Consider the linear parabolic problem (as in Evans) $$\begin{cases} \partial_t u+Lu=f\ \: \mbox{in} \:\Omega \times (0,T]\\ u=0\ \: \mbox{on} \:\partial \Omega \times (0,T]\\ u(\cdot,0)=g \ \: \mbox{in} \:\Omega \end{cases}$$
where $L$ is the operator given by $$Lu=-\mathrm{div}(A(x,t)\nabla u)+b(x,t)\cdot\nabla u+c(x,t)u$$ which we assume to be parabolic with bounded coefficients (and $A$ symmetric).
A weak solution is given by a function $u \in L^2(0,T;H^1_0(\Omega))$ with $u'\in L^2(0,T;H^{-1}(\Omega))$ that satisfies the initial condition and the following equation:
$$\forall 0 < t \leq T,\:\forall v \in H^1_0(\Omega),\: \langle u'(t), v \rangle + a(t;u(t),v)=(f(t),v)_{L^2}$$
(here $a$ is the usual bilinear form, depending on time, associated to $L$)
Consider a Hilbert basis $(w_i)$ of $L^2(\Omega)$ formed by the eigenfunction of the Laplace operator, so that
$$w_i \in H^1_0(\Omega),\:\int_\Omega w_i w_j = \delta_{ij},\:\int_\Omega \nabla w_i\cdot \nabla w_j =||\nabla w_i||^2_{L^2} \delta_{ij}$$
Set $V_m=\overline{\mathrm{span}}(\{ w_i| 1 \leq i \leq m\})$ and denote by $P_m$ the projection on $V_m$.
If $u$ is a weak solution, is it true that $u_m=P_m \circ u$ is a solution of the following approximated problem on $V_m$?
$$\forall 0 < t \leq T,\:\forall v \in V_m,\: (u_m'(t), v)_{L^2} + a(t;u_m(t),v)=(f(t),v)_{L^2}$$
I have been able to prove the following. If $u(t)=\sum_{i=1}^\infty y_i(t) w_i$, then $y_i \in H^1(0,T)$ so that $u_m \in H^1(0,T;V_m)$ with $u_m(t)=\sum_{i=1}^m y_i(t)w_i$ and $u'_m(t)=\sum_{i=1}^m y'_i(t)w_i$. Then one can see that for all $v \in V_m$ $$\langle u'(t),v \rangle=(u'_m(t),v)_{L^2}$$ $$ \int_\Omega A(\cdot,t)\nabla u(t) \cdot \nabla v=\int_\Omega A(\cdot,t)\nabla u_m(t) \cdot \nabla v$$ $$ \int_\Omega c(\cdot,t)u(t)v=\int_\Omega c(\cdot,t)u_m(t)v$$
I don't know if is true that it works also for the $b$ term, that is if for $v \in V_m$ $$\int_\Omega b(\cdot,t)\cdot v \nabla u(t)=\int_\Omega b(\cdot,t) \cdot v \nabla u_m(t)$$ I think it would be sufficient to know that $\int_\Omega w_j \nabla w_i=0$ for all $i \neq j$, but I don't know if it is true and (eventually) how to prove it.