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Solving a Sturm-Liouville BVP in two ways
Athanasios Paraskevopoulos
Athanasios Paraskevopoulos, Hellenic Open University
Posted
2 years ago
Sturm-Liouville Boundary Value problem
Given the following boundary value problem
′
′
X
+
λ
X
0
,
0
<
x
<
1
X
(
0
)
0
,
′
X
(
1
)
+
X
(
1
)
0
E
a
s
i
l
y
w
e
c
a
n
p
r
o
v
e
f
o
r
λ
>
0
,
t
h
e
e
i
g
e
n
v
a
l
u
e
s
o
f
t
h
e
p
r
o
b
l
e
m
s
a
t
i
s
f
y
t
h
e
e
q
u
a
t
i
o
n
λ
-
t
a
n
λ
We are going to solve this problem with two ways
First way
We are going to find the first 10 approximate values and then to find numerically the first 10 eigenvalues
I
n
[
]
:
=
C
l
e
a
r
[
"
G
l
o
b
a
l
`
*
"
]
I
n
[
]
:
=
f
[
x
_
]
:
=
x
I
n
[
]
:
=
h
[
x
_
]
:
=
-
T
a
n
[
x
]
I
n
[
]
:
=
P
l
o
t
[
f
[
x
]
,
{
x
,
0
,
1
2
0
0
}
,
P
l
o
t
R
a
n
g
e
-
>
A
l
l
,
P
l
o
t
L
e
g
e
n
d
s
-
>
"
E
x
p
r
e
s
s
i
o
n
s
"
]
O
u
t
[
]
=
I
n
[
]
:
=
P
l
o
t
[
h
[
x
]
,
{
x
,
0
,
1
2
0
0
}
]
O
u
t
[
]
=
I
n
[
]
:
=
S
h
o
w
[
P
l
o
t
[
{
f
[
x
]
,
h
[
x
]
}
,
{
x
,
0
,
1
2
0
0
}
]
]
O
u
t
[
]
=
From the plot we find the first 10 approximate values
I
n
[
]
:
=
C
o
o
r
d
3
=
{
{
"
4
.
2
5
5
"
,
"
2
.
3
8
8
"
}
,
{
"
2
5
.
0
5
"
,
"
4
.
7
4
8
"
}
,
{
"
6
2
.
2
1
5
"
,
"
8
.
1
8
3
"
}
,
{
"
1
2
5
.
7
9
2
"
,
"
1
1
.
6
5
5
"
}
,
{
"
2
0
1
.
3
6
9
"
,
"
1
4
.
9
4
"
}
,
{
"
2
9
9
.
3
1
8
"
,
"
1
7
.
0
9
"
}
,
{
"
4
1
8
.
4
5
9
"
,
"
2
0
.
3
0
8
"
}
,
{
"
5
5
0
"
,
"
2
3
.
6
9
8
"
}
,
{
"
7
1
5
.
2
9
1
"
,
"
2
6
.
9
3
8
"
}
,
{
"
8
9
1
.
1
8
3
"
,
"
3
0
.
1
2
2
"
}
}
O
u
t
[
]
=
{
{
4
.
2
5
5
,
2
.
3
8
8
}
,
{
2
5
.
0
5
,
4
.
7
4
8
}
,
{
6
2
.
2
1
5
,
8
.
1
8
3
}
,
{
1
2
5
.
7
9
2
,
1
1
.
6
5
5
}
,
{
2
0
1
.
3
6
9
,
1
4
.
9
4
}
,
{
2
9
9
.
3
1
8
,
1
7
.
0
9
}
,
{
4
1
8
.
4
5
9
,
2
0
.
3
0
8
}
,
{
5
5
0
,
2
3
.
6
9
8
}
,
{
7
1
5
.
2
9
1
,
2
6
.
9
3
8
}
,
{
8
9
1
.
1
8
3
,
3
0
.
1
2
2
}
}
No we use the command FindRoot and the approximate values to find numerically the first 10 eigenvalues
I
n
[
]
:
=
x
1
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
4
.
2
5
5
}
]
O
u
t
[
]
=
{
x
4
.
1
1
5
8
6
}
I
n
[
]
:
=
x
2
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
2
5
.
0
5
}
]
O
u
t
[
]
=
{
x
2
4
.
