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Section 6.2 SturmLiouville Theory
97
Since
α
±
= 0, it follows that sinh
α
±
= 0 and this implies that 1

α
2
=0or
α
=
²
1.
We take
α
= 1, because the value

1 does not yield any new eigenfunctions. For
α
= 1, the corresponding solution is
X
=
c
1
cosh
x
+
c
2
sinh
x
=

c
2
cosh
x
+
c
2
sinh
x,
because
c
1
=

αc
2
=

c
2
. So in this case we have one negative eigenvalue
λ
=

α
2
=

1 with corresponding eigenfunction
X
= cosh
x

sinh
x
.
Case III
If
λ
=
α
2
>
0, then the general solution of the diﬀerential equation is
X
=
c
1
cos
αx
+
c
2
sin
αx.
We have
X
±
=

c
1
α
sin
αx
+
c
2
α
cos
αx
. In order to have nonzero solutions, one of
the coeﬃcients
c
1
or
c
2
must be
±
= 0. Using the boundary conditions, we obtain
c
1
+
αc
2
=0
c
1
(cos
α

α
sin
α
)+
c
2
(sin
α
+
α
cos
α
)=0
The ±rst equation implies that
c
1
=

αc
2
and so both
c
1
and
c
2
are
neq
0. From
the second equation, we obtain

αc
2
(cos
α

α
sin
α
c
2
(sin
α
+
α
cos
α

α
(cos
α

α
sin
α
) + (sin
α
+
α
cos
α
sin
α
(
α
2
+1) = 0
Since
α
2
+1
±
= 0, then sin
α
= 0, and so
α
=
nπ
, where
n
=1
,
2
,...
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.
 Fall '11
 StuartChalk

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