\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 24, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/24\hfil Signed radial solutions]
{Signed radial solutions for a weighted $p$-superlinear problem}

\author[S. Herr\'on, E. Lopera \hfil EJDE-2014/24\hfilneg]
{Sigifredo Herr\'on, Emer Lopera}  % in alphabetical order

\address{Sigifredo Herr\'on \newline
Escuela de Matem\'aticas, Universidad Nacional de Colombia Sede
Medell\'in,  Apartado A\'ereo 3840,
 Medell\'{\i}n, Colombia}
\email{sherron@unal.edu.co}

\address{Emer Lopera \newline
Universidad Nacional de Colombia Sede Medell\'in,
Medell\'{\i}n, Colombia}
\email{edlopera@unal.edu.co}

\thanks{Submitted October 1, 2013. Published January 14, 2014.}
\subjclass[2000]{35J92, 35J60}
\keywords{Quasilinear weighted elliptic equations; radial solutions;
\hfill\break\indent one-signed solutions; shooting method}

\begin{abstract}
 We study the existence of one-signed radial solutions for 
 weighted semipositone problems  where $\Delta_p$ operator is
 involved and the nonlinearity is $p$-superlinear at infinity and has
 only two zeros. We establish the existence of at least two
 one-signed solutions when the weight is small enough.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks




\section{Introduction}

Let us consider the existence of one-signed radial solutions for the
boundary-value problem
\begin{equation}
\begin{gathered}
-\Delta_pu=K(\| x\|)  f(u),\quad x\in B_1(0) \\
u=0,\quad \| x\| =1,
\end{gathered}  \label{1}
\end{equation}
where $B_1(0)  \subset\mathbb{R}^N$ is the unit ball and $2\leq p<N$. 

The existence of one-signed radial solutions to \eqref{1} is
equivalent to the existence of one-signed solutions to the ordinary 
differential equation
\begin{equation}
\begin{gathered}
{}[\varphi_p(u'(r))]'+\frac{N-1}{r}\varphi_p(u'(r))
+K(r)  f(u(r))  =0,\quad 0<r<1\\
u'(0)  =0,\quad u(1)  =0,
\end{gathered}  \label{2}
\end{equation}
where $\varphi_p:\mathbb{R}\to\mathbb{R}$ is defined as
$\varphi_p(s)  =|s|^{p-2}s$, $s\neq0$,
$\varphi_p(0)  =0$.

We assume that nonlinearity $f$ satisfies the following hypotheses:
\begin{itemize}
\item[(F1)] $f:\mathbb{R}\setminus\{  0\}  \to \mathbb{R}$ is differentiable, 
$f'(t)  \geq0$ for all $t\leq v_0<0$ and all $t\geq u_0>0$,
where $v_0$ and $u_0$ are the only zeros of $f$.

\item[(F2)] $f(0^{-}) :=\lim_{t\to0^{-}}f(t)$ is a positive
number and $f(0^{+}):=\lim_{t\to0^{+}}f(t)  $ is a negative number.

\item[(F3)] $\lim_{|\alpha|\to\infty}\frac{f(\alpha)  }{\varphi_p(\alpha)  }
=+\infty$.

\item[(F4)] There exist constants $k\in(0,1)  $ and $\theta>N-p$ such that for
all $\delta\geq \theta$
\[
\lim_{|\alpha|\to\infty}\Big(
\frac{\varphi_p(\alpha)  }{f(\alpha)  }\Big)
^{\frac{N}{p}}\Big(\delta F(k\alpha)  -\frac{N-p}{p}\alpha
f(\alpha)\Big)  =+\infty,
\]
where $F(t)  =\int_0^{t}f(s)  ds$.
\end{itemize}

For the weight $K$, hereafter we will assume that
\begin{itemize}
\item[(K1)] $K\in C([0,1],\mathbb{R}^{+})  $ and $K$
is differentiable in $(0,1)$.

\item[(K2)] $r\mapsto N+r\frac{K'(r)  }{K(r)  }>\theta$,  is non-increasing in 
$(0,1)  $ and $\lim_{r\to0^{+}}\frac{K'(r) }{K(r)}$ exists in $\mathbb{R}$.
\end{itemize}
A function satisfying the statements (K1) and (K2) will be called an
admissible weight.

The aim of this article is to prove that under hypotheses
(F1)--(F4) for the nonlinearity and (K1)--(K2) for the weight,
 problem \eqref{2} has at least two solutions, provided that the
weight is small enough. Moreover, one of them is positive and
the other one is negative. In order to prove the existence of
positive solution, we modify  our nonlinearity in the following way
\[
f^{+}(t)  :=\begin{cases}f(t)  &\text{if } t>0\\
f(0^{+})  &\text{if } t=0\\
0 &\text{if } t<0.
\end{cases}
\]
Similar modifications vanishing the positive part  of $f$ give us
 a negative solution.

The case $p=2$ was studied by Castro and Shivaji in \cite{CS}.
Recently, Hakimi and Zertiti, \cite{HZ}, following the ideas in
\cite{CS} obtained existence of positive solutions for a more
general nonlinearity than Castro and Shivaji. Both works
considered a constant weight in the semilinear case. We emphasize that 
in this work we deal with the case $2\leq p<N$ and a nonconstant weight. 
Therefore, it is a generalization of the previous results.

There are several studies related to radial solutions involving $p$-Laplacian 
problems and a lot of techniques have been used. Most of them are interested 
in the existence, nonexistence, and multiplicity of positive radial solutions. 
For instance, in \cite{KLS} the authors considered the equation of 
problem \eqref{1} with $f(u)=u^q, q>p-1$, in exterior domains. 
They employed a global continuation theorem and fixed point index theory 
based on a weighted space as the underlying space. By using this approach 
they obtained multiplicity of positive solutions depending on a certain real 
parameter $\mu$. A one dimensional weighted $p$-Laplacian problem is presented 
in \cite{NT}. Sharp conditions for the existence of solutions with prescribed 
numbers of zeros in terms of the ratio
$f (s)/s^{ p-1}$ at zero and at infinity  were established there. 
Their technique was based on the shooting method together with the qualitative 
theory for half-linear differential equations. Other results can be 
found in \cite{AP, CH, CHV, GMY, Z} and some references therein.

Our main tool for solving problem \eqref{2} is the shooting method. 
Hence, we start considering the auxiliary initial value problem
\begin{equation}
\begin{gathered}
{}[\varphi_p(u'(r))  ]'+\frac{N-1}{r}\varphi_p(u'(r))
+K(r)  f(u(r))  =0,\quad 0<r<1\\
u'(0)  =0,\quad u(0) =\alpha.
\end{gathered}  \label{3}
\end{equation}
It can be shown (see appendix) that this problem has a unique
solution, usually denoted  by $u(\cdot,\alpha,K)  $, which
depends continuously on the initial data $\alpha$.

