2. Scaling property of global radial solutions

Given $\bar{h}>0$ and let $h=h\left( \left\vert x\right\vert \right) $ be a radial solution to (1.6) in $\Omega=B_{1}(0) $ , we have

(2.1) \begin{equation}\left\{\begin{array}{l}h_{rr}+\dfrac{1}{r}h_{r}=\dfrac{1}{\alpha}h^{-\alpha}-p\ \ \text{in}\ \ B_{1}(0),\\ \\[-7pt] 2\int_{0}^{1}rh(r)dr=\bar{h},\\ \\[-7pt] h^{\prime}(1) =0.\end{array}\right. \end{equation}

From the elliptic theory, h is smooth whenever it is positive; hence, we also require that $h^{\prime}(0) =0$ if $h(0) >0$ .

We follow the construction of radial solutions in [Reference Jiang and Ni17]. Fixing $p=\frac{1}{\alpha}$ , we consider the ordinary differential equation

(2.2) \begin{equation}h_{rr}+\frac{1}{r}h_{r}=\frac{1}{\alpha}h^{-\alpha}-\frac{1}{\alpha}\end{equation}

defined on $\left[ 0,\infty\right) $ . It has been shown in [Reference Jiang and Ni17] that for any $\eta>0$ ,

(2.3) \begin{equation}\left\{\begin{array}{l}h_{rr}+\dfrac{1}{r}h_{r}=\dfrac{1}{\alpha}h^{-\alpha}-\dfrac{1}{\alpha},\\ \\[-9pt] h(0) = \eta,\\ \\[-9pt] h^{\prime}(0) = 0\end{array}\right. \end{equation}

has a unique positive solution $h^{\eta}$ defined on $\left[ 0,\infty\right)$ . And when $\eta=0$ , there exists a unique rupture solution $h^{0}$ which is continuous on $\left[ 0,\infty\right) $ such that $h\left( 0\right) =0$ and h is positive and satisfies (2.2) on $\left( 0,\infty\right) $ . We remark here that $h^{0}$ is a weak solution to (2.2) in the sense of distribution even though $\left( h^{0}\right)_{r}(0) =\infty$ . Please see Remark 4.3 in [Reference Jiang and Ni17] for the definition of weak solutions which have higher integrability.

Obviously $h\equiv1$ if $\eta=1$ . When $\eta\geq0$ , $\eta\not =1$ , $h^{\eta}$ oscillates around 1 and there exists an increasing sequence of positive critical radii $\left\{ r_{k}^{\eta}\right\} _{k=1}^{\infty}$ satisfying

\begin{equation*}\lim_{k\rightarrow\infty}r_{k}^{\eta}=\infty,\end{equation*}

such that $\left( h^{\eta} \right) ^{\prime}( r_{k}^{\eta} )=0$ .

Given $\eta\geq0$ , $\eta\not =1$ and a positive integer k, $h^{\eta}(r)$ satisfies the Neumann boundary condition at $r=r_{k}^{\eta}$ . We now define a scaled function

\begin{equation*}h^{\eta,k}(r) =(r_{k}^{\eta})^{-\frac{2}{1+\alpha}}h^{\eta}(r_{k}^{\eta}r)\end{equation*}

and a constant

\begin{equation*}p^{\eta,k}=\frac{1}{\alpha}\left( r_{k}^{\eta}\right) ^{\frac{2\alpha}{1+\alpha}}.\end{equation*}

One can easily verify that $h^{\eta,k}\left( x\right) =h^{\eta,k}(\left\vert x\right\vert ) $ satisfies the elliptic equation

\begin{equation*}\Delta h=\frac{1}{\alpha} h^{-\alpha}-p^{\eta,k}\ \ in\ \ B_{1}(0)\end{equation*}

with Neumann boundary condition

\begin{equation*}\frac{\partial h}{\partial\nu}=0\ on\ \ \partial B_{1}(0).\end{equation*}

We can also calculate the average thickness for $h^{\eta,k}$ ,

\begin{align*}\bar{h}^{\eta,k} & = \frac{1}{|B_{1}(0)|}\int_{B_{1}(0)}h^{\eta,k}(x)dx=\frac{(r_{k}^{\eta})^{-\frac{2}{1+\alpha}}}{|B_{r_{k}^{\eta}}(0)|}\int_{B_{r_{k}^{\eta}}(0)}h^{\eta}(r)dr\\[3pt] & =2(r_{k}^{\eta})^{-\frac{2}{1+\alpha}-2}\int_{0}^{r_{k}^{\eta}}rh^{\eta}(r)dr.\end{align*}

So far we constructed a solution $h^{\eta,k}$ to (2.1) with

\begin{equation*}\bar{h}=\bar{h}^{\eta,k}.\end{equation*}

Actually, all non-constant radial solutions to (2.1) could be obtained in this fashion. Hence, solving (2.1) for given $\bar{h}$ is reduced to find $\eta,k$ so that $\bar{h}=\bar{h}^{\eta,k}$ . So we will analyse the dependence of $\bar{h}^{\eta,k}$ on $\eta$ and k.

Denote $\bar{h}( \eta,k ) =\bar{h}^{\eta,k}$ as a function of $\eta$ and k for averaging thickness. Fixing a positive integer k, from the continuous dependence of ordinary differential equations on the initial data, $\bar{h}(\eta,k ) $ is continuous for $\eta$ in $\left( 0,1 \right) \cup\left(1,\infty\right) $ . As $\eta\rightarrow0^{+}$ , $h^{\eta}$ converges uniformly to the rupture solution $h^{0}$ on $\left[ 0,\infty\right) $ as proved in [Reference Jiang and Miloua16]. Hence, $\bar{h}( \eta,k ) $ is continuous at $\eta=0$ . Moreover, we have

\begin{equation*}\lim_{k\rightarrow\infty}\sqrt{k\pi}\bar{h}( 0,k ) =1.\end{equation*}

Please refer to Theorem 1.6 of [Reference Jiang and Ni17].

Function $\bar{h}( \eta,k ) $ is not well defined when $\eta=1$ . We will discuss the behaviour of $\bar{h}( \eta,k ) $ when $\eta\rightarrow1$ and $\eta\rightarrow\infty$ , respectively, in the next sections. We will first show that

\begin{equation*}\lim_{\eta\rightarrow1}\bar{h}^{\eta,k}=(r_{k}^{\ast})^{-\frac{2}{1+\alpha}},\end{equation*}

where $\left\{ r_{k}^{\ast} \right\} $ is the increasing divergent sequence of the positive critical points of $J_{0}$ , the Bessel’s function of the first kind with order 0 given by (1.7). Hence, $\bar{h}( \eta,k ) $ is a continuous positive function for $\eta\in\left[ 0,\infty\right) $ if we define $\bar{h}( 1,k )=(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}$ . When $\eta\rightarrow\infty$ , we will show in Section 4 that

\begin{equation*}\lim_{\eta\rightarrow\infty}\frac{\bar{h}( \eta,k ) }{\eta^{\frac{\alpha}{1+\alpha}}}=A_{k}\end{equation*}

for some $A_{k}\in\left( 0,\infty\right) $ . That is,

\begin{equation*}\lim_{\eta\rightarrow\infty}\bar{h}( \eta,k ) = \infty.\end{equation*}

Now we are ready to prove our main theorem.

