Ngô Quốc Anh

July 30, 2010

Curvature of a curve

Filed under: Riemannian geometry — Ngô Quốc Anh @ 16:16

Consider a car driving along a curvy road. The tighter the curve, the more difficult the driving is. In math we have a number, the curvature, that describes this “tightness”. If the curvature is zero then the curve looks like a line near this point. While if the curvature is a large number, then the curve has a sharp bend.

Let assume \gamma : I \to \mathbb R^3 be a curve. We also adopt the following notation

r(t)=x(t)\vec i+y(t)\vec j+z(t)\vec k, \quad t \in I.

More formally, if T(t) is the unit tangent vector function then the curvature is defined at the rate at which the unit tangent vector changes with respect to arc length s.

Definition. Curvature is given by

\displaystyle k(t) = \left\|\frac{d}{ds} (T(t)) \right\| = \|r''(s)\|

Example. Let consider an example of a circle of radius r given by

\alpha(t) = (r \sin(t), r \cos(t)).

We then see that

\alpha '(t) = (r \cos t, -r \sin t)

which implies |\alpha'(t)| = r. Thus, s=rt. Our circle is now parameterized by arc length as follows

\displaystyle\alpha(s) = \left(r \sin\left(\frac{s}{r}\right), r \cos \left(\frac{s}{r}\right) \right).

The curvature vector at a given length s is then

\displaystyle \alpha ''(s) = \left( { - \frac{1}{r}\sin \left( {\frac{s}{r}} \right), - \frac{1}{r}\cos \left( {\frac{s}{r}} \right)} \right)

and that

\displaystyle \|\alpha ''(s)\| = \frac{1}{r}.

(more…)

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July 27, 2010

The implicit function theorem: How to prove a continuously dependence on parameters for solutions of ODEs

Filed under: Giải Tích 3, PDEs — Tags: — Ngô Quốc Anh @ 0:25

It is clear that the implicit function theorem plays an important role in analysis. From now on, I am going to demonstrate this significant matter from the theory of differential equations, both ODE and PDE, point of view.

Let us start with the following ODE

-u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}=0

on some domain \Omega \subset \mathbb R^n with \alpha \not\equiv 0 and \beta \not\equiv 0. We assume the existence result on W_+^{2,p} is proved for some p>1. We prove the following

Theorem. The solution u \in W_+^{2,p} depends continuously on (\alpha, \beta) \in L^\infty \times L^\infty.

Proof. Consider the map

\mathcal N : W_+^{2,p} \times (L^\infty \times L^\infty) \to L^p

taking

(u,\alpha,\beta) \mapsto -u''-\alpha^2 u^{-q-1}+\beta^2u^{q-1}.

This map is evidently continuous (since W_+^{2,p} is an algebra). One readily shows that its Fréchet derivative at (u, \alpha, \beta) with respect to u in the direction h is

\mathcal N'[u,\alpha ,\beta ]h = - h'' + \left[ {(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}} \right]h.

The continuity of the map

(u,\alpha,\beta) \mapsto \mathcal N'[u,\alpha ,\beta ]

follows from the fact that W_+^{2,p} is an algebra continuously embedded in C^0(\Omega).

Since \alpha \not\equiv 0 and \beta \not\equiv 0, the potential

V={(q + 1){\alpha ^2}{u^{ - q - 2}} + (q - 1){\beta ^2}{u^{q - 2}}}

is not identically zero. Thus it is well-known that the map

-\Delta +V : W^{2,p} \to L^p

is an isomorphism.

The implicit function theorem then implies that if u_0 is a solution for data (\alpha_0, \beta_0), there is a continuous map defined near (\alpha_0, \beta_0) taking (\alpha, \beta) to the corresponding solution of the ODE. This establishes the conclusion.

For the more details, we prefer the reader to this preprint.

