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I am asking about a passage in the book

http://books.google.co.uk/books?id=3e4h1j9WNlwC&printsec=frontcover&dq=james+robinson+infinite+dimensional&hl=en&sa=X&ei=WlLJUZHUIePU0QXu04CwDw&ved=0CDMQ6AEwAA#v=onepage&q&f=false

please see page 203.

My question is, suppose we have a PDE $$u' + Au = f$$ that we solve by using the Galerkin method. The Galerkin approximation satisfies $$(u_n', w_j) + (Au_n, w_j) = (f,w_j)\tag{1}$$ for $j=1,...,n$, where $w_j$ is the basis. According to this book, we can write this as $$u_n' + Au_n = f\tag{2}$$ but what space is this an equality in? The book seems to suggest $L^2(0,T;V')$, but (1) only holds for the basis functions and hence I can only see the equality holds for $L^2(0,T;V_n')$, where $V_n$ is the span of the first $n$ basis functions.

And then, the author gets an uniform estimate on $\lVert u_n'\rVert_{L^2(0,T;V')}$ by using the fact that $f, Au_n \in L^2(0,T;V')).$ I understand this follows if the equality (2) were to hold in $L^2(0,T;V')$ but if it only holds in $L^2(0,T;V_n')$ does it follow?


I posted a thread about something similar to this previously but I think this is clearer.

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There is no problem in understand equality $(2)$ as an equality in $V'$, in fact remember that $V\subset H\subset V'$ is a Hilbert triple.

If I got the right notation in the book, there they consider $A=-\Delta$, $V=H_0^1$ and $H=L^2$. Now, in one hand, as I have said in first paragraph, $u'_n\in V\subset V'$. On other hand $Au_n\in H\subset V'$, so you can see the equality in $V'$.

Remark: We want to solve the equation $$u'+Au=f\tag{1}$$

Let $P_n$ be the projection onto $V_n$ and applies it in both sides of $(1)$: $$\tag{2} P_nu'+P_n Au=P_nf$$

If we write $u$ in terms of the eigenvector of $A=-\Delta$, i.e $u=\sum \lambda_k w_k$, then $P_nAu=Au_n$ where $u_n=\sum_{k=1}^n\lambda_ kw_k$, hence $(2)$ is equivalently to $$\tag{3} u'_n+Au_n=f_n$$

where $f_n=P_N f$ and $(3)$ is your equation $(1)$. Now you can note that if you solve your equation in $V'_n$, then the same equation is true in $V'$. This is because if you take any elment in $v\in V$ that belongs to $\operatorname{span}\{w_{n+1},...\}$, you have that $$\tag{4}(P_n u',v)=(P_n Au,v)=(P_n f,v)=0$$

Remark 2: I'm gonna show here that $(1)$ of Op is equivalently of my $(2)$.

I - $(1)$ implies $(2)$

Note that $(1)$ implies $$\tag{5}(u'_n,w)+(Au_n,w)=(f,w), \ \forall \ w\in V_n$$

From $(5)$ we conclude that $$\tag{6}(Pu'_n,w)+(APu_n,w)=(f,P_nw),\ \forall w\in V_n$$

Note that (by definition) $(P_nf, w)=(f,P_nw)$, so by applying the same argument as in $(4)$ to the equation $(6)$ we conclude that $$(P_nu'_n)+(P_nAu_n,w)=(P_nf,w),\ \forall w\in V$$

From the last equality we conclude that $(3)$ is true.

II - $(2)$ implies $(1)$

Just apply $w_j$ in both sides and note that $(P_nf,w_j)=(f,P_nw_j)=(f,w_j)$.

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  • $\begingroup$ I agree that $Au_n \in L^2(0,T;V')$, $u_n' \in L^2(0,T;V')$, and $f \in L^2(0,T;V')$. But why is true that $u_n' = Au_n + f$ in $L^2(0,T;V')$? I just don't understand why that equality holds. $\endgroup$ – workl Jun 25 '13 at 12:32
  • $\begingroup$ I am asking why adding $Au_n$ to $f$ equals $u_n'.$ $\endgroup$ – workl Jun 25 '13 at 12:34
  • $\begingroup$ @workl take a look and see if you get convinced now. $\endgroup$ – Tomás Jun 25 '13 at 13:02
  • $\begingroup$ Thanks for your help. I have to say I am still not sure.. remember equation (1) (in your post) holds in a weak sense only so it not that clear to me that you can just apply the projection operator like that. $\endgroup$ – workl Jun 25 '13 at 13:29
  • $\begingroup$ Applying the projection operator to $(1)$ is unnecessary for my conclusion (although it is valid). I have used it only as as heuristic point of view, beucase what I really want to show you was that $(2)$ is equivalently to $(3)$ and this is always true. $\endgroup$ – Tomás Jun 25 '13 at 13:32

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