Question

(Sturm-Liouville) I'm having trouble understanding how in a given problem they get an eigenfunction that they did. I understand how they got their eigenvalue. More info below.

Answer 1

In page 13 of the following PDF, problem III, I get how they got the eigenvalue to be \[\lambda=\bigg(\frac{n \pi}{l}\bigg)^2\]
But I do not understand why they got the sine function as their eigenfunction. Isn't the eigenfunction supposed to be the result of taking your eigenvalue and plugging it back into your original, general solution? This would yield \[y_n=\cos\bigg(\frac{n \pi x}{l}\bigg)\]To me.

Answer 2

I'm thinking this must just be a typo or something, but does what I'm saying make sense?

Answer 3

(?)

Answer 4

I'm literally just asking if this is a typo, could somebody who knows this better than me help me out? I'm 90% sure this is just a mistake, but virtually all it takes is a "yes" or "no" from someone who is familiar with this stuff.

Answer 5

@mathmath333

Answer 6

Thanks, someone just told me that I didn't link the PDF. Somebody mentioning that prior would've been great. http://texas.math.ttu.edu/~gilliam/ttu/ode_pde_pdf/Ch5.pdf

Answer 7

wait wahts the confusion its right

Answer 8

notice the eigen values are just different form from 2 and 3

Answer 9

But where is the additional pi/2 coming from? The argument of the original function being \[y_h=c_1\sin(\alpha x)+c_2\cos(\alpha x)\]Where c_1=0, evaluating the original function at the eigenfunction gives \[y_n =\cos\bigg(\frac{n \pi x}{l}\bigg)\]

Answer 10

*evaluating the original function at the eigenvalue

Answer 11

u cancelled out the wrong part

Answer 12

c2=0

Answer 13

\[y_h=c_1\sin(\alpha x)+c_2\cos(\alpha x)\]\[y'_h=c_1 \alpha \cos(\alpha x)-c_2 \alpha \sin(\alpha x)\]

This is for Part III of the problem, not Parts I or II.

Answer 14

\[y'(0)-c_1 \alpha \cos(0)+c_2 \alpha \sin(0)=0, --> c_1=0.\]

Answer 15

^Equals sign, not minus, and minus, not plus.

Answer 16

they are saying b=0 so u have to start with y'(l) first

Answer 17

y'(0) doesnt give us any useful information in this case

Answer 18

I don't understand what the issues is, I'll put it like this: We agree that the eigenvalue is n^pi^2/l^2, right? We do, don't we?

Answer 19

The way we find an eigenfunction is by taking the eigenvalue and substituting it into the original function. If b (or c_1)=0, then for your original function, you have \[y+h=c_1\cos(\alpha x)\] Right, @dan815 ???

Answer 20

*y_h

Answer 21

\[y_n=c_1\cos(\alpha x)=c_1 \cos\bigg(\frac{n \pi x}{l}\bigg)\]

Answer 22

@bibby

Answer 23

We agree. We agree, right? We agree that \[\lambda =\frac{n^2\pi^2}{l^2}\]

Answer 24

your method is not wrong, its just in this question b=0

Answer 25

The eigenfunction is determined by taking the eigenvalue-which we both agree is the above-and plugging it into the general solution, yes?

Yes, we both agree that b is zero, I'm not disputing that, at all, whatsoever, not at all, dan, that's not what I'm confused about.

Answer 26

ok so

Answer 27

If b is zero, then the original solution must be \[y_h=c_1\cos(\alpha x)\]
Right?

Answer 28

right

Answer 29

so now u do y'(l)

Answer 30

oh i see your ur problem u are confused about their final answer is in sin

Answer 31

thats obviously a mistake xD

Answer 32

Taking the agreed upon eigenvalue-but that doesn't matter, we already know about and agree about the eigenvalue, we get our eigenfunction by substituting our general solution, no derivatives or anything, right? If you have your eigenfunction-HWSD;KJSDFKL;SL;

Answer 33

Alright, I'm done, goodnight. Thank you, Dan.

Answer 34

lol np

Answer 35

Yeah, I just kept earlier over and over asking if it was a typo, that's what I was concerned about.

Answer 36

trying to do too many questions at once

Answer 37

im just skimming over comments

Answer 38

No worries. thanks again.