First, we will give a self-similar solution to the continuity equation (2.4).
Lemma 3.1
For any two
\(C^{1}\)
functions
\(f(z)\geq0\)
and
\(a(t)>0\), define
$$ \rho(r,t)=\frac{f(\frac{r}{a(t)})}{a^{3}(t)},\qquad u(r,t)=\frac{a'(t)}{a(t)}r. $$
(3.1)
Then
\((\rho,u)(r,t)\)
solves the continuity equation (2.4), i.e.,
$$ \rho_{t}+(\rho)_{r}u+\rho u_{r}+ \frac{2\rho u}{r}=0. $$
(3.2)
Here, we can choose \(a(t)\) as a free boundary which satisfies the condition (2.7) or (2.8). So in the following we will determine the form of the function \(f(x)\) and then prove the global existence of the free boundary \(a(t)\). To this end, denoting \(z=\frac{r}{a(t)}\), one can obtain from (2.5) that
$$ \frac{a''(t)}{a^{3}(t)}f(z)z+\gamma f^{\gamma-1}(z)f'(z) \frac {1}{a^{3\gamma+1}(t)}-\frac{3a'(t)}{a^{3\beta+2}(t)}\beta f^{\beta-1}(z)f'(z)=0. $$
(3.3)
Next, we will solve (3.3) according to the free boundary conditions (2.7) or (2.8), respectively.
The continuous boundary condition
Assume that \(\gamma=\beta>\frac{4}{3}\). We require
$$ z=-\gamma f^{\gamma-2}(z)f'(z). $$
(3.4)
Then it follows from (3.3)-(3.4) that
$$\begin{aligned}& f(z)=\biggl[f^{\gamma-1}(1)+\frac{\gamma-1}{2\gamma} \bigl(1-z^{2}\bigr)\biggr]^{\frac{1}{\gamma-1}}, \end{aligned}$$
(3.5)
$$\begin{aligned}& a''(t)-a^{2-3\beta}(t)+3a'(t)a^{1-3\beta}(t)=0. \end{aligned}$$
(3.6)
Using the boundary condition (2.7) yields \(f(z)=[\frac {\gamma-1}{2\gamma}(1-z^{2})]^{\frac{1}{\gamma-1}}\), and then
$$ \rho(r,t)=\frac{[\frac{\gamma-1}{2\gamma}(1-z^{2})]^{\frac{1}{\gamma -1}}}{a^{3}(t)},\qquad u(r,t)=\frac{a'(t)}{a(t)}r. $$
(3.7)
Clearly, if \(a(t)>0\) is a free boundary satisfying the condition (2.7), then it is straightforward to check that \((\rho,u)\) defined by (3.7) is a solution of (2.4)-(2.5), where \(a(t)\) can be determined by
$$ \textstyle\begin{cases} a'(t)=a_{1}+\int_{0}^{t}a^{2-3\beta}(s)\,ds-\frac{3}{2-3\beta}a^{2-3\beta }(t)+\frac{3}{2-3\beta}a_{0}^{2-3\beta},\\ a(0)=a_{0}, \qquad a'(0)=a_{1}, \end{cases} $$
(3.8)
with \(a_{0}>0\) and \(a_{1}\) being the initial location and slope of the free boundary.
Thus, it remains to solve the boundary value problem (3.8). We start with estimates on solutions of (3.8).
