- Implicit function theorem
In the branch of

mathematics calledmultivariable calculus , the**implicit function theorem**is a tool which allows relations to be converted to functions. It does this by representing the relation as thegraph 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 itspartial derivative s, 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 onimplicit function s).**First example**If we define the function "f": $f(x,y)=x^2\; +\; y^2$, then the equation $f(x,y)=1$ cuts out the

unit circle as thelevel set $\{\; (x,y)\; |\; f(x,y)\; =\; 1\; \}$. 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 $pmsqrt\{1-x^2\}$.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 acontinuously differentiable function. We think of**R**^{"n+m"}as thecartesian product **R**^{"n"}×**R**^{"m"}, and we write a point of this product as (**x**,**y**) = ("x_{1}", ..., "x_{n}", "y_{1}", ..., "y_{m}"). "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**) = ("a_{1}", ..., "a_{n}", "b_{1}", ..., "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**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" × "V". In symbols,:$\{\; (mathbf\{x\},\; g(mathbf\{x\}))\; \}\; =\; \{\; (mathbf\{x\},\; mathbf\{y\})\; |\; f(mathbf\{x\},\; mathbf\{y\})\; =\; 0\; \}\; cap\; (U\; imes\; V)$

To state the implicit function theorem, we need the

Jacobian , also called the "differential" or "total derivative", of $f$. This is the matrix ofpartial derivative s of $f$. Abbreviating ("a"_{1}, ..., "a"_{n}, "b"_{1}, ..., "b"_{m}) to (**a**,**b**), the Jacobian matrix is:$egin\{matrix\}(Df)(mathbf\{a\},mathbf\{b\})\; =\; left\; [egin\{matrix\}\; frac\{partial\; f\_1\}\{partial\; x\_1\}(mathbf\{a\},mathbf\{b\})\; cdots\; frac\{partial\; f\_1\}\{partial\; x\_n\}(mathbf\{a\},mathbf\{b\})\backslash \; vdots\; ddots\; vdots\backslash \; frac\{partial\; f\_m\}\{partial\; x\_1\}(mathbf\{a\},mathbf\{b\})\; cdots\; frac\{partial\; f\_m\}\{partial\; x\_n\}(mathbf\{a\},mathbf\{b\})end\{matrix\}\; ight|left.egin\{matrix\}\; frac\{partial\; f\_1\}\{partial\; y\_1\}(mathbf\{a\},mathbf\{b\})\; cdots\; frac\{partial\; f\_1\}\{partial\; y\_m\}(mathbf\{a\},mathbf\{b\})\backslash \; vdots\; ddots\; vdots\backslash frac\{partial\; f\_m\}\{partial\; y\_1\}(mathbf\{a\},mathbf\{b\})\; cdots\; frac\{partial\; f\_m\}\{partial\; y\_m\}(mathbf\{a\},mathbf\{b\})\backslash end\{matrix\}\; ight]\; \backslash \; =\; egin\{bmatrix\}\; X\; |\; Y\; end\{bmatrix\}\backslash end\{matrix\}$

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 acontinuously differentiable function, and let**R**^{n+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**∈**R**^{m}. If the matrix [(∂"f"_{i}/∂"y"_{j})(a,b)] isinvertible , 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::$\{\; (mathbf\{x\},\; g(mathbf\{x\}))\; \}\; =\; \{\; (mathbf\{x\},\; mathbf\{y\})\; |\; f(mathbf\{x\},\; mathbf\{y\})\; =\; mathbf\{c\}\; \}\; cap\; (U\; imes\; V).$**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:$egin\{matrix\}(Df)(a,b)\; =\; egin\{bmatrix\}\; frac\{partial\; f\}\{partial\; x\}(a,b)\; frac\{partial\; f\}\{partial\; y\}(a,b)\backslash end\{bmatrix\}\backslash \; =\; egin\{bmatrix\}\; 2a\; 2b\; end\{bmatrix\}.\backslash end\{matrix\}$

