Hi I am looking at the post discussed about weak solution of biharmonic equation Unique weak solution to the biharmonic equation
I am having trouble verifying statement 2: The bilinear operator is coercive, The claim is $$B(u,u)=\int_|\Delta\,u|^2=\|\Delta u\|_{L^2(\Omega)}^2\ge C\|u\|_{H_0^2(\Omega)}^2$$
I have read the hint (commented by Shuhao Cao) and still couldn't get it. Could any one show me explicitly how the above could be true?
Many thanks!
If we are on $H^2_0$, then both $u$ and it's first order derivatives are zero on the boundary (let's do this in $2$D, since it generalises easily), and so we see that $$\|\Delta u\|_2^2=\int_\Omega u_{xx}^2+2u_{xy}^2+u_{yy}^2=2\int_\Omega u_{xx}^2+u_{yy}^2=2|u|_{H^2}^2,$$ note that the latter equality is deduced by integrating by parts, possibly arguing by approximation.
Now we use poincare inequality, which tells us that $\|u\|_{2}\le C_1\|\nabla u\|_2\le C_2|u|_{H^2}$, and thus $$\|u\|_{H^2_0}^2=\|u\|_2^2+\|\nabla u\|_2^2+|u|_{H^2}^2\le C|u|_{H^2}^2=\frac{1}{2}C\|\Delta u\|_2^2. $$ The rest follows.
I realize that the OP has probably long since moved on from this question, but I thought I'd post an alternate answers for future readers.
Actually, this claim follows from the discussion of elliptic regularity in Evans' book, c.f. Theorem 4 in Section 6.3.2. Indeed, choose $u\in H^2_0(U)$ and put $g=\Delta u$. Note that (trivially) $u$ solves the system $$ \left\{ \begin{array}{lcl} \Delta u = g & & \text{ in } U \\ u = 0 & & \text{ on } \partial U \end{array} \right. $$ Then we have the estimate $$ \|u\|_{H^2(U)}\le c_0(\|g\|_{L^2(U)} + \|u\|_{L^2(U)}), $$ for some constant $c$ depending on $U$. Next, by Theorem 6 in Section 6.2.3 (Evans' book again), $\Delta^{-1}$ is bounded in the sense that $\|u\|_{L^2(U)}\le c_1\|g\|_{L^2(U)}$. Therefore the above estimate simplifies to $$ \|u\|_{H^2(U)}\le c_2\|g\|_{L^2(U)} = c_2\|\Delta u\|_{L^2(U)}. $$ Take $C = c_2^{-2}$ to prove the claim.
