Take $$ u_t(t) + A(t)u(t) = f(t), $$ $$ u(0) = u(T), $$ where $A$ is an linear elliptic operator and the first equation is an equality in $L^2(0,T;V^*)$ for $V \subset H \subset V^*$ Hilbert triple. (there is slight abuse of notation in the equality but never mind)
Under what conditions does a solution to this problem exist? By solution I mean $u \in L^2(0,T;V)$ with $u_t \in L^2(0,T;V^*)$ (or $u_t \in L^2(0,T;H)$ if data is smooth enough). Apart from requiring maybe $A(0) = A(T)$ and $f(0) = f(T)$.
How does one prove this via the Galerkin approach?
Thanks
Step I. Solve the initial value problem $$ \left\{ \begin{array}{lll} X_t(t)=-A(t)X(t), \\ X(s)=I, \end{array} \right. $$ where $I : H\to H$ is the identity. Assume that $K(t,s)$ is the solution - this is a bounded operator from $H$ to $H$, for every $s,t$ - There is both theory and approximations for obtaining $K$, i.e., $$ K(t,s)=\lim_{n\to\infty}\prod_{k=1}^n \bigg(1-\frac{t-s}{n}A\Big(s+k\frac{t-s}{n}\Big)\bigg). $$
Step II. The solution of $$ X_t+A(t)X=f(t), \quad X(0)=X^0, $$ is then expressed as $$ X(t)=K(t,0)\,X^0+\int_0^t K(t,s)\,f(s)\,ds. $$
Step III. Check whether there exists a $X^0$, such that $$ X(T)=X(0) $$ or equivalently $$ K(T,0)\,X(0)+\int_0^T K(T,s)\,f(s)\,ds=X(0). $$ or $$ \big(K(T,0)-I\big)X(0)=-\int_0^T K(T,s)\,f(s)\,ds. $$ This only requires that the operator $K(T,0)-I :H\to H$ possesses an inverse or equivalently $$ 1\not\in \sigma\big(K(T,0)\big). $$