ScalingStacks

5.5. The Poisson equation with prescribed asymptotics [054I]

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 Β· Original author HTML

5.5. The Poisson equation with prescribed asymptotics

In this subsection, we will construct solutions to the Poisson equation on the Calabi space (π’žn,gπ’žn)(\mathcal{C}^{n},g_{\mathcal{C}^{n}}),

(5.200) Ξ”gπ’žn​u=v\Delta_{g_{\mathcal{C}^{n}}}u=v

with controlled asymptotic behavior. As in Section 5.1, we carry out separation of variables. Suppose vv is a smooth function defined on {zβ‰₯1}\{z\geq 1\}. We write

(5.201) u⁑(z,π’š)=βˆ‘k=1∞uk​(z)β‹…Ο†k​(π’š),v⁑(z,π’š)=βˆ‘k=1∞ξk​(z)β‹…Ο†k​(π’š).\displaystyle u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\cdot\varphi_{k}(\bm{y}),\quad v(z,\bm{y})=\sum\limits_{k=1}^{\infty}\xi_{k}(z)\cdot\varphi_{k}(\bm{y}).

So the Poisson equation

(5.202) Ξ”gπ’žn​u=v\Delta_{g_{\mathcal{C}^{n}}}u=v

is reduced to the following inhomogeneous ODE

(5.203) d2​uk​(z)d​z2βˆ’(jk2​n24β‹…zn+n​λk)​znβˆ’2​uk​(z)=znβˆ’1β‹…ΞΎk​(z),zβ‰₯z1.\frac{d^{2}u_{k}(z)}{dz^{2}}-(\frac{j_{k}^{2}n^{2}}{4}\cdot z^{n}+n\lambda_{k})z^{n-2}u_{k}(z)=z^{n-1}\cdot\xi_{k}(z),\quad z\geq z_{1}.

Let 𝒒k​(z)\mathcal{G}_{k}(z) and π’Ÿk​(z)\mathcal{D}_{k}(z) be the growing solution and decaying solution to the corresponding homogeneous equation, which were analyzed in Section 5.2 and 5.3. So applying standard Liouville’ formula, Equation (5.203) has a particular solution

(5.204) uk​(z)≑𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)β‹…(ΞΎk​(r)β‹…rnβˆ’1)​𝑑r+π’Ÿk​(z)𝒲k​(z)β€‹βˆ«z1z𝒒k​(r)β‹…(ΞΎk​(r)β‹…rnβˆ’1)​𝑑r,u_{k}(z)\equiv\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\cdot\Big(\xi_{k}(r)\cdot r^{n-1}\Big)dr+\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z_{1}}^{z}\mathcal{G}_{k}(r)\cdot\Big(\xi_{k}(r)\cdot r^{n-1}\Big)dr,

where 𝒲k\mathcal{W}_{k} is the Wronskian

(5.205) 𝒲k​(z)≑𝒲⁑(𝒒k​(z),π’Ÿk​(z)).\mathcal{W}_{k}(z)\equiv\mathcal{W}\Big(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z)\Big).
Lemma 5.15.

Assume that the function ΞΎk​(z)\xi_{k}(z) satisfies the following property: there are Ξ·0∈(βˆ’Ξ΄b/2,Ξ΄b/2)\eta_{0}\in(-\delta_{b}/2,\delta_{b}/2), a sequence of positive constants 𝔅k>0\mathfrak{B}_{k}>0 such that

(5.206) |ΞΎk​(z)|≀𝔅kβ‹…eΞ·0β‹…zn2.|\xi_{k}(z)|\leq\mathfrak{B}_{k}\cdot e^{\eta_{0}\cdot z^{\frac{n}{2}}}.

Let uk​(z)u_{k}(z) be the particular solution (5.204), then there exists some constant C0>0C_{0}>0 such that the particular solution uku_{k} satisfies the uniform estimate

(5.207) |uk​(z)|≀C0⋅𝔅kβ‹…(Ξ›k)12​nβ‹…eΞ·β‹…zn2|u_{k}(z)|\leq C_{0}\cdot\mathfrak{B}_{k}\cdot(\Lambda_{k})^{\frac{1}{2n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}

for any Ξ·>Ξ·0\eta>\eta_{0}.

