Theory

Implicit functions

When an equation F(x, y) = 0 defines a function y = φ(x), how to compute the derivatives of φ without knowing it, and how the result extends to surfaces and systems.

On this page

The problem
Dini’s theorem

The problem

1. Implicit equation and zero set

Let $F : A \subseteq R^{2} \to R$ . The equation $F (x, y) = 0$ determines the zero set

Z = {(x, y) \in A : F (x, y) = 0},

also called the zero level curve of $F$ . Many plane curves come in this form and not as graphs: the circle $x^{2} + y^{2} - 1 = 0$ , the ellipse $\frac{x ^{2}}{4} + y^{2} - 1 = 0$ , the curve $x^{3} + y^{3} - 2 x y = 0$ . The problem of this chapter is to establish whether, near a point of $Z$ , the set $Z$ coincides with the graph of a function $y = φ (x)$ , and in that case to compute its derivatives.

The curve

x^{3} + y^{3} - 2 x y = 0

: the line

x = \frac{1}{2}

meets it at three points, therefore the curve is not the graph of a function

y = φ (x)

2. Implicitly defined function

Let $(x_{0}, y_{0}) \in Z$ . A function $φ : I \to R$ , with $I$ a neighborhood of $x_{0}$ , is implicitly defined by the equation $F (x, y) = 0$ near $(x_{0}, y_{0})$ if

{φ (x_{0}) = y_{0} F (x, φ (x)) = 0 for every x \in I .

The notion is local: the same equation can define different functions near different points of the curve, and near some points it may define none.

The circle is not a graph, but it is one locally

For $F (x, y) = x^{2} + y^{2} - 1$ the zero set is the unit circle, which is not the graph of any function $y = φ (x)$ : each $x \in (- 1, 1)$ corresponds to two points of the curve. Near the point $(\frac{3}{5}, \frac{4}{5})$ the curve coincides with the graph of $φ (x) = 1 - x^{2}$ ; near $(\frac{3}{5}, - \frac{4}{5})$ with the graph of $- 1 - x^{2}$ . Near $(1, 0)$ no function of $x$ describes the curve, because every neighborhood of $(1, 0)$ contains, for one and the same value $x < 1$ , two points of the circle.

y = \sqrt{1 - x^2}

The upper branch of the circle: near

(\frac{3}{5}, \frac{4}{5})

the curve is the graph of

y = 1 - x^{2}

Remark

In general it is not possible to solve $F (x, y) = 0$ for $y$ algebraically. The equation $y^{5} + y + x = 0$ assigns to each $x$ one and only one $y$ , since the function $y \mapsto y^{5} + y$ is strictly increasing and takes all real values; the solution $y = φ (x)$ , however, cannot be written in terms of elementary functions. Dini’s theorem guarantees existence, local uniqueness and regularity of the implicit function without exhibiting it, and it provides its derivatives.

Dini’s theorem

Dini's theorem (two variables)

Let $F$ be of class $C^{1}$ on an open set $A \subseteq R^{2}$ and let $(x_{0}, y_{0}) \in A$ be such that

{F (x_{0}, y_{0}) = 0 F_{y} (x_{0}, y_{0}) \neq = 0.

Then there exist a neighborhood $I$ of $x_{0}$ , a neighborhood $J$ of $y_{0}$ and a unique function $φ : I \to J$ such that $φ (x_{0}) = y_{0}$ and $F (x, φ (x)) = 0$ for every $x \in I$ . Moreover, $φ$ is of class $C^{1}$ on $I$ and

φ^{'} (x) = - \frac{F _{x} ( x , φ ( x ))}{F _{y} ( x , φ ( x ))} .

The theorem is used in two stages: first the two conditions are checked at the given point, then $φ^{'} (x_{0}) = - \frac{F _{x} ( x _{0} , y _{0} )}{F _{y} ( x _{0} , y _{0} )}$ is computed from the values of the partial derivatives at the point, without knowing $φ$ .

Proof

Let $F_{y} (x_{0}, y_{0}) > 0$ (in the opposite case one argues on $- F$ ). Since $F_{y}$ is continuous, by the sign-preservation property there exists a rectangle $R = [x_{0} - a, x_{0} + a] \times [y_{0} - b, y_{0} + b] \subseteq A$ on which $F_{y} > 0$ . For each fixed $x$ the function $y \mapsto F (x, y)$ is therefore strictly increasing on $[y_{0} - b, y_{0} + b]$ . Since $F (x_{0}, y_{0}) = 0$ , we have $F (x_{0}, y_{0} - b) < 0 < F (x_{0}, y_{0} + b)$ ; by the continuity of $F$ there then exists $δ \in (0, a]$ such that

F (x, y_{0} - b) < 0 < F (x, y_{0} + b) for every x \in I = (x_{0} - δ, x_{0} + δ) .