Lemma 3.2
Let
\(a(t)\in C^{1}[0,T]\)
be a solution to (3.8) for
\(T>0\). Then there exist two uniform positive constants
\(C_{1}\)
and
\(C_{2}>0\), independent of
T, such that
$$ C_{1}(1+t)^{\frac{1}{3(\gamma-1)}}\leq a(t)\leq C_{2}(1+t)^{\frac{6\beta -4}{3(\beta-1)}} . $$
(3.9)
Proof
We first verify the fact that \(a(t)\geq1\) for all \(t\in[0,T]\). Note that (3.7) implies
$$ \rho\bigl(a(t),t\bigr)=0, \qquad u(0,t)=0. $$
(3.10)
We introduce
$$\begin{aligned} H(t) =&\int_{0}^{a(t)}\bigl(r-(1+t)u \bigr)^{2}\rho r^{2}\,dr+\frac{2}{\gamma-1}(1+t)^{2} \int_{0}^{a(t)}\rho^{\gamma}r^{2} \,dr \\ =&\int_{0}^{a(t)}\rho r^{4}dr-2(1+t)\int _{0}^{a(t)}\rho ur^{3}dr+(1+t)^{2} \int_{0}^{a(t)}\rho u^{2}r^{2} \,dr \\ &{}+\frac{2}{\gamma-1}(1+t)^{2}\int_{0}^{a(t)} \rho^{\gamma}r^{2}\,dr. \end{aligned}$$
(3.11)
Due to \(a'(t)=u(a(t),t)\) and (3.10), a direct computation gives
$$\begin{aligned} H'(t) =&\int_{0}^{a(t)} \bigl(\rho_{t}r^{4}-2\rho ur^{3}\bigr) \,dr+(1+t)^{2}\int_{0}^{a(t)}\biggl(\bigl( \rho u^{2}\bigr)_{t}+\frac{2}{\gamma-1}\bigl( \rho^{\gamma}\bigr)_{t}\biggr)r^{2}\,dr \\ &{}+2(1+t)\int_{0}^{a(t)}\biggl(\rho u^{2}r^{2}-(\rho u)_{t}r^{3}+ \frac{2}{\gamma-1}\rho ^{\gamma}r^{2}\biggr)\,dr \\ =&I_{1}+I_{2}+I_{3}. \end{aligned}$$
(3.12)
Equations (2.4) and (3.10) yield
$$I_{1}=-\int_{0}^{a(t)}\bigl(\rho ur^{4}\bigr)_{r}\,dr=0. $$
One has
$$\begin{aligned}& \begin{aligned} I_{2}={}&(1+t)^{2}\int_{0}^{a(t)} \biggl(\bigl(\rho u^{2}\bigr)_{t}+\frac{2}{\gamma-1}\bigl( \rho^{\gamma}\bigr)_{t}\biggr)r^{2}\,dr \\ ={}&(1+t)^{2}\int_{0}^{a(t)}2(\rho u)_{t}ur^{2}-\rho_{t}u^{2}r^{2}+ \frac{2\gamma }{\gamma-1}\rho^{\gamma-1}\rho_{t}r^{2}\,dr \\ ={}&(1+t)^{2}\int_{0}^{a(t)}2u\bigl[- \bigl(\rho u^{2}\bigr)_{r}r^{2}-\bigl( \rho^{\gamma}\bigr)_{r}r^{2}\bigr]-\frac {2\gamma}{\gamma-1} \rho^{\gamma-1}(\rho u)_{r}r^{2} \\ &{}-\frac{2\gamma}{\gamma-1}\rho^{\gamma}u2\gamma+(\rho u)_{r}u^{2}r^{2}-2 \rho u^{3}r+2ur^{2}\biggl(\bigl(2\mu+\rho^{\gamma}\bigr) \biggl(u_{r}+\frac{2u}{r}\biggr)\biggr)_{r}\,dr \\ ={}&(1+t)^{2}\int_{0}^{a(t)}- \frac{2\gamma}{\gamma-1}\bigl(\rho^{\gamma}ur^{2}\bigr)_{r}- \bigl(\rho u^{3}r^{2}\bigr)_{r}+2ur^{2} \biggl(\bigl(2\mu+\rho^{\gamma}\bigr) \biggl(u_{r}+ \frac {2u}{r}\biggr)\biggr)_{r}\,dr \\ ={}&(1+t)^{2}\int_{0}^{a(t)}2ur^{2} \biggl(\bigl(2\mu+\lambda(\rho)\bigr) \biggl(u_{r}+\frac {2u}{r} \biggr)\biggr)_{r}\,dr \\ ={}&(1+t)^{2}\int_{0}^{a(t)}\bigl[\bigl( \rho^{\gamma}\bigr)_{r}\bigl(2uu_{r}r^{2}+4u^{2}r \bigr)+\bigl(2\mu+\rho ^{\gamma}\bigr) \bigl(4uu_{r}r-4u^{2} \bigr)\bigr] \\ ={}&-(1+t)^{2}\int_{0}^{a(t)} \rho^{\gamma}\bigl(2uu_{r}r^{2}+4u^{2}r \bigr)_{r}+(1+t)^{2}\int_{0}^{a(t)} \bigl(2\mu+\rho^{\gamma}\bigr) \bigl(4uu_{r}r-4u^{2} \bigr) \\ ={}&-(1+t)^{2}\int_{0}^{a(t)}\bigl(2 \rho^{\gamma}u_{r}^{2}r^{2}+16 \rho^{\gamma}u^{2}\bigr)\,dr, \end{aligned} \\& \begin{aligned} I_{3}={}&2(1+t)\int_{0}^{a(t)}\rho u^{2}r^{2}-(\rho u)_{t}r^{3}+ \frac{2}{\gamma -1}\rho^{\gamma}r^{2}\,dr \\ ={}&2(1+t)\int_{0}^{a(t)}\frac{5-3\gamma}{\gamma-1} \rho^{\gamma}r^{2}-\biggl(\bigl(2\mu +\rho^{\gamma}\bigr) \biggl(u_{r}+\frac{2u}{r}\biggr)\biggr)_{r}r^{3}dr \\ ={}&2(1+t)\int_{0}^{a(t)}\rho^{\gamma}\bigl(u_{r}r^{3}+2ur^{2}\bigr)_{r}\\ &{}-2(1+t) \int_{0}^{a(t)}\bigl(2\mu+\rho^{\gamma}\bigr) \bigl(2u_{r}r^{2}-2ur\bigr)+\frac{5-3\gamma}{\gamma-1}\rho ^{\gamma}r^{2}\,dr \\ ={}&2(1+t)\int_{0}^{a(t)}3\rho^{\gamma}u_{r}r^{2}+8\rho^{\gamma}ur+\frac {5-3\gamma}{\gamma-1} \rho^{\gamma}r^{2}\,dr. \end{aligned} \end{aligned}$$
Thus, substituting the above estimates into (3.12) and using the Cauchy-Schwarz inequality, one deduces
$$\begin{aligned} H'(t) =&-(1+t)^{2}\int_{0}^{a(t)} \bigl(2\rho^{\gamma}u_{r}^{2}r^{2}+16 \rho^{\gamma}u^{2}\bigr)\,dr+(1+t)\int_{0}^{a(t)}6 \rho^{\gamma}u_{r}r^{2} \\ &{}+16\rho^{\gamma}ur+\frac{5-3\gamma}{\gamma-1}\rho^{\gamma}r^{2}\,dr \\ \leq&\frac{17}{2}\int_{0}^{a(t)} \rho^{\gamma}r^{2}\,dr+\frac{2(5-3\gamma )}{\gamma-1}(1+t)\int _{0}^{a(t)}\rho^{\gamma}r^{2}\,dr, \end{aligned}$$
(3.13)
where one has used
$$\begin{aligned}& 2(1+t)\int_{0}^{a(t)}3\rho^{\gamma}u_{r}r^{2}\,dr\leq2(1+t)^{2}\int _{0}^{a(t)}\rho ^{\gamma}u_{r}^{2}r^{2} \,dr+\frac{9}{2}\int_{0}^{a(t)} \rho^{\gamma}r^{2}\,dr, \\& 16(1+t)\int_{0}^{a(t)}\rho^{\gamma}ur\,dr \leq16(1+t)^{2}\int_{0}^{a(t)}\rho ^{\gamma}u^{2}\,dr+4\int_{0}^{a(t)} \rho^{\gamma}r^{2}\,dr. \end{aligned}$$
Note also that the conservation of total mass implies that
$$ \int_{0}^{a(t)}\rho r^{2}\,dr=\int _{0}^{a_{0}}\rho_{0}r^{2}\,dr=1. $$
In the case \(\gamma\geq\frac{5}{3}\), (3.13) yields
$$ H'(t)\leq\frac{17(\gamma-1)}{2}E_{0},\qquad E_{0}=\int_{0}^{a_{0}}\biggl( \frac{1}{2}\rho _{0}u_{0}^{2}+ \frac{\rho_{0}^{\gamma}}{\gamma-1}\biggr)r^{2}\,dr, $$
(3.14)
and so
$$ H(t)\leq H(0)+\frac{17(\gamma-1)}{2}E_{0}t, $$
(3.15)
and consequently
$$ \int_{0}^{a(t)}\rho^{\gamma}r^{2}\,dr\leq C(1+t)^{-1}. $$
(3.16)
Thus, as a consequence of (3.16) and the conservation of mass, we have, for any \(t>0\),
$$1=\int_{0}^{a(t)}\rho r^{2}\,dr\leq \biggl(\int_{0}^{a(t)}\rho^{\gamma}r^{2}\,dr\biggr)^{\frac {1}{\gamma}}\biggl(\int_{0}^{a(t)}r^{2} \,dr\biggr)^{\frac{\gamma-1}{\gamma}}\leq Ca(t)^{3-\frac{3}{\gamma}}(1+t)^{-\frac{1}{\gamma}}, $$
