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In 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 locus defined by $R(x,y)=0$ will overlap locally with the graph of a function $y=f(x)$ (an explicit function, see article on implicit functions).

First example

The unit circle can be specified as the level curve $f(x,y)=1$ of the function $f(x,y)=x^2+y^2$. Around point A, y can be expressed as a function $y(x)$, specifically $g_1(x)=\sqrt{1-x^2}$. No such function exists around point B.

If we define the function $f(x,y)=x^2+y^2$, then 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 $x\in (-1,1),$ there are two choices of $y$, namely $\pm \sqrt{1-x^2}$.

However, it is possible to represent part of the circle as the graph of a function of one variable. 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.

The purpose of the implicit function theorem is to tell us the existence of functions like $g_1(x)$ and $g_2(x)$, even 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 (x,y) = (x₁, ..., x_n, y₁, ..., y_m). Starting from the given function f, our goal is to construct a function g : Rⁿ → R^m whose graph (x, g(x)) is precisely the set of all (x, y) such that f(x, y) = 0.

As noted above, this may not always be possible. We will therefore fix a point (a,b) = (a₁, ..., a_n, b₁, ..., b_m) which satisfies f(a, b) = 0, and we will ask for a g that works near the point (a, b). 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 satisfies the relation f = 0 on U × V. In symbols,

$\{(𝐱,g(𝐱))\mid 𝐱\in U\}=\{(𝐱,𝐲)\in U\times V\mid f(𝐱,𝐲)=0\}.$

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 (a₁, ..., a_n, b₁, ..., b_m) to (a, b), the Jacobian matrix is

$\begin{array}{ccc}(Df)(𝐚,𝐛)& =& \left[\begin{array}{ccc}\frac{\partial f_1}{\partial x_1}(𝐚,𝐛)& \cdots & \frac{\partial f_1}{\partial x_n}(𝐚,𝐛)\\ ⋮& \ddots & ⋮\\ \frac{\partial f_m}{\partial x_1}(𝐚,𝐛)& \cdots & \frac{\partial f_m}{\partial x_n}(𝐚,𝐛)\end{array}\right|\begin{array}{ccc}\frac{\partial f_1}{\partial y_1}(𝐚,𝐛)& \cdots & \frac{\partial f_1}{\partial y_m}(𝐚,𝐛)\\ ⋮& \ddots & ⋮\\ \frac{\partial f_m}{\partial y_1}(𝐚,𝐛)& \cdots & \frac{\partial f_m}{\partial y_m}(𝐚,𝐛)\end{array}]\\ & =& \left[\begin{array}{ccc}X& |& Y\end{array}\right]\end{array}$

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 (x, y). 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 a, an open set V containing b, and a unique continuously differentiable function g:U → V such that

$\{(𝐱,g(𝐱))|𝐱\in U\}=\{(𝐱,𝐲)\in U\times V|f(𝐱,𝐲)=𝐜\}.$

Regularity

It can be proven that whenever we have the additional hypothesis that f is continuously differentiable up to k times inside U×V, then the same holds true for the explicit function g inside U and

$\frac{\partial g}{\partial x_j}(x)=-{(\frac{\partial f}{\partial y}(x,g(x)))}^{-1}\frac{\partial f}{\partial x_j}(x,g(x))$.

Similarly, if f is analytic inside U×V, then the same holds true for the explicit function g inside U. This generalization is called the analytic implicit function theorem.

The circle example

Let us 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 1×2 matrix, given by

$\begin{array}{ccc}(Df)(a,b)& =& \left[\begin{array}{cc}\frac{\partial f}{\partial x}(a,b)& \frac{\partial f}{\partial y}(a,b)\end{array}\right]\\ & =& \left[\begin{array}{cc}2a& 2b\end{array}\right].\end{array}$

