Implicit function theorem

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]

The unit circle can be specified as the level curve f(x, y) = 1 of the function f(x, y) = x² + y². Around point A, y can be expressed as a function y(x). In this example this function can be written explicitly as ${\displaystyle g_{1}(x)={\sqrt {1-x^{2}}};}$ in many cases no such explicit expression exists, but one can still refer to the implicit function y(x). No such function exists around point B.

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).

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.

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):

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

is invertible, then there exists an open set ${\displaystyle U\subset \mathbb {R} ^{n}}$ containing ${\displaystyle {\textbf {a}}}$ such that there exists a unique continuously differentiable function ${\displaystyle g:U\to \mathbb {R} ^{m}}$ such that ${\displaystyle g(\mathbf {a} )=\mathbf {b} }$, and ${\displaystyle f(\mathbf {x} ,g(\mathbf {x} ))=\mathbf {0} ~{\text{for all}}~\mathbf {x} \in U}$. Moreover, denoting the left-hand panel of the Jacobian matrix shown in the previous section as:

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

the Jacobian matrix of partial derivatives of ${\displaystyle g}$ in ${\displaystyle U}$ is given by the matrix product:[3]

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

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 ${\displaystyle \mathrm {d} F=0}$ 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.

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:

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

giving

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

and

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

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

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

Now the Jacobian matrix of f at a certain point (a, b) [ where ${\displaystyle a=(x'_{1},\ldots ,x'_{m}),b=(x_{1},\ldots ,x_{m})}$ ] is given by

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

where I_m denotes the m × m identity matrix, and J is the m × m matrix of partial derivatives, evaluated at (a, b). (In the above, these blocks were denoted by X and Y. As it happens, in this particular application of the theorem, neither matrix depends on a.) The implicit function theorem now states that we can locally express ${\displaystyle (x_{1},\ldots ,x_{m})}$ as a function of ${\displaystyle (x'_{1},\ldots ,x'_{m})}$ if J is invertible. Demanding J is invertible is equivalent to det J ≠ 0, thus we see that we can go back from the primed to the unprimed coordinates if the determinant of the Jacobian J is non-zero. This statement is also known as the inverse function theorem.

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

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

Since det J = R, conversion back to polar coordinates is possible if R ≠ 0. So it remains to check the case R = 0. It is easy to see that in case R = 0, our coordinate transformation is not invertible: at the origin, the value of θ is not well-defined.

• Inverse function theorem
• Constant rank theorem: Both the implicit function theorem and the inverse function theorem can be seen as special cases of the constant rank theorem.

• Allendoerfer, Carl B. (1974). "Theorems about Differentiable Functions". Calculus of Several Variables and Differentiable Manifolds. New York: Macmillan. pp. 54–88. ISBN 0-02-301840-2.
• Binmore, K. G. (1983). "Implicit Functions". Calculus. New York: Cambridge University Press. pp. 198–211. ISBN 0-521-28952-1.
• Loomis, Lynn H.; Sternberg, Shlomo (1990). Advanced Calculus (Revised ed.). Boston: Jones and Bartlett. pp. 164–171. ISBN 0-86720-122-3.
• Protter, Murray H.; Morrey, Charles B. Jr. (1985). "Implicit Function Theorems. Jacobians". Intermediate Calculus (2nd ed.). New York: Springer. pp. 390–420. ISBN 0-387-96058-9.