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In multivariable calculus, the implicit function theorem^[a] is a tool that allows relations to be converted to functions of several real variables. It does so by representing the relation as the graph of a function. There may not be a single function whose graph can represent 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.

More precisely, given a system of m equations f_i (x₁, ..., x_n, y₁, ..., y_m) = 0, i = 1, ..., m (often abbreviated into F(x, y) = 0), the theorem states that, under a mild condition on the partial derivatives (with respect to each y_i ) at a point, the m variables y_i are differentiable functions of the x_j in some neighborhood of the point. As these functions can generally not be expressed in closed form, they are implicitly defined by the equations, and this motivated the name of the theorem.^[1]

In other words, under a mild condition on the partial derivatives, the set of zeros of a system of equations is locally the graph of a function.

History

Augustin-Louis Cauchy (1789–1857) is credited with the first rigorous form of the implicit function theorem. Ulisse Dini (1845–1918) generalized the real-variable version of the implicit function theorem to the context of functions of any number of real variables.^[2]

First example

If we define the function f(x, y) = x² + y², then the equation f(x, y) = 1 cuts out the unit circle as the level set {(x, y) | f(x, y) = 1}. There is no way to represent the unit circle as the graph of a function of one variable y = g(x) because for each choice of x ∈ (−1, 1), there are two choices of y, namely ${\displaystyle \pm {\sqrt {1-x^{2))))$.

However, it is possible to represent part of the circle as the graph of a function of one variable. If we let ${\displaystyle g_{1}(x)={\sqrt {1-x^{2))))$ for −1 ≤ x ≤ 1, then the graph of y = g₁(x) provides the upper half of the circle. Similarly, if ${\displaystyle g_{2}(x)=-{\sqrt {1-x^{2))))$, then the graph of y = g₂(x) gives the lower half of the circle.

The purpose of the implicit function theorem is to tell us that functions like g₁(x) and g₂(x) almost always exist, even in situations where we cannot write down explicit formulas. It guarantees that g₁(x) and g₂(x) are differentiable, and it even works in situations where we do not have a formula for f(x, y).

Definitions

Let ${\displaystyle f:\mathbb {R} ^{n+m}\to \mathbb {R} ^{m))$ be a continuously differentiable function. We think of ${\displaystyle \mathbb {R} ^{n+m))$ as the Cartesian product ${\displaystyle \mathbb {R} ^{n}\times \mathbb {R} ^{m},}$ and we write a point of this product as ${\displaystyle (\mathbf {x} ,\mathbf {y} )=(x_{1},\ldots ,x_{n},y_{1},\ldots y_{m}).}$ Starting from the given function ${\displaystyle f}$, our goal is to construct a function ${\displaystyle g:\mathbb {R} ^{n}\to \mathbb {R} ^{m))$ whose graph ${\displaystyle ({\textbf {x)),g({\textbf {x))))}$ is precisely the set of all ${\displaystyle ({\textbf {x)),{\textbf {y)))}$ such that ${\displaystyle f({\textbf {x)),{\textbf {y)))={\textbf {0))}$.

As noted above, this may not always be possible. We will therefore fix a point ${\displaystyle ({\textbf {a)),{\textbf {b)))=(a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})}$ which satisfies ${\displaystyle f({\textbf {a)),{\textbf {b)))={\textbf {0))}$, and we will ask for a ${\displaystyle g}$ that works near the point ${\displaystyle ({\textbf {a)),{\textbf {b)))}$. In other words, we want an open set ${\displaystyle U\subset \mathbb {R} ^{n))$ containing ${\displaystyle {\textbf {a))}$, an open set ${\displaystyle V\subset \mathbb {R} ^{m))$ containing ${\displaystyle {\textbf {b))}$, and a function ${\displaystyle g:U\to V}$ such that the graph of ${\displaystyle g}$ satisfies the relation ${\displaystyle f={\textbf {0))}$ on ${\displaystyle U\times V}$, and that no other points within ${\displaystyle U\times V}$ do so. In symbols,

$${\displaystyle \{(\mathbf {x} ,g(\mathbf {x} ))\mid \mathbf {x} \in U\}=\{(\mathbf {x} ,\mathbf {y} )\in U\times V\mid f(\mathbf {x} ,\mathbf {y} )=\mathbf {0} \}.}$$

