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So I am in the section of our book about Strum-Liouville problems and all of the previous questions have had the boundary values equal to a constant (such as $y(10) = 0\,\,\,\,or\,\,\ y(L) = 0)$. But this problem has them set equal to other functions so I am at a bit of a loss. Here is my work so far:

The question in the book says: Find the eigenvalues and eigenfunctions for $$y''+λy=0, \,\,\,\,\,\,y(0) = y(1)\,\,\,\,\,\,\,\,y'(0) = y'(1)$$

I am assuming that $λ>0$ and I am replacing $λ$ with $k^2$ and assuming that $k$ is a non-zero constant.

After going through the characteristic equation I find that $r = \pm ki$.

I then apply Euler's formula because of the imaginary roots and get: $$y(x) = A_1\cos(kx)+iA_2\sin(kx)$$

Now this is where I am not sure if I am heading in the right direction. I figure that I should go ahead and plug in the boundary values and see what happens.

So I end up with: $$y(0) = A_1\cos(0)+iA_2\sin(0) = A_1\cos(k)+iA_2\sin(k) = y(1)$$

since $\cos(0)=1$ and $\sin(0)=0$ so I then simplify: $$y(0) = A_1 = A_1\cos(k)+iA_2\sin(k) = y(1)$$

I could then do some algebra and solve for A$_1$ in terms of A$_2$, but I don't know if that's really going to get me anywhere. Am I heading in the right direction so far?

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What you are doing will eventually work; at the end of the day you will get a linear system and it will have a nontrivial solution when $k$ takes on certain values. But it is usually easier to stay in complex exponential land, allowing your arbitrary constants to perhaps be complex if need be. Doing this in this case you get the following:

$$y(0)=A=y(1)=Ae^{ik} \\ y'(0)=Aik=y'(1)=Aike^{ik}.$$

So you need $e^{ik}=1$; how do you restrict $k$ to obtain this?

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  • $\begingroup$ I don't really see how I could restrict k to get e$^ik$=1 because wouldn't I need to have ik=0, and k is strictly greater than 0. I guess I'm not sure what you're getting at. $\endgroup$ – Eric Oct 21 '15 at 22:03
  • $\begingroup$ @Eric You can have $e^{ik}=1$ without having $ik=0$. Think carefully :) $\endgroup$ – Ian Oct 21 '15 at 22:06
  • $\begingroup$ Are you talking about if we represent e$^ik$ as sines and cosines and what values would make that expression equal 1? $\endgroup$ – Eric Oct 21 '15 at 22:13
  • $\begingroup$ You can do it with sines and cosines if you want. You can also do it with just complex exponentials if you want, if you think the right way. Either way the answer is actually fairly simple. (Consider for instance that in the case with $y(0)=y(1)=0$, you get $k=\pi n$ where $n$ is any positive integer. You should expect something like this.) $\endgroup$ – Ian Oct 21 '15 at 22:22
  • $\begingroup$ This may seem like a silly question, but can we assume that y(0) = y(1) = 0? Are we just assuming this because y"+λy = 0? All the other questions I have had explicitly stated that the boundary values were equal to 0. $\endgroup$ – Eric Oct 21 '15 at 22:52
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Here the differential equation is $$y''+λy=0,~~~~ y(0) = y(1)~~~\text{and}~~~y'(0) = y'(1)\tag1$$

Case I: If $~\lambda~=~0~$, then the given equation becomes $$y''=0\qquad \text{with $~y(0) = y(1)~$and$~y'(0) = y'(1)~$}$$

Solution of the above equation is $~y(x)=~a~x+b\qquad $where $~a,~b~$ are constants.

From the given conditions $~a=0~$and $~b~$is arbitrary.

So eigen function corresponding to the eigen value $~\lambda =0~$ is $~y(x)=b~$, where $~b~$is arbitrary constant.

Case II: If $~\lambda~=-m^2~\lt~0~$ where $~m\in \mathbb{N}~$, then the given equation becomes $$y''-m^2~y=0\qquad \text{with $~y(0) = y(1)~$and$~y'(0) = y'(1)~$}$$

Solution of the above equation is $~y(x)=~a~e^{m~x}+b~e^{-m~x}\qquad $where $~a,~b~$ are constants.

From the given conditions

$$~a+b=a~e^m+b~e^{-m}~$$$$~am-bm=a~m~e^m-b~m~e^{-m}~\implies a-b=a~e^m-b~e^{-m}$$ which gives $~a=b=0~$as$~~m\ne 0~$

So eigen function corresponding to the eigen value $~\lambda \lt 0~$ is $~y(x)=0~$, which is the trivial solution.

Case III: If $~\lambda~=m^2~\gt~0~$ where $~m\in \mathbb{N}~$, then the given equation becomes $$y''+m^2~y=0\qquad \text{with $~y(0) = y(1)~$and$~y'(0) = y'(1)~$}$$

Solution of the above equation is $~y(x)=~a~\cos{(m~x)}+b~\sin{(m~x)}\qquad $where $~a,~b~$ are constants.

Now $~y(0) = y(1)~\implies a=~a~\cos m +b~\sin m $

$y'(0) = y'(1)\implies b=-~a~\sin m +b~\cos m$

which gives either $~a=b=0~$ or, $~\cos m =1~\implies m=2n\pi~$ for $~n\in \mathbb{Z}~$ with $~a,~b~$ are arbitrary.

Now if $~a=b=0~$, we again get the trivial solution.

When $~m=2n\pi~$ for $~n\in \mathbb{Z}~$, the eigen value of the given differential equation $(1)$ is $$~\lambda=m^2=~4~n^2~\pi^2~~~~~\text{where $~n\in \mathbb{Z}~$ }~$$

and the corresponding eigen functions are $$~y(x)~=~a~\cos(2n\pi~x)+b\sin(2n\pi~x)\qquad \text{where $~n\in \mathbb{Z}~~~$and $~a,~b~$ are constants}~.$$

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