Ngô Quốc Anh

August 7, 2012

Almost-Schur lemma

Filed under: Nghiên Cứu Khoa Học, Riemannian geometry — Tags: — Ngô Quốc Anh @ 16:23

Following this note, today we talk about an almost-Schur lemma recently obtained by De Lellis and Topping, see here. If we denote by \mathop {{\text{Ric}}}\limits^ \circ the traceless Ricci tensor, the main theorem of the paper is the following

Theorem. For any integer n \geqslant 3, if (M, g) is a closed Riemannian manifold of  dimension n with nonnegative Ricci curvature, then

\displaystyle \int_M {{{(R - \overline R )}^2}} \leqslant \frac{{4n(n - 1)}}{{{{(n - 2)}^2}}}\int_M {| \mathop {{\text{Ric}}}\limits^ \circ {|^2}}

where \overline R is the average value of the scalar curvature R over M. Moreover equality holds if and only if (M, g) is Einstein.

For a proof of the theorem, recall that the contracted second Bianchi identity tells us that

\displaystyle\delta \text{Ric} + \frac{1}{2}dR = 0

where

\displaystyle {(\delta \text{Ric})_j} = - {\nabla _i}{R_{ij}}.

and hence that

\displaystyle\delta\mathop{\text{Ric}}\limits^\circ = - \frac{{n - 2}}{{2n}}dR.

(more…)

April 4, 2011

An iteration by Stampacchia

Filed under: Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:09

Recently, my friend, CR, has shown me an iteration by Stampacchia. Stampacchia proposed his iteration in a preprint [here] entitled Équations elliptiques du second ordre à coefficients discontinus published in Séminaire Jean Leray in 1963-1964.

Suppose \varphi : [k_0, \infty) \to \mathbb R is a non-negative non-decreasing function satisfying

\displaystyle \varphi (h) \leqslant \frac{c}{(h-k)^\alpha}\big(\varphi(k)\big)^\beta

for any h>k \geqslant k_0 where c, \alpha, \beta are positive given constants.

Then

  • If \beta >1, it holds

    \varphi (k_0+d)=0

    where

    \displaystyle d^\alpha=c\big(\varphi(k_0)\big)^{\beta-1}2^\frac{\alpha\beta}{\beta-1}.

  • If \beta=1, one has

    \displaystyle \varphi (h)\leqslant e e^{-\eta (h-k_0)}\varphi (k_0)

    where

    \displaystyle \eta=(ec)^{-\frac{1}{\alpha}}.

  • If \beta<1 and k_0>0, then

    \displaystyle \varphi (h) \leqslant {2^{\frac{\mu }{{1 - \beta }}}}\left\{ {1 + {c^{\frac{1}{{\beta - 1}}}}{{(2{k_0})}^\mu }\varphi ({k_0})} \right\} = \frac{{{2^{\frac{\mu }{{1 - \beta }}}}}}{{{c^{\frac{1}{{1 - \beta }}}}}}\left\{ {{c^{\frac{1}{{1 - \beta }}}} + {{(2{k_0})}^\mu }\varphi ({k_0})} \right\}

    where

    \displaystyle \mu=\frac{\alpha}{1-\beta}.

(more…)

January 11, 2011

An example of sequence of blow-up solutions with finite limiting mass

Filed under: Nghiên Cứu Khoa Học, PDEs — Ngô Quốc Anh @ 14:35

In this note, we recall an example adapted from an elegant paper due to Y.Y. Li and I. Shafrir published in the Indiana Univ. Math. J. in 1994 [here].

Let us consider the asymptitic behavior of sequences of solutions of

-\Delta u_n=V_n(x)e^{u_n}

on a bounded domain \Omega \subset \mathbb R^2 with V_n a non-negative continuous function. For each solution u_n, we call

\displaystyle \alpha_n := \int_{B_R}V_ne^{u_n}dx

the mass of u_n (over a ball B_R). The terminology limiting mass will be referred to the limit \lim_{n \to \infty} \alpha_n.

For simplicity, we assume V_n \equiv 1. Given m, we are going to construct a sequence of solutions \{u_n\} which blows up exactly at m points, say at a_1,...,a_m \in D where D the unit disc of \mathbb C. Our equation reads as

-\Delta u=e^{u}

in D. Using the Liouville formula for solutions of the above equation, we get

\displaystyle u(z) = \log \frac{{8|f'(z){|^2}}}{{{{(1 + |f(z){|^2})}^2}}}

with f an holomorphic function such that f'(z) \ne 0.

