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Aug 16, 2023 at 6:17 comment added pxchg1200 okay, I see. Are there exists some reference that provided the strict proof of conclusion (*)?
Aug 16, 2023 at 4:32 comment added Terry Tao $w - w_\Omega$ only solves the wave equation in the interior and exterior of $\Omega$; on the boundary $\partial \Omega$ there will be a distributional forcing term.
Aug 16, 2023 at 2:03 comment added pxchg1200 @TerryTao However, according to the arguments above, it seems that we can choose $r_0$ as large as possible, such that $\Omega\times [0, m\cdot d(y,\partial\Omega)]\subset K(x_0,r_0)$ as $m\to+\infty$. This seems very strange to me.
Aug 16, 2023 at 2:01 comment added pxchg1200 @TerryTao Thank you for your comments! But I am still confuse about the restriction $t\leq d(y,\partial\Omega)$. Here is my understanding: After extended $w_{\Omega}$ to the entire Euclidean space, for any fixed $y\in \Omega$, $u=w-w_{\Omega}$ satisfies $\partial_{t}^{2}u-\Delta u=0$ in $\mathbb{R}^n\times (0,+\infty)$, and for any $x_0\in \Omega$, we can find $B(x_0,r_0)$ ($r_0$ large enough) such that $\Omega\times [0,d(y,\partial\Omega)] \subset K(x_0,r_0)$. Then applying the finite speed of propagation property to $u$ in the set $B(x_0,r_0)\times\{t=0\}$. Am I misunderstanding something?
Aug 15, 2023 at 14:54 comment added Terry Tao (Strictly speaking, one needs a version of the finite speed of propagation property that also allows for an inhomogeneous term, but this can be deduced from the version you stated via Duhamel's formula.)
Aug 15, 2023 at 14:51 comment added Terry Tao Apply the finite speed of propagation property to $u = w - w_\Omega$ (where $w_\Omega$ is extended arbitrarily to the entire Euclidean space), and note that $\Omega$ contains $B(y,t)$ whenever $y \in \Omega$ and $t \leq d(y,\partial \Omega)$. (If one is worried about the discontinuities of the extension of $w_\Omega$, one can convolve $u$ by some mollifier (approximation to the identity) if desired, though with the theory of distributions this is not really necessary.)
Aug 15, 2023 at 14:20 history asked pxchg1200 CC BY-SA 4.0