July 24, 2010

Regularity theory for integral equations

Filed under: Giải Tích 6 (MA5205), PDEs — Tags: — Ngô Quốc Anh @ 20:54

My purpose is to derive some regularity result concerning the following integral equation

\displaystyle u(x) = \int_\Omega {\frac{{u(y)}}{{{{\left| {x - y} \right|}^{n - \alpha }}}}dy}

where \Omega \subset \mathbb R^n is open and bounded and 0<\alpha<n. To this purpose, in this entry we first consider the equation

\displaystyle u(x) = \int_\Omega {\frac{{f(y)}}{{{{\left| {x - y}  \right|}^{n - \alpha }}}}dy}

for a suitable choice of f.

The case f \in L^\infty(\Omega). We will prove that u \in C^{1,\beta}(\Omega) for any \beta\in (0,1). Indeed, up to a constant factor, the first derivative of u are given by

\displaystyle {D_i}u(x) = \int_\Omega {\frac{{{x_i} - {y_i}}}{{{{\left| {x - y} \right|}^{n + 2 - \alpha }}}}f(y)dy}.

From this formula,

\displaystyle\left| {{D_i}u({x^1}) - {D_i}u({x^2})} \right| = \mathop {\sup }\limits_\Omega |f|\int_\Omega {\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right|dy} .

By the intermediate value theorem, on the line from x^1 to x^2, there exists some x^3 with

\displaystyle\left| {\frac{{x_i^1 - {y_i}}}{{{{\left| {{x^1} - y} \right|}^{n + 2 - \alpha }}}} - \frac{{x_i^2 - {y_i}}}{{{{\left| {{x^2} - y} \right|}^{n + 2 - \alpha }}}}} \right| \leqslant \frac{C}{{{{\left| {{x^3} - y} \right|}^{n + 2 - \alpha }}}}\left| {{x^1} - {x^2}} \right|.

(more…)

July 21, 2010

Regularity Lifting by Contracting Operators

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 4:47

I just read this method in a book due to W.X. Chen and C.M. Li published by American Institute of Mathematical Sciences this year 2010.

Let V be a topological vector space. Suppose there are two extended norms (i.e. the norm of an element in V might be infinity) defined on V

\|\cdot\|_X, \|\cdot\|_Y :V\to [0,\infty].

Let

X:=\{v \in V: \|v\|_X <\infty\}

and

Y:=\{v \in V: \|v\|_Y <\infty\}.

Theorem (Regularity Lifting I). Let T be a contraction map from X into itself and from Y into itself. Assume that f \in X, and that there exits a function g \in Z := X \cap Y such that f = Tf + g in X. Then f also belongs to Z.

Proof. Firstly, let

\sqrt{\|\cdot\|_X^2+\|\cdot\|_Y^2}

be a norm on Z. We first show that T : Z \to Z is a contraction. Since T is a contraction on X, there exists a constant \theta_1, 0 < \theta_1 < 1 such that

\displaystyle {\left\| {T{h_1} - T{h_2}} \right\|_X} \leqslant {\theta _1}{\left\| {{h_1} - {h_2}} \right\|_X},\quad\forall {h_1},{h_2} \in X.

Similarly, we can find a constant \theta_2, 0 < \theta_2 < 1 such that

\displaystyle {\left\| {T{h_1} - T{h_2}} \right\|_Y} \leqslant  {\theta _2}{\left\| {{h_1} - {h_2}} \right\|_Y},\quad\forall {h_1},{h_2}  \in Y.

Let \theta = \max\{\theta_1,\theta_2\}. Then, for any h_1, h_2 \in Z

\displaystyle {\left\| {T{h_1} - T{h_2}} \right\|_Z} = \sqrt {\left\| {T{h_1} - T{h_2}} \right\|_X^2 + \left\| {T{h_1} - T{h_2}} \right\|_Y^2} \leqslant \theta {\left\| {{h_1} - {h_2}} \right\|_Z}.

Since T : Z \to Z is a contraction, given g \in Z, we can find a solution h \in Z such that h = Th + g. Notice that T : X \to X is also a contraction and g \in Z \subset X, the solution of the equation x = Tx + g must be unique in X. Because both h and f are solutions of the same equation x = Tx + g in X, we deduce that f = h \in Z.