(more…)

June 5, 2010

Why the conformal method is useful in studying the Einstein equations?

Filed under: Nghiên Cứu Khoa Học, PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 19:20

I presume you have some notions about general relativity, especially the Einstein equations

{\rm Eins}_{\alpha\beta}=T_{\alpha\beta}.

As these equations are basically hyperbolic for a suitable metric, it is reasonable to study the Cauchy problems for them. Under the Gauss and Codazzi conditions, we have two constraints called Hamiltonian and Momentum constrains. Cauchy problem is to determine the solvable of these constrains of variables K-the extrinsic curvature and g-the spatial metric. Interestingly, the conformal method says that we can start with an arbitrary metric then we recast the constrain equations into a form which is more amenable to analysis by splitting the Cauchy data. In this method, we try to solve \gamma within the conformal class represented by the initial metric. So, in general, the conformal factor is chosen so that we eventually have a simplest model.

This idea is given via the following theorem.

Theorem. Let \mathcal D =(\gamma, \sigma, \tau,\psi,\pi) be a conformal initial data set for the Einstein-scalar field constraint equations on \Sigma. If

\displaystyle \widetilde \gamma =\theta^\frac{4}{n-2}\gamma

for a smooth positive function \theta, then we define the corresponding conformally transformed initial data set by

\displaystyle\widetilde{\mathcal D} =(\widetilde\gamma, \widetilde \sigma, \widetilde \tau,\widetilde\psi,\widetilde \pi)=(\theta^\frac{4}{n-2}\gamma, \theta^{-2}\sigma, \tau,\psi,\theta^\frac{-2n}{n-2}\pi).

Let W be the solution to the conformal form of the momentum constrain equation w.r.t. the conformal initial data set \mathcal D and let \widetilde W be the solution to the conformal form of the momentum constrain equation w.r.t. the conformal initial data set \widetilde{\mathcal D} (we just assume both exist). Then \varphi is a solution to the Einstein scalar field Lichnerowicz equation for the conformal data \mathcal D with W

\displaystyle \Delta_\gamma \varphi - \mathcal R_{\gamma, \psi}\varphi +\mathcal A_{\gamma, W, \pi}\varphi^{-\frac{3n-2}{n-2}}-\mathcal B_{\tau, \psi}\varphi^\frac{n+2}{n-2}=0

if and only if \theta^{-1}\varphi is a solution to the Einstein scalar field Lichnerowicz equation for the conformal data \widetilde{\mathcal D} with \widetilde W

\displaystyle \Delta_{\widetilde\gamma} (\theta^{-1}\varphi) - \mathcal R_{\widetilde\gamma, \widetilde\psi}(\theta^{-1}\varphi) +\mathcal A_{\widetilde\gamma, \widetilde W, \widetilde\pi}(\theta^{-1}\varphi)^{-\frac{3n-2}{n-2}}-\mathcal B_{\widetilde\tau, \widetilde\psi}(\theta^{-1}\varphi)^\frac{n+2}{n-2}=0.

We refer the reader to a paper due to Yvonne Choquet-Bruhat et al. [here] published in Class. Quantum Grav. in 2007 for details. We adopt this theorem from that paper, however, there is no proof there.

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May 1, 2010

A useful identity in a book due to L. Ahlfors

Filed under: Các Bài Tập Nhỏ, Giải Tích 5, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 3:12

Let \mathbf{x},\mathbf{y} be points in \mathbb R^n. If we denote by \mathbf{x}^\sharp the reflection point of \mathbf{x} with respect to the unit ball, i.e.

\displaystyle \mathbf{x}^\sharp = \frac{\mathbf{x}}{|\mathbf{x}|^2}

we then have the following well-known identity

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

The proof of the above identity comes from the fact that

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}{|^2}|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}} = |\mathbf{y}|\left| {\frac{\mathbf{y}}{{|\mathbf{y}|^2}} - \mathbf{x}} \right|.

