Edelstein-Keshet L. Mathematical Models in Biology
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Selected
Answers
CHAPTER
1
1.
(b)
x
n
- 10 •
2".
2. (b)
x
n
= 1 +
2-(4)-;
3. (b) (i)
A,
+
nA
2
;
6. (a)
A(5)"
+
B(2)«;
(c)
*„
=
(-!)"+
4;
(ii)
(-1)-[A,
+
/iAj;
(b)
A(±)"
+
BH)
n
;
(e)
*„
= 5 +
(-2)".
(iii)
3"[A,
+
«A
2
].
(e)
A(i)»
+
B(-l)«.
10.
11.
13.
14.
18.
7.
8.
9.
(a)
C
n+l
=
C
n
-
pV
n
+
m,
V
n+l
=
aC
n
.
(c)
4aj3
< 1
^>
amount
of
CO
2
lost
per
breath
is
less than
£
(amount
of
CO
2
that
in-
duces
a
unit volume
of
breathing).
4a/3
> 1: A =
|{1
±
Ti}.
7 =
(4«/3
-
1)
1/2
.
|A|
>
1
when
1.
Frequency
Selected
Answers
557
8. (b)
AT,
= 0 or
N
2
=
(\
l/b
-
\)/a.
Ni
stable
iff
A
<
1;
N
2
stable
iff
0 < b(\ -
A~"*)
< 2.
(c)
Nonlinear,
x = 0.
(e)
Nonlinear,
x =
(K
-
&
2
)Ai
Jk,
*
0.
2. (a)
Stable
for |r| < 1. (b)
Unstable.
(c)
Stable.
(d)
Unstable.
4. (a)
CHAPTER
2
1. (a)
Linear,
*„
=
20.
14.
/'(*)
=
kb/(b
+
x)
2
,
/'(O)
= k/b > 1.
16. (b)
C,
+1
=/C,5,;
5r+i
=
S,
-
fC.S,
+ B.
17.
(b)
C
n+1
=
C
-
j8C
n
V
n
+
m,
K
+1
=
aC
n
.
(c)
Need
(wa/3)
< 1 for
stability.
(d)
yes,
for
|*
-
3|
<
2\/2
where
x =
(ma)8)
1/2
.
(h)
C =
HjX
±
VV
+
4A:y}
where
y =
m/fia,
V
=
aC/(K
+ C).
11.
Steady states:
(0, 0) and
558
Selected
Answers
CHAPTER
3
4.
(c)
N
stable
for
11
-
b(\-
l/b
-
1)
| < 1.
5. (b)
1-stable,
2-stable,
3-stable,
4-unstable,
5-stable.
7. (a)
n,+
i
=
An,<T'',
p,+
,
=
n,(l
-
e'"')
for
JV
=
-—-,
P
=
-.
(ac)
a
9.
(b)
a,,
= (1 -
rN/K),
a
n
=
-Na,
a
2
i
-
P/N,
a
22
=
a(N
- P).
10.
(b)
M+,
=
AM
exp[-(a/>,V-
m
],
/%.
=
M(l
-
exp[-(
fl
P
r
)'-"]).
(c)
P =
(In
A/a)%
TV
=
A/>/(A
- 1).
15. (a)
Steady states
A
=
(b)
k =
In/,
b =
(d)
ForF(G,
f/)
=
Qe^~
H
\
G(Q,
H)
= bH
F
Q
(Q,
H)=l,
F
H
(Q,
H)
= -k,
R
X
(Q,
H]
(where
R =
G/tf,
x
=
H/Q).
=
-b
19.
CHAPTER
4
5. (a) r =
intrinsic growth rate.
B =
carrying
capacity,
(d)
For
/
-»
oo,
e
~
n
-»
0 so
N(t)
->
yVofl/A^o
= B.
for
No
small,
A^(f)
«
NoB/Be'"
=
//oe".
6. (b)
(mass
nutrient)/(number
of
bacteria).
8.
ai
=
j,K
mM
=
ratio
of
[emptying time
of
chamber]
to
[(1/ln
2)
times bacterial
doubling
time].
«2
=
c
0
/K
n
=
ratio
of
[stock nutrient concentration]
to
[concentration which produces
a
half-maximal bacterial growth
rate].
10.
11.
Increase
Co, V,
decrease
F. (No
local maximum).
15.
(a)
/'
=
-cot(x):
linear, order
2, not
homogeneous, constant coefficients.