Proof.

We will estimate the two terms in (5.204) individually, and we also divide into several cases.

First consider jk=0j_{k}=0 and k=0k=0. In this case the solutions uku_{k} is given by simple integrals of ΞΎk\xi_{k} and the conclusion is easy to see.

The second case is that kβˆˆβ„€+k\in\mathbb{Z}_{+} and jk=0j_{k}=0. Applying Proposition 5.5, the fundamental solutions 𝒒k​(z)\mathcal{G}_{k}(z) and π’Ÿk​(z)\mathcal{D}_{k}(z) satisfy the uniform estimates

(5.208) 𝒒k​(z)\displaystyle\mathcal{G}_{k}(z) ≀CΞ»k14β‹…z2βˆ’n4β‹…e2​λknβ‹…zn2,\displaystyle\leq\frac{C}{\lambda_{k}^{\frac{1}{4}}}\cdot z^{\frac{2-n}{4}}\cdot e^{2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}},
(5.209) π’Ÿk​(z)\displaystyle\mathcal{D}_{k}(z) ≀CΞ»k14β‹…z2βˆ’n4β‹…eβˆ’2Ξ»knβ‹…zn2.\displaystyle\leq\frac{C}{\lambda_{k}^{\frac{1}{4}}}\cdot z^{\frac{2-n}{4}}\cdot e^{-2\sqrt{\frac{\lambda_{k}}{n}}\cdot z^{\frac{n}{2}}}.

By Lemma 5.3.1, 𝒲k​(z)=𝒲⁑(𝒒k​(z),π’Ÿk​(z))=n2\mathcal{W}_{k}(z)=\mathcal{W}(\mathcal{G}_{k}(z),\mathcal{D}_{k}(z))=\frac{n}{2}. Let us denote Ξ»~k≑2​λkn\tilde{\lambda}_{k}\equiv 2\sqrt{\frac{\lambda_{k}}{n}}, then Ξ»~kβ‰₯2​λ1n=Ξ΄b\tilde{\lambda}_{k}\geq 2\sqrt{\frac{\lambda_{1}}{n}}=\delta_{b}. Now the first integral term in (5.204) has the following bound,

(5.210) 𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)​|ΞΎk​(r)β‹…rnβˆ’1|​𝑑r\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)|\xi_{k}(r)\cdot r^{n-1}|dr
≀\displaystyle\leq C⋅𝔅kΞ»k12β‹…z2βˆ’n4β‹…eΞ»~kβ‹…zn2β‹…βˆ«z∞r3​n4βˆ’12β‹…e(βˆ’Ξ»~k+Ξ·0)β‹…rn2​𝑑r.\displaystyle\frac{C\cdot\mathfrak{B}_{k}}{\lambda_{k}^{\frac{1}{2}}}\cdot z^{\frac{2-n}{4}}\cdot e^{\tilde{\lambda}_{k}\cdot z^{\frac{n}{2}}}\cdot\int_{z}^{\infty}r^{\frac{3n}{4}-\frac{1}{2}}\cdot e^{(-\tilde{\lambda}_{k}+\eta_{0})\cdot r^{\frac{n}{2}}}dr.

By assumption, |Ξ·0|<Ξ΄b2≀λ~k2|\eta_{0}|<\frac{\delta_{b}}{2}\leq\frac{\tilde{\lambda}_{k}}{2}, then

