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

\displaystyle \forall \phi \in A: \int_{\Omega}(u-\tilde{u})\Delta^2 \phi = 0,

and since {\Delta^2} is bijective we have that this is equivalent to

\displaystyle \forall g \in L^2: \int_{\Omega}(u-\tilde{u})g = 0.

Setting {g = u-\tilde{u}} we get that {u-\tilde{u}=0}.

%d bloggers like this: