Elliptic theory on compact manifolds

Question

Maybe this is silly.

On a bounded set $\Omega\subset\mathbb{R}^n$ consider the equation

$$ \Delta u=f \quad\text{ in $\Omega$}$$ $$ u=0\quad\text{ on $\partial\Omega$}.$$

One has the following elliptic estimates

$$ \| u\|_{W^{2,p}}\le C\|f\|_{L^p}. $$

Does one can have the same result if instead of $\Omega\subset\mathbb{R}^n$ one consider the same problem in a compact Riemannian manifold with boundary? I know that one has

$$\|u\|_{W^{2,p}}\le C(\|f\|_{L^p}+\|u\|_{L^p}) $$

for the general case.. is it true that one can get rid of $\| u\|_{L^p}$ considering zero boundary conditions?

This should be true using a covering argument and the result in $\mathbb{R}^n$ but I cannot find anything in literature. Any help will be appreciated.

Answer 1

Consider the case of a sphere. The boundary is empty. A constant $u$ satisfies the equation with $f=0$. You cannot get rid of the $u$ on the right hand side of the inequality in this case.

EDIT: I think this example demonstrates the difficulty of trying to use only local estimates pieced together with partitions of unity to eliminate $u$ on the right hand side of the inequality.

On the other hand, a possible strategy for proving the inequality without $u$ in the right hand side could be the following:

• Note that the Laplace-Beltrami operator with Dirichlet boundary conditions is (essentially) self-adjoint on, say the domain of $u$ in $W^{2,2}$ vanishing at the boundary, and the spectrum is discrete (by compactnesss)

• A maximum principle shows that if the boundary is nonempty (and manifold connected), then $0$ is not an eigenvalue. This is the essential global ingredient.

• The resolvent $\Delta^{-1}$ at $0$ is bounded in $L^2$, which means that we can eliminate $u$ in the right hand side of the inequality for $p=2$.

• We might be able to proceed by interpolation to general $p$.

Answer 2

Previous answer replaced by new one:

Fix a finite relatively open cover $U_1, \dots, U_m$ of $M$, where each $U_i$ is diffeomorphic to either the open ball or the half-ball obtained by intersecting the open ball with a closed half-space. Let $\phi_1, \dots, \phi_N$ be a partition of unity subordinate to this cover.

If $u$ satisfies $$ \Delta u = f\text{ and }u = 0\text{ on }\partial M, $$ then $$ \Delta(\phi_iu) = \phi_if + \Delta\phi_iu + 2\nabla\phi\cdot\nabla u. $$ If the support of $\phi_i$ is contained in an open ball, then $\phi_iu$ vanishes on the boundary. If the support of $\phi_i$ is contained in the half-ball, then $\phi_i u$ vanishes on the boundary of the half-ball, because $u$ is assumed to be zero on $\partial M$. Therefore, by the estimate for the Dirichlet problem and the Gagliardo-Nirenberg inequality, \begin{align*} \|\phi_iu\|_{2,p} &\le C_1(\|f\|_p + \|u\|_p + \|\nabla u\|_p)\\ &\le C_2(\|f\|_p + \|u\|_p + 2\|\nabla^2u\|_p^{1/2}\|u\|_p^{1/2})\\ &\le C_2(\|f\|_p + \|u\|_p + \epsilon\|\nabla^2u\|_p + \epsilon^{-1}\|u\|_p). \end{align*} Set $\epsilon = 1/(2NC_2)$. Adding everything up, we get \begin{align*} \|u\|_{2,p} &=\|\sum_i \phi_iu\|_{2,p} \le \sum_i \|\phi_iu\|_{2,p}\\ &\le C_2N(\|f\|_p + (1+2NC_2)\|u\|_p) + \frac{1}{2}\|u\|_{2,p}. \end{align*} Therefore, $$ \|u\|_{2,p} \le C(\|f\|_p + \|u\|_p). $$

Answer 3

For the specific question of getting rid of the low-order term $\|u\|_p$, what you really need is the compact embedding of $W^{2,p}$ in $L^p$ (Rellich–Kondrachov theorem). Indeed, suppose there is no bound $\|u\|_{2,p}\le C\|f\|_p$, then there is a sequence $u_n$ such that $\|u_n\|_{2,p}=1$ and $\|f_n\|_p\to 0$, where $f_n=\Delta u_n$. By the compact embedding, we can pass to a subsequence so that $u_n$ converges in $L^p$. Combining this with $f_n\to 0$ in $L^p$, we know $u_n$ converges in $W^{2,p}$. Let $u$ be the limit, then $u$ would be a nonzero function whose Laplacian is zero, which is impossible by your boundary conditions.