We have the following equivalent integral formulas to problem \eqref{3}:
\begin{gather}
u(s)  =\alpha-\int_0^{s}\varphi_{p'}\Big(
r^{1-N}\int_0^{r}t^{N-1}K(t)  f(u(
t))  dt\Big)  dr,\quad 0\leq s\leq1,\label{3.a}
\\
\varphi_p(u')  (r)  =-r^{1-N}\int_0
^{r}t^{N-1}K(t)  f(u(t))  dt,\quad 0\leq r\leq1.\label{4}
\end{gather}
Our main result reads as follows.

\begin{theorem}\label{maintheorem}
Under  hypotheses {\rm (F1)--(F4)} and {\rm (K1)--(K2)}, there exists a positive
number $\lambda_0$ such that if $\| K\| _{\infty}<\lambda_0$, then
 problem \eqref{1} has at least two radial solutions. 
Moreover, one of them is positive, decreasing and has negative radial 
derivative in $\| x\| =1$. The second solution is negative, increasing 
and has  positive radial derivative  in $\| x\| =1$.
\end{theorem}

This paper is organized in the following way: 
In Section \ref{preliminary} we present some facts which are useful 
for showing Theorem \ref{maintheorem}. 
Next, in Section \ref{prooftheorem}, we prove the main result and we 
exhibit some examples of functions $f$ and $K$ satisfying conditions 
(F1)--(F4) and (K1)--(K2). Finally, for convenience of the reader, 
we present in the appendix a  Pozohaev type identity and some proofs 
about the existence and uniqueness of \eqref{3}.

\section{Preliminary results}\label{preliminary}

For the rest of this article, we consider problem \eqref{2} with
 nonlinearity $f^+$ instead of $f$. Let $u_1\in (u_0, +\infty)$
be the unique zero of $F^+$ where $F^+(t):=\int_0^tf^+(s)\,ds$. 
Fix an admissible  weight $K$ and set
\begin{gather*}
\eta  :=\min_{r\in[0,1]}K(r)  ;\quad
\bar{\lambda}:=\max_{r\in[0,1]}K(r)  ,\\
\delta_0   :=\min_{r\in[0,1]}\big\{  N+r\frac{K'(r) }{K(r)  }\big\}
 =N+\frac{K'(1)  }{K(1)  }, \\
\delta_1   :=\max_{r\in[0,1]}\big\{  N+r\frac{K'(r)  }{K(r)  }\big\}  =N.
\end{gather*}
For fixed $\alpha\geq u_1/k$, there exists  
$t_0=t_0(\alpha)  \in(0,1)  $ such that $u(t_0,\alpha,K)  =k\alpha$ and
$k\alpha\leq u(t,\alpha,K) \leq\alpha$ for all $t\in[0,t_0)$, where
$u(\cdot,\alpha,K)$ is the unique solution of problem \eqref{3}.
 Moreover, $t_0$ satisfies the  estimate \eqref{5} which, in particular,
shows that $t_0(\alpha)  \to0$ as $\alpha\to+\infty$.
Indeed, as $f^+$ is increasing on $[u_0,+\infty)  $, then for
$t\in[0,t_0]$
\[
t^{N-1}K(t)  f^+(k\alpha)  \leq t^{N-1}K(t)  f^+(u(t,\alpha,K))  
\leq t^{N-1}K(t)  f^+(\alpha)  .
\]
Hence,
\[
-r^{1-N}\int_0^{r}t^{N-1}K(t)  f^+(\alpha) dt\leq\varphi_p(u'(t,\alpha,k))
\leq-r^{1-N}\int_0^{r}t^{N-1}K(t)  f^+(k\alpha)dt,
\]
which in turn, implies that
\[
-\varphi_{p'}\Big(\bar{\lambda}\frac{f^+(\alpha)  }
{N}\Big)  r^{p'-1}\leq u'(t,\alpha,k)
\leq-\varphi_{p'}\Big(\eta\frac{f^+(k\alpha)  }{N}\Big)  r^{p'-1},
\]
for all $t\in[0,t_0]$. Here $p'$ denotes the conjugate of $p$. 
After integration on $[0,t_0]$ we obtain
\[
-\varphi_{p'}\Big(\bar{\lambda}\frac{f^+(\alpha)  }{N}\Big) 
 \frac{t_0^{p'}}{p'}\leq\alpha(k-1)
\leq-\varphi_{p'}\Big(\eta\frac{f^+(k\alpha)  }{N}\Big)  \frac{t_0^{p'}}{p'}.
\]
These inequalities yield the following estimate for $t_0$
\begin{equation}
C_1\Big(\frac{\varphi_p(\alpha)  }{\bar{\lambda}f^+(
\alpha)  }\Big)  ^{1/p}\leq t_0
\leq C_1\Big( \frac{\varphi_p(\alpha)  }{\eta f^+(k\alpha)
}\Big)  ^{1/p}, \label{5}
\end{equation}
where $C_1:=(p'(1-k)  N^{p'-1})^{1/p'}>0$.

