Sliping the laplacian in the sense of distributions

Question

Let $B(r)$ designate the open ball in $\mathbb{R}^{n}$, centered at the origin, $f$ a $C^{\infty}$ and $\phi$ a test function with compact support in $B(R)$. If I take the integral of $f\Delta \phi$ on $B(R)\setminus B(r)$, ($0<r<R$), can I steel slip the laplacina over $f$, i.e., $$\int_{B(R)\setminus B(r)}f(x)\Delta\phi(x)dx=\int_{B(R)\setminus B(r)}\Delta f(x)\phi(x)dx?$$ It seems to me that the answer should be no, because the derivatives of these functions on the boundary of $B(r)$ are not zero and so Green's identity (theorem) doesn't work as we wished. But is there a trick somewhere to make it work?? I badly need it!!!

Answer 1

HINT:

$$\begin{align} f\nabla^2\phi&=\nabla \cdot (f \nabla\phi)-\nabla f \cdot \nabla \phi\\\\ &=\nabla\cdot(f \nabla\phi-\phi\nabla f)+\phi\nabla^2 f \end{align}$$

The integral of the divergence term leads to a boundary integral term over $B(r)$ and $B(R)$.

Finally, let $r\to 0$ and $R\to \infty$.