Linear evolution equation inequality (Evans chapter 7 problem 9) I'm trying to prove an inequality from Evans' PDE book (Chapter 7 Problem 9). It's inequality (54) in $\S7.1.3$ and (59) in $\S7.2.3$.
Problem:
Given $u \in H^2(U) \cap H_0^1(U)$ there exists constants $\beta > 0$, $\gamma \ge 0$ such that
$$
\beta ||u||_{H^2(U)}^2 \le (Lu, -\Delta  u) + \gamma ||u||_{L^2(U)}^2
$$
Hint: Assume $u$ smooth, $u=0$ on $\partial U$. Transform the term $(Lu, -\Delta u)$ by integrating by parts twice and then estimate the boundary terms.
After changing variables locally and using cutoff functions, you may assume the boundary is flat.
My Attempt:
Integration by parts twice
\begin{align*}
(Lu, -\Delta u)
& = -\int_U Lu\, \Delta u dx \\
& = \int_U D Lu \cdot Du\, dx - \int_{\partial U} Lu \frac{\partial u}{\partial \nu} dx \\
& = \int_U \Delta Lu \, u dx + \int_{\partial U} \left(u \frac{\partial Lu}{\partial \nu} - Lu \frac{\partial u}{\partial \nu}\right) dx
\end{align*}
If we assume $u=0$ on $\partial U$ then one of the boundary terms will be zero.
 A: A partial answer.
In the book by Gilbarg and Trudinger it is proved the following (Theorem 8.12, page 186 of the last edition). I simplify the statement a little bit.
$f \in L^2(U)$ and $\varphi \in H^2(U)$ are given. If $L$ is strictly elliptic with smooth coefficients and the boundary of the domain $U$ is regular, then the solution of
$$
\begin{aligned}
 &Lu = f \text{ in } U\\
 &u-\varphi \in H^2(U)
\end{aligned}
$$
satisfies
$$
\|u\|_{H^2(U)}^2 \leq C(\|u\|_{L^2(U)}^2+\|f\|_{L^2(U)}^2+\|\varphi\|_{H^2(U)}^2).
$$
$C$ depends on $L,U$. The theorem is written without squared norms, but you can switch from one to the other easily.
Using this theorem it can be proved that (this is Evans exercise in the case $L = -\Delta$) if $u \in H^2(U) \cap H_0^1(U)$, then
$$
\|u\|_{H^2(U)}^2 \leq C(\|u\|_{L^2(U)}^2+\|-\Delta u\|_{L^2(U)}^2).
$$
Just set $f = \Delta u \in L^2(U)$. A solution of
$$
\begin{aligned}
 &\Delta w = f \text{ in } U\\
 &w = 0 \text { on } \partial U
\end{aligned}
$$
must satisfy
$$
\|w\|_{H^2(U)}^2 \leq C(\|w\|_{L^2(U)}^2+\|-\Delta w\|_{L^2(U)}^2).
$$
but $u$ itself is a solution of that problem. Actually the book by Evans is also a reference for this theorem.
To prove the exercise in general one should prove that
$$
\theta\|-\Delta w\|_{L^2(U)}^2 \leq (Lu,-\Delta u),
$$
where $\theta$ is $L$ ellipticity constant. This can be easily seen in dimension 1, but I did not do the computation in higher dimension. This last point should be checked.
A: We have $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}=\underbrace{\sum_{i,j=1}^{d}\int_{U}a^{ij}\Delta u\mathrm{d}x}_{A}-\underbrace{\int_{U}\left(b\cdot\nabla u\right)\Delta u\mathrm{d}x}_{B}-\underbrace{\int_{U}cu\Delta u\mathrm{d}x}_{C}.$$ We obtain easily by Young's inequality that $B\ge-\left\Vert b\right\Vert _{\infty}\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}+\frac{1}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}\right]$ and that $C\ge-\left\Vert c\right\Vert _{\infty}\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}+\frac{1}{4\varepsilon}\left\Vert u\right\Vert _{L^{2}}^{2}\right]$. Let us now remark that it is enough to prove our inequality for function in $C_{c}^{\infty}(U)$ since $H_{0}^{1}(U)$ is the closure of $C_{c}^{\infty}(U)$ in $H^{1}(U)$ and we have to deal with continuous functions. Thus, for $u\in C_{c}^{\infty}(U)$ we can integrate by parts without obtaining boundary parts and hence  we obtain \begin{alignat*}{1}
A= & \sum_{i,j,k}\int_{U}a^{ij}u_{x_{i}x_{j}}u_{x_{k}x_{k}}\mathrm{d}x=-\sum_{i,j,k}\int_{U}u_{x_{k}}\left(a^{ij}u_{x_{i}x_{j}}\right)_{x_{k}}\mathrm{d}x\\
= & -\sum_{i,j,k}\int_{U}u_{x_{k}}a_{x_{k}}^{ij}u_{x_{i}x_{j}}\mathrm{d}x-\sum_{i,j,k}\int_{U}u_{x_{k}}a^{ij}u_{x_{i}x_{j}x_{k}}\mathrm{d}x\\
= & -\sum_{i,j,k}\int_{U}a_{x_{k}}^{ij}u_{x_{i}x_{j}}u_{x_{k}}\mathrm{d}x+\sum_{i,j,k}\int_{U}\left(u_{x_{k}}a^{ij}\right)_{x_{i}}u_{x_{j}x_{k}}\mathrm{d}x\\
= & \underbrace{-\sum_{i,j,k}\int_{U}a_{x_{k}}^{ij}u_{x_{i}x_{j}}u_{x_{k}}\mathrm{d}x}_{A_{1}}+\underbrace{\sum_{i,j,k}\int_{U}a_{x_{i}}^{ij}u_{x_{j}x_{k}}u_{x_{k}}\mathrm{d}x}_{A_{2}}+\underbrace{\sum_{i,j,k}\int_{U}a^{ij}u_{x_{i}x_{k}}u_{x_{j}x_{k}}\mathrm{d}x}_{A_{3}}.
