Problem Set 3
MH2100
Solutions
1 Questions on understanding the lecture
E±²³´µ¶² · Without doing any calculation, explain why any critical point of the function
±
(
²³´³µ
) = sin
² ·
arctan(
´  ²
) + cos(
´
2
 ²
2
) is degenerate.
S¸¹º»µ¸¼
Since the function is independent of
µ
, all its derivatives with respect to
µ
vanish
and therefore the Hessian matrix looks like
H
±
=
±
²²
±
²´
0
±
²´
±
´´
0
0
0
0
¶
It follows that
H
±
has at least one eigenvalue being 0 so that any critical point of the
function
±
(
) is degenerate.
:)
2 Questions on calculation
E±²³´µ¶² ½ Find critical points and determine their type (degenerate/nondegenerate,
Morse index if there is any) for the following functions:
(a)
±
(
²³´
) = (
²  ´
+ 1)
2
.
(b)
±
(
) =
²´
+
50
²
+
20
´
, where
²³´ >
0.
(c)
±
(
) =
ln(
²
2
+
´
2
).
(d)
±
(
) =
²
+
´
2
4
²
+
µ
2
´
+
2
µ
, where
²³´³µ >
0.
1
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View Full DocumentS±²³´µ±¶
(a)
±
(
²³´
) = (
²  ´
+ 1)
2
. For critical points, we have
±
²
= 2(
²  ´
+ 1) = 0
³
±
´
=

2(
²  ´
+ 1) = 0
³
so both equations are
²  ´
+ 1 = 0, which is a whole straight line consisting of
critical points. Since they are not isolated, they must be degenerate. Since the
value of the function at this line is 0 and
±
(
)
≥
0, we see that this line consists of
degenerate minima.
(b)
±
(
) =
²´
+
50
²
+
20
´
, where
²³´ >
0. For critical points, we have
±
²
=
´ 
50
²
2
= 0
³
±
´
=
² 
20
´
2
= 0
³
so
´
=
50
²
2
and
²
=
20
´
2
=
20
(50
/²
2
)
2
=
²
4
125
. Now we get
²
3
= 125, that is,
²
= 5 and
´
= 2.
Further,
H
±
=
·
100
²
3
1
1
40
´
3
¸
³
H
±
(5
³
2) =
·
4
5
1
1 5
¸
and det
H
±
= 3
>
0, tr
H
±
>
0. Hence (5
³
2) is a local minimum.
(c)
±
(
) =
ln(
²
2
+
´
2
). For critical points, we have
±
²
=
´
ln(
²
2
+
´
2
) +
2
²
²
2
+
´
2
= 0
³
±
´
=
²
ln(
²
2
+
´
2
) +
2
´
²
2
+
´
2
= 0
µ
From
±
²
= 0, we see that either
´
= 0 or ln(
²
2
+
´
2
)+
2
²
2
²
2
+
´
2
= 0. Let’s consider these
two cases separately.
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 Spring '13
 Critical Point, hessian matrix, Morse theory

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