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Sturm-Liouville operator, operating on what?

  1. Sep 23, 2015 #1
    Consider a general Sturm-Liouiville problem

    ##[p(x) y'(x) ]' + [q(x) + \lambda \omega (x) ] y(x) = 0 ##

    ##\quad \Leftrightarrow \quad \hat{L} y(x) = \lambda \omega(x) y(x) \quad \text{with} \quad \hat{L} y(x) = -[p(x) y'(x)]' - q(x) y(x) ##

    where ##p(x), p'(x), q(x), \omega(x)## are real-valued functions, continuous on the interval ##[a,b]##, and ##p(x), \omega(x) \geq 0 \, \forall \,x \in [a,b]##

    with the boundary conditions
    ##\alpha_1 y(a) + \alpha_2 y'(a) = 0 ##
    ##\beta_1 y(b) + \beta_2 y'(b) = 0##

    where ##\alpha_1, \alpha_2 \in \mathbb{R}##, not both zero, and ##\beta_1, \beta_2 \in \mathbb{R}##, not both zero.

    In the various litterature it is claimed that ##\hat{L}## is a linear operator on a Hilbert space of functions defined by the boundary conditions. Can someone give me a rigorous definition of the function space in question?

    For ##\hat{L}## to be a linear operator, surely it must be an endomorphism on some vector space, right? This is where I get confused.

    Suppose the boundary conditions are ##f(a)=f(b)=0##. This supposedly defines a Hilbert space (but for which functions? continuous? square-integrable?). Next, if ##\hat{L}##is an endomorphism, then the function ##\hat{L}f(x)## should be a member of that space also. Then we should have ##\hat{L}f(a) = \hat{L}f(b)##, but I don't see why this would be the case. For example, with ##f(x)## in the space, there's no reason for ##f'(x)## to be in the space.

    I would like a clear definition of the Hilbert space in question, and in what sense ##\hat{L}## truly is an operator (=endomorphism) on that space.

    (Bare in mind that I know very little about functional analysis and Hilbert spaces; I haven't had proper courses in these subjects yet.)

    Thank you.
     
  2. jcsd
  3. Sep 24, 2015 #2

    Svein

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    [itex] \hat{L}[/itex] is a linear operator on the space of twice differentiable functions u(x) on the interval [a, b], subject to certain homogeneous boundary conditions (for example u(a) = 0, u(b) = 0).
    The space is defined above: It is a Hilbert space with the inner product given by [itex] (u, v)=\int_{a}^{b}u(x)v(x)dx[/itex]. Verifying that [itex] \hat{L}[/itex] is a linear operator is not complicated (an integral is a linear operator).
     
  4. Sep 24, 2015 #3
    This I can buy. Perfectly good vector space. Let's call it ##V##.

    Here I get into trouble. Let ##u \in V##. Then ##u## is twice differentiable and ##u(a)=u(b)=0##. We have no other restrictions on ##u##. Now let ##\hat{L}## be defined as above. If ##\hat{L}## were to be an operator, then we must have ##\hat{L} u \in V \,\, \forall \,u \in V##. But ##\hat{L}u = -[pu']' - qu =: v## and I fail to see why this function ##v## should be in ##V##. The function ##v## need not satisfy ##v(a)=v(b)=0##, or even be differentiable?
     
  5. Sep 24, 2015 #4

    Svein

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    Why? I did not state that [itex]\hat{L}: V\rightarrow V[/itex]. In reality [itex]\hat{L}: V\rightarrow L_{2}[a, b] [/itex].
     
  6. Sep 24, 2015 #5
    Oh ok. So ##\hat{L}## is in general not an endomorphism.

    I thought the term "linear operator" on ##V## was the same as endomorphism on ##V##. That's the definition I've seen on wikipedia, and in Axler's linear algebra book, among others.

    Is the same true for "operators" in quantum mechanics? They are not endomorphisms either?

    For finite-dimensional vector spaces, eigenvalues and eigenvectors are defined only for endomorphisms. So I get confused about this in the infinite case. I really need to take a functional analysis course, but that's a few years ahead I'm afraid.
     
  7. Sep 24, 2015 #6

    Svein

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    Well, that depends on how you look at it. Since [itex]V\subset L_{2}[a, b] [/itex] and the inner product is defined on the whole of [itex]L_{2} [/itex], you have [itex] \hat{L}: V\subset L_{2}[a, b]\rightarrow L_{2}[a, b][/itex].
     
  8. Sep 25, 2015 #7
    I see. But ##V \subset L_2[a,b]## only if ##a,b## are finite, right? That is, we get into trouble if the interval is all of ##\mathbb{R}##.
     
