Differential equation with singular boundary conditions

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Homework Help Overview

The discussion revolves around a Sturm-Liouville problem involving the differential equation (x^2 y')' + ax^2y = 0, with boundary conditions y'(0)=0 and y(1)=0. Participants are exploring the implications of a substitution hint provided in the problem statement.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find a general solution using trigonometric functions and considers the implications of boundary conditions at x=0 and x=1. Some participants suggest using the provided substitution to reformulate the equation, while others express confusion about the implications of their findings, particularly regarding the trivial solution.

Discussion Status

The discussion is ongoing, with participants sharing their attempts and questioning the validity of their solutions. There is a recognition of the complexity of the problem, and some guidance has been offered regarding the substitution, though no consensus has been reached on the nature of the solutions.

Contextual Notes

Participants are grappling with the singular boundary condition at x=0 and the implications of the eigenvalue parameter 'a'. There is mention of potential trivial solutions and the need for further exploration of non-trivial solutions.

FunkyDwarf
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Hey guys, just need some hints with this doosey

Homework Statement


We have
(x^2 y')' + ax^2y = 0 where a the eigenvalue (a sturm-lioville problem) (sp?)
with y'(0)=y(1) = 0 and we get the hint to substitute f = y/x.

The Attempt at a Solution



Ok so i get the general solution being a sum of cosines and sines and i fiddled around a LOT with limits and exponentials and small angle approximations etc to try and get around the fact that the function blows up at x = 0 and were supposed to fit a boundary condition there. In all my efforts, the ones that make SOME sense, i get the eigen values = 0 which kind of makes it a trivial problem. I also tried taking advantage of the orthogonality of solutions given we know the weighting function x^2 but that required me to know what at least one eigenvalue was, and when i tried zero i had trouble solving exactly for the second but graphically it appeared to be close to root 2 but not quite.

Im sure I am missing something stupid so if you could, gently, point that out that would be super =)

Cheers
-Z

EDIT: i also tried a series expansion with the first 3 terms for sine and cos but got lost =( i thought about trying a general series solution but wasnt sure
 
Last edited:
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Well, using hint would be a good start! If you let u= y/x, so that y= xu, what does the equation become?
 
With the substitution u=y/x equation becomes
u''+au=0
which has solution
u=c1*cos(kx)+c2*sin(kx) [k=sqrt(a)]
so that
y=xu=c1*x*cos(kx)+c2*x*sin(kx)
So doing y'(0)=c1=0 y(1)=c1*cos(k)+c2*sin(k)=0
i.e. y=0, onlythe trivial solution ?
 
Thats what i got (sorry if it wasnt obvious, yeh i used the hint =) )
 

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