Finding the Set of Eigenvalues

[A.Magnus]

How do I find the set of all eigenvalues of $F^2$, if the problem gives me that $F: X \mapsto X$ is the derivative mapping defined in the space $X$ of all infinitely differentiable real functions?

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

 

[I like Serena]

Hey MaryAnn! (Smile)

Eigenvalues and eigenvectors are defined as solutions for:
$$Ax=\lambda x$$
where $x$ is non-zero.

If $A$ is a matrix we can rewrite it as:
$$Ax - \lambda x = (A-\lambda I)x = 0$$
For non-zero $x$ it has a solution if and only if the determinant $|A-\lambda I|$ is zero.

In this case we have a function (which is effectively an infinite-dimensional vector) instead of a n-dimensional vector, and we have a differential operator instead of a matrix, but the same theory applies:
$$F^2 f(x) = \lambda f(x) \quad\Rightarrow\quad F^2 f(x) - \lambda f(x) = (F^2 - \lambda \operatorname{id}) \circ f(x) = 0$$
However, we can't use the determinant to solve the system now.
Instead we have to solve the corresponding differential equation:
$$f''(x) - \lambda f(x) = 0$$Questions 2-4 are about solving a 2nd order linear differential equation, which is explained for instance here.
Maybe you have notes of your own about solving such differential equations? (Wondering)

 

[A.Magnus]

Hey MaryAnn! (Smile)

Eigenvalues and eigenvectors are defined as solutions for:
$$Ax=\lambda x$$
where $x$ is non-zero.

If $A$ is a matrix we can rewrite it as:
$$Ax - \lambda x = (A-\lambda I)x = 0$$
For non-zero $x$ it has a solution if and only if the determinant $|A-\lambda I|$ is zero.

In this case we have a function (which is effectively an infinite-dimensional vector) instead of a n-dimensional vector, and we have a differential operator instead of a matrix, but the same theory applies:
$$F^2 f(x) = \lambda f(x) \quad\Rightarrow\quad F^2 f(x) - \lambda f(x) = (F^2 - \lambda \operatorname{id}) \circ f(x) = 0$$
However, we can't use the determinant to solve the system now.
Instead we have to solve the corresponding differential equation:
$$f''(x) - \lambda f(x) = 0$$Questions 2-4 are about solving a 2nd order linear differential equation, which is explained for instance here.
Maybe you have notes of your own about solving such differential equations? (Wondering)

Hi ILS again, I am reading your posting with tons of gratitude even as I am typing this response. Thanks for your patient and detail response, your format is exactly what I am needing. Please give me some times to digest it and for the meantime, please take this as my acknowledgment of your gracious help and my gratitude. Thanks again and again. ~MA

PS. Yes, now I do remember the solution to the 2nd order linear differential equation. Thank you.

 

[I like Serena]

Question #4 #5: 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.

From the solutions of the differential equation, we can always find a solution, no matter which $\lambda \in \mathbb R$ we pick.
So all real values are valid eigenvalues for which we can find an (actually infinitely many) eigenfunction(s). (Thinking)

 

[A.Magnus]

From the solutions of the differential equation, we can always find a solution, no matter which $\lambda \in \mathbb R$ we pick.
So all real values are valid eigenvalues for which we can find an (actually infinitely many) eigenfunction(s). (Thinking)

Got it now! Thank you again for the second time for your gracious helping hand. ~MA

 

[A.Magnus]

Hey MaryAnn! (Smile)

Eigenvalues and eigenvectors are defined as solutions for:
$$Ax=\lambda x$$
where $x$ is non-zero.

If $A$ is a matrix we can rewrite it as:
$$Ax - \lambda x = (A-\lambda I)x = 0$$
For non-zero $x$ it has a solution if and only if the determinant $|A-\lambda I|$ is zero.

In this case we have a function (which is effectively an infinite-dimensional vector) instead of a n-dimensional vector, and we have a differential operator instead of a matrix, but the same theory applies:
$$F^2 f(x) = \lambda f(x) \quad\Rightarrow\quad F^2 f(x) - \lambda f(x) = (F^2 - \lambda \operatorname{id}) \circ f(x) = 0$$
However, we can't use the determinant to solve the system now.
Instead we have to solve the corresponding differential equation:
$$f''(x) - \lambda f(x) = 0$$Questions 2-4 are about solving a 2nd order linear differential equation, which is explained for instance here.
Maybe you have notes of your own about solving such differential equations? (Wondering)

One more request if you do not mind at all: Is there anyway you could give me an example of differential operator (appropriate for this problem) for me to visualize? I remotely remember seeing differential operator some years ago, but unfortunately I forget. Perhaps these following are appropriate?

$$D^2 = \frac{d^2}{dx^2}, \quad D^3 = \frac{d^3}{dx^3}.$$

Thank you again for all your help. ~MA

 

[I like Serena]

One more request if you do not mind at all: Is there anyway you could give me an example of differential operator (appropriate for this problem) for me to visualize? I remotely remember seeing differential operator some years ago, but unfortunately I forget. Perhaps these following are appropriate?

$$D^2 = \frac{d^2}{dx^2}, \quad D^3 = \frac{d^3}{dx^3}.$$

Thank you again for all your help. ~MA

Hmm... let's see if I understand what you're asking... (Thinking)

In this particular case we have $F=\frac{d}{dx}$ and $F=\frac{d^2}{dx^2}$.
So $F^2 f = \frac{d^2}{dx^2} f$.
$F$ is applied to a function, and the result is another function (the 2nd derivative).
Another differential operator is $\frac{\partial}{\partial y}$, if we have a function with more than one argument.
Or we could have for instance $L=\frac{d^2}{dx^2} - 2 \frac{d}{dx} - \operatorname{id}$ (a so called elliptical operator). That is, $Lf = f'' - 2f' - f$.

 

[A.Magnus]

Hmm... let's see if I understand what you're asking... (Thinking)

In this particular case we have $F=\frac{d}{dx}$ and $F=\frac{d^2}{dx^2}$.
So $F^2 f = \frac{d^2}{dx^2} f$.
$F$ is applied to a function, and the result is another function (the 2nd derivative).
Another differential operator is $\frac{\partial}{\partial y}$, if we have a function with more than one argument.
Or we could have for instance $L=\frac{d^2}{dx^2} - 2 \frac{d}{dx} - \operatorname{id}$ (a so called elliptical operator). That is, $Lf = f'' - 2f' - f$.

Thank you! ~MA