# Thread: Eigenfunctions of Linear Operator

1. ## Eigenfunctions of Linear Operator

Hello,

Find all the eigenvalues and eigenfunctions of the operator

----------> where the domain D is given by D{f belongs to : f is twice continuously differentiable and f(0)=f'(1)

I can sove differential equation such as
but the above notations are confusing me. Could you please get me started.

Thanks

2. Originally Posted by charikaar
Hello,

Find all the eigenvalues and eigenfunctions of the operator

----------> where the domain D is given by D{f belongs to : f is twice continuously differentiable and f(0)=f'(1) =0 (?)

I can solve differential equation such as
but the above notations are confusing me. Could you please get me started.
A second order differential operator usually needs two initial conditions. I'm guessing that f(0) and f'(1) should both be 0?

In that case, if T is the differential operator given by $\displaystyle Tf = -f''$, and f is an eigenfunction for the operator T, with eigenvalue $\displaystyle \lambda$, then $\displaystyle Tf = \lambda f$ (just as in linear algebra). So you need to solve the differential equation $\displaystyle -f'' = \lambda f$, together with the initial conditions f(0) = f'(1) = 0.

The general solution is $\displaystyle f(t) = A\cos ct + B\sin ct$, where $\displaystyle c = \sqrt{-\lambda}$. The initial condition f(0)=0 tells you that A=0. The condition f'(1)=0 then tells you that $\displaystyle cB\cos c = 0$. So the equation will only have a nonzero solution if $\displaystyle c=0$ or $\displaystyle \cos c=0$. In terms of $\displaystyle \lambda = -c^2$, that implies that $\displaystyle \lambda = 0$ or $\displaystyle \lambda = -(k+\tfrac12)^2\pi^2$, for some integer k. So those are the eigenvalues (an infinite sequence of them, unlike in linear algebra where you only have finitely many eigenvalues for a matrix).