energy estimates Evans PDE chapter 7 Hi I am looking at the proof of theorem 2 energy estimates in Evans PDE. I have some difficulties regarding the estimate for each term.
First for the first term.
Q1 I am a little vague how (23) is obtained. Well, what I see is that
$$2(u'_m,u_m)+2\beta\|u_m\|_{H_0^1(U)}\le 2(u'_m,u_m)+2 B[u_m,u_m;t]+2\gamma\|u_m\|_{L^2(U)}^2$$
Then by (21) the above becomes
$$2(u'_m.u_m)+2\beta\|u_m\|_{H_0^1(U)}\le2(f,u_m)+2\gamma\|u\|_{L^2(U)}^2$$, i.e.
$$\frac{d}{dt}(\|u_m\|_{L^2(U)}^2)+2\beta\|u_m\|_{H_0^1(U)}\le2(f,u_m)+2\gamma\|u\|_{L^2(U)}^2$$
Now, I believe, the following fact (stated in the book)
$$|(f,u_m)|\le\frac{1}{2}\|f\|_{L^2(U)}^2+\frac{1}{2}\|u_m\|_{L^2(U)}^2$$
is used to conclude
$$\frac{d}{dt}(\|u_m\|_{L^2(U)}^2)+2\beta\|u_m\|_{H_0^1(U)}\le C_1\|u_m\|_{L^2(U)}^2+C_2\|f\|_{L^2(U)}^2$$
I guess $C_1$ is $1+2\gamma$ and $C_2$ is 1.
But I am really vague about this, because, the inequality for $|(f,u_m)|$ is not quite the same as the inequality for $(f,u_m)$.
Q2 I do not get the inequality sign on $\eta(0)=\|u_m(0)\|_{L^(U)}\le\|g\|_{L^2(U)}.$ According to (15), i.e. $d_m^k(0)=(g,w_k)$, I have
$$\|u_m(0)\|_{L^2(U)}^2=\|\sum_k (g,w_k)w_k\|_{L^2(U)}^2=(\sum_k (g,w_k)w_k,\sum_j (g,w_j)w_j)=\sum_k\sum_j|(g,w_k)|^2(w_k,w_j)$$
Now since $w_k$ is an orthonormal basis in $L^2$, I think I should end up with an equality rather than inequality. Please clarify.
Next, for the estimate of the 2nd term, I think, all I need is to integrate †he estimate $\max_{t\in[0,T]}\|u_m(t)\|_{L^2(U)}^2$ over $[0,T]$, right?
Finally, for the 3rd term. I cannot see the 3rd last line, i.e.
$$|<u'_m,v>|\le C(\|f\|_{L^2(U)}+\|u_m\|_{H_0^1(U)})$$ is deduced from the previous line.
Basically, my understanding is $$\left|<u'_m,v>\right|\le\left|<f,v^1>-B[u_m,v^1;t]\right|$$...
Please help.


 A: I am not sure how much I can help you, since it has already been 4 months ago. Regarding Q1:
I had a slightly different approach.
$$
\begin{align}
\frac{d}{dt}(\frac{1}{2}||u_m||_{L^2(U)}^2) + B[u_m, u_m,t] &= (f, u_m),\\
&\leq |(f, u_m)|,\\
&\leq \frac{1}{2}||f||_{L^2(U)}^2 + \frac{1}{2}||u_m||_{L^2(U)}^2,\\
\end{align}
$$
Add on both sides $\gamma||u_m||^2_{L^2(U)}$ and multiply everything by $2$ using (22) will give (23). You also mentioned the inequality of $(f,u_m)$ being different then $|(f,u_m)|$, but analogues to what I did; $x\leq |x|$ for all $x$, right?
Q2: Remember $u_m(0):=\sum_{k=1}^m d_m^k(0)w_k$ and $d_m^k(0) = (g, w_k)$, and can be considered a projection of $g$ on the subspace spanned by $\{w_k\}_{k=1}^\infty$. This is an orthogonal projection, hence the inequality.
To see this in your own writing:
$$
\begin{align}
\sum_j |(g, w_k)|^2(w_k, w_j) &= |(g,w_k)|^2,\\
&\leq ||g||^2||w_k||^2 = ||g||^2.
\end{align}
$$
This is due to $w_k$ being orthonormal in $L^2$ and the Cauchy-Schwarz inequality ($|(g,w_k)|\leq||g||||w_k||$).
Next: No, as done in step 3, integrate (23) and after that use the inequality for $\max_{t} ||u_m(t)||^2$ in the first term of (23). Furthermore, the result presented in step 3 is the second term of (23) and (20).
Last: $|<u_m', v>| = |(f, v^1) - B[u_m, v^1;t]|$ and not `$\leq$', this is proved in the first equation (the one above the one you mentioned) on the same page. With help of p.318 you should now be able to get the inequality.
Hopefully you can still do something with this, or better, you already found the answers yourself.
