Let $p \geq 2$ and $B(x_0, d)$ an open ball in $R^n$ ($n \geq 2$). Let $u \in W^{1,p}(B(x_0,d))$ a weak solution of $\Delta_p u = f$ , $u = 0 , \ on \ \partial B(x_0,d)$, that is, $\int_{B(x_0,d)} \nabla u(x). \nabla \varphi (x) \ dx = \int_{B(x_0,d)} f(x) \varphi (x) \ dx $ for all $\varphi \in C^{\infty}_{0}(B(x_0,d))$ where f is a bounded function in $B(x_0,d)$ .
Define $h(x) = \displaystyle\frac{u(x_0 + d x)}{d}, x \in B(0,1)$. My question is : which problem $h$ is weak solution?
Note that if $p=2$ and $u$ is smooth we have $\Delta h (x) = d .f (x_0 + dx)$.
I believe that in the context of my question that $\Delta_p u(x_0 + d. (.)) = d.f(x_0 + d.(.))$ . But I dont know how to prove this....
Someone can give me a help with the above questions ?
thanks in advance
Formally (and by this I mean: applied in smooth functions) the $p$-Laplace operator is defined by $$\Delta_pu=\operatorname{div}(|\nabla u|^{p-2}\nabla u).$$
If $h(x)=\frac{u(x_0+dx)}{d}$, we see from the previous equality that $$\Delta_p h(x)=\frac{\operatorname{div}(|\nabla u(x_0+dx)|^{p-2}\nabla u(x_0+dx)}{d^{p-1}},$$
so if $\Delta_p u=f$, it is expected that $\Delta_p h(x)=d^{p-1} f(x_0+dx)$. You can easily check that the last equality is true, even in the distributional case, by applying the change of variables $x_0+dx=y$ in the integral $$\frac{1}{d^{p-1}}\int_{B(0,1)}|\nabla u(x_0+dx)|^{p-2}\nabla u(x_0+dx)\nabla \varphi(x)dx$$