Heat semigroup representation The 2019 Stack Overflow Developer Survey Results Are InCore for the Laplace operator in a bounded domainExistance of the integral in the domain of generator of the strongly continuous semigroupProving the existence of a solution of the heat equation using semigroup methods$L^p-L^q$ estimates for heat equation - regularizing effectIs the generated semigroup by an elliptic operator be the transition semigroup?why $int_Omega S(t) div(w) = int_Omega div(w)$ ? where $S(t)$ is a heat semigroup with neumann conditionBoundedness of Riemann integral of generator of a semigroupUniform exponential stability of semigroup of operatorsExtending the $C_0$ (heat) semigroup with inverse element?Heat semigroup equality $e^tDeltap(x,x_0, 1)=p(x,x_0,1+t)$?
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Heat semigroup representation
The 2019 Stack Overflow Developer Survey Results Are InCore for the Laplace operator in a bounded domainExistance of the integral in the domain of generator of the strongly continuous semigroupProving the existence of a solution of the heat equation using semigroup methods$L^p-L^q$ estimates for heat equation - regularizing effectIs the generated semigroup by an elliptic operator be the transition semigroup?why $int_Omega S(t) div(w) = int_Omega div(w)$ ? where $S(t)$ is a heat semigroup with neumann conditionBoundedness of Riemann integral of generator of a semigroupUniform exponential stability of semigroup of operatorsExtending the $C_0$ (heat) semigroup with inverse element?Heat semigroup equality $e^tDeltap(x,x_0, 1)=p(x,x_0,1+t)$?
$begingroup$
It is known that the Laplacian operator with Dirichlet boundary condition in $L^2(Omega),$ (with $Omega$ being a open subset of $R^n$) generates a $C_0-$semigroup in $L^2(Omega)$). Moreover, in the case of the whole space $R^n$, the semigroup can be expressed by the means of the Gauss–Weierstrass kernel.
I m wondering if this is also true for a bounded open subset $Omega$ of $R^n$.
functional-analysis operator-theory partial-derivative heat-equation semigroup-of-operators
edited Mar 30 at 23:05
Pedro
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
$endgroup$
add a comment |
$begingroup$
It is known that the Laplacian operator with Dirichlet boundary condition in $L^2(Omega),$ (with $Omega$ being a open subset of $R^n$) generates a $C_0-$semigroup in $L^2(Omega)$). Moreover, in the case of the whole space $R^n$, the semigroup can be expressed by the means of the Gauss–Weierstrass kernel.
I m wondering if this is also true for a bounded open subset $Omega$ of $R^n$.
functional-analysis operator-theory partial-derivative heat-equation semigroup-of-operators
edited Mar 30 at 23:05
Pedro
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
$endgroup$
add a comment |
$begingroup$
It is known that the Laplacian operator with Dirichlet boundary condition in $L^2(Omega),$ (with $Omega$ being a open subset of $R^n$) generates a $C_0-$semigroup in $L^2(Omega)$). Moreover, in the case of the whole space $R^n$, the semigroup can be expressed by the means of the Gauss–Weierstrass kernel.
I m wondering if this is also true for a bounded open subset $Omega$ of $R^n$.
functional-analysis operator-theory partial-derivative heat-equation semigroup-of-operators
edited Mar 30 at 23:05
Pedro
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
$endgroup$
It is known that the Laplacian operator with Dirichlet boundary condition in $L^2(Omega),$ (with $Omega$ being a open subset of $R^n$) generates a $C_0-$semigroup in $L^2(Omega)$). Moreover, in the case of the whole space $R^n$, the semigroup can be expressed by the means of the Gauss–Weierstrass kernel.
