Sturm–Liouville equation with finite number of eigenvalues?
$begingroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ in which $xin(-infty,0)$ and parameter $a>0$. Therefore, the SL coefficient functions $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$. It has a regular singularity $x=0$. This is from some physical modelling and we basically hope for something like homogeneous Dirichlet b.c.
It is solved by making the substitution $y(x)=e^{x/2}x^{-frac{1}{2}+sqrt{(a+frac{1}{2})^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(gamma-x)u'(x)-alpha u(x)=0,$$ in which $alpha=sqrt{(a+frac{1}{2})^2-lambda^2}-a+frac{1}{2}$ and $gamma=1+2sqrt{(a+frac{1}{2})^2-lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
However, seen from this condition for $alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which is seemingly contradicting with the infinite eigenspectrum that SL theory claims.
What is wrong here?
functional-analysis differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
$endgroup$
|
show 3 more comments
$begingroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ in which $xin(-infty,0)$ and parameter $a>0$. Therefore, the SL coefficient functions $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$. It has a regular singularity $x=0$. This is from some physical modelling and we basically hope for something like homogeneous Dirichlet b.c.
It is solved by making the substitution $y(x)=e^{x/2}x^{-frac{1}{2}+sqrt{(a+frac{1}{2})^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(gamma-x)u'(x)-alpha u(x)=0,$$ in which $alpha=sqrt{(a+frac{1}{2})^2-lambda^2}-a+frac{1}{2}$ and $gamma=1+2sqrt{(a+frac{1}{2})^2-lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
However, seen from this condition for $alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which is seemingly contradicting with the infinite eigenspectrum that SL theory claims.
What is wrong here?
functional-analysis differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
$endgroup$
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17
|
show 3 more comments
$begingroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ in which $xin(-infty,0)$ and parameter $a>0$. Therefore, the SL coefficient functions $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$. It has a regular singularity $x=0$. This is from some physical modelling and we basically hope for something like homogeneous Dirichlet b.c.
It is solved by making the substitution $y(x)=e^{x/2}x^{-frac{1}{2}+sqrt{(a+frac{1}{2})^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(gamma-x)u'(x)-alpha u(x)=0,$$ in which $alpha=sqrt{(a+frac{1}{2})^2-lambda^2}-a+frac{1}{2}$ and $gamma=1+2sqrt{(a+frac{1}{2})^2-lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
However, seen from this condition for $alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which is seemingly contradicting with the infinite eigenspectrum that SL theory claims.
What is wrong here?
functional-analysis differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
$endgroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ in which $xin(-infty,0)$ and parameter $a>0$. Therefore, the SL coefficient functions $p(x)=x^2$, $w(x)=1$, and $q(x)=-[(x/2+a)^2+a]$. It has a regular singularity $x=0$. This is from some physical modelling and we basically hope for something like homogeneous Dirichlet b.c.
It is solved by making the substitution $y(x)=e^{x/2}x^{-frac{1}{2}+sqrt{(a+frac{1}{2})^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation $$xu''(x)+(gamma-x)u'(x)-alpha u(x)=0,$$ in which $alpha=sqrt{(a+frac{1}{2})^2-lambda^2}-a+frac{1}{2}$ and $gamma=1+2sqrt{(a+frac{1}{2})^2-lambda^2}$. It has two (1st kind & 2nd kind) independent solutions.
Let's follow some ubiquitous argument when solving eigensystem related to confluent hypergeometric equation. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
However, seen from this condition for $alpha$, obviously we only have a bounded and finite sequence of eigenvalues, which is seemingly contradicting with the infinite eigenspectrum that SL theory claims.
What is wrong here?
functional-analysis differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
functional-analysis differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
edited Jan 7 at 6:46
xiaohuamao
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
258110
258110
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17
|
show 3 more comments
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17
|
show 3 more comments
0
active
oldest
votes
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%2f3062477%2fsturm-liouville-equation-with-finite-number-of-eigenvalues%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3062477%2fsturm-liouville-equation-with-finite-number-of-eigenvalues%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
$begingroup$
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
$endgroup$
– LutzL
Jan 5 at 8:25
$begingroup$
@LutzL But actually assuming imaginary $lambda$ doesn't really give any extra solution to the condition of $alpha$. In any case, I can just use $lambda$ instead and the question remains.
$endgroup$
– xiaohuamao
Jan 5 at 8:35
$begingroup$
If there are only a finite number of eigenfunctions in $L^2$, then there must be continuous spectrum for the operator, provided you have properly defined endpoint conditions that lead to a well-posed selfadjoint problem.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:02
$begingroup$
@DisintegratingByParts Thanks. Yes, I think it is self-adjoint for homogeneous Dirichlet b.c. But a confusing thing is what if we consider this equation on (−∞,+∞) with homogeneous Dirichlet b.c.? The solution shown with $e^{x/2}$ doesn't seem to allow this at +∞. I'm not sure if this is normal. Did I do anything wrong?
$endgroup$
– xiaohuamao
Jan 6 at 7:09
$begingroup$
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
$endgroup$
– DisintegratingByParts
Jan 6 at 7:17