Re: Operators represented as Fourier series
- From: "tanabai" <dinchen2@xxxxxx>
- Date: 12 Nov 2005 10:01:56 -0800
David C. Ullrich schrieb:
> On 11 Nov 2005 16:12:28 -0800, "tanabai" <dinchen2@xxxxxx> wrote:
>
> >I have got the following proble. If I consider the one dimensional
> >Laplace operator T = - d^2/dx^2 acting on the interval (0,1) and
> >additionally supposed to have the boundary condition f(0) = f(1) = 0.
> >Of course the operator have a pure point spectrum,
> >indeed the eigenvalues are (n\pi)^2 and the corresponding
> >eigenfunctions are sin(n\pi x) for n equal or greater than 1.
> >
> >- d^2/dx^2 (sin(n\pi x)) = (n\pi)^2 (sin(n\pi x))
> >
> >Now I want to develop the Fourier expansion of the squar root of the
> >operator - d^2/dx^2 according to (0,1). With no doubt the eigenvalues
> >are (n\pi) , but i do not know what eigenfunctions the square root has.
> >
> >
> >T^{1/2} (sin(n\pi x)) = (n\pi) (sin(n\pi x)) ?
>
> Yes. Let's say T = - d^2/dx^2, and to save typing let's
> write S = T^{1/2} for the positive-definite square root of
> T, and let's write s_n(x) = sin(n\pi x) for positive integers
> n. Then s_n is indeed an eigenvector of S, with eigenvalue
> n pi.
>
> Note that S simply is not a differential operator. There's
> no a priori reason it _should_ be a differential operator,
> and in fact it's simply _not_ one.
>
> >It make sense corresponding to the spectral theorem of unbounded
> >operators, therefore the eigenvalues of T^{1/2} has to be the square
> >root of the eigenvalues of T.
> >
> >But if i take the operator |i d/dx| , which fulfil |id/dx| ^2 =
> >d^2/dx^2, i find the equation
> >
> >|i d/dx (sin(n\pi x)) | = |n\pi| |cos(n\pi x)|
>
> What? You seem to be taking "|i d/dx|" to mean an operator
> such that
>
> |i d/dx| (f) = |i d/dx f|.
>
> But that's not even linear!
>
> Maybe the confusion is caused by some notation in operator
> theory. Let's say that T is a normal operator. You might
> see the notation "|T|" for a certain positive-definite operator.
> But "|T|" does _not_ refer to an operator such that
>
> |T| x = |Tx|.
>
> In general that doesn't even make any sense; elements of
> Hilbert spaces do not have absolute values.
>
> What |T| is is the absolute value of T, "on the spectral
> side": roughly speaking, |T| is the operator such that
> if x is an eigenvector of T with eigenvalue lambda then
> x is also an eigenvector of |T|, with eigenvalue |lambda|.
>
> >but the cosine does not satisfy the boundary condition f(0) = f(1) =
> >0.
> >
> >I do not understand what are the right eigenfunctions of the square
> >root of -d^2/dx^2.
> >It would be very kind, if someone helps me, because i am thinking along
> >time about this problem.
> >
>
> ************************
>
> David C. Ullrich
Thanks a lot, I understand. The only problem remaining is, how can I
prove the first statement
>>From T (sin(n\pi x)) = (n\pi)^2 (sin(n\pi x))
follows that
T^{1/2} (sin(n\pi x)) = (n\pi) (sin(n\pi x)) .
If I use the spectral theorem for a positive selfadjoint unbounded
operators I know that T^{1/2} = \int_0^\infty |lambda| dE(lambda), if
E(lambda) is the spectral measure.
Is there any other theorem to describe the equivalence from above?
And additionally is there a description for the domain of T^{1/2} ?
What can I say about the domain of the square root of -d^2/dx^2, does
T^{1/2} have the boundary conditions of T ?
Thanks a lot.
.
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