How Does Green's Identity Prove Solutions in Poisson's Equation?

[OliviaB]

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 don't know how to start this...

 

[gtoguy97]

Solution: (i) To show that u ≡ 0 on \mathcal{R}, we can use Green's Identity. Green's identity states that for any two functions u and v, \int_\mathcal{R} \bigtriangledown u \cdot \bigtriangledown v \, dx = \int_\mathcal{R} u \bigtriangledown^2 v \, dx - \oint_{\partial R} v \bigtriangledown u \cdot n \, ds Letting u = 1 and v = u, we have \int_\mathcal{R} \bigtriangledown u \cdot \bigtriangledown u \, dx = \int_\mathcal{R} u \bigtriangledown^2 u \, dx - \oint_{\partial R} u \bigtriangledown u \cdot n \, ds Substituting in the equation \bigtriangledown^2 u = p(x) u we have \int_\mathcal{R} \bigtriangledown u \cdot \bigtriangledown u \, dx = \int_\mathcal{R} u^2 p(x) \, dx - \oint_{\partial R} u \bigtriangledown u \cdot n \, ds Since \bigtriangledown \cdot n = 0, the second term is zero. As u > 0 on \mathcal{R}, this implies that \int_\mathcal{R} \bigtriangledown u \cdot \bigtriangledown u \, dx = \int_\mathcal{R} u^2 p(x) \, dx However, as \bigtriangledown u = 0, the left hand side is also zero. This implies that \int_\mathcal{R} u^2 p(x) \, dx = 0, which implies that u ≡ 0 on \mathcal{R}.(ii) To show that u is unique, we can use the uniqueness theorem for the Poisson equation.