Question

(Sturm-Liouville) I'm having trouble understanding the difference in solving Singular SLDE's and Regular SLDE's, and defining that difference. For whatever reason the different sources I've looked at have a kind of obtuse way of talking about the conditions that make a SLDE singular (or maybe I'm looking at stuff above my level and that's why it doesn't make sense)-could somebody explain this?

Answer 1

How is solving a Singular SLDE functionally different?

Answer 2

I don't know much about this, by wiki says
"If the interval is unbounded, or if the coefficients have singularities at the boundary points, one calls L singular. In this case, the spectrum no longer consists of eigenvalues alone and can contain a continuous component"

so it sounds like one difference is integrating over an unbounded region vs some fixed interval.

Answer 3

Right now I'm just scouring Stack Exchange and some other places to look for an understanding to this, a lot of the people typically online who are able to help with this aren't on atm, but they probably will be as the night goes on. I'm checking through these:

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=%22singular+sturm-liouville%22+site:math.stackexchange.com

Just to see if there's a general method for solving SSLDE's.

Answer 4

Solving a singular SLDE like the one we have may be related to using Bessel Functions somehow; are you there?

Answer 5

Eigenvalues and Eigenfunctions of a singular Sturm-Liouville operator using Bessel functions: http://math.stackexchange.com/questions/480935/eigenvalues-and-eigenfunctions-of-a-singular-sturm-liouville-operator-using-bess?rq=1

Sturm-Liouville eigenvalues and the zeroes of a Bessel function: http://math.stackexchange.com/questions/334913/sturm-liouville-eigenvalues-and-the-zeroes-of-a-bessel-function

Answer 6

Hm.. I am just working through some problems right now and when plugging in my singular condition, I get some results that are... kind of interesting. I don't really know how to apply the results of the conditions however.

Answer 7

Yeah, I dunno, we may legitimately want to shelve this for the time being because it looks like the other problems are significantly easier in form.

Answer 8

Shelfing.

Answer 9

Next problem. Gonna see how far I get.

Answer 10

@jim_thompson5910 @FibonacciChick666

Answer 11

I'm sorry this is above my head. Try @ganeshie8 and @hartnn when they come online

Answer 12

Alright, cool; thank you nonetheless for checking it out.

Answer 13

@SithsAndGiggles , please help if possible.

Answer 14

I can't say I've ever worked through a Sturm-Liouville problem before, sorry.

Answer 15

No problem, thank you.

Answer 16

@hartnn You might also be able to help me on this, IDK

Answer 17

do u need help on sturm louiville bessel problem?

Answer 18

Not specifically that, but I'm asking, in general, how are singular sturm-liouville problems solved differently? Are they solved differently? How do you treat it differently from a regular SLDE? The distinction isn't just arbitrary, it must serve a purpose, so presumably there's a difference in the way you solve SLDE's.

Answer 19

sry id have to look more into sturm louiville, why is sturm louiville form so important?

Answer 20

i remember going into for something in physics

Answer 21

Because I'm taking PDE and it's in most of our exams.

Answer 22

One of our exams-the one coming up-is going to have a singular sturm-liouville problem on it.

Answer 23

have u ever watched those sturmlouiville theory videos

Answer 24

Yeah, I haven't looked up anything on singular SLDE's on YouTube and doubt I will find anything useful, but I will, thank you for the suggestion.

Answer 25

Yeah, nothing, it's not suggested in the search results nor do the search results yield any videos on the first page with singular slde's in the title, but I'll poke around a little.

It's pretty absurd for him to ask a question like this of us, but whatever.

Answer 26

No results: https://www.youtube.com/results?search_query=%22singular+sturm-liouville%22

Answer 27

i remember now, it comes up in spherical harmonics and solving for bound states of electrons for hydrogen

Answer 28

im not sure what the different between single and normal sturm form is but

Answer 29

the point of putting something in sturm louville form, means that you have real eigen value solutions + you can determine what function to multiply by to satisfy your orthogonality conditions

Answer 30

like for example in the fourier series
you know how
f(x)= An cos(npix/l) + bnSin(npix/l)
and we know what to multiple by to get An and Bn

Answer 31

similiary we put something in sturm form and see another expansion for your f(x) along another eigen function basis and also we can termine what function we need to multiply by now to get the new constants assiciated with our eigen functions

Answer 32

they are not much different we just have singular points thrown in now right

Answer 33

Yeah, so, if you could summarize what's different about solving them, could you? I'm sorry, it sounds like you've figured it out, but I don't understand. Could you maybe give an example if you have one?

Answer 34

i thought u solve it just like the others... but with the new boundary conditions

Answer 35

for reginal SL we got
a1y+a2y'=0
b1y+b2y'=0
and [a1 a2]^T and [b1 b2]^t not equal to zero

Answer 36

thats regular SL

Answer 37

for Singular the difference is no boundary condition needed

Answer 38

its just bounded at the ends

Answer 39

r(a) =0, no condition needed at x=a
r(b)=0, no condition needed at x=b
now you look for bounded solutions over [a,b]

Answer 40

if u remember normall for the sturm louiville form you have
r(x) > 0 on the open interval (a, b)

Answer 41

and now for the singular case we are given a bound at r(a) and r(b)

Answer 42

so our interval can be over the closed [a,b]

Answer 43

thats about all i know about singular SL

Answer 44

id have to really go thru adjoint function definitions and stuff if im gonna revisit this sturmlouiville

Answer 45

the main concept to take away here is this is a way to make the fourier series more general