(d)
(2y +
2)y'
— y = 0;
nonlinear, order
1,
homogeneous,
nonconstant
coefficients.
(i)
-*•
+ y = - (t
^
0);
linear, order
1,
nonhomogeneous, constant coefficients.
16.
18.
22.
(d)
Steady states
(0, 0), (1, 1);
7(0,
0) =
Selected
Answers
559
where
A =
(«
+
JtuVfei,
B
=
D/k
2
i,
C =
k
n
/(ku
+ s +
*
23
),
E
=
W(*2«
.I"
^
+
*
23
),
F
=
jfcjj/fe.
CHAPTER
5
1-
(b)
(e)
2. (b)
29.
560
Selected
Answers
4.
(b)
(1)
(x,
y) =
(r,
t(t -
1)),
(4) (x, y) =
(cos
r,
sin
i),
(e)
5. (a)
(6) (x, y) = (t,
4f
2
),
(e)
Selected
Answers
7.
12.
15.
561
(g)
6.
(a)
/(I,
1) =
(d)
7(0,
1) =
(a)
neutral center,
(b)
saddle,
(c)
unstable node,
(d)
saddle,
(e)
stable spiral,
(f)
unstable
spiral.
(a)
(b)
(c)
saddle;
7(1,
-1) =
stable node;
•/(-!,
0) =
stable
spiral.
saddle point.
562
20.
(b) K =
K
2
K
4
Il
t
a =
K
3
KJ
0
.
(c)
steady state:
q
=
Selected
Answers
stable.
21. (a)
(b)
(c)
22. (e)
m
=
2,
steady
state:
steady state:
Decaying oscillations
if
(e)
(stable).
(stable).
Selected Answers
563
predation
community:
0,4,2,3}.
Not
qualitatively stable.
14.
17.
18.
23.
24.
2.
3.
4.
6.
9.
(a)
(c)
(b)
(a)
(d)
(b)
(a)
(b)
(c)
(D (a)
(2)
(a)
(3) (a)
(b)
Beverton-Holt
solution:
N
a
e
N
= P
e
n
,
P =
constant,
(steady
state
N = O is
unstable),
not
stabilizing.
(d) not
stabilizing.
AT
=
K
is a
stable steady state.
steady
states:
(0, 0) and
(K\
+
aN
2
)
is the
carrying
capacity
of
species
1.
Thus
the
presence
of
species
2
contributes
positively
to the
carrying capacity
of
species
1.
steady
states:
(0, 0)
(unstable node),
(K
t
,
0), (0,
K
2
)
(saddle points).
(stable
node).
(Last steady state
exists
only
if
#1
=
(a.),
#2
=
detHi
=
a
t
,
del
Hi
-
a\a
2
-
03,
detH
3
=
a
3
(det#
2
).
{1,2}.
(e)
{1,2},
{4,
5}.
not
qualitatively stable;
(e) not
qualitatively stable.
constant.
564
Selected
Answers
CHAPTER
7
29.
32.
2.
4.
6.
7.
12.
16.
19.
20.
21.
23.
(b)
(c)
(b)
(a)
(b)
(c)
(a)
(b)
For A > 0, the
only equilibrium
of
5,1
equations
is (5, /) = (0,
0).
But
7(0,0)
=
(0, 0)
stable.
Steady
states:
(K, 0)
(saddle point),
(J
2
,
fctiz)
where
x
2
=
(stable).
(a)
:c,
=
K
In
c + c =
—Af
+
constant.
(a)
(b)
At
equilibrium
sign pattern
Let*,
=
k
2
=
k
3
-
k
4
= 1.
Steady
state:
x
=
J
=
Selected
Answers
565
CHAPTER
8
Flow
can
escape
to (or
emanate
from)
arbitrarily large
y
values.
Poincare-Bendixson
theory inconclusive.
7. (b) No
limit
cycle,
(c)
limit cycle exists.
10. (a)
Functions have
no
maxima
or
minima, only
an
inflection
point.
13. (b)
Figure
b. (c)
Cannot solve cubic equation.
17.
(c)
Condition
guarantees/
= 0
nullcline intersects
jc-axis
to the right of the
intersec-
tion
of g = 0
with
Jt-axis.
(a)
a(y-l)>l.
(b),
(c) 0<
a(y
- 1) < 1.
4.
(a-c)
No
limit cycle
possible,
(d)
Cannot rule
out
limit
cycle.
18. (a) J =
21. (a)
But/
=
g = 0 at the
nontrivial
steady
state.
6. (b)
2. (b)
= b
>0.