Indeed, by squaring both sides of

\displaystyle |\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = \sqrt {1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}}

we arrive at

\displaystyle |\mathbf{x}|^2\left( {\frac{{|\mathbf{x}|^2}}{{|\mathbf{x}|^4}} - 2\frac{{\mathbf{x} \cdot \mathbf{y}}}{{|\mathbf{x}|^2}} + |\mathbf{y}|^2} \right) = 1 + |\mathbf{x}|^2|\mathbf{y}|^2 - 2\mathbf{x} \cdot \mathbf{y}

which is obviously true. Similarly, the last identity also holds. If we replace \mathbf{y} by -\mathbf{y} we also have

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp }+ \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } + \mathbf{x}} \right|.

Generally, if we consider the reflection point of \mathbf{x} over a ball B_r(0), i.e.

\displaystyle \mathbf{x}^\sharp = \frac{r^2\mathbf{x}}{|\mathbf{x}|^2}

we still have the fact

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } - \mathbf{y}} \right| = |\mathbf{y}|\left| {{\mathbf{y}^\sharp } - \mathbf{x}} \right|.

Indeed, one gets

\displaystyle |\mathbf{x}|\left| {\frac{{{r^2}\mathbf{x}}}{{|\mathbf{x}{|^2}}} - \mathbf{y}} \right| = {r^2}|\mathbf{x}|\left| {\frac{\mathbf{x}}{{|\mathbf{x}{|^2}}} - \frac{\mathbf{y}}{{{r^2}}}} \right| = {r^2}\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|\left| {\frac{{\frac{\mathbf{y}}{{{r^2}}}}}{{{{\left| {\frac{\mathbf{y}}{{{r^2}}}} \right|}^2}}} - \mathbf{x}} \right| = \left| \mathbf{y} \right|\left| {\frac{{{r^2}\mathbf{y}}}{{|\mathbf{y}{|^2}}} - \mathbf{x}} \right|.

Similarly,

\displaystyle |\mathbf{x}|\left| {{\mathbf{x}^\sharp } + y} \right| = |y|\left| {{y^\sharp } + \mathbf{x}} \right|.

Such identity is very useful. For example, in \mathbb R^n (n\geqslant 3) the following holds

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \min \left\{ {\frac{1}{{|{\mathbf{y}}|^{n - 2}}},\frac{1}{r^{n - 2}}} \right\}.

This type of formula has been considered before when n=3 here. For a general case, Lieb and Loss introduced another method in their book published by AMS in 2001. Here we introduce a completely new proof. At first, if |\mathbf{y}|>r by the potential theory, one easily gets

\displaystyle\iint\limits_{{\mathbf{x}} = r} {\frac{{d{\sigma _{\mathbf{x}}}}}{{{{\left| {{\mathbf{x}} - {\mathbf{y}}} \right|}^{n - 2}}}}} = \frac{1}{{|{\mathbf{y}}|^{n - 2}}}.

If |\mathbf{y}|<r, one needs to make use of the reflection point of \mathbf{y} and the above identity to go back to the first case. The point here is |\mathbf{y}^\sharp|>r. The integral is obviously continuous as a function of \mathbf{y}. The above argument is due to professor X.X.W.

April 9, 2010

Kelvin transform: Laplacian

Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:01

For each point x \ne 0, denote x=(x_1,...,x_n) and

\displaystyle \xi = {x^\sharp } = \left( {\frac {{{x_1}}} {{{{\left| x \right|}^2}}},...,\frac {{{x_n}}} {{{{\left| x \right|}^2}}}} \right)

is the inversion of  x with respect to the unit sphere. We have the following identities

\displaystyle\frac {{\partial {\xi _j}}} {{\partial {x_k}}} = \frac {1} {{{{\left| x \right|}^2}}}\left( {{\delta _{jk}} - 2\frac {{{x_j}{x_k}}} {{{{\left| x \right|}^2}}}} \right)

and

\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {\xi _l}}} {{\partial {x_j}}}\frac {{\partial {\xi _l}}} {{\partial {x_k}}}} = \frac {1} {{{{\left| x \right|}^4}}}{\delta _{jk}}.

Thus,

\displaystyle\sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.

Next, think of \xi as a system of orthogonal curvilinear coordinates for x, we deduce that the metric tensor of the Euclidean space in curvilinear coordinates

\displaystyle {g_{j,k}}\left( {{\xi _j},{\xi _k}} \right) = \sum\limits_{l = 1}^n {\frac {{\partial {x_l}}} {{\partial {\xi _j}}}\frac {{\partial {x_l}}} {{\partial {\xi _k}}}} = \frac {1} {{{{\left| \xi \right|}^4}}}{\delta _{jk}}.

