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      f''(x) + A y(x) = 0
    Boundary conditions:  y(0) = y(pi) and  y'(0) = y'(pi)

    Considering the case where A < 0, is it true to say there are no eigenvalues as there is no defining eigenfunction which is not trivial. I used A instead of lambda because I didn't know how to put that symbol in.

    I'm a bit unsure if I have done the evaluation of the boundary conditions correctly. Usually one of my constants cancel out but it wasn't the case this time so I compared coefficients and said that the constants must be zero therefore only resulting in a trivial solution. I just want to make sure that is actually correct. Hope that it's clear what I'm asking and what I've done. If not, ask and I'll try and explain myself a bit better.
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    Your boundary conditions don't look standard. Are you sure they're correct?
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    (Original post by mathprof)
    Your boundary conditions don't look standard. Are you sure they're correct?
    I just saw it as a question on a past exam paper and the boundary conditions were given like that. Also on that note, do you think you could explain what you mean by standard boundary conditions for a sturm-liouville system. I remember briefly reading it in my notes but I didn't really understand so how can you look at the boundary conditions and see they are not standard? And if they are not standard, how do I go about solving the question?
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    By standard (my terminology), I mean boundary conditions of the form

    

A_1y(a)+A_2y'(a)=0

B_1y(b)+B_2y'(b)=0

    Since you saw it somewhere, I'll assume your boundary conditions are correct.

    The d.e. your considering y''+\lambda y=0 has auxillary equation r^2+\lambda =0. Hence r=\pm \sqrt{ -\lambda} . There are 3 cases to consider.

    Case I. \lambda &lt;0 . Put \lambda=-a^2 .

    Here y= c_1e^{ax}+c_2e^{-ax}.

    After applying the boundary conditions, I get a=0 which contradicts our assumption on \lambda. Hence no solution in this case. This is odd!

    Case II. \lambda =0 .

    Here y= c_1x+c_2. Boundary conditions imply here that y= c_2. That is, the solution is not unique. Again, very odd!

    I'm not going to bother with case III because it's clear that this problem is not well-posed. This is why I'm questioning your boundary conditions. The conditions I stated in the beginning lead to a well-posed problem; that is, there exists a unique solution for each choice of \lambda.

    It could be that there's a typo in the exam you're looking at.
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    (Original post by mathprof)
    By standard (my terminology), I mean boundary conditions of the form

    

A_1y(a)+A_2y'(a)=0

B_1y(b)+B_2y'(b)=0

    Since you saw it somewhere, I'll assume your boundary conditions are correct.

    The d.e. your considering y''+\lambda y=0 has auxillary equation r^2+\lambda =0. Hence r=\pm \sqrt{ -\lambda} . There are 3 cases to consider.

    Case I. \lambda &lt;0 . Put \lambda=-a^2 .

    Here y= c_1e^{ax}+c_2e^{-ax}.

    After applying the boundary conditions, I get a=0 which contradicts our assumption on \lambda. Hence no solution in this case. This is odd!

    Case II. \lambda =0 .

    Here y= c_1x+c_2. Boundary conditions imply here that y= c_2. That is, the solution is not unique. Again, very odd!

    I'm not going to bother with case III because it's clear that this problem is not well-posed. This is why I'm questioning your boundary conditions. The conditions I stated in the beginning lead to a well-posed problem; that is, there exists a unique solution for each choice of \lambda.

    It could be that there's a typo in the exam you're looking at.
    Hmm, when I was doing it, I thought it was a bit weird too because I'm usually good at these but this one was a bit of a *****. I assumed I was doing something wrong though. I've just triple checked the question and the boundary conditions are definitely that. It's a question from a past paper. There may be a typo but usually the lecturers point that out. I guess I'll make it a point to email and ask, to be sure.

    The first part asks if any eigenvalues exist A < 0. If so give the corresponding eigenfunctions. Second part goes onto ask to determine whether A = 0 is an eigenvalue and if so, give a corresponding eigenfunction. And then the third part says determine all eigenvalues A > 0 and give corresponding eigenfunctions. From the way the question is phrased, the case A > 0 definitely has eigenvalues so maybe it the other cases are a bit weird, it's meant to be that way? I don't really know.
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    I have thought about this problem some more and I realise that I was too hasty in dismissing the boundary conditions as a typo. In fact, they are correct and are known as periodic boundary conditions.

    In Case I, I forgot about the possibility that c_1=c_2=0 giving the trivial solution. Hence the answer to the first part of the exam question is that there are no eigenvalues in this case because there's only the trivial solution.

    The answer to the second part is that \lambda=0 is an eigenvalue since there's a nontrivial solution. You can take as an eigenfunction y(x)=1 .

    I think Case III leads to two independent eigenfunctions for each eigenvlaue.
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    (Original post by mathprof)
    I have thought about this problem some more and I realise that I was too hasty in dismissing the boundary conditions as a typo. In fact, they are correct and are known as periodic boundary conditions.

    In Case I, I forgot about the possibility that c_1=c_2=0 giving the trivial solution. Hence the answer to the first part of the exam question is that there are no eigenvalues in this case because there's only the trivial solution.

    The answer to the second part is that \lambda=0 is an eigenvalue since there's a nontrivial solution. You can take as an eigenfunction y(x)=1 .

