Laplace equation with the Robin's boundary problem
$begingroup$
$textbf{Problem}$ Let $Omega$ be an open, bounded and connected subset of $mathbb{R}^n$. Suppose that $partial Omega$ is $C^{infty}$. Consider an eigenvalue problem
begin{align*}
begin{cases}
-Delta u=lambda u & textrm{ in } ; Omega \
frac{partial u}{partial nu}=-u & textrm{ on } partial Omega
end{cases}
end{align*}
Define a bilinear operater $(cdot,cdot)_{H^1}$ by
begin{align*}
(u,v)_{H^1}:=int_{Omega} nabla u cdot nabla v ;dx + int_{partial Omega} uv ; dsigma
end{align*}
Show that there exists a constant $theta>0$ independent of $u,v$ such that
begin{align*}
(u,u)_{H^1} geq theta Vert u Vert _{H^1(Omega)}^2
end{align*}
$textbf{Attempt}$
begin{align*}
(u,u)_{H^1}&=int_{Omega} nabla u cdot nabla u ;dx + int_{partial Omega} u^2 ; dsigma \
&=int_{Omega} nabla cdot(unabla u)-uDelta u ; dx +int_{partial Omega} u^2 ; dsigma \
&=int_{partial Omega} u frac{partial u}{partial nu} ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma \
&=-int_{partial Omega} u^2 ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma\
&=lambda Vert u Vert _{L^2(Omega)}^2
end{align*}
I don't know how to get $lambda Vert u Vert_{L^2(Omega)}^2 geq theta Vert u Vert_{H^1(Omega)}^2$...
Any help is appreciated..
Thank you!
analysis pde spectral-theory boundary-value-problem elliptic-equations
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
$endgroup$
add a comment |
$begingroup$
$textbf{Problem}$ Let $Omega$ be an open, bounded and connected subset of $mathbb{R}^n$. Suppose that $partial Omega$ is $C^{infty}$. Consider an eigenvalue problem
begin{align*}
begin{cases}
-Delta u=lambda u & textrm{ in } ; Omega \
frac{partial u}{partial nu}=-u & textrm{ on } partial Omega
end{cases}
end{align*}
Define a bilinear operater $(cdot,cdot)_{H^1}$ by
begin{align*}
(u,v)_{H^1}:=int_{Omega} nabla u cdot nabla v ;dx + int_{partial Omega} uv ; dsigma
end{align*}
Show that there exists a constant $theta>0$ independent of $u,v$ such that
begin{align*}
(u,u)_{H^1} geq theta Vert u Vert _{H^1(Omega)}^2
end{align*}
$textbf{Attempt}$
begin{align*}
(u,u)_{H^1}&=int_{Omega} nabla u cdot nabla u ;dx + int_{partial Omega} u^2 ; dsigma \
&=int_{Omega} nabla cdot(unabla u)-uDelta u ; dx +int_{partial Omega} u^2 ; dsigma \
&=int_{partial Omega} u frac{partial u}{partial nu} ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma \
&=-int_{partial Omega} u^2 ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma\
&=lambda Vert u Vert _{L^2(Omega)}^2
end{align*}
I don't know how to get $lambda Vert u Vert_{L^2(Omega)}^2 geq theta Vert u Vert_{H^1(Omega)}^2$...
Any help is appreciated..
Thank you!