(5.211) 𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)​|ΞΎk​(r)β‹…rnβˆ’1|​𝑑r\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)|\xi_{k}(r)\cdot r^{n-1}|dr ≀\displaystyle\leq C⋅𝔅kΞ»k12β‹…z2βˆ’n4β‹…eΞ»~kβ‹…zn2β‹…e(βˆ’Ξ»~k+Ξ·β€²)β‹…zn2\displaystyle\frac{C\cdot\mathfrak{B}_{k}}{\lambda_{k}^{\frac{1}{2}}}\cdot z^{\frac{2-n}{4}}\cdot e^{\tilde{\lambda}_{k}\cdot z^{\frac{n}{2}}}\cdot e^{(-\tilde{\lambda}_{k}+\eta^{\prime})\cdot z^{\frac{n}{2}}}
≀\displaystyle\leq C⋅𝔅kβ‹…eΞ·β‹…zn2,\displaystyle C\cdot\mathfrak{B}_{k}\cdot e^{\eta\cdot z^{\frac{n}{2}}},

where Ξ·>Ξ·β€²>Ξ·0>0\eta>\eta^{\prime}>\eta_{0}>0. Similarly,

(5.212) π’Ÿk​(z)𝒲k​(z)β€‹βˆ«z0z𝒒k​(r)​|ΞΎk​(r)β‹…rnβˆ’1|​𝑑r≀C⋅𝔅kβ‹…eΞ·β‹…zn2.\displaystyle\frac{\mathcal{D}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z_{0}}^{z}\mathcal{G}_{k}(r)|\xi_{k}(r)\cdot r^{n-1}|dr\leq C\cdot\mathfrak{B}_{k}\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

In the third case jkβˆˆβ„€+j_{k}\in\mathbb{Z}_{+} and Qβ‰₯1Q\geq 1, we need to apply Lemma 5.11. In fact,

(5.213) 𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)​ξk​(r)β‹…rnβˆ’1​𝑑r\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr
≀\displaystyle\leq Cnβ‹…Q14β‹…(jkβ‹…zn)1βˆ’2​α4Ξ“2​(Q+1)β‹…eG^k​(z)𝒲k​(z)∫z∞eF^k​(r)ΞΎk(r)β‹…rnβˆ’1dr\displaystyle C_{n}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)}\xi_{k}(r)\cdot r^{n-1}dr
≀\displaystyle\leq Cn⋅𝔅kβ‹…Q14β‹…(jkβ‹…zn)1βˆ’2​α4Ξ“2​(Q+1)β‹…eG^k​(z)𝒲k​(z)∫z∞eF^k​(r)+Ξ·β€²β‹…rn2dr,\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime}\cdot r^{\frac{n}{2}}}dr,

where Ξ·β€²>Ξ·0\eta^{\prime}>\eta_{0}. We choose any ϡ∈(Ξ΄b/100,Ξ΄b/10)\epsilon\in(\delta_{b}/100,\delta_{b}/10) and denote η′′≑η′+Ο΅\eta^{\prime\prime}\equiv\eta^{\prime}+\epsilon, then by Lemma 5.11,

(5.214) eG^k​(z)𝒲k​(z)β€‹βˆ«z∞eF^k​(r)+Ξ·β€²β‹…rn2​𝑑r\displaystyle\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime}\cdot r^{\frac{n}{2}}}dr
=\displaystyle= eG^k​(z)𝒲k​(z)β€‹βˆ«z∞eF^k​(r)+Ξ·β€²β€²β‹…rn2β‹…eβˆ’Ο΅β€‹rn2​𝑑r\displaystyle\frac{e^{\widehat{G}_{k}(z)}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{\widehat{F}_{k}(r)+\eta^{\prime\prime}\cdot r^{\frac{n}{2}}}\cdot e^{-\epsilon r^{\frac{n}{2}}}dr
≀\displaystyle\leq eF^k​(z)+G^k​(z)+Ξ·β€²β€²β‹…zn2𝒲k​(z)∫z∞eβˆ’Ο΅β‹…rn2dr\displaystyle\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}e^{-\epsilon\cdot r^{\frac{n}{2}}}dr
≀\displaystyle\leq Cnβ‹…eF^k​(z)+G^k​(z)+Ξ·β€²β€²β‹…zn2𝒲k​(z).\displaystyle C_{n}\cdot\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}.

