a solution

*v* to problem (3), (7) satisfying (8),

*cf.* [[

21], Lemma 3.7]. Moreover,

$\frac{{u}_{0,min}(t)}{T-t}\le \frac{v(t)}{T-t}\le \frac{{u}_{0,max}(t)}{T-t},\phantom{\rule{1em}{0ex}}t\in [{t}_{0},T),$

and taking the limit $t\to T$ we obtain^{a} ${u}_{0,min}^{\prime}(T)\ge {v}^{\prime}(T)\ge {u}_{0,max}^{\prime}(T)$, which together with ${u}_{0,min}^{\prime}(T)={u}_{0,max}^{\prime}(T)=0$ gives ${v}^{\prime}(T)=0$.

We now prove that

$|{v}^{\prime}|<S$ on

$[{t}_{0},T]$. The proof is indirect. Let us assume that there exists

$\xi \in [{t}_{0},T)$ such that

$|{v}^{\prime}(\xi )|=S$ and

$|{v}^{\prime}(t)|<S$ for

$t\in (\xi ,T]$. Then

$\chi ({v}^{\prime}(t))=1$ on this interval, and therefore

${v}^{\u2033}(t)=-\frac{a}{t}{v}^{\prime}(t)+\frac{a}{{t}^{2}}v(t)+f(t,v(t))\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [\xi ,T].$

From (8) and (9) we conclude that

$|\frac{a}{{t}^{2}}v(t)+f(t,v(t))|\le \phi (t)\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in [\xi ,T].$

Since

$\begin{array}{rcl}-{v}^{\prime}(t)& =& {v}^{\prime}(T)-{v}^{\prime}(t)={\int}_{t}^{T}{v}^{\u2033}(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ =& -a{\int}_{t}^{T}\frac{{v}^{\prime}(s)}{s}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{\int}_{t}^{T}(\frac{a}{{s}^{2}}v(s)+f(s,v(s)))\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\end{array}$

we have

$|{v}^{\prime}(t)|\le |a|{\int}_{t}^{T}\frac{|{v}^{\prime}(s)|}{s}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{\int}_{t}^{T}\phi (s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

In particular,

$|{v}^{\prime}(t)|\le |a|{\int}_{t}^{T}\frac{|{v}^{\prime}(s)|}{s}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+W\phantom{\rule{1em}{0ex}}\text{for}t\in [\xi ,T].$

By the Gronwall lemma,

$|{v}^{\prime}(t)|\le Wexp(|a|{\int}_{t}^{T}\frac{\mathrm{d}s}{s})=W{\left(\frac{T}{t}\right)}^{|a|},\phantom{\rule{1em}{0ex}}t\in [\xi ,T].$

Therefore

$S=|{v}^{\prime}(\xi )|\le W{\left(\frac{T}{\xi}\right)}^{|a|},$

which contradicts

$W{\left(\frac{T}{\xi}\right)}^{|a|}\le W{\left(\frac{T}{{t}_{0}}\right)}^{|a|}\le S-1.$

Consequently, $|{v}^{\prime}(t)|<S$ for $t\in [{t}_{0},T]$, and so $\chi ({v}^{\prime})$ = 1 on this interval. Thus, *v* is a solution of problem (3), (7).

*Step* 2. *Continuation of the solution* *v*.

It follows from the arguments given in Step 1 that

*v* is a solution of problem (3), (7) on

$[{t}_{0},T]$,

${v}^{\prime}(T)=0$ and

${u}_{0,min}({t}_{0})<v({t}_{0})<{u}_{0,max}({t}_{0})$ because

$v({t}_{0})=Q\in ({u}_{0,min}({t}_{0}),{u}_{0,max}({t}_{0}))$. It is easy to verify that the equality

${\left({t}^{a+2}{\left(\frac{v(t)}{t}\right)}^{\prime}\right)}^{\prime}={t}^{a+1}f(t,v(t))$

(12)

is satisfied for a.e.

$t\in [{t}_{0},T]$. We now integrate the last equality two times over

$[t,T]\subset [{t}_{0},T]$ and have (note that

$v(T)={v}^{\prime}(T)=0$)

$v(t)=t{\int}_{t}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,v(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\phantom{\rule{1em}{0ex}}t\in [{t}_{0},T].$

Let

*u* be a solution of problem (3), (7) on an interval

*J* which is a left-continuation of

*v*. Let us assume that

*u* is not continuable. Let

$\gamma =inf\{t:t\in J\}$. Then

$0\le \gamma <{t}_{0}$ and (12) with

*v* replaced by

*u* holds for a.e.

