### 数学代写|黎曼几何代写Riemannian geometry代考|МАТН6205

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## 数学代写|黎曼几何代写Riemannian geometry代考|The First Dirichlet Eigenvalue Comparison Theorem

Following standard notations and setting (see, e.g., [Cha1] or in this context the seminal survey by Grigoryan in [Gri1]), for any precompact open set $\Omega$ in a Riemannian manifold $M$ we denote by $\lambda(\Omega)$ the smallest number $\lambda$ for which the following Dirichlet eigenvalue problem has a non-zero solution
\left{\begin{aligned} \Delta u+\lambda u &=0 \text { at all points } x \text { in } \Omega \ u(x) &=0 \text { at all points } x \text { in } \partial \Omega \end{aligned}\right.
We shall need the following beautiful observation due to Barta:

Theorem $7.1$ ([B], [Cha1]). Consider any smooth function $f$ on a domain $\Omega$ which satisfies $f_{\left.\right|{\Omega}}>0$ and $f{\mid \text {an }}=0$, and let $\lambda(\Omega)$ denote the first eigenvalue of the Dirichlet problem for $\Omega$. Then
$$\inf {\Omega}\left(\frac{\Delta f}{f}\right) \leq-\lambda(\Omega) \leq \sup {\Omega}\left(\frac{\Delta f}{f}\right)$$
If equality occurs in one of the inequalities, then they are both equalities, and $f$ is an eigenfunction for $\Omega$ corresponding to the eigenvalue $\lambda(\Omega)$.
Proof. Let $\phi$ be an eigenfunction for $\Omega$ corresponding to $\lambda(\Omega)$.
Then $\phi_{\Omega}>0$ and $\phi_{\left.\right|{\Omega}}=0$. If we let $h$ denote the difference $h=\phi-f$, then \begin{aligned} -\lambda(\Omega)=\frac{\Delta \phi}{\phi} &=\frac{\Delta f}{f}+\frac{f \Delta h-h \Delta f}{f(f+h)} \ &=\inf {\Omega}\left(\frac{\Delta f}{f}\right)+\sup {\Omega}\left(\frac{f \Delta h-h \Delta f}{f(f+h)}\right) \ &=\sup {\Omega}\left(\frac{\Delta f}{f}\right)+\inf {\Omega}\left(\frac{f \Delta h-h \Delta f}{f(f+h)}\right) \end{aligned} Here the supremum, $\sup {\Omega}\left(\frac{f \Delta h-h \Delta f}{f(f+h)}\right)$ is necessarily positive since
$$\left.f(f+h)\right|{\Omega}>0$$ and since by Green’s second formula $(6.8)$ in Theorem $6.4$ we have $$\int{\Omega}(f \Delta h-h \Delta f) d V=0 \text {. }$$
For the same reason, the infimum, $\inf _{\Omega}\left(\frac{f \Delta h-h \Delta f}{f(f+h)}\right)$ is necessarily negative. This gives the first part of the theorem. If equality occurs, then $(f \Delta h-h \Delta f)$ must vanish identically on $\Omega$, so that $-\lambda(\Omega)=\frac{\Delta f}{f}$, which gives the last part of the statement.

As already alluded to in the introduction, the key heuristic message of this report is that the Laplacian is a particularly ‘swift actor’ on minimal submanifolds (i.e., minimal extrinsic regular $R$-balls $D_{R}$ ) in ambient spaces with an upper bound $b$ on its sectional curvatures. This is to be understood in comparison with the ‘action’ of the Laplacian on totally geodesic $R$-balls $B_{R}^{b, m}$ in spaces of constant curvature b. In this section we will use Barta’s theorem to show that this phenomenon can indeed be ‘heard’ by ‘listening’ to the bass note of the Dirichlet spectrum of any given $D_{R}$.

## 数学代写|黎曼几何代写Riemannian geometry代考|Isoperimetric Relations

In this and the following two sections we survey some comparison results concerning inequalities of isoperimetric type, mean exit times and capacities, respectively, for extrinsic minimal balls in ambient spaces with an upper bound on sectional curvature. This has been developed in a series of papers, see [Pa] and [MaP1][MaP4].

We will still assume a standard situation as in the previous section, i.e., $D_{R}$ denotes an extrinsic minimal ball of a minimal submanifold $P$ in an ambient space $N$ with the upper bound $b$ on the sectional curvatures.