Proof of Theorem 1. Given any $\bar{h}\in\left(0,\infty\right) $ . Define

\begin{equation*}K=\min\left\{ k\in\mathbb{N}:\bar{h}>(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}\right\} .\end{equation*}

Then we have

\begin{equation*}\bar{h}>(r_{K}^{\ast})^{-\frac{2}{1+\alpha}}\geq(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}\end{equation*}

for any $k\geq K$ . Now $\bar{h}( \eta,k ) $ is a continuous positive function of $\eta$ on $\left( 1,\infty\right) $ with

\begin{equation*}\lim_{\eta\rightarrow1^{+}}\bar{h}^{\eta,k}=(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}\end{equation*}

and

\begin{equation*}\lim_{\eta\rightarrow\infty}\bar{h}( \eta,k ) =\infty.\end{equation*}

Intermediate value theorem implies the existence of $\eta^{k}>1$ , such that

\begin{equation*}\bar{h}=\bar{h}( \eta^{k},k ) .\end{equation*}

Hence, (1.6) admits a radially symmetric solution $\left( h_{k},p_{k}\right) $ where

\begin{equation*}h_{k}(x) =h^{\eta^{k},k}( \left\vert x\right\vert ) =( r_{k}^{\eta^{k})^{-\frac{2}{1+\alpha}}} h^{\eta^{k}}(r_{k}^{\eta^{k}}\left\vert x\right\vert)\end{equation*}

and the pressure

\begin{equation*}p_{k}=p^{\eta_{k},k}=\frac{1}{\alpha}\left( r_{k}^{\eta_{k}}\right)^{\frac{2\alpha}{1+\alpha}}.\end{equation*}

Moreover, $h_{k}$ has exactly k critical points for $r = |x| \in\left( 0,1\right]$ .

3. Behaviour of $\bar{h}( \eta,k ) $ when $\eta\rightarrow1$

To understand the behaviour of $\bar{h}( \eta,k )$ as $\eta\rightarrow1$ , we need to understand the behaviour of $h^{\eta}(r)$ as $\eta\rightarrow1$ . Recall that $h^{\eta}$ is a solution to (2.2) with $h^{\eta}(0) =\eta$ and $\left( h^{\eta}\right) _{r}(0) =0$ . Let $\varepsilon=\eta-1$ and

\begin{equation*}w^{\eta}(r) =\frac{h^{\eta}(r) - 1 }{\varepsilon}.\end{equation*}

Then $w^{\eta}$ is a solution to the differential equation

(3.1) \begin{equation}w_{rr}+\frac{1}{r}w_{r}=\frac{1}{\varepsilon}\left[ \frac{1}{\alpha}\left(1+\varepsilon w\right) ^{-\alpha}-\frac{1}{\alpha}\right]\end{equation}

with initial condition

(3.2) \begin{equation}w(0) = 1,\,w_{r}(0) = 0. \end{equation}

As $\eta\rightarrow1$ , $\varepsilon\rightarrow0$ , formally, (3.1) converges to the Bessel’s differential equation with order 0,

\begin{equation*}w_{rr}^{\ast}+\frac{1}{r}w_{r}^{\ast}+w^{\ast}=0\end{equation*}

with the initial condition $w^{\ast}( 0 ) =1,\,w^{\ast\prime}( 0 ) = 0$ . Such limiting initial value problem has a unique solution $J_{0}$ given by (1.7).

We remark here that $J_{0}$ is oscillating around 0. Denote $r_{k}$ to be the increasing sequence of the critical points of w and $r_{k}^{\ast}$ to be the increasing sequence of the critical points of $J_{0}$ , we have

Proposition 2. As $\eta\rightarrow1$ , the solution $w^{\eta}$ to (3.1) with initial data (3.2) converges uniformly to $J_{0}$ in $\left[ 0,\infty\right) $ . Furthermore, for any positive integer k,

\begin{equation*}\lim_{\eta\rightarrow1}r_{k}=r_{k}^{\ast}.\end{equation*}

Proof. We first show that $w^{\eta}$ is uniformly bounded as $\eta\rightarrow1$ . For simplicity, we will suppress $\eta$ here. Since h is the solution to (2.3), we define energy function

\begin{equation*}e(r)=\frac{1}{2}(h^{\prime}(r))^{2}+F(h(r))\text{ with }F(h)=\frac{1}{\alpha(\alpha-1)}h^{1-\alpha}+\frac{1}{\alpha}h.\end{equation*}

F(h) attains its minimum $\frac{1}{\alpha-1}$ in $\left( 0,\infty\right) $ at $h=1.$ We have

\begin{equation*}\frac{d}{dr}[e(r)]=-\frac{1}{r}(h^{\prime}(r))^{2}\leq0.\end{equation*}

It yields that $F(h(r))\leq e(r)\leq e(0)=F(\eta)$ . Note that as $\eta\rightarrow1$ ,

\begin{align*}& F(1+2(1-\eta))-F(\eta)\\& =[F(1)+\frac{1}{2}F^{\prime\prime}(1)\left[ 2(1-\eta)\right]^{2}]-[F(1)+\frac{1}{2}F^{\prime\prime}(1)(\eta-1)^{2}]+O((\eta-1)^{3})\\& =\frac{3}{2}(1-\eta)^{2}+O((\eta-1)^{3}),\end{align*}

hence for some constant $\delta>0$ , $F(1+2(1-\eta))\geq F(\eta)$ holds whenever $\left\vert \eta-1\right\vert <\delta$ . If $1<\eta<1+\delta$ , then $F(h(r))\leq F( \eta) $ implies $1+2(1-\eta)\leq h(r)\leq\eta$ and if $1-\delta<\eta<1$ , then $\eta\leq h(r)\leq1+2(1-\eta)$ , in both cases

\begin{equation*}-2=\frac{2(1-\eta)}{\varepsilon}\leq w=\frac{h-1}{\varepsilon}\leq\frac{\eta-1}{\varepsilon}=1.\end{equation*}

Thus, $|w\left( r\right) |\leq2$ for any $r>0$ whenever $\left\vert\eta-1\right\vert <\delta$ .

The uniform boundedness of w, as $\eta\rightarrow1$ , implies

\begin{equation*}w_{rr}+\frac{1}{r}w_{r}+w=\frac{1}{\varepsilon}\left[ \frac{1}{\alpha}\left(1+\varepsilon w\right) ^{-\alpha}-\frac{1}{\alpha}\right] +w=O(\varepsilon),\end{equation*}

hence w(x) and $w^{\prime}(x)$ converge uniformly to $J_{0}(x)$ and $J_{0}^{\prime}(x)$ on any bounded interval which implies the convergence of critical points as $\eta\rightarrow1$ since both w and $J_{0}$ are oscillating around 0. From Remark 1, the local maximum and minimum values of $\omega$ at $r_{k}^{\eta}$ form two monotone sequences converging to zero; hence, the local convergence of w(x) to $J_{0}( x ) $ implies the uniform convergence on $\left[ 0,\infty\right) $ .

Since $h^{\eta}=1+\varepsilon w^{\eta}\rightarrow1$ uniformly as $\eta\rightarrow1$ , we have

\begin{equation*}\lim_{\eta\rightarrow1}\bar{h}^{\eta,k}=\lim_{\eta\rightarrow1}\frac{(r_{k}^{\eta})^{-\frac{2}{1+\alpha}}}{|B_{r_{k}^{\eta}}(0)|}\int_{B_{r_{k}^{\eta}}(0)}h^{\eta}(r)dr=(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}.\end{equation*}

Hence, $\bar{h}^{\eta,k}$ is a continuous function in $\eta$ on $\left[0,\infty\right) $ if we define

\begin{equation*}\bar{h}(1,k)=\bar{h}^{1,k}=(r_{k}^{\ast})^{-\frac{2}{1+\alpha}}.\end{equation*}

4. Limiting profile when $\eta\rightarrow\infty$

In this section, we will analyse the behaviour of $\bar{h}(\eta,k)=\bar{h}^{\eta,k}$ as $\eta\rightarrow\infty$ .

