This first integral can be solved explicitly. Rewriting the first integral in terms of the rescaled variables $\frac{\mathfrak{w}}{w_{0}}$ and $\left(\frac{2(n-1)}{n}w_{0}^{n+2}\right)^{-1/n}s$, the ODE simplifies to the form

$$f^{\prime n}=1-f^{-2}.$$

We introduce the function

$$F(x)=\int_{1}^{x}(1-y^{-2})^{-1/n}dy,$$

then

$$F(\frac{\mathfrak{w}}{w_{0}})=\left(\frac{2(n-1)}{n}w_{0}^{n+2}\right)^{-1/n}(s-1).$$

(18)

Inverting $F$ solves $\mathfrak{w}$ as a function of $s$. It is clear that

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

The ODE (14) then implies $\frac{d^{2}\mathfrak{w}}{ds^{2}}>0$, namely the Kähler condition is satisfied. We remark that $F$ can be expressed in terms of the hypergeometric functions, cf. the Appendix.

We then check the initial conditions and the analyticity of the solution near $s=1$.

We have solved the ODE with the prescribed initial condition, on the interval $0<t<1$. For the application to the generalized Calabi ansatz, we need solutions over $0<t<\infty$, and for this purpose the ODE (14) is inadequate. From the ODE (12), it is however a priori clear that $t=1$ is not a singularity.

Recall the ODE has a symmetry under $t\to t^{-1}$ (cf. Remark 3.3). Our strategy to achieve both the $t=0$ and the $t=\infty$ boundary conditions, is to look for symmetric solutions:

$$w(t)=t\tilde{w}(1/t),\quad w(t)=\tilde{w}(t).$$

This is a functional equation on $w(t)$, and it amounts to a matching condition at $t=1$:

$$w(1)=\tilde{w}(1),\quad w^{\prime}(1)=\tilde{w}^{\prime}(1).$$

This is equivalent to

$$w^{\prime}(1)=\frac{1}{2}w(1).$$

(20)

The problem is then to look for $w_{0}>0$ to solve (20). The key is to extract $w^{\prime}(1)$ from the asymptote of $\mathfrak{w}$ as $s\to+\infty$, namely $t\to 1$. The starting point is the identity

$$w^{\prime}(t)=s\frac{d\mathfrak{w}}{ds}-\mathfrak{w}.$$

(21)

which means the value of $w^{\prime}$ is related to the Legendre transform of $\mathfrak{w}$.

The Legendre transform of the convex function $\mathfrak{w}$ is given by

$$\mathfrak{w}^{*}(p)=\sup_{s\in[1,\infty)}sp-\mathfrak{w}(s).$$

In particular, we have

$$\mathfrak{w}^{*}(p)=s\frac{d\mathfrak{w}}{ds}-\mathfrak{w}(s)\qquad\text{ at }p=\frac{d\mathfrak{w}}{ds}$$

We are now going to rewrite the first integral (17) in terms of $\mathfrak{w}^{*}(p)$. First, since $\lim_{s\rightarrow\infty}\mathfrak{w}(s)=\infty$ we see from (17) that

$$\frac{d\mathfrak{w}}{ds}([1,\infty))=[0,p_{*}=\left(\frac{n}{2(n-1)w_{0}^{2}}\right)^{\frac{1}{n}})$$

and so the Legendre transform is defined on this interval. Furthermore, since $\frac{d\mathfrak{w}}{ds}(s=1)=0$ we have $\frac{d\mathfrak{w}^{*}}{dp}(p=0)=1$, and $\mathfrak{w}^{*}(p=0)=-\mathfrak{w}(s=1)=-w_{0}$ by the properties of the Legendre transform. Now from the involution property of the Legendre transform,

$$\mathfrak{w}(s(p))=s(p)p-\mathfrak{w}^{*}(p)=\frac{d\mathfrak{w}^{*}}{dp}p-\mathfrak{w}^{*}(p).$$

