Implicit function theorem

Theorem—If ⁠∂f∂y(x0,y0)≠0, then there exists a unique differentiable function ⁠f(x,φ(x))=0⁠ in a neighbourhood of ⁠f(x,φ(x))=0⁠, one gets ⁠φ(x0)=y0⁠.

Since gx(y)=f(x,y) is strictly monotone in a neighbourhood of φ,ϕ are solutions to the differential equation, then x + y – 1 = 0 cannot be represented as the graph of a function. Around the point B, there is no function of f(x, y) = 1 cuts out the unit circle as the level set "}},"i":0}}]}'/>{(x, y) | f(x, y) = 1}. There is no way to represent the unit circle as the graph of a function of one variable ±1−x2.

However, it is possible to represent part of the circle as the graph of a function of one variable. If we let y = g1(x) provides the upper half of the circle. Similarly, if g1(x) and 2</sub>(''x'')"}},"i":0}}]}'/>g2(x) always exist, even in situations where they cannot be written down by explicit formulas. It guarantees that 1</sub>(''x'')"}},"i":0}}]}'/>g1(x) and 2</sub>(''x'')"}},"i":0}}]}'/>g2(x) are differentiable, and it even works in situations where we do not have a formula for ℝn+m as the Cartesian product f, our goal is to construct a function (𝐱,𝐲) such that f(𝐚,𝐛)=𝟎, and we will ask for a U⊂ℝn containing 𝐛, and a function f=𝟎 on {(𝐱,g(𝐱))∣𝐱∈U}={(𝐱,𝐲)∈U×V∣f(𝐱,𝐲)=𝟎}.

To state the implicit function theorem, we need the Jacobian matrix of (a1,…,an,b1,…,bm) to X is the matrix of partial derivatives in the variables yj. The implicit function theorem says that if V, and ℝn+m have coordinates f(𝐚,𝐛)=𝟎, where U⊂ℝn containing f(𝐱,g(𝐱))=𝟎 for all 𝐱∈U. Moreover, g in f is analytic or continuously differentiable U in order that the same holds true for f(x,y)=x2+y2−1. The matrix of partial derivatives is just a 1 × 2 matrix, given by 2b; the linear map defined by it is invertible if and only if y ≠ 0. For y, that is, b = 0 we have 2xdx+2ydy=0, giving m-dimensional space, parametrised by a set of coordinates h1…hm each being continuously differentiable. These functions allow us to calculate the new coordinates x'1=h1(x1,…,xm),…,x'm=hm(x1,…,xm). One might want to verify if the opposite is possible: given coordinates (x'1,…,x'm,x1,…,xm) are related by f = 0, with (Df)(a,b)=[−1⋯0⋮⋱⋮0⋯−1|∂h1∂x1(b)⋯∂h1∂xm(b)⋮⋱⋮∂hm∂x1(b)⋯∂hm∂xm(b)]=[−Im|J]. where Im denotes the m × m identity matrix, and (x1,…,xm) as a function of x(R, θ) = R cos(θ) and (x, y). When can we go back and convert Cartesian into polar coordinates? By the previous example, it is sufficient to have det J = R, conversion back to polar coordinates is possible if R = 0, our coordinate transformation is not invertible: at the origin, the value of θ is not well-defined.

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(x0,y0)=0, and f:ℝn×ℝm→ℝn such that B⊂ℝm of x0 and y0, respectively, such that, for all y in B, B0⊂ℝm of x0 and y0, such that, for all