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Eigenvalue problem with nonlocal condition

  1. Apr 2, 2014 #1
    Hello guys, suppose we have an eigenvalue problem
    [tex]
    \left\{
    \begin{array}{ll}
    u'' + λu = 0, \quad x \in (0,\pi) \\
    u(0)=0 \quad \\
    \left( \int_0^\pi \! {(u^+)}^2 \, \mathrm{d}x \right)^{\frac{1}{2}} = \left( \int_0^\pi \! {(u^-)}^4 \, \mathrm{d}x \right)^\frac {1}{4}

    \end{array}
    \right.

    [/tex]
    where [tex] u^+, u^-[/tex] is positive, negative part of function respectively.
    I'm having troubles with case when λ> 0. Is there any way how to simplify or express these parts of function ? I've tried some (analytic) brute force methods, tried to simplify that first but still no valuable result.
    Thanks
     
  2. jcsd
  3. Apr 2, 2014 #2

    micromass

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    What if you use that the general solution to

    [tex]u^{\prime\prime} + \lambda u =0[/tex]

    is always of the form

    [tex]u(x) = C_1\sin(\sqrt{\lambda}x) + C_2\cos(\sqrt{\lambda}x)[/tex]

    for ##\lambda>0##. Then from ##u(0) = 0##, you can deduce that

    [tex]C_2=0[/tex]

    So your function has the form

    [tex]u(x) = C_1 \sin(\sqrt{\lambda}x)[/tex]

    Now figuring out what ##C_1## is from the last condition seems tedious, but not all too difficult.
     
  4. Apr 2, 2014 #3

    micromass

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    Hmm, thinking about it, any value of ##C_1## will satisfy the last condition, at least if ##f(x) = \sin(\sqrt{\lambda}x)## satisfies it. So your only job is to find out for which ##\lambda## this is true.
     
  5. Apr 2, 2014 #4
    Thanks for your response.
    The formulation of my problem was quite confusing. First condition is ok, its just straight forward process that leads to sine function, but I'm completely lost with the second one.As you said it's obvious that second condition will hold for any "multiplying" constant.

    These positive / negative parts of function are bit tricky for me. Since number of zeros depends on λ( and then number of positive and negative parts of sine too), I have to split that function somehow and do piecewise integration.Thats the problem. Still seems something is "hidden" for me and I'm sure that will be obvious, but I can't figure it out now.
    I'm not asking for solution, but I would really appreciate some hint ;)
    Thanks
     
    Last edited: Apr 2, 2014
  6. Apr 2, 2014 #5

    HallsofIvy

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    You have already determined, from the equation and the condition that y(0)= 0, that [itex]y= Csin(\sqrt{\lambda} x)[/itex]. Yes, the number of zeros and the number of intervals where y is positive or negative depend upon [itex]\lambda[/itex] but it is not all that difficult to determine. For a given [itex]\lambda[/itex], [itex]cos(\sqrt{\lambda}x)[/itex] will be 0 at [itex](n\pi)/\sqrt{\lambda}[/itex]. It will be positive on the interval from [itex](2n\pi)/\sqrt{\lambda}[/itex] to [itex]((2n+1)\pi)/\sqrt{\lambda}[/itex] and negative on the interval from [itex]((2n+1)\pi)/\sqrt{\lambda}[/itex] to [itex](2(n+1)\pi)/\sqrt{\lambda}[/itex] for any positive integer n.
     
  7. Apr 2, 2014 #6
    Well, not that difficult... Thank you, it really helped.
     
  8. Apr 2, 2014 #7

    AlephZero

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    That shouldn't be a surprise. If the magnitude of an eigenvector is NOT arbitrary, there is something very strange going on!
     
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