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I have a problem with the following part of Evans book PDE. It is in the proof of the improved regularity of weak solution to a second order parabolic equation (Theorem 5, Chapter 7.1, page 361-364).

Here we consider for fixed $T>0$ the equation $$ \left\{ \begin{aligned} u_t+Lu=f &\quad \text{in}\ \ U\times(0,T] \\ u=0 &\quad \text{on}\ \partial U\times[0,T] \\ u=g &\quad \text{on}\ U\times\{t=0\} \end{aligned} \right. $$ where $U$ is an open bounded set in $\mathbb R^n$, $$ Lu = -\sum_{i,j}a^{ij}(x)u_{x_ix_j} + \sum_ib^i(x)u_{x_i} + c(x)u $$ and $\partial t + L$ is uniformly parabolic.

In Theorem 5(ii) (Chapter 7.1, page 361) it is assumed that $$ g \in H_0^1(U),\ f \in H^1(0,T;L^2(U)),\ a^{ij},\ b^i\ \text{and}\ c\ \text{are smooth on}\ \bar U. $$ The proof starts from standard Galerkin approximation. That is, for $m>0$, let $$ u_m(t) = \sum_{k=1}^m d_m^k(t)w_k, \quad \text{s.t.}\ \left\{ \begin{aligned} u'_m + Lu_m = \sum_{k=1}^m \langle f(t),w_k \rangle w_k,\\ u_m(0) = \sum_{k=1}^m \langle g,w_k \rangle w_k, \end{aligned} \right. $$ where $\{w_k\}$ is an orthonormal basis of $L^2(U)$ and an orthognal basis of $H_0^1(U)$. Since $f \in L^2(0,T;L^2(U))$ we have $u_m$ absolutely continuous in $t$ and the equation is satisfied for a.e. $t \in [0,T]$.

By choosing the test function $u'_m$ (here $'$ means the partial differential in time $t$) and applying Gronwall's inequality we deduce in (the first line of) eq(51) that $$ \sup_{[0,T]} \|u'_m(t)\|_{L^2(U)}^2 + \int_0^T \|u'_m\|_{H_0^1(U)}^2dt \le C\big(\|u'_m(0)\|_{L^2(U)}^2 + \|f'\|_{L^2(0,T;L^2(U))}^2\big). $$ Up to here it is okay for me. Then in the last line of (51) the authors obtain further the upper bound $$ C\big(\|f\|_{H^1(0,T;L^2(U))}^2 + \|u_m(0)\|_{H^2(U)}^2\big) $$ by using the weak form of Galerkin equation. I got lost in this step. How can we get this estimate?

Thanks for help!

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1 Answer 1

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We wish to use the equation to show the estimate $$ \lVert u'_m(0) \rVert_{L^2(U)}^2 \leq C\left(\lVert u_m(0) \rVert_{H^2(U)}^2 + \lVert f_m \rVert_{H^1(0,T;L^2(U))}^2\right). $$

The strategy is to consider the equation at $t=0,$ and show the other terms are bounded. This requires some care however, as a-priori the equation only holds almost everywhere in $t.$

To do this, first observe by Section 5.9, Theorem 2 (Calculus on spaces involving time) that we have a continuous embedding $$ H^1(0,T;L^2(U)) \hookrightarrow C([0,T],L^2(U)). $$ Hence for each $k$ the mapping $t \mapsto f_m^k(t) = \langle f_m(t), w \rangle$ is continuous on $[0,T]$ and for each $m$ we have $d_m^k(t)$ satisfies the ODE system $$ (d_m^k)'(t) + \sum_{j=1}^m d_m^j(t) B[w_j,w_k;t] = f_m^k(t). $$ As each $B[w_j,w_k;t]$ is smooth in $t$ (differentiating under the integral sign), by standard ODE theory we deduce that the unique solution $d_m^k(t)$ must be continuously differentiable on $[0,T].$ Therefore the equation holds pointwise on $[0,T],$ and evaluating at $t=0$ we obtain the identiy $$ u_m'(0) = - \sum_{k=1}^m B[u_m(0),w_k;0]w_k + f_m(0). $$ To conclude observe we can control both terms on the right hand side as \begin{align*} \left| B[u_m(0),w_k;t]\right| &\leq C \lVert u_m(0) \rVert_{H^2(U)} \\ \lVert f_m(0) \rVert_{L^2(U)} &\leq C\lVert f_m \rVert_{H^1(0,T;L^2(U))}, \end{align*} where we used the continuous embedding above to estimate the $f_m$ term. Hence putting everything together we get \begin{align*} \lVert u_m'(0) \rVert_{L^2(U)} &\leq \sum_{k=1}^m \left| B[u_m(0),w_k;t]\right| \lVert w_k\rVert_{L^2(U)} + \lVert f_m(0) \rVert_{L^2(U)} \\ &\leq \left(\lVert u_m(0) \rVert_{H^2(U)}^2 + \lVert f_m \rVert_{H^1(0,T;L^2(U))}^2\right), \end{align*} as required.

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