\subsection{Pohozaev Identity}

In this subsection, we present a Pohozaev identity  as well as some
consequences. The proof of this identity is quite standard but it is 
presented in the appendix for the sake of completeness. Let us define 
the Energy associated to  problem \eqref{3} by
\[
E(t,\alpha,K)  :=\frac{|u'(t,\alpha,K)  |^{p}}{p'K(t)  }+F^+(u(t,\alpha,K)).
\]
Also, we define
\[
H(t,\alpha,K)  :=tK(t)  E(t,\alpha,K)+\frac{N-p}{p}\varphi_p(u'(t,\alpha,K))
u(t,\alpha,K)  .
\]
Suppose that $u(\cdot,\alpha,K)  $ is a solution of \eqref{3}.
Then,  a Pohozaev type identity takes place
\begin{equation}\label{pohozaev}
\begin{aligned}
&  t^{N-1}H(t,\alpha,K)  -s^{N-1}H(s,\alpha,K) \\
&  =\int_{s}^{t}r^{N-1}K(r)  \Big[\Big(
N+r\frac{K'(r)  }{K(r)  }\Big)  F^+(u)  -\frac{N-p}{p}f^+(u)  u\Big]dr,
\end{aligned}
\end{equation}
for $0\leq s\leq t\leq1$. We shall use this version of Pohozaev identity 
as follows. For $s=0$ and $t=t_0$, we get
\[
t_0^{N-1}H(t_0,\alpha,K)  =\int_0^{t_0}
r^{N-1}K(r)  \Big[\Big(N+r\frac{K'(r)
}{K(r)  }\Big)  F^+(u)  -\frac{N-p}{p}f^+(u)  u\Big]dr.
\]
Since $f^+$ and $F^+$ are nonnegative and increasing functions on the 
interval $[u_1,\infty)  $, then for all $r\in[0,t_0]$,
\begin{align*}
\Big(N+r\frac{K'(r)  }{K(r)  }\Big)
F^+(u)  -\frac{N-p}{p}uf^+(u) 
&  \geq\delta_0F^+(u)  -\frac{N-p}{p}uf^+(u)\\
&\geq \delta_0F^+(k\alpha)  -\frac{N-p}{p}\alpha f^+(\alpha)  .
\end{align*}
In consequence,
\[
t_0^{N-1}H(t_0,\alpha,K)  \geq\frac{\eta}{N}
\Big(\delta_0F^+(k\alpha)  -\frac{N-p}{p}\alpha f^+(\alpha)\Big)  t_0^N.
\]
From this and \eqref{5} we find that
\begin{equation}
t_0^{N-1}H(t_0,\alpha,K)  
\geq\frac{\eta C_1^N} {N\bar{\lambda}^{N/p}}\Big(\delta_0F^+(k\alpha)  -\frac
{N-p}{p}\alpha f^+(\alpha)\Big)  \Big(\frac{\varphi
_p(\alpha)  }{f^+(\alpha)  }\Big)  ^{N/p}.
\label{7}
\end{equation}
We claim that for each number $\delta\geq\theta$, there is a positive
constant $B_{\delta}$, such that
\begin{equation}
\delta F^+(s)  -\frac{N-p}{p}sf^+(s)  \geq-B_{\delta
},\quad \text{for all }s\in\mathbb{R}. \label{7a}
\end{equation}
In fact, (F4) guarantees existence of $C_{\delta}>0$ satisfying
\[
\delta F^+(s)  -\frac{N-p}{p}sf^+(s)  \geq\delta
F^+(ks)  -\frac{N-p}{p}sf^+(s)  \geq0,
\]
for all $s>C_{\delta}$  and all $s<0$.
On the other hand, if we set $M_{\delta}:=\sup_{s\in[0,C_{\delta}]}|f^+(s)  |$, 
then
\begin{align*}
\delta F^+(s)  -\frac{N-p}{p}sf^+(s)  
&=\int _0^{s}\Big(\delta f^+(t)  -\frac{N-p}{p}f^+(s)\Big)  dt\\
&  \geq-M_{\delta}\int_0^{s}\Big(\delta+\frac{N-p}{p}\Big)dt\\
&=-M_{\delta}s(\delta+\frac{N-p}{p}) \\
&  \geq-M_{\delta}C_{\delta}\Big(\delta+\frac{N-p}{p}\Big)  =:-B_{\delta
}.
\end{align*}

Now, replacing $t_0$ by $s$ in Pohozaev identity \eqref{pohozaev}, 
and using the estimate \eqref{7a} with $\delta_0$ and $\delta_1$ we obtain
\begin{align*}
&t^{N-1}H(t,\alpha,K) \\
&=t_0^{N-1}H(t_0,\alpha,K)  +\int_{A}r^{N-1}K(r)  
\Big[(N+r\frac{K'(r)  }{K(r)  })  F^+(u)  -\frac{N-p}{p}f^+(u)u\Big]dr\\
&\quad  +\int_{B}r^{N-1}K(r) 
\Big[(N+r\frac{K'(r)  }{K(r)  })  F^+(u)-\frac{N-p}{p}f^+(u)  u\Big]dr\\
&\geq t_0^{N-1}H(t_0,\alpha,K)  +\int_{A}
r^{N-1}K(r)  \Big[\delta_0F^+(u)  -\frac{N-p} {p}f^+(u)  u\Big]dr\\
& \quad +\int_{B}r^{N-1}K(r)  \Big[\delta_1F^+(
u)  -\frac{N-p}{p}f^+(u)  u\Big]dr \\
&  \geq t_0^{N-1}H(t_0,\alpha,K)  -B_{\delta_0}
\int_{A}r^{N-1}K(r)  dr-B_{\delta_1}\int_{B}
r^{N-1}K(r)  dr\\
&  \geq t_0^{N-1}H(t_0,\alpha,K)  -2\frac{\bar{\lambda}B_{\delta_1}}{N}.
\end{align*}
Here $A:=\{  r\in[t_0,t]:F^+(u(r,\alpha,K)  \geq0)  \}  $ and 
$B:=\{  r\in[ t_0,t]:F^+(u(r,\alpha,K)  <0)  \}  $.
Then we reach the estimate
\begin{equation}
t^{N-1}H(t,\alpha,K)  \geq t_0^{N-1}H(t_0
,\alpha,K)  -2\bar{\lambda}M_NC_N\Big(1+\frac{N-p}{Np}\Big),   \label{8}
\end{equation}
where we have used the fact that $\delta_1=N$ (it is remarkable  
that $\delta_1$ does not depend on $K$).

\section{Main Result}\label{prooftheorem}

In this section we shall prove our main theorem. 
Before, we will establish three preliminary Lemmas.

\begin{lemma}\label{lemma1}
There exists a positive real number $\lambda_2$ with the
following property. For every admissible weight $K$, with
$\|K\|_{\infty}\equiv\bar{\lambda}<\lambda_2$, there is a real
number $\underline{\alpha}>u_1/k$, such that for all
$\alpha\geq\underline{\alpha}$ and all $t\in[0,1]$,
$|u(t,\alpha,K)  |^{p}+|u'(t,\alpha,K)  |^{p}>0$.
\end{lemma}