Remark. In practice, we usually choose V to be the space of distributions, and X and Y to be function spaces, for instance, X = L^p(\Omega) and Y = W^{1,q}(\Omega). We start from a function f in a lower regularity space X, if we can show that T is a contraction from X to itself and from Y to itself, then we can lift the regularity of f to be in

Z = X \cap Y = L^p(\Omega)\cap W^{1,q}(\Omega).

Applications to PDEs. Now, we explain how the Regularity Lifting Theorem proved in the previous subsection can be used to boost the regularity of week solutions involving critical exponent

\displaystyle -\Delta u =u^\frac{n+2}{n-2}.

Still assume that ­ \Omega is a smooth bounded domain in \mathbb R^n with n \geqslant 3. Let
u \in H^1_0(\Omega) be a weak solution of the above PDE. Then by Sobolev embedding

\displaystyle u \in L^\frac{2n}{n-2}(\Omega).

We can split the right hand side of the PDE in two parts

\displaystyle u^\frac{n+2}{n-2}=u^\frac{4}{n-2}u=a(x)u.

Then obviously a \in L^\frac{n}{2}(\Omega). Hence, more generally, we consider the regularity of the weak solution of the following equation

\displaystyle -\Delta u =a(x)u+b(x).

Theorem. Assume that a,b \in L^\frac{n}{2}(\Omega). Let u \in H_0^1(\Omega) be any weak solution of the foregoing PDE. Then u \in L^p(\Omega) for any 1 \leqslant p<\infty.

The proof of this theorem can be found in the book mentioned above.

Let us go back to the first PDE. Assume that u is a H^1_0(\Omega) weak solution. From the above theorem, we first conclude that u is in L^q(\Omega) for any 1 < q < \infty. Then by a standard regularity result known as the W^{2,p}-regularity for a second order uniformly elliptic operator in divergence form, u is in W^{2,q}(\Omega). This implies that u \in C^{1,\alpha}(\Omega) via Sobolev embedding. Finally, by repeated applications of the Schauder estimates, we derive that u \in C^\infty(\Omega).

July 19, 2010

On the determinant of a matrix

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 20:10

Several days ago, I placed a question on MathLinks asking the relation between \det A and \det(A-\lambda I). The point is how to evaluate

\displaystyle\det\begin{bmatrix}1+|x|^{2}-2x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\  -x_{1}x_{2}&1+|x|^{2}-2x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots  &\vdots &\ddots &\vdots\\  -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-2x_{n}^{2}\end{bmatrix}.

Interestingly, K.M. showed me a new way to attack such a problem but slightly different from the original one. He proved

\displaystyle\det\begin{bmatrix}1+|x|^{2}-x_{1}^{2}&-x_{1}x_{2}&\cdots&-x_{1}x_{n}\\ -x_{1}x_{2}&1+|x|^{2}-x_{2}^{2}&\cdots&-x_{2}x_{n}\\ \vdots &\vdots &\ddots &\vdots\\ -x_{n}x_{1}&-x_{n}x_{2}&\cdots&1+|x|^{2}-x_{n}^{2}\end{bmatrix}=\left(1+|x|^{2}\right)^{n-1}.

Let us discuss the proof of this modified problem.

Let

x=\begin{bmatrix}x_1\\\vdots\\x_n\end{bmatrix}

and let

A=xx^T.

The determinant we are trying to compute is

\displaystyle \det\left((1+|x|^2)I-A\right),

which is the characteristic polynomial of A evaluated at 1+|x|^2.

Now, A is certainly diagonalizable (which doesn’t even matter, but it makes it easier to think about), and we know its eigenvalues. Why do we know its eigenvalues? Because A is a matrix of rank 1, hence nullity n-1, hence n-1 of its n eigenvalues are zero. What is the other eigenvalue? It’s the same as the sum of the eigenvalues, which is the trace of A, which is |x|^2. Put that information together, and we have that the characteristic polynomial of A is

\det(\lambda I-A)=\left(\lambda-|x|^2\right)\lambda^{n-1}=\lambda^n-|x|^2\lambda^{n-1}.

