My question mainly concern about how to construct a elliptic equation with Neumann boundary condition from a minimization problem.

Let $B=B_1 \subset \mathbb{R}^3$ and $E : H^1(B) \to \mathbb{R}$ $$E(u)= \int_{B}|\nabla u|^2+(u^2-1)^2 dx - \int_{\partial B}Q(u)d\mathcal{H}^2$$ We assume that $u_0 \in W^{1,2}$ to be the minimizer of the functional $E$ in the configuration space $$K=\{u\in W^{1,2}(B:\mathbb{R})\}.$$ Since $u_0$ is the critical point of the functional, we let $\xi \in K$, we obtain the equation $$\int_B \nabla u \cdot \nabla \xi + 4(u^2-1)u \xi dx - \int_{\partial B }Q'(u)\xi d\mathcal{H}^2 = 0. $$ If we further require that $\xi$ vanishes on the boundary, we have the EL equation $$\int_B \nabla u \cdot \nabla \xi + 4(u^2-1)u \,\xi dx = 0. $$ Suppose we also have that $u \in H^2(B)$, we have $$\int_B -\Delta u \, \xi + 4(u^2-1)u \xi dx + \int_{\partial B } \dfrac{\partial u}{\partial n}\xi- Q'(u)\xi d\mathcal{H}^2 = 0. $$ We finally obtain the equation $$\Delta u = 4(u^2-1)u \,\text{ in } B \,\text{ and }\, \dfrac{\partial u}{\partial n}=Q'(u) \,\text{ on }\, \partial B.$$

My main goal is to prove the minimizer $u_0$ solve the above equation weakly with the desried Neumann boundary condition. However, my question is how to obtain the $H^2$ bound of $u$? I think we can apply standard estimate to obtain $H^2_{loc}$. If we do not have the fact that $u \in H^2(B)$, we may hard to have the existence of $\dfrac{\partial u}{\partial n}$ on the boundary by trace theorem.

  • $\begingroup$ From the context, it looks like your "$E:\mathbb{R}\to \mathbb{R}$" (third line) should be "$E:H^1(B)\to \mathbb{R}$". $\endgroup$
    – DCM
    May 27, 2019 at 23:16
  • $\begingroup$ Just to make sure I understand correctly: is the goal here to show that the minimising $u_0\in H^1(B)$ is a $H^2(B)$ function satisfying your Neumann problem? $\endgroup$
    – DCM
    May 27, 2019 at 23:22
  • $\begingroup$ @DCM Thanks for pointing out my mistake! $\endgroup$
    – mnmn1993
    May 28, 2019 at 5:12

1 Answer 1


Let $u\in H^1(B)$ be a minimising value of $E:H^1(B)\to \mathbb{R}_+$. Then $E'(u)=0$ in the sense that

$$ E'(u)v = \int_B [\nabla u \cdot \nabla v + 4(u^2-1)uv ] dx - \int_{\partial B} Q'(u)v \hspace{.5pc}d\mathcal{H}^2 = 0 $$

for all $v\in H^1(B)$. This implies, in particular, that

$$ \int_B [-u \Delta v + 4(u^2-1)uv] \hspace{.5pc}dx = 0 $$

for $v\in \mathscr{D}(B)$ (this uses the fact that $v$ and all its derivatives vanish on $\partial B$ if it has compact support in $B$). Having this last equation hold for all $v\in \mathscr{D}(B)$ is exactly the statement that

$$ \Delta u = 4(u^2-1)u \hspace{1pc}\mbox{in $B$ } $$

in the sense of distributions. The interior equation in the distributional sense is therefore just what you get from regarding $E'(u)$ itself as a distribution (i.e. by restricting it to $\mathscr{D}(B))$.

For each compact subset $K$ of $B$, take $\psi_K\in \mathscr{D}(B)$ with $\psi_{|K}=1$. Then considering the action of $E'(u)$ on $\{v(1-\psi_K):v\in C^\infty(\bar B),\mbox{$K\subset B$ compact}\}$ should - I think - also give you a weak form of your boundary condition.

I might be wrong, but I think that it's only once you have your interior equation in a (weak or distributional) form which does a priori involve the assumption that $u\in H^2(B)$ that it's time time to start worrying about regularity. In particular, you may be able to establish that $u\in H^2(B)$ using the fact that it is a $H^1$ solution to your PDE.

  • $\begingroup$ Yes, you are right. For a general $H^1(B)$ function, I think it may be have a well-defined derivative on the boundary and also the $\Delta u$ (ever in weak or distributional sense?). However my main goal is as the title, I do not know if I construct it in the correct way. I also look for another construction if mine is not good enough. $\endgroup$
    – mnmn1993
    May 28, 2019 at 5:16
  • $\begingroup$ I think the general pattern you follow is sound. Assuming $E:H^1(B)\to \mathbb{R}$ is differentiable, a necessary condition for $u\in H^1(B)$ to be a minimiser is that $E'(u)v = 0$ for all $v\in H^1(B)$. Choosing $v\in H^1_0(B)$ gives you your "EL" equation, for example. The key thing is that $u$ has to satisfy the $E'(u)v = 0$ whatever $v\in H^1(B)$ you choose, which is the main mechanism through which you're able to deduce things about its behaviour. $\endgroup$
    – DCM
    May 28, 2019 at 8:46
  • $\begingroup$ In particular, I'd be inclined to see what you get by using Green's identity (thought of in a distributional sense, at least initially) in the equation in your third display, and seeing what drops out of that for different choices of $v$. $\endgroup$
    – DCM
    May 28, 2019 at 8:47
  • $\begingroup$ I think the slickest way to get the interior equation and the boundary equation alone is by choosing test functions $v$ which satisfy a suitable PDE of their own (a PDE designed to make one of the two integrals vanish). $\endgroup$
    – DCM
    May 28, 2019 at 9:15
  • $\begingroup$ I nice analogy here is 'plotting the graph of a function'. $E'(u)$ is a linear function(al) on $H^1(B)$; we learn about its `shape' behaviour by plugging in values and seeing what we get. The behaviour of $E'(u)$ then tells us things about $u$. $\endgroup$
    – DCM
    May 28, 2019 at 9:18

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