Implicit and Inverse Function Theorems, multivariable and vector calculus lesson

The unit circle is not the graph of one function, yet near the point $(0.6, 0.8)$ a small arc of it is the graph of $y = \sqrt{1 - x^2}$ . Near $(1, 0)$ no such description works, because the circle turns vertical there. The implicit function theorem gives a simple test that guarantees the first kind of behavior, using one partial derivative at that point. The inverse function theorem answers the same kind of question for maps that send points to points.

Key terms

C1 regularity on a neighborhood

Definition Let $W$ be an open subset of $\mathbb{R}^k$ and $F \colon W \to \mathbb{R}^m$ . The map $F$ is of class $C^1$ on $W$ when every first partial derivative of every component of $F$ exists at every point of $W$ and is continuous on $W$ . A map is $C^1$ near a point $p$ when it is $C^1$ on some open set containing $p$ . Such a map is differentiable at every point of $W$ , and its derivative is the Jacobian matrix $DF$ .

In plain words The function has partial derivatives everywhere in a small open region around the point, and they change continuously. Both theorems in this lesson need this on a whole neighborhood, not only at one point.

Example Every polynomial, such as $F(x, y) = x^2 + y^2 - 1$ , is $C^1$ on all of $\mathbb{R}^2$ . The function $H(x, y) = x + \lvert y \rvert$ is continuous everywhere, but $\partial H / \partial y$ does not exist at any point with $y = 0$ , so $H$ is not $C^1$ on any open set that meets the $x$ -axis.

Implicit equations

Definition An implicit equation is an equation $F(x, y) = 0$ , where $F$ is defined on an open set of $\mathbb{R}^{n + m}$ with values in $\mathbb{R}^m$ , $x = (x_1, \dots, x_n)$ and $y = (y_1, \dots, y_m)$ . It is a system of $m$ scalar equations in $n + m$ unknowns. Its solution set is the set of all points $(x, y)$ with $F(x, y) = 0$ , and the question is whether the $m$ variables $y$ can be regarded as functions of the $n$ variables $x$ on that set.

In plain words You have as many equations as unknowns you want to solve for, plus some leftover variables that are free to move. The equations tie the unknowns to the free variables without handing you a formula.

Example The single equation $xz + y^2 + z^3 = 3$ has $n = 2$ free variables $x, y$ and $m = 1$ unknown $z$ . The point $(1, 1, 1)$ lies on it, since $1 + 1 + 1 = 3$ . The pair of equations

\begin{gathered} xu + yv = 2, \\ u^2 - v + x - y = 0 \end{gathered}

has $n = 2$ free variables $x, y$ and $m = 2$ unknowns $u, v$ , and $(x, y, u, v) = (1, 1, 1, 1)$ satisfies both.

Local solvability

Definition Let $F(a, b) = 0$ . The equation $F(x, y) = 0$ is locally solvable for $y$ near $(a, b)$ when there are open sets $U \ni a$ in $\mathbb{R}^n$ and $V \ni b$ in $\mathbb{R}^m$ and a function $g \colon U \to V$ with $g(a) = b$ such that, for $(x, y)$ in $U \times V$ ,

F(x, y) = 0 \iff y = g(x).

The function $g$ is called a local branch of the solution set through $(a, b)$ .

In plain words Inside a small box around the point, the solution set is exactly the graph of one function. Points of the solution set outside the box do not matter.

Example For $x^2 + y^2 - 1 = 0$ near $(0.6, 0.8)$ , take $U = (0, 1)$ , $V = (0, 2)$ and $g(x) = \sqrt{1 - x^2}$ . Near $(1, 0)$ there is no local branch $y = g(x)$ , because for $x$ slightly less than $1$ every box around the point contains the two solutions $y = \pm\sqrt{1 - x^2}$ , and for $x$ slightly more than $1$ it contains none.

The implicit function theorem

Definition Let $W \subseteq \mathbb{R}^{n + m}$ be open, let $F \colon W \to \mathbb{R}^m$ be $C^1$ on $W$ , and let $(a, b)$ be a point of $W$ with $F(a, b) = 0$ . Write $D_y F$ for the $m \times m$ matrix of partial derivatives of the components of $F$ with respect to $y_1, \dots, y_m$ . If

\det D_y F(a, b) \ne 0,

then there are open sets $U \ni a$ and $V \ni b$ with $U \times V \subseteq W$ and a unique function $g \colon U \to V$ with $g(a) = b$ such that $F(x, y) = 0$ with $(x, y)$ in $U \times V$ holds exactly when $y = g(x)$ . Moreover $g$ is $C^1$ on $U$ , and $U$ can be chosen so small that $D_y F(x, g(x))$ is invertible for every $x$ in $U$ . For one equation ( $m = 1$ ) the condition is simply that the relevant partial derivative is nonzero, for example $F_z(a, b) \ne 0$ when solving for $z$ .

