Sturm–Liouville equation with finite number of eigenvalues?
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?
differential-equations eigenvalues-eigenvectors boundary-value-problem singularity sturm-liouville
asked yesterday
xiaohuamao
258110
|
show 1 more comment
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?
differential-equations eigenvalues-eigenvectors boundary-value-problem singularity sturm-liouville
asked yesterday
xiaohuamao
258110
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
@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.
– xiaohuamao
yesterday
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.
– DisintegratingByParts
23 hours ago
@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?
– xiaohuamao
23 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago
|
show 1 more comment
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?
differential-equations eigenvalues-eigenvectors boundary-value-problem singularity sturm-liouville
asked yesterday
xiaohuamao
258110
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?
differential-equations eigenvalues-eigenvectors boundary-value-problem singularity sturm-liouville
differential-equations eigenvalues-eigenvectors boundary-value-problem singularity sturm-liouville
asked yesterday
xiaohuamao
258110
asked yesterday
xiaohuamao
258110
edited yesterday
asked yesterday
xiaohuamao
258110
asked yesterday
xiaohuamao
258110
asked yesterday
xiaohuamao
258110
258110
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
@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.
– xiaohuamao
yesterday
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.
– DisintegratingByParts
23 hours ago
@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?
– xiaohuamao
23 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago
|
show 1 more comment
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
@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.
– xiaohuamao
yesterday
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.
– DisintegratingByParts
23 hours ago
@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?
– xiaohuamao
23 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
@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.
– xiaohuamao
yesterday
@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.
– xiaohuamao
yesterday
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.
– DisintegratingByParts
23 hours ago
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.
– DisintegratingByParts
23 hours ago
@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?
– xiaohuamao
23 hours ago
@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?
– xiaohuamao
23 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago
|
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Note that the standard Sturm–Liouville eigenvalue problem is linear in the eigenvalue. In your case that would be equivalent to also allowing imaginary $λ$.
– LutzL
yesterday
@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.
– xiaohuamao
yesterday
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.
– DisintegratingByParts
23 hours ago
@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?
– xiaohuamao
23 hours ago
$x=0$ is a singular point of your equation, which complicates any examination of the equation on $(-infty,infty)$.
– DisintegratingByParts
22 hours ago