  9. Sep 25, 2015 #8

    Svein

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    Well, [itex]L^{2}<-\infty, \infty> [/itex] is defined and exists, but the Sturm-Liouville problem is defined only on a finite interval (see https://en.wikipedia.org/wiki/Sturm–Liouville_theory).
     
  10. Sep 25, 2015 #9

    Orodruin

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    There is still the possibility of having a singular SL problem on an infinite interval. This is also mentioned on the wiki page.
     
  11. Sep 25, 2015 #10

    Svein

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    Yes, but that is too complicated for a first-timer.
     
  12. Sep 25, 2015 #11
    Ok, so thank you Svein for making me understand this a little better. I'm not quite there yet, though.

    In linear algebra (with finite dimensional vector spaces) we define an endomorphism to be self-adjoint if <u,Lv> = <Lu,v>. It doesn't make sense for an operator ##T: U\to V## to be self-adjoint.

    So when we say ##\hat{L}## is self-adjoint, then surely we must regard it as an endomorphism, right?
     
  13. Sep 26, 2015 #12

    Svein

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    Why not? Especially if U⊂V...
     
  14. Sep 26, 2015 #13
    If ##U,V## are finite dimensional inner product spaces and ##T:U\to V## a linear map, then by definition, ##T^\dagger: V \to U##. Thus if ##T=T^\dagger## then we must have ##U=V##.

    Suppose ##U## is a subspace of ##V##, then we may have <v,Tu> = <Tv,u> for all ##u,v \in U## so that ##T^\dagger v = Tv## for all ##v \in U##. But ##T^\dagger v \in U## for all ##v \in V## (and hence for all ##v \in U##) so that ##Tv \in U## for all ##v \in U##. Hence ##T: U \to U##.

    Please prove me wrong or point out my mistake.
     
  15. Sep 29, 2015 #14
    Just a quick note: The space [itex]\{u\in C^2[a,b]\} : u(a)=u(b)=0\}[/itex] with the inner product [itex](u,v)_{L^2} = \int_a^b uv[/itex] is not a Hilbert space (it is not complete). In this case we should work on a so called Sobolev space, in this case the space [itex]H_0^1(a,b) = \{u\in L^2(a,b) : u'\in L^2,u(a)=u(b)=0\} [/itex]. This space is a Hilbert space with the inner product given by
    [tex](u,v)_{H^1} = \int_a^b u'v'[/tex]

    You are right about the spaces being equal. I'll try to explain the definition in the infinite-dimensional case. Suppose you have an operator [itex]T: dom(T)\subset V_1\rightarrow V_2 [/itex] such that [itex]dom(T)[/itex] is dense in [itex]V_1[/itex] ([itex]V_1[/itex] and [itex]V_2[/itex] are Hilbert spaces). Then [itex]dom(T^*) [/itex] is defined as the set of all [itex]v\in V_2[/itex] such that there exists [itex]w\in H_1[/itex] such that

    [tex](v,Tu)_{V_2} = (w,u)_{V_1},\qquad u\in dom(T)[/tex]

    Since [itex]dom(T)[/itex] is dense, [itex]w[/itex] is uniquely determined by [itex]v\in dom(T^*)[/itex]. We can then define the operator [itex]T^*:dom(T^*)\subset V_2 \rightarrow V_1[/itex] by [itex]T^*v := w[/itex]. The pair [itex](dom(T^*),T^*)[/itex] is called the (Hilbert) adjoint of [itex](dom(T),T)[/itex] (the domain is important!!).

    Now, an operator [itex]T: dom(T)\subset H \rightarrow H[/itex] is called symmetric (notice we have only one space [itex]H[/itex]) if
    [tex](Tu,v)=(u,Tv), \qquad u,v\in dom(T).[/tex]

    With this, we finally define an operator [itex](dom(T),T)[/itex] to be self-adjoint if [itex](dom(T),T)=(dom(T^*),T^*)[/itex], that is, if [itex]T[/itex] is symmetric and [itex]dom(T)=dom(T^*)[/itex]. You can prove that if a linear operator [itex]T: H \rightarrow H[/itex] is continuous (as in the finite-dimensional case), then it is self-adjoint if and only if it is symmetric. In conclusion, you had the right idea, the domains must be equal.

    As for the Sturm-Liouville problem, suppose we have the equation
    [tex]-(pu')' + qu = f, \text{ on } (a,b), [/tex]

    with boundary conditions [itex]u(a)=u(b)=0[/itex]. It is possible to prove that for [itex]f\in L^2[/itex] there exists a solution in [itex]H_0^1[/itex] (the space I defined earlier). We can then consider the operator [itex]f\mapsto u[/itex] defined from [itex]L^2[/itex] to [itex]L^2[/itex] (or in [itex]H_0^1[/itex]). This is the operator we prove is self-adjoint (and with that obtain the solution to the eigenvalue problem, for instance). Let me know if you want more details, hope this helps.
     
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