Implicit function theorem

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In multivariable calculus, a branch of mathematics, the implicit function theorem is a tool which allows relations to be converted to functions. It does this by representing the relation as the graph of a function. There may not be a single function whose graph is the entire relation, but there may be such a function on a restriction of the domain of the relation. The implicit function theorem gives a sufficient condition to ensure that there is such a function.

[edit] Example

Consider the unit circle. If we define the function $f$ as $f (x, y) = x 2 + y 2 - 1$ , then the relation $f (x, y) = 0$ cuts out the unit circle. Explicitly, the unit circle is the set ${(x, y) | f (x, y) = 0}$ . There is no way to represent the unit circle as the graph of a function $y = g (x)$ because for each non-zero choice of $x$ , there are two choices of $y$ , namely $\sqrt{1-x^2}$ and $-\sqrt{1-x^2}$ .

However, it is possible to represent part of the circle as a function. If we let $g_1(x) = \sqrt{1-x^2}$ for $- 1 < x < 1$ , then the graph of $y = g 1 (x)$ gives the upper half of the circle. Similarly, if $g_2(x) = -\sqrt{1-x^2}$ , then the graph of $y = g 2 (x)$ gives the lower half of the circle.

It is not possible to find a function which will cut out a neighbourhood of $(1,0)$ or $( - 1,0)$ . Any neighbourhood of $(1,0)$ or $( - 1,0)$ contains both the upper and lower halves of the circle. Because functions must be single-valued, there is no way of writing both the upper and lower halves using one function $y = g (x)$ . Consequently there is no function whose graph looks like a neighbourhood of $(1,0)$ or $( - 1,0)$ . In this case the conclusion of the implicit function theorem fails.

The purpose of the implicit function theorem is to tell us the existence of functions like $g 1 (x)$ and $g 2 (x)$ in situations where we cannot write down explicit formulas. It guarantees that $g 1 (x)$ and $g 2 (x)$ are differentiable, and it even works in situations where we do not have a formula for $f (x, y)$ .<>

[edit] Statement of the theorem

Let $f : \bold R^{n+m} \rightarrow \bold R^m$ be a continuously differentiable function. We think of $\bold R^{n+m}$ as the cartesian product $\bold R^n \times \bold R^m$ , and we write a point of this product as $(x_1, \ldots x_n, y_1, \ldots y_m)$ . $f$ is the given relation. Our goal is to construct a function $g : \bold R^n \rightarrow \bold R^m$ whose graph is the set of all $(x_1, \ldots x_n, y_1, \ldots y_m)$ such that $f(x_1, \ldots x_n, y_1, \ldots y_m) = 0$ .

As noted above, this may not always be possible, so we will fix a point $(a_1, \ldots a_n, b_1, \ldots, b_m)$ which satisfies $f(a_1, \ldots a_n, b_1, \ldots, b_m) = 0$ , and we will ask for a $g$ that works near the point $(a_1, \ldots a_n, b_1, \ldots, b_m)$ . In other words, we want an open set $U$ of $\bold R^n$ , an open set $V$ of $\bold R^m$ , and a function $g : U \rightarrow V$ such that the graph of $g$ equals the relation $f = 0$ on $U \times V$ . In symbols,

$\{ (x_1, \ldots, x_n, g(x_1, \ldots x_n)) \} = \{ (x_1, \ldots x_n, y_1, \ldots y_m) | f(x_1, \ldots x_n, y_1, \ldots y_m) = 0 \} \cap (U \times V)$

To state the implicit function theorem, we need the Jacobian, also called the differential or total derivative, of $f$ . This is the matrix of partial derivatives of $f$ . Abbreviating $(a_1, \ldots, a_n, b_1, \ldots b_m)$ to $(a, b)$ , the Jacobian matrix is

$\begin{matrix} (Df)(a,b) & = & \begin{bmatrix} \frac{\partial f_1}{\partial x_1}(a,b) & \cdots & \frac{\partial f_1}{\partial x_n}(a,b) & \frac{\partial f_1}{\partial y_1}(a,b) & \cdots & \frac{\partial f_1}{\partial y_m}(a,b)\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots\\ \frac{\partial f_m}{\partial x_1}(a,b) & \cdots & \frac{\partial f_m}{\partial x_n}(a,b) & \frac{\partial f_m}{\partial y_1}(a,b) & \cdots & \frac{\partial f_m}{\partial y_m}(a,b)\\ \end{bmatrix}\\ & = & \begin{bmatrix} X & | & Y \end{bmatrix}\\ \end{matrix}$

where $X$ is the matrix of partial derivatives in the $x$ 's and $Y$ is the matrix of partial derivatives in the $y$ 's. The implicit function theorem says that if $Y$ is an invertible matrix, then there are $U$ , $V$ , and $g$ as desired. Writing all the hypotheses together gives the following statement.

Let $f : \bold R^{n+m} \rightarrow \bold R^m$ be a continuously differentiable function, and let $\bold R^{n+m}$ have coordinates $(x_1, \ldots x_n, y_1, \ldots y_m)$ . Fix a point $(a_1, \ldots a_n, b_1, \ldots, b_m) = (a,b)$ . If the matrix $\begin{bmatrix} (\partial f_i / \partial y_j)(a,b) \end{bmatrix}$ is invertible, then there exists an open set $U$ containing $(a_1, \ldots, a_n)$ , an open set $V$ containing $(b_1, \ldots, b_m)$ , and a differentiable function $g : U \rightarrow V$ such that $\{ (x_1, \ldots, x_n, g(x_1, \ldots x_n)) \} = \{ (x_1, \ldots x_n, y_1, \ldots y_m) | f(x_1, \ldots x_n, y_1, \ldots y_m) = 0 \} \cap (U \times V)$ .

[edit] References

Rudin, Walter (1976). Principles of Mathematical Analysis. McGraw-Hill. ISBN 0-07-054235-X.
Spivak, Michael (1965). Calculus on Manifolds. HarperCollins. ISBN 0-8053-9021-9.
Warner, Frank (1983). Foundations of Differentiable Manifolds and Lie Groups. Springer. ISBN 0-387-90894-3.