Ngô Quốc Anh

September 26, 2010

Subharmonic functions

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 1:53

In this entry, we shall discuss a geometric meaning of subharmonic functions. This will help us to easily remember the definition of subharmonic functions.

In mathematics, a harmonic function is a twice continuously differentiable function f : U\to \mathbb R (where U is an open subset of \mathbb R^n) which satisfies Laplace’s equation, i.e.

\displaystyle\frac{\partial^2f}{\partial x_1^2} + \frac{\partial^2f}{\partial x_2^2} + \cdots + \frac{\partial^2f}{\partial x_n^2} = 0

everywhere on U. This is usually written as

\textstyle \Delta f = 0.

In 1D, this condition is about to say that f is harmonic if and only if f is linear. Concerning to the case of functions with one-variable, we have the s0-called convexity saying that function f is convex if and only if the function lies below or on the straight line segment connecting two points, for any two points in the interval. Mathematically, a function f is said to be convex if

\textstyle \Delta f \geqslant 0.

In higher dimension, the notion of linearity and convexity become harmonicity and subharmonicity. Precisely, two points mentioned above become a hyper-surface, for e.g. like a curve in 2D and a straight line becomes a graph of harmonic function. In practice, the closed interval connecting those two points will be replaced by a closed ball. Therefore, we have

Definition. A C^2 function that satisfies \Delta f \ge 0 is called subharmonic. More generally, a function is subharmonic if and only if, in the interior of any ball in its domain, its graph lies below that of the harmonic function interpolating its boundary values on the ball.

Let us consider several examples in 2D.

  • \log functions.

It is well-known that in 2D function \log|z|, where z=(x,y), is harmonic. Therefore, every functions lying below the graph of \log|z| turns out to be subharmonic.

  • \sin functions.

Again, one can easily show that e^x \sin y is harmonic.

September 1, 2010

The inverse of the Laplace transform by contour integration

Filed under: Giải tích 7 (MA4247) — Ngô Quốc Anh @ 17:23

Usually, we can find the inverse of the Laplace transform \mathcal L[\cdot](s) by looking it up in a table. In this entry, we show an alternative method that inverts Laplace transforms through the powerful method of contour integration.

Consider the piece-wise differentiable function f(x) that vanishes for x < 0. We can express the function e^{-cx}f(x) by the complex Fourier representation of

\displaystyle f(x){e^{ - cx}} = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{i\omega x}}\left[ {\int_0^\infty {{e^{ - ct}}f(t){e^{ - i\omega t}}dt} } \right]d\omega }

for any value of the real constant c, where the integral

\displaystyle I = \int_0^\infty {{e^{ - ct}}|f(t)|dt}

exists. By multiplying both sides of first equation by e^{cx} and bringing it inside the first integral

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{ - \infty }^\infty {{e^{(c + i\omega )x}}\left[ {\int_0^\infty {f(t){e^{ - (c + i\omega )t}}dt} } \right]d\omega } .

With the substitution z = c+\omega i, where z is a new, complex variable of integration,

\displaystyle f(x) = \frac{1}{{2\pi }}\int_{c - \infty i}^{c + \infty i} {{e^{zx}}\left[ {\int_0^\infty {f(t){e^{ - zt}}dt} } \right]d\omega }.

The quantity inside the square brackets is the Laplace transform \mathcal L[f](z). Therefore, we can express f(t) in terms of its transform by the complex contour integral

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August 22, 2010

Liouville’s theorem and related problems

Filed under: Giải tích 7 (MA4247), PDEs — Tags: — Ngô Quốc Anh @ 6:35

The following theorem is well-known

Theorem (Liouville). Let \Omega be a simply connected domain in \mathbb R^2. Then all real solutions of

\displaystyle \Delta u +2Ke^u=0

in \Omega where K a constant, are of the form

\displaystyle u=\log\frac{|f'|^2}{\left(1+\frac{K}{4}|f|^2\right)^2}

where f is a locally univalent meromorphic function in \Omega.