which implies
$$ a(t)\geq C(1+t)^{\frac{1}{3(\gamma-1)}}. $$
(3.17)
For \(1<\gamma<\frac{5}{3}\), (3.13) yields
$$\bigl(H(t) (1+t)^{3\gamma-5}\bigr)'\leq C(1+t)^{3\gamma-5}, $$
which yields
$$ H(t)\leq C(1+t)^{\alpha}. $$
(3.18)
Here
$$ \alpha= \textstyle\begin{cases} 1& \mbox{if } \frac{4}{3}< \gamma< \frac{5}{3},\\ 1+\sigma, \quad \sigma>0\ small, &\mbox{if } \gamma=\frac{4}{3},\\ 5-3\gamma, &\mbox{if } 1< \gamma< \frac{4}{3}. \end{cases} $$
(3.19)
As in (3.15)-(3.17), one can show that
$$ a(t)\geq C(1+t)^{\frac{2-\alpha}{3(\gamma-1)}}. $$
(3.20)
Next, we derive an upper bound for \(a(t)\). It follows from (3.8), (3.20) that
$$\begin{aligned} a'(t)&\leq \vert a_{1}\vert +C_{1}(1+t)^{\frac{3\beta-1}{3(\beta-1)}}+C_{2}(1+t)^{\frac {(2-\alpha)(2-3\beta)}{3(\beta-1)}}+C_{3}a_{0}^{2-3\beta} \\ &\leq \vert a_{1}\vert +C(1+t)^{\frac{3\beta-1}{3(\beta-1)}}+C_{2}(1+t)^{\frac {(2-3\beta)}{3(\beta-1)}}+C_{3}a_{0}^{2-3\beta}. \end{aligned}$$
Then
$$a(t)\leq Ct+C(1+t)^{\frac{6\beta-4}{3(\beta-1)}}\leq C(1+t)^{\frac {6\beta-4}{3(\beta-1)}}. $$
This yields (3.9) and completes the proof of Lemma 3.2. □
We are now ready to give the existence and uniqueness of the solution to the boundary value problem (3.8).
Lemma 3.3
There exists a sufficiently small T such that (3.8) has a solution
\(a(t)\), which is unique in
\(C^{1}[0,T]\)
and satisfied with
\(0<\frac{1}{2}a_{0}<a(t)<2a_{0}\).
Proof
The lemma can be proved by a fixed point argument. In fact, set
$$g\bigl(a(t)\bigr)=a_{1}+\int_{0}^{t}a^{2-3\beta}(s) \,ds-\frac{3}{2-3\beta}a^{2-3\beta }(t)+\frac{3}{2-3\beta}a_{0}^{2-3\beta}. $$
Then (3.8) can be rewritten as
$$ \frac{da(t)}{dt}=g\bigl(a(t)\bigr), \qquad a(0)=a_{0},\qquad g\bigl(a(0) \bigr)=a'(0)=a_{1}. $$
Let \(T_{1}\) be a positive small constant to be determined. Define
$$ X=\biggl\{ a(t)\in C^{1}[0,T_{1}], 0< \frac{1}{2}a_{0}< a(t)< 2a_{0},\ \forall t\in [0,T_{1}]\biggr\} . $$
Then, for any \(a_{1}(t)\) and \(a_{2}(t)\in X\), since \(\beta>1\), we have
$$\begin{aligned} \bigl\vert g\bigl(a_{1}(t)\bigr)-g\bigl(a_{2}(t)\bigr) \bigr\vert =&\biggl\vert \int_{0}^{t} \bigl(a_{1}^{2-3\beta}(s)-a_{2}^{2-3\beta }(s)\bigr) \,ds+\frac{3}{2-3\beta}\bigl(a_{2}^{2-3\beta}(t)-a_{1}^{2-3\beta}(t) \bigr)\biggr\vert \\ \leq&\frac{3}{2-3\beta}\biggl(\frac{1}{2}a_{0} \biggr)^{4-6\beta} \vert a_{1}-a_{2}\vert ^{3\beta -2}(t)\\ &{}+\biggl(\frac{1}{2}a_{0}\biggr)^{4-6\beta} \int_{0}^{t}\bigl\vert a_{1}(s)-a_{2}(s) \bigr\vert ^{3\beta-2}\,ds \\ \leq&\biggl(\frac{1}{2}a_{0}\biggr)^{4-6\beta}\biggl( \frac{3}{2-3\beta}+T_{1}\biggr)\sup_{0\leq t\leq T_{1}}\bigl\vert a_{1}(t)-a_{2}(t)\bigr\vert ^{3\beta-2} \\ \leq& L\sup_{0\leq t\leq T_{1}}\bigl\vert a_{1}(t)-a_{2}(t) \bigr\vert , \end{aligned}$$