$t\in J$. The integration now yields

$u(t)=t{\int}_{t}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\phantom{\rule{1em}{0ex}}t\in J,$

(13)

and we claim that

${u}_{0,min}(t)<u(t)<{u}_{0,max}(t)\phantom{\rule{1em}{0ex}}\text{for}t\in (\gamma ,{t}_{0}].$

(14)

We show inequality (14) indirectly. Let us assume that (14) does not hold. Then there exists

$\nu \in (\gamma ,{t}_{0})$ such that

${u}_{0,min}(t)<u(t)<{u}_{0,max}(t)\phantom{\rule{1em}{0ex}}\text{for}t\in (\nu ,{t}_{0}]$

and either $u(\nu )={u}_{0,max}(\nu )$ or $u(\nu )={u}_{0,min}(\nu )$. Assume that $u(\nu )={u}_{0,max}(\nu )$.

Since, by (H

_{3}),

$f(t,u(t))<f(t,{u}_{0,max}(t))$ for a.e.

$t\in [\nu ,{t}_{0}]$ and

$f(t,u(t))\le f(t,{u}_{0,max}(t))$ for a.e.

$t\in [{t}_{0},T]$, we have

$\begin{array}{rcl}u(\nu )& =& \nu {\int}_{\nu}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ <& \nu {\int}_{\nu}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,{u}_{0,max}(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s={u}_{0,max}(\nu ),\end{array}$

which is not possible. The case $u(\nu )={u}_{0,min}(\nu )$ can be discussed analogously. Hence, (14) holds.

Suppose that

$\gamma >0$. Then

$J=(\gamma ,T]$ and since

*u* is bounded on

$(\gamma ,T]$, it follows that

${lim\hspace{0.17em}sup}_{t\to {\gamma}^{+}}|{u}^{\prime}(t)|=\mathrm{\infty}$. The integration of the equality

${u}^{\u2033}(t)=-\frac{a}{t}{u}^{\prime}(t)+\frac{a}{{t}^{2}}u(t)+f(t,u(t))\phantom{\rule{1em}{0ex}}\text{for a.e.}t\in (\gamma ,T]$

over

$[t,T]\subset (\gamma ,T]$ gives

$-{u}^{\prime}(t)=-a{\int}_{t}^{T}\frac{{u}^{\prime}(s)}{s}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{\int}_{t}^{T}(\frac{a}{{s}^{2}}u(s)+f(s,u(s)))\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\phantom{\rule{1em}{0ex}}t\in (\gamma ,T].$

Since

$|\frac{a}{{t}^{2}}u(t)+f(t,u(t))|\le p(t)$ for a.e.

$t\in [\gamma ,T]$, where

$p\in {L}^{1}[\gamma ,T]$, we have

$|{u}^{\prime}(t)|\le |a|{\int}_{t}^{T}\frac{|{u}^{\prime}(s)|}{s}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s+{\int}_{\gamma}^{T}p(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\phantom{\rule{1em}{0ex}}t\in (\gamma ,T].$

Using the Gronwall lemma, we deduce that for

$t\in (\gamma ,T]$,

$|{u}^{\prime}(t)|\le {\left(\frac{T}{t}\right)}^{|a|}{\int}_{\gamma}^{T}p(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\le {\left(\frac{T}{\gamma}\right)}^{|a|}{\int}_{\gamma}^{T}p(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s$

holds. This is a contradiction.

Therefore $\gamma =0$, and then (14) yields ${lim}_{t\to {0}^{+}}u(t)=0$. Consequently $J=[0,T]$, and the assertion of the theorem follows from (13). □

In the next corollary we extend the statement (d) from Lemma 1 to a large set of *A* values.

**Corollary 1** *For each* ${t}_{0}\in (0,T)$ *and each* $A\ge {u}_{0,min}({t}_{0})$ *there exists* $u\in \mathcal{S}$ *satisfying* $u({t}_{0})=A$.

*Proof* The result follows immediately from Lemma 1(d) and Theorem 1. □

**Remark 1** Corollary 1 says that the set

$\mathcal{U}=\{(t,x)\in {\mathbb{R}}^{2}:t\in [0,T],x\ge {u}_{0,min}(t)\}$ is covered by the graphs of functions from

$\mathcal{S}$, that is,

$\mathcal{U}=\{(t,u(t)):t\in [0,T],\phantom{\rule{0.25em}{0ex}}u\in \mathcal{S}\}.$

By Lemma 1(b), (c) we know that functions from $\mathcal{S}$ are uniquely determined by the values −*c* of their derivatives at the right end point $t=T$ only in the case that ${\mathcal{S}}_{c}$ is a singleton set for each $c\ge 0$. Since we cannot uniquely determine all functions from $\mathcal{S}$ via their derivatives at $t=T$, see Lemma 1(e), we discuss their derivatives at the singular point $t=0$.

