ScalingStacks

Second reformulation of the ODE [0230]

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Second reformulation of the ODE

We write for 0<t<10<t<1,

s=11−t∈(1,∞),𝔴⁡(s)=w⁡(t)1−t.s=\frac{1}{1-t}\in(1,\infty),\quad\mathfrak{w}(s)=\frac{w(t)}{1-t}. (13)

Then

d​𝔴d​s=(1−t)2​d​𝔴d​t=(1−t)2​dd​t​(w1−t)=(1−t)​w′+w.\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.
d2​𝔴d​s2=(1−t)2​dd​t​((1−t)​w′+w)=(1−t)3​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

dd​s​(d​𝔴d​s)n−1=(n−1)​d2​𝔴d​s2​(d​𝔴d​s)n−2=𝔴−3.\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

𝔴>0,d​𝔴d​s>0,d2​𝔴d​s2>0.\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=1s=1 is

𝔴=w0+O⁡(t),d​𝔴d​s=b1​t1/(n−1)+O⁡(t),t=1−s−1.\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)

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