Implicit functions

Suppose we have a function of two variables, F(x, y), and we’re interested in its height-c level curve; that is, solutions to the equation F(x, y) = c. For instance, considering F(x, y) = x2 + y2 and c = 1, in which case the level curve we care about is the familiar unit circle. It would be nice if choosing a value for x in the equation F(x, y) = c would immediately determine the value of y — that is, if F(x, y) = c determined y as a function of x. The equation f(x, y) = 1 cuts out the unit circle as the level set {(x, y) | f(x, y) = 1}. There is no way to represent the unit circle as the graph of a function of one variable y = g(x) because for each choice of x ∈ (−1, 1), there are two choices of y, namely ± √(1 − x2). Though, it is possible to have such function locally.

An example in which is possible to define a function implicit

f(x, y) = x + y - 2 = 0

defines a line in the plane, whose points can be equivalentely represented by y = 2 - x, or x = 2 - y.

An equation of the kind F(x, y) = cost, rapresents a level curve: a cross section of the graph F(x, y) taken at a constant value

E = {(x,y) ∈ A: F(x, y) = c}

Some examples:

We pose ourselves the following question:

Given a function F(x, y) at least C1(A) with A open subset of the plane, under which hypothesis f(x, y) = 0, defines implicitely a function y = g(x)?

To state that F(x, y) = 0 defines implicitily a function y= g(x), means that there exists an interval I in which the function g: I → ℝ is such that:

F(x,g(x)) = 0

Theorem 1.0 (Implicit function theorem) - Let A an open subset of ℝ2 and F: A → ℝ, a function C1(A). We suppose that at a given point x0, y0 ∈ A:

F(x0, y0) = 0,   and   Fy(x0, y0) ≠ 0

Then, there exist two intervals centered in x0 and y0 respectively I = (x0 - δ, x0 + δ) and J = (y0 - σ, y0 + σ, with (δ,σ > 0) and a function g: I → J, such that

Proof - Consider the case Fy(x0,y0) > 0. For the continuity of Fy(x,y), there exists a positive number σ such that

Fy(x,y) > 0

for every point (x,y) ∈ A that

|x - x0| ≤ δ,   |y − y0| ≤ σ

The y-value function F(x0,y), is strictly increasing on [y - σ, y0 + σ]. Because F(x0,y0) = 0, it follows that

F(x0,y0 − σ) < 0,   F(x0,y0 + σ) > 0   (3.4)

Applying Corollary 1.0, to functions F(x,y0-σ), F(x, y0+σ) in the variable x, there exists δ > 0 (and < σ) such that

F(x,y0 - σ) < 0,   F(x,y0 + σ) > 0

per ogni x ∈ (x0 − δ, x0 + δ). Pertanto, per ogni x di tale intervallo, la funzione della variabile y ∈ [y0 - σ, y0 + σ]

y → F(x,y)

is strictly increasing and for equations (3.4) has opposite sign values at the extremes of interval [y0 - σ, y0 + σ]. Then, by the The Intermediate Zero Theorem for one dimensional functions: ∀ x in I there exists y ∈ (y0 − σ, y0 + σ) such that F(x, y) = 0: this y value is unique owning to the monotony of the function, and it is indicated with g(x). We just proved that in I × J, J = (y0 − σ, y0 + σ), the equation F(x, y) = 0, holds only if y = g(x).

We show now that g so defined on I is derivable. We have to evaluate the incremental ratio of g

[g(x+h) − g(x)] / h

where 0 < |h| < δ and x ∈ I. Let

γ(t) = F(x + th, g(x) + t[g(x+h) − g(x)])

The composite of differentiable functions is a differentiable function, and it is immediate to note that γ(0) = γ(1) = 0. By Rolle's Theorem there exists t* ∈ (0, 1) such that γ′(t*) = 0. Let a = x + h ⋅ t* and b = g(x) + t* [g(x + h) - g(x)], it follows from the theorem of derivation of composite functions

0 = γ′(t*) = Fx (a, b) h + Fy (a, b)[g(x + h) − g(x)]

This relation allows us to write the incremental ratio of g at a point x as

g'(x) = − Fx(x, g(x)) / Fy(x, g(x)),   ∀x ∈ I

The function g is continuos on I as ita derivative g', since Fx(x, g(x)) and Fy(x, g(x)) are cotinuous (the composite of continuous functions is continuous).□

If F(x0, y0) = 0, and Fx(x0, y0) = 0, it is possible to interchange the roles of x and y and state that in a neighborhood J of y0 there is a unique function x = h(y) defined in J such that:

F(h(y),y) = 0,   ∀y ∈ J

h ∈ C1(J) e

h'(y) = − Fx(x, y) / Fy(x, y),   ∀y ∈ J

In the case of F(x,y) = x2 + y2 given Fy(x,y) = 2y. According to the implicit function theorem, at any point (x*,y*) on the circle except (1,0) and (-1,0), we can locally solve for y as a function of x. Of course there are two possible solutions, y = ± &sqrt;(1 − x2) and the point (x*,y*) will determine which of these it is. At the points (±1, 0) this breaks down, as y can't be defined for x < -1 or x > -1 and the function y = ± &sqrt;(1 − x2) does not even have a one-sided derivative at x = ±1.

There are cases in which it is not possible to write down an analytical equation for g, even though it exists by the hypothesis of Dini's theorem. The following is an example of this situations

Example 1 - Verify that the equation

x ey + y ex = 0   (6.1)

defines implicitily a function y = g(x) in the neighborhood of x = 0.

Solution. We note that f(0,0) = 0. Moreover since fy(0,0) = 1, we can affirm there exists a neighborhood of (0,0), a function y = g(x) defined implicitily by equation (6.1), with y(0) = 0. We are unable to write the function y = g(x). By deriving both members of the equation (6.1) with respect to x

ey + xey ⋅ y' + yex + exy' = 0

from which

y'(x) = (-ey − yex) / (eyx + ex)

by replacing the values x = 0, y(0) = 0, e y'(0)= 0, we obtain y''(0) = -1. If we iterate the step, we can calculate successive derivatives of y(x) in x = 0 and with a Mac Laurin expansion, obtain a good approximation for y(x), even though we don't have an analytical expression for it.

Example 2 - Given the function

f(x,y) = y3 + 2x3 - 2 + xy   (6.1)

Verify that in a neighborhood of P = (1,0) is defined implicitily a function y = g(x). Calculate g'(1).

Solution. We note that f(1,0) = 0. Moreover since fy(1, 0) = 1, possiamo affermare che esiste un intorno I di 1 e una funzione y = g (x) deifinita implicitamente dall'equazione f(x,y) = 0:

since fy(1,0) = 1 and fx(1,0) = 6, g'(1) = -6.

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