Let $\eta>1$ and $h^{\eta}$ be the solution to (2.3). We define the blow-down solution z by $z\left( x\right) =\frac{1}{\eta}h^{\eta}(r)$ with $r=\sqrt{\alpha\eta}x$ . Then we have

\begin{align*}z_{xx}+\frac{1}{x}z_{x} & =\alpha\left( h_{rr}+\frac{1}{r}h_{r}\right)=h^{-\alpha}-1 =\frac{\eta^{-\alpha}}{z^{\alpha}}-1.\end{align*}

Denoting $\varepsilon=\frac{1}{\eta}$ , we have $\varepsilon\rightarrow0^{+}$ as $\eta\rightarrow\infty$ . The blow-down function z is a solution to the initial value problem

(4.1) \begin{equation}\left\{\begin{array}{l}z^{\prime\prime}+\dfrac{1}{x}z^{\prime}=\dfrac{\varepsilon^{\alpha}}{z^{\alpha}}-1\text{ for }x\in\left( 0,\infty\right) ,\\ \\[-9pt] z(0) =1,\ \ \text{and}\ \ z^{\prime}(0) =0.\end{array}\right. \end{equation}

Formally, as $\varepsilon\rightarrow0^{+}$ , (4.1) converges to the limiting equation

(4.2) \begin{equation}\left\{\begin{array}{l}z^{\prime\prime}+\dfrac{1}{x}z^{\prime}=-1,\\ \\[-9pt] z(0) =1,\ \ \text{and}\ \ z^{\prime}(0) =0.\end{array}\right.\end{equation}

which has a unique global solution

\begin{equation*}z(x) =1-\frac{1}{4}x^{2}.\end{equation*}

However, we cannot expect

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}z^{\varepsilon}(x) =1-\frac{1}{4}x^{2}\end{equation*}

since the function $1-\frac{1}{4}x^{2}$ becomes negative when $x>2$ .

Nonetheless, we can establish the following theorem:

Theorem 2. For every $\varepsilon>0$ , let $z^{\varepsilon}(x) $ be the unique solution of the initial value problem (4.1). Then as $\varepsilon\rightarrow0^{+}$ , $z^{\varepsilon}(x) $ converges uniformly to $z_{\ast}(x)$ , the solution of the limiting initial value problem

(4.3) \begin{equation}\left\{\begin{array}{c}z_{\ast}^{\prime\prime}+\frac{1}{x}z_{\ast}^{\prime}=-1,\ \ z_{\ast}>0\ \text{in }\bigcup_{j=0}^{\infty}(a_{j},a_{j+1}).\\ \\[-9pt] z_{\ast}(0) =1,\ \ \text{and}\ \ z_{\ast}^{\prime}(0) =0,\\ \\[-9pt] z_{\ast}( a_{j} ) =0,\ \ z_{\ast}^{\prime}( a_{j}^{+} ) =-z_{\ast}^{\prime}(a_{j}^{-} ),\end{array}\right. \end{equation}

where $a_{0}=0$ , $2=a_{1}<a_{2}<\cdots$ could be inductively computed by solving the initial value problem (4.3).

We will prove Theorem 2 in Section 5 and perform the inductive calculations in Section 6 to obtain the asymptotic behaviour of $a_{k}$ .

The above theorem implies that $z^{\varepsilon}( x ) \ $ converges uniformly to $1-\frac{1}{4}x^{2}$ on $\left[ 0,2\right] $ as $\varepsilon\rightarrow0$ and $\frac{r_{1}^{\eta}}{\sqrt{\alpha\eta}}$ converges to $a_{1}=2$ as $\eta\rightarrow\infty$ . More generally, we have for $k=1,2,3,\cdots$ ,

\begin{equation*}\lim_{\eta\rightarrow\infty}\frac{r_{2k-1}^{\eta}}{\sqrt{\alpha\eta}}=a_{k}\text{ and }\lim_{\eta\rightarrow\infty}\frac{r_{2k}^{\eta}}{\sqrt{\alpha\eta}}=b_{k},\end{equation*}

where $b_{k}$ is the maximum point of $z_{\ast}$ in $\left( a_{k},a_{k+1}\right) $ .

Given a positive integer k and given $\eta>1$ , we have

\begin{align*}\bar{h}^{\eta,k} & =2(r_{k}^{\eta})^{-\frac{2}{1+\alpha}-2}\int_{0}^{r_{k}^{\eta}}rh^{\eta}(r)dr\\& =2(r_{k}^{\eta})^{-\frac{2}{1+\alpha}-2}\alpha\eta^{2}\int_{0}^{\frac{r_{k}^{\eta}}{\sqrt{\alpha\eta}}}sz(s)ds\\& =2\alpha^{-\frac{1}{1+\alpha}}\eta^{\frac{\alpha}{1+\alpha}}\left(\frac{r_{k}^{\eta}}{\sqrt{\alpha\eta}}\right) ^{-\frac{2}{1+\alpha}-2}\int_{0}^{\frac{r_{k}^{\eta}}{\sqrt{\alpha\eta}}}sz(s)ds.\end{align*}

Hence, we have for $k=1,2,3,\cdots$ ,

\begin{equation*}\lim_{\eta\rightarrow\infty}\frac{\bar{h}^{\eta,2k-1}}{\eta^{\frac{\alpha}{1+\alpha}}}=2\alpha^{-\frac{1}{1+\alpha}}a_{k}^{-\frac{2}{1+\alpha}-2}\int_{0}^{a_{k}}sz_{\ast}(s)ds\end{equation*}

and

\begin{equation*}\lim_{\eta\rightarrow\infty}\frac{\bar{h}^{\eta,2k}}{\eta^{\frac{\alpha}{1+\alpha}}}=2\alpha^{-\frac{1}{1+\alpha}}b_{k}^{-\frac{2}{1+\alpha}-2}\int_{0}^{b_{k}}sz_{\ast}(s)ds.\end{equation*}

We remark here that for each positive integer k, $\bar{h}^{\eta,k}\rightarrow\infty$ as $\eta\rightarrow\infty$ .

5. Convergence to the limiting profile

In this section, our goal here is to prove Theorem 2.

Let $\varepsilon\in(0,1) $ , and recall z(x), $x\geq0$ , be the unique solution to (4.1). We need to show that z converges uniformly to $z_{\ast}$ in $\left[0,\infty\right) $ as $\varepsilon\rightarrow0^{+}$ where $z_{\ast}$ is defined by (4.3).

We define an energy function

(5.1) \begin{equation}e(x) =\frac{1}{2}\left( z^{\prime}(x) \right) ^{2}+G(z(x)),\end{equation}

where

\begin{equation*}G(z) =\frac{\varepsilon^{\alpha}}{\alpha-1}z^{1-\alpha}+z.\end{equation*}

It is easy to check that G, defined for $z\in\left( 0,\infty\right) $ , has the following properties:

\begin{equation*}\left\{\begin{array}[c]{l}G( \varepsilon) =\min_{z\in\left( 0,\infty\right) } G(z) =\frac{\alpha}{\alpha-1}\varepsilon,\\ \\[-9pt] G^{\prime}(z) >0\text{ for }z>\varepsilon\text{ and }G^{\prime}(z) <0\text{for }0<z<\varepsilon,\\ \\[-9pt] G^{\prime\prime}(z) >0\text{ for any }z\in\left( 0,\infty\right) ,\\ \\[-9pt] \lim_{z\rightarrow0}G\left( z\right) =\lim_{z\rightarrow\infty}G(z) =\infty.\end{array}\right.\end{equation*}

Since

\begin{equation*}\frac{d}{dx}e(x) =-\frac{1}{x}\left( z^{\prime}(x) \right) ^{2},\end{equation*}

e(x) is monotone decreasing. Hence, for any $x\in\left[ 0,\infty\right)$ ,

\begin{equation*}e(x) \leq e(0) = G(1) =\frac{\varepsilon^{\alpha}}{\alpha-1}+1\end{equation*}

which implies the bounds

\begin{equation*}0<z(x) \leq1\text{, }\left\vert z^{\prime}(x) \right\vert \leq\sqrt{2\left(\frac{\varepsilon^{\alpha}}{\alpha-1}+1\right) }\text{ for any }x\in\left[0,\infty\right) .\end{equation*}

A direct calculation also yields the following simple but useful formulas:

Lemma 1.

