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Topic: special functions problem (Read 2247 times) |
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Marissa
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special functions problem
« on: Mar 26th, 2008, 11:17pm » |
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In order to place the DE into the Sturm-Liouville form, we need to place the DE in it's self-adjoint form. I browsed the Net, and it seems it's only defined for the second order ODE. I first off only know definitions of what self-adjoint is for the matrix case, where a matrix A is equal to it's conjugate-transpose.
Is there a self-adjoint form for all (integer and non-integer) ordinary differential operators? i.e. is there a self-adjoint form for a first-order ODE? third? Fourth?
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Marissa
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Re: special functions problem
« Reply #1 on: Mar 27th, 2008, 8:32am » |
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I'm currently working on Sturm-Liouville problems in second order DE's in math class.
Any help, pls.?
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Michael Dagg
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Re: special functions problem
« Reply #2 on: Mar 28th, 2008, 9:00am » |
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You are asking a good question: What does it mean to be self adjoint
for a differential operator. To explain this properly requires some
advanced topics. The main point is that differential operators act on
infinite dimensional vector spaces. Also, whether or not an operator is
self adjoint depends on which function space it acts on. Some knowledge
of infinite dimensional spaces is required to completely appreciate the
definitions.
The main space of interest is the space of square integrable functions
(called L^2 ). To define this space properly requires the Lebesgue integral.
But, here is the idea: Let's take the operator f --> f'' on the space of square
integrable functions on the real line. A function is square integrable if the
integral of the square of its absolute value over the line is finite. There is
an inner product associated with this space. Given two functions f and g
their inner product is
<f,g> = \int _\infty^\infty fg dx
for the real case. (For complex-valued functions, the product under the
integral must be the conjugate of g). In essence, (but not exactly) self adjoint
for the second derivative operator means
<f'',g> = <f, g''>
By simply using the definition of the inner product and integration by
parts (the key tool) twice, you will see that this formula is true. Note
that the integration by parts gives a minus sign and that integration by
parts twice gives minus times minus = plus. This is the key that makes
second derivatives self adjoint. By this logic, first order differential
operators are not self adjoint but 4th order operators are. The trouble
with all of this is that the derivative is not defined for ALL square
integrable functions. So, the the formula <f'',g> = <f, g''> only holds
for those functions that are twice differentiable with there derivatives
in the space of square integrable functions. This means that the
situation is not exactly the same as for matrices where self adjoint
means essentially that <A v, w> = <v, A w> where the inner product is
the usual inner product of Euclidean space or its complexification.
As I mentioned, the selfadjointness of an operator depends on which
inner product we choose. So, it might be possible to make a first order
operator self adjoint with respect to some strange inner product. But, no
one would care! So, the basic answer to your question is that second
order operators are self adjoint but the others are not.
One reason we like self adjoint is because the spectrum of such an
operator is real.
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| « Last Edit: Mar 28th, 2008, 9:00am by Michael Dagg » |
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Regards,
Michael Dagg
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Marissa
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Re: special functions problem
« Reply #3 on: Mar 28th, 2008, 8:04pm » |
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Thanks.
Definition: Square Integrable Functions
A real valued function f is square integrable on the real interval "I" to the weight function p(x)>0 if
integral|from -oo to +oo| [f(x)]^2 p(x)dx < oo
A Sturm-Liouville Equation is a second order homogeneous DE of the form:
d/dx (p(x)*dy/dx) + [q(x) + lambda*r(x)]y = 0
which is equivalent to the form:
L[y] + lambda*r(x)y = 0
with lamda as a parameter, where L is the self-adjoint operator d/dx (p*d/dx) + q(x).
For existence, r(x) and q(x) are continuous on "I", p(x) is continuously differentiable on "I".
Now, depending on boundary conditions, we get a system that has a countably infinite number of eigenvalues, and a corresponding set of countably infinite eigenvectors. These eigenvalues are all positive, and all corresponding eignvectors orthogonal with weighting function p(x).
If we take a different set of boundary conditions, we get a second infinite set of eigenvalues, as well as corresponding set of eigenfunctions.
Together these two sets would have been the original solutions to the ODE of parameter lambda.
I think that's enough for now and we might just go through and make sure things are well defined. I'd like to get this to a very simple self-adjoint form of a fourth order ODE if we can. I'm a little shaky on the last two, because I'm not 100% familiar with the application of boundary conditions and the recovery of the two forms of the linearly independent solutions to the original DE.
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Michael Dagg
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Re: special functions problem
« Reply #4 on: Mar 28th, 2008, 8:30pm » |
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What's the problem?
That is, you haven't given us one. Seems to me that you are simply
re-typing some definitions from a book or something.
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Regards,
Michael Dagg
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Marissa
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Re: special functions problem
« Reply #5 on: Mar 31st, 2008, 1:15pm » |
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Sorry, I needed to make it clear in my mind.
A linear operator is said to be symmetric on a Hilbert space if:
<Ax,y> = <x,Ay>
and the operator is self-adjoint if it is symmetric everywhere in the domain of x,y.
One thing i'm unsure of is whether or not a Hilbert space is only explicitly defined in cartesian coordinates, with the regular Euclidian inner product. Either way, the self-adjoincy holds for any inner product space. I'm also unsure of what to interpret about an operator, given that it is self -adjoint. The definition of self-adjoincy based on operator symmetry means nothing to me. Possibly some light can be shed on this?
For our second-order ODE:
a(x)y"(x) + b(x)y'(x) + c(x)y(x) = 0
Let L = a(x)d^2/dx^2 + b(x)d/dx + c(x)
then,
<Lf,g> - <f,Lg> = integral|from x1 to x2| p(x)(b-a')(gf'-fg')dx = 0
Such that if b-a' is zero the operator L is in self-adjoint form. There's also the integrating factor for when this isn't the case, but every second order DE can be converted into its self-adjoint form.
Okay now I see why we can't have a first order, sice there is no function r(x) when in the form (ry')' such that only a first-order term pops out.
The simple case of <f'',g>=<f,g''> with the + - as part of an integration by parts wasn't really clueing in.
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Marissa
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Re: special functions problem
« Reply #6 on: Apr 1st, 2008, 1:36pm » |
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Could you possibly shed some lights on this problem?
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TJMann
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Re: special functions problem
« Reply #8 on: Apr 8th, 2008, 2:45pm » |
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interesting discussion but apparently the know-it-cones stop responding.
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jabed937
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Re: special functions problem
« Reply #9 on: Mar 22nd, 2012, 11:00am » |
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very interesting discussion. . Please stay us up to date like this. Thanks for sharing
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