The ODE (7) can be somewhat further simplified:

In terms of the substitution $w=\frac{n+2}{n}v^{n/(n+2)}$, the boundary condition at $t\to 0$ becomes

$$\begin{cases}&w=w_{0}+O(t),\quad w_{0}=\frac{n+2}{n}v_{0}^{n/(n+2)},\\ &w^{\prime}=-w_{0}+b_{1}t^{1/(n-1)}+O(t),\quad b_{1}=av_{0}^{-2/(n+2)},\\ &w^{\prime\prime}=\frac{b_{1}}{n-1}t^{-(n-2)/(n-1)}+O(1).\end{cases}$$

(11)

In the normalization of the ODE

$$w^{\prime\prime}w^{3}(w+(1-t)w^{\prime})^{n-2}=\frac{1}{n-1},$$

(12)

we would have

$$b_{1}^{n-1}w_{0}^{3}=1.$$

Notice $w_{0}$ determines $b_{1}$, which means this boundary condition at $t=0$ comes in a 1-parameter family, instead of the generic 2-parameter family for second order ODEs. The $t=+\infty$ end has a closely related boundary condition via the ODE symmetry (cf. Remark 3.3), which also arises in a 1-parameter family, so one expects the global solutions to the ODE to be isolated.

We write for $0<t<1$,

$$s=\frac{1}{1-t}\in(1,\infty),\quad\mathfrak{w}(s)=\frac{w(t)}{1-t}.$$

(13)

Then

$$\frac{d\mathfrak{w}}{ds}=(1-t)^{2}\frac{d\mathfrak{w}}{dt}=(1-t)^{2}\frac{d}{dt}(\frac{w}{1-t})=(1-t)w^{\prime}+w.$$

$$\frac{d^{2}\mathfrak{w}}{ds^{2}}=(1-t)^{2}\frac{d}{dt}((1-t)w^{\prime}+w)=(1-t)^{3}w^{\prime\prime}.$$

Thus the ODE (12) can be reformulated as

$$\frac{d}{ds}(\frac{d\mathfrak{w}}{ds})^{n-1}=(n-1)\frac{d^{2}\mathfrak{w}}{ds^{2}}(\frac{d\mathfrak{w}}{ds})^{n-2}=\mathfrak{w}^{-3}.$$

(14)

The Kähler condition (cf. Remark 3.2) translates into

$$\mathfrak{w}>0,\quad\frac{d\mathfrak{w}}{ds}>0,\quad\frac{d^{2}\mathfrak{w}}{ds^{2}}>0.$$

(15)

The initial condition at $s=1$ is

$$\mathfrak{w}=w_{0}+O(t),\quad\frac{d\mathfrak{w}}{ds}=b_{1}t^{1/(n-1)}+O(t),\quad t=1-s^{-1}.$$

(16)