Problem 1: We define the function f p =
differential equation p# t = f p t
1
$ p K 1 2 $ p K 5 3 $ p K 7 and consider the
100
.
a. Find the equilibrium solutions for this DE.
Sol'n: By inspection, the zeros of f p are p = 1, 5, 7.
b. Use Maple to calculate and simplify f p $f# p .
Sol'n: First we will define f for Maple:
O f d p/
1
$ p K 1 2$ p K 5 3$ p K 7 ;
100
1
f := p/
pK1 2 pK5
100
3
pK7
(1)
Now that Maple knows the function, we calculate and simplify f p $f # p .
O simplify f p $D f
p ;
3
pK1
5000
3
pK5
5
pK7
p2 K 9 p C 16
(2)
c. Use Maple to solve f p $f # p = 0.
Sol'n: We will use the solve command as follows:
O solve f p $D f
p = 0, p ;
1, 1, 1, 5, 5, 5, 5, 5, 7,
The inflection points are clearly
9
G
2
9
1
C
2
2
17 ,
9
1
K
2
2
17
(3)
17
.
2
d. Use pen/pencil and paper to determine whether solutions with initial conditions in the intervals
determined by the result of part (a) are increasing or decreasing.
Sol'n: The first interval is (0, 1). For p0 2 0, 1 , we have \0 ! p0 ! 1 and so
p0 K 1 ! 0, p0 K 5 ! 0 and p0 K 7 ! 0. Since the exponent on the factor p K 1 is even and the others
dp
are odd, we have
R 0 on this interval and the solutions are increasing on this interval.
dt
The second interval is (1, 5). For p0 2 1, 5 , we have p0 K 1 O 0, p0 K 5 ! 0, and p0 K 7 ! 0. It
dp
follows that
O 0 for initial conditions in this interval.
dt
The third interval is (5, 7). For p0 2 5, 7 , we have p0 K 1 O 0, p0 K 5 O 0, and p0 K 7 ! 0. It
dp
follows at once that
! 0 for initial conditions in this interval.
dt
The fourth interval is 7,N . For p0 O 7, each factor in f p is positive and so
dp
O 0 for initial
dt
conditions in this interval.
e. Determine the concavity of the solutions.
Sol'n: First note that
9
1
C
2
2
17
at 5 digits
6.5616 and
9
1
K
2
2
17
at 5 digits
2.4384. We
proceed as follows:
3
p K 1 3 p K 5 5 p K 7 p K 6.5616
5000
to see that p ## t ! 0 for such p0 . Hence p is concave down on this interval.
For 0 ! p0 ! 1, we use f p f # p =
p K 2.4384
For 1 ! p0 ! 2.4384, we see that f p0 $f # p0 O 0 and so the solutions are concave up for initial
conditions in this interval. Since they are also increasing (see d), they will start out concave up and then
switch to concave down when they increase through p = 2.4384.
For 2.4384 ! p0 ! 5, f p0 $f # p0 ! 0 and so solution curves are concave down here.
For 5 ! p0 ! 6.5616, we see that f p0 $f # p0 O 0 and so solutions are concave up in this interval.
For 6.5616 ! p0 ! 7, we have f p0 $f # p0 ! 0 and so the solution curves are concave down in this
interval.
For p0 O 7, f p0 $f # p0 O 0 and so solution curves are concave up.
f. Use Maple to plot a "phase portrait" using the specified formatting for inflection lines.
Sol'n: We first load the DEtools package:
O with DEtools :
Now we generate the phase plot. (Take note of the syntax and look up the options that are not familiar to
you!)
O DEplot D p t = f p t , p t , t = 0 ..20, p = 0 ..9, p 0 = 0.05, p 0 = 0.1, p 0 = 0.2,
p 0 = 0.5, p 0 = 1.05, p 0 = 1.5, p 0 = 3, p 0 = 4, p 0 = 5.5, p 0 = 5.7, p 0
= 6, p 0 = 6.5, p 0 = 7.1, p 0 = 7.2, p 0 = 7.25, p 0 = 7.35 , linecolor = black,
thickness = 1, arrows = small, dirfield = 40, 40 , stepsize = 0.05 ;
Warning, plot may be incomplete, the following errors(s) were
issued:
cannot evaluate the solution further right of .49095522,
probably a singularity
Warning, plot may be incomplete, the following errors(s) were
issued:
cannot evaluate the solution further right of .18197807,
probably a singularity
9
8
7
6
5
p(t)
4
3
2
1
0
0
5
10
t
15
20