(5.2) \begin{equation}\frac{d}{dx}\left( x^{2}\left( z^{\prime}\right) ^{2}\right) = -2x^{2} ( G\circ z ) ^{\prime} (x), \end{equation}

(5.3) \begin{equation}\frac{d}{dx}\left( xz^{\prime}\right) =x\frac{\varepsilon^{\alpha}}{z^{\alpha}}-x. \end{equation}

Applying the convexity property of G, we have

Lemma 2. Suppose $m<\varepsilon$ , for any $z\in\left( m,\varepsilon\right] $ , we have

(5.4) \begin{equation}\frac{G(m) -G( \varepsilon) }{\varepsilon-m} \leq\frac{G(m) - G(z) }{z-m}\leq-G^{\prime}(m) =\left( \frac{\varepsilon}{m}\right) ^{\alpha}-1.\end{equation}

Suppose $M>\varepsilon$ , for any $z\in\left[ \varepsilon,M\right) $ , we have

(5.5) \begin{equation}\frac{G(M) -G( \varepsilon) }{M-\varepsilon}\leq\frac{G(M) -G(z) }{M-z}\leq G^{\prime}(M) =1-\left( \frac{\varepsilon}{M}\right) ^{\alpha}.\end{equation}

For any $\varepsilon\in\left( 0,1\right) $ , z(x) is oscillating around $\varepsilon$ and the roots to $z(x)=\varepsilon$ could be listed in order as

\begin{equation*}0<x_{1}<y_{1}<x_{2}<y_{2}<\cdots<x_{k}<y_{k}<\cdots\end{equation*}

such that

\begin{align*}z & >\varepsilon\text{ for any }x\in\left( 0,x_{1}\right) \cup\left(\cup_{k=1}^{\infty}\left( y_{k},x_{k+1}\right) \right) ,\\[3pt] z & <\varepsilon\text{ for any }x\in\cup_{k=1}^{\infty}\left( x_{k},y_{k}\right) .\end{align*}

We refer the readers to [Reference Jiang and Ni17] for more details.

Our first step is to show the convergence of z to $z_{\ast}$ on $\left[0,x_{1} \right) $ as $\varepsilon\rightarrow0^{+}$ :

Proposition 3.

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}x_{1} & =a_{1}=2,\\[3pt] \lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( x_{1} ) & = z_{\ast}^{\prime}(a_{1}^{-} ) =-1.\end{align*}

Moreover,

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\sup_{x\in\left[ 0,x_{1}\right]}\left\vert z(x) -z_{\ast}(x) \right\vert =0.\end{equation*}

Proof. Integrating (5.3) from 0 to x, we have

(5.6) \begin{equation}xz^{\prime}(x) =\int_{0}^{x}y\frac{\varepsilon^{\alpha}}{z^{\alpha}}dy-\frac{x^{2}}{2}\geq-\frac{x^{2}}{2} \end{equation}

hence $z^{\prime}(x)\geq-\frac{x}{2}$ . Integrating again, we obtain

\begin{equation*}z(x)\geq1-\frac{x^{2}}{4}\text{ for any }x\in\left[ 0,\infty\right) .\text{}\end{equation*}

Plugging the lower bound for z back into (5.6), we have for any $x\in\left( 0,2\right) $ ,

\begin{equation*}xz^{\prime}(x) =\int_{0}^{x}y\frac{\varepsilon^{\alpha}}{z^{\alpha}}dx-\frac{x^{2}}{2}\leq\int_{0}^{x}y\frac{\varepsilon^{\alpha}}{\left(1-\frac{y^{2}}{4}\right) ^{\alpha}}dy-\frac{x^{2}}{2}\end{equation*}

hence

\begin{equation*}0\leq z^{\prime}(x) +\frac{x}{2}\leq\frac{1}{x}\int_{0}^{x}y\frac{\varepsilon^{\alpha}}{\left( 1-\frac{y^{2}}{4}\right) ^{\alpha}}dy\leq\frac{x\varepsilon^{\alpha}}{2\left( 1-\frac{x^{2}}{4}\right) ^{\alpha}}.\end{equation*}

Fix any $a\in\left( 0,2\right) $ , we have

(5.7) \begin{equation}0\leq z^{\prime}(x) +\frac{x}{2}\leq\frac{a\varepsilon^{\alpha}}{2\left(1-\frac{a^{2}}{4}\right) ^{\alpha}} \end{equation}

holds for any $x\in\left[ 0,a\right] $ . Hence, for any $x\in\left[0,a\right] $

\begin{equation*}1-\frac{x^{2}}{4}\leq z(x) \leq1-\frac{x^{2}}{4}+\frac{a^{2}\varepsilon^{\alpha}}{2\left( 1-\frac{a^{2}}{4}\right) ^{\alpha}}.\end{equation*}

In particular, z converges to $1-\frac{x^{2}}{4}$ uniformly on $\left[ 0,a\right] $ as $\varepsilon\rightarrow0^{+}$ , such fact actually follows directly from the continuously dependence of ordinary differential equations since singularity can be avoided on $\left[ 0,a\right] $ with fixed $a<2$ .

Such convergence implies $x_{1}>1$ for sufficiently small $\varepsilon$ . (5.3) implies $xz^{\prime}(x) $ is decreasing on $\left( 0,x_{1}\right) $ , hence for any $x\in\left[1,x_{1}\right] $ , $xz^{\prime}(x) \leq z^{\prime}(1) <0$ and

\begin{equation*}\ln x_{1}=\int_{1}^{x_{1}}\frac{1}{x}dx\leq\int_{\varepsilon}^{z(1) }\frac{1}{\left\vert z^{\prime}(1) \right\vert }dz=\frac{z(1) -\varepsilon}{\left\vert z^{\prime}(1) \right\vert }.\end{equation*}

Since

\begin{equation*}\,\lim_{\varepsilon\rightarrow0^{+}}z(1) =\frac{3}{4},\,\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}(1) =-\frac{1}{2}\end{equation*}

the above estimate implies that $x_{1}\leq C$ for some constant C independent of $\varepsilon\in\left( 0,1\right) $ .

For any $x\in\left[ 1,x_{1}\right] $ , we have

\begin{align*}0 & \leq z^{\prime}(x)+\frac{x}{2}=\frac{1}{x}\int_{0}^{x}y\frac{\varepsilon^{\alpha}}{z^{\alpha}}dy\leq\int_{0}^{1}y\frac{\varepsilon^{\alpha}}{z^{\alpha}}dy+\int_{1}^{x_{1}}y\frac{\varepsilon^{\alpha}}{z^{\alpha}}dy\\[3pt] & \leq\frac{1}{2}\left( z\left( 1\right) \right) ^{-\alpha}\varepsilon^{\alpha}+\int_{\varepsilon}^{z(1)}\frac{y\varepsilon^{\alpha}}{z^{\alpha}\left\vert \frac{dz}{dy}\right\vert }dz\\[3pt] & \leq\frac{1}{2}\left( z(1)\right) ^{-\alpha}\varepsilon^{\alpha}+\int_{\varepsilon}^{z\left( 1\right) }\frac{y^{2}\varepsilon^{\alpha}}{z^{\alpha}\left\vert z^{\prime}(1)\right\vert }dz\\[3pt] & \leq\frac{1}{2}\left( z(1)\right) ^{-\alpha}\varepsilon^{\alpha}+\frac{x_{1}^{2}}{\left\vert z^{\prime}(1)\right\vert }\int_{\varepsilon}^{\infty}\frac{\varepsilon^{\alpha}}{z^{\alpha}}dz\\[3pt] & =\frac{1}{2}\left( z(1)\right) ^{-\alpha}\varepsilon^{\alpha}+\frac{x_{1}^{2}\varepsilon}{\left\vert z^{\prime}(1)\right\vert \left(\alpha-1\right) }\leq C\varepsilon\end{align*}

for some constant C independent of $\varepsilon\in\left( 0,1\right) $ . Combining the estimate (5.7) with $a=1$ , we conclude for any $x\in\left( 0,x_{1}\right] $