\end{alignat*} If we consider $M:=\sup\left\{ \left\Vert a_{x_{k}}^{ij}\right\Vert :1\le i,j,k\le d\right\} $ (which is finite since $a^{ij}$ are smooth) we get immediately $A_{1}+A_{2}\ge2M\left[\varepsilon\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}+\frac{1}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}\right]$ and also $A_{3}\ge\theta\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}$. All in all we get $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}\ge\left(\theta-\left(2M+\left\Vert b\right\Vert _{\infty}+\left\Vert c\right\Vert _{\infty}\right)\varepsilon\right)\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}-\frac{\left\Vert b\right\Vert _{\infty}+2M}{4\varepsilon}\left\Vert Du\right\Vert _{L^{2}}^{2}-\frac{\left\Vert c\right\Vert _{\infty}}{4\varepsilon}\left\Vert u\right\Vert _{L^{2}}^{2}$$
Now using the inequality $\left\Vert Du\right\Vert _{L^{2}}^{2}\le\eta\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}+C\left(\eta\right)\left\Vert u\right\Vert _{L^{2}}^{2}$ for $\eta=\frac{4\varepsilon^{2}}{\left\Vert b\right\Vert _{\infty}+2M}$ we obtain that $$\left(Lu,-\Delta u\right)_{L^{2}\left(U\right)}\ge\underbrace{\left(\theta-\left(2M+\left\Vert b\right\Vert _{\infty}+\left\Vert c\right\Vert _{\infty}+1\right)\varepsilon\right)}_{\beta\left(\varepsilon\right)}\left\Vert D^{2}u\right\Vert _{L^{2}}^{2}-\underbrace{\left(\frac{\left\Vert c\right\Vert _{\infty}}{4\varepsilon}+C\left(\varepsilon\right)\right)}_{\gamma\left(\varepsilon\right)}\left\Vert u\right\Vert _{L^{2}}^{2}
 $$
so it is enough to choose an $\varepsilon$ small enough such that $\beta(\varepsilon)>0$ and $\gamma(\varepsilon)\ge 0$ and apply Poincare's inequality and we are done.
A: We may consider some simple versions first, for example assume u is smooth with compact support, and 
$$Lu=-\sum_{ij=1}^n (a^{ij}u_i)_j$$
Then use integration by part, now there is no boundary terms:
\begin{align*}
(Lu, \Delta u) &=\int_{U}-\sum_{ij=1}^n (a^{ij}u_i)_j \Delta u dx \\
&= \sum_{ij=1}^{n}\int_{U} D(a^{ij}u_i)\cdot (Du_j) dx \\
&= \sum_{ij=1}^{n}\int_{U} a_{ij}Du_i\cdot Du_j + Da^{ij}u_i Du_j dx \\
&\geq \theta \int_{U} |D^2 u|^2 dx-M \int_{U}\sum_{ij=1}^n |u_i Du_j| dx \\
&\geq C||u||_{H^{2}}^2-M||u||_{H^{1}}||u||_{H^{2}} \\
&\geq C||u||_{H^{2}}^2-M||u||_{L^{2}}^2
\end{align*}
By the $\epsilon$-Cauchy inequality and Poincare inequality.
For the general versions use trace theorem to deal with the boundary term.
A: Would it be possible to see how you did it in dimension 1 ?
I am consistently stuck because of remaing terms. I only can get:
$ \theta ||\Delta u||^2_{L^2} \leq <Lu;-\Delta u> + \text{something annoying} $
Plus, you can refine a little bit what you said. By elliptic regularity (I believe the boundary is supposed to be smooth in this exercise), you directly have:
$ ||u||_{H^2} \leq C||\Delta u||_{L^2} $.
As a result, it would not be surprising that the $||u||_{L^2}$ term actually comes from:
$ \theta ||\Delta u||^2_{L^2} \leq C(<Lu;-\Delta u> + ||u||^2_{L^2}) $.
What do you think about that ?
