ScalingStacks

Proof of Proposition 5.5 . [03IG]

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Proof of Proposition 5.5.

We denote

ψ=d​u2+J𝒞​d​v2=O⁡(eC​δh​z),\psi=du_{2}+J_{\mathcal{C}}dv_{2}=O(e^{C\delta_{h}z}),

Then we have the following expansion as in Section 4.1: let {Λk}k=1\{\Lambda_{k}\}_{k=1} be the spectrum of Y3Y^{3} and {φk}k=1∞\{\varphi_{k}\}_{k=1}^{\infty} are the corresponding eigenfunctions on YY with ℒ∂θ​φk=−1​jk​φk\mathcal{L}_{\partial_{\theta}}\varphi_{k}=\sqrt{-1}j_{k}\varphi_{k},

(5.30) u2=∑kfk​(z)​φk​(zα,θ),v2=∑kgk​(z)​φk​(zα,θ),u_{2}=\sum\limits_{k}f_{k}(z)\varphi_{k}(z_{\alpha},\theta),\ v_{2}=\sum\limits_{k}g_{k}(z)\varphi_{k}(z_{\alpha},\theta),

which implies that

(5.31) d​u2=∑k(fk′​(z)⋅φk⋅d​z+fk​(z)⋅d​φk),d​v2=∑k(gk′​(z)​φk⋅d​z+gk​(z)⋅d​φk).du_{2}=\sum\limits_{k}(f_{k}^{\prime}(z)\cdot\varphi_{k}\cdot dz+f_{k}(z)\cdot d\varphi_{k}),\ dv_{2}=\sum\limits_{k}(g_{k}^{\prime}(z)\varphi_{k}\cdot dz+g_{k}(z)\cdot d\varphi_{k}).

On 𝒞\mathcal{C} by the definition in Section 4.1, we have J𝒞​(z​d​z)=d​θJ_{\mathcal{C}}(zdz)=d\theta, so we have

(5.32) ∑k(fk′​(z)−−1​jk⋅z⋅gk​(z))​φk=ψ(∂z)∑k(z−1⋅gk′​(z)+−1​jk⋅fk​(z))​φk=ψ(∂θ).\displaystyle\begin{split}\sum\limits_{k}\big(f_{k}^{\prime}(z)-\sqrt{-1}j_{k}\cdot z\cdot g_{k}(z)\big)\varphi_{k}&=\psi(\partial_{z})\\ \sum\limits_{k}\big(z^{-1}\cdot g_{k}^{\prime}(z)+\sqrt{-1}j_{k}\cdot f_{k}(z)\big)\varphi_{k}&=\psi(\partial_{\theta}).\end{split}

This implies that for each kk,

(5.33) fk′​(z)−−1​jk⋅z⋅gk​(z)=O⁡(eC​δh​z)z−1⋅gk′​(z)+−1​jk⋅fk​(z)=O⁡(eC​δh​z).\begin{split}f_{k}^{\prime}(z)-\sqrt{-1}j_{k}\cdot z\cdot g_{k}(z)=O(e^{C\delta_{h}z})\\ z^{-1}\cdot g_{k}^{\prime}(z)+\sqrt{-1}j_{k}\cdot f_{k}(z)=O(e^{C\delta_{h}z}).\end{split}

There are three different cases.

If jk≠0j_{k}\neq 0, then we can write fkf_{k} and gkg_{k} are given by a linear combination of one growing solution ℱk\mathcal{F}_{k} and one decaying solution 𝒰k\mathcal{U}_{k}. Using the analysis in Section 4 we know that the asymptotic order of ℱk\mathcal{F}_{k} is ejk​z22e^{\frac{j_{k}z^{2}}{2}} (see Lemma 4.7). The control (5.28) then implies both fkf_{k} and gkg_{k} can only be a multiple of the decaying solution 𝒰k=O⁡(e−jk​z22)\mathcal{U}_{k}=O(e^{-\frac{j_{k}z^{2}}{2}}).

If jk=0j_{k}=0 and λk≠0\lambda_{k}\neq 0, then fkf_{k} and gkg_{k} are given by linear combinations of the exponential functions of the form eλk​ze^{\sqrt{\lambda_{k}}z} and e−λk​ze^{-\sqrt{\lambda_{k}}z}. Let δ¯>0\underline{\delta}>0 be the positive constant given in Proposition 4.10, we use (5.33) and the fact that ψ=O⁡(eC​δh​z)\psi=O(e^{C\delta_{h}z}) to conclude that, if C​δh<δ¯C\delta_{h}<\underline{\delta}, then both fkf_{k} and gkg_{k} must be proportional to the decaying solutions.

If jk=0j_{k}=0 and φk\varphi_{k} is constant, then fkf_{k} and gkg_{k} are linear functions in zz. Now since u2u_{2} and v2v_{2} are harmonic functions on 𝒞\mathcal{C}, by Lemma 4.11, we conclude that

(5.34) |u2|=O⁡(z),|v2|=O⁡(z).|u_{2}|=O(z),\ |v_{2}|=O(z).

This completes the proof of Proposition 5.5.∎

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