where \(L=(\frac{1}{2}a_{0})^{1-3\beta}3^{3(\beta-1)}(\frac{3}{2-3\beta }+T_{1})\) is a constant. We now define a mapping on X by
$$ \mathbb{T}a(t)=a_{0}+\int_{0}^{t} g \bigl(a(s)\bigr)\,ds,\quad \forall t\in[0,T_{1}]. $$
Then \(\mathbb{T}a(t)\in C^{1}[0,T_{1}]\), and, for any \(t< T_{1}\), one can deduce that
$$\begin{aligned} \mathbb{T}a(t) =&a_{0}+\int_{0}^{t} \biggl(a_{1}+\int_{0}^{s}a^{2-3\beta}( \tau)\,d\tau-\frac {3}{2-3\beta}a^{2-3\beta}(s)+\frac{3}{2-3\beta}a_{0}^{2-3\beta} \biggr)\,ds \\ \leq&a_{0}+\biggl(\vert a_{1}\vert -\frac{3}{2-3\beta} \biggl(\frac{1}{2}a_{0}\biggr)^{2-3\beta}\biggr)t+\biggl( \frac {1}{2}a_{0}\biggr)^{2-3\beta}t^{2} \\ \leq&2a_{0} \end{aligned}$$
if
$$ t\leq\frac{\sqrt{(\vert a_{1}\vert -\frac{3}{2-3\beta}(\frac{1}{2}a_{0})^{2-3\beta })^{2}+4a_{0}(\frac{1}{2}a_{0})^{2-3\beta}}-(\vert a_{1}\vert -\frac{3}{2-3\beta}(\frac {1}{2}a_{0})^{2-3\beta})}{2(\frac{1}{2}a_{0})^{2-3\beta}}=T_{2}, $$
and
$$ \mathbb{T}a(t)\geq a_{0}-\biggl(\vert a_{1}\vert - \frac{3}{2-3\beta}a_{0}^{2-3\beta}\biggr)t\geq \frac{1}{2}a_{0} $$
if
$$ t\leq\frac{\frac{1}{2}a_{0}}{\vert a_{1}\vert -\frac{3}{2-3\beta}a_{0}^{2-3\beta}}=T_{3}. $$
Thus, if \(T_{1}\leq\min\{T_{2},T_{3}\}\), then \(\mathbb{T}a(t)\in X\). Furthermore, since
$$ \bigl\vert \mathbb{T}a_{1}(s)-\mathbb{T}a_{2}(s)\bigr\vert =\biggl\vert \int_{0}^{t} g \bigl(a_{1}(s)\bigr)\,ds-\int_{0}^{t} g \bigl(a_{2}(s)\bigr)\,ds\biggr\vert \leq LT_{1}\sup _{0\leq t\leq T_{1}}\bigl\vert a_{1}(t)-a_{2}(t)\bigr\vert , $$
\(\mathbb{T}\) will be a contraction mapping if \((\frac{3}{2-3\beta }+T_{1})T_{1}<(\frac{1}{2}a_{0})^{3\beta-1}3^{3(1-\beta)}\), i.e.,
$$T_{1}< \frac{\sqrt{(\frac{3}{2-3\beta})^{2}+4(\frac{1}{2}a_{0})^{3\beta -1}3^{3(1-\beta)}}-\frac{3}{2-3\beta}}{2}=T_{4}. $$
The above argument shows that \(\mathbb{T}: X\rightarrow X\) is a contraction with the sup-norm for any \(T_{1}=\min\{T_{2},T_{3},T_{4}\}\). By the contraction mapping theorem, there exists a unique \(a(t)\in C^{1}[0,T_{1}]\) such that \(\mathbb{T}a(t)=a(t)\) and then \(a'(t)=g(a(t))\), which yields (3.8). This completes the proof of the lemma. □
Now, Theorem 2.1 follows from Lemma 3.3, the a priori estimates, Lemma 3.2, and the standard continuity argument. This completes the proof of Theorem 2.1.