Therefore,

(5.215) 𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)​ξk​(r)β‹…rnβˆ’1​𝑑r≀Cn⋅𝔅kβ‹…Q14β‹…(jkβ‹…zn)1βˆ’2​α4Ξ“2​(Q+1)β‹…eF^k​(z)+G^k​(z)+Ξ·β€²β€²β‹…zn2𝒲k​(z).\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr\leq C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{Q^{\frac{1}{4}}\cdot(j_{k}\cdot z^{n})^{\frac{1-2\alpha}{4}}}{\Gamma^{2}(Q+1)}\cdot\frac{e^{\widehat{F}_{k}(z)+\widehat{G}_{k}(z)+\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}}{\mathcal{W}_{k}(z)}.

Plugging Lemma 5.10 and Proposition 5.6 into the above inequality,

(5.216) 𝒒k​(z)𝒲k​(z)β€‹βˆ«zβˆžπ’Ÿk​(r)​ξk​(r)β‹…rnβˆ’1​𝑑r\displaystyle\frac{\mathcal{G}_{k}(z)}{\mathcal{W}_{k}(z)}\int_{z}^{\infty}\mathcal{D}_{k}(r)\xi_{k}(r)\cdot r^{n-1}dr ≀\displaystyle\leq Cn⋅𝔅kβ‹…jk1nβ‹…eβˆ’Qβ‹…QQ+1Γ⁑(Q+1)β‹…zβ‹…eΞ·β€²β€²β‹…zn2.\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot\frac{j_{k}^{\frac{1}{n}}\cdot e^{-Q}\cdot Q^{Q+1}}{\Gamma(Q+1)}\cdot z\cdot e^{\eta^{\prime\prime}\cdot z^{\frac{n}{2}}}.
≀\displaystyle\leq Cn⋅𝔅kβ‹…jk1nβ‹…eΞ·β‹…zn2\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot j_{k}^{\frac{1}{n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}
≀\displaystyle\leq Cn⋅𝔅kβ‹…(Ξ›k)12​nβ‹…eΞ·β‹…zn2\displaystyle C_{n}\cdot\mathfrak{B}_{k}\cdot(\Lambda_{k})^{\frac{1}{2n}}\cdot e^{\eta\cdot z^{\frac{n}{2}}}

for any η∈(Ξ·β€²β€²,Ξ·β€²β€²+Ξ΄b100)\eta\in(\eta^{\prime\prime},\eta^{\prime\prime}+\frac{\delta_{b}}{100}), where we used Stirling’s formula for estimating Γ⁑(Q+1)\Gamma(Q+1). Similarly we get the bound for the other term of (5.204).

The fourth case is when jkβ‰₯1j_{k}\geq 1 and Q≀1Q\leq 1. This case is simpler and follows from Corollary 5.12.1 and the argument in the second case.

This completes the proof of the proposition.

∎

Based on the above ODE estimate, we prove the following C0C^{0} and C1C^{1} estimate for the equation to the Poisson equation.

Proposition 5.16.

Let {zβ‰₯1}βŠ‚π’žn\{z\geq 1\}\subset\mathcal{C}^{n} be a subset and let K0β‰₯2​n+1K_{0}\geq 2n+1 be a positive integer. Given any Ξ·0∈(βˆ’Ξ΄b/2,Ξ΄b/2)βˆ–{0}\eta_{0}\in(-\delta_{b}/2,\delta_{b}/2)\setminus\{0\}, if v∈C3​K0,Ξ±({zβ‰₯1})v\in C^{3K_{0},\alpha}(\{z\geq 1\}) for and

(5.217) |v|=O⁑(eΞ·0β‹…z​(𝒙)n2),|v|=O(e^{\eta_{0}\cdot z(\bm{x})^{\frac{n}{2}}}),

then the Poisson equation

(5.218) Ξ”gπ’žn​u=v\Delta_{g_{\mathcal{C}^{n}}}u=v

has a solution u∈C3​K0+2,Ξ±({zβ‰₯1})u\in C^{3K_{0}+2,\alpha}(\{z\geq 1\}) such that for any Ξ·>Ξ·0\eta>\eta_{0}

(5.219) |u⁑(𝒙)|+|βˆ‡u​(𝒙)|≀Cβ‹…eΞ·β‹…zn2,|u(\bm{x})|+|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}},

as z⁑(𝐱)β†’+∞z(\bm{x})\to+\infty, where C>0C>0 is independent of π±βˆˆπ’žn\bm{x}\in\mathcal{C}^{n}.

