I am following a PDE course and I can't understand a step of Evans proof of the improved regularity of weak solution to a second order parabolic equation (Theorem 5, Chapter 7.1):
Let $u\in L^2(0,T,H_0^1(U))$, with $u'\in L^2(0,T,H^{-1}(U))$ be the weak solution of:
$$
\left\{
\begin{aligned}
u_t+Lu=f &\quad \text{in}\ \ U\times(0,T] \\
u=0 &\quad \text{on}\ \partial U\times[0,T] \\
u=g &\quad \text{on}\ U\times\{t=0\}
\end{aligned}
\right.
$$
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 $L$ is uniformly elliptic. Suppose also :
$$
g \in H_0^1(U),\ f \in L^2(0,T;L^2(U)),\ a^{ij},\ b^i\ \text{and}\ c\ \text{are smooth on}\ \bar U
$$
Then in fact:
$$u\in L^2(0,T,H^2(U))\ \cap \ L^\infty(0,T,H_0^1(U)) \qquad u'\in L^2(0,T,L^2(U))$$
The proof starts with the Galerkin approximation.
That is, for $m>0$, let
$$u_m(t) = \sum_{k=1}^m d_m^k(t)w_k, \quad \text{s.t.:}$$
$$(u'_m,w_k)_{L^2(U)} + B[u_m,w_k,t] = (f,w_k)_{L^2(U)} \qquad
d_m^k(0) = (g,w_k)_{L^2(U)}$$
where $\{w_k\}$ is an orthonormal basis of $L^2(U)$ and an orthognal basis of $H_0^1(U)$(e.g. eigenfunctions of laplacian) and
$$B[u,v,t]=\int_U \sum_{i,j}a^{ij}(x)u_{x_i}v_{x_j} + \sum_ib^i(x)u_{x_i}v + c(x)uv \ dx $$
My problem is in step 2 of the proof:
step 2. Combining the above inequalities, we deduce
$$\|u_m'\|_{L^2(U)}^2+\frac{d}{dt}(\frac{1}{2}A[u_m,u_m])\leq \frac{C}{\epsilon}(\|u_m\|_{H_0^1(U)}^2+\|f\|_{L^2(U)}^2)+2\epsilon\|u_m'\|_{L^2(U)}^2$$
Choosing $\epsilon=\frac{1}{4}$ and integrating , we find
$$\int_{0}^{T}\lVert u_m'\rVert_{L^2(U)}^2dt+\sup_{0\leq t\leq T} A[u_m(t), u_m(t)]\leq C(A[u_m(0),u_m(0)])+\int_{0}^{T}\lVert u_m\rVert_{H_0^1(U)}^2+\lVert f\rVert_{L^2(U)}^2dt\leq C(\lVert g\rVert_{H_0^1(U)}^2+\lVert f\rVert_{L^2(0,T;L^2(U))}),$$
according to theorem 2 in $\S$ 7.1.2, where we esimated $\lVert u_m(0)\rVert_{H_0^1(U)}\leq \lVert g\rVert_{H_0^1(U)}$ .
Where
$$A[u,v]=\int_U\sum_{i,j=1}^n a^{ij}u_{x_i}v_{x_j} \ dx$$
I can't find in theorem 2 in $\S$ 7.1.2, the proof of the estimate: $\lVert u_m(0)\rVert_{H_0^1(U)}\leq \lVert g\rVert_{H_0^1(U)}$. It would be easy to prove with the $L^2$ norm using bessel inequality, but with $H_0^1$ norm I don't know how to prove it.
Thank you !
Best Answer
I finally I figured out how to solve the problem. I post my solution to help people who will have my some doubt in future.
Let $\lambda_k$ be the eigenvalue of $-\Delta$, associated to the (weak) eigenfunction $w_k$. Let $\tilde{w_k}=\frac{w_k}{\sqrt\lambda_k}$. Then $\{\tilde{w_k}\}$ are a orthonormal basis of $H_0^1(U)$
Since $w_k$ are weak solutions of $-\Delta u=\lambda_k u$, by definition we have: $$\int_U \nabla w_k \nabla v=\lambda_k\int_U w_kv \qquad \forall v\in H_0^1(U)$$ Thanks to Poincaré inequality $(u,v)_{H_0^1(U)}:=\int_U\nabla u\nabla v$ is an equivalent inner product on $H_0^1(U)$. Taking $v=g$ in the previous equality we get: $$(w_k,g)_{H_0^1(U)}=\lambda_k(w_k,g)_{L^2(U)}$$ Then we have: $$\lVert u_m(0)\rVert_{H_0^1(U)}=\Big\lVert\sum_{k=1}^m (g,w_k)_{L^2(U)}w_k\Big\rVert_{H_0^1(U)}=\Big\lVert\sum_{k=1}^m \frac{1}{\lambda_k}(g,w_k)_{H_0^1(U)}w_k\Big\rVert_{H_0^1(U)}=$$ $$=\Big\lVert\sum_{k=1}^m (g,\tilde{w_k})_{H_0^1(U)}\tilde{w_k}\Big\rVert_{H_0^1(U)}\leq \lVert g\rVert_{H_0^1(U)}$$ where the last inequality is Bessel inequality in $H_0^1(U)$