\begin{proof}
Given a weight $K$ with the properties (K1) and (K2), there is
$\underline{\alpha}>u_1/k$ such that for all
 $\alpha\geq \underline{\alpha}$
\[
\Big(\delta_0F^+(k\alpha)  -\frac{N-p}{p}\alpha f^+(
\alpha)\Big)  \Big(\frac{\varphi_p(\alpha)
}{f^+(\alpha)  }\Big)  ^{N/p}\geq\frac{1}{\eta}.
\]
From \eqref{7} and \eqref{8} we have, for $t\geq t_0$
\begin{equation}\label{lema1-numerada1}
\begin{aligned}
t^{N-1}H(t,\alpha,K)   
&  \geq t_0^{N-1}H(t_0 ,\alpha,K)  -2\bar{\lambda}M_{\delta_1}C_{\delta_1}
\Big(1+\frac{N-p}{Np}\Big) \\
&  \geq\frac{\eta C_1^N}{N\bar{\lambda}^{N/p}}\frac{1}{\eta}-2\bar
{\lambda}M_{N}C_{N}\Big(1+\frac{N-p}{Np}\Big) \\
&  =\bar{\lambda}\Big(\frac{C_1^N}{N\bar{\lambda}^{(N-p)/p}}
-2M_{N}C_{N}(1+\frac{N-p}{Np})\Big).
\end{aligned}
\end{equation}
Now, there exists $\lambda_2>0$ such that
\begin{equation}\label{lema1-numerada2}
 \frac{C_1^N}{N\lambda^{(N-p)/p}}-2M_{N}C_{N}(1+\frac{N-p}{Np})>0,
\end{equation}
for every $\lambda\in (0, \lambda_2)$.
We fix a weight $K$ so that $||K||_\infty:=\bar{\lambda}<\lambda_2$. 
Thus, from \eqref{lema1-numerada1} and \eqref{lema1-numerada2} we have 
$H(t,\alpha, K)  >0$ for $t\in [t_0, 1]$. Then, for all $t\in[t_0,1]$,
$|u(t,\alpha,K)  |^{p}+|u'(t,\alpha,K)  |^{p}>0$. Clearly, 
$|u(t,\alpha,K)  |^{p}+|u'(t,\alpha,K)  |^{p}>0$ holds for $t\in[0,t_0]$.
\end{proof}

\begin{lemma}\label{lemma2}
There exists a real number $\lambda_1\in(0, \lambda_2]$ with the
following property. For every admissible weight $K$, with 
$\bar{\lambda}<\lambda_1$, we have $u(t,\underline{\alpha},K)  \geq u_0$ for all
$t\in[0,1]$.  Here, $\underline{\alpha}$ is a number obtained in
the preceding Lemma.
\end{lemma}

\begin{proof}
Given a weight $K$, set
\[
t_1:=\sup\{  t\in[0,1]:u(r,\underline{\alpha},K)  
\geq u_0\text{ for all }r\in(0,t)  \}  .
\]
We observe that $t_0$ belongs to the previous set. Since $f^+$
is nonnegative on the interval $[u_0,\infty)  $,
we see from \eqref{4}, that
\[
\varphi_p(u'(t,\underline{\alpha},K))
=-t^{1-N}\int_0^{t}r^{N-1}K(r)  f^+(u(r))  dr\leq0,\quad \text{for all }t\in[0,t_1].
\]
Therefore, $u$ is decreasing on $[0,t_1]$. Besides, for all
$t\in[0,t_1]$,
\begin{align*}
|\varphi_p(u'(t,\underline{\alpha},K)) |
&\leq t^{1-N}\int_0^{t}r^{N-1}K(r)  f^+(u(r,\underline{\alpha},K))  dr\\
& \leq\bar{\lambda}f^+(u(0,\underline{\alpha},K)) t^{1-N}\int_0^{t}r^{N-1}dr\\
&\leq\frac{\bar{\lambda}f^+(\underline{\alpha})  }{N}t\\
&\leq\frac{\bar{\lambda}f^+(\underline{\alpha})  }{N}.
\end{align*}
Hence
\[
|u'(t,\underline{\alpha},K)|
\leq\varphi_{p'}\Big(\frac{\bar{\lambda}f^+(\underline{\alpha})  }{N}\Big)  .
\]
Now, fix $\lambda_1\leq\min\{  \lambda_2,\frac{N}{f^+(
\underline{\alpha})  }\varphi_p(\underline{\alpha}-u_0)  \}  $, 
then $\varphi_{p'}(\lambda_1)  \leq\varphi_{p'}(\frac{N}{f^+(\underline
{\alpha})  })  (\underline{\alpha}-u_0)  $. It
follows that if $\bar{\lambda}<\lambda_1$, we have 
$|u'(r,\underline{\alpha},K)  |\leq\underline{\alpha}-u_0$ for all 
$t\in[0,t_1]$. An application of the mean value theorem, allows us to choose 
a real number $\xi\in(0,t_1)  $ such that
\[
u(t_1,\underline{\alpha},K)  -u(0,\underline{\alpha},K)  
=u'(\xi,\underline{\alpha},K)  t_1 \geq-(\underline{\alpha}-u_0)  t_1.
\]
If we assume that $t_1<1$, then $u(t_1,\underline{\alpha},K)>u_0$, 
contradicting the definition of $t_1$. This completes the proof.
\end{proof}


\begin{lemma}\label{lemma3}
For a given admissible weight $K$ with $\bar{\lambda}<\lambda_1$, there
exists $\alpha_1\geq\underline{\alpha}$ such that $u(t,\alpha_1,K)  <0$ 
for some $t\in[0,1]$.
\end{lemma}

