Suppose $g:[0,\infty)\times\mathbb{R}\to\mathbb{R}^2,\;(r,\varphi)\mapsto(r\cos\varphi,r\sin\varphi)$. $X$ is open in $\mathbb{R}^2\backslash H$ where $H=\{(x,y)\in\mathbb{R}^2\;\big|\;x\leq0,\;y=0\}$. Prove \begin{equation*} (\Delta f)\circ g=\frac{\partial^2(f\circ g)}{\partial r^2}+\frac{1}{r}\frac{\partial f\circ g}{\partial r}+\frac{\partial^2 (f\circ g)}{\partial \varphi^2} \end{equation*} on $g^{-1}(X)$, where $\Delta$ is the Laplacian operator.
It suffices to prove the equality holds at each given point $(r,\varphi)$, but I don't know how to expand explicitly $(\triangle f)\circ g(r,\varphi)$.
$\Delta f=\partial_1^2+\partial_2^2 f$.
$\newcommand{\fa}{\forall}$ $\newcommand{\vp}{\varphi}$ $\newcommand{\ptl}{\partial}$ $\newcommand{\fptl}{\frac{\ptl}{\ptl}}$ $\newcommand{\lk}{\left[}$ $\newcommand{\rk}{\right]}$ $\newcommand{\Da}{\Delta}$
$\fa x=(r,\vp)\in g^{-1}(X)$, we have \begin{eqnarray*} \frac{\ptl^2(f\circ g)}{\ptl r^2}(r,\vp) &=&\frac{\ptl}{\ptl r}\lk\fptl{(f\circ g)}{r}(r,\vp)\rk=\fptl{}{r}\lk\ptl_1f\circ g(x)\cos\vp+\ptl_2 f\circ g(x)\sin\vp\rk \\ &=&\ptl_1^2f\circ g(x)\cos^2\vp+2\ptl_{12}^2f\circ g(x)\sin\vp\cos\vp+\ptl_2^2f\circ g(x)\sin^2\vp,\\ \frac{1}{r}\frac{\ptl(f\circ g)}{\ptl r}(r,\vp) &=&\ptl_1f\circ g(x)\frac{\cos\vp}{r}+\ptl_2 f\circ g(x)\frac{\sin\vp}{r}, \\ \frac{1}{r^2}\frac{\ptl^2(f\circ g)}{\ptl \vp^2}(r,\vp) &=&\ptl_1^2f\circ g(x)\sin^2\vp+\ptl_2^2f\circ g(x)\cos^2\vp-\ptl_2 f\circ g(x)\frac{\sin\vp}{r}-\ptl_1f\circ g(x)\frac{\cos\vp}{r}\\ &&-2\ptl_{12}^2 f\circ g(x)\sin\vp\cos\vp. \end{eqnarray*} So $\frac{\ptl^2(f\circ g)}{\ptl r^2}(r,\vp)+\frac{1}{r}\frac{\ptl(f\circ g)}{\ptl r}(r,\vp)+\frac{1}{r^2}\frac{\ptl^2(f\circ g)}{\ptl \vp^2}(r,\vp)=\ptl_1^2f\circ g(r,\vp)+\ptl_2^2f\circ g(r,\vp)=\Da f\circ g(r,\vp)$.