Limit circle/point of ODE with finite eigenvalues
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Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ defined in $(-infty,0]$ or $[0,infty)$ or $(-infty,+infty)$ 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 $x=0$. We basically hope for 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 the Kummer's 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 then follow the standard argument.
Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
Seen from this condition for $alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum as suggested in the comments.
Question
Is it possible to know the limit circle/point classification of this ODE near $0,pminfty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
$endgroup$
|
show 10 more comments
$begingroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ defined in $(-infty,0]$ or $[0,infty)$ or $(-infty,+infty)$ 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 $x=0$. We basically hope for 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 the Kummer's 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 then follow the standard argument.
Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
Seen from this condition for $alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum as suggested in the comments.
Question
Is it possible to know the limit circle/point classification of this ODE near $0,pminfty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
$endgroup$
$begingroup$
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 10 more comments
$begingroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ defined in $(-infty,0]$ or $[0,infty)$ or $(-infty,+infty)$ 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 $x=0$. We basically hope for 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 the Kummer's 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 then follow the standard argument.
Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
Seen from this condition for $alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum as suggested in the comments.
Question
Is it possible to know the limit circle/point classification of this ODE near $0,pminfty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
$endgroup$
Consider the following Sturm–Liouville (SL) eigenvalue problem $$(x^2y')'-[(x/2+a)^2+a]y=-lambda^2y(x),$$ defined in $(-infty,0]$ or $[0,infty)$ or $(-infty,+infty)$ 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 $x=0$. We basically hope for 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 the Kummer's 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 then follow the standard argument.
Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at infinity, the 1st kind is reduced to a polynomial when $-alpha$ is a non-negative integer and eigenvalue $lambda^2$ is attained.
Seen from this condition for $alpha$, we only have a bounded and finite sequence of eigenvalues. Therefore, one probably also expects an additional continuous spectrum as suggested in the comments.
Question
Is it possible to know the limit circle/point classification of this ODE near $0,pminfty$? Is the above solution useful for this purpose? How should one proceed? I also found page 13 of this paper has a limit point/circle classification of the Kummer equation I used. Not sure if relevant.
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
functional-analysis ordinary-differential-equations eigenvalues-eigenvectors spectral-theory sturm-liouville
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
edited Jan 15 at 19:01
xiaohuamao
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
asked Jan 5 at 7:30
xiaohuamaoxiaohuamao
278110
278110
$begingroup$
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 10 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 10 more comments
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$begingroup$
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