Say we have a 3D function, [tex]p(x,y,z)[/tex] and we define it in terms of another function [tex]f(x,y,z)[/tex] via,(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\nabla ^2 p = f.[/tex]

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

[tex]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' [/tex].

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

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

Call the boundary of this domain [tex]\partial \Omega[/tex]. Now say that we havenoboundary conditions specified for [tex]p[/tex] on [tex]\partial \Omega[/tex], but that [tex]f[/tex] is defined on [tex] \Omega[/tex], and is not specified outside of [tex]\Omega[/tex].

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

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

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# Issue with Green's function for Poisson's equation

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