This implies that the so-called Lame coefficients is

\displaystyle {h_j} = \sqrt {{g_{j,j}}\left( {{\xi _j},{\xi _j}} \right)} = \frac {1} {{{{\left| \xi \right|}^2}}}.

(more…)

March 14, 2010

Classification of a system of n first-order PDEs

Filed under: Nghiên Cứu Khoa Học, PDEs — Ngô Quốc Anh @ 11:18

The classification of a system of n first-order PDEs is based on whether there are n directions along which the PDEs reduce to n ODEs. To be more precise, assume that we are given a system of n equations in n unknowns u_1, u_2,...,u_n which we write in matrix form as

\displaystyle \mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u}),

where \mathbf{u}=(u_1,...,u_n)^t, \mathbf{b}=(b_1,...,b_n)^t, and A=(a_{ij}(x,t,\mathbf{u})) is an n \times n matrix.

Now we ask whether there is a family of curves along which the PDEs reduce to a system of ODEs, that is, in which the directional derivative of each u_i occurs in the same direction. We consider a row vector \gamma = (\gamma_1,...,\gamma_n)^t to be determined later. Then

\displaystyle \mathbf{\gamma}^t\mathbf{u}_t + \mathbf{\gamma}^tA(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{\gamma}^t\mathbf{b}(x,t,\mathbf{u}).

We want the above system to have the form of a linear combination of total derivatives of the u_i in the same direction \lambda, that is, we want our system to have the form

\displaystyle \mathbf{m}^t \left( {{{\mathbf{u}}_t} + \lambda {{\mathbf{u}}_x}} \right) = \mathbf{\gamma}^t{\mathbf{b}}

for some \mathbf{m}. Consequently, we require

\displaystyle \mathbf{m}=\gamma, \quad \mathbf{m}^t\lambda=\gamma^tA

or

\displaystyle \gamma^t A=\lambda \gamma^t.

This means that \lambda is an eigenvalue of A and \gamma^t is a corresponding left eigenvector. Note that \lambda as well as \gamma can depend on x, t, and \mathbf{u}. So, if (\lambda, \gamma^t) is an eigenpair, then

\displaystyle \gamma^t \frac{d\mathbf{u}}{dt}=\gamma^t\mathbf{b}

along

\displaystyle \frac{dx}{dt}=\lambda(x,t,\mathbf{u})

and the system of PDEs is reduced to a single ODE along the family of curves, called characteristics, defined by \frac{dx}{dt}=\lambda. The eigenvalue \lambda is called the characteristics direction. Clearly, because there are n unknowns, it would appear that n ODEs are required; but if A has n distinct real eigenvalues, there are n ODEs, each holding along a characteristics direction defined by an eigenvalue. In this case we say that the system is hyperbolic.

Definition. The quasilinear system

\displaystyle \mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u})

is hyperbolic if A has n real eigenvalues and n linearly independent left eigenvectors. Once these eigenvectors are distinct, the system is called stricly hyperbolic.

The system is called elliptic if A has no real eigenvalues, and it is parabolic if A has n real eigenvalues but fewer then n independent left eigenvectors.

No exhaustive classification is made in the case that A has both real and complex eigenvalues. Note that once matrix A has n distinct, real eigenvalues it has n independent left eigenvectors, because distinct eigenvalues have independent eigenvectors.

More general systems of the form

\displaystyle B(x,t,\mathbf{u})\mathbf{u}_t + A(x,t,\mathbf{u})\mathbf{u}_x = \mathbf{b}(x,t,\mathbf{u})

can be considered as well. We refer the reader to a book entitled “An introduction to nonlinear partial differential equations” due to J.D. Logan.

We are now in a position to see why a single first-order quasilinear PDE is hyperbolic. The coefficient matrix for the equation

\displaystyle u_t + c(x,t,u)u_x=b(x,t,u)

is just the real scalar function c(x,t,u) which has the single eigenvalue c(x,t,u) and its corresponding eigenvector 1, a constant function. In this direction, once \frac{dx}{dt}=c(x,t,u) the PDE reduces to the ODE \frac{du}{dt}=b(x,t,u). We refer the reader to the following topic, called characteristic curves, where we consider when the equation has constant coefficients and variable coefficients.