(5.8) \begin{equation}0\leq z^{\prime}(x)+\frac{x}{2}\leq C\varepsilon,\end{equation}

where C is some constant independent of $\varepsilon\in\left( 0,1\right)$ . Integrating from 0 to x, we have for any $x\in\left[ 0,x_{1}\right]$ ,

\begin{equation*}0\leq z(x)-\left( 1-\frac{x^{2}}{4}\right) \leq C\varepsilon x_{1}\leq C\varepsilon.\end{equation*}

In particular, evaluating at $x=x_{1}$ ,

\begin{equation*}0\leq\varepsilon-\left( 1-\frac{x_{1}^{2}}{4}\right) \leq C\varepsilon,\end{equation*}

we deduce

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}x_{1}=2=a_{1}.\end{equation*}

And (5.8) implies

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}(x_{1})=-1=z_{\ast}^{\prime}(a_{1}^{-}).\end{equation*}

Next, we work on intervals $\left( x_{k},y_{k}\right) $ , $k=1,2,3,\cdots$ .

Proposition 4. Let $1<a<b$ and z(x), $x\in\left[ a,b\right] $ be the solution to

(5.9) \begin{equation}\left\{\begin{array}{c}z^{\prime\prime}+\frac{1}{x}z^{\prime}=\frac{\varepsilon^{\alpha}}{z^{\alpha}}-1,\\ \\[-9pt] z(a)=z(b)=\varepsilon.\end{array}\right.\end{equation}

Assume that:

1. (1) $z<\varepsilon\ in\ (a,b).$

2. (2) $\textit{z}\ attains\ its\ unique\ minimum\ \textit{m}\ at\ x_{\min}\in(a,b).$

3. (3) $z^{\prime}<0\ in\ \left[ a,x_{\min}\right)\ and\ z^{\prime}>0\ in\ \left( x_{\min},b\right] .$

Then there exists $\varepsilon_{0}>0$ such that for any $\varepsilon\in\left(0,\varepsilon_{0}\right] $ , $b-a\leq C_{1}\varepsilon$ and

(5.10) \begin{equation}\int_{a}^{b}\frac{(z^{\prime}(x))^{2}}{x}dx\leq C_{2}\varepsilon,\end{equation}

where $\varepsilon_{0}$ , $C_{1}$ , $C_{2}$ are positive constants only depending on $A,B,\alpha$ if $a<A$ , $e(a)<A$ and $\left\vert z^{\prime}(a)\right\vert >B>0$ .

Proof. Integrating (5.2) from $x_{\min}$ to x yields

\begin{equation*}x^{2}\left( z^{\prime}\right) ^{2}=-2\int_{x_{\min}}^{x}y^{2} \left( G \circ z \right) ^{\prime} (y) dy \text{ for any }x\in\left[ a,b\right] .\end{equation*}

Suppose $x\in\left[ a, x_{\min} \right] $ , we deduce

\begin{equation*}2\left( G(m) - G(z) \right) \leq\left\vert \frac{dz}{dx}\right\vert ^{2}\leq\frac{2x_{\min}^{2}}{x^{2}}\left( G(m)-G(z) \right) .\end{equation*}

Evaluating at $x=a$ , we have

\begin{equation*}a^{2}\left( z^{\prime}(a) \right) ^{2}\leq2x_{\min}^{2}\left[ G(m)-G(\varepsilon) \right] .\end{equation*}

Next, applying (5.4),

\begin{align*}x_{\min}-a & =\int_{a}^{x_{\min}}dx\leq\int_{m}^{\varepsilon}\frac{dz}{\sqrt{2\left( G(m)-G(z) \right) }} \leq\frac{\sqrt{2} \varepsilon}{\sqrt{G(m)-G( \varepsilon) }}.\end{align*}

Hence,

\begin{align*}a^{2}\left( z^{\prime}(a) \right) ^{2} & \leq2x_{\min}^{2}\left[G(m)-G(\varepsilon) \right] \\[3pt] & \leq2\left( a+\frac{\sqrt{2} \varepsilon}{\sqrt{G(m) -G(\varepsilon) }}\right) ^{2}\left[ G(m) -G(\varepsilon) \right] \\[3pt] & \leq2\left( a\sqrt{G(m) }+\sqrt{2}\varepsilon\right) ^{2}.\end{align*}

So if

\begin{equation*}\varepsilon_{0}\leq\frac{B}{4}\leq\frac{a\left\vert z^{\prime}(a) \right\vert}{4},\end{equation*}

we have for any $\varepsilon\leq\varepsilon_{0}$ ,

\begin{equation*}G(m) \geq\left( \frac{\left\vert z^{\prime}(a) \right\vert }{\sqrt{2}}-\frac{\sqrt{2}\varepsilon}{a}\right) ^{2}\geq\frac{\left\vert z^{\prime}(a)\right\vert ^{2}}{8}\geq\frac{B^{2}}{8}.\end{equation*}

Hence from the structure of function G, we have $m\leq C\varepsilon^{\frac{\alpha}{\alpha-1}}$ . Now with $\varepsilon_{0}$ sufficiently small, we have for any $\varepsilon\leq\varepsilon_{0}$ , $m\leq\frac{\varepsilon}{2}$ . Hence,

\begin{equation*}x_{\min}-a\leq\frac{\sqrt{2} \varepsilon }{\sqrt{G(m)-G(\varepsilon) }} \leq C_{1} \varepsilon\end{equation*}

since

\begin{equation*}G(m) \geq\frac{B^{2}}{8}\text{ and }G(\varepsilon) =\frac{\alpha}{\alpha-1}\varepsilon.\end{equation*}

And

\begin{align*}\int_{a}^{x_{\min}}\frac{\left( z^{\prime}\right) ^{2}}{x}dx & =\int_{m}^{\varepsilon}\frac{\left\vert z^{\prime}\right\vert }{x}dz\leq\int_{m}^{\varepsilon}\frac{x_{\min}\left( \sqrt{2\left[ G(m)-G(z) \right]}\right) }{x^{2}}dz\\[3pt] & \leq\frac{x_{\min}\sqrt{2G(m)}}{a^{2}}\varepsilon\leq\frac{\left(a+C_{1}\varepsilon\right) \sqrt{2e(a) }}{a^{2}}\varepsilon.\end{align*}

Suppose $x\in\left[ x_{\min},b\right] $ , we have

\begin{equation*}2\frac{x_{\min}^{2}}{x^{2}}\left( G(m)-G(z) \right) \leq\left\vert \frac{dz}{dx}\right\vert ^{2}\leq2\left( G(m)-G(z) \right) .\end{equation*}

Applying (5.4) again,

\begin{equation*}\left\vert \frac{dz}{dx}\right\vert \geq\frac{x_{\min}}{x}\sqrt{2\left(G(m)-G(z) \right) }\geq\frac{x_{\min}}{x}\sqrt{\frac{G(m) -G(\varepsilon)}{\varepsilon-m}2\left( z-m\right) }\end{equation*}

so we have

\begin{align*}\frac{x_{\min}}{b}\left( b-x_{\min}\right) & \leq\int_{x_{\min}}^{b}\frac{x_{\min}}{x}dx\leq\int_{m}^{\varepsilon}\frac{dz}{\sqrt{\frac{2 ( G(m)-G(\varepsilon) ) }{\varepsilon-m}}\sqrt{z-m}}\\& =\frac{\sqrt{2}\left( \varepsilon-m\right) }{\sqrt{G(m)-G(\varepsilon) }}\leq C_{1}\varepsilon.\end{align*}

