The first eigenvalue of Laplacian
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It's well known that for any bounded domain $Sigma$, there exist unique $lambda_Sigma >0$
and nonnegative $varphiin H_0^1(Sigma)$ such that
begin{equation}label{Eq-La-Bel-op}
left{
begin{aligned}
&-Deltavarphi=lambda_Sigmavarphi
quadmbox{in }Sigma, \
&varphi=0 quadquadquadquad~mbox{on } partialSigma,\
end{aligned}
right.
end{equation}
where $Delta$ is the Laplace operator and $lambda_Sigma$ is the first eigenvalue of it.
Question: For any $varepsilon>0$ small, does there eixist some smooth domain $Omega$ such that $SigmasubsetneqOmega$ and
$lambda_{Sigma}leqlambda_{Omega}+varepsilon$?
pde
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up vote
1
down vote
favorite
It's well known that for any bounded domain $Sigma$, there exist unique $lambda_Sigma >0$
and nonnegative $varphiin H_0^1(Sigma)$ such that
begin{equation}label{Eq-La-Bel-op}
left{
begin{aligned}
&-Deltavarphi=lambda_Sigmavarphi
quadmbox{in }Sigma, \
&varphi=0 quadquadquadquad~mbox{on } partialSigma,\
end{aligned}
right.
end{equation}
where $Delta$ is the Laplace operator and $lambda_Sigma$ is the first eigenvalue of it.
Question: For any $varepsilon>0$ small, does there eixist some smooth domain $Omega$ such that $SigmasubsetneqOmega$ and
$lambda_{Sigma}leqlambda_{Omega}+varepsilon$?
pde
Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
It's well known that for any bounded domain $Sigma$, there exist unique $lambda_Sigma >0$
and nonnegative $varphiin H_0^1(Sigma)$ such that
begin{equation}label{Eq-La-Bel-op}
left{
begin{aligned}
&-Deltavarphi=lambda_Sigmavarphi
quadmbox{in }Sigma, \
&varphi=0 quadquadquadquad~mbox{on } partialSigma,\
end{aligned}
right.
end{equation}
where $Delta$ is the Laplace operator and $lambda_Sigma$ is the first eigenvalue of it.
Question: For any $varepsilon>0$ small, does there eixist some smooth domain $Omega$ such that $SigmasubsetneqOmega$ and
$lambda_{Sigma}leqlambda_{Omega}+varepsilon$?
pde
It's well known that for any bounded domain $Sigma$, there exist unique $lambda_Sigma >0$
and nonnegative $varphiin H_0^1(Sigma)$ such that
begin{equation}label{Eq-La-Bel-op}
left{
begin{aligned}
&-Deltavarphi=lambda_Sigmavarphi
quadmbox{in }Sigma, \
&varphi=0 quadquadquadquad~mbox{on } partialSigma,\
end{aligned}
right.
end{equation}
where $Delta$ is the Laplace operator and $lambda_Sigma$ is the first eigenvalue of it.
Question: For any $varepsilon>0$ small, does there eixist some smooth domain $Omega$ such that $SigmasubsetneqOmega$ and
$lambda_{Sigma}leqlambda_{Omega}+varepsilon$?
pde
pde
asked Nov 14 at 14:01
xiaobiaoJia
183
183
Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49
add a comment |
Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49
Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49
add a comment |
1 Answer
1
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up vote
1
down vote
accepted
One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)
- if $Sigma$ is convex and $Omega_n$ are convex sets tending to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
- If $Omega_n$ are uniformly Lipschitz and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$
- If $Omega_nsubset mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
Depending on the regularity and connectedness properties of your set $Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)
- if $Sigma$ is convex and $Omega_n$ are convex sets tending to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
- If $Omega_n$ are uniformly Lipschitz and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$
- If $Omega_nsubset mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
Depending on the regularity and connectedness properties of your set $Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
add a comment |
up vote
1
down vote
accepted
One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)
- if $Sigma$ is convex and $Omega_n$ are convex sets tending to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
- If $Omega_n$ are uniformly Lipschitz and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$
- If $Omega_nsubset mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
Depending on the regularity and connectedness properties of your set $Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)
- if $Sigma$ is convex and $Omega_n$ are convex sets tending to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
- If $Omega_n$ are uniformly Lipschitz and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$
- If $Omega_nsubset mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
Depending on the regularity and connectedness properties of your set $Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
One answer to this question is that you could try to use continuity of eigenvalues with respect to changes of the domain. I haven't looked at the proofs (they aren't in the reference I use anyway), but in Antoine Henrot's book Extremum problems for eigenvalues of elliptic operators, Section 2.3.3 deals with this (and Section 2.3.5 tells you about differentiability). The results are approximately (all sets lying in a fixed compact set)
- if $Sigma$ is convex and $Omega_n$ are convex sets tending to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
- If $Omega_n$ are uniformly Lipschitz and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$
- If $Omega_nsubset mathbb{R}^2$ such that the number of connected components of the complement is uniformly bounded, and converging to $Sigma$ in Hausdorff distance, then $lambda_k(Omega_n)to lambda_k(Sigma)$.
Depending on the regularity and connectedness properties of your set $Sigma$ (and the dimension), this could answer your problems. I am not aware of any necessary conditions for continuity, but of course there might be some.
answered Nov 14 at 14:43
Kusma
3,6881219
3,6881219
add a comment |
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Are you sure this is what you need? Because there is a trivial answer; just take $Omega$ equal to a translate of $Sigma$.
– Giuseppe Negro
Nov 14 at 14:11
A translate of $Sigma$ is not a superset of $Sigma$.
– Kusma
Nov 14 at 14:15
How about $Omega$ is not a planar domain? e.g., $Omega$ is a subdomain of unit sphere. (There is a result: If $Omega$ is a planar domain, then $lambda_{(alphaOmega)}=frac{lambda_{Omega}}{alpha^2}$)
– xiaobiaoJia
Nov 14 at 14:17
OH sorry, that symbol made me think that $Sigma notsubset Omega$.
– Giuseppe Negro
Nov 14 at 14:49
Just scaling your domain only satisfied your condition of $Sigmasubsetneq Omega$ if it is star shaped.
– Kusma
Nov 14 at 14:49