# Regularity of Weak solution for biharmonic problem

Suppose $$\Omega\subset \mathbb R^2$$ be a bounded convex polygonal domain. $$f\in L^2(\Omega)$$ be a force function. Then for the problem \begin{align} -\Delta u&=f\quad\text{in }\Omega\\ u&=0\quad\text{on }\partial\Omega \end{align} The weak problem is to find $$u\in H^1_0(\Omega)$$ such that $$a(u,v)=(f,v)~\forall v\in H^1_0(\Omega)$$ where $$a(u,v)=\int_{\Omega}\nabla u\cdot\nabla v~dx$$. I have seen the regularity results for the weak solution to above defined problem which says that provided the domain is smooth our weak solution will be in $$H^2(\Omega)$$ and $$\lVert u\rVert_{H^2(\Omega)}\lesssim \lVert f\rVert_{L^2(\Omega)}$$ and if $$f\in H^m(\Omega)$$ then $$u\in H^{m+2}(\Omega)$$ with $$\lVert u\rVert_{H^{m+2}(\Omega)}\lesssim \lVert f\rVert_{H^m(\Omega)}$$. All these estimates can be found in standard books like Evans PDE.

Now, I am working on the biharmonic problem and in that too the variatonal inequality concerning biharmonic problem. The continuous problem is defined as \begin{align} \Delta^2 u&=f\quad\text{in }\Omega\\ u=0&=\frac{\partial u}{\partial n}\quad\text{on }\partial \Omega \end{align} For an obstacle $$\chi\in H^3(\Omega)$$ such that $$\chi\leq 0$$ on $$\partial \Omega$$, let a non empty convex set to $$\mathcal K:= \{v\in H^2_0(\Omega)\colon v\geq \chi ~a.e \in \Omega\}.$$ The weak problem is defined as find $$u\in\mathcal K$$ such that $$b(u,v-u)\geq (f,v-u)~\forall v\in\mathcal K$$ where the bilinear form $$b(u,v)=\int_{\Omega}\Delta u \Delta v~dx$$.

Then in one of the paper by Zhang, I have read that $$u\in H^3_{loc}(\Omega)$$ and if the domain is smooth enough $$u\in H^3(\Omega).$$ My query are:

(1). Can someone please let me know about the estimates for the weak solution $$u$$ in terms of $$f$$ and $$\chi$$ as like we have for the obstacle problem.

(2). I once thought that why not we transform our problem into a set of problem namely \begin{align} \Delta u&=v\quad \text{in }\Omega\\ \Delta v&=f\quad \text{in }\Omega \end{align} then using obstacle problem theory $$v\in H^2(\Omega)$$ and hence $$u$$ should be in $$H^4(\Omega)$$ with the estimates that $$\lVert u\rVert_{H^4(\Omega)}\lesssim \lVert f\rVert_{L^2(\Omega)}+\lVert \chi\rVert_{H^3(\Omega)}.$$

Any type of help will be appreciated. Thanks in advance.

Based on the comments leading up to the main question, I think it's worth pointing out that such estimates on the obstacle problem (or the biharmonic obstacle problem) are nontrivial, and do not follow from the corresponding $$L^2$$-based regularity theorems on the unconstrained problem in any meaningful way.

Frehse proved that if $$\chi$$ is locally $$C^{1, 1}$$ and $$f \in H^{-1}$$ (roughly), then $$u$$ is locally $$H^3$$. The idea goes something like this (let's assume $$\chi = f = 0$$ first): ideally we would like to use $$v = u + \epsilon\eta^2 u_{ee}$$ as a test function for the inequality (or competitor in the minimization, if you prefer), where $$\epsilon > 0$$ is tiny, $$e$$ is a fixed unit vector, and $$\eta$$ is a cutoff function. If this was a legal test function, then we would have that $$\int u_{ij} (\eta^2 u_{ee})_{ij} \geq 0$$. Now integrate one of the $$e$$ derivatives by parts to get $$\int \eta^2 (u_e)_{ij} (u_e)_{ij} \leq - \int 2 \eta \eta_e u_{ij}(u_{e})_{ij} + \text{similar terms} \leq C\|u\|_{H^2}(\|u\|_{H^2} + \|\eta u\|_{H^3}).$$ This leads to $$\|\eta u\|_{H^3} \leq C \|u\|_{H^2}$$ and we win.

