Page 140 Fundamentals of Differential Equations R Kent Nagle

Page 140

Fundamentals of Differential Equations

R. Kent Nagle Edward B. Saff Arthur David Snider

The differential equations of the model are similar to those for compartmental analysis dis- cussed in Section 3.2. They involve compartments and parameters. The compartments of the model are r(t) = the population of uninfected CD4+ T cells at time t . 1(t) = the population of infected CD4+ T cells at time V (t) = the population of virus at time t . The parameters (followed by their units) of the model are A(cells-day-1) = constant input source of uninfected cells per day (the human body produces these cells daily in the thymus) . 6(day-1) = normal loss rate constant of uninfected cells (1/ = the average lifespan of (day)-ormal loss ra an uninfected cell in days) (virions-i . day-i-infection rate constant of uninfected cells per infected cell (the rate is of mass action form, i.e., V(r) T(r)) (day-1) = loss rate constant of infected cells (1/ = the average lifespan of an infected cell in days) y(day-1)-loss rate constant of free virus (1/y = the average lifespan of a free virion in days) . N(virions . cell-\') = number of virions produced per day per infected cell (the burst number of an infected cell) .

Solution

Ans-

Solution. First we show N

is closed. Equivalently, M n N

is open. Let x 2 M n N
.
Then d(x; A) > , so choose
0
= d(x; A) . Then D(x;
0
) M n N
. Indeed, if
y 2 D(x;
0
), then, by use of triangle inequality (taking infs over A),
Hence y 2 M n N

d(y; A) d(x; A) d(x; y) >
.
Now we\'ll show A is closed if and only if A = \\fN

0
+
0
= :
j > 0g. If A = \\fN
j > 0g, since
the N
are closed then A is the arbitrary intersection of closed sets, which is closed.
Now suppose that A is closed. One the one hand, since A N

, it follows that A
\\fN

j > 0g. To show the opposite inclusion, suppose x =2 A but x 2 \\fN

j > 0g.
A closed means d(x; A) > 0. Let = d(x; A). Then x =2 N
and this contradicts the
assumption that x 2 \\ fN

j > 0g.
4. End of Chapter 2 Exercise 26. De ne he sequence of numbers a
Show that a
n
a
0
= 1; a
n
= 1 +
1
is a convergent sequence. Find the limit.
Solution. First, note that a
n
1 + a
n1
:
=2
n
by
is bounded below by 0 and above by 2 (a simple induction will
prove these claims). Now we consider the subsequences a
2n
and a
. We\'ll show the former
is monotonically increasing and the latter is monotonically decreasing. Note that
This shows a
n+1
a
a
n+1
n1
a
n1
=
(a
)
(1 + a
n
n
a
n2
)(1 + a
has the same sign as a
n2
n1
)
=
a
n3
(a
n1
2n+1
)
something > 1
a
n3
. Inductively, a
has the
same sign as a
2
a
0
or a
3
a
1
. In fact, the second equality above shows a
n+1
and
a
n
a
n2
have opposite signs. Computing the rst few terms, it follows that a
is monotonically
increasing while a
is monotonically decreasing. Thus each of these subsequences
converges, and they both obey the recurrence relation
2n+1
a
n+1
= 1 +
1
2 +
1
1 + a
n1
Now that we have established that the subsequences converge, it makes sense to take limits
of the above equation. Since both subsequences satisfy the same recurrence relation, they
converge to the same value (so a
n


a
n+1
2n
converges to this value). We easily compute that lim a
=
p
2.
5. End of Chapter 2 Exercise 52. Test the following series for convergence.
(a)
1
X
k=0
e
k
p
k + 1
converges.
. Use the ratio test to get lim
2
k!1

e
e
k
p
k + 1
k+1
p
k + 2

=
1

n1
a
n1
e
< 1, so the series
n