OliviaB
Oct18-10, 09:32 AM
I think this is Poisson's Equation (and inhomogenous). I think I need to use Green's Identity.
Let \mathcal{R} be a bounded region in \mathbb{R}^3, and suppose p(x) > 0 on \mathcal{R}.
(i) If u is a solution of
\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}
show that u \equiv 0 on \mathcal{R}
(ii) If u is a solution of
\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}
show that u is unique (It can be assumed that part (i) is true).
I dont know how to start this...
Let \mathcal{R} be a bounded region in \mathbb{R}^3, and suppose p(x) > 0 on \mathcal{R}.
(i) If u is a solution of
\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}
show that u \equiv 0 on \mathcal{R}
(ii) If u is a solution of
\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}
show that u is unique (It can be assumed that part (i) is true).
I dont know how to start this...