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Leading nonlinear eigenvalue shift in a Sturm-Liouville problem
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Past exam of the mathematics course of the University of Cambridge
/
2023
/
ib
/
Paper 1
/
13A
/
b
/
Solution
Created
2026-09-24
Updated
2026-09-25
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Because the
parameter
enters through
λ
w
y
,
L
(
y
;
λ
0
)
=
L
(
y
;
λ
)
−
(
λ
−
λ
0
)
w
y
=
y
m
+
1
−
ε
m
μ
w
y
.
(1)
Use this
as
f
in part (
a)
. The necessary
orthogonality
condition is
0
=
∫
0
1
y
0
(
y
m
+
1
−
ε
m
μ
w
y
)
d
x
.
(2)
With
y
=
ε
y
0
+
ε
2
y
1
and
μ
=
O
(
1
)
,
y
m
+
1
=
ε
m
+
1
y
0
m
+
1
+
O
(
ε
m
+
2
)
,
w
y
=
εw
y
0
+
O
(
ε
2
)
.
(3)
Consequently
0
=
ε
m
+
1
[
∫
0
1
y
0
m
+
2
d
x
−
μ
∫
0
1
w
y
0
2
d
x
]
+
O
(
ε
m
+
2
)
.
(4)
The normalization makes the
second
integral
one. Divide by
ε
m
+
1
to obtain
μ
=
∫
0
1
y
0
m
+
2
d
x
+
O
(
ε
)
.
(5)
This is the
leading nonlinear eigenvalue shift in a Sturm-Liouville problem
.
Total
articles
:
1