Disk-median curvature expansion (${\mathrm{median}}_{B_h(x)}u=u+\frac{1}{6}{h}^2|\nabla u|\mathrm{div}(\nabla u/|\nabla u|)+o(h^2)$)

At a smooth point with nonzero gradient, rotate coordinates so $\nabla u=G{e}_2$. Write $u=u_0+Gy+\frac{1}{2}(A{x}^2+2Bxy+C{y}^2)+O(|(x,y){|}^3)$. A candidate median $u_0+d{h}^2$ has level boundary $y=(d{h}^2-A{x}^2/2)/G+O(h^3)$ in the disk. The area imbalance is $(2d-A/3)h^3/G+o(h^3)$; equal half-areas force $d=A/6$. Here $A=\Delta u-\nabla u^T(D^{2}u)\nabla u/|\nabla u{|}^2$ is the tangential second derivative, giving the formula. The disk uses uniform area measure, not its boundary measure.