# Implicit function theorem

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") (an explicit function, see article on implicit functions).

First example

If we define the function "f": $f\left(x,y\right)=x^2 + y^2$, then the equation $f\left(x,y\right)=1$ cuts out the unit circle as the level set $\left\{ \left(x,y\right) | f\left(x,y\right) = 1 \right\}$. There is no way to represent the unit circle as the graph of a function $y = g\left(x\right)$ because for each choice of $x in \left(-1,1\right)$, there are two choices of $y$, namely $pmsqrt\left\{1-x^2\right\}$.

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

It is not possible to find a function which will cut out a neighbourhood of $\left(1,0\right)$ or $\left(-1,0\right)$. Any neighbourhood of $\left(1,0\right)$ or $\left(-1,0\right)$ 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 $y = g\left(x\right)$. Consequently, there is no function whose graph looks like a neighbourhood of $\left(1,0\right)$ or $\left(-1,0\right)$. 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 $g_1\left(x\right)$ and $g_2\left(x\right)$ in situations where we cannot write down explicit formulas. It guarantees that $g_1\left(x\right)$ and $g_2\left(x\right)$ are differentiable, and it even works in situations where we do not have a formula for $f\left(x,y\right)$.

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 R"n" &times; R"m", and we write a point of this product as (x,y) = ("x1", ..., "xn", "y1", ..., "ym"). "f" is the given relation. Our goal is to construct a function "g" : R"n"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. As such, we will fix a point (a,b) = ("a1", ..., "an", "b1", ..., "bm") 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 R"n", 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" &times; "V". In symbols,

:$\left\{ \left(mathbf\left\{x\right\}, g\left(mathbf\left\{x\right\}\right)\right) \right\} = \left\{ \left(mathbf\left\{x\right\}, mathbf\left\{y\right\}\right) | f\left(mathbf\left\{x\right\}, mathbf\left\{y\right\}\right) = 0 \right\} cap \left(U imes V\right)$

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, ..., "a"n, "b"1, ..., "b"m) to (a, b), the Jacobian matrix is

:

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" : Rn+m &rarr; Rm be a continuously differentiable function, and let Rn+m have coordinates (x, y). Fix a point ("a"1,...,"a"n,"b"1,...,"b"m) = (a","b) with "f"(a,b)=c, where c&isin; Rm. 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" &rarr; "V" such that::$\left\{ \left(mathbf\left\{x\right\}, g\left(mathbf\left\{x\right\}\right)\right) \right\} = \left\{ \left(mathbf\left\{x\right\}, mathbf\left\{y\right\}\right) | f\left(mathbf\left\{x\right\}, mathbf\left\{y\right\}\right) = mathbf\left\{c\right\} \right\} cap \left(U imes V\right).$

The circle example

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

:

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

Application: change of coordinates

Suppose we have an m-dimensional space, parametrised by a set of coordinates $\left(x_1,ldots,x_m\right)$. We can introduce a new coordinate system by giving m functions $x\text{'}_1\left(x_1,ldots,x_m\right), ldots, x\text{'}_m\left(x_1,ldots,x_m\right)$. These functions allow to calculate the new coordinates $\left(x\text{'}_1,ldots,x\text{'}_m\right)$ of a point, given the old coordinates $\left(x_1,ldots,x_m\right)$. One might want to verify if the opposite is possible: given coordinates $\left(x\text{'}_1,ldots,x\text{'}_m\right)$, can we 'go back' and calculate $\left(x_1,ldots,x_m\right)$? The implicit function theorem will provide an answer to this question. The (new and old) coordinates $\left(x\text{'}_1,ldots,x\text{'}_m, x_1,ldots,x_m\right)$ are related by $f=0$, with:$f\left(x\text{'}_1,ldots,x\text{'}_m,x_1,ldots x_m\right)=\left(x\text{'}_1\left(x_1,ldots x_m\right)-x_1,ldots , x\text{'}_m\left(x_1,ldots, x_m\right)-x_m\right).$Now the Jacobian matrix of "f" at a certain point $\left(a,b\right)$ is given by :Where $1_m$ denotes the $m imes m$ identity matrix, and J is the $m imes m$ matrix of partial derivatives, evaluated at $\left(a,b\right)$. (In the above, these blocks were denoted by X and Y.) The implicit function theorem now states that we can locally express $\left(x_1,ldots,x_m\right)$ as a function of $\left(x\text{'}_1,ldots,x\text{'}_m\right)$ if J is invertible. Demanding J is invertible is equivalent to $det J eq 0$, 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 $\left(R, heta\right)$. We can go to a new coordinate system (cartesian coordinates) by defining functions $x\left(R, heta\right)=R cos heta$ and $y\left(R, heta\right)=R sin heta$. This makes it possible given any point $\left(R, heta\right)$ to find corresponding cartesian coordinates $\left(x,y\right)$. When can we go back, and convert cartesian into polar coordinates? By the previous example, we need $det J eq 0$, with :Since $det J = R$, the conversion back to polar coordinates is only possible if $R eq 0$. This is a consequence of the fact that at that point polar coordinates are not good: at the origin the value of $heta$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 $f:X imes Y o Z$ be Fréchet differentiable. If $\left(x_0,y_0\right)in X imes Y$, $f\left(x_0,y_0\right)=0$, and $ymapsto Df\left(x_0,y_0\right)\left(0,y\right)$ 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 $g:U o V$ such that $f\left(x,g\left(x\right)\right)=0$ and $f\left(x,y\right)=0$ if and only if $y=g\left(x\right)$, for all $\left(x,y\right)in U imes V$.

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 [L. D. Kudryavtsev, "Implicit function" in Encyclopedia of Mathematics,M. Hazewinkel, Ed. Dordrecht, The Netherlands: Kluwer, 1990.] . The following more general form was proven by Kumagai [S. Kumagai, "An implicit function theorem: Comment," "Journal of Optimization Theory and Applications", 31(2):285-288, June 1980.] based on an observation by Jittorntrum [K. Jittorntrum, "An Implicit Function Theorem", "Journal of Optimization Theory and Applications", 25(4), 1978.] .

Consider a continuous function $f : R^n imes R^m ightarrow R^n$ such that $f\left(x_0, y_0\right) = 0$. 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 $y in B$, $f\left(cdot, y\right) : A ightarrow R^n$ 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:$f\left(x, y\right) = 0$has a unique solution:$x = g\left(y\right) in A_0$,where $g$ is a continuous function from $B_0$ into $A_0$.

*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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