Neumann problem for a circle.
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
differential-equations
asked Mar 7 '16 at 19:33
User9523
9701823
add a comment |
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
differential-equations
asked Mar 7 '16 at 19:33
User9523
9701823
add a comment |
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
differential-equations
asked Mar 7 '16 at 19:33
User9523
9701823
We consider the interior neumann problem : $$nabla^{2} u =0:::, ::r<R$$
$$ dfrac{partial u}{partial n}=dfrac{partial u}{partial r}=f(theta)::,::r=R::,:: 0<theta<2pi$$
We first establish the fact which as the solution says is the compatibility condition : $$int_B fds=0$$
which is established using Green's second formula. Now the solution proceeds as follows :
(1) The compatibility condition obtained above can be written as $$R int_0^{2pi} f(theta) dtheta=0$$
(2) In the case of the dirichlet problem of the circle , the solution of the laplace equation is given as : $$u(r,theta) = dfrac{a_0}{2} + sum_{k=1}^{infty}r^{k}(a_k cosktheta + b_k sinktheta)$$
(3) Differentiating with respect to r and applying the boundary conditions we obtain : $$dfrac{partial u}{partial r}(R,theta) = sum_{k=1}^{infty}kR^{k-1}(a_k cosktheta + b_k sinktheta) = f(theta)$$
(4) Now the expression of $f(theta)$ in a series of the form (3) is possible only by the virtue of the compatibility condition since : $$a_0 = dfrac{1}{pi} int_0^{2pi} f(tau) d tau = 0$$
The rest of the proof was okay , can anyone explain the points (1) and (4) ?
Like in (1) are we changing the variable into $theta$ ? If yes , then how is $R$ there ?
And I have no idea about (4). Kindly help !
Thanks in advance !
differential-equations
differential-equations
asked Mar 7 '16 at 19:33
User9523
9701823
asked Mar 7 '16 at 19:33
User9523
9701823
edited Mar 8 '16 at 12:20
asked Mar 7 '16 at 19:33
User9523
9701823
asked Mar 7 '16 at 19:33
User9523
9701823
asked Mar 7 '16 at 19:33
User9523
9701823
9701823
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523
9701823
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%2f1687412%2fneumann-problem-for-a-circle%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
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523
9701823
add a comment |
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523
9701823
add a comment |
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523
9701823
The integral in (1) is transformed into polar co-ordinates by considering the " arc-length " relation. $ds$ here is the arc length. $R$ is the radius of the region. So , using $ theta = dfrac{l}{r}$ , where $l$ is the arc length and $r$ is the radius . Thus we get $ theta = dfrac{ds}{R}$ . Substituting this gives the result.
(4) actually tells the value of the constant $a_0$. That's all.
answered Mar 12 '16 at 10:03
User9523
9701823
answered Mar 12 '16 at 10:03
User9523
9701823
answered Mar 12 '16 at 10:03
User9523
9701823
answered Mar 12 '16 at 10:03
User9523
9701823
9701823
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f1687412%2fneumann-problem-for-a-circle%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