Substitute 1+|x|^2 for \lambda to get the result quoted.

July 15, 2010

The winding number is a special case of the Brouwer degree

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 18:25

In the classical complex analysis, the winding number of a plane curve \Gamma with respect to a point a \notin \mathbb C \setminus \Gamma is defined by

\displaystyle w(\Gamma ,a) = \frac{1}{{2\pi i}}\int_\Gamma {\frac{{dz}}{{z - a}}} .

Let \Omega \subset \mathbb R^n be open and bounded. Let f : \Omega \to \mathbb R^n and p \notin f(\partial \Omega). Notation J_f(a) denotes the Jacobian of f evaluated at a. It is well-known that the Brouwer Degree Theory, usually denoted by \deg, is constructed for continuous function, C^0(\overline\Omega)-class, via the following steps

For the C^1(\overline\Omega)-class: We assume J_f(f^{-1}(p)) \ne 0 then

\displaystyle\deg (f,\Omega ,p) = \begin{cases} \sum\limits_{x \in {f^{ - 1}}(p)} {{\rm sgn} {J_f}(x)} , & {f^{ - 1}}(p) \ne \emptyset ,\\ 0, & {f^{ - 1}}(p) = \emptyset.\end{cases}

For the C^2(\overline\Omega)-class: In case we want to remove the condition J_f(f^{-1}(p)) \ne 0, we then define

\displaystyle\deg (f,\Omega ,p) = \deg (f,\Omega ,p')

where p' is any regular value of f sufficiently closed to p in the sense that

\displaystyle\left\| {p - p'} \right\| < {\rm dist}(p,f(\partial \Omega )).

For the C^0(\overline\Omega)-class: In this case, we define

\displaystyle\deg (f,\Omega ,p) = \deg (g,\Omega ,p)

where g \in C^2(\overline\Omega) is sufficiently closed to f in the sense that

\displaystyle\left\| f-g \right\| < {\rm dist}(p,f(\partial \Omega )).

Now we prove the following fact

Theorem.  Let B(0,1)\subset \mathbb C be the unit ball and \Gamma = \partial B(0,1). Assume f : \overline{B(0,1)} \to \mathbb C is a C^1 function and a \notin f(\Gamma). Then

\displaystyle\deg (f,B(0,1),a) = \frac{1}{{2\pi i}}\int_{f(\Gamma )} {\frac{{dz}}{{z - a}}} .

(more…)

July 13, 2010

What is a curve parametrized by arc length?

Filed under: Riemannian geometry — Ngô Quốc Anh @ 6:58

Let \alpha :I \to \mathbb R^3 be any curve and t_0 \in I. We define the arc length function from t_0 which will denote by S : I \to \mathbb R by

\displaystyle S(t)=\int_{t_0}^t|\alpha'(s)|ds.

Since s \mapsto |\alpha'(s)| is, in general, continuous, the function S is only C^1 and

S'(t)=|\alpha'(t)|.

If we assume \alpha is regular, i.e. \alpha'(t) \ne 0 for any t \in I, then by the Inverse Function Theorem S is differentiable increasing open function. Then if we put

J=S(I)

then S : I \to J is a diffeomorphism between two open intervals. Let \phi : J \to I be the inverse diffeomorphism and let \beta : J \to \mathbb R^3 be the re-parametrization of \alpha given by

\beta = \alpha \circ \phi.

Then this new curve satisfies

\displaystyle \beta '(s) = \alpha '(\phi (s))\phi '(s) = \frac{{\alpha '(\phi (s))}}{{|\alpha '(\phi (s))|}}

and thus |\beta'(s)|=1 for any s \in J. It follows from the above that any regular curve admits a re-parametrization by arc length. So we have

Definition. A curve \alpha :I \to \mathbb R^3 is said to be parametrized by arc length S if |\alpha'(t)| = 1 for all t \in I.

