For the weak formulation of the biharmonic equation on a smooth domain $\Omega$ $$ \Delta^2u=0\;\text{in}\;\Omega\\ u=0, \nabla u\cdot \nu=0\; \text{on}\; \partial\Omega $$ why does one take $H^2_0(\Omega)=\overline{C_c^\infty(\Omega)}^{W^{2,2}}$ as the underlying space? (i.e. $u\in H^2_0$ weak solution iff $\int \Delta u\Delta\phi=0,\;\forall \phi\in H^2_0$)

Isn't $\nabla u=0$ on $\partial\Omega$ for $u\in H^2_0(\Omega)\cap C^1(\Omega)$, which is more than $\nabla u\cdot \nu=0$? If yes, wouldn't $H_0^1(\Omega)\cap H^2(\Omega)$ be the better choice?

  • $\begingroup$ $\nabla u \cdot \nu$ is the normal derivative, hence you have a sort of Neumann condition, which is natural. $\endgroup$ – Siminore Dec 1 '14 at 10:51
  • $\begingroup$ I know that $\nabla u\cdot \nu$ is the normal derivative, but I was wondering if $\nabla u=0$ isn't too much to ask, rather than just $\nabla u\cdot \nu$ $\endgroup$ – Bananach Dec 1 '14 at 10:54

Here is what I thought. I am not very sure but I am happy to discuss with you.

The way we cast $\Delta^2u=0$ into a weak formulation, so that we could use Lax-Milgram, tells us that $H_0^2$ is a suitable space.

Suppose we have a nice solution already, then we test $\Delta^2u=0$ with a $C^\infty$ function $v$ and see what happens.

We have $$\int_\Omega \Delta^2u\,v\,dx = -\int_\Omega \nabla \Delta u \nabla v\,dx+\int_{\partial \Omega}\nabla\Delta u \cdot\nu\,v\,d\sigma(x) = \int_\Omega\Delta u\Delta v \,dx-\int_{\partial \Omega}\Delta u\nabla v\cdot\nu \,d\sigma(x)+\int_{\partial \Omega}\nabla\Delta u \cdot\nu\,v\,d\sigma(x) $$

Hence, we would like to ask $v=0$ and $\nabla v\cdot\nu=0$ on $\partial \Omega$ so that we can obtain a Bilinear operator.

Now we have to choose our underlying space so that we could apply Lax-Milgram. Keep in mind that we need to choose a Banach space $H$, and it has to contain the condition that $v=0$ and $\nabla v\cdot\nu=0$ on $\partial \Omega$. Certainly $H_0^1\cap H^2(\Omega)$ will not work because it can not confirm that $\nabla v\cdot \nu\equiv 0$. Hence the only natural space left for us is $H_0^2(\Omega)$.

I am really not sure that $\{v\in H^2, T[u]=0\text{ and }T[\nabla u]\cdot \nu=0\}$ is a Banach space.

  • $\begingroup$ To your last question, it is, since the trace is continous from $H^1$ to $L^2$ and $\nu$ is $L^2$ (even more, it's smooth on a compact set), thus both conditions are preimages of the singleton zero function $\endgroup$ – Bananach Dec 4 '14 at 9:49
  • $\begingroup$ and yes, $H_0^1\cap H^2$ doesn't suffice, you're right $\endgroup$ – Bananach Dec 4 '14 at 9:50

I realized that $u=0$ on $\partial\Omega$ and $\nabla u\cdot \nu=0$ imply $\nabla u=0$ on $\partial\Omega$ for smooth $u$. So taking $H^2_0$ as space where we seek the weak solution, i.e. intuitively requiring $\nabla u=0$ on $\partial\Omega$, is NOT more than $\nabla u\cdot \nu =0$


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