642:527
SOLUTIONS: ASSIGNMENT 11
FALL 2008
Section 17.7: 7.
Let us first consider
λ >
0 and then write the solution of the ODE as
y
=
A
cos
√
λ x
+
B
sin
√
λ x
.
Then
y
′
=
−
A
√
λ
sin
√
λ x
+
B
√
λ
cos
√
λ x
the boundary
conditions
y
(0)
−
y
(1) = 0 and
y
′
(0) +
y
′
(1) = 0 become
A
−
(
A
cos
√
λ
+
B
sin
√
λ
) = 0
,
B
√
λ
+ (
−
A
√
λ
sin
√
λ
+
B
√
λ
cos
√
λ
) = 0
or
bracketleftbigg
1
−
cos
√
λ
−
sin
√
λ
−
sin
√
λ
1 + cos
√
λ
bracketrightbigg bracketleftbigg
A
B
bracketrightbigg
=
0
.
The determinant vanishes for all
λ
so any
λ
negationslash
= 0 is an eigenvalue; a solution is
A
= sin
√
λ
,
B
= 1
−
cos
√
λ
, so the eigenfunction is
y
= sin
√
λ
cos
√
λ x
+ (1
−
cos
√
λ
) sin
√
λ x
.
In fact this analysis applies for all nonzero
λ
, real or complex;
λ
has two com
plex square roots which we denote
±
√
λ
, and sin
√
λ x
and cos
√
λ x
are defined and
have all the properties needed for the manipulations above: (
d/dx
) sin
√
λ x
= cos
√
λ x
,
(
d/dx
) cos
√
λ x
=
−
sin
√
λ x
, and cos
2
√
λ x
+ sin
2
√
λ x
= 1.
When
λ
= 0 we have
y
=
A
+
Bx
,
y
′
=
B
and so the boundary conditions become
A
−
(
A
+
B
) = 0,
B
+
B
= 0,
so that
B
= 0 but
A
is free, i.e.,
y
= 1 is an eigenfunction.
This is not a regular SturmLiouville system because the boundary conditions are not
separated (each equation gives conditions involving values at both ends of the interval).
It is a bit like the case of periodic boundary conditions studied in Section 17.8; the moral
is that we cannot write down nonseparated boundary conditions arbitrarily and expect to
get a wellbehaved solution.
Section 17.8: 2 (a) See solution in text.
(b) The equation is the same as in 17.7:1 (Assignment 10) above, so we again have the
possibilities (I)–(III), i.e.,
λ >
0,
λ
= 0, or
λ <
0.
Now, however, we have periodic
boundary conditions on the interval
−
1
< x <
5.
(I) If
y
(
x
) =
A
cos
κx
+
B
sin
κx
then
y
′
(
x
) =
−
Aκ
sin
κx
+
Bκ
cos
κx
and
A
and
B
must
satisfy
A
cos(
−
κ
) +
B
sin(
−
κ
) =
A
cos(5
κ
) +
B
sin(5
κ
)
,
−
Aκ
sin(
−
κ
) +
Bκ
cos(
−
κ
) =
−
Aκ
sin(5
κ
) +
Bκ
cos(5
κ
)
,
or
bracketleftbigg
cos(
−
κ
)
−
cos(5
κ
)
sin(
−
κ
)
−
sin(5
κ
)
,
−
sin(
−
κ
) + sin(5
κ
)
cos(
−
κ
)
−
cos(5
κ
)
bracketrightbigg bracketleftbigg
A
B
bracketrightbigg
=
0
.
The determinant of the matrix is, after simplification,
2[1
−
cos(
−
κ
) cos(5
κ
)
−
sin(
−
κ
) sin(5
κ
)] = 2[1
−
cos(6
κ
)]
,
so that the eigenvalue equation is 6
κ
n
= 2
nπ
or
k
n
=
nπ/
3,
n
= 1
,
2
, . . .
. The eigenvalues
are
λ
n
= (
nπ/
3)
2
and since for
κ
=
κ
n
the coefficient matrix is identically zero, both
A
and
B
are arbitrary, i.e., we have two eigenfunctions for each
κ
n
: cos(
nπx/
3) and sin(
nπx/
3).
(II) If
y
(
x
) =
A
+
Bx
then it is easily seen that the boundary conditions imply that
B
= 0
and
A
is arbitrary:
λ
0
= 0 is an eigenvalue with eigenfunction 1.
(III) As in examples done in earlier homework we find that there are no negative eigen
vectors.
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 Fall '08
 Staff
 Math, Cos, Boundary value problem, λ, Boundary conditions, Sturm–Liouville theory

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