Given the following evolution IVP/BVP: $\begin{cases} u_t - \Delta u = f &\text{in } U \times(0,\infty)\\ u=0 &\text{on } \partial U \times (0,\infty)\\ u= g &\text{on } U \times \{t=0\} \end{cases}$,
$g\in L^2(U) $, $f \in L^{\infty}(U_T)$ for each $T>0$ and $U_T:= U\times (0,T]$.
Suppose $\tau > 0$ and $f(x,t) = f(x,t+\tau)$ with $x \in U$, for each $t\geq 0$. Prove that there is a $L^2$ function $g$ whose associated $u$ solving the PDE has the same periodic property as $f$.
Showing uniqueness is pretty straightforward, but I am clueless for existence.
Any hint or suggestion is tremendously appreciated!
This is an expanded version of my comment:
Let $\{\phi_n\}_1^\infty$ and $\{\lambda_n\}_1^\infty$ be the eigenfunctions and eigenvalues of $-\Delta$ on $U$, and project all functions involved onto the orthonormal basis:
\begin{align} u_n(t):=(u(t,x),\phi_n(x)),\\ g_n:=(g(x),\phi_n(x)),\\ f_n(t):=(f(t,x),\phi_n(x)). \end{align} The $f_n$'s are $\tau$-periodic, and by completeness of the basis, it suffices to consider whether the $u_n$'s are $\tau$-periodic as well.
The $u_n$'s solve the initial value problems $$ u_n'(t)+\lambda_n u_n(t)=f_n(t),\\ u_n(0)=g_n, $$ which have the solutions $$ u_n(t)=e^{-\lambda_n t}g_n+\int_0^te^{-\lambda_n(t-s)}f_n(s)ds. $$ We have $\tau$-periodicity if and only if $u_n(t+\tau)=u_n(t)$. Substituting the solution into this expression, extending $f_n$ to $\mathbb R$ using periodicity, and changing variables $s\to s+\tau$ in the integral yields the equation $$ e^{-\lambda_n(t+\tau)}g_n+\int_{-\tau}^te^{-\lambda_n(t-s)}f_n(s)ds=e^{-\lambda_nt}g_n+\int_{0}^te^{-\lambda_n(t-s)}f_n(s)ds, $$ which reduces to \begin{align} (e^{-\lambda_n\tau}-1)g_n&=-\int_{-\tau}^0e^{\lambda_ns}f_n(s)ds\\ &=-e^{-\lambda_n \tau}\int_0^\tau e^{\lambda_n s}f_n(s)ds. \end{align} Since $\lambda_n>0$ for all $n$, we conclude that $$ g_n=\frac{1}{e^{\lambda_n\tau}-1}\int_0^\tau e^{\lambda_n s}f_n(s)ds. $$