I m wondering if this is also true for a bounded open subset $Omega$ of $R^n$.
functional-analysis operator-theory partial-derivative heat-equation semigroup-of-operators
functional-analysis operator-theory partial-derivative heat-equation semigroup-of-operators
edited Mar 30 at 23:05
Pedro
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
edited Mar 30 at 23:05
Pedro
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
edited Mar 30 at 23:05
Pedro
10.9k23475
edited Mar 30 at 23:05
Pedro
10.9k23475
edited Mar 30 at 23:05
Pedro
10.9k23475
10.9k23475
asked Mar 17 at 14:19
RabatRabat
505
asked Mar 17 at 14:19
RabatRabat
505
asked Mar 17 at 14:19
RabatRabat
505
505
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is,
$$
-Delta varphi_n =lambda_n varphi_n, ;;; |varphi_n|_L^2(Omega)=1,\ lambda_1 le lambda_2 le lambda_3 le cdots,
$$
and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^tDeltaf$, which is the solution of $psi_t=Delta psi$ can be expressed in terms of the eigenfunctions of $-Delta$,
$$
H(t)f = sum_n=1^inftye^-sqrtlambda_ntlangle f,varphi_nranglevarphi_n.
$$
There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(Omega)$.
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is,
$$
-Delta varphi_n =lambda_n varphi_n, ;;; |varphi_n|_L^2(Omega)=1,\ lambda_1 le lambda_2 le lambda_3 le cdots,
$$
and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^tDeltaf$, which is the solution of $psi_t=Delta psi$ can be expressed in terms of the eigenfunctions of $-Delta$,
$$
H(t)f = sum_n=1^inftye^-sqrtlambda_ntlangle f,varphi_nranglevarphi_n.
$$
There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(Omega)$.
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
add a comment |
$begingroup$
If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is,
$$
-Delta varphi_n =lambda_n varphi_n, ;;; |varphi_n|_L^2(Omega)=1,\ lambda_1 le lambda_2 le lambda_3 le cdots,
$$
and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^tDeltaf$, which is the solution of $psi_t=Delta psi$ can be expressed in terms of the eigenfunctions of $-Delta$,
$$
H(t)f = sum_n=1^inftye^-sqrtlambda_ntlangle f,varphi_nranglevarphi_n.
$$
There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(Omega)$.
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
add a comment |
$begingroup$
If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is,
$$
-Delta varphi_n =lambda_n varphi_n, ;;; |varphi_n|_L^2(Omega)=1,\ lambda_1 le lambda_2 le lambda_3 le cdots,
$$
and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^tDeltaf$, which is the solution of $psi_t=Delta psi$ can be expressed in terms of the eigenfunctions of $-Delta$,
$$
H(t)f = sum_n=1^inftye^-sqrtlambda_ntlangle f,varphi_nranglevarphi_n.
$$
There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(Omega)$.
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
If you have a bounded region with a smooth boundary, and Dirichlet conditions, then $Delta$ can be represented using an eigenfunction expansion, and the heat semigroup can be expressed in terms of the eigenfunctions of the Laplacian as well. That is,
$$
-Delta varphi_n =lambda_n varphi_n, ;;; |varphi_n|_L^2(Omega)=1,\ lambda_1 le lambda_2 le lambda_3 le cdots,
$$
and the eigenfunctions can be assumed to be mutually orthogonal. And the Heat semigroup $H(t)f=e^tDeltaf$, which is the solution of $psi_t=Delta psi$ can be expressed in terms of the eigenfunctions of $-Delta$,
$$
H(t)f = sum_n=1^inftye^-sqrtlambda_ntlangle f,varphi_nranglevarphi_n.
$$
There would not generally be any simplification of this form. This is a $C^0$ semigroup on $L^2(Omega)$.
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Mar 17 at 16:48
DisintegratingByPartsDisintegratingByParts
60.4k42681
60.4k42681
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
add a comment |
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
In fact I m interested with the continuity of the mapping $tto |S(t)x|_L^infty(Omega), $ and I didn't succeed to do it using this formula.
$endgroup$
– Rabat
Mar 17 at 16:54
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
$begingroup$
@Rabat : I would expect that you would need some additional regularity assumptions on the boundary in order to assert what you want. Sharp cusps on the boundary would affect the eigenfunctions.
$endgroup$
– DisintegratingByParts
Mar 17 at 17:13
add a comment |
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