analysis pde spectral-theory boundary-value-problem elliptic-equations
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
$endgroup$
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22
add a comment |
$begingroup$
$textbf{Problem}$ Let $Omega$ be an open, bounded and connected subset of $mathbb{R}^n$. Suppose that $partial Omega$ is $C^{infty}$. Consider an eigenvalue problem
begin{align*}
begin{cases}
-Delta u=lambda u & textrm{ in } ; Omega \
frac{partial u}{partial nu}=-u & textrm{ on } partial Omega
end{cases}
end{align*}
Define a bilinear operater $(cdot,cdot)_{H^1}$ by
begin{align*}
(u,v)_{H^1}:=int_{Omega} nabla u cdot nabla v ;dx + int_{partial Omega} uv ; dsigma
end{align*}
Show that there exists a constant $theta>0$ independent of $u,v$ such that
begin{align*}
(u,u)_{H^1} geq theta Vert u Vert _{H^1(Omega)}^2
end{align*}
$textbf{Attempt}$
begin{align*}
(u,u)_{H^1}&=int_{Omega} nabla u cdot nabla u ;dx + int_{partial Omega} u^2 ; dsigma \
&=int_{Omega} nabla cdot(unabla u)-uDelta u ; dx +int_{partial Omega} u^2 ; dsigma \
&=int_{partial Omega} u frac{partial u}{partial nu} ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma \
&=-int_{partial Omega} u^2 ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma\
&=lambda Vert u Vert _{L^2(Omega)}^2
end{align*}
I don't know how to get $lambda Vert u Vert_{L^2(Omega)}^2 geq theta Vert u Vert_{H^1(Omega)}^2$...
Any help is appreciated..
Thank you!
analysis pde spectral-theory boundary-value-problem elliptic-equations
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
$endgroup$
$textbf{Problem}$ Let $Omega$ be an open, bounded and connected subset of $mathbb{R}^n$. Suppose that $partial Omega$ is $C^{infty}$. Consider an eigenvalue problem
begin{align*}
begin{cases}
-Delta u=lambda u & textrm{ in } ; Omega \
frac{partial u}{partial nu}=-u & textrm{ on } partial Omega
end{cases}
end{align*}
Define a bilinear operater $(cdot,cdot)_{H^1}$ by
begin{align*}
(u,v)_{H^1}:=int_{Omega} nabla u cdot nabla v ;dx + int_{partial Omega} uv ; dsigma
end{align*}
Show that there exists a constant $theta>0$ independent of $u,v$ such that
begin{align*}
(u,u)_{H^1} geq theta Vert u Vert _{H^1(Omega)}^2
end{align*}
$textbf{Attempt}$
begin{align*}
(u,u)_{H^1}&=int_{Omega} nabla u cdot nabla u ;dx + int_{partial Omega} u^2 ; dsigma \
&=int_{Omega} nabla cdot(unabla u)-uDelta u ; dx +int_{partial Omega} u^2 ; dsigma \
&=int_{partial Omega} u frac{partial u}{partial nu} ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma \
&=-int_{partial Omega} u^2 ; dsigma +int_{Omega} lambda u^2 dx +int_{partial Omega} u^2 ; dsigma\
&=lambda Vert u Vert _{L^2(Omega)}^2
end{align*}
I don't know how to get $lambda Vert u Vert_{L^2(Omega)}^2 geq theta Vert u Vert_{H^1(Omega)}^2$...
Any help is appreciated..
Thank you!
analysis pde spectral-theory boundary-value-problem elliptic-equations
analysis pde spectral-theory boundary-value-problem elliptic-equations
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
asked Dec 10 '18 at 13:56
w.sdkaw.sdka
36919
36919
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22
add a comment |
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is a standard proof by contradiction.
Assume that the inequality does not hold. Then for all $n$ there is $u_nin H^1(Omega)$
such that
$$
|nabla u_n|_{L^2(Omega)}^2 + |u_n|_{L^2(partialOmega)}^2 < n |u_n|_{H^1(Omega)}^2.
$$
This implies $u_nne0$, and we can assume w.l.o.g. $|u_n|_{H^1(Omega)}=1$. Then $nabla u_nto0$ in $L^2(Omega)$ and $uto0$ in $L^2(partialOmega)$ follows immediately.
Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_nrightharpoonup u$ in $H^1(Omega)$. Then $nabla u=0$, and $u$ is a constant. Since $|u|_{L^2(partialOmega)}=0$, it follows $u=0$.
