Implicit function theorem

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In multivariable calculus of mathematics the implicit function theorem says that for a suitable set of equations, some of the variables are defined as functions of the others. There are some natural limitations on this use of a mathematical relation to define implicit functions, which may be seen in trying to use the unit circle as the graph of a function. Firstly, the projection of the circle onto the x-axis is two-to-one on the interval (−1,1); this means that y can only be made a local function of x. Further, at the points (1,0) and (−1,0), the tangent line to the circle is vertical. This means that y cannot be a differentiable function of x at those points.

The implicit function theorem gets around both these difficulties, which represent the typical obstructions. The implicit function is only locally defined, and points at which the first-order behavior would be problematic are outside the scope of the result.

Simplest examples

Consider the function given as

f(x, y, u) = x² + ay + u³ = 0.

Simple algebra tells us that

<math>y = -1/a(x^2 + u^3).</math>

However, this breaks down when a = 0. The implicit function theorem states the conditions under which this kind of expression breaks down; namely, when

<math> \frac{\partial f}{\partial y} = a = 0</math>.

For another example, take the unit-circle function <math>f(x,y) = x^2 + y^2 - 1 = 0</math>. Here, y cannot be written in terms of x when |x| = 1; in the language of the implicit function theorem, this is because the partial derivative of f with respect to y equals zero when |x| = 1.

The implicit function theorem

In general, let

f: R^{m + n} → Rⁿ

be a continuously differentiable function defined in an open subset of R^{m + n} which contains the point (a, b), where

a=<math>(x_1, ..., x_m)</math> ∈ R^m and b=<math>(y_1, ... ,y_n) </math> ∈ Rⁿ. Suppose also that we have the initial equation

f(a, b) = 0.

Let's take the Jacobian of f evaluated at (a, b), D f_{(a, b)}, and split it into [ X | Y ] where the submatrices X, Y are given by

<math> X_{n \times m} = \left({\partial f_i \over \partial x_j}|_{(\mathbf{a},\mathbf{b})}\right)_{1 \leq i \leq n, \; 1 \leq j \leq m}</math>

<math> Y_{n \times n} = \left({\partial f_i \over \partial y_j}|_{(\mathbf{a},\mathbf{b})}\right)_{1 \leq i \leq n, \; 1 \leq j \leq n} </math>

so that D f_{(a, b)} = [ X | Y ]. If Y is invertible (ie., the determinant of Y is nonzero), then f(x, y) = 0 defines y as a function of x near (a, b); more precisely there are open sets A, B such that

a ∈ A ⊆ R^m, b ∈ B ⊆ Rⁿ

and the sets have the following property:

for each x ∈ A, there is a unique y ∈ B such that f(x, y) = 0.

Since the above y is uniquely determined by x with respect to A and B, we can call it g(x), so what we really have is a function

g: A → B

such that for each x ∈ A, f(x, g(x)) = 0, and g(x) is the only element in B with that property. Because f(a, b) = 0, in particular we have g(a) = b; so this function "locally generalizes" the initial equation f(a, b) = 0. Furthermore, the theorem guarantees that the function g is differentiable; in particular, it turns out that

D g(a) = −Y⁻¹X.

An example

Consider the pair of equations

x²u² + xzv + y² = 0

yzu + xyv² - 3x = 0

and let p be the point (x, y, z, u, v) = (3, 3, −3, 0, 1). It is easy to see that the point satisfies the former equations. Lets define a function f: R⁵ → R² by

f(x, y, z, u, v) = (f₁(x, y, z, u, v), f₂(x, y, z, u, v)) = (x²u² + xzv + y², yzu + xyv² - 3x)

It can easily be shown that this function is continuously differentiable. If we set a = (3, 3, −3), b = (0, 1), then we get the initial equation f(p) = f(a, b) = 0. To be able to use the Implicit Function Theorem, we have to show first that the matrix Y (as in the theory) is invertible. Lets see:

The 2×2 matrix

<math> Y = \begin{bmatrix} D_{4}f^{1} & D_{5}f^{1} \\

D_{4}f^{2} & D_{5}f^{2} \end{bmatrix}_{(\mathbf{a}, \mathbf{b})} = \begin{bmatrix} 2ux^2 & xz \\ yz & 2xyv \end{bmatrix}_{(3, 3, -3, 0, 1)} = \begin{bmatrix} 0 & -9 \\ -9 & 18 \end{bmatrix} </math>

It is clear that this matrix is non-singular. So by the Implicit Function Theorem there are open sets A ⊆ R³, B ⊆ R², and a differentiable function g: A → B ⊆ R², call it g = (g₁, g₂) such that for all (x, y, z) ∈ A, the point (x, y, z, g(x, y, z)) = (x, y, z, g₁(x, y, z), g₂(x, y, z)) ∈ R⁵ is a solution of the initial equations. Note that at first this property does not seem so difficult to achieve for a function; the really interesting part is that g is also differentiable.de:Satz von der impliziten Funktion it:Teorema delle funzioni implicite he:משפט הפונקציות הסתומות