Thread: [SOLVED] eigenvalue problem

I want to find the eigenvalues and the eigenfunctions of the operator



so we have the problem



Now I put in the functie y=A*exp(n*x), is it correct to do this?, and get the equation



so Ln=n^2+k

but acording to the solutions it has to be Ln=n^2-k. Where do I make the mistake?
and how do I exactly determine the eigenfunctions

thanks

Originally Posted by charlie123

I want to find the eigenvalues and the eigenfunctions of the operator



so we have the problem



Now I put in the functie y=A*exp(n*x), is it correct to do this?, and get the equation



so Ln=n^2+k

but acording to the solutions it has to be Ln=n^2-k. Where do I make the mistake?
and how do I exactly determine the eigenfunctions

This problem as stated is incomplete. The differential operator is not completely defined unless some boundary conditions are specified. If the boundary condition says that y→∞ as x→∞ then you are correct to take functions of the form A*exp(n*x) as eigenfunctions. More often, the boundary conditions will say for example y=0 when x = 0 and when x = π. In that case, the eigenfunctions will be of the form A*sin(nx) and the eigenvalues will be k–n^2 (still not equal to the given solution n^2–k, but perhaps you are using a different convention about signs).

the boundary conditions are y(0)=0 and y(pi)=0

but how do I know what the the eigenfunction is? Here I can see it of course, but if I didn't know it for instance how do I start solving this problem?
The determining of the eigenvalue is not difficult after that.

thanks for the help

Originally Posted by charlie123

the boundary conditions are y(0)=0 and y(pi)=0

but how do I know what the the eigenfunction is? Here I can see it of course, but if I didn't know it for instance how do I start solving this problem?
The determining of the eigenvalue is not difficult after that.

The differential equation is , where is an eigenvalue. This is a standard SHM equation, and I guess you're supposed to know that the general solution is , where . The boundary condition y(0)=0 tells you that B=0, and the boundary condition y(π)=0 tells you that must be an integer, say n. Therefore .

So you see it's not a question of finding the eigenfunction first and then the eigenvalue. They both emerge together as a result of solving the differential equation together with its boundary conditions.