a nontrivial inequality in the proof of weak solution of biharmonic equation

Question

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!

Answer 1

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.

Answer 2

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.