Mixing
Possible spectra of singular Sturm-Liouville problems
$begingroup$
A Sturm–Liouville (SL) eigenvalue problem with separated boundary condition on $[a,b]$ $$(py')'-qy=-lambda^2wy$$ is regular if $p(x),w(x)>0$ and $p(x),p'(x),q(x),w(x)$ are continuous in the finite interval $[a,b]$. Otherwise, it becomes singular. Let's just take the homogeneous Dirichlet b.c. if necessary.
The regular case exhibits an infinite sequence of discrete eigenvalues and corresponding orthonormal eigenfunctions.
But for the singular case, is there any conclusion and condition on the possibilities of the spectrum? By skimming over this wiki page and other materials, I only vaguely know it's possible to have the following
- only an inifinite sequence of discrete eigenvalues
- only a continuous spectrum
- a finite sequence of discrete eigenvalues
- a discrete/continuous mixture
I'm confused by case 3 and case 4. The questions arise from another post that seems to have a finite sequence of discrete eigenvalues.
- Is only a finite sequence of discrete eigenvalues possible? Or case
3 always implies a complementary continuous spectrum, i.e., case 4? - In case 4 with a finite sequence of discrete eigenvalues, can the ranges (I mean the interval between the upper and lower bounds) of discrete and continuous spectra have overlap? Or always exclusive? If exclusive, could there be region(s) not covered by either of the two?
- how about case 4 with an infinite sequence of discrete eigenvalues? (optional)
Examples are also welcome.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
$endgroup$
add a comment |
$begingroup$
A Sturm–Liouville (SL) eigenvalue problem with separated boundary condition on $[a,b]$ $$(py')'-qy=-lambda^2wy$$ is regular if $p(x),w(x)>0$ and $p(x),p'(x),q(x),w(x)$ are continuous in the finite interval $[a,b]$. Otherwise, it becomes singular. Let's just take the homogeneous Dirichlet b.c. if necessary.
The regular case exhibits an infinite sequence of discrete eigenvalues and corresponding orthonormal eigenfunctions.
But for the singular case, is there any conclusion and condition on the possibilities of the spectrum? By skimming over this wiki page and other materials, I only vaguely know it's possible to have the following
- only an inifinite sequence of discrete eigenvalues
- only a continuous spectrum
- a finite sequence of discrete eigenvalues
- a discrete/continuous mixture
I'm confused by case 3 and case 4. The questions arise from another post that seems to have a finite sequence of discrete eigenvalues.
- Is only a finite sequence of discrete eigenvalues possible? Or case
3 always implies a complementary continuous spectrum, i.e., case 4? - In case 4 with a finite sequence of discrete eigenvalues, can the ranges (I mean the interval between the upper and lower bounds) of discrete and continuous spectra have overlap? Or always exclusive? If exclusive, could there be region(s) not covered by either of the two?
- how about case 4 with an infinite sequence of discrete eigenvalues? (optional)
Examples are also welcome.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
$endgroup$
add a comment |
$begingroup$
A Sturm–Liouville (SL) eigenvalue problem with separated boundary condition on $[a,b]$ $$(py')'-qy=-lambda^2wy$$ is regular if $p(x),w(x)>0$ and $p(x),p'(x),q(x),w(x)$ are continuous in the finite interval $[a,b]$. Otherwise, it becomes singular. Let's just take the homogeneous Dirichlet b.c. if necessary.
The regular case exhibits an infinite sequence of discrete eigenvalues and corresponding orthonormal eigenfunctions.
But for the singular case, is there any conclusion and condition on the possibilities of the spectrum? By skimming over this wiki page and other materials, I only vaguely know it's possible to have the following
- only an inifinite sequence of discrete eigenvalues
- only a continuous spectrum
- a finite sequence of discrete eigenvalues
- a discrete/continuous mixture
I'm confused by case 3 and case 4. The questions arise from another post that seems to have a finite sequence of discrete eigenvalues.
- Is only a finite sequence of discrete eigenvalues possible? Or case
3 always implies a complementary continuous spectrum, i.e., case 4? - In case 4 with a finite sequence of discrete eigenvalues, can the ranges (I mean the interval between the upper and lower bounds) of discrete and continuous spectra have overlap? Or always exclusive? If exclusive, could there be region(s) not covered by either of the two?
- how about case 4 with an infinite sequence of discrete eigenvalues? (optional)
Examples are also welcome.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
$endgroup$
A Sturm–Liouville (SL) eigenvalue problem with separated boundary condition on $[a,b]$ $$(py')'-qy=-lambda^2wy$$ is regular if $p(x),w(x)>0$ and $p(x),p'(x),q(x),w(x)$ are continuous in the finite interval $[a,b]$. Otherwise, it becomes singular. Let's just take the homogeneous Dirichlet b.c. if necessary.
The regular case exhibits an infinite sequence of discrete eigenvalues and corresponding orthonormal eigenfunctions.
But for the singular case, is there any conclusion and condition on the possibilities of the spectrum? By skimming over this wiki page and other materials, I only vaguely know it's possible to have the following
- only an inifinite sequence of discrete eigenvalues
- only a continuous spectrum
- a finite sequence of discrete eigenvalues
- a discrete/continuous mixture
I'm confused by case 3 and case 4. The questions arise from another post that seems to have a finite sequence of discrete eigenvalues.
- Is only a finite sequence of discrete eigenvalues possible? Or case
3 always implies a complementary continuous spectrum, i.e., case 4? - In case 4 with a finite sequence of discrete eigenvalues, can the ranges (I mean the interval between the upper and lower bounds) of discrete and continuous spectra have overlap? Or always exclusive? If exclusive, could there be region(s) not covered by either of the two?