The problem is that it is not clear that $$v$$ is a legitimate test function. First, we do not know it is in $$H^2$$: after all, we are taking four derivatives of $$u$$ and have yet to show that $$u\in H^3$$. This is a typical issue which is resolved by replacing $$u_{ee}$$ with difference quotients like $$w_h(x) = h^{-2}[u(x + he) - 2 u(x) + u(x - he)]$$. The bigger concern is that we do not know that $$v = u + \epsilon \eta^2 w_h \geq \chi (= 0)$$. But here Frehse notices that if $$\epsilon \ll h^2$$, then $$v(x) = (\text{positive number})[u(x + he) + u(x - he)] + \color{red}{(1 - 2 \eta^2(x) \epsilon h^{-2})}u(x)$$ has the red part positive, and as $$u \geq 0$$, this means $$v \geq 0$$. So this test function is actually OK, and we can proceed with our previous argument (notice how the restriction on $$\epsilon$$ small was irrelevant, we just needed it to work for one $$\epsilon > 0$$).

This argument also works for $$\chi$$ convex, and Frehse has some barrier tricks to have it work for $$\chi \in C^{1,1}$$. If $$f$$ is non-zero, you can still run the same argument, since on the right-hand side you will have things like $$\int F_i (\eta^2 w)_i \leq C\|F_i\|_{L^2}\|\eta u\|_{H^3}$$ (for $$f = \text{div} F$$).

Finally, your statement with $$\chi \in H^3$$ follows from this by replacing $$u$$ with $$u - \chi \geq 0$$ and $$\chi$$ with $$0$$. This transforms the right-hand side from $$f$$ to $$f - \Delta^2 \chi \in H^{-1}$$ if $$\chi \in H^3$$.

Regarding question (2): This doesn't work because $$v$$ solves neither an obstacle problem, nor $$\Delta v = f$$. Indeed, we have that $$\Delta^2 u = f$$ when $$u$$ is greater than $$\chi$$, so there $$\Delta v = f$$. But on the interior of $$\{u = \chi\}$$, we instead have $$\Delta v = \Delta \chi$$. And on the interface between these regions, it's pretty unclear what we have. Heuristically (based on e.g. the 1D or radial examples with $$f = 0$$) we expect that $$u$$ has zeroth, first, and second derivatives continuous across the interface, the third derivatives possibly form a jump, and then the fourth derivative is a measure.

This is sort of the correct picture in 2D, as far as it goes: see Caffarelli and Friedman on this topic, who show that $$u \in C^2$$ locally but without any uniform estimate. A more precise characterization of how the third derivatives behave (and how the free boundary looks like) I would say has historically been outside the scope of available methods, but is a topic of current research and perhaps that situation is slowly changing.

• So for sufficiently smooth domain, by Frehse we can at max conclude that $\lVert u\rVert_{H^3(\Omega)}\leq C(\lVert f\rVert_{L^2(\Omega)}+\lVert \chi\rVert_{H^3(\Omega)}).$ Mar 3, 2023 at 3:12
• For a smooth enough domain, this should be true; you can flatten the boundary before running the argument (Frehse in fact treats equations with Lipschitz coefficients, so $C^{1,1}$ should be enough) and then use standard arguments. For just convex domains, you would need to argue more carefully, probably using estimates on the Green's function, but I would guess it's still true. Mar 3, 2023 at 4:09
• Thank you so much for the help. Mar 3, 2023 at 4:12
• can you help me in getting the estimates for $\lVert \Delta^2u\rVert_{L^2(\Omega)}$. what I expected and got is assuming the obstacle $\chi\in H^4(\Omega)$, $$\lVert \Delta^2u\rVert^2_{L^2(\Omega)}\leq C(\lVert f\rVert^2_{L^2(\Omega)}+\lVert \chi\rVert^2_{H^4(\Omega)}).$$ But I am not sure that whether do I need to assume so much regularity on the obstacle to get the estimate on $\Delta^2 u$ and I am not able to find any reference to go through the same. Mar 18, 2023 at 10:50