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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On the Constant Rank Theorem and the Frobenius Theorem for differential equations.

Recently I was reading chapter $4$ (p. $60$) of The Implicit Function Theorem: History, Theorem, and Applications (By Steven George Krantz, Harold R. Parks) on proof's of the equivalence of the Implicit Function Theorem (finite-dimensional vector…
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How to prove Lagrange multiplier theorem in a rigorous but intuitive way?

Following some text books, the Lagrange multiplier theorem can be described as follows. Let $U \subset \mathbb{R}^n$ be an open set and let $f:U\rightarrow \mathbb{R}, g:U\rightarrow \mathbb{R}$ be $C^1$ functions. Let $\mathbf{x}_{0} \in U,…
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A polynomial with nowhere surjective derivative

Let $P:\mathbb {R}^2\rightarrow \mathbb {R}^2$ be a polynomial map. It is given that the Jacobian of $P$ is everywhere not surjective. Must the following be true: There exist polynomial maps $f:\mathbb {R}^2 \rightarrow \mathbb {R}$ , $g:\mathbb…
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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…
Charles Baker
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What's the arc length of an implicit function?

While an explicit function $y(x)$'s arc length $s$ is easily obtained as $$s = \int \sqrt{1+|y'(x)|^2}\,dx,$$ is there any formula for implicit functions given by $f(x,y) = 0$? One can use the implicit differentiation $y'(x) = -\frac{\partial_y…
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Does the implicit function theorem imply Peano existence theorem

In The implicit function theorem written by Krantz & Parks, it's said that the implicit function theorem implies the following existence theorem of ODE: Theorem 4.1.1 If $F(t,x)$, $(t,x)\in\mathbb R\times\mathbb R^N$, is continuous in the…
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Is there a variant of the implicit function theorem covering a branch of a curve around a singular point?

I am looking for a variant of the Implicit function theorem such that it guarantees the analyticity of the implicitly defined function, and works for a single branch of a curve around a singular point? What kind of such variants of the IFT are…
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Invertiblity of the Derivative Matrix ?! (to use Inverse function Theorem)

In the Analysis2 midterm exam, we had the following problem: Let the equation $a_nx^n+\cdots+a_1x+a_0=0$ has $n$ simple real roots (distinct) $\{\alpha_1,\cdots,\alpha_n\}$. Prove that the above equation has still $n$ distinct real roots when the…
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A Lipschitz Implicit Function Theorem.

I look for a reference (book or article) that contains the statement of a version of the implicit function theorem as stated below. This statement I found in notes (with due proof) on the implicit function written by KC Border. In these notes the…
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Prove that $|f|\leq 1$ whenever $|x|\leq 1$.

Let $f :\mathbb{R}^2\rightarrow \mathbb{R}^2 $ be everywhere differentiable such that the Jacobian is not singular at any point in $\mathbb{R}^2$. Assume $|f|\leq 1$ whenever $|x|=1$. Prove that $|f|\leq 1$ whenever $|x|\leq 1$. I think this is…
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Connectedness of a set defined analytically

How can I prove the following set: $$S=\left\{(x,y,z)\in\mathbb{R^3}:x^2+y^2+z^2+\arctan^2(xyz^2)=7\right\}$$ is connected? I thought that I might somehow use the implicit function theorem to prove that $S$ is the union of graphs of continuous…
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$\frac{\partial F}{\partial y}\neq0\implies$ continuous contour line? (Implicit Function Theorem)

I have a function $F(x,y)=z$ and two points $(x_1,y_1),(x_2,y_2)$ s.t. $F(x_1,y_1)=F(x_2,y_2)=c$, $x_1
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If $f∈C^1$ and $\{∇f=0\}$ has Lebesgue measure $0$, then $\{f∈B\}$ has Lebesgue measure $0$ for all Borel measurable $B⊆ℝ$ with Lebesgue measure $0$

Let $d\in\mathbb N$ and $f\in C^1(\mathbb R^d)$. Assume $\left\{\nabla f=0\right\}$ has Lebesgue measure $0$. How can we conclude that $\left\{f\in B\right\}$ has Lebesgue measure $0$ for all Borel measurable $B\subseteq\mathbb R$ with Lebesgue…
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Why is this a safe neighborhood for uniform convergence of the sequence implicitly defining a function?

This discussion pertains to Theorem III-1.4 in C.H. Edwards, Jr.'s Advanced Calculus of Several Variables. I am trying to understand why the neighborhood described in the proof of the implicit function theorem is "safe" for convergence of the…
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Multivariable calculus - find derivative using implicit differentiation

Short simple question which i managed to solve partially. we are given the equation $x^2+y^2-z^2+xz-yz-1=0$. Show using the implicit function theorem that this equation sets in the neighborhood of $(1,0,1)$ $z$ as a function of $x$ and $y$ and find:…
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