Exercise 4.
a) For what
c
∈
R
does there exist a nonzero function
w
: [0
,
2
π
]
→
C
such that:
w
00

c
2
w
= 0
and such that
w
(0) =
w
(2
π
) = 0
?
b) What if
w
instead solves
w
00
+
c
2
w
= 0
(again with the assumption that
w
(0) =
w
(2
π
) = 0
)?
Solution:
Let us first suppose that
c
6
= 0.
From ODE theory, we know that
w
=
a
1
e
ct
+
a
2
e

ct
for some
(complex numbers)
a
1
, a
2
. The condition
w
(0) =
w
(2
π
) = 0 then implies that:
(
a
1
+
a
2
= 0
a
1
e
2
πc
+
a
2
e

2
πc
= 0
From the above two equations, it follows that
a
1
=
a
2
= 0 and so
w
is identically zero. If
c
= 0,
then
w
=
a
1
+
a
2
t
. In this case,
w
(0) = 0 implies that
a
1
= 0 and
w
(2
π
) = 0 implies that
a
2
= 0,
and so
w
is again identically zero. Hence, in a), it is not possible to find such a function
w
.
b) We now consider what happens when
w
00
+
c
2
w
= 0.
Based on part
a
), we need to assume
that
c
6
= 0. In this case, we recall that
w
(
t
) =
a
1
cos(
ct
) +
a
2
sin(
ct
). Since
w
(0) =
a
1
= 0, it follows
that
w
(
t
) =
a
2
sin(
ct
)
.
We then obtain that
w
(2
π
) =
a
2
sin(2
πc
).
Since we want
a
2
6
= 0 (since
otherwise,
w
is identically zero), it follows that we need to have sin(2
πc
) = 0, and hence 2
πc
=
kπ
for some
k
∈
Z
. Consequently,
c
=
k
2
for some
k
∈
Z
\ {
0
}
.
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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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