For fixed $x \in I$ , the continuous function $y \mapsto F (x, y)$ changes sign at the endpoints of $[y_{0} - b, y_{0} + b]$ , hence by the intermediate zero theorem there exists $y \in (y_{0} - b, y_{0} + b)$ with $F (x, y) = 0$ ; such a $y$ is unique by strict monotonicity, and we set $φ (x) = y$ .

y = \sqrt{1 - x^2}

The case

F (x, y) = x^{2} + y^{2} - 1

near

(x_{0}, y_{0}) = (\frac{3}{5}, \frac{4}{5})

: in the rectangle

R

the function is negative on the lower side and positive on the upper side, and for each

x

the zero is unique

For the derivative formula we consider the difference quotient. Setting $P = (x, φ (x))$ and $v = (h, φ (x + h) - φ (x))$ , the points $P$ and $P + v$ belong to the rectangle and $F (P) = F (P + v) = 0$ ; by the mean value theorem applied along the segment there exists a point $Q_{h}$ of the segment such that

0 = F (P + v) - F (P) = F_{x} (Q_{h}) h + F_{y} (Q_{h}) (φ (x + h) - φ (x)),

hence

\frac{φ ( x + h ) - φ ( x )}{h} = - \frac{F _{x} ( Q _{h} )}{F _{y} ( Q _{h} )} .

On $R$ the function $∣ F_{x} ∣$ is bounded and $F_{y}$ has a positive minimum, therefore the difference quotient is bounded and $φ (x + h) \to φ (x)$ as $h \to 0$ : $φ$ is continuous. Consequently $Q_{h} \to (x, φ (x))$ as $h \to 0$ , and since $F_{x}$ and $F_{y}$ are continuous the difference quotient tends to $- \frac{F _{x} ( x , φ ( x ))}{F _{y} ( x , φ ( x ))}$ . Hence $φ$ is differentiable with the stated derivative, which is continuous as a composition of continuous functions, therefore $φ$ is of class $C^{1}$ .

Application to the circle

$F (x, y) = x^{2} + y^{2} - 1$ at the point $(\frac{3}{5}, \frac{4}{5})$ . We have $F (\frac{3}{5}, \frac{4}{5}) = \frac{9}{25} + \frac{16}{25} - 1 = 0$ and $F_{y} (\frac{3}{5}, \frac{4}{5}) = \frac{8}{5} \neq = 0$ , therefore Dini’s theorem applies and

φ^{'} (\frac{3}{5}) = - \frac{F _{x}}{F _{y}} = - \frac{2 \cdot \frac{3}{5}}{2 \cdot \frac{4}{5}} = - \frac{3}{4} .

The value coincides with the derivative of $1 - x^{2}$ at $x = \frac{3}{5}$ , which is known here; the theorem gives the same result even when the explicit form does not exist.

Exchanging the variables

The theorem holds with the roles of $x$ and $y$ exchanged: if $F (x_{0}, y_{0}) = 0$ and $F_{x} (x_{0}, y_{0}) \neq = 0$ , near the point the curve is the graph of a function $x = ψ (y)$ with $ψ^{'} (y) = - \frac{F _{y}}{F _{x}}$ . At the points of the curve where $F_{y} = 0$ and $F_{x} \neq = 0$ the curve therefore remains locally a graph, with respect to the other variable, and the tangent line is vertical. At the points where $F_{x}$ and $F_{y}$ are both zero the theorem provides no information, in either direction: these points are called singular and must be studied directly.

The circle

x^{2} + y^{2} = 1

at the point

(1, 0)

F_{y} = 0

and

F_{x} = 2

, the tangent is vertical and near the point the curve is the graph of

x = 1 - y^{2}

Warning

The condition $F_{y} (x_{0}, y_{0}) \neq = 0$ is sufficient, not necessary. For $F (x, y) = (y - x^{2})^{2}$ the gradient is zero at every point of the zero set, and yet the set is the graph of the function $y = x^{2}$ . When the hypotheses are not satisfied, the existence of the implicit function is not ruled out: it must be decided by studying the equation.

The theory continues with 4 more sections (higher derivatives of the implicit function, tangent line to a level curve, dinis theorem in three variables…), plus all worked examples, the solutions and the PDF.

Unlock now