From: Robin Chapman <rjc@maths.ex.ac.uk>
Subject: Re: Implicit Function Theorem
Date: Wed, 18 Aug 1999 08:24:49 GMT
Newsgroups: sci.math

In article <7pcgg0$i7g$1@nnrp1.deja.com>,
  piersingram@my-deja.com wrote:
> Hi,
> Can anyone fill me in on what this theorem is and how it relates to to
> obtaining an inverse transformation equation, in the context of a
> coordinate transformation.

Let's take the simplest example first. Let F be a function of two real
variables, and consider the equation F(x,y) = 0. If the function F
isn't too wacky this will give some sort of curve in the plane.
This curve might be a graph of a function y = f(x) if you're really
lucky, say for F(x,y) = y - x^2. But usually it isn't, say
F(x,y) = x^2 + y^2 + 1 the curve is empty, while if x^2 + y^2 - 1
gives no y-value if |x| > 1 but two y-values if |x| < 1. But even this
case isn't too bad, if you start with a point P on the curve
x^2 + y^2 = 1 (apart from (+-1,0)) and look at that bit of the
curve near P then that bit is a graph of a function defined
on a a neighbourhood of P. This generalizes; the implicit function
theorem is:

Let F(x,y) be a differentiable function (on an open set in R^2)
with continuous partial derivatives. Let P = (x_0,y_0) be a point
with F(x_0,y_0) = 0 and F_y(x_0,y_0) <> 0. Then there is an open
neighbourhood V of x_0 in R such that there is a unique continuous
function f with the property that f(x_0) =  y_0 and F(x,f(x)) = 0
for all x in V. In addition f has a continuous derivative.

There's a higher dimensional version. Take m continuously
differentiable functions F_1, ..., F_m of n + m variables
x_1,...,x_n,y_1,...,y_m and replace the non-vanishing of
the partial derivative by the non-vanishing of an appropriate
determinant of partial derivatives.

We can relate the implicit function theorem to the inverse function
theorem as follows. Consider the map phi: (x,y) |--> (x, F(x,y)).
The Jacobian determinant of this is non-zero at P, and so there
is an inverse function psi of this defined near (x_0, 0). Obviously
psi(u,v) = (u,G(u,v)) for some function G, and consider
g(x) = G(x,0). Then
(x,F(x,g(x))) = (x,F(x,G(x,0))) = (x, F(psi(x,0)) = phi(psi(x,0))
= (x,0)
and so F(x,g(x)) = 0.

--
Robin Chapman
 http://www.maths.ex.ac.uk/~rjc/rjc.html
  "They did not have proper palms at home in Exeter."
    Peter Carey, _Oscar and Lucinda_


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