Proof.

The proof is constructive, which will be done in two steps.

The first step, as the main part, is to find a solution uu with the prescribed growth (or decay) rate. We will use the method of separation of variables described as follows.

For a fixed slice Y2​nβˆ’1βŠ‚π’žnY^{2n-1}\subset\mathcal{C}^{n}, let {Ξ›k}k=0∞\{\Lambda_{k}\}_{k=0}^{\infty} with Ξ›0=0\Lambda_{0}=0 be the spectrum of Ξ”π’žn\Delta_{\mathcal{C}^{n}} acting on functions. Let {Ο†k}k=0∞\{\varphi_{k}\}_{k=0}^{\infty} be the eigenfunctions satisfying

(5.220) {βˆ’Ξ”π’žn​φk=Ξ›k​φk,β€–Ο†kβ€–L2​(Y2​nβˆ’1)=1.\displaystyle\begin{cases}-\Delta_{\mathcal{C}^{n}}\varphi_{k}=\Lambda_{k}\varphi_{k},\\ \|\varphi_{k}\|_{L^{2}(Y^{2n-1})}=1.\end{cases}

Given a function vv and for any fixed zβ‰₯1z\geq 1, we have the fiberwise L2L^{2}-expansion on Y2​nβˆ’1Y^{2n-1},

(5.221) v⁑(z,π’š)=βˆ‘k=1∞vk​(z)​φk​(π’š).v(z,\bm{y})=\sum\limits_{k=1}^{\infty}v_{k}(z)\varphi_{k}(\bm{y}).

Then we can first construct a formal solution

(5.222) u⁑(z,π’š)=βˆ‘k=1∞uk​(z)​φk​(π’š)u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\varphi_{k}(\bm{y})

to (5.218), which holds in the L2L^{2}-sense for each fixed zβ‰₯1z\geq 1. Here the coefficient functions uk​(z)u_{k}(z) are the particular solutions constructed in Lemma 5.15. The main part is to prove that the above series u⁑(z,π’š)u(z,\bm{y}) converges with higher regularity and hence u⁑(z,π’š)u(z,\bm{y}) is a regular solution to (5.218).

To begin with, we will prove that the series u⁑(z,π’š)u(z,\bm{y}) converges in the C0C^{0}-norm and hence gives a C0C^{0}-function. Combining Lemma 5.13, Lemma 5.15 and the eigenfunction estimate in Lemma 3.32, we have

(5.223) |u⁑(z,π’š)|β‰€βˆ‘k=1∞|uk​(z)|β‹…|Ο†k​(π’š)|≀Cβ€‹βˆ‘k=1∞eΞ·β‹…zn2(Ξ›k)K0βˆ’n2βˆ’12​n.\displaystyle|u(z,\bm{y})|\leq\sum\limits_{k=1}^{\infty}|u_{k}(z)|\cdot|\varphi_{k}(\bm{y})|\leq C\sum\limits_{k=1}^{\infty}\frac{e^{\eta\cdot z^{\frac{n}{2}}}}{(\Lambda_{k})^{K_{0}-\frac{n}{2}-\frac{1}{2n}}}.

