This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 0) = 0. (b) Note that when = 0, (0 ) = 2/ from our expression in part (b) and we also
have (0 0) = 0. It readily follows that
(0 0) doesn’t exist since (0 ) is not
even continuous at = 0.
:)
E
A function (
) is called Morse if all its critical points are nondegenerate.
A function (
) is called harmonic if the equation
+
= 0 holds for all
. Prove
that a harmonic Morse function does not have local maxima or minima, so its critical
points are always saddles.
S
Since
= − , the determinant of the Hessian matrix is − 2 − 2 ≤ 0.
Since all critical points are supposed to be nondegenerate, we have − 2 − 2 < 0 and
therefore all of them are saddle points.
:)
E Let 2− 2 ( + 2 > 0;
= =0 2 2+ 2 )=
0 (a) Find (0 0) and (0 0). (b) Explain why the fact that
S
For ( (0 0) = (0 0) does not contradict Clairaut’s Theorem. (a) As we have ( 0) = (0 ) = 0, it follows that (0 0) = (0 0) = 0.
) = (0 0), by the quotient rule we get (after some calculation)
4 ( )= +4 2 3−
( 2 + 2 )2 Hence
(0 )= −
0 5 5 ( )= / 4 =0 = (− ) 0) = = 5 =− −4 3 2−
( 2 + 2 )2 ;
; =0
=−
=0 and
(0 0) = (0 ) =0 = −1 In the same manner,
(
and
(0 0) = 0) (
5 =0 = =1 =0 = −1 4 :) Hence
(0 0) =
( (0 0) (b) After some calculation, you can verify, using usual diﬀerentiation rules, that for
) = (0 0),
6
+9 4 2−9 2 4− 6
(
)=
(
)=
( 2 + 2 )3 Hence, approaching (0 0) along the axis we have
−
lim (0 →0 ) = lim
→0 6
6 = lim(−1) = −1
→0 while approaching along the axis we have
6 lim
→0 ( 0) = lim
→0 6 = lim 1 = 1
→0 Since these are diﬀerent, we conclude that
(
) (and also
) has no limit as (
)→
(0 0) and hence it is not continuous there. Continuity of the mixed derivatives was a
condition for Clairaut’s theorem, so it is not contradicted by the result from (a). 6...
View Full
Document
 Spring '13

Click to edit the document details