Homework 4 solution
AMATH 383, Autumn 2009
November 30, 2009
1. Periodicity of the Volterra System
(a) try scaling
x
=
x
0
˜
x,
y
=
y
0
˜
y,
t
=
t
0
˜
t
then
d
x
d
t
=
x
0
d˜
x
d
˜
t
d
˜
t
d
t
=
x
0
t
0
d˜
x
d
˜
t
d
y
d
t
=
y
0
t
0
d˜
y
d
˜
t
plug into the equations
d˜
x
d
˜
t
=
rt
0
˜
x

at
0
y
0
˜
x
˜
y
d˜
y
d
˜
t
=
kt
0
(

˜
y
+
bx
0
k
˜
x
˜
y
)
let
rt
0
= 1
at
0
y
0
= 1
bx
0
k
= 1
we have scales for each variables
t
0
=
1
r
x
0
=
k
b
y
0
=
r
a
(b)
d
d
t
A
(
x
(
t
)
,y
(
t
)) =
∂A
∂x
d
x
d
t
+
∂A
∂y
d
y
d
t
=
γ
(1

1
x
)
x
(1

y
) + (1

1
y
)
γy
(
x

1) = 0
(c) as
x
→
0
,
ln
x
→ ∞
, so
A
(
x,y
)
→
y

ln
y
+
γ
(0

(
∞
))
→
+
∞
. Similar argument also
holds for
y
→
0.
D
2
A
=
±
γ
x
2
0
0
1
y
2
¶
is positive deﬁnite, thus is convex
Combining with the previous fact that
A
approaches +
∞
at the boundary of the domain
x >
0
,y >
0,
A
must have a global minimum. The minimum point can be found by solving
DA
= (0
,
0) = (
γ
(1

1
x
)
,
1

1
y
)
⇒
(
x,y
) = (1
,
1)
(d) the minimum value of
A
is
A
(1
,
1) = 1 +
γ
, so for any
c > A
min
= 1 +
γ
,
A
(
x,y
) =
c
is a closed curve, corresponding to a periodic solution (
x
(
t
)
,y
(
t
)). For any positive initial
value of (
x
(0)
,y
(0)) other than (1
,
1),
A
(
x
(
t
)
,y
(
t
)) =
A
(
x
(0)
,y
(0))
> A
min
, thus is periodic.
see also the plots in mathematica for the shape of
A
.
1