Eigenvalues of Linear operator

Feb 2008
184
1
Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
\(\displaystyle d/dt:D\)---------->\(\displaystyle L_2[0,1]\) where the domain D is given by D{f belongs to \(\displaystyle L_2[0,1]\): f is continuously differentiable and f(0)=0)}
I get \(\displaystyle f(x)=e^{\lambda x}\) after solving first order linear differential equation. How do i find \(\displaystyle \lambda\)?
thanks
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Hello,
Find all the eigenvalues and determine with a reason whether L is self-adjoint.
\(\displaystyle d/dt:D\)---------->\(\displaystyle L_2[0,1]\) where the domain D is given by D{f belongs to \(\displaystyle L_2[0,1]\): f is continuously differentiable and f(0)=0)}
I get \(\displaystyle f(x)=e^{\lambda x}\) after solving first order linear differential equation. How do i find \(\displaystyle \lambda\)?
thanks
our operator is just differentiation so, yes, \(\displaystyle \frac{d e^{\lambda x}}{dx}= \lambda e^{\lambda x}\). More generally, if \(\displaystyle \frac{dy}{dx}= \lambda y\), then \(\displaystyle y= Ce^{\lambda 0}\)\(\displaystyle e^{\lambda x}\) is an "eigenvector" and \(\displaystyle \lambda\) is an eigenvalue. As for finding \(\displaystyle \lambda\), you don't! That equation is true for all \(\displaystyle \lambda\)- any real number is an eigenvalue.

As for whether or not it is "self adjoint", the definition of self-adjoint is that <Af, g>= <f, Ag> for any vectors f, g. Here, I presume you are using the "standard" inner product, \(\displaystyle <f, g>= \int_0^1 f(x)g(x)dx\) so that we want \(\displaystyle \int_0^x \frac{df}{dx} g dt\). Use "integration by parts" with u= g and dv= (df/dx)dx. Then du= (dg/dx)dx and v= f.

\(\displaystyle \int u dv= uv- \int vdu= f(x)g(x)- \int f (dg/dx) dx= <f, dg/dx>\) provided f(1)g(1)- f(0)g(0)= 0.
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
I'm curious on one thing. Don't we require that \(\displaystyle f(0) = 0\)?
 
Feb 2008
184
1
Thanks for your help.

\(\displaystyle L_2[0,1]\) is the Hilbert space of complex square integrable functions on [0,1] and D={f belongs to \(\displaystyle L_{2}[0,1]\) : f is continuously differentiable and f(0)=1)}. so \(\displaystyle \lambda\) includes complex numbers also? L is not self-adjoint because there is minus in front of the integral right?

f(0)=1 not zero.