Hence for sufficient small $\varepsilon$ ,

\begin{equation*}b-x_{\min}\leq\frac{C_{1}\varepsilon}{1-\frac{C_{1}\varepsilon}{x_{\min}}}\leq2C_{1}\varepsilon.\end{equation*}

And

\begin{align*}\int_{x_{\min}}^{b}\frac{\left( z^{\prime}\right) ^{2}}{x}dx & \leq\int_{m}^{\varepsilon}\frac{\left\vert z^{\prime}\right\vert }{x}dz\leq\int_{m}^{\varepsilon}\frac{\sqrt{2\left[ G(m)-G(z) \right] }}{x}dz\\[3pt] & \leq\frac{\sqrt{ 2G(m) }}{ x_{\min} }\varepsilon\leq\frac{\sqrt{2e(a)}}{a}\varepsilon.\\[-30pt] \end{align*}

Corollary 1. Suppose

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}x_{k} & =a_{k} \text{ and } \lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( x_{k} ) =z_{\ast}^{\prime}(a_{k}^{-} ) .\end{align*}

Then

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}y_{k} & =a_{k} \text{ and } \lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( y_{k} ) = z_{\ast}^{\prime}(a_{k}^{+} ) .\end{align*}

Moreover

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\sup_{x\in\left[ x_{k},y_{k} \right]}\left\vert z(x)-z_{\ast}(x) \right\vert =0.\end{equation*}

Proof. Since

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\left( y_{k}-x_{k}\right) =0\text{, }\end{equation*}

we have $\lim_{\varepsilon\rightarrow0^{+}}y_{k}=a_{k}$ . Now

\begin{equation*}0\leq e( x_{k} ) -e( y_{k} ) =\int_{x_{k}}^{y_{k}}\frac{\left( z^{\prime}\right) ^{2}}{x}dx\rightarrow0\text{ as } \varepsilon\rightarrow0^{+},\end{equation*}

hence

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\frac{1}{2}\left( \left\vert z^{\prime}(x_{k} ) \right\vert ^{2}-\left\vert z^{\prime}( y_{k} ) \right\vert^{2}\right) =\lim_{\varepsilon\rightarrow0^{+}}\left( e( x_{k} ) -e( y_{k} )\right) =0.\end{equation*}

Since $z^{\prime}( y_{k} ) >0$ , we conclude

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( y_{k} ) =-\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( x_{k} ) =-z^{\prime}_{\ast}( a_{k}^{-} ) =z^{\prime}_{\ast}( a_{k}^{+} ) .\end{equation*}

The convergence

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\sup_{x\in\left[ x_{k},y_{k}\right]}\left\vert z(x) -z_{\ast}(x) \right\vert =0.\end{equation*}

follows from the fact that both z and $z_{\ast}$ converge to 0 in the shrinking to a point interval $\left[ x_{k},y_{k}\right] $ .

Finally, we deal with intervals $\left( y_{k},x_{k+1}\right) $ , $k=1,2,3,\cdots$ .

Proposition 5. Let $1<a<b$ and z(x), $x\in\left[ a,b\right] $ be the solution to

(5.11) \begin{equation}\left\{\begin{array}[c]{c}z^{\prime\prime}+\frac{1}{x}z^{\prime}=\frac{\varepsilon^{\alpha}}{z^{\alpha}}-1,\\ \\[-9pt] z(a) = z(b) = \varepsilon.\end{array}\right.\end{equation}

Assume that:

1. (1) $z>\varepsilon\ in\ (a,b).$

2. (2) $\textit{z}\ attains\ its\ unique\ maximum\ M<1\ at\ x_{\max}\ \in(a,b).$

3. (3) $z^{\prime}>0\ in\ \left[a,x_{\max}\right)\ and\ z^{\prime}<0\ in\ \left( x_{\max},b\right] .$

Then for any $0<\varepsilon\leq\varepsilon_{0}$ , $C_{1}\leq b-a\leq C_{2}$ and

\begin{equation*}\int_{a}^{b}\frac{\varepsilon^{\alpha}}{z^{\alpha}}dx\leq C_{3}\varepsilon,\end{equation*}

where $\varepsilon_{0}$ , $C_{i}$ are constants depending only on $\alpha$ , the upper bound of a and the positive lower bound of $z^{\prime}\left(a\right) $ .

Proof. Integrating (5.3) from a to x, we have

\begin{equation*}xz^{\prime}-az^{\prime}(a) \geq-\frac{1}{2}\left( x^{2}-a^{2}\right) .\end{equation*}

So

\begin{equation*}z^{\prime}\geq\frac{1}{x}\left[ az^{\prime}(a) -\frac{1}{2}\left(x^{2}-a^{2}\right) \right] \geq0\end{equation*}

whenever

\begin{equation*}x\leq\sqrt{a^{2}+2az^{\prime}(a) }.\end{equation*}

Hence,

\begin{equation*}x_{\max}\geq\sqrt{a^{2}+2az^{\prime}(a) },\end{equation*}

and

\begin{align*}M & \geq z\left( \sqrt{a^{2}+2az^{\prime}(a) }\right) \\ \\[-9pt] & \geq\int_{a}^{\sqrt{a^{2}+2az^{\prime}(a) }}\frac{1} {x}\left[ az^{\prime}(a) -\frac{1}{2}\left( x^{2} - a^{2}\right) \right] dx =\frac{a^{2}}{4}u(\frac{2z^{\prime}(a)}{a}) \geq C,\end{align*}

where $u(x)=\left( 1+x\right) \ln(1+x) -x$ is positive and increasing for $x>0$ and C is some constant independent of $\varepsilon$ . We could see that M is uniformly bounded in $\left[ C,1 \right] $ and then we could assume for any $\varepsilon\leq\varepsilon_{0}$ by taking $\varepsilon_{0}$ sufficiently small and any $z\in\left[ \varepsilon,M\right] $

\begin{equation*}\frac{G(M)-G(z)}{M-z}\geq\frac{G(M)-G(\varepsilon)}{M-\varepsilon}\geq\frac{1}{2}.\end{equation*}

Next, integrating (5.2) from $x_{\max}$ to x yields

\begin{equation*}x^{2}\left( z^{\prime}\right) ^{2}=-2\int_{x_{\max}}^{x}y^{2}( G \circ z)^{\prime} (y) dy.\end{equation*}

Suppose $x \in\left[ a,x_{\max} \right] $ , we have

\begin{equation*}2\left( G(M)-G(z) \right) \leq\left\vert \frac{dz}{dx}\right\vert ^{2}\leq\frac{2x_{\max}^{2}}{x^{2}}\left( G(M)-G(z) \right) ,\end{equation*}

hence

\begin{align*}x_{\max}-a & = \int_{a}^{x_{\max}}dx\leq\int_{\varepsilon}^{M}\frac{dz}{\sqrt{2\left( G(M)-G(z) \right) }}\\[3pt] & \leq\int_{\varepsilon}^{M}\frac{dz}{\sqrt{M-z}} \leq2\sqrt{M}.\end{align*}

Now we estimate

\begin{align*}\int_{a}^{x_{\max}}\frac{\varepsilon^{\alpha}}{z^{\alpha}}dx & \leq\int_{\varepsilon}^{M}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{2\left( G(M)-G(z) \right) }}\\ \\[-9pt] & \leq\int_{\varepsilon}^{M}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{M-z}} = \int_{\varepsilon}^{M/2}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{M-z}}+\int_{M/2}^{M}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{M-z}}\\ \\[-9pt] & \leq\frac{\varepsilon}{\left( \alpha-1\right) \sqrt{M/2}}+\frac{2^{\alpha}\varepsilon^{\alpha}}{M^{\alpha}}2\sqrt{M} \leq C_{2}\varepsilon.\end{align*}