Proposition 8.1. We define the following function of $t \in \mathbb{R}{+} \cup{0}$ for every $b \in \mathbb{R}$, for every $q \in \mathbb{R}$, and for every dimension $m \geq 2$ : $$L{q}^{b, m}(t)=q\left(\frac{\operatorname{Vol}\left(S_{t}^{b, m-1}\right)}{m h_{b}(t)}-\operatorname{Vol}\left(B_{t}^{b, m}\right)\right)$$
Then
$$L_{q}^{b, m}(0)=0 \text { for all } b, q, \text { and } m$$
and
$$\operatorname{sign}\left(\frac{d}{d t} L_{q}^{b, m}(t)\right)=\operatorname{sign}(b q) \text { for all } b, q, m, \text { and } t>0 \text {. }$$
Proof. This follows from a direct computation using the definition of $h_{b}(t)$ from equation (3.5) together with the volume formulae (cf. [Gr])
\begin{aligned} \operatorname{Vol}\left(B_{t}^{b, m}\right) &=\operatorname{Vol}\left(S_{1}^{0, m-1}\right) \cdot \int_{0}^{t}\left(Q_{b}(u)\right)^{m-1} d u \ \operatorname{Vol}\left(S_{t}^{b, m-1}\right) &=\operatorname{Vol}\left(S_{1}^{0, m-1}\right) \cdot\left(Q_{b}(t)\right)^{m-1} \end{aligned}

## 数学代写|黎曼几何代写Riemannian geometry代考|A Consequence of the Co-area Formula

The co-area equation (6.4) applied to our setting gives the following
Proposition 9.1. Let $D_{R}(p)$ denote a regular extrinsic minimal ball of $P$ with center $p$ in $N$. Then
$$\frac{d}{d u} \operatorname{Vol}\left(D_{u}\right) \geq \operatorname{Vol}\left(\partial D_{u}\right) \text { for all } u \leq R$$

Proof. We let $f: \bar{D}{R} \rightarrow \mathbb{R}$ denote the function $f(x)=R-r(x)$, which clearly vanishes on the boundary of $D{R}$ and is smooth except at $p$. Following the notation of the co-area formula we further let
\begin{aligned} \Omega(t) &=D_{(R-t)} \ V(t) &=\operatorname{Vol}\left(D_{(R-t)}\right) \text { and } \ \Sigma(t) &=\partial D_{(R-t)} \end{aligned}
Then
\begin{aligned} \operatorname{Vol}\left(D_{u}\right) &=V(R-u) \text { so that } \ \frac{d}{d u} \operatorname{Vol}\left(D_{u}\right) &=-V^{\prime}(t){\left.\right|{i=n-u}} . \end{aligned}
The co-area equation (6.4) now gives
\begin{aligned} -V^{\prime}(t) &=\int_{\partial D_{(R-t)}}\left|\nabla^{P} r\right|^{-1} d A \ & \geq \operatorname{Vol}\left(\partial D_{(R-t)}\right) \ &=\operatorname{Vol}\left(\partial D_{u}\right) \end{aligned}
and this proves the statement.
Exercise 9.2. Explain why the non-smoothness of the function $f$ at $p$ does not create problems for the application of equation (6.4) in this proof although smoothness is one of the assumptions in Theorem 6.1.

## 数学代写黎曼几何代写Riemannian geometry代 考|lsoperimetric Relations

$$L q^{b, m}(t)=q\left(\frac{\operatorname{Vol}\left(S_{t}^{b, m-1}\right)}{m h_{b}(t)}-\operatorname{Vol}\left(B_{t}^{b, m}\right)\right)$$

$$L_{q}^{b, m}(0)=0 \text { for all } b, q, \text { and } m$$

$$\operatorname{sign}\left(\frac{d}{d t} L_{q}^{b, m}(t)\right)=\operatorname{sign}(b q) \text { for all } b, q, m, \text { and } t>0$$

$$\operatorname{Vol}\left(B_{t}^{b, m}\right)=\operatorname{Vol}\left(S_{1}^{0, m-1}\right) \cdot \int_{0}^{t}\left(Q_{b}(u)\right)^{m-1} d u \operatorname{Vol}\left(S_{t}^{b, m-1}\right)=\operatorname{Vol}\left(S_{1}^{0, m-1}\right) \cdot\left(Q_{b}(t)\right)^{m-1}$$

## 数学代写黎曼几何代写Riemannian geometry代考|A Consequence of the Co-area Formula

$$\frac{d}{d u} \operatorname{Vol}\left(D_{u}\right) \geq \operatorname{Vol}\left(\partial D_{u}\right) \text { for all } u \leq R$$

$$\Omega(t)=D_{(R-t)} V(t)=\operatorname{Vol}\left(D_{(R-t)}\right) \text { and } \Sigma(t)=\partial D_{(R-t)}$$

$$\operatorname{Vol}\left(D_{u}\right)=V(R-u) \text { so that } \frac{d}{d u} \operatorname{Vol}\left(D_{u}\right) \quad=-V^{\prime}(t) \mid i=n-u$$

$$-V^{\prime}(t)=\int_{\partial D_{(R-t)}}\left|\nabla^{P} r\right|^{-1} d A \geq \operatorname{Vol}\left(\partial D_{(R-t)}\right)=\operatorname{Vol}\left(\partial D_{u}\right)$$

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