Non-trivial entire harmonic function in plane
$begingroup$
Is it possible to explicitly find a harmonic function $u in C^2(mathbb{R}^2)$ such that
begin{equation}tag{$dagger$}label{eq:dag}
u(x,1) = u(x,-1) = 0
end{equation}
for all $x in mathbb{R}$? Is it possible for $u$ to be a polynomial?
My intuition says that $u$ cannot be a polynomial, but I have been unable to rigorously prove this. In fact, I haven't been able to come up with a function satisfying eqref{eq:dag}.
real-analysis pde harmonic-functions
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
add a comment |
$begingroup$
Is it possible to explicitly find a harmonic function $u in C^2(mathbb{R}^2)$ such that
begin{equation}tag{$dagger$}label{eq:dag}
u(x,1) = u(x,-1) = 0
end{equation}
for all $x in mathbb{R}$? Is it possible for $u$ to be a polynomial?
My intuition says that $u$ cannot be a polynomial, but I have been unable to rigorously prove this. In fact, I haven't been able to come up with a function satisfying eqref{eq:dag}.
real-analysis pde harmonic-functions
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
add a comment |
$begingroup$
Is it possible to explicitly find a harmonic function $u in C^2(mathbb{R}^2)$ such that
begin{equation}tag{$dagger$}label{eq:dag}
u(x,1) = u(x,-1) = 0
end{equation}
for all $x in mathbb{R}$? Is it possible for $u$ to be a polynomial?
My intuition says that $u$ cannot be a polynomial, but I have been unable to rigorously prove this. In fact, I haven't been able to come up with a function satisfying eqref{eq:dag}.
real-analysis pde harmonic-functions
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
$endgroup$
Is it possible to explicitly find a harmonic function $u in C^2(mathbb{R}^2)$ such that
begin{equation}tag{$dagger$}label{eq:dag}
u(x,1) = u(x,-1) = 0
end{equation}
for all $x in mathbb{R}$? Is it possible for $u$ to be a polynomial?
My intuition says that $u$ cannot be a polynomial, but I have been unable to rigorously prove this. In fact, I haven't been able to come up with a function satisfying eqref{eq:dag}.
real-analysis pde harmonic-functions
real-analysis pde harmonic-functions
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
asked Dec 6 '18 at 3:38
ImNotThereRightNow_ImNotThereRightNow_
998
998
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
One such function is
$$
u(x,y)=Rebigl(-i,e^{pi(x+iy)}bigr)=e^{pi,x}sin(pi,y).
$$
If $u$ is such a function, then $u=Re(f(z))$ for some entire function $f$, with $f(x+i)=f(x-i)=0$. Schwarz's reflection principle implies that $f$ must be periodic of period $2,i$, and hence it cannot be a polynomial.
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
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%2f3028002%2fnon-trivial-entire-harmonic-function-in-plane%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
$begingroup$
One such function is
$$
u(x,y)=Rebigl(-i,e^{pi(x+iy)}bigr)=e^{pi,x}sin(pi,y).
$$
If $u$ is such a function, then $u=Re(f(z))$ for some entire function $f$, with $f(x+i)=f(x-i)=0$. Schwarz's reflection principle implies that $f$ must be periodic of period $2,i$, and hence it cannot be a polynomial.
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
One such function is
$$
u(x,y)=Rebigl(-i,e^{pi(x+iy)}bigr)=e^{pi,x}sin(pi,y).
$$
If $u$ is such a function, then $u=Re(f(z))$ for some entire function $f$, with $f(x+i)=f(x-i)=0$. Schwarz's reflection principle implies that $f$ must be periodic of period $2,i$, and hence it cannot be a polynomial.
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
add a comment |
$begingroup$
One such function is
$$
u(x,y)=Rebigl(-i,e^{pi(x+iy)}bigr)=e^{pi,x}sin(pi,y).
$$
If $u$ is such a function, then $u=Re(f(z))$ for some entire function $f$, with $f(x+i)=f(x-i)=0$. Schwarz's reflection principle implies that $f$ must be periodic of period $2,i$, and hence it cannot be a polynomial.
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
$endgroup$
One such function is
$$
u(x,y)=Rebigl(-i,e^{pi(x+iy)}bigr)=e^{pi,x}sin(pi,y).
$$
If $u$ is such a function, then $u=Re(f(z))$ for some entire function $f$, with $f(x+i)=f(x-i)=0$. Schwarz's reflection principle implies that $f$ must be periodic of period $2,i$, and hence it cannot be a polynomial.
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
answered Dec 6 '18 at 10:29
Julián AguirreJulián Aguirre
68.1k24094
68.1k24094
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%2f3028002%2fnon-trivial-entire-harmonic-function-in-plane%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
