Question

(Sturm-Liouville) I'm having trouble in dealing with SLDE's in determining the Eigenfunction. I can generally determine the eigenvalues without too much struggle, but I am failing to understand how to systematically arrive at eigenfunctions for a given eigenvalue. More info/example below.

Answer 1

In this document in Problem 1 (attempting to work through the problem before looking at solution) I was able to get all the way to the Eigenvalue, but I don't understand why the eigenfunction is a sine function. http://www2.math.uu.se/~gaidash/1MA208/Files/Solutions_2.pdf

@Kainui , could you help me out with this?

Answer 2

(Or @freckles )

Answer 3

do you understand how they got their u?

Answer 4

\[u=A \cos(w \ln(x))+B \sin(w \ln(x)) ? \\ \text{ we apply conditions } \\ \text{ first I will find } u' \text{ before doing so } \\ u'=\frac{-Aw}{x}\sin(w \ln(x))+\frac{Bw}{x} \cos(w \ln(x)) \\ \text{ anyways } u(1)=0 \text{ means we have } \\ 0=A \cos(0)+Bsin(0)=A+0 \\ \text{ which means } A=0 \\ \text{ so far we have } u=B \sin(w \ln(x)) \text{ as the solution }\]

Answer 5

we need to apply the other condition

Answer 6

but cos should definitely not be apart of the egienfunction there since A=0

Answer 7

now let's look at the u' think and we only need to look at u'=Bw/x * cos(w ln(x)) since A=0

Answer 8

\[u'(e)=\frac{Bw}{e}\cos(w)=0\]

Answer 9

this means either B=0 or w=0 or cos(w)=0
We don't want B=0 and we don't want w=0
B=0 means we have a trivial solution and we looking for a nontrivial solution
w=0 means we will also have a trivial solution since it will mean lambda is 0
then we would have (xu')'=0 which implies xu'=c
u'=c/x
u=c ln|x|+d
and using the condition u'(e)=0 we have 0=c/e which means c is 0 anyways
so then you have u=d but d is also 0 since u(1)=0

So that means the only thing that we could allow to be 0 is the cos(w) thing
since this would lead to a nontrivial solution

Answer 10

@Mendicant_Bias

Answer 11

Sorry, was in lecture. I'm looking at this now.

Answer 12

ok i will try to clarify anything I said above if you need more clarification on anything

Answer 13

Heh, I think I linked the wrong document, but this is nonetheless helpful and may be nearly the exact same problem

Answer 14

Yeah, the one I intended to link was this one: http://www.ima.umn.edu/~miller/4567midtermsolns2S10.pdf

Answer 15

(Problem 3)

Answer 16

@freckles

Answer 17

So, no Cauchy-Euler, no natural logs involved in the complex roots solution, and slightly different Boundary Conditions. I evaluate it like this: (one second)

Answer 18

\[y''+\lambda y=0, \ \ \ 0<x<\pi, \ \ \ y(0)=0, y'(\pi)=0.\]\[y_h=c_1\cos(\alpha x)+c_2\sin(\alpha x) \]

Answer 19

\[y(0)=c_1\cos(\alpha x) +c_2 \sin(\alpha x)=0, \]\[c_1\cos(0)=0, \ \ \ \mathbf{c_1=0}\]

Answer 20

\[y'(\pi)=[-c_1 \sin(\alpha x)+c_2\cos(\alpha x)]\alpha=0\]

Answer 21

\[y(0)=c_1(\cos( \alpha \cdot 0))+c_2 (\sin( \alpha \cdot 0))=c_1+0 \\ y(0)=c_1 \\ \text{ but since } y(0)=0 \text{ then } c_1=0 \\ \text{ so far we can only have this as a possible nontrivial solution } y=c_2 \sin(ax )\]

Answer 22

\[c_2 \alpha \cos(\alpha \pi)=0, \ \ \ \alpha \pi=\frac{(2n-1)\pi}{2}, \ \ \ \alpha =\frac{(2n-1)}{2}\]

