1.16 Inverse Function Theorem
Theorem 1.16.1 (Inverse function theorem). Let \(f: \mathbb {R}^n \to \mathbb {R}^n\) be in \(C^1(\Omega )\) on an open set \(\Omega \), and let \(\mathbf x_0\in \Omega \). If the Jacobian \(\det J_f(\mathbf x_0)\) is not zero, then there is an open set \(\Omega _1 \subseteq \Omega \) containing \(\mathbf x_0\) on which \(f\) is locally \(C^1\)-invertible. Moreover, if \(g = f^{-1}\) on \(f(\Omega _1)\), then \[J_g(f(\mathbf x_0)) = \big (J_f(\mathbf x_0)\big )^{-1} .\]
The existence of the inverse follows from the implicit function theorem applied to the \(n\) equations \(f_i(\mathbf x) - y_i = 0\), regarded as equations for \(\mathbf x\) in terms of \(\mathbf y\): the Jacobian with respect to \(\mathbf x\) is \(\det J_f(\mathbf x_0)\neq 0\). The formula for the derivative then follows from the general chain rule: differentiating \(g(f(\mathbf x)) = \mathbf x\) gives \(J_g(f(\mathbf x))\,J_f(\mathbf x) = I\), the identity matrix. Note that the Jacobian of the inverse is evaluated at the image point \(f(\mathbf x_0)\).
Example 1.16.2. Show that the polar transformation \(f(r,\theta ) = (r\cos \theta , r\sin \theta )\) is locally invertible at every point with \(r\neq 0\), and find the Jacobian matrix of the inverse.
Solution. \[J_f(r,\theta ) = \begin {pmatrix} \cos \theta & - r\sin \theta \\ \sin \theta & r\cos \theta \end {pmatrix},\qquad \det J_f = r\cos ^2\theta + r\sin ^2\theta = r .\] So \(f\) is locally invertible wherever \(r\neq 0\), and by Theorem 1.16.1 the inverse has Jacobian matrix \[\big (J_f(r,\theta )\big )^{-1} = \frac {1}{r}\begin {pmatrix} r\cos \theta & r \sin \theta \\ -\sin \theta & \cos \theta \end {pmatrix} = \begin {pmatrix} \cos \theta & \sin \theta \\ -\frac {\sin \theta }{r} & \frac {\cos \theta }{r} \end {pmatrix} ,\] the rows giving \(\left (\frac {\partial r}{\partial x}, \frac {\partial r}{\partial y}\right )\) and \(\left (\frac {\partial \theta }{\partial x}, \frac {\partial \theta }{\partial y}\right )\). As a check, \(r = \sqrt {x^2+y^2}\) gives \(\frac {\partial r}{\partial x} = \frac {x}{r} = \cos \theta \). The theorem gives only local invertibility: \(f\) is not one-to-one on the whole half-plane \(r>0\), since \(\theta \) and \(\theta + 2\pi \) have the same image.
Example 1.16.3. Show that \(f(x,y) = (e^x\cos y,\ e^x\sin y)\) has non-zero Jacobian everywhere but is not one-to-one on \(\mathbb {R}^2\).
Solution. \[\det J_f = \begin {vmatrix} e^x\cos y & -e^x\sin y\\ e^x\sin y & e^x\cos y\end {vmatrix} = e^{2x}>0 .\] So \(f\) is locally invertible near every point. But \(f(x, y + 2\pi ) = f(x,y)\), so \(f\) is not one-to-one on the plane. (In complex notation \(f\) is \(z\mapsto e^z\).) A non-vanishing Jacobian guarantees an inverse near each point, not a global one; in one variable the two coincide, but in two or more they do not.
Example 1.16.4. Let \(u = x + y^3\), \(v = x^3 - y\). Find \(\frac {\partial x}{\partial u}\) and \(\frac {\partial y}{\partial v}\) at the point where \((x,y) = (1,1)\).
Solution. \[J = \begin {pmatrix} u_x & u_y\\ v_x & v_y\end {pmatrix} = \begin {pmatrix} 1 & 3y^2\\ 3x^2 & -1\end {pmatrix} = \begin {pmatrix} 1 & 3\\ 3 & -1\end {pmatrix}\ \text {at}\ (1,1),\] with determinant \(-10\neq 0\). By Theorem 1.16.1, \[\begin {pmatrix} x_u & x_v\\ y_u & y_v\end {pmatrix} = J^{-1} = -\frac {1}{10} \begin {pmatrix} -1 & -3\\ -3 & 1\end {pmatrix} = \begin {pmatrix} \frac {1}{10} & \frac {3}{10}\\ \frac {3}{10} & -\frac {1}{10}\end {pmatrix},\] so \(\frac {\partial x}{\partial u} = \frac {1}{10}\) and \(\frac {\partial y}{\partial v} = -\frac {1}{10}\). Note that \(\frac {\partial x}{\partial u}\neq \frac {1}{\partial u/\partial x} = 1\): the reciprocal rule holds for whole Jacobian matrices, not for individual partial derivatives.
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.