We place here three more examples.

Example 1 (The shallow-water equations). The following system

\displaystyle\begin{gathered} {h_t} + u{h_x} + h{u_x} = 0, \hfill \\ {u_t} + u{u_x} + g{h_x} = 0, \hfill \\ \end{gathered}

is trictly hyperbolic.

Example 2. The following system

\displaystyle\begin{gathered} {u_t} - {v_x} = 0, \hfill \\ {v_t} - c{u_x} = 0, \hfill \\ \end{gathered}

is elliptic if c<0 and is hyperbolic if c>0.

Example 3 (The diffusion equations). The following equation

\displaystyle u_t=u_{xx}

may be written as the first-order system

\displaystyle\begin{gathered}u_t-v_x=0, \hfill \\u_x-v = 0, \hfill \\ \end{gathered}

and thus is parabolic.

March 13, 2010

Wave equations: The gradient estimates and a conservation law

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 0:04

Let us consider the following problem

\displaystyle u_{tt} = \Delta u, \quad x \in \mathbb R^n, t>0

with

\displaystyle u(x,0)=f(x), \quad u_t(x,0)=g(x).

Assume f, g \in \mathcal S(\mathbb R^n). Note that the Schwartz space \mathcal S(\mathbb R^n) consists of all indefinitely differentiable functions f : \mathbb R^n \to \mathbb R such that

\displaystyle\mathop {\sup }\limits_{x \in {\mathbb{R}^n}} \left| {{x^\alpha }{D^\beta }f(x)} \right| < \infty

for every multi-index \alpha and \beta. We shall prove that the total energy at time t

\displaystyle E(t) = \frac{1}{2}\int_{{\mathbb{R}^n}} {\left( {u_t^2(x,t) + |{\nabla _x}u(x,t){|^2}} \right)dx}

is constant in time.

Our approach is based on the Fourier transform. Note that the Fourier transform of a Schwartz function \varphi is defined by

\displaystyle\widehat\varphi (\xi ) = \int_{{\mathbb{R}^n}} {\varphi (x){e^{ - 2\pi i\xi \cdot x}}dx} .

It is worth noticing that the Fourier transform maps \mathcal S(\mathbb R^n) to itself. Taking the Fourier transform with respect to space variable x we have

\displaystyle \widehat u_{tt}(\xi,t) = |\xi|^2 \widehat u(\xi,t)

and

\displaystyle \widehat u(\xi,0)=\widehat f(\xi), \quad \widehat u_t(\xi,0)=\widehat g(\xi).

Solving the above initial problem of the ordinary differential equation, we get

\displaystyle\widehat u(\xi ,t) = \widehat f(\xi )\cos \left( {|\xi |t} \right) + \frac{{\widehat g(\xi )}}{{|\xi |}}\sin (|\xi |t)

and therefore

\displaystyle {\widehat u_t}(\xi ,t) = - \widehat f(\xi )|\xi |\sin \left( {|\xi |t} \right) + \widehat g(\xi )\cos (|\xi |t).

As a consequence,

\displaystyle \int_{{\mathbb{R}^n}} {\left( {|{{\widehat u}_t}(\xi ,t){|^2} + |\xi {|^2}|\widehat u(\xi ,t){|^2}} \right)d\xi } = \int_{{\mathbb{R}^n}} {\left( {|\widehat g(\xi ){|^2} + |\xi {|^2}|\widehat f(\xi ){|^2}} \right)d\xi } .

We are now in a position to apply Plancherel’s Theorem to get the desired result.

Theorem (Plancherel). Suppose f \in \mathcal S(\mathbb R^n). Then

\displaystyle\varphi (x) = \int_{{\mathbb{R}^n}} {\widehat\varphi (\xi ){e^{2\pi i\xi \cdot x}}d\xi }.

Moreover

\displaystyle\int_{{\mathbb{R}^n}} {{{\left| {\widehat\varphi (\xi )} \right|}^2}d\xi } = \int_{{\mathbb{R}^n}} {{{\left| {\varphi (x)} \right|}^2}dx}.

Let us denote by E_0 this common value of the total energy. One can show that

\displaystyle\mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {u_t^2(x,t)dx} = \mathop {\lim }\limits_{t \to \infty } \int_{{\mathbb{R}^n}} {|{\nabla _x}u(x,t){|^2}dx} = {E_0}.

