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.


1 Answer 1


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.

  • $\begingroup$ So for sufficiently smooth domain, by Frehse we can at max conclude that $\lVert u\rVert_{H^3(\Omega)}\leq C(\lVert f\rVert_{L^2(\Omega)}+\lVert \chi\rVert_{H^3(\Omega)}).$ $\endgroup$
    – bunny
    Mar 3, 2023 at 3:12
  • 1
    $\begingroup$ 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. $\endgroup$
    – user378654
    Mar 3, 2023 at 4:09
  • $\begingroup$ Thank you so much for the help. $\endgroup$
    – bunny
    Mar 3, 2023 at 4:12
  • $\begingroup$ 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. $\endgroup$
    – bunny
    Mar 18, 2023 at 10:50

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