2.5 ODE reduction [022A]

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2.5 ODE reduction

While we believe the generalized Calabi ansatz has wide applicability, both in the Tian-Yau problem, and in the collapsing polarized degeneration problem as described in [14], solving the NA MA equation (4) is practically quite nontrivial for $m\geq 2$ . We shall now specialize to the proportional line bundle case of section 2.3, and further assume $m=2$ . The NA MA equation becomes a PDE with two independent variables

\det(D^{2}u)(d_{1}\frac{\partial u}{\partial x_{1}}+d_{2}\frac{\partial u}{\partial x_{2}})^{n-2}=\text{const}.

(5)

Motivated by the Calabi ansatz, we wish to look for homogeneous solutions. A preliminary dimensional analysis is useful:

u\sim O(|x|^{\alpha}),\quad Du\sim O(|x|^{\alpha-1}),\quad D^{2}u\sim O(|x|^{\alpha-2}).

Thus we want

2(\alpha-2)+(n-2)(\alpha-1)=0,\quad\alpha=\frac{n+2}{n}.

We try the ansatz

x_{2}=\frac{d_{2}}{d_{1}}tx_{1},\quad u(x_{1},x_{2})=x_{1}^{\frac{n+2}{n}}v(t).

(6)

Routine computation then reduces the NA MA equation to an ODE:

Lemma 2.3.

Under the homogeneous ansatz, the NA MA equation is equivalent to

(vv^{\prime\prime}-\frac{2}{n+2}v^{\prime 2})(\frac{n+2}{n}v+(1-t)v^{\prime})^{n-2}=\text{const}.

(7)

Proof.

We compute

\frac{\partial u}{\partial x_{1}}=(\frac{n+2}{n}v-tv^{\prime})x_{1}^{2/n},\quad\frac{\partial u}{\partial x_{2}}=\frac{d_{1}}{d_{2}}v^{\prime}x_{1}^{2/n},

and the second derivatives

\begin{cases}&\frac{\partial^{2}u}{\partial x_{1}^{2}}=x_{1}^{2/n-1}\{\frac{2(n+2)}{n^{2}}v-\frac{4t}{n}v^{\prime}+t^{2}v^{\prime\prime}\},\\ &\frac{\partial^{2}u}{\partial x_{1}\partial x_{2}}=\frac{d_{1}}{d_{2}}(\frac{2}{n}v^{\prime}-tv^{\prime\prime})x_{1}^{\frac{2}{n}-1},\\ &\frac{\partial^{2}u}{\partial x_{2}^{2}}=(\frac{d_{1}}{d_{2}})^{2}x_{1}^{\frac{2}{n}-1}v^{\prime\prime}.\end{cases}

(8)

Whence

\det(D^{2}u)=(\frac{d_{1}}{d_{2}})^{2}x_{1}^{\frac{4}{n}-2}(\frac{2(n+2)}{n^{2}}vv^{\prime\prime}-\frac{4}{n^{2}}v^{\prime 2}),

d_{1}\frac{\partial u}{\partial x_{1}}+d_{2}\frac{\partial u}{\partial x_{2}}=d_{1}x_{1}^{2/n}(\frac{n+2}{n}v+(1-t)v^{\prime}),

so the NA MA equation becomes

(\frac{2(n+2)}{n^{2}}vv^{\prime\prime}-\frac{4}{n^{2}}v^{\prime 2})(\frac{n+2}{n}v+(1-t)v^{\prime})^{n-2}=\text{const}.

∎

Remark 2.4.

The constant is not essential: it just amounts to rescaling the metric.

Remark 2.5.

The Kähler condition requires $D^{2}u$ to be positive definite, and $d_{1}\frac{\partial u}{\partial x_{1}}+d_{2}\frac{\partial u}{\partial x_{2}}>0$ . These are equivalent to

v^{\prime\prime}>0,\quad(n+2)vv^{\prime\prime}-2v^{\prime 2}>0,\quad\frac{n+2}{n}v+(1-t)v^{\prime}>0.

The first two inequalities imply $v>0$ . These constraints are all quite natural in view of the ODE.

Remark 2.6.

The ODE enjoys a symmetry: under the substitution

v(t)=t^{\frac{n+2}{n}}\tilde{v}(1/t),

we have

(n+2)vv^{\prime\prime}-2v^{\prime 2}=t^{\frac{4}{n}-2}((n+2)\tilde{v}^{\prime\prime}\tilde{v}-2\tilde{v}^{\prime 2}),

\frac{n+2}{n}v+(1-t)v^{\prime}=t^{2/n}\{\frac{n+2}{n}\tilde{v}+(1-t^{-1})\tilde{v}^{\prime}\},

so the function $\tilde{v}$ is another solution of the same ODE. The geometric origin of this symmetry is that the NA MA equation is symmetric in $x_{1},x_{2}$ , up to the minor issue of $d_{1},d_{2}$ which disappears after trivial changes of variables.