Solution to Problem 3

So this problem comes from the functional analysis course I took last semester. Although simple, I thought was pretty cool. In the actual exercise they had broken it down into three parts, which I did not show in the problem post. Here I will show the steps as claims.

Claim 1: The operator ${\Delta^2: A \rightarrow L^2(\Omega)}$ is a bijection

Proof: To show injectivity we show that this operator has trivial kernel. One inclusion is obvious, we just need to show that ${\Delta^2 u = 0~\implies u=0}$ . To see this we recall from elliptic regularity theory for the Laplacian that if ${\Delta \phi = f}$ for some ${f\in H^k(\Omega)}$ then

$\displaystyle \norm{\phi}_{H^{k+2}}\leq C\norm{f}_{H^k}.$

Now back to our question which can be written as ${\Delta(\Delta u)=0}$ , and so applying the estimate we can see that ${\norm{\Delta u}_{H^{2}}\leq 0}$ and so ${\Delta u = 0}$ . Applying the estimate one more time shows that ${u=0}$ . For surjectivity we begin taking some ${f\in L^2(\Omega)}$ and let ${u\in H^1_0(\Omega)}$ be the unique solution to ${\Delta u = f}$ , and by the regularity theory ${u \in H^2(\Omega)}$ . Then let ${v \in H^1_0(\Omega)}$ be the unique solution ${\Delta v = u}$ . It follows that ${v\in A}$ and ${\Delta(\Delta v)=f}$ . (Super cool right?!) $\Box$

Claim 2: If ${\Delta^2 u =f }$ then

$\displaystyle \forall \phi \in A: \int_{\Omega}u\Delta^2 \phi = \int_{\Omega}\phi f.$

Proof: Let ${\phi\in A}$ , multiplying ${\Delta^2 u =f }$ by ${\phi}$ and integrating we get

$\displaystyle \int_{\Omega} \Delta^2 u \phi = \int_{\Omega} f\phi.$

Since ${\phi \in H^1_0(\Omega)}$ we can integrate by parts once with vanishing boundary terms on the left hand side (just replace ${\phi}$ with a sequence of compactly supported smooth functions, integrate by parts using the definition of weak derivative passing limits by dominated convergence) to get

$\displaystyle \int_{\Omega} \Delta(\Delta u) \phi =-\int_{\Omega} \nabla \Delta u \nabla \phi .$

Now, ${\Delta u \in H^1_0}$ (since it is the solution to ${\Delta v =f}$ ) and so we can integrate by parts once more, again with vanishing boundary terms and we get

$\displaystyle \int_{\Omega} f\phi=\int_{\Omega} \Delta u \Delta\phi. \ \ \ \ \ (1)$

Now if we try to integrate by parts again we will run into a dead end, try it! Instead we play the same game with ${\int_{\Omega} u \Delta^2 \phi }$ to get

$\displaystyle \int_{\Omega} u \Delta^2 \phi = \int_{\Omega} \Delta u \Delta\phi . \ \ \ \ \ (2)$

Equating (1) and (2) gives the claim. $\Box$

Now we are ready to finish the problem, which is to prove some regularity result on weak solutions to ${\Delta^2 u =f }$ for ${f\in L^2}$ ; that is ${u\in A}$ . By surjectivity of ${\Delta^2}$ there exists some ${\tilde{u}\in A}$ such that ${\Delta^2 \tilde{u} =f}$ . We just need to conclude now that ${\tilde{u}=u}$ . By claim 2 we have that