Neumann problem with zero average sobolev space
$begingroup$
Let us considère the following Laplace-Neumann problem $-Delta u=0$ with homogenuous boundary condition of type neumann, i:e $frac{{partial u}}{{partial n}} = 0$.
The variational formulation is given as follows $a(u,v)=l(v)$, where $$a(u,v) = int_Omega {nabla unabla v} dx = 0$$ and $l(v)=0$.
Let us cinsidère the space
$$left{ V={u in H^1(Omega), int_Omega v(x)ds=0} right}$$
$l$ is continuous ( is zero), and $a$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.
pde laplacian elliptic-equations variational-analysis
asked Dec 2 '18 at 11:04
GustaveGustave
729211
$endgroup$
add a comment |
$begingroup$
Let us considère the following Laplace-Neumann problem $-Delta u=0$ with homogenuous boundary condition of type neumann, i:e $frac{{partial u}}{{partial n}} = 0$.
The variational formulation is given as follows $a(u,v)=l(v)$, where $$a(u,v) = int_Omega {nabla unabla v} dx = 0$$ and $l(v)=0$.
Let us cinsidère the space
$$left{ V={u in H^1(Omega), int_Omega v(x)ds=0} right}$$
$l$ is continuous ( is zero), and $a$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.
pde laplacian elliptic-equations variational-analysis
asked Dec 2 '18 at 11:04
GustaveGustave
729211
$endgroup$
add a comment |
$begingroup$
Let us considère the following Laplace-Neumann problem $-Delta u=0$ with homogenuous boundary condition of type neumann, i:e $frac{{partial u}}{{partial n}} = 0$.
The variational formulation is given as follows $a(u,v)=l(v)$, where $$a(u,v) = int_Omega {nabla unabla v} dx = 0$$ and $l(v)=0$.
Let us cinsidère the space
$$left{ V={u in H^1(Omega), int_Omega v(x)ds=0} right}$$
$l$ is continuous ( is zero), and $a$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.
pde laplacian elliptic-equations variational-analysis
asked Dec 2 '18 at 11:04
GustaveGustave
729211
$endgroup$
Let us considère the following Laplace-Neumann problem $-Delta u=0$ with homogenuous boundary condition of type neumann, i:e $frac{{partial u}}{{partial n}} = 0$.
The variational formulation is given as follows $a(u,v)=l(v)$, where $$a(u,v) = int_Omega {nabla unabla v} dx = 0$$ and $l(v)=0$.
Let us cinsidère the space
$$left{ V={u in H^1(Omega), int_Omega v(x)ds=0} right}$$
$l$ is continuous ( is zero), and $a$ is continuous and corecive. By lax-Milgram theoremn there exists a unique solution of the variational problem $$a(u,v)=l(v)$$. We can notice that the constant are alos solutions of the problem, but the constants don't satisfy the zero average condition, I don't understand that.
pde laplacian elliptic-equations variational-analysis
pde laplacian elliptic-equations variational-analysis
asked Dec 2 '18 at 11:04
GustaveGustave
729211
asked Dec 2 '18 at 11:04
GustaveGustave
729211
asked Dec 2 '18 at 11:04
GustaveGustave
729211
asked Dec 2 '18 at 11:04
GustaveGustave
729211
asked Dec 2 '18 at 11:04
GustaveGustave
729211
729211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In your case the problem is $-Delta u = 0$ subject to $frac{partial u}{partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $int_{Omega} u ,mathrm{d}x = 0,$ since $a(u,u) = int_{Omega} |nabla u|^2,mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.
So the solution $u equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u equiv k neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $-Delta u = f$, we get $ell(v) = int_{Omega} fv ,mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
$endgroup$
add a comment |
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%2f3022495%2fneumann-problem-with-zero-average-sobolev-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In your case the problem is $-Delta u = 0$ subject to $frac{partial u}{partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $int_{Omega} u ,mathrm{d}x = 0,$ since $a(u,u) = int_{Omega} |nabla u|^2,mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.
So the solution $u equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u equiv k neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $-Delta u = f$, we get $ell(v) = int_{Omega} fv ,mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
In your case the problem is $-Delta u = 0$ subject to $frac{partial u}{partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $int_{Omega} u ,mathrm{d}x = 0,$ since $a(u,u) = int_{Omega} |nabla u|^2,mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.
So the solution $u equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u equiv k neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $-Delta u = f$, we get $ell(v) = int_{Omega} fv ,mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
$endgroup$
add a comment |
$begingroup$
In your case the problem is $-Delta u = 0$ subject to $frac{partial u}{partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $int_{Omega} u ,mathrm{d}x = 0,$ since $a(u,u) = int_{Omega} |nabla u|^2,mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.
So the solution $u equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u equiv k neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $-Delta u = f$, we get $ell(v) = int_{Omega} fv ,mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
$endgroup$
In your case the problem is $-Delta u = 0$ subject to $frac{partial u}{partial n} = 0,$ which only has trivial solutions. Indeed if $u$ is such a solution with zero average $int_{Omega} u ,mathrm{d}x = 0,$ since $a(u,u) = int_{Omega} |nabla u|^2,mathrm{d}x = 0$ we conclude that $u$ is a.e. constant and hence zero as it has zero average.
So the solution $u equiv 0$ is the unique solution subject to the zero average condition. The other constant solutions $u equiv k neq 0$ are also solutions, but they don't satisfy the zero average condition. I assume this is what is meant here.
For a non-homogenous problem $-Delta u = f$, we get $ell(v) = int_{Omega} fv ,mathrm{d}x$ is non-trivial and hence we get non-trivial solutions. However if $u$ is a solution, then so is $u+k$ for any constant $k.$ The zero average condition removes this extra degree of freedom.
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
answered Dec 4 '18 at 20:11
ktoiktoi
2,4061617
2,4061617
add a comment |
add a comment |
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%2f3022495%2fneumann-problem-with-zero-average-sobolev-space%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
