# 
# MATH 3150 SUMMER 99
# PROJECT 1
# GRADIENT FLOWS 
# 
# 
# 
# This Maple project is a short  aid to visualizing gradient flows.
# There are just two problems. 
# 
# The Maple work sheet has a lot of extra junk in it -- namely all my
# comments and explanations. 
# 
# DO  NOT HAND ALL THAT IN.
# 
# Instead,  just copy over the commands you want to another worksheet
# and do the problems there. Label each problem as you go, A.1 etc.,
# otherwise the grader may get lost and be unable to find your marks.
# 
# We need the packages  <plots> and <linalg>, so we load them up now. 
# The commands are ended with colons instead of semicolons in order to
# avoid a lot of extraneous output.
> with(plots):
> with(linalg):
# 
# Heat flow driven by temperature gradients
# 
# Recall that heat flows in the direction  in which the temperature
# decreases most rapidly at a rate proportional to the slope of the
# temperature in that direction.  If T (x,y) is the temperature,this
# translates into mathematics as F = -K *grad (T ) = K*grad (-T), where
# F is the heat flow field and K is the conductivity. Let us illustrate
# this with some plots of the heat flow superimposed on contour plots of
# the temperature. In these problems K is always equal to 1.
# 
# Example: T(x,y) =cos(Pi*y)*sin(2*Pi*x)
# 
# We set this up with T as an expression rather than a function because
# it is a bit more convenient to do it this way in Maple.  The Pi's are
# in there to make it look pretty in a domain which is a 1x1 box.
#  
> T:=cos(Pi*y)*sin(2*Pi*x);

                      T := cos(Pi y) sin(2 Pi x)

# The Maple command for grad is "grad" , with the list [x,y] denoting
# the variables with respect to which  the partials are taken.  Notice
# the - on the T.
> F:=grad(-T,[x,y]);

     F := [-2 cos(Pi y) cos(2 Pi x) Pi, sin(Pi y) Pi sin(2 Pi x)]

# We now make a contour plot of the temperature and a field plot (little
# arrows) of the heat flow  in a 1x1 box.  We follow the commands by :
# so they aren't  displayed.
> Tplot:=contourplot(T,x=0..1,y=0..1):
> Fplot:=fieldplot(F,x=0..1,y=0..1):
# Now we make the display with the two supperimposed  on each other.
> display([Tplot,Fplot]);

# There are hot spots in the lower left and upper right and cold spots
# in the upper left and lower right.  Notice how the heat flows downhill
# and perpindicular to the contours, although if the scales on the x and
# y axis are not the same the arrows may not appear perpindicular. Also
# note that the arrows on the top and bottom edges do not point across
# the  edges, i.e. , there is no heat flow across the edges, they are
# insulated. On the other hand, the left and right edges are at  0
# degrees, so heat flows from the edges to the cold spots, and from the
# hot spots to the edges.
# 
# Question: Is T a steady state solution of the heat equation if the
# temperature is held fixed on the boundaries?
#                     (The Maple function for div is diverge.)
# 
> diverge(F,[x,y]);

                                               2
                     5 cos(Pi y) sin(2 Pi x) Pi

# Answer: No, div(F) is not identically zero and consequently there a
# regions with a net heat flow in or out. This causes the temperature to
# change in these regions.
# 
# OK, here are two similar  problems for you to do. Do the plots on
# x=0..1, y=0..1.
# 
# 
# 
# Problem A.1  T(x,y) = sinh(Pi*y)*sin(Pi*x)
# part 1. Do a field plot of the heat flow superimposed on a contour
# plot of the temperature.
# part 2. Is T a steady state solution of the heat equation if the
# temperature on the boundaries is fixed ?
# 
# Problem A.2 T(x,y)=cosh(Pi*y)*cos(Pi*x)
# part 1. Do a field plot of the heat flow superimposed on a contour
# plot of the temperature.
# part 2. Is T a steady state solution of the heat equation if the
# temperature on the boundaries is fixed ?
# 
# FINIS