I meet some problem in theorem 2(energy estimates) Evans's book chapter 7.
Step 4. Fix any $v\in H_0^1(U)$, with $||v||_{H_0^{U}}\leq 1$, and write $v=v^1+v^2$, where $v^1\in span \{w_k \}$ and $(v^2,w_k)=0 (k=1,\cdots)$. Since the functions $\{w_k\}_{k=0}^{\infty}$ are orthogonal in $H_0^1(U)$, $||v^1||_{H_0^1(U)}\leq ||v||_{H_0^1(U)}\leq 1$. Utilizing (16), we deduce for a.e. $0\leq t\leq T$ that $$(u_m',v^1)+B[u_m,v^1,t]=(f,v^1)$$. Then (14) implies
\begin{equation}
\left\langle u_m',v\right \rangle=(u_m',v)=(u_m',v^1)=(f,v^1)-B[u_m,v^1;t]
\end{equation}
$ \cdots $
Why $\left\langle u_m',v\right \rangle=(u_m',v) $ (the last sentence in step 4), the left side is the pairing of $H^{-1}(U)$ and $H_0^1(U)$ , the right side is inner product in $L^2(U)$.
Could someone give me some advice , thank you!
(14),(16) are equation under below
(14)$$u_m(t):=\Sigma_{k=1}^{m}d_m^k(t)w_k, $$
(16)$$(u_m',w_k)+B[u_m,w_k,t]=(f,w_k) (0\leq t \leq T, k=1, \cdots ,m) $$
Whole theorem is under below.
Best Answer
My guess is the following :
let $\alpha \in H^1_0(Q)$. We can say that $\alpha \in H^{-1}(U)$ with the following construction (using your notation for $()$ and $<>$) :
$$<\alpha|u> := (\alpha |u), \quad \forall u \in H^1_0(U).$$
Indeed, we have
$$|<\alpha|u>| \leq ||\alpha||_{L^2(U)} ||u||_{L^2(U)} \leq \underbrace{||\alpha||_{L^2(U)}}_{:=C} ||u||_{H^1_0(U)}$$ and therefore $\alpha$ can be seen as en element of the dual of $H^1_0(U)$.
Now use that with
$$\alpha=u_m'(t)=\sum_{k=1}^m d_m^{k}{'}(t) w_k \in H^1_0(U)$$ and $v \in H^1_0(U)$.