Weird eigensystem of ODE with regular singularity
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Consider the eigenvalue problem of the following 2nd-order ODE $$(x/2+a)^2y(x)-xy'(x)-x^2y''(x)=lambda^2y(x),$$ in which $yin(-infty,+infty)$ and parameter $a>0$. It has a regular singularity $x=0$ and an irregular singularity at $infty$. This ODE is from some physical modelling and we hence basically hope for something like Dirichlet b.c. at $pminfty$.
It is solved by making the substitution $y(x)=e^{x/2}x^{sqrt{a^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation of $u(x)$ that has two (1st kind & 2nd kind) independent solutions. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial and eigenvalue $lambda$ is attained.
However, it's obvious that $y(x)$ diverges at $+infty$ because of $e^{x/2}$, which doesn't fulfill the b.c. we wanted. But $y(0)=0$ is automatically satisfied, which seems that we can instead use $Y(x)=y(x)theta(-x)$ as the eigenfunction.
My question is whether this $C^0$ continuous solution $Y(x)$ is a valid eigenfunction. Or I missed any other solution? And is it normal to have such an eigenfunction 'cut off' by a singularity?
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions singularity
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
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add a comment |
$begingroup$
Consider the eigenvalue problem of the following 2nd-order ODE $$(x/2+a)^2y(x)-xy'(x)-x^2y''(x)=lambda^2y(x),$$ in which $yin(-infty,+infty)$ and parameter $a>0$. It has a regular singularity $x=0$ and an irregular singularity at $infty$. This ODE is from some physical modelling and we hence basically hope for something like Dirichlet b.c. at $pminfty$.
It is solved by making the substitution $y(x)=e^{x/2}x^{sqrt{a^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation of $u(x)$ that has two (1st kind & 2nd kind) independent solutions. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial and eigenvalue $lambda$ is attained.
However, it's obvious that $y(x)$ diverges at $+infty$ because of $e^{x/2}$, which doesn't fulfill the b.c. we wanted. But $y(0)=0$ is automatically satisfied, which seems that we can instead use $Y(x)=y(x)theta(-x)$ as the eigenfunction.
My question is whether this $C^0$ continuous solution $Y(x)$ is a valid eigenfunction. Or I missed any other solution? And is it normal to have such an eigenfunction 'cut off' by a singularity?
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions singularity
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
$endgroup$
$begingroup$
So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
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– LutzL
Dec 25 '18 at 14:57
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40
add a comment |
$begingroup$
Consider the eigenvalue problem of the following 2nd-order ODE $$(x/2+a)^2y(x)-xy'(x)-x^2y''(x)=lambda^2y(x),$$ in which $yin(-infty,+infty)$ and parameter $a>0$. It has a regular singularity $x=0$ and an irregular singularity at $infty$. This ODE is from some physical modelling and we hence basically hope for something like Dirichlet b.c. at $pminfty$.
It is solved by making the substitution $y(x)=e^{x/2}x^{sqrt{a^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation of $u(x)$ that has two (1st kind & 2nd kind) independent solutions. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial and eigenvalue $lambda$ is attained.
However, it's obvious that $y(x)$ diverges at $+infty$ because of $e^{x/2}$, which doesn't fulfill the b.c. we wanted. But $y(0)=0$ is automatically satisfied, which seems that we can instead use $Y(x)=y(x)theta(-x)$ as the eigenfunction.
My question is whether this $C^0$ continuous solution $Y(x)$ is a valid eigenfunction. Or I missed any other solution? And is it normal to have such an eigenfunction 'cut off' by a singularity?
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions singularity
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
$endgroup$
Consider the eigenvalue problem of the following 2nd-order ODE $$(x/2+a)^2y(x)-xy'(x)-x^2y''(x)=lambda^2y(x),$$ in which $yin(-infty,+infty)$ and parameter $a>0$. It has a regular singularity $x=0$ and an irregular singularity at $infty$. This ODE is from some physical modelling and we hence basically hope for something like Dirichlet b.c. at $pminfty$.
It is solved by making the substitution $y(x)=e^{x/2}x^{sqrt{a^2-lambda^2}}u(x)$, leading directly to a standard confluent hypergeometric equation of $u(x)$ that has two (1st kind & 2nd kind) independent solutions. Requiring nondivergence at $x=0$, the 2nd kind is dropped. Requiring nondivergence at $infty$, the 1st kind is reduced to a polynomial and eigenvalue $lambda$ is attained.
However, it's obvious that $y(x)$ diverges at $+infty$ because of $e^{x/2}$, which doesn't fulfill the b.c. we wanted. But $y(0)=0$ is automatically satisfied, which seems that we can instead use $Y(x)=y(x)theta(-x)$ as the eigenfunction.
My question is whether this $C^0$ continuous solution $Y(x)$ is a valid eigenfunction. Or I missed any other solution? And is it normal to have such an eigenfunction 'cut off' by a singularity?
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions singularity
ordinary-differential-equations eigenvalues-eigenvectors boundary-value-problem eigenfunctions singularity
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
edited Dec 25 '18 at 19:08
xiaohuamao
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
asked Dec 25 '18 at 4:46
xiaohuamaoxiaohuamao
319111
319111
$begingroup$
So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
$endgroup$
– LutzL
Dec 25 '18 at 14:57
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40
add a comment |
$begingroup$
So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
$endgroup$
– LutzL
Dec 25 '18 at 14:57
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40
$begingroup$
So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
$endgroup$
– LutzL
Dec 25 '18 at 14:57
$begingroup$
So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
$endgroup$
– LutzL
Dec 25 '18 at 14:57
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40
add a comment |
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So for the condition at zero you need $λ<a$ and to get polynomial solutions for $u$ the degree $d$ has to satisfy $a-1/2=sqrt{a^2-λ^2}+d$. This does severely restrict the number of eigensolutions of that type. Per general intuition, there should be an infinite sequence of eigenvalues.
$endgroup$
– LutzL
Dec 25 '18 at 14:57
$begingroup$
@LutzL Yes, you're right. This finite sequence of eigenvalues also surprises me.
$endgroup$
– xiaohuamao
Dec 25 '18 at 16:40