# Regulartiy of weak solution to second-order parabolic equation (Evans)

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!

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.