On the other hand, suppose $x\in\left[ x_{\max},b\right] $ , we have

\begin{equation*}\frac{2x_{\max}^{2}}{x^{2}}\left( G(M)-G(z) \right) \leq\left\vert \frac{dz}{dx}\right\vert ^{2}\leq2\left( G(M)-G(z) \right) ,\end{equation*}

hence

\begin{equation*}x_{\max}\ln\frac{b}{x_{\max}} = \int_{x_{\max}}^{b}\frac{x_{\max}}{x}dx \leq\int_{\varepsilon}^{M}\frac{dz}{\sqrt{2\left( G(M)-G(z) \right) }}\leq2\sqrt{M}\end{equation*}

which yields

\begin{equation*}b \leq x_{\max}e^{\frac{2\sqrt{M}}{x_{\max}}} \leq\left( a+2\sqrt{M} \right)e^{\frac{2\sqrt{M}}{a}}.\end{equation*}

We also have

\begin{equation*}b-x_{\max}\geq\int_{\varepsilon}^{M}\frac{dz}{\sqrt{2\left( G(M)-G(z) \right)}} \geq\int_{\varepsilon}^{M}\frac{dz}{\sqrt{2\left( M - z \right) }} =\sqrt{2\left( M - \varepsilon\right) }\end{equation*}

and

\begin{align*}\int_{x_{\max}}^{b}\frac{\varepsilon^{\alpha}}{z^{\alpha}}dx & \leq\int_{\varepsilon}^{M}\frac{x}{x_{\max}}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{2\left( G(M) -G(z) \right) }}\\& \leq\frac{b}{x_{\max}}\int_{\varepsilon}^{M}\frac{\varepsilon^{\alpha}}{z^{\alpha}}\frac{dz}{\sqrt{M-z}} \leq\frac{b}{x_{\max}}C_{2}\varepsilon\leq C_{3}\varepsilon.\\[-40pt]\end{align*}

Corollary 2. Suppose that

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}y_{k} & =a_{k} \text{ and } \lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( y_{k} ) = z_{\ast}^{\prime}(a_{k}^{+} ) .\end{align*}

Then

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}x_{k+1} & =a_{k+1} \text{ and }\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( x_{k+1} ) = z_{\ast}^{\prime}(a_{k+1}^{-} ) .\end{align*}

Moreover,

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\sup_{x\in\left[ y_{k},x_{k+1}\right]}\left\vert z(x) -z_{\ast}(x) \right\vert =0.\end{equation*}

Proof. We define $\tilde{z}$ on $\left[ y_{k},x_{k+1}\right] $ as a solution to

\begin{equation*}\tilde{z}^{\prime\prime}+\frac{1}{x}\tilde{z}^{\prime}=-1\end{equation*}

satisfying $\tilde{z}(y_{k}) = z(y_{k}) $ and $\tilde{z}^{\prime}( y_{k} ) =z^{\prime}( y_{k} )$ . Integrating

\begin{equation*}\left( xz^{\prime}-x\tilde{z}^{\prime} \right) ^{\prime}=x\frac{\varepsilon^{\alpha}}{z^{\alpha}}\end{equation*}

from $y_{k}$ , we have

\begin{equation*}0\leq z^{\prime}-\tilde{z}^{\prime}=\frac{1}{x}\int_{y_{k}}^{x}y\frac{\varepsilon^{\alpha}}{z^{\alpha}(y) }dy\leq\frac{x_{k+1}}{y_{k}}\int_{y_{k}}^{x_{k+1}}\frac{\varepsilon^{\alpha}}{z^{\alpha}(y) }dy\leq C_{1}\varepsilon.\end{equation*}

Integrating again, we obtain

\begin{equation*}0\leq z(x) -\tilde{z}(x) \leq C_{1}\left( x_{k+1}-y_{k}\right)\varepsilon=C_{2}\varepsilon.\end{equation*}

In particular, at $x_{k+1}$ , we have

\begin{equation*}\left( 1-C_{2}\right) \varepsilon\leq\tilde{z}( x_{k+1} ) \leq\varepsilon.\end{equation*}

Let $\tilde{z}^{\ast}$ , defined for $x\geq\min\left( y_{k},a_{k}\right) $ be the solution to

\begin{equation*}z^{\prime\prime}+\frac{1}{x}z^{\prime}=-1\end{equation*}

satisfying

\begin{equation*}z( a_{k} ) =0\text{ and }z^{\prime}( a_{k} ) =z_{\ast}^{\prime}( a_{k}^{+} ).\end{equation*}

Since $x_{k+1}-y_{k}\geq C$ and $\lim_{\varepsilon\rightarrow0^{+}}y_{k}=a_{k}$ , we have $x_{k+1}-a_{k}\geq C$ . The continuously dependence of differential equations with initial data implies

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}\tilde{z}^{\ast}( x_{k+1} ) =0.\end{equation*}

Since

\begin{equation*}\tilde{z}^{\ast}=\left( a_{k}z_{\ast}^{\prime}( a_{k}^{+} ) +\frac{a_{k}^{2}}{2} \right) \ln\frac{x}{a_{k}}-\frac{x^{2}-a_{k}^{2}}{4}\end{equation*}

has a unique root $a_{k+1}$ in $\left( a_{k},\infty\right) $ , we conclude

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}x_{k+1}=a_{k+1}.\end{equation*}

And

\begin{equation*}\lim_{\varepsilon\rightarrow0^{+}}z^{\prime}( x_{k+1} ) =\lim_{\varepsilon\rightarrow0^{+}}\tilde{z}^{\prime}( x_{k+1} ) =\lim_{\varepsilon\rightarrow0^{+}}\tilde{z}^{\ast\prime}( x_{k+1} ) =\tilde{z}^{\ast\prime}(a_{k+1} ) = z_{\ast}^{\prime}( a_{k+1} ) .\end{equation*}

Since

\begin{equation*}\left\vert z(x)-z_{\ast}(x) \right\vert \leq\left\vert z(x)-\tilde{z}(x)\right\vert + \left\vert \tilde{z}(x)-\tilde{z}^{\ast}(x) \right\vert+\left\vert \tilde{z}^{\ast}(x)-z_{\ast}(x) \right\vert\end{equation*}

and all the functions are uniformly small near $a_{k}$ and $a_{k+1}$ , it is easy to check

\begin{align*}\lim_{\varepsilon\rightarrow0^{+}}\sup_{x\in\left[ y_{k},x_{k+1}\right]}\left\vert z(x)-z_{\ast}(x) \right\vert =0.\\[-30pt] \end{align*}

Now we are ready to prove Theorem 2 using the asymptotic behaviour of limit solution $z_{\ast}$ which we will prove in the next section.

6. Asymptotic behaviour of limit solution

From Theorem 2, we have as $\varepsilon\rightarrow0^{+}$ , z(x) converges uniformly on $\left[0,\infty\right) $ to the limit $z_{\ast}(x)$ satisfying (4.3). Now we are going to apply inductive calculations to compute $a_{j}$ and analyse the asymptotic behaviours in the following manner. Similarly as previous, we define the energy function

\begin{equation*}e(x) = \frac{1}{2} \left( z_{\ast}^{\prime}(x) \right) ^{2} + z_{\ast}(x)\end{equation*}

and $e_{j} = e(a_{j})$ . It is easy to check that e(x) is decreasing in x and $e_{j}$ is decreasing in j.