Answer 23

(Solving for eigenvalue where lambda = alpha squared, etc etc, just confused about how to find the eigenfunction for this case)

Answer 24

the eigenfunction is the solution that doesn't give us a trivial case

Answer 25

the trivial case being y(x)=0

Answer 26

(One minute, trying to phrase my questions correctly)

Answer 27

\[y=c_2 \sin(\alpha x) \\ y'=c_2 \cdot \alpha \cos(\alpha x) \\ y'(\pi)=c_2 \alpha \cos(\alpha \cdot \pi) \\ \text{ but } y'(\pi)=0 \text{ so this means we have } \\ c_2 \alpha \cos( \alpha \cdot \pi)=0 \\ \text{ case 1: } c_2=0: \text{ gives } y=0+0=0 \text{ this is a trivial solution } \\ \text{ we do not want to consider that possibility } \\ case 2: \alpha=0: \text{ gives } \lambda=0 \text{ since } \lambda=\alpha^2 \text{ which means you would have the diff equa } \\ y''(x)=0 \text{ with the conditions you gave } y'=c \\ y''=c+dx \\ y'=c \\ y'(\pi)=0 \text{ tells us } c=0 \\ y(0)=0 \text{ tells us } d =0 \\ \text{ so } \alpha=0 \text{ gives a trivial solution \because } y=0+0 =0 \text{ again } \]
so the only other thing that can give us a possible nontrivial solution and also satisfy the equation:
\[c_2 \alpha \cos( \alpha \cdot \pi)=0 \\ \text{ is allowing } \cos( \alpha \cdot \pi)=0 \\ \text{ is the third case } \\ \text{ if } \cos(\alpha \cdot \pi)=0 \text{ then } \alpha \cdot \pi=n \cdot \frac{\pi}{2} \text{ where } n \text{ is an add integer } \\ \alpha \cdot \pi=\frac{n \pi}{2} \text{ implies } \alpha=\frac{ n }{2} \text{ where } n \text{ is odd } \\ \text{ so recall that we had } y=c_2 \sin(\alpha x) \text{ and we just found from the second } \\ \text{ condition that } \alpha \text{ needs to be } \frac{ n}{2} \text{ where again } n \text{ is odd }\]

Answer 28

add=odd
i'm done editing that

Answer 29

Their value for alpha is different than yours, and so is mine if I've read correctly, I'm a little confused as to why alpha isn't \[\alpha=\frac{(2n-1)}{2}\]

Answer 30

actually the alpha is not different

Answer 31

my is n/2 where is odd integer
their's (2k-1)/2 where k is an integer
2k-1 is add for any integer k

Answer 32

omg odd*

Answer 33

Ah, just missed where you wrote that n is odd, got it

Answer 34

probably because i called it add integer
:p

Answer 35

but I think I also referred to as odd integer too in the same paragraph

Answer 36

Alright, so procedurally, you first solve for your eigenvalues after finding the general solution and applying boundary conditions, and then, *after* you have your eigenvalues that work, you plug them back into the general solution along with boundary conditions in order to get your eigenfunction.

Answer 37

yeah i see which of the possible egienvalues gives me eigenfunctions by pluggin in to the orginial
if I don't get y=0 then the possible eigenvalue is an actual eigenvalue
and whatever y equals from that is my eigenfuction

Answer 38

and I didn't do case 3

Answer 39

their case 3

Answer 40

in case 3 they used their conditions for the general from and seen that A=B=0 for lambda in the form of -alpha^2
that is trivial solution having y=0
so there are no egienvalues and therefore no egienfunctions

Answer 41

Yep, got that part. Alright, awesome, thank you so much, I needed clarity on this.

Answer 42

Ok so you are good, for real?
like there were no other possibilities for case 3
e^(u)+e^(v) is never 0 for any v or u because e^u>0 for all u
and e^v>0 for all v
so that means that we can only have A=0
and since B=-A then B=0
which gives us y=0

Answer 43

Yeah, I'm aware, the others made sense to me due to them being trivial, I literally just did not know the proper procedure to find an eigenfunction (but understood how to find eigenvalues).