Issue with Green's function for Poisson's equation

1. Jan 12, 2010

Hoplite

Say we have a 3D function, $$p(x,y,z)$$ and we define it in terms of another function $$f(x,y,z)$$ via,

$$\nabla ^2 p = f.$$

I know that if we are working in $$R^3$$ space (with no boundaries) we can say that,

$$p= \frac{-1}{4\pi}\iiint \limits_R \frac{f(x',y',z')}{\sqrt{(x-x')^2 +(y-y')^2+(z-z')^2}} dx' dy' dz'$$.

But here is my issue: Say that instead of $$R^3$$ , we are working in a twice-infinite channel-shaped domain called $$\Omega$$, defined via,

$$\Omega =\left[ (x,y,z): \quad -\infty < x,y < \infty , \quad -\frac{1}{2} \leq z \leq -\frac{1}{2} \right] .$$

Call the boundary of this domain $$\partial \Omega$$. Now say that we have no boundary conditions specified for $$p$$ on $$\partial \Omega$$, but that $$f$$ is defined on $$\Omega$$, and is not specified outside of $$\Omega$$.

I can see no reason, however, that we couldn't say that $$f=0$$ outside of $$\Omega$$, and that way define $$f$$ on all of $$R^3$$. (But I could be missing something.) I should further specify that $$f=0$$, on the boundary $$\partial \Omega$$, so we could extend the domain of $$f$$ in thids way without sacrificing smoothness. (We can assume that $$f$$ is smooth and finite everywhere on $$\Omega$$.)

So, since $$p$$ is not defined on $$\partial \Omega$$, can I simply treat it as a function defined on $$R^3$$ with no boundaries, and hence solve for $$p$$ using the triple integral above? Or is this illegal, since $$f$$ is defined only on $$\Omega$$?

Last edited: Jan 12, 2010
2. Jan 13, 2010

brahman

i think the problem (must) needs the boundary condition, if not it's ill-posed. The Green function of
$$\nabla ^2 p = f \; , \; \text{ in } \mathbb{R} ^3$$​
is

$$G(x,y,z; \xi , \eta , \theta)= \frac{-1}{4\pi} \frac{1}{ \sqrt{(x-\xi )^2 +(y-\eta )^2+(z-\theta )^2}}$$
and the Green function of the problem
\begin{align} \nabla ^2 p &= f \; , \; \text{ in } \Omega \; \text{(defined by you)} \\ p &= g \; , \; \text{ on } \partial \Omega \end{align}
is

$$G_0 (x,y,z;\xi , \eta , \theta )= \sum_{n= -\infty}^{\infty} [ G(x,y,z; \xi , \eta , k + \theta + 1/2) - G(x,y,z; \xi , \eta , k - \theta - 1/2 ) ]$$
where $$f \in L^2( \mathbb{R} ^3)$$ or $$f \in L^2 ( \Omega )$$.

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