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Let $\mathbb{D} \subset \mathbb{R}^2$ be the unit disk and consider the Dirichlet problem of the Laplace equation $\Delta u = 0$ in $\mathbb{D}$ with $u = 0$ on $\partial \mathbb{D}$. Then we know that the solution to this problem is $u \equiv 0$ and that it minimizes the Dirichlet energy defined as \begin{gather} E[u] := \int_{\mathbb{D}} | \nabla u|^2 \end{gather}

Construct a sequence of functions $\{u_k\} \subset C(\overline{\mathbb{D}})$ satisfying the following conditions

  1. $u_k$ is piecewise smooth and $u_k|_{\mathbb{\partial D}}=0$ $\forall\ k \in \mathbb{N}$
  2. $E[u_k] \to 0$ as $k \to \infty$
  3. $u_k$ diverges as $k \to \infty$ in a set that is dense in $\mathbb{D}$

Then show that $u_k \to 0$ in the Sobolev space $H^1(\mathbb{D})$


Ok, I have constructed a sequence satisfying the aforementioned properties but I cannot understand how can $u_k \to 0$ in $H^1(\mathbb{D})$. I understand that the Sobolev space is equiped with the norm \begin{align} \|u\|_{H^1} &:= \Big( \int_{\mathbb{D}} (|\nabla u |^2 + |u|^2)\Big)^{1/2}\\ \|u\|^2_{H^1} &= \|\nabla u\|^2_{L^2(\mathbb{D})} + \|u\|^2_{L^2(\mathbb{D})} \end{align} and that by Property 2 we have $\|\nabla u_k\|_{L^2(\mathbb{D})} \to 0$ as $k \to \infty$ but still $u_k$ diverges in a dense subset of $\mathbb{D}$ (Property 3) and thus we should have $\|u_k\|_{L^2(\mathbb{D})} \to \infty$ as $k \to \infty$ which means that $u \notin H^1(\mathbb{D})$.

Could you please explain me that? Also, where am I wrong in the above considerations?

Thanks!


Update After @daw's answer I want to clarify that I constructed a sequence diverging in an uncountable dense subset. Does anything on the above declaration of the problem forbids me to do that? Or is it softly implied that the subset must be countable?

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  • $\begingroup$ Imho there is nothing in the question that requires the countability of the subset, where divergence occurs. Could you post your sequence as a separate answer? Would be interesting to see. $\endgroup$ – daw Oct 9 '18 at 10:12
  • $\begingroup$ @daw I am not posting the sequence because I figured out that in fact the energy diverged. Instead I formulated a sequence based on your answer, with logarithmic singularities, that will work. I think that taking $\{q_k\}$ as the Cantor set (instead of $\mathbb{D} \cap \mathbb{Q}$) with the same sequence $u_k$, proves the claim for an uncountable dense subset while ensuring measure zero to prove the convergence in $H^1$. $\endgroup$ – ares Oct 12 '18 at 2:52
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A dense subset still can have measure zero (think of points with rational coordinates), so your conclusion: property (3) implies $\|u_k\|_{L^2}\to\infty$ is not valid. Also it would contradict the Poincare inequality.

The construction of $u_k$ would be something like $$ u_k = \frac1{k^2} \sum_{i=1}^k \phi_i, $$ where $\phi_k \in H^1(\Omega)\setminus L^\infty(\Omega)$, $\|\phi_k\|_{H^1}=1$ with singularity at some $q_k$, where $(q_k)$ is dense in $\mathbb D$.

In order to get an approximation by continuous functions, we can approximate $\phi_k$ by $\phi_{k,n}$ with $\phi_{k,n}\to \phi_k$ and $\|\phi_{k,n}\|_{H^1}\le 2$. Then set $$ u_n = \frac1{n^2} \sum_{i=1}^n \phi_{i,n}, $$

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  • $\begingroup$ I think that this series might not converge. You might want to replace $1/k$ with $1/k^2$? $\endgroup$ – gerw Oct 8 '18 at 6:33
  • $\begingroup$ @daw Thanks you are absolutely right, I was seeing this now, that I can have a countable dense subset. It is that I have constructed the sequence diverging on an uncountable dense subset. So this cannot hold for an uncountable dense subset, right? $\endgroup$ – ares Oct 8 '18 at 6:34
  • $\begingroup$ @daw I haven't accepted the question only because it's still not clear to me. I have slightly updated the question, could you clarify this last point to me? $\endgroup$ – ares Oct 9 '18 at 6:04
  • $\begingroup$ @daw Btw, I just noticed, why you excluded the function space of essentially bounded functions from $H^1$ to pick $\phi_k$? We need $u_k$ to be continuous and thus $\phi_k(q_k)$ must go to infinity for $k\to\infty$ and not for any $k$. What do I miss? $\endgroup$ – ares Oct 12 '18 at 3:43

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