Applying Weyl’s law to the spectrum {Ξ›k}k=1∞\{\Lambda_{k}\}_{k=1}^{\infty},

(5.224) C0βˆ’1​k22​nβˆ’1≀|Ξ›k|≀C0​k22​nβˆ’1,C_{0}^{-1}k^{\frac{2}{2n-1}}\leq|\Lambda_{k}|\leq C_{0}k^{\frac{2}{2n-1}},

where C0>0C_{0}>0 depends only on Y2​nβˆ’1Y^{2n-1} and kk is sufficiently large. Let K0β‰₯2​n+1K_{0}\geq 2n+1, then

(5.225) |u⁑(z,π’š)|≀Cβ‹…eΞ·β‹…zn2β‹…βˆ‘k=1∞1(Ξ›k)3​n2≀Cβ‹…eΞ·β‹…zn2β‹…βˆ‘k=1∞1k3​n2​nβˆ’1≀Cβ‹…eΞ·β‹…zn2.|u(z,\bm{y})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}\cdot\sum\limits_{k=1}^{\infty}\frac{1}{(\Lambda_{k})^{\frac{3n}{2}}}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}\cdot\sum\limits_{k=1}^{\infty}\frac{1}{k^{\frac{3n}{2n-1}}}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

Therefore, u∈C0​(π’žn)u\in C^{0}(\mathcal{C}^{n}) and uu satisfies the C0C^{0}-asymptotic estimate in (5.219).

Based on the above C0C^{0}-regularity, we will apply the standard elliptic regularity on (π’žn,gπ’žn)(\mathcal{C}^{n},g_{\mathcal{C}^{n}}) to show that u∈C2​(π’žn)u\in C^{2}(\mathcal{C}^{n}) is a regular solution to Ξ”gπ’žn​u=v\Delta_{g_{\mathcal{C}^{n}}}u=v. We take the partial sums

(5.226) UN​(z,π’š)β‰‘βˆ‘k=1Nuk​(z)​φk​(π’š),VN​(z,π’š)β‰‘βˆ‘k=1Nvk​(z)​φk​(π’š)\displaystyle U_{N}(z,\bm{y})\equiv\sum\limits_{k=1}^{N}u_{k}(z)\varphi_{k}(\bm{y}),\ V_{N}(z,\bm{y})\equiv\sum\limits_{k=1}^{N}v_{k}(z)\varphi_{k}(\bm{y})

of the expansions

(5.227) u⁑(z,π’š)=βˆ‘k=1∞uk​(z)​φk​(π’š),v⁑(z,π’š)=βˆ‘k=1∞vk​(z)​φk​(π’š).\displaystyle u(z,\bm{y})=\sum\limits_{k=1}^{\infty}u_{k}(z)\varphi_{k}(\bm{y}),\ v(z,\bm{y})=\sum\limits_{k=1}^{\infty}v_{k}(z)\varphi_{k}(\bm{y}).

It is obvious that,

(5.228) Ξ”gπ’žn​UN=VN.\Delta_{g_{\mathcal{C}^{n}}}U_{N}=V_{N}.

For every 𝒙≑(z,π’š)βˆˆπ’žn\bm{x}\equiv(z,\bm{y})\in\mathcal{C}^{n}, we will apply the elliptic regularity on the ball B2​(𝒙)βŠ‚π’žnB_{2}(\bm{x})\subset\mathcal{C}^{n} to obtain the higher regularity of uu.

As a starter, by the same arguments as the above, we have β€–VNβˆ’vβ€–C0​(B2​(𝒙))β†’0\|V_{N}-v\|_{C^{0}(B_{2}(\bm{x}))}\to 0 as Nβ†’βˆžN\to\infty. The proof of the higher order convergence is almost verbatim. In fact, we just need to use β€–vβ€–C2​K0+m\|v\|_{C^{2K_{0}+m}} with m≀K0m\leq K_{0}. Since Ξ”gπ’žβ€‹UN=VN\Delta_{g_{\mathcal{C}}}U_{N}=V_{N}, the standard W2,pW^{2,p}- implies that regularity for every 1<p<∞1<p<\infty,

(5.229) β€–UNβ€–W2,p​(B1​(𝒙))≀Cp,𝒙⋅(β€–VNβ€–C0​(B2​(𝒙))+(β€–UNβ€–C0​(B2​(𝒙)))CLOSE.\|U_{N}\|_{W^{2,p}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}}\cdot(\|V_{N}\|_{C^{0}(B_{2}(\bm{x}))}+(\|U_{N}\|_{C^{0}(B_{2}(\bm{x}))}).