Roughly speaking, instead of using a time variable t to parametrize a curve we use its arc length S. The new equation for the curve in terms of S is called the curve parametrized by arc length.

Example (logarithmic spiral). The curve \alpha : \mathbb R \to \mathbb R^2 given by

\displaystyle \alpha(t)=(ae^{bt}\cos t, a e^{bt} \sin t)

with a>0 and b<0 is called the logarithmic spiral. The case a=2 and b=-\frac{1}{5} is showed below.

The arc length of the logarithmic spiral is given by

\displaystyle S(t) = \frac{{\sqrt {{a^2}(1 + {b^2}){e^{2bt}}} }}{b}.

Solving this equation gives

\displaystyle t = \frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}.

Thus the equation for \alpha parametrized by arc length is given by

\displaystyle \alpha (s) = \left(a{e^{\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}}}\cos \left( {\frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}} \right),a{e^{\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}}}\sin \left( {\frac{1}{b}\log \frac{{bs}}{{a\sqrt {1 + {b^2}} }}} \right)\right).

A lot of formulas become simpler when your curve is parametrized by arc length. For example, the curvature is calculated by taking the derivative of the tangent vector. When the curve is parametrized by arc length, the tangent vector has constant length one. This implies that the derivative of the tangent vector is always normal to the tangent vector, which is necessary to find the curvature.

July 11, 2010

Stereographic projection

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 0:47

In geometry, the stereographic projection, usually denoted by \pi, is a particular mapping (function) that projects a sphere onto a plane. The projection is defined on the entire sphere, except at one point – the projection point. Where it is defined, the mapping is smooth and bijective. It is conformal, meaning that it preserves angles. It is neither isometric nor area-preserving: that is, it preserves neither distances nor the areas of figures.

Intuitively, then, the stereographic projection is a way of picturing the sphere as the plane, with some inevitable compromises. Because the sphere and the plane appear in many areas of mathematics and its applications, so does the stereographic projection; it finds use in diverse fields including complex analysis, cartography, geology, and photography. In practice, the projection is carried out by computer or by hand using a special kind of graph paper called a stereonet or Wulff net.

In Cartesian coordinates  \xi=(\xi_1, \xi_2,...,\xi_{n+1}) on the sphere \mathbb S^n and x=(x_1,x_2,...,x_n) on the plane, the projection \pi : \xi \mapsto x and its inverse \pi^{-1}: x \mapsto \xi are given by the formulas

\displaystyle\xi _i = \begin{cases} \dfrac{{2{x_i}}}{{1 + {{\left| x \right|}^2}}},&1 \leqslant i \leqslant n, \hfill \\ \dfrac{{{{\left| x \right|}^2} - 1}}{{1 + {{\left| x \right|}^2}}},&i = n + 1. \hfill \\ \end{cases}

and

\displaystyle {x_i} = \frac{{{\xi _i}}}{{1 - {\xi _{n + 1}}}}, \quad 1 \leqslant i \leqslant n.

Let us show you an example making use of the projection. We assume v(x) verifies the following PDE

\displaystyle -\Delta v = \frac{n(n-2)}{4}v^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n.

Then the transformed function u(\xi), to be exact u(\pi^{-1}(x)), given by

\displaystyle v(x)=u(\pi^{-1}(x))\left( \frac{2}{1+|x|^2}\right)^\frac{n-2}{2}

satisfies the following PDE

\displaystyle -\Delta_g u + \frac{n(n-2)}{4}u = \frac{n(n-2)}{4}u^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n.

where \Delta_g denotes the Laplace-Beltrami operator with respect to the standard metric g on \mathbb S^n.

Similarly, if function u(\xi) verifying the PDE

\displaystyle -\Delta_g u + \frac{n(n-2)}{4}u = \frac{n(n-2)}{4}u^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n

then a new function v(x) given by

\displaystyle v(x)=u(\pi(x))\left( \frac{2}{1+|x|^2}\right)^\frac{n-2}{2}

will satisfy the following PDE

\displaystyle -\Delta v = \frac{n(n-2)}{4}v^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n.

