Ngô Quốc Anh

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.

(more…)

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.

(more…)

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.

May 2, 2008

Strong Maximum Principle Revisited

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 3:24

Suppose satisfies . If we knew that u \in C^2(\Omega) \cap C^1(\overline\Omega), then we would deduce from the strong maximum principle that there exists an s.t.

u(x) \ge \epsilon \, \mathrm{dist}(x,\partial\Omega) \quad \text{ for any } x \in \Omega.

But assume we only know that . Is the above inequality still true?

Of course not, if we want to use and then unfortunately, the function

u(x) = \exp\left(-\frac 1{1-|x|}\right)

doesn’t work because the inequality is not satisfied:

-u''(x) = u(x) (1-|x|)^{-4} (1-2|x|).

Nevertheless, if , then the function

u(x) = (1-|x|)^s - (1-|x|)^r, \quad \frac 12 < s < 1 < r,

doesn’t satisfy the property that “ for some constant ” while

-\Delta u = \left( s(1-s) + r(r-1)(1-|x|)^{r-s} \right) (1-|x|)^{s-2} + 2\gamma(r-s)\delta \ge 0.

Here is the Dirac measure.

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