# An equality in $L^2(0,T;V')$!? Weak solution to PDE via Galerkin approximations

I have the heat equation $$u' - \Delta u = f$$ as equality in $L^2(0,T;V')$,i.e., $$(u',v) + (\nabla u, \nabla v) = (f,v)$$ for all $v \in L^2(0,T;V)$, where I used the same brackets for duality pairing and inner product for succinctness.

Let $w_j$ be the basis in $V$ and $H$. I read in a book that the finite dimensional (Galerkin) approximations to the PDE $$(u_n',w_j) + (\nabla u_n, \nabla w_j) = (f, w_j), \quad\text{for j=1,...,n}$$ can be written as $$\frac{d}{dt}u_n - \Delta u_n = P_nf\tag{1}$$ where $P_n$ is a projection operator.

In what sense is this latter equation an equality? Presumable not an equality in $L^2(0,T;V')$ because the weak form I wrote above only holds for the basis functions. However, the author later takes an inner product of (1) with an element on $L^2(0,T;V)$, how can he do that?

Well, it is indeed an abuse of notation though, by no means you could say (1) is an equality in $L^2(0,T;V')$, because (1) only holds in a finite dimensional subspace. Multiply (1) by the basis of the finite dimensional subspace $$(u_n',w_j) - (\Delta u_n,w_j) = (P_n f,w_j).$$ Proper assumption about the boundary condition and continuity leads to: $$(u_n',w_j) - (\Delta u_n,w_j) = (u_n',w_j) + (\nabla u_n, \nabla w_j).$$ Hence: $$(P_n f,w_j) = (f,w_j), \quad \text{for } j=1,\ldots,n,$$ which is to say, $P_n f$ is the $L^2$-projection of $f$ onto this finite dimensional subspace.

I didn't see the paper, but I guess the reason to multiply a function in $L^2(0,T;V)$ is to get some a priori error estimate for this approximation by exploiting the fact that $$(u'-u_n',w_j) + \big(\nabla (u-u_n), \nabla w_j\big) = (f-P_n f,w_j).$$

EDIT: question in the comment, instead of claiming (1) holds, rather what we can do is the following \begin{aligned} &(u_n',\phi) + a(u_n,\phi) \\ =& (u_n',P_n\phi) + a(u_n,P_n\phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) \\ =& (P_n f,P_n \phi) + (u_n',\phi - P_n\phi) + a(u_n,\phi-P_n\phi) \\ =& (P_n f,\phi) + \color{blue}{(u_n',\phi - P_n\phi)} + \color{red}{a(u_n,\phi-P_n\phi) }. \end{aligned} Blue term could vanish if we integrate w.r.t. time and integration by parts to move the derivative to $\phi-P_n \phi$. However for the red term to vanish, we must have something like $\Delta u_n = 0$...

EDIT2: If we say $f = g$ in $L^2(0,T;V')$, I don't know what the author did, but according to the Evans' book (see Section 7.1), it is $$\newcommand{\lsub}[2]{{\vphantom{#2}}_{#1}{#2}} \lsub{V'}{\langle f(t,\cdot),v \rangle}_V = \lsub{V'}{\langle g(t,\cdot),v \rangle}_V,$$ for any $v\in V$, at a.e. time $t\in [0,T]$. This is the same with what you wrote in the comment though.

• Thanks for your reply. A thought occurred to me while reading it: since $(u_n', v) = (P_n u_n', v) = (P_n u_n', P_n v)$ by properties of the projection,and likewise with the other terms, maybe it does make sense as an equality in $L^2(0,T;V')$? Because we can take the pairing with any $v \in L^2(0,T;V)$ and use the calculation I just wrote. What do you think? – workl Jun 12 '13 at 15:30
• Once we do the projection, say $$(u_n',P_n v) + (\nabla u_n, \nabla (P_n v)) = (P_n f, P_n v).$$ Still $\{P_n v\}$ is finite dimensional even for all test function $v$, we still can't say this equation holds in $L^2(0,T;V')$. – Shuhao Cao Jun 12 '13 at 15:46
• I agree it's finite dimension but is that the definition of "equality?" As i understand, the equality holds in $L^2(0,T;V')$ if we can take the dual pairing with $v \in L^2(0,T;V)$ and that dual pairing makes sense. Because eg. $(P_nf, v) = (P_nf, P_nv)$, the equation above (1) tells us that $(P_nf, v)$ makes sense, and similarly for the other terms. – workl Jun 12 '13 at 15:58
• So I guess it depends on defn. of equality?? – workl Jun 12 '13 at 15:59
• @workl An equation $A u = f$ being valid in $L^2(0,T;V')$ is equivalent to say $\langle A u,v \rangle = \langle f,v \rangle$, for any $v\in V$. Notice $A u$ and $f$ are in $V'$. In the finite dimensional approximation, $(P_n f,v)$ certainly makes sense, and you can view $P_n f$ in $V'$ as well. But it is definitely not correct to state "(1) holds in $V'$": because $$(u_n',v) + (\nabla u_n,\nabla v) \neq (P_n f,v)$$ for all $v\in V$. Because $$(\nabla u_n,\nabla v)\neq (\nabla u_n,\nabla (P_n v))$$ in general. Did the author claim that? – Shuhao Cao Jun 12 '13 at 19:44