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 onepoint 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}^{a1}{t}^{a1})+t{\int}_{t}^{T}{s}^{a2}\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}Sx2S,\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{aS}{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)}{Tt}\le \frac{v(t)}{Tt}\le \frac{{u}_{0,max}(t)}{Tt},\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 S1.
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}^{a2}\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 leftcontinuation 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}^{a2}\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}^{a2}\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}^{a2}\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}^{a2}\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}^{a1}{t}^{a1})\\ +{\int}_{t}^{T}{s}^{a2}\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}^{a2}\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}^{a2}\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}^{a2}\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}^{a2}\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}^{a2}\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}ta{\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}^{a2}\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}^{a2}\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}^{a2}\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}^{a2}\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. □