Issue with Green's function for Poisson's equation

Click For Summary
SUMMARY

This discussion addresses the application of Green's function for solving Poisson's equation in a twice-infinite channel-shaped domain, denoted as Ω. The user explores the implications of defining the function f(x,y,z) within Ω while considering the boundary conditions on ∂Ω. The conclusion drawn is that without specified boundary conditions for p on ∂Ω, treating p as a function defined on R^3 is problematic and may lead to an ill-posed problem. The correct approach requires acknowledging the boundary conditions to ensure the solution remains valid.

PREREQUISITES
  • Understanding of Green's functions in the context of partial differential equations.
  • Familiarity with Poisson's equation and its applications in mathematical physics.
  • Knowledge of boundary value problems and their significance in solving PDEs.
  • Basic concepts of functional spaces, specifically L² spaces.
NEXT STEPS
  • Study the derivation and properties of Green's functions for various boundary conditions.
  • Explore the implications of boundary conditions in solving Poisson's equation in different domains.
  • Learn about the mathematical framework of L² spaces and their relevance in functional analysis.
  • Investigate the concept of well-posed versus ill-posed problems in the context of PDEs.
USEFUL FOR

Mathematicians, physicists, and engineers working with partial differential equations, particularly those dealing with boundary value problems and Green's functions in various domains.

Hoplite
Messages
50
Reaction score
0
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,

[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 have no boundary 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]?
 
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
[tex]\nabla ^2 p = f \; , \; \text{ in } \mathbb{R} ^3[/tex]​
is

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

[tex]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 ) ][/tex]
where [tex]f \in L^2( \mathbb{R} ^3)[/tex] or [tex]f \in L^2 ( \Omega )[/tex].
 

Similar threads

  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K