Hoplite
- 50
- 0
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?
\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: