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As the analysis of the case ξ > 0 in Eq. (33) was incomplete, the following corrections of Lemma 3, Lemma 4, and Theorem 1 are needed.

Lemma 3: …(2) if ξ > 0,

(2a) with

$a_{1}>-\sqrt{3\xi }a_{0}^{1/3}$
a1>3ξa01/3 the solution a(t) globally exists, such that

\begin{equation}{\mathop{\lim }_{s\rightarrow +\infty }}a(s)=+\infty,\end{equation}
lims+a(s)=+,
(34)

(2b) with

$a_{1}\le -\sqrt{3\xi }a_{0}^{1/3}$
a13ξa01/3⁠, there exists a finite time S, such that
${\mathop{\lim }\limits_{s->S^{-}}}a(s)=0.$
lims>Sa(s)=0.

Lemma 4: …(2) if ξ > 0,

(2a) with

$a_{1}<\sqrt{3\xi }a_{0}^{1/3}$
a1<3ξa01/3 the solution a(t) globally exists, such that

\begin{equation}{\mathop{\lim}_{a(s)-{s\rightarrow +\infty}}}=-\infty,\end{equation}
lima(s)s+=,
(58)

(2b) with

$a_{1}\ge \sqrt{3\xi }a_{0}^{1/3}$
a13ξa01/3⁠, there exists a finite time S, such that
$\mathop{\lim }\limits_{{s->S^{-}}a(s)}=0.$
lims>Sa(s)=0.

Then, we need the corresponding correction for Theorem 1.

Theorem 1:…(1) for σ = −1,….

(b) with ξ > 0 and a0 < 0,

\begin{equation}\rho (t,x)=\frac{f\left( \eta \right) }{a(3t)^{1/3}},\text{ }u(t,x)=\frac{\dot{a}(3t)}{a(3t)}x.\end{equation}
ρ(t,x)=fηa(3t)1/3,u(t,x)=ȧ(3t)a(3t)x.
(19)

If

$a_{1}<\sqrt{3\xi }a_{0}^{1/3}$
a1<3ξa01/3⁠, solution (19) exists globally; If
$a_{1}\ge \sqrt{3\xi }a_{0}^{1/3}$
a13ξa01/3
, solution (19) blows up in a finite time T.

(2) For σ = 1,

(a) with ξ > 0 and a0 > 0,

\begin{equation}\rho (t,x)=\left\lbrace \begin{array}{cc} \frac{f(\eta )}{a(3t)^{1/3}},\text{ for }\eta ^{2}<\frac{\alpha ^{2}}{\xi }\\[4pt]0\text{, for }\eta ^{2}\ge \frac{\alpha ^{2}}{\xi } \end{array} \right. ,u(t,x)=\frac{\dot{a}(3t)}{a(3t)}x.\end{equation}
ρ(t,x)=f(η)a(3t)1/3,forη2<α2ξ0,forη2α2ξ,u(t,x)=ȧ(3t)a(3t)x.
(20)

If

$a_{1}>-\sqrt{3\xi }a_{0}^{1/3}$
a1>3ξa01/3⁠, solution (20) globally exists; If
$a_{1}\le -\sqrt{3\xi }a_{0}^{1/3}$
a13ξa01/3
, solution (20) blows up in a finite time T….