Messages

1516

Misc Math / Re: Mean-value theorem

« on: November 27, 2012, 09:37:58 AM »

Here we integrate over V. Is this V the same as omega? And do we know if the coefficient c in G(x,y) is positive or negative? Thanks!

Yes, should be $\Omega$. No idea what coefficient your are talking about

1517

Misc Math / Re: Mean-value theorem

« on: November 27, 2012, 03:46:53 AM »

I see, thanks.

Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?

Yes, it is exactly correct except $\Sigma \subset \Omega$. Alternatively we can write $\max_{\bar{\Omega}}u \geq \max_{\Sigma}u$ where $\bar{\Omega}=\Omega \cup \Sigma$.

1518

Misc Math / Re: Lec 26 MVT

« on: November 27, 2012, 03:43:54 AM »

Hi Professor and fellow classmates,
in the mean value theorem, it says that the first term in the right-hand expression (7) is 0, but does not state that the Laplacian is 0. Is the Laplacian assumed to be zero in the notes?

Yes--in this place, I put it now explicitly.

And I assume that is how I will prove 2a) of HA8 for a non-zero Laplacian.
Thanks in advance!

It has some sign, so instead of equalities you get inequalities.

1519

Misc Math / Re: Mean-value theorem

« on: November 26, 2012, 04:00:59 PM »

Hi Professor (and fellow classmates),
I am wondering the same question as Thomas; specifically, does it mean this? (See attached)

No, what you are writing is wrong. Function $G(x,y)$ depends on both $x$ and $y$ and we need to differentiate with respect to $x$ not $y$. So in fact I mean is
$$
\nabla_x G(x,y) \cdot \nu(x)
$$
where $\nabla_x$ means gradient with respect to $x$. Here $y$ is considered as a parameter

1520

Misc Math / Re: Lec 26 #3

« on: November 26, 2012, 03:19:45 PM »

In Lec 26, equation (3) involves "antisymmetrizing by u, w (permutting u, w and subtracting from original formula)...

What does antisymmetrizing mean? Or Permutting for that matter? I'm not sure how we move from (2) to (3) to (4)...
 
Can someone kindly clarify?

It means that we have expression $\mathcal{P}(u,w)$ depending on $u$ and $v$. Now we consider expression $\mathcal{P}(w,u)$ so interchange $u$ and $w$.

"Symmetrizing" would mean that we consider $\mathcal{P}(u,w)+\mathcal{P}(w,u)$ (or rather divided by $2$).

"Antisymmetrizing" would mean that we consider $\mathcal{P}(u,w)-\mathcal{P}(w,u)$ (or rather divided by $2$).

1521

Misc Math / Re: Mean-value theorem

« on: November 26, 2012, 03:15:53 PM »

or rather $\nabla G \cdot  \frac{\partial}{\partial x}(\vec{v})$?

No, $\nu_x$ means only that we differentiate with respect to $x$

1522

Misc Math / Re: Mean-value theorem

« on: November 25, 2012, 06:19:55 PM »

Try again

1523

Misc Math / Re: Mean-value theorem

« on: November 25, 2012, 05:26:24 PM »

Hi Professor Ivrii. I have a question on the attachment above. In part b, after dragging out the factor, why the remainder is (partial u)/(partial v) instead of u? Thanks.

Right, fixed and details added
http://www.math.toronto.edu/courses/apm346h1/20129/L26.html#sect-26.4

1524

Misc Math / Re: Mean-value theorem

« on: November 24, 2012, 07:07:57 PM »

b. Gauss identity (you may call it Green in 2D). We proved this before.

c. $\Sigma$ is a sphere of radius $r$ with a center $y$ so $G$ so for $x\in \Sigma$  $G(x,y)=c |x-y|^{2-n}=cr^{2-n}$ is constant and could be moved out of integral.

1525

Technical Questions / Re: LaTeX advices

« on: November 20, 2012, 12:35:02 PM »

hyperref is probably my favourite package

Look at xr-hyper (working in conjugation with hyperref to provide cross-references of different documents).

