Ngô Quốc Anh

October 9, 2017

Jacobian of the stereographic projection not at the North pole

Filed under: Uncategorized — Ngô Quốc Anh @ 18:57

Denote \pi_P : \mathbb S^n \to \mathbb R^n the stereographic projection performed with P as the north pole to the equatorial plane of \mathbb S^n. Clearly when P is the north pole N, i.e. N = (0,...,0,1), then \pi_N is the usual stereographic projection.

As we have already known that, for arbitrary \xi \in \mathbb S^n, the image of \xi is

\displaystyle \pi_P : \xi \mapsto x = P+\frac{\xi-P}{1-\xi \cdot P}.

Here the point x \in \mathbb R^n is being understood as a point in \mathbb R^{n+1} by adding zero in the last coordinate. For the inverse map, it is not hard to see that

\displaystyle \pi_P^{-1} : x \mapsto \xi =\frac{|x|^2-1}{|x|^2+1}P+\frac 2{|x|^2+1}x.

The purpose of this entry is to compute the Jacobian of the, for example, \pi_P^{-1} by comparing the ratio of volumes.

First pick two arbitrary points x, y \in \mathbb R^n and denote \xi = \pi_P^{-1}(x) and \eta = \pi_P^{-1}(y). The Euclidean distance between \xi and \eta is

\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.

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October 3, 2017

Stereographic projection not at the North pole and an example of conformal transformation on S^n

Filed under: Riemannian geometry — Tags: , — Ngô Quốc Anh @ 12:09

Denote \pi_P : \mathbb S^n \to \mathbb R^n the stereographic projection performed with P as the north pole to the equatorial plane of \mathbb S^n. Clearly when P is the north pole N, i.e. N = (0,...,0,1), then \pi_N is the usual stereographic projection.

Clearly, for arbitrary x \in \mathbb S^n, the image of x is

\displaystyle \pi_P : x \mapsto y = P+\frac{x-P}{1-x \cdot P}.

For the inverse map, it is not hard to see that

\displaystyle \pi_P^{-1} : y \mapsto x =\frac{|y|^2-1}{|y|^2+1}P+\frac 2{|y|^2+1}y.

Derivation of \pi_P and \pi_P^{-1} are easy, for interested reader, I refer to an answer in . Let us now define the usual conformal transformation \varphi_{P,t} : \mathbb S^n \to \mathbb S^n given by

\displaystyle \varphi_{P,t} : x \mapsto \pi_P^{-1} \big( t \pi_P ( x) \big)

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February 23, 2017

In a normed space, finite linearly independent systems are stable under small perturbations

Filed under: Giải tích 8 (MA5206) — Ngô Quốc Anh @ 23:21

In this topic, we show that in a normed space, any finite linearly independent system is stable under small perturbations. To be exact, here is the statement.

Suppose (X, \|\cdot\|) is a normed space and \{x_1,...,x_n\} is a set of linearly independent elements in X. Then \{x_1,...,x_n\} is stable under a small perturbation in the sense that there exists some small number \varepsilon>0 such that for any \|y_i\| < \varepsilon with 1 \leqslant i \leqslant n, the all elements of \{x_1+y_1,...,x_n+y_n\} are also linearly independent.

We prove this result by way of contradiction. Indeed, for any \varepsilon>0, there exist n elements y_i \in X with \|y_i\| < \varepsilon such that all elements of \{x_1+y_1,...,x_n+y_n\} are linearly dependent, that is, there exist real numbers \alpha_i with 1 \leqslant i \leqslant n such that

\displaystyle \alpha_1 (x_1+y_1) + \cdots + \alpha_n (x_n+y_n) =0

with

\displaystyle |\alpha_1| + \cdots + |\alpha_n| >0.

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December 11, 2016

Why the equation Δu+1/u^α=0 in R^n has no positive solution?

Filed under: Uncategorized — Ngô Quốc Anh @ 23:50

Following the idea due to Brezis in his note that we had already mentioned once, it seems that we can show that the following equation

\displaystyle \Delta u + \frac 1{u^\alpha}=0

in \mathbb R^n with n\geqslant 2 has no positive C^2-solution. Here we require \alpha>0.

