Sturm Lioville System Problem

[Sarahisme]

Hi guys, i was hopeing someone would be able to guide me through this problem.

 

 

for part (a) i get the transformed equation to be

 

[math] \frac{d^2y}{ds^2}+\frac{4}{s}\frac{dy}{ds} = -\lambda y [/math]

 

i am not 100% sure this is correct, although i have been through the working about 5 times now

 

as for the rest of the question, well, i guess i really need to get part (a) before i can attempt the rest hey? and i don't really get this eigenvalue/vector stuff for functions :S

 

i suppose (a) isnt really my problem, its the rest of the question that is really giving me a hard time!

 

Thanks

-Sarah

[Sarahisme]

to be specific, its parts © & (d) that i don't know how to do, just can't quite figure out anything reasonable, other than a trivial solution! :S

[Sarahisme]

let me see, well i've work out that the general solution is

 

y(x) = Asin(((lambda)^0.5)/x)) + Bcos((((lambda)^0.5)/x))

 

but i don't see how to use y(1) = 0 and y(2) = 0

 

because neither cancels either term.... :S

[Yggdrasil]

for © you shoud find that the p(x) obtained in (b) is zero at the boundaries (x=a and x=b). Therefore, at the boundaries:

 

q(a)y = (lambda)r(a)y

q(b)y = (lambda)r(b)y

 

which has solutions only for certain values of lambda. These lambda are your eigenvalues.

 

Also, if you want to look up more resources on this topic a Strum Liouville system is another name for the self-adjoint form of a differential equation.

[Sarahisme]

for © you shoud find that the p(x) obtained in (b) is zero at the boundaries (x=a and x=b). Therefore' date=' at the boundaries:

 

q(a)y = (lambda)r(a)y

q(b)y = (lambda)r(b)y

 

which has solutions only for certain values of lambda. These lambda are your eigenvalues.

 

Also, if you want to look up more resources on this topic a Strum Liouville system is another name for the self-adjoint form of a differential equation.[/quote']

 

i am little confused.

 

i get for part (b) that

 

p(x) = x^2

q(x) = 0

r(x) = 1/x

 

but p(x) isnt 0 at x=1 or x=2

 

and

 

q(a)y = (lambda)r(a)y

q(b)y = (lambda)r(b)y

 

can't be done because q(x) = 0.

 

also the question asks to find the evals etc. from part (a). i think i'm a bit lost

 

i am guessing all that i have just said is wrong, so please please correct me!

 

-Sarah

[Sarahisme]

i think i need to solve :

 

[math] Asin(\sqrt(\lambda)) + Bcos(\sqrt(\lambda)) =0 [/math]

[math] Asin(\frac{\sqrt(\lambda)}{2}) + Bcos(\frac{\sqrt(\lambda)}{2}) = 0[/math]

 

but i just can't figure out how!

[Sarahisme]

don't worry i've figured it out!

 

thanks for the assistance Yggdrasil