General form of a $2\pi$-periodic $C^2$ function satisfying a certain condition

Question

Suppose $f$ is a $2\pi$-periodic $C^2$ function on $\Bbb R$ satisfying $$ \lambda f(x)-\dfrac{\kappa}{2\pi}\int_{-\pi}^\pi \cos (\psi-x)f(\psi)d\psi=\sigma f''(x),$$ where $\lambda, \kappa$ and $\sigma$ are positive constants. I am asked to find a relation between this constants, and to find the general form of the function $f$, but I have no idea of how to start, because I've never seen this problem before. Any hints?

Answer 1

Remember that

If $f \colon \mathbb{R} \to \mathbb{C}$ a $2\pi$-periodic function and $f \in C^1(-\pi,\pi) $ then $$f(x) = \sum_{n\in \mathbb{Z}} f_n \operatorname{e}^{\operatorname in x} $$ and the Fourier coefficients $f_n$ are defined as $$f_n = \frac{1}{2\pi}\int_{-\pi}^\pi f(x)\operatorname{e}^{-\operatorname in x} \operatorname{d}x $$ The Fourier coefficients of the second derivative are $${f_n}'' = (\operatorname{i}n)^2f_n=-n^2f_n$$

So the equation becomes $$ \begin{split} \sum_{n \in \mathbb Z} -\sigma n^2 f_n \operatorname{e}^{\operatorname i nx} &= \lambda\sum_{n \in \mathbb Z} f_n \operatorname{e}^{\operatorname i nx} -\frac{\kappa}{2\pi}\int_{-\pi}^\pi \cos{(\psi -x)}\sum_{n \in \mathbb Z} f_n \operatorname{e}^{\operatorname i n\psi} \operatorname d \psi\\ &= \sum_{n \in \mathbb Z} \lambda f_n \operatorname{e}^{\operatorname i nx} -\sum_{n \in \mathbb Z} f_n \frac{\kappa}{2\pi}\int_{-\pi}^\pi \cos{(\psi -x)}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi\\ &= \sum_{n \in \mathbb Z} f_n \left\{ \lambda\operatorname{e}^{\operatorname i nx} -\frac{\kappa}{2\pi}\int_{-\pi}^\pi \cos{(\psi -x)}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi\right\}\\ \end{split} $$

So the problem is to evaluate the terms inside the curly brackets. Let's begin with the integral; using the Euler identity

$$\operatorname{e}^{\operatorname{i} z} = \cos z + \operatorname i \sin z$$

we get

$$ \begin{split} \int_{-\pi}^\pi \cos{(\psi -x)}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi &= \int_{-\pi}^\pi \frac{\operatorname{e}^{\operatorname i (\psi - x)} +\operatorname{e}^{-\operatorname i (\psi - x)}}{2}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi\\ &= \frac{1}{2}\int_{-\pi}^\pi \operatorname{e}^{\operatorname i (\psi - x)}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi +\frac{1}{2}\int_{-\pi}^\pi \operatorname{e}^{-\operatorname i (\psi - x)}\operatorname{e}^{\operatorname i n\psi} \operatorname d \psi \\ &= \frac{\operatorname{e}^{-\operatorname i x}}{2}\int_{-\pi}^\pi \operatorname{e}^{\operatorname i \psi(n+1)}\operatorname d \psi +\frac{\operatorname{e}^{\operatorname i x}}{2}\int_{-\pi}^\pi \operatorname{e}^{\operatorname i \psi(n-1)} \operatorname d \psi \\ &= \frac{\operatorname{e}^{-\operatorname i x}}{2}\frac{\operatorname{e}^{\operatorname i \psi(n+1)}}{\operatorname i (n+1)} \Bigg|_{-\pi}^\pi +\frac{\operatorname{e}^{\operatorname i x}}{2}\frac{\operatorname{e}^{\operatorname i \psi(n-1)}}{\operatorname i (n-1)}\Bigg|_{-\pi}^\pi \\ &= \frac{\operatorname{e}^{-\operatorname i x}}{2\operatorname i (n+1)}\left(\operatorname{e}^{\operatorname i \pi(n+1)}-\operatorname{e}^{-\operatorname i \pi(n+1)}\right)+ \frac{\operatorname{e}^{\operatorname i x}}{2\operatorname i (n-1)}\left(\operatorname{e}^{\operatorname i \pi(n-1)}-\operatorname{e}^{-\operatorname i \pi(n-1)}\right)\\ &= \frac{\operatorname{e}^{-\operatorname i x}}{2\operatorname i (n+1)}2\sin{\pi(n+1)}+ \frac{\operatorname{e}^{\operatorname i x}}{2\operatorname i (n-1)}2\sin{\pi(n-1)} = 0 \\ \end{split} \\ $$

Hence

$$ \sum_{n \in \mathbb Z} -\sigma n^2 f_n \operatorname{e}^{\operatorname i nx} = \sum_{n \in \mathbb Z} \lambda f_n \operatorname{e}^{\operatorname i nx} $$ therefore $$ -\sigma n^2 f_n = \lambda f_n $$

But, since $\sigma$ and $\lambda$ are positive constant then $f_n=0$ for every $n \in \mathbb Z$.

Finally, $$ f(x)= \sum_{n \in \mathbb Z} 0 = 0 $$ so $f$ is the null function.

Notice that if $\sigma = \lambda = 0$ then $f_n$ are not forced to be null.