Distributional second-order derivatives of $frac{e^{-|x|}}{4pi |x|}$ to show the solution of $u -Delta u=f$...
$begingroup$
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
$endgroup$
add a comment |
$begingroup$
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
$endgroup$
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
$begingroup$
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
$endgroup$
In Brezis's book "Functional Anlaysis" it is proven that the solutions of the Helmotz equation $u - Delta u=f$ where $f in L^2 (mathbb{Omega})$ belong to $H^2 (mathbb{Omega}) cap H^1 _0 (Omega)$, when $Omega$ is a bounded domain of $mathbb{R}^n$ with $partial Omega$ regular enough. I think he also takes the Dirichlet boundary conditions equal to 0, but the problem I am dealing with takes into account $Omega = mathbb{R}^n$, so it shouldn't be a problem to integrate by parts using sufficently large balls. The thing is, he proves the existence of such a solution by shifting the problem to a minum of a particular functional. He then proves this minumum (that is a weak solution of the equation) belongs to $H^{1}_0 (Omega)$. Then, later on in the book, he proves that under the hypothesis of some regularity for $partial Omega$, the solution is in $H^2(Omega)cap H^1 _0 (Omega)$, and he does that using the so called "method of Nirmberg's translations".
The fact is: I am dealing with the case $Omega= mathbb{R}^3$. I think the same results that were true for regular bounded domains should also be true in this case, I think it should be enough to consider the limit of domains $Omega_N = B(0,N)$ where the proof of Brezis works.
The problem is, I wanted to find the solution explicitly and verify by hand it belongs to $H^{2}(mathbb{R}^3)$, given that the domain is so simple, not using a known theorem. The solution of the equation is given by:
$$u(x)= left( frac{e^{-|y|}}{4 pi |y|} ast f right)(x)$$
But when I try to find a weak second order derivative of $u$ using integration by parts (and I do so by splitting the integration domain in and inside a ball of radius $epsilon$), I get this very annoying term that doesn't allow me to conclude:
$$int_{B(0, epsilon)^c} partial_{x_1}^2 left(frac{1}{4 pi |x|} right) e^{-|x|} phi(x+y) dx=int_{B(0, epsilon)^c} left(frac{3x_1 ^2 - |x|^2}{4 pi |x|^5} right) e^{-|x|} phi(x+y) dx$$
Where $phi in C^{infty}_0 (mathbb{R}^3)$. Now, I understand that something like this was forced to appear. Indeed, I know that $Delta (1/|x|)= delta(x)$, so this isn't a surprise. However, that expression over there does not converge when $epsilon to 0$ (or t least I think so, there should be a singularity like $1/rho$ if you pass to spherical coordinates) and I would like to obtain a weak derivative for $u$ in which I put all the derivatives on the term $e^{-|x|}/(4 pi |x|)$ of the convolution.
EDIT: Turns out I can just cut-off a ball of radius $epsilon$ with a smooth compact supported radial function and then integrating over the angular part first I get either 0 or a constant I don’t remember. Gotta think why I can pass to the limit, but the pointeise convergence is straightforward, and the dominance should follow easiliy, somehow.
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
pde sobolev-spaces distribution-theory calculus-of-variations regularity-theory-of-pdes
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
edited Dec 3 '18 at 18:24
tommy1996q
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
asked Dec 1 '18 at 13:58
tommy1996qtommy1996q
591415
591415
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021373%2fdistributional-second-order-derivatives-of-frace-x4-pi-x-to-show-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3021373%2fdistributional-second-order-derivatives-of-frace-x4-pi-x-to-show-t%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Why wouldn't it converge? The $exp(-lvert xrvert)$ is there to help for large $lvert xrvert$.
$endgroup$
– user10354138
Dec 1 '18 at 14:17
$begingroup$
I mean convergence for $epsilon to 0$. There I have a singularity of order $1/ rho$, passing in spherical coordinates. I'll edit the question, though.
$endgroup$
– tommy1996q
Dec 1 '18 at 14:26