By assumption v∈C3​K0​(π’žn)v\in C^{3K_{0}}(\mathcal{C}^{n}) for K0β‰₯2​n+1K_{0}\geq 2n+1, so it follows that β€–VNβ€–C2​(B2​(𝒙))≀C𝒙\|V_{N}\|_{C^{2}(B_{2}(\bm{x}))}\leq C_{\bm{x}}. Therefore, for every 1<p<∞1<p<\infty,

(5.230) β€–UNβ€–W4,p​(B1​(𝒙))≀Cp,𝒙​(β€–UNβ€–W2,p​(B3/2​(𝒙))+β€–VNβ€–W2,p​(B2​(𝒙)))≀Cp,𝒙.\|U_{N}\|_{W^{4,p}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}}(\|U_{N}\|_{W^{2,p}(B_{3/2}(\bm{x}))}+\|V_{N}\|_{W^{2,p}(B_{2}(\bm{x}))})\leq C_{p,\bm{x}}.

Now it suffices to choose p>2​np>2n, so the Sobolev embedding implies

(5.231) β€–UNβ€–C3,α​(B1​(𝒙))≀Cp,𝒙,α≑1βˆ’2​np,\|U_{N}\|_{C^{3,\alpha}(B_{1}(\bm{x}))}\leq C_{p,\bm{x}},\ \alpha\equiv 1-\frac{2n}{p},

which implies that UNβ†’uU_{N}\to u in the C3C^{3}-norm with respect to gπ’žng_{\mathcal{C}^{n}}. The proof of the first step is done.

We have constructed a solution uu satisfying |u⁑(𝒙)|≀Cβ‹…eΞ·β‹…zn2|u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}. Now we are ready to show that

(5.232) |βˆ‡u​(𝒙)|≀Cβ‹…eΞ·β‹…zn2.|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

This can be accomplished by the elliptic W2,pW^{2,p}-estimate. Since a Calabi space (π’žn,gπ’žn)(\mathcal{C}^{n},g_{\mathcal{C}^{n}}) is collapsed with bounded curvatures as zβ†’+∞z\to+\infty, so there is some constant r0>0r_{0}>0 such that for each π’™βˆˆπ’žn\bm{x}\in\mathcal{C}^{n} satisfying z⁑(𝒙)β‰₯1z(\bm{x})\geq 1, the universal cover (B2​r0​(𝒙)~,𝒙~)(\widetilde{B_{2r_{0}}(\bm{x})},\tilde{\bm{x}}) is non-collapsing. Now we lift the solution uu to this non-collapsing local universal cover, then for any p>1p>1, there exists Cp>0C_{p}>0 such that

(5.233) |u|W2,p​(Br0​(𝒙~))≀Cpβ‹…(|u|Lβˆžβ€‹(B2​r0​(𝒙~))+|​v|Lβˆžβ€‹(B2​r0​(𝒙~)))≀Cpβ‹…eΞ·β‹…zn2.|u|_{W^{2,p}(B_{r_{0}}(\tilde{\bm{x}}))}\leq C_{p}\cdot(|u|_{L^{\infty}(B_{2r_{0}}(\tilde{\bm{x}}))}+|v|_{L^{\infty}(B_{2r_{0}}(\tilde{\bm{x}}))})\leq C_{p}\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

We can choose any p>2​np>2n, then Sobolev embedding gives

(5.234) |u|C1,α​(Br0​(𝒙~))≀Cβ‹…eΞ·β‹…zn2.|u|_{C^{1,\alpha}(B_{r_{0}}(\tilde{\bm{x}}))}\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}}.

In particular,

(5.235) |βˆ‡u​(𝒙)|≀Cβ‹…eΞ·β‹…zn2,|\nabla u(\bm{x})|\leq C\cdot e^{\eta\cdot z^{\frac{n}{2}}},

where α≑1βˆ’2​np\alpha\equiv 1-\frac{2n}{p}. So the proof of the proposition is done.

∎

Original mathematics by the credited authors. Source-backed reader collection; mathematical self-containment is not assessed.