- #1

A.Magnus

- 138

- 0

I found this solution from an online source which is not known for being reliable. Therefore I am not sure if the solution is correct. But if it is, I still have some questions to ask: (My questions are in

*italic*.)

SOLUTION: ~~~~~~~~~~~~~~~~~~~~~~~~~~~

The eigenvalues are given by $F^2 (f(x))=\lambda f(x)$. That is, $f''(x)-\lambda f(x)=0$. Corresponding characteristic equation is $D^2-\lambda=0.$

*Question #1: I know from my class that to find eigenvalues, first I need to set up characteristic equation $|A - \lambda I| = 0$. But here, where are these $F^2 (f(x))=\lambda f(x)$, $f''(x)-\lambda f(x)=0$ and $D^2-\lambda=0$ coming from?*

There are 3 cases:

(1) $\lambda=0$ so that $D^2=0\Rightarrow D=0,\ 0$ and the solutions in this case are $f(x)=a+bx$ where $\lambda=0$ is the eigenvalue.

*Question #2: Where is this $f(x)=a+bx$ coming from?*

(2) $\lambda>0$ so that $D^2=\lambda \Rightarrow D=\pm \sqrt{\lambda}$ and the solutions in this case are $f(x)=ae^{ \sqrt{\lambda }x}+be^{-\sqrt{\lambda }x}$ where $\lambda>0$ is the eigenvalue.

*Question #3: Where is this $f(x) =ae^{ \sqrt{\lambda }x}+be^{-\sqrt{\lambda }x}$ coming from?*

(3) $\lambda<0$ so that $D^2=\lambda\Rightarrow D=\pm i \sqrt{-\lambda}$ where $-\lambda >0$. The solutions in this case are $f(x)=a\cos({ \sqrt{\lambda }x})+b\sin ({\sqrt{\lambda }x})$ where $\lambda<$0 is the eigenvalue.

*Question #4: Again, where is this $f(x)=a\cos({ \sqrt{\lambda }x})+b\sin ({\sqrt{\lambda }x})$ coming from?*

Therefore, the set of all possible eigenvalues is $\lambda \in \mathbb{R}$, that is, set of all real numbers.

*Question #4: I am sorry but I did not see any connection between the case (1), (2) and (3) above and suddenly the conclusion that all possible eigenvalue is the set of all real numbers.*

END OF SOLUTION: ~~~~~~~~~~~~~~~~~~~~~~~~~

As always, thank you fro your gracious helping hand and time spent writing a response to this posting. ~MA