$L^p-L^q$ estimates for heat equation - regularizing effect
Where can I find a proof of the following estimate
$$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.
heat-equation regularity-theory-of-pdes semigroup-of-operators
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
add a comment |
Where can I find a proof of the following estimate
$$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.
heat-equation regularity-theory-of-pdes semigroup-of-operators
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
add a comment |
Where can I find a proof of the following estimate
$$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.
heat-equation regularity-theory-of-pdes semigroup-of-operators
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
Where can I find a proof of the following estimate
$$|S(t)v|_{L^p(Omega)}leq C t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}|v|_{L^q(Omega)}, $$
where $1leq p<q<+infty$, $Omegasubset mathbb{R}^N$ is an open bounded set and ${S(t)}_{tgeq 0}$ is the semigroup generate by the heat equation with Dirichlet boundary condition.
heat-equation regularity-theory-of-pdes semigroup-of-operators
heat-equation regularity-theory-of-pdes semigroup-of-operators
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
edited Feb 29 '16 at 15:31
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
asked Feb 29 '16 at 15:30
Reginaldo Demarque da Rocha
856
856
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
$$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
Since
$$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.
We know that $S(t)v=N_tast v$. From Young's Inequality, we have
$$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$
Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
$$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$
Hence, we have the result.
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
add a comment |
The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):
Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.
It follows from the Kato-Trotter product formula that
$$
0leq e^{tDelta-nchi}fleq e^{tDelta} f
$$
for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
$$
|e^{tDelta_Omega}f|leq e^{tDelta}|f|
$$
for all $fin L^2(Omega)$.
This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.
An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
$$
|f|_{2N/(N-2)}leq C|nabla f|_2,
$$
which can obviously be localized.
answered Nov 26 at 23:43
MaoWao
2,343616
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
$$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
Since
$$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.
We know that $S(t)v=N_tast v$. From Young's Inequality, we have
$$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$
Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
$$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$
Hence, we have the result.
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
add a comment |
Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
$$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
Since
$$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.
We know that $S(t)v=N_tast v$. From Young's Inequality, we have
$$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$
Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
$$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$
Hence, we have the result.
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
add a comment |
Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
$$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
Since
$$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.
We know that $S(t)v=N_tast v$. From Young's Inequality, we have
$$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$
Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
$$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$
Hence, we have the result.
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
Let $N_t:mathbb{R}^Nto mathbb{R}$, $t>0$, be the function defined by
$$N_t(x)=(4pi t)^{-N/2}e^{-|x|^2/4t}.$$
Since
$$int_{mathbb{R}^N} e^{-a|x|^2}dx=left(frac{pi}{a}right)^{N/2},tag{1}label{1}$$
we can see that $N_tin L^1(mathbb{R}^N)$ and $|N_t|_{ L^1(mathbb{R}^N)}=1$.
We know that $S(t)v=N_tast v$. From Young's Inequality, we have
$$|S(t)v|_{ L^p(Omega)}leq |N_tast v|_{ L^p(Omega)}leq |N_t|_{ L^m(Omega)}|v|_{ L^q(Omega)},$$
where $1+frac{1}{p}=frac{1}{m}+frac{1}{q}.$
Now, we just have to estimate $|N_t|_{ L^m(Omega)}$. From eqref{1}, we can see that
$$|N_t|_{ L^m(Omega)}=(4pi t)^{-N/2}left(int_{mathbb{R}^N} e^{-frac{m}{4t}|x|^2}dxright)^{1/m}=(4pi t)^{-N/2}left(frac{pi}{frac{m}{4t}}right)^{N/2m}=C_{m,N}t^{-frac{N}{2}left(1-frac{1}{m}right)}=C_{m,N}t^{-frac{N}{2}left(frac{1}{q}-frac{1}{p}right)}.$$
Hence, we have the result.
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
answered Mar 21 '16 at 10:04
Reginaldo Demarque da Rocha
856
856
add a comment |
add a comment |
The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):
Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.
It follows from the Kato-Trotter product formula that
$$
0leq e^{tDelta-nchi}fleq e^{tDelta} f
$$
for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
$$
|e^{tDelta_Omega}f|leq e^{tDelta}|f|
$$
for all $fin L^2(Omega)$.
This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.
An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
$$
|f|_{2N/(N-2)}leq C|nabla f|_2,
$$
which can obviously be localized.
answered Nov 26 at 23:43
MaoWao
2,343616
add a comment |
The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):
Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.
It follows from the Kato-Trotter product formula that
$$
0leq e^{tDelta-nchi}fleq e^{tDelta} f
$$
for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
$$
|e^{tDelta_Omega}f|leq e^{tDelta}|f|
$$
for all $fin L^2(Omega)$.
This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.
An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
$$
|f|_{2N/(N-2)}leq C|nabla f|_2,
$$
which can obviously be localized.
answered Nov 26 at 23:43
MaoWao
2,343616
add a comment |
The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):
Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.
It follows from the Kato-Trotter product formula that
$$
0leq e^{tDelta-nchi}fleq e^{tDelta} f
$$
for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
$$
|e^{tDelta_Omega}f|leq e^{tDelta}|f|
$$
for all $fin L^2(Omega)$.
This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.
An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
$$
|f|_{2N/(N-2)}leq C|nabla f|_2,
$$
which can obviously be localized.
answered Nov 26 at 23:43
MaoWao
2,343616
The answer by the OP does not directly address their question because it uses the explicit form of the heat kernel on $mathbb{R}^N$, while the question is posed for an arbitrary (bounded) domain $Omega$. Interestingly there is a nice argument to deduce the estimate on $Omega$ from that on $mathbb{R}^N$ (see Davies, Heat Kernels and Spectral Theory, Theorem 2.1.6):
Let $Delta$ be the Laplacian on $L^2(mathbb{R}^N)$, $Delta_Omega$ the Laplacian on $L^2(Omega)$ with Dirichlet boundary conditions, and $chi$ the characteristic function of $mathbb{R}^Nsetminus Omega$. By standard results on semigroup convergence, $e^{t(Delta-nchi)}fto e^{tDelta_Omega}f$ as $nto infty$.
It follows from the Kato-Trotter product formula that
$$
0leq e^{tDelta-nchi}fleq e^{tDelta} f
$$
for all positive $fin L^2(mathbb{R}^N)$. In the limit $Ntoinfty$ it follows that $e^{tDelta_Omega}fleq e^{tDelta}f$ for all positive $fin L^2(Omega)$, which is easily seen to be equivalent to
$$
|e^{tDelta_Omega}f|leq e^{tDelta}|f|
$$
for all $fin L^2(Omega)$.
This shows that the estimate from the question for the semigroup $(e^{tDelta})$ implies the same bound for $(e^{tDelta_Omega})$.
An alternative approach is to notice that the estimate from the question is equivalent to the Sobolev inequality
$$
|f|_{2N/(N-2)}leq C|nabla f|_2,
$$
which can obviously be localized.
answered Nov 26 at 23:43
MaoWao
2,343616
answered Nov 26 at 23:43
MaoWao
2,343616
answered Nov 26 at 23:43
MaoWao
2,343616
answered Nov 26 at 23:43
MaoWao
2,343616
2,343616
add a comment |
add a comment |
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