next up previous
Next: Integration Up: New Functions from old Previous: Inverse functions

Implicit functions

We will prove the implicit function theorem.

Let f (x, y) be a continuous function of two variables so that it is continuous, and differentiable with respect to y when x is kept fixed; in particular, we have expression

f (x, y) = f0(x, y0) + f1(x, y0)(y - y0) + ox(y - y0)

where the subscript in the ox denotes the dependence of the condition on x. The function f1(x, y) is denoted by $ \partial$f /$ \partial$y and is called the partial derivative of f with respect to y (we have already seen this for the case of polynomials). We further assume that f1 is continuous.

Figure 7: The implicit function

Now suppose that f (a, b) = 0 and c = ($ \partial$f /$ \partial$y)(a, b) $ \neq$ 0. We want to find the implicit function h(x) defined by f = 0 (see 7). We do this by showing that for each x near a there is a unique y so that f (x, y) = 0. Equivalently, we need to show that the function g(x, y) = y - f (x, y)/c has a unique fixed point for any chosen x near a.

Contractions give rise to functions with a unique fixed point.

Exercise 49   Let 0 < c < 1 be a constant. Let g(x) be a function so that g send the interval [a, b] to itself and | g(x) - g(y)| < c| x - y| for all x and y in [a, b]. Show that g(x0) = x0 for exactly one point x0 in [a, b].

Consider the function g(x, y) = y - f (x, y)/c as above; we have g(a, b) = b and ($ \partial$g/$ \partial$y) = 0. For each fixed x we would like g(x, y) to be a contraction on some interval around b. As a first step:

Exercise 50   Using the continuity of f1 show that there is an interval [a - r, a + r] around a and an interval [b - s, b + s] around b so that |($ \partial$g/$ \partial$y)(x, y)| $ \leq$ 1/2 for x and y in these respective intervals. (Hint: Write $ \partial$g/$ \partial$y in terms of f1 to show that it is continuous). In particular, by the mean value theorem show that | g(x, y) - g(x, y')| $ \leq$ 1/2(y - y') on these intervals.

Now by the continuity of f (and thus of g) we can choose a smaller r so that | g(x, b) - g(a, b)| $ \leq$ s/2 for x in the interval [a - r, a + r]. Since g(a, b) = b, it follows that g(x, y) = b + (g(x, y) - g(x, b)) + (g(x, b) - g(a, b)) lies in the interval [b - s, b + s]. Applying the above exercise it follows that for every x there is a unique point y so that g(x, y) = y or equivalently f (x, y) = 0. We denote this point y as h(x). This function h is the required implicit function.

We have the identity,

h(x)-h(x')=g(x,h(x))-g(x',h(x'))= \\
\left(g(x,h(x))-g(x',h(x))\right) +

Applying the mean value theorem to g(x', y) we obtain

| h(x) - h(x')| $\displaystyle \leq$ | g(x, h(x)) - g(x', h(x)| + $\displaystyle {\textstyle\frac{1}{2}}$| h(x) - h(x')|

Exercise 51   Use the above inequality and the continuity of g(x, y0) for every fixed y0 to conclude the h(x) is continuous.

Now if we assume in addition that f is differentiable to order k then it follows that h is also differentiable to order k by an entirely similar reasoning to the one in the above exercise.

Exercise 52   If f (x, y) has the form (near the point (a, b))

f(x,y) = f_{1,0}(x-a) + f_{0,1}(y-b) + \\
f_{2,0}(x-a)^2 + 2 f_{1,1}(x-a)(y-b) +
f_{0,2}(y-b)^2 +

with f0, 1 $ \neq$ 0, then show that the implicit function g(x) has the form (near x = a),

g(x) = b - $\displaystyle {\frac{f_{1,0}}{f_{0,1}}}$(x - a) + $\displaystyle {\frac{f_{0,2}f_{1,0}^2 + 2 f_{1,1}f_{1,0}f_{0,1}
- f_{2,0}f_{0,1}^2}{f_{0,1}^3}}$(x - a)2 + o((x - a)2)

next up previous
Next: Integration Up: New Functions from old Previous: Inverse functions
Kapil H. Paranjape 2001-01-20