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We can use a Galerkin method to show that there is a solution to a PDE. So suppose $w_j$ is the basis functions.

I am interested in regularity of solutions. In the book by Evans, he differentiates a weak form involving the finite dimensional approximations (say) $u_m = \sum_{j=1}^mc_j(t)w_j(x)$. So the equation $$(u_m', w_j) + B(u_m, w_j) = f(w_j)$$ he differentiates to get $$(u_m'', w_j) + B(u_m', w_j) = f'(w_j)$$

Then he messes around with this equation and ends up showing that $u_m''$ lies in $L^2(0,T; H^{-1})$.

Questions: 1) How can we differentiate $u_m$ twice? How do we know that $c_j$ is differentiable twice? Even once? 2) Is it enough to show that $u_m''$ lies in the above mentioned space to know that $u_m''$ exists? How do we know that the limit of $u_m''$ as $m \to \infty$ is $u''$??? How to make sense of $u''$??

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$u_m''$ actually is for $x$, not for $t$.So we do not need $c_j$ to be differentiable here. See $u_m = \sum_{j=1}^m c_j(t)w_j(x)$. And $w_j$ should be differentiable. – Yimin Mar 5 '13 at 23:06
@Yimin i don't think so. pretty sure it's differentiation wrt. $t$. – george.s Mar 5 '13 at 23:16
ok, then which chapter is this part in the book? I will look it up later. – Yimin Mar 6 '13 at 0:00
@Yimin try 7.1.3. – george.s Mar 7 '13 at 21:07

1 Answer 1

I believe the $c_j$'s are as smooth as the coefficients of the PDE and the right hand side. If you look at theorem 1, (I will use your notation of $c_k$'s rather than Evans' $d_k$), we have a linear system of ODE $$c^{k'}_m(t)+\sum_{l=1}^me^{kl}(t)c^l_m(t)=f^k(t),$$ for which there exists a unique absolutely continuous function $c_m=(c^1_m,\ldots,c^m_m)$ that satisfies the ODE a.e..

The ODE then tells us that $c^{k'}_m$ is absolutely continuous, (as long as $f$ is nice). We then have the right to differentiate the ODE with respect to $t$, thus obtaining

$$c^{k''}_m(t)=-\sum_{l=1}^me^{kl'}(t)c^l_m(t)+e^{kl}(t)c^{l'}_m(t)+f^{k'}(t),$$ then the second derivative is well behaved as the first derivative, and the derivatives of the coefficients etc.

Provided the problem is in a nice enough setting, this argument can be bootstrapped as much as you like.

We now know that $u_m''$ exists, we then show that $u_m''$ is uniformly bounded in the given space, which gives us weak convergence. If you look at exercise 5 at the end of this chapter in Evans, you will see this notion of convergence is exactly what is needed for the limit of the sequence of the second weak derivative to coincide with the derivative of the limit of the sequence of first weak derivatives, i.e., $$u_m''\rightharpoonup (weak-\lim_{m\to\infty}u_m')'.$$

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