- #1
mathmari
Gold Member
MHB
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Hey! ![Eek! :eek: :eek:](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
I got stuck at the following exercise.
Knowing that:
$"$The eigenvalue problem $Ly=(py')'+qy, a \leq x \leq b$ is a Sturm-Liouville problem when it satisfies the boundary conditions:
$$p(a)W(u(a),v^*(a))=p(b)W(u(b),v^*(b))"$$I have to show that the eigenvalue problem $y''+λy=0$, with boundary conditions $y(0)=0, y'(0)=y'(1)$ is not a Sturm -Liouville problem.
This is what I've done so far:
Let $u, v^*$ solutions of the eigenvalue problem $y''+λy=0$, then:
$u(0)=0, u'(0)=u'(1)$ and $v^*(0)=0, v^{*'}(0)=v^{*'}(1)$.
$W(u(0),v^*(0))=u(0)v^{*'}(0)-u'(0)v^*(0)=0$
$W(u(1),v^*(1))=u(1)v^{*'}(1)-u'(1)v^*(1)=u(1) v^{*'}(0)-u'(0)v^*(1)$
How can I continue? How can I show that this is not equal to $0$?
I got stuck at the following exercise.
Knowing that:
$"$The eigenvalue problem $Ly=(py')'+qy, a \leq x \leq b$ is a Sturm-Liouville problem when it satisfies the boundary conditions:
$$p(a)W(u(a),v^*(a))=p(b)W(u(b),v^*(b))"$$I have to show that the eigenvalue problem $y''+λy=0$, with boundary conditions $y(0)=0, y'(0)=y'(1)$ is not a Sturm -Liouville problem.
This is what I've done so far:
Let $u, v^*$ solutions of the eigenvalue problem $y''+λy=0$, then:
$u(0)=0, u'(0)=u'(1)$ and $v^*(0)=0, v^{*'}(0)=v^{*'}(1)$.
$W(u(0),v^*(0))=u(0)v^{*'}(0)-u'(0)v^*(0)=0$
$W(u(1),v^*(1))=u(1)v^{*'}(1)-u'(1)v^*(1)=u(1) v^{*'}(0)-u'(0)v^*(1)$
How can I continue? How can I show that this is not equal to $0$?