(i) In $\left[0,a_{1}\right] $ , we have

\begin{equation*}z_{\ast}(x) =1-\frac{x^{2}}{4}.\end{equation*}

Hence,

\begin{equation*}a_{1}=2 \text{ and } e_{1}=|z_{\ast}^{\prime}(a_{1})|^{2}=1.\end{equation*}

(ii) In $[a_{1},a_{2}],$

\begin{equation*}(xz_{\ast}^{\prime})^{\prime}=-x\text{ and }z_{\ast}^{\prime}(2^{+})=1.\end{equation*}

Then

\begin{equation*}z_{\ast}(x)=4\ln\frac{x}{2}+\frac{4-x^{2}}{4}.\end{equation*}

Note that $z_{\ast}^{\prime\prime}(x)=-\frac{4}{x^{2}}-\frac{1}{2}<0,$ $z_{\ast}(x)$ is concave down. Therefore, there exists a unique solution $a_{2}\in(2,\infty)$ to

\begin{equation*}4\ln\frac{a_{2}}{2}+\frac{4-a_{2}^{2}}{4}=0.\end{equation*}

That is,

\begin{equation*}a_{2}\approx3.74853 \text{ and } e_{2}=\left\vert z_{\ast}^{\prime}(a_{2})\right\vert ^{2}=\left( \frac{a_{2}}{2}-\frac{4}{a_{2}}\right) ^{2}\approx0.6515.\end{equation*}

(iii) In $[a_{j},a_{j+1}]$ , $j\geq2$ , $z_{\ast}$ is defined by the initial value problem

\begin{equation*}(xz_{\ast}^{\prime})^{\prime}=-x \text{, } z_{\ast}( a_{j} ) =0 \text{ and }z_{\ast}^{\prime}(a_{j}^{+})=\sqrt{e_{j}}.\end{equation*}

For any $x\geq a_{j}$ , integrating twice from $a_{j}$ to x, we obtain

\begin{equation*}z_{\ast}(x) =\left( a_{j}\sqrt{e_{j}}+\frac{a_{j}^{2}}{2}\right) \ln\frac{x}{a_{j}}-\frac{x^{2}-a_{j}^{2}}{4}.\end{equation*}

Since

\begin{equation*}z_{\ast}^{\prime\prime}(x) = -\frac{a_{j}\sqrt{e_{j}}+\frac{a_{j}^{2}}{2}}{x^{2}}-\frac{1}{2}<0\end{equation*}

on $\left( a_{j},\infty\right) $ , there is a unique root $a_{j+1}\in(a_{j},\infty)$ such that

(6.1) \begin{equation}z_{\ast} ( a_{j+1} ) =\left( a_{j}\sqrt{e_{j}}+\frac{a_{j}^{2}}{2}\right)\ln\frac{a_{j+1}}{a_{j}}-\frac{a_{j+1}^{2}-a_{j}^{2}}{4}=0.\end{equation}

And $\sqrt{e_{j+1}}$ is given by

(6.2) \begin{equation}\sqrt{e_{j+1}}=-z_{\ast}^{\prime}( a_{j+1} ) =-\frac{a_{j}}{a_{j+1}}\sqrt{e_{j}} + \frac{ a_{j+1}^{2} -a_{j}^{2} }{ 2 a_{j+1} }.\end{equation}

Next, we consider the asymptotic behaviour of $a_{j}$ and $e_{j}$ as $j\rightarrow\infty$ .

Theorem 3. There exists positive constant A such that as $j\rightarrow\infty$ ,

\begin{equation*}a_{j}\sim Aj^{\frac{3}{4}}\text{ and }\sqrt{e_{j}}\sim\frac{3}{8}Aj^{-\frac{1}{4}}.\end{equation*}

Proof. Let $b_{j}=\frac{\sqrt{e_{j}}}{a_{j}}$ , (6.2) implies

(6.3) \begin{equation}b_{j+1}=\frac{1}{2}-\frac{1}{2}\left( \frac{a_{j}}{a_{j+1}}\right)^{2}\left( 1+2b_{j}\right) . \end{equation}

Since $\sqrt{e_{j}}$ is decreasing and $a_{j}$ is increasing in j, $b_{j}$ is decreasing. If $\lim_{j\rightarrow\infty}b_{j}\not =0$ , then $a_{j}=\frac{\sqrt{e_{j}}}{b_{j}}$ is bounded and hence $\lim_{j\rightarrow\infty}\frac{a_{j}}{a_{j+1}}=1$ . If $\lim_{j\rightarrow\infty}b_{j}=0$ , then (6.3) implies $\lim_{j\rightarrow\infty}\frac{a_{j}}{a_{j+1}}=1$ . Hence, in any case, $\lim_{j\rightarrow\infty}\frac{a_{j}}{a_{j+1}}=1$ which also implies $\lim_{j\rightarrow\infty}b_{j}=0$ . Denote $t_j=(\frac{a_{j+1}}{a_{j}})^{2}-1$ , we have $\lim_{j\rightarrow\infty}t_j=0$ . Now (6.1) could be rewritten into

\begin{equation*}\frac{t_j}{\ln(t_j+1)}=2b_{j}+1.\end{equation*}

By Taylor expansion, we have

\begin{equation*}2b_{j}=\frac{t_j}{2}-\frac{t_j^{2}}{12}+O\left(t_j^{3}\right)\end{equation*}

which yields

\begin{equation*}t_j=4b_{j}+\frac{t_j^{2}}{6}+O\left(t_j^{3}\right) = 4b_{j}+\frac{8}{3}b_{j}^{2}+O\left(b_{j}^{3}\right).\end{equation*}

Therefore,

\begin{align*}\frac{a_{j}}{a_{j+1}} & =\left( 1+t_j\right)^{-\frac{1}{2}}=1-\frac{1}{2}t_j+\frac{3t_j^{2}}{8}+O\left(t_j^{3}\right) \\& =1-2b_{j}+\frac{14}{3}b_{j}^{2}+O(b_{j}^{3}).\end{align*}

Plug the above expansion into (6.3), we have

\begin{align*}b_{j+1} & =\frac{1}{2}-\frac{1}{2}\left( 1-2b_{j}+\frac{14}{3}b_{j}^{2}+O(b_{j}^{3})\right) ^{2}\left( 1+2b_{j}\right) \\& =b_{j}-\frac{8}{3}b_{j}^{2}+O(b_{j}^{3})\end{align*}

which implies

\begin{equation*}\frac{1}{b_{j+1}}=\frac{1}{b_{j}}+\frac{8}{3}+O(b_{j}).\end{equation*}

As $b_{j}$ is decreasing and converges to 0, we conclude

\begin{equation*}\lim_{j\rightarrow\infty}jb_{j}=\frac{3}{8}.\end{equation*}

Next, since

\begin{align*}\frac{b_{j+1}a_{j+1}^{\frac{4}{3}}}{b_{j}a_{j}^{\frac{4}{3}}} &=\frac{b_{j+1}}{b_{j}}\left( \frac{a_{j+1}}{a_{j}}\right) ^{\frac{4}{3}}=\frac{\left( 1-\frac{8}{3}b_{j}+O(b_{j}^{2})\right) }{\left[1-2b_{j}+O(b_{j}^{2})\right] ^{4/3}}\\& =1+O(b_{j}^{2}),\end{align*}

the order of $b_{j}$ implies the limit $\gamma=\lim_{j\rightarrow\infty}b_{j}a_{j}^{\frac{4}{3}}>0$ exists. Hence,

\begin{equation*}\lim_{j\rightarrow\infty}\frac{a_{j}}{j^{\frac{3}{4}}}=\lim_{j\rightarrow\infty}\left( \frac{b_{j}a_{j}^{\frac{4}{3}}}{jb_{j}}\right) ^{\frac{3}{4}}=\left( \frac{8\gamma}{3}\right) ^{\frac{3}{4}}=A\end{equation*}

and finally

\begin{align*}\lim_{j\rightarrow\infty}\frac{\sqrt{e_{j}}}{j^{-\frac{1}{4}}}=\lim_{j\rightarrow\infty}jb_{j}\frac{a_{j}}{j^{\frac{3}{4}}}=\frac{3}{8}A.\\[-40pt] \end{align*}