Extended Global Approximation Theorem
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In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
30919
add a comment |
up vote
1
down vote
favorite
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
30919
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
30919
In Evans,
$textbf{Theorem} $ (Global Approximation Theorem) Assume $U$ is bounded, and $partial U$ is $C^1$. Suppose as well that $u in W^{k,p}(U)$ for some $1leq p < infty$. Then, there exist functions $u_m in C^{infty}(bar{U})$ such that
begin{align*}
u_m rightarrow u quad textrm{ in } W^{k,p}(U)
end{align*}
$textbf{Question}$ Although we change the boundary condition like
begin{align*}
partial U=bigcup_{j=1}^n Gamma_j, quad (textrm{boundary is piecewise } C^{1})
end{align*}
where each $Gamma_j$ for $j=1, cdots, n$ is a $C^1$, $Gamma_j$ and $Gamma_{j^{'}}$ do not intersect except at their endpoints if $jneq j'$, then does the theorem still hold?
Any help is appreciated!!
I want to know references related that...
Thank you!!
analysis pde sobolev-spaces approximation-theory
analysis pde sobolev-spaces approximation-theory
asked Nov 14 at 9:19
w.sdka
30919
asked Nov 14 at 9:19
w.sdka
30919
edited Nov 15 at 16:03
asked Nov 14 at 9:19
w.sdka
30919
asked Nov 14 at 9:19
w.sdka
30919
asked Nov 14 at 9:19
w.sdka
30919
30919
add a comment |
add a comment |
1 Answer
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... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
add a comment |
up vote
0
down vote
up vote
0
down vote
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
... does the theorem still hold?
Smooth boundary is not needed. The theorem is true provided that the domain does not lie on both sides of any part of its boundary, i.e., provided that $U$ satisfies the segment condition.
Reference: Adams book, p. 68.
answered Nov 26 at 15:31
Pedro
10.2k23066
edited Nov 26 at 16:11
answered Nov 26 at 15:31
Pedro
10.2k23066
answered Nov 26 at 15:31
Pedro
10.2k23066
answered Nov 26 at 15:31
Pedro
10.2k23066
10.2k23066
add a comment |
add a comment |
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