Dirichlet problem boundary conditions

  • #1
cianfa72
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TL;DR Summary
About boundary conditions for Dirichlet problem.
The Dirichlet problem asks for the solution of Poisson or Laplace equation in an open region ##S## of ##\mathbb R^n## with a condition on the boundary ##\partial_S##.

In particular the solution function ##f()## is required to be two-times differentiable in the interior region ##S## and continuous on the boundary ##\partial_S##. The boundary condition specifies its value ##u## at each point on it, hence ##f=u## at the boundary ##\partial_S##.

Now my question is: which is the topology w.r.t one asks the function ##f## to be continuous in the closure ##\bar S## ? I believe it is the subspace topology on ##\bar S## from ##\mathbb R^n## as subset.

Is the above correct? Thanks.
 
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  • #2
Id say that more than just topology, it's also an issue of Differential Structure, given you want your function to be (twice)differentiable. Given differentiability implies continuity, I suspect Differential Structure would be enough. I suspect you're dealing with the standard topology, but I'm not 100%.
 
  • #3
WWGD said:
Id say that more than just topology, it's also an issue of Differential Structure, given you want your function to be (twice)differentiable. Given differentiability implies continuity, I suspect Differential Structure would be enough. I suspect you're dealing with the standard topology, but I'm not 100%.
Yes, as far as can I understand, the region where the PDE is solved is an open set ##S## of ##\mathbb R^n## endowed with standard differential structure/topology. Then we have boundary condition on the boundary ##\partial_S##.

The solution ##f## is required to be continuous on the closure ##\bar S##. So the question is: which is the topology on ##\bar S## w.r.t. it is closed anf ##f## is required to be continuous ? My answer: it is the subspace topology from the superset ##\mathbb R^n##.
 
  • #4
Unless otherwise stated, when we speak of [itex]U \subseteq \mathbb{R}^n[/itex] we mean the set together with its local affine structure (so that we can calculate derivatives) and the topology induced by the Euclidean inner product. If [itex]U \neq \mathbb{R}^n[/itex] then the restriction to the subspace topology is implied.

The smooth structure is the maximal atlas which contains the charts [itex](V, \mathrm{id})[/itex] where [itex]V[/itex] is any open subset of [itex]U[/itex]. (There are always at least two such structures, since [tex](V,
p \ni V \to \mathbb{R}^n : q \mapsto ((q_1 - p_1)^3,q_2 - p_2, \dots, q_n - p_n))[/tex] is not smoothly compatible with the identity chart.)
 
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  • #5
pasmith said:
The smooth structure is the maximal atlas which contains the charts [itex](V, \mathrm{id})[/itex] where [itex]V[/itex] is any open subset of [itex]U[/itex]. (There are always at least two such structures, since [tex](V,
p \ni V \to \mathbb{R}^n : q \mapsto ((q_1 - p_1)^3,q_2 - p_2, \dots, q_n - p_n))[/tex] is not smoothly compatible with the identity chart.)
You mean fixed a point ##p =(p_1, p_2 \dots p_n) \in V##, your map [tex]V \ni q \mapsto ((q_1 - p_1)^3,q_2 - p_2, \dots, q_n - p_n))[/tex] is not smoothly compatible with the identity map. However both structures are diffeomorphic (i.e. there exists a diffeomorphism between them).
 
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