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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,

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

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

${\begin{matrix}(Df)(\mathbf {a} ,\mathbf {b} )&=&\left[{\begin{matrix}{\frac {\partial f_{1}}{\partial x_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1}}{\partial x_{n}}}(\mathbf {a} ,\mathbf {b} )\\\vdots &\ddots &\vdots \\{\frac {\partial f_{m}}{\partial x_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m}}{\partial x_{n}}}(\mathbf {a} ,\mathbf {b} )\end{matrix}}\right|\left.{\begin{matrix}{\frac {\partial f_{1}}{\partial y_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1}}{\partial y_{m}}}(\mathbf {a} ,\mathbf {b} )\\\vdots &\ddots &\vdots \\{\frac {\partial f_{m}}{\partial y_{1}}}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m}}{\partial y_{m}}}(\mathbf {a} ,\mathbf {b} )\\\end{matrix}}\right]\\&=&{\begin{bmatrix}X&|&Y\end{bmatrix}}\\\end{matrix}}$

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 (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

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

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

${\frac {\partial g}{\partial x_{j}}}(x)=-\left({\frac {\partial f}{\partial y}}(x,g(x))\right)^{-1}{\frac {\partial f}{\partial x_{j}}}(x,g(x))$.

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

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

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.

Application: change of coordinates

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 $(x'_{1},\ldots ,x'_{m})$ by supplying m functions $h_{1}\ldots h_{m}$. These functions allow to calculate the new coordinates $(x'_{1},\ldots ,x'_{m})$ of a point, given the point's old coordinates $(x_{1},\ldots ,x_{m})$ using $x'_{1}=h_{1}(x_{1},\ldots ,x_{m}),\ldots ,x'_{m}=h_{m}(x_{1},\ldots ,x_{m})$. One might want to verify if the opposite is possible: given coordinates $(x'_{1},\ldots ,x'_{m})$, can we 'go back' and calculate the same point's original coordinates $(x_{1},\ldots ,x_{m})$? 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

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

Now the Jacobian matrix of f at a certain point $(a,b)$ [ where $a=(x'_{1},\ldots ,x'_{m}),b=(x_{1},\ldots ,x_{m})$ ] is given by

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

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. 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 $(x_{1},\ldots ,x_{m})$ 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: 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 $\det J\neq 0$, 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 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 $f:R^{n}\times R^{m}\rightarrow R^{n}$ 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 $y\in B$, $f(\cdot ,y):A\rightarrow R^{n}$ 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/