Find the eigenfunction and eigenvalues of ##\sin\frac{d}{d\phi}##

[Wannabe Physicist]

Homework Statement

Find the eigenfunction and eigenvalues of $\sin\frac{d}{d\phi}$

Relevant Equations

.

Here is what I tried. Suppose $f(\phi)$ and $\lambda$ is the eigenfunction and eigenvalue of the given operator. That is,

$$\sin\frac{d f}{d\phi} = \lambda f$$
Differentiating once,
$$f'' \cos f' = \lambda f' = f'' \sqrt{1-\sin^2f'}$$
$$f''\sqrt{1-\lambda^2 f^2} = \lambda f'$$

I have no idea how to solve this non-linear differential equation. Is this approach even correct? I have also tried expanding the left-hand side of the eigenvalue equation into Taylor expansion of $\sin(f')$. All I get is a function containing higher derivatives of $f$ on one side and $\lambda f$ on the other side and once again I am stuck not knowing how to proceed. Please help

 

[Haborix]

Homework Statement:: Find the eigenfunction and eigenvalues of $\sin\frac{d}{d\phi}$
Relevant Equations:: .

Here is what I tried. Suppose $f(\phi)$ and $\lambda$ is the eigenfunction and eigenvalue of the given operator. That is,

$$\sin\frac{d f}{d\phi} = \lambda f$$
Differentiating once,
$$f'' \cos f' = \lambda f' = f'' \sqrt{1-\sin^2f'}$$
$$f''\sqrt{1-\lambda^2 f^2} = \lambda f'$$

I have no idea how to solve this non-linear differential equation. Is this approach even correct? I have also tried expanding the left-hand side of the eigenvalue equation into Taylor expansion of $\sin(f')$. All I get is a function containing higher derivatives of $f$ on one side and $\lambda f$ on the other side and once again I am stuck not knowing how to proceed. Please help

It isn't pretty and I doubt there is an expression for $f$ in terms of elementary functions, but you could just take arcsine of both sides and directly integrate.

 

[Wannabe Physicist]

Thanks for your response. This is what I did

$$\frac{df}{d\phi} = \sin^{-1}(\lambda f)$$
On integrating the $df$ integral using online integral calculator
$$ \frac{\operatorname{Ci}\left(\arcsin\left(\lambda f\right)\right)}{\lambda} = \phi+C$$

It seems this cannot be expressed in terms of elementary functions, just as you said. I am trying to see if this can be expressed in the form of $f=f(\phi)$. Is it even possible?

 

[dextercioby]

$\sin\left(\frac{d}{d\phi}\right) = \mathfrak{Im} \left( e^{i\frac{d}{dx}}\right)$

Does this help?

 

[Haborix]

Thanks for your response. This is what I did

$$\frac{df}{d\phi} = \sin^{-1}(\lambda f)$$
On integrating the $df$ integral using online integral calculator
$$ \frac{\operatorname{Ci}\left(\arcsin\left(\lambda f\right)\right)}{\lambda} = \phi+C$$

It seems this cannot be expressed in terms of elementary functions, just as you said. I am trying to see if this can be expressed in the form of $f=f(\phi)$. Is it even possible?

It would probably just be an even more horrible function. In the end, you can always just define a function which is the inverse of another, but that's little more than giving something a name. Insight comes when there are known properties of the variously named special functions.

 

[Wannabe Physicist]

It would probably just be an even more horrible function. In the end, you can always just define a function which is the inverse of another, but that's little more than giving something a name. Insight comes when there are known properties of the variously named special functions.

Okay. So I should simply state that the eigenfunctions are all those functions $f(\phi)$ which satisfy $\operatorname{Ci}\left(\sin^{-1}\lambda f\right)/\lambda +C =\phi$ and the corresponding eigenvalue is $\lambda$. Is that right?

 

[Wannabe Physicist]

$\sin\left(\frac{d}{d\phi}\right) = \mathfrak{Im} \left( e^{i\frac{d}{dx}}\right)$

Does this help?

I do not understand how to use this fact actually. I am trying to eyeball some eigenfunction for $e^{i\frac{d}{dx}}$ but no success as of yet.