In geometry, our PDE

\displaystyle \Delta u +2Ke^u=0

says that under the case \Omega=\mathbb R^2, it holds

e^u|dz|^2=f^*g_K

where g_K denotes the standard metric on \mathbb S^2 with constant curvature K. Thus we have

Corollary. All solutions of the PDE in \mathbb R^2 with K>0 and

\displaystyle \int_{\mathbb R^2} e^u<\infty

are of the form

\displaystyle u(x)=\log\frac{16\lambda^2}{\left(4+\lambda^2K|x-x_0|^2\right)^2},\quad \lambda>0, \quad x_0 \in \mathbb R^2.

(more…)

August 17, 2010

Evaluate complex integral via the Fourier transform

Filed under: Giải tích 7 (MA4247) — Tags: — Ngô Quốc Anh @ 5:56

As suggested from this topic, we are interested in evaluating the following complex integral

\displaystyle G(t)=\mathop {\lim }\limits_{A \to \infty } \int\limits_{ - A}^A {{{\left( {\frac{{\sin x}} {x}} \right)}^2}{e^{itx}}dx}.

The trick here is to use the Fourier transform. Thanks to ZY for teaching me this interesting technique.

In \mathbb R, the Fourier transform of function f, denoted by \mathcal F[f], is defined to be

\displaystyle \mathcal F[f](y) = \int_{ - \infty }^\infty {f(x){e^{ - 2\pi ixy}}dx}.

If we apply the Fourier transform twice to a function, we get a spatially reversed version of the function. Precisely,

\displaystyle\begin{gathered} \mathcal{F}\left[ {\mathcal{F}[f]} \right](z) = \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{ - 2\pi iyz}}dy} \hfill \\ \qquad\qquad= \int_{ - \infty }^\infty {\mathcal{F}[f](y){e^{2\pi iy( - z)}}dy} \hfill \\ \qquad\qquad= {\mathcal{F}^{ - 1}}\left[ {\mathcal{F}[f]} \right]( - z) \hfill \\ \qquad\qquad= f( - z) \hfill \\ \end{gathered}

where \mathcal F^{-1} denotes the inverse Fourier transform.

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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}}} .

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April 4, 2010

Growth of an entire function

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247) — Ngô Quốc Anh @ 16:04

The classical Liouville theorem says that

Theorem (Liouville). Every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f(z)| \leqslant M  for all z in \mathbb C is constant.

The aim of this entry to is generalize the constant M, precisely, what happen if we replace M by a polynomial?

Followed by this topic we can prove the following theorem.

Theorem (Generalized Liouville). Assume f is an entire function. If |f(z)| \leqslant A+B|z|^p for all z \in \mathbb C with some positive numbers A,B,p, then f is a polynomial of degree bounded by p.

Proof. Denote the Laurent expansion of f by

\displaystyle f\left( z \right) = \sum\limits_{n = - \infty }^{ + \infty } {{a_n}{z^n}}

where

\displaystyle {a_n} = \frac{1} {{2\pi i}}\int\limits_{\left| z \right| = r < 1} {\frac{{f\left( z \right)dz}} {{{z^{n + 1}}}}}.

Since f is entire, all coefficients a_n with n<0 are zero, i.e.

\displaystyle f\left( z \right) = \sum\limits_{n =0 }^{ + \infty } {{a_n}{z^n}}.

For any integers n>p, one has

\displaystyle\left| {{a_n}} \right| \leqslant \frac{1}{{2\pi }}\int\limits_{\left| z \right| = r < 1} {\left| {\frac{{f\left( z \right)}}{{{z^{n + 1}}}}} \right|dz} = \frac{1}{{2\pi {r^n}}}\int_0^{2\pi } {f(r{e^{i\theta }})d\theta } \leqslant \frac{{A + B{r^p}}}{{2\pi {r^n}}}.

