function Nov10()
% If Tu = f and f(2:n-1) = 0, then
% (1)    2u(1)-u(2)           = f(1)
% (2)    -u(k-1)+2u(k)-u(k+1) = 0
% (3)    -u(n-1)+2u(n)        = f(n)

% Suppose for some k with 2<=k<=n-1 that u(k) = max(u). From (2) we
% 
%                   u(k) = (u(k-1)+u(k+1))/2
%
% It follows that u(1)=...=u(n)