Mathematics of Physics and Engineering

162 The Three Theorems

Combining (3.2.23) and (3.2.24) yields

/// ("V/"^2 + f9) dV = ///(V/ ' VU + 9U) ^ (3-^2 - 25)

G G

By the definition of the dot product, |V/ • Vu| < ||V/|| ||Vu||. Also, for
every two non-negative numbers a, b, 0 < (a — b)^2 = a^2 + b^2 — 2ab, which
means that ab < (a^2 + b^2 )/2. Therefore, (3.2.25) implies

///(ll V/||^2 + fg) dV<\ JJJ \\Vf\\^2 dV + jJj(\\\Vu\\^2 + s«) dV.

G G G

Hence, /(/) < I(u), which is equivalent to (3.2.22).
Conversely, assume that / satisfies (3.2.22). Take any function v that
is twice continuously differentiable in G and is equal to zero outside of an
open set contained in G. For this function v define the function F = F(t)
by F(t) = I(f + tv). By assumption, t = 0 is a local minimum of F. Direct
computations show that F = F(t) is differentiable and

F'(0) = [[[{Vf • Vv + gv) dV. (3.2.26)

G

From the one-variable calculus we know that F'(0) must be equal to zero.
Applying (3.2.19) and keeping in mind that, by assumption, v = 0 on dG,
we conclude that

///

(~Vzf + g)vdV = Q. (3.2.27)

G

Since the function v is arbitrary, this implies V^2 / = g in G and completes
the proof of the theorem. •

EXERCISE 3.2.15.^5 (a) Verify (3.2.26). Hint: replace u in the definition of I
with f + tv and expanding the squares; note that, by construction, f + tv €U for
every v and t. (b) Verify that (3.2.27) indeed implies V^2 / — g in G. Hint-
assume that V^2 / > g at some point P in G. Argue that the same inequality must
then hold in some open ball centered at P and lying completely in G. Then take
a function v that is equal to one in that ball — there is still some space between
the boundary of the ball and the boundary of G where v can come down to zero
— and come to a contradiction with (3.2.27).

Theorem 3.2.7 holds also in M^2 , and, in fact, in every Rn, n > 2.