More general, PDE

\displaystyle -\Delta_g u(\xi) + \frac{n(n-2)}{4}u(\xi) = K(\xi)u(\xi)^\frac{n+2}{n-2} \quad \text{ on } \mathbb S^n

becomes

\displaystyle -\Delta v(x) = K(\pi^{-1}(x))v(x)^\frac{n+2}{n-2} \quad \text{ on } \mathbb R^n.

In conclusion, using the stereographic projection we can transfer some geometric problems on sphere \mathbb S^n to ones in the whole space \mathbb R^n.

July 8, 2010

The Moser-Trudinger inequality


Followed by an entry where the Trudinger inequality had been discussed we now consider an important variant of it known as the Moser-Trudinger inequality.

Let us remind the Trudinger inequality

Theorem (Trudinger). Let \Omega \subset \mathbb R^n be a bounded domain and u \in W_0^{1, n}(\Omega) with

\displaystyle\int_\Omega {{{\left| {\nabla u} \right|}^n}dx} \leqslant 1.

Then there exist universal constants \beta>0, C_1>0 such that

\displaystyle\int_\Omega {\exp (\beta {|u|^\frac{n}{n-1}})dx} \leqslant {C_1}|\Omega |.

The Trudinger inequality has lots of application. For application to the prescribed Gauss curvature equation, one requires a particular value for the best constant \beta_0. In connection with his work on the Gauss curvature equation, J. Moser [here] sharpended the above result of Trungdier as follows

Theorem (Moser). Let \Omega \subset \mathbb R^n be a bounded domain and u \in W_0^{1, n}(\Omega) with

\displaystyle\int_\Omega {{{\left| {\nabla u} \right|}^n}dx} \leqslant 1.

Then there exist sharp constants \beta_0=\beta(n)>0, C_1=C_1(n)>0 given by

\displaystyle \beta_0=n\omega_{n-1}^{\frac{1}{n}-1}

such that

\displaystyle\int_\Omega {\exp (\beta {|u|^\frac{n}{n-1}})dx} \leqslant {C_1}|\Omega|, \quad \forall \beta \leqslant \beta_0.

The constant \beta_0 is sharp in the sense that for all \beta>\beta_0 there is a sequence of functions u_k \in W_0^{1,n}(\Omega) satisfying

\displaystyle\int_\Omega {{{\left| {\nabla u_k} \right|}^n}dx} \leqslant 1

but the integral

\displaystyle\int_\Omega {\exp (\beta {|u_k|^\frac{n}{n-1}})dx}

grow without bound.

For general compact closed manifold (M,g) the constant on the right hand side of the Moser-Trudinger inequality depends on the metric g. Working on a sphere (\mathbb S^2,g_c) with a canonical metric allows us to control the constants.

Theorem (Moser). There is a universal constant C_1>0 such that for all u \in W^{1,2}(\mathbb S^2) with

\displaystyle\int_{\mathbb S^2} {{{\left| {\nabla u} \right|}^n}dv_{g_c}}  \leqslant 1

and

\displaystyle\int_{\mathbb S^2} u dv_{g_c}=0

we have

\displaystyle\int_{\mathbb S^2} \exp(4\pi u^2)  \leqslant {C_1}.

Observe that

\displaystyle 4\pi = \int_{{\mathbb{S}^2}} {d{v_{{g_c}}}} <\int_{{\mathbb{S}^2}} {{e^{4\pi {u^2}}}d{v_{{g_c}}}} \leqslant {C_1}.