Letting r\to \infty we obtain that a_n=0 which implies that f is a polynomial of degree at most p.

Corollary. For a given entire function f, if the following limit

\displaystyle\mathop {\lim }\limits_{|z| \to \infty } \frac{{f(z)}}{{|z{|^p}}}

exists then f is a polynomial of degree at most p.

Question. What happen if

\displaystyle\mathop {\lim }\limits_{|z| \to \infty } |z{|^p}f(z)

exists?

December 17, 2009

Schwarz’s Lemma, Schwarz-Pick theorem, and some applications involving inequalities

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 10:52

In mathematics, the Schwarz lemma, named after Hermann Amandus Schwarz, is a result in complex analysis about holomorphic functions defined on the open unit disk.

Schwarz’s Lemma: Let D=\{z : |z|<1\} be the open unit disk in the complex plane \mathbb C. Let f : D \to \overline D be a holomorphic function with f(0)=0. The Schwarz lemma states that under these circumstances |f(z)| \leq |z| for all z \in D, and |f'(0)| \leq 1. Moreover, if the equality |f(z)|=|z| holds for any z \ne 0, or |f'(0)|=1 then f is a rotation, that is, f(z)=az with |a=1.

This lemma is less celebrated than stronger theorems, such as the Riemann mapping theorem, which it helps to prove; however, it is one of the simplest results capturing the “rigidity” of holomorphic functions. No similar result exists for real functions, of course. To prove the lemma, one applies the maximum modulus principle to the function \frac{f(z)}{z}.

Proof: Let g(z)=\frac{f(z)}{z}. The function g(z) is holomorphic in D (excluding 0) since f(0)=0 and f is holomorphic. Let D_r be a closed disc within D with radius r. By the maximum modulus principle,

\displaystyle |g(z)| = \frac{|f(z)|}{|z|} \leq \frac{|f(z_r)|}{|z_r|} \le \frac{1}{r}

for all z in D_r and all z_r on the boundary of D_r. As r approaches 1 we get |g(z)| \leq 1. Moreover, if there exists a $z_0$ in D such that g(z_0)=1. Then, applying the maximum modulus principle to g, we obtain that g is constant, hence f(z)=kz, where k is constant and |k|=1. This is also the case if |f'(0)|=1.

A variant of the Schwarz lemma can be stated that is invariant under analytic automorphisms on the unit disk, i.e. bijective holomorphic mappings of the unit disc to itself. This variant is known as the Schwarz-Pick theorem (after Georg Pick):

 Schwarz-Pick theorem: Let f : D \to D be holomorphic. Then, for all z_1, z_2 \in D,

\displaystyle\left|\frac{f(z_1)-f(z_2)}{1-\overline{f(z_1)}f(z_2)}\right| \le \frac{\left|z_1-z_2\right|}{\left|1-\overline{z_1}z_2\right|}

and, for all z \in D

\displaystyle\frac{\left|f'(z)\right|}{1-\left|f(z)\right|^2} \le \frac{1}{1-\left|z\right|^2}.

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August 29, 2009

On a polynomials of degree n, having all its zeros in the unit dis

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Tags: — Ngô Quốc Anh @ 21:07

I found a very useful inequality involving a polynomials of degree n, having all its zeros in the unit disk from the following paper, doi:10.1016/j.jmaa.2009.07.049, published in J. Math. Anal. Appl.

Statement. If P(z) is a polynomial of degree n, having all its zeros in the disk |z| \leq 1, then

\left| {zP'(z)} \right| \geq \displaystyle\frac{n} {2}\left| {P(z)} \right|

for |z|=1.