To state the implicit function theorem, we need the Jacobian matrix of ${\displaystyle f}$, which is the matrix of the partial derivatives of ${\displaystyle f}$. Abbreviating ${\displaystyle (a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})}$ to ${\displaystyle ({\textbf {a)),{\textbf {b)))}$, the Jacobian matrix is

$${\displaystyle (Df)(\mathbf {a} ,\mathbf {b} )=\left[{\begin{array}{ccc|ccc}{\frac {\partial f_{1)){\partial x_{1))}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1)){\partial x_{n))}(\mathbf {a} ,\mathbf {b} )&{\frac {\partial f_{1)){\partial y_{1))}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{1)){\partial y_{m))}(\mathbf {a} ,\mathbf {b} )\\\vdots &\ddots &\vdots &\vdots &\ddots &\vdots \\{\frac {\partial f_{m)){\partial x_{1))}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m)){\partial x_{n))}(\mathbf {a} ,\mathbf {b} )&{\frac {\partial f_{m)){\partial y_{1))}(\mathbf {a} ,\mathbf {b} )&\cdots &{\frac {\partial f_{m)){\partial y_{m))}(\mathbf {a} ,\mathbf {b} )\end{array))\right]=\left[{\begin{array}{c|c}X&Y\end{array))\right]}$$

where ${\displaystyle X}$ is the matrix of partial derivatives in the variables ${\displaystyle x_{i))$ and ${\displaystyle Y}$ is the matrix of partial derivatives in the variables ${\displaystyle y_{j))$. The implicit function theorem says that if ${\displaystyle Y}$ is an invertible matrix, then there are ${\displaystyle U}$, ${\displaystyle V}$, and ${\displaystyle g}$ as desired. Writing all the hypotheses together gives the following statement.

Statement of the theorem

Let ${\displaystyle f:\mathbb {R} ^{n+m}\to \mathbb {R} ^{m))$ be a continuously differentiable function, and let ${\displaystyle \mathbb {R} ^{n+m))$ have coordinates ${\displaystyle ({\textbf {x)),{\textbf {y)))}$. Fix a point ${\displaystyle ({\textbf {a)),{\textbf {b)))=(a_{1},\dots ,a_{n},b_{1},\dots ,b_{m})}$ with ${\displaystyle f({\textbf {a)),{\textbf {b)))=\mathbf {0} }$, where ${\displaystyle \mathbf {0} \in \mathbb {R} ^{m))$ is the zero vector. If the Jacobian matrix (this is the right-hand panel of the Jacobian matrix shown in the previous section):

$${\displaystyle J_{f,\mathbf {y} }(\mathbf {a} ,\mathbf {b} )=\left[{\frac {\partial f_{i)){\partial y_{j))}(\mathbf {a} ,\mathbf {b} )\right]}$$

${\displaystyle U\subset \mathbb {R} ^{n))$

${\displaystyle {\textbf {a))}$

${\displaystyle g:U\to \mathbb {R} ^{m))$

${\displaystyle g(\mathbf {a} )=\mathbf {b} }$,

${\displaystyle f(\mathbf {x} ,g(\mathbf {x} ))=\mathbf {0} ~{\text{for all))~\mathbf {x} \in U}$.

$${\displaystyle J_{f,\mathbf {x} }(\mathbf {a} ,\mathbf {b} )=\left[{\frac {\partial f_{i)){\partial x_{j))}(\mathbf {a} ,\mathbf {b} )\right],}$$

${\displaystyle g}$

${\displaystyle U}$

$${\displaystyle \left[{\frac {\partial g_{i)){\partial x_{j))}(\mathbf {x} )\right]_{m\times n}=-\left[J_{f,\mathbf {y} }(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times m}^{-1}\,\left[J_{f,\mathbf {x} }(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times n))$$

Higher derivatives

If, moreover, ${\displaystyle f}$ is analytic or continuously differentiable ${\displaystyle k}$ times in a neighborhood of ${\displaystyle ({\textbf {a)),{\textbf {b)))}$, then one may choose ${\displaystyle U}$ in order that the same holds true for ${\displaystyle g}$ inside ${\displaystyle U}$. ^[4] In the analytic case, this is called the analytic implicit function theorem.

Proof for 2D case

Suppose ${\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} }$ is a continuously differentiable function defining a curve ${\displaystyle F(\mathbf {r} )=F(x,y)=0}$. Let ${\displaystyle (x_{0},y_{0})}$ be a point on the curve. The statement of the theorem above can be rewritten for this simple case as follows:

Theorem — If

$${\displaystyle \left.{\frac {\partial F}{\partial y))\right|_{(x_{0},y_{0})}\neq 0}$$

then in a neighbourhood of the point ${\displaystyle (x_{0},y_{0})}$ we can write ${\displaystyle y=f(x)}$, where ${\displaystyle f}$ is a real function.

