Questions tagged [implicit-function-theorem]

The implicit function theorem gives sufficient conditions to solve a given equation for one or more of the variables as functions of the remaining variables. The basic form of the theorem is that of an existence theorem. However, the contraction mapping proof of the theorem provides an error estimate for a sequence of approximating maps. Sometimes it is also termed the implicit mapping theorem. See http://en.wikipedia.org/wiki/Implicit_function_theorem

The implicit function theorem provides sufficient conditions to solve an equation $G(x,y)=k$ near a point $(a,b)$ for which $G(a,b)=k$ for the $y$-variables as functions of the $x$-variables. In particular, the theorem implies the existence of a function $f$ such that $G(x,f(x))=k$ for $x$ near $a$. The basic idea is simply that if we have $n$-equations in $(m+n)$-unknowns then we may solve for $n$ of the unknowns as dependent variables on the remaining $m$-variables.

Perhaps the most common application is that if $F(x_1, x_2, \dots , x_n )=k$ then we can solve for $x_j$ as a function of $x_1, \dots , x_{j-1},x_{j+1}, \dots , x_n$ near $p\in \mathbb{R}^n$ provided $\frac{\partial F}{\partial x_j}(p)$ is nonzero and $F$ is continuously differentiable ($F \in C^1(p)$) near $p$.

Let us expanding on the general case in more explicit notation. If $G: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable near $(a,b)$ where $a \in \mathbb{R}^m$ and $b \in \mathbb{R}^n$ and $G(a,b)=k$ and the $n \times n$-submatrix of the Jacobian of $G$ corresponding to $y$-derivatives of $G$ is invertible at $(a,b)$ then there exists a function $f: dom(f) \subseteq \mathbb{R}^m \rightarrow \mathbb{R}^n$ which is continuously differentiable near $x=a$ and the solution for $G(x,y)=k$ near $(a,b)$ is given by the graph $y=f(x)$.

An improved version of the implicit function theorem provides a constructive method for which the implicit solution is found as the limit function of a sequence of functions formed by linearizing the equation $G(x,y)=k$ near the initial point $(a,b)$. As typical of such arguments, a fixed point argument is made in concert with the contraction mapping technique. See C.H. Edward's Advanced Calculus of Several Variables for a reasonably complete account of constructive version of the theorem. In particular, Theorem 3.4 of Edward's text states:

Let $G: dom(G) \subseteq \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ be continuously differentiable in an open ball about the point $(a,b)$ where $G(a,b)=k$ (a constant vector in $\mathbb{R}^n$). If the matrix $\tfrac{ \partial G}{\partial y}(a,b)$ is invertible then there exists an open ball $U$ containing $a$ in $\mathbb{R}^m$ and an open ball $W$ containing $(a,b)$ in $\mathbb{R}^m \times \mathbb{R}^n$ and a continuously differentiable mapping $h: U \rightarrow \mathbb{R}^n$ such that $G(x,y)=k$ iff $y=h(x)$ for all $(x,y) \in W$. Moreover, the mapping $h$ is the limit of the sequence of successive approximations defined inductively below $$ h_o(x)=b, \ \ h_{n+1} = h_n(x)-[\tfrac{ \partial G}{\partial y}(a,b)]^{-1}G(x,h_n(x)) \qquad \text{for all $x \in U$.} $$

The implicit function theorem may be used to justify the inverse function theorem and both can be understood as special cases of the more general constant rank theorem. One may consult this Wikipedia article for further examples and discussion.

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Reference Request: Quantitative Implicit Function Theorem

My colleague and I recently wanted to find a form of the Implicit Function Theorem where there were explicit lower bounds on the sizes of the neighborhood of validity (for input and output). The textbooks we have looked at so far do not seem to…
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Implicit Function Theorem in Higher Dimensions

The following function is given: $$f:\Bbb R^3 \rightarrow \Bbb R^2, \left(\begin{matrix}x \\ y \\ z\\ \end{matrix}\right) \mapsto \left(\begin{matrix}-2x^2 + y^2 + z^2 \\ x^2+e^{y-1} - 2y \end{matrix}\right)$$ The first task is to determine whether…
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Does the implicit function theorem hold for discontinuously differentiable functions?