The first integral (17) is rewritten as

$$\frac{n-1}{n}p^{n}+\frac{1}{2}\left(\frac{d\mathfrak{w}^{*}}{dp}p-\mathfrak{w}^{*}(p)\right)^{-2}=\frac{1}{2w_{0}^{2}},$$

namely

$$(p\frac{d}{dp}\mathfrak{w}^{*}-\mathfrak{w}^{*})^{2}\left(\frac{1}{w_{0}^{2}}-\frac{2(n-1)p^{n}}{n}\right)=1.$$

To simplify matters we can rescale. Define

$$y=\frac{1}{p_{*}}p,\quad g(y)=\frac{1}{w_{0}}\mathfrak{w}^{*}.$$

Then the ODE is recast on the interval $y\in[0,1)$ as

$$(y\frac{dg}{dy}-g(y))^{2}(1-y^{n})=1,$$

(22)

subject to the initial conditions

$$g(0)=-1,\quad g^{\prime}(0)=\frac{p_{*}}{w_{0}}=\left(\frac{n}{2(n-1)w_{0}^{2+n}}\right)^{\frac{1}{n}}.$$

This can be integrated explicitly:

We can implement the matching condition $w^{\prime}(1)=\frac{1}{2}w(1)$ as follows. From (21) and the definition of the Legendre transform,

$$w^{\prime}(t=1)=\mathfrak{w}^{*}(p_{*})=w_{0}g(1)=w_{0}\left(\frac{p_{*}}{w_{0}}-\frac{\Gamma(1-\frac{1}{n})\sqrt{\pi}}{\Gamma(\frac{1}{2}-\frac{1}{n})}\right).$$

On the other hand, we have

	$\displaystyle w(t=1)$	$\displaystyle=\lim_{s\rightarrow\infty}\frac{\mathfrak{w}(s)}{s}$
		$\displaystyle=\lim_{p\rightarrow p_{*}}\frac{p(d\mathfrak{w}^{*}/dp)-\mathfrak{w}^{*}}{(d\mathfrak{w}^{*}/dp)}$
		$\displaystyle=\lim_{p\rightarrow p_{*}}p-\frac{\mathfrak{w}^{*}}{(d\mathfrak{w}^{*}/dp)}$

but from the first integral we have $(d\mathfrak{w}^{*}/dp)\rightarrow\infty$ as $p\rightarrow p_{*}$, while $\mathfrak{w}^{*}\rightarrow\mathfrak{w}^{*}(1)<+\infty$. Thus we get

$$w(1)=p_{*}=\left(\frac{n}{2(n-1)w_{0}^{2}}\right)^{1/n}.$$

All together, the matching amounts to solving the equation

$$w_{0}\left(\frac{p_{*}}{w_{0}}-\frac{\Gamma(1-\frac{1}{n})\sqrt{\pi}}{\Gamma(\frac{1}{2}-\frac{1}{n})}\right)=\frac{1}{2}p_{*},$$

Or equivalently

$$\frac{1}{2}\left(\frac{n}{2(n-1)w_{0}^{2}}\right)^{1/n}=w_{0}\frac{\Gamma(1-\frac{1}{n})\sqrt{\pi}}{\Gamma(\frac{1}{2}-\frac{1}{n})}$$

which yields

$\displaystyle w_{0}=\left(\frac{1}{2}\right)^{\frac{n+1}{n+2}}\left(\frac{n}{(n-1)}\right)^{\frac{1}{n+2}}\left(\frac{1}{\sqrt{\pi}}\frac{\Gamma(\frac{1}{2}-\frac{1}{n})}{\Gamma(1-\frac{1}{n})}\right)^{\frac{n}{n+2}}$

(23)

which tends to $\frac{1}{2}$ as $n\rightarrow+\infty$. For our purpose the important fact is that $w_{0}>0$ for $n\geq 3$, which means we have found the initial condition that ensures matching.