In plain words Three things are needed. The function is $C^1$ near the point, the point solves the equation, and the derivative with respect to the unknowns you want is invertible there. Then the solution set near the point is the graph of a smooth function, even if you can never write it down.

Example Take $F(x, y, z) = xz + y^2 + z^3 - 3$ at $(1, 1, 1)$ . It is a polynomial, so it is $C^1$ on $\mathbb{R}^3$ , and $F(1, 1, 1) = 0$ . Since

\begin{gathered} F_z = x + 3z^2, \\ F_z(1, 1, 1) = 4 \ne 0, \end{gathered}

the theorem gives a $C^1$ function $z = g(x, y)$ near $(1, 1)$ with $g(1, 1) = 1$ . For the system above in the unknowns $u, v$ , the relevant Jacobian at $(1, 1, 1, 1)$ is

D_{(u,v)} F = \begin{bmatrix} x & y \\ 2u & -1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & -1 \end{bmatrix},

with determinant $-3 \ne 0$ . So $u$ and $v$ are $C^1$ functions of $(x, y)$ near $(1, 1)$ .

Derivatives of local branches

Definition Under the hypotheses of the implicit function theorem, differentiating the identity $F(x, g(x)) = 0$ with the chain rule gives $D_x F + D_y F \, Dg = 0$ . Since $D_y F$ is invertible along the branch, for every $x$ in $U$

Dg(x) = -\big[ D_y F \big]^{-1} D_x F,

with both matrices evaluated at the point $(x, g(x))$ . Here $Dg$ is the $m \times n$ Jacobian of $g$ and $D_x F$ is the $m \times n$ matrix of partial derivatives of $F$ in $x_1, \dots, x_n$ .

For one equation $F(x, y, z) = 0$ solved for $z$ this reads

\begin{gathered} \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z}, \\ F_z \ne 0. \end{gathered}

In plain words You get the derivatives of the hidden function without solving for it. Differentiate the equations, then solve a small linear system for the rates you want.

Example On $xz + y^2 + z^3 = 3$ near $(1, 1, 1)$ we have $F_x = z$ , $F_y = 2y$ and $F_z = x + 3z^2$ , so at that point

\frac{\partial z}{\partial x} = -\frac{1}{4}, \qquad \frac{\partial z}{\partial y} = -\frac{2}{4} = -\frac{1}{2}.

For the system, differentiate both equations with respect to $x$ , holding $y$ fixed. At $(1, 1, 1, 1)$ this gives

\begin{gathered} 1 + u_x + v_x = 0, \\ 2u_x - v_x + 1 = 0. \end{gathered}

Adding the two equations gives $3u_x + 2 = 0$ , so $u_x = -\frac{2}{3}$ and then $v_x = -\frac{1}{3}$ . The matrix formula gives the same values together with $u_y = 0$ and $v_y = -1$ .

The inverse function theorem

Definition Let $U \subseteq \mathbb{R}^n$ be open, let $f \colon U \to \mathbb{R}^n$ be $C^1$ on $U$ , and let $a$ be a point of $U$ with

\det Df(a) \ne 0.

Then there are open sets $V \ni a$ and $W \ni f(a)$ with $V \subseteq U$ such that $f$ maps $V$ one-to-one onto $W$ , the inverse $f^{-1} \colon W \to V$ is $C^1$ on $W$ , and

D(f^{-1})(f(a)) = \big[ Df(a) \big]^{-1}.

The same formula holds at every point of $W$ , with $a$ replaced by the point of $V$ that $f$ sends there.

In plain words If the derivative at a point is an invertible matrix, then near that point the map itself can be undone smoothly, and the derivative of the undoing map is the inverse matrix.

Example Let $f(u, v) = \langle u^2 - v^2, \, 2uv \rangle$ , which is $C^1$ on $\mathbb{R}^2$ . Its Jacobian has determinant

\det \begin{bmatrix} 2u & -2v \\ 2v & 2u \end{bmatrix} = 4(u^2 + v^2),

which is nonzero away from the origin. At $(2, 1)$ we get $f(2, 1) = (3, 4)$ and

\begin{aligned} D(f^{-1})(3, 4) &= \begin{bmatrix} 4 & -2 \\ 2 & 4 \end{bmatrix}^{-1} \\[0.4em] &= \frac{1}{20} \begin{bmatrix} 4 & 2 \\ -2 & 4 \end{bmatrix}. \end{aligned}

So if $x = u^2 - v^2$ and $y = 2uv$ , the local inverse near $(3, 4)$ has $\partial u / \partial x = \frac{1}{5}$ and $\partial v / \partial x = -\frac{1}{10}$ there.