The key point is to use the Riemann-Lebesgue lemma from harmonic analysis. We refer the reader to a book entitled Fourier Analysis due to Stein and Shakarchi.

March 7, 2010

Heat equations: The gradient estimates

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 16:37

Let us consider the following problem

\displaystyle u_t =u_{xx}, \quad x\in \mathbb R, t>0

and

\displaystyle u(x,0)=f(x),\quad x \in \mathbb R.

We assume further that f \in L^2(\mathbb R). We are interested in the gradient estimate, in fact, we are going to find the bound on

\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx}

and

\displaystyle |{u_x}(x,t)|

in t>0.

So far, once we want to derive any estimates like point-wise, L^2, gradient, the Green function is very important. Again, since the Green function for this heat equation is already known, we can write down everything via the Poisson formula

\displaystyle u(x,t) = \frac{1}{{\sqrt {2\pi t} }}\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} .

A simple calculation shows that

\displaystyle {u_x}(x,t) = \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy}.

Then by using the Holder inequality, we have

\displaystyle\begin{gathered} {\left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)^2} = {\left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right)^2} \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) \hfill \\ \end{gathered}

which implies

\displaystyle |{u_x}(x,t){|^2} \leqslant \frac{1}{{16\pi {t^3}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right).

In order to do next, the following two results are noticed

Proposition 1. The following identities hold

\displaystyle\int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi

and

\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi.

Therefore, it holds that

\displaystyle\int_{ - \infty }^{ + \infty } {{{(x - y)}^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} \mathop = \limits^{z = x - y} \int_{ + \infty }^{ - \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}( - dz)} = \int_{ - \infty }^{ + \infty } {{z^2}{e^{ - \frac{{{z^2}}}{{4t}}}}dz} = 4{t^{\frac{3}{2}}}\sqrt \pi

which is independent of x. Consequently,

\displaystyle\begin{gathered} \int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{16\pi {t^3}}}4{t^{\frac{3}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \hfill \\ \qquad\qquad= \frac{1}{{4{{\sqrt {\pi t} }^{\frac{3}{2}}}}}\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx}. \hfill \\ \end{gathered}

The last double integral can be estimated by the Fubini lemma. In fact,

\displaystyle\int_{ - \infty }^{ + \infty } {\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}{e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right)dx} \leqslant 2{t^{\frac{1}{2}}}\sqrt \pi \int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}

which yields

\displaystyle\int_{ - \infty }^{ + \infty } {|{u_x}(x,t){|^2}dx} \leqslant \frac{1}{{2t}}\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx}.

Remark. If we estimate as follows

\displaystyle\begin{gathered} \left( {\int_{ - \infty }^{ + \infty } {f(y)(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{4t}}}}dy} } \right) = \left( {\int_{ - \infty }^{ + \infty } {f(y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}(x - y){e^{ - \frac{{{{(x - y)}^2}}}{{8t}}}}dy} } \right) \hfill \\ \qquad\qquad\leqslant \left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}} \hfill \\ \end{gathered}

we then have

\displaystyle |{u_x}(x,t)| \leqslant \frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}\left( {\int_{ - \infty }^{ + \infty } {|f(y){|^2}dy} } \right)^\frac{1}{2}{\left( {\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}{\left( {\int_{ - \infty }^{ + \infty } {{{(x - y)}^4}{e^{ - \frac{{{{(x - y)}^2}}}{{2t}}}}dy} } \right)^{\frac{1}{4}}}.

Note that

Proposition 2. The following result holds

\displaystyle\int_{ - \infty }^{ + \infty } {{z^4}{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 3\sqrt 2 {t^{\frac{3}{2}}}\sqrt \pi

and

\displaystyle\int_{ - \infty }^{ + \infty } {{e^{ - \frac{{{z^2}}}{{2t}}}}dz} = 2{t^{\frac{1}{2}}}\sqrt \pi.