\begin{proof}
We argue by contradiction. Suppose that there exists a suitable
weight $K$, such that for all $\alpha\geq\underline{\alpha}$ and all
$t\in[0,1]$, $u(t,\alpha,K)  \geq0$. Without lost
of generality we can assume that $u(t,\alpha,K)  >0$ for all
$t\in[0,1)  $. Let $\bar{t}=\bar{t}(\alpha)  $ be
the supremum of the set
\[
V:=\{  t\in[0,1]:u(\cdot,\alpha,K)  \text{ is
decreasing on }[0,t]\}.
\]
$V$ is a nonempty set because it contains $t_0$. On the other
hand, in view of the inequalities \eqref{7}, \eqref{8} and
hypothesis (F4), we can fix a real number $\alpha_1$ such that
for all $\alpha\geq$ $\alpha_1$ and all $t\in[0,1]$, 
$t^{N-1}H(t,\alpha,K)  >0$. Now, we claim that 
$u'(t,\alpha,K)\neq0$ for all $t\in(0,1]$. For if $u'(t_1,\alpha,K)  =0$ 
for some $t_1\in(0,1]$ then the differential equation in \eqref{2}
would imply that $f^+(u(t_1,\alpha,K))=0$. Hence $u(t_1,\alpha,K) =u_0$, but
\[
0<t_1^{N-1}H(t_1,\alpha,K)  =t_1^NK(t_1)
F^+(u(t_1,\alpha,K))  =t_1^NK(t_1)  F^+(u_0)  <0,
\]
which is a contradiction. We have stated that if 
$\alpha\geq \alpha_1$ then $\bar{t}=1$. Let $v$ be a positive 
solution to problem
\begin{equation}
\begin{gathered}
(\varphi_p(v'))  '+\frac{N-1}{r}\varphi_p(v')  
=-\mu\varphi_p(v)  ,\quad 0<r<\rho:=\frac1{p+1} \\
v(0)  =1,\quad v'(0)  =0,\quad v(\rho)  =0,\quad v(r)  >0\quad
\text{for }0<r<\rho.
\end{gathered}\label{eigenvalue_problem}
\end{equation}
Notice that $v'(\rho) <0$, since
\[
\varphi_p(v'(\rho))  =-\mu\rho^{1-N}\int_0^{\rho}r^{N-1}\varphi_p(v)  dr<0.
\]
Hypothesis (F3) is now used to assure the existence of  $\alpha_0>u_1/k$ 
such that
\begin{equation}
\frac{f^+(x)  }{\varphi_p(x)  }\geq\mu/
{\bar{\lambda}},\quad \text{for all }x\geq\alpha_0. \label{14}
\end{equation}
From the corresponding integral formulas for the solutions $u$ and $v$ 
(cf. \eqref{4} and \eqref{eigenvalue_problem}) we have
\begin{gather}
[r^{N-1}\varphi_p(u')  ]'\varphi_p(v)  =-r^{N-1}K(r)  f^+(u) \varphi_p(v), \label{10}
\\
[r^{N-1}\varphi_p(v')  ]'\varphi_p(u)  =-\mu r^{N-1}\varphi_p(v)
\varphi_p(u)  . \label{11}
\end{gather}
Let $t_1$ be the supremum of the set 
$A:=\{  t\in[0,\rho] :v'u(r)  \leq u'v(r)  \text{ for all }r\in(t,\rho]\} $, 
which is a nonempty set due to the
fact that $0<-v'u(\rho)  +u'v(\rho)$. Certainly
\begin{equation}
v'u(t_1)  =u'v(t_1)  .
\label{12}
\end{equation}
We claim that there exists a real number $t$ on the interval 
$[t_1,\rho]$ that satisfies the inequality $u(t,\alpha,K)  <\alpha_0$. 
The proof of this claim will be carried out arguing
by contradiction. Assume that $u(t,\alpha,K)  \geq \alpha_0$ for all 
$t\in$ $[t_1,\rho]$. Integrating by parts \eqref{10} and \eqref{11}, 
subtracting the resulting equations and taking into account \eqref{12}, 
we see that
\begin{equation} \label{13}
\begin{aligned}
-\rho^{N-1}\varphi_p(v'u)  (\rho)
&=(p-1)  \int_{t_1}^{\rho}r^{N-1}(|
u'v|^{p-2}-|v'u|^{p-2})u'v'dr\\
&\quad  +\int_{t_1}^{\rho}r^{N-1}\Big(\mu-K(r)
\frac{f^+(u)  }{\varphi_p(u)  }\Big)  \varphi_p(uv)  dr.
\end{aligned}
\end{equation}
Since $|u'v|(t)  \leq|uv'|(t)  $ for all $t\in(t_1,\rho) $ and $p-2\geq0$, then
\[
|u'v|^{p-2}(t)  -|uv'|^{p-2}(t)  \leq0.
\]
In consequence, the first term of \eqref{13} is nonpositive. On the other
hand, from \eqref{14} we see that
\[
\mu-\bar{\lambda}\frac{f^+(u)  }{\varphi_p(u)  }\leq0.
\]
Hence, the second term of \eqref{13} is also nonpositive. This is impossible
since $v'u(\rho)  <0$. This proves the claim. Therefore,
$u(t_2,\alpha,K)  =\alpha_0$ for some $t_2\in(0,\rho)$.
Since $\alpha>u_{0\text{ }}$ and $u$ is decreasing, the estimate
\begin{equation}
u(t,\alpha,K)  \leq\alpha_0,\text{ for all }t\in[t_2,1]\label{15}
\end{equation}
holds. Because $F^+$ is increasing on $[u_0,\infty)  $ and
$u(t,\alpha,K)  \geq\alpha k>u_1$ for all $t\in[0,t_0]$
then
\[
E(t,\alpha,K)  \geq F^+(u(t,\alpha,K))\geq F^+(k\alpha)  .
\]
On the other hand, for $t\in(t_0,1]$, since $u(t)  u'(t)  \leq0$, then
\begin{align*}
t^NK(t)  E(t,\alpha,K)
& \geq t^{N-1}H(t,\alpha,K) \\
&  \geq\frac{\eta C_1^N}{N\bar{\lambda}^{N/p}}
\Big(\delta_0F^+(k\alpha)  -\frac{N-p}{p}\alpha f^+(\alpha)\Big)
\Big(\frac{\varphi_p(\alpha)  }{f^+(\alpha)  }\Big)^{\frac{N}{p}}\\
&\quad  -2\bar{\lambda}M_{N}C_{N}\Big(1+\frac{N-p}{Np}\Big)  .
\end{align*}
Thus, $K(t)  E(t,\alpha,K)  \to+\infty$ as
$\alpha\to+\infty$, uniformly in $t\in[0,1]$. Set
$\alpha\geq\alpha_0$ such that
\[
K(t)  E(t,\alpha,K)  \geq\bar{\lambda}F^+(
\alpha_0)  +(p')  ^{p-1}\alpha_0^{p}\quad \text{for all }t\in[0,1].
\]
Relation \eqref{15} implies $F^+(\alpha_0)  -F^+(u(t,\alpha,K))  \geq0$
for all $t$ in $[t_2,1]$, thus
\[
|u'(t,\alpha,K)  |\geq p'\alpha_0,\quad \text{for all }t\in[t_2,1].
\]
Define $\tau=t_2+\frac{1}{p'} $ and  observe that
$(t_2,\tau)  \subseteq(0,1)  $. The mean value theorem applied to $u$
on the interval $[t_2,\tau]$, leads  to the equation
\[
u(\tau,\alpha,K)  -u(t_2,\alpha,K)  =u'(\xi,\alpha,K)  \frac{1}{p'},
\]
for some $\xi\in(t_2,\tau)  $. On the other hand,
$u(t_2,\alpha,K)  =\alpha_0$ and $u'(\xi,\alpha,K)\frac{1}{p'}\leq-\alpha_0$,
thus $u(\tau,\alpha,K) \leq0$, which is  absurd.
\end{proof}


\begin{proposition}\label{the_proposition}
Under hypotheses {\rm (F1)--(F4)} and {\rm (K1)--(K2)}, there exists a
positive number $\lambda_0$ such that if $\|K\| _{\infty}<\lambda_0$ then  
problem \eqref{1} has at least one positive decreasing radial solution, 
with radial negative derivative  in $\| x\| =1$.
\end{proposition}

