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,θ), where r>0 and θ∈(−π,π). Then we have
Δpu(x,y)=(u2r+u2θr2)p−42×((p−1)u2rurr+u3rr+2(p−2)uruθurθr2+u2ruθθr2+urru2θr2−(p−3)uru2θr3+(p−1)u2θuθθr4).
In particular, if u=u(x,y)=u(r), then uθ=uθθ=0, and we get the usual
Δpu(x,y)=(p−1)|ur|p−2urr+|ur|p−2urr.To obtain the claimed expression, let us write all necessary partial derivatives in polar coordinates:
ux=urcosθ−uθsinθr, uy=ursinθ+uθcosθr, uxx=urrcos2θ−urθ2cosθsinθr+uθθsin2θr2+ursin2θr+uθ2cosθsinθr2, uyy=urrsin2θ+urθ2cosθsinθr+uθθcos2θr2+urcos2θr−uθ2cosθsinθr2, uxy=urrcosθsinθ+urθcos2θ−sin2θr−uθθcosθsinθr2−urcosθsinθr−uθcos2θ−sin2θr2.Then, we substitute these expressions into the pointwise formula of the p-Laplacian in 2D (see, e.g., here):
Δpu=div(|∇u|p−2∇u)=|∇u|p−4(|∇u|2Δu+(p−2)u2xuxx+(p−2)u2yuyy+2(p−2)uxuyuxy.).After some standard simplifications, we arrive at the desired expression.