# Phase diagrams for 1st order ODEs are considered and studied.
# 
# Art Belmonte
# Mon, 27/May/96
# Math 308-509 [Maple V Release 4]
# Section 2.6: Population Dynamics and Some Related Problems
# 
# T-62/3-1: 
> with(DEtools):
# dN/dt = F
> F:=n->n * (n-1) * (n-2);

                     F := n -> n (n - 1) (n - 2)

> plot(F(n), n=0..3, y=-1..2, labels=[`n`, `dn/dt`],
>      xtickmarks=3, ytickmarks=2);

# T-62/3-2: NOTE: Capital 'N' appears to be reserved by the system! It
# would appear that within the context of DEplot, N = #differential
# equations. Accordingly, we'll use lower case 'n.'
> unassign('n');
> deq:=diff(n(t), t) = n(t) * (n(t) - 1) * (n(t) - 2);

                    d
             deq := -- n(t) = n(t) (n(t) - 1) (n(t) - 2)
                    dt

> inits:={ [0, 0.2], [0, 0.8], [0, 1.7], [0, 2.1] };

           inits := {[0, .2], [0, .8], [0, 2.1], [0, 1.7]}

> DEplot(deq, n(t), t=-1..3, n=-1..3, inits);

# T-62/3-3: Here are implicit and explicit solutions from dsolve. (It
# this case dsolve COULD explicitly solve for n(t). This may not always
# be the case.) Note that this NONLINEAR equation has three equilibrium
# solutions which are NOT accounted for in the dsolve solutions: n(t) =
# 0, n(t) = 1, and n(t) = 2. This is why we can't speak of a "general"
# solution to a nonlinear differential equation like we can for a linear
# one.
> unassign('n', 't');
> deq; dsolve(deq, n(t));

                 d
                 -- n(t) = n(t) (n(t) - 1) (n(t) - 2)
                 dt


      - 1/2 ln(n(t)) + ln(n(t) - 1) - 1/2 ln(n(t) - 2) + t = _C1

> sol:=dsolve(deq, n(t), explicit);

                2
sol := n(t) = %1 ,

                      2                        2
    n(t) = -%1 (1 - %1  - exp(2 t - 2 _C1) + %1  exp(2 t - 2 _C1))

                                       4
%1 := RootOf((-1 + exp(2 t - 2 _C1)) _Z

                                  2
     + (2 - 2 exp(2 t - 2 _C1)) _Z  + exp(2 t - 2 _C1))

# T-63/9-1: See whiteboard (your written notes) for discussion of
# equilibrium points and stability.
> unassign('n');
> F:=n->n^2 * (n^2-1);

                                   2   2
                        F := n -> n  (n  - 1)

> plot(F(n), n=-2..2, y=-1..2, labels=[`n`, `dn/dt`],
>      xtickmarks=3, ytickmarks=2);

# T-63/9-2
> unassign('n');
> deq:=diff(n(t), t) = n(t)^2 * (n(t)^2 - 1);

                         d             2      2
                  deq := -- n(t) = n(t)  (n(t)  - 1)
                         dt

> inits:=[ [0, 0.2], [0, 0.8], [0, 1.1], [0, -0.3], [0, -1.7] ];

      inits := [[0, .2], [0, .8], [0, 1.1], [0, -.3], [0, -1.7]]

> DEplot(deq, n(t), t=0..3, n=-2..2, inits);

# T-63/9-3: Here is an implicit solution to the D.E. Note that dsolve
# was not able to isolate an explicit solution. Note that this NONLINEAR
# equation has three equilibrium solutions which are NOT accounted for
# in the dsolve solution: N(t) = -1, N(t) = 0, and N(t) = 1. This is why
# we can't speak of a "general" solution to a nonlinear differential
# equation like we can for a linear one.
> deq; dsolve(deq, n(t));

                     d             2      2
                     -- n(t) = n(t)  (n(t)  - 1)
                     dt


           1
        - ---- - 1/2 ln(n(t) - 1) + 1/2 ln(1 + n(t)) + t = _C1
          n(t)

> dsolve(deq, n(t), explicit);

n(t) = exp(RootOf(_Z exp(_Z) - _Z - 2 - ln(exp(_Z) - 2) exp(_Z)

     + ln(exp(_Z) - 2) + 2 t exp(_Z) - 2 t - 2 _C1 exp(_Z) + 2 _C1))

     - 1

> allvalues(", dependent);
Error, (in allvalues/rootseq) cannot evaluate with symbolic
coefficients
>