Prove these two functions are in $H^2(mathbb{R}^n)$
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1
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I want to prove that the equation
$$u -Delta u=f$$
with $f in L^2 (mathbb{R}^n)$ admits a solution in $H^2 (mathbb{R}^n)$ with $n=1,3$.
Taking Fourier transforms and resolving, I get:
$$u(x)=left( frac {e^{-|y|}}{2} ast f right) (x)$$
when $n=1$ and
$$u(x)=left( frac {e^{-|y|}}{4 pi |y|} ast f right) (x)$$
when $n=3$. Do these functions belong to $H^2 (mathbb{R}^n)$? And how do I prove it?
(I don’t know if I should use the $H$ definition with approximations or the $W$ definition with weak derivatives)
EDIT: I just realized (don’t know why it took so much) that the problem I had can be solved by using Young inequality. For some reason I kept thinking I needed the derivative of $frac{e^{-|x|}}{|x|}$ to be in $L^2$
functional-analysis analysis sobolev-spaces
asked Nov 12 at 17:36
tommy1996q
559413
|
show 6 more comments
up vote
1
down vote
favorite
I want to prove that the equation
$$u -Delta u=f$$
with $f in L^2 (mathbb{R}^n)$ admits a solution in $H^2 (mathbb{R}^n)$ with $n=1,3$.
Taking Fourier transforms and resolving, I get:
$$u(x)=left( frac {e^{-|y|}}{2} ast f right) (x)$$
when $n=1$ and
$$u(x)=left( frac {e^{-|y|}}{4 pi |y|} ast f right) (x)$$
when $n=3$. Do these functions belong to $H^2 (mathbb{R}^n)$? And how do I prove it?
(I don’t know if I should use the $H$ definition with approximations or the $W$ definition with weak derivatives)
EDIT: I just realized (don’t know why it took so much) that the problem I had can be solved by using Young inequality. For some reason I kept thinking I needed the derivative of $frac{e^{-|x|}}{|x|}$ to be in $L^2$
functional-analysis analysis sobolev-spaces
asked Nov 12 at 17:36
tommy1996q
559413
Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
1
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25
|
show 6 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that the equation
$$u -Delta u=f$$
with $f in L^2 (mathbb{R}^n)$ admits a solution in $H^2 (mathbb{R}^n)$ with $n=1,3$.
Taking Fourier transforms and resolving, I get:
$$u(x)=left( frac {e^{-|y|}}{2} ast f right) (x)$$
when $n=1$ and
$$u(x)=left( frac {e^{-|y|}}{4 pi |y|} ast f right) (x)$$
when $n=3$. Do these functions belong to $H^2 (mathbb{R}^n)$? And how do I prove it?
(I don’t know if I should use the $H$ definition with approximations or the $W$ definition with weak derivatives)
EDIT: I just realized (don’t know why it took so much) that the problem I had can be solved by using Young inequality. For some reason I kept thinking I needed the derivative of $frac{e^{-|x|}}{|x|}$ to be in $L^2$
functional-analysis analysis sobolev-spaces
asked Nov 12 at 17:36
tommy1996q
559413
I want to prove that the equation
$$u -Delta u=f$$
with $f in L^2 (mathbb{R}^n)$ admits a solution in $H^2 (mathbb{R}^n)$ with $n=1,3$.
Taking Fourier transforms and resolving, I get:
$$u(x)=left( frac {e^{-|y|}}{2} ast f right) (x)$$
when $n=1$ and
$$u(x)=left( frac {e^{-|y|}}{4 pi |y|} ast f right) (x)$$
when $n=3$. Do these functions belong to $H^2 (mathbb{R}^n)$? And how do I prove it?
(I don’t know if I should use the $H$ definition with approximations or the $W$ definition with weak derivatives)
EDIT: I just realized (don’t know why it took so much) that the problem I had can be solved by using Young inequality. For some reason I kept thinking I needed the derivative of $frac{e^{-|x|}}{|x|}$ to be in $L^2$
functional-analysis analysis sobolev-spaces
functional-analysis analysis sobolev-spaces
asked Nov 12 at 17:36
tommy1996q
559413
asked Nov 12 at 17:36
tommy1996q
559413
edited Nov 14 at 15:21
asked Nov 12 at 17:36
tommy1996q
559413
asked Nov 12 at 17:36
tommy1996q
559413
asked Nov 12 at 17:36
tommy1996q
559413
559413
Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
1
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25
|
show 6 more comments
Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
1
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25
Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
1
1
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25
|
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
add a comment |
up vote
2
down vote
accepted
Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
Fourier transforming, we see that the equation is equivalent to
$$
(1+|xi|^2)hat{u}(xi) = hat{f}(xi). $$
In particular,
$$
|u|_{H^2}^2=int_{mathbb R^n} |hat{u}(xi)(1+|xi|^2)|^2, dxi=int_{mathbb R^n} |hat{f}(xi)|^2, dxi, $$
and $hat{f}in L^2(mathbb R^n)$ by Plancherel's theorem, so the integral in the right-hand side is finite. Thus, $uin H^2$.
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
answered Nov 12 at 18:33
Giuseppe Negro
17k328121
17k328121
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
add a comment |
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
I will use this if I can’t find a more direct method. I mean, this is right, but I’ve explicitly calculated the solution, and I would like to have something more “direct”, which uses explicitly the expressions I have found.
– tommy1996q
Nov 12 at 18:40
add a comment |
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Use ast for convolution instead of star.
– Umberto P.
Nov 12 at 18:11
Can you calculate the derivatives directly?
– Umberto P.
Nov 12 at 18:12
Yeah, but I fear those would not be in $L^2$, especially for the second case with $n=3$, because you have a division by $|y|$. Of course I would put all the derivatives on the first term
– tommy1996q
Nov 12 at 18:17
Actually it would be great if it worked with $W^{2,p}$ with $p in (1,2)$, because of the work I’ll have to do later.
– tommy1996q
Nov 12 at 18:21
1
If you need $H^s$ regularity I would work purely in Fourier. A function is $H^2$ iff $(1+|xi|^2)^{s/2}hat{f}in L^2$, use this.
– Giuseppe Negro
Nov 12 at 18:25