Linear differential operator

[Like Tony Stark]

Hi, I have some doubts about the theory of linear differential operators (LDO).
Let's consider the LDO L=(D^2-3), where D=d/dx
Its eigenvalues are the roots of the equation x^2-3=0
Then, the eigenfunctions are the solutions to the differential equation y''-3y=0? Or they are the solutions to y''=(3+λ)y?

And finally, is it invertible? When is a LDO not invertible?

 

[HallsofIvy]

Since you are using "x" in the differential equation it is a bad idea to use it in the eigenvalue equation! Use "\(\displaystyle \lambda\)" instead.

Since the "eigenvalue equation" is \(\displaystyle \lambda^2- 3= 0\) the eigenvalues are \(\displaystyle \lambda= \pm\sqrt{3}\). The eigenfunctions are the two independent solutions to the equation \(\displaystyle y''- 3y= 0\) which are \(\displaystyle y(x)=e^{x\sqrt{3}}\) and \(\displaystyle y(x)= e^{-x\sqrt{3}}\). The "general solution" to that differential equation is \(\displaystyle y(x)= Ae^{x\sqrt{3}}+ Be^{-x\sqrt{3}}\). Any solution to that differential equation is of that form for some constants, A and B.

Given any operator, D, the "eigenvalue problem" is \(\displaystyle Dy= \lambda y\). D is invertible (and finding y such that Dy= f is equivalent to inverting D- \(\displaystyle y= D^{-1}f\)) but since \(\displaystyle (D- \lambda)y= 0\), \(\displaystyle D- \lambda\), for \(\displaystyle \lambda\) an eigenvalue, is not invertible.