AMATH 586 SPRING 2020
HOMEWORK 3 — DUE MAY 8 ON GITHUB BY 11PM
Be sure to do a
git pull
to update your local version of the
amath5862020
repository.
Problem 1:
It is natural to ask if a neighborhood of
z
= 0 can be in the absolute stability region
S
for a LMM. You will show that this cannot be the case. Consider a consistent
and zerostable LMM
r
X
j
=0
α
j
U
n
+
j
=
k
r
X
j
=0
β
j
f
(
U
n
+
j
)
.
Recall the characteristic polynomial
π
(
ξ
;
z
) =
ρ
(
ξ
)

zσ
(
ξ
). Show:
•
Consistency implies that
π
(1; 0) = 0.
•
Stability implies that
ρ
0
(1)
6
= 0.
•
Suppose
ξ
= 1 +
η
(
z
) for
z
near zero so that
π
(
ξ
;
z
) =
π
(1 +
η
(
z
);
z
) = 0.
Compute
η
0
(0). Why does this imply that there must be an interval (0
,
] for
some small
>
0 that does not lie in the absolute stability region
S
.
(a) To have consistency, we need to satisfy (5
.
48), hence
π
(1; 0) =
ρ
(1) =
∑
j
α
j
=
0.
(b) At
z
= 0:
π
(
ξ
; 0) =
ρ
(
ξ
) =
ρ
(1) +
ρ
0
(1)(
ξ

1) +
1
2
ρ
00
(1)(
ξ

1)
2
+
· · ·
.
To have consistency,
ρ
(1) = 0. If
ρ
0
(1) = 0 also, then the above Taylor expan
sion becomes
π
(
ξ
; 0) =
ρ
(
ξ
) =
1
2
ρ
00
(1)(
ξ

1)
2
+
· · ·
.
Hence, we can factor out a repeated root (
ξ

1)
2
at
ξ
= 1 from
π
(
ξ
; 0), which
violates the root condition and makes
π
(
ξ
; 0) unstable at
z
= 0.
(c) Suppose
π
(
ξ
;
z
) =
π
(1 +
η
(
z
);
z
) = 0, then we have
X
j
α
j
(1 +
η
(
z
))
j

z
X
j
β
j
(1 +
η
(
z
))
j
= 0
,
X
j
(
α
j

zβ
j
)(1 +
η
(
z
))
j
= 0
.
Take the derivative with respect to z using implicit differentiation:
X
j

β
j
(1 +
η
(
z
))
j
+ (
α
j

zβ
j
)
j
(1 +
η
(
z
))
j

1
η
0
(
z
) = 0
.
1
Since
η
(0) = 0 (because we know
π
(
ξ
; 0) has a root at
ξ
= 1), we can substitute
η
(0) = 0 and
z
= 0 into the above:
X
j

β
j
+
jα
j
η
0
(0) = 0
.
Hence,
η
0
(0) =
∑
j
β
j
∑
j
jα
j
(
a
)
= 1
,
where (
a
) follows from (5
.
48) for consistency. Since
η
(0) = 0 and
η
0
(0)
>
0,
η
is increasing at
z
= 0 and should be positive when z is a slightly greater than
0. In other words, this would mean that there is a small interval
z
∈
(0
,
], such
that
η
(
z
)
>
0. This would mean that 1 +
η
(
z
) is slightly greater than 1 in this
interval and
π
(1 +
η
(
z
);
z
) = 0 would have a root slightly greater than 1, which
violates the stability condition.
Problem 2:
Recall the test problem
v
000
(
t
) +
v
0
(
t
)
v
(
t
)

β
1
+
β
2
+
β
3
3
v
0
(
t
) = 0
,
where
β
1
< β
2
< β
3
. It follows that
v
(
t
) =
β
2
+ (
β
3

β
2
)cn
2
r
β
3

β
1
12
t,
s
β
3

β
2
β
3

β
1
!
is a solution where cn(
x, k
) is the Jacobi cosine function and
k
is the elliptic modulus.
Some notations use cn(
x, m
) where
m
=
k
2
. The corresponding initial conditions
are
v
(0) =
β
3
, v
0
(0) = 0
, v
00
(0) =

(
β
3

β
1
)(
β
3

β
2
)
6
.