    I think Case III leads to two independent eigenfunctions for each eigenvlaue.
    For case 2, where  \lambda = 0 , I get that  c_1 = 0 and I'm not entirely sure how to determine  c_2 from what I know. How did you get a non-trivial solution?

    And I just want to make sure a part of my reasoning was correct for case 1. I got the trivial solutions but I didn't know if how I justified in my head was mathematically sound.

    For one bit, I got  y(0) = A cosh(\alpha \pi) + Bsinh(\alpha \pi) = A and then concluded A = 0 and B = since there were no cosh or sinh terms on the right.
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    And for \lambda &gt; 0 I got  \lambda = 4k^2 and the corresponding eigenfunction to be  y(x) = Acos\alpha x = A cos (4k^2)x where A =/= 0.

    Just want to make sure that is correct because I don't feel confident on some parts of my working out.
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    (Original post by Preeka)
    For case 2, where  \lambda = 0 , I get that  c_1 = 0 and I'm not entirely sure how to determine  c_2 from what I know. How did you get a non-trivial solution?

    And I just want to make sure a part of my reasoning was correct for case 1. I got the trivial solutions but I didn't know if how I justified in my head was mathematically sound.

    For one bit, I got  y(0) = A cosh(\alpha \pi) + Bsinh(\alpha \pi) = A and then concluded A = 0 and B = since there were no cosh or sinh terms on the right.
    You can't determine c_2, it's arbitrary. The solution is therefore y=c_2 for any choice of c_2. Hence, nontrivial solutions exits. I just picked c_2=1 but anything other choice is fine. Keep in mind that eigenfunctions are sometimes only determined up to a constant multiple (multiplying the d.e. and b.c.s by a constant and you still have a soln with the same eigenvalue.)

    For your second part, I worked with exponentials, not hyperbolic trig functions, and the equations were much simpler to solve. (By the way, the equation you obtained after evaluating at the boundary point doesn't look correct.)
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    (Original post by Preeka)
    And for \lambda &gt; 0 I got  \lambda = 4k^2 and the corresponding eigenfunction to be  y(x) = Acos\alpha x = A cos (4k^2)x where A =/= 0.

    Just want to make sure that is correct because I don't feel confident on some parts of my working out.
    I think there's something wrong here. You should have two independent solutions for each eigenvalue (meaning distinct functions).

    By the way, why do you have a 4 in your expression for the eigenvalue? It serves absolutely no purpose!
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    (Original post by mathprof)
    I think there's something wrong here. You should have two independent solutions for each eigenvalue (meaning distinct functions).

    By the way, why do you have a 4 in your expression for the eigenvalue? It serves absolutely no purpose!
    I can't see where I've made my error. It's most likely an assumption error (I think) so I'm just going to post up some of my working out for  \lambda &gt; 0 .

     

y(x) = Acos(\alpha x) + Bsin(\alpha x)

\Rightarrow y(0) = A

y(\pi) = Acos(\alpha \pi)

y(0) = y(\pi) \Rightarrow cos(\alpha \pi) = 1 

\Rightarrow \alpha = 2k, k\in \mathbb{Z}

     y'(x) = -\alpha Asin(\alpha x) + \alpha B cos(\alpha x)

\Rightarrow y'(0) = \alpha B 

y'(\pi) = -\alpha A sin(\alpha \pi) + \alpha B cos(\alpha \pi)



y'(0) = y'(\pi) \Rightarrow \alpha B cos(\alpha \pi) = \alpha B

\Rightarrow cos(\alpha \pi) = 1

\Rightarrow \alpha = 2k, k\in \mathbb{Z}

     \lambda = (\alpha)^2 \Rightarrow \lambda = 4k^2 and y(x) = Acos(\alpha x) = A cos (4k^2)x, A=/= 0
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    Firstly, ignore my remark about the number 4; I misread your email and didn't realise you were giving me the actual values for \lambda .

    Now for your calculation. I think you made a mistake on the 9th line:

    y'(0)=y'(\pi)\implies \alpha B=-\alpha \sin(\alpha \pi)+\alpha B\cos(\alpha \pi) ;

    not

     \alpha B\cos(\alpha \pi)=\alpha B

    as you have written. You should end up with an eigenspace that's 2-dimensional.

    Everthing else looks correct.
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    (Original post by mathprof)
    Firstly, ignore my remark about the number 4; I misread your email and didn't realise you were giving me the actual values for \lambda .

    Now for your calculation. I think you made a mistake on the 9th line:

    y'(0)=y'(\pi)\implies \alpha B=-\alpha \sin(\alpha \pi)+\alpha B\cos(\alpha \pi) ;

    not

     \alpha B\cos(\alpha \pi)=\alpha B

    as you have written. You should end up with an eigenspace that's 2-dimensional.

    Everthing else looks correct.
    But won't the sin term always equate to zero?
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    Yes, you're quite right. I was looking at general periodic boundary conditions where the function is evaluated at a general point L . When L=\pi , it introduces a simplication. So you're calculation is correct. In this special case, the eigenspace is only 1-dimensional. Keep in mind, however, that in the general case the eigenspace will be 2-dimensional.
 
 
 

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