By compact embedding, $u_nto0$ in $L^2(Omega)$. This implies $uto0$ strongly in $H^1(Omega)$, which is a contradiction to $|u_n|_{H^1}=1$.
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
$endgroup$
add a comment |
$begingroup$
If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}getheta int_Omega |nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line,
$$
(u,u)_{H^1}=frac12int_Omega nabla u cdot nabla u dx + frac12 int_{partialOmega} |u|^2 dsigma + fraclambda2 int_Omega u^2 dx.
$$
Dropping the middle term and using $theta=min(frac12,fraclambda 2)$ you get what you want.
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
$endgroup$
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
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%2f3033943%2flaplace-equation-with-the-robins-boundary-problem%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is a standard proof by contradiction.
Assume that the inequality does not hold. Then for all $n$ there is $u_nin H^1(Omega)$
such that
$$
|nabla u_n|_{L^2(Omega)}^2 + |u_n|_{L^2(partialOmega)}^2 < n |u_n|_{H^1(Omega)}^2.
$$
This implies $u_nne0$, and we can assume w.l.o.g. $|u_n|_{H^1(Omega)}=1$. Then $nabla u_nto0$ in $L^2(Omega)$ and $uto0$ in $L^2(partialOmega)$ follows immediately.
Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_nrightharpoonup u$ in $H^1(Omega)$. Then $nabla u=0$, and $u$ is a constant. Since $|u|_{L^2(partialOmega)}=0$, it follows $u=0$.
By compact embedding, $u_nto0$ in $L^2(Omega)$. This implies $uto0$ strongly in $H^1(Omega)$, which is a contradiction to $|u_n|_{H^1}=1$.
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
$endgroup$
add a comment |
$begingroup$
Here is a standard proof by contradiction.
Assume that the inequality does not hold. Then for all $n$ there is $u_nin H^1(Omega)$
such that
$$
|nabla u_n|_{L^2(Omega)}^2 + |u_n|_{L^2(partialOmega)}^2 < n |u_n|_{H^1(Omega)}^2.
$$
This implies $u_nne0$, and we can assume w.l.o.g. $|u_n|_{H^1(Omega)}=1$. Then $nabla u_nto0$ in $L^2(Omega)$ and $uto0$ in $L^2(partialOmega)$ follows immediately.
Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_nrightharpoonup u$ in $H^1(Omega)$. Then $nabla u=0$, and $u$ is a constant. Since $|u|_{L^2(partialOmega)}=0$, it follows $u=0$.
By compact embedding, $u_nto0$ in $L^2(Omega)$. This implies $uto0$ strongly in $H^1(Omega)$, which is a contradiction to $|u_n|_{H^1}=1$.
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
$endgroup$
add a comment |
$begingroup$
Here is a standard proof by contradiction.
Assume that the inequality does not hold. Then for all $n$ there is $u_nin H^1(Omega)$
such that
$$
|nabla u_n|_{L^2(Omega)}^2 + |u_n|_{L^2(partialOmega)}^2 < n |u_n|_{H^1(Omega)}^2.
$$
This implies $u_nne0$, and we can assume w.l.o.g. $|u_n|_{H^1(Omega)}=1$. Then $nabla u_nto0$ in $L^2(Omega)$ and $uto0$ in $L^2(partialOmega)$ follows immediately.
Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_nrightharpoonup u$ in $H^1(Omega)$. Then $nabla u=0$, and $u$ is a constant. Since $|u|_{L^2(partialOmega)}=0$, it follows $u=0$.
By compact embedding, $u_nto0$ in $L^2(Omega)$. This implies $uto0$ strongly in $H^1(Omega)$, which is a contradiction to $|u_n|_{H^1}=1$.
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
$endgroup$
Here is a standard proof by contradiction.