Indeed, to see this, we first denote

f(t)=-t^{-\alpha}

with t>0. Clearly the function f is monotone increasing in its domain. Using this, we rewrite our equation as follows

\Delta u =f(u).

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September 9, 2016

Benefits of “complete” and “compact” for analysis on Riemannian manifolds

Filed under: Riemannian geometry — Tags: — Ngô Quốc Anh @ 10:08

When working on Riemannian manifolds, it is commonly assumed that the manifold is complete and compact. (The case of non-compactness is also of interest too.) In this entry, let us review the role of completeness and compactness in this setting.

How important the completeness is? Let us recall that for given a Riemannian manifold (M,g), what we have is a nice structure as well as an appropriate analysis on any tangent space T_pM. For a C^1-curve \gamma : [a,b] \to M on M, the length of \gamma is

\displaystyle L(\gamma) = \int_a^b \sqrt{g(\gamma (t)) \langle \partial_t \gamma \big|_t, \partial_t \gamma \big|_t\rangle} dt

where \partial_t \gamma\big|_t \in T_{\gamma (t)}M a tangent vector. (Note that by using curves, the tangent vector \partial_t \gamma\big|_t is being understood as follows

\displaystyle  \partial_t \gamma\big|_t (f) = (f \circ \gamma)'(t)

for any differentiable function f at \gamma(t).) Length of piecewise C^1 curves can be defined as the sum of the lengths of its pieces. From this a distance on M whose topology coincides with the old one on M is given as follows

\displaystyle d_g(x,y) = \inf_\gamma L(\gamma)

where the infimum is taken on all over the set of all piecewise C^1 curves connecting x and y.

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April 20, 2016

Stereographic projection, 6

Filed under: PDEs, Riemannian geometry — Tags: — Ngô Quốc Anh @ 1:08

I want to propose an alternative way to calculate the Jacobian of the stereographic projection \mathcal S. 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 and its inverse 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.

It is well-known that the Jacobian of the stereographic projection \mathcal S: \xi \mapsto x is

\displaystyle \frac{\partial \xi}{\partial x} = {\left( {\frac{2}{{1 + {{\left| x \right|}^2}}}} \right)^n}.

The way to calculate its Jacobian is to compare the ratio of volumes. First pick two arbitrary points x, y \in \mathbb R^n and denote \xi = \mathcal S(x) and \eta = \mathcal S(y).

The Euclidean distance between \xi and \eta is

\displaystyle |\xi -\eta|^2 = \sum_{i=1}^{n+1} |\xi_i - \eta_i|^2 =\sum_{i=1}^n |\xi_i - \eta_i|^2+|\xi_{n+1} - \eta_{n+1}|^2.

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August 9, 2015

The third fundamental lemma in the method of moving spheres

Filed under: Uncategorized — Ngô Quốc Anh @ 4:03

In this post, we proved the following result (appeared in a paper by Y.Y. Li published in J. Eur. Math. Soc. (2004))

Lemma 1. For n \geqslant 1 and \nu \in \mathbb R, let f be a function defined on \mathbb R^n and valued in [-\infty, +\infty] satisfying

\displaystyle {\left( {\frac{\lambda }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) \leqslant f(y), \quad \forall |y - x| > \lambda > 0.

Then f is constant or \pm \infty.

Later, we considered the equality case in this post and proved the following result:

Lemma 2. Let n\geqslant 1, \nu \in \mathbb R and f \in C^0(\mathbb R^n). Suppose that for every x \in \mathbb R^n there exists \lambda(x)>0 such that

\displaystyle {\left( {\frac{\lambda(x) }{{|y - x|}}} \right)^\nu }f\left( {x + {\lambda(x) ^2}\frac{{y - x}}{{|y - x{|^2}}}} \right) =f(y), \quad \forall |y - x| > 0.