1
3
9
3
}
I
n
[
]
:
=
x
3
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
6
2
.
2
1
5
}
]
O
u
t
[
]
=
{
x
6
3
.
6
5
9
1
}
I
n
[
]
:
=
x
4
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
1
2
5
.
7
9
2
}
]
O
u
t
[
]
=
{
x
1
2
2
.
8
8
9
}
I
n
[
]
:
=
x
5
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
2
0
1
.
3
6
9
}
]
O
u
t
[
]
=
{
x
2
0
1
.
8
5
1
}
I
n
[
]
:
=
x
6
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
2
9
9
.
3
1
8
}
]
O
u
t
[
]
=
{
x
3
0
0
.
5
5
}
I
n
[
]
:
=
x
7
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
4
1
8
.
4
5
9
}
]
O
u
t
[
]
=
{
x
4
1
8
.
9
8
7
}
I
n
[
]
:
=
x
8
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
5
5
0
}
]
O
u
t
[
]
=
{
x
3
0
0
.
5
5
}
I
n
[
]
:
=
x
9
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
7
1
5
.
2
9
1
}
]
O
u
t
[
]
=
{
x
7
1
5
.
0
7
7
}
I
n
[
]
:
=
x
1
0
=
F
i
n
d
R
o
o
t
[
f
[
x
]
=
=
h
[
x
]
,
{
x
,
8
9
1
.
1
8
3
}
]
O
u
t
[
]
=
{
x
8
9
2
.
7
3
}
I
n
[
]
:
=
a
=
{
x
1
,
x
2
,
x
3
,
x
4
,
x
5
,
x
6
,
x
7
,
x
8
,
x
9
,
x
1
0
}
O
u
t
[
]
=
{
{
x
4
.
1
1
5
8
6
}
,
{
x
2
4
.
1
3
9
3
}
,
{
x
6
3
.
6
5
9
1
}
,
{
x
1
2
2
.
8
8
9
}
,
{
x
2
0
1
.
8
5
1
}
,
{
x
3
0
0
.
5
5
}
,
{
x
4
1
8
.
9
8
7
}
,
{
x
3
0
0
.
5
5
}
,
{
x
7
1
5
.
0
7
7
}
,
{
x
8
9
2
.
7
3
}
}
We define the asymptotic estimation
I
n
[
]
:
=
λ
[
n
_
]
=
(
(
2
*
n
-
1
)
^
2
*
(
P
i
^
2
)
)
/
4
O
u
t
[
]
=
1
4
2
(
-
1
+
2
n
)
2
π
and we compare with the numerical results
I
n
[
]
:
=
T
a
b
l
e
F
o
r
m
[
T
a
b
l
e
[
{
E
v
a
l
u
a
t
e
[
a
[
[
n
]
]
]
,
N
[
λ
[
n
]
]
}
,
{
n
,
1
,
1
0
}
]
,
T
a
b
l
e
S
p
a
c
i
n
g
-
>
{
3
,
3
}
,
T
a
b
l
e
A
l
i
g
n
m
e
n
t
s
-
>
C
e
n
t
e
r
,
T
a
b
l
e
H
e
a
d
i
n
g
s
-
>
{
T
a
b
l
e
[
n
,
{
n
,
1
,
1
0
}
]
,
{
"
R
o
o
t
"
,
"
A
s
y
m
p
t
o
t
i
c
e
s
t
i
m
a
t
i
o
n
"
}
}
]
O
u
t
[
]
/
/
T
a
b
l
e
F
o
r
m
=
R
o
o
t
A
s
y
m
p
t
o
t
i
c
e
s
t
i
m
a
t
i
o
n
1
x
4
.
1
1
5
8
6
2
.
4
6
7
4
2
x
2
4
.
1
3
9
3
2
2
.
2
0
6
6
3
x
6
3
.
6
5
9
1
6
1
.
6
8
5
4
x
1
2
2
.
8
8
9
1
2
0
.
9
0
3
5
x
2
0
1
.
8
5
1
1
9
9
.
8
5
9
6
x
3
0
0
.
5
5
2
9
8
.
5
5
6
7
x
4
1
8
.
9
8
7
4
1
6
.
9
9
1
8
x
3
0
0
.
5
5
5
5
5
.
1
6
5
9
x
7
1
5
.
0
7
7
7
1
3
.
