Mathematics (or something like that)

This blog on PF is a copy of the posts on my blogspot.com blog. There are pages and content not compatible with PF blogs, so please check out the original! The URL is http://letepsilonbegreaterthanzero.blogspot.com

I will very quickly summarize the main points of the blog here

For the summer of 2012 I am reading through two books, Fourier Analysis by Stein and Shakarchi, and The Road to Reality by Penrose. I am blogging about thoughts I have from these books. I think it would be pretty cool to have feedback here =)

In fall 2012, I will be starting a PhD program with plans of continuing the blog over my course work. If I pass quals in two years (fingers crossed) this blog may change focus.

Anyways, every time I have a new post I will post a link here. I won't have any of the pages or other info on this blog, but at least the bulk of the post will be the same.
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# fourier analysis: nonzero solution of the laplacian

Posted Jun3-12 at 10:42 PM by theorem4.5.9

View full entry at my blogspot.com blog: fourier analysis: nonzero solution of the laplacian

In the last remark of chapter 2, the authors make mention that if $u$ is harmonic on the disc (i.e. $\triangle u = 0$) and converges uniformly to $0$ as $r \rightarrow 1$, then $u$ must be identically $0$. However, if we relax the boundary condition to pointwise convergence, the conclusion no longer holds. Here is some matlab code I wrote up to investigate this problem (the solution is provided in the text).

Here are the two main figures this code creates. As the code describes, the first is a plot of the solution. The solution is unbounded (and hence not continuous on the compact disk), though is finite (i.e. defined) everywhere. Red means the solution is positive, blue negative, green zero (though this is only the boundary, and the $x$-axis. The second plot is the solution on a log-log scale (notice that even this is actually unbounded!). Read the code for more information, or better yet, run it!

% Example of a function z on closed unit disk that
% 1) is not identically equal to zero
% 2) satisfies the laplacian in it's interior
% 3) is zero on the boundary
% 4) limit(r->1) z(r,theta_naught) = 0 for every theta_naught.
%
% The reason the maximum principle does not apply is z is not continuous
% at (r,theta)=(1,0). In fact, the function is not even bounded on
% the closed disk.

r = linspace(0,1,30);
theta = linspace(0,2*pi,100);
[r theta] = meshgrid(r,theta);
z = 2*r.*sin(theta).*(1-r.^2)./(1+r.^2-2*r.*cos(theta)).^2;
% x = r.*cos(theta);
% y = r.*sin(theta);

% only for viewing positive/negative in the plot
for ii = 1:100;
for jj = 1:30;
if z(ii,jj)>0;
c(ii,jj)=1;
else if z(ii,jj)<0
c(ii,jj)=-1;
else c(ii,jj)=0;
end
end
end
end
clear ii jj

% what the solution looks like: red=positive, blue=negative, green=zero
figure;
mesh(r.*cos(theta),r.*sin(theta),z,c)

% scaled for clarity
figure;
mesh(r.*cos(theta),r.*sin(theta),sign(z).*log(log(abs(z)+1)+1)+1)

% the function is unbounded: compute the limit as (r,theta) spirals to (1,0)
syms u R T T_R

u = 2*R*sin(T)*(1-R^2)/(1+R^2-2*R*cos(T))^2;

% to find this value, use solve(diff(U,T),T) and use the third value
T_R = pi - acos(1/4/R*(R^2 - (R^4 + 34*R^2 + 1)^(1/2) + 1));
u = subs(u,T,T_R);
figure;
ezplot(u);
limit(u,R,1,'left')
% which approaches +inf
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