I am wondering how to find a Fourier series solution of the following PDE:
$$u_{t} + i u_{xxx} = 0$$ $$u(0, x) = u_{0}(x)$$
where $t$ is in $(0, \infty)$ and $x$ is in the torus $\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z})$, and we assuming $u_{0}$ is represented by its Fourier series (e.g., $u_0 \in L^2(\mathbb{T})$ ).
I started with supposing $u(t, x) = X(x)T(t)$ and plugged into the PDE to get that $\frac{T'(t)}{T(t)} = \frac{-i X'''(x)}{X(x)} = \lambda$, which gives $T'(t) - \lambda T(t) = 0$, hence $T(t) = C_{\lambda}e^{\lambda t}$.
Then we have $u(t, x) = X(x)T(x) = X(x)C_{\lambda}e^{\lambda t}$. Using the initial condition, we get $u(0, x) = X(x)C = u_{0}(x)$.
Then we suppose there is a general solution of the form $u(t, x) = \sum_{\lambda}X_{\lambda}(x)T_{\lambda}(t) = \sum_{\lambda}\frac{1}{C_{\lambda}}u_{0}(x)C_{\lambda}e^{\lambda t} = \sum_{\lambda}u_{0}(x)e^{\lambda t}$.
At this point I am confused as to how to get a Fourier series solution by setting $u(0,x) = u_0 = \sum_{k}a_{k}e^{i k x}$, where the $a_k$ are the Fourier series coefficients of $u_0$. Since the $\lambda$ might not be an integer, but even if we assume $\lambda_k$ is indexed by the integers, then we are saying $u(0, x) = \sum_{\lambda_k}u_{0}(x) = u_0 = \sum_{k}a_{k}e^{ikx}$, and it is not clear to me how to solve for coefficients for $u(t, x)$.