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 $y=g(x)$ for all points where $y\; eq\; 0$. 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 by giving m functions $x\text{'}\_1(x\_1,ldots,x\_m),\; ldots,\; x\text{'}\_m(x\_1,ldots,x\_m)$. These functions allow to calculate the new coordinates $(x\text{'}\_1,ldots,x\text{'}\_m)$ of a point, given the old coordinates $(x\_1,ldots,x\_m)$. One might want to verify if the opposite is possible: given coordinates $(x\text{'}\_1,ldots,x\text{'}\_m)$, 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 $(x\text{'}\_1,ldots,x\text{'}\_m,\; x\_1,ldots,x\_m)$ are related by $f=0$, with:$f(x\text{'}\_1,ldots,x\text{'}\_m,x\_1,ldots\; x\_m)=(x\text{'}\_1(x\_1,ldots\; x\_m)-x\_1,ldots\; ,\; x\text{'}\_m(x\_1,ldots,\; x\_m)-x\_m).$Now the Jacobian matrix of "f" at a certain point $(a,b)$ is given by :$egin\{matrix\}(Df)(a,b)\; =\; egin\{bmatrix\}\; 1\; cdots\; 0\; frac\{partial\; x\text{'}\_1\}\{partial\; x\_1\}(a,b)\; cdots\; frac\{partial\; x\text{'}\_1\}\{partial\; x\_m\}(a,b)\backslash \; vdots\; ddots\; vdots\; vdots\; ddots\; vdots\backslash \; 0\; cdots\; 1\; frac\{partial\; x\text{'}\_m\}\{partial\; x\_1\}(a,b)\; cdots\; frac\{partial\; x\text{'}\_m\}\{partial\; x\_m\}(a,b)\backslash end\{bmatrix\}\backslash \; =\; egin\{bmatrix\}\; 1\_m\; |\; J\; end\{bmatrix\}.\backslash end\{matrix\}$Where $1\_m$ denotes the $m\; imes\; m$

identity matrix , and J is the $m\; imes\; m$ 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 $(x\text{'}\_1,ldots,x\text{'}\_m)$ 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 theinverse function theorem .**Example: polar coordinates**As a simple application of the above, consider the plane, parametrised by

polar coordinates $(R,\; heta)$. We can go to a new coordinate system (cartesian coordinates ) by defining functions $x(R,\; heta)=R\; cos\; heta$ and $y(R,\; heta)=R\; sin\; heta$. This makes it possible given any point $(R,\; heta)$ to find corresponding cartesian coordinates $(x,y)$. When can we go back, and convert cartesian into polar coordinates? By the previous example, we need $det\; J\; eq\; 0$, with :$J\; =egin\{bmatrix\}\; frac\{partial\; x(R,\; heta)\}\{partial\; R\}\; frac\{partial\; x(R,\; heta)\}\{partial\; heta\}\; \backslash \; frac\{partial\; y(R,\; heta)\}\{partial\; R\}\; frac\{partial\; y(R,\; heta)\}\{partial\; heta\}\; \backslash end\{bmatrix\}=\; egin\{bmatrix\}\; cos\; heta\; -R\; sin\; heta\; \backslash \; sin\; heta\; R\; cos\; hetaend\{bmatrix\}.$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 inBanach space s, 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 $(x\_0,y\_0)in\; X\; imes\; Y$, $f(x\_0,y\_0)=0$, and $ymapsto\; Df(x\_0,y\_0)(0,y)$ 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 $g:U\; o\; V$ such that $f(x,g(x))=0$ and $f(x,y)=0$ if and only if $y=g(x)$, for all $(x,y)in\; U\; imes\; V$.

**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 $f(x\_0,\; y\_0)\; =\; 0$.

**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 $y\; in\; B$, $f(cdot,\; y)\; :\; A\; ightarrow\; R^n$ 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:$f(x,\; y)\; =\; 0$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**

*Wikimedia Foundation.
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