**Lemma 2** *Let* (H

_{1})-(H

_{3})

*hold*.

*Let* $u\in {\mathcal{S}}_{c}$,

$c\ge 0$.

*Then* ${u}^{\prime}(0)=\frac{c}{|a+1|}+{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

(15)

*Proof* It follows from Lemma 1(g) that (6) holds for

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

$u(0)=0$, we have (note that

$a+1<0$)

$\begin{array}{rcl}{u}^{\prime}(0)& =& \underset{t\to 0+}{lim}\frac{u(t)-u(0)}{t}\\ =& \underset{t\to 0+}{lim}(\frac{c{T}^{a+1}}{|a+1|}({T}^{-a-1}-{t}^{-a-1})\\ +{\int}_{t}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s)\\ =& \frac{c}{|a+1|}+{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\end{array}$

and (15) follows. □

**Corollary 2** *Let* $u\in \mathcal{S}$.

*Then* ${u}^{\prime}(0)=-\frac{{u}^{\prime}(T)}{|a+1|}+{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

*Proof* The result follows from Lemma 2, since $u\in {\mathcal{S}}_{-{u}^{\prime}(T)}$. □

**Corollary 3** *Let* $u,v\in \mathcal{S}$, $u\ne v$. *Then* ${u}^{\prime}(0)\ne {v}^{\prime}(0)$.

*Proof* Since

$u\ne v$, there exists

${t}_{0}\in (0,T)$ such that

$u({t}_{0})\ne v({t}_{0})$. We can assume that for instance

$u({t}_{0})<v({t}_{0})$. Then

$u<v$ on

$(0,{t}_{0}]$ and

$u\le v$ on

$({t}_{0},T]$ by Lemma 1(b)(c). Hence,

${u}^{\prime}(T)\ge {v}^{\prime}(T)$ and

$\begin{array}{r}{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \phantom{\rule{1em}{0ex}}<{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,v(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s,\end{array}$

which together with Corollary 2 gives ${u}^{\prime}(0)<{v}^{\prime}(0)$. □

**Corollary 4** *Let* $u\in {\mathcal{S}}_{c}$,

$c\ge 0$.

*Then* ${u}^{\prime}(0)>\frac{c}{|a+1|}.$

(16)

*In particular*, ${u}_{0,min}^{\prime}(0)>0$.

*Proof* Inequality (16) follows from Lemma 2 and the fact that $f(t,u(t))>0$ for a.e. $t\in [0,T]$ by (H_{2}). □

**Corollary 5** *Let* $u\in \mathcal{S}$.

*Then* ${\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=\frac{1}{|a+1|}{\int}_{0}^{T}f(s,u(s))\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

*Proof* The integration of (3) over

$[\epsilon ,T]\subset (0,T]$ gives

${u}^{\prime}(T)-{u}^{\prime}(\epsilon )+a{\int}_{\epsilon}^{T}\frac{{u}^{\prime}(t)}{t}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t-a{\int}_{\epsilon}^{T}\frac{u(t)}{{t}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t={\int}_{\epsilon}^{T}f(t,u(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t.$

Using integration by parts we obtain

${\int}_{\epsilon}^{T}\frac{u(t)}{{t}^{2}}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t=\frac{u(\epsilon )}{\epsilon}+{\int}_{\epsilon}^{T}\frac{{u}^{\prime}(t)}{t}\phantom{\rule{0.2em}{0ex}}\mathrm{d}t,$

and, therefore,

${u}^{\prime}(T)-{u}^{\prime}(\epsilon )-a\frac{u(\epsilon )}{\epsilon}={\int}_{\epsilon}^{T}f(t,u(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t.$

Taking the limit

$\epsilon \to 0$, we have

${u}^{\prime}(T)+|a+1|{u}^{\prime}(0)={\int}_{0}^{T}f(t,u(t))\phantom{\rule{0.2em}{0ex}}\mathrm{d}t.$

On the other hand, it follows from Corollary 2 that

${u}^{\prime}(T)+|a+1|{u}^{\prime}(0)=|a+1|{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.$

Combining the above two equalities yields the result. □

**Lemma 3** *Let* (H_{1})-(H_{3}) *hold*. *Let* $c\ge 0$ *and* ${\mathcal{S}}_{c}$ *be not a singleton set*. *Then for each* $\alpha \in [{u}_{c,min}^{\prime}(0),{u}_{c,max}^{\prime}(0)]$ *there exists a unique* $u\in {\mathcal{S}}_{c}$ *such that* ${u}^{\prime}(0)=\alpha $. *Consequently*, $[{u}_{c,min}^{\prime}(0),{u}_{c,max}^{\prime}(0)]=\{{u}^{\prime}(0):u\in {\mathcal{S}}_{c}\}$.