The stress free boundary condition
First, it follows from the free boundary (2.8) and from (2.4) that
$$ \rho\bigl(a(t)\bigr)=\bigl(\rho_{0}^{\beta-\gamma}(a_{0})+( \gamma-\beta)t\bigr)^{\frac{1}{\beta -\gamma}}. $$
(3.21)
Using the ansatz in (3.1) shows that
$$ a(t)=f^{\frac{1}{3}}(1) \bigl(\rho_{0}^{\beta-\gamma}(a_{0})+( \gamma-\beta )t\bigr)^{\frac{1}{3(\gamma-\beta)}}. $$
(3.22)
Then (3.1) becomes
$$ \rho(r,t)=\frac{f(\frac{r}{a(t)})}{a^{3}(t)},\qquad u(r,t)=\frac{r}{3(\rho _{0}^{\beta-\gamma}(a_{0})+(\gamma-\beta)t)}. $$
(3.23)
Set \(\alpha=f(1)\), and then (3.22) tells us that \(\alpha >0\). Recall the assumption that \(\beta=\frac{1}{2}(\gamma+\frac {1}{3})\), \(\gamma>\frac{5}{3}\). Then (3.3) becomes
$$ \frac{1-\gamma}{2}z+\gamma\alpha^{\frac{1}{3}-\gamma}f^{\gamma -2}(z)f'(z)- \frac{1}{2}\biggl(\gamma+\frac{1}{3}\biggr)\alpha^{\frac{1}{6}- \frac{\gamma}{2}}f^{\frac{3\gamma-11}{6}}(z)f'(z)=0. $$
(3.24)
Denoting \(g(z)=(\frac{f(z)}{\alpha})^{\frac{3\gamma-5}{6}}\) for any \(z\in[0,1]\), then the above equality becomes
$$ \textstyle\begin{cases} g'(z) \{g^{\frac{3\gamma-1}{3\gamma-5}}(z)-\frac{3\gamma+1}{6\gamma } \}=\frac{\alpha^{\frac{2}{3}}(\gamma-1)(3\gamma-5)z}{12\gamma},\\ g(1)=1. \end{cases} $$
(3.25)
Next, we will prove that (3.25) can be solved on \([0,1]\). To this end, we start with a priori estimates and the uniqueness.
Lemma 3.4
For any
\(\gamma>\frac{5}{3}\), let
\(g(z)\)
be a solution to the system (3.25) in
\(C([0,1])\cap C^{1}((0,1])\). Then
$$ \biggl(\frac{3\gamma+1}{6\gamma}\biggr)^{\frac{3\gamma-5}{3\gamma-1}}< g(z)\leq1, $$
(3.26)
for all
\(z\in(0,1]\). Furthermore, such a solution is unique.
Proof
If \(g^{\frac{3\gamma-1}{3\gamma-5}}(z)-\frac{3\gamma+1}{6\gamma}=0\), then (3.25) implies that z must be zero, i.e.
\(g^{\frac{3\gamma-1}{3\gamma-5}}(0)=\frac{3\gamma+1}{6\gamma}\). Namely, if \(z\neq0\), then one has \(g^{\frac{3\gamma-1}{3\gamma-5}}(z)\neq\frac {3\gamma+1}{6\gamma}\).
If for any \(z\in(0,1]\), \(g^{\frac{3\gamma-1}{3\gamma-5}}(z)\in[0,\frac {3\gamma+1}{6\gamma})\), then (3.25) implies that \(g'(z)\leq 0\) and thus
$$1\leq g(z)< \biggl(\frac{3\gamma+1}{6\gamma}\biggr)^{\frac{3\gamma-5}{3\gamma-1}}< 1, $$
which is a contradiction. Thus we can deduce that \(g^{\frac{3\gamma -1}{3\gamma-5}}(z)>\frac{3\gamma+1}{6\gamma}\) for all \(z\in(0,1]\) and together with (3.25) to get \(g'(z)\geq0\) and consequently
$$ \biggl(\frac{3\gamma+1}{6\gamma}\biggr)^{\frac{3\gamma-5}{3\gamma-1}}< g(z)\leq1, \quad \forall z\in(0,1]. $$
(3.27)
It remains to prove the uniqueness.