\begin{proof}
Set $\lambda_0=\lambda_1$ from Lemma \ref{lemma2} and 
\underline{$\alpha$} given by Lemma \ref{lemma1}. 
According to Lemma \ref{lemma3}, there exists 
 $\alpha_1>\underline{\alpha}$ with its corresponding solution being 
negative in some point on $[0,1]$. Let $\tilde{\alpha}$ be the supremum 
of the set $\mathcal{A}$ defined as
\[
\{  \alpha\in[\underline{\alpha},\alpha_1]:u(t,\alpha,K)  \geq0
\text{ for all }t\in[0,1]\}.
\]
This supremum makes sense because of $\underline{\alpha}$ belongs to
$\mathcal{A}$ (which is implied by Lemma \ref{lemma2}). Due to the
continuous dependence of $u$ on $\alpha$, $\mathcal{A}$ is closed.
Thus, $\tilde{\alpha}$ belongs to $\mathcal{A}$ and therefore
$u(t,\tilde{\alpha},K)  \geq0$, for all $t\in[0,1]$. Moreover, 
$\tilde{\alpha}<\alpha_1$. Now, we will see that $u(\cdot,\tilde{\alpha},K)  $ 
is a solution needed.


(i) $u(t,\tilde{\alpha},K)  >0$ for all $t\in[0,1)  $. 
Arguing by contradiction, if $u(\tau,\tilde{\alpha},K)  =0$ for 
some $\tau\in(0,1)  $, then by Lemma \ref{lemma1},
$u'(\tau,\tilde{\alpha},K)  \neq0$. Hence, there exists a
$\tau_1\in(0,1)  $, such that $u(\tau_1,\tilde{\alpha},K)  <0$. 
This is not possible.

(ii) $u(1,\tilde{\alpha},K)  =0$. If $u(1,\tilde{\alpha},K)  >0$, then 
$u(\cdot,\tilde{\alpha},K)>0$ on the compact set $[0,1]$. 
For the continuous dependence in
the initial data, there is some $\alpha$, $\tilde{\alpha}<\alpha<\alpha_1$,
such that $u(\cdot,\alpha,K)  >0$ on $[0,1]$. This
contradicts the definition of $\tilde{\alpha}$.

(iii) $u'(1,\tilde{\alpha},K)  <0$. Due to the
previous steps, $u'(1,\tilde{\alpha},K)  \leq0$. Now, the
fact that $u'(1,\tilde{\alpha},K)  $ is nonzero  follows from Lemma \ref{lemma1}.

(iv) The proof that $u(\cdot,\alpha,K)  $ is decreasing is
contained in the proof of Lemma \ref{lemma3}.

Therefore the proposition is proved.
\end{proof}

\begin{proof}[Proof of theorem \ref{maintheorem}]
The existence of the positive solution is a consequence of previous proposition.
Let
\[
f^{-}(t) :=\begin{cases}
-f(-t)  &\text{if } t>0\\
-f(0^{-}) & \text{if } t=0\\
0 &\text{if } t<0.
\end{cases}
\]
A straightforward application of the Proposition \ref{the_proposition}
 with $f^{-}$ gives us a negative solution with the  desired properties.
\end{proof}

Now, we exhibit some examples of functions $f$ and $K$ satisfying conditions 
(F1)--(F4) and (K1)--(K2). Let $f(t)=t^q-t^{q-1}-1$, for $t>0$ and 
$f(t)=1+(-t)^{q-1}-(-t)^q$ for $t<0$, where $1<q<p<q+1<p^\ast:=\frac{Np}{N-p}$ 
and $K$ any positive constant weight. Choosing $\frac{(N-p)(q+1)}{p}<\theta <N$ 
we can find $k\in (0, 1)$ such that
\begin{equation}\label{example-condition-F4}
\theta\frac{k^{q+1}}{q+1}>\frac{N-p}{p},
\end{equation}
from which follows (F4). If we consider the same nonlinearity $f$ with
$1<q<p<q+1<\frac{(N-1)p}{N-p}$ and $K(r)= \varepsilon e^{-r}$, where 
$\varepsilon>0$ is fixed and we choose $\frac{(N-p)(q+1)}{p}<\theta <N-1$,
we can find $k\in (0,1)$ such that \eqref{example-condition-F4} holds, 
from which follows (F4). 
Since  $N+r\frac{K'(r)}{K(r)}=N-r\geq N-1>\theta$, we obtain (K2). 
Other example is given by $f(t)=t^q\ln t-1$ for for $t>0$ and 
$f(t)=1-(-t)^q\ln (-t)$ for $t<0$, where $2<p<q+1<p^\ast$ and $K$ 
any positive constant weight. Reasoning as before we can show conditions 
{\rm (F1)--(F4)} and (K1)--(K2). Assuming $p<q+1<\frac{(N-1)p}{N-p}$, 
as before $K(r)= \varepsilon e^{-r}$, where $\varepsilon>0$ is fixed, 
is an admissible weight.



\section{Appendix}

This section we establish the Pohozaev identity as well
as the existence of local solution to problem \eqref{3}.

\begin{proposition}[Pohozaev Identity] 
Assume that $u(t,\alpha,K)  $  is a
solution of the initial value problem \eqref{3}, then for all 
$0\leq s\leq t\leq1$, 
\begin{equation} \label{4.3}
\begin{aligned}
&  t^{N-1}H(t,\alpha,K)  -s^{N-1}H(s,\alpha,K)\\
&  =\int_{s}^{t}r^{N-1}K(r)  \Big[\Big(
N+r\frac{K'(r)  }{K(r)  }\Big)  F(
u)  -\frac{N-p}{p}f(u)  u\Big]dr\,.
\end{aligned}
\end{equation}
\end{proposition}

