Exercise 5.
Suppose that
w
: [0
,
+
∞
)
→
R
solves the ODE:
(1)
aw
00
+
bw
0
+
cw
= 0
for some constants
a,b,c
. Furthermore, we assume that
b
≥
0
.
a) Let us deﬁne the
Energy
to be:
E
(
t
) :=
1
2
²
a
(
w
0
(
t
))
2
+
c
(
w
(
t
))
2
³
.
Without solving the ODE
(1)
, show that
E
0
(
t
)
≤
0
.
b) Under the additional assumption that
a >
0
and
c >
0
, show that
w
(0) = 0
and
w
0
(0) = 0
implies
that
w
(
t
) = 0
for all
t
≥
0
.
c) Assume again that
a >
0
and
c >
0
. Show that if
w
1
and
w
2
solve the ODE
(1)
and if
w
1
(0) =
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View Full DocumentPRACTICE HOMEWORK FOR MATH 425, SOLUTIONS
3
w
2
(0)
,w
0
1
(0) =
w
0
2
(0)
, then one can deduce that
w
1
(
t
) =
w
2
(
t
)
for all
t
≥
0
. In this way, we obtain
uniqueness of solutions to
(1)
.
Solution:
a) We use the product rule to calculate:
E
0
(
t
) =
aw
0
w
00
+
cww
0
. We can now use the ODE to
deduce that
w
00
=

bw
0

cw
. Hence:
E
0
(
t
) =
w
0
(

bw
0

cw
) +
cww
0
=

b
(
w
0
)
2
≤
0
since
b
≥
0. In other words,
E
(
t
) is a decreasing function of
t
on [0
,
+
∞
).
b) By assumption
E
(0) =
1
2
±
a
(
w
0
(0))
2
+
c
(
w
(0))
2
²
= 0. Since
a,c >
0, it follows that
E
(
t
) is
nonnegative. Finally, from part a), it follows that
E
(
t
) is a decreasing function on [0
,
+
∞
), hence
E
(
t
) is identically zero on [0
,
+
∞
). In particular, since both
a
and
c
are positive, it follows that
w
(
t
) = 0 for all
t
≥
0.
c) If
w
1
and
w
2
solve the ODE, then so does
w
:=
w
1

w
2
. The function
w
then satisﬁes the
conditions of part b) and the claim follows.
±
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 Spring '09
 Math, Differential Equations, Equations, Integration By Parts, Partial Differential Equations, Complex Numbers, Complex number, Euler's formula, Eθ

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