*Proof* Let

$u\in {\mathcal{S}}_{c}$,

${u}_{c,min}\ne u\ne {u}_{c,max}$. Then it follows from Lemma 1(a) and (H

_{3}) that

$\begin{array}{r}{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,{u}_{c,min}(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \phantom{\rule{1em}{0ex}}<{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,u(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \phantom{\rule{1em}{0ex}}<{\int}_{0}^{T}{s}^{-a-2}\left({\int}_{s}^{T}{\xi}^{a+1}f(\xi ,{u}_{c,max}(\xi ))\phantom{\rule{0.2em}{0ex}}\mathrm{d}\xi \right)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.\end{array}$

Hence, by Lemma 2,

${u}_{c,min}^{\prime}(0)<{u}^{\prime}(0)<{u}_{c,max}^{\prime}(0)\phantom{\rule{1em}{0ex}}\text{for}u\in {\mathcal{S}}_{c},{u}_{c,min}\ne u\ne {u}_{c,max}.$

Since ${\mathcal{S}}_{c}$ is a compact set in ${C}^{1}[0,T]$ by Lemma 1(f), the set $I=\{{u}^{\prime}(0):u\in {\mathcal{S}}_{c}\}\subset [{u}_{c,min}^{\prime}(0),{u}_{c,max}^{\prime}(0)]$ is closed. In fact, let $\{{\beta}_{n}\}\subset I$, ${\beta}_{n}={u}_{n}^{\prime}(0)$, where ${u}_{n}\in {\mathcal{S}}_{c}$, and let ${lim}_{n\to \mathrm{\infty}}{\beta}_{n}=\beta $. Then there exist a subsequence $\{{\beta}_{{k}_{n}}\}$ of $\{{\beta}_{n}\}$ and $u\in {\mathcal{S}}_{c}$ such that ${lim}_{n\to \mathrm{\infty}}{u}_{{k}_{n}}=u$ in ${C}^{1}[0,T]$. In particular, ${\beta}_{{k}_{n}}={u}_{{k}_{n}}^{\prime}(0)\to {u}^{\prime}(0)$ as $n\to \mathrm{\infty}$. Therefore, $\beta ={u}^{\prime}(0)\in I$.

It remains to prove that $I=[{u}_{c,min}^{\prime}(0),{u}_{c,max}^{\prime}(0)]$. Assume that the equality does not hold. Then, from the structure of bounded and closed subsets of ℝ the existence of an open interval $(\alpha ,\beta )\subset [{u}_{c,min}^{\prime}(0),{u}_{c,max}^{\prime}(0)]\setminus I$, $\alpha ,\beta \in I$, follows. Let $\alpha ={u}_{\alpha}^{\prime}(0)$ and $\beta ={u}_{\beta}^{\prime}(0)$, where ${u}_{\alpha},{u}_{\beta}\in {\mathcal{S}}_{c}$. Due to Lemma 1(c), there exists ${t}_{0}\in (0,T)$ such that ${u}_{\alpha}<{u}_{\beta}$ on $(0,{t}_{0}]$, ${u}_{\alpha}\le {u}_{\beta}$ on $({t}_{0},T]$. Choose $\tau \in ({u}_{\alpha}({t}_{0}),{u}_{\beta}({t}_{0}))$. By Corollary 1, there exists $v\in {\mathcal{S}}_{c}$ such that $v({t}_{0})=\tau $. Then ${u}_{\alpha}<v<{u}_{\beta}$ on $(0,{t}_{0}]$ and ${u}_{\alpha}\le v\le {u}_{\beta}$ on $({t}_{0},T]$. Consequently, ${u}_{\alpha}^{\prime}(0)<{v}^{\prime}(0)<{u}_{\beta}^{\prime}(0)$ on $({t}_{0},T]$, that is, ${v}^{\prime}(0)\in (\alpha ,\beta )$, which is not possible. □

Since functions from

$\mathcal{S}$ belong to

$A{C}^{1}[0,T]$, we have

${u}^{\prime}(0)<\mathrm{\infty}$ for each

$u\in \mathcal{S}$. Corollary 4 yields

${u}_{0,min}^{\prime}(0)>0$. Let us denote

$J:=[{u}_{0,min}^{\prime}(0),\mathrm{\infty})\subset (0,\mathrm{\infty}).$

(17)

Lemma 3 implies that functions from $\mathcal{S}$ can be uniquely determined by the values of their derivatives at the singular point $t=0$; see Theorem 2.