\begin{proof}
It is easy to see that our ordinary differential equation can be written as
\begin{equation}\label{appendix18}
[r^{N-1}\varphi_p(u')  ]'=-r^{N-1}K(r)  f(u).
\end{equation}
Multiplying \eqref{appendix18} by $u$ and integrating on $[s, t]$,
by parts,  we obtain
\[
t^{N-1}\varphi_p(u')  u-s^{N-1}\varphi_p(
u')  u=\int_{s}^{t}r^{N-1}\varphi_p(u')  u'dr
-\int_{s}^{t}r^{N-1}K(r)  f(u)  u\,dr.
\]
Then
\begin{equation}
\int_{s}^{t}r^{N-1}|u'|^{p}dr
=b(s,t)  +\int_{s}^{t}r^{N-1}K(r)  f(u)u\,dr,\label{4.2}
\end{equation}
where $b(s,t)  =t^{N-1}\varphi_p(u'(t))  u(t)-s^{N-1}\varphi_p(u'(s))u(s)  $. 
Now, multiplying \eqref{appendix18} by $ru'$ and integrating by parts, we
have
\begin{equation}
t^N\varphi_p(u')  u'-s^N\varphi
_p(u')  u'=\int_{s}^{t}r^{N-1}
\varphi_p(u')  (ru^{\prime\prime}+u')  dr-\int_{s}^{t}r^NK(r)  f(u)
u'dr.\label{4.1}
\end{equation}
From \eqref{appendix18}, we realize that
\[
 \varphi_p(u')  (ru^{\prime\prime}+u')  
=\frac{p-N}{p-1}|u'|^{p}-\frac{K(r)}{p-1}rf(u)  u'.
\]
Then, from \eqref{4.1}, it follows that
\[
a(s,t)  -\int_{s}^{t}r^{N-1}\frac{p-N}{p-1}|
u'|^{p}dr-\int_{s}^{t}r^N\frac{K(r)}{p-1}f(u)  u'dr
=-\int_{s}^{t}r^NK(r)  f(u)  u'dr,
\]
where $a(s,t)  :=t^N\varphi_p(u'(
t))  u'(t)  -s^N\varphi_p(
u'(s))  u'(s)  $.
Therefore,
\[
a(s,t)  +\frac{N-p}{p-1}\int_{s}^{t}r^{N-1}|
u'|^{p}dr=-p'\int_{s}^{t}r^NK(r)  f(u)  u'dr.
\]
This equation and \eqref{4.2} imply that
\[
a(s,t)  +\frac{N-p}{p-1}\Big(b(s,t)
+\int_{s}^{t}r^{N-1}K(r)  f(u)  udr\Big)
=-p'\int_{s}^{t}r^NK(r)  f(u)u'dr.
\]
Now, integrating by parts the right hand side of the last equation we obtain
\begin{align*}
&  a(s,t)  +\frac{N-p}{p-1}b(s,t) \\
& =\frac{p-N}
{p-1}\int_{s}^{t}r^{N-1}K(r)  f(u) udr-p'(t^NK(t)  F(u)-s^NK(s)  F(u))  \\
&\quad  +p'\int_{s}^{t}r^{N-1}(NK(r)+rK'(r))  F(u)  dr.
\end{align*}
Taking into account the definitions of $a(s,t)  $ and $b(s,t)  $, we can see 
that
\begin{align*}
&  t^N|u'|^{p}-s^N|u'|^{p}+\frac{N-p}{p-1}(t^{N-1}\varphi_p(u^{\prime
})  u-s^{N-1}\varphi_p(u')  u)  \\
&  +p'(t^NK(t)  F(u) -s^NK(s)  F(u))  \\
&  =\int_{s}^{t}r^{N-1}\Big[\frac{p-N}{p-1}K(r)
f(u)  u+\frac{p}{p-1}(NK(r)  +rK^{\prime
}(r))  F(u)  \Big]dr.
\end{align*}
Consequently,
\begin{align*}
&  \frac{p-1}{p}t^{N-1}\Big[\Big(t|u'|
^{p}+\frac{N-p}{p-1}\varphi_p(u')  u+p^{\prime
}tK(t)  F(u)\Big)    \\
&    -s^{N-1}\Big(s|u'|^{p}+\frac
{N-p}{p-1}\varphi_p(u')  u+p'sK(
s)  F(u)\Big)  \Big]\\
&  =\int_{s}^{t}r^{N-1}K(r)  \Big[\Big(
N+r\frac{K'(r)  }{K(r)  }\Big)  F(u)  -\frac{N-p}{p}f(u)  u\Big]dr.
\end{align*}
However,
\begin{align*}
&  t^{N-1}\Big(\frac{t}{p'}\{  |u'|^{p}+p'K(t)  F(u)  \}
+\frac{N-p}{p}  \varphi_p(u')  u\Big)  \\
&  -s^{N-1}(\frac{s}{p'}\{  |u' |^{p}+p'K(s)  F(u)  \}
+\frac{N-p}{p}  \varphi_p(u')  u)  \\
&  =t^{N-1}\Big[tK(t)  E(t)  +\frac{N-p}
{p}\varphi_p(u')  u\Big]-s^{N-1}\Big[sK(
s)  E(s)  +\frac{N-p}{p}\varphi_p(u')  u\Big].
\end{align*}
Hence \eqref{4.3} holds.
\end{proof}


\begin{proposition}
Let $p\geq2$, $\alpha>4u_1/3$ and $K$ be a suitable
weight. There exists a positive real number $\varepsilon$ such
that  \eqref{3} has a unique solution $u(\cdot,\alpha,K)  $ on the interval 
$[0,\varepsilon]$.
\end{proposition}

\begin{proof}
For $\varepsilon>0$ and $R=\alpha/4$, set 
$Y:=\{  u\in C([0,\varepsilon]:\mathbb{R}):\| u-\alpha\| _{\infty}\leq R\}  $.
 Define $T:Y\to Y$ by
\[
(Tu)  (s)  :=\alpha-\int_0^{s}
\varphi_{p'}\Big(r^{1-N}\int_0^{r}t^{N-1}K(
t)  f(u(t))  dt\Big)  dr,\quad 0\leq s\leq\varepsilon.
\]
This was suggested by \eqref{3.a}. $T$ is well defined if $\varepsilon$ i
s small enough, since $f$ is Lipschitz continuous on $[3\alpha/4,5\alpha/4]$. 
Now, we will see that $T$ is a contraction. Let $u,v$ be elements
of $Y$ and $s\in\lbrack0,\varepsilon]$. Applying the mean value theorem we
obtain the estimate
\begin{align*}
&  |(Tu)  (s)  -(Tv)  (s)  |\\
&  \leq\int_0^{s}\varphi_{p'}(r^{1-N})
\Big|\varphi_{p'}\Big(\int_0^{r}t^{N-1}K(
t)  f(u)  dt\Big)  -\varphi_{p'}\Big(
\int_0^{r}t^{N-1}K(t)  f(v)  dt\Big) \Big|dr\\
&  \leq(p'-1)  \int_0^{s}\varphi_{p'
}(r^{1-N})  |\xi_{r}|^{p'-2}\Big(
\int_0^{r}t^{N-1}K(t)  |f(u)-f(v)  |dt\Big)  dr,
\end{align*}
where $\xi_{r}$ is a value between the two positive numbers 
$\int _0^{r}t^{N-1}K(t)  f(u)  dt$ and 
$\int_0^{r}t^{N-1}K(t)  f(v) dt$. 
Assume, without loss of generality that
$\int_0^{r}t^{N-1}K(t) f(u)dt\leq\xi_{r}$. Then, we have 
$\frac{\eta}{N}f(\frac{3}{4}\alpha)  r^N\leq\xi_{r}$. Because of
$p'-2\leq0$,
\[
|\xi_{r}|^{p'-2}\leq[\frac{\eta}{N}
r^Nf\big(\frac{3}{4}\alpha\big)  ]^{p'-2}.
\]
Thus,
\[
|(Tu)  (s)  -(Tv)  (
s)  |\leq C_{\alpha}\| u-v\| _{\infty}
\int_0^{s}r^{p'-1}dr<L_{\alpha}\| u-v\|_{\infty}
\]
where $L_{\alpha}<1$ and $C_{\alpha}$ is a suitable constant.
Hence, $T$ is a contraction. Thereupon $T$ has a unique fixed
point, which is the unique local solution of the IVP \eqref{3}.
\end{proof}

