Fourier analysis, second derivative

Question

I'm stuck with a part of a question, I've come to the part where $y''(t) = y(t+\pi)$, the solution says that this is equal to putting $y(t)=C$, where $C$ is a constant, I cannot understand how they do this, any help at this point is greatly appreciated!

Answer 1

Try taking two derivatives of $y(t+\pi)$, you'll get that $y''(t+\pi) = y(t)$, so $y^{(4)}(t) = y(t)$ (using $2\pi$ periodicity). This is a constant coefficient linear homogenous ODE, so assume that $y = re^{\lambda t}$, get that the characteristic equation is $\lambda^4+1 = 0$, so we get that solutions are of the form $\pm 1$, and $\pm i$.

So, we have that $y = c_1 e^{x}+c_2e^{-x}+c_3e^{ix}+c_4e^{-ix}$ is a solution. This isn't yet $2\pi$ periodic (the last two terms are, but the first two aren't). To make it periodic, we can set $y(t+2\pi) = y(t)$. This gives us that: \begin{align*} c_1e^{x+2\pi}+c_2e^{-x-2\pi}+c_3e^{ix+2\pi i}+c_4e^{-ix-2\pi i} & = c_1e^x+c_2e^{-x}+c_3e^{ix}+c_4e^{-ix} \\ c_1e^x(e^{2\pi}-1)+c_2e^{-x}(e^{-2\pi}-1) & = 0 \\ c_1e^{2x} & =c_2\frac{1-e^{-2\pi}}{e^{2\pi}-1} \end{align*} In the above, we have that the only $c_1,c_2$ such that this is true is $c_1 = c_2 = 0$. So, we get that for $y$ to be $2\pi$ periodic, we need $y(x) = c_3e^{ix}+c_4e^{-ix}$.

This means that $y''(x) = -c_3e^{ix}-c_4e^{ix} = -y(x)$ So, if we want $y''(t) = y(t+\pi)$ then we have that $-y(t) = y(t+\pi)$ or $$c_3e^{ix}+c_3e^{ix+i\pi}+c_4e^{-ix}+c_4e^{-ix-i\pi} = c_3e^{ix}(1+e^{i\pi})+c_4e^{-ix}(1+e^{-i\pi})$$ But, we have that $e^{i\pi} = e^{-i\pi} = -1$, so we have that $y = 0$ is our solution.

Finally, I've managed to forget the constant of integration this entire time, so our solution should really be $y = C$.