Several times I found myself looking for an explicit expression of the $p$-Laplacian in polar coordinates. Usual Laplace operator considered in polar coordinates can be very useful if one works with radial domains. So, in some problems it can be quite natural to be interested in the corresponding expression for the $p$-Laplacian. However, such an expression appears to be quite bulky, which makes it complicated to apply. Nevertheless, to find easily this expression in future, I decided to post it here.

Let $u = u(x,y) = u(r,\theta)$, where $r>0$ and $\theta \in (-\pi, \pi)$. Then we have

\begin{align} \notag \Delta_p u(x,y) &= \left(u_r^2 + \frac{u_\theta^2}{r^2} \right)^\frac{p-4}{2} \newline \notag &\times \left( (p-1) u_r^2 u_{rr} + \frac{u_r^3}{r} + \frac{2(p-2) u_r u_\theta u_{r\theta}}{r^2} + \frac{u_r^2 u_{\theta \theta}}{r^2} + \frac{u_{rr} u_\theta^2}{r^2} - \frac{(p-3) u_r u_\theta^2}{r^3} + \frac{(p-1) u_\theta^2 u_{\theta \theta}}{r^4} \right). \end{align}

In particular, if $u = u(x,y) = u(r)$, then $u_\theta = u_{\theta \theta}=0$, and we get the usual

$\Delta_p u(x,y) = (p-1) |u_r|^{p-2} u_{rr} + \frac{|u_r|^{p-2} u_r}{r}.$

To obtain the claimed expression, let us write all necessary partial derivatives in polar coordinates:

$u_{x} = u_r \cos \theta - u_\theta \frac{\sin \theta}{r},$ $u_y = u_r \sin \theta + u_\theta \frac{\cos \theta}{r},$ $u_{xx} = u_{rr} \cos^2 \theta - u_{r\theta} \frac{2 \cos \theta \sin \theta}{r} + u_{\theta \theta} \frac{\sin^2 \theta}{r^2} + u_r \frac{\sin^2 \theta}{r} + u_{\theta} \frac{2 \cos \theta \sin \theta}{r^2},$ $u_{yy} = u_{rr} \sin^2 \theta + u_{r\theta} \frac{2 \cos \theta \sin \theta}{r} + u_{\theta \theta} \frac{\cos^2 \theta}{r^2} + u_r \frac{\cos^2 \theta}{r} - u_{\theta} \frac{2 \cos \theta \sin \theta}{r^2},$ $u_{xy} = u_{rr} \cos \theta \sin \theta + u_{r\theta} \frac{\cos^2 \theta - \sin^2 \theta}{r} - u_{\theta \theta} \frac{\cos \theta \sin \theta}{r^2} - u_r \frac{\cos \theta \sin \theta}{r} - u_{\theta} \frac{\cos^2 \theta - \sin^2 \theta}{r^2}.$

Then, we substitute these expressions into the pointwise formula of the $p$-Laplacian in $2D$ (see, e.g., here):

$\Delta_p u = \text{div}\left(|\nabla u|^{p-2} \nabla u\right) = |\nabla u|^{p-4} \left( |\nabla u|^2 \Delta u + (p-2) u_x^2 u_{xx} + (p-2) u_y^2 u_{yy} + 2(p-2) u_{x}u_y u_{xy}. \right).$

After some standard simplifications, we arrive at the desired expression.