In the same way as we introduce in the entry concerning the Trudinger inequality one can show

Corollary. For

\displaystyle C_2 := \log C_1 +\log\frac{1}{4\pi}

one has

\displaystyle \log \overline\int_{{\mathbb{S}^2}} {{e^{2u}}d{v_{{g_c}}}} \leqslant \left[ {\frac{1}{{4\pi }}\int_{{\mathbb{S}^2}} {{{\left| {\nabla u} \right|}^2}d{v_{{g_c}}}} + 2\overline u } \right] + {C_2}

for all u \in W^{1,2}(\mathbb S^2).

Obviously, C_2 >0 since C_1 >4\pi. It turns out to determine the best constant C_2. This had been done by Onofri known as the Onofri inequality [here].

Theorem (Onofri).Let u \in W^{1,2}(\mathbb S^2) then we have

\displaystyle\log \overline\int_{{\mathbb{S}^2}} {{e^{2u}}d{v_{{g_c}}}} \leqslant \frac{1}{{4\pi }}\int_{{\mathbb{S}^2}} {{{\left| {\nabla u} \right|}^2}d{v_{{g_c}}}} + 2\overline u

with the equality iff

\Delta u +e^{2u}=1.

The proof of the Onofri inequality relies on a result due to Aubin

Theorem (Aubin). For all \varepsilon>0 there exists a constant C_\varepsilon such that

\displaystyle\log \overline\int_{{\mathbb{S}^2}} {{e^{2u}}d{v_{{g_c}}}} \leqslant \left[ {\left( {\frac{1}{2} + \varepsilon } \right)\frac{1}{{4\pi }}\int_{{\mathbb{S}^2}} {{{\left| {\nabla u} \right|}^2}d{v_{{g_c}}}} + 2\overline u } \right] + {C_\varepsilon }

for any u belonging to the following class

\displaystyle S = \left\{ {u \in {W^{1,2}}({\mathbb{S}^2}):\int_{{\mathbb{S}^2}} {{e^{2u}}{x_j}d{v_{{g_c}}}} = 0,j = \overline {1,3} } \right\}.

Source: S-Y.A. Chang, Non-linear elliptic equations in conformal geometry, EMS, 2004.

July 4, 2010

Nash-Moser’s iteration technique

Filed under: PDEs — Tags: — Ngô Quốc Anh @ 23:05

We now continue this topic. Today we discuss a very powerful technique known as the Nash-Moser iteration technique. Let us consider a sub-solution u of the following variational inequality

\displaystyle\int_\Omega {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }\varphi dx} \leqslant 0, \quad \varphi \in H_0^1(\Omega )

where \varphi \geqslant 0. We assume A is of class L^\infty(\Omega) and do satisfy an ellipticity condition. For the sake of simplicity we assume u\geqslant 0. Then by using the test function

\varphi=u^p\eta^2, \quad p>1

with \eta a cut-off function defined as in this topic we arrive at

\displaystyle\int_{B_R} {{A^{\alpha \beta }}(x){\nabla _\alpha }u{\nabla _\beta }({u^p}{\eta ^2})dx} \leqslant 0

which yields

\displaystyle p\int_{B_R} {{A^{\alpha \beta }}(x){u^{p - 1}}{\eta ^2}{\nabla _\alpha }u{\nabla _\beta }udx} + 2\int_{B_R} {{A^{\alpha \beta }}(x){u^p}\eta {\nabla _\alpha }u{\nabla _\beta }\eta dx} \leqslant 0.

Thus by the ellipticity condition

\displaystyle\int_{{B_R}} {{u^{p - 1}}{\eta ^2}{{\left| {\nabla u} \right|}^2}dx} \leqslant \frac{c}{p}\int_{{B_R}} {\left| {\nabla u} \right|{u^{\frac{{p - 1}}{2}}}{u^{\frac{{p + 1}}{2}}}\left| {\nabla \eta } \right|\eta dx}

and thus

\displaystyle\int_{{B_R}} {{u^{p - 1}}{\eta ^2}{{\left| {\nabla u} \right|}^2}dx} \leqslant \frac{c}{{{p^2}}}\int_{{B_R}} {{u^{p + 1}}{{\left| {\nabla \eta } \right|}^2}dx} .

(more…)

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