Jump to content

Sturm Lioville System Problem

Featured Replies

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

 

picture66mz.png

 

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 :P

 

as for the rest of the question, well, i guess i really need to get part (a) before i can attempt the rest hey? :P 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 :)

  • Author

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

  • Author

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

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.

  • Author
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 :P

 

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

 

-Sarah :P

  • Author

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! :(

  • Author

don't worry i've figured it out!

 

thanks for the assistance Yggdrasil :)

Archived

This topic is now archived and is closed to further replies.

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.

Configure browser push notifications

Chrome (Android)
  1. Tap the lock icon next to the address bar.
  2. Tap Permissions → Notifications.
  3. Adjust your preference.
Chrome (Desktop)
  1. Click the padlock icon in the address bar.
  2. Select Site settings.
  3. Find Notifications and adjust your preference.