# Ngô Quốc Anh

## July 21, 2014

### A strong maximum principle due to Tolksdorf and application to the prescribed scalar curvature equation

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 21:08

In this note, we consider a strong maximum principle due to Tolksdorf and application to the prescribed scalar curvature equation. In the context of closed and compact Riemannian manifolds $(M,g)$ of dimension $n$, it can be stated as follows

Theorem (Maximum Principle). Let $(M,g)$ be a compact Riemannian manifold of dimension $n$. Let $u \in C^1(M)$ be such that

$\displaystyle -\Delta_g u \geqslant f(\cdot, u)$

on $M$ where $f$ is a function such that

$\displaystyle \partial_u f \leqslant 0$

(i.e. $f$ is monotone increasing w.r.t the variable $u$) and that

$\displaystyle |f(x,r)| \leqslant C(K+|r|)|r|$

for all $(x,r) \in M \times \mathbb R$ and for some constant $C>0$. If $u \geqslant 0$ in $M$ and $u$ does not vanish identically, then $u>0$ in $M$.

To prove this theorem, we need two auxiliary results. The first result is the weak comparison principle which can be stated as the following.

## March 12, 2013

### The generalized maximum principle

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 9:09

Let us start our notes with a very fundamental maximum principle for any strongly second order elliptic operator. We have

Theorem (Maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle L[u] = \sum\limits_{i,j = 1}^n {{a_{ij}}(x)\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}}} + \sum\limits_{k = 1}^n {{b_k}(x)\frac{{\partial u}}{{\partial {x_k}}}} \geqslant 0$

in a domain $D$ where $L$ is uniformly elliptic. Suppose the coefficients $a_{ij}$ and $b_k$ are uniformly bounded. If $u$ attains a maximum $M$ at a point of $D$, then $u = M$ in $D$.

In order to memorize the above result, let us think about the parabola $y=x^2$ with $x\in[-1,1]$ and $L=\Delta$. In this one-dimentional case, $L[y]=2 \geqslant 0$ which confirms that $y$ only achieves its maximum at $x=\pm 1$.

As you may know the operator $L$ is only assumed to be strongly elliptic which only effects the coefficients $a_{ij}$. Regarding to the coefficients $b_k$, we only assume these are uniformly bounded. However, the uniform ellipticity of the operator L and the boundedness of the coefficients are not really essential as you can check in the proof. Besides, the domain $D$ need not be bounded in this version.

Now for operators of the form $(L + h)$, we still have a result analogous to the above.

Theorem (Maximum principle). Let $u$ satisfy the differential inequality

$\displaystyle (L + h)[u] >0$

with $h <0$, with $L$ uniformly elliptic in $D$, and with the coefficients of $L$ and $h$ bounded. If $u$ attains a non-negative maximum $M$ at an interior point of $D$, then $u = M$.

Clearly, the assumption $h<0$ and $M \geqslant 0$ are crucial as we may face some difficulty as raised in this note. Counterexamples are easily obtained if $h > O$. For example, the function $u = \exp(-r^2)$ has an absolute maximum at $r= 0$.

## February 6, 2012

### A note on the maximum principle

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 18:18

Today, let us discuss a very interesting stuff. Let say $M$ is a compact manifold without boundary of dimention $n \geqslant 3$.

On $M$, we consider the following simple PDE

$\displaystyle -\Delta u + hu = a$

where $h$ and $a$ are smooth functions with $h>0$ and $a \geqslant 0$. Since $h>0$, it is well-known that the operator $-\Delta +h$ is coercive, see here. A standard variation method tells us that there exists a weak solution $u$ to the above PDE. By regularity theorem, $u$ is at least a $C^2$ function, thus, a strong solution (in the classical sense).

Next, we claim that $u \geqslant 0$. To this purpose, assume that the solution $u$ achieves its minimum at some point $x_0 \in M$. In particular, there holds

$-\Delta u (x_0) \leqslant 0$.

This, together with the fact that $h>0$ and $a \geqslant 0$, implies that $u(x_0) \geqslant 0$. Thus, we have shown that $u \geqslant 0$ in $M$.

Once we have the non-negativity of $u$, in view of the strong maximum principle, either $u \equiv 0$ in $M$ or $u>0$. In other words, the solution $u$ cannot achieves its minimum inside the manifold. Since the manifold has no boundary, it is natural to think that the solution $u$ cannot achieve its minimum although $u>0$. This is clear a contradiction to the fact that the manifold is compact and $u$ is of class $C^2(M)$.

So something went wrong but what and why?

In fact, we have made a small mistake. In view of the strong maximum principle, we can only claim that the solution can only achieve its non-positive minimum value on the boundary. Therefore, there are cases so that $u$ may achieve its positive minimum inside $M$. Thus, there is no contradiction here.

In order to see this, let us go back to a proof of the strong maximum principle. Roughly speaking, it starts with the following simple one.

Lemma 1. If $-\Delta u +hu >0$ at any point in $M$, then $u$ cannot have non-positive minimum value in $M$.

Proof. The proof is standard. Assume that at $x_0 \in M$, the function $u$ realizes its minimum, besides, $u(x_0) \leqslant 0$. In particular, $-\Delta u(x_0) \leqslant 0$ and $h(x_0)u(x_0) \leqslant 0$. These force $-\Delta u + hu \leqslant 0$ at $x_0$. A contradiction.

Form the proof above, if $u(x_0)>0$, we cannot get any contradiction. This is why we can claim either $u \equiv 0$ or $u>0$ since $u$ can achieve its positive minimum in $M$.

Notice that, if we don’t have any $h$ (in the operator), the non-positivity can be dropped.

