- #1
OliviaB
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I think this is Poisson's Equation (and inhomogenous). I think I need to use Green's Identity.
Let [itex] \mathcal{R} [/itex] be a bounded region in [itex]\mathbb{R}^3[/itex], and suppose [itex] p(x) > 0 [/itex] on [itex] \mathcal{R}[/itex].
(i) If [itex]u[/itex] is a solution of
[itex]\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}[/itex]
show that [itex]u \equiv 0[/itex] on [itex]\mathcal{R}[/itex]
(ii) If [itex]u[/itex] is a solution of
[itex]\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}[/itex]
show that [itex]u[/itex] is unique (It can be assumed that part (i) is true).
I don't know how to start this...
Let [itex] \mathcal{R} [/itex] be a bounded region in [itex]\mathbb{R}^3[/itex], and suppose [itex] p(x) > 0 [/itex] on [itex] \mathcal{R}[/itex].
(i) If [itex]u[/itex] is a solution of
[itex]\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = 0 \ \x \in \partial \mathcal{R}[/itex]
show that [itex]u \equiv 0[/itex] on [itex]\mathcal{R}[/itex]
(ii) If [itex]u[/itex] is a solution of
[itex]\bigtriangledown^2 u = p(x) u \ \ x \in \mathcal{R} \ \ \bigtriangledown \cdot n = g(x) \ \x \in \partial \mathcal{R}[/itex]
show that [itex]u[/itex] is unique (It can be assumed that part (i) is true).
I don't know how to start this...