Let $v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$. Show that for $x in B(0,R)^c$ we have $v(x) =...
up vote
2
down vote
favorite
Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$
Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.
integration pde harmonic-functions potential-theory
asked Nov 20 at 8:18
h3h325
18910
add a comment |
up vote
2
down vote
favorite
Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$
Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.
integration pde harmonic-functions potential-theory
asked Nov 20 at 8:18
h3h325
18910
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$
Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.
integration pde harmonic-functions potential-theory
asked Nov 20 at 8:18
h3h325
18910
Let $$v(x) = int_{B(0,R)} frac{c}{||x-y||^{n-2}}mathrm{d}y$$ Show that for $x in B(0,R)^c$ we have $$v(x) = c_1||x||^{2-n} + c_0$$
Where $B(0,R) subset mathbb{R}^n$ is the ball centered at $0$ and of radius $R$ and $c_1$ and $c_2$ are constants. I'm afraid my calculus is a bit rusty, I tried changing to polar coordinates but im getting confused with the term $||x-y||$.
integration pde harmonic-functions potential-theory
integration pde harmonic-functions potential-theory
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
edited Nov 23 at 19:53
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
asked Nov 20 at 8:18
h3h325
18910
18910
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:
$$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$
Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$
You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,257313
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',
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%2f3006073%2flet-vx-int-b0-r-fraccx-yn-2-mathrmdy-show-that-for-x%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
up vote
0
down vote
Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:
$$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$
Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$
You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,257313
add a comment |
up vote
0
down vote
Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:
$$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$
Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$
You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,257313
add a comment |
up vote
0
down vote
up vote
0
down vote
Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:
$$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$
Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$
You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,257313
Hint. The change of measures theorem states that if $T:mathrm{X} to mathrm{X}'$ is a homeomorphism and $mu$ is a measure on $mathrm{X}$ then, for the image measure of $mu$ via $T,$ denoted as $T(mu),$ the following relation holds:
$$intlimits_{mathrm{X}'} f circ T^{-1} dT(mu) = intlimits_{mathrm{X}} f dmu$$
Take $mathrm{X} = B(0; R) setminus {0}$ and $mathrm{X}' = (0, R) times mathbf{S}_{n-1}$ with $T:x mapsto (|x|, |x|^{-1} x).$
You may want to observe that if $sigma_{n-1}$ is surface measure on $mathbf{S}_{n-1}$ and $lambda_n$ the Lebuesgue measure on $mathbf{R}^n$ then $T(lambda_n)=lambda_1 otimes sigma_{n-1}.$
answered Nov 23 at 20:13
Will M.
2,257313
answered Nov 23 at 20:13
Will M.
2,257313
answered Nov 23 at 20:13
Will M.
2,257313
answered Nov 23 at 20:13
Will M.
2,257313
2,257313
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%2f3006073%2flet-vx-int-b0-r-fraccx-yn-2-mathrmdy-show-that-for-x%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
