# Ngô Quốc Anh

## March 31, 2009

### Mellin transform with some examples

Filed under: Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 13:11

In mathematics, the Mellin transform is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series, and is often used in number theory and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform, and the theory of the gamma function and allied special functions.

The Mellin transform of a function is

$\displaystyle\mathcal{M}\left\{ {f\left( x \right)} \right\}(s) = \varphi (s) = \int_0^\infty {{x^s}} f(x)\frac{{dx}} {x}$.

The inverse transform is

$\displaystyle {\mathcal{M}^{ - 1}}\left\{ {\varphi \left( s \right)} \right\}(x) = f(x) = \frac{1} {{2\pi i}}\int_{c - i\infty }^{c + i\infty } {{x^{ - s}}} \varphi (s)ds$.

The notation implies this is a line integral taken over a vertical line in the complex plane. Conditions under which this inversion is valid are given in the Mellin inversion theorem. The transform is named after the Finnish mathematician Hjalmar Mellin.

## March 26, 2009

### Solving initial value problem for heat equation via Fourier transform

Filed under: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 21:05

Followed by Solving initial value problem for wave equation via Fourier transform, in this topic I will show you how to solve initial value problem for heat equation via Fourier transform.

Following is the problem

$\displaystyle\left\{\begin{gathered} {u_t} = {c^2}{u_{xx}}, \quad x \in \mathbb{R}, \hfill \\ u(x,0) = \varphi (x). \hfill \\ \end{gathered}\right.$

From the equation by taking Fourier transform to the both sides, we obtain

$\displaystyle\widehat{{u_t}\left( {\eta ,t} \right)} - c^2 {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0$.

This is an ODE, the solution is given by

$\displaystyle\widehat{u\left( {\eta ,t} \right)} = C(\eta)e^{-c^2 \eta^2 t}$.

From the initial date we get

$\displaystyle\widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)}$

which implies that

$\displaystyle\widehat{\varphi \left( \eta \right)} = C(\eta)$.

Thus, we obtain

$\displaystyle\widehat{u\left( {\eta ,t} \right)} = \widehat{\varphi \left( \eta \right)}e^{-c^2 \eta^2 t}$.

Finally, we obtain by taking the inverse Fourier transform

$\displaystyle u\left( {x,t} \right) = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x}}\widehat u\left( {\eta ,t} \right)d\eta } = \frac{1}{{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x - {c^2}{\eta ^2}t}}\hat \varphi \left( \eta \right)d\eta }$.

## March 13, 2009

### Green’s function and differential equations

Filed under: Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:31

In mathematics, a Green’s function is a type of function used to solve inhomogeneous differential equations subject to boundary conditions. The term is also used in physics, specifically in quantum field theory, electrodynamics and statistical field theory, to refer to various types of correlation functions, even those that do not fit the mathematical definition; for this sense, see Correlation function (quantum field theory) and Green’s function (many-body theory).

Green’s functions are named after the British mathematician George Green, who first developed the concept in the 1830s. In the modern study of linear partial differential equations, Green’s functions are largely studied from the point of view of fundamental solutions instead.

Definition and uses

Technically, a Green’s function, $G(x,s)$, of a linear differential operator $L=L(x)$ acting on distributions over a subset of the Euclidean space $\mathbb R^n$, at a point $s$, is any solution of

$LG(x,s)=\delta(x-s)$                         (1)

where $\delta$ is the Dirac delta function.

This technique can be used to solve differential equations of the form

$Lu(x)=f(x)$                                      (2)

If the kernel of $L$ is nontrivial, then the Green’s function is not unique. However, in practice, some combination of symmetry, boundary conditions and/or other externally imposed criteria will give a unique Green’s function. Also, Green’s functions in general are distributions, not necessarily proper functions.

Green’s functions are also a useful tool in condensed matter theory, where they allow the resolution of the diffusion equation, and in quantum mechanics, where the Green’s function of the Hamiltonian is a key concept, with important links to the concept of density of states. The Green’s functions used in those two domains are highly similar, due to the analogy in the mathematical structure of the diffusion equation and Schrödinger equation. As a side note, the Green’s function as used in Physics is usually defined with the opposite sign; that is, $LG(x,s)=-\delta(x-s)$. This definition does not significantly change any of the properties of the Green’s function.

## March 4, 2009

### Show that the closure of a convex set also is convex

Filed under: Các Bài Tập Nhỏ, Giải tích 8 (MA5206), Linh Tinh — Ngô Quốc Anh @ 23:29

Today on the way, my Chinese friend and I have discussed the following question: Let  denote a convex subset of a locally convex topological linear space . Show that the closure  of  also is convex.

I have suggested the following solution.

We define $f: X \times X \times \mathbb{R} \to X$ as following $f: \left( {x,y,\lambda } \right) \mapsto \lambda x + \left( {1 - \lambda } \right)y$. Then  is continuous and $f\left( {K \times K \times \left[ {0,1} \right]} \right) \subset K$ since  is convex. We then obtain that $f\left( {\overline K \times \overline K \times \left[ {0,1} \right]} \right) \subset \overline K$ due to the continuity of , that is $\lambda \overline K + \left( {1 - \lambda } \right)\overline K \in \overline K$ for every . Therefore  is a convex set.

So what I am going to tell you is how correct the solution is? If no, what’s the problem, otherwise, what’s the main point? I will show you a little bit latter. I think I should go for sleep, it’s late now 😦