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Need help-eigenvalue problem

Hey!!! :smile:

Knowing that:
"The eigenvalue problem Ly=(py)+qy,axb 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))p(a)W(u(a),v^*(a))=p(b)W(u(b),v^*(b)),where WW is the wronskian."




I have to show that the eigenvalue problem y+λy=0y''+λy=0 , with boundary conditions y(0)=0,y(0)=y(1)y(0)=0, y'(0)=y'(1) is not a Sturm -Liouville problem.


This is what I've done so far:


Let u,vu, v^* solutions of the eigenvalue problem y+λy=0y''+λy=0 , then:
u(0)=0,u(0)=u(1)u(0)=0, u'(0)=u'(1) and v(0)=0,v(0)=v(1)v^*(0)=0, v^{*'}(0)=v^{*'}(1).


W(u(0),v(0))=u(0)v(0)u(0)v(0)=0W(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)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?
Reply 1
Original post by mathmari
Hey!!! :smile:

W(u(1),v(1))=u(1)v(1)u(1)v(1)=u(1)v(0)u(0)v(1)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?

Could you not find u,vu,v and show it that way? Seems like cheating, but I don't see why you shouldn't do it.
Reply 2
Original post by Smaug123
Could you not find u,vu,v and show it that way? Seems like cheating, but I don't see why you shouldn't do it.


I have found the eigenfunctions of the eigenvalue problem.They are yn(x)=sin(2nπx) y_n(x)=\sin(2 n \pi x) .If we suppose that u u and v v^{*} are solutions of the eigenvalue problem,do I have to take two different n n at the eigenfunction to find u u and v v ? I got stuck right now.. :confused:
Reply 3
Original post by mathmari
I have found the eigenfunctions of the eigenvalue problem.They are yn(x)=sin(2nπx) y_n(x)=\sin(2 n \pi x) .If we suppose that u u and v v^{*} are solutions of the eigenvalue problem,do I have to take two different n n at the eigenfunction to find u u and v v ? I got stuck right now.. :confused:

Hmm, now I'm confused. Because it's not a Sturm-Liouville problem, we aren't guaranteed to have a complete set of eigenfunctions, so we can't proceed by expressing u and v as a sum of eigenfunctions. I'm stumped, I'm afraid :frown:
Reply 4
Original post by Smaug123
Hmm, now I'm confused. Because it's not a Sturm-Liouville problem, we aren't guaranteed to have a complete set of eigenfunctions, so we can't proceed by expressing u and v as a sum of eigenfunctions. I'm stumped, I'm afraid :frown:


Oh ok... How did you mean it that I could find uu and vv? :confused:
Reply 5
Original post by mathmari
Oh ok... How did you mean it that I could find uu and vv? :confused:

In hindsight, I was assuming that any u and v could be expanded in terms of eigenfunctions. That's true for Sturm-Liouville problems, but not necessarily for non-SL ones.
Reply 6
Original post by Smaug123
In hindsight, I was assuming that any u and v could be expanded in terms of eigenfunctions. That's true for Sturm-Liouville problems, but not necessarily for non-SL ones.


Aha! Ok! Do you have any other idea what I could do?

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