To this end, let \(\bar{g}(z)\in C([0,1])\cap C^{1}((0,1])\) be another solution to (3.25) with \(\bar{g}(1)=1\) and \((\frac{3\gamma+1}{6\gamma})^{\frac{3\gamma-5}{3\gamma-1}}< g(z)\leq1\) for all \(z\in(0,1]\).
Define \(w(z)=g(z)-\bar{g}(z)\). Then \(w(z)\) solves the following problem:
$$ \textstyle\begin{cases} \frac{d}{dz} \{w(z)-\frac{\gamma(3\gamma-5)}{(3\gamma+1)(\gamma -1)}\{[w(z)+\bar{g}(z)]^{\frac{6(\gamma-1)}{3\gamma-5}}-\bar {g}(z)^{\frac{6(\gamma-1)}{3\gamma-5}}\} \}=0,\\ w(1)=0,\qquad \bar{g}(1)=1. \end{cases} $$
(3.28)
Set
$$I= \bigl\{ z\in[0,1]|w(\xi)\equiv0, z\leq\xi\leq1 \bigr\} . $$
Here \(I\neq\emptyset\) because of \(1\in I\). Define \(z_{0}=\inf I\) and then \(z_{0}\in[0,1]\). Obviously, the uniqueness of solutions to the system (3.25) will be shown by proving that \(z_{0}\equiv0\) and using a continuity argument.
If not, then \(z_{0}\in(0,1]\), and \(w(z_{0})=0\). For any \(z\in(0,z_{0})\), (3.26) tells us that
$$ \biggl(\frac{3\gamma+1}{6\gamma}\biggr)^{\frac{3\gamma-5}{3\gamma-1}}< g(z)\leq1,\quad \forall z\in(0,z_{0}). $$
(3.29)
Integrating (3.28) over \([z,z_{0}]\) shows
$$ w(z)-\frac{\gamma(3\gamma-5)}{(3\gamma+1)(\gamma-1)}\bigl\{ \bigl[w(z)+\bar {g}(z) \bigr]^{\frac{6(\gamma-1)}{3\gamma-5}}-\bar{g}(z)^{\frac{6(\gamma -1)}{3\gamma-5}}\bigr\} =0. $$
(3.30)
Since for any \(\gamma>\frac{5}{3}\) and \(\frac{6(\gamma-1)}{3\gamma -5}>2\), a Taylor expansion gives
$$ \bigl[w(z)+\bar{g}(z)\bigr]^{\frac{6(\gamma-1)}{3\gamma-5}}-\bigl(\bar{g}(z) \bigr)^{\frac {6(\gamma-1)}{3\gamma-5}}=\frac{6(\gamma-1)}{3\gamma-5}\bigl[\bar {g}(z)\bigr]^{\frac{3\gamma-1}{3\gamma-5}} w(z)+O(1)w^{2}(z) $$
(3.31)
for sufficiently small \(w(z)\). Putting (3.31) into (3.30) and using the fact \(w(z_{0})=0\), one has
$$ \biggl\{ 1-\frac{6\gamma}{3\gamma+1}\bigl[\bar{g}(z)\bigr]^{\frac{3\gamma-1}{3\gamma -5}} \biggr\} w(z)-\frac{\gamma(3\gamma-5)}{(3\gamma+1)(\gamma-1)}O(1)w^{2}(z)=0, $$
(3.32)
for z close to \(z_{0}\). Notice that
$$ 1-\frac{6\gamma}{3\gamma+1}\bigl[\bar{g}(z)\bigr]^{\frac{3\gamma-1}{3\gamma -5}}< 0, \quad \forall z \in(0,z_{0}) $$
by virtue of (3.29). Then one can easily deduce that \(w(z)\equiv0\), \(\forall z\in(z_{0}-\delta,z_{0})\) for some \(\delta>0\). This contradicts to \(z_{0}=\inf I\). Thus \(z_{0}\equiv0\) and the proof of Lemma 3.4 is completed. □
Now we are ready to give an existence result to system (3.25).
