Page 223 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

210 Solutions to the Exercises

7.5 Chapter 5: Minimal surfaces

7.5.1 Generalities about surfaces

Exercise 5.2.1. (i) Elementary.
(ii) Apply (i) with a = v x , b = v y and the definition of E, F and G.
(iii) Since e 3 =(v x × v y ) / |v x × v y |,wehave

he 3 ; v x i = he 3 ; v y i =0 .

Differentiating with respect to x and y, we deduce that

0= he 3 ; v xx i + he 3x ; v x i = he 3 ; v xy i + he 3y ; v x i
= he 3 ; v xy i + he 3x ; v y i = he 3 ; v yy i + he 3y ; v y i
and the result follows from the definition of L, M and N.
Exercise 5.2.2. (i) We have

v x =(−y sin x, y cos x, a) ,v y =(cos x, sin x, 0) ,
(−a sin x, a cos x, −y)
e 3 = p ,
a + y 2
2
v xx =(−y cos x, −y sin x, 0) ,v xy =(− sin x, cos x, 0) ,v yy =0

and hence
a
2 2
E = a + y ,F =0,G =1,L = N =0,M = p
2
a + y 2
which leads to H =0, as wished.
(ii) A straight computation gives
¡ 2 2 ¢ ¡ 2 2 ¢
v x = 1 − x + y , −2xy, 2x ,v y = 2xy, −1+ y − x , −2y ,
¡ 2 2 ¢
2x, 2y, x + y − 1
e 3 = ,
2
2
(1 + x + y )
v xx =(−2x, − 2y, 2) ,v xy =(2y, −2x, 0) ,v yy =(2x, 2y, −2)
and hence
¡ 2 2 ¢ 2
E = G = 1+ x + y ,F =0,L = −2,N =2,M =0
which shows that, indeed, H =0. ¡ ¢
2
Exercise 5.2.3. (i) Since |v x × v y | = w 2 1+ w 02 , we obtain the result.