% chaotic_sqrt.m  :  a variant of apple_quiet.m that 
%   solves the ODE  y' = |y|^(1/2), subject to y(-2) = -1

% Try running: y(-2) = -1, i.e., t_0=-2, y_0=-1.  
% MATLAB seems to prefer the solution:
%
%   y(t) = (-1/4)t^2   for t <= 0, and
%   y(t) = (1/4)t^2    for t >= 0
%
% This is one valid solution of many.  The more general solution is to 
%    take any b >= 0, and to have
%   
%   y(t) = (-1/4)t^2       for t <= 0, 
%   y(t) = 0               for 0 <= t <= b, and 
%   y(t) = (1/4)(t-b)^2    for t >= b
%
% So try:
%   chaotic_sqrt(-2,2,1000,-1);
%   chaotic_sqrt(-2,2,100000,-1);
%
% So far, so good.  You should observe that MATLAB reports:
%   y(t_end)=y(2)=1, and this is the t=1 valule of (1/4)t^2.
% Hence MATLAB seems to prefer the b=0 solution.
%
% Since (1/4)t^2 = 0 at  t=0,  MATLAB should produce the same answer as
%   the above under the initial condition y(0)=0 for the value of y(t_end)
%   at t_end = 2 (or maybe not???...)
%
% Try:
%   chaotic_sqrt(0,2,10000,0);        % hence y_0 = 0
%   chaotic_sqrt(0,2,10000,1e-20);    % hence y_0 = 10^{-20}, which is near 0
%   chaotic_sqrt(0,2,10000,1e-40);    % hence y_0 = 10^{-40}, yet closer to 0
%   chaotic_sqrt(0,2,10000,-1e-1);
%   Explain...
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%  

function result = whatever(t_0,t_end,N,y_0)
                         % the name "whatever" is irrelevant
                         % MATLAB calls this function "chaotic_sqrt"
                         % assuming you have named this file chaotic_sqrt.m
 h = (t_end - t_0)/N;
 t = [t_0 : h: t_end];
  y(1) = y_0;
  for i=1:N
    y(i+1) = y(i) + h * f(t(i),y(i));
  end
  result = y;
  result(N+1)    % this is one thing to observe
end

% Second function: this is where we specify the function f.
% This function is purely local to this program.

function sqrt_y = f(t,y)
  sqrt_y = sqrt( abs(y) );
end