Assume that the inequality does not hold. Then for all $n$ there is $u_nin H^1(Omega)$
such that
$$
|nabla u_n|_{L^2(Omega)}^2 + |u_n|_{L^2(partialOmega)}^2 < n |u_n|_{H^1(Omega)}^2.
$$
This implies $u_nne0$, and we can assume w.l.o.g. $|u_n|_{H^1(Omega)}=1$. Then $nabla u_nto0$ in $L^2(Omega)$ and $uto0$ in $L^2(partialOmega)$ follows immediately.
Since $(u_n)$ is bounded in $H^1$, after extracting a subsequence (denoted the same for simplicity), we have $u_nrightharpoonup u$ in $H^1(Omega)$. Then $nabla u=0$, and $u$ is a constant. Since $|u|_{L^2(partialOmega)}=0$, it follows $u=0$.
By compact embedding, $u_nto0$ in $L^2(Omega)$. This implies $uto0$ strongly in $H^1(Omega)$, which is a contradiction to $|u_n|_{H^1}=1$.
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
answered Dec 10 '18 at 14:30
dawdaw
24.1k1544
24.1k1544
add a comment |
add a comment |
$begingroup$
If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}getheta int_Omega |nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line,
$$
(u,u)_{H^1}=frac12int_Omega nabla u cdot nabla u dx + frac12 int_{partialOmega} |u|^2 dsigma + fraclambda2 int_Omega u^2 dx.
$$
Dropping the middle term and using $theta=min(frac12,fraclambda 2)$ you get what you want.
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
$endgroup$
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
add a comment |
$begingroup$
If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}getheta int_Omega |nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line,
$$
(u,u)_{H^1}=frac12int_Omega nabla u cdot nabla u dx + frac12 int_{partialOmega} |u|^2 dsigma + fraclambda2 int_Omega u^2 dx.
$$
Dropping the middle term and using $theta=min(frac12,fraclambda 2)$ you get what you want.
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
$endgroup$
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
add a comment |
$begingroup$
If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}getheta int_Omega |nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line,
$$
(u,u)_{H^1}=frac12int_Omega nabla u cdot nabla u dx + frac12 int_{partialOmega} |u|^2 dsigma + fraclambda2 int_Omega u^2 dx.
$$
Dropping the middle term and using $theta=min(frac12,fraclambda 2)$ you get what you want.
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
$endgroup$
If you want to show that for $u$ that are solutions of the eigenvalue problem,$$(u,u)_{H^1}getheta int_Omega |nabla u|^2 + |u|^2 dx,$$ you have already done all the hard work. From your equalities, using the first and last line,
$$
(u,u)_{H^1}=frac12int_Omega nabla u cdot nabla u dx + frac12 int_{partialOmega} |u|^2 dsigma + fraclambda2 int_Omega u^2 dx.
$$
Dropping the middle term and using $theta=min(frac12,fraclambda 2)$ you get what you want.
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
answered Dec 10 '18 at 14:12
KusmaKusma
3,7181319
3,7181319
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
add a comment |
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
Is $lambda$ always positive?
$endgroup$
– w.sdka
Dec 10 '18 at 14:15
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
$begingroup$
You have $lambda=frac{(u,u)_{H_1}}{|u|_{L^2}^2}$ (and you assume $u$ nonzero as it is an eigenfunction). That means clearly $lambdage 0$. If $lambda=0$, you get that $int_Omega |nabla u|^2 dx+int_{partialOmega} |u|^2 dsigma=0$ so $u$ is constant, and zero on the boundary, so zero everywhere. So $lambda>0$.
$endgroup$
– Kusma
Dec 10 '18 at 14:29
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%2f3033943%2flaplace-equation-with-the-robins-boundary-problem%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
$begingroup$
What is the connection between the taks 'Show that...' and the eigenvalue problem? As it is stated now, there is no connection, and it looks like the inequality has to be proven for all $uin H^1$.
$endgroup$
– daw
Dec 10 '18 at 14:19
$begingroup$
If I prove the problem, then I'll get every eigenvalue is a positive real number.
$endgroup$
– w.sdka
Dec 10 '18 at 14:22