**Theorem 2** *Let* (H_{1})-(H_{3}) *hold*. *Then there exists a unique* $u\in \mathcal{S}$ *satisfying* ${u}^{\prime}(0)=\alpha $ *if and only if* $\alpha \in J$.

*Proof* We first show

$\{{u}^{\prime}(0):u\in \mathcal{S}\}=J.$

(18)

It follows from Lemma 3 that

$\{{u}^{\prime}(0):u\in \bigcup _{0\le c\le K}{\mathcal{S}}_{c}\}=[{u}_{0,min}^{\prime}(0),{u}_{K,max}^{\prime}(0)]$

for each $K\ge 0$, where we set ${u}_{c,min}=u={u}_{c,max}$ if ${\mathcal{S}}_{c}$ is a singleton set and ${\mathcal{S}}_{c}=\{u\}$. In view of Corollary 4, ${u}^{\prime}(0)>c/|a+1|$ for $u\in {\mathcal{S}}_{c}$. Consequently, (18) follows.

Let us now choose $\alpha \in J$. Then there exists $u\in \mathcal{S}$ satisfying ${u}^{\prime}(0)=\alpha $. The uniqueness of *u* follows from Corollary 3. □

For $\alpha \in J$, we denote by ${u}_{\alpha}$ the unique element of $\mathcal{S}$ satisfying ${u}_{\alpha}^{\prime}(0)=\alpha $.

**Theorem 3** *Let* (H_{1})-(H_{3}) *hold*. *Assume that* $\{{\alpha}_{n}\}\subset J$ *is a convergent sequence and* ${lim}_{n\to \mathrm{\infty}}{\alpha}_{n}=\alpha $. *Then* ${lim}_{n\to \mathrm{\infty}}{u}_{{\alpha}_{n}}={u}_{\alpha}$ *in* ${C}^{1}[0,T]$.

*Proof* Let $K=sup\{-{u}_{{\alpha}_{n}}^{\prime}(T):n\in \mathbb{N}\}$. Then $K<\mathrm{\infty}$ because ${\alpha}_{n}={u}_{{\alpha}_{n}}^{\prime}(0)>-{u}_{{\alpha}_{n}}^{\prime}(T)/|a+1|$ for $n\in \mathbb{N}$ by Corollary 4. Since $\{{u}_{{\alpha}_{n}}\}\subset {\bigcup}_{0\le c\le K}{\mathcal{S}}_{c}$ and ${\bigcup}_{0\le c\le K}{\mathcal{S}}_{c}$ is compact in ${C}^{1}[0,T]$ by Lemma 1(f), the sequence $\{{u}_{{\alpha}_{n}}\}$ is relatively compact in ${C}^{1}[0,T]$. Let $\{{u}_{{\alpha}_{{\ell}_{n}}}\}$ be a subsequence of $\{{u}_{{\alpha}_{n}}\}$ which is convergent in ${C}^{1}[0,T]$, and let ${lim}_{n\to \mathrm{\infty}}{u}_{{\alpha}_{{\ell}_{n}}}=u$. Then $u\in {\mathcal{S}}_{{c}_{0}}$ for a ${c}_{0}\in [0,K]$ and ${u}^{\prime}(0)={lim}_{n\to \mathrm{\infty}}{u}_{{\alpha}_{{\ell}_{n}}}^{\prime}(0)={lim}_{n\to \mathrm{\infty}}{\alpha}_{{\ell}_{n}}=\alpha $. Therefore, $u={u}_{\alpha}$ and hence any subsequence $\{{u}_{{\alpha}_{{\ell}_{n}}}\}$ of $\{{u}_{{\alpha}_{n}}\}$ converging in ${C}^{1}[0,T]$ has the same limit ${u}_{\alpha}$. Consequently, $\{{u}_{{\alpha}_{n}}\}$ is convergent in ${C}^{1}[0,T]$ and ${u}_{\alpha}$ is its limit. □