Proof. Since F is differentiable we write the differential of F through partial derivatives:

$${\displaystyle \mathrm {d} F=\operatorname {grad} F\cdot \mathrm {d} \mathbf {r} ={\frac {\partial F}{\partial x))\mathrm {d} x+{\frac {\partial F}{\partial y))\mathrm {d} y.}$$

Since we are restricted to movement on the curve ${\displaystyle F=0}$ and by assumption ${\displaystyle {\tfrac {\partial F}{\partial y))\neq 0}$ around the point ${\displaystyle (x_{0},y_{0})}$ (since ${\displaystyle {\tfrac {\partial F}{\partial y))}$ is continuous at ${\displaystyle (x_{0},y_{0})}$ and ${\displaystyle \left.{\tfrac {\partial F}{\partial y))\right|_{(x_{0},y_{0})}\neq 0}$). Therefore we have a first-order ordinary differential equation:

$${\displaystyle \partial _{x}F\mathrm {d} x+\partial _{y}F\mathrm {d} y=0,\quad y(x_{0})=y_{0))$$

Now we are looking for a solution to this ODE in an open interval around the point ${\displaystyle (x_{0},y_{0})}$ for which, at every point in it, ${\displaystyle \partial _{y}F\neq 0}$. Since F is continuously differentiable and from the assumption we have

$${\displaystyle |\partial _{x}F|<\infty ,|\partial _{y}F|<\infty ,\partial _{y}F\neq 0.}$$

From this we know that ${\displaystyle {\tfrac {\partial _{x}F}{\partial _{y}F))}$ is continuous and bounded on both ends. From here we know that ${\displaystyle -{\tfrac {\partial _{x}F}{\partial _{y}F))}$ is Lipschitz continuous in both x and y. Therefore, by Cauchy-Lipschitz theorem, there exists unique y(x) that is the solution to the given ODE with the initial conditions. Q.E.D.

The circle example

Let us go back to the example of the unit circle. In this case n = m = 1 and ${\displaystyle f(x,y)=x^{2}+y^{2}-1}$. The matrix of partial derivatives is just a 1 × 2 matrix, given by

$${\displaystyle (Df)(a,b)={\begin{bmatrix}{\dfrac {\partial f}{\partial x))(a,b)&{\dfrac {\partial f}{\partial y))(a,b)\end{bmatrix))={\begin{bmatrix}2a&2b\end{bmatrix))}$$

Thus, here, the Y in the statement of the theorem is just the number 2b; the linear map defined by it is invertible if and only if b ≠ 0. By the implicit function theorem we see that we can locally write the circle in the form y = g(x) for all points where y ≠ 0. For (±1, 0) we run into trouble, as noted before. The implicit function theorem may still be applied to these two points, by writing x as a function of y, that is, ${\displaystyle x=h(y)}$; now the graph of the function will be ${\displaystyle \left(h(y),y\right)}$, since where b = 0 we have a = 1, and the conditions to locally express the function in this form are satisfied.

The implicit derivative of y with respect to x, and that of x with respect to y, can be found by totally differentiating the implicit function ${\displaystyle x^{2}+y^{2}-1}$ and equating to 0:

$${\displaystyle 2x\,dx+2y\,dy=0,}$$

$${\displaystyle {\frac {dy}{dx))=-{\frac {x}{y))}$$

$${\displaystyle {\frac {dx}{dy))=-{\frac {y}{x)).}$$

Application: change of coordinates

Suppose we have an m-dimensional space, parametrised by a set of coordinates ${\displaystyle (x_{1},\ldots ,x_{m})}$. We can introduce a new coordinate system ${\displaystyle (x'_{1},\ldots ,x'_{m})}$ by supplying m functions ${\displaystyle h_{1}\ldots h_{m))$ each being continuously differentiable. These functions allow us to calculate the new coordinates ${\displaystyle (x'_{1},\ldots ,x'_{m})}$ of a point, given the point's old coordinates ${\displaystyle (x_{1},\ldots ,x_{m})}$ using ${\displaystyle x'_{1}=h_{1}(x_{1},\ldots ,x_{m}),\ldots ,x'_{m}=h_{m}(x_{1},\ldots ,x_{m})}$. One might want to verify if the opposite is possible: given coordinates ${\displaystyle (x'_{1},\ldots ,x'_{m})}$, can we 'go back' and calculate the same point's original coordinates ${\displaystyle (x_{1},\ldots ,x_{m})}$? The implicit function theorem will provide an answer to this question. The (new and old) coordinates ${\displaystyle (x'_{1},\ldots ,x'_{m},x_{1},\ldots ,x_{m})}$ are related by f = 0, with

$${\displaystyle f(x'_{1},\ldots ,x'_{m},x_{1},\ldots ,x_{m})=(h_{1}(x_{1},\ldots ,x_{m})-x'_{1},\ldots ,h_{m}(x_{1},\ldots ,x_{m})-x'_{m}).}$$

