Implicit function theorem

In the branch of mathematics called multivariable calculus, 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.

The theorem states that if the equation R(x, y) = 0 (an Implicit function) satisfies some mild conditions on its partial derivatives, then one can in principle solve this equation for y, at least over some small interval. Geometrically, the graph defined by R(x,y) = 0 will overlap locally with the graph of a function y = f(x).

Example

A circle specified by the implicit function $f(x,y)=x^{2}+y^{2}-1$. Around point A y can be expressed as a function of x, specifically $y={\sqrt {1-x^{2}}}$. No such function exists for point B.

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 choice of $x\in (-1,1)$, 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 $y=g_{1}(x)$ provides 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 these two cases, 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 $f(x,y)$.

Statement of the theorem

Let f : R^n+m → R^m be a continuously differentiable function. We think of R^n+m as the cartesian product Rⁿ × R^m, and we write a point of this product as (x₁, ..., x_n, y₁, ..., y_m). f is the given relation. Our goal is to construct a function g : Rⁿ → R^m whose graph (x₁, ..., x_n, g(x₁, ..., x_n)) is precisely the set of all (x₁, ..., x_n, y₁, ..., y_m) such that f(x₁, ..., x_n, y₁, ..., y_m) = 0.

As noted above, this may not always be possible. As such, we will fix a point (a₁, ..., a_n, b₁, ..., b_m) which satisfies f(a₁, ..., a_n, b₁, ..., b_m) = 0, and we will ask for a g that works near the point (a₁, ..., a_n, b₁, ..., b_m). In other words, we want an open set U of Rⁿ, an open set V of R^m, and a function g : U → V such that the graph of g equals the relation f = 0 on U × 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 : R^n+m → R^m be a continuously differentiable function, and let R^n+m have coordinates (x₁,...,x_n, y₁, ..., y_m). Fix a point (a₁,...,a_n,b₁,...,b_m) = (a,b) with f(a,b)=c, where c∈ R^m. If the matrix [(∂f_i/∂y_j)(a,b)] is invertible, then there exists an open set U containing (a₁,..., a_n), an open set V containing (b₁,...,b_m), and a differentiable function g:U → V such that

$\{(x_{1},\ldots ,x_{n},g(x_{1},\ldots ,x_{n}))\}=\{(x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{m})|f(x_{1},\ldots ,x_{n},y_{1},\ldots ,y_{m})=c\}\cap (U\times V).$

Example

Lets go back to the example of the unit circle. In this case $n=m=1$ and $f(x,y)=x^{2}+y^{2}-1$. The matrix of partial derivatives is just a $1\times 2$-matrix, given by

${\begin{matrix}(Df)(a,b)&=&{\begin{bmatrix}{\frac {\partial f}{\partial x}}(a,b)&{\frac {\partial f}{\partial y}}(a,b)\\\end{bmatrix}}\\&=&{\begin{bmatrix}2a&2b\end{bmatrix}}.\\\end{matrix}}$

Thus, here, Y is just a number; the linear map defined by it is invertible iff $b\neq 0$. By the implicit function theorem we see that we can write the circle in the form $y=g(x)$ for all points where $y\neq 0$. For $(-1,0)$ and $(1,0)$ we run into trouble, as noted before.

Example

Suppose we have an m-dimensional space, parametrised by a set of coordinates $(x_{1},\ldots ,x_{m})$. We can introduce a new coordinate system by giving m functions $x'_{1}(x_{1},\ldots ,x_{m}),\ldots ,x'_{m}(x_{1},\ldots ,x_{m})$. These functions allow to calculate the new coordinates $(x'_{1},\ldots ,x'_{m})$ of a point, given the old coordinates $(x_{1},\ldots ,x_{m})$. One might want to verify if the opposite is possible: given coordinates $(x'_{1},\ldots ,x'_{m})$, can we 'go back' and calculate $(x_{1},\ldots ,x_{m})$? The implicit function theorem will provide an answer to this question. The (new and old) coordinates $(x'_{1},\ldots ,x'_{m},x_{1},\ldots ,x_{m})$ are related by $f=0$, with

$f(x'_{1},\ldots ,x'_{m},x_{1},\ldots x_{m})=(x'_{1}(x_{1},\ldots x_{m})-x'_{1},\ldots ,x'_{m}(x_{1},\ldots ,x_{m})-x'_{m}).$

Now the Jacobian matrix of f at a certain point $(a,b)$ is given by

${\begin{matrix}(Df)(a,b)&=&{\begin{bmatrix}1&\cdots &0&{\frac {\partial x'_{1}}{\partial x_{1}}}(a,b)&\cdots &{\frac {\partial x'_{1}}{\partial x_{m}}}(a,b)\\\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\0&\cdots &1&{\frac {\partial x'_{m}}{\partial x_{1}}}(a,b)&\cdots &{\frac {\partial x'_{m}}{\partial x_{m}}}(a,b)\\\end{bmatrix}}\\&=&{\begin{bmatrix}1_{m}&|&J\end{bmatrix}}.\\\end{matrix}}$

Where $1_{m}$ denotes the $m\times m$ identity matrix, and J is the $m\times m$ matrix of partial derivatives, evaluated at $(a,b)$. (In the above, these blocks were denoted by X and Y.) The implicit function theorem now states that we can locally express $(x_{1},\ldots ,x_{m})$ as a function of $(x'_{1},\ldots ,x'_{m})$ if J is invertible. Demanding J is invertible is equivalent to $\det J\neq 0$, thus we see that we can go back from the primed to the unprimed coordinates if the determinant of the Jacobian J is non-zero. This statement is also known as the inverse function theorem.

Example

As a simple application of the above, consider the plane, parametrised by polar coordinates $(R,\theta )$. We can go to a new coordinate system (cartesian coordinates) by defining functions $x(R,\theta )=R\cos \theta$ and $y(R,\theta )=R\sin \theta$. This makes is possible given any point $(R,\theta )$ to find corresponding cartesian coordinates $(x,y)$. When can we go back, and converse cartesian into polar coordinates? By the previous example, we need $\det J\neq 0$, with

$J={\begin{bmatrix}{\frac {\partial x(R,\theta )}{\partial R}}&{\frac {\partial x(R,\theta )}{\partial \theta }}\\{\frac {\partial y(R,\theta )}{\partial R}}&{\frac {\partial y(R,\theta )}{\partial \theta }}\\\end{bmatrix}}={\begin{bmatrix}\cos \theta &-R\sin \theta \\\sin \theta &R\cos \theta \end{bmatrix}}.$

Since $\det J=R$, the conversion back to polar coordinates is only possible if $R\neq 0$. This is a consequence of the fact that at that point polar coordinates are not good: at the origin the value of $\theta$ is not well-defined.

Banach space version

Based on Grave's inverse function theorem it is possible to extend the implicit function theorem to Banach space valued mappings.

Let $X$, $Y$, $Z$ be Banach spaces. Let the mapping $f:X\times Y\to Z$ be Fréchet differentiable. If $(x_{0},y_{0})\in X\times Y$, $f(x_{0},y_{0})=0$, and $y\mapsto Df(x_{0},y_{0})(0,y)$ is a Banach space isomorphism from $Y$ onto $Z$. Then there exist neighborhoods $U$ of $x_{0}$ and $V$ of $y_{0}$ and a Frechet differentiable function $g:U\to V$ such that $f(x,g(x))=0$ and $f(x,y)=0$ if and only if $y=g(x)$, for all $(x,y)\in U\times V$.

See also

• Implicit function theorem