# A PDE problem about $(\partial_x^2 + \partial_t^2)u = 0$ using Fourier series.

I'm trying to solve the following initial value problem (from Folland, pg. 277, exercise $48$a) using Fourier series:

Let $x \mapsto f(x)$ and $x \mapsto u(x, t)$ be periodic functions on $\mathbb{R}$. No regularity is given, so we can just assume whatever is needed as we go. The equation is: $$(\partial_x^2 + \partial_t^2)u = 0.$$ With an initial value: $$u(x, 0) = f(x).$$

What I got after applying the Fourier transform to both sides and solving the corresponding ODE is the following: $$u(x, t) = \sum_{\xi \in \mathbb{Z}} c_1(\xi)e^{2\pi\xi[t + ix]} + c_2(\xi)e^{2\pi\xi[-t + ix]},$$ where $\hat{f}(\xi) = c_1(\xi) + c_2(\xi)$. So far, I think this is ok.

My questions are:

1. Is it possible to go further? Like, is there anything else I can do to make the solution more explicit, such as writing it as a convolution or something?

2. There is a comment in this exercise about Abel means/Abel summation. What does that have to do with anything here?

Thank you!

You will get the general solution of the form $u(x,t)=C_1(x+it)+C_2(x-it)$ , for matching the only condition $u(x,0)=f(x)$ , you can obviously take $C_1(x)=\dfrac{f(x)}{2}+F(x)$ and $C_2(x)=\dfrac{f(x)}{2}-F(x)$ so that you can get $u(x,t)=\dfrac{f(x+it)+f(x-it)}{2}+F(x+it)-F(x-it)$
Note that this solution satisfy on $x,t\in\mathbb{C}$ .