Lemma 3.5
For any
\(\gamma>\frac{5}{3}\), there is a positive function
\(y=g(z)\)
in
\(C([0,1])\cap C^{1}((0,1])\)
satisfying (3.25).
Proof
We can rewrite (3.25) as follows:
$$ \textstyle\begin{cases} g'(z)=G(g(z),z)=\frac{\frac{\alpha^{\frac{2}{3}}(\gamma-1)(3\gamma -5)z}{12\gamma}}{g^{\frac{3\gamma-1}{3\gamma-5}}(z)-\frac{3\gamma +1}{6\gamma}},\\ g(1)=1. \end{cases} $$
(3.33)
We look for a solution to (3.33) such that
$$ g(z)\in C\bigl([0,1]\bigr)\cap C^{1}\bigl((0,1]\bigr), \qquad \biggl( \frac{3\gamma+1}{6\gamma}\biggr)^{\frac {3\gamma-5}{3\gamma-1}}< g(z)\leq1, \quad \forall z\in(0,1]. $$
(3.34)
Set \(\mathbb{R}=\{(z,g(z))|0\leq1-z\leq a, 0\leq1-g(z)\leq b\}\) for small \(a\in(0,1)\), \(b\in(0,1-(\frac{3\gamma+1}{6\gamma})^{\frac {3\gamma-5}{3\gamma-1}})\). Then we can easily deduce that
$$ \bigl\vert G\bigl(g(z),z\bigr)\bigr\vert \leq M, \quad \forall\bigl(z,g(z)\bigr) \in\mathbb{R}, $$
where M is a positive constant only depending on γ, a, b.
Since \(G(g(z),z)\) is continuous in \(\mathbb{R}\), by choosing \(h=\min\{ a,\frac{b}{M}\}\), one can show that the solution to the initial value problem (3.32) exists in the neighborhood \(0\leq1-z\leq h\).
Similarly, we can extend this solution from the left of the neighborhood \(0\leq1-z\leq h\) step by step. Let the maximum interval of the existence of solutions be \((\alpha,1]\) for some \(\alpha\geq0\); with \(g(z)\in C([\alpha,1])\cap C^{1}((\alpha,1])\), we will show that \(\alpha \equiv0\).
If not, then \(\alpha\in(0,1)\). By Lemma 3.4, one has \(g(\alpha)>(\frac{3\gamma+1}{6\gamma})^{\frac{3\gamma-5}{3\gamma-1}}\) and (3.33)1 is well defined in the small neighborhood of \(z=\alpha\) for \(C^{1}\) function \(g(z)\). Thus similar arguments to those above show that the solution of this initial value problem (3.32) in the neighborhood \(\vert \alpha-z\vert \leq h_{0}\) for small \(h_{0}>0\) exists. That is to say, the solution \(g(z)\) can be extended to the interval \([\alpha-h_{0},1]\), which contradicts the fact that \((\alpha ,1]\) is the maximum interval of the existence of solutions. Then we have obtained a \(C([0,1])\cap C^{1}((0,1])\) solution \(g(z)\) to the system (3.25). □
Finally, by virtue of (3.23) and Lemmas 3.4-3.5, we have obtained the global existence of \(y=f(z)\in C([0,1])\cap C^{1}((0,1])\) to (3.24) and the proof of Theorem 2.2 is complete. According to Lemma 3.4, \(f(0)>(\frac{3\gamma+1}{6\gamma})^{\frac{6}{3\gamma-1}}f(1)\), and \(f(z)=\rho(r,t)a^{3}(t)\), we can deduce from (3.21)-(3.22) the following.
Corollary 3.1
The density at the origin of the center has the following estimate:
$$ \biggl(\frac{3\gamma+1}{6\gamma}\biggr)^{\frac{6}{3\gamma-1}} \biggl[\rho_{0}^{\frac {1-3\gamma}{6}}(a_{0})+ \frac{3\gamma-1}{6}t \biggr]^{\frac{-6}{3\gamma -1}}< \rho(0,t)=\frac{f(0)}{f(1)} \biggl[ \rho_{0}^{\frac{1-3\gamma }{6}}(a_{0})+\frac{3\gamma-1}{6}t \biggr]^{\frac{-6}{3\gamma-1}} , $$
and then
$$\rho(0,t)\rightarrow0, \quad \textit{as } t\rightarrow+\infty. $$