MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As a research preface, this question is linked to a problem of increasing magnetism in Ginzburg-Landau equations that I have distilled for the purpose of getting to the bottom of this technical matter.

Suppose $u_n\in H^1(B_1)$ (actually, $u_n$ are smooth), where $B_1$ is the unit ball in $\mathbb{R}^N$, and that we know $||u_n||_{H^1} \leq C$, so that up to a subsequence, $u_n \rightharpoonup u$ in $H^1(B_1)$ weakly. What can we say about the boundedness of the quantity

$\int_{\partial B_1} \nabla u_n(x) \cdot n(x)\;d\mathcal{H}^{N-1}(x)$,

where $n(x)$ is the unit normal to $\partial B_1$?

In particular, is this quantity always finite under these hypotheses, or is there a counterexample that shows this blows up for a bounded subset of $H^1$?

share|cite|improve this question
If you take $u_n(x) := n^{1/2} (|x|-1+\frac{1}{n})_+$, then the $u_n$ are bounded in $H^1$ but the integral blows up like $n^{1/2}$. – Terry Tao Oct 3 '13 at 17:43
up vote 14 down vote accepted

Clearly, what you call $\newcommand{\bn}{\boldsymbol{n}}$ $\nabla u\cdot \bn $ is the normal derivative $\frac{\partial u}{\partial \bn}$. The trace theorem (see e.g. Lions and Magenes, Non-Homogeneous Boundary Value Problems and Applications. I, Thm. 9.4, Chap 1) shows that for $s> \frac{3}{2}$ the restriction map $\newcommand{\pa}{\partial}$

$$R: C^\infty(B_1)\ni u\mapsto (u|_{\pa B_1}, \frac{\pa u}{\pa \bn})\in C^\infty(\pa B_1)\times C^\infty(\pa B_1) $$

extends to a continuous split surjective map

$$ H^s(B_1)\to H^{s-\frac{1}{2}}(\pa B_1) \times H^{s-\frac{3}{2}}(\pa B_1). $$

The result is optimal because Theorem 9.5, Chap. 1 op. cit. shows that if $s\leq \frac{3}{2}$, then for any $\phi\in C^\infty(\pa B_1)$ the linear functional

$$ C^\infty( B_1)\ni u\mapsto \int_{\pa B_1} \frac{\pa u (x)}{\pa \bn} \phi (x) dA(x)\in\mathbb{R} $$

is not continuous in the topology induced by $H^s(B_1)$. In particular, if $s\leq \frac{3}{2}$ there cannot exist a constant $C>0$ such that

$$ \left|\int_{\pa B_1} \frac{\pa u}{\pa \bn} dA\right| \leq C\Vert u\Vert_{H^s(B_1)},\;\;\forall u\in C^\infty(B_1), $$

so that there exists a sequence $u_k\in C^\infty(B_1)$ such that $\Vert u_k\Vert_{H^1(B_1)}\leq 1$ and

$$ \left|\int_{\pa B_1} \frac{\pa u_k}{\pa \bn} dA\right| \to \infty. $$

share|cite|improve this answer

If you restrict a $H^s$-function $f$ to a submanifold of codimension $k$ you get a $H^{s-k/2}$ function.

EDIT: But only if $s$ is large enough (Sobolev lemma threshold). Thus the rest of my answer does not work, and I deleted it.

share|cite|improve this answer

The second order term in your PDE is the Laplacian. Even though in $H^1$ you cannot define the trace on the boundary, the Laplacian helps. I am not sure of which particular problem you are thinking of (many things are called Ginzburg-Landau), but the Pohozaev identity usually helps. See for example 'Vortices in the Magnetic Ginzburg-Landau Model' Here by Sandier and Serfaty -where many references of its use are given. They show for example that for $$ -\Delta u = \frac{u}{\epsilon^2}(1-|u|^2) \textrm{ in }\Omega $$ with $u=g$ on $\partial\Omega$, $|g|=1$ and $\Omega$ star-shaped, there holds (Lemma 5.2) $$ \int_\Omega \frac{(1-|u|^2)^2}{\epsilon^2} + \int_{\partial \Omega} \left|\frac{\partial u}{\partial n}\right|^2 \leq C(\Omega,g), $$ So the normal derivative is $L^2(\partial\Omega)$, and this does not follow from $u\in H^1(\Omega)$.

The Pohozaev identity is to use as a test function $x\cdot \nabla u$ in the case of a ball, that is, a test function that will equal $\partial_n u$ on the boundary. This idea is also known under the name of Morawetz multiplier, and Rellich identity.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.