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Let $U\subset \mathbb{R}^n$ with smooth boundary $\partial U$. And consider the expression

$$\Delta u = f.$$

$$\text{+"convenient boundary conditions"}$$

In my specific case $f\in H^2_0$. Under what (boundary) conditions is it possible to conclude a weak solution which is also in $H^2_0$ of this problem?


The problem is mainly concerned with the subscript zero in $H^2_0$. For instance if I introduce the boundary condition

$$u=0 \text{ on } \partial U\ \text{ (trace-wise),}$$

then for any $f\in L^2$, there exists a weak solution in $H^1_0$. I can regularize the solution to be also in $H^2$, hence it is in $H^1_0\cap H^2$. (I use Evans for referrences concerning to Regularity theorems.) But as far as I know $H^1_0\cap H^2$ is a strictly bigger class of functions then $H^2_0$.

Help is very much appreciated.

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2 Answers 2

Under what (boundary) conditions is it possible to conclude a weak solution to $\Delta u = f$ which is also in $H^2_0$ of this problem?

The answer is that: no boundary conditions can do this.

Rule of thumb: If the imposed boundary condition leads to a well-posed weak formulation (solution exists and is unique), then no matter what this boundary condition is, we can not get a weak solution $u\in H^2_0$ in general.

First notice $H^2_0$ means: $$ H^2_0(U) := \{u\in H^2: \; u\big|_{\partial U} = 0, \; \nabla u\big|_{\partial U} = 0\}. $$ Also note if $u$ is a constant on boundary, then $u$ does not change tangentially along any curve on boundary, so the directional derivative of $u$ along any tangent vector of $\partial U$ is 0. The only thing we wanna deal with is the normal derivative: $$ u\big|_{\partial U} = 0 \text{ and } \nabla u\big|_{\partial U} = 0 \implies u\big|_{\partial U} = 0\text{ and } \frac{\partial u}{\partial n}\Big|_{\partial U} = 0 $$

Now if $u$ is a weak solution of the Poisson equation, and $u\in H^2_0$, this means $$\left\{ \begin{aligned} -\Delta u &= f \quad \text{ in } U, \\ u &= 0 \quad \text{ on } \partial U, \\ \partial u/\partial n &= 0 \quad \text{ on } \partial U, \end{aligned}\right.\tag{$\star$} $$ for some $f$.

However, $(\star)$ is not a well-posed problem. Looking for a function $u\in H^2_0$ is like imposing Dirichlet and Neumann boundary conditions at the same time (a Cauchy boundary problem), which can be proved ill-posed in several ways.

A simple argument can be: if neglecting Neumann boundary at first, Poisson equation together with the Dirichlet boundary has a unique solution for any $f\in L^2(U)$. This means $\nabla u$ can be evaluated a.e. in $U$, and if $\partial u/\partial n|_{\partial U}$ is defined as a limit from the interior to the boundary in the direction of the normal , then $\partial u/\partial n$ is already determined in this case. In general we can't expect $\partial u/\partial n = 0$ coincides with what you obtained from the solution of Dirichlet boundary problem.

A counterexample would be: suppose we have a classical solution which coincides with a strong and weak solution to the Dirichlet boundary problem except a measure zero set, and $f\leq 0$. By maximum principle the maximum is achieved on boundary, $u|_{\partial U}=0> u(x)$ for $x\in U$, then $\partial u/\partial n$ must be strictly greater than 0 pointwisely (see Gilbarg and Trudinger Page 32).

Another argument would be if $\partial u/\partial n|_{\partial U} = 0$, by divergence theorem: $$ \int_U f\,dx = \int_U \mathrm{div}(\nabla u)\,dx = \int_{\partial U} \frac{\partial u}{\partial n} \,dS = 0, $$ which means $f$ has average 0 in $U$. Hence this says that, if $f$ does not have zero average, the $(\star)$ does not have a solution.


Some comment about your terminology, "regularization" doesn't mean what you have written in your question, what you applied is just elliptic regularity.

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I think that you can't solve this equation in $H^2_0$. If a function $u$ is in $H^2_0$ then you have $u=0$ and $\nabla u=0$ at the boundary and these are too much boundary conditions for this equation. Let me think how one can prove the nonexistence.

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It's the same thing that I was fearing. Restricting yourself to $H^2_0$ probably over determines the system. My question is motivated by the following proAblem: Consider two functions $u$ and $v$ in $L^2$, such that for every $\phi\in H^2_0$ it holds that $\int_U u\Delta \phi=\int_U v\Delta \phi$. Can we conclude that $u$ is equal to $v$ almost everywhere? –  Aris May 20 '13 at 18:17
    
Notice that with these boundary conditions the laplacian is self-adjoint. Assume that $u-v$ is smooth. Then you obtain that $u-v$ is harmonic. In fact, u+1 verify that $\int_U (u+1)\Delta \phi=\int_U u\Delta \phi+\int_U\Delta \phi.$ Now you can apply the divergence theorem and you obtain $\int_U\Delta \phi$. Thus, if $U$ is bounded any constant function is in $L2$. Thus, the claim is false, right? –  guacho May 20 '13 at 20:23

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