Suppose that $f\in\mathcal{C}^2$ and $g\in\mathcal{C}^1$. Furthermore, they are such that $g,g'\in\mathcal{PC}(2\pi)$ and $f,f',f''\in\mathcal{PC}(2\pi)$ where $\mathcal{PC}(2\pi)$ is the set of continuous by parts and $2\pi$-periodic functions. Suppose also that $f$ and $g$ satisfy the differential equation
$$f''+\lambda f=g$$
with $\lambda\neq n^2, n=1,2,3,\cdots$. Find the Fourier series of $f$ in terms of the Fourier coefficients of $g$ and prove that it converges everywhere.
This is what I have done so far.
Let's define
$f=\frac{a_0}{2}+\sum_{n=1}^\infty a_n\cos(nt)+b_n\sin(nt)$
$g=\frac{c_0}{2}+\sum_{n=1}^\infty c_n\cos(nt)+d_n\sin(nt)$
Differentiating term by term we obtain
$f''=\sum_{n=1}^\infty -n^2a_n\cos(nt)-n^2b_n\sin(nt)$
Thus, the differential equation becomes
$\frac{\lambda a_0}{2}+\sum_{n=1}^\infty (\lambda-n^2)a_n\cos(nt)+(\lambda-n^2)b_n\sin(nt)=\frac{c_0}{2}+\sum_{n=1}^\infty c_n\cos(nt)+d_n\sin(nt)$
By uniqueness of the Fourier coefficients we conclude that $a_n=\frac{c_n}{\lambda-n^2}$ and $b_n=\frac{d_k}{\lambda-n^2}$.
Is this correct? What I am not sure is if it is valid to differentiate term by term as I did. How can I justify that?
Best Answer
You can differentiate term by term because $f \in C^2$, so the Fourier series of both $f$ and $f'$ are uniformly convergent on the circle.