标签：geq frac maximum principle 原理 极值 partial lambda

Maximum principle in the book geometric analysis

Throughout the book, there is frequent use of the maximum principle, but sometimes it doesn’t seem directly applicable, and the book does not provide an explanation. This troubled me for some time, but now I'm trying to provide something missed.

Let me take an example to explain it.

Corollary 5.8
Let \(M\) be a compact \(m\)-dimensional Riemannian manifold whose boundary is convex in the sense that the second fundamental form is nonnegative with respect to the outward pointing normal vector. Suppose that the Ricci curvature of \(M\) is bounded from below by

\[R_{ij} \geq -(m - 1)R \]

for some constant \(R \geq 0\), and \(d\) denotes the diameter of \(M\). Then there exist constants \(C_1(m)\), \(C_2(m) > 0\) depending on \(m\) alone, such that the first nonzero Neumann eigenvalue of \(M\) satisfies

\[\lambda_1 \geq \frac{C_1}{d^2} \exp(-C_2 d \sqrt{R}). \]

Proof. In view of the proof of Theorem 5.7, it suffices to show that the maximum value for the functional \(Q\) does not occur on the boundary of \(M\). Supposing the contrary that the maximum point for \(Q\) is \(x_0 \in \partial M\), let us denote the outward pointing unit normal vector by \(e_m\), and assume that \(\{e_1, \ldots, e_{m-1}\}\) are orthonormal tangent vectors to \(\partial M\). Since \(Q\) satisfies the differential inequality (5.15), the strong maximum principle implies that

\[e_m(Q)(x_0) > 0. \]

...

In the proof, the strong maximum principle was applied. So let us recall

Lemma (cf. Gilbarg and Trudinger)
Suppose that \(L\) is uniformly elliptic, \(c=0\) and \(Lu \geq 0\) in \(\Omega\). Let \(x_0 \in \partial\Omega\) be such that

• \(u\) is continuous at \(x_0\);
• \(u(x_0)>u(x)\) for all \(x \in \Omega\);
• \(\partial\Omega\) satisfies an interior sphere condition at \(x_0\).

Then the outer normal derivative of \(u\) at \(x_0\), if it exists, satisfies the strict inequality $$\begin{equation}
\frac{\partial u}{\partial \nu} (x_0)>0. \end{equation}$$

If \(c \leq 0\) and \(c/\lambda\) is bounded, the same conclusion holds provided \(u(x_0) \geq 0\), and if \(u(x_0) = 0\) the same conclusion holds irrespective the sign of \(c.\)

Now Q satisfies the differential inequality (5.15), that is

\[\begin{align} \Delta Q \geq & \frac{m}{2(m-1)}|\nabla Q|^2 Q^{-1}+\langle\nabla v, \nabla Q\rangle Q^{-1}\left(\frac{2 \lambda u}{(m-1)(a+u)}-\frac{2(m-2)}{m-1} Q\right) \\ & +\left(\frac{2}{m-1} Q +\frac{4}{m-1} \frac{\lambda u}{a+u} -\frac{2 \lambda a}{a+u} -2(m-1) R \right)Q\\ & +\frac{2}{m-1}\left(\frac{\lambda u}{a+u}\right)^2. \end{align} \]

We need to justify that the maximum principle is applicable.

If

\[\frac{2}{m-1} Q +\frac{4}{m-1} \frac{\lambda u}{a+u} -\frac{2 \lambda a}{a+u} -2(m-1) R \leq 0, \]

then we have

\[Q\leq C_1(m)\lambda+C_2(m)R \]

for some constants denpending on \(m\) alone, which allow us apply the same argument in the proof of Thm 5.7.

So, w.l.o.g., we may assume

\[\frac{2}{m-1} Q +\frac{4}{m-1} \frac{\lambda u}{a+u} -\frac{2 \lambda a}{a+u} -2(m-1) R > 0, \]

near the maximum point \(x_0\).

But now we can "almost" apply the strong maximum principle, since the zero order term is positive.

However, the zero order term involves \(Q\) itself which make it slightly different from the usual linear elliptic equations. I checked some PDE textbooks, from weak maximum principle to strong maximum principle, it seems that they also work for equation (4).

In fact, we don't need any fancier maximum principle!^[1]

Since now we have

\[\begin{aligned} \Delta Q \geq & \frac{m}{2(m-1)}|\nabla Q|^2 Q^{-1}+\langle\nabla v, \nabla Q\rangle Q^{-1}\left(\frac{2 \lambda u}{(m-1)(a+u)}-\frac{2(m-2)}{m-1} Q\right) \\ & +\frac{2}{m-1}\left(\frac{\lambda u}{a+u}\right)^2, \end{aligned} \]

which doesn't involved any zero terms!

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1. Thanks to my colleague Adam's remainder. ↩︎

标签：geq,frac,maximum,principle,原理,极值,partial,lambda
From： https://www.cnblogs.com/crossLH/p/18187936