Now, our goal is to extend this solution to the interval $[0,1]$.

\begin{lemma}
Suppose that $u(\cdot,\alpha,K)  $ is a solution of
 \eqref{3}, restricted to an interval $[0,\varepsilon)$. Assume that
$\lim\sup_{t\to\varepsilon^{-}}|u(t,\alpha ,K)  |<\infty$. Then, there exist a
positive number $\bar{\varepsilon}$ and a function 
$\bar{u}(\cdot,\alpha,K) :[0,\varepsilon+\bar{\varepsilon})  \to\mathbb{R}$,
that solves the IVP and such that $\bar{u}|_{[0,\varepsilon)  }\equiv u$.
\end{lemma}

\begin{proof}
Let $\{ t_n\}  _n$ $\subseteq[0,\varepsilon)  $
be an increasing sequence such that $t_n\to\varepsilon$. Then, for
$n<m$ we have
\begin{align*}
|u(t_n,\alpha,K)  -u(t_{m},\alpha,K)|
&\leq\int_{t_n}^{t_{m}}\Big|\varphi_{p'
}\Big(r^{1-N}\int_0^{r}t^{N-1}K(t)  f(u)  dt\Big)  \Big|dr\\
&  \leq\bar{\lambda}C\int_{t_n}^{t_{m}}\varphi_{p'}(r)  dr\leq C|t_{m}-t_n|,
\end{align*}
for some constant $C$. Here, we have used that $|u(\cdot,\alpha,K)  |$ 
is bounded on $[0,\varepsilon) $, according to the hypothesis,
 $\lim\sup_{t\to \varepsilon^{-}}|u(t,\alpha,K)  |<\infty$.
This chain of inequalities shows that 
$\{  u(t_n,\alpha,K)  \}  _n$ is a Cauchy sequence. Thus, we can define
$u(\varepsilon,\alpha,K)  :=\lim_{n\to\infty}u(t_n,\alpha,K)  $. 
On the other hand
\begin{align*}
|t_n^{N-1}\varphi_p(u'(t_n,\alpha,K))  -t_{m}^{N-1}\varphi_p(u'(
t_{m},\alpha,K))  |
&\leq\int_{t_n}^{t_{m}}|t^{N-1}K(t)  f(u)  dt|\\
&  \leq M|t_{m}-t_n|,
\end{align*}
where $M$ is a suitable constant. Hence, 
$\{t_n^{N-1}\varphi _p(u'(t_n,\alpha,K))  \}  _n$ is convergent and so is
$\{  u'(t_n,\alpha,K) \}  _n$. Then, we can write 
$u'(\varepsilon,\alpha,K)  :=\lim_{n\to\infty}u'(t_n,\alpha,K)  $. 
Now, we consider the problem
\begin{gather*}
{}[\varphi_p(\bar{u}'(r))  ]'+\frac{N-1}{r}\varphi_p(\bar{u}'(r)
)+K(r)  f(\bar{u})  =0,\quad \varepsilon<r<\varepsilon+\tilde{\varepsilon}\\
\bar{u}'(\varepsilon)  =u'(\varepsilon)  ,\quad
\bar{u}(\varepsilon)  =u(\varepsilon),
\end{gather*}
for small $\tilde{\varepsilon}$. Proceeding in the same manner as we did with
the former IVP, we can establish the existence and uniqueness of solution of
this problem in the interval $[0,\varepsilon+\bar{\varepsilon})$. 
Due to the uniqueness property, $\bar{u}|_{[0,\varepsilon)  }\equiv u$.
\end{proof}

\begin{lemma}
The local solution to the IVP can be extended to $[0,1]$.
\end{lemma}

\begin{proof}
We argue by contradiction. As a consequence of the previous Lemma, if
$u(\cdot,\alpha,K)  $ cannot be extended, then it blows up.
Therefore, there exists a sequence $\{t_n\}  _n$ such that
$\lim_{n\to\infty}u(t_n,\alpha,K)  =+\infty$. By
the mean value theorem, without loss of generality, we can assume that
$\lim_{n\to\infty}u'(t_n,\alpha,K)
=+\infty$. From the energy we obtain that
\[
\frac{\partial}{\partial t}E(t,\alpha,K)  
=-\frac{| u'(t,\alpha,K)  |}{p'tK(t)  }\Big(\frac{N-1}{p-1}p+t\frac{K'(t)
}{K(t)  }\Big)  \leq0.
\]
Then 
\[
E(t,\alpha,K)  \leq E(0,\alpha,K)  =F(\alpha) . 
\]
This is a contradiction since $\lim_{n\to \infty}u(t_n,\alpha,K)  =+\infty$.
\end{proof}

\begin{remark} \rm
The continuous dependence of  $u(\cdot,\alpha,K)  $ whit respect to
$\alpha$ is a consequence of the estimate
\[
|u(t,\alpha_1,K)  -u(t,\alpha_2,K)
|\leq|\alpha_1-\alpha_2|+L_{\alpha}\| u(\cdot,\alpha_1,K)  
-u(\cdot,\alpha_2,K)  \| _{\infty},
\]
for all $t\in[0,1]$. Thus
\[
(1-L_{\alpha})  \| u(\cdot,\alpha_1,K)-u(\cdot,\alpha_2,K)  \| _{\infty}
\leq| \alpha_1-\alpha_2|.
\]
\end{remark}


\subsection*{Acknowledgments}
This research was partially supported by DIME, project code 15401.
The authors wish to thank the anonymous referees for their helpful comments.

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