Let $U\in C^2(\mathbb{R}^n\times (0,\infty))$, let $\phi\in C^\infty_c(\mathbb{R}^n)$ and suppose that there is a sufficiently smooth function $\phi_\ast$ such that: $\phi_\ast(x,0)=\phi(x)$,$\forall x\in\mathbb{R}^n$, let $a\in(-1,1)$, i have to prove that, with $X=(x,y)$: $$ \int_{\mathbb{R}^n\times(0,\infty)}y^a\nabla_XU\cdot\nabla_X\phi_\ast\,dX=\int_{\mathbb{R}^n\times(0,\infty)}\text{div}(y^a\phi_\ast\nabla_XU)\,dX,\tag{1}$$ i don't have any idea on how to prove (1). Then, applying the Divergence Theorem i have to prove that, with the hypothesis $\text{div}(y^a\nabla_XU)=0:$ $$ \int_{\mathbb{R}^n\times(0,\infty)}\text{div}(y^a\phi_\ast\nabla_XU)\,dX=-\int_{\mathbb{R}^n\times\{0\}}\phi y^a\partial_yU\,dx, \tag{2}$$ i don't understand how use the Divergence Theorem for obtain (2). Any help would be appreciated.
2026-02-23 16:50:05.1771865405
An application of Divergence Theorem
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