Then for some a \geqslant 0, d>0 and \overline x \in \mathbb R^n

\displaystyle f(x) = \pm a{\left( {\frac{1}{{d + {{\left| {x - \overline x } \right|}^2}}}} \right)^{\frac{\nu }{2}}}.

In this post, we consider the third result which can be stated as follows:

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April 28, 2015

On the simplicity of the first eigenvalue of elliptic systems with locally integrable weight

Filed under: Uncategorized — Ngô Quốc Anh @ 0:52

Of interest in this note is the simplicity of the first eigenvalue of the following problem

\begin{array}{rcl}-\text{div}(h_1 |\nabla u|^{p-2}\nabla u) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}v \quad \text{ in } \Omega\\-\text{div}(h_2 |\nabla v|^{q-2}\nabla v) &=& \lambda |u|^{\alpha-1} |v|^{\beta-1}u \quad \text{ in } \Omega\\u &=&0 \quad \text{ on } \partial\Omega\\v &=&0 \quad \text{ on } \partial\Omega\end{array}

where 1 \leqslant h_1, h_2 \in L_{\rm loc}^1 (\Omega) and \alpha, \beta>0 satisfy

\displaystyle \frac \alpha p + \frac \beta q = 1

with p,q >1. A simple variational argument shows that \lambda exists and can be characterized by

\lambda = \inf_{\Lambda} J(u,v)

where

\displaystyle J(u,v)=\frac \alpha p \int_\Omega h_1 |\nabla u|^p dx + \frac \beta q \int_\Omega h_2 |\nabla v|^q dx

and

\Lambda = \{(u,v) \in W_0^{1,p} (\Omega) \times W_0^{1,q} (\Omega) : \Lambda (u,v) = 1\}

with

\displaystyle \Lambda (u,v)= \int_\Omega |u|^{\alpha-1}|v|^{\beta-1} uv dx.

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April 10, 2015

Existence of antiderivative of discontinuous functions

Filed under: Uncategorized — Ngô Quốc Anh @ 0:17

It is well-known that every continuous functions admits antiderivative. In this note, we show how to prove existence of antiderivative of some discontinuous functions.

A typical example if the following function

f(x)=\begin{cases}\sin \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}

By taking to different sequences x_k = 1/(2k\pi) and y_k = 1/(\pi/2 + 2k\pi) we immediately see that f is discontinuous at x=0. However, we will show that f admits F as its antiderivative.

To this end, we first consider the following function

G(x)=\begin{cases}x^2\cos \frac 1x, & \text{ if } x \ne 0,\\ 0, & \text{ if }x=0.\end{cases}

First we show that G is differentiable. Clearly whenever x \ne 0, we obtain

\displaystyle G'(x)=\sin \frac 1x + 2x \cos \frac 1x.

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April 5, 2015

The set of continuous points of Riemann integrable functions is dense

Filed under: Uncategorized — Ngô Quốc Anh @ 15:03

In this note, we prove that the set of continuous point of Riemann integrable functions f on some interval [a,b] is dense in [a,b]. Our proof start with the following simple observation.

Lemma: Assume that P=\{t_0=a,...,t_n=b\} is a partition of [a,b] such that

\displaystyle U(f,P)-L(f,P)<\frac{b-a}m

for some m; then there exists some index i such that M_i-m_i < \frac 1m where M_i and m_i are the supremum and infimum of f over the subinterval [t_{i-1},t_i].

We now prove this result.

Proof of Lemma: By contradiction, we would have M_i-m_i \geqslant \frac 1m for all i; hence

\displaystyle \frac{b-a}m = \sum_{i} \frac{t_i-t_{i-1}}{m}\leqslant \sum_{i} \big(M_i-m_i\big)(t_i-t_{i-1})=U(f,P)-L(f,P),

which gives us a contradiction.

We now state our main result:

Theorem. Let f be Riemann integrable over [a,b]. Define

\displaystyle \Gamma = \{ x\in [a,b] : f \text{ is continuous at } x\}

Then \Gamma is dense in [a,b].

(more…)

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