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Linear differential operator

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    We have a linear differential operator ##Ly=-y^{''}## working on all ##y## that can be derived at least twice on ##[-\pi ,\pi ]## and also note that ##y(-\pi )=y(\pi )## and ##y^{'}(-\pi )=y^{'}(\pi )##.
    a) Is ##0## eigenvalue for ##L##?
    b) Is ##L## symmetric? (I think the right English expression would be self-adjoint)
    c) Find all positive eigenvalues and corresponding eigenfunctions. What is the dimension of eigenspace?
    d) Does ##L## have any negative eigenvalues?


    2. Relevant equations



    3. The attempt at a solution

    a) No idea how to do this one. Here's my version:

    ##Ly=-y^{''}=\lambda y## which gives me ##y^{''}+\lambda y=0##. Solution of this differential equation is obviously ##y(x)=Acos(\sqrt{\lambda }x)+Bsin(\sqrt{\lambda }x)##

    Using this we find out that the boundary conditions for ##A\neq 0 ## and ##B\neq 0## are fulfilled only when ##\lambda =0##. Therefore the answer to question a) is YES.

    b)

    ##L## is symmetric if ##[PW(y,x)]\mid _a^b =0##. Note that originally the DE is ##P(z)y^{''}+Q(z)y^{'}+R(z)y=0## and where ##W## is Wronskian determinant.

    Obviously ##P=-1##.

    ##[PW(y,x)]\mid _a^b =[-\begin{vmatrix}
    y & z\\
    y^{'} & z^{'}
    \end{vmatrix}]\mid _a^b##

    After short calculus and knowing that ##a=-\pi ## and ##b=\pi ## we find out that the equation above is equal to ##0## and therefore the linear operator ##L## is symmetric.

    c)

    ##Ly=-y^{''}=\lambda^2 y## where I used notation ##\lambda ^2## for eigenvalue instead of ##\lambda ## just because I don't want to write square roots every time.

    Anyhow, solution to this DE is ##y(x)=Acos(\lambda x)+Bsin(\lambda x)## using boundary conditions:

    ##y(\pi )-y(-\pi )=2Bsin(\lambda \pi )=0## and of course for non trivial solutions ##B\neq 0## than:

    ##\lambda =n## for ##n\in \mathbb{Z}^{+}## where i suspect these are all the positive eigenvalues.

    And accordingly, I assume that ##y(x)=Acos(\lambda x)+Bsin(\lambda x)## are eigenfunctions where for each ##\lambda ## the dimension of the eigenspace increases by ##1##, therefore the dimension of eigenspace is ##n##.

    (or do I have to consider ##\lambda =0## here too?)

    d) Hmmm, if ##\lambda <0 ## than the differential equation ##y^{''}+\lambda y=0## changes to ##y^{''}-\lambda y=0##.

    Now ##y## that solves the equation above is ##y(x)=Ae^{\sqrt{\lambda }x}+Be^{-\sqrt{\lambda }x}##.

    Again, taking in mind that ##y(-\pi )=y(\pi )## leaves me with

    ##(B-A)(e^{-\sqrt{\lambda }x\pi }-e^{\sqrt{\lambda }x\pi })=0## which is only true if ##(B-A)=0##.

    Therefore the answer is NO, ##L## does not have negative eigenvalues.
     
  2. jcsd
  3. Apr 29, 2014 #2

    pasmith

    User Avatar
    Homework Helper

    The correct observation is that setting [itex]\lambda = 0[/itex] gives you [itex]y(x) = A \cos 0 + B \sin 0 = A[/itex]. But why not just solve [tex]
    Ly = -y'' = 0
    [/tex] directly to get [itex]y(x) = Cx + D[/itex], where [itex]y(-\pi) = y(\pi)[/itex] requires [itex]C = 0[/itex] but [itex]D[/itex] can be anything?

    [itex]L[/itex] is self-adjoint with respect to an inner product, presumably [itex]\langle f,g \rangle = \int_{-\pi}^{\pi} f(x)g(x)\,dx[/itex] if we're dealing only with real-valued twice-differentiable functions, if and only if [tex]
    \langle Lf, g \rangle = \langle f, Lg \rangle
    [/tex] for all relevant [itex]f[/itex] and [itex]g[/itex]. So you need to show that for all twice-differentiable [itex]f[/itex] and [itex]g[/itex] which satisfy the boundary conditions, you have [tex]
    \int_{-\pi}^{\pi} -f''(x) g(x) \,dx = \int_{-\pi}^{\pi} -f(x) g''(x) \,dx.[/tex]

    Best then to set [itex]\lambda = k^2[/itex] for [itex]k > 0[/itex].

    The eigenvalue is now [itex]\lambda^2[/itex] rather than [itex]\lambda[/itex], so the eigenvalues are [itex]\lambda_n^2 = n^2 > 0[/itex] for [itex]n \in \mathbb{Z}^{+}[/itex].

    Each eigenspace is two-dimensional: the linearly independent eigenfunctions corresponding to the eigenvalue [itex]n^2[/itex] are [itex]\cos nx[/itex] and [itex]\sin nx[/itex].

    If you want to do that, then you need to write [itex]y'' + \lambda y = y'' - |\lambda|y = 0[/itex]. It would be easier to define [itex]\lambda = -k^2[/itex] for [itex]k > 0[/itex].

    So far you've only shown that you must have [itex]B = A[/itex]; you have yet to show that [itex]B = A = 0[/itex].
     
  4. Apr 30, 2014 #3
    That is a lot easier, I agree. So the conclusion is that if ##\lambda =0## we can find a function that satisfies boundary conditions. In this case ##y=D## is a constant function, therefore ##\lambda ## can also be ##0##.

    b) Is ##L## self-adjoint?

    ##<f,g>=\int _{-\pi }^{\pi }f(x)g(x)dx## for real functions.

    ##<Lf,g>=<f,Lg>##

    ##\int _{-\pi }^{\pi }-f^{''}gdx=\int _{-\pi }^{\pi }-fg^{''}dx##

    ##\int _{-\pi }^{\pi }(fg^{''}-f^{''}g)dx=0## That here is still a question. We are trying to prove that the LHS of the equation is 0.

    ##\int _{-\pi }^{\pi }(fg^{''}-f^{''}g)dx=\int _{-\pi }^{\pi }[(g^{'}f)^{'}-f^{'}g^{'}-(f^{'}g)^{'}+f^{'}g^{'}]dx=[g^{'}f-f^{'}g]\mid _{-\pi }^{\pi }##

    ##[g^{'}f-f^{'}g]\mid _{-\pi }^{\pi }=0## for given boundary conditions. Nice, so yes, ##L## is self-adjoint.

    c)

    Ok, I will do that.

    So for ##n## eigenfunctions the eigenspace is ##2n## dimensional? I assume that is because ##sin## and ##cos## are already linearly independent, therefore for each ##\lambda \neq 0## I get two additional linearly independent functions.

    But, is it really ##\lambda _n^2=n^2## ?? Because, if I am not mistaken for [itex]\lambda = k^2[/itex] for [itex]k > 0[/itex] I get

    ##2Bsin(k\pi )=0## therefore ##k=n## and if anything, than ##\lambda _n=k^2=n^2##. Or not?


    d)

    The second part of course comes from second condition, which gives me ##(A+B)(e^{\sqrt{\lambda }\pi }-e^{-\sqrt{\lambda }\pi })=0##

    Now we have both conditions saying that ##A=B## and ##A=-B## which is only possible if ##A=B=0##
     
    Last edited: Apr 30, 2014
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