- how about case 4 with an infinite sequence of discrete eigenvalues? (optional)
Examples are also welcome.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
edited Jan 15 at 1:45
xiaohuamao
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
asked Jan 14 at 22:37
xiaohuamaoxiaohuamao
304111
304111
add a comment |
add a comment |
1 Answer
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Your Case 3 seems inconsistent to me. If the only spectrum you have is discrete, then, because the eigenspaces cannot be more than 2-dimensional, it follows that a finite number of eigenvalues would not lead to a complete system; so in that case there would have to be continuous spectrum as well. Otherwise, the eigenstates would not form a complete system, which they do for well-posed Sturm-Liouvile operators.
The radial equation for the non-relativistic quantum Hydrogen isotope has an interesting spectrum where there is an infinite number of eigenvalues that cluster toward an energy that is an escape energy, and then a continuous spectrum of unbound states that start at the cluster point of the bound states. Expansions require discrete Fourier series types of expansions combined with Fourier integral types of expnsions.
There are Sturm-Liouville systems where there are singular continuous expansion components, meaning that the spectral density measure is a singular continuous measure, and not absolutely continuous with respect to Lebesgue measure. Barry Simon has constructed such cases based on physical cases where the potential is pathological.
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
add a comment |
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1 Answer
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$begingroup$
Your Case 3 seems inconsistent to me. If the only spectrum you have is discrete, then, because the eigenspaces cannot be more than 2-dimensional, it follows that a finite number of eigenvalues would not lead to a complete system; so in that case there would have to be continuous spectrum as well. Otherwise, the eigenstates would not form a complete system, which they do for well-posed Sturm-Liouvile operators.
The radial equation for the non-relativistic quantum Hydrogen isotope has an interesting spectrum where there is an infinite number of eigenvalues that cluster toward an energy that is an escape energy, and then a continuous spectrum of unbound states that start at the cluster point of the bound states. Expansions require discrete Fourier series types of expansions combined with Fourier integral types of expnsions.
There are Sturm-Liouville systems where there are singular continuous expansion components, meaning that the spectral density measure is a singular continuous measure, and not absolutely continuous with respect to Lebesgue measure. Barry Simon has constructed such cases based on physical cases where the potential is pathological.
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
add a comment |
$begingroup$
Your Case 3 seems inconsistent to me. If the only spectrum you have is discrete, then, because the eigenspaces cannot be more than 2-dimensional, it follows that a finite number of eigenvalues would not lead to a complete system; so in that case there would have to be continuous spectrum as well. Otherwise, the eigenstates would not form a complete system, which they do for well-posed Sturm-Liouvile operators.
The radial equation for the non-relativistic quantum Hydrogen isotope has an interesting spectrum where there is an infinite number of eigenvalues that cluster toward an energy that is an escape energy, and then a continuous spectrum of unbound states that start at the cluster point of the bound states. Expansions require discrete Fourier series types of expansions combined with Fourier integral types of expnsions.
There are Sturm-Liouville systems where there are singular continuous expansion components, meaning that the spectral density measure is a singular continuous measure, and not absolutely continuous with respect to Lebesgue measure. Barry Simon has constructed such cases based on physical cases where the potential is pathological.
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
add a comment |
$begingroup$
Your Case 3 seems inconsistent to me. If the only spectrum you have is discrete, then, because the eigenspaces cannot be more than 2-dimensional, it follows that a finite number of eigenvalues would not lead to a complete system; so in that case there would have to be continuous spectrum as well. Otherwise, the eigenstates would not form a complete system, which they do for well-posed Sturm-Liouvile operators.
The radial equation for the non-relativistic quantum Hydrogen isotope has an interesting spectrum where there is an infinite number of eigenvalues that cluster toward an energy that is an escape energy, and then a continuous spectrum of unbound states that start at the cluster point of the bound states. Expansions require discrete Fourier series types of expansions combined with Fourier integral types of expnsions.
There are Sturm-Liouville systems where there are singular continuous expansion components, meaning that the spectral density measure is a singular continuous measure, and not absolutely continuous with respect to Lebesgue measure. Barry Simon has constructed such cases based on physical cases where the potential is pathological.
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
$endgroup$
Your Case 3 seems inconsistent to me. If the only spectrum you have is discrete, then, because the eigenspaces cannot be more than 2-dimensional, it follows that a finite number of eigenvalues would not lead to a complete system; so in that case there would have to be continuous spectrum as well. Otherwise, the eigenstates would not form a complete system, which they do for well-posed Sturm-Liouvile operators.
The radial equation for the non-relativistic quantum Hydrogen isotope has an interesting spectrum where there is an infinite number of eigenvalues that cluster toward an energy that is an escape energy, and then a continuous spectrum of unbound states that start at the cluster point of the bound states. Expansions require discrete Fourier series types of expansions combined with Fourier integral types of expnsions.
There are Sturm-Liouville systems where there are singular continuous expansion components, meaning that the spectral density measure is a singular continuous measure, and not absolutely continuous with respect to Lebesgue measure. Barry Simon has constructed such cases based on physical cases where the potential is pathological.
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
answered Jan 15 at 0:17
DisintegratingByPartsDisintegratingByParts
59.4k42580
59.4k42580
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
add a comment |
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
$begingroup$
Thank you for the nice answer! Is there any guarantee that the continuous spectrum starts at the cluster point or the maxiaml eigenvalue of the discrete part? Or at least no overlap?
$endgroup$
– xiaohuamao
Jan 15 at 5:48
add a comment |
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