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.[3]

The unit circle can be specified as the level curve f(x, y) = 1 of the function ${\displaystyle f(x,y)=x^{2}+y^{2}.}$ 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 ${\displaystyle f(x,y)=x^{2}+y^{2}}$, 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 ${\displaystyle y=g_{1}(x)}$ provides the upper half of the circle. Similarly, if ${\displaystyle g_{2}(x)=-{\sqrt {1-x^{2}}}}$, then the graph of ${\displaystyle y=g_{2}(x)}$ gives the lower half of the circle.

The purpose of the implicit function theorem is to tell us the existence of functions like ${\displaystyle g_{1}(x)}$ and ${\displaystyle g_{2}(x)}$, even in situations where we cannot write down explicit formulas. It guarantees that ${\displaystyle g_{1}(x)}$ and ${\displaystyle g_{2}(x)}$ 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 f, our goal is to construct a function ${\displaystyle g:\mathbb {R} ^{n}\to \mathbb {R} ^{m}}$ whose graph (x, g(x)) is precisely the set of all (x, y) such that f(x, y) = 0.

As noted above, this may not always be possible. We will therefore fix a point (a, b) = (a₁, ..., a_n, b₁, ..., b_m) which satisfies f(a, b) = 0, and we will ask for a g that works near the point (a, b). In other words, we want an open set ${\displaystyle U\subset \mathbb {R} ^{n}}$ containing a, an open set ${\displaystyle V\subset \mathbb {R} ^{m}}$ containing b, and a function g : U → V such that the graph of g satisfies the relation f = 0 on U × V, and that no other points within U × 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 f, which is the matrix of the partial derivatives of f. Abbreviating (a₁, ..., a_n, b₁, ..., b_m) to (a, b), the Jacobian matrix is

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

where X is the matrix of partial derivatives in the variables x_i and Y is the matrix of partial derivatives in the variables y_j. The implicit function theorem says that if Y is an invertible matrix, then there are U, V, and 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 (x, y). Fix a point (a, b) = (a₁, ..., a_n, b₁, ..., b_m) with f(a, b) = 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 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 ${\displaystyle f(\mathbf {x} ,g(\mathbf {x} ))=\mathbf {0} \quad {\text{for all}}\quad \mathbf {x} \in U}$ .

Moreover, the partial derivatives of g in U are given by the matrix product:[4] ${\displaystyle \quad {\frac {\partial g}{\partial x_{j}}}(\mathbf {x} )=-\left[J_{f,\mathbf {y} }(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times m}^{-1}\left[{\frac {\partial f}{\partial x_{j}}}(\mathbf {x} ,g(\mathbf {x} ))\right]_{m\times 1}}$

Suppose ${\displaystyle F:\mathbb {R} ^{2}\to \mathbb {R} }$ is a continuously differentiable function defining a curve ${\displaystyle F(x,y)=0.}$ 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:

If

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

then for the curve around ${\displaystyle (x_{0},y_{0})}$ 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:

${\displaystyle \mathrm {d} F=\operatorname {grad} F\cdot \mathrm {d} {\vec {x}}={\partial F \over \partial x}\mathrm {d} x+{\partial F \over \partial y}\mathrm {d} y.}$

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 ${\displaystyle (x_{0},y_{0}).}$ 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. ∎

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

${\displaystyle (Df)(a,b)=\left[{\frac {\partial f}{\partial x}}(a,b)\ \ {\frac {\partial f}{\partial y}}(a,b)\right]=[2a\ \ 2b]}$

Thus, here, the Y in the statement of the theorem is just the number 2b; the linear map defined by it is invertible iff 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.

• 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.

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• 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.