Thus

\displaystyle |{u_x}(x,t)| \leqslant\underbrace {\frac{1}{{4\sqrt \pi {t^{\frac{3}{2}}}}}{{\left( {\sqrt {2\pi t} } \right)}^{\frac{1}{4}}}{{\left( {3\sqrt {2\pi } {t^{\frac{3}{2}}}} \right)}^{\frac{1}{4}}}}_{\frac{{\sqrt[8]{6}}}{{4t}}}\sqrt {\int_{ - \infty }^{ + \infty } {|f(x){|^2}dx} } = \frac{{\sqrt[8]{6}}}{{4t}}{\left\| f \right\|_{{L^2}(\mathbb{R})}} .

This is a pointwise estimate.

March 2, 2010

On the regularity of solutions to the Yamabe problem

Filed under: Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 21:12

The Yamabe problem has been discussed here. Basically, starting from a metric g, for a given constant R Yamabe wanted to show there always exists a positive function \varphi such that the scalar curvature of metric \overline g defined to be \varphi^\frac{4}{n-2}g equals to R. In terms of PDEs, the scalar curvature satisfies the equation (called Yamabe equation)

\displaystyle \frac{{4\left( {n - 1} \right)}} {{n - 2}}\Delta \varphi + R\varphi = \overline R {\varphi ^{\frac{{n + 2}} {{n - 2}}}}

where \overline R the scalar curvature of metric \overline g.

Clearly, one is also interested in the regularity of solutions to the Yamabe equation. We quote here the result due to N. Trudinger published in Ann. Scuola Norm. Sup. Pisa (3) in 1968 when he was a gradutate student. That paper can be found here.

Let us start with a result from elliptic theory.

Lemma. Let u \in W^{1,2}(M) be a weak, non-negative solution in Mof the linear equation

\displaystyle \Delta u +fu=0

where f \in L^r(M) with r > \frac{n}{2}. Then u is positive and bounded and we have the estimates

\displaystyle\mathop {\sup }\limits_M \left\{ {|u|,\frac{1}{{|u|}}} \right\} \leqslant C\left( {n,{{\left\| f \right\|}_{{L^r}}}} \right)

where C is constant depending only on n and the function f.

Yamabe’s approach. Yamabe tried to solve the subcritical case first. Precisely, he tried to solve the following equation

\displaystyle \frac{{4\left( {n - 1} \right)}}{{n - 2}}\Delta \varphi + R\varphi = \overline R{\varphi ^{q-1}}

where q<\frac{2n}{n-2} = :N. It is clear to see that the above equation can be rewritten as the following

\displaystyle\Delta \varphi + \underbrace {\frac{{n - 2}}{{4\left( {n - 1} \right)}}\left( {\overline R{\varphi ^{q - 2}} - R} \right)}_f\varphi = 0.

Note that the function f is of class L^r where r=\frac{N}{q-2} > \frac{n}{2}. Then once we can prove the existence of weak solution, this solution is positive and bounded. Since r>\frac{n}{2}, elliptic regularity theory then guarantees that \varphi\in C^2(\overline M). By a standard bootstrap argument, one can show that \varphi^{(2)} \in C^2(\overline M). Repeat this process, eventually one has \varphi \in C^\infty(M). At the final stage, Yamabe tried to let q \to N, by a compact embedding which is not true, we claimed that \varphi_q, solution to the above equation, goes to \varphi, the solution we need.

Trudinger’s approach. Trudinger tried to fix this error by assuming the quantity \int_M Rdv_g, called the mean scalar curvature of metric g, is sufficiently small.

A Trudinger’s Regularity Theorem. This is the main part of this entry.

Theorem (Trudinger). Let u \in W^{1,2}(M) be a solution of the Yamabe equation. Then u \in C^\infty(M).

Proof. Clearly, the function u satisfies

\displaystyle\int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}{g^{ij}}{u_i}{\xi _j} + Ru\xi } \right)d{v_g}} = \overline R \int_M {\left( {|u{|^{N - 1}}\xi d{v_g}} \right)}

for all test function \xi \in W^{1,2}(M). We choose an appropriate test function \xi similarly to a method of Serrin published in Acta. Math. in 1965.

Define \overline u=\sup(u,0) and for a fixed \beta>1 defined the functions

\displaystyle G(\overline u ) = \left\{ \begin{gathered} {\overline u ^\beta }\quad \text{ if } \quad\overline u \leqslant l, \hfill \\ {l^{q - 1}}\left( {q{l^{q - 1}}\overline u - (q - 1){l^q}} \right) \quad \text{ if } \quad \overline u > l \hfill \\ \end{gathered} \right.

and

\displaystyle F(\overline u ) = \left\{ \begin{gathered} {\overline u ^q}\quad \text{ if } \quad\overline u \leqslant l, \hfill \\ q{l^{q - 1}}\overline u - (q - 1){l^q}\quad \text{ if } \quad\overline u > l \hfill \\ \end{gathered} \right.

where q=\frac{\beta+1}{2}.

