Math 293 Prelim 1 Solutions and regrades.    10/3/96
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

1) REGRADE POLICY. If you think your prelim has been graded 
improperly:
  a) Don't put any marks on your prelim.  Photocopy the
     relevant page(s) if you need to make marks.
  b) Write a note about why you think points should be
     add or subtracted for a given problem and why.
     Attach this note to your prelim with a paperclip
     (put your name on the note).
  c) turn this in to your TA at your section on
     October 11 or 12 (or some time sooner).


2) The 1 page solutions to prelim 1 passed out in lecture
have an arithmetic error in problem 2b. The mistake happens at the
calculation of x1 and propagates to the end.  Problem 2b should
read
x0 = 5
x1 = 5    + .5 * (2*5    + 1)  =  10.5     (not  5.55)
x2 = 10.5 + .5 * (2*10.5 + 1)  =  21.5     (not 11.6 )
                                  ^^^^ 

3) Rough reasoning for problem bonus problem 4. 
Given y'=sqrt(y), y(0)=0, y(3)=1:  any solutions?

Note that for any point with y>0 there exists a unique solution 
through that point.
For y=0 the uniqueness criteria is not met and there is the
possibility of multiple solutions.

In fact  y=( (x-c)^2 )/4 (*) satisfies the ode for all c (found by
separation of variables).

Also  y=0 satisfies the ODE (found by 'inspection').

Any solution that passes through the point (x,y) = (3,1) is
therefor unique until it hits the x axis.
y = ((x-1)^2)/4 is such a solution but it is not unique
once it touches the x axis at (x,y) = (1,0) where is has a
slope of zero.
The function y=0 satisfies the ODE and goes through both
the point (0,0) and the point (1,0).

By pasting the solutions together, namely
     _
y = |  0            for 0 <=  x  <= 1  (**)
    |  ((x-1)^2)/4  for 1 <=  x  <= 3
     -

you get a curve that is continuous, has a continuous first
derivative, satisfies the ODE everywhere, and goes through the
desired points.

Are there more such solutions? No. In the range 1<x<3 the solution
is unique as already argued. In the range 0 < x <= 1 any solution that
is not on the x axis must be of the form (*), above. Such solutions only
touch the x axis once.  But we need a solution that touches the
x axis at (0,0) and (0,1) so none of these solutions helps.  The only
other possibility for this interval is the solution y=0. So (**) is the
unique solution to the problem.

(One student found this solution.)