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

Application: change of coordinates

Suppose we have an m-dimensional space, parametrised by a set of coordinates $(x_1,\dots ,x_m)$. We can introduce a new coordinate system $(x{\text{'}}_1,\dots ,x{\text{'}}_m)$ by supplying m functions $h_1\dots h_m$. These functions allow to calculate the new coordinates $(x{\text{'}}_1,\dots ,x{\text{'}}_m)$ of a point, given the point's old coordinates $(x_1,\dots ,x_m)$ using $x{\text{'}}_1=h_1(x_1,\dots ,x_m),\dots ,x{\text{'}}_m=h_m(x_1,\dots ,x_m)$. One might want to verify if the opposite is possible: given coordinates $(x{\text{'}}_1,\dots ,x{\text{'}}_m)$, can we 'go back' and calculate the same point's original coordinates $(x_1,\dots ,x_m)$? The implicit function theorem will provide an answer to this question. The (new and old) coordinates $(x{\text{'}}_1,\dots ,x{\text{'}}_m,x_1,\dots ,x_m)$ are related by $f=0$, with

$f(x{\text{'}}_1,\dots ,x{\text{'}}_m,x_1,\dots x_m)=(h_1(x_1,\dots x_m)-x{\text{'}}_1,\dots ,h_m(x_1,\dots ,x_m)-x{\text{'}}_m).$

Now the Jacobian matrix of f at a certain point $(a,b)$ [ where $a=(x{\text{'}}_1,\dots ,x{\text{'}}_m),b=(x_1,\dots ,x_m)$ ] is given by

$\begin{array}{ccc}(Df)(a,b)& =& \left[\begin{array}{cccccc}-1& \cdots & 0& \frac{\partial h_1}{\partial x_1}(b)& \cdots & \frac{\partial h_1}{\partial x_m}(b)\\ ⋮& \ddots & ⋮& ⋮& \ddots & ⋮\\ 0& \cdots & -1& \frac{\partial h_m}{\partial x_1}(b)& \cdots & \frac{\partial h_m}{\partial x_m}(b)\end{array}\right]\\ & =& \left[\begin{array}{ccc}-1_m& |& J\end{array}\right].\end{array}$

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. As it happens, in this particular application of the theorem, neither matrix depends on $a$.) The implicit function theorem now states that we can locally express $(x_1,\dots ,x_m)$ as a function of $(x{\text{'}}_1,\dots ,x{\text{'}}_m)$ if J is invertible. Demanding J is invertible is equivalent to $detJ\ne 0$, 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: polar coordinates

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 it possible given any point $(R,\theta )$ to find corresponding cartesian coordinates $(x,y)$. When can we go back and convert cartesian into polar coordinates? By the previous example, we need $detJ\ne 0$, with

$J=\left[\begin{array}{cc}\frac{\partial x(R,\theta )}{\partial R}& \frac{\partial x(R,\theta )}{\partial \theta }\\ \frac{\partial y(R,\theta )}{\partial R}& \frac{\partial y(R,\theta )}{\partial \theta }\end{array}\right]=\left[\begin{array}{cc}\cos \theta & -R\sin \theta \\ \sin \theta & R\cos \theta \end{array}\right].$

Since $detJ=R$, the conversion back to polar coordinates is only possible if $R\ne 0$. This is a consequence of the fact that at the origin, polar coordinates don't exist: the value of $\theta$ is not well-defined.

Generalizations

Banach space version

Based on the inverse function theorem in Banach spaces, 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 continuously 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 neighbourhoods $U$ 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 $(x,y)\in U\times V$.

Implicit functions from non-differentiable functions

Various forms of the implicit function theorem exist for the case when the function $f$ is not differentiable. It is standard that it holds in one dimension. The following more general form was proven by Kumagai based on an observation by Jittorntrum.

Consider a continuous function $f:R^n\times R^m\to R^n$ such that $f(x_0,y_0)=0$. If there exist open neighbourhoods $A\subset R^n$ and $B\subset R^m$ of $x_0$ and $y_0$, respectively, such that, for all $y\in B$, $f(\cdot ,y):A\to R^n$ is locally one-to-one then there exist open neighbourhoods $A_0\subset R^n$ and $B_0\subset R^m$ of $x_0$ and $y_0$, such that, for all $y\in B_0$, the equation

$f(x,y)=0$

has a unique solution

$x=g(y)\in A_0$,

where $g$ is a continuous function from $B_0$ into $A_0$.

See also

• Constant rank theorem: Both the implicit function theorem and the Inverse function theorem can be seen as special cases of the constant rank theorem.

References

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• Other Sports Official Kull from Drumheller, has hobbies such as telescopes, property developers in singapore and crocheting. Identified some interesting places having spent 4 months at Saloum Delta.
  
  my web-site http://himerka.com/