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Sturm-Liouville Eigenvalue Problem (Variational Method?)

  1. Mar 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Using Sturm-Liouville theory, estimate the lowest eigenvalue ##\lambda_0## of...
    [tex]
    \frac{d^2y}{dx^2}+\lambda xy = 0
    [/tex]
    With the boundary conditions, ##y(0)=y(\pi)=0##

    And explain why your estimate but be strictly greater than ##\lambda_0##


    2. Relevant equations
    ##\frac{d}{dx} \left (p \frac{dy}{dx} \right) + qy + \lambda \rho y = 0 ##
    ##\frac{I}{J}=\frac{\int (p (y')^2 - qy^2) dx}{\int \rho y^2 dx} ##

    3. The attempt at a solution
    I am rather stuck here I think, I am so confused, I am pretty sure our lecturer said we had to use the variatonal method but I cannot see how to do that here.

    But somewhere in my notes it says that the eigenvalues of a SL equation are given by I/J (second equation in the relevant equations section).

    So to make a start that is what I did. Comparing the Sl equation with the equation given to solve, I can see that ##p=1##, ##q=0##, and ##\rho(x) = x## where ##\rho## is the weight function. And I used the trial function that ##y(x)=\sin(x)##

    So I calcuated I/J (lower limit 0, upper limit ##\pi## for the integrals), using integration by parts and trig identities for J and simple trig identities and u-sub for I.

    I found ##I=\pi/2## and ##J=\pi^2/4## so therefore ##I/J=2/\pi## I didnt post all my working because I doubled checked it in an integral calculator. But now I am lost. That cannot just be it.

    I dont understand how to use the variational method here, all examples I have seen using the SL variational method online were to solve problem where a trial function of ##e^{-\alpha x^2} ## satisfied the BC's and the problems required integrating I/J from -'ve to +'ve infinity, and then setting it to zero and solving for ##\alpha## I cant see how I can use that method in this problem.

    Or if that even is what is required. Any help is greatly appreciated, really lost here.
     
  2. jcsd
  3. Mar 23, 2017 #2

    Orodruin

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    Why can't it be it? It is an estimate. You might be able to improve on it by using some sort of trial function that depends on a parameter you can minimise wrt to, but that is another step. How close you get to the actual eigenvalue will depend on how good you are at guessing.
     
  4. Mar 23, 2017 #3
    Thanks for the reply.

    Yeah I know the answer I had was a valid estimate, just I don’t think it is what the assessors will be looking for.

    Do you mean like using ##\sin(\alpha x)## instead and then setting I/J to zero and solving for \alpha? I did attempt that beforehand but I didnt seem to get anywhere usefull but will give it another go if that is what is required.

    Or should I try to improve my trial function?
     
  5. Mar 24, 2017 #4

    Orodruin

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    That function does not satisfy the boundary conditions except for a discrete set of alphas. Your guess must always satisfy the boundary conditions.
     
  6. Mar 24, 2017 #5
    Oh ok, should I try to look for a different
    Ah ok.

    So when you said
    Did you mean a completely different function to Sine? As I am thinking that minimising wrt to a parameter is what I need to do. I cant think of any other function that would satisfy the BC's though,
     
  7. Mar 24, 2017 #6

    Orodruin

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    You cannot think of any other function that satisfies ##y(0) = y(\pi) = 0##??
     
  8. Mar 24, 2017 #7
    I have two suggestions:
    1. Let ##y(x)=sin(nx)##.
    2' Express ##\rho(x)## as a Fourier series.
     
  9. Mar 26, 2017 #8
    It doesn’t matter too much now as the works been submitted. But would need to be able to problems like this for an exam in like 6 week time.

    But I couldnt think of another function that satisfied the BC's and that was suitable.

    I tried ##y(x) = \sin(x) / x ## and the calculation become impossible, or more generally tried ##y(x)=x^n \sin(x)##. And I also tried ##y(x)=e^{ix} + 1## but that leads to a non-real answer.

    I suggested that on post #3, but was told its not valid as it doesnt always satisfy the BC's
     
  10. Mar 26, 2017 #9

    Orodruin

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    No you did not. You suggested ##\sin(\alpha x)## where I interpreted ##\alpha## as a continuous parameter. When ##n## is an integer, it clearly satisfies the boundary conditions. Any linear combination of such functions, for example ##f(x,\alpha) = \cos(\alpha)\sin(x) + \sin(\alpha) \sin(2x)## would be a possible trial function.

    What do you mean by "suitable"? Any function that satisfies the boundary conditions will work. Of course, you might get a better or worse estimate depending on the function you pick.
     
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