Local versus global inverses

Definition A map $f \colon U \to \mathbb{R}^n$ has a global inverse on $U$ when it is one-to-one on all of $U$ . The inverse function theorem gives only local inverses. A $C^1$ map with $\det Df \ne 0$ at every point of an open set need not be one-to-one on that set. In the other direction, $\det Df(a) = 0$ does not rule out an inverse, but if an inverse exists it cannot be differentiable at $f(a)$ .

In plain words An invertible derivative guarantees that the map can be undone near each point separately. Gluing those local inverses into one inverse on the whole domain can fail, because the map may wrap around and hit the same point twice.

Example The polar map $f(r, \theta) = \langle r \cos \theta, \, r \sin \theta \rangle$ on the open set $r > 0$ has $\det Df = r \ne 0$ everywhere, yet $f(1, 0) = f(1, 2\pi) = (1, 0)$ , so it has no global inverse. The squaring map above behaves the same way, since $f(-u, -v) = f(u, v)$ . In one variable, $f(x) = x^3$ is one-to-one with $f'(0) = 0$ , and its inverse $\sqrt[3]{x}$ is not differentiable at $0$ .

Tangent planes to level surfaces

Definition Let $F$ be $C^1$ on an open set of $\mathbb{R}^3$ , let $F(p) = c$ and suppose $\nabla F(p) \ne 0$ . Then some partial derivative of $F$ is nonzero at $p$ , so by the implicit function theorem the level set $F = c$ is, near $p$ , the graph of a $C^1$ function of two of the variables. Its tangent plane at $p = (x_0, y_0, z_0)$ is

\begin{aligned} F_x(p)(x - x_0) &+ F_y(p)(y - y_0) \\ &+ F_z(p)(z - z_0) = 0, \end{aligned}

and $\nabla F(p)$ is a normal vector. If $\nabla F(p) = 0$ the theorem says nothing, and the level set may have no tangent plane there.

In plain words A nonzero gradient both proves that the level set is a smooth surface near the point and hands you its normal vector.

Example For $F(x, y, z) = xz + y^2 + z^3$ at $(1, 1, 1)$ , the gradient is $\langle z, 2y, x + 3z^2 \rangle = \langle 1, 2, 4 \rangle$ . The tangent plane is

(x - 1) + 2(y - 1) + 4(z - 1) = 0,

which simplifies to $x + 2y + 4z = 7$ . Solving for $z$ gives slopes $-\frac{1}{4}$ and $-\frac{1}{2}$ , matching the branch derivatives found earlier. By contrast the cone $x^2 + y^2 - z^2 = 0$ has zero gradient at the origin, and it has no tangent plane at its tip.

Check yourself

Show that $x e^y + y = 1$ defines $y$ as a $C^1$ function of $x$ near $(1, 0)$ , and find $\frac{dy}{dx}$ there.

Show the answer

The function $F(x, y) = x e^y + y - 1$ is $C^1$ on $\mathbb{R}^2$ and $F(1, 0) = 0$ . Its partial $F_y = x e^y + 1$ equals $2 \ne 0$ at $(1, 0)$ , so the theorem applies. Then

\frac{dy}{dx} = -\frac{F_x}{F_y} = -\frac{e^y}{x e^y + 1} = -\frac{1}{2}.

For $f(u, v) = \langle u^2 - v^2, \, 2uv \rangle$ near $(u, v) = (2, 1)$ , find $\partial u / \partial y$ and $\partial v / \partial y$ of the local inverse at $(x, y) = (3, 4)$ . Why is there no local inverse near $(0, 0)$ ?

Show the answer

The second column of the inverse matrix found above gives $\partial u / \partial y = \frac{2}{20} = \frac{1}{10}$ and $\partial v / \partial y = \frac{4}{20} = \frac{1}{5}$ . At the origin the Jacobian determinant $4(u^2 + v^2)$ is $0$ , so the theorem does not apply. In fact no local inverse exists, because every disk around $(0, 0)$ contains a point $(u, v) \ne (0, 0)$ together with $(-u, -v)$ , and these two different points have the same image.

Find the tangent plane to $x^2 + 2y^2 + 3z^2 = 6$ at $(1, 1, 1)$ .

Show the answer

The gradient of $F = x^2 + 2y^2 + 3z^2$ is $\langle 2x, 4y, 6z \rangle = \langle 2, 4, 6 \rangle \ne 0$ . The tangent plane is $2(x - 1) + 4(y - 1) + 6(z - 1) = 0$ , which simplifies to $x + 2y + 3z = 6$ .

Implicit and Inverse Function Theorems

Key terms

C1 regularity on a neighborhood

Implicit equations

Local solvability

The implicit function theorem

Derivatives of local branches

The inverse function theorem

Local versus global inverses

Tangent planes to level surfaces

Check yourself

How this lesson connects