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I am a little confused with the galerkin approximation based proof of existence of weak solution of a linear second order parabolic pde with dirichlet boundary conditions, as stated in Evans' Partial Differential Equations. My questions are probably related to my misunderstanding of the space $H^{-1}(U)$ and representations of elements in that space.

(Pg 356, Eq 27) Why is it important to treat the derivative of the solution, u, with respect to time as a element of $L^{2}(0,T;H^{-1}(U))$? Doesn't the galerkin approximation imply $u^{'}_m(t)$ is also in $H^1_0(U)$. Then doesn't there exist a sub-sequence, $\left\lbrace u^{'}_{m_l} \right\rbrace ^{\inf}_{l=1}$, such that $u^{'}_{m_l}$ converges weakly to $u^{'}$ in $L^{2}(0,T;H^1_0(U))$?

Also, in page 357, while trying to show that the solution, u, satisfies the initial condition, $u(0) = g$, where g is in $L^{2}(U)$, shouldn't we consider $g$ in $H^{1}_{0}( u)$, since u is in $L^{2}(0,T;H^{1}_{0}(U))$. Additionally, while integrating the term $<u',v>$ over time, how is the integration in parts in the time to be understood, since u' is in $L^{2}(0,T;H^{-1}(U))$

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Haudor
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While it is true that $u_m'(t) \in H_0^1(U)$ (since the basis is smooth in space), we only have a bound uniform in $m$ on $u_m'$ in the space $L^2(0,T;H^{-1}(U))$, so we only have the weak convergence in that space. If we had a bound on $u_m'$ in $L^2(0,T;L^2(U))$ or $L^2(0,T;H^1(U))$ then we could say that $u_m' \rightharpoonup u$ with the convergence in $L^2(0,T;L^2(U))$ or $L^2(0,T;H^1(U))$ respectively, but no such bound is given (at this stage).

By assumption $g$ need only be in $L^2$, which is much weaker than requiring it to be in $H^1_0$. Also, if $u \in L^2(0,T;H^1_0(U))$ and $u' \in L^2(0,T;H^{-1}(U))$, it does not follow that $u(t) \in H^1_0(U)$ for each $t$. We only have $u(t) \in L^2(U)$ by the embedding $$C^0([0,T];L^2(U)) \subset \{ u \in L^2(0,T;H^1_0(U)) : u' \in L^2(0,T;H^{-1}(U))\} =: W.$$

The integration by parts formula can be obtained from the result: if $u, v \in W$, then $$\frac{d}{dt}(u(t),v(t))_{L^2(U)} = \langle u'(t), v(t) \rangle + \langle v'(t), u(t) \rangle$$ where the angled brackets represents the duality pairing in $H^1$ and $H^{-1}$.

riem
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  • Thanks, I understood most of it. But initially when we picked $L^{2}(0,T;H^{1}_{0}(U))$ as the space of interest so that the partial derivatives of u with respect to the spatial variables are well defined. But it seems like we ended up with a solution which might not have such a well defined partial derivative for each t? – Haudor Aug 28 '14 at 20:35
  • Well, the solution does lie in $L^2(0,T;H^1_0(U))$ as you said. Remember that this space is a Lebesgue/Bochner space, so if $u \in L^2(0,T;X)$ then all we know is that $u(t) \in X$ for almost every $t \in [0,T]$. This is what you expect since $L^p$ functions are really equivalence classes of functions that agree almost everywhere. So yes it can be the case that $u(t^) \notin X$ for some given $t^$. – riem Aug 28 '14 at 20:45