Issue with Green's function for Poisson's equation

Hoplite
Messages
50
Reaction score
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?
 
Last edited:
Physics news on Phys.org
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 &amp;= f \; , \; \text{ in } \Omega \; \text{(defined by you)} \\<br /> p &amp;= g \; , \; \text{ on } \partial \Omega <br /> \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 ).
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top