Consider the differential equation: $$(2-\cos(\pi x))y''+y=1, -1\le x \le 1, y(-1)=y(1).$$
Let the Fourier series of $y$ be $y(x)=\displaystyle\sum_{n=-\infty}^{n=\infty}\hat{y}_ne^{i\pi n x}$, where $\hat{y}_n=\frac{1}{2}\displaystyle\int_{-1}^{1}y(x)e^{-i\pi n x} dx$. I need to prove that $\hat{y}_n=\hat{y}_{-n}$.
My attempts:
Another way of showing that $\hat{y}_n=\hat{y}_{-n}$, is to show that the imaginary part of $\hat{y}_n$ is zero, or equivalently that $\displaystyle\int_{-1}^{1}y(x)\sin{(\pi n x)} dx=0$.
By substituting the Fourier series of $y$ into the differential equation we get:
$$-2 \pi^2 n^2 \hat{y}_n+\hat{y}_n+\pi^2(\frac{(n-1)^2}{2}\hat{y}_{n-1}+\frac{(n+1)^2}{2}\hat{y}_{n+1})=0$$
which can be used to prove recursively that $\hat{y}_n=\hat{y}_{-n}$, if we know that $\hat{y}_1=\hat{y}_{-1}$ for example.
Would anyone please help me solving this.
Thank you.
The question seems to implicitly assert that the solution of this boundary value problem is unique.
Let $u=y(-x)$, then $u(1)=y(-1)=1=y(1)=u(-1)$, $u''(x)=y''(-x)$ and $$ (2-\cos(πx))u''(x)+u(x)=(2-\cos(π(-x)))y''(-x)+y(-x)=1 $$ This means that also $u$ is a solution and by uniqueness $u=y$.
The Fourier coefficients of real-valued functions satisfy $\hat y_{-n}=\overline{\hat y_n}$. The Fourier coefficients of an even real-valued function are additionally real themselves.