## January 15, 2011

### The Payne’s Maximum Principles

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 15:34

This note, completely based on the elegant paper of L.E. Payne [here], deals primarily with maximum principles for solutions of second order and fourth order elliptic equations. However, some of the results hold for arbitrary sufficiently smooth functions.

Throughout we assume $D$ to be a bounded domain in $\mathbb R^n$ with sufficiently smooth boundary $\partial D$ so that when necessary the governing differential equation will be satisfied on the boundary. In some of the applications we will need $\partial D$ to be a $C^{4+\alpha}$ surface but in most cases this excessive differentiality can be dispensed with.

We shall adopt the summation convention in which repeated Latin indices are to be summed from $1$ to $n$, and we shall use the comma to denote differentiation. The symbol $\partial/\partial \nu$ will be used for the normal derivative directed outward from $D$ on $\partial D$.

Following is the results

1. Inequalities based on the geometry of $D$

We start with the maximum value of the gradient of a function whose normal derivative vanishes on a portion of $\partial D$.

Theorem I. Let $u \in C^2(\overline D)$ have vanishing normal derivative on a portion $\Gamma$ of $\partial D$. Then if the Gaussian curvature o$\Gamma$ is everywhere positive the maximum value of $|{\rm grad} u|^2$ can occur on $\Gamma$ if and only if $u\equiv {\rm constant}$ in $D$.

We also have the following result

Theorem II. Let $u \in C^2(\overline D)$ vanish on a portion $\Gamma_1$ of $\partial D$. Then if the average curvature $K$ is positive at every point of $\Gamma_1$ the maximum value of

$|{\rm grad} u|^2 -2u\Delta u$

cannot occur on $\Gamma_1$ if $u\not \equiv 0$ in $D$.

## August 6, 2010

### An upper bound for solutions via the maximum principle

Filed under: PDEs — Tags: , — Ngô Quốc Anh @ 1:29

It is known that [here] the following PDE

$\displaystyle\begin{cases}\Delta u =e^u, & {\rm in } \, \mathbb R^2,\\\displaystyle\int_{\mathbb R^2}e^u<\infty.\end{cases}$

has no $C^2$ solution. However, this is no longer true if we replace the whole space by a ball of radius $R$, say $B_R(0)$. In this entry, we show that if $u \in C^2(\overline B_R)$ is a solution of

$\Delta u=e^u, \quad {\rm in }\; B_R$

then

$\displaystyle u(0) \leqslant \log 8- 2\log R$.

To this purpose, let us recall the following

The Maximum Principle. Let assume $U\subset \mathbb R^2$ be open and bounded. We consider an elliptic operator $L$ of the form

$\displaystyle Lu = - \sum\limits_{i,j = 1}^2 {{a^{ij}}\frac{{{\partial ^2}u}}{{\partial {x_i}\partial {x_j}}} + \sum\limits_{k = 1}^2 {{b^k}\frac{{\partial u}}{{\partial {x_k}}}} } + cu$

where coefficients are continuous and the standard uniform ellipticity condition holds.

## January 9, 2010

### The maximum principles

Filed under: Nghiên Cứu Khoa Học, PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:02

The maximum principle, in its various forms, is one of the most useful tools for working with elliptic partial differential equations. We gather here a few versions which we use elsewhere in this blog. These versions come from a paper due to James Isenberg published in Class. Quantum Grav. in 1995. In all cases, $\nabla^2$ is the Laplacian (with negative eigenvalues) on a Riemannian manifold $\Sigma^3$ (closed=compact without boundary, or open with boundary, as indicated), and $\psi$ is a real-valued $C^2$ function on $\Sigma^3$.

Version 1. ($\Sigma^3$ closed)

If $\psi$ satisfies either

$\displaystyle\nabla^2 \psi \leq 0, \qquad \nabla^2 \psi=0,$

or

$\displaystyle\nabla^2 \psi \geq 0$

then $\psi$ must be constant.

Hence there is no solution to the equation $\nabla^2 \psi=F(x,\psi)$ with $F(x,\psi) \geq 0$ or $F(x,\psi) \leq 0$ unless in fact $F(x,\psi) =0$.

Version 2. ($\Sigma^3$ closed)

Let $f: \Sigma^3 \to \mathbb R$ be a positive definite ($f(x)>0$) function.

• If $\psi$ satisfies

$\displaystyle\nabla^2\psi+f\psi \geq 0$

then $\psi(x) \geq 0$.

• If $\psi$ satisfies

$\displaystyle\nabla^2\psi+f\psi \geq 0$ (not identically zero)

then $\psi(x) > 0$.

• If $\psi$ satisfies

$\displaystyle\nabla^2\psi+f\psi \leq 0$

then $\psi(x) \leq 0$.

• If $\psi$ satisfies

$\displaystyle\nabla^2\psi+f\psi \leq 0$ (not identically zero)

then $\psi(x) <0$.

Version 3. ($\Sigma^3$ closed)

Let $\mu$ and $\kappa$ be positive constants. If $\psi$ satisfies

$\displaystyle -\nabla^2\psi + \mu\psi \leq \kappa$

then

$\displaystyle \psi(x) \leq \frac{\kappa}{\mu}$.

Version 4. ($\Sigma^3$ open with boundary)

Let $\mu$ be a positive constant, and let $\rho : \partial\Sigma^3 \to \mathbb R$ be a positive definite function. If $\psi$ satisfies either

$\displaystyle -\nabla^2 \psi +\mu\psi=0$ on $\Sigma^3$

and

$\displaystyle\psi = \rho$ on $\partial\Sigma^3$

then $\psi>0$.

We prefer the reader to the book entitled “Maximum Principles in Differential Equations” due to Protter and Weinberger for further discussion and proofs.