0
7
9
1
0
x
8
9
2
.
7
3
8
9
0
.
7
3
2
Second way
Now we are going to solve the same problem with the command DEigenSystem and NDEigenSystem
Let' s start with DEigenSystem
I
n
[
]
:
=
C
l
e
a
r
[
"
G
l
o
b
a
l
`
*
"
]
I
n
[
]
:
=
ℒ
=
-
L
a
p
l
a
c
i
a
n
[
u
[
x
]
,
{
x
}
]
O
u
t
[
]
=
-
′
′
u
[
x
]
I
n
[
]
:
=
=
D
i
r
i
c
h
l
e
t
C
o
n
d
i
t
i
o
n
[
u
[
x
]
=
=
0
,
x
=
=
0
]
;
I
n
[
]
:
=
n
e
w
v
=
N
e
u
m
a
n
n
V
a
l
u
e
[
u
[
x
]
,
x
=
=
1
]
;
I
n
[
]
:
=
e
i
g
e
n
s
y
s
=
D
E
i
g
e
n
s
y
s
t
e
m
[
{
ℒ
+
n
e
w
v
,
}
,
u
[
x
]
,
{
x
,
0
,
1
}
,
1
0
]
;
(
*
E
i
g
e
n
V
a
l
u
e
s
*
)
I
n
[
]
:
=
e
i
g
e
n
s
y
s
[
[
1
]
]
O
u
t
[
]
=
4
.
1
2
…
,
2
4
.
1
…
,
6
3
.
7
…
,
1
2
3
.
…
,
2
0
2
.
…
,
3
0
1
.
…
,
4
1
9
.
…
,
5
5
7
.
…
,
7
1
5
.
…
,
8
9
3
.
…
(
*
E
i
g
e
n
F
u
n
c
t
i
o
n
s
*
)
I
n
[
]
:
=
e
i
g
e
n
s
y
s
[
[
2
]
]
O
u
t
[
]
=
S
i
n
x
4
.
1
2
…
,
S
i
n
x
2
4
.
1
…
,
S
i
n
x
6
3
.
7
…
,
S
i
n
x
1
2
3
.
…
,
S
i
n
x
2
0
2
.
…
,
S
i
n
x
3
0
1
.
…
,
S
i
n
x
4
1
9
.
…
,
S
i
n
x
5
5
7
.
…
,
S
i
n
x
7
1
5
.
…
,
S
i
n
x
8
9
3
.
…
I
n
[
]
:
=
P
l
o
t
[
E
v
a
l
u
a
t
e
[
e
i
g
e
n
s
y
s
[
[
2
]
]
]
,
{
x
,
0
,
1
}
,
P
l
o
t
R
a
n
g
e
-
>
A
l
l
]
O
u
t
[
]
=
Now we are going to use NDEigenSystem
I
n
[
]
:
=
C
l
e
a
r
[
"
G
l
o
b
a
l
`
*
"
]
I
n
[
]
:
=
ℒ
=
-
L
a
p
l
a
c
i
a
n
[
u
[
x
]
,
{
x
}
]
;
I
n
[
]
:
=
=
D
i
r
i
c
h
l
e
t
C
o
n
d
i
t
i
o
n
[
u
[
x
]
=
=
0
,
x
=
=
0
]
;
I
n
[
]
:
=
n
e
w
v
=
N
e
u
m
a
n
n
V
a
l
u
e
[
u
[
x
]
,
x
=
=
1
]
;
I
n
[
]
:
=
e
i
g
e
n
s
y
s
=
N
D
E
i
g
e
n
s
y
s
t
e
m
[
{
ℒ
+
n
e
w
v
,
}
,
u
[
x
]
,
{
x
,
0
,
1
}
,
1
0
,
M
e
t
h
o
d
{
"
P
D
E
D
i
s
c
r
e
t
i
z
a
t
i
o
n
"
{
"
F
i
n
i
t
e
E
l
e
m
e
n
t
"
,
{
"
M
e
s
h
O
p
t
i
o
n
s
"
{
M
a
x
C
e
l
l
M
e
a
s
u
r
e
0
.
0
0
5
}
}
}
}
]
;
(
*
E
i
g
e
n
V
a
l
u
e
s
*
)
I
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We notice that we receive the same result as the first way of our solution
POSTED BY:
Athanasios Paraskevopoulos
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