For graphics (pictures), I would add \usepackage{graphicx}, but it is maybe not so relevant for math.

graphicx is to include existing pictures -- I use it all the time. I am talking about LaTeX drawings (with "frontend" tikz)

Try this: in full LaTeX

Code: [Select]

\documentclass[12pt]{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\fill [red, even odd rule,opacity=.8] (0,2) circle (2.5) circle (3);
\fill [blue, even odd rule,opacity=.8] (1.732,-1) circle (2.5) circle (3);
\fill [green, even odd rule,opacity=.8] (-1.732,-1) circle (2.5) circle (3);
\begin{scope}
\clip  (.9,-.6) circle (.5);
\fill [red, even odd rule,opacity=.8] (0,2) circle (2.5) circle (3);
\end{scope}
\clip (-2.6,1.5) circle (.5);
\fill [red, even odd rule,opacity=.8] (0,2) circle (2.5) circle (3);
\end{tikzpicture}
\end{document}

1526

Technical Questions / LaTeX advices

« on: November 20, 2012, 06:14:04 AM »

We are talking about LaTeX in full, not MathJax

1) Strongly recommended: \documentclass[]{memoir}--very flexible well maintained, can emulate book and article. It allows normalfontsize 14 pt \documentclass[14pt]{memoir} -- note how tests are typed for convenience

2) Strongly recommended: \usepackage{hyperref}--makes better pdf, clickable links (in \label--\ref, \bibitem-\cite, in table of content, index) and also bookmarks

3) Strongly recommended \usepackage{microtype}--better typesetting

4) For graphics (LaTeX drawings) \usepackage{pgf}--the most powerful (another pstricks is also great)

5) For presentations \documentclass{beamer}-- nearest competition powerdot

1527

Home Assignment 7 / Re: Problem 4

« on: November 20, 2012, 06:03:07 AM »

Calvin: I think you should set your constants to zero, as in Lecture 24.

Not really: lecture suggests that we can do it imposing an extra condition $\int _\Omega u \, dS=0$.

A better question: why solution exists? Related: Does solution of the same equation but with b.c. $u_r|_{r=a}=1$ exist?

1528

Home Assignment 7 / Re: Problem 3

« on: November 20, 2012, 05:57:19 AM »

Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!

Yes, simpler definitely to write $n=2m+1$ with $m=0,1,\ldots$ as "even" coefficients are $0$.

1529

Home Assignment 7 / Re: Problem 2

« on: November 20, 2012, 05:48:17 AM »

Calvin: I wasn't really sure that for 2.b) you could have a Bessel function of a complex variable, or if you'd need to do something slightly different?

I read something about this in the textbook (Strauss), but don't have it with me and might not remember correctly. ;P

Actually Bessel functions are analytic functions.


If you consider $n$-dimensional problem and not necessarily spherically symmetric solution then
\begin{equation}
\Delta =\partial_r^2 + (n-1)r^{-1}\partial_r +\r^{-2}\Lambda
\label{eq-1}
\end{equation}
where $\Lambda$ is Laplace-Beltrami on the sphere ($(n-1)$-dimensional sphere $\mathbb{S}^{n-1}$ in $\mathbb{R}^n$).

If we consider equation $\Delta u=0$ and try to solve it by separation of variables we get
\begin{equation}
\frac{r^2 R''+(n-1)r R'}{R}+\frac{\Lambda V}{V}=0
\label{eq-2}
\end{equation}
where $V$ contains spherical variables (except $r$) and therefore
\begin{align}
&\Lambda V=-\lambda V,\label{eq-3}\\
&r^2 R''+(n-1)r R'=\lambda R.\label{eq-4}
\end{align}
One can prove that such solutions $U=VR$ are polynomials of degree $m=0,1,2,\ldots$ with respect to $x$ (assuming solutions are not singular at $0$) and therefore $R=r^m$ (don't care about coefficient) and therefore $\lambda_m= m(m+n-2)$ but their multiplicities are large: the most interesting case is $n=3$ when $\lambda_m=m(m+1)$ of multiplicity $2m+1$ (application to theory of Hydrogen atom).

Solutions of (\ref{eq-3}) are spherical functions which I don't describe. Let us return to $k\ne 0$. In this case separating variables we get
\begin{equation}
\Bigl[\frac{r^2 R''+(n-1)r R'}{R}\mp k^2r^2\Bigr]+\frac{\Lambda V}{V}=0
\label{eq-5}
\end{equation}
and therefore we get (\ref{eq-3}) implying $\lambda=m(m+n-2)$ and instead of (\ref{eq-4}) we get
\begin{equation}
r^2 R''+(n-1)r R' \mp k^2r^2R -\lambda R=0
\label{eq-6}
\end{equation}
which is Bessel equation. By scaling $r_{new}=k r_{old}$ we can get rid off $k$ (and even off sign making complex scaling).

I just want to note that Bessel functions depend on $n$ and $m$. However for odd $n$ (in particular as $n=3$) and $\lambda=0$ they are elementary functions (see problem 1 for $n=3$)

1530

Home Assignment 7 / Re: Problem 1

« on: November 20, 2012, 05:18:40 AM »

http://www.math.toronto.edu/courses/apm346h1/20129/HA7.html#problem-7.1

There is an unfinished business: found solutions satisfy $\Delta u=\pm k^2u0$ as $r>0$ but not necessarily in the origin. Which solutions satisfy this equation in the origin?