Proof. Since all the zeros of P(z) lie in $latex|z| \leq 1$. Hence if z_1, z_2,...,z_n are the zeros of P(z), then |z_j| \leq 1 for all j =1,2,...,n. Clearly,

\displaystyle\Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} = \sum\limits_{j = 1}^n {\Re \frac{{{e^{i\theta }}}}{{{e^{i\theta }} - {z_j}}}} ,

for every point e^{i\theta}, 0 \leq \theta < 2 \pi which is not a zero of P(z). Note that

\displaystyle\sum\limits_{j = 1}^n {\Re \frac{e^{i\theta }}{e^{i\theta } - z_j}}\geq\sum\limits_{j = 1}^n{\frac{1}{2}}=\frac{n}{2}.

This implies

\displaystyle\left| {\frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}}} \right| \geq \Re \frac{{{e^{i\theta }}P'\left( {{e^{i\theta }}} \right)}}{{P\left( {{e^{i\theta }}} \right)}} \geq \frac{n}{2},

for every point e^{i \pi}, 0 \leq \theta < 2\pi. Hence

\left| {zP'(z)} \right| \geq \displaystyle \frac{n}{2}\left| {P(z)} \right|

for |z|=1 and this completes the proof.

August 14, 2009

How to find a conformal mapping between the quadrants and the semidisc

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:34

In the previous topic I show you by the following map

\displaystyle f : z \mapsto \frac{z+1}{z-1}

maps \{z : \Re z < 0\} onto \{w : |w|<1\}, and is conformal. Therefore the map

\displaystyle g : z \mapsto \frac{z+1}{1-z}

maps \{z : \Re z > 0\} onto \{w : |w|<1\}, and is conformal. What I am going to do is to prove that

\displaystyle \Re z > 0\quad \Leftrightarrow \quad \Re \left( {\frac{{z + 1}} {{1-z}}} \right) > 0.

To this purpose, we assume z=x+iy, i.e., \Re z = y. Now by a simple calculation

\displaystyle\frac{{z+1}}{{1-z}}=\frac{{\left({x+1}\right)+iy}}{{\left({1-x}\right)-iy}}=\frac{{\left[{\left({x+1}\right)+iy}\right]\left[{\left({1-x}\right)+iy}\right]}}{{{{\left({1-x}\right)}^{2}}+{y^{2}}}}

which yields

\displaystyle \Re \left( {\frac{{z + 1}} {{1 - z}}} \right) = \frac{{2y}} {{{{\left( {1 - x} \right)}^2} + {y^2}}}.

Having this fact we can easily see that under the map g the first and fourth quadrants maps to upper and lower semidisks, respectively.

An easy way to construct a conformal mapping between upper half plane and the open unit disk

Filed under: Các Bài Tập Nhỏ, Giải tích 7 (MA4247), Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 20:54

Thelocus |z + 1|= | z -1| is the perpendicular bisector of the line segment joining -1 to 1, that is, the imaginary axis. The set |z + 1|< | z -1| is then the set of points z closer to -1 than to 1, that is, the left half-plane \Re z <0. Hence, \Re z <0 if and only if

\displaystyle\frac{|z+1|}{|z-1|}<1.

The map

\displaystyle f : z \mapsto \frac{z+1}{z-1}

maps \{z : \Re z < 0\} onto \{w : |w|<1\}, and is conformal as f' \ne 0. The inverse map is easily seen to be

\displaystyle w \mapsto z=\frac{w+1}{w-1}.

The locus |z + i|= | z - i| is the perpendicular bisector of the line segment joining -i to i, that is, the real axis. The set |z + i|< | z -i| is then the set of points z closer to -i than to i, that is, the lower half-plane \Im z <0. Hence, \Im z <0 if and only if

\displaystyle \frac{|z+i|}{|z-i|}<1.

The map

\displaystyle g : z \mapsto \frac{z+i}{z-i}

maps \{z : \Im z < 0\} onto \{w : |w|<1\}, and is conformal as g' \ne 0. The inverse map is easily seen to be

\displaystyle w \mapsto z=\frac{w+1}{w-1}i.

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