1.16 Inverse Function Theorem
Let \(f: \mathbb {R}^n \longrightarrow \mathbb {R}^n\) be in \(C^1\big (\Omega \big )\) on an open set \(\Omega \) and let \(x_0\) be in \(\Omega \) if the Jacobian determinant \(\big (\det J_{f(x_0)} \neq 0 \big )\) is not zero, then there is an open set \(\Omega _1 \subseteq \Omega \) such that \(x_0 \in \Omega _1\) and \(f\) is locally \(C^1-\) invertible on \(\Omega _1\) moreover if \(g = f^{-1}\) on \(f\big (\Omega \big )\) then \[J_{f^{-1}(x_0)} = J_{g(x_0)} = \Big (J_{f(x_0)}\Big )^{-1}\]
Let \(f: \mathbb {R}^n\) to \(\mathbb {R}^n\) be the polar transformation \(\hspace {0.3cm} x = r\cos \theta \hspace {0.2cm},\hspace {0.2cm} y = r\sin \theta \)
Solution. \[f_1 (r,\theta ) = r \cos \theta \hspace {1cm} f_2(r,\theta ) = r \sin \theta \]
\begin {align*} \frac {\partial (f_1,f_2)}{\partial (r,\theta )} & = \begin {vmatrix} \dfrac {\partial f_1}{\partial r} & \dfrac {\partial f_1}{\partial \theta }\\\\ \dfrac {\partial f_2}{\partial r} & \dfrac {\partial f_2}{\partial \theta }\\ \end {vmatrix}\\\\ & = \begin {vmatrix} \cos \theta & - r\sin \theta \\ \sin \theta & r\cos \theta \\ \end {vmatrix}\\ & = r\\ \end {align*}
\[ J_{f(r,\theta )} = \begin {pmatrix} \cos \theta & - r\sin \theta \\ \sin \theta & r\cos \theta \\ \end {pmatrix}\]
\begin {align*} \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 \\ \dfrac {-\sin \theta }{r} & \dfrac {\cos \theta }{r}\\ \end {pmatrix}\\\\ \end {align*}
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