Wikipedia's statement of the implicit function theorem requires that the original function be continuously differentiable. $\:$ Is it known whether or not that condition can be removed? If it can't be completely removed, can it be replaced…
user57159
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Implicit function Theorem

Do you know some practical application of the Implicit & Inversion Theorem for functions? I wonder if you've encountered a particular practical problem that is not too hard to understand (from physics( mechanichs), astronomy, chemistry etc) in which…
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The converse of the Implicit Theorem

According to the implicit function theorem(on $\mathbb R^2$ for simplicity), if $\displaystyle\frac{\partial f}{\partial y}\ne 0$ at $(x_0, y_0)$, then on a neighborhood of $(x_0, y_0)$, there is a function $g\in C^1$ such that $$y = g(x)$$ There…
ToBY
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A Version of The Implicit Function Theorem

I want to ask if this version of the "implicit function theorem" holds. Let $f:S^1\times S^1\to\mathbb{R}$ be continuous, for every $x\in S^1$, there exists $y\in S^1$, not necessarily unique, such that $f(x,y)=0$. Does there exists $g:S^1\to S^1$,…
Hasan Saad
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What are the conditions for the implicit function theorem to hold?

I learned about the Implicit Function Theorem (IFT) through my economics class. I get the basic idea: given a function $F(x,y,z) = 0$, there are some functions we cannot write one variable in terms of the others. Example: we cannot find $z(x,y)$…
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Implicit Function Theorem in Chain Confusion

Consider the following setup for the application of the IFT. $p.a - C'(y(a,p)) = 0$ determines $y=y(a,p)$ implicitly. With this, we have: $$\frac{\partial y}{\partial a} = \frac{p}{C''(y(a,p))} ~\text{and} \frac{\partial y}{\partial p} =…
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Does the implicit function theorem works on non-open subsets?

I have been looking the implicit function theorem's statement in various websites to solve the equation $$F(a,b)=0,\quad F:\mathbb{R}^{n+1}_+\rightarrow\mathbb{R},\quad a\in\mathbb{R}^{n}_+,b\in\mathbb{R}_+,\quad \mathbb{R}_+=[0,+\infty)$$ and most…
AlephZero
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Implicit theorem in $\mathbb{R}^3$ there is $\xi:(t_0-\varepsilon,t_0+\varepsilon)\to\mathbb{R}$ such that $F(t,\xi(t),\xi'(t))=0$

Let $F:\mathcal{U\subset\mathbb{R}^3}\to \mathbb{R}$ of class $C'$, let $(t_0,x_0,y_0)\in \mathcal{U}$, such that $F(t_0,x_0,y_0)=0$ and $\partial_y F(t_0,x_0,y_0)\neq 0$. Prove that for all $\varepsilon>0$ there is a unique…
Valent
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Continously differentiable function is not injective

I learnt about the implicit function theorem and had to prove the following: Let $F \in C^1(\mathbb{R}^2, \mathbb{R}).$ Show with the implicit function theorem that $F$ is not injective. Proof: Assume $F(x,y) \in C^1(\mathbb{R}^2, \mathbb{R})$ in…
karnan
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Applying implicit function theorem

Suppose I have $2$ functions $\phi$ and $\psi$ and a $C^1$ function $F$ such that $F(\phi , \psi) = 0$ I know that we can write $\phi$ as a function of $\psi$ around a point if $F_\phi \neq 0 $ at that point. (Please mention some conditions I…
ab123
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Implicit function theorem involving $\cos$ function

I have to prove that there exists a unique function $f:\mathbb{R}\rightarrow \mathbb{R}$ that satifies the following $\forall x \in \mathbb{R}$ $x^2f(x)^3+2f(x)=\cos(f(x))$ My attempt: Seeing as I have literally been shown two or three simple…
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Relaxing continuous differentiability in the implicit function theorem

In the statement of the implicit function theorem, we require the function $f$ to be $C^1$. Based on information found here, here, and here, I see that it is possible to relax the $C^1$ requirement and obtain a continuous form of the implicit…
David M.
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Use Implicit Function Theorem to show that J(x,y)=0

Let $u=f(x,y)$ and $v=g(x,y)$ be defined in a region "R". Suppose there exists another non constant function $h$, of class $C^{1}$, s.t $h(u,v)=0$ $\forall (x,y) \in$ R. Prove that $J=\frac{\partial(u,v)}{\partial(x,y)}=0 $, $\forall(x,y)\in$ R.…
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