Question

Sturm-Louiville problem

Answer 1

Find eigenvalues and eigenfunctions

Answer 2

\[Y''+\lambda*Y\]
y(0)=0, y'(3)=0

Answer 3

@Kainui

Answer 4

I was going to re-post it. But if anyone can help me hear that would be great

Answer 5

Yeah, I'm still pretty fresh with SL problems, I should be able to help you out.

Answer 6

i had\[\lambda=((\pi+2n \pi)/6)^2\]

Answer 7

Your diffeq up there isn't equal to anything though, so I'm a little confused.

Answer 8

=0

Answer 9

sorry

Answer 10

only deal with homogenous equations

Answer 11

Ok ok, just making sure you didn't mean to put a = sign where the + sign is.

Answer 12

yea. so i only dealt with the case where\[\lambda>0\]

Answer 13

the other two cases would mean the eigenfunctions would be 0 which weren't really relevant. Thats what i thought

Answer 14

Yeah, you need it to be positive for the theorem to work I believe.

Answer 15

yep!

Answer 16

When I see this though, I immediately see that you can just solve the ODE.

\[\large y=A \sin( \sqrt{\lambda}x)+B \cos( \sqrt{\lambda}x)\]

Then plug y(0)=0 and y'(3)=0 to find the coefficients. From there we sum up all the solutions and then use the orthogonality to solve for the eigenvalues I believe.

Maybe I'm missing something, it's been a while since I've done these and I'm still working it out.

Answer 17

pellet. i didn't do sqrt lambda

Answer 18

Haha yeah I've done that more than I care to admit too.

Answer 19

wait isn't it the same if i let \[\lambda=\]

Answer 20

i just have\[y(x)=Acos(\mu x)+Bsin(\mu x)\]

Answer 21

\[\lambda=\mu^2\]

Answer 22

Yeah that's fine to do it that way, that's what I usually do.

Answer 23

yea. so then with y(0)=0 it implies A=0. so
\[y(x)=Bsin(\mu x)\]

Answer 24

and differentiating that you get
\[y(x) \prime=B \mu \cos(\mu x)\]

Answer 25

Yep you got it.

Answer 26

and then applying the boundary condition
\[0=B \mu \cos(3\mu)\]

Answer 27

and thus
\[B \mu=0\] OR\[\cos(3 \mu)=0\]

Answer 28

but we deal with the second equation since it would be silly if B=0

Answer 29

so
\[3\mu=\pi/2+n \pi \]

Answer 30

Yeah, exactly. You've found your eigenvalue mu_n, you're golden now.

Answer 31

but when i type it in to the online quiz it says its still wrong....

Answer 32

so \[\lambda=\mu^2\]

Answer 33

and you just square mu

Answer 34

Hmm well can you take a screen shot or something? Maybe you're just plugging it in wrong. This is what I'm getting: \[y= \sum_{n=0}^\infty A_n \sin (\sqrt{\lambda_n} x)= \sum_{n=0}^\infty A_n \sin (\frac{\pi(2n+1)}{6}x)\]

Answer 35

Hmm, maybe if you rewrite it as: \[\large \lambda_n=\frac{\pi^2(2n+1)^2}{36}\] it will be better? I don't know, that's really weird to me. I'm gonna take a bathroom break and relook at our work to see if we've messed up somewhere.

Answer 36

No worries! I'll try that.

Answer 37

Yea. Still wrong. Sure we didn't do it wrong if we both got the same answer

Answer 38

Maybe this is where we went wrong, with the cosine. \[3x=\pi(2n-1)\] If they're starting their index at n=1 instead of n=0, then this will make up for the term we're leaving out.

Answer 39

at n=0 \[\mu=\pi/6 \] right

Answer 40

Yeah, but they start out with n=1 then we're essentially starting at (2n+1) which gives 3. Since they're using n=1 with (2n-1) then we have 1 for the first number. I hope that makes sense of why. Just put a negative sign in there and you'll be right I promise.

Answer 41

SUCCESS

Answer 42

let me think about this

Answer 43

ah so our index shift will be n--->n-1 so that we start from n=1 right? AH MAKES SENSE

Answer 44

so we end up have in the numerator


2(n-1)+1
2n-2+1
2n-1

Answer 45

ah awesome

Answer 46

so your eigenfunction would therefore be \[F _{n}=\cos(\lambda x)\]

Answer 47

is that correct?

Answer 48

wait no thats wrong

Answer 49

\[F _{n}=\sin(\sqrt{\lambda}x)\]

Answer 50

ah awesome. i got it out! Thanks alot!

Answer 51

Haha awesome. Yeah, just a subtle little thing always has to mess it up haha.

Answer 52

No kidding right! Thanks alot

Answer 53

could you help me with this question?