Page 65 - Vector Analysis

§2.7 The Implicit Function Theorem 61

(a) G : O Ñ W is one-to-one and onto;

(b) the inverse function G´1 : W Ñ O is of class C r;

(c) ( F (x, ) = ( y))´1.
(DG´1) x, y) (DG)(x,

By Remark 2.62, W.L.O.G. we can assume that O = U ˆ V, where U Ď Rn and V Ď Rm

are open, and x0 P U , y(0 P V. )
Write G´1(u, v) = φ(u, v), ψ(u, v) , where φ : W Ñ U and ψ : W Ñ V. Then

( )( )
(u, v) = G φ(u, v), ψ(u, v) = φ(u, v), F (u, ψ(u, v))

()
which implies that φ(u, v) = u and v =(F (u, ψ()u, v)). Let f (x) = ψ(x, 0). Then u, f (u) P
U ˆ V is the unique point satisfying F u, f (u) = 0 if u P U. Therefore, f : U Ñ V, and

()
F x, f (x) = 0 @ x P U .

() ()
Since G(x0, y0) = (x0, 0) = G x0, f (x0) , (x0, y0), x0, f (x0) P O, and G : O Ñ W is

one-to-one, we must have y0 = f (x0).

By (b) and (c), we have G´1 is of class C 1, and

(DG´1)(u, v) = ( y))´1 .
(DG)(x,

As a consequence, ψ P C 1, and

[ ][ In 0 ]´1
(Duφ)(u, v) (Dvφ)(u, v)
(Duψ)(u, v)
=
(Dvψ)(u, v) [(DxF )(x, y) (DyF )(x, y)
In ]
0

= ( y))´1(DxF ( y))´1 .
´ (Dy (Dy
F )(x, )(x, y) F )(x,

Evaluating the equation above at v = 0, we conclude that

(Df )(u) = (Duψ)(u, 0) = ( )(u, f (u)))´1(DxF ( )
´ (DyF ) u, f (u)

which implies 3. We also note that 4 follows from (b) and 5 follows from 3. ˝

Example 2.69. Let F (x, y) = x2 + y2 ´ 1.

1. If (x0, y0) = (1, 0), then Fx(x0, y0) = 2 ‰ 0; thus the implicit function theorem implies
that locally x can be expressed as a function of y.