The function G(\overline u) is a uniformly Lipschitz continuous function of u and hence belongs to W^{1,2}(M). Likewuse F(\overline u). Observe also that G and F vanish for negative u and that

\displaystyle {\left( {F'(\overline u )} \right)^2} \leqslant qG'(\overline u ), \quad \overline u G(\overline u ) \leqslant {\left( {F(\overline u )} \right)^2}.

Let us now substitute in the variational equation test function \xi=\eta^2G(\overline u) where \eta is an arbitrary, non-negative C^1(M) function. The result is

\displaystyle\frac{{4(n - 1)}}{{n - 2}}\mu \int_M {{\eta ^2}G'(\overline u )u_j^2d{v_g}} \leqslant \int_M {\left( {\frac{{4(n - 1)}}{{n - 2}}\sup |{g^{ij}}|\eta |{\eta _i}| + \sup |R|u{\eta ^2} + \overline R {u^{N - 2}}{\eta ^2}} \right)Gd{v_g}}

and hence

\displaystyle\int_M {{\eta ^2}G'(\overline u )u_j^2d{v_g}} \leqslant C\int_M {\left( {\left( {{\eta ^2} + |{\eta _i}{|^2}} \right)\overline u G(\overline u ) + {\eta ^2}{{\overline u }^{N - 2}}{\eta ^2}\overline u G(\overline u )} \right)d{v_g}}

for some constant C. Clearly, one can estimate further

\displaystyle\int_M {{\eta ^2}{{\left( {F'(\overline u )} \right)}^2}u_j^2d{v_g}} \leqslant Cq\int_M {\left( {\left( {{\eta ^2} + |{\eta _i}{|^2}} \right){{\left( {F(\overline u )} \right)}^2} + {\eta ^2}{{\overline u }^{N - 2}}{\eta ^2}{{\left( {F(\overline u )} \right)}^2}} \right)d{v_g}} .

Let us take \eta now to have compact support in a coordinate patch of M. The integrals above may then be replaced by integrals over a sphere S_R in E^n of radius R where \eta=\eta(x) \in C_0^1(R). We choose R so that

\displaystyle\int_{{S_R}} {|u{|^N}d{v_g}} \leqslant \frac{1}{{2Cq}}.

Then applying the Holder and Sobolev inequalities, we get

\displaystyle {\left\| {\eta F} \right\|_{{L^N}(S)}} \leqslant Cq{\left\| {(\eta + {\eta _i})F} \right\|_{{L^2}(S)}} + \frac{1}{2}{\left\| {\eta F} \right\|_{{L^N}(S)}}

and hence

\displaystyle {\left\| {\eta F} \right\|_{{L^N}(S)}} \leqslant 2Cq{\left\| {(\eta + {\eta _i})F} \right\|_{{L^2}(S)}}.

We choose \beta \in \left( {1,\frac{{n + 2}}{{n - 2}}} \right) so that 2q<N. Hence we may let l \to \infty to obtain the estimate

\displaystyle {\left\| {\eta {{\overline u }^q}} \right\|_{{L^N}(S)}} \leqslant C{\left\| {(\eta + {\eta _i}){{\overline u }^q}} \right\|_{{L^2}(S)}}.

Let S_{\frac{R}{2}} denote the sphere concentric to S of radius \frac{R}{2} and choose \eta=1 on S_{\frac{R}{2}}, |\eta_i| \leq \frac{2}{R} on S_{\frac{R}{2}}. Then we obtain

\displaystyle {\left\| {{{\overline u }^q}} \right\|_{{L^N}(S)}} \leqslant C\left( {1 + \frac{1}{R}} \right).

Replacing u by -u we have a full estimate for u and employing a partition of unity clearly provides a global estimate

\displaystyle {\left\| u \right\|_{{L^r}(S)}} \leqslant C

for some r>N where C will also depend on the local L^N-norm of u. The boundedness of u and subsequently its smoothness are now consequences of the lemma above. The proof is complete.

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