${\displaystyle a=(x'_{1},\ldots ,x'_{m}),b=(x_{1},\ldots ,x_{m})}$

$${\displaystyle (Df)(a,b)=\left[{\begin{matrix}-1&\cdots &0\\\vdots &\ddots &\vdots \\0&\cdots &-1\end{matrix))\left|{\begin{matrix}{\frac {\partial h_{1)){\partial x_{1))}(b)&\cdots &{\frac {\partial h_{1)){\partial x_{m))}(b)\\\vdots &\ddots &\vdots \\{\frac {\partial h_{m)){\partial x_{1))}(b)&\cdots &{\frac {\partial h_{m)){\partial x_{m))}(b)\\\end{matrix))\right.\right]=[-I_{m}|J].}$$

${\displaystyle (x_{1},\ldots ,x_{m})}$

${\displaystyle (x'_{1},\ldots ,x'_{m})}$

Example: polar coordinates

As a simple application of the above, consider the plane, parametrised by polar coordinates (R, θ). We can go to a new coordinate system (cartesian coordinates) by defining functions x(R, θ) = R cos(θ) and y(R, θ) = R sin(θ). This makes it possible given any point (R, θ) to find corresponding Cartesian coordinates (x, y). When can we go back and convert Cartesian into polar coordinates? By the previous example, it is sufficient to have det J ≠ 0, with

$${\displaystyle J={\begin{bmatrix}{\frac {\partial x(R,\theta )}{\partial R))&{\frac {\partial x(R,\theta )}{\partial \theta ))\\{\frac {\partial y(R,\theta )}{\partial R))&{\frac {\partial y(R,\theta )}{\partial \theta ))\\\end{bmatrix))={\begin{bmatrix}\cos \theta &-R\sin \theta \\\sin \theta &R\cos \theta \end{bmatrix)).}$$

Generalizations

Banach space version

Based on the inverse function theorem in Banach spaces, it is possible to extend the implicit function theorem to Banach space valued mappings.^[5][6]

Let X, Y, Z be Banach spaces. Let the mapping f : X × Y → Z be continuously Fréchet differentiable. If ${\displaystyle (x_{0},y_{0})\in X\times Y}$, ${\displaystyle f(x_{0},y_{0})=0}$, and ${\displaystyle y\mapsto Df(x_{0},y_{0})(0,y)}$ is a Banach space isomorphism from Y onto Z, then there exist neighbourhoods U of x₀ and V of y₀ and a Fréchet differentiable function g : U → V such that f(x, g(x)) = 0 and f(x, y) = 0 if and only if y = g(x), for all ${\displaystyle (x,y)\in U\times V}$.

Implicit functions from non-differentiable functions

Various forms of the implicit function theorem exist for the case when the function f is not differentiable. It is standard that local strict monotonicity suffices in one dimension.^[7] The following more general form was proven by Kumagai based on an observation by Jittorntrum.^[8][9]

Consider a continuous function ${\displaystyle f:\mathbb {R} ^{n}\times \mathbb {R} ^{m}\to \mathbb {R} ^{n))$ such that ${\displaystyle f(x_{0},y_{0})=0}$. If there exist open neighbourhoods ${\displaystyle A\subset \mathbb {R} ^{n))$ and ${\displaystyle B\subset \mathbb {R} ^{m))$ of x₀ and y₀, respectively, such that, for all y in B, ${\displaystyle f(\cdot ,y):A\to \mathbb {R} ^{n))$ is locally one-to-one then there exist open neighbourhoods ${\displaystyle A_{0}\subset \mathbb {R} ^{n))$ and ${\displaystyle B_{0}\subset \mathbb {R} ^{m))$ of x₀ and y₀, such that, for all ${\displaystyle y\in B_{0))$, the equation f(x, y) = 0 has a unique solution

$${\displaystyle x=g(y)\in A_{0},}$$

Collapsing manifolds

Perelman’s collapsing theorem for 3-manifolds, the capstone of his proof of Thurston's geometrization conjecture, can be understood as an extension of the implicit function theorem.^[10]