Page 151 - INTRODUCTION TO THE CALCULUS OF VARIATIONS

138 Minimal surfaces

Since E = G and F =0,wehave

L + N
H = ⇒ L + N =2EH . (5.5)
2E
We next prove that hv x ; ∆vi = hv y ; ∆vi =0. Using the equations E = G and
F =0,we have, after differentiation of the first one by x and the second one by
y,
hv x ; v xx i = hv y ; v xy i and hv x ; v yy i + hv y ; v xy i =0 .
This leads, as wished, to hv x ; ∆vi =0 andinasimilarway to hv y ; ∆vi =0.
Therefore ∆v is orthogonal to v x and v y and thus parallel to e 3 ,which means
that there exists a ∈ R so that ∆v = ae 3 . We then deduce that

a = he 3 ; ∆vi = he 3 ; v xx i + he 3 ; v yy i = L + N. (5.6)

Combining (5.5) and (5.6), we immediately get (5.4) and the theorem then fol-
lows.

5.2.1 Exercises
3
Exercise 5.2.1 (i) Let a, b, c ∈ R show that
2 2 2 2
|a × b| = |a| |b| − (ha; bi)
(a × b) × c = ha; ci b − hb; ci a.
√
2
(ii) Deduce that |v x × v y | = EG − F .
(iii) Show that
L = he 3 ; v xx i = − he 3x ; v x i ,
M = he 3 ; v xy i = − he 3x ; v y i = − he 3y ; v x i ,
N = he 3 ; v yy i = − he 3y ; v y i .

Exercise 5.2.2 Show that the surfaces in Example 5.9 and Example 5.10 are
minimal surfaces.

Exercise 5.2.3 Let Σ be a surface (of revolution) given by

v (x, y)=(x, w (x)cos y, w (x)sin y) , x ∈ (0, 1) ,y ∈ (0, 2π) ,w ≥ 0 .
(i) Show that

Z 1 q
2
Area (Σ)= I (w)= 2π w (x) 1+(w (x)) dx.
0
0