3.2 First integral of the ODE [0231]

Original official author HTML, exact retained edition. Historical TeX conversion verdicts remain unchanged. Cited-edition alignment and mathematical self-containment are not assessed.

Complete original source context · Original author HTML

3.2 First integral of the ODE

Lemma 3.4.

If $\mathfrak{w}$ solves the ODE (14) with the initial condition (16), then

\frac{n-1}{n}(\frac{d\mathfrak{w}}{ds})^{n}+\frac{1}{2\mathfrak{w}^{2}}=\frac{1}{2w_{0}^{2}}.

(17)

Proof.

We differentiate

	$\displaystyle\frac{d}{ds}\left(\frac{(n-1)}{n}\left(\frac{d\mathfrak{w}}{ds}\right)^{n}+\frac{1}{2\mathfrak{w}^{2}}\right)$	$\displaystyle=(n-1)\left(\frac{d\mathfrak{w}}{ds}\right)^{n-1}\left(\frac{d^{2}\mathfrak{w}}{ds^{2}}\right)-\mathfrak{w}^{-3}\frac{d\mathfrak{w}}{ds}$
		$\displaystyle=\left(\frac{d\mathfrak{w}}{ds}\right)\left(\frac{d}{ds}\left(\frac{d\mathfrak{w}}{ds}\right)^{n-1}-\mathfrak{w}^{-3}\right)$
		$\displaystyle=0$

Now the result follows by observing that, from the initial conditions we have

\left(\frac{(n-1)}{n}\left(\frac{d\mathfrak{w}}{ds}\right)^{n}+\frac{1}{2\mathfrak{w}^{2}}\right)\bigg|_{s=1}=\frac{1}{2w_{0}^{2}}.

∎

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 $s=1$ .

Corollary 3.5.

The function $\mathfrak{w}(s)$ is a power series in $(s-1)^{n/(n-1)}$ near $s=1$ . To leading orders

\mathfrak{w}=w_{0}+\frac{n-1}{n}w_{0}^{-\frac{3}{n-1}}(s-1)^{\frac{n}{n-1}}+O((s-1)^{\frac{2n}{n-1}}).

(19)

Proof.

Notice near $y=1$ , the function

(1-y^{-2})^{-1/n}=(y-1)^{-1/n}\times\text{Taylor series in $y-1$ with constant term $2^{-1/n}$}.

Upon integration,

F(x)=(x-1)^{(n-1)/n}\times\text{Taylor series in $x-1$ with constant term $2^{-1/n}\frac{n}{n-1}$}.

Raising (18) to the power $n/(n-1)$ , we see

w_{0}^{-\frac{n+2}{n-1}}(s-1)^{n/(n-1)}=\text{Taylor series in $\frac{\mathfrak{w}}{w_{0}}-1$ with first coefficient $\frac{n}{n-1}$}.

Inverting the function, $\frac{\mathfrak{w}}{w_{0}}-1$ is a power series of $(s-1)^{n/(n-1)}$ near $s=1$ . To leading order,

\frac{\mathfrak{w}}{w_{0}}-1=\frac{n-1}{n}w_{0}^{-\frac{n+2}{n-1}}(s-1)^{\frac{n}{n-1}}\left(1+O((s-1)^{\frac{n}{2n-1}})\right),

\frac{d\mathfrak{w}}{ds}\approx w_{0}^{-\frac{3}{n-1}}(s-1)^{1/(n-1)}=b_{1}(s-1)^{1/(n-1)},

which agrees with the initial condition (16). ∎