Sturm-Liouville theory applied to solve Helmholtz equation

[JordanGo]

Homework Statement

Find the eigenfunctions of the Helmholtz equation:
$\frac{d^2y}{dx^2}+k^2y = 0$

with boundary conditions:
$y(0)=0$
$y'(L)=0$

Homework Equations

General Solution:
$y = Asin(kx) + Bcos(kx)$

The Attempt at a Solution

I found that at y(0) that B=0 and that at y'(L) that Akcos(kL)=0.
What do I do with these results?

 

[LCKurtz]

Homework Statement

Find the eigenfunctions of the Helmholtz equation:
$\frac{d^2y}{dx^2}+k^2y = 0$

with boundary conditions:
$y(0)=0$
$y'(L)=0$

Homework Equations

General Solution:
$y = Asin(kx) + Bcos(kx)$

The Attempt at a Solution

I found that at y(0) that B=0 and that at y'(L) that Akcos(kL)=0.
What do I do with these results?

Since you already have B=0, you don't want to have A = 0 to avoid a trivial solution. So you need cos(kL) = 0. What values of kL will do that? That will determine the values of k for which the equation has non-trivial solutions. You may also need to check your original equation when k = 0 to see if that gives an eigenvalue too.

 

[JordanGo]

Ok, so to avoid trivial solutions, we can say that:
$k=\frac{pi}{2L} , k=\frac{3pi}{2L}$
Within the range of [0,2*pi].

But how is knowing a value of k going to help find the eigenfunctions?
If the values of k are the eigenvalues, how do I use them to find the eigenfunctions?

 

[HallsofIvy]

You have already said that B= 0. In order that the solution not be trivial, you must have $k= \pi/(2L)$ or $k= 3\pi/(2L)$. That means that the non-trivial solutions are $Asin(\pi x/(2L))$ and $Asin(3\pi x/(2L))$. Those are the eigenfunctions.

 

[LCKurtz]

And, of course, there is no reason to limit the $kL$ to $[0,2\pi]$. In fact, you best not do that if you want all the eigenvalues and eigenfunctions. And don't forget to check the system when $k=0$.

 

[JordanGo]

Thanks a lot for the help! You guys are great!

One more thing, when you say to check the equation when k=0, you mean use the general equation (knowing A=0) and say that y=B is a solution?

 

[LCKurtz]

Thanks a lot for the help! You guys are great!

One more thing, when you say to check the equation when k=0, you mean use the general equation (knowing A=0) and say that y=B is a solution?

I mean that $y'' + k^2 y = 0$ becomes $y''=0$ with your B.C.s $y(0)=0,\, y'(L)=0$. You don't want to forget to check that because sometimes you get a nontrivial solution which isn't included in the sine-cosine case.

 

[JordanGo]

so when k is zero, we know that y'(L)=0 and y''(x)=0, thus y=C , which is another constant?

 

[LCKurtz]

What is the general solution to y''(x)=0? Make it satisfy y(0)=0 and y'(L)=0.

 

[JordanGo]

Is this Laplace's equation? We haven't seen how to solve it yet... Can you help me? I've searched the internet and it shows plenty of ways to solve it in different coordinate systems, not really sure what to do with it...

 

[LCKurtz]

What is the general solution to y''(x)=0? Make it satisfy y(0)=0 and y'(L)=0.

Is this Laplace's equation? We haven't seen how to solve it yet... Can you help me? I've searched the internet and it shows plenty of ways to solve it in different coordinate systems, not really sure what to do with it...

Are you talking about y''(x)=0? It is just a simple calculus problem. Integrate it twice.

 

[JordanGo]

ok, so the solution to y''=0, y(0)=0 and y'(L)=0 is that either y=0 or y=C (some constant)

 

[LCKurtz]

Are you talking about y''(x)=0? It is just a simple calculus problem. Integrate it twice.

ok, so the solution to y''=0, y(0)=0 and y'(L)=0 is that either y=0 or y=C (some constant)

No comment until you show your work.

 

[JordanGo]

When k=0, the general solution is: y(x)=Ax+B.
First boundary condition says y(0)=0=B, thus B=0, and y'(L)=0=A, thus A=0.
So k=0 is not an eigenvalue.

 

[LCKurtz]

Good. So $y = C \ne 0$ is not an option.