- Implicit function theorem
In the branch of
mathematicscalled 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).
If we define the function "f": , then the equation cuts out the
unit circleas the level set. There is no way to represent the unit circle as the graph of a function because for each choice of , there are two choices of , namely .
However, it is possible to represent part of the circle as a function. If we let for , then the graph of provides the upper half of the circle. Similarly, if , then the graph of gives the lower half of the circle.
It is not possible to find a function which will cut out a neighbourhood of or . Any neighbourhood of or 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 . Consequently, there is no function whose graph looks like a neighbourhood of or . 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 and in situations where we cannot write down explicit formulas. It guarantees that and are differentiable, and it even works in situations where we do not have a formula for .
Statement of the theorem
Let "f" : R"n+m" → R"m" be a
continuously differentiablefunction. We think of R"n+m" as the cartesian productR"n" × 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" × "V". In symbols,
To state the implicit function theorem, we need the
Jacobian, also called the "differential" or "total derivative", of . This is the matrix of partial derivatives of . Abbreviating ("a"1, ..., "a"n, "b"1, ..., "b"m) to (a, b), the Jacobian matrix is
where is the matrix of partial derivatives in the 's and is the matrix of partial derivatives in the 's. The implicit function theorem says that if is an invertible matrix, then there are , , and as desired. Writing all the hypotheses together gives the following statement.
:Let "f" : Rn+m → Rm be a
continuously differentiablefunction, 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∈ 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" → "V" such that::
The circle example
Let us go back to the example of the
unit circle. In this case and . 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. By the implicit function theorem we see that we can write the circle in the form for all points where . For and we run into trouble, as noted before.
Application: change of coordinates
Suppose we have an m-dimensional space, parametrised by a set of coordinates . We can introduce a new coordinate system by giving m functions . These functions allow to calculate the new coordinates of a point, given the old coordinates . One might want to verify if the opposite is possible: given coordinates , can we 'go back' and calculate ? The implicit function theorem will provide an answer to this question. The (new and old) coordinates are related by , with:Now the Jacobian matrix of "f" at a certain point is given by :Where denotes the
identity matrix, and J is the matrix of partial derivatives, evaluated at . (In the above, these blocks were denoted by X and Y.) The implicit function theorem now states that we can locally express as a function of if J is invertible. Demanding J is invertible is equivalent to , 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. We can go to a new coordinate system ( cartesian coordinates) by defining functions and . This makes it possible given any point to find corresponding cartesian coordinates . When can we go back, and convert cartesian into polar coordinates? By the previous example, we need , with :Since , the conversion back to polar coordinates is only possible if . This is a consequence of the fact that at that point polar coordinates are not good: at the origin the value of is not well-defined.
Banach space version
Based on the
inverse function theoremin Banach spaces, it is possible to extend the implicit function theorem to Banach space valued mappings.
Let , , be Banach spaces. Let the mapping be Fréchet differentiable. If , , and is a Banach space isomorphism from onto . Then there exist neighbourhoods of and of and a Frechet differentiable function such that and if and only if , for all .
Implicit functions from non-differentiable functions
Various forms of the implicit function theorem exist for the case when the function 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 such that . If there exist open neighbourhoods and of and , respectively, such that, for all , is locally one-to-one then there exist open neighbourhoods and of and ,such that, for all , the equation:has a unique solution:,where is a continuous function from into .
*Constant rank theorem: Both the implicit function theorem and the
Inverse function theoremcan be seen as special cases of the constant rank theorem.
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